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Show that $64xy=-1$
Let $a=\frac{\pi}{13}$. If $$\begin{aligned} x &= \cos(2a)\cos(5a)\cos(6a)\\ y &= \cos(a)\cos(3a)\cos(9a)\end{aligned}$$ then show that $$64xy = -1$$
I'm trying to use
$$\cos(A)\cos(B)=\frac{\cos(A+B)+\cos(A-B)}{2}$$
then use summation of $\cos(2n+1)$. I need help, please. Thank you.
| As $\cos(\pi-A)=-\cos A$
$\displaystyle xy=(-1)^2\prod_{r=0}^{(13-1)/2-1}\cos(2r+1)a$
We can prove something more generic $$\prod_{r=0}^{n-1}\cos\dfrac{(2r+1)\pi}{2n+1}=\dfrac{(-1)^p}{2^n}\text{ where }p=\left\lfloor\dfrac n2\right\rfloor$$
Proof: Using this , $$\cos(2n+1)x=2^{2n}\cos^{2n+1}x-\cdots+(-1)^n(2n+1)\cos x$$
If $\cos(2n+1)x=-1,(2n+1)x=(2r+1)\pi$ where $r$ is any integer
$x=\dfrac{(2r+1)\pi}{2n+1}$ where $0\le r\le2n$
So, the roots of $$2^{2n}c^{2n+1}+\cdots+(-1)^n(2n+1)c+1=0$$ are $\cos\dfrac{(2r+1)\pi}{2n+1}$ where $0\le r\le2n$
By Vieta's formulas, $$\prod_{r=0}^{2n}\cos\dfrac{(2r+1)\pi}{2n+1}=-1$$
If $2n+1=2r+1\iff n=r,\cos\dfrac{(2r+1)\pi}{2n+1}=-1$
$$\prod_{r=0, r\ne n}^{2n}\cos\dfrac{(2r+1)\pi}{2n+1}=1$$
Again as $\cos(2\pi-B)=+\cos B,$
For $r_1\ne r_2,$
$\cos\dfrac{(2r_1+1)\pi}{2n+1}=\cos\dfrac{(2r_2+1)\pi}{2n+1}\iff\dfrac{(2r_1+1)\pi}{2n+1}+\dfrac{(2r_2+1)\pi}{2n+1}=2\pi\iff r_2=2n-r_1$
$$\implies\left(\prod_{r=0}^{n-1}\cos\dfrac{(2r+1)\pi}{2n+1}\right)^2=1$$
$$\displaystyle\implies\prod_{r=0}^{n-1}\cos\dfrac{(2r+1)\pi}{2n+1}=(-1)^p$$ where $p$ is the number of $r$ such that $\cos\dfrac{(2r+1)\pi}{2n+1}<0\ \ \ \ (1)$
First of all $r<n\implies\dfrac{(2r+1)\pi}{2n+1}<\pi$
So, $(1)$ will occur if $\dfrac\pi2<\dfrac{(2r+1)\pi}{2n+1}<\pi\iff\left\lceil\dfrac{2n-1}4\right\rceil\le r\le n-1$
Observe that $r$ can have $\left\lfloor\dfrac n2\right\rfloor$ distinct values.
Here $2n+1=13\iff n=6,3\le r\le5,p=3$
| {
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Given three non-negatve numbers $a,b,c$. Prove that $1+a^{2}+b^{2}+c^{2}+4abc\geqq a+b+c+ab+bc+ca$ .
Given three non-negatve numbers $a, b, c$. Prove that:
$$1+ a^{2}+ b^{2}+ c^{2}+ 4abc\geqq a+ b+ c+ ab+ bc+ ca$$
Let $t= a+ b+ c$, we have to prove
$$\left(\!\frac{1}{t^{3}}- \frac{1}{t^{2}}+ \frac{1}{t}\!\right)\sum a^{3}+ \left(\!\frac{3}{t^{3}}- \frac{3}{t^{2}}\!\right)\left (\!\sum a^{2}b+ \sum a^{2}c\!\right )+ \left(\!\frac{6}{t^{3}}- \frac{6}{t^{2}}- \frac{3}{t}+ 4\!\right)abc\geqq 0$$
If $0< t< 1$ so
$${\rm LHS}\geqq \left(\frac{3}{t}+ 1\right)\left(\frac{3}{t}- 2\right)^{2}abc\geqq 0$$
If $1< t$ so
$${\rm LHS}= \left(\!\frac{3}{t^{2}}- \frac{3}{t^{3}}\!\right)(\!{\rm Schur.3}\!)+ \frac{1}{t}\left(\!\frac{2}{t}- 1\!\right)^{2}(\!{\rm a.m.}- {\rm g.m.}\!)+ \left(\!\frac{3}{t}+ 1\!\right)+ \left(\!\frac{3}{t}- 2\!\right)^{2}abc\geqq 0$$
(Can you find the way without deviding two cases as above?)
| Partial answer:
Let $a = \frac{1}{2} + a'$, $b = \frac{1}{2} + b'$ and $c = \frac{1}{2} + c'$. Then, from Empy2's hint we have:
$$8a'b'c' + (a'+b')^2 + (b'+c')^2 + (c' + a')^2 ≥ 0$$
which is true if $a', b', c' ≥ 0$. This leaves only the cases when any and only one or all three of $a, b, c$ satisfy $0 < a,b,c < \frac{1}{2}$.
| {
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Why do all primes $n>3$ satisfy $\,309\mid 20^n-13^n-7^n$ Solve the following... $309|(20^n-13^n-7^n)$ in $\mathbb{Z}^+$.
I invested lotof time to it and finally went to WolframAlpha for help by typing...
Solve $309k=20^n-13^n-7^n$ over the integers. It returned the following... Note that all the primes are generated here. Can someone please explain why? Thanks! EDIT: the formulas suggested in the answerare too complicated, but this is so simple!
| $20^6 \equiv 13^6 \equiv 7^6 \equiv 229 \bmod 309$, so $20^n \equiv 13^n + 7^n \mod 309$ if and only if $20^{n+6} \equiv 13^{n+6} + 7^{n+6} \bmod 309$. Since it does work for $n=1$ and $n=5$, but not for $0$, $2$, $3$ or $4$, we find that $20^n \equiv 13^n + 7^n \bmod 309$ if and only if $n \equiv 1$ or $5 \bmod 6$.
All primes except $2$ and $3$ are congruent to $1$ or $5 \bmod 6$. $2$ or $3$ can't divide a number congruent to $1$ or $5 \bmod 6$, so it's not easy for a small number of this form to be composite. The first few composites are $25 = 5^2$, $35 = 5 \cdot 7$, $49 = 7 \cdot 7$, $55 = 5 \cdot 11$, ...
| {
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How to compute a Jacobian using polar coordinates? Consider the transformation $F$ of $\mathbb R^2\setminus\{(0,0)\}$ onto itself defined as
$$
F(x, y):=\left( \frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right).$$
Its Jacobian matrix is
$$\tag{1}
\begin{bmatrix} \frac{y^2-x^2}{(x^2+y^2)^2} & -\frac{2xy}{(x^2+y^2)^2} \\ -\frac{2xy}{(x^2+y^2)^2} & \frac{x^2-y^2}{(x^2+y^2)^2} \end{bmatrix},\quad \text{and its determinant equals}\ \frac{-1}{(x^2+y^2)^2}.$$
The following alternative computation is wrong at (!) and (!!), and I cannot see why.
Let $\phi\colon (0, \infty)\times (-\pi, \pi)\to \mathbb R^2$ be the map $$\phi(r, \theta) =(r\cos \theta, r\sin \theta).$$ Let moreover $$\tag{2}\tilde{F}:=\phi^{-1}\circ F\circ \phi;$$ then, by an easy direct computation, $$\tilde{F}(r, \theta)=\left( \frac1r, \theta\right).$$The Jacobian matrix of $\tilde{F}$ is, thus, $$\tag{!}\begin{bmatrix} \frac{-1}{r^2} & 0 \\ 0 & 1\end{bmatrix} , \quad \text{and its determinant equals }\ \frac{-1}{r^2}.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $\tilde{F}$ are equal. We conclude that the Jacobian determinant of $F$ is $$\tag{!!} \frac{-1}{r^2}=\frac{-1}{x^2+y^2}.$$
The result (!!) is off by a factor of $r^{-2}$ from the correct one, which is given in (1). Equation (!) must also be wrong. Indeed, computing the Jacobian matrix from (2) using the chain rule, and using that
$$
D\phi = \begin{bmatrix} \cos \theta & \sin \theta \\
-r\sin \theta & r\cos \theta\end{bmatrix}$$
and that
$$\tag{!!!}
D(\phi^{-1})= \begin{bmatrix} \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} \\ -\frac{y}{x^2+y^2} & \frac{x}{x^2+y^2}\end{bmatrix},$$
I obtain the result
$$
\begin{bmatrix} \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} \\ -\frac{y}{x^2+y^2} & \frac{x}{x^2+y^2}\end{bmatrix} \begin{bmatrix} \frac{y^2-x^2}{(x^2+y^2)^2} & -\frac{2xy}{(x^2+y^2)^2} \\ -\frac{2xy}{(x^2+y^2)^2} & \frac{x^2-y^2}{(x^2+y^2)^2} \end{bmatrix}\begin{bmatrix} \cos\theta & -r\sin\theta \\ \sin \theta & r\cos \theta\end{bmatrix} = \begin{bmatrix} -\frac1{r^2} & 0 \\ 0 & \frac{1}{r^2}\end{bmatrix},$$
which is different from the matrix in (!), and which gives the correct determinant of $-1/r^4$, as it should be.
Can you help me spot the mistake?
SOLUTION (added at a later time). As answerers pointed out, there is a mistake in (!!!). The correct matrix to be used is
$$ D(\phi^{-1})|_{F\circ \phi(r, \theta)}
= \begin{bmatrix} \frac{\frac{x}{x^2 + y^2}}{\sqrt{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}} & \frac{\frac{y}{x^2 + y^2}}{\sqrt{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}} \\ -\frac{\frac{y}{x^2 + y^2}}{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2} & \frac{\frac{x}{x^2 + y^2}}{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}\end{bmatrix} = \begin{bmatrix}\cos\theta & \sin \theta \\ - r\sin \theta & r\cos\theta \end{bmatrix}.$$
Had I used this matrix, I would have found the correct result for the Jacobian matrix of $\tilde{F}$, which is the equation marked (!). Thus, (!) is actually correct.
My fundamental misunderstanding was the assumption that, because of (2), the Jacobian determinant should be invariant for coordinate changes. This is not true; what follows from (2) is only that
$$
\det D\tilde{F}|_{(r, \theta)}= \det D\phi^{-1}|_{F\circ\phi(r, \theta)}\det D\phi|_{(r, \theta)} \det DF|_{\phi(r, \theta)}.
$$
The first two factors in the right-hand side need not cancel, as I erroneously thought.
| I don't think there is any contradiction here.
Consider the volume form
$$ \omega_{\rm Cart} = dx \wedge dy.$$
Your first calculation shows that the pullback $F^\star(\omega_{\rm Cart})$ is given by
$$ F^\star(\omega_{\rm Cart}) = - \frac{1}{(x^2+y^2)^2}\omega_{\rm Cart}.$$
Now consider the volume form
$$ \omega_{\rm Polar} = dr \wedge d\theta.$$
Your second calculation shows that
$$ F^\star(\omega_{\rm Polar})=-\frac 1 {r^2} \omega_{\rm Polar}. $$
We can use this to recompute $F^\star(\omega_{\rm Cart})$. In view of the fact that
$$ \omega_{\rm Cart} = r \omega_{\rm Polar},$$
we have:
\begin{align}
F^\star(\omega_{\rm Cart}) &= F^\star(r\omega_{\rm Polar}) \\ &= F^\star(r) F^\star(\omega_{\rm Polar}) \\ &= \frac 1 r \left( - \frac 1 {r^2}\omega_{\rm Polar} \right) \\ &= - \frac{1}{r^4} \left(r\omega_{\rm Polar} \right) \\ &= - \frac 1 {r^4} \omega_{\rm Cart}
\end{align}
which is consistent with the first calculation!
As for the application of the chain rule, we have:
$$ (D\bar F)|_{(r, \theta)} = D(\phi^{-1})|_{F\circ \phi(r, \theta)} (DF)|_{\phi(r, \theta)} (D\phi)|_{(r, \theta)}$$
The key point is that you must evaluate $D(\phi^{-1})$ at the point $\left(\frac x { (x^2 +y^2)}, \frac y {(x^2 + y^2)}\right)$, not at the point $(x, y)$.
This is equal to
$$ D(\phi^{-1})|_{F\circ \phi(r, \theta)} = \begin{bmatrix} \frac{\frac{x}{x^2 + y^2}}{\sqrt{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}} & \frac{\frac{y}{x^2 + y^2}}{\sqrt{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}} \\ -\frac{\frac{y}{x^2 + y^2}}{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2} & \frac{\frac{x}{x^2 + y^2}}{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}\end{bmatrix} = \begin{bmatrix}\cos\theta & \sin \theta \\ - r\sin \theta & r\cos\theta \end{bmatrix}$$
which is not the inverse of $(D\phi)|_{(r, \theta)}$.
| {
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Number of Collatz steps for Mersenne numbers I noticed that for all $k \in \mathbb{N} \geq 1$ the following is true (I tested up to $2^{5000}$):
$\text{Collatz_Steps}(2^{2k+1} - 1) + 1 = \text{Collatz_Steps}(2^{2k+2} - 1)$
Where $\text{Collatz_Steps}$ is the number of steps required to reach $1$.
Is there a proof for this?
| Proof sketch:
For a number $2^n-1,$ the first $2n$ steps are the $3x+1$ step and the $x/2$ step in alternating order, until you reach $3^n-1.$
$3^n-1$ is even, so the next step is the $x/2$ step again. We have
$$
\frac{3^n-1}{2} = \sum_{j=0}^{n-1} 3^j
$$
which is odd if $n$ is odd and even if $n$ is even. So for odd $n,$ you get
$$
3\,\cdot \,\frac{3^n-1}{2} +1 = \frac{3^{n+1}-1}{2}
$$
after $2n+2$ steps, while for even $n,$ you get
$$
\frac{3^n-1}{2}
$$
after $2n+1$ steps.
For $n=2k+1,$ you get $\frac{3^{2k+2}-1}{2}$ after $2(2k+1)+2 = 4k+4$ steps.
For $n=2k+2,$ you get $\frac{3^{2k+2}-1}{2}$ after $2(2k+2)+1 = 4k+5$ steps.
| {
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If $f(x)$ Polynomial with real coefficient and $f(0)=1$, $f(2)+f(3)=125$ and $f(x)*f(2x^2)=f(2x^3+x)$ then what is the value of $f(5)$? If $f(x)$ Polynomial with real coefficient and $f(0)=1$, $f(2)+f(3)=125$
and
$f(x)*f(2x^2)=f(2x^3+x)$
then what is the value of $f(5)$?
.
What I tried
$$f(0)=1$$
put x=1 $$f(1)*f(2)=f(3)$$
put x=2 $$f(2)*f(8)=f(18)$$
from this approach, I cant find f(5). Please Suggest a method to solve.
| Here's a somewhat unfulfilling solution.
According to Solve $f(x)f(2x^2) = f(2x^3+x)$, the polynomial $f(x)$ is of the form $f(x)=(x^2+1)^n$ for some $n$.
Using $f(2)+f(3)=125$ we obtain $n=2$, so $f(x)=(x^2+1)^2$. Plugging in $x=5$ we conclude
$$f(5)=(5^2+1)^2=26^2=676.$$
| {
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Solve: $ z^5 = \bar{z} * (- \frac{1}{2} + \frac{\sqrt3}{2} i)$. P.S. if $z = x+y*i$, $\bar{z} = x - y*i$. Solve: $ z^5 = \bar{z} \left(- \frac{1}{2} + \frac{\sqrt3}{2}i\right)$.
P.S. if $z = x+y i$, $\bar{z} = x - y i$.
Seems trivial but I can not solve it. I tried to write $z, \bar{z}$ using $x$ and $y$ and I got nothing. Then I tried by multiplying everything by $z$ and still I got nowhere. Any hint helps.
| You can use polar coordinates. So let $r$ and $\theta$ be such that $z=re^{i\theta}$. Then $\bar{z}=re^{-i\theta}$. Furthermore,
$$-\frac{1}{2}+\frac{\sqrt{3}}{2}i=e^{\frac{2\pi}{3}i}.$$
Thus, we get
$$r^5e^{5i\theta}=re^{-i\theta}\cdot e^{\frac{2\pi}{3}i}\Leftrightarrow r^4e^{6i\theta}=e^{\frac{2\pi}{3}},$$
so that $r^4=1\Leftrightarrow r=1$ and $6i\theta=\frac{2\pi}{3}i\Leftrightarrow \theta=\frac{\pi}{6}$. Hence, $z=e^{\frac{\pi}{6}i}$.
If you want, you can convert this back to a $x+yi$ form by solving $r=\sqrt{x^2+y^2}$ and $\theta=\tan^{-1}\left(\frac{y}{x}\right)$.
| {
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Minimum value of $4$-digit number divided by sum of its digits? If x was a positive 4 digit number and you divide it by the sum of its digits to get the smallest value possible, what is the value of x? For example (1234 = 1234/10)
I got 1099 as my answer however I don't know if this is right or how to prove it.
| Write the $4$-digit number as $x = 1000a+100b+10c+d$, where $a,b,c,d$ are integers from $0$ to $9$ (and $a \neq 0$.)
You want to minimize $A = \dfrac{1000a+100b+10c+d}{a+b+c+d}$, which can be written as
$$
\begin{align*}
A &= \dfrac{999a+99b+9c}{a+b+c+d}+1\\
&\geq \dfrac{999a+99b+9c}{a+b+c+9} + 1 & \text{(since the numerator} > 0\text{, set } d = 9)\\
&= \dfrac{990a+90b-81}{a+b+c+9} + 10\\
&\geq \dfrac{990a+90b-81}{a+b+9+9} + 10 & \text{(since the numerator} > 0\text{, set } c = 9)\\
&= \dfrac{900a-18\times90-81}{a+b+9+9} + 100\\
&= \dfrac{900a-1701}{a+b+18} + 100\\
&\geq \dfrac{900\times 1-1701}{1+b+18} + 100 & \text{(since the } \textbf{denominator} > 0\text{, set } a = 1)\\
&\geq \dfrac{900-1701}{1+0+18} + 100, & \text{(since the numerator} < 0\text{, set } b = 0)\\
\end{align*}
$$
and the minimum attains when $x = 1099$.
| {
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Square root of square as 2/2 power From school I know that
$\sqrt{x^2} = |x|$.
But if we rewrite the above equation in another way
$\sqrt{x^2} = (x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x^1 = x$
then we get another answer.
How is it possible that rewritting an equation changes the answer? And which answer is right, $x$ or $|x|$?
| Let x be a positive number.
So the square of x is $x^2$
Also the square of -x is $(-x)^2 = (-1)^2 x^2 = x^2$.
So we have, $x^2 = (\pm x)^2$
Hence, $\sqrt{x^2} = \sqrt{(\pm x)^2 } = |x|$. As x is positive, |x| = x and as -x is negative |-x| = -(-x) = x
This is simply x because we know in prior that, x is positive.
Now what if x is negative? Then there exists a number x' = -x > 0,
such that $\sqrt{x^2} = \sqrt{(-x')^2} = \sqrt{x'^2}= x' = -x = |x|$.(From the formal case)
Thus in general,$\sqrt{x^2} = |x|$
| {
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Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$
If $$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}= \frac{3\sqrt{3}}{4}$$ for $a,b,c,d>0$
Then
Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$ is
Try: using A.M G.M Inequality
$$1+a^2\geq 2a\;, 1+b^2\geq 2b\;,1+c^2\geq 2c\; 1+d^2\geq 2d$$
$$\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}\geq 4\sqrt{abcd}$$
$$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}\leq \frac{a+b+c+d}{4\sqrt{abcd}}$$
I have edited it, please have a look
Could some help me to solve it , Thanks
| Another way for the proof of the inequality $(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2.$
We obtain: $$(x^2+3)(y^2+3)=x^2y^2+3(x^2+y^2)+9=$$
$$=(xy-1)^2+(x-y)^2+2(x+y)^2+8\geq2((x+y)^2+4).$$
By the same way $$(z^2+3)(t^2+3)\geq2((z+t)^2+4).$$
Id est, by C-S $$\prod_{cyc}(x^2+3)\geq4((x+y)^2+4)(4+(y+z)^2)\geq$$
$$\geq4(2x+2y+2z+2t)^2=16(x+y+z+t)^2.$$
| {
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Probability distribution of a uniform distribution to the third power I have to find a explicit form of probability distribution of $X^3$, if $X \ \mathtt{\sim} \ U[a, b], \ -\infty < a < b < \infty$.
So far I've succesfully done a simpler version, when it's $\ U[0, a], \ \ 0 < a < \infty$:
$F_{X^3}(x) = \mathbb{P}(X^3 \leq x) = \mathbb{P}(X \in [0, \sqrt[3]{x}]) = \int^{\sqrt[3]{x}}_0 \frac{dx}{a} = \frac{\sqrt[3]{x}}{a},$
$f_{X^3}(x) = \frac{dF_{X^3}(x)}{dx} = \dfrac{1}{3ax^{\frac{2}{3}}}$,
where the support is $0 \leq x \leq a^3$.
However, the way of thinking that is above got me lost at the full version of the uniform distribution.
$F_{X^3}(x) = \mathbb{P}(X^3 \leq x) = \mathbb{P}(X \in [-\sqrt[3]{x}, \sqrt[3]{x}]) = \int^{\sqrt[3]{x}}_{-\sqrt[3]{x}} \frac{dx}{b-a} = \frac{2\sqrt[3]{x}}{b-a}$
$f_{X^3}(x) = \frac{dF_{X^3}(x)}{dx} = \dfrac{2}{3x^{\frac{2}{3}} \cdot (b-a)}$, where the support is...?
If the support would be $a^3 \leq x \leq b^3$ and $a < 0$ then the integral
$\int^{b^3}_{a^3} f(x)$ would have no sense for negative values, whereas it should equal $1$.
Or where else did I made a mistake?
| Tip: Don't use $x$ both in the domain's bounds and as the integration's variable. That inevitably leads to confusion. Additionally, we instinctively associate lower and upper case letters as referring to the same thing, so that's the best one to change.
Now if the support for $X$ is $a\leq X\leq b$, then the support for $X^3$ is $a^3\leq X^3\leq b^3$. There are no folds, because $x\mapsto x^3$ is an increasing function over the support for $X$.
$$\begin{align}F_{X^3}(y)&= \mathsf P(X\leq y^{1/3})\\[1ex] &=\mathbf 1_{a^3\leq y\leq b^3} \int_a^{y^{1/3}}\tfrac 1{b-a} \mathrm d x+\mathbf 1_{y>b^3}\\[1ex]&= \phantom{\tfrac{y^{1/3}-a}{b-a}\mathbf 1_{a^3\leq y\leq b^3}+\mathbf 1_{b^3<y}}\\[3ex] f_{X^3}(y)&=\left\lvert\dfrac{\mathrm d ~~}{\mathrm d y}F_{X^3}(y)\right\rvert\\[1ex]&=\phantom{\tfrac{\lvert y^{-2/3}\rvert}{b-a}\mathbf 1_{a^3\leq y\leq b^3}}\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/3210343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is all N that make $2^N + 1$ divisible by 3? When $N = 1$, $2^N + 1 = 3$ which is divisible by $3$.
When $N = 7$, $2^N + 1 = 129$ which is also divisible by $3$.
But when $N = 2$, $2^N + 1 = 5$ which is not divisible by $3$.
A quick lookout can determine that when $N$ is an odd number, $2^N + 1$ is divisible by $3$. If is that correct, how can it be proven? Otherwise, what is the counterexample and the correct $N$?
Can this be made into a more general form, for example $2^N + C$ or something similar?
| If $2^N+1$is divisible by $3$ then in other words $2^N+1$ is a multiple of $3$ .
$2^N+1=3k \implies 2^N=3k-1$ so we want $3k-1$ to be even then we want $3k$ to be odd.
$3k$ is odd only when $k$ is odd.
| {
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"source": "stackexchange",
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Find the limit without using Lhopital this question came out on my analysis exam: Evaluate
$$\lim_{x\to 0}\left(\frac{5^{x^2}+7^{x^2}}{5^x+7^x}\right)^{\frac{1}{x}} $$
I did it using L'hopital rule but is there another way to do this?
| The major part can be handled using squeezing and AM-GM but I need one derivative to calculate the limit
*
*$\color{blue}{(\star)}: \lim_{x\to 0}\left( \frac{5^x + 7^x}{2}\right)^{\frac{1}{x}} = \sqrt{35}$
This is quickly verified when taking the logarithm:
*
*$\frac{\log (5^x + 7^x) - \log 2}{x} \stackrel{x\to 0}{\longrightarrow}f'(0)$ for $f(x)=\log(5^x+7^x)$:
$$f'(0) = \left. \frac{5^x\cdot \log 5 + 7^x\cdot \log 7}{5^x+7^x}\right|_{x=0}=\frac{\log 35}{2} = \log \sqrt{35}$$
Bounding from above:
$$\left(\frac{5^{x^2}+7^{x^2}}{5^x+7^x}\right)^{\frac{1}{x}}\stackrel{AM-GM}{\leq}\left(\frac{5^{x^2}+7^{x^2}}{2\sqrt{5^x\cdot7^x}}\right)^{\frac{1}{x}}\leq \left(\frac{2\cdot7^{x^2}}{2\sqrt{5^x\cdot7^x}}\right)^{\frac{1}{x}} = \frac{7^x}{\sqrt{35}}\stackrel{x \to 0}{\longrightarrow}\frac{1}{\sqrt{35}}$$
Bounding from below:
$$\left(\frac{5^{x^2}+7^{x^2}}{5^x+7^x}\right)^{\frac{1}{x}}\stackrel{AM-GM}{\geq} \left(\frac{2\sqrt{5^{x^2}\cdot 7^{x^2}}}{5^x+7^x}\right)^{\frac{1}{x}} = \left(\sqrt{35} \right)^x \cdot \left(\frac{2}{5^x+7^x}\right)^{\frac{1}{x}}\stackrel{\color{blue}{(\star)} x \to 0}{\longrightarrow}\frac{1}{\sqrt{35}}$$
So, the limit is $\boxed{\frac{1}{\sqrt{35}}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
Here's my working:
Start with $$4+\frac{1}{x}-\frac{1}{x^2}=0$$
Rearranging into standard form:
$$-\frac{1}{x^2}+\frac{1}{x}+4=0$$
Multiply by $-1$ to get a positive leading coefficient $a$:
$$\frac{1}{x^2}-\frac{1}{x}-4=0$$
I'm not sure how to determine my inputs $a,b$ and $c$ with these fractions but I guess $a=\dfrac{1}{x^2}$, $b=\dfrac{1}{x}$ and $c=-4$.
Plugging into quadratic function:
$$x = \frac{-\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
I find this challenging due to the coefficients $a$ and $b$ being fractions.
How can I apply the quadratic formula to $4+\dfrac{1}{x}-\dfrac{1}{x^2}=0$ to arrive at $\dfrac{-1\pm\sqrt{17}}{8}$?
| From
$$4+\left(\frac1x\right)-\left(\frac1x\right)^2=0$$
using the standard formula blindly,
$$\frac1x=\frac{1\pm\sqrt{17}}2$$
and obviously
$$x=\frac2{1\pm\sqrt{17}}.$$
Though this seems to contradict the expected answer, consider
$$\frac2{1\pm\sqrt{17}}=\frac{2(1\mp\sqrt{17})}{1-(\sqrt{17})^2}=\frac{-1\pm\sqrt{17}}8.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$? I need to remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$, and find out what $f(7)$ equals. I am not sure what I've done wrong, but I'm getting 33, which the website I'm using tells me is wrong.
Please help me figure out what I'm doing wrong here.
My Steps:
$$\frac{x^2-11x+28}{x-7} \cdot \frac{x+7}{x+7}$$
$$=\frac{x^3+7x^2-11x^2-77x+28x+196}{x^2-49}$$
$$=\frac{x^3-4x^2-49x+196}{x^2-49}$$
$$=\frac{x^2(x-4)-49(x+4)}{x^2-49}$$
$$=(x-4)(x+4)$$
plug in $7$ for $x$.
$$(7-4)(7+4)=7^2-16=33$$
| To explain why you get the wrong answer, in your second last line, the last term of the numerator is $-49(x-4)$ and not $-49(x+4).$
Faster approach:
$$\frac{x^2-11x+28}{x-7}=\frac{(x-7)(x-4)}{x-7}=x-4$$
Hence to remove the discontinuity at $7$, the right value to redefine at $f(7)$ should be $7-4$.
Factorization should be the way to go, multiplying by the conjugate doesn't serve the purpose here.
| {
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"url": "https://math.stackexchange.com/questions/3213721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can the formula $\frac\pi2=(\frac21)^{1/2}(\frac{2^2}{1\cdot3})^{1/4}(\frac{2^3\cdot4}{1\cdot3^3})^{1/8}\cdots$ prove the irrationality of $\pi$? A less known product formula for $\pi$, due to Sondow, is the following:
$$
\frac{\pi}{2}= \left(\frac{2}{1}\right)^{1/2} \left(\frac{2^2}{1\cdot3}\right)^{1/4} \left(\frac{2^3\cdot4}{1\cdot3^3}\right)^{1/8} \left(\frac{2^4\cdot4^4}{1\cdot3^6\cdot5}\right)^{1/16} \ldots
$$
Is it possible to prove the irrationality of $\pi$ based on this formula, using some sort of convergence acceleration technique leading to a good irrationality measure?
| Dirichlet’s approximation theorem can’t be applied here, as this product formula doesn’t give a sequence of rational approximations due to the fractional exponents.
We can show that this product formula is equivalent to the Wallis product, which does give you a sequence of rational approximations:
Let $a_n(x)$ be the product of the first $n$ terms of
$$\left(\frac{x + 1}{x}\right)^{1/2} \left(\frac{(x + 1)^2}{x(x + 2)}\right)^{1/4} \left(\frac{(x + 1)^3(x + 3)}{x(x + 2)^3}\right)^{1/8} \left(\frac{(x + 1)^4(x + 3)^4}{x(x + 2)^6(x + 4)}\right)^{1/16} \cdots.$$
Then we can see that $a_n(x)$ satisfies this recurrence:
$$a_0(x) = 1, \quad a_{n+1}(x) = \left(\frac{x + 1}{x}\right)^{1/2} \left(\frac{a_n(x)}{a_n(x + 1)}\right)^{1/2}.$$
In the limit as $n \to \infty$, this simplifies to
$$\quad a(x) = \frac{(x + 1)}{x a(x + 1)},$$
and since $\lim_{x \to \infty} a(x) = 1$, we can write
$$a(1) = \frac21 \cdot \frac1{a(2)} = \frac21 \cdot \frac23 \cdot a(3) = \cdots = \frac21 \cdot \frac23 \cdot \frac43 \cdot \frac45 \cdot \frac65 \cdot \frac67 \cdots.$$
However, since the Wallis product doesn’t qualify as lesser-known, I’d imagine that an irrationality measure proof based on it would be well known if it existed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Number of possible integer values of $x$ for which $\frac{x^3+2x^2+9}{x^2+4x+5}$ is integer How many integer numbers, $x$, verify that the following
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5}
\end{equation*}
is an integer?
I managed to do:
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5} = x-2 + \frac{3x+19}{x^2+4x+5}
\end{equation*}
but I cannot go forward.
| Let $t=x+2$, then $$t^2+1\mid 3t+13$$ and thus $$t^2+1\mid t(3t+13)-3(t^2+1)= 13t-3$$
so $$t^2+1\mid 13(3t+13)-3(13t-3) = 178$$
So $$t^2+1\in\{1,2,89,178\}\implies t=\pm 1,0 \implies x\in\{-1,-2,-3\}$$
| {
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"url": "https://math.stackexchange.com/questions/3220135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ Solution
Expanding $(1+x)^{\frac{1}{x}}$ at $x=0$ by Taylor's Formula，we obtain
\begin{align*}
(1+x)^{\frac{1}{x}}&=\exp\left[\frac{\ln(1+x)}{x}\right]=\exp \left(\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(1-\frac{x}{2}+\frac{x^2}{3}+\cdots \right)=e\cdot\exp \left(-\frac{x}{2}+\frac{x^2}{3}+\cdots \right) \\
&=e\left[1+\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e\left(1-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\\
&=e-\frac{ex}{2}+\frac{11}{24}ex^2+\cdots
\end{align*}
Likewise, expanding $e^{(1+x)^{\frac{1}{x}}}$ at $x=0$, we obtain
\begin{align*}
e^{(1+x)^{\frac{1}{x}}}&=(e^e)^{1-\frac{x}{2}+\frac{11}{24}x^2-\cdots}=e^e\cdot (e^e)^{-\frac{x}{2}+\frac{11}{24}x^2+\cdots}\\
&=e^e\cdot\left[1+\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\ln e^e+\frac{1}{2!}\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)^2\ln^2 e^e+\cdots\right]\\
&=e^e \cdot\left[1-\frac{ex}{2}+\frac{1}{24}(11e+3e^2)x^2+\cdots\right]
\end{align*}
Expanding $(1+x)^{\frac{e}{x}}$ at $x=0$, it follows that
\begin{align*}
(1+x)^{\frac{e}{x}}&=\exp\left[\frac{e\ln(1+x)}{x}\right]=\exp \left(e\cdot\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(e-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)=e^e\cdot\exp \left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right) \\
&=e^e\left[1+\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e^e\left[1-\frac{ex}{2}+\frac{1}{24}e(8+3e)x^2+\cdots\right]
\end{align*}
Therefore
\begin{align*}
&\lim_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}\\
=&\lim_{x \to 0}\frac{e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(11+3e)x^2+\cdots\right]-e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(8+3e)x^2+\cdots\right]}{x^2}\\
=&e^e\cdot\frac{1}{8}e\\
=&\frac{1}{8}e^{e+1}
\end{align*}
Please check. Is there any more simpler solution?
| Since both terms, say $A, B$, in numerator tend to same limit $e^e$ we can write $$A-B=B\cdot\frac{\exp (\log A - \log B) - 1}{\log A - \log B} \cdot(\log A - \log B) $$ and thus the numerator can be replaced by $e^e(\log A - \log B) $ or $$e^e\left((1+x)^{1/x}-e\cdot\frac{\log (1+x)}{x}\right)$$ Applying the same technique the above expression can be replaced by $$e^e\cdot e\left(\frac{\log(1+x)}{x}-1-\log\frac{\log(1+x)}{x}\right)$$ or $$e^{e+1}\cdot(u-\log(1+u))$$ where $$u=\frac{\log(1+x)}{x}-1\to 0$$ On the other hand we also know via L'Hospital's Rule or Taylor series that $u/x\to - 1/2$ and therefore the expression $$\frac{u-\log(1+u)}{x^2}=\frac{u-\log(1+u)}{u^2}\cdot\frac{u^2}{x^2}$$ tends to $(1/2)(-1/2)^2=1/8$. The desired limit is thus $e^{e+1}/8$.
In general one should avoid multiplication/division and composition of Taylor series and use the famous ones directly from memory. Often the use of algebraic manipulation combined with standard limits reduces / alleviates the need of any gymnastics with Taylor series.
| {
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"url": "https://math.stackexchange.com/questions/3220990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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The odds in rolling two dice together. Two dice are rolled. I want to see the odds of the following:
$1.$ A sum of $5$.
$2.$ A sum of $8$ or $10$.
$3.$ A sum less than $6$.
$4.$ Not a sum of $7$.
Solution: $1.$ A sum of $5$. $5 = 1+4,2+3,3+2,4+1$. So the odds is $4/36 = 1/9.$
$2.$ A sum of $8$ or $10$. We can express $8 = 2+6,3+5,4+4,5+3,6+2$ and $10 = 4+6,5+5,6+4$. So the odds is $8/36 = 2/9$.
$3.$ A sum less than $6$. We can express $2=1+1$, $3=1+2,2+1$, $4=1+3,2+2,3+1$ and $5 = 1+4,2+3,3+2,4+1$. So the odds is $10/36 = 5/18$.
$4.$ Not a sum of $7$. We can express $7 = 1+6,2+5,3+4,4+3,5+2,6+1$. So the odds is $\frac{36-6}{36} = \frac{30}{36} = \frac56$
Is the solution correct?
| Yes, it states all of the ways the event can happen, then all of the possible events, yielding correct probability.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221120",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How many are the non-negative integer solutions of $ + + + = 16$ where $x < y$?
How many are the non-negative integer solutions of $ + + + = 16$ where $x < y$?
Anyone can explain how to think to approach this type of problem?
The answer is 444.
| Consider cases and find a pattern.
When $x=0$:
$$0+y+z+w=16, y\ge 1, z,w\ge 0 \Rightarrow \\
t+1+z+w=16,t,z,w\ge 0 \Rightarrow \\
t+z+w=15, t,z,w\ge 0 \Rightarrow \\
{15+3-1\choose 3-1}$$
Note: It was used Stars and Bars.
When $x=1$:
$$1+y+z+w=16, y\ge 2, z,w\ge 0 \Rightarrow \\
t+2+z+w=15,t,z,w\ge 0 \Rightarrow \\
t+z+w=13, t,z,w\ge 0 \Rightarrow \\
{13+3-1\choose 3-1}$$
$\vdots$
When $x=7$:
$$7+y+z+w=16, y\ge 8, z,w\ge 0 \Rightarrow \\
t+8+z+w=9,t,z,w\ge 0 \Rightarrow \\
t+z+w=1, t,z,w\ge 0 \Rightarrow \\
{1+3-1\choose 3-1}$$
Hence, it is the sum:
$$\begin{align}\sum_{i=1}^8 {2i-1+3-1\choose 3-1}&=\sum_{i=1}^8 {2i+1\choose 2}=\sum_{i=1}^8 \frac{(2i+1)!}{2!(2i-1)!}=\sum_{i=1}^8 \frac{(2i+1)2i}{2}=\\
&=\sum_{i=1}^8 (2i^2+i)=2\frac{8(8+1)(2\cdot 8+1)}{6}+\frac{8(8+1)}{2}=\\
&=24\cdot 17+36=444.\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan C $ and $ \tan B $
Settings discriminant greater than equal to zero.
$ { (p-1)}^2 - 4p \ge 0 $
This gives $ p \le 3 - 2\sqrt2 $. Or $ p \ge 3 + 2\sqrt2 $
solving both equation
$ A + B + C = \pi $
$ C + B + \frac{\pi}{4} = \pi $
$ C + B = \frac{3\pi}{4} $
Using this to solve both the equation give $ p \in $ real
I found this on Quora.
https://www.quora.com/Let-A-B-C-be-three-angles-such-that-A-frac-pi-4-and-tan-B-tan-C-p-What-are-all-the-possible-value-of-p-such-that-A-B-C-are-the-angles-of-the-triangle
the right method
$ 0 \lt B , C \lt \frac{3\pi}{4} $
Converting tan into sin and cos gives
$ \dfrac {\sin B \sin C}{\cos B \cos C} = p $
Now using componendo and dividendo
$ \frac{\cos (B-C) }{- \cos(B+C) } = \frac{p+1}{p-1} $
We know $ \cos (B+C) = 1/\sqrt2 $
We know the range of $B$ and $C$ $(0, 3π/4)$
Thus the range of $B - C$. $(0, 3π/4 )$
Thus range of $\cos(B+C)$ is $ \frac{ -1}{\sqrt2} $ to $1$
Thus using this to find range gives
$ P \lt 0 $ or $ p \ge 3+ 2\sqrt2 $
| I think that every approach should give correct solution. Different solutions appear if to do mistakes.
Conditions to $\angle B$ and $\angle C$ are WLOG
\begin{cases}
\angle B+\angle C = \dfrac{3\pi}4\\[4pt]
\tan\angle B\tan\angle C = p\\[4pt]
\angle B\in\left(0,\dfrac{3\pi}8\right]\\
\angle C\in\left[\dfrac{3\pi}8,\dfrac{3\pi}4\right).\tag1
\end{cases}
Also, are known relations
$$\tan(a+b) = \dfrac{\tan a + \tan b}{1-\tan a \tan b},\tag{2a}$$
$$\tan\dfrac34\pi = -1,\tag{2b}$$
$$\tan\dfrac38\pi = \frac{1+\cos\frac34\pi}{\sin\frac34\pi}=\sqrt2+1.\tag{2c}$$
Let $x=\tan \angle B$ and $y=\tan \angle C.$
then
$$y = \tan\left(\dfrac{3\pi}4-x\right) = \dfrac{-1-x}{1-x} = \dfrac{x+1}{x-1},\tag3$$
$$xy=p.$$
Therefore,
$$P(p,x) = x^2+x-p(x-1) = 0.\tag4$$
If $\,\underline{x=y=\sqrt2+1}\,$ then
$$P(p,\sqrt2+1) = 4+3\sqrt2 -p\sqrt2 = 0,\\$$
$$p=2\sqrt2+3\tag5.$$
If $\,\underline{x\in (0, \sqrt2+1)}\,$ then
$P(p,z)\,$ has not roots in the interval $\,z\in(-1,0)\,$ and has one root in the interval $\,z\in(0,\sqrt2+1),\,$
\begin{cases}
P(p,-1)P(p,0) > 0\\[4pt]
P(p,0)P(p,\sqrt2+1)<0,
\end{cases}
\begin{cases}
(p+1)p > 0\\[4pt]
p(2\sqrt2+3-p)<0,
\end{cases}
$$p\in(-\infty,0)\cup(2\sqrt2+3,\infty).$$
Taking in account $(5),$ the answer is
$$\color{green}{\boxed{{\phantom{\Big|}\mathbf{p\in(-\infty,0)\cup[2\sqrt2+3,\infty)}.}}\tag6}$$
If the condition $(6)$ is satisfied, then the common solution is
$$(\angle B,\angle C)\in\{(f_\angle(z_1),f_\angle(z_2)), (f_\angle(z_2),f_\angle(z_1))\},\tag7$$
where
$$D=p^2-6p+1 = (p-3-2\sqrt2)(p-3+2\sqrt2),\tag8$$
$$z_{1,2} = \dfrac{p-1\pm\sqrt D}2,\tag9$$
$$f_\angle(z) = \arctan z +\dfrac\pi2(1-\operatorname{sgn} z).\tag{10}$$
Example 1. $\quad p=-\frac32,\quad z\in\{-3,\frac12\},\quad \angle B\approx 27^\circ,\quad \angle C \approx 109^\circ,\quad\angle B+\angle C = 135^\circ =\frac34\pi. $
Example 2. $\quad p=6,\quad z\in\{2,3\},\quad \angle B\approx63^\circ,
\angle C \approx 72^\circ,\quad \angle B +\angle C = 135^\circ -\frac34\pi.$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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The integral solution of $x^{2}-y^{3}=1 (x>1,y>1) $? I know it's a special case of catalan's conjecture.Wiki says its only solution is $(3,2)$.But I cannot work it out.Any help will be appreciated.
| Not finished!
Write $$(x-1)(x+1)=y^3$$
Case 1. $x$ is even then $x+1$ and $x-1$ are relatively prime so $$ x-1 = a^3$$ $$x+1 = b^3$$ where $a<b$ are relatively prime and $ab=y$. Now we have $$2=(b-a)(b^2+ab+a^2)$$
and so:
*
*$b-a=1$ and $b^2+ab+a^2 =2$ so $$ 3a^2+3a+1 =2$$ and this is impossibile.
*$b-a=2$ and $b^2+ab+a^2 =1$ so $$ 3a^2+6a+4=1$$ so $a=-1$ and $b=1$ so $y=-2$ and $x=0$ whivh is not ok again.
Case 2. $x=2z+1$ then $y=2t$ so $$z(z+1)=2t^3$$
Since $z,z+1$ are relatively prime we have
*
*$z=2a^3$ and $z+1=b^3$, so $2a^3=b^3-1= (b-1)(b^2+b+1)$. So $$b-1 = 2p^3$$ $$b^2+b+1=q^3$$ where $p,q$ are relatively prime...
*$z=a^3$ and $z+1=2b^3$, so $2b^3= a^3+1=(a+1)(a^2-a+1)$. So $$a+1 = 2p^3$$ $$a^2-a+1=q^3$$ where $p,q$ are relatively prime...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Evaluate $(\sum_{k=1}^{7} \tan^2(\frac{k\pi}{16})) - \tan^2(\frac{4\pi}{16})$ The question is:
Evaluate
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$
The given answer:$34$
What I've tried:
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+\tan^2\left(\frac{3\pi}{16}\right)+\tan^2\left(\frac{5\pi}{16}\right)+\tan^2\left(\frac{6\pi}{16}\right)+\tan^2\left(\frac{7\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+...+\tan^2\left(\frac{\pi}{2}-\frac{2\pi}{16}\right)+\tan^2\left(\frac{\pi}{2}-\frac{\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\cot^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+\cot^2\left(\frac{2\pi}{16}\right)+\tan^2\left(\frac{3\pi}{16}\right)+\cot^2\left(\frac{3\pi}{16}\right)$$
How do I proceed from here?
| Using $\displaystyle \tan^2(x)+\cot^2(x)=\frac{\sin^4(x)+\cos^4(x)}{\sin^2(x)\cdot \cos^2(x)}=\frac{1-2\sin^2 x\cos^2 x}{\sin^2 x\cos^2 x}.$
So we have $\displaystyle =4\csc^2(2x)-2=2+4\cot^2(2x).$
So our expression is
$\displaystyle 2+4\cot^2(\pi/8)+2+4\cot^2(\pi/4)+2+4\cot^2(3\pi/8)$
$\displaystyle 6+4+4\bigg[\cot^2(\pi/8)+\tan^2(\pi/8)\bigg]$
$\displaystyle =10+4\bigg[2+4\cot^2(\pi/4)\bigg]$
$=10 + 4\bigg[2+4\bigg]=34.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)^2}$.
$(O, R)$ is the circumscribed circle of $\triangle ABC$. $I \in \triangle ABC$. $AI$, $BI$ and $CI$ intersects $AB$, $BC$ and $CA$ respectively at $M$, $N$ and $P$. Prove that $$\large \frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)^2}$$
I have provided my own solution and I would be greatly appreciated if there are any other solutions, perhaps one involving trigonometry. I deeply apologise for the misunderstanding.
| As pointed out in the comments the inequality:
$$\large \frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)}$$
Is not homogeneous and therefore cannot be correct. Take any triangle and any point and even if the given inequality is satisfied fot this configuration then after scaling it by $a$ for sufficiently small $a$ it will stop being correct. But somehow you have managed to convert it to homogeneous inequality:
$$\frac{1}{AI} + \frac{1}{BI} + \frac{1}{CI} \le \frac{2}{R - OI}$$
Which still seems not to be true. And even if it was you obtained it by means of this wrong inequality (no surprise since homogeneous and non-homogeneous inequalities can't be equivalent):
$$\left(\frac{1}{AM} + \frac{1}{BN} + \frac{1}{CP}\right)^2 \le \left(\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP}\right)\left(\frac{1}{AI} + \frac{1}{BI} + \frac{1}{CI}\right)$$
Which is a flawed application of CS inequality. The correct application is:
$$\left(\frac{1}{\sqrt{AM}} + \frac{1}{\sqrt{BN}} + \frac{1}{\sqrt{CP}}\right)^2 \le \left(\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP}\right)\left(\frac{1}{AI} + \frac{1}{BI} + \frac{1}{CI}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3227215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k=4^{-n}~{2n \choose n}.$ I have happened to have proved this sum while attempting to prove another summation. Let
$$S_n=\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k$$
${2k \choose k} $ is the coefficient of $x^0$ in $(x+1/x)^{2k}$.
Consequently, $S_n$ is the coefficient of $x^0$ in $$f(x)= \sum_{k=0}^{2n} {2n \choose k}~\left (x+\frac{1}{x}\right)^{2k}~\left ( \frac{-1}{2}\right)^{k}=\left(1-\left(\frac{x+1/x}{\sqrt{2}}\right)^2 \right)^{2n} = 4^{-n} ~ \left(x^2+\frac{1}{x^2}\right)^{2n}.$$ Finally, the coefficient of $x^0$
in $f(x)$ is $$4^{-n}~{2n \choose n}=S_n.$$
I hope that you will find it interesting and prove it in some other way. Do try!
| We will use
$$
\frac1{1-x}\left(\frac{x}{1-x}\right)^k=\sum_{n=0}^\infty\binom{n}{k}x^n\tag1
$$
and
$$
(1-4x)^{-1/2}=\sum_{k=0}^\infty\binom{2k}{k}x^k\tag2
$$
Extracting the even part of $(1)$
$$
\begin{align}
\sum_{n=0}^\infty\binom{2n}{k}x^{2n}
=\frac12\left[\frac1{1-x}\left(\frac{x}{1-x}\right)^k+\frac1{1+x}\left(-\frac{x}{1+x}\right)^k\right]\tag3
\end{align}
$$
Compute the generating function of the sum we want
$$
\begin{align}
&\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{2k}{k}\binom{2n}{k}\left(-\frac12\right)^kx^{2n}\tag4\\
&=\frac1{2-2x}\sum_{k=0}^\infty\binom{2k}{k}\left(-\frac{x}{2-2x}\right)^k
+\frac1{2+2x}\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{x}{2+2x}\right)^k\tag5\\
&=\frac1{2-2x}\left(1+\frac{4x}{2-2x}\right)^{-1/2}
+\frac1{2+2x}\left(1-\frac{4x}{2+2x}\right)^{-1/2}\tag6\\
&=\frac1{2-2x}\left(\frac{1-x}{1+x}\right)^{1/2}
+\frac1{2+2x}\left(\frac{1+x}{1-x}\right)^{1/2}\tag7\\[6pt]
&=\left(1-x^2\right)^{-1/2}\tag8\\[6pt]
&=\sum_{n=0}^\infty\frac1{4^n}\binom{2n}{n}x^{2n}\tag9
\end{align}
$$
Explanation:
$(4)$: compute the generating function
$(5)$: apply $(3)$
$(6)$: apply $(2)$
$(7)$: simplify
$(8)$: simplify
$(9)$: apply $(2)$
Equating coefficients of $x^{2n}$ gives
$$
\sum_{k=0}^\infty\binom{2k}{k}\binom{2n}{k}\left(-\frac12\right)^k
=\frac1{4^n}\binom{2n}{n}\tag{10}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove $(1 + \frac{3p-3}{p^2-1}) \prod_{\substack{q=3\\q\ \text{prime}}}^{l(p)}(1 + \frac{q+1}{q-1} \frac{1}{p-1})$ goes to 1 $\lim p\to\infty$ Let function $l(p)$ be defined as the largest prime number less than $p$.
For example: $l(7)=5, l(11)=7, l(17)=13$.
Let the function $f(p)$ be defined as follows:
\begin{eqnarray*}
f(p) = \left(1 + \frac{3p-3}{p^2-1}\right) \prod_{\substack{q=3\\q\ \text{prime}}}^{l(p)} \left(1 + \frac{q+1}{q-1} \frac{1}{p-1}\right)
\end{eqnarray*}
Prove that $f(p)$ approaches 1 as $p$ goes to infinity.
I know that $\frac {(3p-3)}{(p^2-1)}$ and $\frac{1}{p-1}$ approach 0 as p gets large, so it makes sense that the $f(p)$ goes to 1.
Here are some example values of $f(p)$.
\begin{eqnarray*}
f(5) &=& \left[(1+ \frac{(3*5 – 3)}{(25-1)} \right] \left[1+ \frac{(3+1)}{(3-1)(5-1)}\right] = 2.25\\
\end{eqnarray*}
\begin{eqnarray*}
f(7) &=& \left[(1+ \frac{(3*7 – 3)}{(49-1)}\right]\left[1+ \frac{(3+1)}{(3-1)(7-1)}\right] \left[1+ \frac{(5+1)}{(5-1)(7-1)}\right] = 2.292
\end{eqnarray*}
$f(11) = 1.955$
$f(13) = 1.947917$
$f(17) = 1.793717$
$f(3137) = 1.154731$
Here is a plot of $f(p)$ up to $p=3137$ and it appears to approach $1$.
| Let $P(n)$ be the set of primes less than $n.$
We have $\lim_{n\to \infty}1+\frac {3n-3}{n^2-1}=1.$ So it suffices to prove that $\lim_{n\to \infty}g(n)=1$ where $$g(n)=\prod_{3\le q\in P(n)}\left(1+\frac {q+1}{q-1}\frac {1}{n-1}\right).$$
Let (as usual) $\pi(n)$ be the number of primes less than $n.$ By elementary means we have $\lim_{n\to \infty}\frac {\pi(n)}{n}=0.$ Hence $$\lim_{n\to \infty}\frac {2(\pi(n)-1)}{n-1}=0.$$
Now if $3\le q<n$ then $1<1+\frac {q+1}{q-1}\frac {1}{n-1}\le 1+\frac {4}{2}\frac {1}{n-1}=1+\frac {2}{n-1}.$
And for $n\ge 4$ the number of terms in the product for $g(n)$ is $\pi(n)-1$. So for $n\ge 4$ we have $$1<g(n)\le \left(1+\frac {2}{n-1}\right)^{\pi(n)-1}=A(n)^{2(\pi(n)-1)/(n-1)}$$ $$\text { where }\quad A(n)=\left(1+\frac {2}{n-1}\right)^{(n-1)/2}. $$
Now $\lim_{n\to \infty}A(n)=e$ (See Footnote) and $\lim_{n\to \infty}2(\pi(n)-1)/(n-1)=0.$
Footnote: For the purposes of this Q it is sufficient that $A(n)\ge 1$ for $n\ge 4 $ and that $\{A(n): 4\le n\in \Bbb N\}$ is a bounded set.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$ if $3\leq aI don't really know if I should use brute force or some kind of theorem, it comes on a calculus past exam and it says:
suppose: $3≤a<b≤8$
prove that $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$
| As Martin R suggested, multiply all sides by conjugate of the middle term (note: $3≤a<b≤8$):
$$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4} \iff \\
\frac{\overbrace{\require{cancel}\cancel{b-a}}^{1}}{6}(\sqrt{1+b} + \sqrt{1+a})≤\overbrace{\cancel{b-a}}^{1}≤\frac{\overbrace{\cancel{b-a}}^{1}}{4}(\sqrt{1+b} + \sqrt{1+a})\iff \\
\text{LHS:} \ \frac{\sqrt{1+b} + \sqrt{1+a}}{6}< \frac{\sqrt{1+b} + \sqrt{1+8}}{6}\le 1 \iff \sqrt{1+b}\le 3 \iff b\le 8 \ \ \color{red}\checkmark\\
\text{RHS:} \ \frac{\sqrt{1+b} + \sqrt{1+a}}{4}>\frac{\sqrt{1+3} + \sqrt{1+a}}{4}\ge 1 \iff \sqrt{1+a}\ge 2 \iff a\ge 3 \ \ \color{red}\checkmark\\
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using complex numbers to find $\cos 5\theta$ I have seen a few examples of finding $\cos 5\theta$ before on here but I was wondering how to just find it by equating the real parts?
So far I have: $$\cos5\theta=\Re(cos\theta+i\sin\theta)^5$$
But how do I know which is the real part and expand from this further? Any hints would be appreciated!
| So here we will use De Moivre's theorem here:
As you stated we will use the fact that $\cos(5\theta)+i\sin(5\theta)=(\cos(\theta)+i\sin(\theta))^5$, so get this just in terms of $\cos(5\theta)$ we can write $\cos(5\theta)=\Re(\cos(\theta)+i\sin(\theta))^5$
So expanding this we have:
$$\cos(5\theta)=\Re(\cos^5\theta+5i\cos^4\theta\sin\theta+10i^2\cos^3\theta \sin^2\theta+10i^3\cos^2 \theta \sin^3\theta +5i^4\cos\theta \sin^4\theta+i^5\sin^5\theta)$$
Since the real parts cannot have an odd power of $\sin\theta$ (since it will not give a real number) we get :
$$\cos(5\theta)=\cos^5\theta+10i^2\cos^3\theta\sin^2\theta+5i^4\cos\theta \sin^4\theta$$
$$\cos(5\theta)=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta \sin^4\theta$$
$$\cos(5\theta)=\cos^5\theta-10\cos^3\theta(1-\cos^2\theta)+5\cos\theta (1-\cos^2\theta)^2$$
$$\cos(5\theta)=\cos^5\theta-10\cos^3\theta+10\cos^5\theta+5\cos\theta-10\cos^3\theta+5\cos^5\theta$$
$$\cos(5\theta)=16\cos^5\theta-20\cos^3\theta+5\cos\theta$$
Hope this helps :)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that generating function coefficients agree with recurrence relation I have the recurrence relation $a_n = 3a_{n-1} + 4a_{n-2}$ with $a_0 = 1, a_1 = 0$. The solution to this is $a_n = \frac{4^n}{5} + \frac{4(-1)^n}{5}$. The generating function is $g(x) = \frac{1-3x}{1-3x-4x^2} = (1-3x)(\sum_{i=0}^{\infty}(-1)^ix^i)(\sum_{k=0}^{\infty}4^kx^k$). I need to show that, for any non-negative integer $n$, the coefficient of $x^n$ is equal to $a_n$ (the solution to the recurrence relation). But how can I find a general formula for this $N$th coefficient, given that there are so many possibilities for what $i$ and $k$ could be?
| Take your generating function, write it as partial fractions:
$\begin{equation*}
g(z)
= \frac{4}{5 (1 + z)} + \frac{1}{5 (1 - 4 z)}
\end{equation*}$
Thus (two geometric series) you have directly:
$\begin{equation*}
[z^n] g(z)
= \frac{4}{5} \cdot (-1)^n + \frac{1}{5} \cdot 4^n
\end{equation*}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x, y \in \mathbb R$: $x^2+y^2=2x^2y^2$ and $(x+y)(1+xy)=4x^2y^2$
Solve the following system of equations. $$\large \left\{ \begin{aligned} x^2 + y^2 &= 2x^2y^2\\ (x + y)(1 + xy) &= 4x^2y^2 \end{aligned} \right.$$
From the system of equations, we have that
$$\left\{ \begin{align*} (x + y)^2 \le 2(x^2 + y^2) = 4x^2y^2\\ 4x^2y^2 = (x + y)(1 + xy) \le \frac{[(x + 1)(y + 1)]^2}{4} \le \frac{(x + y + 2)^4}{4^3} \end{align*} \right.$$
$$\implies (x + y)^2 \le \frac{(x + y + 2)^4}{4^3} \implies |x + y| \le \left|\frac{x + y + 2}{2^3}\right|$$
I don't know what to do next. Please help me solve this problem.
| \begin{align*}
(x+y)^2(1+xy)^2&=(4x^2y^2)^2\\
(x^2+y^2+2xy)(1+xy)^2&=16x^4y^4\\
(2x^2y^2+2xy)(1+xy)^2&=16x^4y^4\\
2xy(1+xy)^3&=16x^4y^4
\end{align*}
So, $xy=0$ or $1+xy=2xy$.
$xy=0$ or $1$.
If $xy=0$, $x=y=0$.
If $xy=1$, $x=y=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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prove $\ln(1+x^2)\arctan x=-2\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{2n+1}x^{2n+1}$ I was able to prove the above identity using 1) Cauchy Product of Power series and 2) integration but the point of posting it here is to use it as a reference in our solutions.
other approaches would be appreciated.
| actually, you can do the product directly, given well-known series
$$\begin{aligned}
\arctan x & = \sum_{n=0}^{\infty} {\frac{(-1)^n x^{2n+1}}{2n+1}}\\
\ln(1+x^2) & = \sum_{n=1}^{\infty} {\frac{(-1)^{n+1} x^{2n}}{n}}
\end{aligned}$$
obviously their product has no even order items, set
$$\arctan x \ln (1+x^2) = \sum_{m=0}^{\infty} {a_{2m+1} x^{2m+1}}$$
for item $x^{2m+1}$, it has pair partitions as $(x,x^{2m}),(x^3,x^{2m-2}),\cdots,(x^{2m-1},x^2)$, thus
$$\begin{aligned}
a_{2m+1}
& = \sum_{n=0}^{m-1} {\frac{(-1)^n}{2n+1} \cdot \frac{(-1)^{m-n+1}}{m-n}} = \sum_{n=0}^{m-1} {\frac{(-1)^{m+1}}{(2n+1)(m-n)}}\\
& = \frac{(-1)^{m+1}}{2m+1} \sum_{n=0}^{m-1} {\frac{2m+1}{(2n+1)(m-n)}} = \frac{(-1)^{m+1}}{2m+1} \sum_{n=0}^{m-1} {\frac{2n+1+2(m-n)}{(2n+1)(m-n)}}\\
& = \frac{(-1)^{m+1}}{2m+1} \left( \sum_{n=0}^{m-1} {\frac1{m-n}} + \sum_{n=0}^{m-1} {\frac2{2n+1}} \right)\\
& = \frac{(-1)^{m+1}}{2m+1} \left( H_{m} + 2\left( \sum_{n=1}^{2m} {\frac1{n}} - \sum_{n=1}^{m} {\frac1{2n}} \right) \right)\\
& = \frac{(-1)^{m+1} (H_{m} + 2H_{2m} - H_{m})}{2m+1} = \frac{(-1)^{m+1} \cdot 2H_{2m}}{2m+1}
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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$(X^2-1)^2+(X^2-1)-12=0$ has 2 solutions only or 3? I am asked to solve for X using substitution:
$(X^2-1)^2+(X^2-1)-12=0$
Let $U = X^2-1$
$U^2+U-12=0$
$U^2+4U-3U-12=0$
$U(U+4)-3(U+4)=0$
$(U+4)(U-3)$
Then:
$U-3=0$
$U=3$
$X^2-1=3$
$X^2=4$
$X=\pm2$
This is the provided solution by my textbook.
However:
$U+4=0$
$U=-4$
$X^2-1=-4$
$X^2=-3$
$X=i\sqrt{3}$
My textbook does not provide $X=i\sqrt{3}$ as a solution. Why is $X=i\sqrt{3}$ not a solution in addition to $\pm2$?
| Probably your textbook is asking for only real solutions and $\pm2$ are only real solutions while $i\sqrt{3}$ is an imaginary solution
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine the stability of an equilibrium point for a given Lagrangian
I have to determine the equilibria and the stability for a Lagrangian given by
\begin{align}
\frac{\dot{x}^2+\dot{y}^2}{2}-V(x,y)
\end{align}
where $V(x,y)=-(x^2+y^2)^4+(2x^2+y^2)^2-1$.
The origin $(0,0)$ is an equilibria, but since the Hessian of the potential energy is indefinite when evaluated at this point, I need an alternative way to move.
I've tried so far to see if $V(x,y)-V(0,0)>0$ in a neighbourhood of the origin, but I can't prove it. The previous point asked me to see if there's any first integral, and I know that $E(x,y,\dot x, \dot y)$ (the energy) is preserved by the "Jacobi function theorem", so maybe I have to use it but I don't know how.
| We have that
$$(2x^2+y^2)^2=x^4+2x^2(x^2+y^2)+(x^2+y^2)^2$$
And if $0<q<1$, we have that $q>q^2$, so we have that (for $0<x^2+y^2<1$)
$$(x^2+y^2)^2 > (x^2+y^2)^4$$
Which means that
$$x^4+2x^2(x^2+y^2)+(x^2+y^2)^2>x^4+2x^2(x^2+y^2)+(x^2+y^2)^2 +(x^2+y^2)^4$$
So
$$(2x^2+y^2)^2-(x^2+y^2)^4 > x^4+2x^2(x^2+y^2)$$
| {
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Use real integral to calculate $\int_R \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$ Problem :
Evaluate $$\int_{-\infty}^{\infty} \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$$
Use only real integral.
What I did :
$$\int_{-\infty}^{\infty} \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$$
$$=2\int_{0}^{\infty} \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$$
$$=2\int_{0}^{\infty} \left( \frac{2\cos(\pi x)}{x^2+2} - \frac{\cos (\pi x)}{x^2 + 1}\right) dx$$
Any easy way to calculate this? or idea like differentiate under the integral sign?
| We can consider a more general function
$$I(t)=2\int_0^\infty\frac{x^2\cos(tx)}{(x^2+1)(x^2+2)}\,\mathrm dx$$
and take its Laplace transform,
$$\begin{align*}
\mathcal{L}\left\{I(t)\right\} &= 2s\int_0^\infty\frac{x^2}{(x^2+s^2)(x^2+1)(x^2+2)}\,\mathrm dx\\
&= 2s\int_0^\infty\left( \frac{1}{(1-s^2)(x^2+1)}+\frac{2}{(s^2-2)(x^2+2)}-\frac{s^2}{(s^4-3s^2+2)(x^2+s^2)}\right )\,\mathrm dx\\
&= \frac{\pi\sqrt{2}}{s+\sqrt{2}}-\frac{\pi}{s+1},\qquad\text{do a bunch of arctan integrals and simplify.}
\end{align*}$$
Now if we use the fact that $\mathcal{L}\left\{e^{-\alpha t}\right\}=\frac{1}{s+\alpha}$, we find
$$I(t)=\pi\sqrt{2}e^{-\sqrt{2}t}-\pi e^{-t}.$$
Plugging in $t=\pi$ gives you what you're looking for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Theorem of Pythagoras - Incorrect Derivation I am trying to solve below question from Coursera Intro to Calculus (link)
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and
$2ab$ units respectively, where $a$ and $b$ are positive real numbers
such that $a$ is greater than $b$. Find an exact expression for the
length of the hypotenuse (in appropriate units).
Below are the choices
$(a - b)^2$
$\sqrt(a^4 + 4a^2b^2 -b^4)$
$a^2 + b^2$
$\sqrt(a^2 + 2ab -b^2)$
$(a + b)^2$
When I attempt to work out the solution (and I even got a 2nd pair of eyes to look at it, but he arrived at the same conclusion), I get this:
$(a^2 - b^2)^2 + (2ab)^2 = x^2$
$a^4 -2a^2b^2 + b^4 + (2ab)^2 = x^2$
$a^4 -2a^2b^2 + b^4 + 4a^2b^2 = x^2$
$a^4 + 2a^2b^2 + b^4 = x^2$
$\sqrt(a^4 + 2a^2b^2 + b^4) = x$
Please help! How to get the correct solution?
| Let $$a' = a^{2} - b^{2}$$ $$ b' = 2ab $$
From the Pythagorean Theorem, $$ c'^{2} = a'^{2} + b'^{2} $$
Substitute both $a'$ and $b'$ into the above equation:
$$ c'^{2} = (a^{2} - b^{2})^{2} + (2ab)^{2} = a^{4} - 2a^{2}b^{2} + b^{4} + 4a^{2}b^{2} = a^{4} + 2a^{2}b^{2} + b^{4}$$
Solve for $c'$ by taking the square root:
$$ c' = \sqrt{a^{4} + 2a^{2}b^{2} + b^{4}} $$
Here, $a^{4} + 2a^{2}b^{2} + b^{4}$ is a perfect square, and $$ \sqrt{a^{4} + 2a^{2}b^{2} + b^{4}} = \sqrt{(a^2 + b^2)^{2}} = a^2 + b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$. This question is taken from book: Exercises and Problems in Calculus, by
John M. Erdman, available online, from chapter 1.1, question $4$.
Request help, as not clear if my approach is correct.
(4) Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$.
Have two conditions, leading to three intervals;
(i) $x \lt \sqrt{-2}$
(ii) $\sqrt{-2} \le x \lt \sqrt{11}$
(iii) $x \ge \sqrt{11}$
(i) no soln.
(ii) $2x^2 = 9 \implies x = \pm\sqrt{\frac 92}$
(iii) no soln.
Verifying:
First, the two solutions must satisfy that they lie in the given interval,
this means $\sqrt{-2} \le x \lt \sqrt{11}\implies \sqrt{2}i \le x \lt \sqrt{11}$.
Am not clear, as the lower bound is not a real one. So, given the two real values of $x = \pm\sqrt{\frac 92}$ need only check with the upper bound.
For further verification, substitute in values of $x$, for interval (ii):
a) For $x = \sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
b) For $x = -\sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
| Perhaps simpler (fewer cases to distinguish):
Note that $|a|=|b|$ (with real numbers $a,b$) means that $a=b$ or $a=-b$.
So here either $x^2+2=x^2-11$, which is absurd, or $x^2+2=-(x^2-11)$, which leads to $2x^2=9$ and so $$x=\pm\frac 32\sqrt 2.$$
Verification: If $x=\pm\frac32\sqrt 2$, then $x^2=\frac 92$ and $|x^2+2|=|\frac{13}2|=\frac{13}2$ and $|x^2-11|=|\frac92-11|=|-\frac{13}2|=\frac{13}2$ as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Show that : $\displaystyle\int_0^{\infty}\frac{\ln(1+x^2)\operatorname{arc\,cot} x}{x}=\frac{π^3}{12}$ I need to prove this :
$I=\displaystyle\int_0^{\infty}\frac{\ln(1+x^2)\operatorname{arc\mkern2mucot} x}{x}=\frac{π^3}{12}$
My try :
$I\displaystyle\int_0^{1}\frac{\ln(1+x^2)\operatorname{arc\mkern2mucot} x}{x}$
$+\displaystyle\int_1^{\infty}\frac{\ln(1+x^2)\operatorname{arc\mkern2mucot} x}{x}$
$y=\frac{1}{x}$ so $dx=-\frac{dy}{y^2}$
$I=\displaystyle\int_0^{1}\frac{\ln(1+x^2)\operatorname{arc\mkern2mucot} x}{x}$
+$\displaystyle\int_0^{1}\frac{(\ln(1+x^2)-2\ln x)\arctan x}{x}$
Now I need use series of $\arctan x=\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{2k+1}$
But I don't know how I complete other integration !!
| The key here is the identity:
$$\mathrm{arccot}x + \arctan x = \arctan \frac{1}{x} + \arctan x = \frac{\pi}{2} \quad \quad \text{forall} \quad x>0$$
Thus,
\begin{align*}
\int_{0}^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x &= \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x + \int_{1}^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x} \, \mathrm{d}x \\
&=\int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x + \int_{0}^{1} \frac{\ln \left ( 1+\frac{1}{x^2} \right ) \arctan x}{x} \, \mathrm{d}x \\
&=\int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x + \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \arctan x}{x}\, \mathrm{d}x - \\
&\quad \quad \quad - \int_0^1 \frac{2 \ln x \arctan x}{x} \, \mathrm{d}x \\
&= \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \left ( \mathrm{arccot} x + \arctan x \right )}{x} \, \mathrm{d}x - \\
& \quad \quad \quad - 2 \int_{0}^{1} \frac{\ln x \arctan x}{x} \, \mathrm{d}x \\
&= \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \left ( \arctan \frac{1}{x} + \arctan x \right )}{x} \, \mathrm{d}x - \\
&\quad \quad \quad - 2 \int_{0}^{1} \frac{\ln x \arctan x}{x} \, \mathrm{d}x \\
&=\frac{\pi}{2} \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{x} \, \mathrm{d}x + \frac{\pi^3}{16} \\
&= \frac{\pi^2}{24} \cdot \frac{\pi}{2} + \frac{\pi^3}{16} \\
&= \frac{\pi^3}{12}
\end{align*}
The latter integrals are rather easy. If needed I can a add a calculation.
Addendum:
For the second integral:
\begin{align*}
\int_{0}^{1} \frac{\ln x \arctan x}{x} \, \mathrm{d}x &= \int_{0}^{1} \frac{\ln x}{x} \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n+1}}{2n+1} \, \mathrm{d}x \\
&=\int_{0}^{1} \ln x \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{2n+1} \, \mathrm{d}x \\
&=\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_{0}^{1}x^{2n} \ln x \, \mathrm{d}x \\
&=-\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\
&= -\beta(3) \\
&=-\frac{\pi^3}{32}
\end{align*}
where $\beta$ is the Dirichlet Beta function.
For the first integral:
\begin{align*}
\int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{x} \, \mathrm{d}x &= \int_{0}^{1} \frac{1}{x} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n}}{n} \, \mathrm{d}x \\
&=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{1} x^{2n-1} \, \mathrm{d}x \\
&= \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} \, \mathrm{d}x \\
&= \frac{\eta(2)}{2}\\
&= \frac{\left ( 1-2^{1-2} \right ) \zeta(2)}{2} \\
&=\frac{\left ( 1-\frac{1}{2} \right ) \zeta(2)}{2} \\
&=\frac{\zeta(2)}{4} \\
&= \frac{\pi^2}{24}
\end{align*}
where $\eta$ is the Dirichlet eta function and $\zeta$ is the Riemann zeta function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Elementary Second partial derivative If $V=\frac{xy} {(x^2+y^2)^2}$ and $x=r\cos\theta$, $y=r\sin\theta$ show that $\frac{\partial^2V} {\partial{r^2}}+\frac{1}{r}\frac{\partial{V}} {\partial{r}}+\frac{1} {r^2} \frac{\partial^2{V}}{\partial{\theta^2}}=0$
My attempt :
$\frac{\partial {V}} {\partial{x}} =y\cdot\frac{y^2-3x^2} {(x^2+y^2)^3}\\$
$\frac{\partial {V}} {\partial{y}} =x\cdot\frac{y^2-3x^2} {(x^2+y^2)^3}$
$\frac{\partial{V} } {\partial{r}}= \frac{\partial {V}} {\partial{x}}\frac{\partial {x}} {\partial{r}}+\frac{\partial {V}} {\partial{y}}\cdot\frac{\partial {y}} {\partial{r}}\\\frac{\partial {V}} {\partial{r}}=\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\}$
$\frac{\partial^2{V} } {\partial{r}^2} =\frac{\partial} {\partial{r}}\biggr\{\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\} \biggr\}\\=$
$\frac{\partial} {\partial{\theta}}\biggr\{\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\} \biggr\} \frac{\partial{\theta} }{\partial{r}}$
At this point it is getting clumsy. Any hint on this.
| We have, by substituting our expressions for $x$ and $y$ $$V=\frac{r^2\cos(\theta)\sin(\theta)}{(r^2\cos^2(\theta)+r^2\sin^2(\theta))^2}=\frac{r^2\cos(\theta)\sin(\theta)}{(r^2(\cos^2(\theta)+\sin^2(\theta)))^2}=\frac{r^2\cos(\theta)\sin(\theta)}{r^4}=\frac{\cos(\theta)\sin(\theta)}{r^2}.$$ We can even utilise a double angle formula $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ to arrive to $$V=\frac{\sin(2\theta)}{2r^2},$$ which will make our derivatives easier.
So now, we compute the partial derivatives $$\frac{\partial V}{\partial r}=-\frac{\sin(2\theta)}{r^3},\quad\frac{\partial^2V}{\partial r^2}=\frac{3\sin(2\theta)}{r^4},$$ $$\frac{\partial V}{\partial\theta}=\frac{\cos(2\theta)}{r^2},\quad\frac{\partial^2V}{\partial\theta^2}=-\frac{2\sin(2\theta)}{r^2}.$$
Then, substituting into the expression $$\frac{\partial^2V} {\partial{r^2}}+\frac{1}{r}\frac{\partial{V}} {\partial{r}}+\frac{1} {r^2} \frac{\partial^2{V}}{\partial{\theta^2}}=\frac{3\sin(2\theta)}{r^4}-\frac{\sin(2\theta)}{r^4}-\frac{2\sin(2\theta)}{r^4}=0,$$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How can I prove this limit converges to the Euler-Mascheroni constant? I'm trying to prove that
$$\lim_{N\to\infty}\;2\left[ \int_0^N \frac{\text{erf}(x)}{x}\,dx - \ln(2N) \right] = \gamma,$$
where $\gamma$ is the Euler-Mascheroni constant.
This looks similar to the definition of $\gamma$,
$$\gamma\equiv \lim_{N\to\infty}\;\left[ \sum_{k=1}^N \frac{1}{k} - \ln N \right] ,$$
so maybe there's a clean proof that manipulates the limit in the first expression into the second limit? I'd prefer proofs that don't involve hypergeometric series as I'm not familiar with them.
| I figured it out thanks to the hint and some further research. The only "extra" result I need is
$$\frac{\Gamma'(z)}{\Gamma(z)}=-\frac{1}{z}-\gamma+z\sum_{k=1}^\infty \frac{1}{k}\cdot\frac{1}{z+k},$$
which is fine with me because Gamelin proves it in Ch. 14 of Complex Analysis.
Starting from my original expression,
$$\lim_{N\to\infty}\;2\left[ \int_0^N \frac{\text{erf}(x)}{x}\,dx - \ln(2N) \right] = \gamma,$$
integrate by parts (set $u=\text{erf}(x)$, $dv = dx/x$), giving
\begin{align}
\\
&= \lim_{N\to\infty}\;2\left[ \ln x\cdot\text{erf}(x)\Big|_0^N - \frac{2}{\sqrt{\pi}}\int_0^N \ln x\cdot e^{-x^2}\,dx - \ln N - \ln 2 \right] \\ \\ \\
&= \lim_{N\to\infty}\;2\left[ - \frac{2}{\sqrt{\pi}}\int_0^N \ln x\cdot e^{-x^2}\,dx - \ln 2 \right] \\ \\ \\
&= -2\left[ \frac{2}{\sqrt{\pi}}\int_0^\infty \ln x\cdot e^{-x^2}\,dx - \ln 2 \right] \\ \\
\end{align}
Now substitute $t=u^2$ to give:
\begin{align}
\\
&= -2\left[ \frac{2}{\sqrt{\pi}} \cdot \frac{1}{4} \int_0^\infty \ln t \cdot t^{-1/2 }e^{-t}\,dt - \ln 2 \right] \\ \\ \\
&= -\frac{1}{\sqrt{\pi}} \int_0^\infty \ln t \cdot t^{-1/2 }e^{-t}\,dt + 2 \ln 2 \\ \\ \\
&= -\frac{1}{\sqrt{\pi}} \frac{d}{dx} \left( \int_0^\infty t^{x-1}e^{-t}\,dt \right) \Bigg|_{x=1/2} + 2 \ln 2 \\ \\ \\
&= -\frac{1}{\sqrt{\pi}} \Gamma'\left(\frac{1}{2}\right) + 2 \ln 2 \\ \\
\end{align}
Now using the result quoted from Gamelin above, we have
\begin{align}
\\
&= -\frac{1}{\sqrt{\pi}} \Gamma\left(\frac{1}{2}\right) \left[ -2-\gamma + \sum_{k=1}^\infty \frac{1}{k}\cdot\frac{1}{2k+1} \right] + 2 \ln 2 \\ \\ \\
&= 2 + \gamma - \left[ \sum_{k=1}^\infty \frac{1}{k}\cdot\frac{1}{2k+1} \right] - 2 \ln 2 \\ \\ \\
&= \gamma
\end{align}
If anyone has a simpler proof, please share.
Appendix:
The sum was harder to evaluate than I expected. Maybe there's a simpler way, but here what I did:
\begin{align}
\sum_{k=1}^\infty \frac{1}{k}\cdot\frac{1}{2k+1} &= \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{2}{2k+1} \right) \\ \\ \\
&= 1 - \sum_{k=1}^\infty \left(\frac{2}{2k+1} - \frac{1}{k+1} \right) \\ \\ \\
&= 1 - 2 \sum_{k=1}^\infty \left(\frac{1}{2k+1} - \frac{1}{2k+2} \right) \\ \\ \\
&= 1 - 2 \sum_{k=3}^\infty \frac{(-1)^{k+1}}{k} \\ \\ \\
&= 1 +2\left( 1 - \frac{1}{2} \right) - 2 \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \\ \\ \\
&= 2 - 2 \ln 2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prob. 5, Sec. 6.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: How to show this function is strictly decreasing using derivative Here is Prob. 5, Sec. 6.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $a > b > 0$ and let $n \in \mathbb{N}$ satisfy $n \geq 2$. Prove that $a^{1/n} - b^{1/n} < (a-b)^{1/n}$. [Hint: Show that $f(x) \colon= x^{1/n} - (x-1)^{1/n}$ is decreasing for $x \geq 1$, and evaluate $f$ at $1$ and $a/b$.]
My Attempt:
We find that for $x > 1$,
$$
\begin{align}
f^\prime(x) &= \frac{1}{n}x^{\frac{1}{n} - 1} - \frac{1}{n}(x-1)^{\frac{1}{n} - 1} \\
&= \frac{1}{n} \left( \frac{x^{1/n}}{x} - \frac{(x-1)^{1/n}}{x-1} \right) \\
&= \frac{1}{nx(x-1)} \left( x^{1/n}(x-1) - x(x-1)^{1/n} \right) \\
&= \frac{x^{1/n}(x-1)^{1/n}}{nx(x-1)} \left( (x-1) - x \right) \\
&= -\frac{1}{nx^{1-\frac{1}{n}} (x-1)^{1-\frac{1}{n}} }.
\end{align}
$$
PS (based on the answer by @auscrypt):
We find that for $x > 1$,
$$
\begin{align}
f^\prime(x) &= \frac{1}{n}x^{\frac{1}{n} - 1} - \frac{1}{n}(x-1)^{\frac{1}{n} - 1} \\
&= \frac{1}{n} \left( \frac{x^{1/n}}{x} - \frac{(x-1)^{1/n}}{x-1} \right) \\
&= \frac{1}{nx(x-1)} \left( x^{1/n}(x-1) - x(x-1)^{1/n} \right) \\
&= \frac{x^{1/n}(x-1)^{1/n}}{nx(x-1)} \left( (x-1)^{1-1/n} - x^{1-1/n} \right) \\
&< 0.
\end{align}
$$
because $1-1/n$ is positive for every $n \in \mathbb{N}$ such that $n \geq 2$ and because $0 < x-1< x$, which implies that
$$ 0 < (x-1)^{1-1/n} < x^{1-1/n},$$
and hence
$$ (x-1)^{1-1/n} - x^{1-1/n} < 0.$$
Thus $f^\prime(x) < 0$ for all $x > 1$. Therefore the function $f$ is strictly decreasing on the interval $[1, +\infty)$. That is, for any $x > 1$ we have $f(x) < f(1)$.
Now if $a > b > 0$, then $a/b > 1$, and so we have
$$ f(a/b) < f(1),$$
that is,
$$ \left(\frac{a}{b}\right)^{\frac{1}{n}} - \left(\frac{a}{b} - 1\right)^{\frac{1}{n}} < 1,$$
which amounts to
$$ \frac{a^{1/n} - (a-b)^{1/n} }{ b^{1/n} } < 1, $$
which implies
$$ a^{1/n} - (a-b)^{1/n} < b^{1/n},$$
and hence
$$ a^{1/n} - b^{1/n} < (a-b)^{1/n}, $$
as required.
Are there any issues with this proof?
| There is an error; the factorisation $x^{1/n}(x-1) - x(x-1)^{1/n} = x^{1/n} (x-1)^{1/n} ((x-1)-x)$ is incorrect and should be $x^{1/n}(x-1) - x(x-1)^{1/n} = x^{1/n} (x-1)^{1/n} ((x-1)^\frac{n-1}{n}-x^\frac{n-1}{n})$.
To correctly prove this, note that since $\frac{1}{n}-1$ is negative we have $x^{\frac{1}{n}-1}$ is decreasing. This implies $(x-1)^{\frac{1}{n}-1} > x^{\frac{1}{n}-1} $ since $x-1<x$, and so $f'(x)$ is clearly negative as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\sum_{k=1}^{n} \frac{1}{k(k+1)}$ I have just started learning sums. I need to evaluate the following sum:
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k(k+1)}$$
$$a_{1} = \frac{1}{2}, a_{2} = \frac{1}{6}, a_{3} = \frac{1}{12}, a_{4} = \frac{1}{20}, a_{5} = \frac{1}{30}, a_{6} = \frac{1}{42}, ...$$
I tried splitting it to partial fractions:
$$\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}$$
$$1 = A(k+1) + Bk$$
$$1 = Ak + A + Bk$$
$$\begin{cases} A = 1 \\ 0 = A + B \quad \Rightarrow B = -1\end{cases}$$
$$\Longrightarrow S_{n} = \sum_{k=1}^{n} \Big( \frac{1}{k} - \frac{1}{k+1} \Big)$$
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k+1}$$
Wolframalpha gives answer $S_{n} = \frac{n}{n+1}$
Help's appreciated. P.S: Are the tags okay?
| "Another way" to write things down, playing with index translation in the discrete sum:
Starting from your last equation:
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k+1}$$
you can translate indices $k=k'+1$ in the first sum:
$$S_{n} = \sum_{k'=0}^{n-1} \frac{1}{k'+1} - \sum_{k=1}^{n} \frac{1}{k+1}$$
then make apparent the "extra" terms:
$$S_{n} = 1+ \sum_{k'=1}^{n-1} \frac{1}{k'+1} - \sum_{k=1}^{n-1} \frac{1}{k+1}-\frac{1}{n+1}$$
and you get your result:
$$S_{n} = 1 -\frac{1}{n+1}$$
| {
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"url": "https://math.stackexchange.com/questions/3256519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int_{-\infty}^{+\infty} \frac { e^{t}( e^{2t} -1)} {t(1+e^{2t})^2 }dt$
How to prove that
$$\int_{-\infty}^{+\infty} \frac { \pi e^{t}( e^{2t} -1)} {4t(1+e^{2t})^2 }d t=C$$
where $C$ is Catalan's constant.
Here is one of my ideas. Let: $$F(a)=\int_{-\infty}^{+\infty} \frac { \pi e^{at}( e^{2t} -1)} {4t(1+e^{2t})^2 }d t$$ for $0<a\leq 1$, then $$\displaystyle F'(a)=\int_{-\infty}^{+\infty} \frac { \pi e^{at}( e^{2t} -1)} {4(1+e^{2t})^2 }d t$$
Now put $x=e^t$ then $$\displaystyle F'(a)=\int_{0}^{+\infty} \frac { \pi x^{a-1}( x^{2} -1)} {4(1+x^{2})^2 }dx$$
put $y=x^2$, then $$\displaystyle F'(a)=\frac {\pi}8 \int_{0}^{+\infty} \frac { y^{\frac a2}( y -1)} {(1+y)^2 }d y =?$$
| $$I=\int_{-\infty}^\infty \frac{e^t(e^{2t}-1)}{t(1+e^{2t})^2}dt\overset{e^t=x}=\int_0^\infty \frac{(x^2-1)}{(x^2+1)^2 \ln x}dx$$
Consider: $$I(a)=\int_0^\infty \frac{(x^a-1)}{(x^2+1)^2 \ln x}dx\Rightarrow I'(a)=\int_0^\infty \frac{x^a}{(x^2+1)^2}dx\overset{x^2=y}=\frac12 \int_0^\infty \frac{y^{a/2}y^{-1/2}}{(y+1)^2}dy$$
$$=\frac12B\left(\frac{a+1}{2},2-\frac{a+1}{2}\right)=\frac12 \Gamma\left(\frac{a+1}{2}\right)\Gamma\left(2-\frac{a+1}{2}\right)=\frac12 \left(\frac{1-a}{2}\right)\frac{\pi}{\sin\left(\frac{\pi(a+1)}{2}\right)}$$
But we have that $I(0)=0$, so integrating back we need to find:
$$I=\frac{\pi}{4}\int_0^2 \frac{1-a}{\sin\frac{\pi(a+1)}{2}}da=\frac1{\pi} \int_\frac{\pi}{2}^\frac{3\pi}{2} \frac{\pi-y}{\sin y}dy =\frac{2}{\pi} \int_{0}^\frac{\pi}{2}\frac{t}{\sin t}dt$$
$$\overset{IBP}=-\frac{2}{\pi}\int_0^\frac{\pi}{2}\ln\left(\tan\frac{t}{2}\right)dt\overset{\tan \frac{t}{2}=e^x}=-\frac{4}{\pi}\int_0^1 \frac{\ln x}{1+x^2}dx$$
$$=-\frac{4}{\pi}\sum_{n=0}^\infty \int_0^1 x^{2n}\ln x dx=\frac{4}{\pi}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}=\frac{4 G}{\pi}$$
| {
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How do I solve quadratic double inequalities? I have two questions involving quadratic double inequalities.
Firstly, what are the steps to get the solution for the following?
$0\le(x+2)^2\le4$
My my thought was to separate the inequality into
$0\le(x+2)^2$ and $(x+2)^2\le4$
Which would then allow me to take the square root of each side of the and.
$0\le x+2$ and $0\ge x+2$ and $x+2 \le 2$ and $x+2 \ge -2$
Which could be simplified to $x+2 \le 2$ and $x+2 \ge -2$
And combined into
$-2 \le x+2 \le 2$
Is this the proper way to think about/solve this problem? Is there a better way to approach it?
Secondly, what strategy could I use to square both sides of
$-2\le x+2\le2$ to get back $0\le(x+2)^2\le4$
Thanks!
| Option:
1) Set $y=x+2$;
$0 \le y^2 \le 4.$
$f(y)=√y$ is an increasing function:
$0\le \sqrt{y^2} \le 2$;
With $ \sqrt{y^2}=|y| $ we get:
$0\le |y| \le 2$;
$-2 \le y\le 2$, or $-2 \le x+2 \le 2$.
2) Rewrite:
$-2 \le y \le 2$ as $|y| \le 2$.
Note $|y| \ge 0$.
Hence:
$|y| \le 2 $ implies
$ |y||y|\le 2 |y| \le 2\cdot 2$, .
$ y^2 \le 4.$
| {
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An identity associated with the centroid of a triangle
Four year ago, I am looking for a proof of my identity as follows:
Let $ABC$ be a triangle, let $G$ be the centroid of $ABC$. Let $P$ be any point on the plane. Let $H, N, O$ on the plane such that: $\overrightarrow{HN}:\overrightarrow{NG}:\overrightarrow{GO}=3:1:2$ , then:
$$PA^2+PB^2+PC^2+PH^2-4PN^2=OA^2+OB^2+OC^2$$
Four special case:
1) When $O$ is the circumcenter, $H$ is the orthocenter, $N$ is Nine point center:
$$PA^2+PB^2+PC^2+PH^2-4PN^2=3R^2$$
2) When $O$ is the circumcenter, $H$ is the orthocenter, $N$ is Nine point center, and $P$ lie on Nine point center.
$$PA^2+PB^2+PC^2+PH^2=4R^2$$
3) when $G \equiv O \equiv H \equiv N$ then $$PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3PG^2$$
4) when $O$ is the circumcenter and $P$ is the Nine point center then
$$NA^2+NB^2+NC^2+NH^2=3R^2$$
| Michael's proof is more elegant, but it is also possible to hack it out with vectors.
Let $G$ be the origin of a cartesian coordinate system, and let boldface lowercase letter denotes the position of the uppercase letter. Then $\mathbf{a}+\mathbf{b}+\mathbf{c}=\vec{0}$, $\mathbf{h}=4\mathbf{n}=-2\mathbf{o}$, so
\begin{align*}\require{cancel}
LHS&=(\mathbf{p}-\mathbf{a})^2+(\mathbf{p}-\mathbf{b})^2+(\mathbf{p}-\mathbf{c})^2+(\mathbf{p}-\mathbf{h})^2-4(\mathbf{p}-\mathbf{n})^2\\
&=2\mathbf{p}\cdot(4\mathbf{n}-\mathbf{a}-\mathbf{b}-\mathbf{c}-\mathbf{h})+\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2+\mathbf{h}^2-4\mathbf{n}^2\\
&=\cancel{2\mathbf{p}\cdot\vec{0}}+\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2+12\mathbf{n}^2\\
RHS &=(\mathbf{a}-\mathbf{o})^2+(\mathbf{b}-\mathbf{o})^2+(\mathbf{c}-\mathbf{o})^2\\
&=\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2-\cancel{2\mathbf{o}\cdot(\mathbf{a}+\mathbf{b}+\mathbf{c})}+3\mathbf{o}^2\\
&=\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2+3\mathbf{o}^2\\
&=\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2+12\mathbf{n}^2.
\end{align*}
| {
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Development of the sine function into the power series I would like to check my solution for the development of the real sine function, $f(x) = \sin x$ into the power series. Here is my solution:
First, we have that $\mathcal{D}_f = \mathbb{R}$. Now, we have to check if the given function is infinitely diferentiable and where:
$$f'(x) = \cos x = \sin (x + \frac{\pi}{2}),$$
$$f''(x) = (\sin(x + \frac{\pi}{2}))' = \cos(x + \frac{\pi}{2}) = \sin (x + \frac{2\pi}{2}),$$
$$f'''(x) = (\sin(x + \frac{2\pi}{2}))' = \cos(x + \frac{2\pi}{2}) = \sin(x + \frac{3\pi}{2}),$$
$$\vdots$$
$$f^{(n)}(x) = \sin(x + \frac{n\pi}{2}).$$
So, we, have that given function is infinitely diferentiable on the set $\mathbb{R}$.
Now, notice the point $x_0 = 0$, in which we are going to try to develop the given function.
Because $f$ is infinitely diferentiable, from that follows that $f$ is also continuous with all of its derivatives up to the $n$ - th order, in the arbitrarily neighborhood $U$ of the point $x_0 = 0$, and it has the derivative of the $n + 1$ - st order in $U$. Now, we have that $f$ fulfills the conditions of the theorem from here:
Checking my understanding of the process of developing function into power series
Now, we have that the $f$ can be represented with Maclaurin series:
$$f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + R_n(x) =$$
$$= \sum\limits_{n = 0}^{\infty} \frac{\sin \frac{n \pi}{2}}{n!} x^n + R_n(x).$$
Next, I need to examine if $\lim\limits_{n \to \infty} R_n(x) = 0$:
We can notice that $\forall n \in \mathbb{N}$, $\forall x \in \mathbb{R}$, $|f^{(n)}(x)| \le 1$.
Further, we can notice that for $\forall h \in \mathbb{R}$, $f$ is infinitely diferentiable in $[x_0 - h, x_0 + h]$ and that there is a constant $M = 1$, such that $\forall n \in \mathbb{N}$ and $\forall x \in [x_0 - h, x_0 + h]$ it is $|f^{(n)}(x)| \le M$.
So, We have that conditions of the lema from the question I already linked, are fulfilled. Because of that, we can conclude that $\lim\limits_{n \to \infty} R_n(x) = 0$ for $\forall x \in [x_0 - h, x_0 + h]$, or in this case $\forall x \in \mathbb{R}$.
Conclulsion:
We can conclude that the function $f(x) = \sin x$, can be developed into the power series
$$\sum\limits_{n = 0}^{\infty} \frac{\sin \frac{n \pi}{2}}{n!} x^n$$
in the neighborhood of the point $x_0 = 0$.
Edit:
I tried to simplify the power series I got, to try and get the shape suggested in the answer by Chris Custer. I think I did it:
$$\sum\limits_{n = 0}^{\infty} \frac{\sin \frac{n \pi}{2}}{n!} x^n = \frac{0 x^0}{0!} + \frac{1 x^1}{1!} + \frac{0 x^2}{2!} + \frac{(-1) x^3}{3!}+\frac{0 x^4}{4!} + \frac{1 x^5}{5!} + \frac{0 x^6}{6!} + \frac{(-1) x^7}{7!} + \cdots =$$
$$= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!} + \frac{x^{13}}{13!} - \frac{x^{15}}{15!} + \cdots = $$
$$= (-1)^0 x^{2\cdot 0 + 1} + (-1)^1 \frac{x^{2\cdot 1 + 1}}{(2\cdot 1 + 1)!} + (-1)^2 \frac{x^{2\cdot 2 + 1}}{(2\cdot 2 + 1)!} + (-1)^3 \frac{x^{2\cdot 3 + 1}}{(2\cdot 3 + 1)!} + (-1)^4 \frac{x^{2\cdot 4 + 1}}{(2\cdot 4 + 1)!} + (-1)^5 \frac{x^{2\cdot 5 + 1}}{(2\cdot 5 + 1)!} + \cdots = $$
$$= \sum\limits_{n = 0}^{\infty} (-1)^n \frac{x^{2n + 1}}{(2n + 1)!}$$
Please, could you tell me if my solution is correct, and if it isn't could you tell me where I made a mistake?
| Take this as a supplement, relying on the (omnipotent) power series of the exponential function
$$\exp( z ) \:=\:\sum_{k=0}^\infty \frac{z^k}{k!}\:=\:1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + \ldots$$
in the complex plane, which is given instead of developed here.
Papa Rudin's Prologue(*) is dedicated to it and defines the exponential function by the above series (as do many others), being valid for all $\,z\in\mathbb C\:\!$, in other words, the radius of convergence is $\infty\,$.
Pursuing this definition line one obtains $\,\cos\,$ and $\,\sin\,$ as the real and imaginary part of $\,\exp(ix)$ for $x\in\mathbb R$:
$$\exp(ix)\:=\:\sum_{k=0}^\infty \frac{(ix)^k}{k!}
\;=\;\underbrace{\sum\nolimits_{k=0}^\infty\: (-1)^k\frac{x^{2k}}{(2k)!}}_{\cos x} \:+\:i\:\underbrace{\sum\nolimits_{n=0}^\infty\:(-1)^n\frac{x^{2n+1}}{(2n+1)!}}_{\sin x}$$
This corresponds to the result you have accomplished above.
Furthermore, the "neighbourhood of the point $_0=0$" you mention is the largest possible, namely all of $\mathbb R$.
* Walter Rudin: "Real and Complex analysis"
| {
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Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+7=(1-\sqrt{x+2})^2$ # square both sides
(Use perfect square formula on right hand side $a^2-2ab+b^2$)
$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$ # lhs radical is removed, rhs use perfect square formula
$3x+7=1+2(\sqrt{x+2})+x+2$ # simplify
$3x+7=x+3+2\sqrt{x+2}$ # keep simplifying
$2x+4=2\sqrt{x+2}$ # simplify across both sides
$(2x+4)^2=(2\sqrt{x+2})^2$
$4x^2+16x+16=4(x+2)$ # now that radical on rhs is isolated, square both sides again
$4x^2+12x+14=0$ # a quadratic formula I can use to solve for x
For use int he quadratic function, my parameters are: a=4, b=12 and c=14:
$x=\frac{-12\pm\sqrt{12^2-(4)(4)(14)}}{2(4)}$
$x=\frac{-12\pm{\sqrt{(144-224)}}}{8}$
$x=\frac{-12\pm{\sqrt{-80}}}{8}$
$x=\frac{-12\pm{i\sqrt{16}*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{4i*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{-4\sqrt{5}}}{8}$ #since $4i*i\sqrt{5}$ and i^2 is -1
This is as far as I get:
$\frac{-12}{8}\pm\frac{4\sqrt{5}}{8}$
I must have gone of course somewhere further up since the solution is provided as x=-2.
How can I arrive at -2?
|
$\sqrt{3x+7}-\sqrt{x+2}=1$
$3x+7=(1 \color{red}{\mathbf{ \,-\, }}\sqrt{x+2})^2$ # square both sides
You want: $3x+7=(1 \color{blue}{\mathbf{ \, + \,}}\sqrt{x+2})^2$
Note: not only $x=-2$ solves this equation, also $x=-1$.
| {
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How to determine whether $\Bbb P ( B_{t_1} \in [x - c , x + c ], \ldots , B_{t_n} \in [x -c , x + c ])$ is decreasing in $x$? Let $\phi_t (z) := \frac{1}{\sqrt{2\pi t}} e^{-\frac {z^2} {2t}}$. By intuition $\omega : [0,\infty ) \to [0,1]$
\begin{align}
\omega (x) &:= \Bbb P ( B_{t_1} \in [x - c , x + c ], \ldots , B_{t_n} \in [x -c , x + c ])\\ &=\int_{x-c}^{x+c} \phi_{t_1} (y_1) \int_{x-c}^{x+c} \phi_{t_2-t_1} (y_2 - y_1) \ldots \int_{x-c}^{x+c} \phi_{t_n -t_{n-1}} (y_n - y_{n-1}) \text d y_n \ldots\text d y_1
\end{align}
should be decreasing in $x$. For $n=1$ this is easily done by derivating in $x$. Does anyone see a better way for this case here?
| $\newcommand{\PRx}[2]{\Bbb{P}_{#1} (#2)}$
Let $c>0$ and $G\subset (0,\infty)$ a finite set of time points. The function $\omega : \Bbb R \to [0,1]$ given by $\omega (x) := \Bbb P_x ({\vert{B_s}\vert \leq c\ \forall s \in G})$ is radial nonincreasing.
The strategy is induction over $\vert G \vert$. For $\vert G \vert =1$ the derivative of $\omega (x)$ is $\phi_t (x+c) - \phi_t (x-c) \leq 0$. Let $G$ be arbitrary, $s = \min G$ and write $\tilde{G} := \{t-s : t \in G\setminus \{s\} \}$. Then
\begin{align*}
\omega (x) = \int_{-c}^c \phi_s (y-x) \Bbb P_y ({\vert{B_r}\vert\leq c \ \forall r\in \tilde{G}}) \text d y = \int_{-c}^c \phi_s (y-x) \tilde{\omega} (y) \text d y
\end{align*}
where $\tilde{\omega}(y) := \PRx{y}{\vert{B_r}\vert\leq c \ \forall r\in \tilde{G}}$ is radial nonincreasing by assumption. Let $x\geq 0 $. Then by differentiation of parameter integrals
\begin{align*}
\omega ' (x) &= \int_{-c}^c \frac{\partial}{\partial x} \phi_s (y-x) \tilde{\omega} (y) \text d y = \int_{-c}^c \frac{y-x}{s} \phi_s (y-x) \tilde{\omega} (y) \text d y\\
&= \int_{-c-x}^{c-x} \frac{y}{s} \phi_s (y) \tilde{\omega} (y+x) \text d y
\end{align*}
which is clearly non-positive if $x\geq c$. In the other case rewrite above as
\begin{align*}
&\int_0^{c-x} \frac{y}{s} \phi_s (y) \tilde{\omega} (y+x) \text d y + \int_{-c-x}^0 \frac{y}{s} \phi_s (y) \tilde{\omega} (y+x) \text d y\\
&= \int_0^{c-x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x+y) \text d y - \int_0^{c+x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x -y) \text d y\\
&= \int_0^{c-x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x +y) \text d y - \int_0^{c-x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x -y) \text d y -\int_{c-x}^{c+x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x -y) \text d y\\
&= \int_0^{c-x} \frac{y}{s} \phi_s (y) ( \tilde{\omega} (x +y) - \tilde{\omega} (x -y) )\text d y -\int_{c-x}^{c+x} \frac{y}{s} \phi_s (y) \tilde{\omega} (x -y) \text d y\\
&\leq 0
\end{align*}
since $\tilde{\omega} (x +y) - \tilde{\omega} (x -y) \leq 0$ due to the fact that $\vert{x-y}\vert \leq \vert{x+y}\vert$. Therefore the assertion follows by induction.
| {
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Evaluate $\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......$
Prove that $$\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......=12e-5$$
$$
e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+......
$$
I have no clue of where to start and I am not able to find the general expression of the nth term.
| Your series is apparently
$$\sum_{n=1}^{\infty} \frac{3n^2+n+5}{n!}$$
| {
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Sum of fractions. Let $a_1 < a_2 <a_3 < a_4$ be positive integers such that $\sum_{i=1}^{4} \frac {1}{ a_i}$=$ 11/6$. Then $a_4-a_2=?$
I've tried to find the $a_i$'s but couldn't.
Is there a general way to solve such equations?
| Note that if $2\leq a_1<a_2<a_3<a_4$ then
$$\frac{77}{60}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\geq \sum_{i=1}^{4} \frac {1}{ a_i}=\frac{11}{6}$$
which is a contradiction. Hence $a_1=1$.
Now we are left with $2\leq a_2<a_3<a_4$ such that
$$\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}=\frac{5}{6}.$$
Similarly if $3\leq a_2<a_3<a_4$ then
$$\frac{47}{60}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\geq \frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}=\frac{5}{6}.$$
which is a contradiction. Hence $a_2=2$.
Then we have to solve
$$\frac{1}{a_3}+\frac{1}{a_4}=\frac{1}{3}$$
with $3\leq a_3<a_4$. With $a_3=3$ and $a_3\geq 6$ we have again a contradiction, so $a_3$ can be $4$ or $5$.
Can you take it from here?
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Solutions to a system of three equations with Pythagorean triples Is there any solution to this system of equations where $x,y,z,s,w,t\in\mathbb{Z}$, none are $0$.
\begin{align*}
x^2+y^2=z^2\\\
s^2+z^2=w^2\\\
x^2+t^2=w^2
\end{align*}
EDIT: Thank you zwim for the answer. Maybe I should explain where this came from. Well these six numbers correspond to another four numbers $a,b,c,d\in\mathbb{Z}$ in such a way that if you take any two numbers among $a,b,c,d$ their difference produces a perfect square. There is six pairs so it corresponds to finding six perfect squares that satisfy the system of equations above. In the end we have a set of numbers that all differ between each other by a perfect square. I now wonder if there is a set of five numbers with such a property.
| Time... began to write, and I will add a few words....
Do there exist four distinct integers such that the sum of any two of them is a perfect square?
This is equivalent to solving the following system of equations:
$$\left\{\begin{aligned}& b+a=x^2 \\&b+c=y^2\\&b+f=z^2\\&a+c=e^2\\&a+f=j^2\\&c+f=p^2\end{aligned}\right.$$
Let: $F,T,R,D$ - any asked us integers.
For ease of calculation, let's make a replacement.
$$q=(8F^2+4FT-T^2)R^2+2(T+2F)RD-D^2$$
$$k=(8F^2+8FT+2T^2)R^2+2(T+2F)RD$$
$$s=-T^2R^2+2(T+2F)RD-D^2$$
$$t=(8F^2+12TF+3T^2)R^2+2(T+2F)DR-D^2$$
Then the solutions are of the form:
$$x=s^2+k^2-t^2+2(t-k-s)q$$
$$y=t^2+k^2-s^2+2ks-2tk$$
$$z=s^2+k^2-t^2$$
$$e=t^2+k^2+s^2-2kt-2ts$$
$$j=t^2+s^2-k^2+2ks-2ts$$
$$p=3s^2+3k^2+3t^2-6kt-6st+8ks+2(t-k-s)q$$
$$b=\frac{x^2+y^2-e^2}{2}$$
$$a=\frac{e^2+x^2-y^2}{2}$$
$$c=\frac{e^2+y^2-x^2}{2}$$
$$f=\frac{2z^2+e^2-x^2-y^2}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A First Course in Mathematical Analysis, David Alexander Brannan, ch 1, problem 11 Self studying Brannan's book. For some problem, I am not seeing how they got the solution.
Theorem 3. Arithmetic Mean-Geometric Mean Inequality:
For any positive real numbers $a_{1},a_{2},\dots,a_{n}$ we have
$$
(a_{1}a_{2}\dots a_{n})^{\frac{1}{n}}\leq\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}.
$$
Problem 11 Use theorem 3 with the $n+1$ positive numbers
$1,1+\frac{1}{n},1+\frac{1}{n},\dots,1+\frac{1}{n}$ to prove that,
for any positive integer $n$
$$
(1+\frac{1}{n})^{n}\leq(1+\frac{1}{n+1})^{n+1}.
$$
Book solution:
\begin{align*}
((1+\frac{1}{n})^{n})^{\frac{1}{n+1}}&\leq \frac{1}{n+1}\times(n+1+n\times\frac{1}{n})\\
&= \frac{n+2}{n+1}\\
&= 1+\frac{1}{n+1};
\end{align*}
taking the $(n+1)$th power of this last inequality, by the Power
Rule, we deduce that
$$
(1+\frac{1}{n})^{n}\leq(1+\frac{1}{n+1})^{n+1}.
$$
I am not seeing how Brannan got the RHS of the first line of his solution. I am having trouble seeing how it maps to theorem 3. I tried a few different things and am (embarrassingly) stuck. Can someone please explain this pedantically? Thanks.
Update:
Here is my updated answer based on input from @Lord Shark the Unknown
(for the RHS expansion and subsequent simplification except for a
few steps at the end that I filled in). From @peek-a-boo's answer
I am gathering that (for reasons I don't understand) we are able to
expand theorem 3 to include an $(n+1)$th term, though the definition
of theorem 3 doesn't provide for one as it also doesn't provide for
the LHS having a power of $\frac{1}{n+1}$, but
rather a power of $n$. I'd like to know why we can make those changes
to theorem 3 and have it still hold (is it because of induction or ?). Anyway, here is my own updated format and solution based on their input:
First raise both sides of the inequality to the power $\frac{1}{n+1}$
to get the inequality into a form similar to theorem 3 so that we
can use theorem 3's RHS. Then:
\begin{align*}
((1+\frac{1}{n})_{1}(1+\frac{1}{n})_{2}\cdots(1+\frac{1}{n})_{n})^{\frac{1}{n+1}}\leq & \frac{1}{n+1}\times(1+(1+\frac{1}{n})+(1+\frac{1}{n})+\cdots+(1+\frac{1}{n})) \\
= & \frac{1}{n+1}\times(1+n(1+\frac{1}{n}))\\
= & \frac{1}{n+1}\times(1+n+1)\\
= & \frac{n+2}{n+1}\\
= & \frac{n}{n+1}+\frac{2}{n+1}\\
= & 1-\frac{1}{n+1}+\frac{2}{n+1}\\
= & 1+\frac{1}{n+1}
\end{align*}
| The AM of $1$ and $n$ copies of $1+\frac1n$ is
$$\frac1{n+1}\left(1+\left(1+\frac1n\right)+
\cdots+\left(1+\frac1n\right)\right)=\frac1{n+1}
\left(1+n\left(1+\frac1n\right)\right)=\frac1{n+1}\left(1+n+1\right)
=\frac{n+2}{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}.$
Prove that $$S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\frac{e}{2}-1.$$
$$
S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}\\
=\frac{1}{2}\bigg[\frac{1}{3.0!}+\frac{1}{4.1!}+\frac{1}{5.2!}+\frac{1}{6.3!}+\dots\bigg]=\frac{1}{2}\bigg[\Big(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots\Big)-\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big)\bigg]\\=\frac{e}{2}-\frac{1}{2}\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big)
$$
How do I proceed further as I am stuck with the last infinite series.
| Hint: Take two derivation of both side of
$$\dfrac{e^x-1-x-\dfrac12x^2}{x}=\sum_{n=2}^{\infty}\dfrac{x^n}{(n+1)!}$$
then let $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to factor a fourth degree polynomial I'm working on a math problem but I am having a hard time figuring out the method used by my textbook to make this factorization:
$$x^4 + 10x^3 + 39x^2 + 70x + 50 = (x^2 + 4x + 5)(x^2 + 6x + 10)$$
I've tried to see if this equation can be factored by grouping or by long division to no avail. Any help would be greatly appreciated.
| For all real $k$ we obtain:
$$x^4+10x^3+39x^2+70x+50=$$
$$=(x^2+5x+k)^2-25x^2-k^2-10kx-2kx^2+39x^2+70x+50=$$
$$=(x^2+5x+k)^2-((2k-14)x^2+(10k-70)x+k^2-50).$$
Now, we'll choose $k$ such that we'll get a difference of squares.
For which we need $$25(k-7)^2-(2k-14)(k^2-50)=0$$ or
$$(k-7)(2k^2-25k+75)=0$$ or
$$(k-7)(k-5)(2k-15)=0.$$
We see that only $k=7.5$ is valid and we obtain:
$$$x^4+10x^3+39x^2+70x+50=(x^2+5x+7.5)^2-(x^2+5x+6.25)=$$
$$=(x^2+5x+7.5)^2-(x+2.5)^2=(x^2+4x+5)(x^2+6x+10).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 7,
"answer_id": 5
} |
$bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b)$ My book has used the equality $$bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b)$$ But its proof is not given.
When I open the LHS, I get $b^2c-bc^2+c^2a-ca^2+a^2b-ab^2$. I do not know how to proceed next.
| $bc(b-c)+ca(c-a)+ab(a-b)=c[b(b-c)+a(c-a)]+ab(a-b)=c[b^2-bc+ac-a^2]+ab(a-b)=c[b^2-a^2+ac-bc]+ab(a-b)=c[(b-a)(b+a)-c(b-a)]-ab(b-a)=c(b-a)(b+a-c)-ab(b-a)=(b-a)[c(b+a-c)-ab]=-(a-b)(bc+ac-c^2-ab)=-(a-b)[c(b-c)-a(b-c)]=-(a-b)(c-a)(b-c)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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minimum value of of $(5+x)(5+y)$
If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$
Then find the minimum value of $(5+x)(5+y)$
What I try
$$(5+x)(5+y)=25+5(x+y)+xy$$
$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$
I am finding $f(x,y)=22+5(x+y)+(x+y)^2$
How do I solve it? Help me please
| Hint: From the equation $$y^2+xy+x^2-3=0$$ we get
$$y_{1,2}=-\frac{x}{2}\pm\sqrt{3-\frac{3}{4}x^2}$$ so you will obtain the two functions
$$f_1(x)=(5+x)\left(5-\frac{x}{2}+\sqrt{3-\frac{3}{4}x^2}\right)$$
or
$$f_2(x)=(5+x)\left(5-\frac{x}{2}-\sqrt{3-\frac{3}{4}x^2}\right)$$
to consider.
Doing this we get $$(5+x)(5+y)\geq 16$$, the equal sign holds if $$x=y=-1$$ as Mr. Rozenberg stated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$
I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if someone has a more straightforward solution.
So applying the sum formula for sine and doing simple algebra we have:
$$\lim_{x\to 0} \frac{\cos{2} \,(\sin{x^2}-\sin{x})}{x} - \frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
The first limit is easy to evaluate and is equal to $-\cos{2}$. However, the second limit is harder, as it follows:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
I came across a solution by using the following sum-to-product identity:
$$\cos{A}-\cos{B}=-2\sin{\Big(\frac{A+B}{2}\Big)} \sin{\Big(\frac{A-B}{2}\Big)}$$
Setting $A=x^2$ and $B=x$, we have that
$$\cos{x^2}-\cos{x}=-2\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}$$
This is my only point of concern whether I applied the identity correctly. The rest of it flows more easily:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} = \lim_{x\to 0} \frac{-2\sin{2}\,\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}}{x}$$
$$= -2 \sin{2} \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} \lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}\Big)} $$
The first limit can be solved as it follows:
$$\lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} = \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)} \Big(\frac{x^2+x}{2}\Big)}{x \Big(\frac{x^2+x}{2}\Big)} = 1 \cdot \lim_{x \to 0} \frac{x^2 + x}{2x} = \frac{1}{2} $$
The second limit is equal to zero
$$\lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}}\Big)=0$$
| The second limit can be evaluated as follows:
$\quad \quad \quad \lim_{x\to0} \frac{sin2(cosx^2 - cosx)}{x}$
$\quad \quad \quad \lim_{x\to0} sin2 \cdot (\frac{cosx^2 - cosx}{x})$
$\quad \quad \quad \lim_{x\to0} sin2 \cdot (\frac{cosx^2}{x} - \frac{cosx}{x})$
$\quad \quad \quad \lim_{x\to0} sin2 \cdot (\frac{1}{x} \cdot (cosx^2 - cosx))$
$\quad \quad \quad sin2 \cdot \lim_{x\to0} \frac{1}{x} \cdot \lim_{x\to0} (cosx^2-cosx)$
By the multiplication rule for limits
$\quad \quad \quad sin2 \cdot \lim_{x\to0} (\frac{1}{x} \cdot (cosx^2-cosx))$
Substituting $x = 0$ for the latter expression
$\quad \quad \quad sin2 \cdot \lim_{x\to0} (\frac{1}{x} \cdot 0)$
$\quad \quad \quad sin2 \cdot \lim_{x\to0} (0)$
The second part is the limit of a constant, which is just the constant, therefore
$\quad \quad \quad sin2 \cdot 0$
Therefore, the entire second limit evaluates to $0$, and the answer to your question is the first limit; $-cos(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx$ I'm currently working on a definite integral and am hoping to find alternative methods to evaluate. Here I will to address the integral:
\begin{equation}
I_n = \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx
\end{equation}
Where $n \in \mathbb{N}$. We first observe that when $n = 2k + 1$ ($k\in \mathbb{Z}, k \geq 0$) that,
\begin{equation}
I_{2k + 1} = \int_0^\frac{\pi}{2}\ln^{2k + 1}\left(\tan(x)\right)\:dx = 0
\end{equation}
This can be easily shown by noticing that the integrand is odd over the region of integration about $x = \frac{\pi}{4}$. Thus, we need only resolve the cases when $n = 2k$, i.e.
\begin{equation}
I_{2k} = \int_0^\frac{\pi}{2}\ln^{2k}\left(\tan(x)\right)\:dx
\end{equation}
Here I have isolated two methods.
Method 1:
Let $u = \tan(x)$:
\begin{equation}
I_{2k} = \int_0^\infty\ln^{2k}\left(u\right) \cdot \frac{1}{u^2 + 1}\:du = \int_0^\infty \frac{\ln^{2k}\left(u\right)}{u^2 + 1}\:du
\end{equation}
We note that:
\begin{equation}
\ln^{2k}(u) = \frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0}
\end{equation}
By Leibniz's Integral Rule:
\begin{align}
I_{2k} &= \int_0^\infty \frac{\frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0}}{u^2 + 1}\:du = \frac{d^{2k}}{dy^{2k}} \left[ \int_0^\infty \frac{u^y}{u^2 + 1} \right]_{y = 0} \nonumber \\
&= \frac{d^{2k}}{dy^{2k}} \left[ \frac{1}{2}B\left(1 - \frac{y + 1}{2}, \frac{y + 1}{2} \right) \right]_{y = 0} =\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \Gamma\left(1 - \frac{y + 1}{2}\right)\Gamma\left( \frac{y + 1}{2} \right) \right]_{y = 0} \nonumber \\
&=\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \frac{\pi}{\sin\left(\pi\left(\frac{y + 1}{2}\right)\right)} \right]_{y = 0} = \frac{\pi}{2}\frac{d^{2k}}{dy^{2k}} \left[\operatorname{cosec}\left(\frac{\pi}{2}\left(y + 1\right)\right) \right]_{y = 0}
\end{align}
Method 2:
We first observe that:
\begin{align}
\ln^{2k}\left(\tan(x)\right) &= \big[\ln\left(\sin(x)\right) - \ln\left(\cos(x)\right) \big]^{2k} \nonumber \\
&= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right)
\end{align}
By the linearity property of proper integrals we observe:
\begin{align}
I_{2k} &= \int_0^\frac{\pi}{2} \left[ \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right) \right]\:dx \nonumber \\
&= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \int_0^\frac{\pi}{2} \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right)\:dx \nonumber \\
& = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j F_{n,m}(0,0)
\end{align}
Where
\begin{equation}
F_{n,m}(a,b) = \int_0^\frac{\pi}{2} \ln^n\left(\cos(x)\right)\ln^{m}\left(\sin(x)\right)\:dx
\end{equation}
Utilising the same identity given before, this becomes:
\begin{align}
F_{n,m}(a,b) &= \int_0^\frac{\pi}{2} \frac{d^n}{da^n}\big[\sin^a(x) \big] \cdot \frac{d^m}{db^m}\big[\cos^b(x) \big]\big|\:dx \nonumber \\
&= \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[ \int_0^\frac{\pi}{2} \sin^a(x)\cos^b(x)\:dx\right] = \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{1}{2} B\left(\frac{a + 1}{2}, \frac{b + 1}{2} \right)\right] \nonumber \\
&= \frac{1}{2}\frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right]
\end{align}
Thus,
\begin{equation}
I_{2k} = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \frac{1}{2}\frac{\partial^{2k }}{\partial a^j \partial b^{2k - j}}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right]_{(a,b) = (0,0)}
\end{equation}
So, I'm curious, are there any other Real Based Methods to evaluate this definite integral?
| Slightly different way - use well known results
$$\int_0^{\frac{\pi}{2}}\tan^ay\;dy=\frac{\pi}{2}\frac{1}{\sin \frac{\pi}2(a+1) } $$
( this integral is considered in this site probably many times.)
$$\frac{1}{\sin x}=\frac{1}{x}+\sum _{n=1}^\infty (-1)^n\left ( \frac{1}{x-n\pi}+ \frac{1}{x+n\pi}\right )$$
and differentiate with respect to $a$ as many times as necessary.
| {
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"url": "https://math.stackexchange.com/questions/3294446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
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Compute $\gcd(a+b, 2a+3b)$ if $\gcd(a,b) = 1$ A question from a problem set is asking to compute the value of $\gcd(a+b, 2a+3b)$ if $\gcd(a+b) = 1$, or if it isn't possible, prove why.
Here's how I ended up doing it:
$\gcd(a,b) = 1$ implies that for some integers $x$, and $y$, that $ax+by = 1$.
Let $d = gcd(a+b, 2a+3b)$. This implies:
$\implies \text{d is divisible into }2(2a+3b) - 4(a+b) = 2b\cdots (1)$
$\implies \text{d is divisible into} 6(a+b) - 2(2a+3b) = 2a\cdots (2)$
Statement $(1)$ implies that $d$ divides $2by$ for some integer $y$
Statement $(2)$ implies that $d$ divides $2ax$ for some integer $x$
This implies that $d$ is divisible into $2(ax+by)$, which implies:
$\gcd(a+b, 2a+3b) =\text{ either 1 or 2}$
Thus the result is not generally determinable as it takes $2$ possible values.
Are my assumptions and logic correct? If not, where are the errors?
Thank you!
| It's simple if you use matrices. Suppose a positive integer $d$ divides both $a + b$ and $2a + 3b$. This can be restated as $d$ dividing both entries of the vector
\begin{equation}\begin{bmatrix} 1 & 1 \\ 2 & 3\end{bmatrix}\begin{bmatrix} a \\ b\end{bmatrix}\end{equation}
Multiplying on the left by the inverse matrix, which has integer coefficients, $d$ must therefore also divide each entry of
\begin{equation}\begin{bmatrix} 3 & -1 \\ -2 & 1\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 2 & 3\end{bmatrix}\begin{bmatrix} a \\ b\end{bmatrix}\end{equation}
\begin{equation}=\begin{bmatrix} a \\ b\end{bmatrix}\end{equation}
Thus $d$ divides both $a$ and $b$ and hence is $1$. Thus $gcd(a+b,2a + 3b) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find polynomials $p(x) $ such that $(x+10)p(2x)=(8x-32)p(x+6) $ when $p(1)=210$ Hi I have polynomials as below form.
$(x+10)p(2x)=(8x-32)p(x+6)$, when $p(1)=210$
I try to assign some $x$ eg. $0.5$ , $-6$ but can not find the correct relation. Any advice or guidance on this would be greatly appreciated, Thanks.
| Let $S$ be the set of zeroes of $p$.
If $x \in S$, $x \ne -4$ then $$\underbrace{(x+4)}_{\ne 0}p(2x-12) = 8(x-10)p((x-6)+6) = 8(x-10)p(x) = 0$$
so $2x-12 \in S$.
Notice that $8 \in S$ since plugging in $x = 4$ gives $14p(8) = 8(4-4)p(10) = 0$.
Therefore $4 = 2\cdot 8 -12 \in S$ and then $-4 = 2\cdot 4 - 12 \in S$ so $\{8,-4,4\} \subseteq S$.
@Vasily Mitch noticed that $\deg p = 3$ so the only solution is $S = \{8,-4,4\}$ or $$p(x) = a(x-8)(x-4)(x+4)$$
for some $a \in \mathbb{R}$. From $p(1) = 210$ it follows $a = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability to get a card in a card game Yesterday, while playing a card game, we had an argument about probabilities.
The game settings (for 4 players): there are 2 red cards ($R$) and 3 blue cards ($B$). At the beginning of a round, those 5 cards are shuffled and one is taken out. Then the card are distributed one by one to the 4 players.
*
*What is the probability that 2 red card are present?
*What is the probability that only 1 red card is present?
*What is the probability that only 1 red card is present knowing that I have a blue card?
$P(2R)$
For this one, there is $\frac{2}{5}$ chances that a red card has been taken out and $\frac{3}{5}$ that a blue card has been taken out. $P(2R) = \frac{3}{5}$.
$P(1R) = \frac{2}{5}$ in a similar way.
This problem can also be considered the other way around. We got a set of 5 cards, and we pick them one by one until we have a set of 4, thus leaving one out.
For the first card, you have a $\frac{2}{5}$ chances to pick a red card. For the second card, you have a $\frac{1}{4}$ chances to pick a red card. Thus, you have a $\frac{1}{10}$ chance to pick 2 reds with the 2 first card.
If you do that for all combinations (with a tree to avoid mistakes), you get $\frac{1}{10}$ for each of the 6 branch ending with 2 reds. Thus, in total, you have $\frac{6}{10} = \frac{3}{5}$ to have 2 reds in play.
Next the conditional probability: $P_B(1R) = \frac{P(B\bigcap1R)}{P(B)}$.
$P(B)$ is the probability that I get a blue card. Thus, I think it is
$P(B) = \frac{3}{4}*P(3B, 1R) + \frac{2}{4}*P(2B, 2R) = \frac{3}{4}*\frac{2}{5} + \frac{2}{4}*\frac{3}{5} = \frac{3}{5}$.
Next, and that's the one I am unsure about: $P(B\bigcap1R) = P(B)*P(1R) = \frac{3}{5}*\frac{2}{5} = \frac{6}{25}$.
Why am I unsure? If I recall correctly, this multiplication can be done if both events are independent. Can we say that those 2 events are independent? What if they are not?
And thus: $P_B(1R) = \frac{6}{15}$
How wrong am I? Thanks for the help :)
|
What is the probability that two red cards are present?
As you observed, the probability that two red cards are present is equal to the probability that a blue card was removed, which is
$$\Pr(\text{two red cards are present}) = \Pr(\text{a blue card is removed}) = \frac{3}{5}$$
as you found
As for your failed attempt, observe that there are $\binom{5}{2}$ ways to pick two positions for the red cards. Two red cards can be found in the first four positions in $\binom{4}{2}$ ways. Hence, the probability that two red cards are in the first four positions is
$$\Pr(\text{two red cards are present}) = \frac{\dbinom{4}{2}}{\dbinom{5}{2}} = \frac{6}{10} = \frac{3}{5}$$
What was your mistake?
You did not take into account the probability of selecting a blue ball?
Your first calculation is correct. The probability that red balls are in positions 1 and 2 is
$$\Pr(\text{red, red}) = \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{10}$$
When you calculated the probability that red balls are in positions 1 and 3, you did not take into account the probability that a blue ball was picked second.
$$\Pr(\text{red, blue, red}) = \frac{2}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} = \frac{1}{10}$$
The fact that both calculations yielded $\frac{1}{10}$ is not a coincidence. Let's do those two calculations again, this time writing out all five terms.
\begin{align*}
\Pr(\text{red, red, blue, blue, blue} & = \frac{2}{5} \cdot \frac{1}{4} \cdot \frac{3}{3} \cdot \frac{2}{2} \cdot \frac{1}{1} = \frac{3!2!}{5!} = \frac{1}{10}\\
\Pr(\text{red, blue, red, blue, blue} & = \frac{2}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} \cdot \frac{2}{2} \cdot \frac{1}{1} = \frac{3!2!}{5!} = \frac{1}{10}
\end{align*}
In fact, the probability that the two red balls appear in any two positions of the sequence is always $1/10$ since there are $\binom{5}{2} = 10$ equally likely ways to place the red balls in the sequence of two red and three blue balls.
What is the probability that only one red card is present?
As you observed, the probability that only one red card is present is equal to the probability that a red card is removed, which is
$$\Pr(\text{only one red card is present}) = \Pr(\text{a red card is removed}) = \frac{2}{5}$$
as you found.
What is the probability that only one red card is present given that you have a blue card?
Since you are equally likely to have any of the cards, the probability that you have a blue card is $3/5$.
Clearly, your blue card could not have been removed. Therefore, the card that is removed must be one of the other four cards, of which two are red. Since a red card must be removed if only one red card is present, the probability that you only one red card is present given that you have a blue card is
$$\Pr(\text{only one red card is present} \mid \text{you have a blue card}) = \frac{2}{4} = \frac{1}{2}$$
which is the answer to your question.
What was your mistake?
If we denote the event only one card is present by $1R$ and the event that you have a blue card by $B$, then the probability that you have a blue card and only one red card is present is
$$\Pr(1R \cap B) = \Pr(B)\Pr(1R \mid B) = \frac{3}{5} \cdot \frac{1}{2} = \frac{3}{10}$$
How could you have seen this directly?
If you write the sequences of three blue and two red cards in which a blue card is in the first position (representing the card you hold), there are three ways for a red card to be in the fifth position, namely those in which the other red card is in the second, third, or fourth positions. Hence, three of the ten sequences of three blue and two red cards have you holding a blue card with only red card present in the game.
| {
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1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) above it We can observe the following -
$1 - 1$
$2 - 2 $
$3 - 2$
$4 - 4$
$5 - 4$
. . .
$8 - 8$
$9 - 8$
. . .
We can notice that $ 4^{th}$ term is 4 and similarly, the $ 8^{th}$ term is 8.
So the $ 1025^{th}$ term must be 1024 as $ 1024^{th} $ term starts with 1024.
So the value of $ 1025^{th}$ term is $ 2^{10} $ .
Is there any other method to solve this question?
| The first term is $2^0=1$; the next $2^1=2$ terms are equal to $2^1=2$; the next $2^2=4$ terms are equal to $4$ and so on. Since
$$
2^0+2^1+2^2+\dots+2^n=2^{n+1}-1
$$
the term at place $1023$ is $512$. The next $2^{10}$ terms are equal to $1024$.
| {
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Modular arithmetic. $x \equiv 1 \pmod {m^k}$ implies $x^m \equiv 1 \pmod {m^{k+1}}$ Show that for $k \gt 0$ and $m \ge 1$, $x \equiv 1 \pmod {m^k}$ implies $x^m \equiv 1 \pmod {m^{k+1}}$
This question has already been asked in SE (Show that for $k \gt 0$ and $m \ge 1$, $x \equiv 1 \pmod {m^k}$ implies $x^m \equiv 1 \pmod {m^{k+1}}$.), but I think it´s not really answered. ( the hint given was enough but I didn’t realize it before)
I tried with Fermat´s little theorem but I get nowhere.
note: If $a \equiv b \pmod m$, then $a \cdot t \equiv b \cdot t \pmod {m \cdot t}$ with $t \gt 0$ (don´t know if this is useful)
Any help would be appreciated. Thanks.
| We have $x^m - 1 = (x-1) (x^{m-1} + x^{m-2} + \cdots + 1)$. Now, by hypothesis, $m^k \mid x-1$. On the other hand, since $k > 0$, we also have $x \equiv 1 \pmod{m}$, so $x^{m-1} \equiv 1 \pmod{m}$, $x^{m-2} \equiv 1 \pmod{m}$, ..., $1 \equiv 1 \pmod{m}$. Therefore, $x^{m-1} + x^{m-2} + \cdots + 1 \equiv 1 + 1 + \cdots + 1 = m \equiv 0 \pmod{m}$, so $m \mid x^{m-1} + x^{m-2} + \cdots + 1$.
| {
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$a, b$ are the integer part and decimal fraction of $\sqrt7$ find integer part of $\frac{a}{b}$
$a, b$ are the integer part and decimal fraction of $\sqrt7$
find integer part of $\frac{a}{b}$
using calculator :
$\sqrt 7$ = 2.645
$\frac{a}{b} = \frac{2}{0.6} = \frac{20}{6} = 3.333$
integer part of $\frac{a}{b} = 3$
how to do it without calculator? if i don't know $\sqrt7$ part
| Note that $$ 2.5<\sqrt 7<3$$
$$ \frac {a}{b} = \frac {2}{\sqrt 7 -2} = \frac {2(\sqrt 7+2)}{3}$$
$$3= \frac {2( 2.5+2)}{3} < \frac {2(\sqrt 7+2)}{3} <\frac {2( 3+2)}{3}<4$$
Thus the integer part of $\frac {2}{\sqrt 7 -2}$ is $3$
| {
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Find the Range of $y=\frac{x+a}{x^2+bx+c^2}$ Given that $x^2-4cx+b^2 \gt 0$ $\:$ $\forall$ $x \in \mathbb{R}$ and $a^2+c^2-ab \lt 0$
Then find the Range of $$y=\frac{x+a}{x^2+bx+c^2}$$
My try:
Since $$x^2-4cx+b^2 \gt 0$$ we have Discriminant
$$D \lt 0$$ $\implies$
$$b^2-4c^2 \gt 0$$
Also
$$x^2+bx+c^2=(x+a)^2+(b-2a)(x+a)+a^2+c^2-ab$$
Hence
$$y=\frac{1}{(x+a)+b-2a+\frac{a^2+c^2-ab}{x+a}}$$
But Since $a^2+c^2-ab \lt 0$ We can't Use AM GM Inequality
Any way to proceed?
| If a number $p$ is in the range of the $y$ you have given, it means $\frac{x+a}{x^2+bx+c^2} = p$ for some $x$.
Or, in other words, that equation has a real solution.
=> $px^2+bpx+pc^2 = x+a$ For a real value of $x$
=> the discriminant of $px^2+(bp-1)x+pc^2-a$ is non-negative.
After some rearrangement we have any $p$ where the following inequality stands true is in the range of y.
=> $p^2(b^2-4c^2) + p(4a-2b) + 1 \geq 0$
The discriminant upon expansion turns out to be $16(a^2+c^2-ab)$ which is given to be strictly negative. As you have noted, the leading coefficient is positive. This means the given expression is always positive so every value of $p$ satisfies the condition. So, $p$ takes all real values. This means the range of the given function is $(-\infty, \infty)$.
| {
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Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
I would like a little help, so that I can finish solving this exercise.
So far, I've got
$$
\frac{\frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h)^2}{(x-a-h)^2(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h)^2}{(x-a-h)^2(x-a)^2}}{ \frac{h}{1}}
= \frac{(x-a)^2 - (x-a-h)^2}{h(x-a-h)^2(x-a)^2}
$$
In the numerator maybe I can continue with a difference of squares, but I'm a little confused.
| hint
Begin by putting
$$x-a=b$$
then
$$\frac{1}{(b-h)^2}-\frac{1}{b^2}=$$
$$\frac{b^2-(b-h)^2}{b^2(b-h)^2}=$$
$$\frac{h(2b-h)}{b^2(b-h)^2}.$$
And if we divide by h, one find
$$\frac{2(x-a)-h}{(x-a)^2(x-a-h)^2}.$$
| {
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Minimizing $9 \sec^2{x} + 16 \csc^2{x}$ Find the minimum value of $$9 \sec^2{x} + 16 \csc^2{x}$$
My turn :
Using AM-GM
$$9\sec^2{x} + 16\csc^2{x} \geq 2 \sqrt{144 \sec^2{x} \csc^2{x}}$$
$$9 \sec^2{x} + 16\csc^2{x} \geq 24 \sec{x} \csc{x} $$
But the equality sign holds iff
$$9 \sec^2{x} = 16\csc^2{x}$$
Then $$ \tan{x} = \frac{4}{3}$$
Then the minimum value is $$24 \times \frac{5}{4} \times \frac{5}{3} = 50$$
Is there any mistake with the solution ?
| @Michael Rozenberg has already given an answer using Cauchy-Schwarz inequality. Here's my answer using the AM-GM inequality.
Write $\sec^2 x = 1 + \tan^2 x$ and $\csc^2 x = 1 + \cot^2 x$. Then
$$9\sec^2 x + 16\csc^2 x = 9(1+\tan^2 x) + 16(1+ \cot^2 x) = 25 + 9\tan^2 x + \frac{16}{\tan^2 x}.$$
Then by AM-GM inequality
$$9\tan^2 x + \frac{16}{\tan ^2 x} \geq 2\sqrt{9\tan^2 x\cdot\frac{16}{\tan^2 x}}$$
$$=2\sqrt{9\cdot 16}=2\cdot3\cdot4 = 24.$$
Thus
$$9\sec^2 x + 16\csc^2 x \geq 25+ 24=49.$$
| {
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How many different matrices $A$ and $B$ are possible such that product $AB$ is defined? Two matrices $A$ and $B$ have in total $6$ different elements (none repeated). How many different matrices $A$ and $B$ are possible such that product $AB$ is defined?
My attempts
$1^{st} Attempt$
Considering that there are $6$ different elements in both $A$ and $B$ matrices, i.e., there are total $12$ elements.
$2^{nd} Attempt$
| This answer was revised in response to the posting of the solution. I initially overlooked that the product of a matrix with two elements and a matrix with four elements could be formed in four ways. The method in the posted solution is preferable to the one written below.
Your strategy is correct. However, saying that matrices $A$ and $B$ have in total six elements means that adding the number of elements in matrix $A$ to the number of elements in matrix $B$ yields $6$.
We can partition $6$ in the following ways:
\begin{align*}
6 & = 1 + 5\\
& = 2 + 4\\
& = 3 + 3
\end{align*}
Thus, we can form products if we can find a matrix with one element that can be multiplied by a matrix with five elements, a matrix with two elements that can be multiplied by a matrix with four elements, or a matrix with three elements that can be multiplied by a matrix with three elements. In each case, we can satisfy the requirement that the number of columns in $A$ is equal to the number of rows in $B$ in two ways.
\begin{array}{c c c}
A & B & AB\\ \hline
1 \times 1 & 1 \times 5 & 1 \times 5\\
5 \times 1 & 1 \times 1 & 5 \times 1\\
1 \times 2 & 2 \times 2 & 1 \times 2\\
2 \times 2 & 2 \times 1 & 2 \times 1\\
4 \times 1 & 1 \times 2 & 4 \times 2\\
2 \times 1 & 1 \times 4 & 2 \times 4\\
1 \times 3 & 3 \times 1 & 1 \times 1\\
3 \times 1 & 1 \times 3 & 3 \times 3
\end{array}
Since there are $8$ possible ways to form the product and the matrices $A$ and $B$ can be populated in $6!$ different ways by $6$ distinct elements in each case, the number of different matrices $A$ and $B$ that can be constructed so that the product is defined is $8 \cdot 6!$.
| {
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The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two partss. Find the area of smaller part. Question: The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two parts. Find the area of smaller part.
This question is from Basic Mathematics. Please explain how I can solve it according to class 11th student.
My work is below.
$\implies y = \frac{x^2}2 \\
\implies 2y = x^2 \\
\implies -x^2 +2y = 0 \ \ \ \ (1)$
and
$x^2 + y^2 = 8 \ \ \ \ (2) $
Adding (1) and (2)
$y^2 + 2y = 8 \\
(y-2)(y+4) =0 \\
\therefore y = 2 or -4$
But when I plot it in the plain with help of https://www.desmos.com/ value of y can only be 2 for the point of intersections.
I'm unable to understand how I can eliminate -4 from the results.
| I can substitute: $(x^2/2)^2+x^2=8$. I obtain: $x^4+4x^2-32=0$. Let $t=x^2$: $t^2+4t-32=0$. In other words: $(t+8)(t-4)$. $x^2=-8$ is impossible; $x^2=4$ has two solutions: $x=2$ or $x=-2$
| {
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How to find all positive integers whose square ends in $444$? The question is from the $1995$ British Mathematical Olympiad.
I’ve figured out that $38^2$ ends with a $444$. So $N^2 \equiv 38^2 \pmod{ 1000}$. Consequentially $N^2 - 38^2 \equiv 0 \pmod{ 1000}$. Using difference of two squares we see that $(N-38)(N+38) \equiv 0 \pmod{ 1000}.$
Now I’m stuck. I’m not too sure if factoring was the right way to go. Also feel free to post any other solutions that begin with with a different route.
| $$N^2\equiv444\pmod{1000}\implies N^2\equiv4\pmod{10}$$
$\implies N=10m\pm2$
$N^2=(10m\pm2)^2=4\pm40m+100m^2\equiv4\pm40m\pmod{100}$
We need $4\pm40m\equiv44\pmod{100}$
$4+40m\equiv44\pmod{100}\implies40m\equiv40\pmod{100}\iff100|40(m-1)\iff5|2(m-1)$
$m=5r+1$(say)
$N=10(5r+1)+2=50r+12$
$N^2=(50r+12)^2=2500r^2+1200r+144\equiv444\pmod{1000}$
$2500r^2+1200r\equiv300\pmod{1000}$
$\iff25r^2+12r\equiv3\pmod{10}$
$\iff5r^2+2r\equiv3\pmod{10}$
Clearly, $r$ can not be even $r=2t+1$(say)
$5r^2+2r-3=5(2t+1)^2+2(2t+1)-3\equiv4t+4\pmod{10}\iff5|2(t+2)\implies5|(t+2)$
$t+2=5s$(say)
$r=2(5s-2)+1=10s-3$
$N=50(10s-3)+12=500s-138,0\le500s-138<1000\implies1\le s\le2$
Similarly for $N=10m-2$ which can also be derived from $N=10m+2$
as for $a$ is a solution, so will be $1000-a$
| {
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Show that $z+\frac{1}{z} = 2\cos(\theta)$ and $z^n+\frac{1}{z^n} = 2\cos (n\theta)$. Hence express $\cos^6 (\theta)$ Show that $z+ \frac{1}{z} = 2\cos (\theta)$ and $z^n+\frac{1}{z^n} = 2\cos (n\theta)$
Hence express $\cos^6(\theta)$
So I have nailed down the proof:
Setting the value of $z=\cos(\theta) + i\sin(\theta)$, $$z^n+\frac{1}{z^n}$$$$=z^n+z^{-n}$$
$$=(\cos\theta+i\sin\theta)^n+(\cos\theta+i\sin\theta)^{-n}$$
Using d-Moivre's theorem,
$$=\cos(n\theta)+i\sin(n\theta)+\cos(-n\theta)+i\sin(-n\theta)$$
$$=\cos(n\theta)+i\sin(n\theta)+\cos(n\theta)-i\sin(n\theta)$$
$$=2\cos(n\theta)$$
I'm stuck from here since, I assume we somehow need to use the theorem.
and equate $\cos^6$ to something based on $z^n+\frac{1}{z^n}$
| I assume you want to express $\cos^6(x)$ in terms of a linear combination of terms from $\{\cos(kx)\}_{k=0}^6$, as this is a standard question. By the binomial theorem,
$$2^6\cos^6(x)=(z+z^{-1})^6=(z^6+z^{-6})+6(z^4+z^{-4})+\binom{6}{2}(z^2+z^{-2})+\binom{6}{3}$$
Which reduces to
$$2^5\cos^6(x)=\cos(6x)+6\cos(4x)+15\cos(2x)+10$$
The overarching point is that we can form a recursive relation between $\cos^nx$ and $\cos^{n-2}(x)$ to prove any power of cosine is expressible as a linear combination of $\{\cos(kx)\}_{k=0}^n$. The polynomial $T_n(x)$ for which $T_n(\cos(x))=\cos(nx)$ is known as the Chebyshev polynomial of the first kind.
| {
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How to get the value of $A + B ?$ I have this statement:
If $\frac{x+6}{x^2-x-6} = \frac{A}{x-3} + \frac{B}{x+2}$, what is the value of $A+B$ ?
My attempt was:
$\frac{x+6}{(x-3)(x+2)} = \frac{A(x+2) + B(x-3)}{(x-3)(x+2)}$
$x+6=(x+2)A + B(x-3)$: But from here, I don't know how to get $A + B$, any hint is appreciated.
| Put $x=-2$ and $x=3$ to get $B=-\frac{4}{5}$ and $A=\frac{9}{5}$ respectively.
| {
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$34!=295232799cd96041408476186096435ab000000$, find $a$ ,$b$, $c$, and $d$ There was a number theory question that I have to do for homework.
$34!=295232799cd96041408476186096435ab000000$, find $a$ ,$b$, $c$, and $d$
I know $b=0$ because $10^7\big|34!$ only. But how can I find other variables?
Remark:
No electronic calculation device can be used for calculation. (I need to write the full solution.)
Can someone help me?
| Sum of digits is divisible by $9$, which gives
$$a+c+d\equiv5(\mod9).$$
$$2-9+5-2+3-2+7-9+9-c+d-9+6-0+4-1+4-0+8-4+7-6+1-8+6-0+9-6+4-3+5-a\equiv0(\mod11),$$
which gives
$$a+c-d\equiv-1(\mod11).$$
From here we can get $a+c=2$ and $d=3$.
Now, since $a\in\{2,6\}$, we obtain $a=2$, $c=0$ and we are done!
| {
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Partial Fractions: Why does this shortcut method work? Suppose I want to resolve $1/{(n(n+1))}$ into a sum of partial fractions. I solve this by letting $1/{(n(n+1))} = {a/n} + {b/(n+1)}$ and then solving for $a$ and $b$, which in this case gives $a=1$ and $b=-1$.
But I learnt about a shortcut method. It says suppose $1/{(n(n+1))} = {a/n} + {b/(n+1)}$, then find $a$ by finding the value which makes its denominator in the RHS equal to $0$ and computing the LHS with the $0$ term (or $a$'s denominator in RHS) removed so we get $a = {1/(0+1)} = 1$ [as $n=0$], and we get $b = {1/(-1)} = -1$ [as $n+1=0$].
Another example, if I am not clear, suppose $$\frac{1}{n(n+1)(n+2)} = \frac{a}{n} + \frac{b}{n+1} + \frac{c}{n+2};$$ then
$$
\begin{eqnarray}
a &=& \frac{1}{(0+1)(0+2)}=\frac{1}{2}, \\
b &=& \frac{1}{(-1)(-1+2)}=-1, \\
c &=& \frac{1}{(-2)(-2+1)}=\frac{1}{2}.
\end{eqnarray}
$$
Why does this shortcut method work?
| I'll just start with $a$ as an example. Multiply both sides by $n$ to get
$$\frac 1{(n+1)(n+2)} = a + \frac {bn}{n+1} + \frac {cn}{n+2}$$
Since this is true for all $n$, plug in $0$. The $b$ and $c$ terms become $0$, leaving $a$ in the formula described above:
$$\frac 1{(0+1)(0+2)} = a$$
| {
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Can i factor this expression: $x^3+y^3+z^3$ I have the following numerical expression, which is exactly equal to $1$
Text version:
(-(7 2^(2/3))/38 + 3^(2/3)/19 + (23 5^(2/3))/95 + (5 6^(1/3))/19 + 3/38 (5^(1/3) 6^(2/3)) - 10^(1/3)/19 - 3/190 (3^(1/3) 10^(2/3)) - (4 15^(1/3))/19 - 6/95 (2^(1/3) 15^(2/3)))*(2^1/3+3^1/3+5^1/3)
Image version:
according to this fact, $x^3+y^3+z^3$ would be factorized at all and first factor of it should be $x+y+z$ if we set
$
x^3=2,y^3=3,z^3=5
$
If so then how to factor $x^3+y^3+z^3$ ?
PS. I tried to divide $x^3+y^3+z^3$ by $x+y+z$ and failed (got remainder).
| By Eisenstein's criterion and Gauss' lemma, $x^3+y^3+z^3$ is irreducible in $k[x,y,z]$ for any field $k$ not of characteristic $3$. So, for example, it is irreducible in $\mathbb C[x,y,z]$.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Modular arithmetic $(2n+1)x \equiv -7 \pmod 9$ Find a solution $(2n+1)x \equiv -7 \pmod 9$
I’m sure this is trivial but I still have doubts about it.
I know the equation has solution for certain $n \in \mathbb {Z}$. Actually I have tried a few and got a similar results (with Diophantine equations ). I wonder if there’s general solution for the equation without changing the n for an integer.
Thanks in advance.
| $9$ is not prime so it has $0$ divisors and you cant solve $3x \equiv k\pmod 9$ unless $k$ is a multiple of $3$.
Basically if $\gcd(m, n) = 1$ there will always be a solution (and only one solution) to $mx \equiv 1\pmod n$. We can notate that solution as $m^{-1}$. ( So for example $5^{-1} = 2\pmod 9$ because $2*5 \equiv 1 \pmod 9$.
So for any $mx \equiv k \pmod n$ we can do $m^{-1}mx \equiv m^{-1}k\pmod n$ and so $x \equiv m^{-1}k\pmod n$. So in your example if $2n +1 = 5$ we could solve $5x\equiv -7\pmod 9$ so $2*5x \equiv x \equiv 2*(-7)\equiv -14\equiv -5 \equiv 4\pmod 9$. (And indeed $4*5 \equiv -7\pmod 9$).
But if $\gcd(m,n) \ne 1$ this does not follow unless $k$ is a multiple of $\gcd(m,n)$. But if $k$ is a multiple of $\gcd(m,n)$ we can solve.
.....
To put this all in perspective these are actually just a restatement of Bezouts lemma.
$mx \equiv k \pmod n$ is solveable if and only if there if is an integer $w$ so that $mx + wn = k$ which is solveable if and only if $k$ is a multiple of $\gcd(m,n)$.
So to solve $(2n + 1)x \equiv -7\pmod 9$: as $-7$ is not a multiple of an factor of $9$ other than $1$, this will only be solvable if $\gcd(2n+1, 9)= 1$.
So we may if and only if $2n+1$ is not a multiple of $3$. Or in other words if and only if $2n+1 \not \equiv 0\pmod 3$ or $2n\not \equiv -1\pmod 3$ or $n \not \equiv 1\pmod 3$.
..... so final answer .....
For there to be solutions we can't have $n\equiv 1\pmod 3$. In other words we cant have $n\equiv 1,4,7\pmod 9$.
So we can have solutions if $n \equiv 0,2,3,5,6,8 \pmod 9$.
In those case $2n+1 \equiv 1,2,4,5,7,8\pmod 9$.
We can find $(2n+1)^{-1}\mod 9$ for those values.
$1*1 = 1; 2*5\equiv 1; 4*7\equiv 1; 5*2\equiv 1; 7*4\equiv 1; 8*\equiv 1\pmod 9$ so $(2n+1)^{-1}\equiv 1,5,7,2,4,8\pmod 9$ when $n \equiv 0,2,3,5,6,8\pmod 9$ respectively.
So solution to $(2n+1)x \equiv -7 \equiv 2 \pmod 9$ is $x\equiv (2n+1)^{-1}*2 \pmod 9$.
So if $n \equiv 0,2,3,5,6,8\pmod 9$ then $x \equiv (2n+1)^{-1}*2 \equiv 1*2,5*2,7*2,2*2,4*2, 8*2 \equiv 2,1,5,4,8,7\pmod 9$ respectively.
| {
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Calculating value of series by taking the difference Apologies if this is a basic question!
I'm trying to understand a solution to a problem I was solving. The author suggests a trick to calculate expected value by multiplying the expected value series by 0.5 (line 2) and taking the difference (line 3):
$E(X) = 0.5^1 + 2 \cdot0.5^2 + 3\cdot 0.5^3...\\$
$0.5E(X) = 0.5^2 + 2 \cdot0.5^3 + 3\cdot 0.5^4...\\$
$0.5E(X) = 0.5^1 + 0.5^2 + 0.5^3...$
My question: how did he calculate the difference on line 3?
Thanks for your help.
| We have $$E = 0.5^1 + 2\cdot0.5^2 + 3\cdot0.5^3 + 4\cdot0.5^4+\cdots$$
$$0.5E = 0.5^2 + 2\cdot0.5^3 + 3\cdot0.5^4+\cdots$$
Combining terms with equal powers of $0.5$,$$E - 0.5E = 0.5^1 + 0.5^2(2-1) + 0.5^3(3-2) + 0.5^4 (4-3) \cdots$$
$$\implies 0.5E = 0.5^1 + 0.5^2+0.5^3\cdots$$
| {
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Prove that $\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx = \frac{\pi}{4}\log(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}$ I have to prove that
$$
I=\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}{4}\log(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}
$$
I know that
$$ I_{+}=\int_{0}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}{8}\log(2)
$$
and I can write
$$
I= \int_{0}^{1}\frac{\log(1+x)}{1+x^2}dx+\int_{0}^{1}\frac{\log(1-x)}{1+x^2}dx
$$
So I have only to evaluate $$I_{-}=\int_{0}^{1}\frac{\log(1-x)}{1+x^2}dx
$$
Using the expansion of the logarithm valid in the range of integration I can also write
$$I_{-}=-\sum_{n\geq1}\frac{1}{n}\int_{0}^{1}\frac{x^n}{1+x^2}dx
$$
but then I can go any further because I can't find a simple solution that depend only on $n$ for the rational (and solvable) integral under the sum sign .
Maybe is simple to considerate instead of $ I_{-} $ the integral $\int_{0}^{1}\frac{\log(1-x^2)}{1+x^2}dx=I$ because it seems to me that $\int_{0}^{1}\frac{x^{2n}}{1+x^2}dx= \sum_{k=0}^{n-1} (\frac{(-1)^{n+k}}{2k+1} + \frac{\pi}{4}(-1)^n)$ but again I can go further.
Any help would be appreciated.
Thanks
| Yet another way is to use Feynman's trick of differentiating under the integral sign. Let
$$I(a) = \int_{-1}^1 \frac{\ln (1 + ax)}{1 + x^2} \, dx.$$
Note that $I(0) = 0$ and we require $I(1)$. Now
\begin{align}
I'(a) &= \int_{-1}^1 \frac{x}{(1 + ax)(1 + x^2)} \, dx\\
&= \frac{a}{1 + a^2} \int_{-1}^1 \frac{dx}{1 + x^2} + \frac{1}{1 + a^2} \int_{-1}^1 \frac{x}{1 + x^2} \, dx - \frac{a}{1 - a^2} \int_{-1}^1 \frac{dx}{1 + ax}\\
&= \frac{\pi a}{2(1 + a^2)} + \frac{1}{1 + a^2} \ln \left (\frac{1 - a}{1 + a} \right )
\end{align}
Thus
\begin{align}
I(1) &= \frac{\pi}{4} \int_0^1 \frac{2a}{1 + a^2} \, da + \underbrace{\int_0^1 \ln \left (\frac{1 - a}{1 + a} \right ) \frac{da}{1 + a^2}}_{a \, \mapsto \, (1 - a)/(1 + a)}\\
&= \frac{\pi}{4} \ln 2 + \int_0^1 \frac{\ln a}{1 + a^2} \, da\\
&= \frac{\pi}{4} \ln 2 + \sum_{n = 0}^\infty (-1)^n \int_0^1 a^{2n} \ln x \, da\\
&= \frac{\pi}{4} \ln 2 + \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\int_0^1 a^{2n + s} \, da \right ]_{s = 0}\\
&= \frac{\pi}{4} \ln 2 + \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\frac{1}{2n + s + 1} \right ]_{s = 0}\\
&= \frac{\pi}{4} \ln 2 - \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)^2}\\
&= \frac{\pi}{4} \ln 2 - \mathbf{G}
\end{align}
| {
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Computation Of Integrals Computer the Integral: $$\int\frac{2x+1}{(x-1)(x-2)}dx$$
Now using partial fraction we can write $$\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$$, So we get $$\frac{2x+1}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$$ Now for all $x$ not equal to $1, 2$ we can cancel out the denominator to get $$2x+1=A(x-2)+B(x-1)$$ Now to find $A$ and $B$ how can we put $x=1$ and $x=2$ in this identity as this identity is valid if and only if $x$ is not equal to $1, 2$
| You have
$$2x+1=A(x-2)+B(x-1) \tag{1}\label{eq1}$$
As for why substituting values of $x = 1$ and $x = 2$ works, this is because \eqref{eq1} is an identity and, thus, must be true for all values of $x$. With the original equation involving the denominators, due to continuity for all $x \neq 1,2$, the numerators must still match values at $x = 1,2$. Thus, by using $x = 2$ first, you eliminate the $A$ parameter so you only have the $B$ parameter, to get $B = 5$. Likewise, using $x = 1$ eliminates the $B$ parameter, leaving an equation in just $A$ to solve to get $-A = 3 \implies A = -3$. However, this method doesn't always work well in more complicated sets of equations (e.g., where higher powers are involved, there are considerably more variables being used so you can't isolate just one of them, etc.), which is why I present the more general method below.
Although it's sometimes more work, you can also expand and collect the terms of the same powers of $x$ together. In this case, \eqref{eq1} becomes
$$2x + 1 = (A + B)x - 2A - B \implies (2 - (A + B))x + (1 + 2A + B) = 0 \tag{2}\label{eq2}$$
Thus, you get for the coefficient of $x$ to be $0$ that
$$A + B = 2 \tag{3}\label{eq3}$$
and for the constant term to be $0$ that
$$-2A - B = 1 \tag{4}\label{eq4}$$
Adding the $2$ equations gives $-A = 3 \implies A = -3$. Thus, from \eqref{eq3}, you then get $B = 2 + 3 = 5$. This, of course, matches what was originally determined by using $x = 1$ and $x = 2$.
| {
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Prove $ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{3} \le (x^{9} + y^{9} + z^{9})[ (1+x)^{3/2} + (1+y)^{3/2} + (1+z)^{3/2} ]^{2} $ Prove
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{3} \le (x^{9} + y^{9} + z^{9})[ (1+x)^{3/2} + (1+y)^{3/2} + (1+z)^{3/2} ]^{2} $$
if $x,y,z \ge 0$
I made this problem using Holders inequality, notice that
$$ (1+x)x^{3} + (1+y)y^{3} + (1+z)z^{3} \le \left[ (x^{3})^{3} + (y^{3})^{3} + (z^{3})^{3} \right]^{1/3} \left[ (1+x)^{3/2} + (1+y)^{3/2} + (1+z)^{3/2} \right]^{2/3} $$
then taking cube both sides we get
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{3} \le (x^{9} + y^{9} + z^{9})[ (1+x)^{3/2} + (1+y)^{3/2} + (1+z)^{3/2} ]^{2} $$
My question is, will this inequality be obvious to be solved without Holders inequality? if so then what is the alternative solution?
| Directly citing the Hölder inequality is the shortest solution. If one wants to reduce this to some somewhat more elementary arguments, one can use the inequality of the arithmetic and cubic mean.
For non-negative $a_k$ and positive $b_k$ we get, using the weighted mean value inequalities, that is, Jensen's inequality for a discrete measure, and using weights $w_k=b_k^3$,
\begin{align}
\frac{\sum a_kb_k^2}{\sum b_k^3}
=\frac{\sum w_k\frac{a_k}{b_k}}{\sum w_k}
\le\sqrt[3]{\frac{\sum w_k\left(\frac{a_k}{b_k}\right)^3}{\sum w_k}}
=\sqrt[3]{\frac{\sum a_k^3}{\sum b_k^3}}
\\~\\
\implies \left(\sum a_kb_k^2\right)^3 \le \sum a_k^3\left(\sum b_k^3\right)^2
\end{align}
Now set $a_1=x^3$, $b_1=\sqrt{1+x}$ etc.
| {
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equation simplification. $(5y-1)/3 + 4 =(-8y+4)/6$ Simplification of this equation gives two answers when approched by two different methods.
Method 1 Using L.C.M( least common multiple)
$(5y-1)/3 + 4 =(-8y+4)/6$
$(5y-1+12)/3 = (-8y+4)/6$
$5y-11 = (-8y+4)/2$
$(5y-11)2= (-8y+4)$
$10y-22 = -8y+4$
$18y=26$
$y = 26/18=13/9$
Method 2 multiplying every term by 3
$3(5y-1)/3 + 4*3 = 3(-8y+4)/6$
$5y-1 + 12 = (-8y+4)/2$
$2(5y-1 + 12) = -8y+4$
$10y-2+24 = -8y+4$
$18y + 22 = 4$
$18y = -18$
$y = -1$
The correct method is method 2 and the correct answer is y = -1
Why is method 1 is incorrect? Could anyone explain why the answer is wrong when using the L.C.M( method 1)?
| Your first method multiplies $(5y-1+12)/3$ by $3$ to give $5y-11$ when it should give $5y+11$
| {
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Line$ 3x-4y+k$ touches a circle $x^2+y^2-4x-8y-5 = 0$ at $(a,b)$, then find $k+a+b =?$
The line $3x-4y+\lambda=0, (\lambda > 0)$ touches the circle $x^2+y^2-4x-8y-5 = 0$ at $(a,b)$. Find the value of $\lambda+a+b$.
I tried solving in the following manner:
Clearly the circle has center at $(2,4)$ with a radius of $5$.
So, the perpendicular distance from the center onto the given line must be the radius (Radius is perpendicular to tangent)
So, using the formula for perpendicular distance
$$\frac{|3\times2-4\times4+\lambda|}{(3^2+4^2)^{1/2}}= 5$$
$$\implies\frac{|3\times2-4\times4+\lambda|}{5} = 5$$
$$\implies|\lambda-10| = 25$$
From this, I get $\lambda = 35,-15$, but since in the question it is given $\lambda>0$, I'm taking $\lambda = 35$
Now, the line is $3x-4y+35 = 0$
Given, it is the tangent at $(a,b)$
Writing equation of tangent at (a,b), I get,
$$ax+by+(-2)(x+a)+(-4)(y+b)-5 = 0$$
$$\implies (a-2)x+(b-4)y+-2a-4b-5 = 0$$
which should be identical to the line $$3x-4y+35= 0$$
So, I write
$$\frac{(a-2)}{3} = \frac{b-4}{-4}=\frac{-(2a+4b+5)}{35}$$
Solving these, I obtained that a = -1 and b = 8
So $\lambda+a+b = 42$
But the answer says 20. Is there a problem with my above procedure?
| We can take advantage of the fact that the line touches the circle at a single point:
$$y=\frac34x+\frac{\lambda}{4} \Rightarrow \\
x^2+\left(\frac34x+\frac{\lambda}{4}\right)^2-4x-8\left(\frac34x+\frac{\lambda}{4}\right)-5 = 0 \Rightarrow \\
25x^2+2(3\lambda-80)x+\lambda^2-32\lambda-80=0 \quad (1)$$
The quadratic equation has a single solution when $D=0$:
$$(3\lambda-80)^2-25(\lambda^2-32\lambda-80)=0 \Rightarrow \\
\lambda_1=-15,\lambda_2=35.$$
When we plug $\lambda=35>0$ into $(1)$:
$$25x^2+50x+1225-1120-80=0 \Rightarrow \\
25x^2+50x+25=0 \Rightarrow \\
x^2+2x+1=0 \Rightarrow \\
(x+1)^2=0 \Rightarrow \\
x=-1=a \Rightarrow y=8=b$$
Hence:
$$\lambda+a+b=35-1+8=42.$$
| {
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Find all the integer pairs $(x,y)$ that satisfy the equation $7x^2-40xy+7y^2=(|x-y|+2)^3$ Find all the integer pairs $(x,y)$ that satisfy the equation
$7x^2-40xy+7y^2=(|x-y|+2)^3$
it is clear the equation is symmetric therefore you can assume w.l.o.g that $x\ge y$ which makes the new equation equal to
$(x-y+2)^3=7x^2-40xy+7y^2$
which becomes
$0=x^3+x^2(-1-3y)+x(12+28y+3y^2)+(8+y^3-y^2-12y)$
but I don't know what to do next , help, suggestions and solutions would be appreciated.
I think Lagrange multipliers can be used but there is a preferred solution that doesn't use calculus.
Taken from the 2019 SAIMC https://chiuchang.org/imc/wp-content/uploads/sites/2/2019/08/SAIMC-2019_Keystage-3_Individual_Final.x17381.pdf
| First notice that $7x^2-40xy+7y^2\le\frac{27}{2}|x-y|^2$ and equality holds only if $x+y=0$. (You can show this directly by expanding)
Now, let $t=|x-y|$ then $(t+2)^3\le\frac{27}{2}t^2$ and $t\ge 0$.
Since $(t+2)^3-\frac{27}{2}t^2=(t-4)^2(t+\frac{1}{2})$, this holds only if $t=4$, and even when $t=4$, you get equality and this implies $x+y=0$.
Thus, $|x-y|=4$ and $x+y=0$ is the only candidate.
Therefore, the only possible pairs are $(2,-2)$ and $(-2,2)$. (They are actually integer pairs, but even for real number pair we get the same)
| {
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find min of $a+b$ given sum of $a\geq b\geq c\geq d$ is 9 and square sum is 21 Suppose $a\geq b\geq c\geq d>0$ and all are real numbers, and $a+b+c+d=9,a^2+b^2+c^2+d^2=21$, how to find the minmum of $a+b$?
What I attempted:
I can show $b\geq 1.5$ and $a\leq 3$.for $r\geq 0$, I consider $\sum(a-r)^2=\sum a^2-2r\sum a+4r^2=21-18r+4r^2$, then $a\leq \sqrt{21-18r+4r^2}+r$ which implies $a\leq 3$. Also I guess the min shoul be 5 and there are two solutions: 3,2,2,2 and $2.5,2.5,2.5,1.5$. But I cannot prove it.
New attempts:
*
*$6-2a=(a-3)^2+\sum(b-2)^2$
*$2d-3=\sum(a-2.5)^2+(d-1.5)^2$
then $2(d-a)+3\geq 1$ which implies that $d\leq c\leq b\leq a\leq d+1$
| Your solution is right and very nice!
The equality occurs for $x=4$ and $$(18x-x^2-2y-39)(2y-x^2)=0.$$
For $2y-x^2=0$ we obtain $y=8$, $c=d=2$, $a=3$ and $b=2$.
For $18x-x^2-2y-39=0$ we obtain $y=8.5$, $c=2.5$, $d=1.5$ and $a=b=2.5.$
We got all cases of equality occurring, as you said.
| {
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Eliminating parameter $\beta$ from $x=\cos 3 \beta + \sin 3 \beta$, $y = \cos \beta - \sin \beta$
Based on the given parametric equations:
$$\begin{align}
x &=\cos 3 \beta + \sin 3 \beta \\
y &= \cos \beta \phantom{3}- \sin \beta
\end{align}$$
Eliminate the parameter $\beta$ to prove that $x-3y+2y^3=0$.
What I got so far:
$$\cos 3 \beta + \sin 3 \beta = ( \cos \beta - \sin \beta)(1+4\sin\beta\cos\beta)$$
Which trigonometric identity should I use to proceed?
| Hint: Write your system in the form
$$x=\sqrt{2}\sin\left(\frac{\pi}{4}-\beta\right)(1+2\sin(2\beta))$$
$$y=\sqrt{2}\sin\left(\frac{\pi}{4}-\beta\right)$$
then you will get $$x=y(1+2\sin(2\beta))$$ and you can eliminate $\beta$
| {
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Show that $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{3x-4}{x^2+5}$ is injective I'm working on proving that $f: \Bbb{R} \rightarrow \Bbb{R}$ defined by $f(x) = \frac{3x-4}{x^2+5}$ is injective. However I'm stuck.
Assuming that $f(x_1) = f(x_2)$:
$$f(x_1) = f(x_2)$$
$$\frac{3x_1-4}{x_1^2+5} = \frac{3x_2-4}{x_2^2+5}$$
$$(3x_1-4)(x_2^2+5) = (3x_2-4)(x_1^2+5)$$
$$3x_1x_2^2 + 15x_1 - 4x_2^2 - 20 = 3x_2x_1^2 + 15x_2 - 4x_1^2 - 20$$
$$3x_1x_2^2 + 15x_1 - 4x_2^2 = 3x_2x_1^2 + 15x_2 - 4x_1^2$$
How do I get to $x_1 = x_2$ from here on out?
| $$f(x) = \frac{3x-4}{x^2+5}$$
$$f(0)=-\dfrac 45 = f\left(-\dfrac{15}{4}\right)$$
You would have had an easier time solving
\begin{align}
\dfrac{3x-4}{x^2+5} &= k \\
3x-4 &= kx^2 + 5k \\
kx^2 - 3x +(5k+4) &= 0 \\
x &= \dfrac{3 \pm \sqrt{9-16k-20k^2}}{2k}
\end{align}
So $x$ is double valued for all
$$-\dfrac 25 - \dfrac{\sqrt{61}}{10} < k < -\dfrac 25 + \dfrac{\sqrt{61}}{10}$$
except it is single-valued at $k=0$
| {
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Please explain the formula for the sum of the cubes and the difference: $a^3 - b^3$ and $a^3 + b^3$? I have not yet fully mastered the formula for accelerated multiplication. I was able to understand everything except the last two. I would like to deal with them. Why have : $$ a^3 + b^3 = (a+b)(a^2 -ab + b^2)^* $$ and $$ a^3 - b^3 = (a-b)(a^2 + ab + b^2) $$ Can someone paint me a formula in detail. Why is this the way it works.
| Let $x=a/b$. We want to factor $b^3(x^3\pm1)$. Note that $1$ is a root of $x^3-1$, so $x-1$ is a factor of $x^3-1$. You could find the quotient by polynomial long division. Also, $-1$ is a root of $x^3+1$, so $x+1$ is a factor of $x^3+1$, and again you could find the quotient by polynomial long division.
(From this, it is easy to see why $a^n+b^n$ has that kind of factorization only for $n$ odd.)
| {
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"timestamp": "2023-03-29T00:00:00",
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proving that $a + b \sqrt {2} + c \sqrt{3} + d \sqrt{6} $ is a subfield of $\mathbb{R}$ The question is given below:
My questions are:
1- How can I find the general form of the multiplicative inverse of each element?
2-How can I find the multiplicative identity?
3-Is the only difference between the field and the subfield definition is that in the case of a subfield every nonzero element has an additive identity but in the field every element not only nonzero ones?
Could anyone help me in understanding these questions, please?
EDIT:
I have found this solution on the internet:
My question: is that a fully acceptable answer to the question? I guess yes.
| Regarding 1: Note that if we multiple your general element by its conjugate with respect to $\sqrt{3}$,
$$ (a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6})(a + b\sqrt{2} - c\sqrt{3} - d\sqrt{6}) \\ = a^2 + 2b^2 - 3c^2 -6d^2 + (2ab-6cd)\sqrt{2} \text{,} $$
which we can simplify further by multiplying with this new expression's conjugate with respect to $\sqrt{2}$, $a^2 + 2b^2 - 3c^2 -6d^2 - (2ab-6cd)\sqrt{2}$. We find
$$ (a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6})(a + b\sqrt{2} - c\sqrt{3} - d\sqrt{6})(a^2 + 2b^2 - 3c^2 -6d^2 - (2ab-6cd)\sqrt{2}) \\
= a^4-4 a^2 b^2-6 a^2 c^2-12 a^2 d^2+48 a b c d+4 b^4-12 b^2 c^2-24 b^2 d^2+9 c^4-36c^2 d^2+36 d^4 \text{.} $$ Now divide both sides by $(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6})$ and $(a^4-4 a^2 b^2-6 a^2 c^2-12 a^2 d^2+48 a b c d+4 b^4-12 b^2 c^2-24 b^2 d^2+9 c^4-36c^2 d^2+36 d^4)$ to find an expression for the multiplicative inverse of $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$.
Notice that this is entirely analogous to what we do with complex numbers. For $a+b\mathrm{i}$, we multiply by its conjugate (with respect to $\mathrm{i}$),
$$ (a + b\mathrm{i})(a - b\mathrm{i}) = a^2 + b^2 \text{.} $$
Then $\frac{a - b \mathrm{i}}{a^2 + b^2} = \frac{1}{a + b \mathrm{i}}$.
Regarding 2: What's the multiplicative identity in $\mathbb{R}$? Is it possible for the multiplicative identity of a subfield to be a different number? Suppose $s$ in a subfield satisfies $sx = x$ for every $x$ in that subfield, then that equation is also true for all those elements in the field. If we write $1 \in \mathbb{R}$ as the multiplicative identity in $\mathbb{R}$, then $sx = 1x$ is an equivalent equation in $\mathbb{R}$ and $(s-1)x = 0$. It is a familiar property of $\mathbb{R}$ that this forces either $s-1 = 0$ or $x = 0$. Since this holds for at least one non-zero $x$ (because subfields are fields, so they have at least two elements), it must be the case that $s-1 = 0$, so $s = 1$.
Regarding 3: Every element in any field (or subfield) has an additive inverse. For a field $F$ and an element $x \in F$, its additive inverse is $-x$. For your element $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$, its additive inverse is $-a -b\sqrt{2} -c\sqrt{3} -d\sqrt{6}$ for every choice of $a,b,c,d$. Even zero has an additive inverse. (It's zero.)
| {
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Integrating $\int_a^b\left[ \left(1 - \frac{a}{x} \right)\left(\frac{b}{x}-1 \right) \right]^\frac{1}{2}dx$ We are interested in the following integration:
$$ \int_a^b \left[ \left(1 - \frac{a}{x} \right)\left(\frac{b}{x}-1 \right) \right]^{1/2} dx. $$
I try to use substitution with:
$$u = \left(1-\tfrac{a}{x}\right)\left(\tfrac{b}{x} -1\right) = (a+b)x^{-1} - 1 - (ab)x^{-2}.$$
$$ du = \left( (-a-b)x^{-2} + (2ab)(x^{-3}) \right)dx $$
or
$$ dx = du \left( \left( (-a-b)x^{-2} + (2ab)(x^{-3}) \right)^{-1} \right)$$
Now the integration becomes something ugly. I try integration by parts on the ugly part, and it becomes even uglier. Obviously, I don't see some easier ways. I would appreciate some advices or the solution!
Edit: It may be easier to see the problem as
$$ \int_a^b \left[ (a-x)(x-b)x^{-2} \right]^{1/2} dx .$$
| Rescale the integration range with the following variable change and the shorthand $q$
$$u = \frac{x-a}{b-a}, \>\>\>\>\>\>q=\frac{b}{a}-1$$
to simplify the original integral
$$ I = \int_a^b \left[ \left(1 - \frac{a}{x} \right)\left(\frac{b}{x}-1 \right) \right]^{1/2} dx=aq^2\int_0^1 \frac{\sqrt{u(1-u)}}{1+qu} du\tag{1}$$
Then, let $u=\sin^2 t$ to rewrite (1) as
$$ I = 2a\int_0^{\pi/2} \frac{q^2\sin^2 t\cos^2 t}{1+q\sin^2 t}dt$$
Decompose the integrand,
$$I = 2a\int_0^{\pi/2}\left( 1+q\cos^2 t-\frac{1+q}{1+q\sin^2 t}\right) dt$$
Carry out the two integrals,
$$I_1=\int_0^{\pi/2}\left( 1+q\cos^2 t\right) dt=\frac{\pi}{2}\left(1+\frac{q}{2}\right)$$
$$I_2=\int_0^{\pi/2} \frac{1+q}{1+q\sin^2 t}dt=\int_0^{\infty} \frac{(1+q)ds}{1+(1+q)s^2 }=\frac{\pi}{2}\sqrt{1+q}$$
where $s=\tan t$ is used in evaluating $I_2$. Thus,
$$I= 2a(I_1- I_2)= \frac{\pi}{2}(b+a-2\sqrt{ab})$$
| {
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If $a$, $b$, $c$ are sides of a triangle, prove $2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$
$a$, $b$, $c$ are sides of a triangle, prove:
$$2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$$
What I have tried:
$$
⇔2\sum (a+b)ab\geq \sum a^3+9abc
$$
so I can't use
$$\sum a^3+3abc\geq \sum ab(a+b)$$
| WLOG $$c=\max\{a,b,c\}\implies a=x+u,b=x+v\text{ and }c=x+u+v (x>0, u,v\ge 0)$$
Then your inequality is equivalent to
$$x(u^2-uv+v^2)+2(u+v)(u-v)^2\ge 0*\text{true}*$$
| {
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Convolution of $f(x)={1 \over 2} \chi_{[-1,1]}*\chi_{[-5,5]}$ Convolution of $f(x)={1 \over 2} \chi_{[-1,1]}*\chi_{[-5,5]}$. This is my first exercise, it's basically an interpolation between the two functions. So here my results:
Calling $u={1 \over 2} \chi_{[-1,1]}$ and $v=\chi_{[-5,5]}$ I can calculate the different values:
$u*v(0)={ 3 \over 2}$
$u*v({1 \over 2})={ 3 \over 2}$
$u*v(-{1 \over 2})={ 3 \over 2}$
So for the borders, we have:
*
*$0$ for $x<-5,x>5$
*$1$ for $x \geq -5,x \leq -{ 3\over 4}$
*$-x+{ 3 \over 4}$ for $-{ 3\over 4} \leq x \leq -{ 1 \over 4}$
*${ 3 \over 2}$ for $-{ 1 \over 4} \leq x \leq { 1 \over 4}$
*$x-{ 3 \over 4}$ for ${ 1 \over 4} \leq x \leq { 3 \over 4}$
*$1$ for $x \geq { 3 \over 4},x\leq5$
I hope it's right (I can't draw the graphic here); if there is some error or not rigorous passage, please let me know I want to be capable to solve this at the best of possibilities
| $f(x)=\frac{1}{2}\int_{-\infty}^{+\infty} X_{[-5,5]}(y)X_{[-1,1]}(x-y)dy$
$$X_{[-1,1]}(x-y)=1 \Longleftrightarrow -1 \leq x-y \leq 1 $$$$\Longleftrightarrow x-1\leq y \leq x+1 \Longleftrightarrow y \in[x-1,x+1]\Longleftrightarrow X_{[x-1,x+1]}(y)=1$$
Thus $X_{[-1,1]}(x-y)=X_{[x-1,x+1]}(y)$
Also $G(y)=X_{[-5,5]}(y)X_{[x-1,x+1]}(y)=X_{[-5,5] \cap [x-1,x+1]}(y)$
$1.$ If $|x| \leq 4$ then $[x-1,x+1] \cap [-5,5]=[x-1,x+1]$ thus $G(y)=X_{[x-1,x+1]}(y) \Longrightarrow f(x)=1$
$2.$ If $x-1>5$ the the intervals are disjoint so the integral is zero,so $f(x)=0$
$3.$ If $x+1<-5$ again as in $2$ we have $f(x)=0$
$4.$ If $x=-6,6$ the intersection of the intervals is a singleton set so $f(x)=0$
$5.$ If $-6<x<-4$ then $x+1<-3$ and $x-1<-5$ and $-5<x+1<-3 $ and the intersection of the intervals is $[-5,x+1]$ thus $f(x)=\int_{-5}^{x+1}dy=\frac{x+6}{2}$
$6.$ If $4<x<6$ then $x+1>5$ and $3<x-1<5$ then the intersection is the interval $[x-1,5]$ thus $f(x)=\int_{x-1}^{5}dy=\frac{6-x}{2}$
| {
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Proving a result by making discriminant zero
If the roots of given Quadratic equation
$$a(b-c)x^2 +b(c-a)x + c(a-b)=0$$
are equal, prove the following:
$$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$$.
MY approach:
Method 1: put Discriminant=0 and get stuck.
Method2:add $acx$ on both sides and get $(x-1)$ as factor ,so other root is also 1 and hence the result.
But if someone can prove the result by making discriminant zero (method 1), that would be more rigorous.
Thank you.
| Following your first approach, we have that the discriminant is
$$\begin{align}
\Delta&=b^2(c−a)^2-4a(b−c)c(a−b)\\
&=b^2(c−a)^2+4ac(b^2-b(a+c)+ac)\\
&=b^2(c+a)^2+(2ac)^2-2b(a+c)(2ac)\\
&=(b(a+c)-2ac)^2
\end{align}.$$
and by letting $\Delta=0$, it is easy to show that $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$.
As regards your second approach, since you already noted that $x=1$ is a root, we have a double root if and only if the following polynomial identity holds (polynomial factorization is unique):
$$a(b-c)x^2 +b(c-a)x + c(a-b)=0=a(b-c)(x-1)^2.$$
Now by letting $x=0$ we find $$c(a-b)=a(b-c)$$ which implies $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$.
| {
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Prove that $\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$ How to show
$$\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$$
I tried hypergeometric expansion, yielding $\, _2F_1\left(\frac{1}{2},\frac{3}{4};\frac{7}{4};-1\right)$. Can this be evaluated analytically? Any help will be appreciated.
| As I said in my comment I believe by making the substitution:
$$x^2=\tan(t)$$
you get:
$$\int_0^1\frac{x^2}{\sqrt{x^4+1}}dx=\frac{1}{2}\int_0^{\pi/4}\sin(t)^{1/2}\cos(t)^{-3/2}dt=\frac{1}{4}\int_0^{1/\sqrt[4]{2}}u^{-1/4}(1-u)^{-5/4}du=B\left(2^{-1/4};\frac34,-\frac14\right)$$
| {
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The sum of infinite series $(1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4..................$ Initially, I broke this series as $1/2 = 1-1/2, 2/3 = 1-1/3, 3/4 = 1-1/4$, then I got two series as it is
$$= (1/5)^2 + (1/5)^3 + (1/5)^4 +.......-[1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
After solving the first part, we reached at this point
$$= 1/20 - [1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
Please help someone in solving this question. I am very grateful to you.
| It is given:
$$\frac12\cdot \left(\frac15\right)^2+\frac23\cdot \left(\frac15\right)^3+\frac34\cdot \left(\frac15\right)^4+\cdots+\frac{n}{n+1}\cdot \left(\frac15\right)^{n+1}+\cdots$$
Consider the function:
$$f(x)=\frac12\cdot x^2+\frac23\cdot x^3+\frac34\cdot x^4+\cdots+\frac{n}{n+1}\cdot x^{n+1}+\cdots$$
We want to find $f\left(\frac15\right)$. (Why?)
Take derivative from both sides:
$$f'(x)=\color{blue}{x+2x^2+3x^3+\cdots +nx^n+\cdots}$$
Consider another function:
$$g(x)=1+x+x^2+x^3+\cdots+x^n+\cdots=\frac1{1-x}, \quad \left(x=\frac15<1\right)$$
Take derivative from both sides:
$$g'(x)=1+2x+3x^2+\cdots+nx^{n-1}+\cdots=\frac1{(1-x)^2}$$
Multiply both sides by $x$:
$$xg'(x)=\color{blue}{x+2x^2+3x^3+\cdots+nx^n+\cdots}=\frac{x}{(1-x)^2}$$
Note that $f'(x)=xg'(x)$, so:
$$f'(x)=\frac{x}{(1-x)^2}$$
Now we integrate both sides:
$$\int f'(x)dx=\int \frac{x}{(1-x)^2}dx \Rightarrow\\
f(x)=\int \frac{x-1+1}{(1-x)^2}dx=-\int \frac1{1-x}dx+\int \frac1{(1-x)^2}dx =\\
\ln (1-x)+\frac1{1-x}+C$$
Note that $f(0)=0$:
$$f(0)=\ln (1-0)+\frac1{1-0}+C=0 \Rightarrow C=-1$$
So:
$$f(x)=\ln (1-x)+\frac1{1-x}-1$$
At last we plug $x=\frac15$:
$$f\left(\frac15\right)=\ln \left(1-\frac15\right)+\frac1{1-\frac15}-1=\ln \frac45-\frac14.$$
| {
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Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
My attempt:
$\gcd\left(9,5,4\right) = 1 = 9r+5s+4\times0$
$9 = 1\times5+4$
$5 = 1\times4+1$
so $1 = 5-(9-5) = 2\times5 - 1\times9$
thus $r=-1$ and $s = 2$
so $x = 2\times5\times8 -1\times9\times1=80 - 9=71\mod 180$
But the problem is that $71\equiv8\mod9$ and $71\equiv1\mod5$, but $71 \ne 2 \mod 4$. What is wrong?
| I am using the method for solving this system of equations as outlined here.
Here, we have $a_1 = 2, a_2 = 8, a_3 = 1$. Additionally, $n_1 = 4, n_2 = 9, n_3 = 5$. Now, $N = n_1 \cdot n_2 \cdot n_3 = 180$.
Hence, $y_1 = 45, y_2 = 20, y_3 = 36$. Now, we are supposed to find the values of $z_i$ for $i = 1, 2, 3$. I'll outline the method for finding $z_1$, and the others follow similarly.
Now, $z_1 \equiv 45^{-1} \mod 4$. Now, what is the value of $45^{-1}$? It is that value of $x \in \mathrm{Z_4}$ such that $$45x \equiv 1 \mod 4$$ We can easily see that $x = 1$. So, we get: $z_1 \equiv 1 \mod 4$, which gives $z_1 = 1$.
Can you take it from here?
| {
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Circles and squares and polygon
If i replace the square with circle, i can very easily find the area of circle and get some approximation for the area of the square.
Even the square is doable, but i was thinking that can we find the area of the regular polygon which is made to sit in place of the square ?
Since, i don't have the answer to that problem, i don't know if approximating that with circle would work.
Would a regular polygon of arbitrarily large number of sides fit into that space and can we find its area (i know only upto undergrad. maths)?
I am trying to solve it by coordinate geometry but i am not getting as many equations as many variables i have.
$a^2$+b(b-7)=49=$d^2$+c(c-7)=($e^2$+$f^2$)=($g^2$+$h^2$)=(f-b)(g-c)49/(e-a)(h-d)=$(f-b)^2$+$(e-a)^2$+49-$(g-c)^2$-$(h-d)^2$=$(a-c)^2$+$(b-d)^2$+49-$(e-g)^2$-$(f-h)^2$
Thank you.
| A solution for a regular polygon with one edge a chord of the large circle and the vertices of the opposite edge on the smaller circles
Consider coordinates with $O$ as origin and the line of symmetry as the $x$-axis. Let the edge length of the regular polygon be $2L$ and let the distance between opposite edges be $D$.
$D$ is $2L$ for a square, $2\sqrt 3L$ for a hexagon and $(2+2\sqrt 2)L$ for an octagon.
The large circle has equation $x^2+y^2=49$.
The upper small circle has equation $x^2-\frac{7}{\sqrt 2}x+y^2-\frac{7}{\sqrt 2}y=0$.
The line $y=L$ intersects the upper circle at $(x,L)$ and the large circle at $(x+D,L)$. Eliminating $x$ from the two equations for these points gives us a polynomial equation for $L$ in terms of $D$.
(1) $x^2+2Dx+D^2+L^2=49$.
(2) $x^2-\frac{7}{\sqrt 2}x+L^2-\frac{7}{\sqrt 2}L=0$.
First eliminate $x^2$.
$(7+2\sqrt 2 D)x=49\sqrt 2-\sqrt 2 D^2-7L$
Substituting into Equation 1 then gives us the required polynomial. We shall now do this for the square.
$(7+4\sqrt 2 L)x=49\sqrt 2-7L-4\sqrt 2 L^2$
$(7+4\sqrt 2 L)(x+D)=4\sqrt 2 L^2+7L+49\sqrt 2$
$(4\sqrt 2 L^2+7L+49\sqrt 2)^2=(49-L^2)(4\sqrt 2 L+7)^2$
$64L^4+112\sqrt 2 L^3-686L^2-2058\sqrt 2 L+2401=0$
The relevant solution is $L=0.7274$ and so the length of a side is approximately $1.455$.
| {
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Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$ I'm supposed to prove that the product of three successive natural numbers the middle of which is the square of a natural number is divisible by $60$. Here's my attempt.
My Attempt:
$$\text{P}=(n^2-1)n^2(n^2+1)=n(n-1)(n+1)[n(n^2+1)]$$
It is now enough to prove that $n(n^2+1)$ is divisible by $10$. But for $n=4$, $4(17)\ne10\lambda$ but for $n=4$, $\text{P}$ is $4080=60\cdot68$ which means apart from just being a multiple of $6$, $n(n-1)(n+1)$ is actually helping $n(n^2+1)$ with a $5$ to sustain divisibility by $60$.
How to tackle this? Thanks
| Use the Chinese Remainder theorem to show this product is congruent to $0\bmod 3, 4$ and $5$. You'll determine first what the squares are, modulo these numbers.
A sketch for the case of the modulus $4$:
Every number $n\equiv 0,1,2$ or $3\: (\equiv -1)\bmod 4$. So
$$n^2\equiv 0^2=0,\: 1^2=1, 2^2\equiv 0\quad\text{or}\quad 3^2\equiv(-1)^2=1.$$
As a conclusion, either $n^2$ or $n^2-1\equiv 0\mod 4$.
| {
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$\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. $\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. What is the product of all possible value of $p$?
Note that $p$ could be a complex number.
I tried some basic Vieta's formulas, couldn't find easy way to simplify...
| $$(x^3-1)^3=(x^2-px)^3$$
$$(x^3)^3-1-3x^3(x^3-1)=(x^3)^2-p^3(x^3)-3px^3(x^3-1)$$
Replace $x^3+1$ with $y$
$$(y-1)^3-3(y-1)(y-2)=(y-1)^2-p^3(y-1)-3p(y-1)(y-2)$$
$$y^3+(\cdots)y^2+(\cdots)y-1-6+1-p^3+6p=0$$
Now apply Vieta's formula
| {
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Find all solutions of $a,b,c,d$ if $a+b+c+d=12$ and $abcd = 27 +ab+ac+ad+bc+bd+cd$ Find all solutions of $a,b,c,d>0$ if
$$a+b+c+d=12$$
$$abcd = 27 +ab+ac+ad+bc+bd+cd$$
Attempt:
$a + b + c + d = 12 \implies 3 \ge \sqrt[4]{abcd} \implies 81 \ge abcd$ by AM-GM, eqyality when $a=b=c=d$. Also
$$ abcd - 27 = ab + ac + ad + bc + bd + cd \ge 6 \sqrt[6]{a^{3} b^{3} c^{3} d^{3}} = 6 \sqrt{a b c d} $$
$$ (abcd - 27)^{2} \ge 36 a b c d$$
dari AM-GM, equality when $ab=ac=ad=bc=bd=cd$. then we must have
$$ (abcd - 27)^{2} = 36 \times 81 \implies abcd = 54 + 27 = 81 $$
atau
$$ (abcd - 27)^{2} = 36 \times 81 \implies abcd = -54 + 27 = -27 (\text{not possible}) $$
So $abcd = 81$. Since $a=b=c=d$ then $a=b=c=d=3$.
Is this the only solution? if it is, why?
| You are fine : essentially, you got $$a+b+c+d \geq 4 \sqrt[4]{abcd} \implies 81 \geq abcd$$
from the first line. Then from the next few, you got :
$$
ab+bc+cd+da+ac+bd \geq 6 \sqrt[6]{a^3b^3c^3d^3} = 6 \sqrt{abcd}
$$
Combining this with $ab+bc+cd+da+ac+bd = abcd - 27$ gives that $abcd - 27 \geq 6 \sqrt{abcd}$, which can be rearranged to $(\sqrt{abcd} - 3)^2 \geq 36$. Now from here and the fact that $\sqrt{abcd}$ must be positive we get $\sqrt{abcd} \geq 9$. Combining with our first observation we get that equality is attained in AM-GM which gives $a=b=c=d=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx$
Find $\displaystyle\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx$
My approach is as follow
$\cos^2x=t$;
$\sin2x\ dx=-dt$
Therefore,
\begin{align}
\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx&=\int \frac{\sin2x+\frac{2\tan x\cos^2x}{\cos^2x}}{\cos^6x+6\cos^2x+4}dx\\
&=\int \frac{\sin2x+\frac{\sin2x}{\cos^2x}}{\cos^6x+6\cos^2x+4}dx\\
&=\int \frac{-(1+\frac{1}{t})}{t^3+6t+4}dt
\end{align}
I am not able to proceed from here
| Let $$I=-\int \frac{1+1/t}{t^3+6t+4} dt=\int \frac{1+t}{12t(t^3+6t+4)} dt=\int \left(\frac{1}{4t}-\frac{1}{12} \frac{3t^3+6}{(t^3+6t+4)}\right) dt$$ $$=\frac{\ln t}{4}-\frac{1}{12} \ln
~(t^3+6t+4)+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding the Taylor Series Expansion using Binomial Series, then obtaining a subsequent Expansion. Hi all doing some prep work for a course and really struggling to wrap my head around some of the revision questions. Some help would be really appreciated. There are two halfs, part one is as follows:
Consider the function:
$$ f(x) =\frac{1}{1-x} $$
Find the Taylor series expansion by expanding as a binomial series.
My solution is using $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...$
In this case, $n = -1$ and $x = -1$.
So by substituting I end up with $$ 1 + x + x^2 + x^3 + ... $$
So the Taylor Series Expansion in this case is:
$$\sum_{n=0}^\infty x^n $$
Ok so far, now the second half.
Using your result for f (x) obtain the Taylor series for:
$$ g(x) =\frac{x^2}{1+x^2} $$
From what I understand reading, I need to convert f(x) into g(x) then apply whatever mathematical steps made to the terms in the series, which I did:
$$ f(x) =\frac{1}{1-x} * \frac{1-x}{1} = \frac{1}{1}$$
$$ \frac{1}{1} * \frac {1}{x^2+1} = \frac{1}{x^2+1}$$
$$ \frac{1}{x^2+1} * x^2 = \frac{x^2}{x^2+1} = g(x)$$
I'll leave out the individual steps, but applying these to the terms of the series, I get:
$$ \frac{x^2}{x^2+1} - \frac{x^3}{x^2+1} + \frac{x^3}{x^2+1} + \frac{x^4}{x^2+1} + \frac{x^4}{x^2+1}$$
Which is as far as I'm comfortable going, since this really doesn't feel right at all for a solution. I'd really appreciate if someone could point me in the right direction or let me know if I'm made some catastrophic error. I'm really not sure how to write the final terms in sigma notation either. Feel free to give me a slap on the wrist if I've made any silly errors.
| $$g(x) =\frac{x^2}{1+x^2}=\frac{x^2+1-1}{1+x^2}=1-\frac{1}{1+x^2}$$ make $x^2=-t$ to face $f(t)$, simplify and, at the end, replace $t$ by $-x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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