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Find $g(x|y=\frac{1}{2})$, the conditional pdf of $X$ given $Y = \frac{1}{2}$ (Need confirmation) Let X and Y be continuous random variables having the joint pdf $$f(x,y) = 8xy , 0\leq{y}\leq{x}\leq{1}$$ I found that the marginal pdf of Y is $f_2(y) = 4y - 4y^3$. Does $g(x|y=\frac{1}{2}) = \frac{f(x,\frac{1}{2})}{f_2(\frac{1}{2})} = \frac{8x}{3}$ or $g(x|y=\frac{1}{2}) = \frac{f(x,y)}{f_2(\frac{1}{2})} = \frac{16xy}{3}$. A little confused here with the definition. Sorry for the tiny fractions, I don't know how to enlarge them.
|
$$f_Y(y) = \int_{y}^{1} 8xy \: \text{dx} = 4y - 4y^3$$
$$ f_Y(y) =
\begin{cases}
4y - 4y^3, & 0 \leq y \leq 1 \\
0, & \text{otherwise}
\end{cases}
$$
$$g(x \mid y=\frac{1}{2}) = \frac{f_{X,Y}(x,y=\frac{1}{2})}{f_Y(y=\frac{1}{2})} = \frac{8x}{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1771281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Find the Fourier transform of $\sin x^2$. I've tried it by applying integratrion by parts, but I'm not getting the answer correct. Its answer is
$$\frac{1}{\sqrt{2}}\,\sin\left(\frac{k^2}{4} +\frac{\pi}{4}\right).$$
Please help in this.
|
Let be
$$
\sin(ax^2)=\frac{\mathrm e^{iax^2}-\mathrm e^{iax^2}}{2i}\qquad\text{and}\qquad\cos(ax^2)=\frac{\mathrm e^{iax^2}+\mathrm e^{iax^2}}{2}
$$
anf use the Fourier transform defined as $$\mathcal F\left\{f(x)\right\}=F(k)=\int_{-\infty}^{\infty} f(x)\mathrm e^{-i2\pi k x}\,\mathrm d x$$
The Fourier transform can be found by using the Fourier Transform of the Gaussian. Observing that
$$\mathcal F\left\{\mathrm e^{-\beta x^2}\right\}=\sqrt{\frac{\pi}{\beta}}\mathrm e^{-\frac{pi^2 k^2}{\beta}}$$
for $\beta=ia$ for $a>0$
$$\mathcal F\left\{\mathrm e^{-i ax^2}\right\}=\sqrt{\frac{\pi}{ai}}\mathrm e^{-\frac{\pi^2 k^2}{ia}}=\sqrt{\frac{\pi}{a}}\mathrm e^{-i\frac{\pi}{4}}\mathrm e^{-\frac{\pi^2 k^2}{ia}}=\sqrt{\frac{\pi}{a}}\mathrm{exp}\left(i\left[\frac{\pi^2 k^2}{a}-\frac{\pi}{4}\right]\right)$$
for $a<0$, $\beta=-|a|i$
$$
\mathcal F\left\{\mathrm e^{-i ax^2}\right\}=
\sqrt{\frac{\pi}{-|a|i}}\mathrm e^{-\frac{\pi^2 k^2}{i|a|}}
=\sqrt{\frac{\pi}{|a|}}\mathrm e^{i\frac{\pi}{4}}\mathrm e^{-i\frac{\pi^2 k^2}{|a|}}
=\sqrt{\frac{\pi}{|a|}}\mathrm{exp}\left(-i\left[\frac{\pi^2 k^2}{|a|}-\frac{\pi}{4}\right]\right)
$$
that is
$$
\mathcal F\left\{\mathrm e^{-i ax^2}\right\}=\begin{cases}
\sqrt{\frac{\pi}{a}}\mathrm{exp}\left(i\left[\frac{\pi^2 k^2}{|a|}-\frac{\pi}{4}\right]\right) & \text{for }a>0\\
\sqrt{\frac{\pi}{|a|}}\mathrm{exp}\left(-i\left[\frac{\pi^2 k^2}{|a|}-\frac{\pi}{4}\right]\right) & \text{for }a<0
\end{cases}
$$
So we have, for positive $a$,
$$
\mathcal F\left\{\mathrm e^{-iax^2}\right\}=\sqrt{\frac{\pi}{a}}\mathrm{exp}\left({i\left[\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right]}\right)
$$
and
$$
\mathcal F\left\{\mathrm e^{iax^2}\right\}=\mathcal F\left\{\mathrm e^{-i(-a)x^2}\right\}=\sqrt{\tfrac{\pi}{|-a|}}\mathrm{exp}\left({-i\left[\tfrac{\pi^2k^2}{|a|}-\tfrac{\pi}{4}\right]}\right)=\sqrt{\frac{\pi}{a}}
\mathrm{exp}\left({-i\left[\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right]}\right)
$$
So we have, for positive $a$,
\begin{align}
\mathcal F\left\{\sin(ax^2)\right\}&=\frac{\mathcal F\left\{\mathrm e^{iax^2}\right\}-\mathcal F\left\{\mathrm e^{-iax^2}\right\}}{2i}\\
&=\frac{1}{2i}\left[\sqrt{\frac{\pi}{a}}\mathrm e^{-i\left(\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right)}-\sqrt{\frac{\pi}{a}}\mathrm e^{i\left(\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right)}\right]\\
&=-\sqrt{\frac{\pi}{a}}\sin\left(\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right)
\end{align}
and observing that $\sin(-t)=-\sin(t)$, we have
$$
\mathcal F\left\{\sin(ax^2)\right\}=\begin{cases}
-\sqrt{\frac{\pi}{a}}\sin\left(\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right) & \text{for }a>0\\
\sqrt{\frac{\pi}{|a|}}\sin\left(\frac{\pi^2k^2}{|a|}-\frac{\pi}{4}\right) & \text{for }a<0
\end{cases}
$$
In the same way we have
$$
\mathcal F\left\{\cos(ax^2)\right\}=\begin{cases}
\sqrt{\frac{\pi}{a}}\cos\left(\frac{\pi^2k^2}{a}-\frac{\pi}{4}\right) & \text{for }a>0\\
\sqrt{\frac{\pi}{|a|}}\cos\left(\frac{\pi^2k^2}{|a|}-\frac{\pi}{4}\right) & \text{for }a<0
\end{cases}
$$
Thus
\begin{align}
\mathcal F\left\{\sin(x^2)\right\}=-\sqrt{\pi}\sin\left(\pi^2k^2-\frac{\pi}{4}\right)
\end{align}
Using the Fourier transform defined as $$\mathcal F\left\{f(x)\right\}=\hat f(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)\mathrm e^{-i\xi x}\,\mathrm d x=\frac{1}{\sqrt{2\pi}}F\left(\frac{\xi}{2\pi}\right)$$
we find
$$
\mathcal F\left\{\sin(x^2)\right\}=-\frac{1}{\sqrt{2}}\sin\left(\frac{\xi^2}{4}-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\cos\left(\frac{\xi^2}{4}+\frac{\pi}{4}\right)
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
finding the number of factors $2^{15}\times3^{10}\times5^6$ The number of factors of $2^{15}\times3^{10}\times5^6$ which are either perfect square or perfect cubes(or both)
I don't know how to start this even!
Plz solve this!
|
Case 1: factors that are perfect squares
\begin{align}
A & =\{2^0,2^2,2^4,2^6,2^8,2^{10},2^{12},2^{14}\} \\
B & =\{3^0,3^2,3^4,3^6,3^8,3^{10}\} \\
C & =\{5^0,5^2,6^4,5^6\}
\end{align}
Choose 1 from each set. We have $8\times6\times4=192$ choices.
Case 2: factors that are perfect cubes
\begin{align}
A& =\{2^0,2^3,2^6,2^9,2^{12},2^{15}\} \\
B & =\{3^0,3^3,3^6,3^9\} \\
C & =\{5^0,5^3,5^6\}
\end{align}
Choose 1 from each set. We have $6\times4\times3=72$ choices.
Therefore, we have $72+192-3\times2\times2=252$ factors in total.
Note: subtract $3\times2\times2$ from the total because some sixth powers have been counted twice.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove
$$x^2y+y^2z+z^2x < \frac12$$
This inequality has been verified by Mathematica. $\frac12$ is not the best bound. I try to do AM-GM for this one but not yet success. The condition $x+y^2+z^3$ is very weird.
|
This answer is incomplete.
Let $A=x+y^2$, $B=y^2+z^3$ and $C=z^3+x$.
Claim: $x^2y \leq \dfrac{A}{2}x^{3/2}$,
$y^2z \leq \dfrac{B^{2/3}}{4^{2/3}}y^{4/3}$, $z^2x \leq \dfrac{C^{4/3}}{4^{2/3}}x^{1/3}$.
Assuming the claim, we note that $A+B+C=2$ and we need to prove that the given sum is less than $\dfrac{A+B+C}{4}$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1775498",
"timestamp": "2023-03-29T00:00:00",
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|
Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
$x,y,z >0$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$
Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame.
Update 1: In this forum, somebody said that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as "coffin problems" in my country. The official solution is very elementary and elegant.
Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with "pencil and papers." I think the approach by Peter Scholze in here may help.
Update 3: Michael has tried to apply Peter Scholze's method but not found the solution yet.
Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the "King of Inequality".
|
Not sure, if I missed out anything here. Take a look.
For non negative, $X,Y,Z$,
We can perhaps use Titu's inequality (a mix of Holder and CS), sometimes called Titu's screw lemma (https://en.wikipedia.org/wiki/Nesbitt%27s_inequality).
\begin{equation}
\sum_{k=1}^{n}{\frac{x_{k}^{2}}{a_{k}}} \ge \frac{\left(\sum_{k=1}^{n}{x_{k}}\right)^{2}}{\sum_{k=1}^{n}{a_{k}}}
\end{equation}
With $n\to3$ terms, $x_{1}\to X^{2},x_{2} \to Y^{2}, x_{3} \to Z^{2}$ and $a_{1} \to A, a_{2}\to B, a_{3} \to C$, we will have
\begin{eqnarray*}
\frac{\left(X^2\right)^{2}}{A}+\frac{\left(Y^2\right)^{2}}{B}+\frac{\left(Z^2\right)^{2}}{C} &\ge& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{A+B+C} \\
\end{eqnarray*}
With
\begin{eqnarray*}
A &=& \alpha X^{3} +\beta Y^{3} \\
B &=& \alpha Y^{3} +\beta Z^{3} \\
C &=& \alpha Z^{3} +\beta X^{3}
\end{eqnarray*}
where,
\begin{eqnarray*}
A+B+C &=& (\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right)
\end{eqnarray*}
\begin{eqnarray}
\frac{X^4}{A}+\frac{Y^4}{B}+\frac{Z^4}{C} &=&\frac{\left(X^2\right)^{2}}{A}+\frac{\left(Y^2\right)^{2}}{B}+\frac{\left(Z^2\right)^{2}}{C}\\
&\ge& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{A+B+C} \\
&=& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{(\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right)} \\
&\overset{(p)}{\ge}& \frac{\left(X^{3}+Y^{3}+Z^{3}\right)\left(X+Y+Z\right)}{(\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right)} \\
&=& \frac{\left(X+Y+Z\right)}{(\alpha+\beta)}
\end{eqnarray}
QED.
Here $(p)$ is from the fact that,
\begin{eqnarray*}
(X^2+Y^2+Z^2)^{2} -\left(X^{3}+Y^{3}+Z^{3} \right) (X+Y+Z) &=& XY(X-Y)^{2}+YZ(Y-Z)^{2}+ZX(Z-X)^{2} \\
&\ge& 0
\end{eqnarray*}
Here $\alpha=8$ and $\beta=5$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove $x^{\frac85}+y^{\frac85}+z^{\frac85} \geqslant 3$ $x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove
$$x^{\frac85}+y^{\frac85}+z^{\frac85} \geqslant 3$$
1) The equality occurs only at $x,y,z=1$. Let's assume $F=x^n+y^n+z^n$, I noticed that $n=1$ then $F \leqslant 3$ and $n=2$ then $F \geqslant 3$. I believe $n=\frac85=1.6$ is a very sharp inequality.
2) I try trig substitution but cannot remove the radical . $n=\frac85$ is too much for algebraic manipulation.
|
Let $x=2\cos\alpha$ and $y=2\cos\beta$, where $\{\alpha,\beta\}\subset\left[0,\frac{\pi}{2}\right]$.
Hence, $z=2\cos\gamma$, where $\alpha+\beta+\gamma=\pi$ and $\gamma\in\left[0,\frac{\pi}{2}\right]$.
Let $f(x)=\left(\cos x\right)^{\frac{8}{5}}$.
Since $f''(x)=\frac{-8(1+4\cos2x)}{25\sqrt[5]{\cos^2x}}\geq0$ for all $x\in\left[\frac{\pi}{3},\frac{\pi}{2}\right)$, by Vasc's RCF Theorem
it remains to prove our inequality for $\beta=\alpha$ and $\gamma=\pi-2\alpha$ or
for $y=x$ and $z=2-x^2$, where $0\leq x\leq\sqrt2$.
Id est, it remains to prove that $2x^{\frac{8}{5}}+(2-x^2)^{\frac{8}{5}}\geq3$, which is obvious.
|
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|
Prove that $\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$
Prove that $$\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2).$$
I was thinking of using mathematical induction for this. That is,
We prove by induction on $n$. The case $n=1$ holds trivially since $2 \leq 2$. Now assume the result holds for some $m$. Then by assumption we know that $$\displaystyle \sum_{k=1}^{m+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+m}\bigg ) \leq \ln(2m) + 2 -\ln(2)+\dfrac{1}{m+1}+\dfrac{2}{2m+1}. $$
We must relate this somehow to $\ln(2(m+1)) + 2 -\ln(2)$.
|
First note, that we can rewrite the RHS of your inequality as $\ln(n) +2$.
We proceed by induction. We compute
$$ \sum_{k=1}^{n+1} \left( \frac{1}{k} + \frac{2}{k+n+1}\right)
= \frac{1}{n+1} + \frac{2}{2n+2} + \sum_{k=1}^n \frac{1}{k} + \sum_{k=1}^{n} \frac{2}{k+n+1}$$
shifting the index yields
$$ =\frac{2}{n+1} + \sum_{k=1}^n \frac{1}{k} + \sum_{k=2}^{n+1} \frac{2}{k+n}
= \sum_{k=1}^n \frac{1}{k} + \sum_{k=1}^{n+1} \frac{2}{k+n}
= \sum_{k=1}^n\left( \frac{1}{k} + \frac{2}{k+n}\right) + \frac{2}{2n+1}$$
applying the induction hypothesis gives
$$ \leq \ln(n) + 2 + \frac{2}{2n+1} = \ln(n+1) +2 + \left( \frac{2}{2n+1} + \ln(n) - \ln(n+1) \right).$$
Hence, we are left to show that
$$ \frac{2}{2n+1} - \ln\left(1 + \frac{1}{n}\right) = \frac{2}{2n+1} + \ln(n) - \ln(n+1) \leq 0. $$
This is already done in a previous answer.
|
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|
How to remember sum to product and product to sum trigonometric formulas? They are:
\begin{align}
\cos(a)\cos(b)&=\frac{1}{2}\Big(\cos(a+b)+\cos(a-b)\Big) \\[2ex]
\sin(a)\sin(b)&=\frac{1}{2}\Big(\cos(a-b)-\cos(a+b)\Big) \\[2ex]
\sin(a)\cos(b)&=\frac{1}{2}\Big(\sin(a+b)+\sin(a-b)\Big) \\[2ex]
\cos(a)\sin(b)&=\frac{1}{2}\Big(\sin(a+b)-\sin(a-b)\Big) \\[2ex]
\cos(a)+\cos(b)&=2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right) \\[3ex]
\cos(a)-\cos(b)&=-2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right) \\[3ex]
\sin(a)+\sin(b)&=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right) \\[3ex]
\sin(a)-\sin(b)&=2\cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)
\end{align}
I have found nice mnemonics that helped me to remember the reduction formulae and others but I can't find a simple relationship between the formulas above. Can you help?
|
How about just restating the LHS. For example, you could restate $\cos a\sin b$ as
$$\frac{\sin a\cos b +\cos a\sin b + \cos a\sin b - \sin a\cos b}{2}$$ and just figure it out from there. For Example, Let's start off with $\cos a\sin b$ and try to derive $\frac{1}{2}[\sin(a+b) - \sin(a-b)]$
$$
\begin{align}
\cos a \sin b &= \frac{1}{2}\bigg[2\cos a\sin b\bigg] \\
&= \frac{1}{2}\bigg[\cos a\sin b + \cos a\sin b\bigg] \\
&= \frac{1}{2}\bigg[\cos a\sin b + \cos a\sin b + 0\bigg] \\
&= \color{red}{\frac{1}{2}\bigg[\cos a\sin b + \cos a\sin b + (\sin a \cos b - \sin a\cos b)\bigg]} \\
&= \frac{1}{2}\bigg[(\cos a\sin b + \sin a \cos b) +(\cos a\sin b - \sin a\cos b)\bigg] \\
&= \frac{1}{2}\bigg[(\cos a\sin b + \sin a \cos b) -(\sin a\cos b - \cos a\sin b )\bigg] \\
&= \frac{1}{2}\bigg[(\cos a\sin b + \sin a \cos b) -(\sin a\cos (-b) + \cos a\sin (-b) )\bigg] \\
&=\frac{1}{2}\bigg[\sin(a+b) - \sin(a-b)\bigg]
\end{align}
$$
Usually I just remember/figure out the red line.
|
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|
Show that $\int_0^\infty\frac{1}{1+x^2+x^{\pi}+x^{\pi-2}}dx = \int_0^\infty\frac{1}{1+x^2+x^{e}+x^{e-2}}dx $ Vladimir Reshetnikov's Identity
(1)
$$\int_0^\infty\frac{1}{1+x^2}\cdot\frac{1}{1+x^{\pi}}dx =\int_0^\infty\frac{1}{1+x^2}\cdot\frac{1}{1+x^{e}}dx $$
Note that $(1+x^2)(1+x^{\pi})=1+x^2+x^{\pi}+x^{2+\pi}$ and
$(1+x^2)(1+x^e)=1+x^2+x^e+x^{2+e}$
**An imitation of Vladimir Reshetnikov's identity **
(2)
$$\int_0^\infty\frac{1}{1+x^2+x^{\pi}+x^{\pi-2}}dx = \int_0^\infty\frac{1}{1+x^2+x^{e}+x^{e-2}}dx $$
The problem with this denominator $(1++x^2+x^{\pi}+x^{\pi-2})$ we can't factorised it
The proved of Identity (1) was offered by Dr.MV
I was hoping to follow the method of Dr.MV of prove of (1) and use it in (2) but its denominator wasn't factorisable, so I am stuck at this point. Can anybody offer their own way of proving identity (2)
My try:
Let (2) be writen in the general form $$I=\int_0^\infty\frac{1}{1+x^2+x^a+x^{a-2}}dx $$
I change (2) into $$I=\int_0^\infty\frac{1}{(1+x^{-2})(1+x^a)+x^2-x^{-2}}dx$$ from here I still think it is more harder then before, anyway I am stuck. **I need help **
|
Let we do something more general. Let $a>0$ and
$$ I(a) = \int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^a)}. $$
Well, $I(a)$ is constant. Don't you believe it? Replace $x$ with $\frac{1}{z}$ to get:
$$ I(a) = \int_{0}^{+\infty}\frac{z^a\,dx}{(1+z^2)(1+z^a)}\,dz $$
from which:
$$ I(a) = \frac{1}{2}\int_{0}^{+\infty}\frac{1+x^a}{(1+x^2)(1+x^a)}\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{dx}{1+x^2}=\color{red}{\frac{\pi}{4}}.$$
You integrals are just $I(e-2)$ and $I(\pi-2)$, and they both equal $\frac{\pi}{4}$.
|
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|
Find $\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms Find $S=\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms
I first multiplied and divided $S$ with $1\cdot3\cdot5$
$$\frac{S}{15}=\frac{1}{1\cdot3\cdot5\cdot7}+\frac{1\cdot3}{1\cdot3\cdot5\cdot7\cdot9}+\frac{1\cdot3\cdot5}{1\cdot3\cdot5\cdot7\cdot9\cdot11}+\cdots$$
Using the expansion of $(2n)!$
$$1\cdot3\cdot5\cdots(2n-1)=\frac{(2n)!}{2^nn!}$$
$$S=15\left[\sum_{r=1}^{20}\frac{\frac{(2r)!}{2^rr!}}{\frac{(2(r+3))!}{2^{r+3}(r+3)!}}\right]$$
$$S=15\cdot8\cdot\left[\sum_{r=1}^{20}\frac{(2r)!}{r!}\cdot\frac{(r+3)!}{(2r+6)!}\right]$$
$$S=15\sum_{r=1}^{20}\frac{1}{(2r+5)(2r+3)(2r+1)}$$
How can I solve the above expression? Or is there an simpler/faster method?
|
Hint:
$\frac{1}{(2r+5)(2r+3)(2r+1)}=\frac{1}{4}\left(\frac{1}{(2r+3)(2r+1)}-\frac{1}{(2r+5)(2r+3)}\right)$
|
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Let g(x) be a non negative continuous function on R such that $g(x) +g(x+\frac{1}{3})=5$.. Problem :
Let g(x) be a non negative continuous function on R such that $g(x) +g(x+\frac{1}{3})=5$ then calculate the value of integral $\int^{1200}_0 g(x) dx$
My approach :
$g(x) +g(x+\frac{1}{3})=5$.....(1)
put x = x +$\frac{1}{3}$
we get $g(x+\frac{1}{3})+g(x+\frac{2}{3})=5$...(2)
Subtracting (2) from (1) we get
$g(x) = g(x+\frac{2}{3})$
$\Rightarrow $ g(x) is periodic with period $\frac{2}{3}$
So we can write the integral as
$1800 \int^{2/3}_0 g(x)dx $
now what to do further please suggest , will be of great help. thanks
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I don't know whether this solution is the fastest but I am pretty sure it is the most accurate one.
Step 1: multiply the differential infinitesimal (dx) on $g(x) +g(x+\frac{1}{3})=5$
$(g(x) +g(x+\frac{1}{3}))dx= 5dx$
Because of the multiplication's disturbution law:
$g(x)dx +g(x+\frac{1}{3})dx= 5dx$
Step 2: integral two side with from low limit(0) to upper limmit(1/3):
$\int_0^{\frac{1}{3}}g(x)dx + g(x+\frac{1}{3})dx = \int_0^{\frac{1}{3}} 5 dx$
By the integration's indentitiy:
$\int_0^{\frac{1}{3}}g(x)dx + \int_0^{\frac{1}{3}}g(x+1/3)dx = \int_0^{\frac{1}{3}} 5 dx$
substitude x for x+1/3
$\int^{2/3}_0 g(x)dx = \int_0^{\frac{1}{3}}g(x)dx + \int_\frac{1}{3}^{\frac{2}{3}}g(x)dx = \int_0^{\frac{1}{3}} 5 dx = \frac{5}{3}$
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solve for $x$: $\frac{\sin(x)}{x}=\frac{5}{6}$ Is it possible to solve for x the following equation without root finding:
$$\frac{\sin(x)}{x}=\frac{5}{6}$$
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Equations which mix polynomial and trigonometric terms do not show analytical solutions (this is already the case for $x=\cos(x)$) and numerical methods should be used.
However, some rather good approximations can be made and, for your curiosity, I give you the links to two questions of mine (here and here).
Using the beautiful $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician, your equation becomes $$\frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}=\frac 56 x$$ which reduces to a simple quadratic $$-\frac{10 x^2}{3}+\left(\frac{10 \pi }{3}-16\right) x-\frac{25 \pi ^2}{6}+16 \pi=0$$ for which the solution to keep is $$x=\frac{1}{40} \left(-96+20 \pi +\sqrt{9216+3840 \pi -1600 \pi ^2}\right)\approx 1.02288 $$ while the exact solution, obtained using Newton method, would be $\approx 1.02674$.
Now, knowing that the solution is close to $1$, we could approximate $\sin(x)$ using its simplest Padé approximant built at $x=1$; this should give $$\sin(x)=\frac{\frac{1}{2} (x-1) (2 \cos (1)+\sin (1) \tan (1))+\sin (1)}{1+\frac{1}{2} (x-1)
\tan (1)}$$ which, again, reduces to a quadratic the solution of which being $\approx 1.02676$.
Another way could be to use the Taylor series of $\sin(x)$ around $x=\frac \pi 3$ which is $$\sin(x)=\frac{\sqrt{3}}{2}+\frac{1}{2} \left(x-\frac{\pi }{3}\right)-\frac{1}{4} \sqrt{3}
\left(x-\frac{\pi }{3}\right)^2-\frac{1}{12} \left(x-\frac{\pi
}{3}\right)^3+O\left(\left(x-\frac{\pi }{3}\right)^4\right)$$ Limiting to first order, the solution would then be $$x_{(1)}=\frac{1}{2} \left(3 \sqrt{3}-\pi \right)\approx 1.02728$$ Limiting to second order, the solution would then be $$x_{(2)}=\frac{1}{3} \left(-\frac{2}{\sqrt{3}}+\pi +\sqrt{\frac{2}{3} \left(29-5 \sqrt{3}
\pi \right)}\right)\approx 1.02674$$
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Prove that $\frac{1}{7\sqrt{2}} \leq \int_{0}^1 \frac{x^6}{\sqrt{1+x^2}}dx \leq \frac{1}{7}$
Prove that $\displaystyle \dfrac{1}{7\sqrt{2}} \leq \int_{0}^1 \dfrac{x^6}{\sqrt{1+x^2}}dx \leq \dfrac{1}{7}$.
My book says let $f(x) = \dfrac{1}{\sqrt{1+x^2}}$ and $g(x) = x^6$. Then $\displaystyle \int_0^1 \dfrac{x^6}{\sqrt{1+x^2}}dx = \dfrac{1}{\sqrt{1+\xi^2}} \int_{0}^1 x^6 dx$ where $0 \leq \xi \leq 1$. Thus $$\dfrac{1}{7\sqrt{2}} \leq \dfrac{1}{\sqrt{1+\xi^2}} \leq \dfrac{1}{7}.$$
I don't understand how they are getting all these results. The problem did say to use other results in the book, but I didn't see any that related. For reference it is in chapter 13 of Michael Spivak's calculus book.
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We have $$\frac{1}{\sqrt 2} \le \frac{1}{\sqrt{1+x^2}} \le 1$$ for all $x \in [0,1]$ since $0 \le x^2 \le 1$. Indeed, then $1 \le 1+x^2 \le 2$ and so $$\frac{1}{2} \le \frac{1}{1+x^2} \le 1.$$ Thus $$\frac{1}{\sqrt 2} \le \frac{1}{\sqrt{1+x^2}} \le 1.$$
Then just multiply by $x^6$ and integrate.
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How to calculate $\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$? Evaluate this limit
$$\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$$
What's the method?
The answer is $1000$.
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This answer is pretty useless for the results per se, much shorter proofs have been given. Honestly, it took me some time to write it, and I could non just abandon it in the forest.
[EDIT] Its intent is to play with standard inequalities, that are useful to keep in mind, and keep them as long as possible, without using de l'Hôpital rule, which I revere for the insight. However it requires some care, and probably should mention Johan Bernoulli in its name. Keeping inequalities may provide you with convergence rate estimates.
The function (call it $f(x)$) is even. Borrowing from @Paramanand Singh notation, it can be factorized as:
$$f(x) = \frac{1 - \left(\frac{\sin x }{x}\right)^n}{x^2 \left(\frac{\sin x }{x}\right)^n}\,.$$
We use standard inequalities close enough to $0^+$:
$$ 1-ny \le (1-y)^n \le 1-ny + \frac{n(n+1)}{2} y^2$$
for $n>0$ and
$$ x - \frac{x^3}{6} \le \sin x \le x - \frac{x^3}{6} + \frac{x^5}{5!} $$
hence
$$ 1 - \frac{x^2}{6} \le \frac{\sin x }{x} \le 1 - \frac{x^2}{6} + \frac{x^4}{5!}\,.$$
From the last one, we see that $\left(\frac{\sin x }{x}\right)^n \to 1$, so we forget about it and study $g(x) = f(x)\left(\frac{\sin x }{x}\right)^n$.
By replacing above $y$ by $ \frac{x^2}{6} $ or $ \frac{x^2}{6} - \frac{x^4}{5!}$ , we have:
$$ n x^2 \left(\frac{1}{6} - \frac{x^2}{5!}\right) - \frac{n(n+1)x^4}{2}\left(\frac{1}{6} - \frac{x^2}{5!}\right) ^2\le 1 - \left(\frac{\sin x }{x}\right)^n \le n\left(\frac{x^2}{6} \right)$$
hence
$$\frac{n}{6}+a_nx^2+b_n x^4+c_nx^6\le g(x) \le \frac{n}{6}$$
with constants $a_n,b_n,c_n $ which you can compute explicitly, especially if you want some rate of convergence, for instance I got $a_n = -\frac{n(5n+8)}{360}$. And for your question of course the limit is $1000 = 6000/6$.
Here is a visual simulation of the bounds for function $g(x)$, only for $n=4$ for numerical reasons:
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Finding the limit of $\lim_{n\rightarrow\infty}(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+...[\text{upto n times}])$ The question is as follows-
Evaluate the limit.
$$\lim_{n\rightarrow\infty}(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+...[\text{upto n times}])$$
I have no idea on how to solve this limit.
Thanks for any help,response or hint!!
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Use formula:
$$\frac{1}{n^3-n}=\frac1{(n-1)n(n+1)}=\frac12 \left(\frac1{n(n-1)}-\frac1{n(n+1)} \right)$$
Then
$$\lim_{n\rightarrow\infty}\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+...+\frac{1}{n^3-n}\right)=$$
$$=\lim_{n\rightarrow\infty}\frac12\left(\frac{1}{2\cdot1}-\frac{1}{2\cdot3}+\frac{1}{3\cdot2}-\frac{1}{3\cdot4}+\frac{1}{4\cdot3}-...-\frac{1}{n(n+1}\right)=$$
$$=\lim_{n\rightarrow\infty}\frac12\left(\frac12-\frac{1}{n(n+1)} \right)=\frac14$$
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Find $k_{max}$ if such $\frac{2(a^2+kab+b^2)}{(k+2)(a+b)}\ge \sqrt{ab}$ Let $a,b>0$ then we have
$$\color{crimson}{\dfrac{2(a^2+kab+b^2)}{(k+2)(a+b)}\ge \sqrt{ab}}$$ Find $k_{\max}$
Everything I tried has failed so far.
Here is one thing I tried, but obviously didn't work.
Consider the Special case $a=b$
then
$$\color{crimson}{\dfrac{2(a^2+kab+b^2)}{(k+2)(a+b)}=\dfrac{2(2a^2+ka^2)}{2(k+2)a}=a=RHS}$$Thanks in advance
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The inequality can be rewritten as $$2(a-b)^2\ge (k+2)(\sqrt{a}-\sqrt{b})^2$$ which is equivalent when $a\neq b$ to $$2(\sqrt{a}+\sqrt{b})^2\ge (k+2)\sqrt{ab}$$ or $$2(\sqrt{a}-\sqrt{b})^2\ge (k-6)\sqrt{ab}$$ This shows that the inequality always holds for $k=6$. For any $k>6$, we may take $a=1, b=1+\epsilon$ for sufficiently small $\epsilon>0$, and the inequality becomes false. So $k=6$ is best.
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show that $\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $ Let $a,b,c$ are positive numbers,if $$a+b+c=1$$
show that $$\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $$
I am tried proving it but failed.Any hints will be appreciated.
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We shall use the sigma sign ($\sum$) for cyclic sum.
By changing every number 1 into $a+b+c$, what we are trying to prove is equivalent to :
$$\sum \frac {1}{a+b} \ge \sum \frac{2}{(a+b)+(c+a)}$$
Let
$$a+b=x$$
$$b+c=y$$
$$c+a=z$$
We have to prove :
$$\sum \frac {1}{x} \ge \sum \frac{2}{x+y}$$
We have $$ \frac 1x + \frac 1y \ge \frac 4 {x+y}$$
$$ \frac 1y + \frac 1z \ge \frac 4 {y+z}$$
$$ \frac 1z + \frac 1x \ge \frac 4 {z+x}$$
By adding all three side of all three inequalities above, we get
$$\sum \frac {2}{x} \ge \sum \frac{4}{x+y}$$
Dividing each side by 2, we get what we have to prove.
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Inequality problem $\frac 1x + \frac 1y + \frac 1z > 5$ prove Prove that:
$$\frac 1x + \frac 1y + \frac 1z > 5$$
where $x+y+z=1$, $x, y, z$ are real numbers not equal $0$
and $x\neq y \neq z $
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Note: This somewhat longish answer contains several proofs of the inequality. I'm leaving things in the order in which they accreted, but I recommend skipping to the boldface "added yet later" section for the simplest of the proofs here. I also highly recommend Macavity's answer as the simplest of all.
Using AGM twice, we have
$${1\over x}+{1\over y}+{1\over z}\ge3\sqrt[3]{1\over xyz}={3\over\sqrt[3]{xyz}}\ge{3\over\left(x+y+z\over3\right)}={9\over x+y+z}=9\gt5$$
Added later: Here is a second, completely different proof, which gets the requested inequality (with a $5$) without getting the stronger inequality with a $9$.
The requirement that $x+y+z=1$ with $x,y,z\gt0$ means that $(\sqrt x,\sqrt y,\sqrt z)$ lies on the unit sphere.. This means we can parameterize the variables, using spherical coordinates, as
$$\begin{align}
x&=\sin^2\theta\cos^2\phi\\
y&=\sin^2\theta\sin^2\phi\\
z&=\cos^2\theta
\end{align}$$
Then
$$\begin{align}
{1\over x}+{1\over y}+{1\over z}
&={1\over\sin^2\theta\cos^2\phi}+{1\over\sin^2\theta\sin^2\phi}+{1\over\cos^2\theta}\\
&={\sin^2\phi+\cos^2\phi\over\sin^2\theta\sin^2\phi\cos^2\phi}+{1\over\cos^2\theta}\\
&={1\over\sin^2\theta\sin^2\phi\cos^2\phi}+{1\over\cos^2\theta}\\
&={4\over\sin^2\theta(2\sin\phi\cos\phi)^2}+{1\over\cos^2\theta}\\
&={4\over\sin^2\theta\sin^22\phi}+{1\over\cos^2\theta}\\
&\ge4+1\\
&=5
\end{align}$$
Added yet later: Here is a third, fairly simple proof.
The conditions on $x$, $y$, and $z$ imply $0\lt x,y,z\lt1$, so we can write $x=1-u$, $y=1-v$, and $z=1-w$ with $0\lt u,v,w\lt1$. The equation $x+y+z=1$ translates into $u+v+w=2$. Thus
$$\begin{align}
{1\over x}+{1\over y}+{1\over z}
&={1\over1-u}+{1\over1-v}+{1\over1-w}\\
&=(1+u+u^2+\cdots)+(1+v+v^2+\cdots)+(1+w+w^2+\cdots)\\
&\gt(1+u)+(1+v)+(1+w)\\
&=3+(u+v+w)\\
&=5
\end{align}$$
Remarks: If you include Macavity's extremely simple proof, we now have three proofs of the requested inequality that don't give anything stronger without additional work. It's worth noting that Macavity's approach can be strengthened by $1$ with just a small twist: If we order $0\lt x\le y\le z\lt1$, then $x+y+z=1$ implies $x\le{1\over3}$ and $y\le{1\over2}$, which gives
$${1\over x}+{1\over y}+{1\over z}\gt3+2+1=6$$
It might be entertaining to see if there are any approaches that naturally (whatever "naturally" means) give the inequality with a $7$ but not an $8$, or an $8$ but not the $9$. Update: Macavity has given proofs that do this in comments below.
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Tricky Multivariable Inequality Given 100 positive real numbers $x_1, x_2, \cdots, x_n$ that satisfy
$x_1^2+x_2^2+\cdots+x_n^2>10000$ and
$x_1+x_2+\cdots x_n\le 300$,
prove that there exist three numbers from this set such that the sum of these three numbers is larger than 100.
I have tried using Lagrange Multipliers and smoothing but it appears not to work. An "equality case" is $(100/3,...$(9 times)$..., 100/3,0,0,0...)$
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I shall prove the following statement, from which the problem is solved by setting $B=300$ and $n=100$.
Let $S=\{x_1,x_2,\ldots,x_n\}$ be a set of $n \geq 3$ non-negative numbers and let $t=\sum_{i=1}^{n}x_i^2$ and
$s=\sum_{i=1}^{n} x_i$.
Suppose that $s \geq \dfrac{B^2}{9}$ and $t \leq B$. Then there exist three numbers in $S$ whose sum is at least
$\dfrac{B}{3}$.
Proof:
Let $A=\dfrac{B^2}{9}$, $p_i=\dfrac{x_i}{s}$.
We have: $\sum_{i=1}^{n}{p_i}=1$ and $\sum p_ix_i= \dfrac{t}{s} \geq \dfrac{A}{B}$.
Hence we can find a number $x$ in the set of value at least $\dfrac{A}{B}$.
Repeating this argument, we can find two more numbers $y,z$ such that:
$y \geq \dfrac{A-x^2}{B-x}$ and $z \geq \dfrac{A-x^2-y^2}{B-x-y}$.
We need to prove:
$x+y+z \geq 3\dfrac{A}{B}$,
for which I'll use the following claims.
Claim 1: Let $A,B> 0$ such that $B^2>A$ and let $f(x)=x+\dfrac{A-x^2}{B-x}$ and $g(x)=x+2\dfrac{A-x^2}{B-x}$.
Then $f$ is increasing when $x \leq B-\sqrt{\dfrac{B^2-A}{2}}$ and $g$ is increasing when $x \leq 2B-\sqrt{3B^2-2A}$.
In particular, if $B^2=9A$, then $f,g$ are both increasing when $x \leq \dfrac{B}{3}$.
This claim is proved by differentiating $f,g$ and some algebra.
Claim 2: Let $A,B >0$, $A \leq B^2 \leq \dfrac{4A}{3}$ and $\sqrt{A} \geq x \geq \dfrac{A}{B}$;
then $f(x)=x+\dfrac{A-x^2}{B-x} \geq 2\dfrac{A}{B}$, with the minimum at $x=\dfrac{A}{B}$.
Proof of Claim 2: If $x \geq 2\dfrac{A}{B}$, we are done because the second term is non-negative,
so let $x \leq 2\dfrac{A}{B}$. Using claim 1 for $f$ and some algebra, we see that $f$ is increasing in this range of $x$;
hence the minimum is at $x=\dfrac{A}{B}$.
Now we prove the original desired statement that $x+y+z \geq 3\dfrac{A}{B}$.
If $x \geq \dfrac{B}{3}=3\dfrac{A}{B}$, we are done, so let's assume that $x \leq \dfrac{B}{3}$.
Consider $y+ \dfrac{A-x^2-y^2}{B-x-y}$: this is at least $2\dfrac{A-x^2}{B-x}$ by applying Claim 2 and
checking that its preconditions are met.
Thus $x+y+z \geq x+2\dfrac{A-x^2}{B-x}$.
Again using Claim 1 for $g$, we see that $g$ is increasing in the range of $x$ and hence its minimum is at $x=\dfrac{A}{B}$.
This gives $x+y+z \geq 3\dfrac{A}{B}=\dfrac{B}{3}$, which completes the proof.
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Parametrize the "alpha curve" I wanted to check if my solution would be correct for the question:
Q. Parametrize the "alpha curve"
$y^2$=$x^3$+$x^2$
A. y=tx
$(tx)^2$=$x^3$+$x^2$
$t^2$$x^2$=$x^3$+$x^2$
$t^2$$x^2$-$x^2$=$x^3$
$x^2$($t^2$-1)=$x^3$
x=$t^2$-1
Substitute x=$t^2$-1 into $y^2$=$x^3$+$x^2$
$y^2$=$x^3$+$x^2$
= $x^2$(x+1)
=($t^2$-1)$^2$($t^2$-1+1)
y=${\sqrt{(t^2-1)^2(t^2)}}$
=$t(t^2-1)$
So, the parametrization of the alpha curve is
$t\mapsto(t^2-1, t(t^2-1)$
and the double point (0,0) is obtained twice for t=${\sqrt{1}}$
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In polar coordinates,
$$r^2\sin^2(\theta)=r^3\cos^3(\theta)+r^2\cos^2(\theta),$$
then
$$r=\frac{\sin^2(\theta)-\cos^2(\theta)}{\cos^3(\theta)}=\frac{1-2\cos^2(\theta)}{\cos^3(\theta)}.$$
Then numerator cancels for the four angles $\theta=\pm\frac\pi4,\pm\frac{3\pi}4$, corresponding to a double point at the origin, the curve being tangent to the main bissectors. The denominator cancels for $\theta=\frac\pi2,\frac{3\pi}2$, giving points at infinity.
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Error in solving $\int \sqrt{1 + e^x} dx$ . I want to solve this integral for $1 + e^x \ge 0$
$$\int \sqrt{1 + e^x} dx$$
I start by parts $$\int \sqrt{1 + e^x} dx = x\sqrt{1 + e^x} - \int x \frac{e^x}{2\sqrt{1 + e^x}} dx $$
Substitute $\sqrt{1 + e^x} = t \implies dt = \frac{e^x}{2\sqrt{1 + e^x}} dx$
So I remain with $$x\sqrt{1 + e^x} - \int \ln((t-1)(t+1)) dt = x\sqrt{1 + e^x} + 2t +(-t+1)\ln(t-1)+ (-t+1)\ln(t+1) $$
that is
$$(x+2)\sqrt{1 + e^x} + (-\sqrt{1 + e^x} + 1) \ln(+\sqrt{1 + e^x} - 1) + (-\sqrt{1 + e^x} + 1) \ln(+\sqrt{1 + e^x} + 1) + C $$
But the solution should be $$2\sqrt{1 + e^x} + \ln(+\sqrt{1 + e^x} - 1) - \ln(+\sqrt{1 + e^x} + 1) $$
Where is the mistake hidden?
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Where is the mistake hidden?
In the following part :
So I remain with $$x\sqrt{1 + e^x} - \int \ln((t-1)(t+1)) dt = x\sqrt{1 + e^x} + 2t +(-t+1)\ln(t-1)+ (-t+1)\ln(t+1) $$
It should be the following (a sign mistake) :
$$x\sqrt{1 + e^x} - \int \ln((t-1)(t+1)) dt = x\sqrt{1 + e^x} + 2t +(-t+1)\ln(t-1)+ (-t\color{red}{-}1)\ln(t+1)\color{red}{+C} $$
that is
$$x\sqrt{1 + e^x} + 2\sqrt{1 + e^x} +(-\sqrt{1 + e^x}+1)\ln(\sqrt{1 + e^x}-1)$$$$+ (-\sqrt{1 + e^x}-1)\ln(\sqrt{1 + e^x}+1)+C$$
$$=(x+2)\sqrt{1+e^x}-\sqrt{1+e^x}\ (\ln (\sqrt{1+e^x}-1)+\ln(\sqrt{1+e^x}+1))+\ln(\sqrt{1+e^x}-1)$$$$-\ln(\sqrt{1+e^x}+1)+C$$
$$=2\sqrt{1+e^x}+\ln(\sqrt{1+e^x}-1)-\ln(\sqrt{1+e^x}+1)+C$$
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Show that if $n\equiv 3, 6 \pmod9 $ then $n$ is not a sum of two squares Show that if $n\equiv 3, 6 \pmod9 $ then $n$ is not a sum of two squares.
I started by: Assume $n=a^2+b^2$ a sum of two squares. Then $a^2,b^2\equiv 0,1,4,7 \pmod9$, and no combination these numbers can yield $3$ or $6$ so that $a^2+b^2\equiv 3,6 \pmod9$.
But then I would need to show the first result, but I don't know any results (and shouldn't need to apply) results for quadratic residues modulo a composite number. Otherwise, maybe I need to use the result that $n$ is a sum of two squares if $n \not\equiv 3\pmod4$.
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It suffices to show that if $3\mid a^2+b^2$, then $3\mid a$ or $3\mid b$. There is a separate question about this: Prove that if $3\mid a^2+b^2$ then $3\mid a$ and $3\mid b$. (And you might find something also in the questions which are linked there.) Unfortunately, that question is not in a very good state. It was edited, some of the answers address the original version, some of the answers address the current version of the question and some of the answers address a common generalization of both the original and the new version.
For this reason, I will summarize here things which are said in some answers to other posts.
Claim 1. If $3\mid a^2+b^2$, then $3\mid a$ and $3\mid b$.
Proof. By contrapositive. Let us assume $3\nmid a$ or $3\nmid b$.
If $3\nmid a$, then we have $a^2\equiv 1\pmod 3$. For any $b$ we have $b^2\equiv 0,1\pmod 3$
This implies $a^2\equiv b^2\equiv 1 \pmod 3$. And now we get
$$a^2+b^2 \equiv 1,2 \pmod 3$$
which means $3\nmid a^2+b^2$. $\square$
Claim 2. If $n\equiv 3,6 \pmod 9$, then $n$ is not sum of two squares.
Proof. By contradiction. Assume that $n=a^2+b^2$ and $n\equiv 3,6 \pmod 9$. Clearly, this implies $3\mid a^2+b^2$ and by Claim 1 we get $3\mid a$ and $3\mid b$. But this implies $a^2=(3k)^2=9k^2$, i.e. $9\mid a^2$. We get $9\mid b^2$ by the same argument. Together we get $9\mid a^2+b^2$, which means
$$a^2+b^2\equiv 0 \pmod 9,$$
a contradiction. $\square$
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Evaluation of $\int_{0}^{1}\frac{\ln x}{x^2-x-1}dx$
Evaluation of $\displaystyle \int_{0}^{1}\frac{\ln x}{x^2-x-1}dx$
$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2-x-1}dx=\int_{0}^{\infty}\frac{\ln(x)}{\left(x-\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2}dx$
Now Put $\displaystyle \left(x-\frac{1}{2}\right)=\frac{\sqrt{5}}{2}\sec \theta\;,$ Then $\displaystyle dx = \frac{\sqrt{5}}{2}\sec \theta \tan \theta$
So $$ I= -\frac{2}{\sqrt{5}}\int_{-\frac{1}{2}}^{\infty}\frac{\ln(\sqrt{5}\sec \theta+1)-\ln(2)}{\tan \theta}\cdot \sec \theta d\theta$$
So $$I = \frac{2}{\sqrt{5}}\int_{-\frac{1}{2}}^{\infty}\frac{\ln(2)+\ln(\cos \theta)-\ln(\sqrt{5}+\cos \theta)}{\sin \theta} d\theta$$
Now How can I solve after that, Help Required, Thanks
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So, by the comment of the OP, the integral we have to study is $$\int_{0}^{1}\frac{\log\left(x\right)}{x^{2}-x-1}dx.$$ We note that $$\begin{align} I= & \int_{0}^{1}\frac{\log\left(x\right)}{x^{2}-x-1}dx \\ = & \int_{0}^{1}\frac{\log\left(x\right)}{\left(x-\frac{1-\sqrt{5}}{2}\right)\left(x-\frac{1+\sqrt{5}}{2}\right)}dx \\ = & -\frac{2}{\sqrt{5}}\left(\int_{0}^{1}\frac{\log\left(x\right)}{2x+\sqrt{5}-1}dx-\int_{0}^{1}\frac{\log\left(x\right)}{2x-\sqrt{5}-1}dx\right) \\ = & -\frac{2}{\sqrt{5}}\left(I_{1}-I_{2}\right), \end{align}$$ say. Let us analyze $I_{1}$. Integrating by part we get $$I_{1}=\int_{0}^{1}\frac{\log\left(x\right)}{2x+\sqrt{5}-1}dx=\frac{1}{\sqrt{5}-1}\int_{0}^{1}\frac{\log\left(x\right)}{\frac{2x}{\sqrt{5}-1}+1}dx=-\frac{1}{2}\int_{0}^{1}\frac{\log\left(\frac{2x}{\sqrt{5}-1}+1\right)}{x}dx
$$ $$=\frac{1}{2}\textrm{Li}_{2}\left(-\frac{2}{\sqrt{5}-1}\right)
$$ where $\textrm{Li}_{2}(x)$ is the Dilogarithm function. In a similar way we can find that $$I_{2}=\int_{0}^{1}\frac{\log\left(x\right)}{2x-\sqrt{5}-1}dx=\frac{1}{2}\textrm{Li}_{2}\left(\frac{2}{\sqrt{5}+1}\right)
$$ so $$I=\frac{\textrm{Li}_{2}\left(\frac{2}{\sqrt{5}+1}\right)-\textrm{Li}_{2}\left(-\frac{2}{\sqrt{5}-1}\right)}{\sqrt{5}}
$$ and since holds $$\textrm{Li}_{2}\left(\phi^{-1}\right)=\frac{1}{10}\pi^{2}-\log^{2}\left(\phi\right),\,\textrm{Li}_{2}\left(-\left(\phi-1\right)^{-1}\right)=-\frac{1}{10}\pi^{2}-\log^{2}\left(\phi\right)
$$ where $\phi=\frac{\sqrt{5}+1}{2}
$ is the golden ratio we have $$I=\frac{\pi^{2}}{5\sqrt{5}}.$$
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To show that the variables in the system are same in magnitude I am stuck with this interesting problem,
If for non-negative integers $a, b, \text{and} c$, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ are both integers then show that $|a|=|b|=|c|$.
I could only make a few observations
1) Because the system is symmetric wrt. $a, b, c$ as the two expressions swap, when any two variables are swapped. So, we can assume that $a \geq b+1$ and $a \geq c+2$ as assuming equality of any two makes the problems trivial.
2) Adding and subtracting the two equations give $\frac{(a+b)(b+c)(c+a)}{abc}$ and $\frac{(a-b)(b-c)(c-a)}{abc}$ are both integers.
Also, $\frac{(ab+bc+ca)(a+b+c)}{abc}$ is an integer.
Also, seeing that the expressions involved in 2) are close to the elementary symmetry polynomials, I considered the polynomial $P(x)=(x-a)(x-b)(x-c)$
I think the above observations are useful for the problem but I have failed to build on them so someone please help me. Please!! By the way, I am in high school and don't know any college level stuff.
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Let $m=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $n=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$. Then $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are the roots of $x^3-mx^2+nx-1=0$.
Hence we have $a^3-ma^2b+nab^2-b^3=0$, where $a,b,m,n$ are all integers, with $a,b$ positive. Let $p^r$ be the highest power of $p$ dividing $a$, and $p^s$ be the highest power of $p$ dividing $b$. Suppose $r>s$. Then the highest power of $p$ dividing $a^3-ma^2b+nab^2$ is at least $r+2s$, whereas the highest power of $p$ dividing $b^3$ is $3s<r+2s$. Contradiction. So we must have $r\le s$. Similarly $s\le r$, so $r=s$. This is true for any prime $p$, so we must have $a=b$.
Similarly, $b=c$.
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Prove $\ln(2^5)-\pi=8\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{e^{n\pi}+1}-\frac{1}{e^{2n\pi}+1}\right)$ We took this idea from Simon Plouffe see here
$$\ln(2^5)-\pi=8\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{e^{n\pi}+1}-\frac{1}{e^{2n\pi}+1}\right)$$
Can anyone prove this identiy?
We found this identity via a sum calculator by varying Simon Plouffe identities
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We can calculate the sum $$\sum_{n = 1}^{\infty}\frac{1}{n(e^{nx} + 1)}$$ as follows.
Let $q = e^{-x}$ and then we have
\begin{align}
F(q) &= \sum_{n = 1}^{\infty}\frac{1}{n(e^{nx} + 1)}\notag\\
&= \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 + q^{n})}\notag\\
&= \sum_{n = 1}^{\infty}\frac{1}{n}\sum_{k = 1}^{\infty}(-1)^{k - 1}q^{kn}\notag\\
&= \sum_{k = 1}^{\infty}(-1)^{k - 1}\sum_{n = 1}^{\infty}\frac{q^{kn}}{n}\notag\\
&= \sum_{k = 1}^{\infty}(-1)^{k}\log(1 - q^{k})\notag\\
&= \sum_{k \text{ even}}\log(1 - q^{k}) - \sum_{k \text{ odd}}\log(1 - q^{k})\notag\\
&= 2\sum_{k \text{ even}}\log(1 - q^{k}) - \sum_{k = 1}^{\infty}\log(1 - q^{k})\notag\\
&= 2\sum_{k = 1}^{\infty}\log(1 - q^{2k}) - \sum_{k = 1}^{\infty}\log(1 - q^{k})\notag\\
&= 2f(q^{2}) - f(q)\notag
\end{align}
where we have from the linked answer (the $f(q)$ of this answer is same as $-a(q)$ of the linked answer, also note that the linked answer proves another formula of Plouffe for $\pi$) $$f(q) = \sum_{k = 1}^{\infty}\log(1 - q^{k}) = \frac{\log k}{12} + \frac{\log k'}{3} + \frac{\log 2}{3} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{24K}\tag{1}$$ and $$f(q^{2}) = \frac{\log(kk')}{6} + \frac{\log 2}{6} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{12K}\tag{2}$$ and $$f(q^{4}) = \frac{1}{12}\log(k^{4}k') - \frac{\log 2}{6} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{6K}\tag{3}$$ From the above equations we get $$F(q) = \frac{\log k}{4} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{8K}\tag{4}$$ and $$F(q^{2}) = \frac{\log k}{2} - \frac{\log 2}{2} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{4K}\tag{5}$$ The sum in your question is $$8(F(q) - F(q^{2}))$$ where $q = e^{-\pi}$ and this is clearly equal to $$8\left(-\frac{\log k}{4} + \frac{\log 2}{2} - \frac{\pi K'}{8K}\right)$$ and we note that $k = 1/\sqrt{2}$ and $K' = K$ for $q = e^{-\pi}$. Hence the desired sum is $$8((5/8)\log 2 - (\pi/8)) = 5\log 2 - \pi$$
Most of the sums which involve the expressions of type $e^{n\pi}$ I try to express them as function of $q = e^{-\pi}$ and hope that the result series / product in $q$ can be related to the elliptic integrals $K$ and modulus $k$. If this is possible (like here) then the problem of dealing with infinite series/product is transformed into dealing with finite expressions consisting of $k, K, \pi$ and simple algebraic manipulations on them. This may not work always (or may not be convenient every time), but appears simpler to me when it works compared to general techniques for summing infinite series.
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Solution of integral $\int \frac{\sin (x)}{\sin (5x) \sin (3x)}\,\mathrm dx.$ Find the following integral:
$$\int \frac{\sin (x)}{\sin (5x) \sin (3x)}\,\mathrm dx.$$
I don't know how to deal with the $\sin (x)$ in the numerator. If it had been $\sin (2x)$ then we could have used $\sin (2x)= \sin (5x-3x)$. How to deal given integral? Could someone help me with this?
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I think it's a little easier if you leave in in terms of Chebyshev polynomials of the second kind, $U_n(\cos x)=\frac{\sin(n+1)x}{\sin x}$. Then the integral looks like
$$\begin{align}\int\frac{\sin x}{\sin5x\sin3x}dx&=\int\frac{\sin x}{\sin^2xU_4(\cos x)U_2(\cos x)}dx\\
&=\int\frac{-dv}{(1-v^2)U_4(v)U_2(v)}\end{align}$$
Where we have made the substitution $v=\cos x$. Then you can use $\sin(n+1)x=2\sin nx\cos x-\sin(n-1)x$ to run these out far enough:
$$\begin{array}{rl}\sin2x&=2\sin x\cos x\\
\sin3x&=4\sin x\cos^2x-\sin x\\
\sin4x&=8\sin x\cos^3x-4\sin x\cos x\\
\sin5x&=16\sin x\cos^4x-12\sin x\cos^2x+\sin x\end{array}$$
So now we are up to
$$\int\frac{\sin x}{\sin5x\sin3x}dx=\int\frac{dv}{64\left(v^2-1\right)\left(v^2-\frac14\right)\left(v^4-\frac34v^2+\frac1{16}\right)}$$
The roots of that last factor are given by $\sin\left(5\cos^{-1}v\right)=0=\sin n\pi$, so $v=\cos\frac{n\pi}5$, or $$v\in\left\{\frac{\phi}2,\frac1{2\phi},-\frac1{2\phi},-\frac{\phi}2\right\}$$
Where $\phi=\frac{\sqrt5+1}2$, just like @Jack D'Aurizio said. So we have
$$\begin{align}&\frac1{64\left(v^2-1\right)\left(v^2-\frac14\right)\left(v^4-\frac34v^2+\frac1{16}\right)}\\
&=\frac1{64(v-1)(v+1)(v-\frac12)(v+\frac12)(v-\frac{\phi}2)(v-\frac1{2\phi})(v+\frac1{2\phi})(v+\frac{\phi}2)}\\
&=\frac A{v-1}+\frac B{v+1}+\frac C{v-\frac12}+\frac D{v+\frac12}+\frac E{v-\frac{\phi}2}+\frac F{v-\frac1{2\phi}}+\frac G{v+\frac1{2\phi}}+\frac H{v+\frac{\phi}2}\end{align}$$
We can knock these partial fractions out pretty quick with L'Hopital's rule. For example
$$\begin{align}\lim_{v\rightarrow\frac{\phi}2}\frac{v-\frac{\phi}2}{64\left(v^2-1\right)\left(v^2-\frac14\right)\left(v^4-\frac34v^2+\frac1{16}\right)}&=\frac1{64\left(\frac{\phi^2}4-1\right)\left(\frac{\phi^2}4-\frac14\right)\left(4\frac{\phi^3}{8}-\frac34(2)\frac{\phi}2\right)}\\
&=-\frac1{5\phi}=\lim_{v\rightarrow\frac{\phi}2}\frac{E(v-\frac{\phi}2)}{v-\frac{\phi}2}=E\end{align}$$
The symmetries mean that we only need one of a pair of algebraic conjugates or additive inverses, so really only a total of $3$ numerators need be separately computed. Eventually after getting the other coefficients and integrating and expressing in terms of $x$, we get
$$\begin{align}\int\frac{\sin x}{\sin5x\sin3x}dx&=-\frac1{5\phi}\ln\left|\frac{\cos x-\frac{\phi}2}{\cos x+\frac{\phi}2}\right|+\frac{\phi}5\ln\left|\frac{\cos x+\frac1{2\phi}}{\cos x-\frac1{2\phi}}\right|\\
&+\frac13\ln\left|\frac{\cos x-\frac12}{\cos x+\frac12}\right|+\frac1{30}\ln\left|\frac{\cos x-1}{\cos x+1}\right|+C\end{align}$$
I checked it by comparison with numeric quadrature, so the only errors left are the typos. The use of Chebyshev polynomials of the second kind left the denominator in a nicer factored form.
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Solving a trig equation that is quadratic? I have to solve for $x$ given
$$\tan^2 x = 2 + \tan x\;\;\;\;\;\;0≤x≤2\pi$$
I brought it all to one side and set it all equal to zero like:
$$\tan^2 x - \tan x - 2 = 0$$
What am i supposed to do from there to solve for x?
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You began with:
$ \tan^2 x = 2 + \tan x $.
As you mentioned, bring all terms to the L.H.S.:
$ \tan^2 x - \tan x - 2 = 0 $
Now let:
$ y = \tan z $.
As a result:
$ y^2 - y - 2 = 0 $
Use the quadratic formula to solve for $y$.
$ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} $
$ = \frac{1 \pm \sqrt{9}}{2} $
$ = \frac{1 \pm 3}{2} $
$ = \frac{1 + 3}{2}, \frac{1 - 3}{2} $
$ = \frac{4}{2}, \frac{-2}{2} $
$ = 2, -1 $
Since $y = 2, -1 $, and $ y = \tan z $:
$ \tan z = 2, -1 $
From there:
$ z = \arctan 2, \arctan-1 $
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If $a, b, c >0$ prove that $ [(1+a)(1+b)(1+c)]^7 > 7^7a^4b^4c^4 $. I solved it using AM, GM inequalities and reached to $[(1+a)(1+b)(1+c)]^7 > 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.
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$\dfrac{1 + a + b + c + ab + bc + ac + abc}{7} > \dfrac{a + b + c + ab + bc + ac + abc}{7} \geq \sqrt[7]{a^4b^4c^4}$
where the second inequality follows from the AM-GM inequality.
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Integrating trig function I'm stuck at this problem:
$$ \int{\sqrt{(\sin^2 x)^2 + (2\sin x \cos x)^2}dx} = \int{\sqrt{\sin^2 x \sin^2 x + 4\sin^2 x \cos^2 x} dx}$$
I tried a few trig identities: $\sin^2 x = \frac{1-\cos 2x}{2} $ and $ \cos^2 x = \frac{1+\cos 2x}{2} $ and $\sin^2 x + \cos^2 x = 1$. Keep hitting dead end. Any tips?
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$$\begin{align} &\int{\sqrt{\sin^2x \sin^2x + 4\sin^2 x \cos^2 x} dx} \\ &
=\int{\sqrt{\sin^2x( \sin^2x + 4\cos^2 x)} dx} \\ &
=\begin{cases}\int{\sin x\sqrt{( \sin^2x + 4\cos^2 x)} dx} & & & &, \sin x \geq 0 \\ -\int{\sin x\sqrt{( \sin^2x + 4\cos^2 x)} dx} & & & &, \sin x < 0\end{cases} \\ &
=\begin{cases}\int{\sin x\sqrt{(1 + 3\cos^2 x)} dx} & & & & & &, \sin x \geq 0 \\ -\int{\sin x\sqrt{(1 + 3\cos^2 x)} dx} & & & & & &, \sin x < 0\end{cases} \\ &
=\begin{cases}-\frac{1}{\sqrt3}\int{\sin x\sqrt{(1 + 3\cos^2 x)} d(\sqrt3\cos x)} & , \sin x \geq 0 \\ \frac{1}{\sqrt3}\int{\sin x\sqrt{(1 + 3\cos^2 x)} d(\sqrt3\cos x)} & , \sin x < 0\end{cases}\end{align}$$
Now say $z=\sqrt3\cos x$. Then the integration becomes
$$=\begin{cases}-\frac{1}{\sqrt3}\int{\sin x\sqrt{1 + z^2} dz} & , \sin x \geq 0 \\ \frac{1}{\sqrt3}\int{\sin x\sqrt{1 + z^2} dz} & , \sin x < 0\end{cases}$$
I hope you can finish it now, using the necessary formula. The formula can be deduced using integration by parts and proper substitutions and is denoted in the linked page as no. 8.
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Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
I think we'll have to use number theory to do it. Simply solving the equations won't do.
If we divide the second equation by the first, we get:
$$x^2 - xy + y^2 = 1 + z.$$
Also, since they are integers $z^2 \ge z \implies -z^2 \le -z$. This implies
$$x + y = 1 - z \ge 1 - z^2 = x^3 + y^3.$$
This shows that atleast one of $x$ and $y$ is negative with the additive inverse of the negative being larger than that of the positive.
I have tried but am not able to proceed further. Can you help me with this?
Thanks.
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Substituting the first in the second gives
$$x^3+y^3+(x+y)^2-2(x+y)=0$$
so $x+y=0$ (giving $z=1$) or
$$x^2-xy+y^2+x+y-2=0,$$
that is,
$$\left(x+\frac12-\frac y2\right)^2+\frac34(y+1)^2-3=0$$
so $(y+1)^2\leq4$, which leaves to check $y\in\{-3,-2,-1,0,1\}$.
All solutions are given by
$$\begin{align*}(x,y,z)\in\{&(a,-a),\;a\in\mathbb Z,&&(z=1)\\
&(-2,-3),&&(z=6)\\
&(-3,-2),(0,-2),&&(z=6,z=3)\\
&(-2,0),(1,0),&&(z=3,z=0)\\
&(0,1)&&(z=0)\}\end{align*}$$
Perhaps some explanation how I got $\left(x+\frac12-\frac y2\right)^2+\frac34(y+1)^2-3=0$. This is called Completing the square:
Starting from $x^2-xy+y^2+x+y-2=0$ we first get rid of the linear term in $x$. Using $x^2+x=(x+\frac12)^2-\frac14$ we find:
$$\left(x+\frac12\right)^2-\frac14-xy+y^2+y-2=0.$$
Let $X=x+\frac12$. We have
$$X^2-\frac14-Xy+\frac y2+y^2+y-2=0.$$
Now we want to get rid of the mixed term (for the moment we don't care about additional terms in $y$ or constant terms). Using $X^2-Xy=(X-\frac y2)^2-\frac{y^2}4$ we find:
$$\left(X-\frac y2\right)^2-\frac{y^2}4-\frac14+\frac y2+y^2+y-2=0.$$
Now we're left only with terms in $y$: $\frac34y^2+\frac32y-\frac94$. Using $y^2+2y=(y+1)^2-1$ we find:
$$\frac34y^2+\frac32y-\frac94=\frac34(y+1)^2-\frac34-\frac94.$$
So finally,
$$\left(X-\frac y2\right)^2+\frac34(y+1)^2-3=0;\qquad X=x+\frac12.$$
Note: Using this technique, any (inhomogeneous) binary quadratic equation
$$ax^2+bxy+cy^2+\text{linear and constant terms}=0$$
with nonzero discriminant $D=b^2-4ac$ can be rewritten in the form
$$U^2-DV^2=c$$
where $U$ is a linear (better: affine) function of $x$ and $y$, and $V$ is an affine function of $y$. If $D<0$ (as was the case here), the equation clearly has only finitely many solutions. It can be shown that if $D>0$ it has either $0$ or $\infty$ solutions (in that case we call it a Pell-type-equation or something).
If $D=0$ things get ugly.
Geometrically, these correspond to finding integer points on an ellipse if $D<0$, a hyperbola if $D>0$ and a parabola or a union of at most two lines if $D=0$.
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Find the volume of the region enclosed by $x^2+y^2+z^2=2$ and $x^2+y^2=z$.
Find the volume of the region enclosed by $x^2+y^2+z^2=2$ and $x^2+y^2=z$.
I tried to solve the problem above by doing a change of variables to the spherical coordinate system, that is,
$x= \rho \cos \theta \sin \varphi$
$y = \rho \sin \theta \sin \varphi$
$z= \rho \sin \varphi$
But I'm struggling with the range of each variable, $\rho, \theta$ and $\varphi$. What I did is set $\theta$ from $0$ to $2 \pi$, $\varphi$ from $\pi /4$ to $3 \pi /4$ and $\rho$ from $0$ to $\sqrt{2}$, so I get
$$v= \int_0^{\sqrt{2}}\left( \int_0^{2 \pi} \left( \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \rho ^2 \sin \varphi d\varphi \right) d\theta \right)d \rho = \frac{8 \pi}{3}$$
Is this result right? If not, can you suggest me how to proceed? Thanks in advance!
EDIT: After some thinking, I've arrived to the following:
Using Tom-Tom suggestion in the comment that I should split the figure in two parts and change to cylimdral coordinates, let's call the two volumes $S_1$ and $S_2$, where $S_1$ is the upper part (from $z=1$ to $z=\sqrt{2}$) and $S_2$ the rest of it.
For $S_1$, let's first fix the variable $\rho$ which in this case goes from $0$ to $1$. Now, $z$ goes from $0$ to the sphere defined by $x^2+y^2+z^2=2$, which in cylimdral coordinates is $z^2= 2- \rho^2$, or $z=\sqrt{2-\rho^2}$. And finall, $\theta$, which moves from $0$ to $2 \pi$. So then,
$$v(S_1)=\int_{S_1} 1 = \int_0^{2\pi} \int_0^1 \int_1^{\sqrt{2-\rho^2}} \rho \ dz \ d\rho \ d\theta = 2\pi \int_0^1 \rho (\sqrt{2-\rho^2}-1)\ d\rho = 2\pi \left(\frac{\sqrt{8}}{3}-\frac{5}{6}\right)$$
Mow, for $S_2$, we fix the variable $z$ first, that in this case goes from $0$ to $1$. $\rho$ moves from $0$ to the paraboloid defined by $z=x^2+y^2$, which in cylindral coordinates is $\rho = \sqrt{z}$ and $\theta$ goes fro $0$ to $2\pi$ as before. So
$$v(S_2)=\int_{S_2} 1= \int_0^1 \int_0^{2\pi} \int_0^{\sqrt{z}} \rho \ d\rho \ d\theta \ dz = \frac{\pi}{2}$$
And since $S_1$ and $S_2$ intersect on a circle, whic has null measure in $\mathbb{R}^3$,
$$v(S_1 \cup S_2)= v(S_1) + v(S_2)$$
Is these reasoning correct or I'm stil messing it up with the change of coordinates?
|
I can't understand why this solid is getting chopped up into two pieces. We know the $\theta$ goes all the way around, from $0$ to $2\pi$, then the boundary conditions can be solved as $z=r^2$, $r^2+z^2=r^2+r^4=2$, so $r^4+r^2-2=(r^2+2)(r^2-1)=(r^2+2)(r+1)(r-1)=0$ leads to $r=1$ as the only positive solution. So we know the limits on $r$ are from $0$ to $1$, and the limits on z are due to the paraboloid on the bottom as $z=r^2$ and the sphere on top as $z=\sqrt{2-r^2}$ so the volume is
$$\begin{align}V&=\int_0^{2\pi}\int_0^1\int_{r^2}^{\sqrt{2-r^2}}dz\,r\,dr\,d
\theta=2\pi\int_0^1\left[\sqrt{2-r^2}-r^2\right]r\,dr\\
&=2\pi\left[-\frac13(2-r^2)^{3/2}-\frac14r^4\right]_0^1=2\pi\left[-\frac13+\frac{2\sqrt2}3-\frac14\right]=\frac{\pi}6\left(8\sqrt2-7\right)\end{align}$$
Which is the same answer as everyone else got, but with fewer integrals.
|
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|
Range of function $f(x) = \sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$
Range of function $f(x) = \sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$
$\bf{My\; Try::}$ For $\min$ of $f(x)$
$$\left(\sqrt{13-x}+\sqrt{x}\right)^2=13-x+x+2\sqrt{x}\sqrt{13-x}= 13+2\sqrt{x}\sqrt{13-x}\geq 13$$
Now $$\sqrt{x+27} + \sqrt{13-x}+\sqrt{x} \geq \sqrt{27} + \sqrt{13}$$
and equality hold at $x=0$
Now How can i calculate $\max$ of $f(x)\;,$ Help required, Thanks
|
Hint:by Cauchy-Schwarz inequality $$121=[(x+27)+3(13-x)+2x][1+\dfrac{1}{3}+\dfrac{1}{2}]\ge f^2(x)$$
|
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|
Prove that $\int_{0}^{1}{\ln(x) \over 2-x}\mathrm dx={\ln^2(2)-\zeta(2)\over 2}$ $$I=\int_{0}^{1}{\ln(x) \over 2-x}\,\mathrm{d}x={\ln^2(2)-\zeta(2)\over 2}$$
$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}$$
Using binomial series here
$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}={1\over2}+{x\over 4}+{x^2\over8}+{x^3\over 16}+\cdots=\sum_{k=0}^{\infty}{x^k\over2^{k+1}}$$
$$I=\sum_{k=0}^{\infty}{1\over 2^{k+1}}\int_{0}^{1}x^k\ln(x)\,\mathrm{d}x$$
$$\int_{0}^{1}x^k\ln(x)\,\mathrm{d}x=-{1\over (1+k)^2}$$
$$I=-\sum_{k=0}^{\infty}{1\over 2^{k+1}(1+k)^2}$$
So this is where I got to, unable to determine this infinite sum. Do anyone know how to prove that,
$$-\sum_{k=0}^{\infty}{1\over 2^{k+1}(1+k)^2}={\ln^2(2)-\zeta(2) \over 2}$$
|
On the path of user1952009,
Perform the change of variable $y=1-x$,
$\displaystyle I=\int_0^1 \dfrac{\ln(1-x)}{1+x}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$\begin{align}
I&=\int_0^1 \dfrac{\ln\left(\tfrac{2x}{1+x}\right)}{1+x}dx\\
&=\int_0^1 \dfrac{\ln 2}{1+x}dx+\int_0^1 \dfrac{\ln x}{1+x}dx-\int_0^1 \dfrac{\ln(1+x)}{1+x}dx\\
&=\ln 2\Big[\ln(1+x)\Big]_0^1+\int_0^1 \dfrac{\ln x}{1+x}dx-\Big[\dfrac{1}{2}(\ln(1+x))^2\Big]_0^1\\
&=\dfrac{1}{2}(\ln 2)^2+\int_0^1 \dfrac{\ln x}{1+x}dx
\end{align}$
$\begin{align}\int_0^1 \dfrac{\ln x}{1+x}dx&=\int_0^1 \ln x\left(\sum_{n=0}^{\infty} (-1)^n x^n\right)dx\\
&=\sum_{n=0}^{\infty} \left(\int_0^1 (-1)^n x^n\ln x dx\right)\\
&=-\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(1+n)^2}\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^2}\\
&=\sum_{n=1}^{\infty} \dfrac{1}{(2n)^2}-\sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2}\\
&=\dfrac{1}{4}\zeta(2)-\left(\zeta(2)-\dfrac{1}{4}\zeta(2)\right)\\
&=-\dfrac{1}{2}\zeta(2)
\end{align}$
therefore,
$\boxed{I=\dfrac{1}{2}(\ln 2)^2-\dfrac{1}{2}\zeta(2)}$
|
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|
Extreme values of $\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}$ Let $a,b,c$ be side lengths of a triangle. What are the minimum and maximum of $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}?$$
When $a=b=c$, the value is $9$. In addition, we can write $a=x+y,b=y+z,c=z+x$ since they are side lengths of a triangle. The expression becomes
$$\frac{8(x+y+z)(xy+yz+xz)}{(x+y)(y+z)(z+x)}.$$
|
We will prove that
$$8 \leqslant \frac{8(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)} \leqslant 9$$
For the Left Hand Side:
$$(x+y+z)(xy+yz+zx) - (x+y)(y+z)(z+x)=xyz \geqslant 0$$
For the Right Hand Side
\begin{align*}
\ & 9(x+y)(y+z)(z+x)-8(x+y+z)(xy+yz+zx)
\\ &=x^2y+x^2z+xy^2+xz^2+y^2z+yz^2-6xyz
\\ &\geqslant 0
\end{align*}
which is true according to AM-GM.
So the minimum value is $8$ , achieve when $x=0$ and cyclic permutation
The maximum value is $9$, achieve at $x=y=z$
|
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|
Find all pairs (x, y) of real numbers such that $16^{x^2+y} + 16^{x+y^2}=1$ Find all pairs (x, y) of real numbers such that
$$16^{x^2+y} + 16^{x+y^2}=1$$
|
A very nice problem!
By application of inequality $a+b\ge2\sqrt{ab}$ we have
$$16^{x^2+y}+16^{x+y^2}\ge 2\sqrt{16^{x^2+y}\cdot 16^{x+y^2}}=2\cdot 4^{x^2+x+y^2+y}$$
Since $t^2+t\ge-\frac{1}{4}$ holds for all $t\in\mathbb{R}$, for all $x,y\in\mathbb{R}$ we have
$$16^{x^2+y}+16^{x+y^2}\ge2\cdot 4^{x^2+x+y^2+y}\ge 2\cdot 4^{-\frac{1}{2}}=1$$
with equality only in case $x=y=-\frac{1}{2}$, so this is the only solution.
|
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|
Solve the equation $|2x-1| -|x+5| = 3$
Problem : Solve the equation $|2x-1| - |x+5| = 3$
In my attempt to solve the problem, I only manage to get one of the solutions.
Attempted Solution
$$\begin{equation}
\begin{split}
|x|-|y| & \leq |x-y| \\
\implies |2x-1| - |x+5| & \leq |(2x-1)-(x+5)| \\
\implies 3 &\leq |x-6| \\
\implies 3 &= |x-6| \\
\implies (x = 9) \ & \lor \ (x= 3 \ (\text{Reject})) \\
\implies x &= 9
\end{split}
\end{equation}
$$
However plugging the correct solution set to this equation is $x \in \{-\frac{7}{3}, 9\}$. Why does my attempted solution, not reach $-\frac{7}{3}$, as one of the possible solutions? Furthermore, how would you go about solving a problem of this nature?
|
Remember that $|x|=\begin{cases}x&\text{if}~x\geq 0\\ -x&\text{if}~x<0\end{cases}$
Now, we attempt to rewrite the original expression without absolute value signs. To do so, we ask ourselves when do we need to change the sign for each? On what regions will we need to change the sign of $|2x-1|$? The sign of $|x+5|$?
$2x-1<0$ when $x<\frac{1}{2}$.
$x+5<0$ when $x<-5$
We see then that the expression can be written without absolute value signs if broken into the following three regions: case 1:$x<-5$, case 2: $-5\leq x<\frac{1}{2}$, and case 3: $\frac{1}{2}\leq x$
$|2x-1|-|x+5| = \begin{cases} -(2x-1)-(-(x+5))&\text{when}~x<-5\\
-(2x-1)-(x+5)&\text{when}~-5\leq x<\frac{1}{2}\\
(2x-1)-(x+5)&\text{when}~\frac{1}{2}\leq x\end{cases}$
Setting each of these equal to three and solving for $x$ individually, and then checking that such a value for $x$ lies within its respective region will finish the problem.
|
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|
Proof of the second symmetric derivative Prove that if $f''(a)$ exists, then $$f''(a)=\lim_{h\to0}\frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.$$
I really have no idea on this one. Am I supposed to apply the mean value theorem?
|
Let us make the problem more general trying to approximate the second derivative based on three points $x+ah$, $x+bh$, $x+ch$.
By Taylor, we have $$f(x+kh)=f(x)+h k f'(x)+\frac{1}{2} h^2 k^2 f''(x)+\frac{1}{6} h^3 k^3
f^{(3)}(x)+\frac{1}{24} h^4 k^4
f^{(4)}(x)+O\left(h^5\right)$$ Now, consider $$F=Af(x+ah)+Bf(x+bh)+Cf(x+ch)$$ and apply the above formula for each of the terms and group terms; you should get
$$F=f(x) (A+B+C)+h f'(x) (a A+b B+c C)+\frac{1}{2} h^2 f''(x) \left(a^2 A+b^2 B+c^2
C\right)+\frac{1}{6} h^3 f^{(3)}(x) \left(a^3 A+b^3 B+c^3
C\right)+\frac{1}{24} h^4 f^{(4)}(x) \left(a^4 A+b^4 B+c^4 C\right)+O\left(h^5\right)$$ What we then want is to only get the term with $f''(x)$; so we need to cancel the previous terms. Then, this implies $$A+B+C=0\tag 1$$ $$a A+b B+c C=0\tag 2$$ $$\frac{1}{2} \left(a^2 A+b^2 B+c^2
C\right)=1\tag 3$$ So, three linear equations in $A,B,C$ to be solved. This gives $$A=\frac{2}{(a-b) (a-c)}\qquad B=\frac{2}{(b-a) (b-c)}\qquad C=\frac{2}{(c-a) (c-b)}$$ For your case $a=1,b-1,c=0$ which gives $A=1,B=1,C=-2$.
So, we have $$f(x+h)+f(x-h)-2f(x)=h^2 f''(x)+\frac{1}{12} h^4 f^{(4)}(x)+O\left(h^5\right)$$
The same procedure can apply to the derivative of any order.
|
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|
Decomposition of a rational function in partial fractions I am trying to decompose the following rational function: $\dfrac{1}{(x^2-1)^2}$ in partial fractions (in order to untegrate it later).
I have notices that $(x^2-1)^2 = (x+1)^2(x-1)^2$
Therefore $\exists A, B, C, D$ s.t: $\dfrac{1}{(x^2-1)^2} = \dfrac{1}{(x-1)^2(x+1)^2} = \dfrac{Ax+B}{(x+1)^2}+\dfrac{Cx+D}{(x-1)^2}$
We then have $Ax+B =\left. \dfrac{1}{(x-1)^2} \right\vert _{x=-1} \implies B-A=1/4$
Same thing for Cx+D: $Cx+D =\left. \dfrac{1}{(x+1)^2} \right\vert _{x=1} \implies C+D=1/4 $
How do I find A, B, C, D from here?
|
Hint: substitute $x=i $ to get two additional equations one for the real part and one for the imaginary part
|
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|
Intersection of plane How would I do this question.
Find a plane that contains the point A(3,1,−1) and touches the cylinder with radius 3 whose axis is the line p : x = 0, y = z.
It was on a test I did a few days ago and has been bugging me since as I don't even know where to approach this from. Any help would be much appreciated.
Thanks.
|
Assume the plane is $\pi:ax+by+cz+d=0$. The vector $(0,1,1)$ is on the cylinder axis so it is perpendicular to the normal of the plane:
$$(0,1,1)\cdot(a,b,c)=0\implies c=-b$$
Also, $A(3,1,-1)$ is on the plane so: $$3a+b-c+d=0 \implies b=-\frac{1}{2}d-\frac{3}{2}a$$
We know $(0,0,0)$ is on the axis of the cylinder so the distance between the plane and the point is 3. plugging this into the formula for distance between a plane and a point we get:
$$\frac{|d|}{\sqrt{a^2+b^2+c^2}}=3 \\ d^2 = 9a^2+9b^2+9c^2\\d^2=18(\frac{1}{2}d+\frac{3}{2}a)^2+9a^2\\99a^2+(54d)a+7d^2=0$$
solving this for $a$ gives us:
$$a_{1,2}=\frac{-54d\pm\sqrt{54^2d^2-4\cdot99\cdot7d^2}}{2\cdot99}=\frac{-54d\pm12d}{198}$$
$$a_1=-\frac{7}{33}d,\, a_2=-\frac{1}{3}d$$
Note: Remember that after squaring an equation we have to substitute the solutions in it and check that they really are solutions. In this case both solutions are correct.
$$$$
Case 1:
Substituting $a_1$ in equations for $b,c$ gives us $b = -\frac{2}{11}d,\, c= \frac{2}{11}d$. Plugging this into $\pi$ gives us: $$\pi:-(\frac{7}{33}d)x-(\frac{2}{11}d)y+(\frac{2}{11}d)z+d=0\\$$ $$\pi:7x+6y-6z-33=0\\$$
Case 2:
Substituting $a_2$ in equations for $b,c$ gives us $b = 0,\, c= 0$. Plugging this into $\pi$ gives us: $$\pi:-(\frac{1}{3}d)x + d=0\\$$ $$\pi:x-3=0\\$$
So the two planes are $\,7x+6y-6z-33=0,\, x-3=0$.
|
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|
Integrating the following $\int \sqrt{\tan x+1}\,dx$
Question: Integrate the following, $$\int\sqrt{\tan x+1}\;dx.$$
Wolfram Alpha returns a non-elementary answer. Can someone please spot the mistake I have made here:
First consider this integral:
$$\int \frac{1}{(x+1)\sqrt{x+3}} \, dx = -\sqrt{2}\tanh^{-1}\frac{\sqrt{x+3}}{\sqrt{2}} + c$$
Wolfram Alpha confirms that result.
Then, we have
$$I=\int \sqrt{\tan x+1} \, dx, \quad \tan x=u+2,
\quad dx=\frac{du}{\sec^{2}x}=\frac{du}{(u+2)^{2}-1}=\frac{dx}{(u+3)(u+1)}$$
So this transforms the integral to the first integral on this post, which we can evaluate. Then after evaluation and resubstitution I get:
$$I=-\sqrt{2}\tanh^{-1}\frac{\sqrt{\tan x+1}}{\sqrt{2}}+c$$
However differentiating this with Wolfram Alpha gives me a messy trigonometric expression which doesn't seem to be equal (I tested some values in both expressions and get different answers). I also estimated the area under the integral between some values and also obtained different answers using the closed form. Any ideas why?
EDIT: I used the wrong identity. Nevertheless, we can still use this method to integrate sqrt(tanhx integrals). E.g:
$$I=\int \sqrt{\tanh x+1} \, dx, \tanh x=u+2,\quad
-dx = \frac{du}{\operatorname{sech}^2 x} = \frac{du}{(u+2)^2-1} = \frac{dx}{(u+3)(u+1)}$$
To obtain:
$\int \sqrt{\tanh x+1} \, dx = I=\sqrt{2}\tanh^{-1} \dfrac{\sqrt{\tanh x+1}}{\sqrt{2}}+c$
|
This integral is not non-elementary. Following from @Harry Peter's hint, we have
$$ \begin{align} \frac{2u^2}{u^4 - 2u^2 + 2} &= \frac{2}{u^2 + \dfrac{2}{u^2} - 2} \\ &= \frac{1 - \dfrac{\sqrt{2}}{u^2}}{\left(u+\dfrac{\sqrt{2}}{u}\right)^2-2\sqrt{2}-2} + \frac{1 + \dfrac{\sqrt{2}}{u^2}}{\left(u - \dfrac{\sqrt{2}}{u}\right)^2 + 2\sqrt{2} - 2 }
\end{align} $$
Then you can substitute $s = u + \dfrac{\sqrt{2}}{u}$ and $t = u - \dfrac{\sqrt{2}}{u}$, respectively. The first integral is in terms of logarithms and the second is in terms of arctangent. As you can see, this can be expressed in terms of elementary functions.
|
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|
vector of eigenvalues is an eigenvector When is it the case that the vector $\begin{bmatrix} \lambda_1 \\ \lambda_2 \\ ... \end{bmatrix}$ of eigenvalues of a matrix is in fact an eigenvector of that matrix?
|
To make our lives easier, let's try $2\times 2$:
$$\left \{ \begin{align*}
\begin{pmatrix} a & b \\ c & d \end{pmatrix}
\begin{pmatrix} u \\ v \end{pmatrix} = u
\begin{pmatrix} u \\ v \end{pmatrix} \\[5pt]
\begin{pmatrix} a & b \\ c & d \end{pmatrix}
\begin{pmatrix} v \\ u \end{pmatrix} = v
\begin{pmatrix} v \\ u \end{pmatrix}
\end{align*} \right.$$
On solving, we have
\begin{align*}
\begin{pmatrix} a & b \\ c & d \end{pmatrix} &=
\begin{pmatrix} u & v \\ v & u \end{pmatrix}
\begin{pmatrix} u & 0 \\ 0 & v \end{pmatrix}
\begin{pmatrix} u & v \\ v & u \end{pmatrix}^{-1} \\[5pt] &=
\begin{pmatrix}
\frac{u^2+uv+v^2}{u+v} & -\frac{uv}{u+v} \\
\frac{uv}{u+v} & \frac{uv}{u+v}
\end{pmatrix}
\end{align*}
For $u\begin{pmatrix} v \\ u \end{pmatrix}$ and
$v\begin{pmatrix} u \\ v \end{pmatrix}$ convention:
\begin{align*}
\begin{pmatrix} a & b \\ c & d \end{pmatrix} &=
\begin{pmatrix} v & u \\ u & v \end{pmatrix}
\begin{pmatrix} u & 0 \\ 0 & v \end{pmatrix}
\begin{pmatrix} v & u \\ u & v \end{pmatrix}^{-1} \\[5pt] &=
\begin{pmatrix}
\frac{uv}{u+v} & \frac{uv}{u+v} \\
-\frac{uv}{u+v} & \frac{u^2+uv+v^2}{u+v}
\end{pmatrix}
\end{align*}
For $n$ distinct assigned eigenvalues, there are $n!$ permutations of eigenvectors, so there are $\displaystyle \, n! \binom{n!}{n}$ of possible matrices.
|
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|
How to prove $|a_{2n}-a_{n}|<\frac{10}{27}$ Let sequence $$a_{1}=1,a_{n+1}=\dfrac{2}{2a_{n}+1}$$
show that
$$\left|a_{2n}-a_{n}\right|<\dfrac{10}{27}$$
and the constant $\frac{10}{27}$ A smaller number instead
$$a_{n+1}+\dfrac{1+\sqrt{17}}{4}=2(1+\sqrt{17})\cdot\dfrac{a_{n}+\frac{1+\sqrt{17}}{4}}{4(2a_{n}+1)}\tag{1}$$
$$a_{n+1}-\dfrac{-1+\sqrt{17}}{4}=-2(-1+\sqrt{17})\cdot\dfrac{a_{n}-\frac{-1+\sqrt{17}}{4}}{4(2a_{n}+1)}\tag{2}$$
$\dfrac{(1)}{(2)}$,that
$$\dfrac{a_{n+1}+\dfrac{1+\sqrt{17}}{4}}{a_{n+1}-\dfrac{-1+\sqrt{17}}{4}}=\dfrac{1+\sqrt{17}}{1-\sqrt{17}}\cdot\left(\dfrac{a_{n}+\dfrac{1+\sqrt{17}}{4}}{a_{n}-\dfrac{-1+\sqrt{17}}{4}}\right)$$
so we I get ugly even more can't solve this problem.can you some recommended or solve this?
|
The function $f(x)=\frac{1}{x+1/2}$ maps $[2/3,1]$ to $[2/3,6/7] \subset [2/3,1]$, since $1/3 < 10/27$ the result follows.
|
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Evaluate the limit $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$
Evaluate the limit $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$, without using a Riemann sum
$\bf{My\; Try:}$ Using the graph of $\displaystyle f(x) = \frac{1}{\sqrt{x}}\;,$ we get
$$\int_{1}^{n+1}\frac{1}{\sqrt{x}}dx <\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<1+\int_{1}^{n}\frac{1}{\sqrt{x}}dx$$
So we get $$2\sqrt{n+1}-2<\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<2\sqrt{n}-1$$
So $$\lim_{n\rightarrow \infty}\frac{2\sqrt{n+1}-2}{\sqrt{n}}=2<\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<\lim_{n\rightarrow \infty}\frac{2\sqrt{n}-1}{\sqrt{n}} = 2$$
So using the Sandwich Theorem, we get $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)=2$
My question is can we solve it by using any other method? If yes then please explain it here.
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By Cesaro-Stolz
$$\ldots = \lim_{n \to \infty} \frac{\sum_{k=1}^{n+1} \frac{1}{\sqrt{k}}- \sum_{k=1}^{n} \frac{1}{\sqrt{k}} }{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}(\sqrt{n+1} + \sqrt{n})}{n+1 - n} = 2 $$
Also, there is another approach using the squeeze theorem that parallels your argument without appealing to the integral.
Since $\sqrt{k} + \sqrt{k-1} < 2\sqrt{k} < \sqrt{k} + \sqrt{k+1}$ we find the bounds
$$2(\sqrt{k+1} - \sqrt{k})=\frac{2}{\sqrt{k+1} + \sqrt{k}} \leqslant \frac{1}{\sqrt{k}} \leqslant \frac{2}{\sqrt{k} + \sqrt{k-1}} = 2(\sqrt{k} - \sqrt{k-1}).$$
As sums of the LHS and RHS are telescoping, we have
$$2(\sqrt{n+1} - 1) \leqslant \sum_{k=1}^n \frac{1}{\sqrt{k}} \leqslant 2\sqrt{n},$$
and
$$2(\sqrt{1+1/n} - 1/\sqrt{n}) \leqslant \frac{1}{\sqrt{n}}\sum_{k=1}^n \frac{1}{\sqrt{k}} \leqslant 2.$$
Now apply the squeeze theorem.
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"timestamp": "2023-03-29T00:00:00",
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|
Suppose $-a\sin(s) - b\cos(s) = 0$, then $a^2 + b^2 = 1$?
Suppose $-a\sin(s) - b\cos(s) = 0$ and given that $a^2 + b^2 \le 1$, then $a^2 + b^2 = 1$?
I am having trouble getting the above identity. I vaguely recall that
$$a\sin(s) + b\cos(s)=\sqrt{a^2+b^2}\sin(s+\theta)$$
where $$\tan(\theta) = \frac{b}{a}$$
But I still don't see why it's obvious?
|
Let us see the identity you vaguely recall:
$$a\sin(s) + b\cos(s)=\sqrt{a^2+b^2}\sin(s+\theta)$$
with $\tan\theta=\frac ba$. To get it, we factor out the square root, then set $\sin\theta=\frac{b}{\sqrt{a^2+b^2}}$ and $\cos\theta=\frac{a}{\sqrt{a^2+b^2}}$, which is licit since the squares of those numbers sum to 1. Now we have:
$$a\sin s+b\cos s=\sqrt{a^2+b^2}(\cos\theta\sin s+\sin\theta\cos s),$$
and $\sin(s+\theta)$ expands to just that using addition formula.
So the identity at question top now becomes:
$$\sqrt{a^2+b^2}\sin(s+\theta)=0.$$
I have, of course, changed signs, which doesn't alter the equation. This can never be true for all $s$ unless $a=b=0$, in which case obviously $a^2+b^2=0\neq1$.
Another answer suggests the problem might have a $1$ at the start of the equations, and that the identity works only for some $t$. Then we would have, for some $t$, that:
$$1-\sqrt{a^2+b^2}\sin(t+\theta)=0\iff\sin(t+\theta)=\frac{1}{\sqrt{a^2+b^2}}.$$
$a^2+b^2\leq1$, so the RHS is at least 1, but the LHS is at most 1 in absolute value, so for the equation to hold you need $a^2+b^2=1$.
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_0^\infty\frac{1-e^{-x}(1+x )}{x(e^{x}-1)(e^{x}+e^{-x})}dx$
\begin{equation}
\int_0^\infty\frac{1-e^{-x}(1+x )}{x(e^{x}-1)(e^{x}+e^{-x})}dx
\end{equation}
My colleague got this problem from his friend but he didn't know the answer so he asked my help. Unfortunately, after hours of tired effort I was unable to crack this integral. I was unable to find a way to evaluate it from online search either. I used to be good at solving this kind of problem but now I feel so embarrassed by my stupidity. I'm stuck and I badly need your help. It's a humbling request to ask people here being so kind to help me out. Thank you.
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I answer my own OP instead of improving it as a proof to user @mickep that I did give a try to this problem but I didn't want to post some useless efforts like "I tried substitution $x=\tan y$ then I failed... miserably". I consider putting this kind of effort is a a complete joke as shown in some posts with integration tag.
Okay, here is my try. Following user @Anastasiya-Romanova秀's suggestion, I use the substitution $y=e^{-x}$ and the original integral becomes
\begin{equation}
\int_0^1\frac{y(y-1)-y^2\log y}{(1-y)(1+y^2)\log y}dy=-\int_0^1\frac{y}{1+y^2}\left(\frac{y}{1-y}+\frac{1}{\log y}\right)dy
\end{equation}
Now, we consider the following parametric integral
\begin{equation}
I(a):=\int_0^1\frac{y^{a-1}}{1+y^2}\left(\frac{y}{1-y}+\frac{1}{\log y}\right)dy
\end{equation}
Differentiating $I(a)$ we get
\begin{align}
I'(a)& =\int_0^1\left(\frac{\log y}{1-y}+1\right)\frac{y^{a-1}}{1+y^2}dy\\
&=\int_0^1\frac{y^{a}\log y}{1-y^4}dy+\int_0^1\frac{y^{a+1}\log y}{1-y^4}dy+\int_0^1\frac{y^{a-1}}{1+y^2}dy\\
&=\sum_{k=0}^\infty\int_0^1\left(y^{4k+a}\log y+y^{4k+a+1}\log y+(-1)^ky^{2k+a-1}\right)dy\\
&=\sum_{k=0}^\infty\left(-\frac{1}{(4k+a+1)^2}-\frac{1}{(4k+a+2)^2}+\frac{(-1)^k}{2k+a}\right)\\
&=\frac{1}{16}\left(-\psi_1\!\left(\frac{a+1}{4}\right)-\psi_1\!\left(\frac{a+2}{4}\right)+4\psi\!\left(\frac{a+2}{4}\right)-4\psi\!\left(\frac{a}{4}\right)\right)
\end{align}
where I use the following relation
\begin{equation}
\sum_{k=0}^\infty\frac{(-1)^k}{(z+k)^{m+1}}=\frac1{(-2)^{m+1}m!}\!\left(\psi_m\left(\frac{z}{2}\right)-\psi_m\!\left(\frac{z+1}{2}\right)\right)
\end{equation}
Hence
\begin{equation}
I(a)=\int_{\infty}^aI'(a)\ da=-\frac{1}{4}\psi\!\left(\frac{a+1}{4}\right)-\frac{1}{4}\psi\!\left(\frac{a+2}{4}\right)+\log\Gamma\!\left(\frac{a+2}{4}\right)-\log\Gamma\!\left(\frac{a}{4}\right)
\end{equation}
Hence our considered problem is $\color{red}{-I(2)}$ which confirms user @ClaudeLeibovici's result.
\begin{align}
\int_0^\infty\frac{1-e^{-x}(1+x )}{x(e^{x}-1)(e^{x}+e^{-x})}dx&=\frac{1}{4}\psi\!\left(\frac{3}{4}\right)+\frac{1}{4}\psi\!\left(1\right)+\log\Gamma\!\left(\frac{1}{2}\right)\\[10pt]
&=\frac{1}{8}\left(\pi+\log\left(\frac{\pi ^4}{64}\right)-4\gamma\right)
\end{align}
where I use special value of the digamma function
\begin{equation}
\psi\!\left(\frac{1}{4}\right)=-\frac{\pi}{2}-3\log2-\gamma
\end{equation}
and the reflection formula
\begin{equation}
\psi\!\left(1-x\right)-\psi\!\left(x\right)=\pi\cot\pi x
\end{equation}
Done!
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|
Integrate $\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx$ $\displaystyle\int \dfrac {x^{2}} {\sqrt {x^{2}-16}}dx$
Effort 1:
Let be $x=4\sec u$
$dx=4.\sin u.\sec^2u.du$
Then integral;
$\displaystyle\int \dfrac {\sec^2u \; .4.\sin u.\sec^2u.du} {\sqrt {16\sec^2u-16}}=\displaystyle\int \sec^3.du$
After I didn't nothing.
Effort 2:
$\displaystyle\int \dfrac {x^{2}dx} {\sqrt {x^{2}-16}}$
Let's doing integral by parts;
$du=\dfrac{x}{\sqrt{x^2-16}}$
$v=x$
$\displaystyle\int \dfrac {x^{2}dx} {\sqrt {x^{2}-16}}=x.\sqrt{x^2-16}-\displaystyle\int\sqrt{x^2-16}dx$
We have $\quad \displaystyle\int\sqrt{x^2-16}dx$
let be $\quad x=4\sec j$
$dx=4\dfrac{\sin j}{\cos^2 j}dj$
and;
$\displaystyle\int\sqrt{x^2-16}\;dx=16.\displaystyle\int\dfrac{\sin j}{\cos j}\dfrac{\sin j}{\cos^2 j}dj=16.\displaystyle\int \sec j.tan^2j.dj$
After I didn't nothing.
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Note that
\begin{align}
\int \frac{x^2}{\sqrt{x^2-16}} dx &= \int \frac{(x^2-16)+16}{\sqrt{x^2-16}}dx \\
&=\int \sqrt{x^2-16}dx + 16 \int \frac{dx}{\sqrt{x^2-16}} \\
&= \sqrt{x^2-16} x - \int \frac{x^2}{\sqrt{x^2-16}}dx+ 16 \int \frac{dx}{\sqrt{x^2-16}}dx
\end{align}
Hence
$$\int \frac{x^2}{\sqrt{x^2-16}} = \frac{1}{2} x \sqrt{16-x^2} + 8 \int \frac{dx}{\sqrt{x^2-16}}dx. $$
One verifies that
$$\int \frac{dx}{\sqrt{x^2-16}} = \ln\left(\sqrt{x^2-16}+x\right)$$
Therefore
$$\int \frac{x^2}{\sqrt{x^2-16}} dx = \frac{1}{2} x \sqrt{16-x^2} + 8 \ln\left(\sqrt{x^2-16}+x\right)+C$$
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|
Sum of combinatorics sequence $\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1}$ I need to find sum like $$\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1},\qquad \text{ for even } n$$
Example:
Find the sum of $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$
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Hint:
Consider the difference between:
$$ \sum_{k=0}^{20}\binom{20}{k}1^k = (1+1)^{20}\quad\text{and}\quad \sum_{k=0}^{20}\binom{20}{k}(-1)^k = (1-1)^{20}.$$
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"timestamp": "2023-03-29T00:00:00",
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|
What is the meaning of $X^{2}$? Can you let me know the meaning of $X^{2}$ where X is a random variable? (Please, give me an example in real life and explain the meaning of $X^{2}$?)
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Here's a simple example. Suppose I give you an unusual six-sided die. It's not loaded, so any of the six sides is equally likely to be rolled, but the faces are numbered $$1, 1, 1, 2, 2, 3.$$ So, for any given roll of the die, you assign the random variable $X$ to the outcome you observe. Then you would say $$\begin{align*} \Pr[X = 1] &= \frac{3}{6} = \frac{1}{2}, \\ \Pr[X = 2] &= \frac{2}{6} = \frac{1}{3}, \\ \Pr[X = 3] &= \frac{1}{6}. \end{align*}$$
But what if you were interested not in the number value of the roll, but the square of the value rolled? You'd first notice that the support of $X$ is the set $\{1, 2, 3\}$; that is to say, the only possible outcomes you can obtain from a single die roll are $\{1, 2, 3\}$. But the support of $X^2$ is $\{1, 4, 9\}$, because if $X = 1$, then $X^2 = 1$; and if $X = 2$, then $X^2 = 4$; and if $X = 3$, then $X^2 = 9$. And because we know the value of $X^2$ when we know $X$ itself, it is intuitively obvious that $$\begin{align*} \Pr[X^2 = 1] &= \Pr[X = 1] = \frac{1}{2}, \\ \Pr[X^2 = 4] &= \Pr[X = 2] = \frac{1}{3}, \\ \Pr[X^2 = 9] &= \Pr[X = 3] = \frac{1}{6}. \end{align*}$$ In this situation, we are dealing with a discrete random variable $X$ with finite support, and it so happens that the relationship between $X$ and $X^2$ is one-to-one: that is to say, $X^2$ is uniquely determined from $X$, but also, $X$ is uniquely determined from $X^2$. So for example, if you rolled the die and told me that the square of the value you obtained is $X^2 = 9$, I'd tell you that you must have seen $X = 3$.
But now suppose I give you a different die, and although it is also unloaded, the numbers on its faces are $$-1, -1, 1, 1, 2, 2.$$ Then you let $Y$ be the random variable for the value you obtain when rolling this die; you'd state $$\begin{align*} \Pr[Y = -1] &= \frac{1}{3}, \\ \Pr[Y = 1] &= \frac{1}{3}, \\ \Pr[Y = 2] &= \frac{1}{3}. \end{align*}$$ But because $(-1)^2 = 1^2 = 1$, the support of $Y^2$ is only $\{1, 4\}$. You would say $$\begin{align*} \Pr[Y^2 = 1] &= \Pr[Y = 1] + \Pr[Y = -1] = \frac{2}{3}, \\ \Pr[Y^2 = 4] &= \Pr[Y = 2] = \frac{1}{3}. \end{align*}$$ In this case, if you rolled the die and only told me that $Y^2 = 1$, I would not be able to tell whether the value you rolled was $1$ or $-1$, because both of these outcomes would give the same squared value.
When we deal with continuous random variables, or other transformations of random variables besides squaring, the basic idea is much the same as in the discrete-valued example above. For a general random variable $X$ following some parametric probability distribution, and some function $f$ that is well-defined on the support of $X$, the expression $f(X)$ is also a random variable, whose support is the image of the support of $X$ under $f$. For instance, if the support is $$X \in \{ x \in \mathbb R : -1 \le x \le 1 \},$$ and $f(x) = 3x + 5$, then $$f(X) \in \{y \in \mathbb R : 2 \le y \le 8\}.$$
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"timestamp": "2023-03-29T00:00:00",
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|
Find $3^{333} + 7^{777}\pmod{ 50}$ As title say, I need to find remainder of these to numbers. I know that here is plenty of similar questions, but non of these gives me right explanation. I always get stuck at some point (mostly right at the beginning) and don't have idea how to start.
Thanks in advance.
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First, let us compute $\varphi(50)=\varphi(2\cdot 25)=\varphi(2)\varphi(25)=20$. By Euler's theorem, if $a$ is relatively prime to 50, $a^{20}\equiv 1 \pmod{50}$. Therefore
$$3^{333}+7^{777}\equiv 3^{16*20+13}+7^{38*20+17}=1^{16}*3^{13}+1^{38}*7^{17}\pmod{50}.$$
Thus, we have reduced the problem to computing $3^{13} \pmod{50}$ and $7^{17}\pmod{50}$. There are a number of ways to proceed from here, but one that I like is repeated squaring. I will demonstrate the procedure for one of the terms.
Write $13$ in base $2$ as $13=8+4+1=1+2(2(1+2))$. Then $3^{13}=3*((3^{1+2})^2)^2$. We then compute $3^3=27$, $27^2\equiv 29\pmod{50}$, $29^2\equiv (-21)^2\equiv 41 \equiv -9 \pmod{50}$, and finally $3*(-9)\equiv -27 \pmod{50}$. With this method, we can compute $a^b \pmod{n}$ in less than $2\log_2 \varphi(n)$ steps where each step is either a squaring or a multiplication by $a$, and we never have any intermediate terms bigger than $n^2$.
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|
Solutions to $\frac1{\lfloor x\rfloor}+\frac1{\lfloor 2x\rfloor}=\{x\}+\frac13$
Find all solutions to $$\dfrac{1}{\lfloor x\rfloor}+\dfrac{1}{\lfloor 2x\rfloor}=\{x\}+\dfrac{1}{3}$$
$$$$
Unfortunately I have no idea as to how to go about this. On rearranging, I got $$3\lfloor 2x\rfloor = 3\lfloor x\rfloor\{x\}-2\lfloor x\rfloor$$
I'm not sure about what to do with the $3\lfloor 2x\rfloor $ term; I'd prefer to resolve it in terms of $\lfloor x\rfloor $ but am not able to. All that struck me was using the identity for $\lfloor nx\rfloor, n\in \Bbb Z$. However on first glance, it did not strike me as particularly useful.$$$$
I would be grateful for any help. Many thanks!
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I decided to do @mathlove's case $2$.
CASE $2$
We see that $n \gt 0$.
$$\dfrac{1}{n} + \dfrac{1}{2n + 1}= \delta + \frac 13$$
$$\dfrac{3n+1}{2n^2+n} - \dfrac 13 = \delta$$
$$ \delta = \frac{-2n^2 + 8n + 3}{6n^2 + 3n}$$
\begin{align}
0 \le \frac{-2n^2 + 8n + 3}{6n^2 + 3n} \lt 1 \\
0 \le -2n^2 + 8n + 3 \lt 6n^2 + 3n \\
\end{align}
\begin{align}
0 &\le -2n^2 + 8n + 3 \\
2n^2 - 8n - 3 &\le 0 \\
n^2 - 4n &\le \frac 32 \\
(n - 2)^2 &\le \frac{11}{2} \\
(n - 2)^2 &\le 4 \\
n &\in \{1,2,3,4\}
\end{align}
\begin{align}
-2n^2 + 8n + 3 &\lt 6n^2 + 3n \\
8n^2 - 5n - 3 &\gt 0 \\
256n^2 - 160n &\gt 96 \\
(16n - 5)^2 &\gt 96 \\
16n - 5 &\gt 10 \\
16n &\gt 15 \\
n &\ge 1
\end{align}
$n= -1 \implies \delta = -\frac 73$
$n=1 \implies \delta = 1$
$n=2 \implies \delta = \frac{11}{30} \implies x = 2\frac{11}{30}$
$n=3 \implies \delta = \frac 17 \implies x = 3\frac 17$
$n=4 \implies \delta = \frac{1}{36} \implies x = 1\frac{1}{36}$
$n=5 \implies \delta = - \frac{7}{165}$
So $n \in \{ 2\frac{11}{30}, 3\frac 17, 1\frac{1}{36}\}$
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$25$ men are employed to do a work $25$ men are employed to do a work, which they could finish it in $20$ days but the drop off by $5$ men at the end of every $10$ days. In what time will the work be completed?
My Attempt
In $20$ days, $25$ men can do $1$ work.
In $1$ day, $25$ men can do $\frac {1}{20}$ work.
In $1$ day, $1$ man can do $\frac {1}{20\times 25}$ work.
Now how can I proceed further?
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As you've said in one day a man can complete $\frac{1}{500}$ of the job.
The amount of work done every 10 days is then given by $\frac{1}{500}\times10\times(30-5t)$ where $t$ is how many lots of 10 days have passed (starting from $t=1$).
Adding this from $t=1$ to $t=n$ gives:
$$\sum_{t=1}^n\frac{1}{500}\times10(30-5t)=\sum_{t=1}^n\frac{3}{5}-\frac{1}{10}t$$
$$=\frac{3}{5}n-\frac{1}{20}n(n+1)$$
$$=\frac{11}{20}n-\frac{1}{20}n^2$$
Then we need to solve this equal to $1$ to complete the job.
$$1=\frac{11}{20}n-\frac{1}{20}n^2$$
$$n^2-11n+20=0$$
$$n=\frac{11\pm\sqrt{11^2-4\times1\times20}}{2}$$
$$n=\frac{11\pm\sqrt{41}}{2}$$
$$n\approx2.3,8.7$$
So the job will get completed in between 20 and 30. To work out exactly we need to do the individual sums.
First ten days: $\frac{1}{500}\times10\times25=\frac{1}{2}$
Second ten days: $\frac{1}{500}\times10\times20=\frac{2}{5}$, total so far: $\frac{9}{10}$.
Remaining: $\frac{1}{10}$. Days needed for this: $\frac{1}{500}\times15\times n=\frac{1}{10}$ gives $n=\frac{10}{3}$ so $3$ and $\frac{1}{3}$ days.
So the total is $23$ and $\frac{1}{3}$ days.
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|
If $9 ≥ 4x + 1$, which inequality represents the possible range of values of $12x + 3?$ If $9 ≥ 4x + 1$, which inequality represents the possible range of values of $12x + 3$?
I've been trying to do SAT prep, and I came across this question. It allowed me to show an explanation and it still didn't make any sense to me.
"If we look closely, we see that $12x+3=3(4x+1)$."
What did they do to get $12x+3=3(4x+1)?$
The answer choices:
A. $12x+3≥17$
B. $12x+3≤17$
C. $12x+3≥27$
D. $12x+3≤27$
Thank you for your time.
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$\begin{array}{llr}12x+3 &= 1\cdot (12x + 3)&\text{due to properties of 1}\\
&=3\cdot\frac{1}{3}\cdot(12x+3)&\text{by replacing 1 with}~3\cdot\frac{1}{3}\\
&=3\cdot(\frac{1}{3}\cdot 12x + \frac{1}{3}\cdot 3)&\text{by distributivity of multiplication over addition}\\
&=3\cdot(4x+1)&\text{by evaluating}~\frac{1}{3}\cdot 12~\text{and}~\frac{1}{3}\cdot 3\end{array}$
Going in reverse is faster:
$3(4x+1)=3\cdot 4x + 3\cdot 1 = 12x+3$, again by the distributivity property of multiplication over addition.
In general $a(b+c)=ab+ac$
|
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|
Why does an argument similiar to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$ show that $2+4+8+...=-2$ See how to prove $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$
$x=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
$2x=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
Then:
$x=1$
Now I use the same argument to prove $2+4+8+...=-2$
$x=2+4+8+...$
$2x=4+8+16+...$
Then:
$x=-2$
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You can't subtract divergent series.
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|
Minimizing $\cot^2 A +\cot^2 B + \cot^2 C$ for $A+B+C=\pi$
If $A + B + C = \pi$, then find the minimum value of $\cot^2 A +\cot^2 B + \cot^2 C$.
I don't know how to solve it. And can you please mention the used formulas first.
What I can see is that if one of the angles $A$, $B$, $C$ is small, then the value $\cot^2A$ or $\cot^2B$ or $\cot^2C$ will be big. So I want to make angles big (more precisely, close to $\pi/2$, where cotangent is zero), but the condition $A+B+C=\pi$ prevents me from making all three of them very big .
I can see that if $A=B=C=\frac\pi3$, then I get $\cot A=\cot B=\cot C=\frac1{\sqrt3}$ and $\cot^2A+\cot^2B+\cot^2C=1$. But I do not know whether this is indeed minimum. (According to WolframAlpha this is the minimum. However, I would like to see some proof of this fact.)
|
Put $x=\cot A,y=\cot B$. Using the standard formulae we have $z=\cot C=-\cot(A+B)=\frac{1-xy}{x+y}$. So we want to minimize $$f(x,y)=\frac{(x^2+y^2)(x+y)^2+(xy-1)^2}{(x+y)^2}$$ where $x,y$ can take any real values.
If we want to minimize $(x^2+y^2)(x+y)^2+x^2y^2-2xy+1$ subject to $x+y=k$, then using Lagrange multipliers we find $4x^3-2y+6x^2y+6xy^2+2y^3=\lambda=-2x+2x^3+6yx^2+6xy^2+4y^3$ and so $x=y$ or $x^2+xy+y^2=0$. Note that this result is independent of $k$. Note also that $f(x,y)$ is large for $|x|$ or $|y|$ large, so the minimum will occur at a stationary point.
If $x=y$ then we want to minimize $2x^2+\frac{(x^2-1)^2}{4x^2}{4x^2}$. Differentiating, this has a minimum at $x=\frac{1}{\sqrt3}$ that makes $A=B=C=\frac{\pi}{3}$ and the minimum value 1.
If $x^2+xy+y^2=0$, then $-xy=x^2+y^2\ge0$. But $0\le(x+y)^2=x^2+xy+y^2+xy=xy$, so we must have $xy=0$ and hence also $x^2+y^2=0$ and so $x=y=0$, which clearly does not yield a minimum.
----------Added later, courtesy of Martin Sleziak ---------
I am not particularly happy with the argument above, which seems like using a hammer to crack a nut. The OP in this earlier question In a $\triangle ABC,$ Evaluation of minimum value of $\cot^2 A+\cot^2 B+\cot^2 C$ provided a much simpler argument:
By AM/GM we have $\cot^2A+\cot^2B\ge 2\cot A\cot B$, and similarly $\cot^2B+\cot^2C\ge 2\cot B\cot C$ and $\cot^2C+\cot^2A\ge 2\cot C\cot A$ so we have $$\cot^2A+\cot^2B+\cot^2C\ge \cot A\cot B+\cot B\cot C+\cot C\cot A$$ But now by the formula at the start we have $\cot C(\cot A+\cot B)=1-\cot A\cot B$ which gives immediately $$\cot^2A+\cot^2B+\cot^2C\ge1$$
Instead of using AM/GM we could also use the rearrangement inequality.
If you like that argument upvote his question rather than this answer!
|
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|
Area of the triangle formed by circumcenter, incenter and orthocenter Lets say we have $\triangle$$ABC$ having $O,I,H$ as its circumcenter, incenter and orthocenter. How can I go on finding the area of the $\triangle$$HOI$.
I thought of doing the question using the distance (length) between $HO$,$HI$ and $OI$ and then using the Heron's formula, but that has made the calculation very much complicated. Is there any simple way to crack the problem?
|
Given any triangle $\triangle ABC$, we will
abuse notation and use the same letter to represent
both a vertex and the angle at that vertex. Let
*
*$a, b, c$ be the side lengths $|BC|$, $|CA|$ and $|AB|$.
*$L = a + b + c$ be the perimeter.
*$c_X, s_X, t_X$ be $\cos X, \sin X, \tan X$ for any angle $X \in \{ A, B, C \}$.
*$R$ be the circumradius.
Let $I, O, H$ be the incenter, circumcenter and orthocenter.
Their barycentric coordinates are given by
$$\begin{array}{ccccc}
\alpha_I : \beta_I : \gamma_I &=& \sin A : \sin B : \sin C &=& t_A c_A : t_B c_B : t_C c_C\\
\alpha_O : \beta_O : \gamma_O &=& \sin 2A : \sin 2B : \sin 2C &=& 2t_A c_A^2 : 2t_B c_B^2 : 2t_C c_C^2\\
\alpha_H : \beta_H : \gamma_H &=& \tan A : \tan B : \tan C &=& t_A : t_B : t_C
\end{array}$$
Let $\mathcal{A}_0$ and $\mathcal{A}$ be the area of $\triangle ABC$ and $\triangle IOH$, their ratio is given by
$$\frac{\mathcal{A}}{\mathcal{A}_0} =
\left|\det\begin{bmatrix}
\alpha_I & \beta_I & \gamma_I\\
\alpha_O & \beta_O & \gamma_O\\
\alpha_H & \beta_H & \gamma_H
\end{bmatrix}\right|
= \frac{\mathcal{N}}{\delta_I\delta_O\delta_H}
$$
where $\;\begin{cases}
\delta_I &= \sin A + \sin B + \sin C\\
\delta_O &= \sin 2A + \sin 2B + \sin 2C\\
\delta_H &= \tan A + \tan B + \tan C
\end{cases}
\;$ and
$$\begin{align}\mathcal{N}
&= \left|\det\begin{bmatrix}
t_A c_A & t_B c_B & t_C c_C\\
2t_A c_A^2 & 2t_B c_B^2 & 2 t_C c_C^2\\
t_A & t_B & t_C
\end{bmatrix}\right|
= 2t_A t_B t_C
\left|\det\begin{bmatrix}
c_A & c_B & c_C\\
c_A^2 & c_B^2 & c_C^2\\
1 & 1 & 1
\end{bmatrix}\right|\\
&= 2t_A t_B t_C |(c_A - c_B)(c_B - c_C)(c_C - c_A)|
\end{align}
$$
Since $A + B + C = \pi$, $\delta_H = t_A + t_B + t_C = t_A t_B t_C$.
Together with the relations,
$$\begin{cases}
\mathcal{A}_0 &= \frac12 R^2(\sin 2A + \sin 2B + \sin2C)\\
L &= 2R(\sin A + \sin B + \sin C)
\end{cases}
\quad\implies\quad
\begin{cases}
\delta_O = \frac{2\mathcal{A}_0}{R^2}\\
\delta_I = \frac{L}{2R}
\end{cases}
$$
We find
$$\frac{\mathcal{A}}{\mathcal{A}_0} = \frac{2}{\delta_I\delta_O}|(c_A - c_B)(c_B - c_C)(c_C - c_A)|
= \frac{2R^3}{\mathcal{A}_0L}|(c_A - c_B)(c_B - c_C)(c_C - c_A)|
$$
Notice
$$c_A - c_B = \frac{-a^2 + b^2 + c^2}{2bc} - \frac{a^2-b^2+c^2}{2ac}
= \frac{(b-a)L(L-2a)}{2abc}$$
and similar expressions for $c_B - c_C, c_C - c_A$, we find
$$\frac{\mathcal{A}}{\mathcal{A}_0}
= \left(\frac{2R^3}{\mathcal{A}_0L}\cdot\frac{L^3(L-2a)(L-2b)(L-2c)}{8a^3b^3c^3}\right)|(a-b)(b-c)(c-a)|
$$
Recall the Heron's formula and a beautiful relation between $\mathcal{A}_0$ and $R$:
$$16\mathcal{A}_0^2 = L(L-2a)(L-2b)(L-2c)\quad\text{ and }\quad
4\mathcal{A}_0 R = abc
$$
What's inside the parenthesis above can be simplified as
$$\frac{2R^3}{\mathcal{A_0}L}\cdot\frac{L^2\cdot16\mathcal{A_0}^2}{8(4\mathcal{A}_0R)^3}
= \frac{L}{16A_0^2} = \frac{1}{(L-2a)(L-2b)(L-2c)}$$
This leads to a reasonably simple ratio one can use to compute the area $\mathcal{A}$.
$$\frac{\mathcal{A}}{\mathcal{A}_0} = \frac{|(a-b)(b-c)(c-a)|}{(-a+b+c)(a-b+c)(a+b-c)}$$
|
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|
Proof from Matrix I have tried upto where I could. please anyone help me to complete my proof..
Help much appreciated
|
I therefore ends your calculations:
$\left(
\begin{array}{cc}
1 & \frac{\sin \frac \alpha 2}{\cos \frac \alpha 2} \\
-\frac{\sin \frac \alpha 2}{\cos \frac \alpha 2} & 1
\end{array}
\right) \left(
\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}
\right) =\left(
\begin{array}{cc}
\cos \alpha +\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\sin \alpha &
-\sin \alpha +\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\cos \alpha
\\
-\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\cos \alpha +\sin \alpha
& \cos \alpha +\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\sin \alpha
\end{array}
\right) $ ;
so developing the expressions, in the last matrix, using trigonometric forms,
$\sin \alpha =2\sin \frac \alpha 2\cos \frac \alpha 2$
$\cos \alpha =2\cos ^2\frac \alpha 2-1$
$\cos ^2\frac \alpha 2+\sin ^2\frac \alpha 2=1$,
the result is obtained as
$\left(
\begin{array}{cc}
1 & -\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha } \\
\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha } & 1
\end{array}
\right) $
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to arrive at Ramanujan's nested radicals? Ramanujan found that $\sqrt[3]{\cos\left(\frac {2\pi}{7}\right)}+\sqrt[3]{\cos\left(\frac {4\pi}{7}\right)}+\sqrt[3]{\cos\left(\frac {8\pi}{7}\right)}=\sqrt[3]{\frac {1}{2}\left(5-3\sqrt[3]{7}\right)}$ on the last page of Ramanujan's second notebook.
He also found that $\sqrt[3]{\sec\left(\frac {2\pi}{9}\right)}+\sqrt[3]{\sec\left(\frac {4\pi}{9}\right)}-\sqrt[3]{\sec\left(\frac {\pi}{9}\right)}=\sqrt[3]{6\sqrt[3]{9}-6}$
My question: How do we find the denesting with only the knowledge on the nested radical? Example: How would you find the denestings for $\sqrt[3]{\frac {1}{2}\left(5-3\sqrt{7}\right)}$ without knowledge on $\sqrt[3]{\cos\left(\frac {2\pi}{7}\right)}+\sqrt[3]{\cos\left(\frac {4\pi}{7}\right)}+\sqrt[3]{\cos\left(\frac {8\pi}{7}\right)}$?
|
Those identities are not difficult to prove once established. For instance, $\alpha=\cos\left(\frac{2\pi}{7}\right),\beta=\cos\left(\frac{4\pi}{7}\right),\gamma=\cos\left(\frac{6\pi}{7}\right)=\cos\left(\frac{8\pi}{7}\right)$ are algebraic conjugates, roots of the polynomial:
$$ p(x) = 8x^3+4x^2-4x-1.$$
If follows that $\alpha^{1/3},\beta^{1/3},\gamma^{1/3}$ are roots of the polynomial:
$$ q(x) = 8x^9-4x^6-4x^3-1 $$
and it is not difficult to compute, from $q(x)$, a polynomial that vanishes over $\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}$.
With some god-like inspiration, one may notice that
$$ r(x) = 4x^9-30x^6+75x^3+32 $$
is a factor of such polynomial. Then the conjecture
$$ \alpha^{1/3}+\beta^{1/3}+\gamma^{1/3} = \sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}} $$
is natural, since $r(x)$ is the minimal polynomial of the RHS, and it is straightforward to check.
However, how Ramanujan actually saw that is a mystery to me. Probably the term genius was forged for a reason.
|
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|
how to solve $x^{113}\equiv 2 \pmod{143}$ I need to solve $x^{113} \equiv 2 \pmod{143}$
$$143 = 13 \times 11$$
I know that it equals to $x^{113}\equiv 2 \pmod{13}$ and $x^{113}\equiv 2 \pmod{11}$
By Fermat I got
1) $x^{5} \equiv 2 \pmod{13}$
2) $x^{3} \equiv 2 \pmod{11}$
Now I'm stuck..
|
Brute-force approach:
We want to find an $x$ such that $$x^{113} \equiv 2 \pmod{143}$$
We split $\mathbb{Z}_{143}$ into $\mathbb{Z}_{13} \times \mathbb{Z}_{11}$ since $13\cdot 11 = 143$ and $11,13$ are prime. Then, we compute the result of the congruences in $\pmod{13}$ and $\pmod{11}$, that is we solve
$$x^{113} \equiv 2 \pmod{13}$$
$$x^{113} \equiv 2 \pmod{11}$$
On which you already correctly applied Fermat, so you can reduce the exponent always $\pmod{p-1}$ with the prime modulus. Now you solve
$$x^{5} \equiv 2 \pmod{13}$$
$$x^{3} \equiv 2 \pmod{11}$$
By hand, to get
$$x \equiv 6 \pmod{13}$$
$$x \equiv 7 \pmod{11}$$
CRT goes like this: You know two congruences of $x$ in two moduli, that is
$$x \equiv a \pmod{p}$$
$$x \equiv b \pmod{q}$$
Where $p,q$ are relatively prime to each aother. From this, we can deduce the congruence of $x$ modulo $p\cdot q$ with
$$x \equiv a\cdot (q^{-1} \text{ mod }p)\cdot q + b\cdot (p^{-1} \text{ mod } q)\cdot p \pmod{pq} $$ (Follows from https://en.wikipedia.org/wiki/Chinese_remainder_theorem)
Through @Merlin's answer, we know that in this case
$$x \equiv 6 \pmod{13}$$
$$x \equiv 7 \pmod{11}$$
We calculate the needed inverses by using the extended eucledian algorithm (EEA):
$$13^{-1} \equiv 6 \pmod{11} $$
$$11^{-1} \equiv 6 \pmod{13} $$
We reconstruct $x$:
\begin{align} x &\equiv a\cdot (q^{-1} \text{ mod }p)\cdot q + b\cdot (p^{-1} \text{ mod } q)\cdot p \pmod{pq}\\
&\equiv 6\cdot6\cdot11 + 7\cdot6\cdot 13 \equiv 942\pmod{11\cdot13} \\
&\equiv 84 \pmod{143}
\end{align}
We check that this is indeed a solution by direct computation with wolfram alpha:
So yes, this checks out.
|
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|
When is $\sum_{i=1}^n a_i^{-2}=1$?
For which natural numbers $n$ do there exist $n$ natural numbers $a_i\ (1\le i\le n)$ such that $\displaystyle\sum_{i=1}^n a_i^{-2}=1$?
I didn't see an easy way of solving this. There is a solution for $n=1$ since we just take $a_i = 1$ and a solution for $n=4$ if we take $a_i = 2$. How do we find all possible values of $n$?
|
If $n$ works, then $n+3$ and $n+8$ works.
Proof. Take a term, $a_i$, and replace that with $4$ copies of $2a_i$. Take a term, $a_i$, and replace that with $9$ copies of $3a_i$. Both work.
So, since $1$ works, all numbers $\equiv1\pmod 3$ work.
Since $9$ works ($a_i=3\forall i$), all numbers $\equiv0\pmod 3$ greater than or equal $9$ work.
Since $1$ works, $1+8+8=17$ works, all numbers $\equiv2\pmod 3$ greater than or equal $17$ work.
Now these are the exceptions we must check: $2, 3, 5, 6, 8, 11, 14$.
Clearly $2, 3$ are impossible as $\frac{1}{a_i^2}\leq\frac{1}{4}$ for $a_i>2$, making $\sum\frac{1}{a_i^2}<1$.
For $5$, one of the $\frac{1}{a_i^2}\geq\frac{1}{5}$, so $a_i=2$ for some $i$. We have $4$ numbers adding to $\frac{3}{4}$. One of the $\frac{1}{a_i^2}\geq\frac{3}{16}>\frac{1}{9}$, so $a_i=2$ for another $i$. We have $3$ numbers adding to $\frac{1}{2}$. One of the $\frac{1}{a_i^2}\geq\frac{1}{6}>\frac{1}{9}$, so $a_i=2$ for another $i$. We have $2$ numbers adding to $\frac{1}{4}$. One of the numbers $\geq\frac{1}{8}$, so impossible.
$6$ is actually possible. $1=3\times\frac{1}{4}+2\times\frac{1}{9}+1\times\frac{1}{36}$
$8$ is actually possible. $1=2\times\frac{1}{4}+4\times\frac{1}{9}+2\times\frac{1}{36}$, so $8+3=11$ and $11+3=14$ are possible.
So now, the updated set of possible numbers is all integers, except for $2, 3, 5$.
|
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|
Determine matrix of linear map Linear map is given through:
$\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $
$\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$
Determine matrix $A$ linear map.
Here I have solution but I dont understand how to get it.
$A=\begin{pmatrix} -3 & -3 \\ -2 & 4 \end{pmatrix} $
|
One possibility is "brute force" solution: you know $A\pmatrix{3 \\ -2} = \pmatrix{-3 \\ -14}, A\pmatrix{3 \\ 0} = \pmatrix{-9 \\ -6}$, that's four equations for four unknown elements, solvable in a number of ways. This approach is inelegant and quickly becomes hard for bigger dimensionality, but is perfectly doable.
The other is that you know matrix of $\phi$ in basis $b' = (\pmatrix{3 \\ -2}, \pmatrix{3 \\ 0})$ (it is $A' = \pmatrix{-3 & -9 \\ -14 & -6}$) and you want to learn what matrix it has in basis $b_0 = (\pmatrix{1 \\ 0}, \pmatrix{0 \\ 1})$. That would be $A'*\pmatrix{3 & 3 \\ -2 & 0}^{-1} = A'*\pmatrix{0 & -{1 \over 2} \\ 1 \over 3 & 1 \over 2} = \pmatrix{-3 & -3 \\ -2 & 4}$.
|
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|
Evaluation of Irrational Integral
Evaluation of $$\int\frac{x^4}{(1-x^4)^{\frac{3}{2}}}dx$$
$\bf{My\; Try::}$ Let $$I = \int\frac{x^4}{(1-x^4)^{\frac{3}{2}}}dx = -\frac{1}{4}\int x\cdot \frac{-4x^3}{(1-x^{4})^{\frac{3}{2}}}dx$$
Using Integration by parts, We get
$$I =\frac{x}{2(1-x^4)^{\frac{1}{2}}}-\int\frac{1}{(1-x^4)^{\frac{1}{2}}}dx$$
Now How can I solve after that, Help required, Thanks
|
$$
\int\frac{1}{\sqrt{1-x^4}}\,dx=\int\frac{1}{\sqrt{1+x^2}}\frac{1}{\sqrt{1-x^2}}\,dx
$$
Let $u=\arcsin x$. Then $du=1/\sqrt{1-x^2}\,dx$, and we get
$$
\int\frac{1}{\sqrt{1+\sin^2 u}}\,du=F(-1\,|\, u)+C=F(-1\,|\,\arcsin x)+C
$$
where $F$ denotes an elliptic integral of first kind. Also, see wiki for notational variations.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Deriving formula from Fourier series: $\frac{\pi^2}{12} = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ The equation/formula $$ \frac{\pi^2}{12} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$$ is to be derived.
I know that the Fourier expansion of $f(x)=x$ for $x \in (-\pi,\pi)$ is $$f(x)=x=\sum_{n=1}^{\infty} \frac{2\,(-1)^{n+1}}{n}\,\sin{nx}$$ and there was an example of how for $x=\frac{\pi}{2}$ you'd get $\frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}- \cdots = \sum_{l=0}^\infty \frac{(-1)^{l}}{2l+1}$, but here it's not the square of $n$. I got a Fourier series for $$f_1(x)=x^2=\sum_{n=1}^{\infty} \frac{4\,(-1)^n}{n^2}\,\cos{nx}$$ but then the exponent of the $(-1)$ isn't the same.
I'm obviously missing something, so any hints are appreciated.
|
This does not directly answer the question posed in the OP regarding the use of Fourier Series to prove that
$$\frac{\pi^2}{12}=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} \tag 1$$
But, I thought it might be instructive to present an approach that relies only on the Basel Problem
$$\frac{\pi^2}{6}=\sum_{n=1}^\infty \frac{1}{n^2} \tag 2$$
which was proven by Euler without use of Fourier Series. To that end, we proceed.
Note that we can write the series in $(1)$ as
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}-\frac{1}{(2n)^2}\right) \tag 3$$
Alongside this result, we can write the series in $(2)$ as
$$\begin{align}
\frac{\pi^2}{6}&=\sum_{n=1}^\infty \frac{1}{n^2} \\\\
&=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}+\frac{1}{(2n)^2}\right) \\\\
&=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}-\frac{1}{(2n)^2}\right)+2\sum_{n=1}^\infty\frac{1}{(2n)^2} \\\\
&=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}-\frac{1}{(2n)^2}\right)+\frac12\sum_{n=1}^\infty\frac{1}{n^2} \\\\
&=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}-\frac{1}{(2n)^2}\right)+\frac{\pi^2}{12}\\\\
\frac{\pi^2}{12}&=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}-\frac{1}{(2n)^2}\right)\tag 4
\end{align}$$
Substituting $(4)$ into $(3)$ yields the coveted result.
|
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|
If $x= m-m^2-2$ then find $x^4+3x^3+2x^2-11x+6$ where m is a cube root of unity If $$x= m-m^2-2$$ then find $$x^4+3x^3+2x^2-11x+6$$ where $m$ is a cube root of unity.
My try:
Since $ m+ m^2+1=0$ the value of $x$ is $-1$.
Let $f(x)=x^4+3x^3+2x^2-11x+6$
then $ f(-1)=5$ So the answer is $5$.
Am I correct? In my book it says that answer is $1$. Please help me to identify my mistake.
|
If $m$ is a primitive cube root of unity, $m^2+m+1=0$ is correct, but that just gives
$$ x = m-m^2-2 = -2\pm i\sqrt{3}. $$
Given $p(z)=z^4+3z^3+2z^2-11z+6$, we have:
$$ q(z) = p(z-2) = z^4-5z^3+8z^2-15z+28 $$
and
$$ p(x) = q(\pm i\sqrt{3}) =\color{red}{13}.$$
|
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|
Find a matrix $B$ such that $B^3 = A$
$$A=\begin{pmatrix} 1 & -1 \\ -2 & 1 \end{pmatrix}$$
Find a matrix $B$ such that $B^3$ = A
My attempt:
I found $\lambda_1= 1+{\sqrt 2}$ and $\lambda_2= 1-{\sqrt 2}$
I also found their corresponding eigenvectors $\vec v_1 =\begin{pmatrix} \frac{-\sqrt 2}{2} \\ 1 \end{pmatrix}$ and $\vec v_2 = \begin{pmatrix} \frac{\sqrt 2}{2} \\ 1 \end{pmatrix}$
I know the Power function of a matrix formula $A=PDP^{-1}$
Because it's the cubed root I'm looking for I don't know how to get the cubed root of the eigenvaules and keep the maths neat. Is there another way to solve this problem or an I going the wrong way about doing it ?
|
you find a matrix $P=\left(
\begin{array}{cc}
\frac{-\sqrt{2}}2 & \frac{\sqrt{2}}2 \\
1 & 1
\end{array}
\right) $ diagonalizes $ A $ that is
$P^{-1}AP=\left(\begin{array}{cc}
1+\sqrt{2} & 0 \\
0 & 1-\sqrt{2}
\end{array}\right) $,
then we look
for a simple matrix $ M =PBP^{-1} $ such that it is diagonal and $
M^3 = D$ an simple solution is $M=\left(
\begin{array}{cc}
\sqrt[3]{\left( 1+\sqrt{2}\right) } & 0 \\
0 & -\sqrt[3]{\left( -1+\sqrt{2}\right) }
\end{array}
\right) $ and so $B=PDP^{-1}$ is a solution. Precisely $$B= \left(
\begin{array}{cc}
\frac 12\sqrt[3]{\left( 1+\sqrt{2}\right) }-\frac 12\sqrt[3]{\left( -1+\sqrt{%
2}\right) } & -\frac 14\sqrt{2}\sqrt[3]{\left( 1+\sqrt{2}\right) }-\frac 14%
\sqrt{2}\sqrt[3]{\left( -1+\sqrt{2}\right) } \\
-\frac 12\sqrt{2}\sqrt[3]{\left( 1+\sqrt{2}\right) }-\frac 12\sqrt{2}\sqrt[3%
]{\left( -1+\sqrt{2}\right) } & \frac 12\sqrt[3]{\left( 1+\sqrt{2}\right) }%
-\frac 12\sqrt[3]{\left( -1+\sqrt{2}\right) }
\end{array}
\right) $$
|
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|
How to sketch the region on the complex plane? I am going through a basic course on complex analysis. I have a problem in understanding the following.
E $\subset\mathbb{C}$ is defined as $$E := \{z\in\mathbb{C}:\vert z+i \vert = 2\vert z\vert \}$$ I want to know if this set is connected, closed, bounded but I do not know how to sketch this set or visualize it.
|
I would write $z=x+\mathrm i y$ and then see what equation I got.
\begin{eqnarray*}
|z+\mathrm i| &=& 2|z| \\ \\
|(x+\mathrm iy)+\mathrm i| &=& 2|x+\mathrm iy| \\ \\
|x+(1+y)\mathrm i|&=& 2|x+\mathrm iy| \\ \\
\sqrt{x^2+(1+y)^2} &=& 2\sqrt{x^2+y^2} \\ \\
x^2+(1+y)^2 &=& 4(x^2+y^2) \\ \\
x^2+1+2y+y^2 &=& 4x^2+4y^2 \\ \\
1 &=& 3x^2 + 3y^2 -2y \\ \\
\frac{1}{3} &=& x^2 + y^2 - \frac{2}{3}y \\ \\
\frac{1}{3} &=& x^2 + \left(y - \frac{1}{3}\right)^2 - \frac{1}{9} \\ \\
\frac{4}{9} &=& x^2 + \left(y - \frac{1}{3}\right)^2
\end{eqnarray*}
You have a circle, centre $(0,\frac{1}{3})$ with radius $\frac{2}{3}$.
|
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|
When is $2^m3^n +1$ the square of some integer?
Find all pairs of natural numbers $(m, n)$ for which $2^m3^n +1$ is the square of some integer.
The powers of $2$ modulo $10$ cycle as $2,4,8,6,\ldots$ and the powers of $3$ as $3,9,7,1,\ldots$. The square of am integer needs to end in $0,1,4,5,6,$ or $9$. Thus, $2^m3^n$ ends in $4$ or $8$. Therefore, if $m \equiv 1 \pmod{4}$, then $n \equiv 2 \pmod{4}$; if $m \equiv 2 \pmod{4}$ then $n \equiv 3,0 \pmod{4}$; if $m \equiv 3 \pmod{4}$, then $n \equiv 1,0 \pmod{4}$; if $m \equiv 0 \pmod{4}$, then $n \equiv 1,2 \pmod{4}$. Do I preform casework from here or is there an easier way?
|
$2^m3^n+1=k^2$ for some $k\in\mathbb Z^+$ is true if and only if $2^m3^n=(k+1)(k-1)$.
Natural numbers may or may not include $0$ (see Wikipedia or MathWorld). I'll assume $0\in\mathbb N$.
By Euclidean algorithm: $$\gcd(k+1,k-1)$$
$$=\gcd((k+1)-(k-1),k-1)$$
$$=\gcd(2,k-1)$$
$$=\gcd(2,k+1)$$
If $m=0$, then $\gcd(k+1,k-1)=1$, so, since also $k+1>k-1$, we get $k+1=3^n$ and $k-1=1$, so $k=2$, $n=1$. We get the solution $(m,n)=(0,1)$.
If $m\ge 1$, then $2\mid (k+1)(k-1)$, so by Euclid's Lemma either $2\mid k+1$ or $2\mid k-1$, i.e. either $\gcd(2,k+1)=2$ or $\gcd(2,k-1)=2$. In any case, $\gcd(k+1,k-1)=2$, so $$2^{m-2}3^n=\left(\frac{k+1}{2}\right)\left(\frac{k-1}{2}\right),$$
where $\gcd\left(\frac{k+1}{2},\frac{k-1}{2}\right)=1$, so there are two cases:
$1)\ \ \ $ $\frac{k+1}{2}=2^{m-2}$ and $\frac{k-1}{2}=3^n$. Then $2^{m-2}=3^n+1>1$, so $m-2\ge 1$. If $n=0$, then $(m,n)=(3,0)$. Let $n\ge 1$. Then $(-1)^{m-2}\equiv 1\pmod{3}$, so $m-2=2t$ for some $t\in\mathbb Z^+$, so $$\left(2^t+1\right)\left(2^t-1\right)=3^n$$ $$\gcd\left(2^t+1,2^t-1\right)=\gcd\left(2,2^t-1\right)=1,$$
also $2^t+1>2^t-1$, so $2^t+1=3^n$ and $2^t-1=1$, so $t=1$, $n=1$, so $m=4$. We get the solution $(m,n)=(4,1)$.
$2)\ \ \ $ $\frac{k+1}{2}=3^n$, $\frac{k-1}{2}=2^{m-2}$. Then $3^n=2^{m-2}+1>1$, so $n\ge 1$, so $m-2\ge 1$. If $m-2=1$, we get $(m,n)=(3,1)$. If $m-2\ge 2$, we get $(-1)^{n}\equiv 1\pmod{4}$, so $n=2h$ for some $h\in\mathbb Z^+$, so
$$\left(3^h+1\right)\left(3^h-1\right)=2^{m-2}$$
$$\gcd\left(3^h+1,3^h-1\right)=\gcd\left(2,3^h-1\right)=2$$
Therefore, since $3^h+1>3^h-1$ and $m\ge 4$, we get $3^h+1=2^{m-3}$ and $3^h-1=2$, so $h=1$, $m=5$, so $n=2$, so $(m,n)=(5,2)$.
Answer: $(m,n)=(0,1),(3,0),(4,1),(3,1),(5,2)$.
|
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|
Roots of $y=x^3+x^2-6x-7$ I'm wondering if there is a mathematical way of finding the roots of $y=x^3+x^2-6x-7$?
Supposedly, the roots are $2\cos\left(\frac {4\pi}{19}\right)+2\cos\left(\frac {6\pi}{19}\right)+2\cos\left(\frac {10\pi}{19}\right)$, $2\cos\left(\frac {2\pi}{19}\right)+2\cos\left(\frac {14\pi}{19}\right)+2\cos\left(\frac {16\pi}{19}\right)$ and $2\cos\left(\frac {8\pi}{19}\right)+2\cos\left(\frac {12\pi}{19}\right)+2\cos\left(\frac {18\pi}{19}\right)$.
Anything helps. I don't think substituting $x$ with something like $t+t^{-1}$ will work.
|
By the translation $x\to x-\frac13$, you cancel the square term.
$$(x-\frac13)^3+(x-\frac13)^2-6(x-\frac13)-7=x^3-\frac{19x}3-\frac{133}{27}.$$
Then multiplying by $27$ and replacing $x\to3x$, you get
$$x^3-57x-133=0.$$
Next, setting $x=2\sqrt{19}\cos(\theta)$ (with the aim to let $4\cos^3(\theta)-3\cos(\theta)=\cos(3\theta)$ appear),
$$152\sqrt{19}\cos^3(\theta)-114\sqrt{19}\cos(\theta)-133=38\sqrt{19}(4\cos^3(\theta)-3\cos(\theta))-133=0$$
so that
$$\cos(3\theta)=\frac{133}{38\sqrt{19}}.$$
Retrieving the three roots from here is straightforward.
This solution is due to François Viète. You can see it as an instance of the famous "angle trisection" problem.
Another approach is by decomposing $x=u+v$. You can develop the equation as
$$u^3+3uv(u+v)+v^3-57(u+v)-133=0.$$
Then setting $3uv=57$, two terms cancel out and you end up with the system
$$\begin{cases}uv=19\iff u^3v^3=6859\\u^3+v^3=133.\end{cases}$$
This is a sum/product problem and
$$u^3,v^3=\frac{133\pm\sqrt{133^2-4\cdot6850}}2.$$
But the story ends here, because these solutions are complex numbers, which is harmless in itself, but you need to take their cubic roots and this requires the use of... trigonometric functions. This is called the casus irreductibilis as radicals don't suffice.
|
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|
Prove inequality $\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}>1$
For any $n\ge2, n \in \mathbb N$ prove that
$$\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}>1$$
My work so far:
1) $$\sqrt{n+1}-\sqrt{n}>\frac1{2\sqrt{n+0.5}}$$
2) $$\sqrt{n+1}-\sqrt{n}<\frac1{2\sqrt{n+0.375}}$$
|
Another proof is this:
Note that
$$
2 = \sqrt{\frac{4n}{n}} = \frac{1}{\sqrt{n}}\sum_{j=0}^{4n-1}\sqrt{j+1}-\sqrt{j}
$$
where the RHS can be expressed as
$$
\frac{1}{\sqrt{n}}\left(\sum_{j=1}^{2n}(\sqrt{2j}-\sqrt{2j-1})+\sum_{j=1}^{2n-1}(\sqrt{2j-1}-\sqrt{2j-2})\right)
$$
Using that the function $f:(0,+\infty)\to \mathbb{R}$ given by $f(x)=\sqrt{x+1}-\sqrt{x}$ is strictly decreasing, we deduce, for all $j\in \{1,...,2n\}$,
$$
\sqrt{2j-1}-\sqrt{2j-2}>\sqrt{2j}-\sqrt{2j-1}
$$
hence
$$
2\sum_{j=1}^{2n}\left(\sqrt{\frac{2j-1}{n}}-\sqrt{\frac{2j-2}{n}}\right)>2
$$
|
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|
Let $a, b, c>0$, such that $a+b+c=1$, prove that $\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\ge\frac{9}{4}$ Let $a, b, c>0$, such that $a+b+c=1$, prove that:
$$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\ge\frac{9}{4}$$
|
Hint: $$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}=$$
$$=\frac{a}{(1-a)^2}+\frac{b}{(1-b)^2}+\frac{c}{(1-c)^2}$$
Let $f(t)=\frac t{1-t^2}$
Use Jensen's Inequality
|
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|
Express the function $ f $ without using absolute value signs $\left|\frac{x-2}{x+3}\right|e^{\left|x-2\right|}$? Good evening to everyone:
This is the equation $$ f(x) = \left|\frac{x-2}{x+3}\right|e^{\left|x-2\right|} $$ What I've tried is: $$ \frac{x-2}{x+3}\ge 0 => x-2 \ge 0 => x \ge 2$$ Then $$ \frac{-x+2}{-x-3} < 0 => -x+2 <0 => x > 2$$ These 2 combined give the result $$ \left|\frac{x-2}{x+3}\right| = \frac{-x+2}{-x-3} $$ for $x>2 $ and $$x-2 \ge 0 => x \ge 2$$ and $$ -x+2<0 => -x<-2 => x>2$$ therefore the result is $$ \left|x-2\right| =-x+2$$ for $x>2$. So the equation becomes $$ f(x) = \frac{-x+2}{-x-3}e^{-x+2} $$ for $ x>2 $ .But my teacher says it's not right. Can you clarify it for me please. Thanks for any possible response.
Edit: The real answers are: $$\frac{x-2}{x+3}e^{2-x}\:$$ for $ x<3$ $$ \frac{2-x}{x+3}e^{2-x}\: $$ for $-3 < x \le 2 $ and $$ \frac{x-2}{x+3}e^{x-2} $$ for $ x > 2 $
|
$$\frac{x-2}{x+3}\ge 0\stackrel{\text{Mult. by}\;(x+3)^2}\iff (x-2)(x+3)\ge0\;,\;\;x\neq-3\iff$$
$$x<-3\;\;\text{or}\;\;x\ge2$$
and then
$$\left|\frac{x-2}{x+3}\right|e^{|x-2|}=\begin{cases}\frac{x-2}{x+3}e^{x-2},&x\ge2\\{}\\\frac{x-2}{x+3}e^{-(x-2)},&x<-3\end{cases}$$
And thus
$$-3<x<2\implies \left|\frac{x-2}{x+3}\right|e^{|x-2|}=-\frac{x-2}{x+3}e^{-(x-2)}$$
|
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|
If $x^2+y^2 \equiv 0\pmod{p}$, then $p \equiv 1 \pmod{4}$
Prove that if $x^2+y^2 \equiv 0\pmod{p}$ where $p$ is a prime and $x,y$ are not both divisible by $p$, then $p \equiv 1 \pmod{4}$.
I tried using that $x^2 \equiv -y^2 \pmod{p}$ and conjectured that $-1$ must a quadratic residue modulo $p$, but I am not sure how that would help.
|
Since $p \not \mid y$, there exists integer $y'$ such that $yy' \equiv 1 \pmod p$. Then multiply both side of the congruence with $y'$ to get that:
$$(xy')^2 \equiv -(yy')^2 \equiv - 1 \pmod p$$
So hence $-1$ is a quadratic residue modulo $p$. To prove that $p \equiv 1 \pmod 4$ from this note that:
$$1 \equiv x^{p-1} \equiv (x^2)^{\frac{(p-1)}{2}} \equiv (-1)^{\frac{(p-1)}{2}} \pmod p$$
This implies that $\frac{(p-1)}{2}$ is even and hence $p \equiv 1 \pmod 4$
|
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|
Evaluation of $\int_{0}^{\sqrt{2}-1}\frac{\ln(1+x^2)}{1+x}dx$
Evaluation of $$\int_{0}^{\sqrt{2}-1}\frac{\ln(1+x^2)}{1+x}dx$$
$\bf{My\; Try:::}$ Let $$I(a) = \int_{0}^{\sqrt{2}-1}\frac{\ln(1+ax^2)}{1+x}dx$$
Now $$I'(a) = \int_{0}^{\sqrt{2}-1}\frac{x^2}{(1+ax^2)(1+x)}dx = \frac{1}{a}\int_{0}^{\sqrt{2}-1}\frac{(1+ax^2)-1}{(1+ax^2)(1+x)}dx$$
So $$I'(a) = \frac{\ln 2}{2a}-\frac{1}{a^2}\int_{0}^{\sqrt{2}-1}\frac{1}{(x^2+\frac{1}{a})(1+x)}dx$$
So $$I'(a) = \frac{\ln 2}{2a}-\frac{1}{a^2(a+1)}\int_{0}^{\sqrt{2}-1}\left[\frac{1-x}{x^2+\frac{1}{a}}-\frac{1}{1+x}\right]dx$$
Now How can i solve after that, Help required, Thanks
|
Since the number $\sqrt{2}-1$ is invariant under the transformation $x \mapsto \frac{1-x}{1+x}$, it is natural to make that substitution. Doing so, we find:
$$\begin{align} I &= \int_0^{\sqrt{2}-1} \frac{\ln(1+x^2)}{1+x} dx
\\&= \int_{\sqrt{2}-1}^1 \cfrac{\ln(1+x^2)+\ln\left(\frac2{(1+x)^2}\right)}{1+x}dx
\\&=\frac12 I + \frac12 \int_{\sqrt{2}-1}^1 \frac{\ln(1+x^2)}{1+x} dx +\frac12 \int_{\sqrt{2}-1}^1 \frac{\ln2 - 2 \ln(1+x)}{1+x} dx
\\&=\frac12 \int_0^1 \frac{\ln(1+x^2)}{1+x} dx -\frac18 \ln^2 2
\\&= \frac{\ln^2 2}{4}-\frac{\pi^2}{96}.\end{align}$$
You can check the evaluation $$\int_0^1 \frac{\ln(1+x^2)}{1+x} dx =\frac34 \ln^2 2- \frac{\pi^2}{48}$$ here, for example.
|
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|
What are the differences between: $\sqrt{(-3)^2}$, $\sqrt{-3^2}$ and $(\sqrt{-3})^2$. First, is $\sqrt{-3}$ is equal to $-3$ or is it imaginary?
What is the difference between:
*
*$\sqrt{(-3)^2}$
*$\sqrt{-3^2}$
*$(\sqrt{-3})^2$
Can I write $(\sqrt{-3})^2 = -3$?
And, given the rule that $\sqrt{a^n}$ is equal to ($\sqrt{a})^n$, can I say that $\sqrt{-3^2}=-3$?
|
$\sqrt{(-3)^2}=\sqrt{9}=3$,
$\sqrt{-3^2}=\sqrt{-9}=3i$,
$(\sqrt{-3})^2=(\sqrt{3}i)^2=-3$
The rule is true only in the case $a>0$.
|
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|
Compute $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\theta$ I''m stuck in a exercise in complex analysis concerning integration of rational trigonometric functions. I have the solution but I don't understand a specific part. Here it goes:
We want to find $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\theta$.
Let $z=e^{i\theta}$ so that $d\theta=dz/iz$, $\cos \theta = \frac12 (z+z^{-1})$ and $\sin \theta = \frac {1}{2i}(z-z^{-1})$. We have
$$\begin{align}
I&=\int_0^{2\pi} \frac{\sin^2\theta}{5+3\cos \theta}d\theta\\\\
&=\oint_C \frac{-\frac14(z-z^{-1})^2}{5+3\frac12 (z+z^{-1})}\frac{dz}{iz}\\\\
&=\oint_C \frac{\frac i4 (z^2-2+z^{-2})}{5z+ \frac32 z^2 + \frac32}dz\\\\
&=\oint_C \frac{\frac i2 (z^2-2+z^{-2})}{3z^2 + 10z + 3}dz\\\\
&=\frac i2 \oint_C \frac{z^4-2z^2+1}{z^2(3z^2 + 10z + 3)}dz \quad (1)
\end{align}$$
where $C$ is the unit circle in the complex $z$-plane.
If I understand the last step correctly we multipliy numerator/denominator by $z^2$ in order to get rid of the negative power of $z$.
Now for the part that I don't understand in the solution. It is said that
$$(1) = \frac i2 \oint_C \frac{z^4-2z^2+1}{z^23(z+3)(z+\frac13)}dz = \frac i6 \oint_C \frac{z^4-2z^2+1}{z^2(z+3)(z+\frac13)}dz$$
Where does this $3$ come from in the denominator? If I factor out $3z^2 + 10z + 3$ I simply get $(z+3)(z+\frac13)$.
What is even stranger to me is that, after the computation of residues, the author is getting the correct solution of $\frac{2\pi}{9}$ (I verified with Wolfram) while I'm getting $\frac{2\pi}{3}$ so I'm missing a factor of $\frac13$ (i.e. I'm missing that $3$ in the denominator).
Can someone help me with this?
|
$$I=\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\theta=\int_{0}^{\pi}\sin^2(\theta)\left(\frac{1}{5+3\cos\theta}+\frac{1}{5-3\cos\theta}\right)\,d\theta$$
hence:
$$ I = 10 \int_{0}^{\pi}\frac{\sin^2\theta}{25-9\cos^2\theta}\,d\theta = 20\int_{0}^{\pi/2}\frac{\cos^2\theta}{25-9\sin^2\theta}d\theta $$
and by setting $\theta=\arctan t$ we get:
$$ I = 20\int_{0}^{+\infty}\frac{dt}{(1+t^2)(25+16t^2)}=\frac{20}{9}\int_{0}^{+\infty}\left(\frac{1}{1+t^2}-\frac{16}{25+16t^2}\right)\,dt=\color{red}{\frac{2\pi}{9}}. $$
|
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|
Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$
Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$
Attempt:
$\star$ denotes $3\cdot 2^n+2\cdot 3^n$
$$\text{for }n=1:\quad\star\equiv 5\not\equiv 1\\
\text{for }n=2:\quad\star\equiv 5\not\equiv 1\\
\text{for }n=3:\quad\star\equiv 3\not\equiv 1\\
\text{for }n=4:\quad\star\equiv 0\not\equiv 1\\
\text{for }n=5:\quad\star\equiv \color{red}1\\
\text{for }n=6:\quad\star\equiv 5\not\equiv 1\\
\vdots$$
this is how should I aprroach this? or maybe there is smarter way?
|
As $2^3\equiv1,3^3\equiv-1\pmod7$
let us start with $n=3a,3a+1,3a+2$
Case$\#1:$ $n=3a$
$$1\equiv3\cdot2^{3a}+2\cdot3^{3a}\equiv3+2(-1)^a$$
$\iff2(-1)^a\equiv-2\iff(-1)^a\equiv-1$ as $(2,7)=1$
$\implies a$ is odd $=2b+1$(say) $\implies n\equiv3\pmod6$
Case$\#2:$ $n=3a+1$
$$1=3\cdot2^{3a+1}+2\cdot3^{3a+1}\equiv6(1+(-1)^a)$$
$\iff1+(-1)^a\equiv6^{-1}\equiv6$ as $6^2\equiv1\pmod7$
$\iff(-1)^a\equiv5\pmod7$ which is impossible as $5\not\equiv\pm1\pmod7$
Case$\#3:$ $n=3a+2$
Left for you!
|
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|
Factorization of $x^n + y^n$, what sort of coefficients show up? We know that$$a^2 + b^2 = (a + bi)(a - bi).$$ What are the complete factorizations of $a^3 + b^3$, $a^4 + b^4$, $\ldots$ , $a^k + b^k$, etc.? What sort of coefficients show up?
|
You could always write $a^n+b^n = a^n - c^n$ where $c = \xi_n b$, where $\xi_n = e^{\pi i/n}$ (so that $\xi_n^n = -1$). Then fixing $a$, the roots of $a^n - c^n$ are solutions to $a^n = c^n$ and so there are $n$ solutions for $c$; namely, the numbers in the form of $\zeta_n^k a$ where $0\leq k\leq n-1$ and $\zeta = e^{2\pi i/n}$ so that $\zeta_n^n = 1$. So the original polynomial factors completely over $\Bbb{C}$ as
$$(c - a)(c - \zeta_na)(c - \zeta_n^2a)...(c - \zeta_n^{n-1}a)$$
Such $\zeta_n$ is called an $n^{th}$ root of unity. There are always $n$ $n^{th}$ roots of unity, namely, the distinct powers of $\zeta_n$.
|
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|
If A and B are positive real numbers and each of the equations: $x^2+ax+2b=0$ and $x^2+2bx+a=0$ has real roots, what is the smallest value of A+B Problem
In the equation:$x^2+ax+2b=0$ and $x^2+2bx+a=0$, we have to figure out the sum of a and b by using the following identity:
$P(x)=Q(x)*D(x)+R(x)$
where P(x) is the equations and q(x) is the quotient and d(x) is the divisor and r(x) is the remainder.
My though process
my initial stance was that because these two equations are both equal to zero, then they must equal each other so i set
$x^2+ax+2b=x^2+2bx+a$
which simplifies into:
$ax+2b=2bx+a$
then i knew that that didn't go anywhere so then i set x=0 in the first equation and thought that B was zero but that can't be true
So then i assigned R and S as the roots to the equations where i figured that in general given a simple quadratic equation: $ax^2+bx+c$ with the roots r and s, $r+s=\frac{-b}{a}$ and $rs=\frac{c}{a}$
and going back to the real equation we can set
$r+s=\frac{-a}{1}$ in the equation $x^2+ax+2b=0$ and $rs=\frac{2b}{1}$
and with the same thing to the other equation I got:
$r+s=\frac{-2b}{1}$ and $rs=\frac{a}{1}$
as stated these two equation can be set equal to each other because they are both equal to 0.
This means that $\frac{-a}{1}$=$\frac{-2b}{1}$
and $\frac{2b}{1}$=$\frac{a}{1}$
and thats where i got stuck.
|
HINT:
We need discriminant $\ge0,$
$\implies a^2\ge8b, 4b^2\ge4a\iff b^2\ge a$
$a^4\ge64a\iff a(a-4)(a^2+4a+4^2)\ge0\iff a(a-4)\ge0$ as $a^2+4a+4^2=(a+2)^2+2^2\ge2^2>0$
$\implies$ either $a\le0$ or $a-4\ge0\iff a\ge4$
But $a$ is positive.
|
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|
Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Attempt:
$$\gcd(4n-5,60-12n)=(4n-5,-8n+55)=(4n-5,-4n+50)=(4n-5,45)$$
$$n=1: (4-5,45)=1\quad \checkmark\\
n=2: (3,45)=3\quad \times\\
n=3: (7,45)=1\quad \checkmark\\
n=4: (11,45)=1\quad \checkmark\\
n=5: (15,45)=15\quad \times\\
n=6: (19,45)=1\quad \checkmark\\
n=7: (23,45)=1\quad \checkmark\\
n=8: (27,45)=9\quad \times\\
\vdots$$
So the answer is that it can't be reduced for $n=1,3,4,6,7,..$ i.e
$$\bigg\{n\bigg|n\notin \begin{cases}a_1=2\\a_n=a_{n-1}+3\end{cases}\bigg\}$$
I want to verify that my solution is correct
|
We have that $\gcd(pq + r,p) = \gcd(p,r)$ and since $12n - 60 = 3(4n-5)+45$, we have that $\frac{4n-5}{60-12n}$ can't be reduced if and only if $\gcd(4n-5,45) = 1$ and since $45=3^2\cdot 5$, this happens if and only if $3\nmid 4n-5$ and $5\nmid 4n-5$.
Now,
$4n-5\equiv 0 \pmod 3\iff 4n \equiv 8 \pmod 3 \iff n\equiv 2\pmod 3$ (last equivalence is due to $\bar 4$ being invertible in $\mathbb Z/3\mathbb Z$)
$4n-5\equiv 0\pmod 5\iff 4n\equiv 0\pmod 5 \iff n\equiv 0\pmod 5$ (last equivalence is due to $\bar 4$ being invertible in $\mathbb Z/5\mathbb Z$)
Thus, $\frac{4n-5}{60-12n}$ can't be reduced if and only if $n\not\equiv 2\pmod 3$ and $n\not\equiv 0\pmod 5$.
|
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|
How to find this function, and what method to use? The function is $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}$ and they are asking us to find out what $f(-x)$ is?
|
Observe that $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}=(x-\frac{1}{x})^3+3\cdot x\cdot \frac{1}{x}\cdot(x-\frac{1}{x})=(x-\frac{1}{x})^3+3(x-\frac{1}{x})$
Hence, we can say that $$f(z)=z^3+3z$$ where $z$ is any real variable.
So we have that $$\color{red}{f(-x)}=\color{blue}{-x^3-3x}$$
|
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|
Maximize $k=x^2+y^2$ Subject to $x^2-4x+y^2+3=0$ Question
Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. Find the maximum and minimum values of $x^2+y^2$.
My work
Let $k=x^2+y^2$
Therefore, $x^2-4x+y^2+3=0$ ---> $k-4x+3=0$ .
What do I do next? How do I find an expression in terms of $k$ that I can maximize?
|
$F(x,y,\lambda)=x^2+y^2+\lambda (x^2-4 x+y^2+3)$ is lagrange multiplier. We need to solve system of equations $$F'_x=0$$ $$F'_y=0,$$ $$x^2-4 x+y^2+3=0$$ or $$2x+2\lambda x-4\lambda=0,$$ $$2y+2\lambda y=0,$$ $$x^2-4 x+y^2+3=0.$$ We have $(x,y,\lambda)\in\{(1,0,1),(3,0,-3)\}$. For $\lambda=1$ is $d^2 F>0$ so $(x,y)=(1,0)$ is point of minimum and $(x,y)=(3,0)$ is point of maximum, since $d^2 F<0$ for $\lambda=-3$.
|
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|
Factorial sum estimate $\sum_{n=m+1}^\infty \frac{1}{n!} \le \frac{1}{m\cdot m!}$ Prove that:
$$\displaystyle \sum_{n=m+1}^\infty \dfrac{1}{n!} \le \dfrac{1}{m\cdot m!}$$
I have tried induction on $m$ but it does not work very well. Any suggestion?
|
$$\sum_{n=m+1}^{\infty} \frac{1}{n!}=\frac{1}{m!} \left(\frac{1}{m+1}+\frac{1}{(m+1)(m+2)}+\frac{1}{(m+1)(m+2)(m+3)}+\ldots \right)$$ $$ \leq \frac{1}{m!} \left(\frac{1}{m+1}+\frac{1}{(m+1)^2}+\frac{1}{(m+1)^3}+\ldots \right)=\frac{1}{m!}\frac{1}{m+1}\frac{1}{1-\frac{1}{m+1}}=$$ $$\frac{1}{m!}\frac{1}{m+1-1}=\frac{1}{m!}\frac{1}{m}$$
|
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|
Prove $\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$ This series represents the sum of reciprocal pentagonal numbers (multiplied by $3$). I got its value from Wolfram Alpha:
$$\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$$
My attempt: Turn the series into an integral:
$$\sum_{n=1}^\infty \frac{x^{3n}}{n}=-\ln(1-x^3)$$
$$\sum_{n=1}^\infty \frac{x^{3n-2}}{n}=-\frac{\ln(1-x^3)}{x^2}$$
$$\sum_{n=1}^\infty \frac{1}{n(3n-1)}=-\int_0^1 \frac{\ln(1-x^3)}{x^2} dx$$
I don't know how to solve this integral.
We can also transform the common term into two parts:
$$\frac{1}{n(3n-1)}=\frac{3}{3n-1}-\frac{1}{n}$$
But this gives us two diverging series, so it doesn't help.
|
We can do the indefinite integral. Start by the obvious guess
$$\frac{d}{dx} \frac{-\log(1-x^3)}{x} = \frac{\log(1-x^3)}{x^2}+\frac{3x}{1-x^3}$$
Now we have to wipe out that second term.
$$
\frac{3x}{1-x^3}=\frac{1}{1-x}+\frac{x-1}{1+x+x^2} \\
\int\frac{1}{1-x}= -\log(1-x)
$$
leaving the term $\frac{x-1}{1+x+x^2}$ to deal with. A reasonable try would be subtracting off
$$
\frac{d}{dx} (\log{\sqrt{1+x+x^2}}) = \frac{x+\frac12}{1+x+x^2}
$$
And we are left with
$$\frac12\int\frac1{1+x+x^2}dx$$
If you know that $$\frac{d}{dx} \tan^{-1}(a+b*x) = \frac{b}{1+(a+bx)^2}$$ and set that equal to $\zeta/(1+x+x^2)$ you can find $a, b$ and the multiplicative constant, to get
$$\frac{d}{dx}\left( \frac{\tan^{-1}(\frac{1+2x}{\sqrt{3}})} {\sqrt{3}} \right) =\frac12\int\frac1{1+x+x^2}dx$$
I may have a factor of $3$ error in this chain. At any rate, you can now see that the $\sqrt{3}\pi$ in the answer comes from an arctan of $\frac{1+2\cdot 1}{\sqrt{3}}$, which is $\pi/3$.
Glue all these parts together and you have your indefinite integral.
|
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|
Find $x$ in $1!+2!+\ldots+100!\equiv x \pmod{19}$ Here I come from one more (probably again failed) exam. We never did congruence with factorials; there were 3 of 6 problems we never worked on in class and they don't appear anywhere in scripts or advised literature.
First is this one:
Find $x$ in $1!+2!+\ldots+100! \equiv x \pmod {19}$
I got that every $x$ after $19!$ is $0$.
But I am left with $1!+2!+\ldots+18!$, and my calculator can't calculate $18!$ without exponential, so there must be some simpler way, but I have never learned that.
|
$18! \equiv -1 \pmod{19}$ by Wilson's Theorem. Then you can find the modular inverse of $18$ modulo $19$, multiply both sides by it and you will get $17! \equiv 1 \pmod{19}$. You can continue simularly for the big numbers that are left.
Also you can go step by step in the multiplication. For example for $13!$ you can start:
$$13! \equiv 1 \cdot 2 \cdot ... \cdot 13 \equiv 2 \cdot 3 \cdot ... \cdot 13 \equiv 6 \cdot 4 \cdot ... \cdot 13 \equiv 5 \cdot 5 \cdot ... \cdot 13 \equiv 6 \cdot 6 \cdot ... \cdot 13 \equiv 17 \cdot 7 \cdot ... \cdot 13 $$
$$\equiv 5 \cdot 8 \cdot ... \cdot 13 \equiv 2 \cdot 9 \cdot ... \cdot 13 \equiv 1 \cdot 10 \cdot ... \cdot 13$$
$$ \equiv 10 \cdot 11 \cdot 12 \cdot 13 \equiv 15 \cdot 12 \cdot 13 \equiv 9 \cdot 13 \equiv 3 \pmod {19}$$
|
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|
Find the highest point of intersection
Find the highest point of intersection of the sphere $x^2+y^2+z^2=30$ and the cone $x^2+2y^2-z^2=0$.
Am I supposed to use the Lagrange multiplier for this?
EDIT:
So this is what I've tried...
$z^2=-x^2-y^2+30$ and $z^2=x^2+2y^2$. Then $-x^2-y^2+30 = x^2+2y^2$.
This gives us $3y^2=-2x^2+30$, which is then: $y^2=-\frac23x^2+10$. Substitute $y^2$ in to the sphere equation, we have:
$z^2 = -x^2 - (-\frac23x^2+10)+30 = -\frac13x^2+20$, so the maximum $z$ is $\sqrt{20}$...?
If this is true, then calculating the rest of the point, we have $(0,\sqrt{10},\sqrt{20})$....?
|
As suggested by John Molokach: solving for $z^2$ yields
$$
2x^2+3y^2=30,
$$
so the intersection of both curves projected in the $x,y$ plane is the ellipse
$$
\begin{cases}
x=\sqrt{15}\cos t\\
y=\sqrt{10}\sin t
\end{cases}
$$
Replacing these equations in the cone yields
$$
z=\sqrt{15+5\sin^2t}
$$
which has maximum $z_{MAX}=2\sqrt{5}$ with elementary derivative analysis.
|
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How do we prove that $4(3\sqrt2-4)=\prod_{n=1}^{\infty}\left({e^{2\pi(2n-1)}-1\over e^{2\pi(2n-1)}+1}\right)^8?$ How do we prove that
$$4(3\sqrt2-4)=\prod_{n=1}^{\infty}\left({e^{2\pi(2n-1)}-1\over e^{2\pi(2n-1)}+1}\right)^8\tag1$$
Rewrite as, to keep it simple
Let $a=e^{2\pi(2n-1)}$
$$4(3\sqrt2-4)=\prod_{n=1}^{\infty}\left(a-1\over a+1\right)^8\tag2$$
Take the log
$${1\over 8}\ln(12\sqrt2-16)=\sum_{n=1}^{\infty}\ln\left({a-1\over a+1}\right)\tag3$$
We have this series
$$-\ln\left({x-1\over x+1}\right)={2\over x-1}-{2^2\over 2(x-1)^2}+{2^3\over 3(x-1)^3}-\cdots\tag4$$
$$-{1\over 8}\ln(12\sqrt2-16)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}{(-1)^{m-1}2^m\over m(a-1)^m}\tag5$$
How do I evaluate this double series?
|
This one is similar to this answer of your previous question and you should have guessed it. Based on the definition of Ramanujan's class invariants
\begin{align}G_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 + e^{-\pi\sqrt{n}})(1 + e^{-3\pi\sqrt{n}})(1 + e^{-5\pi\sqrt{n}})\cdots\tag{1}\\
g_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 - e^{-\pi\sqrt{n}})(1 - e^{-3\pi\sqrt{n}})(1 - e^{-5\pi\sqrt{n}})\cdots\tag{2}\end{align}
the desired product is equal to $(g_{4}/G_{4})^{8}$. From the linked answer we have $$G_{1} = 1, g_{1} = 2^{-1/8}$$ and then $$g_{4}=2^{1/4}g_{1}G_{1} = 2^{1/4}2^{-1/8} = 2^{1/8}$$ Thus $g_{4}^{8} = 2$. Again we have $$g_{4}^{8}G_{4}^{8}(G_{4}^{8} - g_{4}^{8}) = \frac{1}{4}$$ so that $$2x(x - 2) = \frac{1}{4}$$ where $x = G_{4}^{8}$. Thus $$8x^{2} - 16x - 1 = 0$$ or $$x = G_{4}^{8} = \frac{4 + 3\sqrt{2}}{4}$$ and then the desired product is $$\frac{8}{4 + 3\sqrt{2}} = 4(3\sqrt{2} - 4)$$ We have used the following identities for Ramanujan Class Invariants: $$g_{4n} = 2^{1/4}g_{n}G_{n},\, (g_{n}G_{n})^{8}(G_{n}^{8} - g_{n}^{8}) = \frac{1}{4}\tag{3}$$ Note that the first equation $g_{4n} = 2^{1/4}g_{n}G_{n}$ follows directly by multiplying equations $(1)$ and $(2)$ and the second equation in $(3)$ follows by noting that $G_{n} = (2kk')^{-1/12}, g_{n} = (2k/k'^{2})^{-1/12}$ and $k^{2} + k'^{2} = 1$.
|
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|
Is it possible to have a power series for arctan(x) centered at 1? My Calculus 2 professor referenced that such a series is impossible, but why? I understand how to properly find the power series of arctan(x) centered at 0.
|
Since
$$
\begin{align}
\tan(\arctan(1+x)-\arctan(1))
&=\frac{(1+x)-1}{1+(1+x)\cdot1}\\
&=\frac x{2+x}\tag{1}
\end{align}
$$
we have
$$
\begin{align}
\arctan(1+x)
&=\frac\pi4+\arctan\left(\frac{x}{2+x}\right)\\
&=\frac\pi4+\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\left(\frac x{2+x}\right)^{2k+1}\tag{2}
\end{align}
$$
Expanding $(2)$ using the binomial theorem, the coefficient of $x^n$, for $n\ge1$, is
$$
\begin{align}
&\frac{\left(-\frac12\right)^n}n\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}(-1)^{k+1}\binom{n}{2k+1}\\
&=\frac{\left(-\frac12\right)^n}n\sum_{k=0}^n\frac{(-i)^k-i^k}{2i}\binom{n}{k}\\
&=\frac{\left(-\frac12\right)^n}n\frac{(1-i)^n-(1+i)^n}{2i}\\
&=\frac{\left(-\frac1{\sqrt2}\right)^n}n\frac{\left(\frac{1-i}{\sqrt2}\right)^n-\left(\frac{1+i}{\sqrt2}\right)^n}{2i}\\
&=-\frac{\left(-\frac1{\sqrt2}\right)^n}n\sin\left(\frac{n\pi}4\right)\tag{3}
\end{align}
$$
Therefore,
$$
\arctan(1+x)=\frac\pi4-\sum_{n=1}^\infty\frac{\sin\left(\frac{n\pi}4\right)}n\left(-\frac{x}{\sqrt2}\right)^n\tag{4}
$$
Note that the series in $(2)$ converges for $-1\lt x\lt\infty$ and $(4)$ matches Jack D'Aurizio's answer.
|
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|
Conjectured value of $\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}$ I was curious whether this integral has a closed form expression :
$$\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln
x}\right)\frac{\mathrm{d}x}{x^2+1}$$
The integrand has a singularity at $x=1$, but it's removable. And as $x \to \infty$, the integrand behaves like $\frac{1}{x \ln^{2}x}$. So the integral clearly converges.
Although I have not been able to derive its closed form, I think, by reverse symbolic calculators, up to 20 digits it could be
$$I=\frac{4G}{\pi}$$
where $G$ is Catalan's constant. Is it true or is it completely fabulous?
EDIT. NOTE :
For better search to this integral I have renamed the title from Conjectured value of logarithmic definite integral, which is ambiguous and did not say anything, to the current one with integral explicitly written.
|
Though using the residue method is somewhat straightforward, but not everyone can understand it. So, here is a residue-free method:
Split the integral into two terms where each term is in the interval $0<x<1$ and $1<x<\infty$, then use the substitution $x\mapsto\frac{1}{x}$ to the second term. We will get
$$
\left[\int_{0}^{1}+\int_{1}^{\infty}\right]\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}=\int_{0}^{1}\frac{(x-1)^2}{x\ln^2 x}\cdot\frac{\mathrm{d}x}{x^2+1}\tag1
$$
Now, for $a\ge-1$ , one may consider the following integral
$$
I(a)=\int_{0}^{1}x^a\cdot\frac{(x-1)^2}{\ln^2 x}\cdot\frac{\mathrm{d}x}{1+x^2}\tag2
$$
and the desired integral is $I(-1)$. Since $0<x<1$, one may observe that $I(\infty)\to0$ as $a\to\infty$.
\begin{align}
I''(a)&=\int_{0}^{1}\frac{x^a(x-1)^2}{1+x^2}\ \mathrm{d}x\\[10pt]
&=\int_{0}^{1}\sum_{k=0}^\infty(-1)^k\ x^{2k+a}\ (x^2-2x+1)\ \mathrm{d}x\\[10pt]
&=\sum_{k=0}^\infty(-1)^k\left(\frac{1}{2k+a+3}-\frac{2}{2k+a+2}+\frac{1}{2k+a+1}\right)\\[10pt]
&=\frac{1}{4}\left[\psi\left(\frac{a+5}{4}\right)-2\psi\left(\frac{a+4}{4}\right)+2\psi\left(\frac{a+2}{4}\right)-\psi\left(\frac{a+1}{4}\right)\right]\\[10pt]
I'(a)&=\ln\Gamma\left(\frac{a+5}{4}\right)-\ln\Gamma\left(\frac{a+1}{4}\right)+2\ln\Gamma\left(\frac{a+2}{4}\right)-2\ln\Gamma\left(\frac{a+4}{4}\right)\\[10pt]
I(a)&=4\left[\psi\left(-2,\frac{a+5}{4}\right)-\psi\left(-2,\frac{a+1}{4}\right)+2\psi\left(-2,\frac{a+2}{4}\right)-2\psi\left(-2,\frac{a+4}{4}\right)\right]\tag3\\[10pt]
\end{align}
Hence
$$
I(-1)=4\left[\psi\left(-2,1\right)-\psi\left(-2,0\right)+2\psi\left(-2,\frac{1}{4}\right)-2\psi\left(-2,\frac{3}{4}\right)\right]=\frac{4G}{\pi}
$$
Wolfram Alpha confirms it. One may also use the special values of generalized polygamma function and its related relation with derivative of Hurwitz Zeta Function: $$\psi(-2,x)=\zeta'(-1,x)-\frac{x^2}{2}+\frac{x}{2}-\frac{1}{12}$$
|
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|
Minimal polynomial and possible Jordan forms
Let $A$ be an $8\times 8$ complex matrix with characteristic polynomial $$p_A(x)=(x-1)^4(x+2)^2(x^2+1)$$ and minimal polynomial $$m_A(x)=(x-1)^2(x+2)^2(x^2+1).$$ Determine all possible Jordan canonical forms of $A$. Also, what is the dimension of the eigenspace for $\lambda = 1$ for each case?
I haven't learned minimal polynomial yet, so I wanted to check if I am on the right track.
From the characteristic polynomial, the eigenvalues are $i,-i,-2,1$. From the minimal polynomial, the block sizes are $1,1,2$ for eigenvalues $i,-i,-2$ respectively. For $\lambda = 1$, the largest block size is 2, so the possible block sizes are $2,2$, which has eigenspace dim of 2 ($J_1$). Or the sizes are $2,1,1$, which has eigenspace dim of 3 ($J_2$). Writing the full Jordan matrix:
$J_1=\begin{pmatrix}1&1&0&0&0&0&0&0\\0&1&0&0&0&0&0&0\\0&0&1&1&0&0&0&0\\0&0&0&1&0&0&0&0\\0&0&0&0&-2&1&0&0\\0&0&0&0&0&-2&0&0\\0&0&0&0&0&0&i&0\\0&0&0&0&0&0&0&-i\end{pmatrix}$, $J_2=\begin{pmatrix}1&1&0&0&0&0&0&0\\0&1&0&0&0&0&0&0\\0&0&1&0&0&0&0&0\\0&0&0&1&0&0&0&0\\0&0&0&0&-2&1&0&0\\0&0&0&0&0&-2&0&0\\0&0&0&0&0&0&i&0\\0&0&0&0&0&0&0&-i\end{pmatrix}$
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You identified the blocks and eigenspace dimensions correctly. The blocks for a given form, however, can be permuted. Thus, your first matrix represents 5!/2! = 60 forms, and your second matrix, 6!/2! = 360 forms for a total of 420 forms.
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Proof about Pythagorean triples Show that if $(x,y,z)$ is a Pythagorean triple, then $10\mid xyz$
Proof
First, if $x$, $y$, $z$ are all odd, then so are $x^2$, $y^2$, $z^2$, so $x^2+y^2$ is even, which means that $x^2+y^2 \neq z^2 $. Hence, at least one of $x$, $y$, $z$ is even, so $2\mid xyz$ (clear).
Next, for any $n \in Z$, if $5$ doesn't divide $n$, then $(n^2)^2=n^4 \equiv 1\pmod{5}$ (as you can check or quote Euler's theorem), and therefore $n^2 \equiv \pm 1\pmod{5}$. Now, if $5$ doesn't divide $xy$, then $x^2 \equiv \pm 1\pmod{5}$ and $y^2 \equiv \pm 1\pmod{5}$, so
$$x^2+y^2 \equiv -2,0,2 \pmod{5}$$
Therefore if $x^2+y^2=z^2$ and $5$ doesn't divide $xy$ then $z^2=x^2+y^2 \equiv -2,0,2\pmod{5}$, so $5\mid z^2$ (why? it looks weird to me because $z^2$ can be also congruent to $2$ and $-2$) and hence $5\mid z$ (otherwise $z^2 \equiv \pm 1\pmod{5}$). It follows that $x^2+y^2=z^2$, then either $5\mid xy$ or $5\mid z$ (how does this follow? if $5$ doesn't divide $z$ then how can $5$ divide $xy$), so in any case $5\mid xyz$
Finally, we can conclude that if $x^2+y^2=z^2$ then $2\mid xyz$ and $5\mid xyz$, so $10\mid xyz$ (intuitively it looks right but I can't prove it!)
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No problem to give a "theoretical" proof. Just for fun let me work with a table in $\mathbb F_5^2$.
By the quite known parametrization of the Pythagorean triple, $xyz$ is even so we must prove in
$\mathbb F_5$ that
$$xyz=mn(m^2-n^2)(m^2+n^2)=0\qquad (1)$$
If $m=n$ and if $mn=0$ it is clear so we prove $(1)$ with $m\ne n$ and $mn\ne 0$ so we need to verify $(1)$ with $12$ couples $(m,n)$
$$\begin{array}{|c|c|}\hline (1,1) & (1,2) & (1,3) &(1, 4) \\\hline (2,1) & (2,2) & (2,3) &(2,4)\\\hline (3,1) &(3,2) & (3,3) &(3,4)\\\hline (4,1)&(4,2)&(4,3)&(4,4)\\\hline\end{array}$$
If $(m,n)\in \{(1,2),(1,3),(2,1),(2,4),(3,1),(3,4),(4,2),(4,3)\}$ then $m^2+n^2=0$
If $(m,n)\in\{(1,4),(2,3),(3,2),(4,1)\}$ then $m^2-n^2=0$
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How to find $ \tan \left(\frac{x}{2}\right) $ knowing that $\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $ Good evening to everyone. I don't know how to find $ \tan \left(\frac{x}{2}\right) $ knowing that $$\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $$ and x$\in (0,\frac{\pi}{3})$ Here's what I've tried:
$$\tan \left(\frac{x}{2}\right) = \frac{1-\cos \left(x\right)}{\frac{7}{5}- \cos \left(x\right)}$$ But I don't know what to do from here. Can someone explain to me how to solve this? Thanks for any answers.
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\begin{align}
\cos x +\sin x &= \frac 75 \\
\cos^2 x + 2 \cos x \sin x + \sin^2 x &= \frac{49}{25} \\
2 \cos x \sin x &= \frac{24}{25} \\
\hline
\cos^2 x - 2 \cos x \sin x + \sin^2 x &= 1 - 2 \cos x \sin x \\
(\cos x - \sin x)^2 &= \dfrac{1}{25} \\
\cos x - \sin x &= \pm \frac 15 \\
\hline
(\cos x, \sin x) &\in \left\{ \left(\frac 45, \frac 35\right),
\left(\frac 35, \frac 45\right) \right\} \\
\hline
\tan \frac x2 &= \frac{\sin x}{1 + \cos x} \\
\tan \frac x2 &\in \left\{\frac 39, \frac 48\right\} \\
\tan \frac x2 &\in \left\{\frac 13, \frac 12\right\}
\end{align}
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How to prove by induction that $3^{3n}+1$ is divisible by $3^n+1$ for $(n=1,2,...)$ So this is what I've tried:
Checked the statement for $n=1$ - it's valid.
Assume that $3^{3n}+1=k(3^n+1)$ where $k$ is a whole number (for some n). Proving for $n+1$: $$3^{3n+3}+1=3^33^{3n}+1=3^3(3^{3n}+1)-26=3^3k(3^n+1)-26=3^3k(3^n+1)-3^3+1=3^3[k(3^n+1)-1]+1$$ and I'm stuck. Any help please?
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It's true for $n=1$. Assume it holds for $n$, i.e: $3^{3n} + 1 = k(3^n +1)$ then consider $$3^{3(n+1)} +1 = 27 \cdot 3^{3n} + 1 = 27 (3^{3n} +1) - 26 = 27k(3^n +1) - 26$$
Let's get it into a more amenable form $$\begin{equation}3^{3(n+1)} +1 = 9k3^{n+1} + 27k - 26 = 9k(3^{n+1}+1) + 18k - 26 \end{equation} \tag{1}$$
So we want to show that $3^{n+1} + 1$ always divides $18k - 26$ where $$k = \frac{3^{3n} + 1}{3^n + 1} = 3^{2n} - 3^n + 1$$
Equivalently we want to show that $3^{n+1} + 1$ divides $18\cdot 3^{2n} - 18 \cdot 3^n - 8 $. But: $$18\cdot 3^{2n} - 18 \cdot 3^n - 8 = 2(3^{n+1} - 4)(3^{n+1} + 1) \quad \quad (\star)$$
It is then clear that $3^{n+1} + 1$ divides $18k - 26$ since $2(3^{n+1} - 4)$ is an integer.
And hence, since $3^{3(n+1)} + 1$ is the sum of two terms (see $(1)$), each divisible by $3^{n+1} + 1$, where the divisibility holds due to a relation we exploited from the inductive hypothesis, we are done inductively.
If $(\star)$ seems a bit magical, simply look at $f(x) = 18x^2 - 18x - 8 = 2(3x-4)(3x+1)$ and substitute in $x = 3^n$.
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If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$
*
*$\sin x + \sin y = 1$
*$\cos x + \cos y = 0$
Any valid pair of $(x, y)$ is fine, as the restrictions on the board in the image below are obscured.
I got the question from chapter 26 of a comic called Yamada-kun.
How can I solve this equation?
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There are identities for sum of sin and cos:
$$\sin(x)+\sin(y) = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)_.$$
$$\cos(x) + \cos(y) = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right).$$
Using the first equation tells us that $\cos\left(\frac{x-y}{2}\right)\neq 0.$ Therefore by the second equation, $\cos\left(\frac{x+y}{2}\right) = 0.$
This may or may not be useful in finishing the problem.
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Compute $1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \cdots + n \cdot \frac {1}{2^n} + \cdots $ I have tried to compute the first few terms to try to find a pattern but I got
$$\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}$$
but I still don't see any obvious pattern(s). I also tried to look for a pattern in the question, but I cannot see any pattern (possibly because I'm overthinking it?) Please help me with this problem.
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First consider the partial sum by showing
$$ \sum_{n=1}^k \frac{n}{2^n} = 2^{-k}(-k+2^{k+1}-2). $$
Now $k\to \infty$ gives the result $2$.
Edit: Proof of the partial sum by induction
$k=1$:
$$\sum_{n=1}^1 \frac{1}{2} = 1/2 = 2^{-1}(-1+2^{2}-2) \quad \checkmark$$
Let $\sum_{n=1}^k \frac{n}{2^n} = 2^{-k}(-k+2^{k+1}-2)$ be true for any $k\geq1$.
Induction step:
$$ \sum_{n=1}^{k+1} \frac{n}{2^n} = \sum_{n=1}^k \frac{n}{2^n} + \frac{k+1}{2^{k+1}} = 2^{-k}(-k+2^{k+1}-2) + \frac{k+1}{2^{k+1}} = \frac{2(-k+2^{k+1}-2)+k+1}{2^{k+1}} = \frac{-k+2^{k+2}-3}{2^{k+1}} = 2^{-(k+1)}(-(k+1)+2^{k+2}-2) $$
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If $ax^2 +bxy+cy^2+5x-2y+3$ divided by $x-y+1$ has remainder $0$, determine $a$, $b$, and $c$.
If $ax^2 +bxy+cy^2+5x-2y+3$ divided by $x-y+1$ has remainder $0$, determine $a$, $b$, and $c$.
I do not know how to approach this problem and would appreciate advice how to proceed.
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Since there is no remainder the original equation will be the product of $x-y+1$ with another unknown equation. In order to create the $x^2$, $y^2$, and $xy$ terms we expect the form to be $dx+ey+f$.
$$ax^2+bxy+cy^2+5x−2y+3 = (x−y+1)(dx+ey+f)$$
$$ax^2+bxy+cy^2+5x−2y+3 = dx^2+(e-d)xy-ey^2+(d+f)x+(e-f)y+f$$
working right to left
$$f=3$$
$$e-f=-2, e=1$$
$$d+f=5, d=2$$
$$c=-e, c=-1$$
$$b=e-d, b=-1$$
$$a=d, a=2$$
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Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$ $$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.
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Hint. Observe that
$$
(\sqrt{5}-1)^2=6-2\sqrt{5},\quad (\sqrt{5}+1)^2=6+2\sqrt{5}.
$$
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Simplify the expression $\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +\binom{n+k}{k}$ Simplify the expression $\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +\binom{n+k}{k}$
My attempt: Using the formula $\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$
$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots \binom{n+k-1}{k-1}+\binom{n+k}{k}$
=$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +\binom{n+k-1}{k-1}+(\binom{n+k-1}{k}+\binom{n+k-1}{k-1})$
=$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +2\binom{n+k-1}{k-1}+\binom{n+k-1}{k}$
I can again use the same formula for the term $2\binom{n+k-1}{k-1}$, and in the next to step to the term $3\binom{n+k-2}{k-2}$.
But I don't think this way the expression will get simplified.
Any help is appreciated.
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First show that $\large{n\choose r}={{n-1}\choose {r-1}}+{{n-1}\choose r}$,
from which we get $\large{{n+r+1}\choose r}={{n+r}\choose r}+{{n+r}\choose {r-1}}$
By the same rule, $\large{{n+r}\choose {r-1}}={{n+r-1}\choose {r-1}}+{{n+r-1}\choose {r-2}}$
$\large{{n+r-1}\choose {r-2}}={{n+r-2}\choose {r-2}}+{{n+r-3}\choose {r-3}}$
$..$ $\quad \quad \quad \quad ..$ $ \quad \quad \quad ..$
$..$ $\quad \quad \quad \quad ..$ $ \quad \quad \quad ..$
$..$ $\quad \quad \quad \quad ..$ $ \quad \quad \quad ..$
$\large{{n+3}\choose {2}}={{n+2}\choose {2}}+{{n+2}\choose {1}}$
$\large{{n+2}\choose {1}}={{n+1}\choose {1}}+{{n+1}\choose {0}}$
Adding, we get $\large{n\choose 0}+{{n+1}\choose 1}+...+{{n+r-1}\choose {r-1}}+{{n+r}\choose r}={{n+r+1}\choose r}$
Alternatively, we can fix any $r$ of the $(n+r+1)$ objects given. Label them $A_1,A_2,...A_r$. Now our choices of $r$ objects from the $(n+r+1)$ objects may or may not contain any or all of the set $\{A_1,A_2,...,A_r\}$.
Case I: It does not contain $A_1$
This will happen in $\large{{n+r}\choose r}$ ways as the $r$ things have to be chosen from the remaining $(n+r)$ things.
Case II: It contains $A_1$ but does not contain $A_2$.
This will happen in $\large{{n+r-1}\choose {r-1}}$ ways, because having chosen $A_1$ and rejectd $A_2$, we have only $(n+r-1)$ things to choose from and we need only $(r-1)$.
$...$
Case r: It contains $A_1,A_2,...,A_{r-1}$ but does not contain $A_r$.
This will happen in $\large {{n+1}\choose 1}$ ways.
Case $(r+1)$: It contains $A_1,A_2,...,A_r$.
This will happen in $\large{n\choose 0}=1$ way.
Hence, $\displaystyle\sum_{k=0}^{r}{{n+k}\choose k}={{n+r+1}\choose r}$
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How to formally use Taylor expansions for $n$th derivatives and generating functions? When deriving Catalan numbers, the generating function takes on this form:
$$C(x) = \frac{1}{2} (1 - \sqrt{1-4x}) = \frac{1}{2} (1 - f(x))$$
where $f(x) = \sqrt{1-4x}$
How does one formally show what it evaluates to? I can do it somewhat "informally" like so:
$$f(x) = \sqrt{1-4x} = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$
$$f(x) = \frac{f^{(0)}(0)}{0!} x^0 + \frac{f^{(1)}(0)}{1!} x^1 + \frac{f^{(2)}(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + ...$$
Now at this point I can literally start computing the successive derivatives of $f(x)$ and then plug in $0$ for each $x$ and look for patterns and "eyeball" a closed-form for it.
But is there a way at this point to formally show what the $n$th derivative is? Or is this more of an art than a science that changes depending on the function?
I could eyeball the $n$th derivative formula as well:
$f^{0}(x) = (1 - 4x)^{\frac{1}{2}}$
$f^{1}(x) = (\frac{1}{2})(-4)(1 - 4x)^{-\frac{1}{2}}$
$f^{2}(x) = (-\frac{1}{2})(\frac{1}{2})(-4)(-4)(1 - 4x)^{-\frac{3}{2}}$
$f^{3}(x) = (-\frac{3}{2})(-\frac{1}{2})(\frac{1}{2})(-4)(-4)(-4)(1 - 4x)^{-\frac{5}{2}}$
$f^{4}(x) = (-\frac{5}{2})(-\frac{3}{2})(-\frac{1}{2})(\frac{1}{2})(-4)(-4)(-4)(-4)(1 - 4x)^{-\frac{7}{2}}$
At this point I can eyeball the pattern and solve it manually. But again, what if I couldn't do that? Is there a more direct and methodical way to this?
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May be, you could use the generalized binomial expansion $$(1+x)^a=\sum_{n=0}^\infty \binom{a}{n} x^n$$ which makes $$f(x)=\sqrt{1-4x}=\sum_{n=0}^\infty (-1)^n\binom{\frac 12}{n} (4x)^n=1+\sum_{n=1}^\infty (-1)^n\binom{\frac 12}{n} (4x)^n$$ and then $$C(x) = \frac{1}{2} (1 - \sqrt{1-4x})=\frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}\binom{\frac 12}{n} (4x)^n$$
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Which answer is correct? Finding the limit of a radical as $x$ approaches infinity. When I take $$\lim_{x \to -∞} \sqrt{x^2+7x}+x,$$ I multiply by the conjugate over the conjugate to get
$$\lim_{x \to -∞}\frac{7x}{\sqrt{x^2+7x}-x},$$ and multiply by either $\frac{\frac{1}{x}}{\frac{1}{x}}$ or $\frac{\frac{1}{-x}}{\frac{1}{-x}}$ to get an undefined answer or $\frac{-7}{2}.$
My teacher's solution involves multiplying by $\frac{\frac{1}{-x}}{\frac{1}{-x}}:$
$$=\lim_{x \to -∞}\frac{-7}{\sqrt{x^2/x^2+7x/x^2}+1}$$
$$=-\frac{7}{\sqrt{1+0}+1}$$
$$=\frac{-7}{2}$$
However, I multiplied
by $\frac{\frac{1}{x}}{\frac{1}{x}}$ and got the following:
$$\lim_{x \to -∞}\frac{7}{\sqrt{x^2/x^2+7x/x^2}-1}$$
$$\frac{7}{\sqrt{1+0}-1}$$
$$\frac{7}{0}$$
Which is undefined.
Why does multiplying by what is essentially $1$ cause different answers in general, and in particular for evaluating limits?
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Let's see what happens to $$f(x) = \sqrt{x^2 + 7x} + x$$ when $x$ takes on actual negative values. When $x = -100$, we get $$f(-100) = 10 \sqrt{93} - 100 \approx -3.56349.$$ When $x = -10000$, we get $$f(-10000) = 100 \sqrt{9993} - 10000 \approx -3.50061.$$ So, from a numerical standpoint, this seems to suggest that the limit should exist. When we rationalize the numerator, i.e. $$f(x) = \frac{7x}{\sqrt{x^2 + 7x} - x},$$ we are still okay, but the next step, division by $x$, requires care because when we write $$\frac{7}{\frac{1}{x}\sqrt{x^2 + 7x} - 1} = \frac{7}{\sqrt{1 + 7/x^2} - 1},$$ we inadvertently commit a sign error: this is because if $x < 0$, we no longer have $$\frac{1}{x} = \sqrt{\frac{1}{x^2}}.$$ The LHS is negative; the RHS is positive. Instead, when $x < 0$, we should have $$\frac{1}{x} = -\sqrt{\frac{1}{x^2}}.$$ So to preserve the sign in the denominator, we must write instead $$\frac{7}{\frac{1}{x}\sqrt{x^2 + 7x} - 1} = \frac{7}{-\sqrt{1 + 7/x^2} - 1},$$ and now the correct limiting behavior is retained.
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Minimum value of $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$ If $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$
Find the minimum value of the function
I tried using the AMGM inequality and differentiation but didn't know how to solve it any ideas?
This is from a math competition. ( I would like to see the most efficient way as I think differentiation in a math competition is not that efficient)
Using Mogjals comment and using the AM-GM inequality , setting $A = \sqrt{x^2 + (1-x)^2}$ and $B=\sqrt{(1-x)^2 +(1+x)^2}$
then
$A+B \geq 2\sqrt{AB}$
$$ \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2} \geq 2\sqrt{2}\sqrt{x^2+1}\sqrt{2x^2-2x+1}$$
With equality if and only $A=B$ so $x=-\frac{1}{2}$
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Let $A=(0,1)$, $B=(1,1)$, and $C=(-x,x)$ as in the picture. Then
$$AC=\sqrt{x^2+(1-x)^2}, BC=\sqrt{(1+x)^2+(1-x)^2}.$$
Let $D$ be symmetric to $A$ about $x+y=0$ (trajectory of $C$). Apparently
$$AC+BC\ge BD=AE+BE.$$
Minimal is attained at $C=E$. I leave you figure the coordinates of $E$.
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Is it possible to $\int \sqrt{\cot x}$ by hand $$\int \sqrt{\cot x}{dx}$$
$$\int \sqrt{\frac{\cos x}{\sin x}}{dx} $$
Using half angle formula
$$\int \sqrt{\frac{1-\tan^2 \frac{x}{2}}{2\tan \frac{x}{2}}}{dx}$$
But I am not getting any lead from here .I think it is not possible to integrate $\sqrt{\cot x}$ by hand .
I calculated the result with the help of integral calculator
$$\dfrac{\ln\left(\left|\tan\left(x\right)+\sqrt{2}\sqrt{\tan\left(x\right)}+1\right|\right)-\ln\left(\left|\tan\left(x\right)-\sqrt{2}\sqrt{\tan\left(x\right)}+1\right|\right)+2\arctan\left(\frac{2\sqrt{\tan\left(x\right)}+\sqrt{2}}{\sqrt{2}}\right)+2\arctan\left(\frac{2\sqrt{\tan\left(x\right)}-\sqrt{2}}{\sqrt{2}}\right)}{2^\frac{3}{2}}$$
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Let $$I = \int \sqrt{\cot x}dx = \frac{1}{2}\int 2\sqrt{\cot x}dx = \frac{1}{2}\int \left[\left(\sqrt{\cot x}+\sqrt{\tan x}\right)+\left(\sqrt{\cot x}-\sqrt{\tan x}\right)\right]dx$$
Now Let $$J = \int \left(\sqrt{\cot x}+\sqrt{\tan x}\right)dx = \sqrt{2}\int\frac{\cos x+\sin x}{\sqrt{\sin 2x}}dx = \sqrt{2}\int\frac{(\sin x-\cos x)'}{\sqrt{1-(\sin x-\cos x)^2}}dx$$
So we get $$J = \sqrt{2}\sin^{-1}(\sin x-\cos x)+\mathcal{C_{1}}$$
Similarly Let $$K = \int \left(\sqrt{\cot x}-\sqrt{\tan x}\right)dx = \sqrt{2}\int\frac{\cos x-\sin x}{\sqrt{\sin 2x}}dx = \sqrt{2}\int\frac{(\sin x+\cos x)'}{\sqrt{(\sin x+\cos x)^2-1}}dx$$
So we get $$K = \sqrt{2}\ln \left|(\sin x+\cos x)+\sqrt{\sin 2x}\right|+\mathcal{C_{2}}$$
So we get $$I = \int \sqrt{\cot x}dx = \frac{1}{\sqrt{2}}\sin^{-1}(\sin x-\cos x)+\frac{1}{\sqrt{2}}\ln \left|(\sin x+\cos x)+\sqrt{\sin 2x}\right|+\mathcal{C}$$
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1873601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
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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.