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Proof by induction $\sum_{k=1}^{2^n} \frac1k \ge 1 + \frac{n}2 $ Show that for all $n \in N , n \ge 1$ :
$\sum_{k=1}^{2^n} \frac1k \ge 1 + \frac{n}2 $
Here is what I got:
I. $A(1)$: $\sum_{k=1}^{2} \frac1k = 1.5 \ge 1.5 \;\; \implies A(1) \text{ is correct} $
II. $A(n)$ is correct: $\sum_{k=1}^{2^n} \frac1k \ge 1 + \frac{n}2 $
III. Is $A(n+1)$ correct? $\sum_{k=1}^{2^n+1} \frac1k \ge 1 + \frac{n+1}2 $
IV. Now I need the proof but I don't get the solution for it. Trying for hours - so hope someone can give me an answer with a reasonable explanation. Thank you very much!
|
Base case:
Let $n=1$. Then $\sum_{k=1}^{2^1} \frac{1}{k} \geq 1 + \frac{1}{2}$ by evaluation / inspection.
Induction step (examining the $n+1$ case against the $n$ case):
$(\sum_{k=1}^{2^{n+1}} \frac{1}{k}) - (\sum_{k=1}^{2^n} \frac{1}{k}) = \sum_{k=2^n+1}^{2^{n+1}} \frac{1}{k}$
$(1+\frac{n+1}{2}) - (1+\frac{n}{2}) = \frac{1}{2}$
We only need to prove that $(\sum_{k=2^n+1}^{2^{n+1}} \frac{1}{k}) \geq \frac{1}{2}$.
Note that there are $2^n$ terms in the lefthand summation, where all values of $\frac{1}{k}$ are $\geq \frac{1}{2^{n+1}}$.
Therefore $(\sum_{k=2^n+1}^{2^{n+1}} \frac{1}{k}) \geq (2^n)(\frac{1}{2^{n+1}}) \geq \frac{1}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Knowing that m and n are two positive integers, solve the equation m/n = n.m (ex.: 5/2 = 2.5) Knowing that $m$ and $n$ are two positive integers, find all the solutions for the equation:
$\frac{m}{n} = n.m$
For example, the pair $(m=5, n=2)$ is a solution because $\frac{5}{2} = 2.5$ while $(m=294, n=17)$ comes close to a solution but it is not because $\frac{294}{17} = 17.29411...$ .
|
$m/n = n.m= n + m/10^k$ so $10^kn^2 + mn - 10^km$
So by quadratic equation $n = \frac{-m + \sqrt{m^2 + 4m*10^{2k}}}{2*10^k} $
Notice:
m has k digits.
So n = $\frac{-m + \sqrt{m^2 + 4m*10^{2k}}}{2*10^k}$ has $\lceil k/2 \rceil$ digits. But m/n $\approx$ n then has ${\lceil k/2 \rceil}$ digits, so $m$ has between $\lfloor k/2 \rfloor$ and ${\lceil k/2 \rceil}$ digits. But m has $k $ digits. So k = 1 or 2 and m has one or 2 digits and n has 1 digit. $nm$ is a multiple of $10^k$ but m/n has only 1 digit so k = 1 and m has 1 digit.
$n = \frac{-m + \sqrt{m^2 + 400}}{20} $
So $m^2 + 400$ is a perfect square. Only possibility is m =5 so n = 2.
|
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|
Determine the complex contour integral $\oint \limits_{C} \frac{2}{z^3+z}dz$ without using Residue Theorems
Without residue theory, determine $$\oint \limits_{C} \frac{2}{z^3+z}dz$$ if $C: \big|~z~-~\frac{i}{2}~\big|=1$ is positively oriented.
We first find that our integrand has three distinct singular points $z=\{-i,~0,~i\}$
Now if we draw a sketch of $C$, we see that only two of these singular points are within $C$.
We can thus rewrite our integral as $$\oint \limits_{C} \frac{2}{z^3+z}dz = 2 \oint \limits_C \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz$$
Now we may define two closed, piecewise smooth curves $C_1, C_2$ around each singular point in $C$ as follows
We then have that \begin{align}\oint \limits_{C} \frac{2}{z^3+z}dz &= 2 \oint \limits_C \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz \\ &= 2 \bigg[ \underbrace{\oint \limits_{C_1} \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz}_{\displaystyle I_1} + \underbrace{\oint \limits_{C_2} \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz}_{\displaystyle I_2} \bigg] \end{align}
Now let $f(z) = \frac{1}{z(z+i)}$ then, from Cauchy Integral Formula, we know that \begin{align}I_1 &=\oint \limits_{C_1} \frac{f(z)}{(z-i)}dz \\ &= 2\pi i ~f(i) \\ &=-\pi i\end{align}
Now let $g(z) = \frac{1}{(z+i)(z-i)}$, then from Cauchy Integral Formula, we know that
\begin{align}I_2 &= \oint \limits_{C_2} \frac{g(z)}{z}dz \\ &= 2\pi i ~ g(0) \\ &= 2\pi i\end{align}
So finally we have that \begin{align}\oint \limits_C \frac{2}{z^3 + z}dz = 2\big( -\pi i + 2\pi i\big) = 2 \pi i\end{align}
Does this seem correct?
|
It's correct. Essentially what you've done is exactly how the proof of the residue theorem goes anyway - you surround each isolated pole with a small contour and apply Cauchy's theorem to each individual pole. The result is the sum of the residues, multiplied by $2\pi i$, which is the residue theorem. Now you know why the residue formula is what it is!
|
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|
Calculate the limit $\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$ $$\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$$
My attempt
\begin{align}
\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)} &= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-n+1)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{1 + (-2n)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}
\end{align}
since $\left(\frac{n^2+1}{n^2} \right) \to 1$ when $n\to\infty$
$$\lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)} = \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}$$
Also I tried to make similar simplifications into the braсkets but nothing happens and no proof that limit $= 1$ or whatever.
And here is a rule for the task. This is a limit of a sequence so no usage of functional simplifications and derives are allowed. If you have really beautiful solution for the task then post it anyway.
|
For all $n$, we have $0 < \frac{n^2}{n^2 + 1} < 1.$
For $n > 1$, we have $\frac{n-1}{n+1} > 0.$
Therefore, for $n > 1$,
$$ \left(\frac{n^2}{n^2 + 1}\right)^{(n-1)/(n+1)} < 1.$$
On the other hand, $\frac{n^2 + 1}{n^2} > 1$ for all $n$, and for $n > -1$ we have $\frac{2}{n+1} > 0$, so
\begin{align}
\left(\frac{n^2}{n^2 + 1}\right)^{(n-1)/(n+1)}
&= \left(\frac{n^2}{n^2 + 1}\right) \cdot
\left(\frac{n^2}{n^2 + 1}\right)^{-2/(n+1)} \\
&= \left(\frac{n^2}{n^2 + 1}\right) \cdot
\left(\frac{n^2 + 1}{n^2}\right)^{2/(n+1)} \\
&> \frac{n^2}{n^2 + 1}. \\
\end{align}
In short, for any $n > 1$,
$$ \frac{n^2}{n^2 + 1} < \left(\frac{n^2}{n^2 + 1}\right)^{(n-1)/(n+1)} < 1.$$
Finding the limit should now be simple enough.
|
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|
Trigonometry equation, odd-function. So I have the following equation:
$\sin\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{4}\right)=0$
It should be solved using the fact that Sin is an odd function, I can not really get the gripp of how and what I need to do? Any sugestions?
|
Sorry that this approach does not use the oddness of the Sine function, after I finishing writing this it was too late for me to turn back and utilize that property.
Using trigonometric addition formulae as mentioned in the comments:
$$\sin(A \pm B)= \sin A\cos B\pm\cos A\sin B$$
$$\cos(A \pm B)= \cos A\cos B\mp\sin A\sin B$$
Write $$\sin\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{4}\right)=0$$
as $$\sin(x)\cos\left(\frac{\pi}{6}\right)-\cos(x)\sin\left(\frac{\pi}{6}\right)+\cos(x)\cos\left(\frac{\pi}{4}\right)-\sin(x)\sin\left(\frac{\pi}{4}\right)$$
$$=\frac{\sqrt{3}}{2}\sin(x)-\frac{1}{2}\cos(x)+\frac{1}{\sqrt{2}}\cos(x)-\frac{1}{\sqrt{2}}\sin(x)$$
$$=\cos(x)\left(\frac{1}{\sqrt{2}}-\frac{1}{2}\right)+\sin(x)\left(\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\right)$$
$$\implies\cos(x)\left(\frac{2-\sqrt{2}}{2\sqrt{2}}\right)+\sin(x)\left(\frac{\sqrt{6}-2}{2\sqrt{2}}\right)=0$$
Since
$$\cos(x)\ne 0$$
$$\implies\left(\frac{2-\sqrt{2}}{2\sqrt{2}}\right)+\tan(x)\left(\frac{\sqrt{6}-2}{2\sqrt{2}}\right)=0$$
$$\implies\tan(x)\left(\frac{\sqrt{6}-2}{2\sqrt{2}}\right)=\left(\frac{\sqrt{2}-2}{2\sqrt{2}}\right)$$
$$x=\tan^{-1}\left(\frac{\sqrt{2}-2}{\sqrt{6}-2}\right)$$
$$x=\tan^{-1}\left(\frac{(\sqrt{2}-2)(\sqrt{6} + 2)}{(\sqrt{6}-2)(\sqrt{6} + 2)}\right)=\tan^{-1}\left(\frac{(\sqrt{2}-2)(\sqrt{6} + 2)}{2}\right)$$
$$=\tan^{-1}{(\sqrt{2}+\sqrt{3}-\sqrt{6}-2)}+k\pi$$
where $k \in \mathbb{N}$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Finding a sequence satisfying this recurrence relation? I just don't even know where to start with this,
Find a sequence $(x_n)$ satisfying the recurrence relation:
$2x_n$$_+$$_2$ = $3x_n$$_+$$_1$ + $8x_n$ + $3x_n$$_-$$_1$ Where n is a natural number and
$x_0$ = -1,
$x_1$ = 3 and
$x_2$ = 3
Thanks in advance!
|
Let $2x_{n+2} = 3 \, x_{n+1} + 8 \, x_{n} + 3 \, x_{n-1}$, with $x_{0} = -1,
x_{1} = 3$ and $x_{2} = 3$ for which letting $x_{n} = p^{n}$ the equation
$$2 \, p^{3} - 3 \, p^{2} - 8 \, p - 3 = 0$$
is obtained. This equation can be factored to $(p-3)(p+1)(2p+1) = 0$ and leads to the roots $p \in \{ 3, -1, -1/2 \}$. From this the general form of $x_{n}$ is
$$x_{n} = a_{0} \, 3^{n} + a_{1} \, (-1)^{n} + a_{2} \, \left( - \frac{1}{2} \right)^{n}.$$
Applying the initial conditions yields:
\begin{align}
-1 &= a_{0} + a_{1} + a_{2} \\
3 &= 3 \, a_{0} - a_{1} - \frac{a_{2}}{2} \\
3 &= 9 \, a_{0} + a_{1} + \frac{a_{2}}{4}
\end{align}
which leads to $a_{0} = \frac{1}{2}$, $a_{1} = - \frac{3}{2}$, $a_{2} = 0$.
The resulting sequence is generated by
\begin{align}
x_{n} = \frac{3}{2} \, \left( 3^{n-1} + (-1)^{n-1} \right)
\end{align}
|
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|
Simultaneous equations How can I find the solution $(3, 3, \frac{-3}{2})$ from the following system of equations:
$x+(y-1) z = 0$
$(x-1) z+y = 0$
$x (y-1)-y+2 z = 0$
I have done eq1 - eq2 to find the other solutions. How would i get to the solution $(3, 3, \frac{-3}{2})$
|
$x+(y-1) z = 0$ : (eq1)
$(x-1)z+y = 0$ : (eq2)
$x (y-1)-y+2 z = 0$ : (eq3)
Rearranging eq1 and we have: $x=-(y-1)z$. Subbing this into eq2 and eq3 gives:
$(-(y-1)z-1)z+y=0$ : (eq4)
$-(y-1)z(y-1)-y+2z=0$ : (eq5)
Rearranging eq5 we get: $z=\frac{y}{2-(y-1)^2}$
Subbing this into eq4 gives:
$$\left(-(y-1)\frac{y}{2-(y-1)^2}-1\right)\frac{y}{2-(y-1)^2}+y=0$$
$$\left(-\frac{y(y-1)}{-y^2+2y+1}-1\right)\frac{y}{-y^2+2y+1}+y=0$$
$$(-y(y-1)+y^2-2y-1)y+y(-y^2+2y+1)^2=0$$
$$y(-y-1+y^4-4y^3+2y^2+4y+1)=0$$
$$y(y^4-4y^3+2y^2+3y)=0$$
$$y^2(y^3-4y^2+2y+3)=0$$
$$y^2(y-3)(y^2-y-1)=0$$
Hence: $y=0$, $y=3$, $y=\frac{-1\pm\sqrt{5}}{2}$
Subbing back into $x$ and $z$ gives the four solutions:
$$(x,y,z) = (0,0,0),\left(3,3,-\frac{3}{2}\right),\left(\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2},1\right), \left(\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2},1\right)$$
|
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|
Show that T is a rotation through an angle, and find the angle $T\left[\begin{array}{c}x \\ y \end{array}\right] = \frac{1}{\sqrt{2}}\left[\begin{array}{c}x+y \\ -x+y \end{array}\right]$
This is my attempt, but I'm stuck. Can someone continue?
$T\left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{cc} \cos θ & -\sin θ \\ \sin θ & \cos θ \end{array}\right] \left[\begin{array}{c}x \\ y \end{array}\right]$
$= \left[\begin{array}{c} x\cos θ - y\sin θ \\ x\sin θ + y\cos θ \end{array}\right]$
|
You want to find a rotation matrix that has the same effect as the matrix you have. You can then easily find the angle of rotation. We can write it like this to make it clearer (I've just re-written your first equation as a
matrix product rather than the way you have it):
$$
\left( \begin{array}{ccc}
\cos\theta & -\sin \theta \\
\sin\theta & \cos \theta \end{array} \right) \left(
\begin{array}{c}
x\\
y\\
\end{array}
\right)= \left( \begin{array}{ccc}
\frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array} \right) \left(
\begin{array}{c}
x\\
y\\
\end{array}
\right)
$$
so you can see we need to find the angle $\theta$ such that:
$$ \cos\theta=\frac{1}{\sqrt{2}} \;\;\text{ and } \;\; -\sin\theta=\frac{1}{\sqrt{2}}\\
\sin\theta=-\frac{1}{\sqrt{2}} \;\;\text{ and } \;\; \cos\theta=\frac{1}{\sqrt{2}}
$$
which is a consistent set of equations with a solution: $$\theta=\frac{\pi}{4}$$
So we can then interpret it as a rotation by an angle of $\pi/4$ radians or $45^\circ$.
Hopefully this makes it a little clearer how I got the right hand side of the first equation:
$$ T \left(\begin{array}{c}
x\\
y\\
\end{array}
\right) =\frac{1}{\sqrt{2}}\left(\begin{array}{c}
x+y\\
-x+y\\
\end{array}
\right)=\frac{1}{\sqrt{2}}\left( \begin{array}{ccc}
1 & 1 \\
-1 & 1 \end{array} \right)\left(\begin{array}{c}
x\\
y\\
\end{array}
\right)=\left( \begin{array}{ccc}
\frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array} \right) \left(
\begin{array}{c}
x\\
y\\
\end{array}
\right)$$
|
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|
How to rationalize denominator? Suppose $c$ is not a complete square integer, ${a_0},{a_1} \in \mathbb{Q}$, we have
$$
\frac{1}{{{a_0} + {a_1}\sqrt c }}
= \frac{{{a_0} - {a_1}\sqrt c }}{{a_0^2 - a_1^2c}}.
$$
We need to show ${{a_0} + {a_1}\sqrt c } = 0$ iff ${a_0} = {a_1} = 0$. I know it's not hard.
Suppose $c$ is not a complete cube integer, ${a_0},{a_1},{a_2} \in \mathbb{Q}$, how can we deal with
$$
\frac{1}{{{a_0} + {a_1}\sqrt[3]{c} + {a_2}\sqrt[3]{{{c^2}}}}}
$$
similarly?
Any help will be appreciated.
|
I found a way. Let ${p^3} = c$, we write
$\left\{ \begin{gathered}
{a_0} + {a_1}p + {a_2}{p^2} = S \hfill \\
{a_2}c + {a_0}p + {a_1}{p^2} = pS \hfill \\
{a_1}c + {a_2}cp + {a_0}{p^2} = {p^2}S \hfill \\
\end{gathered} \right.$
So
$\left| {\begin{array}{*{20}{c}}
{{a_0} - S}&{{a_1}}&{{a_2}} \\
{{a_2}c - pS}&{{a_0}}&{{a_1}} \\
{{a_1}c - {p^2}S}&{{a_2}c}&{{a_0}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{{a_0}}&{{a_1}}&{{a_2}} \\
{{a_2}c}&{{a_0}}&{{a_1}} \\
{{a_1}c}&{{a_2}c}&{{a_0}}
\end{array}} \right| - S\left| {\begin{array}{*{20}{c}}
1&{{a_1}}&{{a_2}} \\
p&{{a_0}}&{{a_1}} \\
{{p^2}}&{{a_2}c}&{{a_0}}
\end{array}} \right| = 0$
And
$\frac{1}{S} = \left| {\begin{array}{*{20}{c}}
1&{{a_1}}&{{a_2}} \\
p&{{a_0}}&{{a_1}} \\
{{p^2}}&{{a_2}c}&{{a_0}}
\end{array}} \right|{\left| {\begin{array}{*{20}{c}}
{{a_0}}&{{a_1}}&{{a_2}} \\
{{a_2}c}&{{a_0}}&{{a_1}} \\
{{a_1}c}&{{a_2}c}&{{a_0}}
\end{array}} \right|^{ - 1}}$
But I cannot see why $\left| {\begin{array}{*{20}{c}}
{{a_0}}&{{a_1}}&{{a_2}} \\
{{a_2}c}&{{a_0}}&{{a_1}} \\
{{a_1}c}&{{a_2}c}&{{a_0}}
\end{array}} \right| \ne 0$.
|
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|
Find the length of $\sqrt x$. Let $f(x) = \sqrt x$. Find the length of the curve for $0\le x \le a$.
So I know the formula is:
$$\int_0^a \sqrt{(1+(f')^2(x)} \ dx = \ldots = \int_0^a \sqrt{1+\frac{1}{4x}} \ dx$$
Now, how do I evaulate this integral?
I tried the subtitution $t = \sqrt {1+\frac{1}{4x}}$ but it became somewhat complicated.
|
$$\int\sqrt{1+\frac{1}{4x}}\space\text{d}x=$$
$$\int\sqrt{1+\frac{x+\frac{1}{4}}{x}}\space\text{d}x=$$
Substitute $u=\frac{x+\frac{1}{4}}{x}$ and $\text{d}u=\left(\frac{1}{x}-\frac{x+\frac{1}{4}}{x^2}\right)\space\text{d}x$:
$$-\frac{1}{4}\int\frac{\sqrt{u}}{(1+u)^2}\space\text{d}u=$$
Substitute $s=\sqrt{u}$ and $\text{d}s=\frac{1}{2\sqrt{u}}\space\text{d}u$:
$$-\frac{1}{2}\int\frac{s^2}{(1-s^2)^2}\space\text{d}s=$$
$$-\frac{1}{2}\int\frac{s^2}{(s^2-1)^2}\space\text{d}s=$$
$$-\frac{1}{2}\int\left(-\frac{1}{4(s+1)}+\frac{1}{4(s+1)^2}+\frac{1}{4(s-1)}+\frac{1}{4(s-1)^2}\right)\space\text{d}s=$$
$$\frac{1}{8}\int\frac{1}{s+1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s+1)^2}\space\text{d}s-\frac{1}{8}\int\frac{1}{s-1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$
Substitute $p=s+1$ and $\text{d}p=\text{d}s$:
$$\frac{1}{8}\int\frac{1}{p}\space\text{d}p-\frac{1}{8}\int\frac{1}{(s+1)^2}\space\text{d}s-\frac{1}{8}\int\frac{1}{s-1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$
$$\frac{\ln|p|}{8}-\frac{1}{8}\int\frac{1}{(s+1)^2}\space\text{d}s-\frac{1}{8}\int\frac{1}{s-1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$
Substitute $w=s+1$ and $\text{d}w=\text{d}s$:
$$\frac{\ln|p|}{8}-\frac{1}{8}\int\frac{1}{w^2}\space\text{d}w-\frac{1}{8}\int\frac{1}{s-1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$
$$\frac{\ln|p|}{8}+\frac{1}{8w}-\frac{1}{8}\int\frac{1}{s-1}\space\text{d}s-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$
Substitute $v=s-1$ and $\text{d}v=\text{d}s$:
$$\frac{\ln|p|}{8}+\frac{1}{8w}-\frac{1}{8}\int\frac{1}{v}\space\text{d}v-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$
$$\frac{\ln|p|}{8}+\frac{1}{8w}-\frac{\ln|v|}{8}-\frac{1}{8}\int\frac{1}{(s-1)^2}\space\text{d}s=$$
Substitute $q=s-1$ and $\text{d}q=\text{d}s$:
$$\frac{\ln|p|}{8}+\frac{1}{8w}-\frac{\ln|v|}{8}-\frac{1}{8}\int\frac{1}{q^2}\space\text{d}q=$$
$$\frac{\ln|p|}{8}+\frac{1}{8w}-\frac{\ln|v|}{8}+\frac{1}{8q}+\text{C}=$$
$$\frac{\ln|s+1|}{8}+\frac{1}{8(s+1)}-\frac{\ln|s-1|}{8}+\frac{1}{8(s-1)}+\text{C}=$$
$$\frac{\ln\left|\sqrt{u}+1\right|}{8}+\frac{1}{8\left(\sqrt{u}+1\right)}-\frac{\ln\left|\sqrt{u}-1\right|}{8}+\frac{1}{8\left(\sqrt{u}-1\right)}+\text{C}=$$
$$\frac{\ln\left|\sqrt{\frac{x+\frac{1}{4}}{x}}+1\right|}{8}+\frac{1}{8\left(\sqrt{\frac{x+\frac{1}{4}}{x}}+1\right)}-\frac{\ln\left|\sqrt{\frac{x+\frac{1}{4}}{x}}-1\right|}{8}+\frac{1}{8\left(\sqrt{\frac{x+\frac{1}{4}}{x}}-1\right)}+\text{C}$$
|
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|
Double integral with Polar coordinates - hard example Calculate using polar coordinates:
$$\iint_{D}^{} (x^2+y^2)^\frac{1}5 \ dx \ dy $$
where D is the region inside the circle with radius 1.
Working:
D: $ \ x^2+y^2=1 \\ $ so
$ 0 \leq r \leq 1 \ \ , $ $0 \leq \theta \leq \pi \ $
and $x=rcos\theta \ ,$ $y=rsin\theta $ so
$$\int_{0}^{1} \int_{0}^{\pi} r^2(cos^2\theta + sin^2\theta)^\frac{1}5 \ d\theta \ dr $$
But here is where I get stuck! Isn't $ (cos^2\theta + sin^2\theta) = 1$? So its just
$$\int_{0}^{1} \int_{0}^{\pi} r^2 \ d\theta \ dr $$
I am really confused as what to do! Any help is appreciated
|
According to your calculations it should go on as:
$$\iint_{D}^{} (x^2+y^2)^\frac{1}5 \ dx \ dy $$
$$=\int_{0}^{1} \int_{0}^{2\pi} (r^2\cos^2\theta + r^2\sin^2\theta)^\frac{1}{5} \cdot r d\theta \ dr $$
$$=\int_{0}^{1} \int_{0}^{2\pi} r^\frac{2}{5} \cdot r d\theta \ dr$$
$$=\int_{0}^{1} r^\frac{7}{5} dr\int_{0}^{2\pi} d\theta $$
$$=\frac{5}{12} \cdot 2\pi$$
$$=\frac{5\pi}{6}$$
NOTE: The limits for $\theta$ must be $0$ to $2\pi$ and not $0$ to $\pi$.
|
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|
Find number of words of length n over alphabet {A, B, C} where any nonterminal A must be followed by B. Let $a_n$ be the number of words of length n over the alphabet {A, B, C} such that any nonterminal A has to be immediately followed by B. Find $a_n$.
Here's what I know:
If the word starts with $C$ or $B$, any valid sequence of length $n-1$ will work to complete the word.
If the word starts with $A$, it must be followed with $B$, then any valid sequence of length $n-2$, so I get a recurrence relation of: $$a_n = 2a_{n-1} + a_{n-2}.$$
How could I solve for $a_n$?
|
This can be done recursively. We have $a_1 = 3$ and $a_2 = 7$ by inspection.
Then for $n>2$, we see that there are three possible ways to start a sequence: $AB$, $B$, or $C$. Depending on which one we choose, we recurse into a smaller sequence of known size, where the counting process is repeated all over again.
This suggests the recurrence $a_n = 2a_{n-1} + a_{n-2}$ with $a_1 = 3$ and $a_2 = 7$.
To get a closed form, note that the characteristic polynomial is $x^2 - 2x - 1 = 0$, or $(x-1)^2-2 = 0$, which has roots $1 + \sqrt{2}$ and $1 - \sqrt{2}$. Therefore:
$a_n = A(1 + \sqrt{2})^n+B(1 - \sqrt{2})^n$ for some unknown $A,B$.
But we already know $a_1$ and $a_2$:
$3 = A(1 + \sqrt{2})^1+B(1 - \sqrt{2})^1$
$7 = A(1 + \sqrt{2})^2+B(1 - \sqrt{2})^2$
Solving this system yields $A = \frac{1}{2}+\frac{1}{\sqrt{2}}, B = \frac{1}{2}-\frac{1}{\sqrt{2}}$. Therefore:
$$a_n = (\frac{1}{2}+\frac{1}{\sqrt{2}})(1 + \sqrt{2})^n+(\frac{1}{2}-\frac{1}{\sqrt{2}})(1 - \sqrt{2})^n$$
Simplifying:
$$a_n = \frac{1}{2} ((1+\sqrt{2})^{n+1}+(1-\sqrt{2})^{n+1})$$
|
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|
How do you prove that $\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x$? I have the task to prove that
$$
\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x ,\left|x\right|\le 1
$$
I do not have any ideas from where I should start.
Can anyone help me solve it?
|
Let $\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-x}})=y$
$\implies-\dfrac\pi2\le y\le\dfrac\pi2$
and $\dfrac12\sqrt{2-\sqrt {2-x}}=\sin y$
$\implies2-\sqrt {2-x}=(2\sin y)^2$
Using $\cos2A=2\cos^2A-1=1-2\sin^2A,$
$\sqrt {2-x}=2-4\sin^2y=2\cos2y$
$\implies x=2-(2\cos2y)^2=-2\cos4y$
$\implies\cos4y=-\dfrac x2$
$\arccos\left(-\dfrac x2\right)=\begin{cases}2\pi+4y &\mbox{if } -2\pi\le4y<-\pi \\-4y &\mbox{if } -\pi\le4y<0\\ 4y &\mbox{if } 0\le4y\le\pi \\
2\pi-4y & \mbox{if }\pi<4y\le2\pi \end{cases}$
Now $\arccos\left(-\dfrac x2\right)=\dfrac\pi2-\arcsin\left(-\dfrac x2\right)$
$\implies\arccos\left(-\dfrac x2\right)=\dfrac\pi2+\arcsin\dfrac x2$ as $\arcsin(-u)=-\arcsin u$
|
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|
Solving the integral $\int_{-1}^1 2\sqrt{2-2x^2}\,dx$ I'm working on a triple integral and have managed to get it to a certain point:
$$\int_{-1}^1 2\sqrt{2-2x^2}dx $$
When I check this with WolframAlpha it gives the answer $\pi\sqrt{2}$, which is the right answer to the problem.
I know I should do a trig substitution to solve from here, so I used $x=\sin t$, which gives me
$$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{2\cos^2t}\,dt $$
but this has the answer $4\sqrt{2}$.
I'm probably making a glaringly obvious mistake, but if someone could help me out, I'd greatly appreciate it!
|
$$\int 2\sqrt{2-2x^2}\, dx=2\sqrt{2}\int \sqrt{1-x^2}\, dx$$
For the expression to make sense, we must have $1-x^2\ge 0$, i.e. $-1\le x\le 1$, so exists $t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ such that $x=\sin t$. Then $dx=\cos t\, dt$.
$$=2\sqrt{2}\int\cos^2 t\, dt=2\sqrt{2}\int \frac{1+\cos(2t)}{2}\, dt$$
$$=\sqrt{2}\left(\int dt+\frac{1}{2}\int \cos(2t)\, d(2t)\right)$$
$$=\sqrt{2}\left(t+\frac{1}{2}\sin(2t)\right)+C=\sqrt{2}\left(\arcsin x+\sin t\cos t\right)+C$$
$$=\sqrt{2}\left(\arcsin x+x\sqrt{1-x^2}\right)+C$$
$$\int_{-1}^12\sqrt{2-2x^2}\, dx=\sqrt{2}\left(\arcsin (1)+1\sqrt{1-1^2}\right)-\sqrt{2}\left(\arcsin (-1)+(-1)\sqrt{1-(-1)^2}\right)$$
$$=\sqrt{2}\left(\arcsin (1)-\arcsin (-1)\right)=\sqrt{2}\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)=\pi\sqrt{2}$$
|
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|
Limit of a fraction with a square root
Find $$\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}$$ (without L'Hopital)
Where is the following wrong? (The limit is 6.)
\begin{align}\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{9-x^2-5}}= \\
& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{-x^2+4}}=\\
& = \lim_{x \to 2}\sqrt{(2-x)(2+x)}=0.
\end{align}
|
Let $f(x) = x^2, g(x) = \sqrt {x^2 + 5}.$ The expression equals
$$\frac{(f(x) - f(2))/(x-2)}{(g(x) - g(2))/(x-2)}.$$
By definition of the derivative, as $x\to 2$ this $\to f'(2)/g'(2),$ which is easy to compute.
|
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|
Confused about multinomials. Can we write $\binom{n}{a,b,c}=\binom{n}{a}\binom{n-a}{b}\binom{n-a-b}{c}$ if $a+b+c \le n$? Can we write $\binom{n}{a,b,c}=\binom{n}{a}\binom{n-a}{b}\binom{n-a-b}{c}$ if $a+b+c \le n$?
The definition for multinomial says $a+b+c=n$ must hold or else $\binom{n}{a,b,c}=0$.
I found that if $a+b+c \ge n$ we get $\binom{n}{a,b,c}=0$, but if $a+b+c \le n$, then $\binom{n}{a,b,c}=\binom{n}{a,b,c, n-(a+b+c)}$
|
No, for any $(a,b,c)\in\Bbb N^3$ and $a+b+c\leq n$ then,
$$\dbinom{n}{a}\dbinom{n-a}{b}\dbinom{n-a-b}{c}=\dbinom{n\qquad\qquad\qquad}{a,b,c, n-a-b-c}\;(n-a-b-c)!$$
Unlike the binomial coefficient, the usual convention is not to leave the last lower term of the multinomial coefficient implicit. The sum of the lower terms is required to equal the upper term.
If you are using non-standard notation you should mention this somewhere (in your work).
|
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|
Finding $\binom n0+\binom n3+\binom n6+\cdots $ How to get
$$\binom n0 + \binom n3 + \binom n6 + \cdots$$
MY ATTEMPT
$$(1+\omega)^n = \binom n0 + \binom n1 \omega^1 + \binom n2 \omega^2 + \cdots$$
$$(1+\omega^2)^n = \binom n0 + \binom n1 \omega^2 + \binom n2 \omega^4 + \cdots $$
$$(1 + 1)^n = 2^n = \binom n0 + \binom n1 + \binom n2 + \cdots$$
$$(1+\omega)^n + (1+\omega^2)^n + (1 + 1)^n = 3 \left(\binom n0 + \binom n3 + \binom n6 + \cdots\right)$$
But how to solve LHS? I got the required equation in RHS
|
$$(1+\omega)^n+(1+\omega^2)^n+2^n\\
=(-\omega^2)^n+(-\omega)^n+2^n\\
=(-1)^n(\omega^{2n}+\omega^n)+2^n\\
$$
i) $n=3m$:
$$(-1)^n(\omega^{2n}+\omega^n)+2^n=2\cdot(-1)^n+2^n$$
ii) $n=3m+1$ or $3m+2$:
$$(-1)^n(\omega^{2n}+\omega^n)+2^n=(-1)^{n+1}+2^n$$
|
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|
Finding $\sqrt{(14+6\sqrt 5)^3}+\sqrt{(14-6\sqrt 5)^3}$ Find $$\sqrt{(14+6\sqrt{5})^3}+ \sqrt{(14-6\sqrt{5})^3}$$
A.$72$
B.$144$
C.$64\sqrt{5}$
D.$32\sqrt{5}$
How to cancel out the square root?
|
The square of that sum equals:
$$(14+6\sqrt{5})^3+(14-6\sqrt{5})^3+2\sqrt{(14^2-36\cdot 5)^3}$$
that is:
$$ 2\cdot 14^3 +2\cdot 3\cdot 14\cdot 36\cdot 5+2\sqrt{16^3} = 2^8\cdot 3^4$$
so the original sum equals $2^4\cdot 3^2 = \color{red}{144}.$
|
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|
Indefinite integral with substitution For my engineering math course I got a couple of exercises about indefinite integrals. I ran trought all of them but stumbled upon the following problem.
$$\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx $$
We can write $1+x-2x^2$ as $(1-x)(2x+1)$
So I got:
$$
\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx = \int \frac{1-x}{\sqrt{(1-x)(2x+1)}}\,dx
$$
We can also replace $1-x$ in the denominator with $\sqrt{(1-x)^2}$
$$
\int \frac{1-x}{\sqrt{(1-x)(2x+1)}}\,dx = \int \frac{\sqrt{(1-x)^2}}{\sqrt{(1-x)(2x+1)}}\,dx
$$
If we simplify this fraction we get:
$$
\int \frac{\sqrt{1-x}}{\sqrt{2x+1}}\,dx
$$
Next we apply the following substitutions
$$
u = -x
$$
so : $-du = dx$
We can rewrite the integral as following:
$$-\int \frac{\sqrt{1+u}}{\sqrt{1-2u}}\,du$$
Then we apply another substitution:
$\sqrt{1+u} = t $ so $ \frac{1}{2\sqrt{1+u}} = dt $
We rewrite: $ \sqrt{1+u} $ to $\frac{1}{2}t^2 \,dt $
We can also replace $\sqrt{1-2u} $ as following:
$$\sqrt{-2t^2+3}=\sqrt{-2(1+u)+3}=\sqrt{1-2u}$$
With al these substitutions the integral has now the following form:
$$-\frac{1}{2}\int \frac{t^2}{\sqrt{-2t^2+3}}\,dt$$
Next we try to ''clean'' up the numerator:
$$-\frac{1}{2} \int \frac{t^2}{\sqrt{\frac{1}{2}(6-t^2)}} \, dt$$
$$-\frac{\sqrt{2}}{2} \int \frac{t^2}{\sqrt{6-t^2}} \, dt$$
And that's where I got stuck. I can clearly see that an arcsin is showing up in the integral but don't know how to get rid of the $t^2$.
|
Substitute $t = \sqrt{6} \sin(u)$, so $dt = \sqrt{6} \cos(u) du$ and the integrand becomes (up to losing the $-\frac{1}{\sqrt{2}}$ at the start)
$$\sqrt{6} \int \sqrt{6} \sin^2(u) du$$
I think you can do that!
|
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|
If $x_{n+1}= \frac{x_n^2+x_n+1}{x_n+1}$ find$ \sum_{n=1}^{p}\frac{1}{1+x_n}$ If $x_{n+1}= \frac{x_n^2+x_n+1}{x_n+1}$ find $ \sum_{n=1}^{p}\frac{1}{1+x_n}$,
given $x_1=1/2$
The original question was for p=2014, but I want to find the general solution.
I have tried using $x^2+x+1=(x- \omega)(x-\omega^2)$ and breaking $1/x_{n+1}$ into factors, but yielded no results.
I tried expressing $\frac{1}{1+x_{n+1}}$ in terms of $x_n$ but still no result.
Any help/hints on how to proceed will be highly appreciated.
|
Observe that $x_{n + 1} = x_n + \frac {1} {1 + x_n}$ so $\frac {1} {1 + x_n} = x_{n + 1} - x_n$ so $$\sum_{n = 1}^{p} \frac {1} {1 + x_n} = \sum_{n = 1}^{p} (x_{n + 1} - x_n) = x_{p + 1} - x_1 = x_{p + 1} - \frac {1} {2}.$$
|
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|
Question on indefinite integrals I have to integrate:
$$I_2 = \int \frac{e^{2x} - e^{x} + 1}{(e^x\cos(x) - \sin(x))\cdot \left(e^x\sin(x) + \cos(x)\right)} \text{d}x$$
I simply can't understand from where to begin with. Please help me in solving this problem.
|
I consulted Moor and he suggested an approach involving differentiating the denominator.
Let $f(x) = (e^x \cos (x) - \sin(x))$ and $g(x) = (e^x \sin x + \cos x)$. Then, note that, by the product rule $(fg)'=f'g +g'f = (e^x\cos(x)-e^x\sin(x) - \cos(x))(e^x \sin (x) + \cos (x)) + (e^x\sin (x) + e^x \cos(x)-\sin(x))(e^x \cos(x) - \sin(x))$
but this doesn't cancel out nicely to get $e^{2x}-e^x+1$. In fact, expanding gives
$$(fg)' = (e^{2x} \sin(x) \cos(x) + e^x \cos^2(x) - e^{2x} \sin^2 (x) - \cos^2(x) - 2e^x \sin(x)\cos(x)) \\
+(e^{2x}\sin(x)\cos(x) - e^x \sin^2(x) + e^{2x} \cos^2 (x) + \sin^2(x) - 2e^x \sin (x) \cos (x))$$
Note that $fg'$ will cancel with a lot of $f'g$ terms if $fg'$ is negative. In particular, note that $f'(x)g(x) - f(x)g'(x) = -(e^{2x} - e^x + 1)$.
So, our integral looks like
\begin{align*}
\int \frac{e^{2x} - e^{x} + 1}{(e^x\cos(x) - \sin(x))\cdot \left(e^x\sin(x) + \cos(x)\right)} \, dx &=
-\int \frac{f'(x)g(x) - f(x)g'(x)}{f(x)g(x)}\,dx \\
&= - \int \left(\frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)} \right) \,dx \\
&= - \ln(f(x)) + \ln(g(x)) + C \\
&= \ln\left(\frac{g(x)}{f(x)}\right) + C \\
&= \ln\left(\frac{e^x \sin(x) + \cos(x)}{e^x \cos(x) - \sin(x)} \right) +C
\end{align*}
|
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|
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Update (2015-12-01) after your answers:
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $
Contradiction
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
|
Using the definition of a Ring, it is pretty simple to show that
$$\forall a, 0\times a=a\times 0=0$$
Thus, If we were to define division (only in the case where $c=ab=ba$):
$$\forall a, a=\frac{0}{0}$$
Which can be true if and only if your ring is the Zero ring
$$\{0\}$$
Therefore, when $\frac{0}{0}$ exists, it is actually $0$ (but there aren't any other numbers in the ring)
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|
2 problems related to the number 2015
*
*Let $p=\underbrace{11\cdots1}_\text{2015}\underbrace{22\cdots2}_\text{2015}$. Find $n$, where $n(n+1) = p$
*Prove that $\frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{2015^2} < \frac{2014}{2015}$
For 1, I tried dividing in various ways until I got a simpler expression, but no result. For 2, I know that $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$, but the proofs i found are over the elementary level the problem is aimed for. I also proved that it's smaller than $\frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{1024^2} < 1$, but that obviously is missing $\frac{1}{2015}$
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$1)$ $n=\frac{10^{2015}-1}{3}=3...3$ with $2015$ $3's$ :
$$\frac{10^{2015}-1}{3}\times \frac{10^{2015}+2}{3}=\frac{10^{4030}+10^{2015}-2}{9}=10^{2015}\times \frac{10^{2015}-1}{9}+2\times \frac{10^{2015}-1}{9}=10^{2015}R_{2015}+2\times R_{2015}$$
$2)$ follows from $\frac{1}{2^2}+...+\frac{1}{2015^2}<\sum_{j=2}^{\infty} \frac{1}{j^2}=\frac{\pi^2}{6}-1<\frac{2014}{2015}$
|
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|
Summation of Greatest Integer $ \sum_{k=1}^{n} \left \lfloor{\sqrt{k}}\right \rfloor $ I want to know the concept behind summing up a Greatest Integer function like $\sum_{k=1}^{n} \left \lfloor{x}\right \rfloor $ can be evaluated by simply writing the series but something like $$ \sum_{k=1}^{n} \left \lfloor{\sqrt{k}}\right \rfloor $$ This is creating problem by simply writing the series it came out to be $$ S=1\times3 +2\times4+3\times5+4\times6+... $$
The sum of this can be written as $$ \sum_{r=1}^{n} (r)(2r+1) $$ but then i am not able to relate $r$ and $k$ to each other.
|
For any given $m\geq0$, we have $\lfloor\sqrt{k}\rfloor=m$ if and only if $m^2\leq k<(m+1)^2$. So the number of times $m$ occours in the sum is
$$(m+1)^2-m^2=2m+1.$$
Setting $\ell:=\lfloor\sqrt{n}\rfloor$ this indeed leads to the simplification you already found:
$$\sum_{k=0}^n\lfloor\sqrt{k}\rfloor=\ell\cdot(n+1-\ell^2)+\sum_{m=0}^{\ell-1}m\cdot(2m+1),$$
where the term $\ell\cdot(n+1-\ell^2)$ comes from the fact that we're not summing over all integers $k$ for which $\lfloor\sqrt{k}\rfloor=\ell$, but just the first $n+1-\ell^2$ such integers. Now we can write the sum out as
\begin{eqnarray*}
\ell\cdot(n+1-\ell^2)+\sum_{m=0}^{\ell-1}m\cdot(2m+1)
&=&\ell\cdot(n+1-\ell^2)+2\sum_{m=0}^{\ell-1}m^2+\sum_{m=0}^{\ell-1}m\\
&=&\ell\cdot(n+1-\ell^2)+\frac{1}{3}\cdot\ell(\ell-1)(2\ell-1)+\frac{1}{2}\ell(\ell-1)\\
&=&-\frac{1}{3}\ell^3-\frac{1}{2}\ell^2+\left(n+\frac{5}{6}\right)\ell\\
&=&n\ell-\frac{1}{6}\ell\left(\ell-1\right)\left(2\ell+5\right)
\end{eqnarray*}
|
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|
Evaluate the integral $\int \sqrt{(x-a)(b-x)}$ I'm trying to figure out how to evaluate the following integral:
$$\int \sqrt{(x-a)(b-x)} \, dx $$
I've tried various trig substitutions, but can't seem to get anywhere. This is an exercise in Apostol's Calculus Volume 1 (Section 6.22, Exercise 46). The solution provided in the text is
$$\frac{1}{4} |b-a|(b-a) \arcsin \sqrt{\frac{x-a}{b-a}} + \frac{1}{4} \sqrt{(x-a)(b-x)} (2x-(a+b)) + C.$$
|
$$\sqrt{(x-a)(b-x)} = \sqrt{\left(\frac{a+b}{2}\right)^2 - ab - \left(x^2 - (a+b)x + \left(\frac{a+b}{2}\right)^2 \right)}$$
$\left(\frac{a+b}{2}\right)^2 - ab = \left(\frac{a-b}{2}\right)^2 $, call it $k^2$ for some $k$. The remaining term is $\left(x - \frac{a+b}{2}\right)^2$, and you may make the substitution $ u = x - \frac{a+b}{2}$. Consequently you get
$$\int \sqrt{k^2 - u^2} \mathrm{d}u$$
which can be solved with the substitution $u = k \sin \theta$.
|
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|
If $ c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ prove that it is Isosceles Triangle
In a $\triangle ABC\;,$ If $\displaystyle c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ Then how can we prove that $\triangle ABC$
is an Isoceles $\triangle.$
$\bf{My\; Try::}$ Using $\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\;,$ We get
$$k\sin C\left[k\sin A+k\sin B\right]\cdot \cos \frac{B}{2} = k\sin B\left[k\sin A+k\sin C\right]\cdot \cos \frac{C}{2} $$
So we get $$\sin C\left[\sin \left(\frac{A+B}{2}\right)\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\sin \left(\frac{A+C}{2}\right)\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
Now Using $A+B+C=\pi\;,$ We get $\displaystyle \frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$ and $\displaystyle \frac{A+C}{2}=\frac{\pi}{2}-\frac{B}{2}$
So we get $$\sin C\left[\cos \frac{C}{2}\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\cos \frac{B}{2}\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
So we get $$\sin C\cdot \cos \left(\frac{A-B}{2}\right)=\sin B\cdot \cos \left(\frac{A-C}{2}\right)$$
Now if we put $B=C\;,$ Then these two are equal.
My question is how can we prove it.
Help me, Thanks
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using the sine rule in the form:
$$
b = 4R\sin\frac{B}2\cos\frac{B}2 \\
c = 4R\sin\frac{C}2\cos\frac{C}2
$$
we obtain:
$$
(a+c)\sin\frac{B}2 = (a+b)\sin\frac{C}2
$$
which, together with the original relation:
$$
c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}
$$
gives
$$
b \tan\frac{B}2 = c\tan\frac{C}2
$$
but the function $x\tan\frac{x}2$ is monotonic on $[0,\pi)$ hence the result
|
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|
Evaluation of $\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$ Evaluate
$$\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$$
My main concern is finding the indefinite integral as once i have that the rest is fairly straight forward. Please give a detailed answer with reference to why you made each substitution (what indicated that said substitution would work etc.)
My initial substitution was $x= \sin \theta$ which tidies it up a bit
$$\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \space dx$$
Also the answer from Wolfram Alpha is;
$$ = \frac{1}{4}\bigg( log(1-2 x^2)+2 \tanh^{-1}\bigg(\frac{x}{\sqrt{1-x^2}}\bigg)+2 \sin^{-1}x\bigg)+constant$$
I need to be able to answer questions like this in under 15 mins so any help is appreciated
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BIG HINT:
$$\int\frac{1}{x+\sqrt{1-x^2}}\space\text{d}x=$$
Substitute $x=\sin(u)$ and $\text{d}x=\cos(u)\space\text{d}u$.
Then $\sqrt{1-x^2}=\sqrt{1-\sin^2(u)}=\cos(u)$ and $u=\arcsin(x)$:
$$\int\frac{\cos(u)}{\sin(u)+\cos(u)}\space\text{d}u=$$
$$\int\frac{\sec^3(u)}{\sec^3(u)}\cdot\frac{\cos(u)}{\sin(u)+\cos(u)}\space\text{d}u=$$
$$\int\frac{\sec^2(u)}{\sec^2(u)+\sec^2(u)\tan(u)}\space\text{d}u=$$
Prepare to substitute $s=\tan(u)$. Rewrite $\frac{\sec^2(u)}{\sec^2(u)+\sec^2(u)\tan(u)}$ using $\sec^2(u)=1+\tan^2(u)$:
$$\int\frac{\sec^2(u)}{1+\tan(u)+\tan^2(u)+\tan^3(u)}\space\text{d}u=$$
Substitute $s=\tan(u)$ and $\text{d}s=\sec^2(u)\space\text{d}u$:
$$\int\frac{1}{s^3+s^2+s+1}\space\text{d}s=$$
$$\int\left(\frac{1-s}{2(s^2+1)}+\frac{1}{2(s+1)}\right)\space\text{d}s=$$
$$\int\frac{1-s}{2(s^2+1)}\space\text{d}s+\int\frac{1}{2(s+1)}\space\text{d}s=$$
$$\frac{1}{2}\int\frac{1-s}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$
$$\frac{1}{2}\int\left(\frac{1}{s^2+1}-\frac{s}{s^2+1}\right)\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$
$$-\frac{1}{2}\int\frac{s}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$
Substitute $p=s^2+1$ and $\text{d}p=2s\space\text{d}s$:
$$-\frac{1}{4}\int\frac{1}{p}\space\text{d}p+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$
$$-\frac{\ln\left|p\right|}{4}+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$
$$-\frac{\ln\left|p\right|}{4}+\frac{\arctan\left(s\right)}{2}+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$
Substitute $w=s+1$ and $\text{d}w=\space\text{d}s$:
$$-\frac{\ln\left|p\right|}{4}+\frac{\arctan\left(s\right)}{2}+\frac{1}{2}\int\frac{1}{w}\space\text{d}w=$$
$$-\frac{\ln\left|p\right|}{4}+\frac{\arctan\left(s\right)}{2}+\frac{\ln\left|w\right|}{2}+\text{C}$$
|
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|
Diophantine equation $x^2 + y^2 = z^3$ I have found all solutions to the Diophantine equation $x^2 + y^2 = z^3$ when
$z$ is odd. I am having some difficulty finding the solutions when $z$ is even. I am asking for a proof that provides the solutions where $z$ is even. I want the proof to be elementary and use only Number theory and perhaps Calculus or basic ideas about groups and rings.
|
Unfortunately, there isn't (apparently) one complete polynomial parameterization to
$$x^2+y^2 = z^k\tag1$$
when $k>2$. For $k=2$, the complete solution is,
$$x,\,y,\,z = (a^2-b^2)s,\; (2ab)s,\; (a^2+b^2)s$$
where $s$ is a scaling factor. Using complex numbers $a+b i$, one can generalize the method. For $k=3$, it is
$$x,\,y,\,z = (a^3 - 3a b^2)s^3,\; (3a^2 b - b^3)s^3,\; (a^2+b^2)s^2\tag2$$
but you can no longer find rational $a,b,s$ for certain solutions. For example,
$\hskip2.7in$ $9^2+46^2 = 13^3\quad$ Yes
$\hskip2.7in$ $58^2+145^2=29^3\quad$ No
(You can click on the Yes/No links for Walpha output.) A related discussion can be found in this post while an alternative method is described here. For the case $k=3$, if $a^2+b^2=c^3$, then an infinite more can be found as,
$$(a u^3 + 3 b u^2 v - 3 a u v^2 - b v^3)^2 + (b u^3 - 3 a u^2 v - 3 b u v^2 +
a v^3)^2 = c^3(u^2+v^2)^3\tag3$$
which should provide some solutions not covered by $(2)$.
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|
Limit of function $x^2-x\cot\left(\frac{1}{x}\right)$ How To compute $\lim_{x \to \infty} x^2 -x \cot(1/x)$? Wolfram says it is $\frac{1}{3}$ and I know it is supposed to converge to a number other than 0 but I keep getting infinity.
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WithOut L'Hospital:
i will guess that the limit exist so :
$$l=\lim_{y\to 0}\frac{1-y \cot y}{y^2}$$
$$l=\lim_{y\to 0}\frac{\sin y-y \cos y}{y^2\sin y}$$
$$l=\lim_{y\to 0}\frac{\sin y-y \cos y}{y^3}$$
$y\to 2y$
$$l=\frac{1}{4}\lim_{y\to 0}\frac{\sin y \cos y-y (1-2\sin^2y)}{y^3}$$
$$l=\frac{1}{4}\lim_{y\to 0}\frac{\sin y \cos y-y+2y\sin^2y-y\cos^2y+y\cos^2y}{y^3}$$
$$l=\frac{1}{4}\lim_{y\to 0}\frac{\sin y \cos y-y\cos^2y+y\cos^2y-y+2y\sin^2y}{y^3}$$
$$l=\frac{1}{4}\lim_{y\to 0}(\frac{\cos y(\sin y -y\cos y)}{y^3}+\frac{y\cos^2y-y}{y^3}+\frac{2y\sin^2y}{y^3})$$
$$l=\frac{1}{4}(l-1+2)$$
$$l=\frac{1}{3}$$
|
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|
Length of a side of a triangle given the angles and the area We have $\triangle ABC$ with the following measures:
$A = 65^\circ$,
$B= 75^\circ$,
$\text{area}= 88\,m^2$.
How can I determine the longest side?
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We want to find $AC=b$.
We have
$$88=\frac 12ac\sin75^\circ\tag1$$
By the law of sines, we have
$$\frac{a}{\sin 65^\circ}=\frac{b}{\sin 75^\circ}=\frac{c}{\sin 40^\circ}$$
So,
$$a=\frac{\sin 65^\circ}{\sin 75^\circ}b,\quad c=\frac{\sin 40^\circ}{\sin 75^\circ}b\tag2$$
From $(1)(2)$,
$$88=\frac 12\cdot\frac{\sin 65^\circ}{\sin 75^\circ}b\cdot \frac{\sin 40^\circ}{\sin 75^\circ}b\sin75^\circ$$
Hence,
$$(\text{the longest side})=b=\sqrt{\frac{176\sin 75^\circ}{\sin 65^\circ\sin 40^\circ}}$$
|
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|
Finding $\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$ $$\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$$
I'm trying to understand trigonometric substitution better, because I never could get a good handle on it. All I know is that this integral is supposed to reduce to the integral of some power of cosine. I tried $x^2=\tan\theta$, but I ended up with $\sin\theta\cos^3\theta$ as my integrand. Can someone explain how to compute this?
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Via contour integration $\oint_C \frac{z^{\alpha-1}}{{1+z}} dz$ it was shown that
$$\int_0^\infty \frac{y^{\alpha-1}}{1+y} dy = \frac{\pi}{\sin \pi \alpha}.$$
Let $y=x^4$ so that $dy = 4x^3 dx.$
$$\int_{-\infty}^\infty \frac{x^2}{1+x^4} dx =\frac{1}{4} \int_{-\infty}^\infty \frac{y^{-1/4}}{1+y} dy = \frac{1}{2} \int_0^\infty \frac{y^{-1/4}}{1+y}dy=\frac{1}{2} \frac{\pi}{\sin \frac{3\pi}{4}}=\frac{\pi}{\sqrt{2}}.$$
|
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|
Find all incongruent solutions of $x^8\equiv3\pmod{13}$. Find all incongruent solutions of $x^8\equiv3\pmod{13}$.
I know that $2$ is a primitive root of $13$ and that $2^4\equiv3\pmod{13}$, so we want to solve $x^8\equiv2^4\pmod{13}$.
Now, $\gcd(8,\phi(13))=\gcd(8,12)=4$ divides the exponent of $2$, which is $4$, so $x^8\equiv3\pmod{13}$ has exactly $4$ incongruent solutions modulo $13$.
I was able to find on my calculator (using brute force) that these solutions are $4,6,7,$ and $9$ (i.e. $\pm4,\pm6$), but how would I go about finding them without using a calculator?
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$$x^8-3\equiv x^8-16\equiv \left(x^2-2\right)\left(x^2+2\right)\left(x^4+4\right)\pmod{13}$$
$x^2\equiv 2\pmod{13}$ and $x^2\equiv -2\pmod{13}$ are both unsolvable (by Quadratic Reciprocity), because $13\equiv 5\pmod{8}$.
$$x^4+4\equiv x^4-9\equiv \left(x^2+3\right)\left(x^2-3\right)\pmod{13}$$
$$x^2+3\equiv x^2-36\equiv (x+6)(x-6)\pmod{13}$$
$$x^2-3\equiv x^2-16\equiv (x+4)(x-4)\pmod{13}$$
|
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|
How do you calculate this limit without using L'Hopital or Taylor? I need to calculate this limit $$\lim_{x\rightarrow 0}\frac{xe^x-(e^x-1)}{x^2}$$ but only with elementary methods, so no L'Hopital/Taylor.
I've done quite a bit of manipulation but nothing seems to work, could you give me a hint?
|
Clearly we have $$\frac{xe^{x} - (e^{x} - 1)}{x^{2}} = \frac{xe^{x} - x - (e^{x} - 1 - x)}{x^{2}} = \frac{e^{x} - 1}{x} - \frac{e^{x} - 1 - x}{x^{2}}\tag{1}$$ and we know that $(e^{x} - 1)/x \to 1$ as $x \to 0$ hence it follows from the equation $(1)$ that our job is done if we can calculate the limit $$\lim_{x \to 0}\frac{e^{x} - 1 - x}{x^{2}} = L\tag{2}$$ and the answer to original question would be $1 - L$.
The limit $L$ can be easily (very very easily) calculated using L'Hospital's Rule or Taylor series. But since these methods are forbidden we need to invoke some definition of $e^{x}$. The simplest approach seems to be to use the defining equation $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{3}$$ Let's first handle the case when $x \to 0^{+}$. By the binomial theorem we have
$$\left(1 + \frac{x}{n}\right)^{n} = 1 + x + \dfrac{1 - \dfrac{1}{n}}{2!}x^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}x^{3} + \cdots$$
and hence
\begin{align}
L &= \lim_{x \to 0}\frac{e^{x} - 1 - x}{x^{2}}\notag\\
&= \lim_{x \to 0}\lim_{n \to \infty}\dfrac{\left(1 + \dfrac{x}{n}\right)^{n} - 1 - x}{x^{2}}\notag\\
&= \lim_{x \to 0}\lim_{n \to \infty}\dfrac{1 - \dfrac{1}{n}}{2!} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}x + \cdots\notag\\
&= \lim_{x \to 0}\lim_{n \to \infty}\phi(x, n)\text{ (say)}\tag{4}
\end{align}
If $x \to 0^{+}$ then we can see that $$\frac{1}{2} - \frac{1}{2n}\leq \phi(x, n) \leq \frac{1}{2} + \frac{x}{3!} + \frac{x^{2}}{4!}\cdots \leq \frac{1}{2} + \frac{x}{2\cdot 3} + \frac{x^{2}}{2\cdot 3\cdot 3} + \cdots = \frac{1}{2} + \frac{x}{6 - 2x}\tag{5}$$ for $0 < x < 3$. Further note that $\phi(x, n)$ is increasing as $n$ increases and by above equation $(5)$ it is bounded above. Hence $\lim_{n \to \infty}\phi(x, n) = \phi(x)$ exists for $0 < x < 3$. From equation $(5)$ it follows that $$\frac{1}{2} \leq \phi(x) \leq \frac{1}{2} + \frac{x}{6 - 2x}$$ and then by Squeeze Theorem $\phi(x) \to 1/2$ as $x \to 0^{+}$. It now follows from equation $(4)$ that $L = 1/2$ (as far $x \to 0^{+}$ is concerned).
The case $x \to 0^{-}$ is easy (try it and it will surprise you!). Thus $L = 1/2$ and desired limit $(1 - L) = 1/2$.
|
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|
Find the product xy. Given
$$
\log (x) + \frac{\log (xy^8)}{(\log x)^2+(\log y)^2} = 2\\
\log (y) + \frac{\log \left(\frac{x^8}{y}\right)}{(\log x)^2+(\log y)^2} = 0
$$
Find the product $xy$ if both $x$ and $y$ are real.
After applying basic log identities, I tried equating value of $ \large\frac{1}{(\log x)^2}+\frac{1}{(\log x)^2} $ but I am not getting any fruitful result.
|
hint put $\log(x)=a\,\log(y)=b$ this leads to two equations with two unknowns.
They are:
(I) $a+\frac{a+8b}{a^2+b^2}=2$ and
(II) $b+\frac{(8a-b)}{a^2+b^2}=0.$
Mutiplying (I) by $b$ and (II) by $a$ we get $ab+\frac{ab+8b^2}{a^2+b^2}=2b$ and $ab+\frac{8a^2-ab} {a^2+b^2}=0$, adding them up, we get $2ab+8=2b$ and $ab+4=b$ thus $a = \frac{b-4}{b}$. Solving equation (II) and substituting value for $a$ we have $b^3-\frac{16}{b} = 0$, thus $b=\pm 2$ thus $a=3,-1$ therefore $$(x , y) = \left(1000, \dfrac{1}{100}\right) || \left(\dfrac{1}{10} , 100\right)$$. Finally, we have, $$xy=10$$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show $\sin(\frac{\pi}{3})=\frac{1}{2}\sqrt{3}$ I have to show that
$$\sin\left(\frac{\pi}{3}\right)=\frac{1}{2}\sqrt{3}$$
and
$$\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$$
Should I use the exponential function?
|
Without using triangles:
We use the following core formulas to prove this identity:
*
*Pythagorean identity: $cos^2(x) + sin^2(x) = 1$
*Angle transformation formula: $sin(x - y) = sin(x)cos(y) - cos(x)sin(y)$
*Double-angle formula: $sin(2x) = 2sin(x)cos(x)$
To show: $sin(\frac{\pi}{3}) = \frac{\sqrt3}{2}$
We start by showing the following identity: $sin(\pi - x) = sin(x)$
$sin(\pi - x) \stackrel{2.}{=} sin(\pi)cos(x) - cos(\pi)sin(x)$
$\stackrel{sin(\pi) = 0}{\Rightarrow}$
$\stackrel{cos(\pi) = -1}{\Rightarrow}$
$= sin(x)$
We use $x = \frac{\pi}{3}$ and get: $sin(\frac{2\pi}{3}) = sin(\frac{\pi}{3})$
Now: $sin(2\frac{\pi}{3}) \stackrel{3.}{=} 2sin(\frac{\pi}{3})cos(\frac{\pi}{3})$
$\stackrel{/sin(\frac{\pi}{3})}{\Rightarrow}$
$\stackrel{/2)}{\Rightarrow}$
$\frac{1}{2} = cos(\frac{\pi}{3})$
Finally we use 1. to reach the desired result:
$cos^2(\frac{\pi}{3}) + sin^2(\frac{\pi}{3}) = 1 \Rightarrow sin^2(\frac{\pi}{3}) = 1 - cos^2(\frac{\pi}{3}) \stackrel{\frac{1}{2} = cos(\frac{\pi}{3})}{\Rightarrow} sin^2(\frac{\pi}{3}) = \frac{3}{4} \stackrel{(\sqrt)}{\Rightarrow} sin(\frac{\pi}{3}) = \frac{\sqrt3}{2}$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find $\int _{0}^{2\pi}{1\over a\sin t+b \cos t +c}$ where $\sqrt{a^2+b^2}=1Find $\int _{0}^{2\pi}{1\over a\sin t+b \cos t +c}$ where $\sqrt{a^2+b^2}=1<c$. I am lead to believe I should be using curves but I really don't understand what curves to choose and how to properly use them. I would appreaciate some help on the issue.
|
Suppose we seek to evaluate
$$\int_{0}^{2\pi} \frac{1}{a\sin x + b\cos x + c} \; dx$$
where $a^2+b^2 = 1 < c.$
Introduce $z=\exp(ix)$ so that $dz=iz \; dx$ to get
$$\int_{|z|=1}
\frac{1}{a(z-1/z)/2/i + b(z+1/z)/2 + c}
\frac{dz}{iz}
\\ = \int_{|z|=1}
\frac{1}{az^2/2-a/2 + ibz^2/2+ib/2 + ciz}
\; dz.$$
Call this function (the integrand) $f(z).$
The two poles are at
$$\rho_{0,1} = \frac{-ci\pm \sqrt{a^2+b^2-c^2}}{a+bi}.$$
This is
$$\rho_{0,1} = \frac{-ci\pm \sqrt{1-c^2}}{a+bi}.$$
With the principal branch of the square root we thus have
(recall that $c\gt 1$)
$$|\rho_{0,1}| = |-c\pm\sqrt{c^2-1}|$$
Therefore $\rho_1$ is definitely not inside the contour but for
$\rho_0$ we have $c-\sqrt{c^2-1} \lt 1$ since $c-1\lt \sqrt{c^2-1}$ is
equivalent to $c^2-2c+1\lt c^2-1$ or $2\lt 2c$ and $c\gt 1.$
Next differentiate the denominator of our function $f(z)$ to get for
the residue at $z=\rho_0$
$$\left.\frac{1}{az+ibz+ci}\right|_{z=\rho_0}
= \left.\frac{1}{(a+ib)z+ci}\right|_{z=\rho_0}
= \frac{1}{-ci+\sqrt{1-c^2}+ci}
\\ = \frac{1}{\sqrt{1-c^2}}.$$
With $c\gt 1$ a real number this is
$$\frac{1}{i\sqrt{c^2-1}}.$$
We thus get for the result
$$2\pi i \times \mathrm{Res}_{z=\rho_0} f(z)
= \frac{2\pi}{\sqrt{c^2-1}}.$$
|
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|
Rational solutions $(a,b)$ to the equation $a\sqrt{2}+b\sqrt{3} = 2\sqrt{a} + 3\sqrt{b}$ Find all rational solutions $(a,b)$ to the equation
$$a\sqrt{2}+b\sqrt{3} = 2\sqrt{a} + 3\sqrt{b}.$$
I can see that we have the solutions $(0,0), (2,0), (0,3), (3,2), (2,3)$, and I suspect that there are no more.
I tried to do the thing where you square both sides, rearrange terms, square both sides again, but it got messy.
Edited: Also, I think I recall that all distinct square roots of square-free numbers are linearly independent over the rationals. Is this true? This might lead to a more direct way of proving it.
|
We can use the interesting
Proposition. The square roots of the square-free naturals are $\Bbb Q$-linearly independant.
Write $a=c^2n$, $b=d^2m$ with $c,d\in\Bbb Q$ and $n,m$ square-free (note that $a,b$ cannot be negative anyway). Then we have
$$ a\sqrt 2+b\sqrt 3-2c\sqrt n-3d\sqrt m=0.$$
By the proposition, the following conclusions can be made casewise:
*
*$a=b=0$.
*$a=0$, $b\ne 0$. Then $b\sqrt 3-3d\sqrt m=0$ implies $m=3$ and $3d=b$, so $b=3$
*$b=0$, $a\ne 0$. Then $a\sqrt 2-2c\sqrt n=0$ implies $n=2$ and $2c=a$, so $a=2$.
*$a\ne 0,b\ne 0$. Then either $n=2$, $m=3$, $2c=a$, $3d=b$; this means $a=2$, $b=3$. Or $n=3$, $m=2$, $2c=b$, $3d=a$; this means $3b^2=12c^2=4a$ and $2a^2=18d^2=9b$, so $72b=16a^2=9b^4$, $b^3=8$, $b=2$ and $a=3$.
|
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|
Calculating the convolution of piecewise constant functions Let $f(x) = \frac{1}{2}$ on $[-1,1]$. Find $f*f*f$
$(f*g)(x)=\int\limits_{-\infty}^\infty f(t)g(x-t)\,dt$.
So $(f*f)(x)=\begin{cases} \frac{1}{4}x+\frac{1}{2} & -2\le x \le 0 \\ \frac{-1}{4}x+\frac{1}{2} & 0 < x \le 2 \\ 0 & \text{otherwise} \end{cases}$
This is almost identical to this question
Which this answer in this questions does now clearly explain how to find the limits of integration. I figured it out for $f*f$, but I am now lost in trying to calculate $f*f*f$.
I know that $(f*f*f)(x)$ will be in three pieces on $[-3,-1],[-1,1],$ and $[1,3]$, but I do not know how to set up my integrals.
|
The convolution is $f*f*f(w)=\int_{x}\int_{y} f(x)f(y)f(w-x-y) dydx$. We seek the integration limits, firstly for $x$ given $w$ (with $-3\lt w\lt 3$), then for $y$ given $w,x$.
Limits for $x$ given $w$:
*
*$-1\lt x\lt 1$
*$w-2\lt x\lt w+2\qquad \text{(lower bound comes from $x=w-y-z\;$ with $y=z=1$)}$
$\qquad\qquad\qquad\qquad\qquad\text{(upper bound comes from $x=w-y-z\;$ with $y=z=-1$)}$
So the limits for $x$: $\qquad\max\{-1,w-2\}\lt x\lt\min\{1,w+2\}$
Note:
*
*$-1\lt w-2$ iff $w\gt 1$
*$1\lt w+2$ iff $w\gt -1$
Hence, we split the interval $w\in [-3,3]$ at points $-1,1$.
$\\$
Limits for $y$ given $w,x$:
*
*$-1\lt y\lt 1$
*$w-x-1\lt y\lt w-x+1\qquad \text{(lower bound comes from $y=w-x-z\;$ with $z=1$)}$
$\qquad\qquad\qquad\qquad\qquad\qquad\text{(upper bound comes from $y=w-x-z\;$ with $z=-1$)}$
So the limits for $y$: $\qquad\max\{-1,w-x-1\}\lt y\lt\min\{1,w-x+1\}$
$\\$
For interval $w\in(-3,-1)$:
*
*$-1\lt x\lt w+2$
$\\$
*
*$\max\{-1,w-x-1\} = -1$
*$\min\{1,w-x+1\} = w-x+1$
$\qquad\therefore\quad -1\lt y\lt w-x+1$
$\\$
For interval $w\in(-1,1)$:
*
*$-1\lt x\lt 1$
$\\$
*
*$\max\{-1,w-x-1\} = -1 \text{ only when $x\gt w$}$
*$\min\{1,w-x+1\} = 1 \text{ only when $x\lt w$}$
So if $-1\lt x\lt w$ then $w-x-1\lt y\lt 1$.
And if $w\lt x\lt 1$ then $-1\lt y\lt w-x+1$.
$\\$
For interval $w\in(1,3)$:
*
*$w-2\lt x\lt 1$
$\\$
*
*$\max\{-1,w-x-1\} = w-x-1$
*$\min\{1,w-x+1\} = 1$
$\qquad\therefore\quad w-x-1\lt y\lt 1$
$\\$
So our integrals are:
For $-3\lt w\lt -1$:
\begin{align}
f*f*f(w) &= \int_{x=-1}^{w+2}\int_{y=-1}^{w-x+1} f(x)f(y)f(w-x-y)\;dydx \\
&= \left(\dfrac{w+3}{4} \right)^2
\end{align}
For $-1\lt w\lt 1$:
\begin{align}
f*f*f(w) &= \int_{x=-1}^{w}\int_{y=w-x-1}^{1} f(x)f(y)f(w-x-y)\;dydx \\
&\qquad + \int_{x=w}^{1}\int_{y=-1}^{w-x+1} f(x)f(y)f(w-x-y)\;dydx \\
&= \dfrac{3-w^2}{8}
\end{align}
For $1\lt w\lt 3$:
\begin{align}
f*f*f(w) &= \int_{x=w-2}^{1}\int_{y=w-x-1}^{1} f(x)f(y)f(w-x-y)\;dydx \\
&= \left(\dfrac{w-3}{4} \right)^2
\end{align}
It might help to see the limits of integration by looking at how the plane $x+y+z=w$ moves through the cube $[-1,1]^3$ as $w$ varies from $-3$ to $3$.
|
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|
Solve the given Differential Equation Solve the non-linear first order differential equation
$$\frac{dy}{dx}=\frac{x+2y-5}{2x+xy-4} $$
I tried substituting $x=X+h$ and $y=Y+k$ but the $xy$ term is creating problem.
How to solve it?
Any suggestion is appreciated.
|
$\dfrac{dy}{dx}=\dfrac{x+2y-5}{2x+xy-4}$
$(x+2y-5)\dfrac{dx}{dy}=(y+2)x-4$
This belongs to an Abel equation of the second kind.
Let $u=x+2y-5$ ,
Then $x=u-2y+5$
$\dfrac{dx}{dy}=\dfrac{du}{dy}-2$
$\therefore u\left(\dfrac{du}{dy}-2\right)=(y+2)(u-2y+5)-4$
$u\dfrac{du}{dy}-2u=(y+2)u-2y^2+y+6$
$u\dfrac{du}{dy}=(y+4)u-2y^2+y+6$
Let $s=y+4$ ,
Then $\dfrac{du}{dy}=\dfrac{du}{ds}\dfrac{ds}{dy}=\dfrac{du}{ds}$
$\therefore u\dfrac{du}{ds}=su-2(s-4)^2+s-4+6$
$u\dfrac{du}{ds}=su-2s^2+17s-30$
Let $t=\dfrac{s^2}{2}$ ,
Then $s=\pm\sqrt{2t}$
$\dfrac{du}{ds}=\dfrac{du}{dt}\dfrac{dt}{ds}=s\dfrac{du}{dt}$
$\therefore su\dfrac{du}{dt}=su-2s^2+17s-30$
$u\dfrac{du}{dt}=u-2s+17-\dfrac{30}{s}$
$u\dfrac{du}{dt}-u=\mp2\sqrt{2t}+17\mp\dfrac{30}{\sqrt{2t}}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf
|
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|
I want to show that $\sum_{n=1}^\infty \sin{\left(n\frac{\pi}{2}\right)}\cdot\frac{n^2+2}{n^3+n} $ converges or diverges. I want to show that $$\sum_{n=1}^\infty \sin{\left(n\frac{\pi}{2}\right)}\cdot\frac{n^2+2}{n^3+n} $$ converges (absolutly?) or diverges.
My idea was: $n=2k+1$ and then it becomes: $$\sum_{n=1}^\infty (-1)^n\cdot\frac{(2k+1)^2+2}{(2k+1)^3+2k+1} $$
Then I somehow try to show whether the limit is $0$ and whether it is decreasing. But I am not sure how I would go about that?
|
The sum converges by the alternating series test.
It turns out that the sum may be evaluated using complex analysis, i.e., the residue theorem. First simplify the sum
$$\sum_{n=0}^{\infty} \sin{\left (n \frac{\pi}{2} \right )} \frac{n^2+2}{n^3+n} $$
as
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \left [1+\frac1{ (2 k+1)^2 + 1} \right ] = \frac{\pi}{4} + \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \frac1{ (2 k+1)^2 + 1}$$
Note that the sum on the RHS is equal to
$$\frac12 \sum_{k=-\infty}^{\infty} \frac{(-1)^k}{2 k+1} \frac{1}{ (2 k+1)^2 + 1} $$
We may evaluate this sum using a result that is a consequence of the residue theorem. I will not prove it here, just state the result, which is if $f$ is sufficiently "well-behaved", then
$$\sum_{k=-\infty}^{\infty} (-1)^k \, f(k) = -\sum_n \operatorname*{Res}_{z=z_n} \left [\pi \, \csc{(\pi z)}\, f(z) \right ]$$
where the $z_n$ are non-integer poles of $f$ in the complex plane.
In our case,
$$f(z) = \frac1{2 z+1} \frac1{(2 z+1)^2+1} = \frac18 \frac1{z+\frac12} \frac1{\left (z+\frac12 \right )^2 + \frac14}$$
We have three poles: $z_1=-1/2$, $z_2 = -1/2 + i 1/2$, and $z_3=-1/2-i 1/2$. Thus, the doubly infinite sum is equal to
$$-\pi \left [\csc{\left (-\frac{\pi}{2} \right )} \frac12 + \csc{\left (-\frac{\pi}{2} + i \frac{\pi}{2} \right )} \left (-\frac14 \right ) + \csc{\left (-\frac{\pi}{2} - i \frac{\pi}{2} \right )} \left (-\frac14 \right ) \right ] $$
which simplifies to
$$\frac{\pi}{2} \left [1-\operatorname{sech}{\left (\frac{\pi}{2} \right )} \right ] $$
Putting everything altogether, we have
$$\sum_{n=0}^{\infty} \sin{\left (n \frac{\pi}{2} \right )} \frac{n^2+2}{n^3+n} = \frac{\pi}{2} - \frac{\pi}{4}\, \operatorname{sech}{\left (\frac{\pi}{2} \right )} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
show that the given series converges Prove that the series:
$$
\sum _{n=1} ^\infty \frac{(-1)^{\lfloor n/3 \rfloor}}{n}
$$
I tried to use Leibniz rule but I did not succeed. Any suggestions? Thanks for helping!
|
\begin{align}
\sum _{n=1}^{\infty} \frac{(-1)^{\lfloor n/3\rfloor}}{n} &= \frac{3}{2}+\sum _{n=1}^{\infty/3} \left[ \frac{(-1)^{n}}{3n} + \frac{(-1)^{n}}{3n+1} +\frac{(-1)^{n}}{3n+2} \right]\\
&=\frac{3}{2}+\sum _{n=1}^{\infty} (-1)^n \left(\frac{1}{3 n}+\frac{1}{3 n+1} + \frac{1}{3 n+2}\right)\\
\end{align}
By the Alternating series test, the three series converge, see Paul's for further details. Below, we deduce their summation, using integration techniques.
\begin{align}
\hphantom{\sum _{n=1}^{\infty} \frac{(-1)^{\lfloor n/3\rfloor}}{n}} & \\&=
\frac{3}{2}+\left[-\frac{\log (8)}{9} \right] +
\left[ \frac{\pi \sqrt{3} + \log (8)}{9} - 1 \right] +
\left[ \frac{\pi \sqrt{3} - \log (8)}{9} - \frac{1}{2}\right] \\
&= \frac{2\pi \sqrt{3} - \log (8)}{9}.
\end{align}
|
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|
Factoring Multivariable polynomials Determine a constant $k$ such that the polynomial
$$ P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) $$
is divisible by $x+y+z$.
Im having trouble utilizing the multivariable factor theorem for this problem>
|
Write these symmetric expressions in terms of $s = x+y+z$, $q = x y + x z + y z$, $p=x y z$. We get
\begin{eqnarray}
x^5 + y^5 + z^5 &=& s^5 - 5 q s^3 + 5 q^2 s + 5 p s^2 - 5 p q\\
x^3 + y^3 + z^3 &=& s^3 - 3 q s + 3 p \\
x^2 + y^2 + z^2 &=& s^2 - 2 q
\end{eqnarray}
Therefore
$$x^5 + y^5 + z^5 + k (x^3 + y^3 + z^3)(x^2 + y^2 + z^2) \equiv - 5 pq + k (3p) \cdot (-2 q)= -(5 + 6k) pq \mod s$$
Hence $k = -\frac{5}{6}$. Check:
$$x^5 + y^5 + z^5 -\frac{5}{6} (x^3 + y^3 + z^3)(x^2 + y^2 + z^2)
= \frac{1}{6}s^2 \cdot (s^3 - 5 q s + 15 p) $$
|
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|
Calculus - limit of a function: $\lim\limits_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$ How do you compute the following limit without using the l'Hopital rule?
If you were allowed to use it, it becomes easy and the result is $\sqrt{3}\over 3$ but without it, I am not sure how to proceed. $$\lim_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$$
|
Proceed as follows:
$$\lim_{x\to\frac{\pi}{3}} \frac{\sin(x-\pi/3)}{1-2\cos x}$$
$$=\lim_{t\to0} \frac{\sin t}{1-2\cos (t+\frac{\pi}{3})}$$
$$=\lim_{t\to0} \frac{\sin t}{1-2\cos t\cos \frac{\pi}{3}+2\sin t \sin\frac{\pi}{3}}$$
$$=\lim_{t\to0} \frac{\sin t}{(1-\cos t)+2\sin t \sin\frac{\pi}{3}}=\frac{1}{2\sin\frac{\pi}{3}}=\frac{1}{\sqrt3}$$
Maybe a little bit of reasoning: $1-\cos t\approx t^2/2$ for small $t$, which is insignificant compared to $\sin t$. If unsure, maybe use $1-\cos t = 2\sin^2 \frac{t}{2}$ and put everything else into half angles to. In that case, you get
$$=\lim_{t\to0} \frac{2\sin \frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}+4\sin\frac{t}{2}\cos\frac{t}{2} \sin\frac{\pi}{3}}=\lim_{t\to0} \frac{2\cos\frac{t}{2}}{2\sin\frac{t}{2}+4\cos\frac{t}{2} \sin\frac{\pi}{3}}=\frac{2}{4\sin\frac{\pi}{3}}=\frac{1}{\sqrt3}$$
|
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|
When does $nx^4+4x+3=0$ have real roots?
Find all positive integers $n$ such that the equation $nx^4+4x+3=0$ has real roots.
I think the answer must also include the cases with $2$ real roots. But my main question is, how do I start? Thanks.
|
We have
\begin{align}
nx^4+4x+3 & = (n-1)x^4 + x^4+4x+3 = (n-1)x^4 + (x+1)^2((x-1)^2+2)\\
& = (n-1)x^4 + (x^2-1)^2 + 2(x+1)^2
\end{align}
Hence, for $n \geq 2$, the function is a sum of squares and is always positive, since all three of them cannot be simultaneously zero and therefore has no real roots.
For $n=1$, we see that $x=-1$ is the only possible real root.
|
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|
Is it possible to find $g^k(x)$? Given $g(x)=\frac{x}{2}+\frac{1}{x}$, is it possible to find an expression for $g^k(x)=(g\underbrace{\circ \cdots \circ}_k g)(x)$, where $k$ is some positive whole number?
For example, given $h(x)=\frac{x}{2}$, the expression for $h^k(x)$ is $h^k(x)=\frac{x}{2^k}$.
|
Looking at
$$
g(x)=\frac{x}{2}+\frac{1}{x} = \frac{x^2 + 2}{2x}
$$
and cranking some iterations through the Maxima CAS, I get
$$
g^2(x) =
{{x^4+12\,x^2+4}\over{4\,x^3+8\,x}} \\
g^3(x) =
{{x^8+56\,x^6+280\,x^4+224\,x^2+16}\over{8\,x^7+112\,x^5+224\,x^3+
64\,x}} \\
g^4(x) =
{{x^{16}+240\,x^{14}+7280\,x^{12}+64064\,x^{10}+205920\,x^8+256256
\,x^6+116480\,x^4+15360\,x^2+256}\over{16\,x^{15}+1120\,x^{13}+17472
\,x^{11}+91520\,x^9+183040\,x^7+139776\,x^5+35840\,x^3+2048\,x}}
$$
This gives no simple pattern I am able to spot, just more or less
complicated rational functions with nominator degree $2^k$ and denominator degree $2^k - 1$, and the proper coefficients to go asymptotically to $h^k$.
Here are plots of $\text{id}, g, g^2, g^3, g^4, g^5$:
(Larger version)
This is slightly more interesting, as we see that for $x > 0$ the series $g^k$ seems to iterate against a constant function, at value of the positive fixed point $x_+^*$ of $g$. (Similar for the negative arguments against $x_-^*$).
Going for the fixed points:
$$
x = g(x) = \frac{x^2 + 2}{2x} \Rightarrow \\
2 x^2 = x^2 + 2 \Rightarrow \\
x^2 = 2 \Rightarrow \\
x = \pm \sqrt{2}
$$
|
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Method of completing squares with 3 variables I want to use the method "completing squares" for this term:
$x^2-2xy +y^2+z^2*a+2xz-2yz$
The result should be $(x-y+z)^2 +(a-1)*z^3$
Is there a "recipe" behind how to do this? Hope someone could help
|
Aras' answer is really good. But you might be wondering how one develops intuition and what happens if one doesn't immediately see $x^2 - 2xy + y^2$ is a square. (Part of developing intuition is that this does become obvious.)
This'll get messy. But
So you have:
$\color {blue}{x^2}-2xy +y^2+z^2*a+2xz-2yz$
And you have an $x^2$ so we'll "complete" x first so take all the x terms and put them nearby.
$\color {blue}{x^2}\color {purple}{-2xy +2xz}+y^2+z^2*a-2yz$
We know that completing the square, the expression will need to start something like this: $x^2 \pm 2*x*something + ....$ so we factor out the $2x$
$\color {blue}{x^2 + 2x}\color {purple}{(z - y)}+y^2+z^2*a-2yz$
Now we know to "complete the square" we must do $x^2 \pm 2x*something \color{green}{+ something^2} \color {red}{- something^2} +...$
So
$\color {blue}{x^2 + 2x(z - y)} \color{green}{ + (z -y)^2} \color{red}{- (z -y)^2}+y^2+z^2*a-2yz=$
$\color {blue}{(x + (y -z))^2} \color{red}{- (z^2 - 2zy + y^2) }+y^2+z^2*a-2yz$
So let's clean up a little.
$\color {blue}{(x + y -z)^2} \color{red}{- z^2 + 2zy - y^2 }+y^2+z^2*a-2yz$
$\color {blue}{(x + y -z)^2} -z^2 +z^2*a$
So now we need to deal with just the $ ... - z^2 + z^2*a$. Well we factor out common terms.
$\color {blue}{(x + y -z)^2} +z^2(-1 +a)$
$\color {blue}{(x + y -z)^2} +z^2(a-1)$
|
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What is the probability that $x^4-y^4$ is divisible by 5? Two numbers $x$ and $y$ are chosen at random without replacement from the set $\{1,2,3...,5n\}$.
What is the probability that $x^4-y^4$ is divisible by $5$?
I divided the numbers into groups of 5 $(1,2,3,4,5),(6,7,8,9,10),...$.The probability in the first group would itself be the answer.But how to find that?
|
Hint: If $x \equiv 0\pmod{5}$, then $x^4 \equiv 0\pmod{5}$. If $x \not\equiv 0\pmod{5}$, then $x^4 \equiv 1\pmod{5}$ by Fermat's Little Theorem.
Alternatively, if you don't know modular arithmetic, you can do the following to get the same result.
If $x = 5q+r$ for some integers $q$ and $r$ with $0 \le r \le 4$, then we have $x^4 = (5q+r)^4$ $= 625q^4+500q^3r+150q^2r^2+20qr^3+r^4$ $= 5(125q^4+100q^3r+30q^2r^2+4qr^3)+r^4$ where $r^4$ is one of $\{0, 1, 16, 81, 256\}$. Thus, if $x$ is divisible by $5$, then so is $x^4$, and if $x$ is not divisible by $5$, then $x^4$ is one more than a multiple of $5$.
Hence, $x^4-y^4$ is divisible by $5$ iff $x$ and $y$ are both divisible by $5$ or both not divisible by $5$.
Here is how to finish the problem in case you are still stuck:
There are $5n(5n-1)$ total ways to choose $x$ and $y$ without replacement. There are $n(n-1)$ ways to choose $x$ and $y$ without replacement such that both are divisible by $5$, and there are $4n(4n-1)$ ways to choose $x$ and $y$ without replacement such that both are not divisible by $5$. Hence, the probability that $x^4-y^4$ is divisible by $5$ is $\dfrac{n(n-1)+4n(4n-1)}{5n(5n-1)} = \dfrac{17n-5}{25n-5}$.
|
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How to compute the area of the portion of a paraboloid cut off by a plane? How to compute:
The area of that portion of the paraboloid $x^2+z^2=2ay$ which is cut off by the plane $y=a$ ?
I think I have to compute $\iint f(x,z) dx dz$ , where $f(x,z)=\sqrt{(x^2+z^2)/2a}$ ; but I can't figure out what should be the limits of integration . Please help . Thanks in advance .
|
So we have the paraboloid $x^2+z^2=2ay$ and the plane $y=a$. Let me just rotate the whole shebang: this is the same as $x^2+y^2=2az$ cut by $z=a$. Just to ease my 3D imagination :).
The picture shows the paraboloid for $a=\frac12$, so it is $x^2+y^2=z$. The section we must find the area of is evidently a circle. So all we need is the radius.
Let us rewrite the equation: $z=\frac{x^2+y^2}{2a}$. Admitting, of course, $a\neq0$. Naturally, $x^2+y^2=r^2$, so $z=\frac{r^2}{2a}$. But $z=a$, so $a=\frac{r^2}{2a}$, which means $r=\sqrt2a$. Hence, the area of the section will be $\pi r^2=2\pi a^2$.
For $a=0$, the equation becomes $x^2+y^2=2\cdot0z=0$, so it is a line. The section is a point, since the cutting plane is not a plane the line is contained in, hence the area of the section is 0.
Edit
Seems I misunderstood the problem. We are supposed to find the area of the yellow part, not the area of the cross-section. Let us then try parametrizing that surface. If we take the $z=a$ plane as $z=0$, the paraboloid will have graph $z=\frac{x^2+y^2}{2a}-a$. So we just need to integrate that over the circle with center 0 and radius the one found before, which was $\sqrt2a$. Hence, the area will be:
\begin{align*}
-A={}&\int\limits_{B(0,\sqrt2a)}\left(\frac{x^2+y^2}{2a}-a\right)dxdy=-a\times\pi(\sqrt2a)^2+\int\limits_{B(0,\sqrt2a)}\frac{x^2+y^2}{2a}dxdy={} \\
{}={}&-2\pi a^3+\int\limits_0^{2\pi}\left(\int\limits_0^{\sqrt2a}\frac{r^2}{2a}rdr\right)d\theta=2\pi\left(-a^3+\int\limits_0^{\sqrt2a}\frac{r^3}{2a}dr\right)=2\pi\left(-a^3+\frac{r^4}{8a}\Big|_{0}^{\sqrt2a}\right)={} \\
{}={}&2\pi\left(\frac{4a^4}{8a}-0-a^3\right)=\pi(a^3-2a^3)=-\pi a^3.
\end{align*}
Naturally, we should have expected a negative result from integrating a negative function, which is why I said $-A$ is the integral. $A$ is an area, so it must be positive, and will be $A=\pi a^3$. I wonder if it is a coincidence that I get the same result as the other "volume" answer or it isn't. If the answerer would elaborate on their purported equivalence…
Edit
Sorry, I must be thinking of something else. The above is indeed the volume. To find the surface, I must compute a surface integral. Let me parametrize the surface:
$$\sigma(x,y)=\left(x,y,a-\frac{x^2+y^2}{2a}\right),$$
with $(x,y)\in B(0,\sqrt2a)$. OK, I rotated it to have $z>0$. The area won't change. The surface integral will be the integral over the ball $B(0,\sqrt2a)$ of $\|\partial_x\sigma(x,y)\times\partial_y\sigma(x,y)\|$, where, $\partial$ is the component-wise partial derivative and $\|\cdot\|$ is the standard Euclidean norm. Let us get down to the calcs then. The derivatives:
\begin{align*}
\partial_x\sigma(x,y)={}&\left(1,0,\frac{1}{2a\sqrt{x^2+y^2}}2x\right), \\
\partial_y\sigma(x,y)={}&\left(0,1,\frac{1}{2a\sqrt{x^2+y^2}}y\right).
\end{align*}
The vector product:
\begin{align*}
\partial_x\sigma(x,y)\times\partial_y\sigma(x,y)={}&\left|\begin{array}{ccc} i & j & k \\ 1 & 0 & \frac{x}{2a\sqrt{x^2+y^2}} \\ 0 & 1 & \frac{y}{2a\sqrt{x^2+y^2}} \end{array}\right|={} \\
{}={}&i\left(-\frac{x}{2a\sqrt{x^2+y^2}}\right)-j\left(\frac{y}{2a\sqrt{x^2+y^2}}\right)+k=\left(\begin{array}{c} -\frac{x}{2a\sqrt{x^2+y^2}} \\ -\frac{y}{2a\sqrt{x^2+y^2}} \\ 1 \end{array}\right)={} \\
{}={}&(-\nabla\sigma_3(x,y),0)+(0,0,1).
\end{align*}
Since those two vectors are orthogonal, the norm of their sum is the sum of the norms, which means:
$$\left\|\partial_x\sigma(x,y)\times\partial_y(x,y)\right\|=\|\nabla\sigma_3(x,y)\|+1.$$
This seems to be a general formula for areas of surfaces of graphs of cartesian functions, i.e. whenever you want the area of a surface $\sigma(x,y)=(x,y,f(x,y))$ the area is the integral over the domain of $\sigma$ of $\|\nabla f(x,y)\|+1$. In our case, the gradient is normalized to $\frac{1}{2|a|}$, hence we will get an area of $(\frac{1}{2|a|}+1)\pi(\sqrt2a)^2$. Which is pretty different from what the other answer says, so I prompt the answerer to point out any mistake of mine :).
|
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How to handle indices with fractional degree? An algebra problem ate my head!!!
$x$ and $y$ are positive real numbers such that
$$\sqrt{x^2 + \sqrt[3]{x^4 y^2}} + \sqrt{y^2 + \sqrt[3]{x^2 y^4}} = 512.$$
Find $x^{2/3} + y^{2/3}$.
It would be a great help if anybody helps me in solving this problem. I tried taking conjugates and all but I didn't get any answer.
thank you
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Hint: Notice that $\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^3}}$ $= \sqrt{x^2+x^{4/3}y^{2/3}}+\sqrt{y^2+x^{2/3}y^{4/3}}$
$= \sqrt{x^{4/3}(x^{2/3}+y^{2/3})}+\sqrt{y^{4/3}(y^{2/3}+x^{2/3})}$ $= x^{2/3}\sqrt{x^{2/3}+y^{2/3}}+y^{2/3}\sqrt{y^{2/3}+x^{2/3}}$
$= (x^{2/3}+y^{2/3})\sqrt{x^{2/3}+y^{2/3}}$ $= (x^{2/3}+y^{2/3})^{3/2}$.
Can you finish the problem from here?
|
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Why does this method to solve a quadratic equation for $x$ omit $x=0$? Here is a simple quadratic equation:
$$9x^2 - 36x = 0$$
We proceed as following:
\begin{align*}
9x^2 & = 36x\\
9x & = 36\\
x & = 4
\end{align*}
So, we get $x=4$.
But, here's another way:
\begin{align*}
9x^2 - 36x & = 0\\
x(9x-36) & = 0\\
\end{align*}
Therefore, $\boldsymbol{x=0}$ or $x=4$.
What's the problem in the first method that doesn't allow $x$ to be $0$?
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To solve an equation, you have to establish a sequence of logical equivalences.
In your first method, you only established a sequence of logical implications. This is the reason why you lost the solution $x=0$.
Remark: In order to solve an equation, an sequence of implications can fails for two reasons:
*
*You lost one (o more than one) "right solution".
*You find one (or more then one) "wrong solution".
*
*Example 1 (provided by you):
$$\begin{align}
&9x^2-36x=0\\
\Rightarrow\quad&9x^2=36x\\
\Rightarrow\quad&9x=36\qquad \text{if } x\neq 0\\
\Rightarrow\quad&x=4\qquad \text{if } x\neq 0\\
\end{align}$$
Here, we lost the right solution $x=0$.
*Example 2:
$$\begin{align}
&x^2+1=0\\
\Rightarrow\quad&(x^2+1)(x^2-1)=0(x^2-1)\\
\Rightarrow\quad&x^4-1=0\\
\Rightarrow\quad&x^4=1\\
\Rightarrow\quad&x=1\\
\end{align}$$
Here, we find the wrong solution $x=1$.
*Example 3 (provided by you too):
$$\begin{align}
&9x^2-36x=0\\
\Leftrightarrow\quad&x(9x-36)=0\\
\Leftrightarrow\quad&x=0\text{ or }9x-36=0\\
\Leftrightarrow\quad&x=0\text{ or }x=4\\\end{align}$$
Here, we lost nothing right and find nothing wrong.
*Example 4:
$$\begin{align}
&9x^2-36x=0\\
\Leftrightarrow\quad&9x^2=36x\\
\Leftrightarrow\quad&9x=36 \text{ if } x\neq 0\qquad \text { or } \qquad x=0\\
\Leftrightarrow\quad&x=4\text { or }x=0\\
\end{align}$$
Here, we solved the equation because we established a sequence of equivalences.
|
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Nesbitt's Inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ I'm reading a book which focus in inequality.
I'm stuck in this question. Let $a,b,c$ be positive real numbers. (Nesbitt's inequality) Prove the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$
So the first step of solution given is $\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{a+c}{c+b}+\frac{c+b}{a+c}+\frac{b+a}{a+c}+\frac{a+c}{b+a}\geq2+2+2=6$
I don't know how to proceed from the question to the first step of solution. Can anyone explain?
|
The first step of the solution says that $X+\frac1X+Y+\frac1Y+Z+\frac1Z\geq2+2+2$.
where $X=\frac{a+b}{b+c}$ and so on.
Do you know that $X+\frac1X$ is always $2$ or more whenever $X>0$?
|
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Solving functional equation $f(x+y)=f(x)+f(y)+xy$ We are given $f(0)=0$. Then when $x+y=0$:
$$0=f(-y)+f(-x)+xy$$
Can I now use $x=0$ and obtain:
$$0=f(-y)?$$ Is this correct? Is there a better way to solve this equation?
|
I show that the only solution is $f = \frac{1}{2}x^2 + bx$ (assuming $f$ continuous).
My strategy was to transform $f(x+y) = f(x) + g(y) + xy$ into a linear equation.
Write $f(x+y) = f(x) + g(y) + xy$ as $f(x+y) - \frac{1}{2}(x+y)^2 = (f(x) - \frac{1}{2}x^2) + (g(y) - \frac{1}{2}y^2)$. Substitute $g(z) = f(z) - \frac{1}{2}z^2$. Then we have:
$g(x+y) = x+y$, which is linear, so $g(x) = bx$ (assuming continuity).
Thus $f(z) - \frac{1}{2}z^2 = bz$, so $f(z) = \frac{1}{2}z^2 + bz$.
|
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Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$ It's required to prove that
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
I managed to go about out it two ways:
*
*Show it is equivalent to $\mathsf{true}$:
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
$$\Longleftrightarrow\sin x(1+\cos x+\sin x)\equiv(1+\cos x)(1-\cos x+\sin x)$$
$$\Longleftrightarrow\sin x+\cos x\sin x+\sin^2 x\equiv1-\cos x+\sin x+\cos x-\cos^2 x+\sin x \cos x$$
$$\Longleftrightarrow\sin^2 x\equiv1-\cos^2 x$$
$$\Longleftrightarrow\cos^2 x +\sin^2 x\equiv1$$
$$\Longleftrightarrow \mathsf{true}$$
*Multiplying through by the "conjugate" of the denominator:
$${\rm\small LHS}\equiv\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} $$
$$\equiv\frac{1+\cos x + \sin x}{1 - (\cos x - \sin x)} ~~\cdot ~~\frac{1+(\cos x - \sin x)}{1 +(\cos x - \sin x)}$$
$$\equiv\frac{(1+\cos x + \sin x)(1+\cos x - \sin x)}{1 - (\cos x - \sin x)^2}$$
$$\equiv\frac{1+\cos x - \sin x+\cos x + \cos^2 x - \sin x \cos x+\sin x + \sin x \cos x - \sin^2 x}{1 - \cos^2 x - \sin^2 x + 2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- \sin^2 x}{2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- 1 + \cos^2 x}{2\sin x \cos x}$$
$$\equiv\frac{2\cos x (1+\cos x)}{2\cos x(\sin x)}$$
$$\equiv\frac{1+\cos x}{\sin x}$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\equiv {\rm\small RHS}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$$
Both methods of proof feel either inelegant or unnecessarily complicated. Is there a simpler more intuitive way to go about this? Thanks.
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Since $1-\cos^2 x = \sin^2 x$, we have $f(x) := \dfrac{1+\cos x}{\sin x} = \dfrac{\sin x}{1-\cos x}$. Therefore,
\begin{align*}\dfrac{1+\cos x + \sin x}{1-\cos x + \sin x} &= \dfrac{f(x)\sin x + f(x)(1-\cos x)}{1-\cos x + \sin x} \\ &= \dfrac{f(x)[1-\cos x + \sin x]}{1-\cos x + \sin x} \\ &= f(x) \\ &= \dfrac{1+\cos x}{\sin x}.\end{align*}
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Prove that $xy+yz+zx+\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge 2x^2+2y^2+2z^2$ for every $x,y,z$ strictly positive I checked the inequality for $y=z$. In this particular case, after some simplifications, the inequality becomes:
$$
2x^3+y^3\ge 3x^2y,
$$
which is true, according to the arithmetic-geometric mean inequality applied to the numbers $x^3,\ x^3$ and $y^3$. I have no idea, at least for now, on how to proceed in the general case. Please give me a hint.
|
Use Schur inequality
$$a^4+b^4+c^4+abc(a+b+c)\ge \sum_{cyc}(a^3b+ab^3)\ge \sum_{cyc}2a^2b^2$$
Let
$$a=xy,b=yz,c=xz$$
then is your inequality
|
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Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?
|
After checking this for k=1, k=2 you need to follow this:
Let's say $A_1=(1+2+...+k)^2$; $B_1=1^3+...k^3$ and $A_2=(1+2+...+ (k+1))^2$
$B_2=1^3+...(k+1)^3$
Then to use induction it is needed to prove that $A_2-A_1=B_2-B_1$ or:
$$(1+2+...+ (k+1))^2-(1+2+...+k)^2=1^3+...(k+1)^3-1^3+...k^3$$
And it is equal to:
$$(k+1)(1+2+...+k)+(1+2+...+k+1)=(k+1)^3$$
$$(k+1)(\frac{k(k+1)}{2}+\frac{(k+1)(k+2)}{2})=(k+1)^3$$
$$(k+1)^2\frac{2k+2}{2}=(k+1)^3$$
$$(k+1)^3=(k+1)^3$$
This is the prove.
|
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Find the real ordered pairs $(x,y)$ satisfying $x^2+x=y^4+y^3+y^2+y.$ Find the real ordered pairs $(x,y)$ satisfying $x^2+x=y^4+y^3+y^2+y.$
$x^2+x=y^4+y^3+y^2+y$
$4x^2+4x=4y^4+4y^3+4y^2+4y$
$4x^2+4x+1=4y^4+4y^3+4y^2+4y+1$
$(2x+1)^2=4y^4+4y^3+4y^2+4y+1$
I am stuck here.Is there a general method to solve such type of equations?Please help me.
|
Under the assumption that the problem asks for integer solutions, (which is probably the case)
We have $$(2y^2+y)^2<(2x+1)^2=4y^4+4y^3+4y^2+4y+4<(2y^2+y+1)^2$$ for $y > 3$ or $y <-1$, so we must have $-1 \le y \le 3$. Now bash.
|
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$\lim \frac{\cos{x}}{x^2}$ as x goes to infinity $\displaystyle \lim_{x \to \infty} \frac{\cos{x}}{x^2} =\lim_{x \to \infty} \frac{\frac{d}{dx} \cos{x}}{\frac{d}{dx}x^2} = -\frac{1}{2}\lim_{x \to \infty}\frac{\sin{x}}{x}$. But $-\frac{1}{x} \le \frac{\sin{x}}{x} \le \frac{1}{x}$ so $\displaystyle -\lim_{x \to \infty}\frac{1}{x} \le \lim_{x \to \infty}\frac{\sin{x}}{x} \le \lim_{x \to \infty}\frac{1}{x} \iff 0 \le \lim_{x \to \infty}\frac{\sin{x}}{x} \le 0 \iff \lim_{x \to \infty}\frac{\sin{x}}{x} = 0.$
Therefore $\displaystyle \lim_{x \to \infty} \frac{\cos{x}}{x^2} = 0.$
Is the above correct?
|
You don’t need to use the derivatives: For every $x > 0$ we have
$$
-\frac{1}{x^2} \leq \frac{\cos(x)}{x^2} \leq \frac{1}{x^2}.
$$
Because $\lim_{x \to \infty} \frac{1}{x^2} = 0$ it follows that $\lim_{x \to \infty} \frac{\cos(x)}{x^2} = 0$.
|
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|
Find all real numbers $x,y > 1$ such that $\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$
Find all real numbers $x,y > 1$ such that $$\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$$
Attempt
We can rewrite this as $x^2(x-1)+y^2(y-1) = 8(x-1)(y-1)$. Then I get a multivariate cubic, which I find hard to find all solutions to.
|
As suggested by matt, let $a=x-1$, $b=y-1$ and apply the inequality $r+s\ge 2\sqrt{rs}$ for $r,s>0$ with equality if and only if $r=s$. Then $a,b>0$ and
$$\frac{(a+1)^2}{b}+\frac{(b+1)^2}{a}=\left(\frac{a^2}{b}+\frac{1}{b}\right)+\left(\frac{b^2}{a}+\frac{1}{a}\right)+2\left(\frac{a}{b}+\frac{b}{a}\right)$$
$$\ge \left(2\sqrt{\frac{a^2}{b}\cdot \frac{1}{b}}\right)+\left(2\sqrt{\frac{b^2}{a}\cdot \frac{1}{a}}\right)+2\left(2\sqrt{\frac{a}{b}\cdot \frac{b}{a}}\right)$$
$$=2\left(\frac{a}{b}+\frac{b}{a}\right)+4\ge 2\left(2\sqrt{\frac{a}{b}\cdot \frac{b}{a}}\right)+4=8$$
For equality to hold, we must have $\frac{a^2}{b}=\frac{1}{b}$ and $\frac{b^2}{a}=\frac{1}{a}$, i.e. $a=b=1$, i.e. $x=y=2$ (and indeed equality holds).
|
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If $x=2+i$, $gcd(a,b,c)=1$, and $ax^4+bx^3+cx^2+bx+a=0$, then what is $|c|$?
Suppose
$$a(2+i)^4 + b(2+i)^3 + c(2+i)^2 + b(2+i) + a = 0,$$
where $a,b,c$ are integers whose greatest common divisor is $1$.
Determine $|c|$.
So I first simplified the exponents and combined like terms.
I received $$a(-6+24i)+b(4+12i)+c(3+4i)=0.$$
I don't really know how to progress. What is the answer $|c|$?
|
We use your calculation. Note that as the other answers show there would have been nicer ways to proceed.
From the equation that you obtained we get $-6a+4b+3c=0$ and $24a+12b+4c=0$. Divide the second expression through by $4$, and add the first expression. We get $7b+4c=0$.
Thus $4c=-7b$ and therefore $c=7k$, $b=4k$ for some integer $k$. Substituting in the first equation we get $6a=5k$. It follows that $k=6m$ for some integer $m$. Thus $c=42m$, $b=-24m$, and $a=5m$.
By relative primality we get $|m|=1$, and therefore $|c|=42$.
|
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Strange Mean Inequality This problem was inspired by this question.
$\sqrt [ 3 ]{ a(\frac { a+b }{ 2 } )(\frac { a+b+c }{ 3 } ) } \ge \frac { a+\sqrt { ab } +\sqrt [ 3 ]{ abc } }{ 3 } $
The above can be proved using Hölder's inequality.
$\sqrt [ 3 ]{ a(\frac { a+b }{ 2 } )(\frac { a+b+c }{ 3 } ) } =\sqrt [ 3 ]{ (\frac { a }{ 3 } +\frac { a }{ 3 } +\frac { a }{ 3 } )(\frac { a }{ 3 } +\frac { a+b }{ 6 } +\frac { b }{ 3 } )(\frac { a+b+c }{ 3 } ) } \ge \sqrt [ 3 ]{ (\frac { a }{ 3 } +\frac { a }{ 3 } +\frac { a }{ 3 } )(\frac { a }{ 3 } +\frac { \sqrt { ab } }{ 3 } +\frac { b }{ 3 } )(\frac { a }{ 3 } +\frac { b }{ 3 } +\frac { c }{ 3 } ) } (\because \text{AM-GM})\\ \ge \frac { a+\sqrt { ab } +\sqrt [ 3 ]{ abc } }{ 3 } (\because \text{Holder's inequality)}$
However, I had trouble generalizing this inequality to
$\sqrt [ n ]{ \prod _{ i=1 }^{ n }{ { A }_{ i } } } \ge \frac { \sum _{ i=1 }^{ n }{ { G }_{ i } } }{ n } $
when ${ A }_{ i }=\frac { \sum _{ j=1 }^{ i }{ { a }_{ i } } }{ i } $
and ${ G }_{ i }=\sqrt [ i ]{ \prod _{ j=1 }^{ i }{ { a }_{ i } } } $ as I could not split the fractions as I did above.
|
This result was conjectured by professor Finbarr Holand, and then it was proved by
K. Kedlaya in an article in the American Mathematical Monthly that could be found here, and then it was generalized by Professor Holand in an article that could be found here.
|
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|
Denoting the sum of $n$ odd numbers $1+3+5+\cdots+(2n-1) = 1+3+5+\cdots+(2n+1)$? Let's say we are denoting the sum of $n$ odd numbers.
Then in symbols $1+3+5+\cdots+(2n-1)$.
If we substitute $(k+1)$ for $n$. $2n-1=2k+1$
So $1+3+5+...+2k+1$
Then can we use $1+3+5+\cdots+(2n+1)$ instead of $1+3+5+\cdots+(2n-1)$?
Logically, I think they are the same, but when I think of the number of terms, $1+3+5+\cdots+(2n+1)$ has one more term than $1+3+5+\cdots+(2n-1)$.
I also think $1+3+5+\cdots+(2n+1)$ is equal to $1+3+5+\cdots+(2n-1)+(2n+1)$, so it would be contradiction. But I don't know how to explain $1+3+5+\cdots+(2n-1)=1+3+5+\cdots+2k+1$ is a contradiction.
|
If you're writing $1+3+5+\cdots+45$, for instance, you could describe that as
$1 + 3 + 5 + \cdots + (2n-1)$ for $1 \le n \le 23$, and there are 23 terms.
You could also describe it as
$1 + 3 + 5 + \cdots + (2n+1)$, but this time we have $0 \le n \le 22$, and there are still only 23 terms: even though the last value of $n$ is 22, this time we're also counting $n=0$.
|
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|
$\frac{2^{2n}}{2n} \le {2n \choose n}$? Show that $\frac{2^{2n}}{2n} \le {2n \choose n}$, $n\in\mathbb{N}$.
What I can see easily is that $\frac{2^{2n}}{2n + 1} \le {2n \choose n}$, since
$(1+1)^{2n} = \sum_{m=0}^{2n}{2n \choose m} = 2^{2n}$. This sum contains $2n+1$ terms with the largest being ${2n \choose n}$. So ${2n \choose n}$ is larger than $\frac{2^{2n}}{2n + 1}$. However I have found the statement $\frac{2^{2n}}{2n} \le {2n \choose n}$, which is not obvious to me.
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The first and last terms are $\left( \begin{array}{c} 2n \\ 0 \end{array} \right) = \left( \begin{array}{c} 2n \\ 2n \end{array} \right) = 1$.
So with $n \ge 1$, the sum of these two terms is $\le \left( \begin{array}{c} 2n \\ n \end{array} \right)$ and we only have $2n$ terms in the sum now.
|
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Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$ Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$
My try
My book gives as a hint to move everything to the left hand side of the inequality and then factor and see what I get in the long factorization process and to lookout for squares.
So that's what I have tried:
\begin{array}
((a^7+b^7)(a^2+b^2) &\ge (a^5+b^5)(a^4+b^4) \\\\
(a^7+b^7)(a^2+b^2)-(a^5+b^5)(a^4+b^4) &\ge 0 \\\\
(a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)(a^2+b^2)-(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)(a^4+b^4) &\ge 0 \\\\
(a+b)\left[(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)((a+b)^2-2ab)-(a^4+b^4)(a^4-a^3b+a^2b^2-ab^3+b^4)\right] &\ge 0
\end{array}
Now it's not clear what I have to do next.I am stuck.
Note: My book doesn't teach any advanced technique for solving inequality as AM-GM ,Cauchy inequality etc..
|
\begin{align*}
(a^7 + b^7)(a^2 + b^2) - (a^5 + b^5)(a^4 + b^4) &= a^7 b^2 + a^2 b^7 - a^5 b^4 - a^4 b^5 \\
&= a^2 b^2 \big(a^5 + b^5 - a^3 b^2 - a^2 b^3\big) \\
&= a^2 b^2 \big(a^2 (a^3 - b^3) + b^2 (b^3 - a^3)\big) \\
&= a^2 b^2 (a^2 - b^2)(a^3 - b^3)
\end{align*}
Now regardless of how $a$ and $b$ are related to each other, the two difference terms above have the same sign.
|
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|
General values of $\theta$ for which $(\sin\theta)x^2+(2\cos\theta)x+\frac{\cos\theta+\sin\theta}{2}$ is the square of a linear function. Find the general values of $\theta$ for which the quadratic function $(\sin\theta)x^2+(2\cos\theta)x+\frac{\cos\theta+\sin\theta}{2}$ is the square of a linear function.
As the quadratic function is the square of a linear function,so its discriminant should be zero.
$(2\cos\theta)^2=4(\sin\theta)\frac{\cos\theta+\sin\theta}{2}$
$2\cos^2\theta=\sin^2\theta+\sin\theta\cos\theta$
$(\cos^2\theta-\sin^2\theta)=(\sin\theta\cos\theta-\cos^2\theta)$
$(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)=(-\cos\theta)(\cos\theta-\sin\theta)$
Either $\cos\theta-\sin\theta=0$ or $\tan\theta=-2$
So i found general solution as $n\pi+\frac{\pi}{4}$ or $n\pi-\arctan(2),$ where $n$ is an integer. But the answer given in the book is $2n\pi+\frac{\pi}{4}$ or $(2n+1)\pi-\arctan(2)$,where $n$ is an integer.
I dont know where i am wrong.
|
Since it is the square of a linear function, it has to be positive. So the first coefficient $\sin \theta$ has to be positive. This gives the criteria that $\theta$ is in the first or second quadrant. Hence the answer $2n\pi+\cdots$
|
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|
Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$
It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?
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$x^3=a^2,y^3=b^2,z^3=c^2 \implies x^3+y^3+z^2 +3xyz \ge 2(\sqrt{(xy)^3}+\sqrt{(yz)^3}+\sqrt{(xz)^3})$
we have $x^3+y^3+z^3 +3xyz \ge xy(x+y)+yz(y+z)+xz(x+z)$
$xy(x+y)\ge 2xy\sqrt{xy}=2\sqrt{(xy)^3} \implies xy(x+y)+yz(y+z)+xz(x+z)\ge 2(\sqrt{(xy)^3}+\sqrt{(yz)^3}+\sqrt{(xz)^3})$
|
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prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$
If $x<y<z$ are real numbers, prove that $$(x-y)^3+(y-z)^3+(z-x)^3 > 0.$$
Attempt
We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of using rearrangement but then I would get $\geq$ instead of $>$.
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Because it is only the differences in the variables that matter, we can subtract the smallest value ($x$) from each term:
$-y^3 + (y-z)^3 + z^3 > 0$.
Let $y = d_1$ and $z-y = d_2$ and note $z = d_1 + d_2$. Then:
$-d_1^3 - d_2^3 + (d_1 + d_2)^3 > 0$.
Simplify:
$3d_1 d_2^2 + 3 d_1^2 d_2 >0$.
or
$3(d_1 d_2)(d_1 + d_2) > 0$.
|
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|
Value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Find the value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Now the trivial method is to put $X=5+2\sqrt{-4}$ in the polynomial and calculate but this is for $2$ marks only and that takes a hell lot of time for $2$! So I was thinking may be there is some trick or other technique to get the result quicker . Can anybody help please $?$
Thank you .
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$X+5=4i$ squaring you get $x^2+10x+41=0$ now just divide the given polynomial with this equation so solving $$\frac{x^4+9x^3+35x^2-x+4}{x^2+10x+41}$$ you get it equal to $(160)$
|
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Prove that if $a,b,$ and $c$ are positive real numbers then $(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}.$
Prove that if $a,b,$ and $c$ are positive real numbers then $$(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}.$$
This looks like a simple question. We can apply AM-GM twice to get $(a+b)(a+c) \geq 4a\sqrt{bc}$. Then how do I use that fact to get $(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}$?
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we get after squaring and factorizing $$ \left( {a}^{2}+ab+ac-cb \right) ^{2}$$ and this is nonnegative
|
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|
Evaluating integral $\int\frac{e^{\cos x}(x\sin^3x+\cos x)}{\sin^2x}dx $ $$\int\frac{e^{\cos x}(x\sin^3x+\cos x)}{\sin^2x}dx $$
The usual form $\int e^x(f(x)+f'(x))dx $ does not apply here. What substitution should I make ?
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The integral is of the form
\begin{equation*}
\int \left( x\sin x+\frac{\cos x}{\sin ^{2}x}\right) e^{\cos x}dx=\int
h(x)e^{g(x)}dx.
\end{equation*}
This form recalls the well-known formula
\begin{equation*}
\int \left( f^{\prime }(x)+g^{\prime }(x)f(x)\right)
e^{g(x)}dx=f(x)e^{g(x)}+C.
\end{equation*}
Its proof maybe found at
Proof for formula $\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$
So we are done if we find a function $f(x)$ such that
\begin{equation*}
h(x)=f^{\prime }(x)+g^{\prime }(x)f(x).
\end{equation*}
In what follows, I will show that $f(x)=-x-\csc x,$ and therefore
\begin{equation*}
\int \left( x\sin x+\frac{\cos x}{\sin ^{2}x}\right) e^{\cos x}dx=\left(
-x-\csc x\right) e^{\cos x}+C.
\end{equation*}
$\color{red}{\bf Problem:}$ We want to write $x\sin x+\frac{\cos x}{\sin ^{2}x}$ as $
f^{\prime }(x)+g^{\prime }(x)f(x)$ where $g(x)=\cos x,$ $g^{\prime
}(x)=-\sin x$ and $f(x)$ is to be determined.
First, it is easy to see that
\begin{equation*}
x\sin x+\frac{\cos x}{\sin ^{2}x}=\frac{\cos x}{\sin ^{2}x}+(-\sin x)(-x)=
\frac{\cos x}{\sin ^{2}x}+g^{\prime }(x)(-x).
\end{equation*}
If we put $f_{1}(x)=(-x),$ then
\begin{equation*}
f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)=\left( -1\right) +(-\sin x)\left(
-x\right) =-1+x\sin x
\end{equation*}
This suggests to add and to subtract the term $-1$ as follows
\begin{equation*}
x\sin x+\frac{\cos x}{\sin ^{2}x}=x\sin x+\frac{\cos x}{\sin ^{2}x}%
-1+1=\left( -1+x\sin x\right) +\left( 1+\frac{\cos x}{\sin ^{2}x}\right) ,
\end{equation*}
Now let us find $f_{2}(x)$ such that :
\begin{equation*}
\left( 1+\frac{\cos x}{\sin ^{2}x}\right) =f_{2}^{\prime }(x)+g^{\prime
}(x)f_{2}(x)=f_{2}^{\prime }(x)-(\sin x)f_{2}(x).
\end{equation*}
It is easy to see that
\begin{equation*}
\frac{\cos x}{\sin ^{2}x}+1=\frac{\cos x}{\sin ^{2}x}+(-\sin x)\left( \frac{
-1}{\sin x}\right) =\frac{\cos x}{\sin ^{2}x}+g^{\prime }(x)\left( \frac{-1}{
\sin x}\right)
\end{equation*}
Then if we put $f_{2}(x)=\left( \frac{-1}{\sin x}\right) $ it follows that
\begin{equation*}
f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)=\left( \frac{-1}{\sin x}\right)
^{\prime }-(\sin x)\left( \frac{-1}{\sin x}\right) =\frac{\cos x}{\sin ^{2}x}
+1.
\end{equation*}
It follows that
\begin{eqnarray*}
\left( x\sin x+\frac{\cos x}{\sin ^{2}x}\right) &=&\left( -1+x\sin x\right)
+\left( \frac{\cos x}{\sin ^{2}x}+1\right) \\
&& \\
&=&\left( f_{1}^{\prime }(x)+g^{\prime }(x)f_{1}(x)\right) +\left(
f_{2}^{\prime }(x)+g^{\prime }(x)f_{2}(x)\right) \\
&& \\
&=&\left( (f_{1}(x)+f_{2}(x))^{\prime }+g^{\prime
}(x)(f_{1}(x)+f_{2}(x))\right)
\end{eqnarray*}
then, it suffices to take
\begin{equation*}
f(x)=f_{1}(x)+f_{2}(x)=-x-\frac{1}{\sin x}=-x-\csc x.\ \ \
\color{red}
\blacksquare
\end{equation*}
|
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|
Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$ I've worked out a proof, but I was wondering about alternate, possibly more elegant ways to prove the statement. This is my (hopefully correct) proof:
Starting from the identity $2^m = \sum_{k=0}^m \binom{m}{k}$ (easily derived from the binomial theorem), with $m = 2n$:
$2^{2n} = 4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{2n-1} + \binom{2n}{2n}$
Applying the property $\binom{m}{k} = \binom{m}{m-k}$ to the second half of the list of summands in RHS above:
$4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n-1} + \binom{2n}{n} + \underbrace{\binom{2n}{n-1} +\cdots \binom{2n}{1} + \binom{2n}{0}}_{\binom{m}{k} = \binom{m}{m-k} \text{ has been applied}}$
Rearranging the above sum by alternately taking terms from the front and end of the summand list in RHS above (and introducing the term $\binom{2n}{-1} = 0$ at the beginning just to make explicit the pattern being developed):
$4^n = (\binom{2n}{-1} + \binom{2n}{0}) + (\binom{2n}{0} + \binom{2n}{1}) + \cdots + (\binom{2n}{n-1} + \binom{2n}{n})$
Finally, using the property $\binom{m}{k} + \binom{m}{k-1} = \binom{m+1}{k}$ on the paired summands, we get the desired result:
$4^n = \binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n}$
|
Let $A$ be the powerset (i.e., the set of subsets of) of $\{1,\ldots,2n\}$. Let $B$ the set of all subsets of $\{1,\ldots,2n+1\}$ at most $n$ elements. Then "clearly" $|A|=2^{2n}=4^n$ and $|B|=\sum_{k=0}^n{2n+1\choose k}$.
Define the following map $f\colon B\to A$:
$$f(S)=\begin{cases}S&\text{if }2n+1\notin S\\\{1,\ldots,2n+1\}\setminus S&\text{if }2n+1\in S\end{cases} $$
and define $g\colon A\to B$ by
$$g(S)=\begin{cases}S&\text{if }|S|\le n\\\{1,\ldots,2n+1\}\setminus S&\text{if }|S|>n\end{cases} $$
Finally, verify that $f$ and $g$ are inverse of each other, hence $|A|=|B|$.
|
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Minimum value of $\frac{(1 + x + x^2)(1 + y + y^2)}{xy}$
What is the minimum value of $$\frac{(1 + x + x^2)(1 + y + y^2)}{xy},~~(x \neq 0)$$
Should we find the minimum value of each quadratic?
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$$\frac{(1+x+x^2)(1+y+y^2)}{xy} = \frac{1+x+x^2}{x} \cdot \frac{1+y+y^2}{y}$$
By AM–GM inequality $(1+x+\frac1x)(1+y+\frac1y)≥3*3=9$
|
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Given $a+b+c=3$ .Prove $ \sum \limits_{cyc} \frac {1}{a^2+b^2+2} \le \frac 34$ Yesterday I found this on the Internet:
Give 3 non-negative numbers $a,b,c$ that $a+b+c=3$. Prove
$$ \sum _{cyc} \frac {1}{a^2+b^2+2} \le \frac 34 $$
I have tried to solve this using AM-GM:
From AM-GM we got:
$$a^2+b^2+2\ge2(a+b)$$
$$\Rightarrow \frac{1}{a^2+b^2+2} \le \frac12\frac1{a+b}$$
$$\Rightarrow \sum _{cyc} \frac {1}{a^2+b^2+2} \le \frac12\sum_{cyc}\frac{1}{a+b}$$
We have to prove $$\frac12\sum_{cyc}\frac{1}{a+b}\le\frac34$$
or $$\sum_{cyc}\frac{1}{a+b}\le\frac32$$
However the above statement seem to be false. If $a = 0 , b = 1, c = 2$:
$$\sum_{cyc}\frac{1}{a+b} = 1+\frac13+\frac12=\frac{11}6\gt \frac32$$
Anyone know the solutions?
|
It's suffice to show the following inequality $$ \sum\limits_{sic}{\frac{a^2+b^2}{a^2+b^2+2}} \ge \frac{3}{2} $$ By using Cauchy, we have $$ LHS \ge \frac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a^2+b^2+c^2\right)+6} \ge \frac{\sqrt{3\left(a^2b^2+b^2c^2+c^2a^2\right)}+2\left(a^2+b^2+c^2\right)}{\left(a^2+b^2+c^2\right)+3} $$ Notice that $ \sqrt{3\left(a^2b^2+b^2c^2+c^2a^2\right)} \ge ab+bc+ca $, thus, the last term is greater than $$ \frac{2\left(a^2+b^2+c^2\right)+\left(ab+bc+ca\right)}{\left(a^2+b^2+c^2\right)+3} = \frac{3}{2} $$ The conclusion follows. Note that $ a+b+c=3 $ implies $ a^2+b^2+c^2+2ab+2bc+2ca=9 $.
|
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|
Converting $(1+...+n)^2*(n+1)^3$ to $(2+...+2n)^2$ I'm currently going through Calculus by Spivak by myself, and came across a proof by induction requiring to prove $1^3+...+n^3 = (1+...+n)^2$
Naturally, to prove this, I need to somehow convert $(1+...+n)^2+(n+1)^3$ to $(2+...+2n)^2$.
After quite a bit of thinking, I'm still not sure how to do this. I think i may be forgetting about some property of squares that we're supposed to be using.
Note: Please only provide a hint, not the complete answer.
Edit: I was mistakenly taking $(1+...+n)^2*(n+1)^3$ rather than $(1+...+n)^2+(n+1)^3$. Corrected.
|
Naturally, to prove this, i need to somehow convert $(1+...+n)^2*(n+1)^3$ to $(2+...+2n)^2$.
Unfortunately, "$(1+...+n)^2*(n+1)^3 = (2+...+2n)^2$" is not true for all $n$. Once you simplify the summations, the left side will be a polynomial with degree $2 \cdot 2 + 3 = 7$ while the right side will be a polynomial with degree $2 \cdot 2 = 4$.
The induction step is to assume that $$1^3+2^3+\cdots+n^3 = (1+2+\cdots+n)^2$$
for some integer $n$, and then prove that $$1^3+2^3+\cdots+n^3+(n+1)^3 = (1+2+\cdots+n+(n+1))^2.$$
As Arnie Dris suggested, the identity $1+2+\cdots+n = \dfrac{n(n+1)}{2}$ is useful.
Note that \begin{align*}& (1+2+\cdots+(n+1))^2 - (1+2+\cdots+n)^2 \\ &= \left[\dfrac{(n+1)(n+2)}{2}\right]^2-\left[\dfrac{n(n+1)}{2}\right]^2 \\ &=\dfrac{(n+1)^2}{4}\left[(n+2)^2-n^2\right] \\ &= \dfrac{(n+1)^2}{4}[4n+4] \\ &= (n+1)^3\end{align*}
Do you see how to complete the proof using this?
|
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|
Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$? Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$?
The justification is that $\lim_\limits{x\to 0}\frac{\sin(x)}{x} = 1$
But, I am not seeing the connection.
L'Hospital's rule? Is there a double angle substitution happening?
|
As user Daniel has written you can express the ratio $\frac{\sin 6x}{\sin 2x}$ in a way which makes it amenable to the use of standard limit $$\lim_{x\to 0} \frac{\sin x}{x} = 1\ .$$ Notice that $$\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\sin 2x}{2x} = \lim_{x\to 0} \frac{\sin 6x}{6x}\ .$$ As long as the argument $x$ is not equal to zero (and $\sin 2x \neq 0$) you can prolong the ratio $\frac{\sin 6x}{\sin 2x}$ as follows.
$$\frac{\sin 6x}{\sin 2x} = \frac{\sin 6x}{6x} \cdot \frac{6x}{2x} \cdot \frac{2x}{\sin 2x} = \underbrace{\frac{\sin 6x}{6x}}_{\to 1} \cdot \frac{6}{2} \cdot \underbrace{\frac{1}{\frac{\sin 2x}{2x}}}_{\to\frac{1}{1}} \to 1 \cdot 3 \cdot 1 = 3 \ .$$ Hence you obtain the result $$\lim_{x\to 0}\frac{\sin 6x}{\sin 2x} = 3 \ .$$
|
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|
Find the value of $A/B$ for this definite integration. $$\int_{0}^{\infty}e^{-\sqrt 3 x^{\frac{2}{3}}}\sin(x^{\frac{2}{3}})\mathrm{d}x$$
Given that the integral above is equal to $\frac{3\pi^A}{B}$ for rational numbers $A$ and $B$, find the value of $A/B$.
I thought of taking $x^{2/3}$ as t but that didn't help. Any other suggestion?
|
Letting $t=x^{\frac13}$ and noting
$$ (-\frac{1}{8} e^{-\sqrt{3} t^2} (\sqrt{3} \sin(t^2)+\cos(t^2)))'=te^{-\sqrt 3 t^2}\sin(t^2)$$
then we have
\begin{eqnarray}
&&\int_{0}^{\infty}e^{-\sqrt 3 x^{\frac{2}{3}}}\sin(x^{\frac{2}{3}})\mathrm{d}x\\
&=&3\int_{0}^{\infty}e^{-\sqrt 3 t^2}\sin(t^2)t^2\mathrm{d}t\\
&=&-\frac{3}{8}\int_{0}^{\infty}td(e^{-\sqrt{3} t^2} (\sqrt{3} \sin(t^2)+\cos(t^2)))\\
&=&\frac{3}{8} \int_{0}^{\infty}e^{-\sqrt{3} t^2} (\sqrt{3} \sin(t^2)+\cos(t^2)))dt.
\end{eqnarray}
Noting that
$$ \int_{0}^{\infty}e^{-\sqrt{3} t^2+it^2}=\int_{0}^{\infty}e^{-(\sqrt{3}-i) t^2}=\frac{\sqrt{\pi}}{2\sqrt{\sqrt{3}-i}}=\frac{\sqrt\pi}{2}\left(\frac{1}{4}(1+\sqrt{3})+i\frac{1}{4}(\sqrt{3}-1)\right) $$
we have
$$ \int_{0}^{\infty}e^{-\sqrt{3} t^2}\cos(t^2)dt=\frac{\sqrt\pi}{8}(1+\sqrt{3}), \int_{0}^{\infty}e^{-\sqrt{3} t^2}\sin(t^2)dt=\frac{\sqrt\pi}{8}(\sqrt{3}-1).$$
Thus
\begin{eqnarray}
&&\int_{0}^{\infty}e^{-\sqrt 3 x^{\frac{2}{3}}}\sin(x^{\frac{2}{3}})\mathrm{d}x\\
&=&\frac{3}{8}\left(\sqrt{3} \frac{\sqrt\pi}{8}(\sqrt{3}-1)+\frac{\sqrt\pi}{8}(1+\sqrt{3}))\right)\\
&=&\frac{3\sqrt\pi}{16}.
\end{eqnarray}
|
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|
Second derivative of $x^3+y^3=1$ using implicit differentiation I need to find the $D_x^2y$ of $x^3+y^3=1$ using implicit differentiation
So,
$$
x^3 + y^3 =1 \\
3x^2+3y^2 \cdot D_xy = 0 \\
3y^2 \cdot D_xy= -3x^2 \\
D_xy = - {x^2 \over y^2}
$$
Now I need to find the $D_x^2y$.
I am pretty sure that means the second derivative.
How would I do it to find the second derivative? apparently, it is supposed to be$$-{2x \over y^5}$$
|
If
$x^n + y^n =1
$
then
$nx^{n-1} +ny^{n-1}y' = 0$
or
$y'
=-\frac{x^{n-1}}{y^{n-1}}
$.
Differentiating again,
$\begin{array}\\
0
&=(n-1)x^{n-2}+y^{n-1}y''+(n-1)y^{n-2}(y')^2\\
&=(n-1)x^{n-2}+y^{n-1}y''+(n-1)y(\frac{x^{n-1}}{y^{n-1}})^2\\
&=y^{n-1}y''+(n-1)x^{n-2}(1+\frac{x^n}{y^{2n-3}})\\
&=y^{n-1}y''+(n-1)x^{n-2}\frac{x^n+y^{2n-3}}{y^{2n-3}}\\
\end{array}
$
so
$y''
=-\frac{(n-1)x^{n-2}\frac{x^n+y^{2n-3}}{y^{2n-3}}}{y^{n-1}}
=-\frac{(n-1)x^{n-2}(x^n+y^{2n-3})}{y^{3n-4}}
$
If $n=3$,
$y'
=-\frac{x^2}{y^2}
$
and
$y''
=-\frac{2x(x^3+y^3)}{y^5}
=-\frac{2x}{y^5}
$
since
$x^3+y^3 = 1$.
|
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|
Prove that $\frac{1}{a+ab}+\frac{1}{b+bc}+\frac{1}{c+ca} \geq \frac{3}{2}.$
Let $a,b,$ and $c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{1}{a+ab}+\dfrac{1}{b+bc}+\dfrac{1}{c+ca} \geq \dfrac{3}{2}.$$
I thought about substituting in $abc = 1$ to get $$\dfrac{1}{a+\dfrac{1}{c}}+\dfrac{1}{b+\dfrac{1}{a}}+\dfrac{1}{c+\dfrac{1}{b}} = \dfrac{c}{ac+1}+\dfrac{a}{ab+1}+\dfrac{b}{bc+1}.$$ Then I am not sure what inequality to apply.
|
Notice that the inequality is equivalent to
$(abc+1)(\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)})+3\ge 6$
This is equivalent to proving that
$\frac{1+abc+a+ab}{a(1+b)}+\frac{1+abc+b+bc}{b(1+c)}+\frac{1+abc+c+ca}{c(1+a)}\ge6$.
However, $\frac{1+abc+a+ab}{a(1+b)}+\frac{1+abc+b+bc}{b(1+c)}+\frac{1+abc+c+ca}{c(1+a)}=\frac{1+a}{a(1+b)}+\frac{1+b}{b(1+c)}+\frac{1+c}{c(1+a)}+\frac{b(c+1)}{b+1}+\frac{c(a+1)}{c+1}+\frac{a(b+1)}{a+1} \ge 6 (\because AM-GM)$.
Therefore, our proof is done.
|
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|
Prove that $ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$ Let $a,b,c,x,y,z$ be positive real numbers such that $x+y+z=1$. Prove that
$$ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$$.
my try:
$2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{\frac{2(a+b+c)}{3}}$
But this is not the right choice because
$ax+by+cz\le{\frac{a+b+c}{3}}$ is not always true.
|
Let $d = \sqrt{2(ab+bc+ca)}$ and $t = \sqrt{2(xy+yz+xz)}$. It is easy to check
$$
\begin{cases}
a^2 + b^2 + c^2 + d^2 &= a^2+b^2+c^2 + 2(ab+bc+ca) = (a+b+c)^2\\
x^2 + y^2 + z^2 + t^2 &= x^2+y^2+z^2 + 2(xy+yz+xz) = (x+y+z)^2 = 1
\end{cases}
$$
Apply Cauchy Schwarz to the two 4-vectors $(a,b,c,d)$ and $(x,y,z,t)$,
we immediately get
$$\text{LHS} = ax+by+cz + dt \le \sqrt{(a^2+b^2+c^2+d^2)(x^2+y^2+z^2+t^2)} = a+b+c = \text{RHS}$$
|
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|
A line is drawn through the point $A(1,2)$ to cut the line $2y=3x-5$ in $P$ and the line $x+y=12$ in $Q$. If $AQ=2AP$, find $P$ and $Q$. A line is drawn through the point $A(1,2)$ to cut the line $2y=3x-5$ in $P$ and the line $x+y=12$ in $Q$. If $AQ=2AP$, find the coordinates of $P$ and $Q$.
I found the lengths of the lines $AQ$ and $AP$ in terms of $x$ and $y$ and used $AQ=2AP$ to get this equation $11x^2-40x-16=0$
Working: $AP=\sqrt{(x-1)^2+(\frac{3}{2}x-\frac{9}{2})^2}=\sqrt{\frac{1}{4}(13x^2-62x+85)}$
$AQ=\sqrt{(x-1)^2+(-x+10)^2}=\sqrt{2x^2-22x+101}$
$AQ=2AP\Rightarrow \sqrt{2x^2-22x+101}=2\sqrt{\frac{1}{4}(13x^2-62x+85)}$
$\Rightarrow 2x^2-22x+101=13x^2-62x+85\Rightarrow 11x^2-40x-16=0$
Using the quadratic formula I found $x=4,x=-\frac{4}{11}$
I then substituted these values into the two equations in the question to get the coordinates $(4,\frac{7}{2}),(-\frac{4}{11},-\frac{15}{11})$ or $(4,8),(-\frac{4}{11},\frac{136}{11})$
But the answers in the book are $(4,\frac{7}{2}),(7,5)$ or $(\frac{2}{5},-\frac{19}{10}),(\frac{11}{5},\frac{49}{5})$
I don't know where I went wrong in my method.
Edit: Found some errors in my method and calculation.
Using $x=4$ in $2y=3x-5$ gives $P(4,\frac{7}{2})$
Line with points $AP$ has equation $y-2=\frac{\frac{7}{2}-2}{3}(x-1)\Rightarrow y=\frac{1}{2}x+\frac{3}{2}$
$Q$ lies on the same line. Therefore, the intersection between $y=\frac{1}{2}x+\frac{3}{2}$ and $y+x=12$ is the point $Q$. Working: $\frac{1}{2}x+\frac{3}{2}=-x+12\Rightarrow x=7$ Therefore, $y=-7+12=5$. Point Q has coordinates $(7,5)$.
But $x=-\frac{4}{11}$ is not the other $x$ coordinate of $P$ according to the answers in the book. I have no idea where I went wrong at this point.
|
Here is a method you might prefer.
We can write $$P(2s+1, 3s-1)$$ where $s$ is a parameter. Clearly the parametrization is not unique, and I have chosen this for convenience.
Similarly we can write $$Q(t, 12-t)$$
Now $$\overrightarrow{AP}=\left(\begin{matrix}2s\\3s-3\end{matrix}\right)$$ and $$\overrightarrow{AQ}=\left(\begin{matrix}t-1\\10-t\end{matrix}\right)$$
Then $$\overrightarrow{AQ}=\pm2\overrightarrow{AP}\Rightarrow\left(\begin{matrix}t-1\\10-t\end{matrix}\right)=\pm2\left(\begin{matrix}2s\\3s-3\end{matrix}\right)$$
With the $+$ sign we get $s=1.5, t=7$ giving $$P(4, 3.5), Q(7, 5)$$
With the $-$ sign we get $s=-\frac{3}{10}, t=\frac{11}{5}$, giving $$P(\frac 25,-\frac{19}{10}), Q(\frac{11}{5},\frac{49}{5})$$
|
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|
Find the last $4$ digits of $2016^{2016}$
Find the last $4$ digits of $2016^{2016}$.
Technically I was able to solve this question by I used Wolfram Alpha and modular arithmetic so the worst I had to do was raise a $4$ digit number to the $9$th power. I would do $2016^2 \equiv 4256^2 \equiv \cdots$
and then continue using the prime factorization of $2016$. I am wondering if there is a better way to solve this.
|
Allow me give a different answer from the three ones above and easy to achieve with a calculator, please. This direct procedure can be applied to similar calculations.
One can simply use one or both properties, $A^{abc}=(A^a)^{bc}\equiv(B)^{bc}\pmod{M}$ and $A^n=A^r\cdot A^s\equiv(B)\cdot (C)\pmod{M}$ where $r+s=n$, iterating the process according to convenience of exponents.$$******$$
$2016^{2016}=(2^5\cdot3^2\cdot7)^{2016}=(2^{2^5\cdot3^2\cdot5\cdot7})(3^{2^6\cdot3^2\cdot7})(7^{2^5\cdot3^2\cdot7})$
Calculating separately each of the three factors, we have
$2^{2^5\cdot3^2\cdot5\cdot7}\equiv(8368)^{6\cdot6\cdot 8}\equiv (9024)^{6\cdot 8}\equiv (8976)^8\equiv6176\pmod{ 10^4} $
$3^{2^6\cdot3^2\cdot7}\equiv(3203)^{8\cdot8\cdot3}\equiv(3761)^{8\cdot3}\equiv(8881)^3\equiv1841\pmod{ 10^4} $
$7^{2^5\cdot3^2\cdot7}\equiv(4007)^{4\cdot4\cdot6}\equiv(5649)^{4\cdot4}\equiv(2401)^4\equiv9601\pmod{ 10^4} $
Hence $$2016^{2016}\equiv6176\cdot1841\cdot9601\equiv\color{red}{3616}\pmod{ 10^4} $$
|
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|
Prove that $\frac{a^3}{(1+b)(1+c)}+\frac{b^3}{(1+a)(1+c)}+\frac{c^3}{(1+a)(1+b)} \geq \frac{3}{4}.$
Let $a,b,$ and $c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{a^3}{(1+b)(1+c)}+\dfrac{b^3}{(1+a)(1+c)}+\dfrac{c^3}{(1+a)(1+b)} \geq \dfrac{3}{4}.$$
Attempt
We have $$\dfrac{a^3}{(1+b)(1+c)}+\dfrac{b^3}{(1+a)(1+c)}+\dfrac{c^3}{(1+a)(1+b)} = \dfrac{a^3}{1+b+c+bc}+\dfrac{b^3}{1+a+b+ac}+\dfrac{c^3}{1+a+b+ab} = \dfrac{a^4}{a+ab+ac+1}+\dfrac{b^3}{b+ab+bc+1}+\dfrac{c^4}{c+bc+ac+1}.$$
I get stuck here.
|
From the AM-GM inequality :
$$\frac{a^3}{(1+b)(1+c)}+\frac{1+b}{8}+\frac{1+c}{8} \geq 3\sqrt[3]{\frac{a^3}{(1+b)(1+c)} \cdot \frac{1+b}{8}\cdot\frac{1+c}{8}}=\frac{3a}{4}$$
Now do this for the others and add them all to get :
$$\frac{a^3}{(1+b)(1+c)}+\frac{b^3}{(1+a)(1+c)}+\frac{c^3}{(1+a)(1+b)}\geq \frac{a+b+c}{2} -\frac{3}{4}\geq \frac{3}{4}$$ because : $$a+b+c \geq 3\sqrt[3]{abc}=3$$ from AM-GM.
|
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|
Find the least value of $4\csc^{2} x+9\sin^{2} x$
Find the least value of $4\csc^{2} x+9\sin^{2} x$
$a.)\ 14 \ \ \ \ \ \ \ \ \ b.)\ 10 \\
c.)\ 11 \ \ \ \ \ \ \ \ \ \color{green}{d.)\ 12} $
$4\csc^{2} x+9\sin^{2} x \\
= \dfrac{4}{\sin^{2} x} +9\sin^{2} x \\
= \dfrac{4+9\sin^{4} x}{\sin^{2} x} \\
= 13 \ \ \ \ \ \ \ \ \ \ \ \ (0\leq \sin^{2} x\leq 1)
$
But that is not in options.
I look for a short and simple way.
I have studied maths upto $12$th grade.
|
Without nice tricks, you can use calculus. Consider
$$
f(x)=\frac{4}{\sin^2x}+9\sin^2x
$$
defined over $(0,\pi)$. Note that the general function is periodic of period $\pi$.
Then
$$
f'(x)=-\frac{8\cos x}{\sin^3x}+18\sin x\cos x
=2\frac{\cos x}{\sin^3x}(9\sin^4x-4)
=2\frac{\cos x}{\sin^3x}(\sqrt{3}\sin x-\sqrt{2})(\sqrt{3}\sin x+\sqrt{2})(3\sin^2x+2)
$$
In the given interval, $\sin x>0$; so the derivative vanishes for $x=\arcsin\sqrt{2/3}$, $x=\pi/2$ and $x=\pi-\arcsin\sqrt{2/3}$.
Since $\lim_{x\to0^+}f(x)=\lim_{x\to\pi^-}f(x)=\infty$, the function has a minimum at $\arcsin\sqrt{2/3}$ and $x=\pi-\arcsin\sqrt{2/3}$ (it has the same value at these points). Clearly
$$
f(\arcsin\sqrt{2/3})=4/(2/3)+9\cdot\frac{2}{3}=12
$$
|
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|
Occurrence of 5 consecutive tails before occurrence of 2 consecutive heads In repeated tossing of a fair coin find the probability that $5$ consecutive tails occur before occurrence of $2$ consecutive heads.
My attempt:
I tried to find the probability of non-occurrence of two consecutive heads in $n$ throws.
Let $a_{n}$ be the number of possibilities in which $2$ consecutive heads do not occur in $n$ throws.
I managed to find the recursion formula.
$a_{1}=2$
$a_{2}=3$
$a_{n}=a_{n-1}+a_{n-2}$
But I am not able to get a closed form of $a_{n}$.
Once $a_{n}$ gets determined it may be possible then to find probability of occurrence of $5$ consecutive heads
|
Here is a slightly different proof which may be of interest although
it is not as elegant as the one by @leonbloy.
Suppose we treat the problem of $t$ tails before $h$ heads.
Encoding this in a generating function with $u$ marking sequences of
tails of length at least $t$ and $v$ sequences of heads of length at
least $h$ and finally $w$ marking the final occurrence of $h$ heads
and introducing
$$G_t(z) = z+z^2+\cdots +z^{t-1}+uz^t\frac{1}{1-z}
\quad\text{and}\quad
G_h(z) = z+z^2+\cdots +z^{h-1}+vz^h\frac{1}{1-z}$$
we obtain
$$H(z) = (1+G_t(z))
\left(\sum_{k\ge 0} G_h(z)^k G_t(z)^k\right)
\left(1+z+\cdots+z^{h-1} + wz^h + z^{h+1}\frac{1}{1-z}\right).$$
Observe that when we remove the three markers $u,v$ and $w$ we obtain
$$Q(z) = \frac{1}{1-z}
\left(\sum_{k\ge 0} \frac{z^k}{(1-z)^k} \frac{z^k}{(1-z)^k}\right)
\frac{1}{1-z}
\\ = \frac{1}{(1-z)^2} \frac{1}{1-z^2/(1-z)^2}
= \frac{1}{(1-z)^2-z^2} = \frac{1}{1-2z}$$
which is good news because it means we have enumerated all $2^n$
possible bit strings of length $n.$
Now extracting coefficients we are interested in the series on $w$
which yields
$$H_1(z) = z^h (1+G_t(z))
\left(\sum_{k\ge 0} G_h(z)^k G_t(z)^k\right)$$
The next step is to discard those terms that have $v\ge 1$ (meaning an
internal occurrence of $h$ heads) which yields on setting $v=0$
$$H_2(z) = z^h (1+G_t(z))
\left(\sum_{k\ge 0}
\left(z\frac{1-z^{h-1}}{1-z}\right)^k G_t(z)^k\right).$$
Finally we need to compute
$$H_3(z) =
\left. H_2(z)\right|_{u=1} - \left. H_2(z)\right|_{u=0}$$
to remove those terms not containing a run of at least $t$ tails.
This yields
$$H_3(z) = z^h \frac{1}{1-z}
\left(\sum_{k\ge 0}
\left(z\frac{1-z^{h-1}}{1-z}\right)^k
\left(\frac{z}{1-z}\right)^k\right)
\\ - z^h \frac{1-z^t}{1-z}
\left(\sum_{k\ge 0}
\left(z\frac{1-z^{h-1}}{1-z}\right)^k
\left(z\frac{1-z^{t-1}}{1-z}\right)^k\right).$$
This finally produces
$$H_3(z) = z^h\frac{1}{1-z}
\frac{1}{1-z^2 (1-z^{h-1})/(1-z)^2}
\\ - z^h\frac{1-z^t}{1-z}
\frac{1}{1- z^2(1-z^{h-1})(1-z^{t-1}))/(1-z)^2}
\\ = z^h
\frac{1-z}{(1-z)^2-z^2 (1-z^{h-1})}
\\ - z^h (1-z^t)
\frac{1-z}{(1-z)^2- z^2 (1-z^{h-1})(1-z^{t-1})}
\\ = z^h
\frac{1-z}{1 - 2z + z^{h+1}}
- z^h (1-z^t)
\frac{1-z}{1 - 2z + z^{h+1} + z^{t+1} - z^{h+t}}.$$
We obtain the probability by setting $z=1/2$
which yields
$$\frac{1}{2^{h+1}} 2^{h+1}
- \frac{1}{2^{h+1}} \left(1-\frac{1}{2^t}\right)
\frac{1}{1/2^{h+1}+1/2^{t+1}-1/2^{h+t}}
\\ = 1
- \frac{2^t-1}{2^{h+t+1}}
\frac{1}{1/2^{h+1}+1/2^{t+1}-1/2^{h+t}}
\\ = 1
- (2^t-1)
\frac{1}{2^t+2^h-2}
= \frac{2^t+2^h-2-(2^t-1)}{2^t+2^h-2}
\\ = \frac{2^h-1}{2^t+2^h-2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1617646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
To evaluate the limits $\lim\limits_{n \to \infty} \{\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3+n^3}\}$ To me it seems like that we need to manipulate the given sum into the Riemann sum of some function. First writing in the standard summation form;
$$\{\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3+n^3}\}=\sum\limits_{k=0}^n \frac{k^2}{k^3+n^3}$$
I have tried factoring out $\frac{1}{n}$ from the above sum. But I cant seem to get into the right form. Any help!
|
You can proceed from here as $$\lim_{n\to \infty}\sum\limits_{k=0}^n \frac{k^2}{k^3+n^3}= \lim_{n\to\infty}\sum\limits_{k=0}^n \;\frac{1}{n}\frac{\frac{k^2}{n^2}}{\frac{k^3}{n^3}+1}= \lim_{h\to 0} \;h\sum_{k=1}^n \frac{{(kh)}^2}{1 + {(kh)}^3}= \int_0^1 \frac{x^2}{1+x^3} \mathrm{d}x\;.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1617952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
If $\omega$ is an imaginary fifth root of unity, then $\log_2 \begin{vmatrix} 1+\omega +\omega^2+\omega^3 -\frac{1}{\omega} \\ \end{vmatrix}$ =
If $\omega$ is an imaginary fifth root of unity, then $$\log_2 \begin{vmatrix}
1+\omega +\omega^2+\omega^3 -\frac{1}{\omega} \\
\end{vmatrix} =$$
My approach :
$$\omega^5 = 1 \\ \implies 1+\omega +\omega^2 +\omega^3 + \omega^4 =0$$
Therefore, \begin{align}\log_2 |1+\omega +\omega^2+ \omega^3 -\frac{1}{\omega}| &=\log_2 |1+\omega +\omega^2+ \omega^3 -\omega^4|\\& =\log_2|-2\omega^4|\\ &=\log_2 2 +\log_2 \omega^4 \end{align}
Now how to solve further; please suggest.
|
Since
$$
\begin{align}
1+\omega+\omega^2+\omega^3+\omega^4
&=\frac{1-\omega^5}{1-\omega}\\[3pt]
&=0
\end{align}
$$
we have
$$
\begin{align}
1+\omega+\omega^2+\omega^3-\frac1\omega
&=-\omega^4-\frac1\omega\\
&=-\frac{\omega^5+1}\omega\\
&=-\frac2\omega
\end{align}
$$
Therefore
$$
\begin{align}
\log_2\left|1+\omega+\omega^2+\omega^3-\frac1\omega\right|
&=\log_2\left|-\frac2\omega\right|\\
&=\log_2(2)\\[6pt]
&=1
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1619460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \ \ \ d.)\ 7 $
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta \\
=\sin^{2} \theta +\dfrac{1}{\sin^{2} \theta }+\cos^{2} \theta+\dfrac{1}{\cos^{2} \theta }+\tan^{2} \theta+\dfrac{1}{\tan^{2} \theta } \\
\color{blue}{\text{By using the AM-GM inequlity}} \\
\color{blue}{x+\dfrac{1}{x} \geq 2} \\
=2+2+2=6 $
Which is not in options.
But I am not sure if I can use that $ AM-GM$ inequality in this case.
I look for a short and simple way .
I have studied maths upnto $12$th grade .
|
Using standard trigonometric identities, we see this is $1+2\tan^2\theta+1+2\cot^2\theta+1$.
Now we can use AM/GM to show that $2\tan^2\theta+2\cot^2\theta\ge 4$, and the value $4$ is attained at $\pi/4$.
Remark: Your AM/GM argument is enough to identify the right answer of this multiple choice question. For as you saw the minimum is $\ge 6$, and there is only one choice which is $\ge 6$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1620239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Compute $\lim_{n \to +\infty} n^{-\frac12 \left(1+\frac{1}{n}\right)} \left(1^1 \cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\frac{1}{n^2}}$
How to compute
$$\displaystyle \lim_{n \to +\infty} n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} \left(1^1\cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\dfrac{1}{n^2}}$$
I'm interested in more ways of computing limit for this expression
My proof:
Let $u_n$be that sequence we've:
\begin{eqnarray*}
\ln u_n &=& -\frac{n+1}{2n}\ln n + \frac{1}{n^2}\sum_{k=1}^n k\ln k\\
&=& -\frac{n+1}{2n}\ln n + \frac{1}{n^2}\sum_{k=1}^n k\ln \frac{k}{n}+\frac{1}{n^2}\sum_{k=1}^n k\ln n\\
&=& \frac{1}{n^2}\sum_{k=1}^n k\ln \frac{k}{n}\\
&=& \frac{1}{n}\sum_{k=1}^n \frac{k}{n}\ln \frac{k}{n}\\
&\to&\int_0^1 x\ln x\,dx = -1/4
\end{eqnarray*}
Therefore the limit is $e^{-\frac{1}{4}}$
|
We have $$\exp \left (- \frac {1} {2} \left (1 + \frac {1} {n}\right) \log n + \frac {1} {n^2} (1 \log 1 + 2 \log 2 + \cdots + n \log n)\right) = \exp \left (- \frac {1} {2} \left (1 + \frac {1} {n}\right) \log n + \frac {1} {2} \log n - \frac {1} {4} + \frac {\log n} {2n} + o \left (\frac {1} {n}\right) \right) = \exp \left (-\frac {1} {4} + o \left (\frac {1} {n}\right) \right),$$ using the fact that $$1 \log 1 + 2 \log 2 + \cdots + n \log n = \frac {1} {2} n^2 \log n - \frac {1} {4} n^2 + \frac {1} {2} n \log n + o (n).$$ By letting $n \to \infty$, we have $$\exp \left (-\frac {1} {4} + o \left (\frac {1} {n}\right) \right) \to \exp (-\frac {1} {4}).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1624690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Complex Numbers - Quartic Find two distinct real roots of the equation $z^4-3z^3+5z^2-z-10$, and hence solve this equation completely.
The problem is how do you find the two distinct real roots?
|
$$z^4-3z^3+5z^2-z-10=0\Longleftrightarrow$$
$$(z-2)(z+1)((z-2)z+5)=0$$
So, we've got three equations:
*
*$$z-2=0\Longleftrightarrow z=2$$
*$$z+1=0\Longleftrightarrow z=-1$$
*$$(z-2)z+5=0\Longleftrightarrow$$
$$z^2-2z+5=0\Longleftrightarrow$$
$$z=\frac{-\left(-2\right)\pm\sqrt{\left(-2\right)^2-4\cdot(1)\cdot(5)}}{2\cdot1}\Longleftrightarrow$$
$$z=\frac{2\pm\sqrt{4-20}}{2}\Longleftrightarrow$$
$$z=\frac{2\pm\sqrt{-16}}{2}\Longleftrightarrow$$
$$z=\frac{2\pm i\sqrt{16}}{2}\Longleftrightarrow$$
$$z=\frac{2\pm 4i}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1624862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the maximum value of $(12\sin x-9\sin^{2} x)$
The maximum value of $(12\sin x-9\sin^{2} x)$
is equal to
$a.)\ 3 \\
\color{green}{b.)\ 4} \\
c.)\ 5 \\
d.)\ \text{none of these}$
As
$-1\leq \sin x\leq 1 ,\\
12\sin x-9\sin^{2} x \\
=12-9=3 \\
$
But the answer given is $4.$
I am looking for a short and simple way.
I have studied maths up to $12$th grade.
Note : I can't use calculus.
|
As you noted $\sin(x)$ takes on all values from $-1 \dots 1$.
So consider $f(y) = 12y - 9y^2$, where $y$ is in the range $-1 \dots 1$.
What is the maximum of $f$? The $y^2$ coefficient is negative, so the maximum occurs at the vertex of the parabola.
The vertex of a parabola $f(x) = ax^2 + bx + c$ occurs at $x = \frac{-b}{2a}$, so the vertex of $12y - 9y^2$ is at $y=\frac{-12}{2 \times -9} = \frac{2}{3}$.
Note that $y=\frac{2}{3}$ is in the range of $-1 \dots 1$, so there is some $x$ such that $y = \sin(x) = \frac{2}{3}$.
So the maximum is $f\left(\frac 23\right) = 12\times\frac 23 - 9\times\left(\frac 23\right)^2 = 8 - 4 = 4$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1625692",
"timestamp": "2023-03-29T00:00:00",
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|
Bounding the remainder I need to find the 3rd order Taylor polynomials and bound the remainder term at $(0,0)$.
The function is $$f(x,y)=\cos(x)\sin(y)$$
Here is what I did: first, I found the taylor expansions of sin and cos functions and multiply them, and then I get the following remainder term
$$R_3(x)-\frac{x^2}{2!}R_3(y)+yR_3(x)-\frac{y^3}{3!}R_3(x)+R_3(y)R_3(x)$$
I need to be using the inequality $R_{n,a}(x)\le \frac{M}{(n+1)!}|x-a|^{(n+1)}$
How do I proceed?
This is a pretty standard problem so I would be thankful if someone could give a methodological answer.
Thanks in advance!
|
Lets refer to a set of notes for Multivariable Taylor and MV Taylor.
Lets do it the long way first.
*
*$(a, b) = (0, 0)$
*$f(x, y) = \cos x \sin y$
*$(x + y)^1 = x + y$, so we need partials $f_x, f_y$
*$f_{x}(x, y) = -\sin x \cos y \rightarrow f_x(0, 0) = 0$
*$f_{y}(x, y) = -\cos x \cos y \rightarrow f_y(0, 0) = 1$
*$(x + y)^2 = x^2 + 2 xy + y^2$, so we need partials $f_{xx}, f_{xy}, f_{yy}$
*$f_{xx}(x, y) = -\cos x \sin y \rightarrow f_{xx}(0, 0) = 0$
*$f_{xy}(x, y) = -\sin x \cos y \rightarrow f_{xy}(0, 0) = 0$
*$f_{yy}(x, y) = -\cos x \sin y \rightarrow f_{yy}(0, 0) = 0$
*Notice how all the second order terms were zero? This will be a pattern.
*$(x + y)^3 = x^3 + 3 x^2 y + 3 x y^2 + y^3$, so we need partials $f_{xxx}, f_{xxy}, f_{xyy}, f_{yyy}$
*$f_{xxx}(x, y) = \sin x \sin y \rightarrow f_{xxx}(0, 0) = 0$
*$f_{xxy}(x, y) = -\cos x \cos y \rightarrow f_{xxy}(0, 0) = -1$
*$f_{xyy}(x, y) = \sin x \sin y \rightarrow f_{xyy}(0, 0) = 0$
*$f_{yyy}(x, y) = -\cos x \cos y \rightarrow f_{yyy}(0, 0) = -1$
*$(x + y)^4 = x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + y^4$, so we need partials $f_{xxxx}, f_{xxxy}, f_{xxyy}, f_{xyyy}, f_{yyyy}$
*However, we already detected a pattern that these will all be zero. This means that this error term is not enough and we need one more.
*$(x + y)^5 = x^5 + 5 x^4 y + 10x^3y^2+10x^2y^3+5 xy^4 + y^5$
We are now ready to calculate the linear approximation and the remainder term.
We have:
$$L(x, y) = L_1 (x, y) + \dfrac{1}{2!} L_2(x, y) + \dfrac{1}{3!} L_3(x, y)$$
*
*Using $(x + y)^1$
*$~~L_1(x, y) = f_x(a, b) (x - a) + f_y(a, b) (y - b) = 0 + y$
*Using $(x + y)^2$
*$~~L_2(x, y) = f_{xx}(a, b)(x-a)^2 + 2 f_{xy}(a, b)(x-a)(y-b) + f_{yy}(a,b)(y-b)^2 = 0$
*Using $(x + y)^3$
*$~~L_3(x, y) = f_{xxx}(a, b)(x-a)^3 + 3 f_{xxy}(a, b)(x-a)^2(y-b) +3 f_{xyy}(a,b)(x-a)(y-b)^2 + f_{yyy}(a, b)(y-b)^3 = 0 + 3(-1)x^2 y + 0 + (-1)y^3 = -3x^2y - y^3$
$$L(x, y) = y + \dfrac{1}{3!}(-3x^2 y - y^3) = y - \dfrac{x^2 y}{2} - \dfrac{y^3}{6}$$
*
*We already know $L_4 = 0$, so need one more term for the remainder, from $(x + y)^5$ we have:
*$R_5(x, y) = f_{xxxxx}(a, b)(x-a)^5 + 5f_{xxxxy}(a, b)(x-a)^4(y-b) + 10 f_{xxxyy}(a, b)(x-a)^3(y-b)^2 + 10 f_{xxyyy}(a, b)(x-a)^2(y-b)^3+5_{xyyyy}(a, b)(x-a)(y-b)^4+f_{yyyyy}(a, b)(y-b)^5$
As for bounding that error, we have that the max of the products for each of the partials of cosine and sine terms is $M = 1$, thus we can write:
$$|R_5(x, y)| \le \dfrac{M}{5!}(|x^5| + 5|x^4y| + 10|x^3 y^2| + 10 |x^2y^3|+ 5|x y^4| + |y^5|) = \dfrac{1}{120}(|x|+|y|)^5$$
Now, if they provided a region for $(x, y)$, we can easily calculate the max error.
There is a much easier approach to arrive at this solution.
The third order McLaurin series expansion for $\cos x = 1-\frac{x^2}{2}$ and for $\sin y = y-\frac{y^3}{6}$. This gives (see notes for dropped term):
$$L(x, y) = \cos x \sin y= \left(1-\frac{x^2}{2}\right) \left(y-\frac{y^3}{6}\right) = y -\frac{x^2 y}{2}-\frac{y^3}{6}$$
Compare that to the result above.
For the error term, we have the expansion for $(x + y)^5$. (When using this, be careful to know the max of the function because you need that value (see the sets of notes I linked above for examples) as I showed above for this particular example.
|
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|
Proving Trig Identities (Complex Numbers)
Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n + {1\over z^n} = 2\cos(n\theta) $$
Hence prove that $\cos^6(\theta)$ $=$ $\frac 1{32}$$(\cos(6\theta)$ + $6\cos(4\theta)$ + $15\cos(2\theta)$ + $10$)
I learnt to prove the first part in another post linked here.
The second part is where I am confused because there is a 'hence'
so I thought of taking 2 approaches:
either $$ z^6 + \frac 1{z^6} $$
or $$ z^6 $$ and equating real parts.
I will start with my first approach
$$ z^6 + \frac 1{z^6} $$
$$ (\cos(x)+i\sin(x))^6 + \frac 1{(\cos(x)+i\sin(x))^6} $$
$$ (\cos(x)+i\sin(x))^6 + {(\cos(x)+i\sin(x))^{-6}} $$
$$ \cos(6x) + i\sin(6x) + \cos(-6x)+ i \sin(-6x) $$
$$ 2\cos(6x) $$
Which is no where near what I am suppose to prove..
So with my second approach (expanding and equation real parts)
$$ z^6 $$
$$ (\cos(x) + i \sin(x))^6 $$
Using pascals
$$ \cos^6(x) + i*6\cos^5(x)\sin(x) + i^2*15\cos^4(x)\sin^2(x) + i^3*20\cos^3(x)\sin(x) + i^4*15\cos^2(x)\sin^4(x) + i^5*6\cos(x)\sin^5(x) + i^6 * \sin^6(x) $$
Simplifying
$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) +i(6\cos^5(x)\sin(x)-20\cos^3(x)\sin(x) + 6cos(x)\sin^5(x)$$
Now considering only real
$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) $$
At this point I'm confused , am I on the right approach?
|
This is simply because for such a $z$, $\;\frac1{z^n}=\bar z^n=\overline{z^n}$, and
$$z^n+\overline{z^n}=2\operatorname{Re}(z^n)=2\cos n\theta\qquad\quad\text{by De Moivre's formula.}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1627410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Prove that one of the elements can't be in the interval $(0,1)$ Let be $a,b,c\in\mathbb{R}$ so that the sum of two of them is never equal to $1$. Prove that atleast on of $\frac{ab}{a+b-1},\frac{bc}{b+c-1},\frac{ca}{c+a-1}$ can't be in the interval $(0,1)$.
I aproached it with contradiction, but can't get sth that is not true.
|
Let $0<\frac{ab}{a+b-1}<1$, $0<\frac{ac}{a+c-1}<1$ and $0<\frac{bc}{b+c-1}<1$.
Hence, $0<\frac{a^2b^2c^2}{\prod\limits_{cyc}(a+b-1)}<1$.
In another hand, $0<1-\frac{ab}{a+b-1}<1$, $0<1-\frac{ac}{a+c-1}<1$ and $0<1-\frac{bc}{b+c-1}<1$, which gives
$0<\frac{(1-a)(b-1)}{a+b-1}<1$, $0<\frac{(1-c)(a-1)}{a+c-1}<1$ and $0<\frac{(1-b)(c-1)}{b+c-1}<1$, which gives
$0<\frac{-\prod\limits_{cyc}(a-1)^2}{\prod\limits_{cyc}(a+b-1)}<1$, which is a contradiction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1627539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Summation up to $n$ terms : $\sum r\cdot (r+1)^2$ Summation up to $n$ terms : $$\sum_{r=1}^{n} r\cdot (r+1)^2$$
My attempt :
$$\sum_{r=1}^{n} r\cdot (r+1)^2=\sum_{r=1}^{n} r^3+2\sum_{r=1}^{n} r^2+\sum_{r=1}^{n} r$$
$$\sum_{r=1}^{n} r\cdot (r+1)^2=\sum_{r=1}^{n} r^3+2\sum_{r=1}^{n} r^2+\sum_{r=1}^{n} r$$
But for this method I should know the answers of $\sum r^3$ , $\sum r^2$ and $\sum r$.
Is there another method to find the sum without using $\sum r^3$ , $\sum r^2$ and $\sum r$ ?
|
Let $f(r)=a+br+cr^2+dr^3+er^4$.
Then $$f(r+1)-f(r)$$
$$=a+b(r+1)+c(r^2+2r+1)+d(r^3+3r^2+3r+1)+e(r^4+4r^3+6e^2+4e+1)-(a+br+cr^2+dr^3+er^4)$$
$$=b+2cr+c+3dr^2+3dr+d+4er^3+6er^2+4er+e$$
$$=(b+c+d+e)+(2c+3b+4e)r+(3d+6e)r^2+4er^3$$
We need this to equal $r(r^2+2r+1)=r+2r^2+r^3$.
So $4e=1$ etc...
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\sum_{n=1}^{\infty }\int_{0}^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^{2}}\,\mathrm{d}x$ I have some trouble in evaluating this series
$$\sum_{n=1}^{\infty }\int_{0}^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^{2}}\mathrm{d}x$$
I tried to calculate the integral first, but after that I found the series become so complicated.
Besides, I found maybe the series equals to
$$\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}\zeta \left ( 2k+\dfrac{3}{2} \right )}{2k+\dfrac{3}{2}}$$
this is definitely a monster to me. So I want to know is there a good way to solve this integral-series.
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Let
$$I(n) = \int_{0}^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^{2}}\mathrm{d}x.$$
Substitution:
$$x = t^2,\quad \mathrm{d}x = 2t\mathrm{d}t:$$
$$I(n) = \int\limits_{0}^{\frac{1}{\sqrt{n}}}\frac{2t^2}{1+t^{4}}\mathrm{d}t.$$
$$I(n) = \int\limits_{0}^{\frac{1}{\sqrt{n}}}\frac{2}{t^2 + \dfrac{1}{t^2}}\mathrm{d}t.$$
$$I(n) = \int\limits_{0}^{\frac{1}{\sqrt{n}}}\frac{1-\dfrac1{t^2}}{\left(t + \dfrac{1}{t}\right)^2-2}\mathrm{d}t + \int\limits_{0}^{\frac{1}{\sqrt{n}}}\frac{1+\dfrac1{t^2}}{\left(t - \dfrac{1}{t}\right)^2+2}\mathrm{d}t.$$
$$I(n) = \dfrac{1}{2\sqrt{2}}\left.\ln\left|\frac{t + \dfrac{1}{t}-\sqrt 2}{t + \dfrac{1}{t}+\sqrt 2}\right|\right|_0^{\frac1{\sqrt n}} + \dfrac1{2\sqrt2}\left.\arctan\dfrac{ {t - \dfrac{1}{t}}}{\sqrt2}\right|_0^{\frac1{\sqrt n}}.$$
$$I(n) = \dfrac{1}{2\sqrt{2}}\ln\frac{n - \sqrt {2n} + 1}{n + \sqrt {2n} + 1} + \dfrac1{2\sqrt{2}}\left(\dfrac{\pi}{2}-\arctan\dfrac{n - 1}{\sqrt{2n}}\right), \quad I(1) = \dfrac{\pi}{4\sqrt2}.$$
or
$$I(n) = \dfrac{1}{2\sqrt{2}}\ln\frac{1 -\dfrac{\sqrt{2n}}{n + 1}}{1 +\dfrac{\sqrt{2n}}{n + 1}} + \dfrac1{2\sqrt{2}}\arctan\dfrac{\sqrt{2n}}{n - 1}, \quad I(1) = \dfrac{\pi}{4\sqrt2}.$$
UPD
Maclaurin series is as follows:
$$
I(n) = \dfrac{\sqrt n}{n+1}\left(1+\dfrac13\dfrac{2n}{(n+1)^2}+\dfrac15\dfrac{(2n)^2}{(n+1)^4}+\dots++\dfrac1{2k+1}\dfrac{(2n)^k}{(n+1)^{2k}}+\dots\right)$$$$+\dfrac12\,\dfrac{\sqrt n}{n-1}\left(1-\dfrac13\dfrac{2n}{(n-1)^2}+\dfrac15\dfrac{(2n)^2}{(n-1)^4}-\dots+\dfrac1{2k+1}\dfrac{(-2n)^k}{(n-1)^{2k}}+\dots\right),
$$
$$I(1) = \dfrac{\pi}{4\sqrt2}.$$
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2} \, \mathrm{d}\theta$ I have some trouble in how to evaluate this integral:
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right)
\,\mathrm{d}\theta
$$
I think it maybe has another form
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right)
\,\mathrm{d}\theta
=
\sum_{n=1}^{\infty}{1 \over n^{2}}
\left[\psi\left(n + {1 \over 2}\right) - \psi\left(1 \over 2\right)\right]
$$
|
Obviously we have
$$\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2}\mathrm{d}\theta =4\int_{0}^{\pi /2}x\ln \tan x\mathrm{d}x$$
then use the definition of Lobachevskiy Function(You can see this in table of integrals,series,and products,Eighth Edition by Ryzhik,page 900)
$$\mathrm{L}\left ( x \right )=-\int_{0}^{x}\ln\cos x\mathrm{d}x,~ ~ ~ ~ ~ -\frac{\pi }{2}\leq x\leq \frac{\pi }{2}$$
Hence we have
\begin{align*}
\int_{0}^{\pi /2}x\ln\tan x\mathrm{d}x &= x\left [ \mathrm{L}\left ( x \right )+\mathrm{L}\left ( \frac{\pi }{2}-x \right ) \right ]_{0}^{\pi /2}-\int_{0}^{\pi /2}\left [ \mathrm{L}\left ( x \right )+\mathrm{L}\left ( \frac{\pi }{2}-x \right ) \right ]\mathrm{d}x\\
&= \left ( \frac{\pi }{2} \right )^{2}\ln 2-2\int_{0}^{\pi /2}\mathrm{L}\left ( x \right )\mathrm{d}x
\end{align*}
use
$$\mathrm{L}\left ( x \right )=x\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k-1}}{k^{2}}\sin 2kx$$
(Integrate the fourier series of $\ln\cos x$ from $0$ to $x$.)
we can calculate
\begin{align*}
\int_{0}^{\pi /2}\mathrm{L}\left ( x \right )\mathrm{d}x&=\frac{1}{2}\left ( \frac{\pi }{2} \right )^{2}\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k-1}}{k^{2}}\int_{0}^{\pi /2}\sin 2kx\mathrm{d}x \\
&= \frac{\pi ^{2}}{8}\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{1}{\left ( 2k-1 \right )^{3}}
\end{align*}
So
\begin{align*}
\int_{0}^{\pi/2}x\ln\tan x\mathrm{d}x &=\frac{\pi ^{2}}{4}\ln 2-2\left [ \frac{\pi ^{2}}{8}\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{1}{\left ( 2k-1 \right )^{3}} \right ] \\
&=\sum_{k=1}^{\infty }\frac{1}{\left ( 2k-1 \right )^{3}}\\
&=\sum_{k=1}^{\infty } \frac{1}{k^{3}}-\sum_{k=1}^{\infty }\frac{1}{\left ( 2k \right )^{3}}=\frac{7}{8}\zeta \left ( 3 \right )
\end{align*}
Hence the initial integral is
$$\boxed{\Large\color{blue}{\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2}\mathrm{d}\theta=\frac{7}{2}\zeta \left ( 3 \right )}}$$
in addition,as you mentioned
$$\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2}\mathrm{d}\theta=\color{red}{\sum_{n=1}^{\infty }\frac{1}{n^{2}}\left [ \psi \left ( n+\frac{1}{2} \right )-\psi \left ( \frac{1}{2} \right ) \right ]=\frac{7}{2}\zeta \left ( 3 \right )}$$
or
$$\sum_{n=1}^{\infty }\frac{1}{n^{2}}\psi \left ( n+\frac{1}{2} \right )=\frac{7}{2}\zeta \left ( 3 \right )-\left ( \gamma +2\ln 2 \right )\frac{\pi ^{2}}{6}$$
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"timestamp": "2023-03-29T00:00:00",
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|
Choosing a substitution to evaluate $\int \frac{x+3}{\sqrt{x+2}}dx$ Is there any other value you can assign to the substitution variable to solve this integral?
$$\int \frac{x+3}{\sqrt{x+2}}dx$$
Substituting $u = x + 2$:
$$du = dx; u +1 = x+3 ,$$
and we get this new integral that we can then split into two different ones:
$$\int \frac{u + 1}{\sqrt{u}}du = \int \frac{u}{\sqrt{u}}du + \int \frac{1}{\sqrt{u}}du .$$
We can substitute again $s = \sqrt u$ and get two immediate integrals:
$$s = \sqrt{u}; \quad ds = \frac{1}{2\sqrt{u}}du; \quad 2s^2 =u .$$ Substituting back $u$ to $s$ and $x$ to $u$ we get this result,
$$s^2 + \ln{\left | \sqrt{u} \right |} = u + \ln{\left | \sqrt{u} \right |} = x+2+\ln{\left | \sqrt{x+2} \right |},$$ which doesn't look quite to be right. What am I doing wrong? I'm pretty unsure about the second substitution, $2s^2 = u$. Is it correct?
|
let's make it easier than that!
Use this:
$$x + 2 = t^2 ~~~~~~~~~~~ x+3 = t^2 + 1 ~~~~~~~ \text{d}x = 2t\ \text{d}t$$
Obtaining
$$I = \int\frac{t^2 + 1}{t}\ 2t\ \text{d}t = 2\int t^2 + 1\ \text{d}t = \frac{2}{3}t^3 + 2t$$
Coming back to $x$, having $t = \sqrt{x+2}$ and you'll have
$$I = \frac{2}{3}\sqrt{x+2}(x+5)$$
|
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"url": "https://math.stackexchange.com/questions/1634375",
"timestamp": "2023-03-29T00:00:00",
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|
Bounded sequence $a_n=\sqrt{4+2 \sqrt{4+\cdots+2 \sqrt{4+2 \sqrt{4+2 \sqrt{4+4}}}}}$ Let
$$a_n=\sqrt{4+2 \sqrt{4+\cdots+2 \sqrt{4+2 \sqrt{4+2 \sqrt{4+4}}}}}$$
the sign $\sqrt{}$ occurs $n$ times.
a) Prove, that $a_n< \sqrt{5}+1$ for all $n$.
b) Find $\lim_{n\rightarrow \infty } a_n$
Author O.Kukush
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You can prove (a) by mathematical induction. $a_1$ is less than $1+\sqrt5$ by direct comparison as $a_1 = \sqrt8 < 3$ while $1+\sqrt5>3$. Let $a_n < 1+\sqrt5$. Then $4+2a_n < 6+2\sqrt5$ and taking square root on both sides (that both are positive is trivial) we obtain $a_{n+1} < 1+\sqrt5$.
It also follows that $(a_n)$ is strictly growing. You can check that by comparing $a_{n+1}^2 = 4+2a_n$ with $a_n^2$ knowing that (a) is true.
From the monotonicity and boundedness it follows that $(a_n)$ has a limit and this number, let's call it $a$, needs to be a fixed point of $\sqrt{4+2a} = a$. The only solution larger than $0$ is $1+\sqrt5$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $p$ is a prime other than 2 or 5, prove that $p$ must be one of the forms $10k + 1$, $10k + 3$, $10k + 7$, or $10k + 9$ If $p$ is a prime other than 2 or 5, prove that $p$ must be one of the forms $10k + 1$, $10k + 3$, $10k + 7$, or $10k + 9$
-The section we are covering is on the division algorithm, although I am unsure as to how to utilize it in this problem. Any help is appreciated.
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So you wish to express $10k + c$ as $bq + r$ with $0 \leq r < b$. If $r = 0$ and $10k + c \neq b$, that means $10k + c$ and $b$ share a common prime factor, and if $b$ is the prime then $10k + c$ is composite.
Try it with $b = 2$, which you know is a prime number. Then:
*
*$10k + 0 = 2(5k) + 0$
*$10k + 2 = 2(5k + 1) + 0$
*$10k + 4 = 2(5k + 2) + 0$
*$10k + 6 = 2(5k + 3) + 0$
*$10k + 8 = 2(5k + 4) + 0$
Since $r = 0$ in each of these cases and we know 2 is prime, this must mean that these $10k + c$ are composite.
Very similarly you take care of $10k + 5$ with $b = 5$.
This leaves only $10k + 1$, $10k + 3$, $10k + 7$ and $10k + 9$ as potentially prime.
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"timestamp": "2023-03-29T00:00:00",
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|
This simple algebra question \begin{cases}
3x + y = \frac{12}{y^2}\\
3y + x = \frac{15}{x^2}
\end{cases}
$x+y = ?$
I tried adding side by side but I found only $4x+4y = \dfrac{15}{x^2} + \dfrac{12}{y^2}$ so I couldn't find anything. Can someone solve it?
|
*
*Multiply the first equation by $y^2$: $3xy^2+y^3 = 12$
*Multiply the second equation by $x^2$: $3x^2y+x^3=15$
*Add them up to get $3xy^2+y^3+3x^2y+x^3=(x+y)^3=27.$
Hence, $x+y=3$.
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"timestamp": "2023-03-29T00:00:00",
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|
What is the number of ordered triplets $(x, y, z)$ such that the LCM of $x, y$ and $z$ is ...
What is the number of ordered triplets $(x, y, z)$ such that the LCM of $x, y$ and $z$ is $2^33^3$ where $x, y,z\in \Bbb N$?
What I tried :
At least one of $x, y$ and $z$ should have factor $2^3$ and at least one should have factor $3^3$. I then tried to figure out the possible combinations but couldn't get the correct answer.
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We use Inclusion/Exclusion.
First we find the number of (positive) triples in which each entry divides $2^33^3$. At each of $x$, $y$, $z$ we have $(4)(4)$ choices, for a total of $16^3$.
We want to subtract the number of such triples in which each entry divides $2^23^3$. There are $12^3$ such triples. There are also $12^3$ such triples in which each element divides $2^33^2$.
But we have subtracted once too many times the $9^3$ triples in which each entry divides $2^23^2$.
So the total is $16^3-2\cdot 12^3+9^3$.
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.