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Solving limit without L'Hôpital's rule: $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}$ How can I solve this limit without L'Hôpital's rule?
$$\begin{align}\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}&=\lim\limits_{x \to 0} \frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\&=\lim\limits_{x \to 0} \frac{\frac{\tan x-\sin x}{x^3}}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}\end{align}$$
I can't proceed anymore from here.
|
Hint:
$$\tan x- \sin x=\frac{x^3}{3}-\frac{x^3}{3!}+......$$
|
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|
Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
Here's my working:
Start with $$4+\frac{1}{x}-\frac{1}{x^2}=0$$
Rearranging into standard form:
$$-\frac{1}{x^2}+\frac{1}{x}+4=0$$
Multiply by $-1$ to get a positive leading coefficient $a$:
$$\frac{1}{x^2}-\frac{1}{x}-4=0$$
I'm not sure how to determine my inputs $a,b$ and $c$ with these fractions but I guess $a=\dfrac{1}{x^2}$, $b=\dfrac{1}{x}$ and $c=-4$.
Plugging into quadratic function:
$$x = \frac{-\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
I find this challenging due to the coefficients $a$ and $b$ being fractions.
How can I apply the quadratic formula to $4+\dfrac{1}{x}-\dfrac{1}{x^2}=0$ to arrive at $\dfrac{-1\pm\sqrt{17}}{8}$?
|
Note that$$4+\frac1x-\frac1{x^2}=\frac1{x^2}\left(4x^2+x-1\right).$$So, solve the equation $4x^2+x-1=0$.
|
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|
Solving second-order differential equations. I have a few problems that I'm trying to work through. Want to see if these few are correct.
*
*$$3y'' + 4y' - 3y = 0$$
auxiliary equation is: $$3r^2 + 4r -3 = 0$$ where $a = 3$, $b = 4$, $c = -3$
can't really find roots by factoring so gonna use quadratic:
$$r = \frac{-4 \pm \sqrt{16 - 4(3)(-3)}}{6}$$
$$r = \frac{-4 \pm \sqrt{52}}{6}$$
$$r = \frac{-2}{3} \pm \frac{\sqrt{52}}{6}$$
$$r = \frac{-2}{3} \pm \frac{2\sqrt{13}}{3}$$
so there are 2 real roots. So the general solution is:
$$y = c_1e^{r_1x} + c_2e^{r_2x}$$
where $r_1 = \frac{-2}{3} + \frac{2\sqrt{13}}{3}$
where $r_2 = \frac{-2}{3} - \frac{2\sqrt{13}}{3}$
*$$9y'' + 4y = 0$$
auxiliary equation (could have used quadratic): $$9r^2 + 4 = 0$$
$$9r^2 = -4$$
$$r^2 = -4/9$$
$$r = \pm \frac{2}{3}i$$
so the two roots are:
$$r_1 = 0 + \frac{2}{3}i$$ and $$r = 0 - \frac{2}{3}i$$
where $\alpha = 0$ and $\beta = \frac{2}{3}$
and so the general solution is:
$$y = e^0(c_1cos\frac{2}{3}x + c_2sin\frac{2}{3}x)$$
*$$y = y''$$
$$y'' - y = 0$$
$$r^2 - 1 = 0$$
$$r^2 = 1$$
$$r = \pm 1$$
two real roots so:
general solution is: $$y = c_1e^x + c_2e^{-x}$$
*$$y'' + 2y = 0$$
$$r^2 + 2 = 0$$
$$r^2 = -2$$
$$r = \pm \sqrt{-2} = \pm \sqrt{2}i$$ and so $\alpha = 0$ and $\beta = \sqrt{2}$ and so $e^{\alpha x} = 1$
so
$$y = c_1cos\sqrt{2}x + c_2sin\sqrt{2}x$$
do these look right?
|
For this type of problems you can also follow the table given below.
I think, this table will help you to deal with all linear differential equations with constant coefficients.
|
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|
Largest circle in basin of attraction of the origin. We're given dynamical system:
$$
\dot x = -x + y + x (x^2 + y^2)\\
\dot y = -y -2x + y (x^2 + y^2)
$$
Question is what's the largest constant $r_0$ s.t. circle $x^2+y^2 < r_0^2$ lies in the origins basin of attraction.
So far with relatively easy algebra I've got:
$$ \dot r = \frac{r}{2}(-2-\sin(2 \phi)+2r^2) \\
\dot \phi = -(1+\cos^2(\phi))
$$
Which immediately shows $r_0 \geq \sqrt{1/2}$. How to show that there is no better bound?
|
After solving the system
$$
\dot r = \frac{r}{2}(-2-\sin(2 \phi)+2r^2) \\
\dot \phi = -(1+\cos^2(\phi))
$$
we have
$$
\left\{
\begin{array}{rcl}
r & = & \frac{\sqrt{3} \sqrt{3-\cos \left(2 \sqrt{2} \left(t-2 c_1\right)\right)}}{\sqrt{3 e^{2 t} c_2-\cos \left(2 \sqrt{2} \left(t-2
c_1\right)\right)+\sqrt{2} \sin \left(2 \sqrt{2} \left(t-2 c_1\right)\right)+9}} \\
\phi & = & -\tan ^{-1}\left(\sqrt{2} \tan \left(\sqrt{2} t-2 \sqrt{2} c_1\right)\right) \\
\end{array}
\right.
$$
Regarding the $r$ behavior the line of maxima after considering $c_1 = c_2 = 0$ is given by
$$
r_{max} = \frac{\sqrt{3} \sqrt{3-\cos \left(2 \sqrt{2} t\right)}}{\sqrt{\sqrt{2} \sin \left(2 \sqrt{2} t\right)-\cos \left(2 \sqrt{2} t\right)+9}}
$$
with $\min r_{max} = \sqrt{1-\frac{1}{\sqrt{13}}}\approx 0.850088$
|
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|
Solving a set of multivariate polynomials; four equations, four unknowns, maximum degree 2 I am currently trying to solve a set of four polynomial expressions for a geometric purpose. in short they take on this form:
$A_0=\frac{w_0^2+w_1 w_0+w_2 w_0+w_3 w_0+27 w_0+6 w_1+6 w_2+6 w_3+117}{\left(w_0+w_1+w_2+w_3+27\right){}^2}$
$A_1=\frac{w_1^2+w_0 w_1+2 w_3 w_1+24 w_1+4 w_0+11 w_3+87}{\left(w_0+w_1+w_2+w_3+27\right){}^2}$
$A_2=\frac{w_2^2+w_0 w_2+2 w_1 w_2+24 w_2+4 w_0+11 w_1+87}{\left(w_0+w_1+w_2+w_3+27\right){}^2}$
$A_3=\frac{w_3^2+w_0 w_3+2 w_2 w_3+24 w_3+4 w_0+11 w_2+87}{\left(w_0+w_1+w_2+w_3+27\right){}^2}$
The values of $A_0,A_1,A_2,A_3$ are assumed known, and the purpose is to invert these equations to achieve $w_0,w_1,w_2,w_3$.
I initially solved for $w_0$ in terms of all knowns, but for the remaining $w$ values, the formulae that result are dependent on $w_0$ for their result. This may lead to issues down the road especially if there are no real positive candidate roots for $w_0$.
The framework I laid out mostly involves solving each of the four equations for each of the four unknowns, leading to a total of 16 equations, and substituting different ones into one another until only one unknown is left. Furthermore, the current equations so far have consistently relied on the solutions of cubic and quartic equations, which can be unwieldy and could provide incorrect or spurious results.
My question is if I am taking this the wrong way and/or skipping key assumptions, and if there is a simpler approach that could be used to solve all the $w_1,w_2,w_3$ solely in terms of $A_0,A_1,A_2,A_3$. Does this set of equations remind you of any mathematical construct in particular that has had
a proven solution? How would you go about to attack this problem?
|
No matter how you attack this, generally, you will have one variable the other three depend on, another variable that the remaining two depend on, and of these last two, one will depend on the other. (I say "generally" because very, very rarely, the variables decouple. Consider $x^2 - x - y + 1 = 0$. This (surprise!) factors as $(x-1)(y-1) = 0$, so $x$ and $y$ turn out to be independent. However, this does not happen for $x^2 - x - y + 2 = 0$, or for any other small perturbation of the first equation. This decoupling is a rare coincidence.)
(There was much typing, copying, pasting, and editing in the following. A typo' could easily have slipped into this answer.)
I see several repeated subexpressions that are linear in the $w_i$, meaning we can solve for the subexpressions, then multiply by a matrix to get the $w_i$. Let $x = w_0 + w_1 + w_2 + w_3 + 27$. Then your system is \begin{align*}
A_0 x^2 &= w_0x +6 w_1 +6 w_2 +6 w_3 + 117 \\
A_1 x^2 &= (w_1+4)x - w_1 w_2 + w_1 w_3 - 7 w_1 - 4 w_2 + 7 w_3 - 21 \\
A_2 x^2 &= (w_2+4)x + w_1 w_2 - w_2 w_3 + 7 w_1 - 7 w_2 - 4 w_3 - 21 \\
A_3 x^2 &= (w_3+4)x - w_1 w_3 + w_2 w_3 - 4 w_1 + 7 w_2 - 7 w_3 - 21
\end{align*}
Adding weighted subsets of these together (which subsets and weights is evident from the left-hand sides), and setting $y = w_1+w_2+w_3+5$, \begin{align*}
A_0 x^2 &= w_0x + 6y + 87 \\
A_1 x^2 &= (w_1+4)x +(-w_2 + w_3 - 7)y +w_2^2 - w_3^2 + 8 w_2 + 9 w_3 + 14 \\
A_2 x^2 &= (w_2+4)x + (w_1 - w_3 - 7)y -w_1^2 + w_3^2 + 9 w_1 + 8 w_3 + 14 \\
A_3 x^2 &= (w_3+4)x +(-w_1 + w_2 - 7)y + w_1^2 - w_2^2 + 8 w_1 + 9 w_2 + 14
\end{align*}\begin{align*}
(A_0 + A_1 + A_2 + A_3)x^2 &= x(x-15) + 2y + 44 \tag{9} \\
(A_1 + A_2 + A_3)x^2 &= (x-4)y + 7x - 43 \\
(2A_0 + (x-4)(A_0 + A_1 + A_2 + A_3)) x^2 &= x^3 - 19x^2 + 90 x - 90 \tag{11}
\end{align*}
Equation $(11)$ is a cubic polynomial in $x$, constants, and the $A_i$s, so we can solve it for $x$, obtaining either one or three real roots. For each candidate $x$, use equation $(9)$ to get its corresponding $y$. Notice that $x - y - 22 = w_0$, so you can already determine whether $w_0$ is a positive number.
For explicitness, equation $(11)$ is, taking $A = A_0+A_1+A_2+A_3$,
$$ (A-1)x^3 + (-4A + 2A_0+19)x^2 - 90x + 90 = 0 \text{.} $$
|
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|
Different methods give different answers Solve $ \sec x- 1 = (\sqrt 2 - 1) \tan x $ Solve $ \sec x- 1 = (\sqrt 2 - 1) \tan x $
Case 1)
Square both the sides and using $ \sec ^2 x = 1+ \tan^2 x. $ And solving the quadratic we get answer $\tan x = 1$ or $\tan x = 0$.
Putting them back in also solves the equation.
Thus $x$ is either $n \pi $ or $x = n \pi + \frac{\pi}{4}, n \in \mathbb{Z}$
Case 2)
$ \sec x- 1 = (\sqrt 2 - 1) \tan x $
$ \frac{1-\cos x}{\cos x} = (\sqrt 2 - 1) \frac{\sin x}{\cos x} $
$ 2\sin^2 \frac{x}{2}= (\sqrt 2 - 1) 2 \sin \frac{x}{2} \cos \frac{x}{2}$
Thus solution is $\sin \frac{x}{2} = 0$ and $\tan \frac{x}{2} = \frac{\pi}{8}$
Thus $x$ is either $x = 2n \pi$
Or $x = 2n\pi + \frac{π}{4}$
These two answer are not same. So something is wrong.
Even though below question look similar to this one. None of the concepts in its answers really help this question.
Different ways gives different results - solving $\tan 2a = \sqrt 3 $
|
Squaring both sides of an equation can introduce extraneous solutions. Substitution demonstrates that if $n$ is odd, then $x = n\pi$ is not a solution as $-2 \neq 0$ and that $x = n\pi + \frac{\pi}{4}$ is not a solution as the LHS is negative while the RHS is positive. On the other hand, direct substitution shows that your second method produces correct solutions.
|
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|
Proof that $ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $ for $p>5$ Proof that $$ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $$
for $p>5$ and $p$ is prime.
$\newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)}$
My try
Let show that
$$p^2 - 1 \equiv 0 \Mod{30} \vee p^2 - 1\equiv 18 \Mod{30}$$
Let check $$p^2 -1 = (p-1)(p+1) $$
We know that (for example from here) that this is dividable by $2$ and by $3$ so by $6$
Let consider $5$ cases:
$$\exists_k p=5k \rightarrow \mbox{false because p is prime}$$
$$\exists_k p=5k+1 \rightarrow p^2 - 1 = 5k(5k+2) \rightarrow \mbox{ dividable by 6 and by 5 so we have rest 0}$$
$$\exists_k p=5k+2 \rightarrow p^2 - 1 = (5k+1)(5k+3) \rightarrow \mbox{ ??? }$$
$$\exists_k p=5k+3 \rightarrow p^2 - 1 = (5k+2)(5k+4) \rightarrow \mbox{ ??? }$$
$$\exists_k p=5k+4 \rightarrow p^2 - 1 = (5k+3)(5k+5) \rightarrow \mbox{ dividable by 6 and by 5 so we have rest 0}$$
I have stucked with $???$ cases...
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Any prime $> 5$ is congruent to $1$ mod $2$, to $1$ or $2$ mod $3$, and to $1, 2, 3$ or $4$ mod $5$.
Its square is congruent to $1$ mod $2$, to $1$ mod $3$, and to $1$ or $4$ mod $5$. The numbers that are congruent to $1$ mod $2$, to $1$ mod $3$ and to $1$ mod $5$ are congruent to $1$ mod $30$, while those that are congruent to $1$ mod $2$, to $1$ mod $3$ and to $4$ mod $5$ are congruent to $19$ mod $30$ (see Chinese Remainder Theorem).
|
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|
Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ Solution
Expanding $(1+x)^{\frac{1}{x}}$ at $x=0$ by Taylor's Formula，we obtain
\begin{align*}
(1+x)^{\frac{1}{x}}&=\exp\left[\frac{\ln(1+x)}{x}\right]=\exp \left(\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(1-\frac{x}{2}+\frac{x^2}{3}+\cdots \right)=e\cdot\exp \left(-\frac{x}{2}+\frac{x^2}{3}+\cdots \right) \\
&=e\left[1+\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e\left(1-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\\
&=e-\frac{ex}{2}+\frac{11}{24}ex^2+\cdots
\end{align*}
Likewise, expanding $e^{(1+x)^{\frac{1}{x}}}$ at $x=0$, we obtain
\begin{align*}
e^{(1+x)^{\frac{1}{x}}}&=(e^e)^{1-\frac{x}{2}+\frac{11}{24}x^2-\cdots}=e^e\cdot (e^e)^{-\frac{x}{2}+\frac{11}{24}x^2+\cdots}\\
&=e^e\cdot\left[1+\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\ln e^e+\frac{1}{2!}\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)^2\ln^2 e^e+\cdots\right]\\
&=e^e \cdot\left[1-\frac{ex}{2}+\frac{1}{24}(11e+3e^2)x^2+\cdots\right]
\end{align*}
Expanding $(1+x)^{\frac{e}{x}}$ at $x=0$, it follows that
\begin{align*}
(1+x)^{\frac{e}{x}}&=\exp\left[\frac{e\ln(1+x)}{x}\right]=\exp \left(e\cdot\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(e-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)=e^e\cdot\exp \left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right) \\
&=e^e\left[1+\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e^e\left[1-\frac{ex}{2}+\frac{1}{24}e(8+3e)x^2+\cdots\right]
\end{align*}
Therefore
\begin{align*}
&\lim_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}\\
=&\lim_{x \to 0}\frac{e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(11+3e)x^2+\cdots\right]-e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(8+3e)x^2+\cdots\right]}{x^2}\\
=&e^e\cdot\frac{1}{8}e\\
=&\frac{1}{8}e^{e+1}
\end{align*}
Please check. Is there any more simpler solution?
|
Looks good, and thumbs up for taking the time to provide your own work!
$$\lim\limits_{x \to 0}\frac{e^{(1+x)^{1/x}}-(1+x)^{e/x}}{x^2}$$
Not sure if this is any better, but taking $t=\left(1+x\right)^{1/x}$ turns the numerator into $e^t-t^e$ with $t \to e$ when $x \to 0$. Its first non-zero term in the series expansion (around $t=e$) is:
$$\frac{1}{2} e^{e - 1} \left(t - e\right)^2 \tag{$\star$}$$
Now borrowing your expansion of $t=\left(1+x\right)^{1/x}$ around $x=0$:
$$(1+x)^{\frac{1}{x}}=\color{blue}{e-\frac{ex}{2}}+\frac{11}{24}ex^2+\ldots$$
And plugging this only up to order $1$ (blue) into $(\star)$ gives:
$$\frac{1}{2} e^{e - 1} \left(\color{blue}{e-\frac{ex}{2}} - e\right)^2 =\frac{1}{8} e^{e + 1} x^2$$
For $(\star)$, with $f(t)=e^t-t^e$ you have:
*
*$f'(t)=e^t-et^{e-1} \implies f'(e)=0$
*$f''(t)=e^t-e(e-1)t^{e-2} \implies f''(e)=e^{e-1}$
So:
$$\begin{align}f(t) & = f(e)+f'(e)(t-e)+\frac{1}{2}f''(e)(t-e)^2 + \ldots \\
& = 0 + 0 + \frac{1}{2} e^{e - 1} \left(t - e\right)^2 + \ldots\end{align}$$
|
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|
Why is it that exponents that are divisors of φ(N) capable of generating the identity element as a power when N is prime? Let $p=x$ where $a^x \equiv 1\pmod N$. When $N$ is prime I can check whether $p=N-1$ and if it is, there are a full set of remainders. However, if $p<N-1$, I need to keep checking. For example, if $N=11$ and $a=3$ the following powers are generated:
$3^1 \pmod {11} =3 \\
3^2 \pmod {11} =9 \\
3^3 \pmod {11} =5 \\
3^4 \pmod {11} =4 \\
3^5 \pmod {11} =1 \\
3^6 \pmod {11} =3 \\
3^7 \pmod {11} =9 \\
3^8 \pmod {11} =5 \\
3^9 \pmod {11} =4 \\
3^{10} \pmod {11} =1 $
$3^{10} \pmod {11}$ doesn't represent the smallest exponent where $a^x \equiv 1\pmod N$ because equivalent powers exist (e.g. $3^6 \equiv 3^1 \pmod {11}$. The least positive exponent where $a^x \equiv 1\pmod N$ is $5$. So $p<N-1$. What I'm trying to figure out is why only divisors of $φ(N)$ are capable of $a^x \equiv 1\pmod N$, why can't $7$ be a possible exponent that returns $1$?
|
Let $N$ be a prime (as you specify) and let $(a,N) = 1$.
Let $\omega$ be the smallest positive integer such that
$$ a^{\omega} \equiv 1 \pmod{N}.$$
Then $\omega$ is called the order of $a$ mod $N$.
Since $\omega | \phi(N)$, it follows that $\phi(N) = k\omega$, for some positive integer $k$. Hence,
$$a^{\phi(N)} \equiv a^{k\omega} \equiv (a^{\omega})^{k}\equiv (1)^{k}\equiv 1 \pmod{N}.$$
In fact, we see that by taking any integer $s \in \lbrace 1, 2, \ldots, k \rbrace$, then
$$ a^{s\omega} \equiv (a^{\omega})^{s} \equiv (1)^{s} \equiv 1 \pmod{N}.$$
And so, we obtain a congruence equivalent to $1$ whenever the exponent is a multiple of the order of $a$ modulo $N$ (or equivalently, when the exponent divides the value of Euler's $\phi$-function).
Remark: There is no need to require that $N$ be prime; the above holds for composite $N$ as well.
|
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|
Evaluate $\sum_{n=1}^{\infty}{\frac{1}{n 2^{n-1}}}$ I need to find $$S = \sum_{n=1}^{\infty}{\frac{1}{n 2^{n-1}}}$$
Attempt:
$$f'(x) = \sum_{n=1}^{\infty}\frac{x^n}{2^n} = \frac{x}{2-x}$$
Which is just evaluating geometric series
$$f(x) = \sum_{n=1}^{\infty}\frac{x^{n+1}}{(n+1)2^{n}}$$
Now, by finding antiderivative of $\frac{x}{2-x}$
$$f(x) = -x-2\ln(x-2)$$
Finding sum should be just $$S = 1 + f(1)$$
But $f(x)$ is undefined as a real-valued function for $x \leq 2$
|
$$S=2\sum_{n=1}^{\infty} \frac{2^{-n}}{n}=2\sum_{k=1}^{\infty}~ 2^{-n} \int_{0}^{1} x^{n-1} ~dx=2 \int_{0}^{1}\sum_{n=1}^{\infty} \frac{dx}{x}\left(\frac{x}{2}\right)^n =\int_{0}^{1}\frac{2}{x} \frac{x/2}{1-x/2} ~dx=2 \ln 2. $$
|
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|
How to find the sum: $\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) $ $$S_n=\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) = \frac{1}{r-s}\left( \frac{1}{s+1}+\frac{1}{s+2}\cdots+\frac{1}{r}\right)$$
I fail to see how this:
$$S_n = \frac{1}{r-s}\left( \frac{1}{s+1}-\frac{1}{r-1}+\frac{1}{s+2}-\frac{1}{r+2}+\cdots\right)$$
converges to the right answer. Can someone please give a detailed explanation?
$r,s$ are integers and $r>s>0$
|
Consider the partial sum
$$S_p=\sum_{n=0}^p \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right)=\frac{1}{r-s}\left(H_{p+s+1}-H_{p+r+1}+H_r-H_s \right)$$ and use the asymptotics of harmonic numbers to get
$$S_p=\frac{H_r-H_s}{r-s}-\frac{1}{p}+\frac{r+s+3}{2
p^2}+O\left(\frac{1}{p^3}\right)$$ Then, the limit.
|
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"timestamp": "2023-03-29T00:00:00",
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|
A Problem Concerning the Series Problem
Let $\{a_n\},\{b_n\}$ be two sequences such that $a_n>0(\forall n \geq 1)$, $\sum\limits_{n=1}^{\infty}b_n$ is absolutely convergent, and $$\frac{a_n}{a_{n+1}}\leq 1+\frac{1}{n}+\frac{1}{n\ln n}+b_n(\forall n \geq 2).\tag{1}$$
Prove
*
*$\dfrac{a_n}{a_{n+1}}< \dfrac{n+1}{n}\cdot \dfrac{\ln(n+1)}{\ln
n}+b_n(\forall n \geq 2)\tag{2};$
*$\sum\limits_{n=1}^{\infty}a_n$ is divergent.
Proof for 1
It suffices to show
\begin{align*}
&~~~1+\frac{1}{n}+\frac{1}{n\ln n}<\dfrac{n+1}{n}\cdot \dfrac{\ln(n+1)}{\ln
n}\tag{3}\\
\Leftrightarrow &~~~n\ln n+\ln n+1< (n+1)\ln(n+1)\\
\Leftrightarrow &~~~\ln n^{n+1}+1< \ln(n+1)^{n+1}\\
\Leftrightarrow &~~~\ln\left(1+\frac{1}{n}\right)^{n+1}> 1\\
\Leftrightarrow &~~~\left(1+\frac{1}{n}\right)^{n+1}> e.\tag{4}
\end{align*}
Define
$$x_n:=\left(1+\frac{1}{n}\right)^{n+1},n=1,2,\cdots$$
By AM-GM inequality, we obtain
\begin{align*}
\frac{1}{x_n}&=\frac{1}{\left(1+\dfrac{1}{n}\right)^{n+1}}=\left(1-\frac{1}{n+1}\right)^{n+1}\\
&=\left(1-\frac{1}{n+1}\right)\cdot\left(1-\frac{1}{n+1}\right)\cdots\left(1-\frac{1}{n+1}\right)\cdot 1\\
&< \left[\dfrac{\left(1-\dfrac{1}{n+1}\right)+\left(1-\dfrac{1}{n+1}\right)+\cdots+\left(1-\dfrac{1}{n+1}\right)+1}{n+2}\right]^{n+2}\\
&= \left[\dfrac{(n+1)\left(1-\dfrac{1}{n+1}\right)+1}{n+2}\right]^{n+2}\\
&=\frac{1}{\left(1+\dfrac{1}{n+1}\right)^{n+2}}\\
&=\frac{1}{x_{n+1}},
\end{align*}
which implies $x_n > x_{n+1}$, namely $\{x_n\}$ is a decreasing sequence. Moreover, we can find
$$\lim_{n \to \infty}x_n=e.$$
Hence
$$\inf_{x \in \mathbb{N^+}} x_n=e,$$
which implies
$$x_n>e,$$which is desired.
AM I RIGHT? HOW TO TACKLE 2?
|
Step 1 is correct though the simple estimate $\log{(1+\frac{1}{n})} > \frac{1}{n} - \frac{1}{2n^2} \ge \frac{1}{n+1}$ immediately shows (4)
Now let $x>A>0, |y|<\frac{x}{2}$, then $\frac{1}{x+y}=\frac{1}{x}+z, |z| < \frac{2|y|}{A^2}$ since obviosuly $z=-\frac{y}{x(x+y)}$ and $x(x+y)>\frac{x^2}{2}> \frac{A^2}{2}$ by our assumptions.
Let $c_n=\frac{1}{n\ln n}, n \ge 2, \frac{c_n}{c_{n+1}}>1, \Sigma c_n=\infty$
Using the absolute convergence of $b_n$ we can find $N$ s.t. $|b_n|<\frac{1}{8}, n>N$, so by the above we can write $\frac{1}{\frac{c_n}{c_{n+1}}+b_n}=\frac{c_{n+1}}{c_n}+d_n, |d_n|<2|b_n|<\frac{1}{4}, \Sigma{|d_n|}=B<\infty$,all for $n>N$ of course.
In particular relation (2) becomes:
$\frac{a_{n+1}}{a_n}> \frac{c_{n+1}}{c_n}+d_n, n>N$
Taking logarithms we get
$\log a_{n+1} - \log a_{n}>\log c_{n+1} - \log c_{n}+\log(1+\frac{c_n}{c_{n+1}}d_n) \ge \log c_{n+1} - \log c_{n} - 2|d_n|, n>N$,
using $\frac{c_n}{c_{n+1}}|d_n| < 2|d_n| < \frac{1}{2}, \log{(1+x)}\ge -2|x|, 0 \le |x|<\frac{1}{2}$
summing now from say $N+1$ to $N+p$, we get:
$\log a_{N+p} - \log a_{N+1} > \log c_{N+p} - \log c_{N+1} -2B$ or
$a_{N+p}>(\frac{a_{N+1}}{c_{N+1}e^{2B}})c_{N+p}=Kc_{N+p}, p\ge 1$.
Combining with the divergence of $\Sigma{c_n}$ we are done!
Note that we proved the following result:
$a_n>0, \frac{a_n}{a_{n+1}}< \frac{c_n}{c_{n+1}}+b_n$, where $c_n>0, \Sigma{c_n}=\infty, \Sigma{|b_n|}<\infty, 0<C_1<\frac{c_n}{c_{n+1}}<C_2$ implies $\Sigma{a_n}=\infty$, where here we have $C_1=1, C_2=2$ but obviously we can easily adjust the proof for general $C_1, C_2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inserting random numbers from 1 to $n^2$ in a matrix of size $n \times n$ I have two matrices of size nxn with random numbers that are in range of $1$ to $n^2$.
I'm trying to calculate the probability of :
*
*the numbers 1 and 9 are present in the same indices in the two matrices,for example($n=4$):
\begin{bmatrix}
14& 10 & 2 & 1 \\
11& \textbf{1} & 16 & 15 \\
4 & 9& 13 & \textbf{9}\\
6 & 3& 5 & 7\\
\end{bmatrix}
\begin{bmatrix}
13& 4& 2 & 7 \\
15& \textbf{1} & 11& 10\\
6 & 16& 5& \textbf{9}\\
12 & 8& 3 & 14\\
\end{bmatrix}
*A pair (one or more) of two numbers that are sequential that exist in the same index of the two matrices.
for example(n=4):
\begin{bmatrix}
14& 10 & 2 & 1 \\
11& 1 & 16 & 15 \\
4 & 9& 13 & \textbf{6}\\
6 & 3& \textbf{5} & 7\\
\end{bmatrix}
\begin{bmatrix}
13& 4& 2 & 7 \\
15& 3 & 11& 10\\
6 & 16& 2& \textbf{6}\\
12 & 8& \textbf{5} & 14\\
\end{bmatrix}
*Row values of the first matrix are equal to the row values of the second matrix (not necessarily in the same index values) ,example(n=4):
\begin{bmatrix}
14& 10 & 2 & 8 \\
11& 1 & 16 & 15 \\
4 & 12& 9& 6\\
13 & 3& 5 & 7\\
\end{bmatrix}
\begin{bmatrix}
10& 2& 8 & 14 \\
16& 11 & 15& 1\\
6 & 9& 4& 12\\
13 & 3& 7 & 5\\
\end{bmatrix}
|
I'll solve the first part of the problem and leave the rest for you to solve
First place 1 and 9 at any two locations out of $n^2$ available locations. Then fill the rest of $2n^2-4$ with any number between 1 and $n^2$. So the total number of desired possibilities $=^{n^2}\!\!\!P_{2}(n^2)^{2n^2-4}$.
The number of possible matrix pairs without any restrictions is $(n^2)^{2n^2}$.
Hence the probability is $\frac{n^2-1}{n^6}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I order these numbers without a calculator?
*
*Classify the following numbers as rational or irrational. Then place them in order on a number line:
$$\pi^2, -\pi^3, 10, 31/13, \sqrt{13}, 2018/2019, -17, 41000$$
I know $\pi$ is irrational so $\pi^2$ and $-\pi^3$ are irrational. 10, 31/30, 2018/2019, -17, 41000 are all rational because then can be written in the form $\frac{a}{b}, a,b \in \mathbb{Z}, b\neq 0$
I'm not sure how I can look at $\sqrt{13}$ and determine if it's rational or irrational.
For ordering them I am approximating $\pi$ as 3, so
$\pi^2 \approx 9$ and
$-\pi^3 \approx -27$.
$31/13 \approx 30/10 = 3$
$\sqrt{9}<\sqrt{13}<\sqrt{16}$ so $3<\sqrt{13}<4$
$2018/2019 \approx 1$
If I were to order these, I would say:
$-\pi^3, -17, 2018/2019, 31/13, \sqrt{13}, \pi^2, 41000$
*Put these numbers in order from least to greatest (no calculator):
$$10^8, 5^{12}, 2^{24}$$
All I can think of is that $2^{10} \approx 10^3$ so
$2^{24} = (2^{10})^{2}* 2^{4} \approx (10^3)^2 *2^4 = 10^6 *16 <10^{8}$
So $2^{24} < 10^8$
Not sure how $5^{12}$ fits in.
|
$\sqrt{13}$ is irrational, by user.
The order of your first question should be $-\pi^3,-17,2018/2019,31/13,\pi^2,10,41000$.
$-\pi^3<-3^3 = -27<-17$, so we can compare two negative numbers.
We now compare $\pi^2$ and $10$, others are easy to compare. Note that $\pi = 3.1415926\dots$ and so $\pi<3.15$. Now $3.15^2 = 9.9225<10$, so $\pi^2<3.15^2<10$.
For your second question, just divide them respectively.
$\frac{10^8}{5^{12}} = \frac{2^85^8}{5^{12}} = \frac{2^8}{5^4} = \frac{4^4}{5^4}<1$, and $\frac{10^8}{2^{24}} = \frac{2^85^8}{2^{24}} = \frac{5^8}{2^{16}} = \frac{5^8}{4^8}>1$. Thus $2^{24}<10^8<5^{12}$.
|
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|
How to prove De Moivre's theorem inductively It is given that $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ where $n\in Z^+$.
I can show it works for $n=1$ but I am stuck in showing it inductively. I have got as far as below but are stuck in the rearranging:
$$(\cos\theta + i\sin\theta)^{k+1}= (\cos k\theta + i\sin k \theta)(\cos\theta + i\sin\theta)$$
How can I rearrange the left to look like the right?
|
By induction,
$(\cos \theta + i\sin \theta )^{k+1} = (\cos \theta + i\sin \theta )^k.(\cos \theta + i\sin \theta ) = (\cos k\theta + i\sin k\theta )(\cos \theta + i\sin \theta )$
$(\cos \theta + i\sin \theta )^{k+1} = \cos k\theta\cos \theta + i\sin k\theta\cos \theta + i\sin \theta \cos k\theta - \sin k\theta\sin \theta$
$(\cos \theta + i\sin \theta )^{k+1} = \cos((k+1)\theta) + i\sin((k+1)\theta) $
For $k=n \in \mathbb{N}$
$$(\cos \theta + i\sin \theta )^n = \cos n\theta + i\sin n\theta$$
We can also prove this by simple exponentiation rule and Euler's form of complex numbers.
$z = \cos \theta + i\sin \theta = e^{i\theta}$
As, $e^{ix} = \cos x+ i\sin x\cdots(i)$
$z^n = (e^{i\theta})^n$
As, $(e^a)^b= e^{ab}$
$z^n = (e^{i\theta})^n =e^{in\theta}$
From (i), we can write,
$$z^n = (\cos \theta + i\sin \theta )^n = \cos n\theta + i\sin n\theta$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
convergence of a fibonacci-like sequence I posted a question earlier on finding a formula for the sequence
$$t_1, t_2, t_1+t_2, t_1+2t_2,....$$
This is the question I posted earlier
I want to show that as $n\rightarrow \infty$, $\frac{t_{n+1}}{t_n} \rightarrow \phi$
The relationship is $T_n = A_nF_{n-2} +B_n F_{n_1}$
Where $A_1=1, A_2=0, A_{n+2} = A_{n+1}+A_{n}$ and
$B_1=0, B_2=1, B_{n+2} = B_{n+1}+B_{n}$
Since $A_n=F_{n-2}$ and $B_n = F_{n-1}$ then
$T_n = t_1 F_{n-2}+t_nF_{n-1}$
I'm wondering if this proof will work. I know that this sequence is like Fibonacci and it converges to the golden ratio...
suppose as $n \rightarrow \infty$, $F_{n+1}/F_n$ converges to a limit $L$.
Then: $L = \lim_{n\rightarrow \infty} \frac{F_{n+1}}{F_n} = \lim_{n\rightarrow \infty} 1 +\frac {1}{L}$
So I solve $L=1+\frac{1}{L} \implies L=\frac{1\pm\sqrt{5}}{2}$
We take the positive root so the answer is $\phi$
The reason i am confused is because this is for convergence $\frac{F_{n+1}}{F_n}$ but my sequence isn't exactly this.
|
From linear recurrence theory, as characteristic polynomial of the sequence is $x^2 - x - 1$ and it's roots are $\frac{1 \pm \sqrt{5}}{2}$, we have $t_n = A \left(\frac{1 + \sqrt{5}}{2}\right)^n + B \left(\frac{1 - \sqrt{5}}{2}\right)^n$ where $A$ and $B$ can be found from $t_1$ and $t_2$. If $A \neq 0$ then $\lim\limits_{n \to \infty} \frac{t_{n + 1}}{t_n} = \lim\limits_{n\to \infty} \left[\frac{1 + \sqrt{5}}{2} + \frac{B}{A}(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})^n\right] / \left[1 + \frac{B}{A}(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})^n\right] = \frac{1 + \sqrt{5}}{2}$ (because $|\frac{1 - \sqrt{5}}{1 + \sqrt{5}}| < 1$), otherwise it is $\frac{1 - \sqrt{5}}{2}$.
And if $A = 0$ then $t_1$ and $t_2$ have different signs - so if they are both positive, $A \neq 0$.
|
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How many solutions are there to $x_{1} + x_{2} + x_{3} + x_{4} = 15$ How many solutions are there to (I think they mean non-negative integer solutions)
$x_{1} + x_{2} + x_{3} + x_{4} = 15$
where
$1\leq x_{1} \leq 4$
$2 \leq x_{2} \leq 5$
$7\leq x_{3}$
$2\leq x_{4}$
My "Solution"
I use the method shown in this video: https://www.youtube.com/watch?v=Y0CYHMqomgI&list=PLDDGPdw7e6Aj0amDsYInT_8p6xTSTGEi2&index=7
$$N(\overline{C_{1}}\overline{C_{2}}\overline{C_{3}}\overline{C_{4}}) = $$
$$ N - (N_{C1}+N_{C2}+N_{C3}+N_{C4}) + N_{C1C2}+ N_{C1C3}+ N_{C1C4}+ N_{C2C3}+ N_{C2C4}+N_{C3C4} - N_{C1C2C3} - N_{C1C2C4} - N_{C1C3C4} - N_{C2C3C4} + N_{C1C2C3C4}$$
$$N=\binom{18}{15}$$
Now this part (find NC1) i am unsure about. when $$1\leq x_{1} \leq 4$$
$$N_{C1}$$
$$x_{1} + x_{2} + x_{3} + x_{4} = 15$$ $$1\leq x_{1} \leq 4$$
$$ x_{1}{}' + x_{2} + x_{3} + x_{4} = 15 -4 = 11$$ $$x_{1}{}' \geq 0$$
$$N_{C1} =\binom{14}{11}$$
Is this the right way to find $$N_{C1} $$ ?
|
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\sum_{x_{1} = 1}^{4}\sum_{x_{2} = 2}^{5}
\sum_{x_{3} = 7}^{\infty}\sum_{x_{4} = 2}^{\infty}
\bracks{z^{15}}z^{x_{1} + x_{2} + x_{3} + x_{4}}}
\\[5mm] = &\
\sum_{x_{1} = 0}^{3}\sum_{x_{2} = 0}^{3}
\sum_{x_{3} = 0}^{\infty}\sum_{x_{4} = 0}^{\infty}
\bracks{z^{15}}z^{\pars{x_{1} + 1} + \pars{x_{2} + 2} + \pars{x_{3} + 7} + \pars{x_{4} + 2}}
\\[5mm] = &\
\bracks{z^{3}}\sum_{x_{1} = 0}^{3}z^{x_{1}}
\sum_{x_{2} = 0}^{3}z^{x_{2}}
\sum_{x_{3} = 0}^{\infty}z^{x_{3}}\sum_{x_{4} = 0}^{\infty}z^{x_{4}} =
\bracks{z^{3}}\pars{z^{4} - 1 \over z - 1}^{2}
\pars{1 \over 1 - z}^{2}
\\[5mm] = &\
\bracks{z^{3}}\pars{1 - z}^{-4} = {-4 \choose 3}\pars{-1}^{3} =
-{-\pars{-4} + 3 - 1 \choose 3}\pars{-1}^{3}
\\[5mm] = &\ {6 \choose 3} = \bbx{20}
\end{align}
|
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|
The coefficient of $x^n$ in the expansion of $\frac{2-3x}{1-3x+2x^2}$ is? The coefficient of $x^n$ in the expansion of $\frac{2-3x}{1-3x+2x^2}$ is?
Working: $1 – 3x + 2x^2 = (1 – x)(1 – 2x)
=> (1 – x)^{-1} = 1 + (-1)(-x) + {(-1)(-1 -1)/2!}(-x)^2 + {(-1)(-1-1)(-12)/3!}(-x)^3 + … = 1 + x + x^2 + x^3 + …. $
$(1 – 2x)^{-1} = 1 + (-1)(-2x) + {(-1)(-1 – 1)/2!}(-2x)^2 + {(-1)(-1-1)(1-2)/3!}(-2x)^3 + … = 1 + 2x + (2x)^2 + (2x)^3 + …..$
Coefficient of $x^n$ in $(1 – x)^{-1}(1 – 2x)^{-1} = 2^n + 2^{n-1} + 2^{n-2} + …. + 2 + 1 = 2^{n+1} – 1$
Coefficient of $x^{n-1}$ in $(1 – x)^{-1}(1 – 2x)^{-1}$ = $2^n – 1$
Coefficient of $x^n$ in $(2 – 3x)(1 – x)^{-1}(1 – 2x)^{-1} = 2^{n+2} – 2 – 3(2n – 1)$ = $2^{n+2} – 3(2)^n + 1 = (4)2^n – (3)2^n + 1 = 2^n + 1.$
Is there a shorter, quicker (and easier) method to do this?
|
$\dfrac{2-3x}{1-3x+2x^2}=\dfrac{1}{1-x}+\dfrac1{1-2x}$
The coefficient of $x^n$ is $1^n+2^n=2^n+1$
|
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|
How to compute gcd of two polynomials efficiently I have two polynomials $A=x^4+x^2+1$
And $B=x^4-x^2-2x-1$
I need to compute the gcd of $A$ and $B$ but when I do the regular Euclidean way I get fractions and it gets confusing, are you somehow able to use a SylvesterMatrix to find the gcd or am I probably doing something wrong?
I don’t know how to format properly yet so apologies
|
$$ \left( x^{4} + x^{2} + 1 \right) $$
$$ \left( x^{4} - x^{2} - 2 x - 1 \right) $$
$$ \left( x^{4} + x^{2} + 1 \right) = \left( x^{4} - x^{2} - 2 x - 1 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 2 x^{2} + 2 x + 2 \right) $$
$$ \left( x^{4} - x^{2} - 2 x - 1 \right) = \left( 2 x^{2} + 2 x + 2 \right) \cdot \color{magenta}{ \left( \frac{ x^{2} - x - 1 }{ 2 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ x^{2} - x - 1 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{2} - x + 1 }{ 2 } \right) }{ \left( \frac{ x^{2} - x - 1 }{ 2 } \right) } $$
$$ \left( x^{2} - x + 1 \right) \left( \frac{ 1}{2 } \right) - \left( x^{2} - x - 1 \right) \left( \frac{ 1}{2 } \right) = \left( 1 \right) $$
$$ \left( x^{4} + x^{2} + 1 \right) = \left( x^{2} - x + 1 \right) \cdot \color{magenta}{ \left( x^{2} + x + 1 \right) } + \left( 0 \right) $$
$$ \left( x^{4} - x^{2} - 2 x - 1 \right) = \left( x^{2} - x - 1 \right) \cdot \color{magenta}{ \left( x^{2} + x + 1 \right) } + \left( 0 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( x^{2} + x + 1 \right) } $$
$$ \left( x^{4} + x^{2} + 1 \right) \left( \frac{ 1}{2 } \right) - \left( x^{4} - x^{2} - 2 x - 1 \right) \left( \frac{ 1}{2 } \right) = \left( x^{2} + x + 1 \right) $$
.....
|
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Limit of powers of $3\times3$ matrix Consider the matrix
$$A = \begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$$
What is $\lim_{n→\infty}$$A^n$ ?
A)$\begin{bmatrix} 0 & 0 & 0\\ 0& 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$
B)$\begin{bmatrix} \frac{1}{4} &\frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\end{bmatrix}$
C)$\begin{bmatrix} \frac{1}{2} &\frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\end{bmatrix}$
D)$\begin{bmatrix} 0 &\frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}$
E) The limit exists, but it is none of the above
The given answer is D). How does one arrive at this result?
|
By this question, we know that
\begin{equation}
A^n=
\begin{pmatrix}
2^{-n} & n\cdot 2^{-n-1} - 2^{-n-1} + \frac12 & {1-\frac{n+1}{2^n}\over2}\\
0 & {2^{-n}+1\over2} & {1-2^{-n}\over2} \\
0 & {1-2^{-n}\over2} & {2^{-n}+1\over2}
\end{pmatrix}.
\end{equation}
It is thus clear that $\lim_{n\to\infty} A^n = \begin{pmatrix} 0 &\frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\end{pmatrix}$.
|
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|
Calculate $\lim _{n\rightarrow +\infty} \sum _{k=0}^{n} \frac{\sqrt{2n^2+kn-k^2}}{n^2}$
Calculate $$\lim _{n\rightarrow +\infty} \sum _{k=0}^{n} \frac{\sqrt{2n^2+kn-k^2}}{n^2}$$
My try:
$$\lim _{n\rightarrow +\infty} \sum _{k=0}^{n} \frac{\sqrt{2n^2+kn-k^2}}{n^2}=\lim _{n\rightarrow +\infty} \frac{1}{n} \sum _{k=0}^{n} \sqrt{2+\frac{k}{n}-(\frac{k}{n})^2}=\lim _{n\rightarrow +\infty} \frac{1}{n} \sum _{k=0}^{n} f(\frac{k}{n})=\int^{1}_{0} \sqrt{2+x-x^2} dx=\int^{1}_{0} \sqrt{-(x-\frac{1}{2})^2+\frac{9}{4}} dx=\int^{\frac{1}{2}}_{-\frac{1}{2}} \sqrt{\frac{9}{4}-u^2} du$$In this sollution:
$$f:[0,1]\rightarrow \mathbb R, f(x)=\sqrt{2+x-x^2}$$
$$u=x-\frac{1}{2}, du=dx$$Unfortunatelly I don't know what I can do with $\int^{\frac{1}{2}}_{-\frac{1}{2}} \sqrt{\frac{9}{4}-u^2} du$ because my only idea is integration by substitution. But if I use for example $s=u^2$ then I have $ds=2udu$ so $du=\frac{ds}{2u}$ so I did not get rid of $ u $ which is problematic.Can you help me how to bypass this problem?
|
Hint. Just make the change of variable
$$
u=\frac 32 \sin x,\qquad du = \frac 32 \cos x \,dx,
$$ giving
$$
\int^{\frac{1}{2}}_{-\frac{1}{2}} \sqrt{\frac{9}{4}-u^2} _,du=\frac 94\int^{\arcsin\frac{1}{3}}_{-\arcsin\frac{1}{3}} \cos^2 x \,dx=\frac 98\int^{\arcsin\frac{1}{3}}_{-\arcsin\frac{1}{3}}\left(1+ \cos 2 x\right) dx.
$$ The conclusion is then standard.
|
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|
Find all roots of the equation :$(1+\frac{ix}n)^n = (1-\frac{ix}n)^n$ This question is taken from book: Advanced Calculus: An Introduction to Classical Analysis, by Louis Brand. The book is concerned with introductory real analysis.
I request to help find the solution.
If $n$ is a positive integer, find all roots of the equation :
$$(1+\frac{ix}n)^n = (1-\frac{ix}n)^n$$
The binomial expansion on each side will lead to:
$$(n.1^n+C(n, 1).1^{n-1}.\frac{ix}n + C(n, 2).1^{n-2}.(\frac{ix}n)^2 + C(n, 3).1^{n-3}.(\frac{ix}n)^3+\cdots ) = (n.1^n+C(n, 1).1^{n-1}.\frac{-ix}n + C(n, 2).1^{n-2}.(\frac{-ix}n)^2 + C(n, 3).1^{n-3}.(\frac{-ix}n)^3+\cdots )$$
$n$ can be odd or even, but the terms on l.h.s. & r.h.s. cancel for even $n$ as power of $\frac{ix}n$. Anyway, the first terms cancel each other.
$$(C(n, 1).1^{n-1}.\frac{ix}n + C(n, 3).1^{n-3}.(\frac{ix}n)^3+\cdots ) = (C(n, 1).1^{n-1}.\frac{-ix}n + C(n, 3).1^{n-3}.(\frac{-ix}n)^3+\cdots )$$
As the term $(1)^{n-i}$ for $i \in \{1,2,\cdots\}$ don't matter in products terms, so ignore them:
$$(C(n, 1).\frac{ix}n + C(n, 3).(\frac{ix}n)^3+\cdots ) = (C(n, 1).\frac{-ix}n + C(n, 3).(\frac{-ix}n)^3+\cdots )$$
$$2(C(n, 1).\frac{ix}n + C(n, 3).(\frac{ix}n)^3+\cdots ) = 0$$
$$C(n, 1).\frac{ix}n + C(n, 3).(\frac{ix}n)^3+\cdots = 0$$
Unable to pursue further.
|
Hint:
Put $$z=\frac{1+i\frac{x}{n}}{1-i\frac{x}{n}}$$ then $z$ will be a $n$-root of unity and solve for $x:$ $$z= \frac{1+i\frac{x}{n}}{1-i\frac{x}{n}}=\exp{\left(i\frac{2k\pi}{n}\right)},\quad k\in\{0,1,...,n-1\}$$
|
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|
Calculate limit in use of integrals. Calculate limit in use of integrals:
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k+n}{3k^2+n^2+1} $$
Solution:
$$\sum_{k=1}^{n} \frac{k+n}{3k^2+n^2+1} = \frac{1}{n} \sum_{k=1}^{n} \frac{\frac{k}{n}+1}{3(k/n)^2+1 + n^{-2}} = \\
\frac{1}{n}\cdot \sum_{k=1}^{n} \frac{\frac{k}{n}+1}{3(k/n)^2+1 +n^{-2}} $$
I want use there approximation by Riemann integrals. But I don't know how to deal with $n^{-2}$
|
For any $\varepsilon>0$ and all $n>1/\sqrt\varepsilon$, we have
$$
\frac{k+n}{3k^2+n^2}\geq
\frac{k+n}{3k^2+n^2+1}\geq
\frac{k+n}{3k^2+(1+\varepsilon)n^2}.
$$
So we have
$$
\int_0^1\frac{x+1}{3x^2+1}\,\mathrm{d}x\geq\lim_{n\to\infty}\frac{k+n}{3k^2+n^2+1}\geq\int_0^1\frac{x+1}{3x^2+(1+\varepsilon)}\,\mathrm{d}x
$$
and now take $\varepsilon\to 0$.
|
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|
How do I see if $a^3+b^3=c^3+d^3$ has any solutions where $ 1 \le a,b,c,d \in \mathbb{Z} \le 1000$ and $a \ne b \ne c \ne d$? How do I see if $a^3+b^3=c^3+d^3$ has any solutions where $ 1 \le a,b,c,d \in \mathbb{Z} \le 1000$ and $a \ne b \ne c \ne d$ ?
I know I can write a program to brute force this and find out, but is there a way I can determine this through algebra?
I thought I could use the Difference of Powers formula:
$(a-c)(a^2+ac+c^2)=a^3-c^3=d^3-b^3=(d-b)(d^2+db+b^2)$.
At this point I am stuck.
|
You are asking about Taxicab Numbers, of which the most famous is
$$1729 = 12^3 + 1^3 = 10^3 + 9^3.$$
There is no (known) algebraic method to solve this equation. All of our current understand of these types of numbers come from computer calculation. There is also an OEIS entry dedicated to them.
If you're curious, we also have found numbers which can be written with $3$+ pairs:
\begin{align*}
87539319 &=167^3 + 436^3 \\
&= 228^3 + 423^3 \\
&= 255^3 + 414^3 \\
6963472309248 &= 2421^3 + 19083^3 \\
&= 5436^3 + 18948^3 \\
&= 10200^3 + 18072^3 \\
&= 13322^3 + 16630^3. \\
\end{align*}
|
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|
Theorem of Pythagoras - Incorrect Derivation I am trying to solve below question from Coursera Intro to Calculus (link)
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and
$2ab$ units respectively, where $a$ and $b$ are positive real numbers
such that $a$ is greater than $b$. Find an exact expression for the
length of the hypotenuse (in appropriate units).
Below are the choices
$(a - b)^2$
$\sqrt(a^4 + 4a^2b^2 -b^4)$
$a^2 + b^2$
$\sqrt(a^2 + 2ab -b^2)$
$(a + b)^2$
When I attempt to work out the solution (and I even got a 2nd pair of eyes to look at it, but he arrived at the same conclusion), I get this:
$(a^2 - b^2)^2 + (2ab)^2 = x^2$
$a^4 -2a^2b^2 + b^4 + (2ab)^2 = x^2$
$a^4 -2a^2b^2 + b^4 + 4a^2b^2 = x^2$
$a^4 + 2a^2b^2 + b^4 = x^2$
$\sqrt(a^4 + 2a^2b^2 + b^4) = x$
Please help! How to get the correct solution?
|
You are done. just multiply out $(a^2 + b^2)^2$ and check that you get $x^2$.
|
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|
Prove for all positive a,b,c that $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$ Prove for all positive a,b,c $$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$$
My Try
I tried taking common denominator of the expression,
$\frac{a^2b+ab^2+b^2c+c^2b+ac^2+a^2c}{abc}$
How to proceed? Is there a way to write them as perfect squares to get the least value? a Hint is much appreciated. Thanks!
|
Using Am-Gm for 6 terms we get: $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\geq 6\sqrt[6]{\frac{a}{b} \frac{b}{a} \frac{b}{c} \frac{c}{b} \frac{c}{a}\frac{a}{c}}= 6$$
|
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|
Calculate the minimum value of $\sum_\mathrm{cyc}\frac{a^2}{b + c}$ where $a, b, c > 0$ and $\sum_\mathrm{cyc}\sqrt{a^2 + b^2} = 1$.
$a$, $b$ and $c$ are positives such that $\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} = 1$. Calculate the minimum value of $$\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b}$$
I have not come up with any ideas to solve the problem yet. I will probably in the near future but right now, I can't.
|
I have an alternative solution, by CS:
$$A=\sum\limits_{cyc} \frac{a^2}{b+c}=\sum\limits_{cyc} \frac{a^4}{a^2b+a^2c}\geq \frac{\left(\sum\limits_{cyc}a^2\right)^2}{\sum\limits_{sym}a^2b} $$
By Jensen:
$$1=\sum\limits_{cyc}\sqrt{a^2+b^2}\leq 3\sqrt{\frac{2}{3}(a^2+b^2+c^2)}$$
So
$$A\geq \frac{\left(\sum\limits_{cyc}a^2\right)^2}{(\sum\limits_{sym}a^2b )* 1}\geq \frac{1}{\sqrt{6}}\frac{\left(\sum\limits_{cyc}a^2\right)^{\frac{3}{2}}}{\sum\limits_{sym}a^2b}$$
If you plug in $a=b=c=\frac{1}{3\sqrt{2}}$ you will get $A=\frac{1}{2\sqrt{2}}$
We will show that $\frac{1}{\sqrt{6}}\frac{\left(\sum\limits_{cyc}a^2\right)^{\frac{3}{2}}}{\sum\limits_{sym}a^2b}\geq \frac{1}{2\sqrt{2}}$
Or
$$4\left(\sum\limits_{cyc}a^2\right)^{3}\geq 3\left(\sum\limits_{sym}a^2b\right)^2$$
Which is true after foiling by Muirhead and Schur
|
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|
Let $A, B$ and $C$ be the angles of an acute triangle. Show that: $\sin A+\sin B +\sin C > 2$. Let $A, B$ and $C$ be the angles of an acute triangle. Show that: $\sin A+\sin B +\sin C > 2$.
I started from considering
$$\begin{align}\sin A+\sin B+\sin (180^o-A-B) &= \sin A+\sin B+\sin(A+B)
\\&=\sin A+\sin B+\sin A\cos B + \cos A\sin B
\\&=\sin A(1+\cos B)+\sin B(1+\cos A).\end{align}$$
How to proceed?
|
Note that since $\sin(x)$ is concave and $\left ( \frac{\pi}{2} , \frac{\pi}{2}, 0 \right ) \succ (A, B, C)$, then by Karamata's Inequality
$$\sin(A) + \sin(B) + \sin(C) > \sin \left ( \frac{\pi}{2} \right ) + \sin \left ( \frac{\pi}{2} \right ) + \sin(0) = 2.$$
|
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|
The value of $I=\int_{|z|=2}z^2\sin\frac{1}{z}dz$? I want to get the value of $I=\int_{|z|=2}z^2\sin\frac{1}{z}dz$.
My idea is to let $t=\frac{1}{z}$, then
$$I=\int_{|z|=2}z^2\sin\frac{1}{z}dz=\int_{|t|=\frac{1}{2}}\frac{\sin t}{t^4}dt=2\pi i \times-\frac{1}{6}=-\frac{\pi i}{3}$$ Is this right?
|
Your result has the wrong sign because you did not take into account that the substitution $t=-\frac{1}{z}$ reverts the orientation of the circle.
You could also apply the residue theorem directly to
$$
z^2 \sin \frac 1z = z^2 \left( \frac 1z - \frac{1}{3!z^3} + \ldots \right) = z -\frac{1}{6z} + \ldots
$$
so that
$$\operatorname{Res}(z^2 \sin \frac 1z, 0) = -\frac 16$$ and therefore
$$
I = 2 \pi i \operatorname{Res}(z^2 \sin \frac 1z, 0) = - \frac{\pi i}{3} \,.
$$
|
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|
Find the sum of $\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$ $$S(x)=\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$$
The question is divided into three parts:
1. Determine its radius of convergence
2. By using the power series of $\frac{1}{1-x}$, show that for all x $\in$ ]-1,1[ , we get $\ln(1-x)=-\sum_{n}^\infty \frac{1}{n}x^n$
3. Find the value of the sum S(x)=$\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$
I was able to solve 1 and 2 but I found some difficulty in 3. I will write how I solved the first two questions and if possible I would like to know if I have made a mistake.*
1. Using the ratio test with $U_n=\frac{1}{n(n+2)}$
$$\frac{U_{n+1}}{U_n}=\frac{n(n+2)}{(n+1)(n+3)}=\frac{n^2+2n}{n^2+4n+3}$$
$$L=\lim\limits_{n \to \infty}(\frac{n^2+2n}{n^2+4n+3})=\lim\limits_{n\to \infty}(\frac{\frac{1}{n^2}(1+\frac{2}{n})}{\frac{1}{n^2}(1+\frac{4}{n}+\frac{3}{n^2})})=1$$
Radius of converge R=1/L $\rightarrow$ R = 1
2.
$$\frac{1}{1-x}=-\frac{dy}{dx}\ln(1-x)$$
and since it is known that $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ for $x \in]-1,1[$ (too lazy to write proof) we can take the integral of $x^n$ to get $\ln(1-x)=-\sum_{n}^\infty \frac{1}{n}x^n$
$$-\int_{0}^{x} x^n = -\frac {x^{n+1}}{n+1} \rightarrow -\sum_{n=0}^\infty\frac {x^{n+1}}{n+1}= -\sum_{n=1}^\infty\frac {x^n}{n}$$
3.
$$S(x)=\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$$
I took
$$U_n=\frac{1}{n(n+2)}=\frac{A}{n}+\frac{B}{n+2}$$
I got
$$U_n=\frac{1}{n(n+2)}=\frac{1}{2n}+\frac{-1}{2(n+2)}$$
$\therefore$
$$S(x)=\sum_{n=1}^\infty (\frac{1}{2n}+\frac{-1}{2(n+2)})x^n$$
First, I started to solve $\sum_{n=1}^\infty (\frac{1}{2n})x^n$, and by using the proof of question n.2 I got: $$\sum_{n=1}^\infty (\frac{1}{2n})x^n=\frac{1}{2}\sum_{n=1}^\infty (\frac{1}{n})x^n=\frac{1}{2}(-ln(1-x))=ln(\frac{1}{\sqrt{1-x}})$$
However, I was not able to solve the second part: $$\sum_{n=1}^{\infty}\frac{-1}{2(n+2)}x^n$$
How can I find the sum of the second part to find S(x)? I appreciate the help and I would be glad to know if I have made any mistakes too.
|
Note that $$\sum \limits_{n=1}^{\infty} \frac{-1}{2(n+2)}x^n = \frac{1}{x^2}\sum \limits_{n=1}^{\infty} \frac{-1}{2(n+2)}x^{n+2} = -\frac{1}{x^2}(\frac{1}{2}x + \frac{1}{4}x^2 +\sum \limits_{n=1}^{\infty} \frac{1}{2n}x^n).$$
But you have already found $\sum \limits_{n=1}^{\infty} \frac{1}{2n}x^n$, so the rest is straightforward.
|
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|
Simplifying $\sum_{cyc}\tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)$. I get $0$, but the answer is $\pi$. So the question is
$$ \tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)+\tan^{-1}\left(\sqrt{\frac{y(x+y+z)}{xz}}\right)+\tan^{-1}\left(\sqrt{\frac{z(x+y+z)}{yx}}\right) =\ ? $$
So my take on the question is to rewrite it as
$$ \tan^{-1}\left(x\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(y\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(z\sqrt{\frac{(x+y+z)}{yzx}}\right) $$
Then say $$\frac{x+y+z}{yzx}= a^2.$$
We get
$$ \tan^{-1}\left( \frac{a((x+y+z)-a^2xyz)}{1-a^2(xy+yz+zx)}\right)$$
And since $ (x+y+z) = a^2xyz $ , this is just equal to $\tan^{-1}(0)= 0 $ but the answer given is $\pi.$
|
Like Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$\tan^{-1}\sqrt{\dfrac{x(x+y+z)}{yz}}+\tan^{-1}\sqrt{\dfrac{y(x+y+z)}{zx}}$$ $$=\begin{cases} \tan^{-1}\left(\dfrac{\sqrt{\dfrac{x(x+y+z)}{yz}}+\sqrt{\dfrac{y(x+y+z)}{zx}}}{1-\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}}\right) &\mbox{if } \sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}<1 \ \ \ \ (1) \\
\pi+\tan^{-1}\left(\dfrac{\sqrt{\dfrac{x(x+y+z)}{yz}}+\sqrt{\dfrac{y(x+y+z)}{zx}}}{1-\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}}\right) & \mbox{if } \sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}>1 \ \ \ \ (2) \end{cases} $$
Now $R=\dfrac{\sqrt{\dfrac{x(x+y+z)}{yz}}+\sqrt{\dfrac{y(x+y+z)}{zx}}}{1-\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}}=\sqrt{\dfrac{x+y+z}{xyz}}\cdot\dfrac{|z|(|x|+|y|)}{|z|-|x+y+z|}$
If $|x+y|,|z|,|x+y+z|\ge0,$ $$R=-\sqrt{\dfrac{x+y+z}{xyz}}\cdot z=-\sqrt{\dfrac{z(x+y+z)}{xy}}$$
Again $(2)$ will hold true if $\sqrt{\dfrac{x(x+y+z)}{yz}}\cdot\sqrt{\dfrac{y(x+y+z)}{zx}}>1 \iff (x+y+z)^2>z^2$
which is true if $x,y,z>0$
Finally $\tan^{-1}(-u)=-\tan^{-1}u$
|
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|
Proving $-\frac{1}{a}<\int_a^b \sin(x^2) dx<\frac{1}{a}$ I have encountered a question:
Prove
$$-\frac{1}{a}<\int_a^b \sin(x^2) dx<\frac{1}{a}$$
There are plenty of solutions to $\int_0^{\infty} \sin(x^2) dx$ online, but there seems to be no solution to the boundary of $\int_a^b \sin(x^2) dx$.
Could anyone help me with this please? I tried to calculate the integral directly, but I cannot cancel out b and get a boundary only with $a$.
Applying inequality to the integrand $\sin(x^2)<x^2$ does not work either.
|
By change of variables $u=x^2$, we have $\displaystyle{\int_{a}^{b}\sin(x^2)\,{\rm d}x=\frac{1}{2}\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}\,{\rm d}u}$.
Now, by integration by parts we have
$$\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}\,{\rm d}u=-\int_{a^2}^{b^2}\frac{1}{\sqrt{u}}{\rm d}(\cos u)=\frac{\cos(a^2)}{a}-\frac{\cos(b^2)}{b}-\frac{1}{2}\int_{a^2}^{b^2}\frac{\cos u}{u^{3/2}}{\rm d}u$$
Now, note that $$\left|\frac{\cos(a^2)}{a}\right|\leq \frac{1}{a},$$ $$\left|\frac{\cos(b^2)}{b}\right|\leq \frac{1}{b}$$ and $$\frac{1}{2}\left|\int_{a^2}^{b^2}\frac{\cos u}{u^{3/2}}{\rm d}u\right|< \frac{1}{2}\int_{a^2}^{b^2}\frac{1}{u^{3/2}}{\rm d}u=\frac{1}{a}-\frac{1}{b}.$$
Hence, $\displaystyle{\left|\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}{\rm d}u\right|<\frac{1}{a}+\frac{1}{b}+\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{2}{a}}$.
We conclude $$\left|\int_{a}^{b}\sin(x^2)\,{\rm d}x\right|=\frac{1}{2}\left|\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}{\rm d}u\right|<\frac{1}{a} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$
How to evaluate $$\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$$
I tried to integrate by parts, but no way so far, help me, thanks.
|
different approach to evaluate $\displaystyle\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx$ :
from here , we have $\displaystyle\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1-x)}{x}\ dx=\frac{\pi^3}{16}\tag{1}$
and from here , we have $\displaystyle \ 3\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=0\tag{2}$
by combining $(1)$ and $(2)$, we obtain that $\displaystyle\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=3\int_0^1\frac{\arctan x\ln(1-x)}{x}\ dx+\frac{3\pi^3}{32}\tag{3}$
we have
\begin{align}
\int_0^1 \frac{\arctan x\ln(1-x)}{x}\ dx&=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1 x^{2n}\ln(1-x)\ dx\\
&=-\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}=-\text{Im}\sum_{n=1}^\infty\frac{i^nH_n}{n^2}\\
\end{align}
and using the generating function with $\ x=i$ $$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$
we get $\ \displaystyle\int_0^1 \frac{\arctan x\ln(1-x)}{x}\ dx=\frac{\pi}{16}\ln^22+\frac12G\ln2+\text{Im}\operatorname{Li}_3(1-i)\tag{4}$
plugging $(4)$ in $(3)$, we get $$\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2+3\text{Im}\operatorname{Li}_3(1-i)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Ordinary Differential Equation getting two different answers Solve $y(y^2-2x^2)dx+x(2y^2-x^2)dy=0$ and find a particular curve passing through $(1,2)$
My attempt:
1st Solution: Rewrite as
$y(y^2dx+x2ydy)-x(x^2dy+y2xdx)=0$
$\implies xy^2d(xy^2)-x^2yd(x^2y)=0$ (multiplying $xy$)
$\implies (xy^2)^2-(x^2y)^2 = c\,\,\,$ Or $x^2y^2(y^2-x^2) = c$
$\implies x^2y^2(y^2-x^2) = 12 $
2nd Solution: Rewrite as
$\frac{dy}{dx}=\frac{2\frac yx-(\frac yx)^3}{2(\frac yx)^2-1}$
Let $y=tx\implies \frac{dy}{dx}=t+x\frac{dt}{dx}=\frac{2t-t^3}{2t^2-1}$
$\implies x\frac{dt}{dx}=\frac{3(t-t^3)}{2t^2-1}$
$\implies \int\frac{2t^2-1}{t^3-t}dt+\int\frac 3x dx=0$
$\implies \int \left(\frac 2t+\frac{1}{(t-1)}+\frac{1}{(t+1)}\right)dt+\int\frac 6x dx = 0 $
$\implies \ln|t^2(t^2-1)x^6| = ln c $
$\implies |t^2(t^2-1)x^6| = c $
$\implies |x^2y^2(y^2-x^2)| = 12 $
$\implies x^2y^2(y^2-x^2) = \pm 12 $
Where I am doing wrong in second or in first solution!
|
I got $$\frac{y}{x}\left(\frac{y^2}{x^2}-2\right)-\left(2\frac{y^2}{x^2}-1\right)y'(x)=0$$ now let $$\frac{y}{x}=u$$ and then we get
$$-u^3-u=u'x(2u^2-1)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $(\sqrt{3}+i)^{14}+(\sqrt{3}-i)^{14}$ Evaluate $(\sqrt{3}+i)^{14}+(\sqrt{3}-i)^{14}$
I tried by using the De'moivers theorem but I didn't get proper value I get a mess value 4..but I am not sure about the answer can anyone please tell me
|
Observe that
$$
(\sqrt{3} + i)^{14} = ((\sqrt{3} + i)^2)^7
$$
Also, $ (\sqrt{3} + i)^2 = 2 + i 2\sqrt{3} = 2 (2 e^{j\theta}) = 4 e^{j\theta}$, where $ \theta = \pi / 3 $.
Similarly, for the second term we have that $ (\sqrt{3} - i)^2 = 4 e^{-j\theta} $.
You problem reduces to finding
$$
M = 4^7 \cdot e^{j 7\theta} + 4^7 \cdot e^{-j 7\theta}
= 4^7 \cdot 2 \cdot (\frac{e^{j 7\theta} + e^{-j 7\theta}}{2})
= 4^7 \cdot \cos(7\theta)
= 4^7 \cdot 1
= 16384
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find maximum and minimum of $\sin x + \sin y$ I am working on my scholarship exam practice but I am stuck on finding the minimum. Pre-university maths background is assumed.
When $x + y = \frac{2\pi}{3}, x\geq0, y\geq0$, the maximum of
$\sin x+\sin y$ is ....., and the minimum of that is .....
Let me walk you through what I have got.
$\sin x+\sin y = 2\sin (\frac{x+y}{2})\cos (\frac{x-y}{2})$
By substituting $x + y = \frac{2\pi}{3}$ into the sine function, we have
$\sin x+\sin y = 2\sin (\frac{2\pi}{3\cdot2})\cos (\frac{x-y}{2})$
$\sin x+\sin y = \sqrt{3}\cos (\frac{x-y}{2})$
To find the maximum and minimum, we know that
$-1 \leq\cos (\frac{x-y}{2})\leq1$
$-\sqrt{3} \leq\sqrt{3}\cos (\frac{x-y}{2})\leq\sqrt{3}$
Hence, the maximum is $\sqrt{3}$ which is correct and in accordance with the answer key.
However, it seems that the minimum equals to $-\sqrt{3}$ is incorrect. The answer key provided is $\frac {\sqrt{3}}{2}$. Could you please elucidate how I can get to this answer? My guess is something to do with the condition $x\geq0$ and $y\geq0$ given by the question.
|
For the minimum, note that since $x,y\ge0\implies y\le\dfrac{2\pi}3$, we have $$\dfrac{x-y}2=\dfrac{x+y-2y}2=\dfrac{\dfrac{2\pi}3-2y}2=\frac\pi3-y$$ so $$\sin x+\sin y = \sqrt{3}\cos\frac{x-y}{2}=\sqrt3\cos\left(\frac\pi3-y\right)\ge\begin{cases}\sqrt3\cos\left(\frac\pi3-0\right)\\\sqrt3\cos\left(\frac\pi3-\frac{2\pi}3\right)\end{cases}=\frac{\sqrt3}2.$$
|
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|
Does $\int_{0}^{\infty} \sin(x) \cdot\sin(x^2)\,dx$ converge? My try: Let $x = \sqrt{t}$, then $dx = \frac{1}{2\sqrt{t}}dt$. We get the following integral: $\int_{0}^{\infty}\frac{\sin(t)\sin(\sqrt{t})}{2\sqrt{t}}dt$. Now I tried to use Dirichlet's test: The function $g(x) = 1/2\sqrt{t}$ has limit $0$ at infinity and it is monotonically decreasing. Now if I could show that the function $F(b) = \int_{0}^{b} \sin(t)\sin(\sqrt{t})$ is bounded, that would mean the integral converges by Dirichlet's test, but I don't know how to prove it. Suggestions?
|
As usual, trig identities are the answer:
\begin{align}
\int_0^\infty \sin(x^2)\sin(x)dx =& \frac{1}{2}\int_0^\infty\left[ \cos(x^2-x)-\cos(x^2+x)\right]dx \\ =& \frac{1}{2}\int_0^\infty\left( \cos\left[\left(x-\frac{1}{2}\right)^2 -\frac{1}{4}\right]-\cos\left[\left(x+\frac{1}{2}\right)^2 -\frac{1}{4}\right]\right)dx
\\ = &\frac{1}{2}\left[\int_{-1/2}^\infty\cos\left(x^2 -\frac{1}{4}\right)dx-\int_{1/2}^\infty\cos\left(x^2 -\frac{1}{4}\right)dx\right]
\end{align}
and these integrals both converge, since the integrals of $\sin(x^2)$ and $\cos(x^2)$ both converge, and $\cos(x^2-1/4) = \cos(1/4)\cos(x^2)+\sin(1/4)\sin(x^2)$. Thus, the original integral converges as well.
The difference of integrals can be simplified to
\begin{align}
\int_0^\infty \sin(x^2)\sin(x)dx = &\int_0^{1/2}\cos\left(x^2-\frac{1}{4}\right)dx
\\= &\cos\left(\frac{1}{4}\right)\int_0^{1/2}\cos(x^2)dx + \sin\left(\frac{1}{4}\right)\int_0^{1/2}\sin(x^2)dx
\\ = &\sqrt{\frac{\pi}{2}}\left[\cos\left(\frac{1}{4}\right)C\left(\frac{1}{\sqrt{2\pi}}\right) + \sin\left(\frac{1}{4}\right)S\left(\frac{1}{\sqrt{2\pi}}\right)\right],
\end{align}
where $S$ and $C$ are the Fresnel sine and cosine integrals, respectively.
|
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|
$\sqrt{2} x^2 - \sqrt{3} x +k=0$ with solutions $\sin\theta$ and $\cos\theta$, find k
If the equation $\sqrt{2} x^2 - \sqrt{3} x +k=0$ with $k$ a constant has two solutions $\sin\theta$ and $\cos\theta$ $(0\leq\theta\leq\frac{\pi}{2})$, then $k=$……
My approach is suggested below but I am not sure how to continue.
Since $\sin\theta$ and $\cos\theta$ are two solutions of the equation,
Then we have,
$\sqrt{2} \sin^2\theta - \sqrt{3} \sin\theta +k=0$ .....Equation (1)
$\sqrt{2} \cos^2\theta - \sqrt{3} \cos\theta +k=0$ .....Equation (2)
Add (2) to (1),
$\sqrt{2} (\sin^2\theta + \cos^2\theta) - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$
$\sqrt{2} - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$
The answer key provided is $\frac{\sqrt{2}}{4}$. I think I am probably on the right track here but not sure how I should proceed with $\sin\theta$ and $\cos\theta$ next. Please help.
|
By the Viete we obtain: $$1=\left(\sqrt{\frac{3}{2}}\right)^2-\sqrt2k.$$
Can you end it now?
|
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|
Given remainders from other polynomial divisions. Find the remainder in a polynomial division Let $F(x)$ be a polynomial.
If $F(x)$ is divided by $(x-1)^2$ the remainder will be $x+1$
and if $F(x)$ is divided by $x^2$ the remainder will be $2x+3$.
What is the remainder if $F(x)$ is divided by $x^2(x-1)$?
My solution :
$F(x) = (x-1)^2 P(x) + x+1$, substitute $x=1$ and get $F(1) = 2$.
Differentiate it and get $F'(x) = (x-1)^2 P'(x)+P(x)(2)(x-1)+1$ then $F'(1)=1$.
And do the same to the other equation $( F(0)=3, F'(0) = 2 )$.
Let the remainder of $F(x)$ divided by $x^2(x-1)$ be $R(x)$. $( R(x) = ax^2+bx+c)$
so $F(x) =x^2(x-1)A(x) + R(x)$
$F(0) = R(0) = 3$
$F(1) = R(1) = 2$
$F'(x) = (x)^2(x-1) A'(x)+x^2A(x)+2x(x-1)A(x) + R'(x)$
$F'(0) = R'(0) = 2$
from $R(0) = 3$ then $c=3$
from $R'(0) = 2$ then $R'(x) = 2ax+b$ then $R'(0) = b = 2$
$R(x) = ax^2+2x+3$ from $R(1) = 2$, will get $a+5 = 2 , a=-3$
so the remainder will be $-3x^2+2x+3$
but one said that the answer can be $-3x^2+2x+1$ too, so what's correct answer.
|
Your answer is correct.
The second answer cannot be also true, as $(-3x^2 + 2x + 3) - (-3x^2 + 2x + 1) =2 $ is not divisible by $(x-1)x^2$.
Here is another elementary solution:
*
*$(1):F(x) = (x-1)^2p(x) + x+1$
*$(2):F(x) = x^2q(x) + 2x+3$
*Setting $p(x) =x^2f(x)+ax+b$ you get
$$(x-1)^2p(x) + x+1 = x^2q(x) -2bx+ax+b+x+1 $$
$$\stackrel{(2)}{\Rightarrow} b+1 = 3, 1+a-2b = 2\Rightarrow \boxed{b=2, a= 5}$$
$$\Rightarrow p(x) =x^2f(x)+5x+2$$
Plugging this into $(1)$, you need only consider
$$(x-1)^2(5x+2) + x+1 = 5x^3-8x^2+2x+3 $$
$$= 5(x^3 - x^2) \boxed{- 3x^2+2x+3}$$
|
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|
Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$ Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$
Start by squaring both sides
$$\frac{-4x^2-x+14}{(x-1)^2}<0$$
Factoring and multiplied both sides with -1
$$\frac{(4x-7)(x+2)}{(x-1)^2}>0$$
I got
$$(-\infty,-2)\cup \left(\frac{7}{4},\infty\right)$$
Since $x\leq2$ then
$$(-\infty,-2)\cup \left(\frac{7}{4},2\right]$$
But the answer should be $(-\infty,1)\cup \left(\frac{7}{4},2\right]$. Did I missed something?
|
Define $f(x)=\frac{3\sqrt{2-x}}{x-1}-2$. Being continuous on its domain $(-\infty,2]\setminus\{1\}$,the function may change its sign only at its zero $7/4$ or at its singularity, namely at $1$. Now check the sign of $f$ in the corresponding intervals $(-\infty,1)$, $(1,7/4)$ and $(7/4,2]$; you want $f(x)<0$.
|
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|
Find remainder of division of $x^3$ by $x^2-x+1$ I am stuck at my exam practice here.
The remainder of the division of $x^3$ by $x^2-x+1$ is ..... and that of $x^{2007}$ by $x^2-x+1$ is .....
I tried the polynomial remainder theorem but I am not sure if I did it correctly.
By factor theorem definition, provided by Wikipedia,
the remainder of the division of a polynomial $f(x)$ by a linear polynomial $x-r$ is equal to $f(r)$.
So I attempted to find $r$ by factorizing $x^2-x+1$ first but I got the complex form $x=\frac{1\pm\sqrt{3}i}{2}=r$.
$f(r)$ is then $(\frac{1+\sqrt{3}i}{2})^3$ or $(\frac{1-\sqrt{3}i}{2})^3$ which do not sound right.
However, the answer key provided is $-1$ for the first question and also $-1$ for the second one. Please help.
|
Another way if you're familiar with modular arithmetic is to work modulo $x^2-x+1$, in which case we have $x^2\equiv x-1$ and thus
$$x^3\equiv xx^2\equiv x(x-1)\equiv x^2-x\equiv (x-1)-x\equiv -1.$$ This can be extended to your other question by noting that $x^{2007}=\left(x^3\right)^{669}$.
|
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|
Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+7=(1-\sqrt{x+2})^2$ # square both sides
(Use perfect square formula on right hand side $a^2-2ab+b^2$)
$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$ # lhs radical is removed, rhs use perfect square formula
$3x+7=1+2(\sqrt{x+2})+x+2$ # simplify
$3x+7=x+3+2\sqrt{x+2}$ # keep simplifying
$2x+4=2\sqrt{x+2}$ # simplify across both sides
$(2x+4)^2=(2\sqrt{x+2})^2$
$4x^2+16x+16=4(x+2)$ # now that radical on rhs is isolated, square both sides again
$4x^2+12x+14=0$ # a quadratic formula I can use to solve for x
For use int he quadratic function, my parameters are: a=4, b=12 and c=14:
$x=\frac{-12\pm\sqrt{12^2-(4)(4)(14)}}{2(4)}$
$x=\frac{-12\pm{\sqrt{(144-224)}}}{8}$
$x=\frac{-12\pm{\sqrt{-80}}}{8}$
$x=\frac{-12\pm{i\sqrt{16}*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{4i*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{-4\sqrt{5}}}{8}$ #since $4i*i\sqrt{5}$ and i^2 is -1
This is as far as I get:
$\frac{-12}{8}\pm\frac{4\sqrt{5}}{8}$
I must have gone of course somewhere further up since the solution is provided as x=-2.
How can I arrive at -2?
|
Hint: Writing your equation in the form
$$\sqrt{3x+7}=1+\sqrt{x+2}$$
squaring gives
$$3x+7=1+x+2+2\sqrt{x+2}$$ so
$$x+2=\sqrt{x+2}$$ squaring again:
$$(x+2)^2=x+2$$
Can you finish?
|
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|
Determine fractional transformation which maps $\mathbb{R}$ to $\mathbb{R}$ and $S^1$ to $S^1$ I want to determine all fractional transformation $F:\mathbb{C}\to\mathbb{C}$ which maps $\mathbb{R}$ on $\mathbb{R}$ and the unit circle on the unit circle. Let $\begin{pmatrix}a&b\\c&d
\end{pmatrix}$ be such a linear transformation.
For $x\in\mathbb{R}$, we must have that $\frac{ax+b}{cx+d}\in\mathbb{R}.$ Then $$\frac{ax+b}{cx+d}=
\frac{\overline{ax+b}}{\overline{cx+d}}=\frac{\overline{a}x+\overline{b}}{\overline{c}x+\overline{d}}.
$$
Can we conclue from this that $\begin{pmatrix}a&b\\c&d\end{pmatrix}=\lambda\begin{pmatrix}\overline{a}&\overline{b}\\\overline{c}&\overline{d}\end{pmatrix}$ for some $\lambda\in\mathbb{C}$ met $|\lambda|=1$. Then also for $y\in S^1$, we have that $$\bigg|\frac{ay+b}{cy+d}\bigg|=1$$
how can I continue?
|
As you have noted in your comment, $F$ either fixes $1$ and $-1$ or swaps them.
Replacing $F$ by $-F$ if necessary, we can assume $F$ fixes $1$ and $-1$. Then we have
$$
\frac{a\cdot 1 + b}{c \cdot 1 + d} = 1 \quad\quad \frac{a\cdot -1 + b}{c \cdot -1 + d} = -1
$$
So:
$
a + b = c + d
$
and
$
b - a = c - d
$
Adding these equations gives $2b = 2c$ and subtracting them gives $2a = 2d$.
If $b = 0$, $F(z) = z$, otherwise we can divide through by $b$ and so assume $b = 1$. Thus the matrix of $F$ is either
$$
\begin{pmatrix}1&0\\
0 & 1\end{pmatrix} \quad\mbox{or} \quad
\begin{pmatrix}a&1\\
1 & a\end{pmatrix}
$$
for some $a$. However, $F$ maps the unit circle to itself, and as $F$ maps circles to circles (or lines), this will hold iff $|F(i)| = 1$, i.e., iff
$
|ai + 1| = |i + a|
$
which holds iff
$
(ai+1)(\overline{ai+1}) = (i + a)(\overline{i + a})
$
but
$$
\begin{align*}
(ai+1)(\overline{ai+1}) &= a\overline{a} + (a - \overline{a})i + 1
\\
(i + a)(\overline{i + a}) &= a\overline{a} + (\overline{a} - a)i + 1
\end{align*}
$$
So $|F(i)| = 1$ iff $a - \overline{a} = 0$, i.e., iff $a \in \Bbb{R}$.
We conclude that the linear transformation $F$ maps $\Bbb{R}$ to $\Bbb{R}$ and $S^1$ to $S^1$ iff its matrix is either
$$
\pm \begin{pmatrix}1&0\\
0 & 1\end{pmatrix} \quad\mbox{or} \quad
\pm \begin{pmatrix}a&1\\
1 & a\end{pmatrix}
$$
for some $a \in \Bbb{R}$. Note that this is degenerate when $a = 1$. For $a \neq 1$ it maps $\Bbb{R}$ onto $\Bbb{R}$ and $S^1$ onto $S^1$
|
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|
Prime factor inversion Define a function $f(n)$ for $n \in \mathbb{N}$ to "invert" the prime
factorization of $n$ in the following sense. Let me start with an example.
If $n= 3564 = 2^2 \cdot 3^4 \cdot 11^1$,
then $f(n) = 2^2 \cdot 4^3 \cdot 1^{11} = 256$,
inverting the base primes and their exponents.
In general, if the prime
factorization of $n$ is $p_1^{m_1} \cdot \cdots \cdot p_k^{m_k}$,
then $f(n) = m_1^{p_1} \cdot \cdots \cdot m_k^{p_k}$.
Continuing the above example:
\begin{eqnarray}
f(n) &= f(3564 = 2^2 \cdot 3^4 \cdot 11^1) & = 2^2 \cdot 4^3 \cdot 1^{11} = 256\\
f^2(n)&=f(256=2^8) &= 8^2 =64 \\
f^3(n)&=f(64 = 2^6) &= 6^2 = 36 \\
f^4(n)&=f(36 = 2^2 \cdot 3^2) &= 2^2 \cdot 2^3 = 32\\
f^5(n)&=f(32 = 2^5) &= 5^2 = 25\\
f^6(n)&=f(25 = 5^2) &= 2^5 = 32
\end{eqnarray}
and we have fallen into a $2$-cycle. My questions concern the iterated behavior of $f(n)$.
*
*Q1. If $n= \Pi p_i$, $p_i$ primes, then $f(n) = 1$.
If $n= \Pi p_i^{p_i}$, $p_i$ primes, then $f(n) = n$, a fixed point.
Are any other $n$ fixed points or $n$ that eventually lead to $1$-cycles?
*Q2. If $n= \Pi p_i^{q_i} \cdot q_j^{p_j}$, all of $p_i,p_j,q_i,q_j$ primes, then $f^2(n)=n$, a $2$-cycle.
Can you see a characterization of those $n$ that eventually fall into a $2$-cycle?
*Q3. Are there any $n$ that eventually fall into $k$-cycles
for $k \ge 3$?
|
Q1: Yes, for example $p^q q^p$, where $p$ and $q$ are primes, is a fixed point (or products that consist of these). More generally you should be able to use permutations of order $2$, i.e products of disjoint transpositions.
|
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|
Integrating quadratics in denominator I'm following a book on Calculus that introduces partial fraction expansion. They discuss common outcomes of the partial fraction expansion, for example that we are left with an integral of the form:
$$
\int \frac{dx}{x^2+bx+c}
$$
And then we can use complete the square and $u$-substitution:
$$
x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right) = u^2+\alpha^2
$$
where $u=x+\frac{b}{2}$ and $\alpha=\frac{1}{2}\sqrt{4c-b^2}$. The book says: "... this is possible because $4c-b^2>0$."
Eagerly I tried an example, using the quadratic $x^2-8x+1$.
Then let $a=1, b=-8, c=1$ and:
$$
x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right) = u^2 + \alpha^2
$$
where $u = x+b/2 = x-4$ but we run into a problem:
$$\alpha=\frac{1}{2}\sqrt{4c-b^2} = \frac{1}{2}\sqrt{(4)(1) - (-8)^2} = \frac{1}{2}\sqrt{4-64}$$
So $\alpha$ doesn't satisfy $4c-b^2>0$. Maybe I'm missing something obvious? Or is the book missing a caveat that this method doesn't always work. Because in the book they make it sound like "... this is possible because $4c-b^2>0$." is always true.
|
For integrating the Quadratics in the denominator, there are two cases,
$b^2-4c>0:$
$x^2+bx+c= (x+\frac{b}{2})^2-\frac{b^2-4c}{4} = u^2 - \beta^2 \Rightarrow $ This will give you: $\frac{1}{2 \beta} \ln \frac{u- \beta} {u+ \beta}; (\beta> 0)$
$4c-b^2>0:$
$x^2+bx+c= (x+\frac{b}{2})^2+\frac{4c-b^2}{4} = u^2 + \alpha^2 \Rightarrow $ This will give you: $\frac{1}{\alpha} \arctan \frac{u} {\alpha}; (\alpha >0)$
Your denominator x^2-8x+1 belongs to FIRST case.
|
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|
Form groups of people using combinatorics I have this problem:
How many groups of 5 people can be formed between 4 boys and 7 girls
if there should be at least 2 girls included?
My attempt was:
Ways of select two girls of a total of $7$ is: $\binom{7}{2} = 21$
Ways to select the $3$ remaining people, out of a total of $4$ boys and $5$ girls, is: $\binom{9}{3} = 84$
Therefore, $21 \cdot 84 = 1764$, but according to the guide my answer is incorrect.
What is wrong with my development? Thanks in advance.
|
There are four cases, with 2, 3, 4, or 5 girls in the 5-group.
The solution is:
$$ {4 \choose 3}{7 \choose 2} +
{4 \choose 2}{7 \choose 3} +
{4 \choose 1}{7 \choose 4} +
{4 \choose 0}{7 \choose 5} = 455
$$
What you have counted is :
$$ {4 \choose 3}{7 \choose 2} {\color{red} {2 \choose 2}} +
{4 \choose 2}{7 \choose 3} {\color{red} {3 \choose 2}} +
{4 \choose 1}{7 \choose 4} {\color{red} {4 \choose 2}} +
{4 \choose 0}{7 \choose 5} {\color{red} {5 \choose 2}} = 1764
$$
meaning that between the choosen girls there are two special, i.e. the two ones selected in the beggining to ensure the quota.
|
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|
Solve $x^{3} = 6+ 3xy - 3 ( \sqrt{2}+2 )^{{1}/{3}} , y^{3} = 9 + 3xy(\sqrt{2}+2)^{{1}/{3}} - 3(\sqrt{2}+2)^{{2}/{3}}$ Solve the system of equations for $x,y \in \mathbb{R}$
$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}} $
$ y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}$
I just rearranged between those equations and get $ \frac{y^{3}-9}{x^{3} -6} = (\sqrt{2}+2)^{\frac{1}{3}}$ then I don't know how to deal with it.
Please give me a hint or relevant theorem to solve the equation.
Thank you, and I appreciate any help.
Furthermore I get an idea
how about we subtract two equation and get
$y^{3}-x^{3} = 3 + 3xy((\sqrt{2}+2)^{\frac{1}{3}} -1) - (3(\sqrt{2}+2)^{\frac{2}{3}} - 3(\sqrt{2}+2)^{\frac{1}{3}})$
$(y-x)(x^2+xy+y^2)= 3[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)]$
$y-x = 3$ and $x^2 +xy+y^2 =[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $
or,
$y-x = [1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $ and $x^2 +xy+y^2 = 3$
Am I on the right track?
|
$$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}$$
$$y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}} $$
Let $A=\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}$, then re-write the equations as
$$x^{3} = 6+ 3xy - 3A\tag{1}$$
$$y^{3} = 9 + 3xyA - 3A^2\tag{2} $$
Solving (1) for $3xy$
$$ 3xy=x^3+3A-6 \tag{3}$$
Substituting (3) into (2) gives
$$ y^3=Ax^3-6A+9 \tag{4}$$
Cubing (3) and combining with (4) gives
$$ 27x^3(Ax^3-6A+9)=(x^3+3A-6)^3 $$
Let $x^3=u,\,6A-9=B$, and $3A-6=C$. Then equation (5) becomes
$$ 27u(Au-B)=(u+C)^3 $$
$$ u^3+(3C-27A)u^2+(3C^2+27B)u+C^3=0 $$
Solving for $c$ and the cube roots of $c$ there is only one real solution $(3.555,4.074)$ [for which I cheated and used desmos].
|
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|
Calculating $H'(x)$ given $H(x) = \int_{x^3 + 1}^{x^2 + 2x} e^{-t^2} dt$ Given $\displaystyle H(x) = \int_{x^3 + 1}^{x^2 + 2x} e^{-t^2} dt$, we want to find $H'(x)$.
First, we rewrite $H(x)$ as follows:
$$\begin{align}
&= \int_0^{x^2 + 2x} e^{-t^2} dt + \int_{x^3 + 1}^0 e^{-t^2} dt \qquad &\text{Properties of integrals} \\
&= \int_0^{x^2 + 2x} e^{-t^2} dt - \int_{0}^{x^3 + 1} e^{-t^2} dt \qquad &\text{Definition of backwards integrals} \tag{1}
\end{align}
$$
Next, we'll define $\displaystyle F(x) = \int_0^x e^{-t^2} dt$.
We know its derivative is $F'(x) = e^{-x^2}$, by the Fundamental Theorem of Calculus.
Next, we'll define new functions for the two integrals in $(1)$:
$$\begin{align*}
H_1(x) &= \int_0^{x^2 + 2x} e^{-t^2} dt &\qquad H_2(x) &= \displaystyle\int_{0}^{x^3 + 1} e^{-t^2} dt \\
&= F(x^2 + 2x)& &=F(x^3 + 1)
\end{align*}$$
We use the chain rule to find their derivatives:
$$ H_1'(x) = e^{-(x^2 + 2x)^2} (2x + 2) \qquad H_2'(x) = e^{-(x^3 + 1)^2} (3x) $$
Therefore,
$$H'(x) = e^{-(x^2 + 2x)^2} (2x + 2) - e^{-(x^3 + 1)^2} (3x)$$
Is my calculation correct?
|
Use Lebnitz rule:
$$\frac{d}{dx} \int_{L(x)}^{U(x)} f(t) dt= U'(x) f(U(x)) -L'(x) f(L(x))$$
So in your case you get $$H'(x)=(2x+2) e^{-(x^2+2x)^2}-3x^2 e^{-(x^3+1)^2}.$$
|
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|
Let $x, y, z$ be real numbers. If $x + y + z = 1$ and $x^2 + y^2 + z^2 = 1$, then what is the minimum value of $x^3 + y^3 + z^3$? Through manipulation, I got as far as $x^3 +y^3 + z^3 = 1 + 3xyz$
I tried solving it using AM - GM inequality but it only gave me the maximum value.
GM - HM does not to help either.
|
The intersection of sphere and plane is a smooth curve, one can apply standard optimization methods.
With the Lagrange functional
$$
L(x,y,z,\lambda,\mu)=x^3+x^3+z^3+λ(x^2+y^2+z^2-1)+μ(x+y+z-1)
$$
one gets the equilibrium conditions
\begin{align}
3x^2+2λx+μ&=0\\
3y^2+2λy+μ&=0\\
3z^2+2λz+μ&=0
\end{align}
As now $x,y,z$ are solutions of the same quadratic polynomial, two of them have to be equal, wlog. $y=z$. Then from the original equations $x=1-2y$ so that
$$
(1-2y)^2=x^2=1-2y^2\\
6y^2-4y=0\\
y=z=0\lor y=z=\frac23
$$
which gives $x=0$ in the first case with cubic sum $1$ and in the second case $x=-\frac13$ with cubic sum $\frac{5}{9}$. These are the extremal values that the cubic sum function takes on the circle given by the constraints.
|
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|
Radius of convergence of $\sum_{n=0}^\infty a_n x^n$, with $a_{n+2} = \frac{n(n+1) a_{n+1} - a_n}{(n+2)(n+1)}, a_2 = -a_0/2$ Problem
Find the radius of convergence of the power series
$$
\sum_{n=0}^\infty a_n x^n
$$
where $a_n$'s are defined by the following recurrence relation
$$
\begin{aligned}
a_{n+2} &= \frac{n(n+1) a_{n+1} - a_n}{(n+2)(n+1)}, n\ge 1 \\[8pt]
a_2 &= -a_0/2
\end{aligned}
$$
with arbitrary $a_0, a_1$.
Try1
I have tried to directly applying the ratio test, by dividing the above recurrence relation by $a_{n+1}$,
$$
r_{n+1} = \frac{n}{n+2} - \frac{1}{(n+2)(n+1)} \frac{1}{r_n}
$$
where $r_n := a_{n+1}/a_n$. We can observe that
$$
r_n r_{n+1} = \frac{n}{n+2} r_n - \frac{1}{(n+2)(n+1)}
$$
but I cannot proceed to find the expression for
$$
\lim_{n \to \infty} \vert r_n \vert
$$
Try2
Since the recurrence relation depends on the arbitrary choice pf $a_0, a_1$, let us proceed by letting $a_0:= 0$. We have
$$
\begin{aligned}
a_2 &= a_0 = 0 \\[7pt]
a_3 &= -\frac{1}{6} a_1 \\[7pt]
a_4 &= \frac{2}{4} a_3 = -a_1/12 \\[7pt]
a_5 &= \frac{3}{5} a_4 - \frac{1}{5\cdot 4}a_3 = -\frac{1}{24} a_1 \\[7pt]
\end{aligned}
$$
where I cannot find any simple rules. Likewise, by letting $a_1:= 0$, we have
$$
\begin{aligned}
a_2 &= -a_0/2 \\[7pt]
a_3 &= \frac{1}{3} a_2 = -\frac{1}{6}a_0 \\[7pt]
a_4 &= \frac{2}{4} a_3 - \frac{1}{4\cdot 3}a_2 = -\frac{1}{24} a_0\\[7pt]
a_5 &= \frac{3}{5} a_4 - \frac{1}{5\cdot 4}a_3 = -\frac{1}{60} a_0 \\[7pt]
\end{aligned}
$$
where again I have failed to find any rules. So, I cannot find the interval that the following composition is valid.
$$
\sum_{n=0}^\infty a_n x^n = a_0 \left[ 1 - x^2/2 - x^3/6 - x^4/24 - x^5/60 + \cdots \right] + a_1 \left[ x - x^3/6 - x^4/12 - x^5/24 + \cdots\right]
$$
Any help will be appreciated.
|
A possible way to approach the problem is to consider
$$F(x):=\sum_{n\ge0}a_nx^n$$
as a purely formal series (so without thinking about convergence radii etc) and try to understand what function does it actually represent. Playing around with the recursive formula for the coefficients, you can derive (up to mistakes on my part)
$$\frac{d^2}{dx^2}F(x) = -F(x)+x\frac{d^2}{dx^2}F(x)\ .$$
From this, you can try to solve the differential equation to find a closed formula for $F(x)$, derive the Taylor coefficients from it, and thus determine the $a_n$. Having done this, you can finally determine the convergence radius.
|
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|
Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$
I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if someone has a more straightforward solution.
So applying the sum formula for sine and doing simple algebra we have:
$$\lim_{x\to 0} \frac{\cos{2} \,(\sin{x^2}-\sin{x})}{x} - \frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
The first limit is easy to evaluate and is equal to $-\cos{2}$. However, the second limit is harder, as it follows:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
I came across a solution by using the following sum-to-product identity:
$$\cos{A}-\cos{B}=-2\sin{\Big(\frac{A+B}{2}\Big)} \sin{\Big(\frac{A-B}{2}\Big)}$$
Setting $A=x^2$ and $B=x$, we have that
$$\cos{x^2}-\cos{x}=-2\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}$$
This is my only point of concern whether I applied the identity correctly. The rest of it flows more easily:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} = \lim_{x\to 0} \frac{-2\sin{2}\,\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}}{x}$$
$$= -2 \sin{2} \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} \lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}\Big)} $$
The first limit can be solved as it follows:
$$\lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} = \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)} \Big(\frac{x^2+x}{2}\Big)}{x \Big(\frac{x^2+x}{2}\Big)} = 1 \cdot \lim_{x \to 0} \frac{x^2 + x}{2x} = \frac{1}{2} $$
The second limit is equal to zero
$$\lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}}\Big)=0$$
|
Hint:
You can rewrite this fraction as
$$\frac{\sin(x^2+2)-\sin(x+2)}{x}=\frac{\sin(x^2+2)-\sin(2)}{x}-\frac{\sin(x+2)-\sin(2)}{x},$$
which is the difference of two rates of variation. Can take it from there?
|
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|
Find modulus and argument of $\zeta = {\frac {1-\cos4(\theta) + i \sin4(\theta)} {\sin2(\theta)+ 2i\cos^2(\theta)}}$ A guideline on which identity to use would be greatly appreciated, as $1-\cos2A=2\sin^2A$ identity isn't giving me the correct answer I think.
Given that:
$$\zeta = {\frac {1-\cos4(\theta) + i \sin4(\theta)} {\sin2(\theta)+ 2i\cos^2(\theta)}} $$
To prove:
$$|\zeta| = 2\sin(\theta) \qquad\text{and}\qquad \arg(\zeta) = \theta $$
My work so far;
Proof:
$$\zeta = {\frac {\sqrt{(1-\cos4\theta)^2 + \sin^24(\theta)}} {\sqrt{\sin^22(\theta)+ 4\cos^4(\theta})}} $$
I'm not sure about the above, if there's an easy explanation or method to go through would be great as questions like these are giving me a bit of trouble.
|
Method 1: Using the $1-\cos2A=2\sin^2A$ identity
For $|\zeta| = 2\sin(\theta)$:
$$\zeta = {\frac {|1-\cos4(\theta) + i \sin4(\theta)|} {|\sin2(\theta)+ 2i \cos^2(\theta)|}} $$
$$\zeta = {\frac {\sqrt{(1-\cos4\theta)^2 + \sin^24(\theta)}} {\sqrt{\sin^22(\theta)+ 4\cos^4(\theta})}} $$
$$\zeta = {\frac {\sqrt{(1-2\cos4\theta + \cos^24\theta+\sin^24\theta}} {\sqrt{4\sin^2\theta\cos^2\theta+ 4\cos^4(\theta})}} $$
$$\zeta = {\frac {\sqrt{(2-2\cos4\theta}} {\sqrt{4\cos^2\theta(\sin^2\theta+ \cos^2(\theta}))}} $$
$$\zeta = {\frac {\sqrt{2}\sqrt{1-\cos4\theta}} {2\cos\theta}}
\begin{cases}
\mbox{$1-\cos2A=2\sin^2A$ identity}
\end{cases} $$
$$\zeta = {\frac {\sqrt{2}\sqrt{2\sin^22\theta}} {2\cos\theta}} $$
$$\zeta = {\frac {2\sin2\theta} {2\cos\theta}} $$
$$\zeta = {\frac {2\sin\theta\cos\theta} {\cos\theta}} = 2\sin\theta.$$
For $\arg(\zeta) = \theta$:
$$\arg(\zeta) = \arg(1-\cos4\theta+i\sin4\theta)-\arg(\sin2\theta+2i\cos^2\theta)$$
$$\arg(\zeta) = \arctan(\frac{\sin4\theta}{1-\cos4\theta})-\arctan(\frac{2\cos^2\theta}{\sin2\theta})$$
$$\arg(\zeta) = \arctan(\frac{2\sin2\theta\cos2\theta}{2\sin^22\theta})-\arctan(\frac{2\cos\theta\cos\theta}{2\sin\theta\cos\theta})$$
$$\arg(\zeta) = \arctan(\cot2\theta)-\arctan(\cot\theta)$$
$$\arg(\zeta) = \arctan[\tan(90°-2\theta)]-\arctan[\tan(90°-\theta)])$$
$$\arg(\zeta) = 90°-2\theta-90°+\theta$$
$$\arg(\zeta) = -\theta.$$
Method 2: Using De Moivre's theorem,
$$\zeta = {\frac {1-\cos4(\theta) + i \sin4(\theta)} {\sin2(\theta)+ 2i \cos^2(\theta)}} $$
$$\zeta = {\frac {2\sin^22\theta + i 2\sin2\theta\cos2\theta} {2\sin\theta\cos\theta+ 2i \cos^2\theta}} $$
$$\zeta = {\frac {2\sin2\theta(\sin2\theta + i\cos2\theta)} {2\cos\theta(\sin\theta+ i \cos\theta)}} $$
$$\zeta = {\frac {2\sin\theta\cos\theta(\sin2\theta + i\cos2\theta)} {\cos\theta(\sin\theta+ i \cos\theta)}} * {\frac{-i}{-i}}$$
$$\zeta = {\frac {2\sin\theta(\cos2\theta - i\sin2\theta)} {\cos\theta - i \sin\theta}}$$
$$\zeta = {\frac {2\sin\theta[(\cos-2\theta + i\sin-2\theta)]} {\cos(-\theta) + i \sin(-\theta)}}$$
$$\zeta = {\frac {2\sin\theta(\cos\theta+i\sin\theta)^{-2}} {(\cos\theta+i\sin\theta)^{-1}}}
\begin{cases}
\mbox{De Moivre's Theorem}
\end{cases} $$
$$\zeta = 2\sin\theta[\cos(-\theta)+i\sin(-\theta)]$$
$$|\zeta| = 2\sin(\theta) \qquad\text{and}\qquad \arg(\zeta) = -\theta $$
|
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|
1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) above it We can observe the following -
$1 - 1$
$2 - 2 $
$3 - 2$
$4 - 4$
$5 - 4$
. . .
$8 - 8$
$9 - 8$
. . .
We can notice that $ 4^{th}$ term is 4 and similarly, the $ 8^{th}$ term is 8.
So the $ 1025^{th}$ term must be 1024 as $ 1024^{th} $ term starts with 1024.
So the value of $ 1025^{th}$ term is $ 2^{10} $ .
Is there any other method to solve this question?
|
The number $n$ first appears in the sequence at position $n$ until position $2n-1$. So, the number $1024$ appears in the sequence at position $1024$ until $2047$. Therefore the number will be $1024$.
|
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|
Solve inequality Given positive numbers $a,b,c$ satisfying $a^2+b^2+c^2=1$, prove the following inequality
$$\frac{a}{\sqrt{1-bc}} + \frac{b}{\sqrt{1-ac}}+\frac{c}{\sqrt{1-ab}}\le\frac{3}{\sqrt{2}}$$
Thanks
I have tried using CS, try to make use of $a+b+c\leq\sqrt3$, $abc\leq\frac{1}{3\sqrt3}$, but got nowhere –
|
hint:
$\dfrac{a}{\sqrt{2-2bc}} \le \dfrac{a}{\sqrt{1+a^2}}$
if you can prove $f(x)=\sqrt{\dfrac{x}{1+x}}$ is concave function, then the problem is solved.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given $(x-1)^3+3(x-1)^2-2(x-1)-4=a(x+1)^3+b(x+1)^2+c(x+1)+d$, find$(a,b,c,d)$
Given $(x-1)^3+3(x-1)^2-2(x-1)-4=a(x+1)^3+b(x+1)^2+c(x+1)+d$,
find$(a,b,c,d)$
my attempt:
$$(x+1)=(x-1)\frac{(x+1)}{(x-1)}$$
but this seems useless?
I want to use synthetic division but I don't know how
|
In case you don't know how to solve this most elegantly, there is still the straightforward possibility by expanding both sides. The LHS is given by
$$
x^3-5x,
$$
whereas the RSH is
$$
ax^3 + (3a + b)x^2 + (3a + 2b + c)x + a + b + c + d.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\omega = \frac{1 + \sqrt3 i}{2}$ , $ \omega^5 = ? $
$\omega = \frac{1 + \sqrt3 i}{2} $, $ \omega^5 = ? $
$\omega^3 = 1$ by definition?
So, $\omega^5 = \omega^2$
But why do i get wrong answer?
|
Because $\omega=\cos\left(\frac\pi3\right)+i\sin\left(\frac\pi3\right)$ and therefore it is not true that $\omega^3=1$. Actually, $\omega^3=-1$. And$$\omega^5=-\omega^2=-\cos\left(\frac{2\pi}3\right)-i\sin\left(\frac{2\pi}3\right)=\frac{1-i\sqrt3}2.$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to show the matrix has Rank $\le 5$
I want to show that the following matrix has Rank $\le 5$.
The matrix is
\begin{bmatrix}
2&1&1&1&0&1&1&1\\
1&2&1&1&1&0&1&1\\
1&1&2&1&1&1&0&1\\
1&1&1&2&1&1&1&0\\
0&1&1&1&2&1&1&1\\
1&0&1&1&1&2&1&1\\
1&1&0&1&1&1&2&1\\
1&1&1&0&1&1&1&2
\end{bmatrix}
I found that there is a submatrix in the matrix which has rank $ =4$ given by
$[2,1,1,1],[1,2,1,1],[1,1,2,1],[1,1,1,2]$.
I need to show the given matrix has at least 3 zero rows in order to show that Rank $\le 5$..
But I dont know how to show it.
Can someone help.
|
By inspection, the vectors $(1,-1,1,-1,1,-1,1,-1)$, $(1,0,-1,0,1,0,-1,0)$, and $(0,1,0,-1,0,1,0,-1)$ are all in the nullspace, and are linearly independent, so the rank is at most $5$.
|
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|
Find $n$ if $\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$ Find $n$ if $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$$
In this video they show a shortcut and say $n=-1/2$ without any explanation.
Key observation here is that the geometric mean of $9$ and $4$ is $6$.
It seems numerator and denominator are partial sums of geometric series, but I don't know how to proceed. Any help?
|
$$\dfrac{9(9^n) + 4(4^n)}{9^n + 4^n} = 6 $$ divide numerator and denominator by $9^n$:
$$ \dfrac{9 + 4\left(\frac{4}{9}\right)^n}{1 + \left(\frac{4}{9}\right)^n} = 6$$
Let $\left(\frac{4}{9}\right)^n = a$:
$$ \dfrac{9+ 4a}{1 + a} = 6 $$
$$a = \frac{3}{2} $$
$$ \left(\frac{4}{9}\right)^n =\left(\frac{2}{3}\right)^{2n} = \left(\frac{2}{3}\right)^{-1}$$
$$ n = -\frac{1}{2}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Maximizing area of rectangle inscribed in circle sector of radius 2 Question:
A rectangle is inscribed in a circle sector. The top two corners of the rectangle lies on the radius of the circle sector and the bottom two corners lie on the arc of the circle sector.
The radius is 2 and angle is $\frac{2\pi}{3}$.
Find the maximum area of the rectangle that has these properties and are located exactly like this using calculus-based optimization.
Attempted solution:
Let the shorter side of the rectangle be labelled x and the longer be labelled y. The distance from the center of the circle to the top of the rectangle is b.
Now, create a dashed-line triangle from the center of the circle to the bottom right of the rectangle. Call the top angle $\alpha$.
From this, we can get a value for $\frac{y}{2}$, since this is the side opposite the angle:
$$\sin \alpha = \frac{\frac{y}{2}}{2} \Rightarrow y = 4 \sin \alpha$$
Using the same triangle but with the adjacent side x + b:
$$\cos \alpha = \frac{x+b}{2} \Rightarrow x = 2 \cos \alpha - b$$
Then we use half of the top triangle above the rectangle to get a value for b in terms of y:
$$\tan \frac{2\pi}{6} = \frac{\frac{y}{2}}{b} \Rightarrow b = \frac{\frac{1}{2}y}{\tan \frac{\pi}{3}}$$
Combining these to get a value for x:
$$x = 2 \cos \alpha - b = 2 \cos \alpha - \frac{\frac{1}{2}y}{\tan \frac{\pi}{3}} = 2 \cos \alpha - \frac{2 \sin \alpha}{\tan \frac{\pi}{3}}$$
The area becomes:
$$A(\alpha) = yx = 4 \sin \alpha (2 \cos \alpha - \frac{2 \sin \alpha}{\tan \frac{\pi}{3}})$$
Taking the derivative:
$$A'(\alpha) = 8(-sin^2 \alpha + cos^2 \alpha - 2 \sin \alpha \cos \alpha)$$
Setting to zero:
$$A'(\alpha) = 0 \Rightarrow a = \frac{\pi}{2} - \frac{3\pi}{8}$$
Putting this into the formula for A:
$$A( \frac{\pi}{2} - \frac{3\pi}{8}) = 4(\sqrt{2} -1)$$
However, this does not match the expected answer of $\frac{4}{\sqrt{3}}$.
What went wrong? What are some productive ways to finish this off?
|
$A'(\alpha) = 8(-\sin^{2}{\alpha} + \cos^{2}{\alpha} - \frac{2}{\sqrt{3}} \sin{\alpha}\cos{\alpha})$. Setting this to zero gives the family of solutions $\alpha = \frac{1}{6}(3 \pi n + \pi)$, $n \in \mathbb{Z}$. Taking $n = 0$ gives $A(\pi/6) = \frac{4}{\sqrt{3}}$.
|
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|
Minimize $\left\lVert \mathbf{x} \right\rVert_1 - \left\lVert \mathbf{x} - \overline x \,\mathbf{1}_n\right\rVert_1$ over the unit $n$-cube I want to solve the following minimization problem
$$\min_{\mathbf{x} \in [0, 1]^n} \left\lVert \mathbf{x} \right\rVert_1 - \left\lVert \mathbf{x} - \overline x \,\mathbf{1}_n\right\rVert_1$$
where $\mathbf{x} = (x_1,x_2,\dots,x_n)$ and
$$\overline x := \frac{1}{n} \sum_{i=1}^{n} x_i = \frac{1}{n} \mathbf{1}_n^\top \mathbf{x}$$
I can prove $- \left\lVert \mathbf{x} - \overline x \,\mathbf{1}_n\right\rVert_1$, subject to $x_i \in [0, 1]$, takes the minimum value when half $x_i=0$ and the other half $x_i=1$. However, I have no idea how to find the minimum after adding the $1$-norm $\left\lVert \mathbf{x}\right\rVert_1$.
|
We need to minimize the following expression
$$
S = \sum_{i=1}^n (x_i - |x_i - \overline{x}|)
$$
where $x_i \in [0,1]$.
We can rewrite it as
$$
S = \sum_{i\in I_{>}} (x_i - (x_i - \overline{x})) + \sum_{i \in I_{<}} (x_i - (\overline{x} - x_i)).
$$
Here
$$
I_{>} = \{i\in \{1,2,\ldots n\}: x_i \ge \overline{x} \},
$$
$$
I_{<} = \{i\in \{1,2,\ldots n\}: x_i < \overline{x} \}.
$$
Hence
$$
S = \overline{x} (|I_{>}| - |I_{<}|) + 2 \sum_{i\in I_{<}} x_i,
$$
where $|A|$ is a number of elements in the set $A$.
Our goal is to make this expression as small as possible. Let $|I_{>}| =k$. Then $|I_{<}| = n-k$ and
$$
S = \overline{x} (2k - n) + 2 \sum_{i \in I_{<}} x_i.
$$
First of all, $2k -n$ must be negative in order for $S$ to have the smallest value. It's also easy to see that we should take all $k$ values of $x_i$ for $i$ from $I_{>}$ to be equal to $1$, since in that case $\overline{x}$ will be bigger and $\overline{x}(2k-n)$ will be smaller.
So,
$$
S = \frac{k+ \sum_{i \in I_{<}}x_i}{n} (2k-n) + 2 \sum_{i \in I_{<}} x_i = \frac{k(2k-n)}{n} + \sum_{i \in I_{<}} x_i \bigg(\frac{2k}{n} + 1\bigg).
$$
The previous expression have the smallest value when all $x_i = 0$ for $i \in I_{<}$.
Hence,
$$
\min S = \min_{1 \le k \le (n-1)} \Big\{\frac{k(2k-n)}{n} \Big\} = \min_{1 \le k \le (n-1)} \Big\{ \frac{2k^2}{n} -k \Big\}.
$$
Next consider a function
$$
f(x) = \frac{2x^2}{n} - x
$$
and determine it's minimum using calculus:
$$
f^{'}(x) = \frac{4x}{n} - 1 = 0 \implies x = \frac{n}{4}.
$$
So, for $n = 4l$ we get $k = l$ and the answer is
$$
\frac{2l^2}{n} - l = -\frac{n}{8}.
$$
For other values of $n$ the answer will be either $\lceil n/4 \rceil$ or $\lfloor n/4 \rfloor$.
Let's consider other values of $n$.
Case $n = 4l+1$. Then $\lfloor n/4 \rfloor = l$ and $\lceil n/4 \rceil = l +1$
$$
\frac{2l^2}{4l+1} - l = -\frac{2l^2+l}{4l+1}
$$
$$
\frac{2(l + 1)^2}{4l+1} - (l+1) = -\frac{2l^2+l - 1}{4l+1}
$$
This shows that in this case $k = \lfloor n/4 \rfloor$ gives the right answer.
Case $n = 4l+2$.
$$
\frac{2l^2}{4l+2} - l = -\frac{2l^2+2l}{4l+2}
$$
$$
\frac{2(l + 1)^2}{4l+2} - (l+1) = -\frac{2l^2+2l}{4l+2}
$$
This shows that in this case $k = \lfloor n/4 \rfloor$ and $k =\lceil n/4 \rceil$ both give the right answer.
Finally case $n = 4l+3$.
$$
\frac{2l^2}{4l+3} - l = -\frac{2l^2+3l}{4l+3}
$$
$$
\frac{2(l + 1)^2}{4l+3} - (l+1) = -\frac{2l^2+3l+1}{4l+3}
$$
This shows that in this case $k =\lceil n/4 \rceil$ gives the right answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$a+b+c=-3\sqrt{ac}$
If $a,b,c$ be in Geometic Progression, and $b-c,c-a,a-b$ in Harmonic Progression then prove that $a+b+c=-3\sqrt{ac}$.
My attempt: $$c-a=\frac{2(b-c)(a-b)}{b-c+a-b}$$$$(c-a)^2+2(b-c)(a-b)=0$$$$c^2+a^2-2ac+2(ab-b^2-ac+bc)=0$$$$a^2+c^2-2b^2-4ac+2ab+2bc=0$$$$a^2+c^2-6ac+2b(a+c)=0(b^2=ac)$$$$a^2+c^2+2ac+2b(a+c)=8ac$$$$(a+c)(a+c+2b)=8ac$$$$(a+c)(\sqrt{a}+\sqrt{c})^2=8(\sqrt{ac})^2(b=\sqrt{ac})$$
|
Let $$a,b,c \equiv k,kr,kr^2 \text{ be in GP } $$
Then $$c-a = \frac{2(b-c)(a-b)}{b-c+a-b} \implies- (kr^2-k)^2 = 2(kr-kr^2)(k-kr) $$
$$\implies-(r^2-1)^2 = 2(r)(1-r)(1-r) \implies-(r-1)^2(r+1)^2 = 2r(r-1)^2$$
$\color{red}{r\neq1} $ as we have a GP
$$ -(r+1)^2 = 2r \text{ or } r^2+1 =- 4r$$
Now $$a+b+c = k(1+r+r^2) = k(-4r+r) = -3kr$$ and $-3\sqrt{ac} = -3\sqrt{k^2r^2} = -3|kr| =\begin{cases} -3kr \iff kr>0 \\ 3kr \iff kr<0 \end{cases}$
So if $kr>0$ we have
$$a+b+c = -3\sqrt{ac} = -3kr$$
Edit : There's a possibility that $k$ and $r$ can both be negative. So, I've used $kr>0$
|
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|
Trying to find Pythagorean triples for a given ratio of $\frac{C}{A}$ I want to solve a ratio equation for $n$ in terms of $m$ and $R$ hoping, given a Ratio and a range of $m$ values, that finding an integer $n$ will give me the $m,n$ needed to generate a Pythagrean triple where $\frac{C}{A}$ has that ratio.
Given $A=m^2-n^2\qquad C=m^2+n^2$
$$\frac{C}{A}=\frac{m^2+n^2}{m^2-n^2}=R\implies Rm^2-Rn^2=m^2+n^2\implies (R+1)n^2+0n-m^2(R-1)=0$$
$$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{\pm\sqrt{4*(R+1)*m^2(R-1)}}{2(R+1)}=\frac{2m\sqrt{R^2-1}}{2(R^2-1)}=\frac{m\sqrt{R^2-1}}{R^2-1}$$
If I take the example of $f(2,1)=3,4,5$, $R=\frac{5}{3}$ and $\frac{2\sqrt{\frac{25}{9}-1}}{\frac{25}{9}-1}=\frac{2\sqrt{\frac{16}{9}}}{\frac{16}{9}}=\frac{\frac{8}{3}}{\frac{16}{9}}=\frac{8*9}{3*16}=\frac{72}{48}=1.5$
Where is my mistake?
|
In
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{\pm\sqrt{4*(R+1)*m^2(R-1)}}{2(R+1)}=\frac{2m\sqrt{R^2-1}}{2(R^2-1)}=\frac{m\sqrt{R^2-1}}{R^2-1}$$
Going from the second to the third part, you introduced an extra factor of $R-1$ into the denominator, i.e., it went from $2(R+1)$ to $2(R^2-1)$. Thus, with handling for both both positive & negative radicals, the correct result is
$$n = \pm\frac{m\sqrt{R^2-1}}{R+1} \tag{1}\label{eq1}$$
As such, with your example, using $m = 2$, $R = \frac{5}{3}$ in \eqref{eq1} gives $n = \pm 1$. Also, with your calculated results, multiplying $1.5 = \frac{3}{2}$ by $R - 1 = \frac{5}{3} - 1 = \frac{2}{3}$ (to undo dividing by $R-1$ in your calculations) gives $1$, i.e., the correct value of $n$.
|
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|
Evaluating $\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx$ My first instinct was to evaluate the indefinite form of the integral, which I did by substituting $x=\tan t$, therefore yielding
\begin{align}
\int \frac{x\ln x}{(1+x^2)^2} \,dx
&= \int \frac{\tan t \sec^2 t \ln\tan t}{(1+\tan^2 t)^2} \,dt
&& \text{by substitution} \\
&= \int \tan t \cos^2 t \ln \tan t \,dt \\
&= \int \sin t \cos t \ln \tan t \,dt \\
&= -\frac{1}{2} \cos^2 t \ln \tan t + \frac{1}{2} \int \cot t \, dt
&& \text{by parts} \\
&= -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t + k
\end{align}
I run into a wall when I introduce the limits of the integral, since I get
\begin{align}
\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx
&= \bigg[ -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t \bigg] ^\frac{\pi}{2} _0
\end{align}
I'm not too sure how to evaluate the limit of the final equation as $t \rightarrow 0$. I feel like the solution is something very trivial, but I can't quite put my finger on what I'm forgetting.
|
Different approach:
\begin{align}
I&=\int_0^\infty\frac{x\ln x}{(1+x^2)^2}\ dx\overset{x^2=y}{=}\frac14\int_0^\infty\frac{\ln y}{(1+y)^2}\ dy\overset{1+y=z}{=}\frac14\int_1^\infty\frac{\ln(z-1)}{z^2}\ dz\\
&\overset{\large z\ \mapsto\ \frac{1}{z}}{=}\frac14\int_0^1\ln\left(\frac{1-z}{z}\right)\ dz=\frac14\underbrace{\int_0^1\ln(1-z)\ dz}_{\large1-z\ \mapsto\ z}-\frac14\int_0^1\ln z\ dz\\
&=\frac14\int_0^1\ln z\ dz-\frac14\int_0^1\ln z\ dz=0
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate $\int \frac{dx}{1-\sin^4x}$
Calculate $$\int \frac{dx}{1-\sin^4x}$$
My try:
\begin{align}
\int \frac{dx}{1-\sin^4x}&=\int \frac{(1-\sin^2x)+(1+\sin^2x)dx}{1-\sin^4x}
\\&=\int \frac{dx}{1-\sin^2x}+\int \frac{dx}{1+\sin^2x}
\\&=\tan x+\int \frac{dx}{1+\sin^2x}
\end{align}
How to deal with the second one?
|
$$\dfrac{dx}{a\sin x\cos x+b\sin^2x+c\cos^2x}$$
Divide numerator and denominator
by $\cos^2x$ and set $\tan x=u$
Or by $\sin^2x$ and set $\cot x=v$
Here $(1)+\sin^2x=(\cos^2x+\sin^2x)+\sin^2x=?$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determine the maximum possible volume, excluding the volume of the legs. A table is to be constructed by gluing together 68 cubes of dimension $1\times 1\times 1$. All four legs and the rectangular top will be formed by the cubes. The four legs must be the same length and must be one cube thick, and the top is one cube thick, as well. What is the maximum possible volume of the space between the table's top and the floor, excluding the space taken up by the table's legs?
My solution:
Let $a$ be the area of the top and $c$ be the length of a leg.
$$
\begin{align}
a + 4c &= 68 \cr
a &= 68 - 4c
\end{align}
$$
We try to find the maximum value of ac, and then subtract 4c:
$$ac = (68 - 4c)c = -4c^2 + 68c$$
Completing the square gets $-4(c - 17/2)^2 + 289$.
The lowest value of $ac$ is when $c= 17/2$, and the lowest value is $289$. Then we subtract $289 - 4c = 289 - 34 = 255$.
Therefore, the maximum volume when excluding the volume of the legs is $255$.
However, the actual answer is $256$ as the maximum volume when excluding the legs. Where did I go wrong?
|
Noting that subtracting $68-4=64$, we can see that, if the legs are only $1$ cube high, that leaves $8\times 8\times 1=64$ for the table top. The space under the table (minus the legs) is $64-(4\times 1\times 1)=60$ so the volume from the table top to the floor is $2*64-4=124$.
If we let the legs be $4\times 2=8$, the table top is $5\times 12=60$, volume: $3*60-8=172$.
If we let the legs be $4\times 3=12$, the top is $68-12=54$ volume: $4\times 54-12=204$
If we let the legs be $4\times 4=16$, the top is $68-16=52$, volume: $5\times 52-16=244$
If we let the legs be $4\times 5=20$, the top is $68-20=48$, volume: $6\times 48-20=268$
If we let the legs be $4\times 6=24$, the top is $68-24=44,$ volume: $7\times 44-24=284$
If we let the legs be $4\times 7=28$, the top is $68-28=40,$ volume: $8\times 40-28=292$
If we let the legs be $4\times 8=32$, the top is $68-32=36,$ volume: $9\times 36-32=292$
If we let the legs be $4\times 9=36$, the top is $68-36=32,$ volume: $10\times 32-36=284$
If we let the legs be $4\times 10=40$, the top is $68-40=28,$ volume: $11\times 28-40=268$
It appears $292$ is the maximum volume with outer dimensions of $8\times 5\times 8$ or $9\times 6\times 6$
|
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|
Find the number of polynomials $P(x)$ with coefficients $0,1,2,3$ such that $P(2) = n$.
Find the number of polynomials $P(x)$ with coefficients $0,1,2,3$ such that $P(2) = n$.
Only what I currently know is that I should consider polynomials with the smallest degree $k$, where $k$ is first integer such that
$$ 3 \cdot (1 + 2 + 4 + ... + 2^k) = 3 \cdot \frac{2^{k+1}-1}{1} \ge n $$
$$ 2^{k+1} \ge \frac{n}{3}+1 $$
$$ k \ge \log_2(\frac{n}{3}+1) - 1 $$
But what should be done next? I don't see a pattern here.
|
Let your polynomial be $P(x) = \sum_{k=0}^n c_i x^i$.
Let $A_k$ be the sets of nonnegative integers $i$ such that $c_i = k$, $k=0,1,2,3$. Thus $$
\eqalign{n &= \sum_{i \in A_1} 2^i + 2 \sum_{i \in A_2} 2^i + 3 \sum_{i \in A_3} 2^i\cr
&= \sum_{i \in A_1 \cup A_3} 2^i + 2\sum_{i \in A_2 \cup A_3} 2^i}$$
Given any nonnegative integers $a,b$ such that $n = a + 2 b$, we can find $A_0, A_1, A_2, A_3$ such that $a = \sum_{i \in A_1 \cap A_3} 2^i$ and $b = \sum_{i \in A_2 \cup A_3} 2^i$, namely $A_0$ is the set of $i$ such that
the base-2 digit in the "$2^i$" place for both $a$ and $b$ are $0$, $A_1$ where the digit is $1$ for $a$ and $0$ for $b$, $A_2$ where the digit is $0$ for $a$ and $1$ for $b$, $A_3$ where the digit is $1$ for both.
Now count how many ways there are to write $n=a+2b$...
|
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|
$\epsilon - N$ proof of $\sqrt{4n^2+n} - 2n \rightarrow \frac{1}{4}$ I have the following proof for $\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$ and was wondering if it was correct. Note that $\sqrt{4n^2+n} - 2n = \frac{n}{\sqrt{4n^2+n} + 2n}$.
$$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| \\
= \left|\frac{2n - \sqrt{4n^2+n}}{4(\sqrt{4n^2+n} + 2n)}\right|=\left|\frac{\sqrt{4n^2+n} - 2n}{4(\sqrt{4n^2+n} + 2n)}\right|\\
= \left|\frac{n}{4(\sqrt{4n^2+n} + 2n)^2}\right| \leq \left|\frac{n}{4(4n)^2}\right| = \left|\frac{n}{64n^2}\right| \\
= \left|\frac{1}{64n}\right| < \epsilon \\
\implies n>\frac{1}{64\epsilon}$$
|
You did the scratch work correctly. But I wouldn't call this a proof (of course it contains all ingredients of a good proof!).
You didn't introduce $\epsilon$. Of course, everyone knows what you mean but one should write it out if one wants to be fully rigorous.
So, your proof should read:
Let $\epsilon >0$. Let $N$ be an integer greater than --insert your choice for $N$ here--.
If $n \geq N$, we have
--insert your estimations here--
Thus, by definition of limit, the result follows. QED
|
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|
Convert a double integral into polar coordinates If $a$ is a positive number, compute
$$
\int_0^{2a}\int_{-\sqrt{2ay-y^2}}^0 \sqrt{x^2+y^2} \, dx \, dy.
$$
I wanted to use polar coordinates to compute this double integral. However, the lower $x$-limit
$$
x=-\sqrt{2ay-y^2}
$$
is equivalent to saying that $x$ satisfies
$$
x^2+(y-a)^2=a^2.
$$
I am still not sure how we can change this integral to polar coordinates...
|
Let $x=r\cos\theta,y=r\sin\theta$. Since $x\leq 0$, $\frac{\pi}2\leq \theta\leq\pi$. Since $x^2+y^2\leq 2ay$, $r^2\leq 2ar\sin\theta$ thus $r\leq 2a\sin\theta$. Hence
\begin{align*}
\int_0^{2a}\int_{-\sqrt{2ay-y^2}}^0 \sqrt{x^2+y^2} \, dx \, dy&=\int\limits_{x^2+y^2\leq 2ay\\x\leq 0}\sqrt{x^2+y^2} \, dx \, dy\\&=\int_{\frac{\pi}2}^\pi \,d\theta\int_0^{2a\sin\theta}r^2\,dr\\&=\frac{8a^3}{3}\int_{\frac{\pi}2}^\pi(\sin\theta)^3\,d\theta.
\end{align*}
I believe that you can move on now.
|
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|
Value of a and P(x) when P(x) is a rational number when satisfies a certain equation This is the question as I still don't have permission to post picture, but it is basically when P(x) = x^3+x^2+ax+1, when a is a rational number, P(X) is also rational number for every x that satisfy x^2+2x-2=0
Consider the integral expression in $x$ $$ P= x^3 + x^2 + ax +1 $$ where $a$ is a rational number. At $a=a_0$ the value of $P$ is a rational number for any $x$ which satisfies the equation $x^2 + 2x -2 = 0$, and n this case the value of $P$ is $P_0$.
Find the values of $a_0$ and $P_0$
|
Hint: By a direct computation, $x^2+2x-2=0$ has two solutions, namely
$$
x=\pm \sqrt{3}-1.
$$
For $x=\sqrt{3}-1$ we have
$$
P=x^3+x^2+ax+1=(a+4)\sqrt{3}-a-5.
$$
Hence for $a=-4$, this is rational. For $x=-\sqrt{3}-1$ we have
$$
P=-(a+4)\sqrt{3}-a-5.
$$
|
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|
How to find the area of a region bounded by a simple closed curve? I have the following equation:
$$
\frac{p}{(a-x)^2+y^2}+\frac{1-p}{(b-x)^2+y^2}=1 \text{ where } 0\leq p\leq 1
$$
Which represent a simple close curve. Obviously, when $p=0,p=1$ or $a=b$ we recover a unit circle. However the shape of the curve is more interesting in the other cases. Here is an example for $p=0.2$, $a=1$ and $b=0$:
The area inside the curve is supposed to represent an uniform distribution, thus I suspect the area to be equal to one (unless I did a mistake!). However I don't know how to compute it. My attempt was to shift to polar coordinates and compute a double integral. However I have trouble determining the boundaries of these integrals.
Any hints or general advice are appreciated !
|
By clearing denominators, we see that our curve is a quartic $p(x, y) = 0$. For generic values of $p, a, b$, it is elliptic and so does not admit a rational parameterization. Probably the areas can be computed in terms of elliptic functions.
In the special case $p = \frac{1}{2}$, the curve is symmetric not only about the $x$-axis but also the line $x = \frac{1}{2}(a + b)$, and by translating our coordinate system we may as well assume $b = -a$, $a > 0$. The curve then can be written in the simple form
$$x^4 + 2 x^2 y^2 + y^4 - (2 a^2 + 1) x^2 + (2 a^2 - 1) y^2 + a^2 (a^2 - 1) = 0 .$$
This is a biquadratic in, say, $y$, so we can solve for $y$ in terms of $x$ using two square root extractions:
$$y = \pm F(x), \qquad F(x) := \frac{1}{2} \sqrt{2 - 4 (a^2 + x^2) + 2 \sqrt{16 a^2 x^2 + 1}} ,$$
and using the fourfold symmetry of the curve we find that the area it encloses is
$$4 \int_{x_0}^{x_1} F(x) \,dx .$$
If $b > 1$ then $F$ has two positive roots, and we set $x_0$ to be the smaller root and $x_1$ the larger. If $0 \leq b \leq 1$ then $F$ has one positive root, and we take $x_1$ to be that root and take $x_0 = 0$.
Substituting $x = \frac{t}{2 a (1 - t^2)}$ gives an integrand without a nested radical:
$$\int_{t_0}^{t_1} \frac{(t^2 + 1) \sqrt{-4 a^4 t^4 + (8 a^4 - 4 a^2 - 1) t^2 + 4 a^2 (1 - a^2)} \, dt}{(1 - t^2)^3} .$$
(NB if $x_0 = 0$ then $t_0 = 0$.) Generically the occurrence of the square root of quartic polynomial in an algebraic expression means that, at best, explicitly writing its antiderivative generically requires the use of elliptic functions, and trying a few parameter values for $a$ confirms this.
If the quartic has a multiple root, however, we can pull an even number of factors out of the radical, leaving at worst the square root of quadratic polynomial, and such integrals can often be evaluated using elementary techniques. The discriminant of the quartic is a constant multiple of $(8 a^2 + 1)^2 a^6 (a^2 - 1)$, so the only positive value of $a$ for which it has a multiple root is $a = 1$. For this parameter value the curve has genus $0$ and has a self-intersection at the origin, and the integral simplifies to
$$\int_0^{\sqrt{3} / 2} \frac{(t^2 + 1) t \sqrt{3 - 4 t^2} \,dt}{(1 - t^2)^3} = \frac{2}{3} \pi + \sqrt{3} .$$
To illustrate how complicated the explicit expressions can be even for the case $p = \frac{1}{2}, b = -a$ for nonspecial values of $a$, Maple returns for $a = 2$ that the area is
$${\frac {\left[6 \sqrt {2} \left( (136 + 8 \sqrt {
3}\sqrt {11} )\,\Pi \left( 1,2\,{\frac {
\sqrt {3}\sqrt {11}}{111+\sqrt {3}\sqrt {11}}}, \omega \right) - (135 + 9\,\sqrt {3}\sqrt {11})K \left( \omega
\right) +32\,\Pi \left( 1,2\,{\frac {\sqrt {3}\sqrt {11}
}{\sqrt {3}\sqrt {11}+15}}, \omega \right) \right) \right] }
{\sqrt {111+\sqrt {3}\sqrt {11}} \left( \sqrt {3}\sqrt {11}+15
\right) }},$$
where $K$ is the complete elliptic integral of the first kind, $\Pi$ is the incomplete elliptic integral of the third kind (but NB Maple's convention for these functions is different from Wikipedia's), and $\omega := {\frac {\sqrt {2}\sqrt [4]{3}\sqrt [4]{11}}{\sqrt {111+\sqrt {3}\sqrt {11}}}}$.
Numerically integrating, by the way, we find that the area of the interior of the example curve with parameters $a = 1, b = 0, p = \frac{1}{5}$ is $3.56910\ldots$, not $1$ as intended.
|
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|
Express a vector as a linear combination of two other vectors. Let $A = \begin{bmatrix}
1 & -1 & 1 \\
1 & 0 & 2 \\
-1 & 2 & 0
\end{bmatrix}$ and $B = \begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}$. $A$ row reduces to $B$ but you are asked not to verify this.
1. Find a basis for the column space $\text{col}(A)$.
From $B$, it is clear that the first and second columns of the matrix are pivot columns. Thus, the first and second columns of $A$ form a basis, i.e.,
\begin{equation*}
\left\{\begin{pmatrix}
1 \\
1 \\
-1 \\
\end{pmatrix}, \begin{pmatrix}
-1 \\
0 \\
2
\end{pmatrix}\right\}.
\end{equation*}
2. Determine the vector of $\text{col}(A)$ which is nearest to the vector $\begin{bmatrix}
2 \\
-2 \\
0
\end{bmatrix}$.
The vectors that span $C = \text{col}(A)$ are all linearly independent. We first find the projection matrix given by $P = C(C^TC)^{-1}C^T$. Lets first calculate $C^TC$
\begin{equation*}
C^TC = \begin{bmatrix}
1 & 1 & -1 \\
-1 & 0 & 2
\end{bmatrix}
\begin{bmatrix}
1 & -1 \\
1 & 0 \\
-1 & 2
\end{bmatrix} = \begin{bmatrix}
3 & -3 \\
-3 & 5
\end{bmatrix}.
\end{equation*}
This means that
\begin{equation*}
(A^TA)^{-1} = \frac{1}{6}\begin{bmatrix}
5 & 3 \\
3 & 3
\end{bmatrix}.
\end{equation*}
Hence,
\begin{equation*}
\begin{split}
P = A(A^TA)^{-1}A^T &= \frac{1}{6}\begin{bmatrix}
1 & -1 \\
1 & 0 \\
-1 & 2
\end{bmatrix}
\begin{bmatrix}
5 & 3 \\
3 & 3
\end{bmatrix}
\begin{bmatrix}
1 & 1 & -1 \\
-1 & 0 & 2
\end{bmatrix} \\
&= \frac{1}{6}\begin{bmatrix}
1 & -1 \\
1 & 0 \\
-1 & 2
\end{bmatrix}
\begin{bmatrix}
2 & 5 & 1 \\
0 & 3 & 3
\end{bmatrix} \\
&= \frac{1}{6}\begin{bmatrix}
2 & 2 & -2 \\
2 & 5 & 1 \\
-2 & 1 & 5
\end{bmatrix}.
\end{split}
\end{equation*}
So the projection of $(2,-2,0)$ onto $C$ is
\begin{equation*}
\frac{1}{6}\begin{bmatrix}
2 & 2 & -2 \\
2 & 5 & 1 \\
-2 & 1 & 5
\end{bmatrix}
\begin{bmatrix}
2 \\
-2 \\
0
\end{bmatrix} = \begin{bmatrix}
0 \\
-1 \\
-1
\end{bmatrix}.
\end{equation*}
3. Express the vector $\begin{bmatrix}
2 \\
-2 \\
0
\end{bmatrix}$ in $\mathbb{R}^3$ as a linear combination of two vectors, one in $\text{col}(A)$ and the other in $\text{col}(A)^{\perp}$.
Not sure how to do this one any help would be great!
|
Just subtract $\begin{bmatrix} 0 \\ -1 \\ -1 \end{bmatrix}$ from $\begin{bmatrix} 2 \\ -2 \\ 0 \end{bmatrix}$. That vector must be in $\text{col}(A)^{\perp}$, by properties of projections. Indeed, you can check its perpendicularity to your basis for col$(A)$ by computing the dot product.
|
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|
Help differentating $f(x) = \sqrt\frac{x^2-1}{x^2+1}$ The equation I'm trying to differentiate is, $ f(x) = \sqrt\frac{x^2-1}{x^2+1}$ and I know the answer is meant to be
$$=\frac{\frac{x\sqrt {x^2+1}}{\sqrt {x^2-1}}-\frac{x\sqrt {x^2-1}}{\sqrt {x^2+1}}}{x^2+1}$$
But when I do the working out I get this
$$=\frac{(x^2-1)^\frac{1}{2}}{(x^2+1)^\frac{1}{2}}$$
$$=\frac{\frac{1}{2}(x^2-1)^\frac{-1}{2}\cdot2x\cdot(x^2+1)^\frac{1}{2}-(x^2-1)^\frac{1}{2}\cdot\frac{1}{2}(x^2+1)^\frac{-1}{2}\cdot2x}{x^2+1}$$
simplify
$$=\frac{x(x^2-1)^\frac{-1}{2}\cdot(x^2+1)^\frac{1}{2}-(x^2-1)^\frac{1}{2}\cdot x(x^2+1)^\frac{-1}{2}}{x^2+1}$$
$$=\frac{\frac{\sqrt {x^2+1}}{x\sqrt {x^2-1}}-\frac{\sqrt {x^2-1}}{x\sqrt {x^2+1}}}{x^2+1}$$
As you can see two of my $x$'s are in the wrong location, and I just can't figure out what I'm doing wrong. Any help as to what steps I'm doing wrong or missing would be much appreciated.
|
Just a small trick.
When you face expressions which contains products, quotients, powers, logarithmic differentiation makes life easier.
$$y= \sqrt\frac{x^2-1}{x^2+1}\implies \log(y)=\frac 12 \log(x^2-1)-\frac 12 \log(x^2+1)$$ Differentiate both sides
$$\frac {y'} y=\frac 12 \frac {2x}{x^2-1}-\frac 12 \frac {2x}{x^2+1}=\frac{2x}{(x^2-1)(x^2+1)}$$
$$y'=y\times\frac {y'} y=\sqrt\frac{x^2-1}{x^2+1}\times\frac{2x}{(x^2-1)(x^2+1)}$$ and simplify.
|
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|
Evaluating $\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$ Problem:
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$$
$$\lim_{x \to -\infty} \sqrt{\frac{1}{x^6}}=0$$ so...
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}=\frac{1}{0}$$
The answer is $-1$ and I know how to get that answer. Where is the mistake in this method though?
|
Yes, $\lim_{x\to-\infty}\sqrt{\frac{1}{x^6}}=0$. But $\lim_{x\to-\infty}\sqrt{x^6+4}=\infty$.
|
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|
If $a+b+c = 4, a^2+b^2+c^2=7, a^3+b^3+c^3=28$ find $a^4+b^4+c^4$ and $a^5+b^5+c^5$ I have tried to solve it but cannot find any approach which would lead me to the answer
|
Use $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
and $$a^3+b^3+c^3-3bc=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}$$ to find $ab+bc+ca=u, abc=v$(say)
So, $a,b,c$ are the roots of $$t^3-4t^2+ut-v=0$$
Use Newton's Sums
|
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|
Computing a circle from a given set with a tangent line condition I was asked to compute the circles' equations from the set
$$x^2 + y^2 -3x + (k-6)y + (9-3k)=0$$
that fulfill the following request: the circles must be tangent to the line
$$x+y-3=0$$
I started to compute
$$
\begin{cases}
x^2 + y^2 -3x + (k-6)y + (9-3k)=0\\
x+y-3=0
\end{cases}
$$
$$
\begin{cases}
x^2 + y^2 -3x + (k-6)y + (9-3k)=0\\
x=3-y
\end{cases}
$$
obtaining
$$
\begin{cases}
2y^2 +(k-3)y +(9-3k)=0\\
x=3-y
\end{cases}
$$
Now, in order to obtain a tangent line to the circle, I must put the condition $\Delta=0$, yielding
$$(k-3)^2-8(9-3k)=0$$
and so
$$k_1=3, \quad k_2=-24$$
So I have the circles, by replacing $k$:
$$x^2 + y^2 -3x -3y =0$$
and
$$x^2 + y^2 -3x -30y + 81=0$$
while the solutions reported were $$x^2+y^2-3x-5y+8=0$$ and $$x^2+y^2-3x-13y+32=0$$
Could anyone tell me where I am getting wrong?
Thanks in advance.
|
You have occoured in an error when you say: "obtaining: " $$\begin{cases}
2y^2 +(k-3)y +(9-3k)=0\\
x=3-y
\end{cases}$$
You should have obtained: $$\begin{cases}
2y^2 +(k-9)y +(9-3k)=0\\
x=3-y
\end{cases}$$
From here: $(k-9)^2-8(9-3k)=0$ and $k^2+6k+9=0$, so $k=-3$. The equation of the circle is: $x^2+y^2-3x-9y+18=0$.
Here the graph:
The solution $x^2+y^2-3x-5y+8=0$ is impossible because $9-3k=8$ ($k=\frac{1}{3}$) and $k-6=5$ ($k=11$) that is impossible.
|
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|
How can we not use Muirhead's Inequality for proving the following inequality? There was a question in the problem set in my math team training homework:
Show that $∀a, b, c ∈ \mathbb{R}_{≥0}$ s.t. $a + b + c = 1, 7(ab + bc + ca) ≤ 2 + 9abc.$
I used Muirhead's inequality to do the question (you can try out yourself):
By Muirhead's inequality, $$\begin{align}7(ab+bc+ca)&=7(a+b+c)(ab+bc+ca)\\&=21abc+6\big(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\big)+\sum_{sym}a^2b\\&\le21abc+6\big(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\big)+\sum_{sym}a^3\\&=2(a+b+c)^3+9abc\\&=2+9abc\end{align}$$As $(3,0,0)$ majorizes $(2,1,0)$.
Is above proof correct? Also, can we find a proof not using Muirhead's inequality?
Any help is appreciated!
|
A proof by SOS:
$$2+9abc-7(ab+ac+bc)=2(a+b+c)^3+9abc-7(a+b+c)(ab+ac+bc)=$$
$$=\sum_{cyc}(2a^3+6a^2b+6a^2c+4abc+3abc-7a^2b-7a^2c-7abc)=$$
$$=\sum_{cyc}(a^3-a^2b-ab^2+b^3)=\sum_{cyc}(a-b)^2(a+b)\geq0.$$
Also, $uvw$ kills it immediately.
|
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|
tangent inequality in triangle Let $a$, $b$ and $c$ be the measures of angles of a triangle (in radians).
It is asked to prove that
$$\tan^2\left(\dfrac{\pi-a}{4}\right)+\tan^2\left(\dfrac{\pi-b}{4}\right)+\tan^2\left(\dfrac{\pi-c}{4}\right) \ge 1$$
When does equality occur ?
My try :
Letting $u:= \tan\left(\dfrac{\pi-a}{4}\right)$ and $v:= \tan\left(\dfrac{\pi-b}{4}\right)$ the inequality reduces to proving
$$u^2+v^2+\dfrac{(1-uv)^2}{(u+v)^2} \ge 1\quad\quad (*)$$ ($$u,v\in (0,1)$$)
( using $a+b+c=\pi$ and the formula for $\tan(x+y)$ and the fact that $\tan\left(\dfrac{\pi}{2}-x\right)=\dfrac{1}{\tan x}$ )
I'm having trouble in proving that last inequality.
Any suggestions are welcome.
Thanks.
Edit : is the following reasoning to prove the inequality (*) sound ?
(*) is obvious when $u^2 + v^2 \ge 1$ so we only have to deal with the case $u^2 + v^2 \le 1$ which we assume true in what follows.
(*) $\iff (u^2 + v^2)(u + v)^2 \ge (u+v)^2+(1-uv)^2$
Setting $x:= u^2 + v^2$ and $a:=uv$ we get
(*) $\iff x^2+(2a-1)x \ge a^2+4a-1$
Now some calculus $(x^2+(2a-1)x)' = 2x + 2a-1$
the function $\phi:x\mapsto x^2+(2a-1)x$ then has a minimum at $\dfrac 12 - a$ which is $\phi\left(\dfrac 12 - a\right) = -a^2+a-\dfrac 14$
It then suffices to have $-a^2+a-\dfrac 14 \ge a^2+4a-1$
This last ineq is equivalent to
$8a^2+12a-3 \le 0$
which, in turn, is equivalent to
$a \in \left[\dfrac{-6-\sqrt{60}}{8}, \dfrac{-6+\sqrt{60}}{8}\right]$
Recall that we're working under the assumption $u^2 +v^2 \le 1$, that yields in particular that $uv \le \dfrac 12$.
since $ \dfrac{-6-\sqrt{60}}{8} \le 0 \le a := uv \le \dfrac 12 \le \dfrac{-6+\sqrt{60}}{8}$, we're done.
Thanks for taking time to check the correctness of the above proof.
|
$$\left(\tan^2\frac{\pi-x}{4}\right)''=\frac{2-\sin\frac{x}{2}}{8\cos^4\frac{\pi-x}{4}}>0.$$
Thus, by Jensen
$$\sum_{cyc}\tan^2\frac{\pi-\alpha}{4}\geq3\tan^2\frac{\pi-\frac{\alpha+\beta+\gamma}{3}}{4}=1.$$
|
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|
Inverse of rational function $y= \frac{3-x}{1+x^2}$ I have the function $$y= \frac{3-x}{1+x^2}$$ and I want to find the inverse of this function.
I know that
$$x= \frac{1 \pm \sqrt{1-4y(3-y)}}{2y}$$
My question is how do I find the domain where the function is
$$x= \frac{1- \sqrt{1-4y(3-y)}}{2y}$$
and
$$x= \frac{1+ \sqrt{1-4y(3-y)}}{2y}$$
respectively?
|
Domain of $x$ is the range of $y$, and vice versa.
You have a square root for $x=f(y)$, so you should solve the condition for the expression within that square root is bigger than or equal $0$. Or:
$1-12y+4y^2\ge 0$
Solving $1-12y+4y^2=0$ yields $(1/2)(3+2\sqrt{2})$ and $(1/2)(3-2\sqrt{2})$. These two are real values, the coefficient a of the original quadratic expression is $4$, or $4>0$. Hence, the expression will have a negative value between the 2 roots.
Therefore, for this, $y<(1/2)(3+2\sqrt{2})$ and $y>(1/2)\sqrt{3-2\sqrt{2}}$ will make the square root of $f(y)$ "meaningful", thus, that is the domain of $x$. And yes, the two $x=f(y)$ share the same domain
However, based on the comments, I suggest you to double-check on your math again, just to make sure the $x=f(y)$ is written correctly. The way to process after that is the same though.
|
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|
Asymptotes of hyperbola by considering x tends to infinity I saw this derivation of equations of the asymptotes of hyperbola and it goes like this...
For a standard hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, rearranging the terms we get
$y=\pm\frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}}$
So as $x\rightarrow\infty$, $\frac{a^2}{x^2}\rightarrow 0$ and hence $y\rightarrow\pm\frac{b}{a}x$.
Therefore the equations of the asymptotes are $y=\pm\frac{b}{a}x$.
I am not very sure if this derivation is correct. Would appreciate if you can share your opinion. Like if there is any step above that is not true in general.
|
I think you are right to be suspicious of this method.
Let's apply it to the hypberbola given by the equation
$$ \frac{x^2}{2^2} + x -\frac{y^2}{2^2}=1 .$$
Solving for $y$ as in the method in the question, we get
$$ y = \pm x \sqrt{1 + \frac 4x - \frac 4{x^2}},$$
and as $x\to\infty,$ we find that $\sqrt{1 + \frac 4x - \frac 4{x^2}}\to 1,$
so this method derives the asymptotes $y = x$ and $y = -x.$
But the actual asymptotes are $y = 2 + x$ and $y = 2-x.$
The pitfall in this method can be seen by applying it to any straight line.
For the equation $y = mx + b,$ We factor $x$ out of the right side to obtain
$$ y = x \left(m + \frac bx\right), $$
and then $\left(m + \frac bx\right)\to m$ as $x\to\infty,$
so the method yields $y = mx.$
What the method is actually finding is the directions from the origin to the points at infinity on the curve, which gives the slopes of the asymptotes but not the $x$- or $y$-intercepts.
It will coincidentally give the correct result when the asymptote happens to pass through the origin, as we can predict will happen with
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ due to symmetry,
but that's an extra fact that has to be shown and it works only in that special case.
However, the method does find the slope of each asymptote.
We can then find the $y$-intercept by taking the difference between the curve and a line through the origin with the same slope as the asymptote.
Taking the hyperbola $ \frac{x^2}{2^2} + x -\frac{y^2}{2^2}=1$ again,
taking the solution $y = \sqrt{x^2 + 4x -4}$ and
comparing it with the line $y = x,$
we find that
$$ \lim_{x\to\infty} (y_\mathrm{\,hyperbola} - y_\mathrm{\,line})
= \lim_{x\to\infty} \sqrt{x^2 + 4x -4} - x = 2.$$
Hence the difference between the curve $y = \sqrt{x^2 + 4x -4}$
and the line $y = 2 + x$ as $x\to\infty$ is zero.
This correctly predicts that $y = 2 + x$ is an asymptote.
Aretino's answer shows how this method is correctly applied to the hyperbola in the question.
|
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|
Probability Question 3.17 in Wackerly, Mendenhall, and Scheaffer, 5th Ed.
Problem 3.17. Two construction contracts are to be randomly assigned to one or more of three firms:
I, II, and III. Any firm may receive both contracts. If each contract will yield a profit of
$\$90,000$ for the firm, find the expected profit for firm I. If firms I and II are actually owned
by the same individual, what is the owner's expected total profit?
My Answer. Let $Y$ be the number of contracts assigned to firm I. Then this problem is equivalent to
indistinguishable balls put into distinguishable urns, where the urns can be empty. That means
the number of ways to assign $2$ indistinguishable contracts to $3$ distinguishable firms is
$$\binom{2+3-1}{2}=\binom{4}{2}=6. $$
It follows that the distribution for $Y$ is as follows:
$$
\begin{array}{c|c|c|c}
y &0 &1 &2 \\ \hline
P(Y=y) &1/2 &1/3 &1/6
\end{array}
$$
This means that the expected profit, given by $E(\$90000Y),$
is
$$E(\$90000Y)=\$90000\left[0/2+1/3+2/6\right]=\$60000. $$
Now if one owner owns two firms, let $X$ be a random variable describing how many contracts either
company I or II got. This essentially coalesces the two companies I and II in terms of our analysis.
We have the following probability distribution for $X:$
$$
\begin{array}{c|c|c|c}
x &0 &1 &2 \\ \hline
P(X=x) &1/3 &1/3 &1/3
\end{array}
$$
It follows that the expected value for the total profit is given by
$$E(\$90000X)=\$90000[0/3+1/3+2/3]=\$90000. $$
My Question. The answer in the back of the book, however, is $\$120000.$ Why is that? Is there an error in my reasoning? Thanks in advance for your time!
|
There's a problem with your reasoning: each combination is not just as likely to occur. If each contract has a probability of $\frac{1}{3}$ to go to one of the firms, than the probabilities are as follows:
$$P(0) = \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}$$
$$P(1) = \frac{2}{3} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{2}{3} = \frac{4}{9}$$
$$P(2) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$$
We find:
$$E[X] = \frac{4}{9} \cdot 0 + \frac{4}{9} \cdot 90000 + \frac{1}{9} \cdot 180000 = 60000$$
You could also arrive here by dividing the total expected profit for all companies by the number of companies, or $180000$ divided by three. If one individual owns both companies, we can simply multiply by two and arrive at $120000$. Alternatively:
$$P(0) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$$
$$P(1) = \frac{2}{3} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{2}{3} = \frac{4}{9}$$
$$P(2) = \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}$$
We then find:
$$E[X] = \frac{1}{9} \cdot 0 + \frac{4}{9} \cdot 90000 + \frac{4}{9} \cdot 180000 = 120000$$
|
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|
All possible antiderivatives for Integration by Parts Just had a quick inquiry with regards to the formula for Integration by Parts. If I'm not mistaken, the formula states that
$$\int f'(x)g(x) = f(x)g(x)- \int f(x)g'(x)$$
However, in the case that I try to substitute an antiderivative with a valid constant, the formula does not appear to work. I attempted to use a simple product such as $7x(x^2)$, using $7x$ as $f'(x)$ and $x^2$ as $g(x)$ respectively. I found that using $\frac{7}{2}x^2+5$ for the antidervative for $7x$ does not work in the formula, while $\frac{7}{2}x^2$ without the constant does indeed work. I am sure that I am missing something, however, why do both of these solutions not work, even though both are valid antiderivatives?
|
Say you pick $f(x) = \frac{7}{2}x^2 + 5$, $g(x) = x^2$, and are doing
$$\int f'(x)g(x)\,dx = \int (7x)x^2\,dx = \int 7x^3\,dx = \frac{7}{4}x^4 + C.$$
Now, if you try using integration by parts with the anti-derivative you pick, you have
$$\begin{align*}
\int f'(x)g(x)\,dx &= f(x)g(x) - \int g'(x)f(x)\,dx\\
&= \left(\frac{7}{2}x^2+5\right)(x^2) - \int 2x\left(\frac{7}{2}x^2+5\right)\,dx\\
&= \frac{7}{2}x^4 + 5x^2 - \int (7x^3 + 10x)\,dx\\
&= \frac{7}{2}x^4 + 5x^2 - \left( \frac{7}{4}x^4 + 5x^2 + D\right)\\
&= \frac{7}{2}x^4 - \frac{7}{4}x^4 + 5x^2 -5x^2 - D\\
&= \frac{7}{4}x^4 - D.
\end{align*}$$That is, the same answer, up to a constant.
So long as you use the same antiderivative in both instances of $f(x)$ on the right hand side, it will work out. Recall that if $f(x)$ is one antiderivative, then every antiderivative is of the form $f(x)+D$, with $D$ a constant. So you would get:
$$\begin{align*}
(f(x)+D)g(x) &- \int (f(x)+D)g'(x)\,dx\\&= f(x)g(x) + Dg(x) -\int f(x)g'(x)\,dx - \int Dg'(x)\,dx\\
&= f(x)g(x) + Dg(x) - \int f(x)g'(x)\,dx - D\int g'(x)\,dx\\
&= f(x)g(x) + Dg(x) - \int f(x)g'(x)\,dx - D(g(x)+E)\\
&= f(x)g(x) + Dg(x) - Dg(x) - \int f(x)g'(x)\,dx - DE\\
&= f(x)g(x) - \int f(x)g'(x)\,dx
\end{align*}$$
(because that final constant gets "absorbed" into the indefinite integral).
|
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How to evaluate $\int_{1-\sqrt{1-t^2}}^{1+\sqrt{1-t^2}}(y^2-2y+t^2)dy$, where t is constant I need to evaluate the following one. Can't understand the method in my textbook.
$$\int_{1-\sqrt{1-t^2}}^{1+\sqrt{1-t^2}}(y^2-2y+t^2)dy$$
My textbook is to let $\alpha=1-\sqrt{1-t^2}$, $\beta=1+\sqrt{1-t^2}$, so it becomes
$$\int_\alpha^\beta\color{blue}{(y-\alpha)(y-\beta)}dy\color{red}{=\frac16(\alpha-\beta)^3}$$
And from here it can be solved normally.
How can I come up the idea to make it into "the blue part"?
And is there a formula that can make "the red part"?
|
The author set up the limits so that
\begin{align*}
\alpha + \beta & = 1 - \sqrt{1 - t^2} + 1 + \sqrt{1 - t^2}\\
& = 2\\
\alpha\beta & = (1 - \sqrt{1 - t^2})(1 + \sqrt{1 - t^2})\\
& = 1 - (1 - t^2)\\
& = t^2
\end{align*}
Therefore,
\begin{align*}
(y - \alpha)(y - \beta) & = y^2 - (\alpha + \beta)y + \alpha\beta\\
& = y^2 - 2y + t^2
\end{align*}
and evaluating the definite integral yields
\begin{align*}
\int_\alpha^\beta (y - \alpha)(y - \beta)~dy & = \int_\alpha^\beta [y^2 - (\alpha + \beta)y + \alpha\beta]~dy\\
& = \left[\frac{1}{3}y^3 - \frac{1}{2}(\alpha + \beta)y^2 + \alpha\beta y\right]\bigg|_\alpha^\beta\\
& = \frac{1}{6}\left[2y^3 - 3(\alpha + \beta)y^2 + 6\alpha\beta y\right]\bigg|_\alpha^\beta\\
& = \frac{1}{6}\left[2\beta^3 - 3(\alpha + \beta)\beta^2 + 6\alpha\beta^2 - (2\alpha^3 - 3(\alpha + \beta)\alpha^2 + 6\alpha^2\beta)\right]\\
& = \frac{1}{6}\left[2\beta^3 - 3\alpha\beta^2 - 3\beta^3 + 6\alpha\beta^2 - 2\alpha^3 + 3\alpha^3 + 3\alpha^2\beta - 6\alpha^2\beta\right]\\
& = \frac{1}{6}(-\beta^3 + 3\alpha\beta^2 - 3\alpha^2\beta + \alpha^3)\\
& = \frac{1}{6}(\alpha - \beta)^3
\end{align*}
|
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How to compare $\pi, e\cdot 2^{1/3}, \frac{1+\sqrt{2}}{\sqrt{3}-1}$ This is in the GRE exam where we are supposed to answer fast so I think there might be some trick behind this to allow us to do that. But so far the best I can do is to write $\frac{1+\sqrt{2}}{\sqrt{3}-1}=\frac{1+\sqrt{6}+\sqrt{2}+\sqrt{3}}{2}$ and compute the nominator with the value of square root 2 and 3 memorized. And as to $e\cdot 2^{1/3}$, I just don't see how to compare it to other two items without take cubic and compute. This whole process is very time consuming.
I have seen some tricks to compare say $2^\pi,\pi^2$. But the technique does not seem to apply here.
|
Here's a dirty decimal arithmetic method that presumes knowledge only of the bounds $1.41 < \sqrt{2} < 1.42$, $1.73 < \sqrt{3} < 1.74$---which you probably know if you're taking the GRE subject test---and the not-too-obscure fact $e^3 > 20$: Since $\sqrt 6 = \sqrt 2 \sqrt 3$ multiplying gives $2.43 < \sqrt{6} < 2.47$ Then, using the rationalization $$\frac{1 + \sqrt 2}{\sqrt 3 - 1} = \frac{1}{2} (1 + \sqrt 2 + \sqrt 3 + \sqrt 6) ,$$ and substituting the decimal values gives
$$\pi < 3.29 < \frac{1 + \sqrt 2}{\sqrt 3 - 1} < 3.32 .$$ Now, $3.32 < \frac{10}{3}$, so $$\left(\frac{1 + \sqrt 2}{\sqrt 3 - 1}\right)^3 < \left(\frac{10}{3}\right)^3 < 40 = 2 \cdot 20 < (\sqrt[3]{2} e)^3,$$
establishing
$$\color{#bf0000}{\boxed{\pi < \frac{1 + \sqrt{2}}{\sqrt{3} - 1} < \sqrt[3]{2} e}} .$$
Alternatively, here's a version that uses only estimates using fractions with small denominators (which themselves follow from the decimal bounds above): Since $$\frac{7}{5} < \sqrt{2} < \frac{10}{7} \qquad \textrm{and} \qquad \frac{12}{7} < \sqrt{3} < \frac{7}{4} ,$$ we have
$$\frac{1 + \sqrt{2}}{\sqrt{3} - 1} > \frac{1 + \frac{7}{5}}{\frac{7}{4} - 1} = \frac{16}{5}.$$ (Of course we can verify the bounds on $\sqrt{2}, \sqrt{3}$ without knowing anything about the numbers' decimal representations---just square all of the numbers, which reduces the problem to comparing rational numbers.) This is $3.2 > \pi$, but we can avoid decimal representations using $\frac{16}{5} > \frac{22}{7} > \pi$.
On the other hand,
$$\frac{1 + \sqrt{2}}{\sqrt{3} - 1} < \frac{1 + \frac{10}{7}}{\frac{12}{7} - 1} = \frac{17}{5} .$$
Since $e^3 > 20$, we have $(\sqrt[3]{2} e)^3 > 40$, but
$$\left(\frac{17}{5}\right)^3 < 40,$$ giving the order
$$\color{#bf0000}{\boxed{\pi < \frac{1 + \sqrt{2}}{\sqrt{3} - 1} < \sqrt[3]{2} e}} .$$
See this this follow-up question that discusses methods for deriving the inequality $e^3 > 20$ by hand.
|
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|
Question about the particular part in a non homogeneous recurrence Question about the particular part in the following non homogeneous recurrence :
$$a_n - 6a_{n-1} + 9a_{n-2} = n * 3^n $$
I have the following particual part : $$ a_n = n * 3^n$$
Now the solution of the homogenous part is $$x_1 = 3, x_2 = 3$$ and is of the form $$a_n = (An * B)* 3^n$$
What im struggling with is understanding how this helps me solve for the particual part. The solution to the particual part is: $$ a_n = (Cn^3 + Dn^2) * 3^n$$ how i got to it was $$n^2 * (Cn + D) * 3^n$$ now the part i dont understand is why is there $$(Cn + D)$$ and not just $ Cn * 3^n$
Edited
|
OK, now with the full recurrence. For the homogeneous part $a_{n + 2} - 6 a_{n + 1} + 9 a_n = 0$ we have a characteristic equation $r^2 - 6 r + 9 = 0$, which is just $(r - 3)^2 = 0$. You have two equal roots, so the solution to the homogeneous part is $a_n^h = (c_1 n + c_2) \cdot 3^n$.
The forcing function is $n \cdot 3^n$, for which the guess of a particular part would ordinarily be $(A n + B) \cdot 3^n$. But this is a solution to the homogeneous recurrence, you have to go up to two more (one for the $3^n$, one for the factor $n$), i.e., guess $a_n^p = (A n^3 + B n^2) \cdot 3^n$ (lower terms are part of the homogeneous solution, they would cancel out). Substitute in your recurrence:
$\begin{align*}
(A (n + 2)^3 + B (n + 2)^2) \cdot 3^{n + 2}
- 6 (A (n + 1)^3 + B (n + 1)^2) \cdot 3^{n + 1}
+ 9 (A n^3 + B n^2) \cdot 3^n
&= (54 A n + 54 A + 18 B) \cdot 3^n
\end{align*}$
Comparing to your recurrence you see:
$\begin{align*}
54 A
&= 1 \\
54 A + 18 B
&= 0
\end{align*}$
so $A = 1/54, B = - 1/18$. The full general solution is:
$\begin{align*}
a_n
&= a_n^h + a_n^p \\
&= \left(\frac{n^3}{54} - \frac{n^2}{18} + c_1 n + c_2\right) \cdot 3^n
\end{align*}$
|
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What is the method to factor $x^3 + 1$? In the solution to a problem, it's stated that
We see that $x^3+1=(x+1)(x^2-x+1)$.
Why is this, and what method can I use for similar problems with different coefficients?
The full problem is
Find the remainder when $x^{81}+x^{48}+2x^{27}+x^6+3$ is divided by $x^3+1$.
|
$x^3+1=(x+1)(x^2-x+1)$ is something one just knows, similar to how one knows that
$
x^2+2x+1=(x+1)^2
$, or $
x^2-1=(x+1)(x-1)
$
If you don't already know it, note that $(-1)^3+1=0$, which implies that $x-(-1)$ is a factor of $x^3+1$. Finding the second factor can be done through long division.
|
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|
Finding generating series from equation How can I solve $f(x)=(1+x+x^2)f(x^2)$, where $f$ is a generating series of a set? I just need the technique, not necessarily the solution.
|
$f(x) = a_0 + a_1 x + a_2 x^2 + \cdots$
$f(x^2) = a_0 + a_1 x^2 + a_2 x^4 + \cdots$
so $(1 + x + x^2) f(x^2) = a_0 + a_1 x + a_2 x^2 + \cdots + a_0 x + a_1 x^2 + a_2 x^3 + \cdots + a_0 x^2 + a_1 x^3 + a_2 x^4 + \cdots$
Given $f(x) = (1 + x + x^2) f(x^2)$ we equate the following powers of $x$:
$x^0$ therefore: $a_0 = a_0$
$x^1$ therefore: $a_1 = a_1 + a_0$
But these two equations shows that $a_0 = 0$ ($\checkmark$).
$x^2$ therefore: $a_2 = a_2 + a_1 + \underbrace{a_0}_{=0}$
Thus $a_1 = 0$ ($\checkmark$).
Can you continue?
|
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|
Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Note: $x$ and $y$ are obtuse angles.
My attempt that is not simple is as follows.
Expand both known constraints, so we have
\begin{align}
\cos x \cos y &=4/15\\
\sin x \sin y &=-1/15
\end{align}
Eliminate $x$ using $\sin^2 x +\cos^ 2 x=1$, we have
$$
225 \sin^4 y -210 \sin^2 y +1=0
$$
with its solution
$\sin^2 y = \frac{7\pm4\sqrt3}{15}$.
Then, $\cos^2 y = \frac{4(2\mp\sqrt3)}{15}$.
\begin{align}
\sin^2(2y) &= 4\cos^2 y\sin^2 y\\
&= 4 \times \frac{4(2\mp\sqrt3)}{15}\times \frac{7\pm4\sqrt3}{15} \\
\sin 2 y & = - \frac{4}{15}\sqrt{(2\mp\sqrt3)(7\pm4\sqrt3)}
\end{align}
$\sin 2y$ must be negative.
Edit
Thank you for your effort to answer my question. However, the existing answers seem to be more complicated than my attempt above.
By the way, I am confused in deciding which the correct pair among $(2\mp\sqrt3)(7\pm4\sqrt3)$ is.
|
Try to solve the first equation for $x$ and plug this in the second equation .
I got this for $y$:
$$\cos ^{-1}\left(\frac{1}{3}\right)=2
y+\cos
^{-1}\left(\frac{1}{5}\right)$$
|
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|
Calculate $\sum_{k=1}^n k2^{n-k}$ I want to Calculate $\sum_{k=1}^n k2^{n-k}$. Here's my attempt:
$$\begin{align}
&\sum_{k=1}^n k2^{n-k} = \sum_{k=1}^n k2^{n+1}2^{-k-1} =2^{n+1}\sum_{k=1}^n k2^{-k-1} & &\text{(1)} \\
&k2^{-k-1} = -\frac{d}{dk}2^{-k} & &\text{(2)}
\end{align}$$
Plugging $(2)$ to $(1)$,
$$\begin{aligned}
2^{n+1}\sum_{k=1}^n k2^{-k-1} &= 2^{n+1}\sum_{k=1}^n-\frac{d}{dk}2^{-k} \\
&= -2^{n+1}\sum_{k=1}^n\frac{d}{dk}2^{-k} \\
&= -2^{n+1}\frac{d}{dk}\sum_{k=1}^n2^{-k} \\
&= -2^{n+1}\frac{d}{dk}(1-2^{-n}) \\
&= -2^{n+1}\cdot 0 \\
&= 0 \end{aligned}$$
Obviously, I don't know much about summation and derivative. Can you please let me know where I am doing wrong and give me a hint for this question? Thanks!
|
Let $S=\sum_{k=1}^n k 2^{n-k}$
$ S = 2^{n-1} + 2(2^{n-2}) + 3(2^{n-3}) + \cdots + n(2^0)$
$ {S \over 2} = \quad\quad\quad1(2^{n-2}) + 2(2^{n-3}) + \cdots + (n-1)(2^0) + n(2^{-1})$
$\begin{align} (1-{1\over2})S
&= (2^{n-1} + 2^{n-2} + 2^{n-3} + \cdots + 1) - n(2^{-1})\cr
{S\over2}&= (2^n-1) - n(2^{-1})\cr\cr
S &= 2(2^n-1) - n
\end{align}$
|
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"url": "https://math.stackexchange.com/questions/3348824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Integration simple question I want to integrate $$\int\left(\frac{y}{2}\right)^\frac{2}{3}dy$$
Doing this I get $$\cfrac{3}{5}\left(\cfrac{y}{2}\right)^\cfrac{5}{3}$$
But the answer is $$\cfrac{6}{5}\left(\cfrac{y}{2}\right)^\cfrac{5}{3}$$
What step am I missing
|
$$\int (\frac{y}{2})^\frac{2}{3} dy = \frac{1}{2^\frac{2}{3}} \int y^\frac{2}{3} dy = \frac{1}{2^\frac{2}{3}} \frac{3}{5} (y)^\frac{5}{3} = \frac{2 \cdot 3}{2^\frac{5}{3} \cdot 5} (y)^\frac{5}{3} = \frac{6}{5}(\frac{y}{2})^\frac{5}{3}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Eigen Decomposition Check I am following the wiki entry on eigen dicomposition with the following matrix:
$$A = \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}$$
I wish to find a diagonalizing matrix T.S.
$$T^{-1}AT=\Lambda$$
$$AT=T\Lambda$$
where,
$$\Lambda = \begin{pmatrix}
x & 0 \\
0 & y \\
\end{pmatrix}$$
$$T = \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}$$
I am following with the wiki:
$$A\begin{pmatrix}a \\c \\\end{pmatrix}=x\begin{pmatrix}a \\c \\\end{pmatrix}$$
$$A\begin{pmatrix}b \\d \\\end{pmatrix}=y\begin{pmatrix}b \\d \\\end{pmatrix}$$
and the eigenvalues are:
$$\left|A-I\lambda\right| = \left|\begin{pmatrix}-\lambda & 1 \\1 & -\lambda \\\end{pmatrix}\right|=(\lambda^2-1)=(\lambda-1)(\lambda+1)$$
Hence, $\lambda_{1,2}=\pm 1$
Now I am finding the eigenvectors with:
$$A\begin{pmatrix}a \\c \\\end{pmatrix}=\begin{pmatrix}0 & 1\\1 & 0 \\\end{pmatrix}\begin{pmatrix}a\\c \\\end{pmatrix}=\begin{pmatrix}c \\a \\\end{pmatrix}=1\begin{pmatrix}a \\c \\\end{pmatrix}$$
Hence, $a=c$. And,
$$A\begin{pmatrix}b \\d \\\end{pmatrix}=\begin{pmatrix}0 & 1\\1 & 0 \\\end{pmatrix}\begin{pmatrix}b\\d \\\end{pmatrix}=\begin{pmatrix}d \\b \\\end{pmatrix}=-1\begin{pmatrix}b \\d \\\end{pmatrix}$$
Hence, $b=-d$.
So I take $T$ to be:
$$T = \begin{pmatrix}
a & b \\
a & -b \\
\end{pmatrix}=\begin{pmatrix}
1 & 1 \\
1 & -1 \\
\end{pmatrix}$$
with:
$$T^{-1} = \frac{1}{det(T)}\begin{pmatrix}
1 & -1 \\
-1 & -1 \\
\end{pmatrix}=\frac{1}{-1-1}\begin{pmatrix}
1 & -1 \\
-1 & -1 \\
\end{pmatrix}=\frac{1}{-2}\begin{pmatrix}
1 & -1 \\
-1 & -1 \\
\end{pmatrix}=\frac{1}{2}\begin{pmatrix}
-1 & 1 \\
1 & 1 \\
\end{pmatrix}$$
Now, when I try to calculate $T^{-1}AT$ I get:
$$\Lambda =\frac{1}{2}\begin{pmatrix}
-1 & 1 \\
1 & 1 \\
\end{pmatrix}\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}\begin{pmatrix}
1 & 1 \\
1 & -1 \\
\end{pmatrix}=\frac{1}{2}\begin{pmatrix}
1 & -1 \\
1 & 1 \\
\end{pmatrix}\begin{pmatrix}
1 & 1 \\
1 & -1 \\
\end{pmatrix}=\frac{1}{2}\begin{pmatrix}
0 & -2 \\
2 & 0 \\
\end{pmatrix}=\begin{pmatrix}
0 & -1 \\
1 & 0 \\
\end{pmatrix}$$
Where have I gone wrong here?
|
As a good check you can compute $TT^{-1}$, and you'll find you don't get the identity matrix. So the mistake must be in your computation of $T^{-1}$.
It looks to me like you've set it up wrong. Seems you're trying to use the formula $$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$$ but you've mixed up some of the terms.
|
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|
if $4n^2-1=3m^2$ has a positive integer solution, show that $2n-1$ always square Let $n\in \mathbb{N}$, if $4n^2-1=3m^2$ has a positive integer solution, show that $2n-1$ is a perfect square.
For example $n=1$, it is clear.
and $n=13$ because $$ 4\cdot 13^2-1=3\cdot 15^2$$ and $$2n-1=25=5^2$$
|
An Experimental approach:
There are infinitely many integers like n such that:
$2n-1=k^2$; $k∈N$
They make following series:
$n=13, 41, 61, 181, 221 . . .$
therefore we may write:
$4\times 13^2-1=3\times 15^2$
$4\times 41^2-1=3\times 47.34^2$
$4\times 61^2-1=3\times 70.4^2$
$4\times 181^2-1=3\times 209^2$
$4\times 221^2-1=3\times 255.19^2$
.
.
For $4 . n^2-1=3 .m^2$ The condition $2n-1=k^2$ is necessary but not sufficient. That is if equation ($4 . n^2-1=3 .m^2$) has integer solution then ($2n-1$) is definitely is a square.
|
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|
Are there integer solutions such that the sum of the roots of x and y equal the root of a prime p? How could one prove whether or not integer pairs (x, y) exist such that $\sqrt{x} + \sqrt{y} = \sqrt{p}$, where p is prime? `
|
A slightly less elementary approach. Consider the polynomial whose roots are the pairwise sums of the roots of the minimal polynomials of $\sqrt x$ and $\sqrt y$: $$(z - \sqrt x - \sqrt y)(z - \sqrt x + \sqrt y)(z + \sqrt x - \sqrt y)(z + \sqrt x + \sqrt y) \\
= ((z - \sqrt x)^2 - y) ((z + \sqrt x)^2 - y) \\
= (z^2 + x - y - 2z\sqrt x) (z^2 + x - y + 2z\sqrt x) \\
= (z^2 + x - y)^2 - 4z^2x \\
= z^4 - 2(x+y) z^2 + (x-y)^2$$
If $\sqrt p$ is a root then its minimal polynomial $z^2 - p$ is a factor. Since we have two monic polynomials, that implies that $p$ is a factor of $(x-y)^2$. The tail of Elliot's argument wraps it up.
We could instead have gone on to do a long division, finding that the remainder is $$p^2 - 2(x+y)p + (x-y) ^2 =0$$ as in Fimpellizieri's answer.
|
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|
Formulas for Sequences Removing Multiples of 2, 3, and 5 First off, I am a programmer so please excuse if some of the terms I use are not the correct mathematical terms. I was working on devising a function to improve one of my prime number generation algorithms. With this in mind, I first set out to find the formulas for a sequence removing multiples of 2 and 3:
\begin{array}{c|c}
x&y\\ \hline
0&5\\ \hline
1&7\\ \hline
2&11\\ \hline
3&13\\ \hline
4&17\\ \hline
5&19\\ \hline
6&23\\ \hline
7&25\\ \hline
8&29\\ \hline
\vdots&\vdots
\end{array}
The equations that I came up for for this sequence are as follows:
$$y = 3x + 5 - x \bmod 2$$
$$x = \left\lfloor\frac{y - 5 + [y \bmod 3 \neq 0]}{3}\right\rfloor$$
After this, I tried to do the same for a sequence removing multiples of 2, 3, and 5:
\begin{array}{c|c}
x&y\\ \hline
0&7\\ \hline
1&11\\ \hline
2&13\\ \hline
3&17\\ \hline
4&19\\ \hline
5&23\\ \hline
6&29\\ \hline
7&31\\ \hline
8&37\\ \hline
9&41\\ \hline
10&43\\ \hline
11&47\\ \hline
12&49\\ \hline
13&53\\ \hline
\vdots&\vdots
\end{array}
While I think I found an equation to get $y$ from a value of $x$, I cannot find a way to get the value of $x$ from a given value $y$.
$$y = 4x + 7 - 2\left\lfloor\frac{1}{8}x\right\rfloor - 2\left[\{2, 3, 6\} \ \text{contains}\ (x \bmod 8)\right] - 4\left[\{4, 5, 7\} \ \text{contains}\ (x \bmod 8)\right]$$
$$x =\ ?$$
I am wondering if an equation that produces the corresponding value of $x$ for a given value of $y$ for the aforementioned sequence exists, and if indeed it does, what the equation is.
|
The first one is:
$$y=5+6\bigg\lfloor \frac{x}{2} \bigg\rfloor + 2(x\mod2)$$
Although arguably $x(0)=1$, so:
$$y=1+6\bigg\lfloor \frac{x+1}{2} \bigg\rfloor - 2(x\mod2)$$
The second, with $x(0)=1$, is:
$$y=15+30\bigg\lfloor \frac{x}{8} \bigg\rfloor +4(x\mod8-3.5)-\frac{(x\mod8-3.5)}{|x\mod8-3.5|}\cdot2(\bigg\lfloor\frac{(|x\mod8-3.5|+0.5)}{2}\bigg\rfloor\mod2)$$
|
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|
Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$ I'm supposed to prove that the product of three successive natural numbers the middle of which is the square of a natural number is divisible by $60$. Here's my attempt.
My Attempt:
$$\text{P}=(n^2-1)n^2(n^2+1)=n(n-1)(n+1)[n(n^2+1)]$$
It is now enough to prove that $n(n^2+1)$ is divisible by $10$. But for $n=4$, $4(17)\ne10\lambda$ but for $n=4$, $\text{P}$ is $4080=60\cdot68$ which means apart from just being a multiple of $6$, $n(n-1)(n+1)$ is actually helping $n(n^2+1)$ with a $5$ to sustain divisibility by $60$.
How to tackle this? Thanks
|
$$
\frac{n^2(n^2-1)(n^2+1)}{60} = 2n\binom{n+2}5 + 2\binom{n+1}4 + \binom{n+1}3.
$$
|
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|
Solving Diophantine Equations Using Inequalities Prove that if $x$ and $y$ are positive integers such that $xy$ divides $x^2+y^2+1$ then prove that the quotient is $3$.
Now as $x^2+y^2 +1 > 2xy$ we have $$\frac{x^2+y^2+1}{xy} \geq 3$$ So now we have to prove that $$\frac{x^2+y^2+1}{xy} \leq 3$$
Now if we prove that for positive integers $x$, $y$ and $z$ we have $x^2+y^2+z^2 \leq 3xyz$ then for $z=1$ our problem gets solved but I don't know how to prove it.
|
Suppose $x y \mid (x^2 + y^2 + 1)$ and without loss of generality assume that $x < y$.
Then $(x, y)$ is a pair of positive integers with $x < y$ such that $x \mid (y^2 + 1)$ and $y \mid (x^2 + 1)$. Notice that this implies that $\gcd (x, y) = 1$.
Therefore there exist positive integers $w, z$ such that $x^2 + 1 = w y$ and $y^2 + 1 = x z$. If $x > 1$, then
$$w = \frac {x^2 + 1} y < \frac {x^2 + x} y = \frac {x(x + 1)} y \le \frac {x y} y = x$$
Also, from $1 = wy - x^2$ and $1 = x z - y^2$, we have
$$wy - x^2 = x z - y^2 \quad\implies\quad y(w + y) = x(x + z) \quad\implies\quad w + y = \frac {x (x + z)} y$$
Therefore
$$w^2 + 1 = w^2 + wy - x^2 = w(w + y) - x^2 = w \, \frac {x (x + z)} y - x^2 = x \left ( w\, \frac {x + z} y - x \right )$$
where we know that $\frac {x + z} y$ is an integer because $\frac {x (x + z)} y$ is an integer and $\gcd(x, y) = 1$.
We have found another pair $(w, x)$ with $w < x$ such that $w \mid (x^2 + 1)$ and $x \mid (w^2 + 1)$. Moreover,
$$\frac {w^2 + x^2 + 1} {w x} = \frac {w^2 + wy} {wx} = \frac {w + y} x = \frac {wy + y^2} {xy} = \frac {x^2 + y^2 + 1} {x y}$$
so the ratio is the same. If $w > 1$, then we can find another pair $(v, w)$ with $v < w$ such that $v \mid (w^2 + 1)$ and $w \mid (v^2 + 1)$, and so on. Iterating this procedure, we must end at a pair $(1, b)$, which satisfies the property if and only if $b \mid (1^2 + 1) = 2$, so either $b = 1$ or $b = 2$. Since
$$\frac {1^2 + 1^2 + 1} {1 \cdot 1} = 3 \qquad \frac {1^2 + 2^2 + 1} {1 \cdot 2} = \frac 6 2 = 3$$
we have proved that $3$ is the only possible value for the ratio.
|
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|
Find all natural numbers $n$ such that $n+1$ divides $3n+11$ Following the example of my teacher:
Find all natural numbers $n$ such that $n-2$ divides $n+5$.
$$n+5 = n-2+7$$
As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:
*
*$n-2=-7 \Leftrightarrow n = -5 $
*$n-2=-1 \Leftrightarrow n = 1 $
*$n-2=7 \Leftrightarrow n = 3 $
*$n-2=-1 \Leftrightarrow n = 9 $
So $S = \{ 1, 3, 9 \}$
I decompose $n+1$ the exact same way:
$$n+1 = 3n+11 - 2(n+5)$$
But I'd get stuck as $2(n+5)$ since only $-2$, $-1$, $1$ and $2$ are divisors, which don't satisify the equation as I'd hoped:
*
*$3n+11 = -2 \Leftrightarrow n =\frac{-13}{3}$
*$3n+11 = -1 \Leftrightarrow n = -4$
*$3n+11 = 1 \Leftrightarrow n =\frac{-10}{3}$
*$3n+11 = 1 \Leftrightarrow n = -3$
Any clues?
|
You can use polynomial long division, which is useful if you have a more complex expression:
$$
\require{enclose}
\begin{array}{rll}
3 && \\[-3pt]
n+1 \enclose{longdiv}{\ 3n+11}\kern-.2ex \\[-3pt]
\ \ - \underline{\ (3n+3)} && \\[-3pt]
8 && \\[-3pt]
\end{array}
$$
Therefore $\frac{3n+11}{n+1} = 3 + \frac{8}{n+1}$. Can you continue?
|
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|
How is this Taylor expansion computed? I am reading a paper, where we consider the following function,
$$F(r) = \frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)$$
where $b>0$ is a constant such that $r>2/b.$ After stating this definition the author writes that the taylor expression of $F$ is,
$$F(r) = b\frac{r^2}{4}-\frac{\ln(r)}{b} -\frac{1}{2b} -\frac{\ln(2b)}{b} + \mathcal{O}(b^{-3}r^{-2}).$$
I know how to compute taylor series, but I am not sure what the taylor series is centered around. I have computed the derivatives upto order $2$.
$$F'(r) = \frac{1}{2}\sqrt{b^2r^2-4}, \quad F''(r) = \frac{b^2r}{2\sqrt{b^2r^2-4}}.$$
Any explanations regarding the expansion would be much appreciated.
|
$$F(r) = \frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)\tag{1}$$
Expand the square-root term with $1/(b^2r^2)$ as the small variable,
$$\sqrt{b^2r^2-4}=br\sqrt{1-\frac{4}{b^2r^2}}=br \left( 1-\frac{2}{b^2r^2} \right) \tag{2}=br - \frac{2}{br} $$
Similarly, expand the log term as
$$\ln(\sqrt{b^2r^2-4} + br) =\ln \left[br \left(1+ \sqrt{1-\frac{4}{b^2r^2}} \right) \right]$$
$$=\ln \left[br \left( 2-\frac{2}{b^2r^2}\right) \right] = \ln(2br) + \ln \left( 1-\frac{1}{b^2r^2}\right) = \ln(2b) + \ln r - \frac{1}{b^2r^2}\tag{3}$$
Plug the two Taylor expansions (2) and (3) into (1) to get
$$F(r) = b\frac{r^2}{4}-\frac{\ln(r)}{b} -\frac{1}{2b} -\frac{\ln(2b)}{b} $$
|
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Real Analysis Inf and Sup question I am hung up on this question for real analysis ( intro to anaylsis ).
Find $\inf D$ and $\sup D$
$$\mathrm{D}=\left\{\frac{m+n\sqrt{2}}{m+n\sqrt{3}} :m,n\in\Bbb{N}\right\}$$
I have spent enough time staring at this thing that I know the $\sup D=1$ and $\inf D=\frac{\sqrt{2}}{\sqrt{3}}$.
for $\sup D$:
$$m+n\sqrt{2}<m+n\sqrt{3}\implies\frac{m+n\sqrt{2}}{m+n\sqrt{3}}<1$$
so $1$ is an upper bound for $D$, and then for the confirmation that 1 is the least upper bound I can prove by contradiction that $\sup D$ cannot be less than $1$, because I could always find a $d \in D$ such that $$\sup D<d<1$$, which is the contradiction since no $d \in D$ can be greater than $\sup D$.(proof omited)
So my problem is with $\inf D$. I am having trouble establishing that $\frac{\sqrt{2}}{\sqrt{3}}$ is a lower bound. I am just not seeing it. The intuition is that if $m$ is small and $n$ is large than the fraction $\frac{\sqrt{2}}{\sqrt{3}}$ dominates the expression, however it will always be slightly greater than $\frac{\sqrt{2}}{\sqrt{3}}$. Analytically I am just not able to show it.
Any help would be greatly appreciated
|
In general if $\alpha$ is a lower bound of $X$ and $Y$ is a nonempty subset of $X$ with $\alpha = \text{inf}(Y)$, then $\alpha = \text{inf}(X)$.
By fleablood's opening argument, we know that $\alpha = \frac{\sqrt 2}{\sqrt 3}$ is a lower bound for $D$.
Let $\mathrm{E}=\left\{\frac{1+n\sqrt{2}}{1+n\sqrt{3}} :n\in\Bbb{N}\right\}$.
We are going to show that the infimum of $E$ is equal to $\alpha$.
We have
$\tag 1 \frac{1+n\sqrt{2}}{1+n\sqrt{3}}= \frac{1}{1+n\sqrt{3}} + \frac{\sqrt{2}}{\frac{1}{n}+\sqrt{3}}$
Let $\varepsilon \gt 0$.
Find an $n$ such that both
$\quad \big| \frac{1}{1+n\sqrt{3}} \big| \lt \frac{\varepsilon}{2}$
$\quad \big| \frac{\sqrt{2}}{\frac{1}{n}+\sqrt{3}} -\alpha \big|\lt \frac{\varepsilon}{2}$
are true.
Then, using $\text{(1)}$ and substitution, associativity and the triangle inequality,
$\quad \big| \frac{1+n\sqrt{2}}{1+n\sqrt{3}} - \alpha \big| = \big| (\frac{1}{1+n\sqrt{3}} + \frac{\sqrt{2}}{\frac{1}{n}+\sqrt{3}}) -\alpha \big| = \big| \frac{1}{1+n\sqrt{3}} + (\frac{\sqrt{2}}{\frac{1}{n}+\sqrt{3}}) -\alpha) \big| \le$
$\quad \quad \big| \frac{1}{1+n\sqrt{3}} \big| + \big|(\frac{\sqrt{2}}{\frac{1}{n}+\sqrt{3}}) -\alpha) \big| \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$
we conclude that there are numbers in $E$ that are arbitrarlly close to the lower bound $\alpha$ of $E$. But then $\text{inf}(E) = \alpha$, as was to be shown.
|
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|
Factorizing $x^5+1$ as a product of linear and quadratic polynomials. I am encountering some trouble with this question:
Factorize $$x^5+1$$ as a product of real linear and quadratic polynomials.
I know that if we subtract 1 from $x^5+1$, we get that $x^5 = -1$, but I am unsure where to go from here.
Can anyone help with this? Thanks.
|
If you remember the formula for the sum of a geometric progression, you will have
$$\begin{align} & 1 + x + x^2 + \cdots + x^{n-1} = \frac{1-x^n}{1-x}\\
\iff & x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1)
\end{align}
$$
When $n = 2k+1 $ is odd, substitute $x$ by $-x$ and move the minus sign around, you will get
$$x^{2k+1} + 1 = (x+1)(x^{2k} - x^{2k-1} + x^{2k-2} + \cdots - x + 1)$$
In particular, for $n = 5$, this gives you
$$x^5 + 1 = (x+1)(x^4 - x^3 + x^2 - x + 1)$$
This sort of factorization is something you should know. If not, you know it now.
Let's look at the coefficients of $x^k$ in second factor, they are $1, -1, 1, -1, 1$.
Did you notice it is palindromic, i.e stay the same if you reverse the order. For this sort of polynomial. If $r \ne 0$ is a root, so does $\frac1r$. When the number of this list of coefficients is odd, you can simplify the expression by grouping them in powers of $x+\frac1x$.
Let's see what happens to the factor. We have
$$\begin{align} x^4 - x^3 + x^2 - x + 1
&= x^2(x^2 - x + 1 - x^{-1} + x^{-2})\\
&= x^2((x^2 + 2 + x^{-2}) - (x + x^{-1}) - 1)\\
&= x^2((x + x^{-1})^2 - (x + x^{-1}) - 1 )\\
&= x^2\left[\left(x + x^{-1} -\frac12\right)^2 - \frac{5}{4}\right]\\
\end{align}
$$
Another identity you should remember is for positive $a$,
$$y^2 - a = (y + \sqrt{a})(y - \sqrt{a})$$
Apply this identity to above expression, you get
$$x^4 - x^3 + x^2 - x + 1
= x^2\left(x + x^{-1} -\frac{1+\sqrt{5}}{2}\right)
\left(x + x^{-1} -\frac{1-\sqrt{5}}{2}\right)$$
Move the $x$ factor around to get rid of the $x^{-1}$ and combine with the $x+1$ factor, one obtain
$$x^5 + 1 = (x+1)\left(x^2 - \frac{1+\sqrt{5}}{2} x + 1 \right)
\left(x^2 - \frac{1-\sqrt{5}}{2} x + 1 \right)$$
This is the decomposition of $x^5+1$ into product of linear and quadratic factors you are seeking.
|
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Given three positive numbers $a,b,c$. Prove that $\sum\limits_{sym}\frac{a+b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$ .
(A problem due to Mr. Le Khanh Sy). Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{sym}\frac{a+ b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$$
I'm eagerly interested in learning one method which assumes $c\not\equiv {\rm mid}(\!a, b, c\!)$. But if $c\equiv {\rm mid}(\!a, b, c\!)$:
$$2\sqrt{(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)(\!a+ b+ c\!)}\leqq c(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)+ \frac{a+ b+ c}{c}= \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}$$
We need to prove
$$\begin{align} \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}\leqq \frac{a+ b}{c}+ \frac{b+ c}{a} & + \frac{c+ a}{b}\Leftrightarrow 1+ \frac{c^{2}}{ab}\leqq \frac{c}{a}+ \frac{c}{b}\Leftrightarrow \\ & \Leftrightarrow (\frac{c}{a}- 1)(\frac{c}{b}- 1)\leqq 0\Leftrightarrow \frac{(c- a)(c- b)}{ab}\leqq 0 \end{align}$$
Who can teach me what would we do if $c\not\equiv {\rm mid}(\!a, b, c\!)$ ? I am goin' to set a bounty, thank u so much
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By the Vornicu Schur, we have
$$\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}} \right)\left( {\frac{a}{c} + \frac{c}{b} + \frac{b}{a}} \right) - \left( {a + b + c} \right)\left( {\frac{a}{{bc}} + \frac{b}{{ca}} + \frac{c}{{ab}}} \right) = \sum \frac{(a-b)(a-c)}{a^2} \geqslant 0.$$
Theforere we will show that
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geqslant 2\sqrt{\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}} \right)\left( {\frac{a}{c} + \frac{c}{b} + \frac{b}{a}} \right)},$$
equivalent to
$$\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}} \right)+\left( {\frac{a}{c} + \frac{c}{b} + \frac{b}{a}} \right) \geqslant 2\sqrt{\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}} \right)\left( {\frac{a}{c} + \frac{c}{b} + \frac{b}{a}} \right)}.$$
It's $x+y \geqslant 2\sqrt{xy}.$ The proof is completed.
Note. The inequality follows from the famous inequality
$$(x+y+z)^2 \geqslant 4(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C).$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3373080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
}
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