Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proving $\sum_{k=0}^{\infty} F_{mk}z^k=\frac{F_mz}{1-z(F_{m-1}+F_{m+1})+(-1)^mz^2}$ I have read in a few places that$$\sum_{k=0}^{\infty} F_{mk}z^k=\frac{F_mz}{1-z(F_{m-1}+F_{m+1})+(-1)^mz^2}$$where $F_i$ denotes the $i$-th Fibonacci number. The series is a generalization of the more known $$\sum_{k=0}^{\infty} F_{k}z^k=\frac{1}{1-z-z^2}$$The latter is relatively easy to prove, as done here, but I havent been able find a proof of the former.
| Try using Binet’s formula for the Fibonacci numbers:
$$F_n=\frac{1}{\sqrt{5}}\bigg(\frac{1+\sqrt{5}}{2}\bigg)^n-\frac{1}{\sqrt{5}}\bigg(\frac{1-\sqrt{5}}{2}\bigg)^n$$
Combining this with the formula for the sum of a geometric series, we have that your sum is equal to
$$\begin{align}
\sum_{k=0}^\infty F_{mk}z^k &= \frac{1}{\sqrt{5}}\sum_{k=0}^\infty (\phi^{mk}-(-\phi)^{-mk})z^k \\
&= \frac{1}{\sqrt{5}}\bigg(\frac{1}{1-\phi^m z} - \frac{1}{1-(-\phi)^{-m}z}\bigg) \\
&= \frac{1}{\sqrt{5}}\frac{1-(-\phi)^{-m} z-1+\phi^{m}z}{(1-\phi^m z)(1-(-\phi)^{-m}z)} \\
&= \frac{1}{\sqrt{5}} \frac{\phi^m z-(-\phi)^{-m}z}{1-(\phi^m+(-\phi)^{-m})z+(-1)^m z^2} \\
&= \frac{F_m z}{1-(F_{m-1}+F_{m+1})z+(-1)^m z^2}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$ Evaluate $$\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$$
I tried substitutions like $u=\frac{\pi}{4}-x$, and trig identities like $\cos^2{x}=1-\sin^2{x}$ after getting a common denominator.
$$4\int_0^{\frac{\pi}{4}} \frac{\sin^2{(5x)\cos^2{x}-\cos^2{(5x)}\sin^2{x}}}{\sin^2{(2x)}} \mathop{dx}$$
Where should I go from here? Any help is appreciated!
| Use the Chebyshev formulas: $$\int_0^{\frac{\pi}{4}} \frac{( \color{red}{\sin{(5x)} \color{black}{)^2}}}{\sin^2{x}} -\frac{( \color{blue}{\cos{(5x)} \color{black}{)^2}}}{\cos^2{x}} \,{dx}$$
$$=\int_0^{\frac{\pi}{4}}\frac{\left( \color{red}{ 16
\sin^5 (x)- 20 \sin^3( x) + 5\sin (x) } \right)^2}{\sin^2{x}} -\frac{ \left(\color{blue}{16 \cos^5 (x) -20 \cos^3(x) + 5 \cos(x) } \right)^2}{\cos^2{x}} \,{dx}$$
$$=\int_0^{\frac{\pi}{4}} {\left( 16
\sin^4 (x)- 20 \sin^2( x) + 5\right)^2} -{\left(16 \cos^4 (x) -20 \cos^2(x) + 5 \right)^2}\,{dx}$$
$$=\int_0^{\frac{\pi}{4}} {\left( 16
\sin^4 (x)- 20 \sin^2( x) + 5\right)^2} -{\left(16 \cos^4 (x) -20 \cos^2(x) + 5 \right)^2}\,{dx}$$Multiply it out; when you do, it cleans up rather nicely:
$$
= \int _0 ^{\pi/4} 8 (2 \cos(2 x) + \cos(6 x)) \,dx
$$
$$
=\left. 8\cdot \left( \sin(2x)+\frac{1}{6}\sin(6x) \right)\right|_0^{\pi/4} = 8\cdot \frac{5}{6}=\frac{20}{3}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Implicit differantiation Let z and w be differantiable functions of x and y and they satisfy the following equations.
$$ xw^3+yz^2+z^3=-1$$ $$ zw^3-xz^3+y^2w=1$$
Find $\frac{\partial z}{\partial x}$ and its value at $(x,y,z,w)=(1,-1,-1,1)$.
I don't know what to do with two implicit functions. So I thought maybe I can write $w$ in the first equation as a function of x,y and z such as :
$$ w = \left( -\frac{1+yz^2+z^3}{x} \right) ^{1/3}$$
Then I can put it into the second equation:
$$ z\left( \frac{-(1+yz^2+z^3)}{x} \right) - xz^3+y^2\left( \frac{-(1+yz^2+z^3)}{x} \right)^{1/3}=1$$
And here I can calculate the partial derivative as following:
$$F(x,y,z) = z\left( \frac{-(1+yz^2+z^3)}{x} \right) - xz^3+y^2\left( \frac{-(1+yz^2+z^3)}{x} \right)^{1/3} $$
$$\frac{\partial z}{\partial x} = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}} $$
Then I found the result as:
$$ \frac{\partial z}{\partial x} = \frac{\frac{z(1+yz^2+z^3)}{x^2}-z^3+\frac{y^2(1+yz^2+z^3)^{1/3}}{3x^{4/3}}}{\frac{-(4z^3+3yz^2+1)}{x}-3xz^2-\frac{y^2\left(\frac{z(3z+2y)}{3(z^3+yz^2+1)^{2/3}} \right)}{3x^{4/3}}}$$
But I'm guessing that I missed something here by eliminating $w$. Could you please give me your opinions? Thank you.
| Since you have
\begin{align}
&x w^3 + y z^2 + z^3 = -1 \\
&z w^3 - x z^3 + y^2 w = 1,
\end{align}
take the derivative with respect to $x$ for both equation, you have
\begin{align}
&x (3 w^2 \frac{\partial w}{\partial x}) + w^3 + y (2 z \frac{\partial z}{\partial x})
+ 3 z^2 \frac{\partial z}{\partial x} = 0 \\
&z (3 w^2 \frac{\partial w}{\partial x}) + (\frac{\partial z}{\partial x}) w^3 -
x (3 z^2 \frac{\partial z}{\partial x}) - z^3 + y^2 \frac{\partial w}{\partial x} = 0,
\end{align}
then by substituting $(x, y, z, w) = (1, -1, -1, 1)$, you have
\begin{align}
& 3 \frac{\partial w}{\partial x} + 1 + 2 \frac{\partial z}{\partial x}
+ 3 \frac{\partial z}{\partial x} = 0 \\
-&3 \frac{\partial w}{\partial x} + \frac{\partial z}{\partial x} -
3 \frac{\partial z}{\partial x} + 1 + \frac{\partial w}{\partial x} = 0.
\end{align}
Obviously, the solution for $\frac{\partial z}{\partial x}$ is
$$
\frac{\partial z}{\partial x}|_{(x,y,z,w)=(1,-1,-1,1)} = -\frac{5}{4}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a quick (hopefully elementary) way to prove that $6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0$ has only one solution? I have the Diophantine equation $$6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0.$$ Numerical calculations suggest this has only one positive integer solution, namely $(b,c)=(2,3)$. Is there a quick way to prove or disprove that?
| The equation is quadratic in $c$.
If $b, c$ are positive integers, we have
$$c = \frac{18b + \sqrt{3[(2b^2+1)^3 - 54]}}{6b^2 + 3}.$$
So, $(2b^2+1)^3 - 54 = 3m^2$ for some positive integer $m$.
Since $3 | 54$ and $3 | 3m^2$, we know that $3 | (2b^2 + 1)$. Let $x = \frac{2b^2+1}{3}$ and $y = \frac{m}{3}$.
We have $x^3 - 2 = y^2$. Since $x$ is a positive integer, we know that $y$ is also a positive integer.
The equation $x^3 - 2 = y^2$ is well-known.
The only positive integer solutions to $x^3 - 2 = y^2$ are $(x, y) = (3, 5)$.
So, $(b, c) = (2, 3)$ is the only positive integer solution.
See:
[1] Theorem 3.4 (page 7), https://kconrad.math.uconn.edu/blurbs/gradnumthy/mordelleqn1.pdf
[2] Solving the diophantine equation $y^{2}=x^{3}-2$
[3] https://en.wikipedia.org/wiki/Mordell_curve
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$ (I need a review of my resolution please :) ) Find the limit:
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$$
This is what I have, i'm not sure about my answer (I'm just learning limits).
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \lim_{x \to 1} \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1} $$
And I know that cos is a bounded limit, also its $L=-1$.
The other limit:
$$ \lim_{x \to 1} \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1} $$
$$ \lim_{x \to 1} \sqrt{\frac{(x-1) \cdot (x-1)}{(x-1)\cdot (x+1)}} \cdot \frac{1-x}{x^2+x-1} $$
Simplifying:
$$ \lim_{x \to 1} \sqrt{\frac{(x-1) \cdot (x-1)}{(x-1)\cdot (x+1)}} \cdot \frac{1-x}{x^2+x-1} $$
$$ \lim_{x \to 1} \sqrt{\frac{(x-1)}{(x+1)}} \cdot \frac{1-x}{x^2+x-1} $$
Evaluating:
$$ \lim_{x \to 1} \sqrt{\frac{(1-1)}{(1+1)}} \cdot \frac{1-1}{1^2+1-1} $$
$$ = \sqrt{0} \cdot \frac{0}{1} $$
$$\fbox {= 0}$$
So since I have a bounded limit and the other limit function is zero, the whole limit is $0$.
So;
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1} = 0$$
| Too long for comments.
Your work is fine but I think that there is a faster solution.
When I see a problem such as $\lim_{x \to a} f(x)$, my first reaction is to let $x=y+a$ and consider $\lim_{y \to 0} g(y)$.
If we do it for your problem, we have
$$\lim_{x \to 1} \cos(\pi x) \, \sqrt{\frac{(x-1)^2}{(x^2-1)}} \, \frac{1-x}{x^2+x-1}=\lim_{y \to 0}\cos (\pi y)\frac{y }{y^2+3 y+1}\sqrt{\frac{y}{y+2}}$$ which gives $1\times 0\times 0=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $\sum\limits_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$? I am trying to prove this. I used the telescoping method, but the problem is I need the first fraction to be $\frac{1}{4(n+1)}$ and I also tried to relate it to alternating Harmonic series which didn't work. Any hint would be greatly appreciated.$$\sum_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$$
| May be interesting to consider the most general case of
$$S_p=\sum _{n=1}^{p } \left(\frac{1}{a n+b}-\frac{1}{c n+d}\right)$$ Using the digamma function,
$$S_p=\frac{c \psi \left(\frac{b}{a}+p+1\right)-a \psi \left(\frac{d}{c}+p+1\right)-c \psi
\left(\frac{b}{a}+1\right)+a
\psi\left(\frac{d}{c}+1\right)}{a c}$$ Expanded as series for large values of $p$
$$S_p=\frac{(c-a) \log (p)-c \psi \left(\frac{a+b}{a}\right)+a \psi
\left(\frac{c+d}{c}\right)}{a
c}+O\left(\frac{1}{p}\right)$$ which can converge only if $c=a$. In such a case
$$S_p=\frac{\psi \left(\frac{a+d}{a}\right)-\psi
\left(\frac{a+b}{a}\right)}{a}+O\left(\frac{1}{p}\right)$$
For the particular case where $d=0$, $\psi(1)=-\gamma$ and
$$S_p=-\frac 1a\left(\psi \left(1+\frac{b}{a}\right)+\gamma \right)+O\left(\frac{1}{p}\right)$$ and if $a=k b$ the term in parentheses has simple expressions up to ... $k=4$.
| {
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Find $n$ such that $1-a c^{n-1} \ge \exp(-\frac{1}{n})$ I am trying to find the integer $n$ such that
\begin{align}
1-a c^{n-1} \ge \exp(-\frac{1}{n})
\end{align}
where $a>0$ and $c \in (0,1)$.
I know that finding it exactly is difficult. However, can one find good upper and lower bounds it.
It tried using lower bound $\exp(-x) \le 1-x+\frac{1}{2}x^2$. However, it didn't really work.
| You could chain you're inequalities in the following way:
$$1 - ac^{n - 1} \geq exp\left (\frac{-1}{n} \right) \geq 1 - \frac{1}{n}$$
Then
$$-ac^{n - 1} \geq -\frac{1}{n}$$
$$\iff ac^{n - 1} \leq \frac{1}{n} $$
Now, we can move all of the constants on one side and all of the variables to the other:
$$\frac{a}{c} \leq \frac{1}{n c^n}$$
Clearly, the left side is constant. On the other hand, we know that because $c \in (0,1),$ then
$$\frac{1}{n c^n} < \frac{1}{c^n}$$
Therefore, we can chain in the following way:
$$\frac{a}{c} \leq \frac{1}{n c^n} < \frac{1}{c^n}$$
$$\iff \ln(\frac{a}{c} ) < n ln(\frac{1}{c})$$
$$\iff \frac{\ln(\frac{a}{c} )}{\ln(\frac{1}{c})} < n$$
| {
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Solution verification: $ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $ This limit is not too difficult but I was just wondering if my work/solution looked good?
Thanks so much for your input!!
$$ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $$
$$ 2 x - 6 = 2 x \left( 1 - \frac 6 { 2 x } \right) $$
$$ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } =
\lim _ { x \to 3 } \frac { 2 x \left( 1 - \frac 6 { 2 x } \right) } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \frac { x \left( 1 - \frac 6 { 2 x } \right) } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \frac { x - 3 } { \sqrt x - \sqrt 3 } $$
By rationalizing the denominator:
$$ \frac { x - 3 } { \sqrt x - \sqrt 3 } = \sqrt x + \sqrt 3 $$
$$ 2 \cdot \lim _ { x \to 3 } \frac { x - 3 } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \left( \sqrt x + \sqrt 3 \right) $$
By plugging in $ x = 3 $:
$$ 2 \cdot \lim _ { x \to 3 } \left( \sqrt x + \sqrt 3 \right) =
2 \left( \sqrt 3 + \sqrt 3 \right) = 4 \sqrt 3 $$
| Yes, the solution is correct. The only minor stylistic change I would personally consider is not going into as much detail when factoring out the 2 in the second line, and mention that you're multiplying your fraction by $\frac{\sqrt{x}+\sqrt{3}}{\sqrt{x}+\sqrt{3}}.$
| {
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Prove that $TK=TO$
Given $\triangle ABC$ such that $\angle A=90^\circ$ inscribed in circle with center $O$. Let $D$ be the feet perpendicular from $A$ to $BC$ and $M$ be the mid-point of $BD$. Draw the line $AM$ and let it intersect the circumcircle at $X$. Let $K$ be the point on $AX$ such that $OK//XC$. Lastly, denote $T$ as the intersection of the perpendicular from $AX$ at $K$ to $XC$. Prove that $TK=TO$
I do some angle chasing but I haven't use anything that the problem given like the perp and midpoint for example, as I dun know how I could apply it. BTW, my approach is to prove that $\angle{BCA}=\angle{TOC}$ or perhaps prove that $\triangle ABX$ is similar to $\triangle TOC$. Please help
| Our solution relies on removing all the "annoying" points; essentially, $T$ and $K$ do not have many properties that we can use, so we attempt to rid them from our equations.
As you noted, we only need to have $\triangle ABX \sim \triangle TOC$, and then we are done. Since $\angle TCO = \angle BAX$, we only need to prove that $\frac{TC}{XA} = \frac{OC}{AB}$, or that $$2 \cdot TC \cdot AB = 2 \cdot OC \cdot XA = BC \cdot XA = BA \cdot CX + BX \cdot AC$$ by Ptolemy's theorem. Notice that $CX - CT = XT$, so we now want to prove that $$AB \cdot TC = BA \cdot XT + BX \cdot AC$$ or that $AB \cdot CX = 2 \cdot BA \cdot XT + BX \cdot AC$. Now, $\triangle XKT \sim \triangle ABC$, so we have $XT \cdot AB = XK \cdot BC$, and our equation now becomes $$AB \cdot CX = 2 \cdot XK \cdot BC + BX \cdot AC$$ and we have successfully removed the point $T$ from our equation. Now, to remove $K$, we notice that $\triangle OMK \sim \triangle CMX$, so $\frac{KX}{OC} = \frac{MX}{MC}$ and therefore substituting for $XK$ we now want to prove $$AB \cdot CX = BC^2 \frac{MX}{MC} + BX \cdot AC$$ We now make our final change to this equation, and then apply some trigonometry to finish the problem. By Power of a Point, we have $BM \cdot MC = AM \cdot MX$, or that $\frac{MX}{MC} = \frac{BM}{AM} = \frac{MD}{AM} = \sin \angle MAD$. Thus, we have that we want to prove $$AB \cdot CX = BC^2\sin \angle MAD + BX \cdot AC$$
Now, notice that $AB = BC \cos \angle ABC$, $CX = BC \cos \angle BCX$, $BX = BC \sin \angle BCX$, and $AC = BC \sin \angle ABC$, so we wish to prove that $$BC^2(\cos \angle ABC\cos \angle BCX - \sin \angle BCX\sin \angle ABC) = BC^2 \sin \angle MAD$$ However, it is well-known that for any angles $x$ and $y$, $\cos x \cos y - \sin x \sin y = \cos{(x+y)}$, so all we wish to show now is that $$\cos{(\angle ABC + \angle BCX)} = \sin \angle MAD$$ which is true since $\angle ABC + \angle BCX = \angle ABM + \angle BAX = \angle AMD$, so, since all our steps are reversible, we are done.
| {
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Primes Powers and Mods The Question is below:
For which primes $p$ is $(p − 1)^p + 1$ a power of $p$?
I think the answer is $2$ and $3$, none of the others work.
Here is what I have:
let $p^k-1=(p-1)^p$. Then we have $$p^{k-1}+...+p+1 \equiv0 \pmod{p-1}.$$
Please also include a proof of your answer
| If $(p-1)^p+1=p^k$ then
\begin{align*}
k&=\log_p \left(1+(p-1)^p\right)\\
&=\log_p \left(p^p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)\right)\\
&=p+\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)
\end{align*}
clearly
$$\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)<0$$
since $\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p<1$
and so it suffices to show that
$$\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)>-1$$
for $p>3$ since then $p+\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)$ cannot be an integer for $p>3$. Differentiating $\left(1-\frac{1}{x}\right)^x$ we get that it is an increasing function and so for $p\geq5$ we get that
$$\left(1-\frac{1}{x}\right)^x\geq\left(1-\frac{1}{5}\right)^5>.3$$
for $p\geq5$ we also clearly have that $\frac{1}{p}\leq \frac{1}{5}=.2$: it is thus that
\begin{align*}
\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p &> \left(1-\frac{1}{p}\right)^p\\
&>.3\\
&>.2\\
&\geq \frac{1}{p}
\end{align*}
taking logs base $p$ of both sides gets that
$$\log_p \left(\frac{1}{p^p}+\left(1-\frac{1}{p}\right)^p\right)>-1$$
and our proof is complete.
| {
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Closed form sought for $a_1 = a_2 = 1, a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i $ where $n>2$ I've been working through a problem that I've got as far as getting a recursive answer to. I was hoping to turn this into more of a "closed form" answer, but haven't really gotten anywhere. I'm hoping that someone can help with this, though anything would be greatly appreciated.
The recursive answer I have is a sequence of real numbers given by $$\begin{gather}
a_1 = a_2 = 1 \\
a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i \qquad (n > 2)
\end{gather}$$
The first few non-trivial members of this sequence are
*
*$a_3 = \frac{5}{3}$
*$a_4 = 2$
*$a_5 = \frac{37}{15}$
*$a_6 = \frac{26}{9}$
*$a_7 = \frac{349}{105}$
I've tried to express these in terms of $a_1$ and $a_2$ and and constants and arrived at
*
*$a_3 = 1 + \frac{2}{3} a_1$
*$a_4 = 1 + \frac{2}{4} a_1 + \frac{2}{4} a_2$
*$a_5 = (1 + \frac{2}{5}) + (\frac{2}{5} + \frac{2^2}{3\cdot5} ) a_1 + \frac{2}{5} a_2$
*$a_6 = (1 + \frac{2}{6} + \frac{2}{6}) + (\frac{2}{6} + \frac{2^2}{3 \cdot 6} + \frac{2^2}{4 \cdot 6}) a_1 + (\frac{2}{6} + \frac{2^2}{4 \cdot 6}) a_2$
*$a_7 = (1 + \frac{2}{7} + \frac{2}{7} + \frac{2}{7} + \frac{2^2}{5 \cdot 7}) + (\frac{2}{7} + \frac{2^2}{3\cdot7} + \frac{2^2}{4 \cdot 7} + \frac{2^2}{5 \cdot 7} + \frac{2^3}{3 \cdot 5 \cdot 7}) a_1 + (\frac{2}{7} + \frac{2^2}{4 \cdot 7} + \frac{2^2}{5 \cdot 7}) a_2$
I am not seeing a pattern developing here.
I also rearranged the above noting that $a_1 = a_2 = 1$ and got
*
*$a_3 = 1 + \frac{2}{3} $
*$a_4 = 1 + 2 (\frac{2}{4})$
*$a_5 = 1 + 3 (\frac{2}{5}) + \frac{2^2}{3\cdot5}$
*$a_6 = 1 + 4 (\frac{2}{6}) + \frac{2^2}{3 \cdot 6} + 2 (\frac{2^2}{4 \cdot 6})$
*$a_7 = 1 + 5 (\frac{2}{7}) + \frac{2^2}{3 \cdot 7} + 2 (\frac{2^2}{4 \cdot 7}) + 3 (\frac{2^2}{5 \cdot 7}) + \frac{2^3}{3 \cdot 5 \cdot 7}$
Here I do notice a couple of things
*
*The expression for $a_n$ begins with "$1 + (n-2) \frac{2}{n}$".
*The remaining terms of the expression look like "$k \dfrac{2^{i+1}}{b_1 \cdots b_{i} \cdot n}$" where each $b_j$ is between $3$ and $n-2$ and consecutive numbers cannot appear among them. The $k$ seems to be determined by the smallest number among the $b_j$, but this is more of a guess than anything right now.
These observations are not really helping me much at all.
| I wrote Benedict W. J. Irwin's simplified recurrence in the form
$$
na_n = 1 + 2a_{n - 2} + (n - 1)a_{n - 1} ,
$$
which gave me for the generating function $G(x) = \sum\nolimits_{n = 1}^\infty {a_n x^n }$ the ODE
$$
\frac{1}{{1 - x}} + 2xG(x) + (x - 1)G'(x) = 0.
$$
The particular solution we are looking for is
$$
G(x) = \frac{{1 - e^{ - 2x} }}{{2(x-1)^2}}.
$$
You should be able to obtain a formula for the $a_n$'s from this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Showing that $\sin^2x\cdot\sin^22x\cdot\sin^24x\cdot\sin^28x\cdots\sin^22^nx\leq\frac{3^n}{4^n}$
Show that $$\sin^2x\cdot\sin^22x\cdot\sin^24x\cdot\sin^28x\cdots\sin^22^nx\leq\frac{3^n}{4^n}$$
I understand the result of an arithmetic sequence $(\sin1^\circ)(\sin3^\circ)(\sin5^\circ)…(\sin89^\circ)$, how about the geometric sequence case?
| We first prove that
$$ (\sin x)^4(\sin 2x)^2 \leq \left(\frac{3}{4}\right)^3. $$
Indeed, applying the double angle formula $\sin 2x = 2\sin x\cos x$ and substituting $t = \sin^2 x$, we have
$$ (\sin x)^4(\sin 2x)^2 = 4t^3(1-t) $$
and the right-hand side is maximized at $t = \frac{3}{4}$ with the value $(3/4)^3$ as desired. Now, returning to the original problem, the above inequality yields
\begin{align*}
&(\sin x)^2 (\sin 2x)^2 \dots (\sin 2^n x)^2 \\
&= \Biggl[ (\sin x)^2 (\sin 2^n x)^4 \prod_{k=0}^{n-1} (\sin 2^k x)^4 (\sin 2^{k+1}x)^2 \Biggr]^{1/3} \\
&\leq \Biggl[ \prod_{k=0}^{n-1} \left(\frac{3}{4}\right)^3 \Biggr]^{1/3} \\
&= \left(\frac{3}{4}\right)^n
\end{align*}
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
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Is the inequality true for all $n\geq 2$? Let $x,y,z>0$. I am wondering if the following inequality is true?
$$\sum_{cyc}\frac{x^n}{y^2+yz+z^2}\geq\frac{x^{2n-2}+y^{2n-2}+z^{2n-2}}{x^n+y^n+z^n},\qquad n\geq 2$$
If not, is it known for which $n$ it is true?
$\displaystyle(1)\qquad\sum_{cyc}\dfrac{x^2}{y^2+yz+z^2}\overset{AGM}{\geq} \frac{2}{3}\sum_{cyc}\frac{x^2}{y^2+z^2}\overset{Nesbitt}{\geq}1=\frac{x^2+y^2+z^2}{x^2+y^2+z^2}$
$\displaystyle(2)\qquad\sum_{cyc}\dfrac{x^3}{y^2+yz+z^2}=\sum_{cyc}\dfrac{3x^4}{3xy^2+3xyz+3xz^2}\overset{AGM}{\geq}\sum_{cyc}\frac{3x^4}{3x^3+3y^3+3z^3}=\frac{x^4+y^4+z^4}{x^3+y^3+z^3}$
$\displaystyle(3)\qquad\sum_{cyc}\dfrac{x^4}{y^2+yz+z^2}\overset{AGM}{\geq}\frac{2}{3}\sum_{cyc}\dfrac{x^4}{y^2+z^2}\overset{(*)}{\geq}\frac{x^6+y^6+z^6}{x^4+y^4+z^4}$
$(*)\iff\forall$ $a,b,c>0$, $\,\,\displaystyle \frac{2}{3}\sum_{cyc}\dfrac{a^2}{b+c}\geq\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$
$\iff \displaystyle \frac{2}{3}\left(\sum_{cyc}\dfrac{a^2}{b+c}-\frac{a+b+c}{2}\right)\geq\frac{a^3+b^3+c^3}{a^2+b^2+c^2}-\frac{a+b+c}{3}$
$\iff\displaystyle\frac{(a+b+c)}{3(a+b)(b+c)(c+a)}\sum_{cyc}(a+b)(a-b)^2\geq\frac{1}{3(a^2+b^2+c^2)}\sum_{cyc}(a+b)(a-b)^2$
$\iff\displaystyle\left((a+b+c)(a^2+b^2+c^2)-(a+b)(b+c)(c+a)\right)\sum_{cyc}(a+b)(a-b)^2\geq0$
where
$$(a+b+c)(a^2+b^2+c^2)-(a+b)(b+c)(c+a)$$
$$\ge(a+b+c)(ab+bc+ca)-(a+b)(b+c)(c+a)=abc>0$$
Done!
| A proof for $n=4$.
We need to prove that:
$$\sum_{cyc}\frac{x^4}{y^2+yz+z^2}\geq\frac{x^6+y^6+z^6}{x^4+y^4+z^4}.$$
Now, by AM-GM $$\sum_{cyc}\frac{x^4}{y^2+yz+z^2}\geq\sum_{cyc}\frac{x^4}{y^2+\frac{y^2+z^2}{2}+z^2}=\frac{2}{3}\sum_{cyc}\frac{x^4}{y^2+z^2}.$$
Id est, it's enough to prove that
$$\sum_{cyc}\frac{x^2}{y+z}\geq\frac{3}{2}\cdot\frac{x^3+y^3+z^3}{x^2+y^2+z^2}$$ for positives $x$, $y$ and $z$.
Now, by C-S $$\sum_{cyc}\frac{x^2}{y+z}=\sum_{cyc}\frac{x^6}{x^4y+x^4z}\geq\frac{(x^3+y^3+z^3)^2}{\sum\limits_{cyc}(x^4y+x^4z)}.$$
Thus, it's enough to prove that $$2(x^3+y^3+z^3)(x^2+y^2+z^2)\geq3\sum\limits_{cyc}(x^4y+x^4z)$$ or
$$\sum_{cyc}(2x^5-3x^4y-3x^4z+2x^3y^2+2x^3z^2)\geq0$$ or
$$\sum_{cyc}(x^5-3x^4y+2x^3y^2+2x^2y^3-3xy^4+y^5)\geq0$$ or
$$\sum_{cyc}(x+y)(x-y)^4\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Evaluating limit of the function at $\frac{\pi}{2}$ I'm trying to solve this
$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}}{(x-\frac{\pi}{2})^3}$$
I have tried using the L'Hôpital's rule
But I'm stuck at
$$\lim_{x \to \frac{\pi}{2}} \frac{-\sin{x}}{3(x-\frac{\pi}{2})^2}$$
Since the above equation is not in the $\frac{0}{0}$ , $\frac{\infty}{\infty}$
or $\frac{anything}{\infty}$ form
Then I tried expanding the $\cos{x}$ as taylor series at $x=\frac{\pi}{2}$. Which on simplifying I am left with
$$ \lim_{x \to \frac{\pi}{2}} \frac{-1}{(x - \frac{\pi}{2})^2} + \frac{1}{6} - \frac{1}{120} (x - \frac{\pi}{2})^2 + \frac{(x - \frac{\pi}{2})^4}{5040} - ... $$
and I'm stuck again. How do I proceed ahead?
| Since we have an indeterminate form of type $(0/0)$, we can apply the l'Hopital's rule:
$$\color{blue}{\lim_{x \to \frac{\pi}{2}} \frac{\cos{\left(x \right)}}{\left(x - \frac{\pi}{2}\right)^{3}}} = \color{magenta}{\lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx}\left(\cos{\left(x \right)}\right)}{\frac{d}{dx}\left(\left(x - \frac{\pi}{2}\right)^{3}\right)}}$$
Hence we have:
$$\lim_{x \to \frac{\pi}{2}}\left(- \frac{\sin{\left(x \right)}}{3 \left(x - \frac{\pi}{2}\right)^{2}}\right)=\lim_{x \to \frac{\pi}{2}}\left(- \frac{4 \sin{\left(x \right)}}{3 \left(\pi - 2 x\right)^{2}}\right)$$
After if $x\to \pi/2$ we have $\text{limited function}/0\to +\infty$ being $3 \left(\pi - 2 x\right)^{2}\geq0$, but with minus sign we have:
$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{\left(x \right)}}{\left(x - \frac{\pi}{2}\right)^{3}} = -\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate: $\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx.$
Prove that:$$\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx=\frac{2\ln c}{\sqrt{4c^2-b^2}}\cot^{-1}\left(\frac{b}{\sqrt{4c^2-b^2}}\right),$$ where $4c^2-b^2>0, c>0.$
We have:
\begin{align}
4(x^2+bx+c^2)&=(2x+b)^2-\left(\underbrace{\sqrt{4c^2-b^2}}_{=k\text{ (say)}}\right)^2\\\\
&=(2x+b+k)(2x+b-k).
\end{align}
Thus,
\begin{align}
I&=4\int_0^{\infty}\frac{\ln x}{(2x+b+k)(2x+b-k)}\,dx\\\\
&=\frac4{2k}\int_0^{\infty}\left(\frac1{2x+b-k}-\frac1{2x+b+k}\right)\ln x\, dx\\\\
&=\frac2k\left[\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b-k}\,dx}_{=I_1}-\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b+k}\,dx}_{=I_2}\right]
\end{align}
For $I_1:$ Letting $2x+b-k=t$ yields
\begin{align}
I_1&=\int_{b-k}^{\infty} \frac{\ln(t-b+k)-\ln 2}t\,dt\\\\
&=\int_{b-k}^{\infty}\frac{\ln(t-b+k)}t\, dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t
\end{align}
Now $\ln(t-b+k)=\ln(k-b)+\ln\left(\frac{t}{k-b}+1\right).$ So,$$I_1=\ln(k-b)\int_{b-k}^{\infty}\frac{dt}t+\int_{b-k}^{\infty} \frac{\ln\left(\frac t{k-b}+1\right)}t\,dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t.$$
If we let: $\frac t{k-b}=-u.$ Then $$I_1=\ln\left(\frac{k-b}2\right)\int_{b-k}^{\infty} \frac{dt}t+\int_1^{\infty}\frac{\ln(1-u)}u\,du.$$
It's getting messier and messier. Also it seems that $I_1$ and hence $I$ diverges. Please tell me, am I heading towards something useful? It has become difficult for me to handle this integral. Please show me a proper way.
| Hint : Substitute $xy=c^2\to x=\frac{c^2}{y}$ then the integral becomes $$ I= \int_0^{\infty}\frac{\ln(c^2)-\ln y}{y^2+by+c^2}dy\\2I= \int_0^{\infty}\frac{\ln(c^2)}{y^2+by+c^2}dy$$ and since $$ y^2+by+c^2= \left(y+\frac{b}{2}\right)^2+\left(c^2-\frac{b^2}{4}\right)$$ and using elementary integral formula of $$\int_{0}^{\infty}\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)$$ we have our desired closed form.
| {
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"url": "https://math.stackexchange.com/questions/3754243",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Simplify $\sqrt{8-\sqrt{63}}$ I simplified the expression into $$\sqrt{8-3\cdot \sqrt{7}}$$ but my tutor said it wasn't the answer he was looking for. Can someone help me?
| $$ 8-3\sqrt{7} = a^2 + b^2 - 2ab $$
Let $3\sqrt{7}= 2ab$
$$ab = 1.5\sqrt{7}$$
$$b = \frac{1.5\sqrt{7}}{a}$$
$$a^2 + b^2 = 8$$
$$a^2 +\frac{15.75}{a^2} = 8$$
$y = a^2$
$$y + \frac{15.75}{y} = 8$$
$$y^2 + 15.75 = 8y$$
$$y^2 + 15.75-8y = 0$$
Solve and get
$$y = \frac{7}{2}$$
$$y = \frac{9}{2}$$
Case 1
$$y = \frac{7}{2}$$
$$a = \pm \sqrt \frac{7}{2}$$
$$b = \pm 1.5\sqrt{2}$$
Just remember the sign of a and b are opposite
$$a -b = \sqrt \frac{7}{2}-1.5\sqrt{2}= \frac{\sqrt{7} -3}{\sqrt{2}} $$or
$$\frac{ 3-\sqrt{7}}{\sqrt{2}}$$
Case 2 :
$$y = \frac{9}{2}$$
$$a = \pm\frac{3}{\sqrt2}$$
$$b = \pm \frac{1.5\sqrt{14}}{3}=\pm\frac{ \sqrt{14}}{2}=\pm\frac{ \sqrt{7}}{\sqrt{2}} $$
which is the identical solution as above
| {
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"url": "https://math.stackexchange.com/questions/3754592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem with proving inequalities Question:
Prove that if $x,y,z$ are positive real numbers such that $x+y+z=a$ then $(a-x)(a-y)(a-z)>\frac8{27}a^3$ is not true.
My Approach:
$$\frac{a-x}{2}=\frac{y+z}2$$
$$\frac{a-y}{2}=\frac{x+z}2$$
$$\frac{a-z}{2}=\frac{x+y}2$$
Using $AM>GM$ we get $$\frac{x+y+z}{3}>\root 3 \of {xyz}$$ Cubing both sides and multiplying by $8$, $$\frac{8a^3}{27}>8xyz$$
Also, by $AM>GM$, $$(\frac{y+z}2)(\frac{x+z}2)(\frac{x+y}2)>8xyz$$
Now, how do I find the relation between $(\frac{y+z}2)(\frac{x+z}2)(\frac{x+y}2)$ and $\frac{8a^3}{27}$?
| Let $x=y\rightarrow0^+$ and $z\rightarrow a$.
Thus, the left side is closed to $0$, but the right side is greater than $0$,
which says that our inequality is not true.
| {
"language": "en",
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"source": "stackexchange",
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"answer_id": 3
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Estimating $\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$. I'm trying to solve this:
Which of the following is the closest to the value of this integral?
$$\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$$
(A) 1
(B) 1.2
(C) 1.6
(D) 2
(E) The integral doesn't converge.
I've found a lower bound by manually calculating $\int_{0}^{1} \sqrt{1+\frac{1}{3}} \ dx \approx 1.1547$. This eliminates option (A). I also see no reason why the integral shouldn't converge. However, to pick an option out of (B), (C) and (D) I need to find an upper bound too. Ideas? Please note that I'm not supposed to use a calculator to solve this.
From GRE problem sets by UChicago
| We know that,
${\displaystyle\int}\sqrt{\dfrac{1}{3x}+1}\,\mathrm{d}x$=$=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{\sqrt{3}}}}{\displaystyle\int}\sqrt{\dfrac{1}{x}+3}\,\mathrm{d}x$
Substitute $u=\sqrt{\dfrac{1}{x}+3}$ and $\dfrac{\mathrm{d}u}{\mathrm{d}x} = -\dfrac{1}{2\sqrt{\frac{1}{x}+3}x^2}$,i.e $\mathrm{d}x=-2\sqrt{\dfrac{1}{x}+3}x^2\,\mathrm{d}u$
${\displaystyle\int}\sqrt{\dfrac{1}{x}+3}\,\mathrm{d}x$=$-\class{steps-node}{\cssId{steps-node-2}{2}}{\displaystyle\int}\dfrac{u^2}{\left(u^2-3\right)^2}\,\mathrm{d}u$
$={\displaystyle\int}\left(\dfrac{\class{steps-node}{\cssId{steps-node-5}{u^2-3}}}{\left(u^2-3\right)^2}+\dfrac{\class{steps-node}{\cssId{steps-node-6}{3}}}{\left(u^2-3\right)^2}\right)\mathrm{d}u$
$={\displaystyle\int}\dfrac{1}{u^2-3}\,\mathrm{d}u+\class{steps-node}{\cssId{steps-node-7}{3}}{\displaystyle\int}\dfrac{1}{\left(u^2-3\right)^2}\,\mathrm{d}u$
Perform partial fraction decomposition:
$={\displaystyle\int}\left(\dfrac{1}{2\sqrt{3}\left(u-\sqrt{3}\right)}-\dfrac{1}{2\sqrt{3}\left(u+\sqrt{3}\right)}\right)\mathrm{d}u$ + ${\displaystyle\int}\dfrac{1}{u+\sqrt{3}}\,\mathrm{d}u$
On solving this further we get
$\dfrac{\sqrt{\frac{1}{3x}+1}}{3\left(\frac{1}{3x}+1\right)-3}+\dfrac{\ln\left(\sqrt{\frac{1}{x}+3}+\sqrt{3}\right)}{6}-\dfrac{\ln\left(\left|\sqrt{\frac{1}{x}+3}-\sqrt{3}\right|\right)}{6}+C$
I.e,
$\dfrac{6\sqrt{\frac{1}{3x}+1}x+\ln\left(\sqrt{\frac{1}{x}+3}+\sqrt{3}\right)-\ln\left(\left|\sqrt{\frac{1}{x}+3}-\sqrt{3}\right|\right)}{6}+C$
$\boldsymbol{\int\limits^{1}_{0}{f(x)}\,\mathrm{d}x =}$=${1\over6}[4 \sqrt(3) + \ln(7 + 4 \sqrt(3)]$
Approximation:
$1.593686504020857$,
i.e $1.6$.
The answer is option $(C)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
} |
On the series $\sum_{n=-\infty}^{\infty} \frac{\cos nx}{\Gamma\left ( a+n+1 \right ) \Gamma \left ( a-n+1 \right )}$ For the values of $a$ for which the following series makes sense, prove that
$$\sum_{n=-\infty}^{\infty} \frac{\cos nx}{\Gamma\left ( a+n+1 \right ) \Gamma \left ( a-n+1 \right )} = \frac{\left ( 2 \cos \frac{x}{2} \right )^{2a}}{\Gamma\left ( 2a+1 \right )}$$
I ran into this series recently and I have no idea how to attack it. How would you go to prove the identity?
| Indeed, following Jack's comment we have
\begin{align*}
\sum_{n=-\infty}^{\infty} \frac{\cos nx}{\Gamma\left ( a+n+1 \right ) \Gamma \left ( a-n+1 \right )} &= \sum_{n=-\infty}^{\infty} \frac{\cos nx}{\left ( a+n \right )! \left ( a-n \right )!} \\
&=\frac{1}{\left (2a \right )!}\sum_{n=-\infty}^{\infty} \binom{2a}{a+n} \cos nx \\
&=\frac{1}{\left (2a \right )!}\mathfrak{Re} \left ( \sum_{n=-\infty}^{\infty} \binom{2a}{a+n} e^{inx} \right ) \\
&= \frac{\left ( 2 \cos \frac{x}{2} \right )^{2a}}{\left ( 2a \right )!} \\
&= \frac{\left ( 2 \cos \frac{x}{2} \right )^{2a}}{\Gamma\left ( 2a+1 \right )}
\end{align*}
from the extended binomial theorem.
| {
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"url": "https://math.stackexchange.com/questions/3756433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What method can I use to compute the limit of this series? Let
$$
\begin{cases}
a_1 &=1 \\a_{n+1}&=\frac{1}{a_1+a_2+\cdots +a_n}-\sqrt{2}
\end{cases}
$$
Then $\sum_{i=1}^{\infty}{a_i}=?
$
I assume that the limit exists, and then I get the limit is equal to $\frac{\sqrt{2}}{2}$
$\sum_{k=1}^{n}{a_k=A_n,}A_{n+1}-A_n=\frac{1}{A_n}-\sqrt{2}.$take the limit of both sides can get $\frac{\sqrt{2}}{2}$.
$$A_n+\frac{1}{A_n}-\frac{3}{\sqrt{2}}=\frac{\left( \sqrt{2}A_n-1 \right) \left( A_n-\sqrt{2} \right)}{\sqrt{2}A_n}=\frac{\sqrt{2}A_{n+1}-1}{\sqrt{2}}$$
$$\frac{\left( \sqrt{2}A_n-1 \right) \left( A_n-\sqrt{2} \right)}{A_n}=\sqrt{2}A_{n+1}-1=\left( \sqrt{2}A_n-1 \right) \left( 1-\frac{\sqrt{2}}{A_n} \right)$$
All I have to do is prove that $\left( \frac{\sqrt{2}}{A_n} \right)$ has a bound greater than 0 and less than 1, but I didn't prove that bound.
| We recall that $\ A_n= \displaystyle \sum_{i=1}^n a_i$.
We have: $\ A_1 = 1 \ \text{ and } \ \forall n \in \mathbb N \ , \ A_{n+1}=A_n+\dfrac{1}{A_n}-\sqrt{2}$
Let $\ f(x)=x+\dfrac{1}{x}-\sqrt{2}$.
We have:
$f\circ f(x)-\dfrac{\sqrt{2}}{2} = \left(x-\dfrac{\sqrt{2}}{2}\right) \dfrac{ (\sqrt{2}-x)(1+\sqrt{2}-x)(x+1-\sqrt{2})}{x(x^2-\sqrt{2}x+1)}$
$f\circ f(x)-x = - \left( x-\dfrac{\sqrt{2}}{2}\right)^3 \dfrac{2\sqrt{2}}{x^2-\sqrt{2}x+1}$
So: $\ \forall x \in \left[ \dfrac{\sqrt{2}}{2},1\right] \ , \ \dfrac{\sqrt{2}}{2} \leqslant f(x)\leqslant x \leqslant 1$
Then, by induction: $\ \forall n \in \mathbb N \ , \ \dfrac{\sqrt{2}}{2} \leqslant A_{2n+1}\leqslant 1$
And: $\forall n \in \mathbb N \ , \ A_{2n+3} \leqslant A_{2n+1}$
$(A_{2n+1})_{n\in \mathbb N}$ is decreasing and bounded. It is convergent vers a real $\ell$ and $\ell = f\circ f)(\ell)$. So $\ \ell=\dfrac{\sqrt{2}}{2}$.
Now $(A_{2n+2})_{n\in\mathbb N} =(f(A_{2n+1})_{n\in \mathbb N})$ converges towards $f(\ell)=\ell$.
We conclude that $(A_n)_{n\in\mathbb N^*}$ converges towards $\ell=\dfrac{\sqrt{2}}{2}$
| {
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Sufficient to show the cases when $x = 0$, $y=0$, and $(x,y) \ne (0,0)$?
For a fixed $k \in \mathbb{N}$, define $f_k: \mathbb{R}^2 \rightarrow \mathbb{R}$ by:
$$
f_k(x,y)=
\begin{cases}
\dfrac{x^2(x+y^2)}{x^2+y^{2k}} &, (x,y)\neq (0,0)\\
0 &, (x,y)=(0,0)\\
\end{cases}
$$
Show that $f_1$ is not differentiable at $(0,0)$, but $f_k$ is differentiable at $(0,0)$ for each $k\geq 2$.
(Hint: At some point it may help to separately consider the cases $|x|\geq |y|^k$ and $|x|\leq |y|^k$.)
So I was able to determine that if $f_k$ is differentiable, the derivative is $(1,0)$. So after some algebra, I see that I have to show that
$\left|\dfrac{x^2y^{2} - x y^{2k}}{(x^2 + y^{2k})\sqrt{x^2+y^2}}\right| \to 0$ as $(x,y) \to (0,0)$.
When evaluating this limit, is it sufficient to show the cases when $x = 0$, $y=0$, and $(x,y) \ne (0,0)$? Why or why not?
| So far you have shown that
$$\lim\limits_{(x,y)\to(0,0)}|\frac{f_k(x,y)-f_k(0,0)-\frac{\partial f_k}{\partial x}(0,0)x-\frac{\partial f_k}{\partial y}(0,0)y}{\sqrt{x^2+y^2}}|=\lim\limits_{(x,y)\to(0,0)}|\frac{x^2y^2-xy^{2k}}{(x^2+y^{2k})\sqrt{x^2+y^2}}|$$
The function is differentiable at $(0,0)$ if and only if the above limit is zero.
Note that for the case $k=1$ the function $f_1$ indeed isn't differentiable since along the path $x=y$ we have
$$\lim\limits_{(y,y)\to(0,0)}|\frac{x^2y^2-xy^2}{(x^2+y^2)\sqrt{x^2+y^2}}|=\lim\limits_{(y,y)\to(0,0)}|\frac{y^4-y^3}{2\sqrt2y^3}|=\frac{1}{2\sqrt{2}}\neq0$$
Now suppose that $k\geq2$. Then, using the standard inequality $|\frac{x-y}{2}|\leq\sqrt{\frac{x^2+y^2}{2}}$, we have
$$|\frac{x^2y^2-xy^{2k}}{(x^2+y^{2k})\sqrt{x^2+y^2}}|\leq\sqrt{2}\cdot\sqrt{\frac{x^4y^4+x^2y^{4k}}{(x^2+y^{2k})^2(x^2+y^2)}}\leq\sqrt{2}\cdot\sqrt{\frac{x^2y^4(x^2+y^{2k})}{(x^2+y^{2k})^2(x^2+y^2)}}\leq\sqrt{2}|y|$$
and so
$$\lim\limits_{(x,y)\to(0,0)}|\frac{x^2y^2-xy^{2k}}{(x^2+y^{2k})\sqrt{x^2+y^2}}|\leq\lim\limits_{y\to0}\sqrt{2}|y|=0,$$
i.e. $f_k$ is differentiable at $(0,0)$ for $k\geq2$.
$$$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does $e^x\cdot e^X$ equal $e^{x+X}$? I know that they equal each other, but when I'm trying to prove it, something doesn't match. Please mind the difference between the two equations, one is a lowercase $x$ and the other is an uppercase $x.$ I know that the formula to get $e^x$ is $\frac{x^n}{n!}$. So I apply on $e^x$and it becomes $1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots$ and the same for $e^X$. This image shows what I'm thinking, and what happens when I multiply them on each other. But I want to go more deeper and show the full equation of the numbers shown, for example, showing $\frac{1}{6}x^3$ and others as well. If I did, there would be a total of 16 numbers shown that are arranged in order of degrees. So I try to prove it like this picture. But the problem shows when I try to prove it more. I tried proving it more, but $\frac{1}{6}(x+X)^3$ won't work. Than what should I do to make it work? Or what formula should I use?
| \begin{align}
& \left( \sum_{j=0}^\infty \frac{a^j}{j!} \right) \left( \sum_{k=0}^\infty \frac{b^k}{k!} \right) \\[10pt]
= & \Big(\bullet\Big) \left( \sum_{k=0}^\infty \frac{b^k}{k!} \right) = \sum_{k=0}^\infty \left( \bullet \cdot \frac{b^k}{k!} \right) = \sum_{k=0}^\infty \left(\left( \sum_{j=0}^\infty \frac{a^j}{j!} \right) \frac{b^k}{k!} \right) \\[8pt]
= {} & \sum_{k=0}^\infty \sum_{j=0}^\infty \left( \frac{a^j}{j!} \cdot \frac{b^k}{k!} \right).
\end{align}
This sum includes all the cases where $j+k=0,$ and all cases where $j+k=1$, and all cases where $j+k=2,$ and so on. What are the cases where $j+k=4,$ for example? They are these:
$$
\frac{a^0b^4}{0!4!} + \frac{a^1 b^3}{1!3!} + \frac{a^2b^2}{2!2!} + \frac{a^3 b^1}{3!1!} + \frac{a^4b^0}{4!0!}
$$
That is the same as
\begin{align}
& \frac 1 {4!} \left( \frac{4!}{0!4!} a^0b^4 + \frac{4!}{1!3!} a^1 b^3 + \frac{4!}{2!2!} a^2b^2 + \frac{4!}{3!1!} a^3 b^1 + \frac{4!}{4!0!} a^4b^0 \right) \\[8pt]
= {} & \frac 1 {4!} (a+b)^4.
\end{align}
And similarly for other numbers than $4.$ Thus the sum is
$$
\sum_{\ell=0}^\infty \frac 1 {\ell!} (a+b)^\ell = e^{a+b}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
How to evaluate $\int_0^{\pi/2} \frac{\sin x}{\sin^{2n+1}x +\cos^{2n+1}x} dx$? I have an exercise to evalute the following integral for all $n\geq 1 $
$$I(n)=\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin^{2n+1} x+\cos^{2n+1} x}dx$$
I attempted to find the closed form for the integral above in the following manner, where I used the integral identity $\int_a^bf(x)=\int_a^b f(a+b-x)dx$. $$I(\bar{n})=\int_0^{\frac{\pi}{2}}\frac{\cos x}{\cos^{2n-1} x+\sin ^{2n-1} x}dx$$ adding $I(n)$ and $I(\bar{n})$ its reduces to $$\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\cos x +\sin x}{\cos^{2n+1}x +\sin^{2n+1}x}dx$$ using the algebraic identity $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2+\cdots +b^{n-1})$ for odd integers $n$, I get $$\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos^{2n}x-\cos^{2n-1}\sin x+\cdots +\sin^{2n}x}dx $$
I'm now stuck here. How can I continue now?. Thanks in advance.
| Utilize the decomposition
$$\frac{\sin x +\cos x}{\sin^{2n+1}x +\cos^{2n+1}x}
=\frac{2^{n+1}}{2n+1}\sum_{k=1}^n\frac{(-1)^{k+1}\sin\frac{a_k}2 \cos^{n-1}a_k}{\csc a_k+\cot a_k \sin 2x}
$$
with $a_k=\frac{2\pi k}{2n+1}$ and integrate piecewise to arrive at
\begin{align}
\int_0^{\pi/2} \frac{\sin x}{\sin^{2n+1}x +\cos^{2n+1}x} dx
=\frac{2^{n+1}}{2n+1}\sum_{k=1}^n (-1)^{k+1}\ a_k \sin\frac{a_k}2 \cos^{n-1}a_k
\end{align}
The general result above produces
\begin{align}
\int_0^{\pi/2} \frac{\sin x}{\sin^{3}x +\cos^{3}x} dx
&= \frac{2\pi}{3\sqrt3}\\
\int_0^{\pi/2} \frac{\sin x}{\sin^{5}x +\cos^{5}x} dx
&= \frac{2\pi}{5} \sqrt{1+\frac{2}{\sqrt 5}}\\
\int_0^{\pi/2} \frac{\sin x}{\sin^{7}x +\cos^{7}x} dx
&= \frac{4\pi}7\left(\sin\frac\pi7+\sin\frac{3\pi}7\right)\\
\int_0^{\pi/2} \frac{\sin x}{\sin^{9}x +\cos^{9}x} dx
&= \frac{4\pi}{9\sqrt3}\left(1+4\cos\frac{\pi}9\right)\\
\int_0^{\pi/2} \frac{\sin x}{\sin^{11}x +\cos^{11}x} dx
&= \ \cdots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Other way to evaluate $\int \frac{1}{\cos 2x+3}\ dx$? I am evaluating
$$\int \frac{1}{\cos 2x+3} dx \quad (1)$$
Using Weierstrass substitution:
$$ (1)=\int \frac{1}{\frac{1-v^2}{1+v^2}+3}\cdot \frac{2}{1+v^2}dv =\int \frac{1}{v^2+2}dv \quad (2) $$
And then $\:v=\sqrt{2}w$
$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)^2+2}\sqrt{2} dw$$$$= \frac{1}{2} \int \frac{1}{\sqrt{2}\left(w^2+1\right)}dw$$$$ = \frac{1}{2\sqrt{2}}\arctan \left(w\right) + C$$$$= \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C$$
Therefore,
$$\int \frac{1}{\cos 2x+3} dx = \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C $$
That's a decent solution but I am wondering if there are any other simpler ways to solve this (besides Weierstass). Can you come up with one?
| $$\int \frac{1}{\cos2x+3}dx=\int \frac{1}{\frac{1-\tan^2x}{1+\tan^2x}+3}dx$$
$$=\int \frac{1+\tan^2x}{2\tan^2x+4}dx$$
$$=\frac12\int \frac{\sec^2x\ dx}{\tan^2x+2}$$
$$=\frac12\int \frac{d(\tan x)}{(\tan x)^2+(\sqrt2)^2}$$
$$=\frac12\frac{1}{\sqrt2}\tan^{-1}\left(\frac{\tan x}{\sqrt2}\right)+C$$
$$=\bbox[15px,#ffd,border:1px solid green]{\frac{1}{2\sqrt2}\tan^{-1}\left(\frac{\tan x}{\sqrt2}\right)+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Use linearisation of a certain function to approximate $\sqrt[3]{30}$ Background
Find the linearisation of the function
$$f(x)=\sqrt[3]{{{x^2}}}$$
at
$$a = 27.$$
Then, use the linearisation to find
$$\sqrt[3]{30}$$
My work so far
Applying the formula
$${f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) + f^\prime\left( a \right)\left( {x – a} \right),}$$
where
$${f\left( a \right) = f\left( {27} \right) }={ \sqrt[3]{{{{27}^2}}} }={ 9.}$$
Then, the derivative using the power rule:
$${f^\prime\left( x \right) = \left( {\sqrt[3]{{{x^2}}}} \right)^\prime }={ \left( {{x^{\frac{2}{3}}}} \right)^\prime }={ \frac{2}{3}{x^{\frac{2}{3} – 1}} }={ \frac{2}{3}{x^{ – \frac{1}{3}}} }={ \frac{2}{{3\sqrt[3]{x}}}.}$$
then
$${f^\prime\left( a \right) = f^\prime\left( {27} \right) }={ \frac{2}{{3\sqrt[3]{{27}}}} }={ \frac{2}{9}.}$$
Substitute this in the equation for $L(x)$:
$${L\left( x \right) = 9 + \frac{2}{9}\left( {x – 27} \right) }={ 9 + \frac{2}{9}x – 6 }={ \frac{2}{9}x + 3.}$$
Then, to use this linearisation to find
$$\sqrt[3]{30}$$
I perform the following $\Delta x = x – a = 30 – 27 = 3$ as the condition is $x =30$ and the staring point is $a=27$
As the the derivative of this particular function is given by $f\left( x \right) = \sqrt[\large 3\normalsize]{x}$
$${f’\left( x \right) = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } } = {{\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } } = {\frac{1}{3}{x^{ – \large\frac{2}{3}\normalsize}} } = {\frac{1}{{3\sqrt[\large 3\normalsize]{{{x^2}}}}},}$$
and its value at point $a$ is equal to
$${f’\left( {a} \right) = \frac{1}{{3\sqrt[\large 3\normalsize]{{{{27}^2}}}}} } = {\frac{1}{{3 \cdot {3^2}}} = \frac{1}{{27}}.}$$
Thus, getting the solution
$${f\left( x \right) \approx f\left( {a} \right) + f’\left( {a} \right)\Delta x,\;\;}\Rightarrow {\sqrt[\large 3\normalsize]{{30}} \approx \sqrt[\large 3\normalsize]{{27}} + \frac{1}{{27}} \cdot 3 } = {3 + \frac{1}{9} } = {\frac{{28}}{9} \approx 3,111.}$$
Is my process correct so far? Or, did I go wrong in the second part? Also, as $a=27$ is from the original linearisation, this would be brought into the linearisation approximation for $\sqrt[3]{30}$?
| Just to show you something a little bit different, we can do this with the binomial theorem.
$(a+b)^k = a^k + k a^{k-1}b + \frac {k(k-1)}{2} a^{k-2}b^2 + \cdots$
You learned this in algebra / pre-calculus with integers. It actually works for all real numbers.
$(27 + 3)^\frac 13 = 27^\frac 13 + \frac 13 (27^{-\frac 23})(3) - \frac 19 (27^{-\frac 53})(3^2)+\cdots$
$3 + \frac 19 - \frac 1{3^5}+ \cdots$
The first 2 terms would be all that you would use for a linear approximation, but for additional precision you can extend.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to solve this ODE: $x^3dx+(y+2)^2dy=0$? I am trying to solve $$ x^3dx+(y+2)^2dy=0 \quad( 1)$$
Dividing by $dx$, we can reduce the ODE to seperate variable form, i.e
$$ (1) \to (y+2)^2y'=-x^3 $$
Hence,
$$ \int (y(x)+2)^2y'(x) dy = \int -x^3dx = - \frac{x^4}{4} + c_1$$
This LHE seems to be easy to solve using integration by parts:
$$ \int (y(x)+2)^2y'(x) dy = y(x)(y(x)+2)^2 - \int y^3dy- \int 4y^2dy + \int4ydy \iff$$
$$ \iff - \frac{y^4}{4} + -\frac13 y^3 + 4y^2 + 8y + c_2$$
But then solving the ODE for $y(x)$ is a struggle:
$$ \iff - \frac{y^4}{4} + -\frac13 y^3 + 4y^2 + 8y = - \frac{3}{4}x + C$$
Any ideas on how I can solve this?
| $$x^3dx+(y+2)^2dy=0$$
$$x^3dx=-(y+2)^2dy$$
$$x^3=-(y+2)^2y'$$
Integratation gives:
$$\int x^3dx=-\int (y+2)^2 y'dx$$
$$\int x^3dx=-\int (y+2)^2 dy$$
Substitute $u=y+2$
$$\int x^3dx=-\int u^2du$$
$$\dfrac {x^4}4+\dfrac {u^3}{3}=C$$
$$\dfrac {x^4}4+\dfrac {(y+2)^3}{3}=C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove that $(a b+b c+c a-1)^{2} \leq\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)$.
Let $a$, $b$, and $c$ be real numbers. Prove that $$(a b+b c+c a-1)^{2} \leq\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)\,.$$
In solution of this author take Let $a=\tan x, b=\tan y, c=\tan z$ with $-\frac{\pi}{2}<x, y, z<\frac{\pi}{2}$ but i did not understand the reason behind letting that $-\frac{\pi}{2}<x, y, z<\frac{\pi}{2}$ ,
i mean if we just take $a=\tan x, b=\tan y, c=\tan z$ then is something wrong,
I do not want solution,just want to clear this step
| Since $$(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2,$$ we obtain:
$$(a^2+1)(b^2+1)(c^2+1)=((a+b)^2+(ab-1)^2)(c^2+1)=$$
$$=((a+b)c+ab-1)^2+(a+b-(ab-1)c)^2\geq(ab+ac+bc-1)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
Initially, this seemed like one could work it out with $AM-GM$, but it doesn't seem so.
From $AM-GM$ one gets that $$\frac{1^2+2^2+ \dots+n^2}{n} \geqslant \sqrt[\leftroot{-1}\uproot{2}n]{1^2\cdot2^2\dots\cdot n^2}
$$
is this of any help here?
Remark. Thanks to Favst, the source of the problem is Problem 1 of 1994 British Mathematical Olympiad Round 2
| This question has already been posted on this website. See my solution via Pell's equation here, where I wrote that the answer is $337.$ The problem appeared on the 1994 British Mathematical Olympiad Round 2
Edit: as suggested by Batominovski, I am copying my old solution here:
A while ago, I found this problem as the 1994 British Mathematical Olympiad - Round 2, Problem 1, but the solution is mine. Here it is.
The equation is $$m^2=\frac{1}{n}\sum_{k=1}^{n}{k^2}=\frac{1}{n}\cdot \frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{6}.$$
With some manipulation, this is equivalent to $$(4n+3)^2-48m^2=1,$$ which can be solved by Pell's equation. The fundamental solution for $D=48$ in Pell's equation $x^2-Dy^2=1$ is $(x,y)=(7,1),$ so all solutions are parameterized by $$x_t + y_t \sqrt{48}=(7+\sqrt{48})^t.$$ We want to find the first solution $t>1$ for which $x_t\equiv 3\pmod{4}.$ While $t=2$ does not work, $t=3$ yields $$1351+195\sqrt{48}.$$ Since $1351=337\cdot 4+3,$ the answer is $337.$
We can check that $$\frac{(337+1)(2\cdot 337+1)}{6}=3^2\cdot 5^2\cdot 13^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Fundamental group of Klein Bottle It is well know that the fundamental group of the Klein Bottle $G$ is defined by
$$G=BS(1,-1)=\langle a,b: bab^{-1}=a^{-1}\rangle.$$
I know, for example that $BS(1,2)$ can be defined as the group
$$BS(1,2)=\langle A,B\rangle $$
where
$$A=\left(
\begin{array}{cc}
1 & 1 \\
0 & 1 \\
\end{array}
\right), B=\left(
\begin{array}{cc}
2 & 0 \\
0 & 1 \\
\end{array}
\right).$$
These matrices satisfy the equation $BAB^{-1}=A^{2}$ and are free : there is not an integer $k$ such that $A^{k}=I$ or $B^{k}=I$. This implies that we obtain an "explicit description" of $BS(1,2)$ as the group generated by $A$ and $B$.
I know that the matrices
$$a=\left(
\begin{array}{cc}
1 & 1 \\
0 & 1 \\
\end{array}
\right), b=\left(
\begin{array}{cc}
-1 & 0 \\
0 & 1 \\
\end{array}
\right)$$
satisfy the relation $bab^{-1}=a^{-1}$ but $b^{2}=I$. This implies that $BS(1,-1)$ is not generated by $a$ and $b$.
My question is: is there an "explicit description" for $G=BS(1,-1)$ with matrices or maybe another couple of objects?
| Finally , $G$ can be described as the group of $2\times 2$ matrices generated by two matrix $A,B$ such that $BAB^{-1}=A^{-1}$ and $A^{k}\neq I$, $B^{k}\neq I$ for all $k$. Let $A=\left(
\begin{array}{cc}
a & b \\
c & d\\
\end{array}
\right)$ and $B=\left(
\begin{array}{cc}
\lambda & 0 \\
0 & \mu \\
\end{array}
\right)$ (to simplify computations). Working whith the equation $BAB^{-1}=A^{-1}$ and assuming that $ad-bc=1$ and $b,c\neq 0$ we have that
$A=\left(
\begin{array}{cc}
a & b \\
c & a\\
\end{array}
\right)$ and $B=\left(
\begin{array}{cc}
\lambda & 0 \\
0 & -\lambda \\
\end{array}
\right)$
This family of matrices satisfy the relation $BAB^{-1}=A^{-1}$.
If $A=\left(
\begin{array}{cc}
2 & 3 \\
1 & 2\\
\end{array}
\right)$ and $B=\left(
\begin{array}{cc}
2 & 0 \\
0 & -2 \\
\end{array}
\right)$, then $BAB^{-1}=A^{-1}$ and $A^{k}\neq I$, $B^{k}\neq I$, for all $k$
Hence $G\equiv \langle A,B\rangle$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
What is the pdf of $\frac{|x-y|}{(x+y)(2-x-y)}$ when $x,y$ are i.i.d uniform on $[0,1]$? If $x,y$ are i.i.d uniform random variables on $[0,1]$.
I know that the PDF of $|x-y|$ is:
$$f(z) = \begin{cases}
2(1-z) & \text{for $0 < z < 1$} \\
0 & \text{otherwise.}
\end{cases}$$
I know that the PDF of $x+y$ is
$$f(t)=f(z) = \begin{cases}
z & \text{for $0 < z < 1$} \\
2-z & \text{for $1 \le z < 2$} \\
0 & \text{otherwise.}
\end{cases}$$
I am trying to calculate the PDF of $\frac{|x-y|}{(x+y)(2-x-y)}$
What I'm trying to do is to calculate first the PDF of $(x+y)(2-x-y)$ by doing the following: since $(x+y)$ and $(2-x-y)$ have the same PDFs, the PDF of the product should be:
$$f_{(x+y)^2}(z)=\int_{0}^1 f_{x+y}(t) f_{x+y}(z/t) \frac{1}{|t|}\,dt=\int_{0}^1z/t \,dt=z-z\,\log z$$
if $0<z<1$ and
$$f_{(x+y)^2}(z)=\int_{1}^2 f_{x+y}(t) f_{x+y}(z/t) \frac{1}{|t|}\,dt=$$
$$=\int_{1}^2 \frac{(2-t)(2-z/t)}{t} \,dt=-2 - z + z \log 2 + \log 16$$
if $1<z<2$ and $0$ elsewhere.
However, this is not a density function because its area is not 1. I cannot find the mistake.
| Let $$Z = \frac{|X-Y|}{(X+Y)(2-X-Y)}.$$ We first determine the support of $Z$ for $X, Y \sim \operatorname{Uniform}(0,1)$. A quick glance shows that the minimum value is $0$; the maximum is attained at $(X,Y) \in \{ (1,0), (0,1) \}$; therefore, $Z \in [0,1]$. Then we want $$\Pr[Z \le z] = \int_{x=0}^1 \int_{y=0}^1 \mathbb 1\left(\frac{|x-y|}{(x+y)(2-x-y)} \le z \right) \, dy \, dx.$$ Since the the integrand is symmetric in $x$ and $y$, as well as symmetric about the line $x + y = 1$ (since replacing $(x,y)$ with $(1-y, 1-x)$ gives the same function), we can restrict our attention to the triangular region with vertices $(0, 0), (1/2, 1/2), (1, 0)$. The transformation $$u = x-y, \quad v = x+y$$ turns this region into $(0,0), (0,1), (1,1)$ in $(u,v)$-space, and the integral becomes $$\Pr[Z \le z] = 4 \int_{v=0}^1 \int_{u=0}^v \mathbb 1 \left( \frac{u}{v(2-v)} \le z \right) \cdot \frac{1}{2} \, du \, dv, \tag{1}$$ where the $4$ comes from the symmetry argument above, and the $1/2$ is the Jacobian of the transformation from $(x,y)$ to $(u,v)$. We consider the locus of points satisfying the equation $$u = z v(2-v)$$ as comprising part of the boundary of the region of integration, the other parts being the aforementioned triangle. As this is clearly a parabola with vertex at $(u,v) = (z,1)$, axis of symmetry $u = 1$, and passing through $(0,0)$, we note that when $1/2 < z \le 1$, there is a nontrivial point of intersection with the line $u = v$ at $$u = v = 2 - \frac{1}{z}.$$ Therefore, we must consider these cases separately. When $0 \le z \le 1/2$, the integral $(1)$ is simply the area of the parabolic region given by $$2 \int_{v=0}^1 \int_{u=0}^{zv(2-v)} \, du \, dv = 2 \int_{v=0}^1 zv(2-v) \, dv = \frac{4z}{3}.$$ When $1/2 < z \le 1$, we instead evaluate $$\begin{align}
\Pr[Z \le z]
&= 2 \left( \int_{v=0}^{2 - 1/z} \int_{u=0}^v \, du \, dv + \int_{v= 2-1/z}^1 \int_{u=0}^{zv(2-v)} \, du \, dv \right) \\
&= (2-1/z)^2 + 2 \int_{v=2 - 1/z}^1 zv(2-v) \, dv \\
&= (2-1/z)^2 + 2\frac{-1 + 3z + 2z^3}{3z^2} \\
&= \frac{1-6z+12z^2-4z^3}{3z^2}.
\end{align}$$
Therefore, the full CDF is given by $$F_Z(z) = \begin{cases} 0, & z < 0 \\
\frac{4z}{3}, & 0 \le z \le 1/2 \\
\frac{1-6z+12z^2-4z^3}{3z^2}, & 1/2 < z \le 1 \\
1, & 1 < z. \end{cases}$$
Consequently, the density is
$$f_Z(z) = \begin{cases} \frac{4}{3}, & 0 < z < 1/2 \\ \frac{2(-1 + 3z - 2z^3)}{3z^3}, & 1/2 < z < 1. \end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \frac{A}{\pi} +B $ $$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \frac{A}{\pi} +B $$ if $\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ find $\frac{A^2}{B^2}$
My Attempt
$$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \sum_{n=1}^\infty csch^2(i\omega\pi n)= 4\sum_{n=1}^\infty \big(e^{\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big) }-e^{-\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big)}\big) ^{-2} $$ $$\sum_{n=1}^\infty \big(e^{\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big) }-e^{-\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big)}\big) ^{-2}= \big(ie^{\pi\frac{\sqrt{3}}{2}}+ie^{-\pi\frac{\sqrt{3}}{2}}\big)^{-2}+ \big(-e^{\pi\sqrt{3}}+e^{-\pi\sqrt{3}}\big)^{-2} +\big(-ie^{3\pi\frac{\sqrt{3}}{2}}-ie^{-3\pi\frac{\sqrt{3}}{2}}\big)^{-2}+ \big(e^{2\pi\sqrt{3}}-e^{-2\pi\sqrt{3}}\big)^{-2} +...$$$$= \sum_{n=0}^\infty \big(ie^{(4n+1)\pi\frac{\sqrt{3}}{2}}+ie^{-(4n+1)\pi\frac{\sqrt{3}}{2}}\big)^{-2} +\sum_{n=0}^\infty \big(-e^{(2n+1)π√3}+e^{-(2n+1)π√3}\big)^{-2} +\sum_{n=0}^\infty \big(-ie^{(3+4n)\pi\frac{\sqrt{3}}{2}}+-ie^{-(4n+3)\pi\frac{\sqrt{3}}{2}}\big)^{-2}+
\sum_{n=0}^\infty \big(e^{(2n)π√3}-e^{-(2n)π√3}\big)^{-2} $$ $$\sum_{n=0}^\infty \big(-e^{(4n+1)\pi\sqrt{3}}-2-e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=0}^\infty \big(e^{2(2n+1)\pi\sqrt{3}}-2+e^{-2(2n+1)\pi\sqrt{3}}\big)^{-1} + \sum_{n=0}^\infty \big(e^{(3+4n)\pi\sqrt{3}}-2+e^{-(3+4n)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=1}^\infty \big(e^{4n\pi\sqrt{3}}-2+e^{-4n\pi\sqrt{3}}\big)^{-1} $$ I have found the sums numerically and $\sum_{n=0}^\infty \big(-e^{(4n+1)\pi\sqrt{3}}-2-e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=0}^\infty \big(e^{2(2n+1)\pi\sqrt{3}}-2+e^{-2(2n+1)\pi\sqrt{3}}\big)^{-1} + \sum_{n=0}^\infty \big(e^{(3+4n)\pi\sqrt{3}}-2+e^{-(3+4n)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=1}^\infty \big(e^{4n\pi\sqrt{3}}-2+e^{-4n\pi\sqrt{3}}\big)^{-1} \approx -0.00429$
How can I evaluate this analytically?
| Here is an approach using the residue theorem.
The residues of $\newcommand{\res}{\operatorname*{Res}}f(z)=\pi\cot\pi z\csc^2\omega\pi z$ at its poles are:
$$\res_{z=0}f(z)=\frac{1-\omega}{3};\quad\res_{z=n}f(z)=\csc^2\omega\pi n;\quad\res_{z=n/\omega}f(z)=-\omega\csc^2\omega\pi n\quad(n\in\mathbb{Z}_{\neq 0})$$ (easy to obtain using power series; we keep in mind $\omega^3=1\implies 1/\omega=-1-\omega$).
Now consider $\int_{\Gamma}f(z)\,dz$ where $\Gamma$ is the boundary of the parallelogram
$\hspace{3cm}$
$$\big\{z : \max\{|\Re z|,|\Re(\omega z)|\}\leqslant N+1/2\big\}$$
and take $N\to\infty$. On $AB$ or $CD$ we have $z=\pm\frac1\omega\left(N+\frac12+it\right)$, with "$+$" on $CD$ and "$-$" on $AB$, and $t\in\mathbb{R}$; using $\lim\limits_{y\to-\infty}\cot(x+iy)=i$ (uniformly w.r.t. $x\in\mathbb{R}$), we see that both $\int_{AB}$ and $\int_{CD}$ tend to $$-\frac\pi\omega\int_{-\infty}^\infty\frac{dt}{\cosh^2\pi t}=-\frac2\omega$$ as $N\to\infty$. The integrals along $BC$ and $DA$ both tend to $0$, hence the residue theorem gives $$\frac{1-\omega}{3}+2(1-\omega)\sum_{n=1}^\infty\csc^2\omega\pi n=-\frac{2}{\omega\pi i}\implies\color{blue}{\sum_{n=1}^\infty\csc^2\omega\pi n=\frac{1}{\pi\sqrt3}-\frac16}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Show that the elements of the sequence are divisible by $2^n$ I am trying to prove the following:
Consider the sequence defined by $A_{n+2}=6A_{n+1}+2A_n, A_0=2, A_1=6$. Show that $2^n|A_{2n-1}$ but $2^{n+1}\nmid A_{2n-1}$.
The first terms of this sequence are 2, 6, 40, 252, 1592, 10056, 63520.
In fact, the maximal exponent of $2$ that divides $A_n$ seems to follow a pattern with a period of 4:
1, 1, 3, 2, 3, 3, 5, 4, 5, 5, 7, 6...
But I haven't been able to prove this.
| Using induction, it's easy to prove that $A_n=(3+\sqrt{11})^n+(3-\sqrt{11})^n$ (roots of the characteristic polynomial of the corresponding recurrence $t^2-6t-2$ are $3\pm\sqrt{11}$).
Define
$$
B_n:=A_{2n+1}~\text{for}~n\geq 0.
$$
Then,
$$
B_n=(3+\sqrt{11})^{2n+1}+(3-\sqrt{11})^{2n+1}.
$$
It's easy to check that $B_0=6$, $B_1=252$ and
$$
B_{n+2}=40B_{n+1}-4B_n
$$
(because $(3\pm\sqrt{11})^2=20\pm 6\sqrt{11}$ are roots of $t^2-40t+4$).
Now, define $C_n:=\frac{B_n}{2^{n+1}}$. Then, we can rewrite previous equality as
$$
2^{n+3}C_{n+2}=40\cdot 2^{n+2}C_{n+1}-4\cdot 2^{n+1} C_n.
$$
or
$$
C_{n+2}=20C_{n+1}-C_n.
$$
Also, $C_0=3$ and $C_1=63$.
It's clear from the recurrence relation for $\{C_n\}$ that all $C_n$ are integers. Moreover, each $C_n$ is odd (because $C_0,C_1$ are odd and from the recurrence relation: $C_{n+2}\equiv C_n\pmod 2$).
Therefore, $A_{2n+1}=2^{n+1}\cdot C_n$ for some odd $C_n$. Thus, $2^{n+1\mid A_{2n+1}}$ and $2^{n+2}\not\mid A_{2n+1}$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove an inequality for positive real numbers Prove that :-
$$\frac{x^2}{y}+ \frac{y^2}{z}+\frac{z^2}{x} \geq x+y+z$$
Where $x,y,z$ are positive real numbers
My attempt :-
L.H.S = $\frac{x^3 z +x y^3 + y z^3 }{xyz} $
We need to show that
$x^3 z + x y^3 + y z^3 \geq xyz (x+y+z) $
I tried the AM-GM
$3(x^3 z + x y^3 + y z^3) \geq (xyz)^{\frac{4}{3}}$
But i could not go on any more !
| We have by Cauchy-Schwarz-Bunyakovsky inequality:
$$ (a^2+b^2+c^2)(x^2+y^2+z^2)\geq (ax+by+cz)^2$$
or writen like this:
$$(a+b+c)(x+y+z)\geq (\sqrt{ax}+\sqrt{by}+\sqrt{cz})^2$$
we have a folloving:
$$(y+z+x)(\frac{x^2}{y}+ \frac{y^2}{z}+\frac{z^2}{x}) \geq (x+y+z)^2$$
and thus a conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $k$ in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$ Find the value of 'k' in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$
My approach is as follow
$\sin {1^\circ} = \sin {179^\circ}$
$T = {\sin ^2}{1^\circ}{\sin ^2}{3^\circ}..{\sin ^2}{89^\circ}$
$\left( {\frac{{1 - \cos {2^\circ}}}{2}} \right) = {\sin ^2}{1^\circ}$
$T = \left( {\frac{{1 - \cos {2^\circ}}}{2}} \right)\left( {\frac{{1 - \cos {6^\circ}}}{2}} \right)\left( {\frac{{1 - \cos {{10}^\circ}}}{2}} \right)....\left( {\frac{{1 - \cos {{178^\circ}}}}{2}} \right)$
$2 + \left( {n - 1} \right)4 = 178 \Rightarrow n = 45$
$T = \frac{1}{{{2^{45}}}}\left( {1 - \cos {2^\circ}} \right)\left( {1 - \cos {6^\circ}} \right)\left( {1 - \cos {{10}^\circ}} \right)....\left( {1 - \cos {{178^\circ}}} \right)$
$\left( {1 - \cos {{178}^\circ}} \right) = \left( {1 + \cos {2^\circ}} \right)$
$T = \frac{1}{{{2^{45}}}}\left( {1 - {{\cos }^2}{2^\circ}} \right)\left( {1 - {{\cos }^2}{6^\circ}} \right)\left( {1 - {{\cos }^2}{10^\circ}} \right)....\left( {1 - {{\cos }^2}{{86}^\circ}} \right)\left( {1 - \cos {{90}^\circ}} \right)$
$T = \frac{{{{\sin }^2}{2^\circ}.{{\sin }^2}{6^\circ}.{{\sin }^2}{{10}^\circ}......{{\sin }^2}{{86}^\circ}}}{{{2^{45}}}}$
Not able to proceed from here
| In your approach, there's a shorter way to arrive at T, but anyway taking off from there :
$$ T = \frac{{{{\sin }^2}{2^\circ}.{{\sin }^2}{6^\circ}.{{\sin }^2}{{10}^\circ}......{{\sin }^2}{{86}^\circ}}}{{{2^{45}}}} \\
= \frac{({\sin }{2^\circ}.{\sin }{6^\circ}.{\sin }{{10}^\circ}......{\sin }{{86}^\circ})^2}{{{2^{45}}}} \\
= \frac{({\cos }{4^\circ}.{\cos }{8^\circ}....{\cos }{{84}^\circ}.{\cos }{{88}^\circ})^2}{{{2^{45}}}} $$
Now let $p= {\cos }{4^\circ}.{\cos }{8^\circ}....{\cos }{{84}^\circ}.{\cos }{{88}^\circ} \\ $
And let $q = {\sin }{4^\circ}.{\sin }{8^\circ}....{\sin }{{84}^\circ}.{\sin }{{88}^\circ}$
So $pq = {(1/2^{22})}.{{\sin }{8^\circ}.{\sin }{16^\circ}...{\sin }{{176}^\circ}}\\
= {(1/2^{22})}.{\sin }{4^\circ}.{\sin }{8^\circ}....{\sin }{{84}^\circ}.{\sin }{{88}^\circ} {(\because \sin(180-x)=\sin x)}\\
= {(1/2^{22})}q $
So $p={(1/2^{22})}$
Therefore $T={(1/2^{89})}$
So $k=89$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Probability of exactly $2$ sixes in $3$ dice rolls where $2$ dice have $6$ on $2$ faces? Three dice are rolled. One is fair and the other two have 6 on two faces.
Find the probability of rolling exactly 2 sixes.
My textbook gives an answer of $\frac{20}{147}$ but I get an answer of: $$\frac{1}{6}\frac{2}{6}\frac{4}{6}+\frac{1}{6}\frac{4}{6}\frac{2}{6}+\frac{5}{6}\frac{2}{6}\frac{2}{6}=\frac{8}{216}+\frac{8}{216}+\frac{20}{216}=\frac{36}{216}=\frac{1}{6}$$
I just want to know where I am going wrong or could the textbook be mistaken ?
| Note that the official answer is correct if you make the (unnatural) assumption that the two non-standard dice have seven sides (two of which show $6$).
In that case the answer is $$\frac 27\times \frac 27\times \frac 56+2\times \frac 27\times \frac 57\times \frac 16=\frac {20}{147}$$
To be sure, this was arrived at by reverse engineering, not by any sensible reading of the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Definite integration $\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx$ How do I integrate $$\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx\quad ?$$
The actual integral that I encountered is:
$$\int_{-\infty}^\infty dx \left(\frac{N}{\cosh(\frac{\pi }{c}(x-1))}+\frac{1}{\cosh(\frac{\pi}{c}x)} \right) 2 \tan^{-1}\left(\frac{2x-2}{c} \right)$$ where c is a constant with $$\Re c>0$$ Not sure if these two terms makes it easier.
I was trying to solve just the last term, but I couldn't make any progress. Numerical integration gives $\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx= -1.01334 $. Any hint on how to do it analytically?
| Assume $a>0$ and $b \in \mathbb{R}$.
Let's first make the substitution $u = ax+b$ to get $$\int_{-\infty}^{\infty} \frac{\arctan (ax+b)}{\cosh(\pi x)} \, \mathrm dx = \int_{-\infty}^{\infty} \frac{\arctan u}{a\cosh \left(\pi \left(\frac{u-b}{a} \right) \right)} \, \mathrm du.$$
Following the general approach that Iaroslav V. Blagouchine uses in the paper Rediscovery of Malmsten’s integrals, their evaluation by contour integration methods and some related results, let's integrate the function $$\frac{\log \Gamma \left(\frac{z}{2ia}+\frac{1}{2a} \right)}{a\cosh\left(\pi \left(\frac{z-b}{a} \right) \right)}, $$ where $\log \Gamma (z)$ is the log-gamma function, around an infinitely wide rectangular contour in the upper half of the complex plane of height $2ia $ (which is the period of the denominator).
(The branch cut for the log-gamma function in the numerator runs down the imaginary axis from $z=-i$, and the denominator grows much faster than the numerator as $\Re(z) \to \pm \infty$.)
Integrating around the contour, and using the property $\log(x) + \log \Gamma(x) = \log \Gamma(x+1)$, we get $$\begin{align} &\int_{-\infty}^{\infty} \frac{\log \Gamma \left(\frac{x}{2ia }+\frac{1}{2a} \right)}{a\cosh\left(\pi \left(\frac{x-b}{a} \right) \right)} \, \mathrm dx -\int_{-\infty}^{\infty} \frac{\log \Gamma \left(\left(\frac{x}{2ia }+\frac{1}{2a} \right)+1 \right)}{a\cosh\left(\pi \left(\frac{x-b}{a} \right) \right)} \, \mathrm dx \\ &= -\int_{-\infty}^{\infty} \frac{\log \left(\frac{x}{2ia}+\frac{1}{2a} \right)}{a\cosh\left(\pi \left(\frac{x-b}{a} \right) \right)} \, \mathrm dx \\ &= 2 \pi i \left(\operatorname{Res} \left[f(z), b+ \frac{ia}{2} \right] + \operatorname{Res} \left[f(z), b+ \frac{3ia}{2}\right] \right) \\ &= 2 \pi i \left(\frac{1}{\pi i} \, \log \Gamma\left(\frac{1}{4}+ \frac{1}{2a} -\frac{ib}{2a} \right) - \frac{1}{\pi i} \, \log \Gamma \left(\frac{3}{4}+ \frac{1}{2a} - \frac{ib}{2a} \right)\right) \\&= 2 \left( \log \Gamma\left(\frac{1}{4}+ \frac{1}{2a} - \frac{ib}{2a} \right) - \log \Gamma \left(\frac{3}{4}+ \frac{1}{2a}- \frac{ib}{2a} \right) \right). \end{align}$$
Then equating the imaginary parts on both sides of the equation, we get $$\int_{-\infty}^{\infty} \frac{\arctan x}{a\cosh\left(\pi \left(\frac{x-b}{a} \right) \right)} \, \mathrm dx = 2 \Im \left( \log \Gamma\left(\frac{1}{4}+ \frac{1}{2a} - \frac{ib}{2a} \right) - \log \Gamma \left(\frac{3}{4}+ \frac{1}{2a} - \frac{ib}{2a} \right)\right). $$
By the Schwarz reflection principle, the result can also be expressed as $$2 \Im \left( \log \Gamma\left(\frac{3}{4}+ \frac{1}{2a} + \frac{ib}{2a} \right) - \log \Gamma \left(\frac{1}{4}+ \frac{1}{2a} + \frac{ib}{2a} \right)\right),$$ which agrees with pisco's answer.
| {
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"url": "https://math.stackexchange.com/questions/3774068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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The values of the parameters for which $\frac{ab}{c}+\frac{a(a+1)b(b+1)}{2!c(c+1)}+\frac{a(a+1)(a+2)b(b+1)(b+2)}{3!c(c+1)(c+2)}+\dots$ converges. Determine the values of the parameters for which $\frac{ab}{c}+\frac{a(a+1)b(b+1)}{2!c(c+1)}+\frac{a(a+1)(a+2)b(b+1)(b+2)}{3!c(c+1)(c+2)}+\dots$ converges.
Then $\frac{a_{n+1}}{a_n}=\frac{(a+n)(b+n)}{(n+1)(c+n)}$, I guess then I should use Raabe's test, but I don't know how to write $\frac{(a+n)(b+n)}{(n+1)(c+n)}$ in form of $1-\frac{p}{n}$. Any help? Thanks.
| $$
\frac{(a+n)(b+n)}{(c+n)(1+n)} = 1 - \frac{1+c-a-b}{n} + O(1/n^2)
$$
as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of the Inscribed Angle Theorem using vectors I am trying to prove the inscribed angle theorem using vectors. I fixed the dots $A=(\cos\theta,\sin\theta)$, and $B=(\cos\varphi,\sin\varphi)$, and I took another point $C=(\cos\psi,\sin\psi)$ in the biggest arc $AB$.
My idea was to calculate $\dfrac{\langle A-C,B-C\rangle}{\lvert{A-B}\rvert\lvert{B-C\rvert}}$, what according to my calculations is $$\dfrac{1+\cos(\theta-\varphi)-\cos(\psi-\theta)-\cos(\varphi-\psi)}{2\sqrt{1+\cos(\theta-\psi)\cos(\varphi-\psi)-\cos(\theta-\psi)-\cos(\varphi-\psi)}}.$$
My main difficulty here is the square root, which I can't get rid of. Does someone know how to proceed from here?
Or maybe to solve the problem with vectors with a different approach?
| You expanded your terms all the way. Take a few steps back:
\begin{align}
\langle A-C, B-C\rangle &= \langle (\cos\theta-\cos\psi,\sin\theta-\sin\psi),(\cos\phi-\cos\psi,\sin\phi-\sin\psi)\rangle\\
&= (\cos\theta-\cos\psi)(\cos\phi-\cos\psi)+(\sin\theta-\sin\psi)(\sin\phi-\sin\psi)\\
&= -2\sin\frac{\theta-\psi}2\sin\frac{\theta+\psi}2 \cdot(-2)\sin\frac{\phi-\psi}2\sin\frac{\phi+\psi}2 \\&\quad+ 2\cos\frac{\theta+\psi}2\sin\frac{\theta-\psi}2 \cdot 2\cos\frac{\phi+\psi}2\sin\frac{\phi-\psi}2\\
&= 4\sin\frac{\theta-\psi}2\sin\frac{\phi-\psi}2\left(\sin\frac{\theta+\psi}2\sin\frac{\phi+\psi}2 + \cos\frac{\theta+\psi}2\cos\frac{\phi+\psi}2\right)\\
&= 4\sin\frac{\theta-\psi}2\sin\frac{\phi-\psi}2\cos\left(\frac{\theta+\psi}2-\frac{\phi+\psi}2\right)\\
&= 4\sin\frac{\theta-\psi}2\sin\frac{\phi-\psi}2\cos\frac{\theta-\phi}2
\end{align}
\begin{align}
|A-C|^2|B-C|^2 &= ((\cos\theta-\cos\psi)^2+(\sin\theta-\sin\psi)^2)((\cos\phi-\cos\psi)^2+(\sin\phi-\sin\psi)^2)\\
&= (2-2\cos\theta\cos\psi - 2\sin\theta\sin\psi)(2-2\cos\phi\cos\psi - 2\sin\phi\sin\psi)\\
&= 4(1-\cos(\theta-\psi))(1-\cos(\phi-\psi))\\
&= 4\cdot 2\sin^2\frac{\theta-\psi}2\cdot 2\sin^2\frac{\phi-\psi}2\\
&= 16 \sin^2\frac{\theta-\psi}2\sin^2\frac{\phi-\psi}2
\end{align}
and hence
$$\cos\measuredangle ACB = \frac{\langle A-C, B-C\rangle}{|A-C||B-C|} = \cos\frac{\theta-\phi}2$$
which means that the angle $\measuredangle ACB$ does not depend on $C$ and is equal to half the central angle: $$\measuredangle ACB = \frac12(\theta-\phi) = \frac12 \measuredangle AOB.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + 2$ = $(x - 1)(x - 2)$
and I can see $(x - 1)^2 \equiv 1$ $($mod $x - 2)$ . We also have :-
$$\frac{(x - 1)^{100}}{(x - 1)(x - 2)} = \frac{(x - 1)^{99}}{(x - 2)}.$$
We have :- $(x - 1)^{98} \equiv 1$ $($mod $x - 2).$ $\rightarrow (x - 1)^{99} \equiv (x - 1)$ $($mod $x - 2)$. Now for the case of $(x - 2)^{200}$ we have :-
$$\frac{(x - 2)^{200}}{(x - 1)(x - 2)} = \frac{(x - 2)^{199}}{(x - 1)}.$$
We have :- $(x - 2) \equiv (-1)$ $($mod $x - 1)$ $\rightarrow (x - 2)^{199} \equiv (-1)$ $($mod $x - 1)$.
Adding all these up we have :- $(x - 1)^{100} + (x - 2)^{200} \equiv (x - 2)$ $($mod $x² - 3x + 2)$ .
On checking my answer with wolfram alpha , I found the remainder to be $1$, so I messed up in some step .
Can anyone help me?
| Since $(x - 1)^{99} \equiv (x - 1)\;\mod (x - 2)$, we get that
$(x - 1)^{100}\equiv (x-1)^2\;\mod (x-2)(x-1). \quad(*)$
Since $(x-2)^{199}\equiv -1\;\mod (x-1)$, we get that
$(x-2)^{200}\equiv -(x-2)\;\mod (x-1)(x-2). \quad(**)$
So, by adding $(*)$ and $(**)$, it follows that
$(x - 1)^{100} + (x - 2)^{200} \equiv (x-1)^2-(x-2)\\\mod (x-1)(x-2),$
that is
$(x - 1)^{100} + (x - 2)^{200}\equiv x^2-3x+3\mod (x^2-3x+2)$.
Hence,
$(x - 1)^{100} + (x - 2)^{200} \equiv 1\;\mod (x^2-3x+2)$.
| {
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Given $\sinh x$, find the exact value of $\cosh x, \cos 2x$ and $\tan 2x$ Given $\sinh x = 8/14$, find the exact value of $\cosh x, \cos 2x$ and $\tan 2x$.
I have been getting two answers which has made me confused. I keep getting $\sqrt{65/7}$ or $\sqrt{4/7}.$
That's what I got: $$\cosh^2 x - \sinh^2 x = 1$$ $$\cosh^2 x = 1 + \sinh^2 x$$ $$\cosh^2 x = 1 +\left(\frac{8}{14}\right)^2$$ $$\cosh^2 x=\frac{65}{49}$$ $$\cosh (x) = \sqrt{\frac{65}{49}}$$ $$\cosh x = \frac{\sqrt{65}}{7}$$
| You are right! Just use $$\cosh^2 x = 1+\sinh^2 x.$$
Now it only amounts to keeping track of what $\sinh x$ is and carefully doing the calculation. Is it $\sinh^2 x = \left(\frac{8}{13}\right)^2$ or is it $\sinh^2 x = \left(\frac{8}{14}\right)^2$?
UPDATE: With $\sinh x = \frac{4}{7}$:
$$\cosh^2 x = 1+ \left(\frac{4}{7}\right)^2 = 1+ \frac{16}{49}$$
so
$$\cosh x = \frac{\sqrt{65}}{7}.$$
| {
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Elegant way of finding the least perimeter of triangle A triangle $ABC$ has positive integer sides, $\angle A = 2\angle B$ and $\angle C > \pi/2$ , then the minimum length of the perimeter of $ABC$ is?
We have $\angle A = 2\angle B$
$\Rightarrow \sin A=\sin 2B=2 \sin B \cos B $
$\sin C=\sin(\pi-3B)=\sin(3B)=3\sin B-4\sin^3B$
Using $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
$a=2b\cos B $
$c=b(3-4\sin^2 B)$
From above two equations $a^2=b(c+b) $
NOTE: similar questions have been asked earlier but I want to do this question analysing with $a^2=b(c+b)$ and no fancy inequalities
| Found an another solution:
we have $a^2=b(c+b)$
A triangle of smallest perimeter means $gcd(a,b,c)=1$
In fact $gcd(b,c)=1$ since any common factor of $b,c$ would be a factor of $a$ as well.
A perfect square $a^2$ is being expressed as the product of two relatively prime integers $b$ and $c$.
it must be the case that both $b$ & $b + c$ are perfect squares. Thus for some integer $m$ & $n$ with
$gcd(m,n)= 1$ we have
$b=m^2$ and $b+c=n^2$,$a=mn$
$2\ cosB=\frac{n}{m}=\frac{a}{b}$
As $\angle C >\frac{\pi}{2} \Rightarrow 0<\angle B<\frac{\pi}{6}$
$\Rightarrow \sqrt{3}<2\ cosB=\frac{n}{m}<2$
The smallest value of $(m, n)$ that satisfies the above mentioned conditions are $4$ and $7$ $\blacksquare$
| {
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Determine $\sqrt{1+50\cdot51\cdot52\cdot53}$ without a calculator? I've tried a lot of things but failed to do it, I've calculated the result inside the square root which is $7027801$ using substitution and factoring but $\sqrt{7027801}$ isn't possible to simplify.
| Including $50\cdot51\cdot 52\cdot 53$ inside the square root suggests that you should choose a value for $x$. Suppose you choose the highest value and set $x=53$. Then
\begin{align}1+x(x-1)(x-2)(x-3)&=1+x(x-3)\cdot(x-1)(x-2)\\&
=1+(x^2-3x)(x^2-3x+2)
\end{align}
Now let $y=x^2-3x$. The above equation becomes
$$1+y(y+2)=y^2+2y+1=(y+1)^2$$
therefore
$$\sqrt{(y+1)^2}=y+1=x^2-3x+1={53}^2-3(53)+1=2651$$
| {
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If $xyz=32$, find the minimal value of If $xyz=32;x,y,z>0$, find the minimal value of $f(x,y,z)=x^2+4xy+4y^2+2z^2$
I tried to do by $A.M.\geq M.G.$:
$\frac{x^2+4y^2+2z^2}{2}\geq\sqrt{8x^2y^2z^2}\to x^2+4y^2+2z^2\geq32$
But how can I maximaze 4xy?
| Your application of AM-GM is wrong. The statement for AM-GM states that for positive integers $a_1, a_2, \dots, a_n$:
$$\frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1a_2\dots a_n}$$
with equality at $a_1 = a_2 = \dots = a_n$. Observe that
$$\frac{x^2+2xy+2xy+4y^2+z^2+z^2}{6} \geq \sqrt[6]{16x^4y^4z^4} = 16 \implies x^2 + 4xy + 4y^2 + 2z^2 \geq 96.$$
Equality occurs when $x^2 = 2xy = 4y^2 = z^2 \implies (x,y,z)=(4,2,4)$.
| {
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The coefficient of $x^{n}$ in the expansion of $(2-3 x) /(1-3 x+$ $\left.2 x^{2}\right)$ is The coefficient of $x^{n}$ in the expansion of $\frac{(2-3 x)}{(1-3 x+\left.2 x^{2}\right)}$ is
$(a) \quad(-3)^{n}-(2)^{n / 2-1}$
(b) $2^{n}+1$
$(c) 3(2)^{n / 2-1}-2(3)^{n}$
(d) None of the foregoing numbers.
Now, $1-3 x+2 x^{2}=(1-x)(1-2 x)$
Now, $(1-x)^{-1}$
$=1+x+x^{2}+x^{3}+\ldots$
Now, $(1-2 x)^{-1}$
$=1+2 x+(2 x)^{2}+(2 x)^{3}+\ldots \ldots$
Coefficient of $x^{n}$ in $(1-x)^{-1}(1-2 x)^{-1}=2^{n}+2^{n-1}+2^{n-2}+\ldots .+2+1=$
$2^{n+1}-1$
What to do next??
Any shortcut or objective approach for this type of problems would be highly appreciated.
| You're on the right track. Consider splitting the fraction up using partial fractions.
\begin{align*}
\frac{2-3x}{1-3x+2x^2} &= \frac{2-3x}{(1-2x)(1-x)}\\ &=\frac{1}{1-2x} + \frac{1}{1-x}\\&=\sum_{n=0}^\infty 2^nx^n + \sum_{n=0}^\infty x^n\\&=\sum_{n=0}^\infty \left (2^n+1\right )x^n
\end{align*}
Hence, the coefficient of $x^n$ is $2^n+1$.
| {
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Method of summation for third order difference series: $2+12+36+80+150\dots$ I am unable to understand how one derived the formula for the $n$-th term $= an^3 +bn^2 + cn + d$, where the degree of the polynomial depends on the step at which we get a constant A.P. Here its at $2$nd step so degree $=2+1=3$.
But how do we derive this?
| $2+12+36+80+150\dots = (1+1) + (4 + 8) + (9 + 27)+(16+64)+(25+125)+...=\sum (k^2+k^3)$
$$\sum _{k=1}^n \left(k^3+k^2\right)=\sum _{k=1}^n k^3+\sum _{k=1}^n k^2=\frac{1}{4} n^2 (n+1)^2+\frac{1}{6} n (n+1) (2 n+1)=$$
$$=\frac{1}{12} n (n+1) (n+2) (3 n+1)$$
| {
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How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$ Here is a new challenging problem:
Show that
$$I=\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx=2\ln(2)G-\frac{\pi}{8}\ln^2(2)-\frac{5\pi^3}{32}+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
My attempt:
With Weierstrass substitution we have
$$I=2\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1-x^2}{1+x^2}\right)dx\overset{x\to \frac{1-x}{1+x}}{=}4\int_0^1\frac{\frac{\pi}{4}-\arctan x}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx$$
$$=\pi\underbrace{\int_0^1\frac{1}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx}_{I_1}-4\underbrace{\int_0^1\frac{\arctan x}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx}_{I_2}$$
By setting $x\to \frac{1-x}{1+x}$ in the first integral we have
$$I_1=\frac12\int_0^1\frac{1}{x}\ln\left(\frac{1-x^2}{1+x^2}\right)dx$$
$$=\frac14\int_0^1\frac{1}{x}\ln\left(\frac{1-x}{1+x}\right)dx=\frac14\left[-\text{Li}_2(x)+\text{Li}_2(-x)\right]_0^1=-\frac38\zeta(2)$$
For the second integral, write $\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$
$$I_2=\frac12\int_0^1\frac{\arctan x}{1-x}\ln\left(\frac{2x}{1+x^2}\right)dx+\frac12\int_0^1\frac{\arctan x}{1+x}\ln\left(\frac{2x}{1+x^2}\right)dx$$
The first integral is very similar to this one
$$\int_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{192}\,\pi^{3} +
2\Im\left\{%
\text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}$$
So we are left with only $\int_0^1\frac{\arctan x\ln(1+x^2)}{1+x}dx$ as $\int_0^1\frac{\arctan x\ln x}{1+x}dx$ is already nicely calculated by FDP here. Any idea?
I noticed that if we use $x\to\frac{1-x}{1+x}$ in $\int_0^1\frac{\arctan x\ln(1+x^2)}{1+x}dx$ we will have a nice symmerty but still some annoying integrals appear.
In $I$, I also tried the Fourier series of $\ln(\cos x)$ but I stopped at $\int_0^{\pi/2} \frac{x\cos(2nx)}{\sin x}dx$. I would like to see different approaches if possible.
Thank you.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\mc}[1]{\mathcal{#1}}
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\begin{align}
I & \equiv \int_{0}^{\pi/2}x{\ln\pars{\cos\pars{x}}
\over \sin\pars{x}}\,\dd x
\\[5mm] & =
\bbox[5px,#ffd]{2\ln\pars{2}\,\mrm{G} - {\pi \over 8}\ln^{2}\pars{2} - {5\pi^{3} \over 32} + 4\,\Im\pars{\mrm{Li}_3\pars{1 + \ic \over 2}}}:\ {\Large ?}\label{1}\tag{1}
\end{align}
$\ds{\mrm{G}}$ is the Catalan Constant and
$\ds{\mrm{Li}_{s}}$ is the polylogarithm.
\begin{align}
I & \equiv \bbox[5px,#ffd]{\int_{0}^{\pi/2}x{\ln\pars{\cos\pars{x}}
\over \sin\pars{x}}\,\dd x}
\\[5mm] & =
\left. \Re\int_{x\ =\ 0}^{x\ =\ \pi/2}\bracks{-\ic\ln\pars{z}}{\ln\pars{\bracks{z + 1/z}/2}
\over \pars{z - 1/z}/\pars{2\ic}}\,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] & =
\left. -2\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\ln\pars{z}\,
\ln\pars{1 + z^{2} \over 2z}
\,{\dd z \over 1 - z^{2}}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] & =
2\,\Im\int_{1}^{0}\bracks{\ln\pars{y} + {\pi \over 2}\,\ic}\,
\bracks{\ln\pars{1 - y^{2} \over 2y} - {\pi \over 2}\,\ic}
\,{\ic\,\dd y \over 1 + y^{2}}
\\[5mm] & =
-2\int_{0}^{1}\bracks{\ln\pars{y}\ln\pars{1 - y^{2} \over 2y} + {\pi^{2} \over 4}}\,
\,{\dd y \over 1 + y^{2}}
\\[5mm] & =
-2\
\overbrace{\int_{0}^{1}{\ln\pars{y}\ln\pars{1 - y} \over 1 + y^{2}}\,\dd y}^{\ds{I_{1}}}\ -\ 2\
\overbrace{\int_{0}^{1}{\ln\pars{y}\ln\pars{1 + y} \over 1 + y^{2}}\,\dd y}^{\ds{I_{2}}}
\\[2mm] & + 2\ln\pars{2}\
\underbrace{\int_{0}^{1}{\ln\pars{y} \over 1 + y^{2}}\,\dd y}
_{\ds{I_{3}}}\ +\ 2\
\underbrace{\int_{0}^{1}{\ln^{2}\pars{y} \over 1 + y^{2}}\,\dd y}
_{\ds{I_{4}}}\ -\
\underbrace{{\pi^{2} \over 2}\int_{0}^{1}{\dd y \over 1 + y^{2}}}
_{\ds{\pi^{3} \over 8}}
\\ & =
-2I_{1} -2I_{2} + 2\ln\pars{2}\, I_{3} +2I_{4} - {\pi^{3} \over 8}
\label{2}\tag{2}
\end{align}
Those integrals are well known or/and very -laboriously- doable:
\begin{equation}
\left\{\begin{array}{rcl}
\ds{I_{1}} & \ds{=} &
\ds{-\,{\pi \over 32}\,\ln^{2}\pars{2}} - {\pi^{3} \over 128} +
\Im\pars{\mrm{Li}_{3}\pars{1 + \ic \over 2}}
\\[2mm]
\ds{I_{2}} & \ds{=} &
\ds{\phantom{-}2\mrm{G}\ln\pars{2} +
{3\pi \over 32}\,\ln^{2}\pars{2}} + {11\pi^{3} \over 128} -
3\,\Im\pars{\mrm{Li}_{3}\pars{1 + \ic \over 2}}
\\[2mm]
\ds{I_{3}} & \ds{=} & \ds{-\,\mrm{G}}
\\[2mm]
\ds{I_{4}} & \ds{=} & \ds{\phantom{-}{\pi^{3} \over 16}}
\end{array}\right.\label{3}\tag{3}
\end{equation}
(\ref{2}) and (\ref{3}) lead to the coveted result (\ref{1}).
| {
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Use generating functions to solve the non-homogenous recurrence relation Let $a_0=0, a_1=2,$ and $a_2=5$. Use generating functions to solve the recurrence equation: $$a_{n+3} = 5a_{n+2} - 7a_{n+1}+3a_n + 2^n$$ for $n\geq0$.
This is a book problem from Applied Combinatorics. I am really confused about tackling $2^n$ part of the recurrence relation using generating functions.
Edit:
I know I need to convert the recurrence into series and I have broken it down, but am struggling with getting it into a proper form to do partial fractions. These are the equations I have managed to get.
If we let $A(x) = \sum_{n \geq 0} a_n x^n$ be the generating function for $a_n$ then after the calculations I got:
$$A(x)\cdot(1-5x+7x^2-3x^3)= 12x^3 - 9 x^2 + \frac{2x}{1-2x}$$
After simplifying:
$$A(x) = \frac{12x^3 - 9 x^2 + \frac{2x}{1-2x}}{1-5x+7x^2-3x^3}$$
$$= \frac{24 x^4 - 30 x^3 + 9 x^2 - 2 x}{(1-2x)(x-1)^2(3x-1)}$$
Then, the partial fraction decomposition is:
$$A(x) = -\frac{8}{1-2x} + \frac{13}{4}\frac{1}{1-3x} + \frac{37}{4}\frac{1}{1-x} - \frac{1}{2} \frac{1}{(1-x)^2} - 4$$
I have tried to plug in the values, but something doesn't seem right. Please let me know where I would have gone wrong.
| You made a mistake somewhere in the generating function derivation (hard to tell where since you did not include this part), I've got
\begin{align}
A(x)&=2x+5x^2+\sum_{n \geq 3}a_{n}x^n\\
&=2x+5x^2+5\sum_{n \geq 3}a_{n-1}x^n-7\sum_{n \geq 3}a_{n-2}x^n+3\sum_{n \geq 3}a_{n-3}x^n+\sum_{n \geq 3}2^{n-3}x^n\\
&=2x+5x^2+5x\sum_{n \geq 2}a_{n}x^n-7x^2\sum_{n \geq 1}a_{n}x^n+3x^3\sum_{n \geq 0}a_{n}x^n+x^3\sum_{n \geq 0}2^{n}x^n\\
&=2x+5x^2+5x(A(x)-2x)-7x^2(A(x)-0)+3x^3A(x)+x^3\cdot \frac{1}{1-2x}
\end{align}
which solves to
\begin{align}
A(x)&=\frac{x(11x^2-9x+2)}{(1-2x)(1-3x)(x-1)^2}\\
&=\frac{2}{(x-1)^2}-\frac{3}{2}\frac{1}{1-x}-\frac{1}{1-2x}+\frac{1}{2}\frac{1}{1-3x}.
\end{align}
Check your solution, hopefully you can finish it from here.
| {
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Complex Matrix is Orthogonal if and only if... Let D be a 2x2 matrix with entries in the complex numbers. Prove that D is orthogonal if and only if, it is of the form:
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}
or
\begin{pmatrix}
a & b\\
b & -a
\end{pmatrix}
Proof. I've already proved that if D is equal to those forms then, it implies that D is an orthogonal matrix. But how can I prove this? If D is orthogonal, then it must be of the form:
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}
or
\begin{pmatrix}
a & b\\
b & -a
\end{pmatrix}
Update: a and b must satisfy that $a^2+b^2=1$
| Let a 2x2 matrix $A$ be orthogonal i.e $AA^t = A^tA = E$. Notice that
$$ E= \left (\begin{array}{cc}a & b \\ c & d\end{array} \right )\left (\begin{array}{cc}a & c \\ b & d\end{array} \right ) \Rightarrow \|(a,b)\| = \|(c,d)\| = 1,\ (a,b)\perp (c,d) $$
Conversely, suppose a 2x2 matrix $A$ has orthonormal columns $(a,b)^t$ and $(c,d)^t$. Then
$$ A^tA = \left (\begin{array}{cc}a & b \\ c & d\end{array} \right )\left (\begin{array}{cc}a & c \\ b & d\end{array} \right ) = \left ( \begin{array}{cc}a^2 + b^2 & 0 \\ 0 & c^2+d^2 \end{array} \right ) = E $$
Likewise $AA^t = E$.
For the complex case we say a square matrix $A$ is unitary, if $A^HA=AA^H = E$, where $A^H := \overline{A}^t$. Orthonormality of column vectors remains an equivalent condition.
| {
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Limit using Taylor expansion : which term do we expand? I want to check the limit $\displaystyle{\lim_{n\rightarrow +\infty}\text{exp}\left (\frac{n}{2}\ln\left (\frac{n^2-2n+1}{n^2+1}\right )\right )}$ using the Taylor expansion.
I have done the following:
$$\lim_{n\rightarrow +\infty}\text{exp}\left (\frac{n}{2}\ln\left (\frac{n^2-2n+1}{n^2+1}\right )\right )=\lim_{n\rightarrow +\infty}\text{exp}\left (\ln\left (\frac{n^2-2n+1}{n^2+1}\right )^{\frac{n}{2}}\right )=\lim_{n\rightarrow +\infty}\left (\frac{n^2-2n+1}{n^2+1}\right )^{\frac{n}{2}}=\lim_{n\rightarrow +\infty}\left (\frac{(n-1)^2}{n^2+1}\right )^{\frac{n}{2}}=\lim_{n\rightarrow +\infty}\frac{(n-1)^n}{\left (n^2+1\right )^{\frac{n}{2}}}$$
For which term do we have to write the Taylor expansion?
| Expand the logarithm
$$
\ln\left(\frac{n^2-2n+1}{n^2+1}\right)=\ln\left (1-\frac{2n}{n^2+1}\right)=-\frac{2n}{n^2+1}+\ldots
$$
| {
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If $a+\sqrt{a}=b+\sqrt{b}$ is $a=b$? If $a+\sqrt{a}=b+\sqrt{b}$, does this automatically mean that $a=b$?
I first tried to square both sides but that seemed to get me nowhere.
$$a-b=\sqrt{b}-\sqrt{a}$$
Can we just conclude that $a$ has to be equal to $b$ to make this expression to be true?
| $a-b=\sqrt{a}-\sqrt{b} \implies (\sqrt{a}-\sqrt{b})(-1+\sqrt{a}+\sqrt{b})=0\implies \sqrt{a}=\sqrt{b}$ or $1=\sqrt{a}+\sqrt{b}$. So it might not be the case that $a = b$.
I realize my answer above is for a different problem. So back to this one. We have $a - b = \sqrt{b} - \sqrt{a} \implies a - b +\sqrt{a}-\sqrt{b} = 0 \implies (\sqrt{a}-\sqrt{b})(1+\sqrt{a}+\sqrt{b}) = 0 \implies \sqrt{a} = \sqrt{b} \implies a = b$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial sums over rational functions I recently came across the result that
$$\sum_{n=2}^\infty \frac{n^4-n^3+n+1}{n^6-1} = \frac{1}{2}$$
I am wondering how one could proof this, generally how one could evaluate a sum over rational functions.
If I plug the sum into Wolfram Alpha it gives
$$\frac{3k^4-k-2}{6k(k+1)(k^2+k+1)}$$
as the $k$-th partial sum. Taking the limit as $n \to \infty$, this would in fact proof the upper equality.
Sadly, I could not wrap my head around how to get to Wolfram Alphas partial sum result. If anyone has an idea let me know. Any tips are appreciated.
| In a word: the terms telescope in a nice way.
Using partial fractions, we have
$$
\frac{n^4-n^3+n+1}{n^6-1} = \frac{1}{3}\left(\frac{n - 2}{n^2 - n + 1}- \frac{n-1}{n^2 + n + 1} + \frac{1}{n - 1} - \frac{1}{n + 1}\right)
$$We need to be a bit careful because the harmonic series diverges, so we should only group terms with opposite signs together. The last two terms form a well-known telescoping series:
$$
\frac{1}{3}\sum_{n=2}^{m}\left(\frac{1}{n - 1} - \frac{1}{n + 1} \right)= \frac{1}{3}\left(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1}\right)\longrightarrow \frac{1}{2}
$$The first two terms telescope as well, with the second term eating the one before it:
$$
\frac{1}{3}\sum_{n=2}^{m}\left(\frac{n - 2}{n^2 - n + 1}- \frac{n-1}{n^2 + n + 1} \right)= -\frac{1}{3}\cdot \frac{m-1}{m^2+m+1}\longrightarrow 0
$$This sort of thing won't work in general but I won't look a gift horse in the mouth.
| {
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"url": "https://math.stackexchange.com/questions/3799725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $\sum_{cyc}\frac{(a-1)(c+1)}{1+bc+c}\geq 0$ for positive $a$, $b$, $c$ with $abc=1$. I recently saw the following inequality,
$$\frac{(a-1)(c+1)}{1+bc+c}+\frac{(b-1)(a+1)}{1+ca+a} + \frac{(c-1)(b+1)}{1+ab+b} \geq 0 \tag1$$
for all $abc=1$ and $a,b,c \in \mathbb{R}_+ \setminus \{0 \}$.
To prove this inequality, I thought to try to look at each term, by knowing that $abc=1$, and as such we could express the three terms of our inequalities in terms of three/two variables and eliminate the denominator.
First, we shall look at the expression:
$$\frac{(a-1)(c+1)}{1+bc+c} = \frac{(a-1)(\frac{1}{ab}+1)}{1+ b \frac{1}{ab}+\frac{1}{ab}} = \frac{\frac{(a-1)(ab+1)}{ab}}{1+ \frac{1}{a}+\frac{1}{ab}} = \frac{(a-1)(ab+1)}{ab(1+ \frac{1}{a}+\frac{1}{ab})} = \frac{(a-1)(ab+1)}{ab+ {b}+{1}}$$
One can see that the denominator now is the same as the denominator of the third term of our inequality. Thus, we shall also bring the denominator of the second term to the form $ab+b+1$. We have
$$\frac{(b-1)(a+1)}{1+ca+a}=\frac{(b-1)(a+1)}{1+\frac{1}{ab}a+a} = \frac{(b-1)(a+1)}{1+\frac{1}{b}+a} = \frac{(b-1)(a+1)b}{b(1+\frac{1}{b}+a)} = \frac{(b-1)(ab+b)}{b+1+ab} = \frac{(b-1)(ab+b)}{ab+b+1}$$
Now we will sum our terms now with the same denominator,
\begin{align*}
I &= \frac{(a-1)(ab+1)}{ab+ {b}+{1}} + \frac{(b-1)(ab+b)}{ab+b+1} +\frac{(c-1)(b+1)}{1+ab+b} \\ &= \frac{(a-1)(ab+1)+(b-1)(ab+b)+(c-1)(b+1)}{ab+ {b}+{1}}
\end{align*}
with
$$I=\sum_{cyc}\frac{(a-1)(c+1)}{1+bc+c} $$
Thus
$$I = \frac{a^2b+a-ab-1+ab^2+b^2-ab-b+bc+c-b-1}{ab+ {b}+{1}}$$
So
$$I = \frac{a^2b+ab^2 -2(ab+b+1)+b^2+bc+a+c}{ab+ {b}+{1}}$$
Now we must prove that
$$\frac{a^2b+ab^2 -2(ab+b+1)+b^2+bc+a+c}{ab+ {b}+{1}} \geq 0$$
Since $a,b,c$ are positive real numbers, $ab+b+1 \in \mathbb{R}_+^*$, so we now are left to prove
$$a^2b+ab^2 -2(ab+b+1)+b^2+bc+a+c \geq 0$$
$$a^2b+ab^2+b^2+bc+a+c \geq 2(ab+b+1)$$
Note. Here, AM-GM would be a good idea to use. I am currently thinking about a way to use it.
If anybody has a solution or a hint, it'd be much appreciated.
| Hint: A common trick in an inequality like this is to substitute:
$$a = \frac xy, b = \frac yz, c = \frac zx$$
which transforms your final inequality to:
$$(x^3z+y^3x+z^3y) + (x^2y^2+z^2x^2+y^2z^2)\geq 2xyz(x+y+z).$$
But the above is easy: you can suitably use AM-GM and prove that two expressions in parentheses is at least $xyz(x+y+z).$ I will leave the AM-GM part for you to find the coefficients.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this proof of $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ incomplete? So, for any angle $\alpha$ :
$$\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = \dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha} = \dfrac{\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}}{\dfrac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}= \dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}$$
Now, $\cos\alpha = \cos\Big(2\cdot\dfrac{\alpha}{2}\Big) = \dfrac{1-\tan^2\dfrac{\alpha}{2}}{1+\tan^2\dfrac{\alpha}{2}}$
Now, using the componendo and dividendo rule, we get :
$$\dfrac{\cos\alpha+1}{\cos\alpha-1} = \dfrac{2}{-2\tan^2\dfrac{\alpha}{2}} = \dfrac{-1}{\tan^2\dfrac{\alpha}{2}} \implies \tan^2\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{1+\cos\alpha}$$
$$\implies \tan^2\dfrac{\alpha}{2} = \dfrac{(1-\cos\alpha)(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)} = \Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)^2$$
$$\implies \Bigg|\tan\Big(\dfrac{\alpha}{2}\Big)\Bigg| = \Bigg|\dfrac{1-\cos\alpha}{\sin\alpha}\Bigg|$$
Now, only if $\mathrm{sign}\Big(\tan\dfrac{\alpha}{2}\Big) = \mathrm{sign}\Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)$ is true, we can say that $\tan\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{\sin\alpha}$
So, I think that without proving that, the proof will be incomplete but my Math textbook doesn't prove it.
So, is it necessary to prove it? If not, why not?
Thanks!
| Yes, it is necessary to show.
As $\tan$ has a periodicity $\pi$, it's enough to check the signs for $\dfrac{\alpha}{2}$ in the ranges, $\left[0,\dfrac{\pi}{4}\right], \left[\dfrac{\pi}{4},\dfrac{\pi}{2}\right),\left(\dfrac{\pi}{2},\dfrac{3\pi}{4}\right] $ and $\left[\dfrac{3\pi}{4},\pi\right]$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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What is $3^{99} \pmod{100}$? I saw this post on how to solve $3^{123}\pmod{100}$ using Euler's Totient Theorem.
How about for $3^{99}\pmod{100}$? It seems more complicated because applying Euler's Totient Theorem gets us $3^{40}\equiv 1\pmod{100}$. This means $3^{80}\equiv 1 \pmod{100}$, which isn't enough, because we still need to find $3^{19}\pmod{100}$.
Now, when the terms are listed, a pattern does emerge.
$$\begin{array}{|c|c|c|c|}
\hline
3 & 9 & 27 & 81 \\ \hline
43 & 29 & 87 & 61 \\ \hline
83 & 49 & 47 & 41\\ \hline
23 & 69 & 07 & 21 \\ \hline
03 & 89 & 67 & 01 \\ \hline
\end{array}$$
And $3^{19}$ ends in $67$.
But how can I find this in other methods, besides bashing? Is there some sort of theorem that I can use?
| You can do this using the Chinese remainder theorem and by calculating some powers.
$3^{99} = (3^3)^{33} \equiv 2^{33} \pmod{25} = (2^7)^4 \times 2^5 \equiv 3^4 \times 2^5 = 3^3 \times 3 \times 2^5 \equiv 3 \times 2^6 \equiv 17$.
$3^{99} \equiv (-1)^{99} \equiv 3 \pmod{4}$.
So we need the unique number satisfying those constraints in $[0, 100)$, which is $67$.
Your way comes out much the same: $3^{19}$ mod $4$ is $3$ again, and $3^{19}$ mod $25$ is $(3^3)^6 \times 3$ which is $2^6 \times 3$ which is again $17$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating $\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx$ without using Beta function and Euler sum. Is it possible to show that
$$\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx=-\frac12\zeta(4)$$
without using the Beta function
$$\text{B}(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
and the generalized Euler sum
$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k),\quad q\ge 2\ ?$$
By integration by parts I found
$$\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx=\color{blue}{\int_0^1\frac{\ln^2(1-x)\ln x}{x}dx}$$
Setting $1-x\to x$ gives the blue integral again.
This integral seems tough under such restrictions. All approaches are welcome.
thanks.
| \begin{align}J&=\int_0^1 \frac{\ln^2 x\ln(1-x)}{1-x}dx\\
&\overset{\text{IBP}}=\underbrace{\left[\left(\int_0^x \frac{\ln^2 t}{1-t}dt-\int_0^1 \frac{\ln^2 t}{1-t}dt\right)\ln(1-x)\right]_0^1}_{=0}+\\&\int_0^1 \frac{1}{1-x}\left(\underbrace{\int_0^x \frac{\ln^2 t}{1-t}dt}_{u(t)=\frac{t}{x}}-\int_0^1 \frac{\ln^2 t}{1-t}dt\right)dx\\
&=\int_0^1\int_0^1 \left(\frac{x\ln^2(ux)}{(1-x)(1-ux)}-\frac{\ln^2 u}{(1-x)(1-u)}\right)dudx\\
&=\int_0^1 \int_0^1 \left(\frac{\ln^2 x}{(1-x)(1-u)}+\frac{2\ln x\ln u}{(1-x)(1-u)}-\frac{\ln^2(ux)}{(1-u)(1-ux)}\right)dudx\\
&=2\zeta(2)^2+\int_0^1\frac{1}{1-u}\left(\int_0^1 \frac{\ln^2 x}{1-x}dx-\underbrace{\int_0^1 \frac{\ln^2(ux)}{1-ux}dx}_{z(x)=ux}\right)du\\
&=2\zeta(2)^2+\int_0^1\frac{1}{u(1-u)}\left(u\int_0^1 \frac{\ln^2 z}{1-z}dz-\int_0^u \frac{\ln^2 z}{1-z}dz\right)du\\
&\overset{\text{IPP}}=2\zeta(2)^2-\int_0^1 \ln\left(\frac{u}{1-u}\right)\left(\int_0^1 \frac{\ln^2 z}{1-z}dz-\frac{\ln^2 u}{1-u}\right)du\\
&=2\zeta(2)^2+\int_0^1 \frac{\ln\left(\frac{u}{1-u}\right)\ln^2 u}{1-u}du=\underbrace{2\zeta(2)^2}_{=5\zeta(4)}+\underbrace{\int_0^1 \frac{\ln^3 u}{1-u}du}_{=-6\zeta(4)}-J
\end{align}
Therefore,
\begin{align}\boxed{J=-\frac{1}{2}\zeta(4)}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Factoristaion of $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1$ My question revolves around the polynomial
$$x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1$$
I know that it can be factorised into
$$(x^2+x+1)(x^6+x^3+1)$$
but what method can we employ to obtain this result in the first place? Any method that works will be of great help to me.
Thank you for your help.
| Method 1: Use what you know about polynomials of the form $x^n-1$. For instance, your polynomial is $\frac{x^9-1}{x-1}$. So if you can factor the numerator, many of those factors would be inherited to your polynomial. And indeed, we have
$$
x^9-1=(x^3)^3-1=((x^3)^2+x^3+1)(x^3-1)\\
=(x^6+x^3+1)(x^2+x+1)(x-1)
$$
Put this into the fraction above, and the factorisation follows.
Method 2: Inspection. There are nine terms. They can be grouped three-by-three:
$$
(x^8+x^7+x^6)+(x^5+x^4+x^3)+(x^2+x+1)\\
=x^6(x^2+x+1)+x^3(x^2+x+1)+(x^2+x+1)
$$
and the factorisation follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of series $\sum_{n=1}^\infty\left[\left(1+\frac{1}{n}\right)^{n+1}-a\right]\sin{n}$ How to deduce if series is convergent (depending on parameter $a$):
$$
\sum_{n=1}^\infty\left[\left(1+\frac{1}{n}\right)^{n+1}-a\right]\sin{n}?
$$
If $a=e,$ we have that $\lim_{n\to\infty}a_n=0,$ but it is not sufficient to show that series converges.
I'm stuck with this today, not at my best. Any help is welcome. Thanks in advance.
| If $a \neq e$, then $$\left[\left(1+\frac{1}{n}\right)^{n+1}-a\right]\sin{n}$$
does not tend to $0$ so the series cannot converge.
If $a=e$, you have
$$\left[\left(1+\frac{1}{n}\right)^{n+1}-e\right]\sin{n} = \left[e^{(n+1)\ln\left( 1 + \frac{1}{n}\right)}-e\right]\sin{n}=e\left[e^{(n+1)\ln\left( 1 + \frac{1}{n}\right)-1}-1\right]\sin{n}$$
Now
$$(n+1)\ln\left( 1 + \frac{1}{n}\right)-1 = (n+1)\left( \frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} + o\left( \frac{1}{n^3}\right)\right) - 1$$ $$= \frac{1}{2n} - \frac{1}{6n^2} + o\left( \frac{1}{n^2}\right)$$
So
$$\left[\left(1+\frac{1}{n}\right)^{n+1}-e\right]\sin{n} =e\left[ \frac{1}{2n} - \frac{1}{24n^2} + o\left( \frac{1}{n^2}\right)\right]\sin{n}$$
Finally, the series
$$\sum \frac{\sin(n)}{n}$$
converges (by Dirichlet's test) ; and the series
$$\sum \frac{\sin(n)}{n^2}$$
is absolutely convergent. So your series converges.
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $\sum_{cyc}\frac{{a^2}{b}}{c}\ge a^2+b^2+c^2$ prove $$\sum_{cyc}\frac{{a^2}{b}}{c}\ge a^2+b^2+c^2$$ where $a,b,c>0$ and $a\ge b\ge c$
My try it seemed qite simple but i couldnt apply the rearrangement inequality directly. so i tried manipulating the inequality.
the inequality can be written as $$a^2(b-c)+b^2(c-a)+c^2(a-b)\ge 0$$ .It seemed like 'schurs' inequality could be uused but i couldn't procceed.Also i tried using the weighted a-m g-m method. Could anyone give me a hint (i want to solve the problem myself).
source: Excursion in mathematics(Modak)
| Your first step is wrong:
We need to prove that:
$$\sum_{cyc}(a^3b^2-a^3bc)\geq0$$ and it indeed gives a proof:
$$\sum_{cyc}(2a^3b^2-2a^3bc)\geq0$$ or
$$\sum_{cyc}(a^3b^2+a^3c^2-2a^3bc)\geq\sum_{cyc}(a^3c^2-a^3b^2)$$ or
$$\sum_{cyc}a^3(b-c)^2\geq(ab+ac+bc)(a-b)(b-c)(c-a),$$ which is obvious.
We can get $$\sum_{cyc}(a^3c^2-a^3b^2)=(ab+ac+bc)(a-b)(b-c)(c-a)$$ by the following way.
For $a=b$, $a=c$ and $b=c$ we obtain identity, which says that $$\sum_{cyc}(a^3c^2-a^3b^2)=P(a,b,c)(a-b)(b-c)(c-a),$$ where $P$ is a cyclic homogeneous polynomial of second degree.
Id est $$P(a,b,c)=\sum_{cyc}(ka^2+mab).$$
Now, $k=0$ because, otherwise there is a problem on $\infty$.
Also, let $c=0$.
We obtain: $$b^3a^2-a^3b^2=mab(a-b)b(-a)$$ or
$$a^2b^2(b-a)=ma^2b^2(b-a),$$ which gives $m=1$.
Factoring of some Schur's polynomials:
$$\sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c),$$
$$\sum_{cyc}(a^3b-a^3c)=(a+b+c)(a-b)(a-c)(b-c),$$
$$\sum_{cyc}(a^4b-a^4c)=(a^2+b^2+c^2+ab+ac+bc)(a-b)(a-c)(b-c),$$
$$\sum_{cyc}(a^3b^2-a^3c^2)=(ab+ac+bc)(a-b)(a-c)(b-c),$$
$$\sum_{cyc}(a^5b-a^5c)=$$
$$=(a^3+b^3+c^3+a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+abc)(a-b)(a-c)(b-c),$$
$$\sum_{cyc}(a^4b^2-a^4c^2)=(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc)(a-b)(a-c)(b-c),...$$
| {
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"url": "https://math.stackexchange.com/questions/3805200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$ If $a+b+c=1$ and $a,b,c>0$ then prove that $$\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$$
My try : We have to prove: $$ab+bc+ca+36abc(ab+bc+ca)\ge 21abc$$ or after homogenising we get :
$$\sum a^4b+3\sum a^3b^2+6\sum a^2b^2c\ge 14\sum a^3bc$$
I don't know what to do next. I think SOS may be helpful but find it difficult to factorize. Any other methods are also welcome but, if possible, can someone help me continue from here?
| Suppose $t = \frac{a+b}{2}$ and $c = \max \{a,b,c\}.$ Let
$$f(a,b,c) = \frac{1}{abc}+36 - \frac{21}{ab+bc+ca}.$$
We have
$$f(a,b,c) -f(t,t,c) = \frac{1}{abc}-\frac{4}{c(a+b)^2}+\frac{84}{(a+b)(a+b+4c)}-\frac{21}{ab+bc+ca}$$
$$ = \frac{(a-b)^2}{a+b}\left(\frac{1}{abc(a+b)}-\frac{21}{(ab+bc+ca)(a+b+4c)}\right) \geqslant 0,$$
because
$$(ab+bc+ca)(a+b+4c)=(a+b+c)(ab+bc+ca)(a+b+4c) $$
$$ \geqslant 9abc (a+b+4c) > 21abc(a+b).$$
Thefore
$$f(a,b,c) \geqslant f(t,t,c) = f\left(\frac{1-c}{2},\frac{1-c}{2},c\right)=\frac{4(3c^2-3c+1)(3c-1)^2}{c(3c+1)(c-1)^2} \geqslant 0.$$
The proof is completed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Using rnfequation to show that $\left(\frac{\frac{-1+\sqrt{5}}{2}+\sqrt{\frac{-\sqrt{5}-5}{2}}}{2}\right)^5=1$, in PARI/GP I want to understand how can the code below shows that $\left(\frac{\frac{-1+\sqrt{5}}{2}+\sqrt{\frac{-\sqrt{5}-5}{2}}}{2}\right)^5=1$
{
V=rnfequation(T^2-5,x^2-Mod((-T-5)/2,T^2-5),1);
a=V[2];
b=x-V[3]*a;
lift(((a-1+2*b)/4)^5)
}
recall :
rnfequation(nf,pol,{flag = 0})
given a number field nf as output by nfinit (or simply a polynomial) and a polynomial pol with coefficients in nf defining a relative extension L of nf, computes the absolute equation of L over Q.
If flag is non-zero, outputs a 3-component row vector [z,a,k], where z is the absolute equation of L over Q, as in the default behaviour, a expresses as an element of L a root alpha of the polynomial defining the base field nf, and k is a small integer such that theta = beta+kalpha where theta is a root of z and beta a root of pol.
| By using the property that each side of a regular decagon inscribed in a circumference is the golden part of the radius, we get that
$\cfrac{\text{side}}{\text{radius}}=\cfrac{-1+\sqrt{5}}{2}\;$ and $\;\sin 18^\circ=\cfrac{\text{side / 2}}{\text{radius}}=\cfrac{-1+\sqrt{5}}{4}\;.$
Moreover,
$\cos 18^\circ=\sqrt{1-\sin^218^\circ}=\sqrt{1-\left(\cfrac{-1+\sqrt{5}}{4}\right)^2}=\\=\sqrt{\cfrac{5+\sqrt{5}}{8}}=\cfrac{1}{2}\sqrt{\cfrac{\sqrt{5}+5}{2}}\;.$
Since $\;\cfrac{\cfrac{-1+\sqrt{5}}{2}+\sqrt{\cfrac{-\sqrt{5}-5}{2}}}{2}=\sin 18^\circ+i\cos 18^\circ\;,\;$ then, by applying De Moivre’s formula, we obtain,
$\left(\cfrac{\cfrac{-1+\sqrt{5}}{2}+\sqrt{\cfrac{-\sqrt{5}-5}{2}}}{2}\right)^5=\left(\sin 18^\circ+i\cos 18^\circ\right)^5=\\=\sin 90^\circ+i\cos 90^\circ=1\;.$
| {
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"timestamp": "2023-03-29T00:00:00",
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$|z-1|+|z-2|+|z-3|=6$ in the Argand Plane In the argand plane $$C:|z-1|+|z-2|+|z-3|=6 ~~~(1)$$ represents a bounded curve which is a rounded blob mimicking an "ellipse".
Using the the inequality $$|Z_1+Z_2+Z_3| \le |Z_1|+|Z_2|+|Z_3|, ~~~~(2)$$ we can see that $$6=|z-1|+|z-2|+|z-3|\ge |3z-6| \implies |z-2| \le 2. ~~~(3)$$
The circle $C_1: |z-2|=2$ touches and remains outside $C$.
Next, it will be interesting to use some other inequality or something to find the equation of a bounded curve $C_2$ which touches and remains inside $C$. Eventually, $C$ will touch $C_1$ and $C_2$ and it will be enclosed between them.
The question is to find the curve $C_2.$
Edit:
For $C_2$, the RMS-AM inequality gives:
$$\sqrt{\frac{|z-1|^2+|z-2|^2+|z-3|^2}{3}} \ge \frac{|z-1|+|z-2|+|z-3|}{3}=2~~~~(4)$$
The equality in above doesn't holds as $|z-1|=|z-2|=|z-3|$ cannot be met
$$\implies(x-2)^2+y^2 >\sqrt{\frac{10}{3}}=1.82574~~~~(5)$$
This circle $|z-2|=\sqrt{\frac{10}{3}}$ cannot be the required $C_2$ as it cannot touch $C$ in (1).
| Our claim is that the circle
$$C_2:|z-2|=\sqrt\frac{44}{3}-2$$
lies entirely inside $C$. It is not hard to show that this value is the unique real $r$ for which $2\pm ir$ are on $C$.
We can calculate that
$$|z-2|^2=\frac{|z-1|^2+|z-3|^2}{2}-1.$$
We have that
$$\frac{|z-1|^2+|z-3|^2}{2}\geq\left(\frac{|z-1|+|z-3|}{2}\right)^2,$$
with equality if and only if $|z-1|=|z-3|$, so
$$|z-1|+|z-3|\leq 2\sqrt{|z-2|^2+1},$$
again with equality if and only if $|z-1|=|z-3|$. So, for a point $z$ on $C$,
$$6=|z-1|+|z-2|+|z-3|\leq |z-2|+2\sqrt{|z-2|^2+1};$$
since the right side is increasing in $|z-2|$, this implies that $|z-2|$ is at least the real $r$ for which $6=r+2\sqrt{r^2+1}$, with equality reached if and only if $z=2\pm ir$. It can be calculated that $r=\sqrt\frac{44}3-2$ is the positive value of $r$ that satisfies $6=r+2\sqrt{r^2+1}$, so we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the $26^{th}$ digit of a $50$ digit number divisible by $13$. $N$ is a $50$ digit number (in the decimal scale). All digits except the $26^{th}$ digit (from the left)
are $1$. If $N$ is divisible by $13$, find the $26^{th}$ digit.
This question was asked in RMO $1990$ and is very similar to this question and the same as this question but it is not solved by the approach used by me whereas I want to verify my approach.
My approach:
Suppose $N=111\cdots a\cdots111$ and $N\equiv 0\pmod {13}$
Now $N=10^{49}+10^{48}+\ldots+a10^{24}+\ldots+10+1=(10^{49}+10^{48}\ldots+10+1)+(a-1)10^{24}$
$N=\dfrac{10^{50}-1}{9}+(a-1)10^{24}$
Now $10^{12}\equiv 1\pmod {13}\Rightarrow 10^{24}\equiv 1\pmod {13}$ by fermat's little theorem.
Thus $(a-1)10^{24}\equiv (a-1) \pmod{13}\Rightarrow \dfrac{10^{50}-1}{9}\equiv 1-a\pmod{13}$ since $N\equiv 0\pmod{13}$
$10^{24}\equiv 1\pmod{13}\Rightarrow 10^{48}\equiv 1\pmod{13}$ or $10^{50}-1\equiv -5 \pmod{13}$
Now $10^{50}-1\equiv -5\pmod {13}\Rightarrow 9(1-a)\equiv -5\pmod{13}$
$a=3$ clearly satisfies the above conditions
$\therefore$ The $26^{th}$ digit from the left must be $3$.
Please suggest what is incorrect in this solution and advice for alternative solutions.
THANKS
| Another way is to use the trick from Wikipedia (that doesn't solve your solution)
Taking $N$ from the right, and applying the sequence $(1, −3, −4, −1, 3, 4)$ as instructed on the page (multiply the digits from the right by the given numbers in sequence), we get
$0$ for the 6 first digits from the right ($1-3-4-1+3+4=0$), repeating the sequence, $0$ up to digit 24 (from right), we still have $0$
Then, the next $6$ are our $a$ and $5\times 1$, or
$$a-3-4-1+3+4\\=a-1$$
We did $30$ digits, $20$ to go. The next $18$ will give $0$, the last $2$ give $1-3$, thus the whole sum is $$a-1-2=a-3$$
The only digit that would have $a-3\equiv 0\pmod {13}$ is $$\bbox[5px,border:2px solid #ba9]{a=3}$$
| {
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Sequence and polynomial power relationship I recently encountered this relationship between polynomial powers and a certain associated sequence and I am seeking any help or idea that might answer why the relationship is true.
Let $P(x)$ be a polynomial, say for instance $P(x)=1+3x+2x^2$. Consider the consecutive powers of $P(x)$ and arrange the numerical coefficients in order of appearance. For the given $P(x)$ we have for up to fifth power:
$P(x)^1=1+3x+2x^2$
$P(x)^2=1 + 6 x + 13 x^2 + 12 x^3 + 4 x^4$
$P(x)^3=1 + 9 x + 33 x^2 + 63 x^3 + 66 x^4 + 36 x^5 + 8 x^6$
$P(x)^4=1 + 12 x + 62 x^2 + 180 x^3 + 321 x^4 + 360 x^5 + 248 x^6 + 96 x^7 +
16 x^8$
$P(x)^5=1 + 15 x + 100 x^2 + 390 x^3 + 985 x^4 + 1683 x^5 + 1970 x^6 +
1560 x^7 + 800 x^8 + 240 x^9 + 32 x^{10}$.
Also, consider the sequence defined by
$a_{m,n}=a_{(m-1),n}+3a_{(m-1),(n-1)}+2a_{(m-1),(n-2)}$ with $a_{1,1}=1,a_{1,2}=3$ and $a_{1,3}=2$.
Observe that the sequence above completely determines the entries in the expansion of the polynomial power. For instance the number $63$ in the third power of $P(x)$ is equal to $63=12+3(13)+2(6)$.
I am wondering why is this TRUE. Thanks for your help and suggestion.
| In the product $$(1 + 6 x + 13 x^2 + 12 x^3 + 4 x^4)(2x^2+3x+1)$$ if you try to compute the coefficient of $x^3$, note that $x^3$ can be formed in the following ways
$12x^3$ from the first bracket, $1$ from the second bracket
$13x^2$ from the first bracket, $3x$ from the second bracket
$6x$ from the first bracket, $2x^2$ from the second bracket
that is, you can form the power $3$ of $x$, in the following ways $$3+0,\ 2+1,\ 1+2$$ where the first term in these sums is the power of $x$ you are taking from the first bracket and the second term is that you are taking from the second bracket.
Hence, the coefficient of $\texttt{x^3}$ in $P(x)^3$ is
$(\text{coeff. of x^2 in P(x)}=2)\times\text{coeff. of x in }P(x)^2$
$+(\text{coeff. of x^1 in P(x)}=3)\times\text{coeff. of x^2 in }P(x)^2$
$+(\text{coeff. of x^0 in P(x)}=1)\times\text{coeff. of x^3 in }P(x)^2$
$= 1\cdot a_{2,n} + 3\cdot a_{2,n-1} + 2\cdot a_{2,n-2}$
Basically, $a_{m,n}$ is the coefficient of $x^{n}$ in $P(x)^m$.
So, what your recurrence relation is doing is framing the distributive law of multiplication over addition in an interesting way. (This representation will be helpful to you in understanding how generating functions aid counting in combinatorics.)
| {
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Evaluating $I=\int\frac{\sin(x)+\cos(x)}{\sin^4(x)+\cos^2(x)}~dx$ I'm interested in the following problem:
Evaluate the indefinite integral$$I=\int\frac{\sin(x)+\cos(x)}{\sin^4(x)+\cos^2(x)}~dx$$
Here's what I did:
We note that
$$I = \int\frac{\sin(x)}{\sin^4(x)+\cos^2(x)}~dx+\int\frac{\cos(x)}{\sin^4(x)+\cos^2(x)}~dx$$Now we evaluate each part separately. Let $\cos(x)=u$ and thus,
$$I_1:=\int\frac{\sin(x)}{\sin^4(x)+\cos^2(x)}~dx=\int\frac{-du}{(1-u^2)^2+u^2}=-\int\frac{du}{1+u^4-u^2}$$For the second part, we substitute $\sin(x)=v$ and so,
$$I_2:=\int\frac{\cos(x)}{\sin^4(x)+\cos^2(x)}~dx = \int\frac{dv}{v^4+1-v^2}=\int\frac{du}{u^4+1-u^2}$$
Now as, $I=I_1+I_2$, we conclude
$$I=\int 0~ du = \text{Constant}$$However, we get a different answer by WolframAlpha.
So my question where am I wrong in above process?
Any help will be highly appreciated.
| You can factorize
$$
u^4-u^2+1=(u^2+\sqrt{3} u+1)(u^2-\sqrt{3} u+1),
$$
so that the partial fraction decomposition is
$$
\frac{1}{u^4-u^2+1}=\frac{1}{4 \sqrt{3}}\left[\frac{(2 u+\sqrt{3})+\sqrt{3}}{u^2+\sqrt{3} u+1}-\frac{(2 u-\sqrt{3})-\sqrt{3}}{u^2-\sqrt{3} u+1}\right],
$$
and we have
$$
\int\frac{1}{u^4-u^2+1}du=\frac{1}{4 \sqrt{3}}\log \left(\frac{u^2+\sqrt{3} u+1}{u^2-\sqrt{3} u+1}\right)+\frac{1}{2} \left[\arctan\left(2 u+\sqrt{3}\right)+\arctan\left(2u-\sqrt{3}\right)\right]+C.
$$
The two integrals become
$$
-\frac{1}{4 \sqrt{3}}\log \left(\frac{\cos ^2x+\sqrt{3} \cos
x+1}{\cos ^2x-\sqrt{3} \cos x+1}\right)-\frac{1}{2} \left[\arctan \left(2 \cos x+\sqrt{3}\right)+\arctan\left(2 \cos x-\sqrt{3}\right)\right]+C,
$$
and
$$
+\frac{1}{4 \sqrt{3}}\log \left(\frac{\sin ^2x+\sqrt{3} \sin x+1}{\sin ^2x-\sqrt{3} \sin x+1}\right)+\frac{1}{2} \left[\arctan\left(2 \sin
x+\sqrt{3}\right)+\arctan\left(2 \sin x-\sqrt{3}\right)\right]+C.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the power series of $\frac{3x+4}{x+1}$ around $x=1$. I'm trying to find the power series of
$$
\frac{3x+4}{x+1}
$$
around $x=1$.
My idea was to use the equation
$$
\left(\sum_{n\ge0}a_n (x-x_0)^n\right)\left(\sum_{k\ge0}b_k (x-x_0)^k\right) = \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-x_0)^n \tag{1}
$$
taken from here, to solve this problem.
My attempt
I can write the numerator of the expression as
$$
3x +4 = 3(x-1)+7 = \sum_{n\ge0}a_n (x-1)^n
$$
where
$$
a_n=\begin{cases}
7 & n=0 \\
3 & n=1 \\
0 & n>1
\end{cases}
$$
On the other hand, given $|x-1| <2$ we get
\begin{align*}
\frac{1}{x+1} = \frac{1}{2} \cdot \frac{1}{1-\left(\frac{1-x}{2}\right)} = \frac{1}{2} \sum_{k\ge0} \left(\frac{1-x}{2}\right)^n = \sum_{k\ge0}\underbrace{\frac{-1}{(-2)^{k+1}}}_{\color{blue}{b_k}} (x-1)^k
\end{align*}
And then, using equation $(1)$ we get
\begin{align*}
\frac{3x+4}{x+1} &= \left(\sum_{n\ge0}a_n (x-1)^n\right)\left(\sum_{k\ge0}b_k (x-1)^k\right)\\
&= \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-1)^n\\
&= \sum_{n\ge0}\left(a_0 b_n + a_1 b_{n-1} \right)(x-1)^n\\
&= \sum_{n\ge0}\left(7 \frac{-1}{(-2)^{n+1}} + 3 \frac{-1}{(-2)^{n}} \right)(x-1)^n\\
&= \sum_{n\ge0}\left( \frac{-7 +3(-1)(-2)}{(-2)^{n+1}} \right)(x-1)^n\\
&= \sum_{n\ge0} \frac{-1}{(-2)^{n+1}} (x-1)^n\\
\end{align*}
And this seems to imply that $\frac{3x+4}{x+1} = \frac{1}{x+1}$, at least for $|x-1| <2$, which is clearly not true.
I've gone over the steps, but I don't see where my mistake is. Could anyone tell me where my solution went wrong? Thank you!
| Hint: $f(x)=\frac{3x+4}{x+1}=3+\frac{1}{x+1}$ and from here we obtain $f^{(n)}(x)=\frac{(-1)^n n!}{(x+1)^{n+1}}, n \geqslant 1.$
Now, knowing derivative, Taylor series will be .. can you finish?
Addition:
It should be
$$\frac{3x+4}{x+1}=\frac{7}{2}-\frac{1}{4}(x-1)+ \cdots$$
and
$$\frac{1}{x+1}=\frac{1}{2}-\frac{1}{4}(x-1)+ \cdots$$
And about mistake - it is 3-d line of series multiplication. There should be
$$a_0b_0+(a_0b_1+b_0a_1)(x-1)+\cdots$$
| {
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Is there a way to solve the equation $\sin x = x\ln x$ analytically?
Is there a way to solve the equation $\sin x = x\ln x$ numerically or analytically?
The only way I have been able to solve this is using a graphic calculator like Desmos, but is there another way to solve this?
| This is a transcendental equation; this means no hope for a closed form solution and numerical methods are reaquired. So, considering that you looks for the zero's of function $$f=\sin (x)- x\log(x)$$ for which $x=0$ is a trivial solution. By inspection, the solution is between $1$ and $2$. Being lazy (myself), make a Taylor expansion around $x=\frac \pi 2$; this would give
$$f=\left(1-\frac{1}{2} \pi \log \left(\frac{\pi }{2}\right)\right)+\left(x-\frac{\pi
}{2}\right) \left(\log \left(\frac{2}{\pi }\right)-1\right)-\frac{(2+\pi )
\left(x-\frac{\pi }{2}\right)^2}{2 \pi }+\frac{2 \left(x-\frac{\pi
}{2}\right)^3}{3 \pi ^2}+\left(\frac{1}{24}-\frac{2}{3 \pi ^3}\right)
\left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^5\right)$$ Now, using series reversion
$$x=\frac{\pi }{2}-\frac{f-1+\frac{1}{2} \pi \log \left(\frac{\pi }{2}\right)}{1+\log
\left(\frac{\pi }{2}\right)}-\frac{(2+\pi ) \left(f-1+\frac{1}{2} \pi \log
\left(\frac{\pi }{2}\right)\right)^2}{2 \pi \left(1+\log \left(\frac{\pi
}{2}\right)\right)^3 }+O\left(\left(f-1+\frac{1}{2} \pi \log \left(\frac{\pi
}{2}\right)\right)^3\right)$$
Making $f=0$ leads to $x=1.74843$ while the "exact" solution (obtained by Newton method) is $1.75268$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all real solutions $x$ for the equation $x^{1/2} − (2−2x)^{1/2} = 1$ This is what the answer says:
Note that the equation can be rewritten as $\sqrt{x} − \sqrt{2 − 2x} = 1$,
and the existence of such real $x$ implies that $x$ is larger than or equal to $0$ and $x$ is less than or equal to $1$, since we implicitly assume that $\sqrt{x}$ and $\sqrt{2 − 2x}$ are real as well. Then
$\sqrt{x} − \sqrt{2 − 2x} < \sqrt{1} = 1 \text{ if } x < 1$, and
$\sqrt{x} − \sqrt{2 − 2x} = 1 \text{ if } x = 1$.
Thus, $x = 1$ is the only real solutions of the equation $\sqrt{x} − \sqrt{2 − 2x} = 1$.
I don't understand how
$\sqrt{x} − \sqrt{2 − 2x} < \sqrt{1} = 1 \text{ if } x < 1$.
I understand everything else.
Can someone please explain? Thank you.
| Notice that $\sqrt{x}-\sqrt{2-2x} \leq \sqrt{x}$. Then if $x<1$ we have $\sqrt{x}<\sqrt{1}=1$. It follows that $\sqrt{x}-\sqrt{2-2x} <1$ for $x<1$. So if $x<1$ it is not a solution. If $x=1$ we have $\sqrt{x}-\sqrt{2-2x}=\sqrt{1}-\sqrt{2-2}=1 $. So $x=1$ is a solution
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving infinite nested square roots of 2 converging to finite nested radical Can anyone explain to solve the identity posted by my friend $$2\cos12°= \sqrt{2+{\sqrt{2+\sqrt{2-\sqrt{2-...}}} }}$$ which is an infinite nested square roots of 2. (Pattern $++--$ repeating infinitely)
Converging to finite nested radical of $2\cos12° = \frac{1}{2}\times\sqrt{9+\sqrt5+\sqrt{(30-6\sqrt5)}}$
The finite nested radical, I was able to derive $\cos12° = \cos(30-18)°$ as follows
$$\cos30°\cdot\cos18° + \sin30°\cdot\sin18°$$
$$= \frac{√3}{2}\cdot\frac{\sqrt{2+2\cos36°}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2-2\cos36°}}{2}$$
Where $\cos18° = \frac{\sqrt{2+2\cos36°}}{2}$ (by Half angle cosine formula) and
$\sin18° = \frac{\sqrt{2-2\cos36°}}{2}$ (solving again by half angle cosine formula)
$2\cos36° =\frac{ \sqrt5 +1}{2}$ which is golden ratio
$\frac{\sqrt3}{2}\cdot\frac{\sqrt{10+2\sqrt5}}{4}+ \frac{1}{2}\cdot\frac{\sqrt{5}-1}{4} = \frac{\sqrt{30+6\sqrt5}}{8}+ \frac{\sqrt5-1}{8}$
Further steps finally lead to the finite nested radical
Method actually I tried to solve infinite nested square roots of 2 is as follows.
$2\cos\theta = \sqrt{2+2\cos2\theta}$ and
$2\sin\theta = \sqrt{2-2\cos2\theta}$
Now simplifying infinite nested square roots of 2, we will get the following as simplified nested radical
$$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2\cos12°}}}}$$
Simplifying step by step as follows
$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\sin6°}}}$ then
$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\cos84°}}}$ (by $\sin\theta = \cos(90-\theta)$
$2\cos12° = \sqrt{2+\sqrt{2+2\sin42°}}$
$2\cos12° = \sqrt{2+\sqrt{2+2\cos48°}}$
$2\cos12° = \sqrt{2+2\cos24°}$
$2\cos12° = 2\cos12°$
We are back to $\sqrt1$
Actually this is how I got stuck!
But for infinite nested square roots of 2(as depicted), if I run program in python I am able to get good approximation ( Perhaps if we run large number of nested square roots in python we get more number of digits matching the finite nested radical), because I'm not able get anywhere solving such a kind of infinite cyclic nested square roots of 2.
Dear friends, is there anyway to find the solution by any other means like solving infinite nested square roots
Thanks in advance.
| Somehow I got the answer from my subsequent post (after a long homework for cyclic infinite nested square roots of 2)
Sivakumar Krishnamoorthi (https://math.stackexchange.com/users/686991/sivakumar-krishnamoorthi), Solving cyclic infinite nested square roots of 2 as cosine functions, URL (version: 2020-09-26): https://math.stackexchange.com/q/3841605
$2\cos48°$ or $2\cos\frac{4}{15}π$ is cyclic infinite nested square roots of 2 of form $cin\sqrt2[2-2+]$ i.e. $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+...}}}}$
According to half angle cosine formula within next 2 steps we get $2\cos12°$ or $2\cos\frac{π}{9}$ as $cin\sqrt2[2+2-]$ i.e. cyclic infinite nested square roots of 2 as $\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-...}}}}$
Therefore $2\cos12°$ as a multiple of 3 it can be solved as nested radicals as in LHS of my question and on RHS it is cyclic infinite nested square roots of 2 which is also equivalent. As both the result belong to same $\cos \frac{π}{15}$ cyclic infinite nested square roots of 2 on RHS and finite nested radical are the same in terms of results.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ \lim_{x \to 0}{\frac{\sin( \pi \cos x)}{x \sin x} }$ I have to evaluate the following limit
$$ \lim_{x \to 0}{\frac{\sin( \pi \cos x)}{x \sin x} }$$
My solution is:
$$ \lim_{x \to 0}{\frac{\sin( \pi \cos x)}{x \sin x} }=\lim_{x \to 0}{\frac{\pi \cos x}{x \cdot x} }=+\infty$$
But the correct result is $\frac{\pi}{2}$. I can't understand where I'm making mistakes.
| By some elementary inequalities, near $x=0$ we have
$$
\cos(x)=1-x^2/2+O(x^4)$$In particular, near zero we have
$$
\frac{\sin(\pi\cos(x))}{x\sin(x)} = \frac{\sin(\pi(1-x^2/2)+ O(x^4))}{x\sin(x)}
$$Use the sine-reflection and angle-addition:
$$
\frac{\sin(\pi(1-x^2/2)+O(x^4))}{x\sin(x)} = \frac{\sin(\pi/2 \cdot x^2+O(x^4))}{x\sin(x)}
$$
$$
= \frac{\sin(\pi/2\cdot x^2)\cos(O(x^4))}{x\sin(x)}+ \frac{\cos(\pi/2 \cdot x^2)\sin(O(x^4))}{x\sin(x)}
$$Using $\lim_{\theta\to 0}\frac{\sin(\theta)}{\theta}=1$, the second term vanishes in the limit, and $\cos(O(x^4))\to 1$. So we have
$$
\lim_{x\to 0}\frac{\sin(\pi/2 \cdot x^2)\cdot 1}{x\sin(x)} = \lim_{x\to 0}\frac{\sin(\pi/2 \cdot x^2)}{x^2}\cdot \frac{x}{\sin(x)} =\frac{\pi}{2}\cdot 1 =\frac{\pi}{2}
$$
| {
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Proving $\sum_{i=1}^n (1-\frac{1}{(i+1)^2}) = \frac{n+2}{2n+2}$ using induction. My textbook has the following question:
Prove the follwing statement using induction for all natural numbers $n$
$$(1- \frac{1}{4})+(1- \frac{1}{9})+.......+(1- \frac{1}{(n+1)^2})=\frac{n+2}{2n+2}$$
So, I check both the sides for $n=1$. In that case LHS=RHS=$\frac{3}{4}$.
Now I assume the statement to be true for $n=k$ which gives
$$
\sum_{i=1}^k (1-\frac{1}{(i+1)^2}) = \frac{k+2}{2k+2}
$$
Now I evaluate the original statement for $n=k+1$ which leaves us with the
$$ LHS= \frac{3k^3+16k^2+26k+14}{2(k+1)(k+2)^2} $$
And we have to prove this LHS to be equal to RHS which is
$$
RHS= \frac{k+3}{2k+4}
$$
But these (new) LHS and RHS don't seem to be equal. And hence I am not able to complete the proof.
How should I proceed?
A solution without induction is also welcome.
Book: Comprehensive Algebra VOL-1
Author: Vinay Kumar
Publisher: McGraw Hill Education.
| So the claim is
$$\sum_{i=1}^{n} (1-\frac{1}{(i+1)^2}) = \frac{n+2}{2n+2}$$
Base case is true, now assuming that it is true for all naturals less than $n+1$ we will show that it is true for $n+1$.
$$\sum_{i=1}^{n+1} (1-\frac{1}{(i+1)^2}) = \frac{n+3}{2n+4}$$
is to be proved
Instead of working with summation,we will estimate difference.
Obviosuly
$$\frac{n+3}{2n+4}- \frac{n+2}{2n+2} = 1-\frac{1}{(n+2)^2}$$
if the formula is correct (and also, using inductive hypothesis)
But
$$\frac{n+3}{2n+4}- \frac{n+2}{2n+2} = \frac{1}{2} \left( \frac{(n+3)(n+1)-(n+2)^2}{(n+2)(n+1)} \right) = $$
$$ = -\frac{1}{2(n+1)(n+2)}$$
which is obviously wrong
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rate of convergence for a sequence (Preferably without Taylor series) I am trying to solve the following problem:
Knowing that the sequence $(a_{n})$ with:
$$a_{n+1}=\frac{1}{2}(a_{n}+\frac{3}{a_{n}})$$
converges to $\sqrt{3}$, find it's rate of convergence.
After doing some searching, I found this formula from wikipedia:
$$\lim\limits_{n \to \infty} \frac{|a_{n+1}-L|}{|a_{n}-L|} = μ$$
And I think that our L is $\sqrt{3}$.
Do I need to find the value of $a_{n}$ to find the rate of convergence (μ)? And how do I find $a_{n}$ ?
UPDATE: I can simply use the formula above but I need to make my limit approach to $\sqrt{3}$ because we have $a_{n} \to \sqrt{3}$:
$$\lim\limits_{x \to \sqrt{3}} \frac{|\frac{1}{2}(x+\frac{3}{x})-\sqrt{3}|}{|x-\sqrt{3}|}$$
But my problem is that this limit is resulting in a indeterminate form because of $\frac{0}{0}$
How can I solve this limit without expanding series?
UPDATE 2 - ANSWER:
Using @user's approach we can write our limit as:
$$\lim\limits_{x \to \sqrt{3}} \frac{\frac{1}{2}(x+\frac{3}{x})-\sqrt{3}}{x-\sqrt{3}}=\frac{x^2-2\sqrt 3x+3}{2x(x-\sqrt{3})}=\frac{(x-\sqrt{3})^2}{2x(x-\sqrt{3})}=\frac{x-\sqrt{3}}{2x}\to 0$$
and then the sequence converges Q-superlinearly to $\sqrt 3$. Look at here.
| @ClaudeLeibovici notes this iteration is by the Newton-Raphson method; so, under mild conditions (which apply here), the convergence is quadratic (i.e. the order of convergence is $2$) so $\mu=0$. @user's work makes this easy to check. With $x:=\tfrac{a-\sqrt{3}}{a+\sqrt{3}}$ we have$$a_n-\sqrt{3}=2\sqrt{3}x^{2^n}\underbrace{\frac{1}{1-x^{2^n}}}_{\sim1}\implies\frac{a_{n+1}-\sqrt{3}}{(a_n-\sqrt{3})^2}\sim\frac{1}{2\sqrt{3}},$$where we've used the convergence requirement $\lim_{n\to\infty}x^{2^n}=0$.
To address the update, note that$$\begin{align}\lim{y\to\sqrt{3}}\frac{(y+3/y)/2-\sqrt{3}}{y-\sqrt{3}}&=\lim{z\to0}\frac{z-\sqrt{3}+3/(z+\sqrt{3})}{2z}\\&=\lim{z\to0}\frac{z}{2(z+\sqrt{3})}\\&=0.\end{align}$$Again, we can prove something stronger with a squared denominator:$$\lim{z\to0}\frac{z-\sqrt{3}+3/(z+\sqrt{3})}{2z^2}=\lim{z\to0}\frac{1}{2(z+\sqrt{3})}=\frac{1}{2\sqrt{3}},$$as in the above calculations.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $A,B,C$ are collinear, prove $\vec{A}\times \vec{B} + \vec{B}\times \vec{C} + \vec{C}\times \vec{A} =\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$ Prove that if $A, B$ and $C$ are collinear, then
$\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A} =\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$.
So far, I know that $\vec{AB}, \vec{BC}, \vec{AC}$ are scalar multiples and their cross product is zero. I'm not sure how to apply this to prove $\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A}$ though.
| Since $A$, $B$ and $C$ are collinear, you can write $C$ as $A+\lambda(B-A)\require{cancel}$. So\begin{align}A\times B+B\times C+C\times A&=A\times B+B\times(A+\lambda(B-A))+(A+\lambda(B-A))\times A\\&=\cancel{A\times B+B\times A}+\lambda B\times(B-A)+\cancel{A\times A}+\lambda(B-A)\times A\\&=-\lambda B\times A+\lambda B\times A\\&=0.\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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In triangle ABC a point X is taken on AC and a point Y is taken on BC if AY and BX meet at O This question is from pre collage mathematics.
The question goes on like this:
In $\triangle ABC$ a point $X$ is taken on $\overline{AC}$ and a point $Y$ is taken on $\overline{BC}$. If $\overline{AY}$ and $\overline{BX}$ meet at $O$, find the area of $\triangle CXY$ if the areas of triangles $OXA$, $OAB$ and $OBY$ are $x,y,z$ respectively.
My answer is:
$$\frac{xz(x+y)(y+z)}{y(y^2-zx)}$$
| $\frac{A(ABY)}{A(ACY)}=\frac{A(BXY)}{A(CXY)}=\frac{BY}{CY}$
$\frac{y+z}{x+A(OXY)+A(CXY)}=\frac{z+A(OXY)}{A(CXY)}$
$A(CXY)\times(y+z)=(x+A(OXY)+A(CXY))\times(z+A(OXY))$
$y\times A(CXY)=xz+x\times A(OXY)+z\times A(OXY)+(A(OXY))^2+A(CXY)\times A(OXY)$
$A(CXY)=\frac{xz+x\times A(OXY)+z\times A(OXY)+(A(OXY))^2}{y- A(OXY)}=\frac{xz+\frac{x^2z}{y}+\frac{xz^2}{y}+\frac{x^2z^2}{y^2}}{y-\frac{xz}{y}}=\frac{y^2+xy+yz+xz}{y(y^2-xz)}=\frac{xz(x+y)(y+z)}{y(y^2-xz)}$
and using this
$\frac{A(ABO)}{A(AXO)}=\frac{A(OBY)}{A(OXY)}=\frac{BO}{XO}$
that is
$\frac{y}{x}=\frac{z}{A(OXY)}$
then
${A(OXY)}=\frac{xz}{y}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $x^5-x^3+x-2=0$, find $\lfloor x^6\rfloor$.
If $\alpha$ is a real root of the equation $x^5-x^3+x-2=0$, find the value of $\lfloor\alpha^6\rfloor$, where $\lfloor x\rfloor$ is the least positive integer not exceeding $x$.
My approach is to bound the value of $\alpha^6=\alpha^4-\alpha^2+2\alpha$.
First I proved the function $f(x)=x^5-x^3+x-2$ is monotone increasing by derivative. Then I argue that $1<\alpha<\frac32$ as $f(1)<0$ and $f(\frac32)>0$.
Then I tried to create an upper and lower bound for $\alpha$, as such
$$\alpha^6=\alpha^4-\alpha^2+2\alpha<\frac94\alpha^2-\alpha^2+2\alpha=\frac54\alpha^2+2\alpha<\frac{45}{16}+3=\frac{93}{16}$$
and
$$\alpha^6=\alpha^4-\alpha^2+2\alpha>\alpha^3-\alpha^2+2\alpha=\alpha^2(\alpha-1)+2\alpha>2$$
Now we know that $\lfloor\alpha\rfloor\in\{2,3,4,5\}$. But I cannot proceed any further.
Any idea, help, hint, or answer is appreciated. Thank you.
| $$x^5-x^3+x-2=x^5+x^2-(x^2-x+1)-(x^3+1)=$$
$$=(x^2-x+1)(x^3+x^2-x-2).$$
Now, we see that our equation has an unique real root $1<\alpha<2$ and from here $$[\alpha]=1.$$
Now, about your new problem.
Easy to see that for our root $\alpha$ we have $$1.205<\alpha<1.206,$$
which gives $$3<\alpha^6<4$$ and from here: $$[\alpha^6]=3.$$
| {
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"url": "https://math.stackexchange.com/questions/3832120",
"timestamp": "2023-03-29T00:00:00",
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Positive integers $(a, b, c)$ are a primitive Pythagorean triple
Show that if $a = m^2 - n^2$ , $b = 2mn$, $c = m^2 + n^2$ , where $m$, $n$ are relatively prime, not both odd, and $m>n$, then $(a, b, c)$ is a primitive Pythagorean triple.
This is part one of a proof I am required to do.
I know that if $m$ and $n$ are not both odd, then they can be written as $2k+1$ and $2l$, respectively, or as $2k$ and $2l$. I plugged in the given values for a, b, and c into the equation $a^2 + b^2 = c^2$ and got $m^4 + 2m^2n^2 + n^4$, but this is as far as I can get, however. I know that to show $a,b,c$ are primitive I need to show their GCD is $1$, but I don't know how to do this. Can someone show me where to start?
| Line up your ducks. And then shoot them.
Does
$(m^2 - n^2)^2 + (2mn)^2 {? \over=} (m^2+n^2)^2$
$m^4 - 2m^2n^2 + n^4 + 4m^2n^2 {? \over=} m^4 + 2m^2n^2 + n^4$
$m^4 + 2m^2n^2 {? \over=} m^4 + 2m^2n^2 + n^4$?
The answer is... yes, it does.
So $m^2-n^2, 2mn, m^2 + n^2$ are a pythogorean triple.
====
But are the a primative triplet? That is:
And are $m^2 - n^2$ and $2mn$ relatively prime if $m,n$ are and they are not both odd?
If $p$ is a prime divisor that divides $2mn$ then either
*
*$p|2$ so $p=2$.
But $m,n$ are relatively prime so they are not both even and they are not both odd so $m^2 -n^2$ is odd and so $p\not \mid m^2 - n^2$.
*$p|m$
But $m,n$ are relatively prime $p\not \mid n$. So $p|m^2$ but not $n^2$ so $p \not \mid m^2 -n^2$.
*$p|n$
Same argument. $p\not \mid m$ so $p|n^2$ but not $m^2$ and therefor $p\not \mid m^2 - n^2$.
so no prime factor of $2mn$ is a facctor of $m^2 - n^2$ so $m^2-n^2$ and $2mn$ are relatively prime.
So $m^2-n^2, 2mn, m^2+n^2$ is a primitive pythagorean triplet.
| {
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"timestamp": "2023-03-29T00:00:00",
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To Prove $\frac{1}{b}+\frac{1}{c}+\frac{1}{a} > \sqrt{a}+\sqrt{b}+\sqrt{c}$ The three sides of a triangle are $a,b,c$, the area of the triangle is $0.25$, the radius of the circumcircle is $1$.
Prove that $1/b+1/c+1/a > \sqrt{a}+\sqrt{b}+\sqrt{c}$.
what I've tried:
$$\frac{1}{4} = \frac{1}{2}ab\sin C \Rightarrow ab=\frac{1}{2}\sin C \\c=2\sin C \Rightarrow \frac{1}{c}=\frac{1}{2}*\sin C $$
so, $$\frac{1}{c}=ab \Rightarrow abc=1 \Rightarrow \sqrt{abc}=1$$
now the problem becomes
$$ab+bc+ac > \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}},$$
with $0<a\leq b\leq c\leq 2$, and $a+b>c$.
But even so I don't know how to prove it.
Any help or hint is appreciated. Thank you.:)
| The following relation:
$$(ab+bc+ca)(\frac{1}{b}+\frac{1}{c}+\frac{1}{a})\geq(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$$ is easily obtained by C-S inequality.
Since you arleady know $abc=1$ , you can easily notice that $$ab+bc+ca=\frac{1}{b}+\frac{1}{c}+\frac{1}{a}$$
and your question becomes much easier at this point.
What you should be careful about is that when $ab+bc+ca=\sqrt{a}+\sqrt{b}+\sqrt{c}$ , you get $a=b=c=1$ , which apparently doesn't fit the condition in the original question.
Therefore,
$$ab+bc+ca=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>\sqrt{a}+\sqrt{b}+\sqrt{c}$$
is proven.
| {
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"timestamp": "2023-03-29T00:00:00",
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Characteristic polynomials for a class of simple matrices Let $M_{k}$ be the kxk-matrix with entries $m(i,j,k)=0$ for $i+j<k-1$ and $m(i,j,k)=1$ else.
For example $M_{3}=\begin{bmatrix}0 & 0 & 1\\0&1&1\\ 1&1&1\end{bmatrix}.$
Let $F_{n}(x)$ denote the Fibonacci polynomials defined by $F_{n}(x)= F_{n-1}(x)+x F_{n-2}(x)$ with initial values $ F_{0}(x)= F_{1}(x)=1.$
Computations suggest that the characteristic polynomial of $M_{k}^2$ is $F_{2k}(-x)=\sum_{j=0}^k\binom{2k-j}{j}(-x)^k.$
Is there a simple proof of this result?
| The squares $N_k = M_k^2$ are very nice matrices: we have (indexing from $1$ to $k$)
$$N_k(i, j) = \text{min}(i, j)$$
hence, for example,
$$N_4 = \left[ \begin{array}{cc} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \end{array} \right].$$
Define the polynomials $P_k(x) = \det (N_k - xI)$, and (it turns out we will need this) also define $Q_k(x)$ to be the determinant of $N_k - \text{diag}(x, x, \dots 0)$. Subtracting the second-to-last-row from the last row of $N_k - xI$ gives last row $(0, 0, \dots x, 1-x)$, and Laplace expanding along this row gives
$$P_k(x) = (1 - x) P_{k-1}(x) - x Q_{k-1}(x).$$
The same row operation for $N_k - \text{diag}(x, x, \dots 0)$ gives last row $(0, 0, \dots x, 1)$, and Laplace expanding along this row gives
$$Q_k(x) = P_{k-1}(x) - x Q_{k-1}(x).$$
This gives
$$\left[ \begin{array}{c} P_k(x) \\ Q_k(x) \end{array} \right] = \left[ \begin{array}{cc} 1-x & -x \\ 1 & -x \end{array} \right] \left[ \begin{array}{c} P_{k-1}(x) \\ Q_{k-1}(x) \end{array} \right]$$
which, together with the initial conditions $P_0(x) = 1, Q_0(x) = 0$ (which you can check is consistent with $P_1(x) = 1-x, Q_1(x) = 1$), gives
$$\left[ \begin{array}{c} P_k(x) \\ Q_k(x) \end{array} \right] = \left[ \begin{array}{cc} 1-x & -x \\ 1 & -x \end{array} \right]^n \left[ \begin{array}{cc} 1 \\ 0 \end{array} \right].$$
So $P_k(x)$ satisfies a linear recurrence relation with characteristic polynomial the characteristic polynomial of $\left[ \begin{array}{cc} 1-x & -x \\ 1 & -x \end{array} \right]$, which is
$$\lambda^2 - (1-2x) \lambda + x^2.$$
This gives $P_0(x) = 1, P_1(x) = 1 - x$, and
$$P_k(x) = (1 - 2x) P_{k-1}(x) - x^2 P_{k-2}(x).$$
This matches up to your Fibonacci polynomials by induction, although note that Wikipedia uses different initial conditions and puts the $x$ in a different place than you and I don't know which convention is standard. To see this concretely write
$$F_{2k}(-x) = F_{2k-1}(-x) - x F_{2k-2}(-x)$$
$$F_{2k-1}(-x) = F_{2k-2}(-x) - x F_{2k-3}(-x)$$
$$F_{2k-2}(-x) = F_{2k-3}(-x) - x F_{2k-4}(-x)$$
Writing the third equation as $F_{2k-3}(-x) = F_{2k-2}(-x) + x F_{2k-4}(x)$ and substituting twice to remove the odd terms gives
$$F_{2k}(-x) = (1 - 2x) F_{2k-2}(-x) - x^2 F_{2k-4}(-x)$$
as desired.
More abstractly, $F_k(-x)$ satisfies a recurrence relation with characteristic polynomial $\lambda^2 - \lambda + x$ so it can be expressed as a sum of $k^{th}$ powers of the two roots of this polynomial. This means $F_{2k}(-x)$ satisfies a recurrence relation with characteristic polynomial the polynomial whose roots are the squares of $\lambda^2 - \lambda + x$. In general, if $\lambda^2 - b \lambda + c$ has roots $r, s$, the polynomial with roots $r^2, s^2$ is $\lambda^2 - (b^2 - 2c) \lambda + c^2$ and applying this transformation to $\lambda^2 - \lambda + x$ gives $\lambda^2 - (1 - 2x) \lambda + x^2$ as expected.
| {
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"timestamp": "2023-03-29T00:00:00",
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Angle between the curves without differentiation $y^2=4(x+1)$ and $x^2=4(y+1)$
Are the two curves, the answer online uses calculus and we have not been taught that.
can this be solved any other way?
Thanks in advance.
| Write the product of the equations of the parabolas around the intersection points $(2\pm 2\sqrt 2, 2\pm 2\sqrt2):$ $-16[(x-(2\pm 2\sqrt{2}))^2\pm \sqrt{2}(x-(2\pm 2\sqrt{2}))(y-(2\pm 2\sqrt{2}))+(y-(2\pm 2\sqrt{2}))^2]+$ higher order terms, and use the $\tan \theta = {2 \sqrt{h^2-ab} \over a+b},ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0:$ $$\tan \theta={2 \sqrt{(\sqrt{\pm 2})^2-1} \over 1+1}=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Integrate $\int_0^3 \int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{81-x^2-y^2}}(1) dzdydx $ How to integrate using cylindrical coordinates the following function?
$$\int_0^3 \int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{81-x^2-y^2}}(1) dzdydx $$
My problem is not with the function itself but with arranging the respective boundaries with the variables.
Is it correct $\int_0^3 \int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{81-x^2-y^2}}(1) dzdydx \Rightarrow \int_0^3 \int_0^{\sqrt{9-r^2cos^2\theta}}\int_0^{\sqrt{81-r^2}}(1) r dzd\theta dr $
When I got to the integration I got very confused, as I mentioned, with my arrangement of the boundaries and the change in area $dzd\theta dr $ since that in the second integral $\int_0^\sqrt{9-r^2cos^2\theta}$ I repeat the variable $r$. Thanks for the help.
| The boundary is a cylinder with a cap, that lies on the first octant. The projection on the $x$-$y$ plane is the positive part of $\frac14$ of a circle with radius $3$ (lying on the first quadrant). How do you find such area using polar coordinates?
$$A=\int_0^{3}\int_0^{\sqrt{9-x^{2}}}dxdy=\int_{0}^{\pi/2}\int_0^3rdrd\theta.$$
On the other hand, to find the upperbound of $z$, simply recall that $x^2+y^2=r^2$, so:
$$\begin{cases}
z=\sqrt{81-(x^2+y^2)}\\
x^2+y^2=r^2
\end{cases}
\implies z=\sqrt{81-r^2}$$
therefore, yielding
$$\begin{aligned}V&=\int_0^3\int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{81-x^2-y^2}}dzdydx=\int_{0}^{\pi/2}\int_0^3\int_0^{\sqrt{81-r^2}}rdzdrd\theta\\&=\frac{\pi}{2}\int_0^3r\sqrt{81-r^2}drd\theta=-\frac\pi2\frac{1}{3} \left(81-r^2\right)^{3/2}\Big|_0^3=\frac{9}{2} \left(27-16 \sqrt{2}\right) \pi
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$ z,w\in\mathbb{C},|z|=|w|=R\gt0$. Show that $\left(\frac{z+w}{R^2+zw}\right)^2+\left(\vcenter{\frac{z-w}{R^2-zw}}\right)^2\ge\frac1{R^2}$ Let $ z, w \in \mathbb{C} $ be such that $ |z| = |w| = R > 0 $. Show that
$ \left(\frac{z + w}{R^2 + zw}\right)^2 + \left(\frac{z - w}{R^2 - zw}\right)^2 \geq \frac{1}{R^2} $
Well, i could only proof that
$ u=\left(\frac{z + w}{R^2 + zw}\right) $ , $ v = \left(\frac{z - w}{R^2 - zw}\right) \in \mathbb{R} $
By showing that
$ u = \overline{u}$ and $ v = \overline{v}$
Which leads to
$ u^2 + v^2 = |u|^2 + |v|^2 $
But i can't find a way to compute $|u|$ or $|v|$
| Written as $\left(\frac{z+w}{|z||w|+zw}\right)^2+\left(\frac{z-w}{|z||w|-zw}\right)^2\ge\frac1{|z||w|}$, the inequality scales nicely, so we can assume wlog that $R=|z|=|w|=1$.
Set $z=e^{i\alpha}$ and $w=e^{i\beta}$. If $(zw)^2\ne1$, then $\cos^2(\alpha+\beta)\ne1$. Furthermore,
$$
\begin{align}
&\left(\frac{z+w}{1+zw}\right)^2+\left(\frac{z-w}{1-zw}\right)^2\\
&=2\,\frac{z^2+w^2+z^4w^2+z^2w^4-4z^2w^2}{1-2z^2w^2+z^4w^4}\tag1\\
&=2\,\frac{\bar{w}^2+\bar{z}^2+z^2+w^2-4}{\bar{z}^2\bar{w}^2-2+z^2w^2}\tag2\\
&=2\,\frac{2\cos(2\alpha)+2\cos(2\beta)-4}{2\cos(2\alpha+2\beta)-2}\tag3\\
&=2\,\frac{1-\cos(\alpha+\beta)\cos(\alpha-\beta)}{1-\cos^2(\alpha+\beta)}\tag4\\[6pt]
&\gt1\tag5
\end{align}
$$
Explanation:
$(1)$: expand and combine
$(2)$: multiply by $\frac{\bar{z}^2\bar{w}^2}{\bar{z}^2\bar{w}^2}$
$(3)$: $z^2+\bar{z}^2=2\cos(2\alpha)$, $w^2+\bar{w}^2=2\cos(2\beta)$
$\phantom{\text{(3):}}$ $z^2w^2+\bar{z}^2\bar{w}^2=2\cos(2\alpha+2\beta)$
$(4)$: $\cos(2\alpha)+\cos(2\beta)=2\cos(\alpha+\beta)\cos(\alpha-\beta)$
$\phantom{\text{(4):}}$ $\cos(2\alpha+2\beta)=2\cos^2(\alpha+\beta)-1$
$(5)$: apply $(6)$ below
Since $\frac{1-xy}{1-x^2}$ is monotonic in $y$ for each $x\in(-1,1)$, we have for $y\in[-1,1]$
$$
\frac12\lt\frac1{1+|x|}\le\frac{1-xy}{1-x^2}\le\frac1{1-|x|}\tag6
$$
| {
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Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$.
I tried to substitute some basic values like $-1,0,1$ and try to find the roots of the function but couldn't.
Then I graphed the function on desmos and this is the graph.
So from this, we can say that $x^{12}-x^9+x^4-x+1>0$ for all values of $x$.
But I want to know how to find the required values of $x$ without graphing
| Hint: Break it into cases:
If $x \ge 1$, then $x^{12} \ge x^9$ and $x^4 \ge x$.
If $x \le 0$, then $x^{12} \ge 0$ and $x^9 \le 0$ and $x^4 \ge 0$ and $x \le 0$.
If $0 < x < 1$, then $x^{12} > 0$ and $x^4 > x^9$ and $1 > x$.
Can you finish each of these cases?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\frac{dg}{dθ}$ at $(r,θ)=(2\sqrt{2},\frac{π}{4})$ where $g(x,y)=\frac1{x+y^2}$ using chain rule?
Okay so my first step is to find the partial derivatives:
$$\frac{\partial \:}{\partial \:x}\left(\frac{1}{x+y^2}\right)=-\frac{1}{\left(x+y^2\right)^2}$$
$$\frac{\partial \:}{\partial \:y}\left(\frac{1}{x+y^2}\right)=-\frac{2y}{\left(x+y^2\right)^2}$$
Then I multiply each partial derivative with its respective equal and add it together:
$$(fx)(32r\cos(\theta)) + (fy)(3r(\sin(\theta))$$
Then I sub in X and Y in the partial derivatives. And then plug in $r$ and $\theta$:
$$-\left(\frac{1}{\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:}{4}\right)+\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi }{4}\right)\right)^2\right)^2}\right)\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:}{4}\right)\right)+-\left(\frac{2\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)}{\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:\:}{4}\right)+\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)^2\right)^2}\right)\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)$$
And I show my answer in the screenshot above. But for some reason the software is telling me it's wrong. I've also tried not multiplying the partial derivatives by the $32rcos(\theta)$ thing but it's still giving me the wrong answer.
What am I doing wrong? Thank you.
| We have that
$$\frac{\partial g}{\partial \theta}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial \theta}=-\frac{1}{\left(x+y^2\right)^2}(-32 r \sin \theta)-\frac{2y}{\left(x+y^2\right)^2}(3r\cos \theta)=$$
with $x=64$ and $y=6$ then
$$=-\frac{1}{100^2}(-64)-\frac{12}{100^2}(6)=-\frac{8}{100^2}$$
| {
"language": "en",
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Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$ This question appeared in one of the national exams (MCQs) in Saudi Arabia.
In this exam;
*
*Using calculators is not allowed,
*The student have $72$ seconds on average to answer one question.
PROBLEM:
Compare $a=(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $b=(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$.
CHOICES:
A) $a>b$
B) $a<b$
C) $a=b$
D) Given information is not enough
Using algebra to evaluate each expression is easy, and the correct choice is $A$, but that will take a long time.
Any suggestion to solve this problem in a short time? THANKS.
| The sum of all four factors is the same in both cases. To maximize the product, we want the factors to be as close together as possible.
The factors $(2 + \frac12)(5 + \frac15)$ are closer to their average than $(2 + \frac15)(5 + \frac12)$, so $(2+\frac12)(5+\frac15) > (2+\frac15)(5+\frac12)$.
Similarly, $(3 + \frac13)(4+\frac14) > (3 + \frac14)(4 + \frac13)$.
We could also compare each of these pairs by multiplying them out. But we don't have to multiply out everything. When we expand $(2 + \frac12)(5 + \frac15)$ and $(2 + \frac15)(5 + \frac12)$, the terms $2\cdot 5$ and $\frac12 \cdot \frac15$ will be common between them. However, $2 \cdot \frac15 + \frac12 \cdot 5 > 2 \cdot \frac12 + \frac15 \cdot 5$, which is not hard to see: just $\frac12 \cdot 5$ is $2.5$ on the left, and the right is $2$.
The same thing happens with the other pair: $3 \cdot \frac14 + \frac13 \cdot 4 > 3 \cdot \frac13 + \frac14 \cdot 4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of hitting probabilities not equal to 1? I have a Markov Chain with transition matrix
\begin{equation}
P=
\begin{pmatrix}
0 & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4}\\
\frac{1}{5} & 0 & 0 & 0 & \frac{4}{5} \\
0 & 0 & 1 & 0 & 0\\
\frac{5}{6} & 0 & \frac{1}{6} & 0 & 0\\
0 & 0 & 0 & 0 & 1
\end{pmatrix}
\end{equation}
The states are labelled $0,1,2,3,4$. States $2$ and $4$ are absorbing. I am interested in calculating the probabilities that I reach state $2$ and $4$ respectively. Initially, I did the following.
$$$$
Define
\begin{equation}
f=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\left(\frac{1}{4}\frac{1}{5}\right)^{i}\left(\frac{1}{4}\frac{5}{6}\right)^{j}=\frac{1}{1-\frac{1}{20}}\frac{1}{1-\frac{5}{24}}=\frac{480}{360}
\end{equation}
$f$ is the sum of probabilities that the chain loops between states $0$ and $1$ $i$ times and $0$ and $3$ $j$ times for all possible values of $i$ and $j$. Then
\begin{equation}
P(2)=\frac{1}{4}f+\frac{1}{4}\frac{1}{6}f=\frac{140}{361}
\end{equation}
Where the first term is the probability of accessing state $2$ from state $0$ and the second term is the probability of accessing state $2$ from state $3$.
\begin{equation}
P(4)=\frac{1}{4}f+\frac{1}{4}\frac{4}{5}f=\frac{216}{361}
\end{equation}
Where the first term is the probability of accessing state $4$ from state $0$ and the second term is the probability of accessing state $4$ from state $1$.
The problem with this is that these 2 probabilities do not sum to 1. Does anyone know why?
| Nvm, I found the answer. I was not taking into account the ordering of the loops. eg looping between states $0$ and $1$ followed by $0$ and $3$ results in a different chain as compared to looping between $0$ and $3$ first then $0$ and $1$. $f$ should be changed as following
\begin{equation}
f=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\left(\frac{1}{20}\right)^{i}\left(\frac{5}{24}\right)^{j}\frac{(i+j)!}{i!j!}
\end{equation}
Which is a binomial expansion and can be simplified to
\begin{equation}
f=\sum_{n=0}^{\infty}\left(\frac{1}{20}+\frac{5}{24}\right)^{n}
\end{equation}
which can be evaluated as an infinite geometric sum.
| {
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Integrate $\int ((x^2-1)(x+1))^{-2/3} \, dx$ using $u = \tan^{-1}(x)$ The problem asks to solve $\int ((x^2-1)(x+1))^{-2/3} \, dx$ using the u-substitution $u = \tan^{-1}(x)$.
I was able to solve the integral using the $u$-substitution $v = \frac{x-1}{x+1}$ (see below); however, I was not able to make any progress when $u = \tan^{-1}(x)$ except for finding that $du = \frac{1}{1+x^2} dx$. Any hints would be greatly appreciated.
$$\int ((x^2-1)(x+1))^{-2/3} \, dx = \int \frac{1}{\Big(\frac{x-1}{x+1}\Big)^{2/3}(x+1)^2} \, dx = \frac{1}{2}\int {v}^{-2/3} \, dv = \frac{3}{2}\Big(\frac{x-1}{x+1}\Big)^{1/3}+c$$
| I think I have figured out a messy way to get the solution. Let $u = \tan^{-1}(x)$.
$\begin{align*}
\int \frac{1}{(x^2-1)^{2/3}(x+1)^{2/3}} \, dx &= \int \frac{1+\tan^2(u)}{(\tan^2(u)-1)^{2/3}(\tan(u)+1)^{2/3}} \, du \\
&= \int \frac{\sec^2(u)}{(\tan^2(u)-1)^{2/3}(\tan(u+\frac{\pi}{4})(1-\tan(u)))^{2/3}} \, du\\
&= \int \frac{1}{(\tan(u+\frac{\pi}{4}))^{2/3}(\cos(u)-\sin(u))^2\Big(\frac{\cos(u)+ \sin(u)}{\cos(u)-\sin(u)}\Big)^{2/3}} \, du\\
&= \int \frac{1}{(\cos(u)-\sin(u))^2(\tan(u+\frac{\pi}{4}))^{4/3}} \, du\\
&= \int \frac{\sec^2(u+\frac{\pi}{4})}{2(\tan(u+\frac{\pi}{4}))^{4/3}} \, du\\
&= -\frac{3}{2}(\tan(u+\frac{\pi}{4}))^{-1/3} + c \\
&= \frac{3}{2}\Big(\frac{x-1}{x+1}\Big)^{1/3}+c\\
\end{align*}$
Please let me know if you find an easier solution!
| {
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Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, $x < 0, a = 1$ The textbook answer is $\frac {1}{3}$, I went through all the steps, but couldn't interpret it.
Below were my steps.
Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, x < 0, a = 1
First I tried to find the inverse, rearrange the equation to $$x = y - \frac{2}{y} \implies x = \frac{y^{2}-2}{y}$$
Rearrange, $xy=y^{2}-2 \implies y^{2}-xy-2 = 0$
Then, I tried to use the equation $\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$, and I got:
$y = \frac{x \pm \sqrt{x^{2}+8}}{2}=f^{-1}$
Next, I tried to find $(f^{-1}){'}$, for both $+$ and $-$
$\frac{1}{2} \cdot (x' + ((x^{2}+8)^{\frac{1}{2}})')$
Apply chain rule:
$((x^{2}+8)^{\frac{1}{2}})'$ = $\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x$
Then do the same for -
And I got $(f^{-1})'$ = $\begin{cases} \frac{1}{2}\cdot(1+\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x) \\ \frac{1}{2}\cdot(1-\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x) \end{cases}$
When a = 1,
$(f^{-1})'=\begin{cases} \frac{\sqrt{9}+1}{2\sqrt{9}} = \frac{2}{3} \\ \frac{\sqrt{9}-1}{2\sqrt{9}} = \frac{1}{3} \end{cases}$
I need some help. It says when $x < 0$, but both of my answers are positive. How do I interpret it?
| The domain of $f$ is the interval $(-\infty,0)$.
Thus, $g:=f^{-1}$ should be such that its range is $(-\infty,0)$.
This way, one can see that
$$
g(s) = \frac{s-\sqrt{s^2+8}}{2}
$$
is the proper solution for the inverse of $f$,
(because we always have $g(s)<0$), but
$g(s) = \frac{s+\sqrt{s^2+8}}{2}$ would not be, because its values
can be positive (for example we have $g(0)>0$).
Now that the proper inverse function is known, we can see from your calculations
that $\frac13$ is the correct answer and $\frac23$ is wrong.
It says when $x < 0$, but both of my answers are positive. How do I interpret it?
While the output of the function $g=f^{-1}$ should be negative,
its derivative $g'=(f^{-1})'$ can be positive, because these are different objects.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a|(b+c)$ and $\gcd(b,c)=1$, prove that $\gcd(a,b)=1$ and $\gcd(a,c)=1$.
If $a|(b+c)$ and $\gcd(b,c)=1$, prove that $\gcd(a,b)=1$ and $\gcd(a,c)=1$.
I started with:
Suppose $a|(b+c)$ and $\gcd(b,c)=1$.
This means that $ak=b+c$, for some integer $k$.
And $1|b$ and $1|c$.
I know I can solve this using the theorem that if $\gcd(a,b)=1$ then there exists integers $u$ and $v$ such that $au+bv=1$.
But I was wondering how else I can solve this, without using this theorem.
| One thing I like to do is figure if $k$ is a common (positive) divisor of $a$ and $b$, and $a|b+c$ then $k|b+c$. But we also have $k|b$ so $k| (b+c)-b = c$. So $k$ is a common (positive) divisor of $b,c$. But $\gcd(b,c) = 1$ so the only common (positive) divisor of $b,c$ is one so $k=1$. So the only common divisor of $a$ and $b$ is $1$. So $\gcd(a,b)=1$.
And doing similar to prove $\gcd(a,c) =1$ is nearly entirely the same. If $d$ is a common divisor of $a$ and $c$ then $d|b+c$ and $(b+c)-c=b$ and so $d$ is a common divisor of $b$ and $c$ which are relatively prime.
....
Or maybe even easier: $a|b+c$. So $\gcd(a,b)|a$ so $\gcd(a,b)|b+c$ but $\gcd(a,b)|b$ so $\gcd(a,b)|(b+c)-b=c$. So $\gcd(a,b)|b$ and $\gcd(a,b)|c$ so $\gcd(a,b)|\gcd(b,c)=1$. So $\gcd(a,b) = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $f(x) = |x+1|-|x-1|$, find $f \circ f\circ f\circ f ... \circ f(x)$ (n times). Let $f(x) = |x+1|-|x-1|$, find $f \circ f\circ f\circ f ... \circ f(x)$ (n times). I don't know where to start... Should I use mathematical induction? But what should be my hypothesis? Should I calculate for n = 1, n = 2?
| First note
$$f(x) = |x+1| - |x-1|\\
= \begin{cases}
-2, & x< -1\\
2x, & x\in [-1, 1] \\
2, & x>1\\
\end{cases}$$
Proposition: $$\color{red}{f^{n+1}(x) = \left|2^nx+1\right| - \left| 2^nx-1\right|}$$.
Note: this is exactly equivalent to
$$f^{n+1}(x) = \begin{cases}
-2, & x< -\frac1{2^n}\\
2^{n+1}x, & x\in [-\frac1{2^n}, \frac1{2^n}] \\
2, & x>\frac1{2^n}\\
\end{cases}$$
Clearly true for $n=0$. Assuming it holds for $n$, the inductive step is
$$f^{n+2}x = f\circ f^{n+1} (x)= \begin{cases}
f(-2), & x < -\frac1{2^n}\\
f(2^{n+1}x), &x\in [-\frac1{2^n}, \frac1{2^n}] \\
f(2), & x>\frac1{2^n}\\
\end{cases}\\
=\begin{cases}
-2, & x < -\frac1{2^n}\\
-2, &x\in [-\frac1{2^n}, -\frac1{2^{n+1}}) \\
f(2^{n+1}x), &x\in [-\frac1{2^{n+1}}, \frac1{2^{n+1}}] \\
2, &x\in (\frac1{2^{n+1}}, \frac1{2^n}] \\
2, & x>\frac1{2^n}\\
\end{cases}$$
$$=\begin{cases}
-2, & x < -\frac1{2^{n+1}} \\
2^{n+2}x, &x\in [-\frac1{2^{n+1}}, \frac1{2^{n+1}}] \\
2, &x > \frac1{2^{n+1}}\\
\end{cases}\\
= |2^{n+1}x+1| - |2^{n+1}x-1|$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this integral $I = \int\dfrac{\cos^3x}{\sin x + \cos x}dx$? $\displaystyle\int\dfrac{\cos^3x}{\sin x + \cos x}dx$
I added $J =\displaystyle \int\dfrac{\sin^3x}{\sin x + \cos x}dx$
then $I + J = \displaystyle\int\dfrac{\cos^3x + \sin^3x}{\sin x + \cos x}dx = x + \dfrac{1}{2}\cos2x + C$
but I can't find how to solve $I-J$
And is that the true way to solve it?
Please help!
| $$I-J = \int \frac{ \cos^3 x - \sin^3 x}{ \sin x + \cos x} dx = \int \frac{ (\cos x - \sin x)( \cos^2 x + \sin^2 x + \sin x \cos x )}{ \sin x + \cos x} dx$$
Substitute $$ \sin x + \cos x = t$$
$$t^2 = 1 - 2 \sin x \cos x$$
Or,
$$ \sin x \cos x = \frac{1-t^2}{2}$$
$$ I-J= \int \frac{(1+ ( \frac{1-t^2}{2}))}{t} dt$$
Can you finish?
| {
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Stuck on Mathematical Induction Proof I have the following question: Prove with mathematical induction that $3^n+4^n\le 5^n$ for all $n\ge 2$.
$$\text{Assume true: }3^n+4^n \le 5^n \text{. Prove that $3^{n+1}+4^{n+1} \le 5^{n+1}$} \\
= 3\cdot3^n+4\cdot4^n \\ =3\cdot3^n+4^n(3+1)\\
=3\cdot3^n+3\cdot4^n+4^n \\
=3(3^n+4^n)+4^n \\
\le 3(5^n)+4^n \\
(5-2)(5^n)+4^n \\
5^{n+1} - 2\cdot5^n+4^n\\
5^{n+1} - 2\cdot(3^n+4^n) +4^n$$
I am stuck on. I see now that I double the $5^n$, and this leads me nowhere. Where can I go from here to solve this? I have tried playing with the algebra, but I am not getting anywhere with this last step.
| From where you left off, we have
$$
5^{n+1}-2(3^n+4^n)+4^n=5^{n+1}-2\cdot3^n - 4^n\leq 5^{n+1}
$$
and your proof is finished.
| {
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"timestamp": "2023-03-29T00:00:00",
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Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$?
Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$.
Here's my progress. Let $x = \sqrt[4]{2}$. Then our expression can be written as $x^4/(x^4 - x)$, which simplifies to $x^3/(x^3 - 1)$. Multiply top and bottom by $(x^3 + 1)$ to get $x^3(x^3 + 1)/(x^6 - 1)$. Multiply top and bottom by $(x^6 + 1)$ to get$$x^3(x^3 + 1)(x^6 + 1)/(x^{12} - 1) = x^3(x^3 + 1)(x^6 + 1)/7 = {1\over7}(8 + 4\sqrt[4]{2} + 2 \sqrt{2} + 2^{3/4}).$$However, Wolfram Alpha also tells me that we can write this as$${1\over{14}}\Big(16 + 4\sqrt{2} + 7\sqrt{{{64}\over{49}} + {{72{\sqrt2}}\over{49}}}\Big)$$But how do I derive that? Seems impossible!
| We can use rationalization by a double step
$$\frac2{2 - \sqrt[4]{2}} \cdot \frac{2 +\sqrt[4]{2}} {2 + \sqrt[4]{2}} \cdot \frac{4+\sqrt {2}} {4+ \sqrt{2}}=\frac{(2 +\sqrt[4]{2})(4+\sqrt {2})}7$$
| {
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How do I prove that if: $\cos^3(x) + \sin^3(x) = 1$ then: $\cos(x) = 0 ; \sin(x)=1$ or $\cos(x)=1 ; \sin(x)=0$ How do I prove that if:
$$\cos^3(x) + \sin^3(x) = 1$$
then:
$$\cos(x) = 0 ; \sin(x)=1 \text{ or } \cos(x)=1 ; \sin(x)=0?$$
Starting from the first expression, I couldn't figure out how to reach the conclusion. I replaced 1 by $\cos^2(x) + \sin^2(x) $ hoping to factor it but to no avail.
| Write $u = \cos x$, $v=\sin x$. From $u^2 + v^2=1$ we get $u$, $v$ $\le 1$ so $u^3\le u^2$, $v^3 \le v^2$, and adding up we get $u^3 + v^3\le 1$. We have equality if and only if we have equality in the previous equalities, therefore $(u,v)=(1,0)$ or $(0,1)$.
It is interesting to sketch the curves $u^2+v^2=1$ and $u^3+v^3=1$. From the above, the function $u^3+v^3$ has maximum value $1$ on $u^2+v^2=1$, so the two curves are tangent at $(1,0)$ and $(0,1)$. In other words, the multiplicities of the intersection at $(1,0)$, $(0,1)$ are $2$. Bézout's theorem tells us there should be two more intersection points. They are complex, $(-1+\frac{i}{\sqrt{2}},-1-\frac{i}{\sqrt{2}})$ and its conjugate $(-1-\frac{i}{\sqrt{2}},-1+\frac{i}{\sqrt{2}})$
| {
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There are no rationals $r, s$ such that $\sqrt{3} = r + s\sqrt{2}$. There are no rationals $r, s$ such that $\sqrt{3} = r + s\sqrt{2}$.
Our professor gave us lengthy and a bit of a complex proof. I tried to construct my own, more simple proof. Could you please verificate it ?
Suppose there are $r,s$ such that$\sqrt{3} = r + s\sqrt{2}$. Then $\sqrt{3}-r= s\sqrt{2} \to \frac{\sqrt{3}}{\sqrt{2}} - \frac{r}{\sqrt{2}} = s$.
$\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} = \frac{1}{2} * \sqrt{6}$ Clearly this number is irrational
$\frac{r}{\sqrt{2}} = \frac{r\sqrt{2}}{2} = \frac{r}{2} * \sqrt{2}$ Again this is irrational.
The only way to substract irrationals and receive an integer is if they are equal. Clearly, this is impossible in our case. Hence, this is contradiction.
Actually could you please explain why difference of irrationals is equal to irrational or $0$.
UPD: I now realize my proof was wrong. I constructed a new one.
$\sqrt{3} - r = s\sqrt{2}$. Squaring this equation we get $3-2r\sqrt{3} + r^2 = 2s^2 \to 3-2s^2+r^2=2r\sqrt{3}$ Clearly, $2r\in Q $ and so RHS is irrational.
Suppose $r=0$. Then $3-2s^2=0 \to s=\sqrt{\frac{3}{2}}$ Contradiction, since s is rational.
Suppose $r\neq 0$ Then there is contradiction since LHS is rational and RHS is irrational. Therefore there does not exist such rational r, s.
| The proof is wrong .
For example $2+\sqrt{3}$ and $-2+\sqrt{3}$ are irrtionlas and give an integer that is not zero on subtraction
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit of the finite series $\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$ The problem is to find the limit of:
$$\ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$$
A the series is finite, it looks as if it would be required to find the sum of the series - however, I have to find the limit. It resembles for me:
$$\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n^2-kn }{(k+n)^3-2kn^2-2k^2n}=\\
\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n(n-k) }{(k+n)^3-2kn(n+k)}=\\
\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n(n-k) }{(k+n)((k+n)^2-2kn)}$$
but I don't know what to do next and how to solve it. I would appreciate your help.
Edit: Does it has something in common with Riemann sum?
| Attempts
Using the integral test, we can show the convergence since
$$I=\int \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}\,dk=\int \left(\frac{n}{k^2+n^2}+\frac{1}{k+n}\right)\,dk=$$
$$I=\log (k+n)-\tan ^{-1}\left(\frac{n}{k}\right)$$
Integrating between $k=1$ and $k=\lfloor n + \sqrt{n}\rfloor$ and simplifying, we have
$$\tan ^{-1}\left(\frac{n \left(\left\lfloor n+\sqrt{n}\right\rfloor
-1\right)}{\left\lfloor n+\sqrt{n}\right\rfloor +n^2}\right)+\log \left(\frac{\left\lfloor n+\sqrt{n}\right\rfloor +n}{n+1}\right)$$ which is asymptotic to $\frac \pi 4+\log(2)\approx 1.47855$.
Integrating between $k=0$ and $k=\lfloor n + \sqrt{n}\rfloor$ and simplifying, we have
$$\log \left(\frac{\left\lfloor n+\sqrt{n}\right\rfloor +n}{n}\right)-\tan
^{-1}\left(\frac{n}{\left\lfloor n+\sqrt{n}\right\rfloor }\right)+\frac{\pi }{2}$$ which shows the same asymptotic value.
On the other hand, we can write the summation as
$$-\frac{1}{2} i H_{\left\lfloor n+\sqrt{n}\right\rfloor -i n}+\frac{1}{2} i H_{i
n+\left\lfloor n+\sqrt{n}\right\rfloor }+H_{n+\left\lfloor n+\sqrt{n}\right\rfloor
}-H_n-\frac{1}{2 n}+\frac{1}{2} \pi \coth (\pi n)$$ and using the asymptotics of harmonic numbers, the same result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Riemann zeta function in the critical strip Hello wonderful people
It seems that the Riemann zeta function in the critical strip may be given by:
\begin{gather}
\zeta(s)=\frac{1}{s-1}+1-s \int\limits_{1}^{+\infty} \frac{x-[x]}{x^{s+1}} \mathrm{d}x
\end{gather}
Does someone have a nice and pedagogical proof for a public of engineers?
Thx in advance
| If a "public of engineers" wants a much more detailed proof, see below.
We assume $Re(s) > 1$ until indicated otherwise.
\begin{align*}
\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} &= \sum_{n=1}^{1} \frac{n}{n^s}
+ \sum_{n=2}^{\infty} \frac{1}{n^s}
= \sum_{n=1}^{1} \frac{n}{n^s} + \sum_{n=2}^{\infty} \frac{n - (n-1)}{n^s} \\
&= \sum_{n=1}^{1} \frac{n}{n^s} + \sum_{n=2}^{\infty} \frac{n}{n^s}
- \sum_{n=2}^{\infty} \frac{n-1}{n^s}
= \sum_{n=1}^{\infty} \frac{n}{n^s} - \sum_{n=2}^{\infty} \frac{n-1}{n^s} \\
&= \sum_{n=1}^{\infty} \frac{n}{n^s} - \sum_{n=1}^{\infty} \frac{n}{(n+1)^s}
= \sum_{n=1}^{\infty} n\left[ \frac{1}{n^s} - \frac{1}{(n+1)^s}\right] \\
& = s \sum_{n=1}^{\infty} n \int_n^{n+1} x^{-s-1}\,dx.
\end{align*}
Since $[x] = n$ for any $x$ in the interval $[n,n+1)$, we have
\begin{align*}
&= s \sum_{n=1}^{\infty} \int_n^{n+1} [x]x^{-s-1}\,dx
= s \int_1^{\infty} [x]x^{-s-1}\,dx \\
&= s \left[ \int_1^{\infty} x^{-s}\,dx \right] - s \int_1^{\infty} \{x\} x^{-s-1}\,dx
\quad\text{(because $[x] = x − \{x\}$)}.\\
&= s \left[ \frac{x^{-s+1}}{-s+1} \Bigg|_1^{\infty} \right] - s \int_1^{\infty} \{x\} x^{-s-1}\,dx,
\end{align*}
allowing the following simplification
\begin{align*}
\zeta(s) &= \frac{s}{s-1} - s\int_{1}^{\infty} \frac{\{x\} }{x^{s+1}}\,dx
\quad Re(s) > 1.
\end{align*}
Because $0 \leq \{x\} \leq 1$, the last integral converges and is holomorphic on $Re(s) > 0$. But that means the full equation is meromorphic on $Re(s) > 0$, and thus provides an analytic continuation of $\zeta(s)$ on the half plane $Re(s) > 0$. The $s/(s-1)$ term gives a simple pole at $s = 1$ with residue $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
In this linear equation system, find the $a,b,c$ values such that...
Consider this linear equations system:
\begin{align*} \begin{pmatrix} a & 1 & 1\\ 1 & a & 1 \\ 1 & 1 & 1
\end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}=
\begin{pmatrix} b\\ c\\ 2 \end{pmatrix} \end{align*}
*
*Determine the $a,b,c$ values such that the system has a unique solution.
*To those $a,b,c$ values such that the system has a non unique solution, find all the possible solutions.
We have that:
\begin{align*}
ax+y+z&=b\\x+ay+z&=c\\x+y+z&=2
\end{align*}
Then:
\begin{align*}
\begin{pmatrix}
a & 1 & 1 & \vdots & b\\
1 & a & 1 & \vdots & c\\
1 & 1 & 1 & \vdots & 2
\end{pmatrix}\sim \cdots \sim
\begin{pmatrix}
1 & 0 & 0 & \vdots & \frac{b-2}{a-1}\\
0 & 1 & \frac{1}{a} & \vdots & \frac{c(a-1)-(b-2)}{a(a-1)}\\
0 & 0 & 1 & \vdots & \frac{a \left [ 2(a-1)-(b-2) \right ]}{(a-1)^{2}}
\end{pmatrix}
\end{align*}
But I don't know how to continue to solve 1 and 2. What am I doing wrong or how would you solve it? I would really appreciate your help.
| How I would solve the first question: A quick check shows that $\det A=(a-1)^2$, so there is a unique solution unless $a=1$.
A hands-on approach without any theory works too: Note that $x+y+z=2$ and so
$$b-2=(ax+y+z)-(x+y+z)=(a-1)x,$$
$$c-2=(x+ay+z)-(x+y+z)=(a-1)y,$$
which shows that if $a\neq1$ then
$$x=\frac{b-2}{a-1}\qquad\text{ and }\qquad y=\frac{c-2}{a-1},$$
and hence
$$z=2-x-y=\frac{2a-b-c+2}{a-1}.$$
So if $a\neq1$ there is a unique solution, for any value of $b$ and $c$.
As for the second question: If $a=1$ then it is necessary and sufficient that $b=c=2$ for a solution to exist. The solutions are all $x$, $y$ and $z$ such that $x+y+z=2$. That is, all triplets of the form
$$(x,y,2-x-y).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3862452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Functional equation involving integral
Find $f(x)$ given that it satisfies the equation $f(x) = x^2 + \int_0^{\frac{\pi}{2}} f(t) \sin(x+1) dt$
My attempt:
Let $\int_0^{\pi/2}f(t)dt=c$
So
$$f(x)=x^2+c\sin(x+1)$$
Integrate w.r.t. $x$ from $0$ to $\pi/2$
$$\int_0^{\pi/2}f(x)dx=\int_0^{\pi/2}x^2dx+c\int_0^{\pi/2}\sin(x+1)dx$$
$$c=\frac{x^3}{3}\bigg|_0^{\frac{\pi}{2}}-c(\cos(x+1))\bigg|_0^{\frac{\pi}{2}}$$
$$c=\frac{\pi^3}{24}+c(\sin(1)+\cos(1))$$
$$\implies c=\frac{\pi^3}{24(1-\sin1-\cos1)}$$
Plugging it back we obtain:
$$f(x)=x^2+\frac{\pi^3\sin(x+1)}{24(1-\sin1-\cos1)}$$
This however doesn't seem to match with the given answer:
$$f(x) = x^2-\frac {2(3\pi^2-4\pi-8)}{\pi^2-4}\sin x-\frac {\pi^3-16}{\pi^2-4} \cos x$$
I have tried expanding $\sin(x+1)$ in my expression but it didn't lead to anything.
Are these expressions equivalent? If yes, how do I inter-convert them?
EDIT
The following solution was given:
| $$f(x) = x^2 + \int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt$$
differentiate twice wrt $x$
$$f''(x) = 2-\int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt$$
but
$$\int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt=f(x)-x^2$$
thus we get
$$f''(x)=2-f(x)+x^2$$
$$f(x) = x^2+a \cos (x)+b \sin (x)$$
Compute now
$$\int_0^{\frac{\pi}{2}} f(t) \sin(x+t) dt=\frac{1}{4} \left(\left(\pi a+2 b+\pi ^2-8\right) \sin (x)+(2 a+\pi (b+4)-8) \cos (x)\right)$$
we know that $f(x)-x^2=a \cos (x)+b \sin (x)$ so we have
$$a \cos (x)+b \sin (x)=\frac{1}{4} \left(\left(\pi a+2 b+\pi ^2-8\right) \sin (x)+(2 a+\pi (b+4)-8) \cos (x)\right)$$
which leads to the system of two linear equations
$$
\left\{
\begin{array}{l}
(2 a+\pi (b+4)-8)=4a \\
\pi a+2 b+\pi ^2-8=4b \\
\end{array}
\right.
$$
$$a= -\frac{\pi ^3-16}{\pi ^2-4},\;b= -\frac{2 \left(3\pi^2-4\pi-8\right)}{\pi ^2-4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3863931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm.
In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm. Suppose the lines $AD$ and $BC$ intersect at right angles and the lines $AC$ and $BD$ when extended at point $Q$ form an angle of $45^\circ$. Compute the area of $ABCD$.
What I Tried :- Here is the picture :-
Now to find the area of $ABCD$, I just need to find its height, but I cannot find it.
I can see that $\Delta AOB \sim \Delta COD$. So :-
$$\frac{AB}{CD} = \frac{AO}{OD} = \frac{BO}{OC} = \frac{2}{5}$$
So I assumed $AO = 2x$ , $BO = 2y$ , $CO = 5y$ , $DO = 5x$.
Now in $\Delta AOB$, by Pythagorean Theorem :-
$AO^2 + OB^2 = AB^2$
$\rightarrow 4x^2 + 4y^2 = 16$
$\rightarrow x^2 + y^2 = 4$
Also $\Delta QAB \sim \Delta QDC$. So:-
$$\frac{QA}{AC} = \frac{QB}{BD}$$
I get $AC$ and $BD$ by Pythagorean Theorem again, which gives me :-
$$\frac{QA}{\sqrt{4x^2 + 25y^2}} = \frac{QB}{\sqrt{25x^2 + 4y^2}}$$
I don't know how to proceed next, as this result only gives me that $\left(\frac{QA}{OB}\right)^2 = \frac{21y^2 + 16}{21x^2 + 16}$ . Also I couldn't think of any way to use the $45^\circ$ angle, except that I can figure out that the triangle is cyclic.
Can anyone help?
| Let $OC = a$, $OD = b$. So $OA=\frac{2}{5}OC$, $OB = \frac{2}{5} OD$.
(Note you have swapped labels $C$ and $D$ in figure)
Also let $AD=3x$, $BC=3y$, so that $QA=2x$, $QB=2y$.
We have $a^2+b^2=100$
By Pythagoras,
$$ (OA^2+OD^2)+(OB^2+OC^2)=9(x^2+y^2) $$
$$ \Rightarrow x^2+y^2=116/9 $$
By cosine-rule in $\triangle QAB$,
$$ 4x^2+4y^2-4\sqrt{2}xy=4^2 $$
$$\Rightarrow xy=\dfrac{40\sqrt{2}}{9}$$
So
$$
\begin{align}
[ABCD] &= (1-\dfrac{4}{25})[QDC] \\
&=\dfrac{21}{25}(\frac{1}{2}\cdot QD\cdot QC\cdot\sin 45^{\circ})\\
&=\dfrac{21}{25}(\frac{1}{2}\cdot 5x\cdot5y\cdot\frac{1}{\sqrt{2}}) \\
&=\boxed{\dfrac{140}{3}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3864897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Proving that $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}].$ Here is the question I want to answer:
Find polynomials $f(x), g(x) \in \mathbb{Q}[x]$ such that $\sqrt{2} = f(\sqrt{2} + \sqrt{3})$ and $\sqrt{3} = g(\sqrt{2} + \sqrt{3}).$ Deduce the equality of fields: $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}].$
Here is what I managed to prove so far:
I managed to write $\sqrt{2}$ as a function in $\sqrt{2} + \sqrt{3}$ and $\sqrt{3}$ as a function in $\sqrt{2} + \sqrt{3}$ as can seen below:
$\textbf{Finding f(x).}$
Since $(\sqrt{2} + \sqrt{3})^2 = 5 + 2 \sqrt{2.3}.$ Then we have that \begin{align*}
(\sqrt{2} + \sqrt{3})^2 (\sqrt{2} + \sqrt{3}) &= (5 + 2 \sqrt{2.3}) (\sqrt{2} + \sqrt{3} )\\
&= 5 \sqrt{2} + 5 \sqrt{3} + 4 \sqrt{3} + 6 \sqrt{2} \\
&= 11 \sqrt{2} + 9 \sqrt{3} \\
&= 9 (\sqrt{2} + \sqrt{3}) + 2 \sqrt{2}
\end{align*}
Then if we put $x = \sqrt{2} + \sqrt{3},$ we will get that $$x^3 = 9x + 2 \sqrt{2}.$$ And so we have that $$\sqrt{2} = \frac{x^3}{2} - \frac{9x}{2} = f(x).$$\
$\textbf{Finding g(x).}$
\begin{align*}
(\sqrt{2} + \sqrt{3})^3
&= 11 \sqrt{2} + 9 \sqrt{3}\\
&= 11 \sqrt{2} + 9 \sqrt{3} + 2 \sqrt{3} - 2 \sqrt{3}\\
&= 11 (\sqrt{2} + \sqrt{3}) - 2 \sqrt{3}
\end{align*}
Then if we put $x = \sqrt{2} + \sqrt{3},$ we will get that $$x^3 = 11x - 2 \sqrt{3}.$$ And so we have that $$\sqrt{3} = \frac{- x^3}{2} + \frac{11x}{2} = g(x).$$\
And I understand that this proves this inclusion $\mathbb{Q}[\sqrt{2},\sqrt{3}] \subseteq \mathbb{Q}[\sqrt{2} + \sqrt{3}] $
My question is how can I prove the other inclusion $\mathbb Q(\sqrt2+\sqrt3)\subseteq \mathbb Q(\sqrt2,\sqrt3)$? could anyone help me in showing this please pointing out for me what exactly the definition of $\mathbb Q(\sqrt2,\sqrt3)$ for me.
| You are asking about definition for $\Bbb{Q}(\sqrt{2},\sqrt{3})$.
Let me add some more.
Suppose $F$ is a field of a subfield $K$ and $\alpha$ is an element of $K$. Then the collection of all subfields of $K$ containing both $F$ and $\alpha$ is nonempty(why!).
Since the intersection of subfields is again a subfield, it implies that there is a unique smallest subfiles containing both $F$ and $\alpha$. Similarly, you can replace $\alpha$ by a collection $\alpha,\beta,\dots$ of elements of $K$.
We conclude a definition from above,
Definition Let $K$ be an extension of the filed $F$ and let $a,b,c,\dots\in K$ be collection of elements in $K$. Then the smallest field containing both $F$ and the elements $a,b,c\dots$ denoted $F(a,b,c,...)$ is called the field generated by $a,b,c\dots$ over $F$.
Clearly $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest field containing $\Bbb{Q},\sqrt{2}$ and $\sqrt{3}$.
i.e., $\sqrt{2},\sqrt{3}\in \Bbb{Q}(\sqrt{2},\sqrt{3})$.
Note that $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is a field. So by the closure property w.r.t. addition,
$\sqrt{2}+\sqrt{3}\in\Bbb{Q}(\sqrt{2},\sqrt{3})$
$\Rightarrow \Bbb{Q}(\sqrt{2}+\sqrt{3})\subseteq \Bbb{Q}(\sqrt{2},\sqrt{3})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3866399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question about asymptotes.
Let $f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2}.$ Give a polynomial
$g(x)$ so that $f(x) + g(x)$ has a horizontal asymptote of $0$ as $x$
approaches positive infinity.
I've tried using that if the degree of the denominator is bigger than the degree of the numerator, the horizontal asymptote is $y = 0$, but I couldn't get anywhere. How should I solve it?
| Perform the Euclidean division of the numerator by the denominator:
$$x^4+x^3+x^2+1=(x^2+3)(x^2+x-2) -3x+7. $$
We deduce
$$f(x)=3(x^2+3)+3\frac{-3x+7}{x^2+x-2}.$$
The fraction tends to $0$ as $x\to\infty$, hence we can take $\:g(x)=-3(x^2+3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Fractions in Questions and Answers
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