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Evaluating $\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}$. I was wondering if it was possible to evaluate
$$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}}}$$
I let the expression equal $x>0$ and wrote $$x=\sqrt{9-5\sqrt{3-x}}$$ However, there is not just one value $x$ can take; $x=2$ or $x=3$.
How do I find out which one it is, or does this infinite-nested radical converge at all? Perhaps it merely oscillates between $2$ and $3$, but I am not entirely sure. Any help or hints would be much appreciated.
Thank you in advance.
The ellipsis means "and so on". It measures the following: $$\sqrt{9-5}$$ $$\sqrt{9-5\sqrt{3-\sqrt{9-5}}}$$ $$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5}}}}}$$ $$\vdots$$
Incidentally, I did not refuse to clarify the meaning. I am only active on Math.SE for so long. Whatever requests that occur can only be followed up the moment I am active, can see them and have time to act.
| The answer is $2$. While $3$ is also a fixed point, it is unstable because if we let $x=3-\epsilon$ for some small $\epsilon$, and iterate $x\leftarrow \sqrt{9-5\sqrt{3-x}}$, it will diverge away from $3$.
$\sqrt{9-5\sqrt{3-x}}$">
If you look at the graph, you will find that the slope approaches $\infty$ as $x\to 3$. The derivative of $\sqrt{9-5\sqrt{3-x}}$ is $\frac5{4\sqrt{9-5\sqrt{3-x}}\sqrt{3-x}}$. When $x\to 3$, the $\sqrt{3-x}$ in the denominator will approach $0$, which means the derivative approaches $\infty$ as $x\to 3$. Therefore, the fixed point is unstable and will very quickly diverge away from $3$. Plugging $2$ into the equation gives $\frac58$, which is less than $1$. Therefore, the fixed point is stable.
In conclusion:
$$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}}}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to find the minimum of $abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$ when $ab+bc+cd+da+ac+bd=6$ If $a,b,c,d\geq0$ are such that $ab+bc+cd+da+ac+bd=6$, then what is the minimum value of $$f(a,b,c,d)=abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$$ ?
My attempts: I think the minimum $5$ is obtained at $a=b=c=d=1$. I tried the Hölder inequality: $$f(a,b,c,d)\geq abcd + (1+\sqrt{a b c d})^2,$$ but the last term is $1<5$ for $d=0$.
Also, direct AM-GM on the two terms of $f$ did not work either. What to do?
Remark: This problem is from AoPS. There, it is also asked what happens for $ab+bc+cd+da+ac+bd=7$, which seems even harder.
| We have the following identity:
$$\prod_{cyc} (a^2+1) = \left(\sum_{cyc}a-\sum_{cyc}abc\right)^2+\left(\sum_{sym}ab-abcd-1\right)^2$$
So
$$\prod_{cyc} (a^2+1) \geq \left(\sum_{sym}ab-abcd-1\right)^2$$
and this implies
$$1+abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)} \geq ab+bc+cd+da+bd+ca$$
Now, we can see that the minimum is indeed $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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prove $ 1 < \varphi = \frac{1 + \sqrt{5}}{2} < 2$ Prove $ 1 < \varphi = \frac{1 + \sqrt{5}}{2} < 2$
Just want to see if my "reasoning is sound.
1) Showing $1 < \frac{1 + \sqrt{5}}{2}$
Consider $\frac{1}{2}$:
$$\frac{1}{2} < \frac{1}{2} + \frac{1}{2} < \frac{1}{2} + \frac{\sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2} $$
2) Showing $\frac{1 + \sqrt{5}}{2} < 2$
We know $$\sqrt{5} < 3 \\ \Rightarrow 1 + \sqrt{5} < 4 \\ \Rightarrow \frac{1 + \sqrt{5}}{2} < 2$$
Therefore: $ 1 < \varphi = \frac{1 + \sqrt{5}}{2} < 2$
| What about:
$4 < 5 < 9$
$2< \sqrt 5 < 3$
$3 < 1 + \sqrt{5} < 4$
$\frac 32 < \frac {1+\sqrt{5}}2 < 2$.
$1 < \frac 32 < \frac{1+\sqrt{5}} 2 < 2$.
.....
I suppose maybe a more direct way of solving would be
$n < \frac {1+\sqrt 5}2 < n+1 \iff$
$2n < 1+\sqrt 5 < 2n + 2\iff$
$2n-1 < \sqrt 5 < 2n+1 \iff$
$(2n-1)^2 < 5 < (2n+1)^2$.
So our task is to: Find to consecutive odd numbers so that $5$ is between the squares. Those odd numbers are $1$ and $1^2 < 5$ and $3$ and $3^2 > 5$ and so $2n -1 = 1$ and $2n+1 = 3$
And so $n =1$.
| {
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"url": "https://math.stackexchange.com/questions/3540601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The radius of a circle, having minimum area, which touches the curve $y = 4 - x^2$ and the lines, $y = |x|$ is The radius of a circle, having minimum area, which touches the curve $y = 4 - x^2$ and the lines, $y = |x|$ is
My attempt is as follows:-
Let the circle be $x^2+y^2+2gx+2fy+c=0$, let the line $y=x$ touching circle at $(\alpha,\alpha)$
$$y\alpha+x\alpha+g(x+\alpha)+f(y+\alpha)+c=0$$
$$x(\alpha+g)+y(\alpha+f)+g\alpha+f\alpha+c=0$$
Comparing it with $y=x$
$$\dfrac{1}{\alpha+f}=\dfrac{-1}{\alpha+g}$$
$$\alpha+g=-\alpha-f=2\alpha+g+f=0\tag{1}$$
$$g\alpha+f\alpha+c=0$$
$$g+f=-\dfrac{c}{\alpha}$$
$$-2\alpha=-\dfrac{c}{\alpha}$$
$$2\alpha^2=c\tag{2}$$
Comparing it with $y=-x$, let the line $y=-x$ touching circle at $\left(\beta,-\beta\right)$
$$-y\beta+x\beta+g(x+\beta)+f(y-\beta)+c=0$$
$$x(g+\beta)+y(f-\beta)+g\beta-f\beta+c=0$$
$$\dfrac{1}{f-\beta}=\dfrac{1}{g+\beta}$$
$$f-g=2\beta$$
$$g\beta-f\beta+c=0$$
$$\beta(g-f)=-c$$
$$2\beta^2=c$$
So we got $\alpha=\pm\beta$
Let the parametric point on $y=4-x^2$ be $(\gamma,4-\gamma^2)$
$$y-(4-\gamma^2)=-2\gamma(x-\gamma)$$
$$y-4+\gamma^2=-2\gamma x+2\gamma^2$$
$$y-4=-2\gamma x+\gamma^2$$
Let's write the equation of tangent to circle at point $(\gamma,4-\gamma^2)$
$$y(4-\gamma^2)+x\gamma+g(x+\gamma)+f(y+4-\gamma^2)+c=0$$
$$y(4-\gamma^2+f)+x(\gamma+g)+g\gamma+f(4-\gamma^2)+c=0$$
$$\dfrac{1}{4-\gamma^2+f}=\dfrac{2\gamma}{\gamma+g}$$
Its getting too long, any other way of doing this question?
|
Given the symmetry with respect to the $y$-axis,let the equation of the circle be
$$x^2+(y-b)^2=r^2$$
Note that the origin, the center of the circle and the touch point with one of the lines $y=|x|$ form a 45-45 right triangle, which gives
$$ b = \sqrt2 r\tag 1$$
Also, match the gradients at the touch point with the curve $y=4-x^2$,
$$ y' = -2x = -\frac{x}{y-b}$$
which leads to $y = \frac12 + b$. Plug it into $y=4-x^2$ to get $x$-coordinates of the touch points $x=\pm\sqrt{\frac72-b}$. Then, the distance of the center to the touch points is
$$(\frac72 -b) + \frac14 = r^2=\frac{b^2}2$$
where (1) is used in the last step. solve to obtain $b = \sqrt{\frac{17}2}-1$
and, in turn, the radius
$$r = \frac12(\sqrt{17}-\sqrt2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $a_n = \frac{1}{1^3\cdot 1}+\frac{1}{1^3\cdot 2+2^3\cdot 1}+\cdots+\frac{1}{1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1}$ Let
$$a_n = \frac{1}{1^3\cdot 1}+\frac{1}{1^3\cdot 2+2^3\cdot 1} + \cdots +\frac{1}{1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1}$$
Does this sequence converge to a simple number?
My thought was to compute each denominator:
$$1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1=(n+1)\sum_{k=1}^n k^3-\sum_{k=1}^nk^4$$
and this are known, I find $\frac{1}{60}n(n+1)(2n+1)(3n^2+6n+1)$. But can we find maybe a closed formula for $a_n$ after this?
| Continuing in a different way, you can express the limit in terms of the digamma function as follows:
$$
\mathop {\lim }\limits_{n \to + \infty } 30\left[ {\sum\limits_{k = 1}^n {\frac{1}{k}} + \sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} + \sum\limits_{k = 1}^n {\frac{1}{{k + 2}}} - \frac{3}{2}\sum\limits_{k = 1}^n {\frac{1}{{k + 1 + \sqrt {2/3} }}} - \frac{3}{2}\sum\limits_{k = 1}^n {\frac{1}{{k + 1 - \sqrt {2/3} }}} } \right]
\\
= \mathop {\lim }\limits_{n \to + \infty } 30\left[ {\frac{1}{2} - \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \frac{3}{2}\sum\limits_{k = 1}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 + \sqrt {2/3} }}} \right]} + \frac{3}{2}\sum\limits_{k = 1}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 - \sqrt {2/3} }}} \right]} } \right]
\\
= \mathop {\lim }\limits_{n \to + \infty } 30\left[ {6 + \frac{1}{2} - \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \frac{3}{2}\sum\limits_{k = 0}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 + \sqrt {2/3} }}} \right]} + \frac{3}{2}\sum\limits_{k = 0}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 - \sqrt {2/3} }}} \right]} } \right]
\\
= 30\left[ {6 + \frac{1}{2} + \frac{3}{2}\left( {\psi \left( {\sqrt {2/3} } \right) + \gamma } \right) + \frac{3}{2}\left( {\psi \left( { - \sqrt {2/3} } \right) + \gamma } \right)} \right]
\\
= 195 + 90\gamma + 45\left( {\psi \left( {\sqrt {2/3} } \right) + \psi \left( { - \sqrt {2/3} } \right)} \right)
\\
= 195 + 90\gamma + 45\pi \cot (\pi \sqrt {2/3} ) + 45\sqrt {\frac{3}{2}} + 90\psi (\sqrt {2/3} ).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the minimum value of $x$ s.t. $\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$ Let $x,y\in \mathbb{R}$ such that $$\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$$. Find the minimize value of $x$.
[Edit by Michael Rozenberg] I tried to use AM-GM to retire the radical but failed:
$$27=\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}\geq2\sqrt[4]{\left(\frac{x+y}{2}\right)^3\left(\frac{x-y}{2}\right)^3}=$$
$$=2\sqrt[4]{\frac{(x^2-y^2)^3}{64}}=\sqrt[4]{\frac{(x^2-y^2)^3}{4}}.$$
Help me
| Note that $x(y)$ is an even function of $y$, with the domain $y\le |x|$. So, just examine $0\le y \le x $. Evaluate,
$$x'(y) =\frac{-y}{\left(\sqrt{\frac{x+y}{2}} +\sqrt{\frac{x-y}{2}}\right)^2}<0$$
i.e. $x(y)$ strictly decreases for $0< y \le x $. Therefore, the minimum is at $y=x$, and by symmetry also at $y=-x$. Plug $y=\pm x$ into
$$\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$$
to obtained $x_{min} = 9$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given a group and defined operation. Can't find and prove the that there exist an inverse. Let $G = \{x \in \mathbb{R} \mid x \ne - 1\}$, and let
$x * y = x + y + xy$, where $xy$ is a product of $x$ and $y$. We need to show that this is the group.
I had not difficulties with closure axiom, associative law, and identity element, but I can't find the inverse $h \in G$ such that $h*g = e = g*h$, for every $g \in G$.
| Since $0$ is also the identity under
$x \ast y = x + y + xy, \tag 1$
$y$ is the inverse of $x$ if
$x + y + xy = 0; \tag 2$
it is also clear that
$y = 0 \Longleftrightarrow x = 0; \tag 3$
so if
$x \ne 0, \tag 4$
then
$y \ne 0, \tag 5$
and thus
$xy \ne 0; \tag 6$
therefore, (2) yields
$\dfrac{1}{y} + \dfrac{1}{x} + 1 = 0, \tag 7$
that is,
$\dfrac{1}{y} = -\dfrac{1}{x} - 1 = -\dfrac{1 + x}{x}, \tag 8$
or
$y = -\dfrac{x}{1 + x}. \tag{10}$
We can easily check this result; indeed, we have
$x -\dfrac{x}{1 + x} - \dfrac{x^2}{1 + x} = \dfrac{x + x^2}{1 + x} - \dfrac{x}{1 + x} - \dfrac{x^2}{1 + x}$
$= \dfrac{x + x^2 - x - x^2}{1 + x} = \dfrac{0}{1 + x^2} = 0. \tag{11}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac b{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}$ when $a,b,c>0$ I want to prove that for $a,b,c>0$ we have
$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}=
\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac{b}{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}.$$
My first attempt: By Cauchy-Schwarz we have $$\left(\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}\right)^2\le3\sum_{cyc}\frac{a^2}{a^2+3b^2+3c^2}$$ so we only need to prove that the right-hand side is always less than $\frac{9}{7}$, but this is false. Failed
Second attempt: By Cauchy-Schwarz
$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}=\sum_{cyc} \frac 1{\sqrt{1+3\frac{b^2}{a^2}+3\frac{c^2}{a^2}}}\le\sum_{cyc} \frac{\sqrt 7}{1+3\frac{b}{a}+3\frac{c}a}$$
so it remains to prove that $$\sum_{cyc} \frac{a}{1+3b+3c}\le\frac37$$ but this is wrong for example for $a=1,b=1,c=2$. Failed
Third attempt: Let $S=3(a^2+b^2+c^2)$. We need to prove $$\sum_{cyc} \frac{a}{\sqrt{S-2a^2}}\le \frac37.$$ But $x\mapsto \frac{x}{\sqrt{S
-2x^2}}$ is convex so Jensen has the wrong direction...
| We need to prove that:
$$\sqrt{\frac{a}{a+3b+3c}}+\sqrt{\frac{b}{b+3c+3a}}+\sqrt{\frac{c}{c+3a+3b}}\leq\frac{3}{\sqrt7}$$
Let's normalize with $a+b+c=3$. Then the inequality is equivalent with:
$$\sqrt{\frac{a}{9-2a}}+\sqrt{\frac{b}{9-2b}}+\sqrt{\frac{c}{9-2c}}\leq \frac{3}{\sqrt{2}}$$
Without loss of generality suppose that $a\le b\le c$. Then we have $a+b\leq 2$ and we will prove:
$$\sqrt{\frac{a}{9-2a}}+\sqrt{\frac{b}{9-2b}} \leq \sqrt{\frac{2(a+b)}{9-a-b}}$$
Squaring twice, this is equivalent with:
$$\frac{(a-b)^2[729+81(a+b)^2-486(a+b)-16ab(a+b)]}{(9-2a)^2(9-2b)^2(9-a-b)^2}\geq 0$$
We have $16ab(a+b)\leq 16(a+b)$ and
$$729+81x^2-502x\geq 0, \text{ when }x \leq 2$$
It remains to prove that:
$$\sqrt{\frac{14(3-c)}{6+c}}+\sqrt{\frac{7c}{9-2c}}\leq 3$$
Squaring twice this is equivalent with:
$$\frac{81(c-1)^2 (12 - 5 c)^2}{(9 - 2 c)^2 (6 + c)^2}\geq 0$$
which gives the two equality cases.
| {
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"url": "https://math.stackexchange.com/questions/3546034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Using partial fraction to power that less than 1 I need help to check this answer since, unfortunately, the worksheet doesn't attach the key.
Integrate:
$$\int \frac{1}{\sqrt{x} - \sqrt[3]{x}}dx$$
So, I tried use a substitution that allows the variables to become polynomials. I rewrote the expression to
$$\int \frac{1}{x^\frac{5}{6}\left(x^\frac{-2}{6}- x^\frac{-3}{6}\right)} \, dx$$
when I set $$u = x^\frac{1}{6} \text{ and } \frac{dx}{x^\frac{5}{6}}=6 \, du$$ then
$$\int \frac{6 \, du}{\frac{1}{u^2}-\frac{1}{u^3}} = \int \frac{6 u^3 \, du}{u-1}.$$
This becomes
$$6 \int \left( u^2 + u + 1 + \frac{1}{u-1} \right) \, du = 6 \left(\frac{u^3}{3} + \frac{u^2}{2} + u + \ln\lvert u-1\rvert \right) + c.$$
and last part is to substitute back $u = x^\frac{1}{6}$.
I don't know whether this right or not. This is first time I am dealing with partial factions that have less than one power. Please correct it if this work is wrong. Thanks in advance.
| An alternate view it to do a power expansion and the cancel powers. In this example $u^3$ can be seen as:
$$u^3 = ( (u-1) + 1)^3 = (u-1)^3 + 3 \, (u-1)^2 + 3 \, (u-1) + 1$$
and
$$\frac{u^3}{u-1} = (u-1)^2 + 3 \, (u-1) + 3 + \frac{1}{u-1}.$$
Integration yields:
\begin{align}
I &= \int \frac{u^3 \, du}{u-1} \\
&= \int (u-1)^2 \, du + 3 \, \int (u-1) \, du + 3 u + \int \frac{du}{u-1} \\
&= \frac{(u-1)^3}{3} + \frac{3 \, (u-1)^2}{2} + 3 \, u + \ln(u-1) + c_{0} \\
6 I &= 2 (u-1)^3 + 9 (u-1)^2 + 18(u-1) + 6 \, \ln(u-1) + c_{1}
\end{align}
This leads to the form
$$\int \frac{dx}{\sqrt{x} - \sqrt[3]{x}} = 2 (\sqrt[6]{x}-1)^3 + 9 (\sqrt[6]{x}-1)^2 + 18(\sqrt[6]{x}-1) + 6 \, \ln(\sqrt[6]{x}-1) + c_{1}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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value of $\tan B\tan C$ in a triangle
In a triangle $ABC,$ equation of side $BC$ is $2x-y=3$ and circumcenter and orthocenter of triangle are $(2,4)$ and $(1,2)$ respectively, Then $\tan B\tan C=$
what i try
In a triangle, Let $BC$ be base side and $H(1,2)$ be orthocenter of triangle and $O(2,4)$ be orthocenter of triangle.
and we know that Image of orthocenter $H(1,2)$ of triangle $ABC$ with respect to side $BC$ lie on circumference of Circumcirlce of $\triangle ABC$
for image point coordinate
using formula
$\displaystyle \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=-\frac{(ax_{1}+by_{1}+c)}{a^2+b^2}$
above image of point $P(x_{1},y_{1})$ w. r to line $ax+by+c=0$
$$\frac{x-1}{2}=\frac{y-2}{-1}=-\frac{2(1)-2-3}{2^2+1^2}$$
Coordinate of any point Image point which lie on circumcircle is $$(x,y)=\bigg(\frac{11}{5},\frac{7}{5}\bigg)$$
so circumradius of circle is $$\sqrt{\bigg(2-\frac{11}{5}\bigg)^2+\bigg(4-\frac{7}{5}\bigg)^2}=\frac{\sqrt{170}}{5}$$
How do i solve it Help me please
| If $K$ is the intersection of altitude $AH$ with $BC$, and $H'$ the reflection of $H$ about $BC$, then:
$$
\tan B\cdot\tan C={AK\over BK}\cdot{AK\over CK}={AK^2\over AK\cdot KH'}=
{AK\over KH'}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solving degree 3 equations Solve for $x$,
\begin{cases}4x^3+3x^2y+y^3=8\\
2x^3-2x^2y+xy^2=1\end{cases}
I tried substitution of $x$, but it got very complex.
Is there a simpler way to do this?
| We obtain:
$$4x^3+3x^2y+y^3=8(2x^3-2x^2y+xy^2)$$ or
$$12x^3-19x^2y+8xy^2-y^3=0$$ or
$$12x^3-12x^2y-7x^2y+7xy^2+xy^2-y^3=0$$ or
$$(x-y)(12x^2-7xy+y^2)=0$$ or
$$(x-y)(3x-y)(4x-y)=0.$$
Can you end it now?
I got the following answer:
$$\left\{(1,1),\left(\frac{1}{\sqrt[3]5},\frac{3}{\sqrt[3]5}\right), \left(\frac{1}{\sqrt[3]{10}},\frac{4}{\sqrt[3]{10}}\right)\right\}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many ways to select $n$ balls using generating function Given an infinite number of red, orange and yellow balls. In how many ways can you select $n$ balls if order doesn't matter and the number of red balls must be at least twice as big as the number of orange balls?
This is what I tried so far:
If $x_1$ is the number of red balls, $x_2$ the number of orange balls and $x_3$ the number of yellow balls, then we know:
$$x_1 \geq 2x_2,$$
$$x_1 + x_2 + x_3 = n.$$
From $x_1 \geq 2x_2$ follows $x_1 - 2x_2 \geq 0$, and if we create a new variable $x_4 = x_1 -2x_2$ then we find $x_4 \geq 0$.
Since $x_4 = x_1 -2x_2$ it follows that $x_1 = x_4 + 2x_2$. Now we find $x_1 + x_2 + x_3 = x_4 + 2x_2 + x_2 + x_3 = x_4 + 3x_2 + x_3 = n$ with $x_4 \geq 0$.
For the individual generating functions I found:
$$A_3(x) = A_4(x) = 1 + x + x^2 + ... = \frac{1}{1-x},$$
$$A_2(x) = 1 + x^3 + x^6 + .... = \frac{1}{1-x^3},$$
so for the total generating function I found:
$$A_x(x) = \frac{1}{(1-x)^2(1-x^3)}.$$
I want to rewrite this to the form $\sum_{n=0}^{\infty} ... x^n$, but I am not sure how to do that.
| $a$ is the number of orange balls
$2a+b$ is the number of red balls
$c$ is the number of yellow balls
We want
$$
3a+b+c=n\tag1
$$
The generating function is
$$
\begin{align}
\overbrace{\ \frac1{1-x^3}\ }^{3a}\overbrace{\ \ \frac1{1-x}\ \ }^{b}\overbrace{\ \ \frac1{1-x}\ \ }^{c}
&=\overbrace{\sum_{k=0}^\infty\binom{-1}{k}\left(-x^3\right)^k}^{\frac1{1-x^3}}\overbrace{\sum_{j=0}^\infty\binom{-2}{j}(-x)^j}^{\left(\frac1{1-x}\right)^2}\\
&=\sum_{k=0}^\infty x^{3k}\sum_{j=0}^\infty(j+1)x^j\tag2
\end{align}
$$
Using the Cauchy product formula, we get the coefficient for $x^n$ to be
$$
\begin{align}
\sum_{k=0}^{\left\lfloor\frac{n+1}3\right\rfloor}(n-3k+1)
&=(n+1)\left\lfloor\frac{n+4}3\right\rfloor-\frac32\left\lfloor\frac{n+4}3\right\rfloor\left\lfloor\frac{n+1}3\right\rfloor\\
&=\left\lfloor\frac{n+4}3\right\rfloor\left(n+1-\frac32\left\lfloor\frac{n+1}3\right\rfloor\right)\\[6pt]
&=\frac{(n+4)(n+1)}6+\frac32\left(\left\{\frac{n+1}3\right\}-\left\{\frac{n+1}3\right\}^2\right)\\
&=\bbox[5px,border:2px solid #C0A000]{\left\lfloor\frac{(n+2)(n+3)}6\right\rfloor}\tag3
\end{align}
$$
| {
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"url": "https://math.stackexchange.com/questions/3552690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Modular equation solving I have a modular equation that is pretty strange to solve for me.
Basically it is nothing special, but that modulo is really embarrassing for me.
Equation is this :
$\mod(A,x-B)=-1$ , $x \in \Bbb N$
How can I find the $x$ that make this equation true ? Of course there are more than one solution, and it would like to be great to find a general formula to get all of them...
Thanks in advance
| $\mod(A,x-B) = -1$ means that there is an integer $k$ so that $A = -1 + k(x-B)$ or that $A+1= k(x-B)$
If we solve for $x$ we get.....
$A + 1 = kx - kB$
$kx = kB + A + 1$.
So $x = \frac {kB + A+1}k= B + \frac {A+1}k$ where $k$ will allow that to be an integer.
In other words if $k$ is any factor $A+1$
Then $x = B + \frac {A+1}k$ will be a solution.
In for example of $A = 15$ and $B = 16$ ....
$A + 1 = 16$. $16$ has many factors. we can have $k=1, 2, 4, 8, 16$ and we can have
$x = B+\frac {A+1}k = 16+\frac {16}k = 17, 18, 20,24, 32$
And indeed $\mod(15, x-16)=-1$ means $15\equiv -1\mod (1,2,4,8,16)$
Everything is equiv $0\pmod 1$ (because $1$ divides everything) so $15\equiv -1 \equiv 0 \pmod 1$.
$15\equiv -1 \pmod 2$ as both $15$ and $-1$ or odd.
$15\equiv -1\pmod 4$ because $15 = -1 + 4*4$.
$15\equiv -1\pmod 8$ becuase $15 =-1 + 8*2$
And $15\equiv -1\pmod 16$ because $15 = -1 + 16*1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$z^2 = -1$ Why what i do is wrong? I want to calculate $z^2 = -1$ i think i should get to $z = i$.
$-1 = -1+0i = 1(\cos(2\pi k)+i\sin(2\pi k)), k \in Z$
$z^2 = r^2(\cos(2 \theta) + i\sin(2\theta))$
So i get:
$$r^2(\cos(2 \theta) + i\sin(2\theta)) = 1(\cos(2\pi k)+i\sin(2\pi k))$$
Namely:
$r = 1, 2 \theta = 2 \pi k \Rightarrow \theta = \pi k$
So for $k = 0$ i get $z_0 = 1$ for $k=1: z_1 = -1$
What i do wrong?
| Note that
$$-1 = -1+0i = 1[\cos(\pi+2\pi k)+i\sin(\pi+2\pi k)], \>k \in Z$$
Then,
$$z^2=r^2[\cos(2 \theta) + i\sin(2\theta)] = 1[\cos(\pi+2\pi k)+i\sin(\pi+2\pi k)]$$
and you get $r =1$ and $\theta = \frac\pi2 + k\pi$. As a result,
$$z_k = \cos(\frac\pi2+\pi k)+i\sin(\frac\pi2+\pi k)$$
So, for $k=0$ and $k=1$, you have respectively,
$$z_0= \cos\frac\pi2 + i\sin\frac\pi2 = i;\>\>\>\>\>z_1= \cos\frac{3\pi}2 + i\sin\frac{3\pi}2 =-i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Minimum value of $f(a,b) = a^2+ab+b^2-3a-6b+11$
Minimum value of
$f(a,b) = a^2+ab+b^2-3a-6b+11$ for all $a,b\in \mathbb{R}$
what i try
Let $$k=a^2+ab+b^2-3a-6b+11$$
$$k=a^2+(b-3)a+b^2-6b+11$$
$\displaystyle k=\bigg[a^2+(b-3)a+\frac{(b-3)^2}{4}+b^2-6b+11-\frac{(b-3)^2}{4}\bigg]$
$$\displaystyle k = \bigg[\bigg(a+\frac{b-3}{2}\bigg)^2+3b^2-18b+35\bigg]$$
$$k=\bigg[\bigg(a+\frac{b-3}{2}\bigg)^2+3(b-3)^2+8\bigg]\geq 8$$
but answer is $20$
Help me please
| Note that as $a \to \pm \infty$ or $b\to \pm \infty$, the function $f$ diverges positively so the boundary will not lead to the infinimum. Using the standard technique,
\begin{align*}
f_a=2a+b-3=0\implies b=3-2a
\end{align*}
\begin{align*}
f_b=2b+a-6=0\implies a=6-2b.
\end{align*}
Hence, the critical point occurs when $b=3-2(6-2b)=4b-9\implies b=3.$ Then it easily follows that $a=0$. Plugging in, $f(0,3)=3^2-6\cdot 3+11=2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
An indeterminate limit form of infinity/infinity I am trying to solve the limit:
$$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)$$
I was trying to find a way to bring it into a fraction form to apply L'Hospital's rule, and I tried using
$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$
But it made it even more complex and after applying L'Hospital's rule I got stuck with all the terms. Is there a smarter way to evaluate it?
| First I would factor $x^{1/3}$ out of the second term:
$$
x^{5/3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)
= x^{2}\left(\left(1+ \frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^{1/3}-1\right)
$$
Then I would substitute $t = \frac{1}{x}$:
$$
\lim_{x\to\infty}x^{2}\left(\left(1+ \frac{1}{x}\sin\left(\frac{1}{x}\right)\right)^{1/3}-1\right)
= \lim_{t\to 0^+} \frac{(1+t \sin t)^{1/3} - 1}{t^2}
$$
At this point we could use Taylor's Theorem, or the Binomial series, or L'Hôpital's rule. The limit of the quotient of the derivatives is:
\begin{align*}
\lim_{t\to 0^+} \frac{\frac{1}{3}(1+t\sin t)^{-2/3}(t \cos t + \sin t)}{2t}
&= \frac{1}{6}\cdot 1 \cdot \lim_{t\to 0^+} \left(\cos t + \frac{\sin t}{t}\right)
\\&= \frac{1}{6}(1+1) = \frac{1}{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Minimum value when $abc+ab+4bc+9ca=144$ If $a,b,c$ are non-negative real numbers such that $abc+ab+4bc+9ca=144$, find the minimum value of $a+b+c$.
I tried with Lagrange multipliers. I got the system:
$bc+b+9c=ca+a+4c=ab+4b+9a$
Replacing in the condition, I found four solutions, but only one $(4,0,4)$ is non-negative. So the minimum value is $8$. My question is, can this be done without Lagrange Multipliers?
| Let $a+b+c<8,$ $a=kx$, $b=ky$ and $c=kz$ such that $k>0$ and $x+y+z=8.$
Thus, $$k(x+y+z)<8,$$ which gives $0<k<1.$
Thus, $$144=k^3xyz+k^2(xy+4yz+9zx)<xyz+xy+4yz+9zx,$$ which is a contradiction because we'll prove now that
$$xyz+xy+4yz+9zx\leq144.$$
Indeed, we need to prove that: $$xyz+\frac{(x+y+z)(xy+4yz+9zx)}{8}\leq\frac{144(x+y+z)^3}{512}$$ or
$$9(x+z)(x-z)^2+y(23x+11z)(x-z)+y^2(9y+23x+11z)\geq0,$$ for which it's enough to prove that
$$y^2(23x+11z)^2-36y^2(x+z)(23x+11z)\leq0,$$ which is obvious.
Id est, $$a+b+c\geq8.$$
The equality occurs for $a=c=4$ and $b=0,$ which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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if $x$ is odd, show that $x^3+x$ has a remainder 2 when divided by 4 I did part of this question but am stuck and don't know how to continue
I let $x= 2k +1$
Also noticed that $x^3+x = x(x^2+1)$
therefore
$4m+2 = 2k+1((2k+1)^2+1)$
I simplified this and ended up with
$4m+2 = 8k^3+12k^2+8k+2$
I don't know how to continue from and prove that $x^3+x$ has remainder 2 when divided by 4
| $x = 2k + 1; \tag 1$
$x ^3 = 8k^3 + 12k^2 + 6k + 1; \tag 2$
$x ^3 + x = 8k^3 + 12k^2 + 6k + 1 + (2k + 1)$
$ = 8k^3 + 12k^2 + 8k + 2 = 4(2k^3 + 3k^2 + 2k) + 2, \tag 3$
that is,
$x^3 + x \equiv 2 \mod 4, \tag 4$
which by the Euclidean division implies the remainder of $x^3 + x$ when divided by $4$ is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
What does this series converge to? $\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)$ What does the following expression converge to?
$$\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)$$
(It looks like this problem)
$\displaystyle\sum_{k=0}^n a_{2k+1}+a_{2k+2}-a_{k+1}$ ;
$\displaystyle\sum_{k=0}^n a_{k+1}$ does not converge.
| Another way:
$$\sum_{k=0}^{+\infty}\left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=$$
$$=\sum_{k=0}^{+\infty}\left(\frac{1}{4k+1}-\frac{1}{4k+2}+\frac{1}{4k+3}-\frac{1}{4k+4}\right)-\sum_{k=0}^{+\infty}\left(\frac{1}{4k+2}-\frac{1}{4k+4}\right)=$$
$$=\ln2-\frac{\ln2}{2}=\frac{\ln2}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Proving the determinant of a $3\times 3$ matrix is given by $2s^3(s-a)(s-b)(s-c)$. Prove:
$$\begin{vmatrix}a^2&(s-a)^2&(s-a)^2\\(s-b)^2&b^2&(s-b)^2\\(s-c)^2&(s-c)^2&c^2\end{vmatrix}=2s^3(s-a)(s-b)(s-c),\;\;s=\frac{a+b+c}{2}$$
My attempt:
Let $c_1,c_2,c_3$ be columns and $r_1,r_2,r_3$ be rows.
$c_3-c_2:$
$$\begin{vmatrix}a^2&(s-a)^2&0\\(s-b)^2&b^2&s(s-2b)\\(s-c)^2&(s-c)^2&-s(s-2c)\end{vmatrix}$$
$c_2-c_1:$
$$\begin{vmatrix}a^2&s(s-2a)&0\\(s-b)^2&-s(s-2b)&s(s-2b)\\(s-c)^2&0&-s(s-2c)\end{vmatrix}$$
$c_2\cdot\frac{1}{s},c_3\cdot\frac{1}{s}$
$$s^2\begin{vmatrix}a^2&(s-2a)&0\\(s-b)^2&-(s-2b)&(s-2b)\\(s-c)^2&0&-(s-2c)\end{vmatrix}$$
Then, by LaPLace on the $3^{\text{rd}}$ column, I got:
$$-(s-2b)(s-2a)(s-c)^2-(s-2c)\left(a^2(2b-s)-(s-2a)(s-b)^2\right)$$
Which got too complicated. May I ask how to end this task?
| COMMENT.-Put $S_a=s-a$ and analogues you have
$$\begin{vmatrix}a^2&S_a^2&S_a^2\\S_b^2&b^2&S_b^2\\S_c^2&S_c^2&c^2\end{vmatrix}=2s^3S_aS_bS_c$$ By property of determinants one has
$$LHS=\begin{vmatrix}a^2&S_a^2&0\\S_b^2&b^2&S_b^2-b^2\\S_c^2&S_c^2&S_c-c^2\end{vmatrix}=2S_a^2S_b^2S_c^2-a^2S_b^2S_c^2-b^2S_a^2S_c^2-c^2S_a^2S_b^2$$ With this the verification of the identity is straightforward but tedious.
An easier way is to consider $a,b,c$ as the sides of a triangle so you have to prove
$$LHS=R^6\begin{vmatrix}4(\sin \alpha)^2&(-\sin \alpha+\sin\beta+\sin \gamma)^2&(-\sin \alpha+\sin\beta+\sin \gamma)^2\\(\sin \alpha-\sin\beta+\sin \gamma)^2&4(\sin \beta)^2&(\sin \alpha-\sin\beta+\sin \gamma)^2\\(\sin \alpha+\sin\beta-\sin \gamma)^2&(\sin \alpha+\sin\beta-\sin \gamma)^2&4(\sin \gamma)^2\end{vmatrix}$$ where $R$ is the radius of the circumscribed circle.
You have for the $RHS$ $$s=R(\sin \alpha+\sin \beta+\sin \gamma)\\s-a=R(-\sin \alpha+\sin\beta+\sin \gamma)\\s-b=R(\sin \alpha-\sin\beta+\sin \gamma)\\s-c=R(\sin \alpha+\sin\beta-\sin \gamma)$$ This way is easier without doubt.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can a Sum of distinct squares ever equal power of two? Does there exist $2^t,\ t\in\mathbb{Z}_+$ which can be express as Sum of two or more distinct square number.
Or
Can it be shown that
$$\begin{split}2^t &\ne \sum a_i^2 = a_1^2+ a_2^2+\cdots+a_n^2\end{split}$$
Where $n\ge 2$ and $\{a_i,t\}\in\mathbb{Z}_+$ and $a_i \ne a_j$ for $1\le i,j \le n$
Example: $2^6=64=7^2+3^2+2^2+1^2+1^2$ here $1^2$ repeat two times so this is not allowed.
My incomplete attempt for arithmetic squares
Edit: check related new post, Can a sum arithmetic square ever equal to power of two?
Let $n,u,d\in\mathbb{Z}_+$
$$\begin{split}\sum_{q=0}^u (n+qd)^2 &=n^2+(n+d)^2+(n+2d)^2+\cdots+(n+ud)^2\\ &=n^2(u+1)+\frac{(u+1)u}{2}(2nd+d^2)+\frac{(u+1)u(u-1)}{3}d^2 \end{split}$$
Let
$$\begin{split}2^t &=\sum_{q=0}^u (n+qd)^2 \\ \implies 3\cdot 2^{t+1}&=6n^2(u+1)+3(u+1)u(2nd-d^2)+(u+1)u(u-1)2d^2 \\ &= (u+1)(6n^2+3u(2nd+d^2)+u(u-1)2d^2)\\ &(in\ case,\ u+1= 3) \\
\implies 2^t&= 3n^2+3(2nd+d^2)+2d^2\\ &= n^2+(n+1)^2+(n+2d)^2 \end{split}$$
Now we need to simplify for case, $6n^2+3u(2nd+d^2)+u(u-1)2d^2=3\cdot2^x$ and $u+1=2^y$ where $x+y=t+1$ but I'm stuck here. Thank you.
Related post:
Can a sum of consecutive $n$th powers ever equal a power of two?
| It would be extremely surprising if the answer were no. Consider just the first $n$ squares. There are $2^n$ ways to construct distinct sums of these, and each sum adds up to something between $0$ and $\frac16(2n^3-3n^2+n)$ which is approximately $\frac{n^3}3$.
But $2^n$ is vastly larger than $\frac{n^3}3$. Consider even a small number like $n=80$. On the one hand we have $2^{80}$ which is the number of atoms in the universe and on the other we have the everyday number 173,880.
What are the chances that any of the $\frac{n^3}3$ possible sums $S$ is so unusual that none of the $2^n$ possible selections of squares adds up to $S$? There would have to be a very good reason, and it would not be a subtle one.
Instead, we should guess that the opposite is true: except for a few very small exceptions, every power of 2 should be representable as a sum of distinct squares in a great many ways.
| {
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Find $a,b$ and $c$ if $(1+\sqrt[3]{2})^{-1}$ in the form of $a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$ Given that
$$ (1+\sqrt[3]{2})^{-1} =a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$$
find the value of rationals $a,b,c.$
Solution I tried: I tried to rationalize it, but I'm not getting the answer:
$$\frac{1}{(1+\sqrt[3]{2})}\times \frac{(1-\sqrt[3]{2})}{(1-\sqrt[3]{2})}$$
$$\frac{(1-\sqrt[3]{2})}{1-2^{\frac{2}{3}}}$$
so doing so not getting answer; also, I tried to expand it, but we have condition that $(1+x)^n$ where $n$ is in fraction can be expandable only when $x < 1$, but here the cube root of $2$ is not less than $1.$
Thank you
| So we have $\frac 1{1 + \sqrt[3]2}$ and we want to get
$\frac 1{1 + \sqrt[3]2}\frac {something}{something}=\frac{something}{something\ with\ no\ radicals}$
Taking a page from doing this for square roots where, from $a + \sqrt b$ we realize $(a+\sqrt b)(a-\sqrt b) = a^2 - b$, which works because $(m-n)(m+n) = m^2 -n^2$.
If we use the idea $(m\pm k)(m^{n-1} \mp m^{n-2}k + ......) = m^n \pm k^n$.
So if we can see $(1+\sqrt[3]{2})(1 - \sqrt[3]{2} + \sqrt[3]{2}^2) = 1 + \sqrt[3]2^3 = 1+2=3$
And $\frac 1{1 + \sqrt[3]2}=\frac 1{1 + \sqrt[3]2}\frac {1-\sqrt[3]2 + \sqrt[3]2^2} {1-\sqrt[3]2 + \sqrt[3]2^2}= \frac {1-\sqrt[3]2 + \sqrt[3]2^2}3=\frac 13 -\frac 13\sqrt[3]2 + \frac 13\sqrt[3]{2^2}$
Or $a = c =\frac 13$ and $b =-\frac 13$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of ordered pairs of n-length binary strings with edit distance n-1 An edit operation is a single character insert, delete or substitution. The edit distance between two strings is the minimum number of edit operations needed to transform one string into the other one.
How many ordered pairs of binary strings of length $n$ are there with
edit distance exactly $n-1$?
For $n = 1,\dots, 10$ the exact numbers are $2, 8, 26, 54, 92, 138, 192, 254, 324, 402, 488, 582$ which is not in the OEIS. For $n=3$ the full list of pairs is:
000 011
000 101
000 110
001 010
001 100
001 111
010 001
010 100
010 101
010 111
011 000
011 101
011 110
100 001
100 010
100 111
101 000
101 010
101 011
101 110
110 000
110 011
110 101
111 001
111 010
111 100
For comparison, the number of ordered pairs of binary strings of length $n$ which have edit distance exactly $n$ is $2n$.
| Feeding these numbers to Wolfram|Alpha yields these iterated differences:
\begin{array}
2 & & 8 & & 26 & & 54 & & 92 & & 138 & & 192 & & 254 & & 324 & & 402\\
& 6 & & 18 & & 28 & & 38 & & 46 & & 54 & & 62 & & 70 & & 78 & \\
& & 12 & & 10 & & 10 & & 8 & & 8 & & 8 & & 8 & & 8 & & \\
& & & -2 & & 0 & & -2 & & 0 & & 0 & & 0 & & 0 & & & \\
& & & & 2 & & -2 & & 2 & & 0 & & 0 & & 0 & & & & \\
& & & & & -4 & & 4 & & -2 & & 0 & & 0 & & & & & \\
& & & & & & 8 & & -6 & & 2 & & 0 & & & & & & \\
& & & & & & & -14 & & 8 & & -2 & & & & & & & \\
& & & & & & & & 22 & & -10 & & & & & & & & \\
& & & & & & & & & -32 & & & & & & & & &
\end{array}
That seems to indicate that except for the first three terms the sequence is given by a quadratic polynomial. Trying again without the first three terms yields the polynomial $2\left(2n^2+13n+12\right)$. Since you seem to have code for producing initial terms, maybe you can produce a few more to check this. In view of the result for Average number of strings with edit distance exactly 2, it seems plausible that the result would be a polynomial. If you need a proof, some of the ideas I applied in the answer to that question may prove helpful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum $9a+25b+49c$ when $ab+bc+ca+abc=4$ If $a,b,c\ge 0$ such that $ab+bc+ca+abc=4$, find the minimum value of $9a+25b+49c$.
I know that $a+b+c\ge 3$, but I don't this is good to use here. So I tried with Lagrange multipliers:
$$L(x,y,z)=9a+25b+49c+\lambda(ab+bc+ca+abc-4)$$
With the partial derivative I found:
$$\frac{b+c+bc}{9}=\frac{a+c+ac}{25}=\frac{a+b+ab}{49}$$
and with $ab+bc+ca+abc=4$, I found minimum $59$ at $(3,1,1/7)$. My question is, can it be done with traditional ways? I tried to prove $ab+bc+ca+abc\le 4$ when $9a+25b+49c=59$ (with idea from this question: Minimum value when $abc+ab+4bc+9ca=144$), but I got lost after expanding.
| A similar idea as the accepted answer after rewriting the condition as:
$$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1$$
and using Cauchy-Schwarz:
$$
\begin{aligned}
\frac{a+1}{a+2}&=\frac{1}{b+2}+\frac{1}{c+2}\\
&=5\cdot\frac{1}{5(b+2)}+7\cdot\frac{1}{7(c+2)}\\
&\geq \frac{144}{25b+49c+148}
\end{aligned}
$$
Therefore
$$25b+49c \geq \frac{144}{a+1}-4$$
and using AM-GM:
$$
\begin{aligned}
9a+25b+49c &\geq 9a+\frac{144}{a+1}-4\\
&=9(a+1)+\frac{144}{a+1}-13\\
& \geq 2\sqrt{9(a+1)\cdot \frac{144}{a+1}}-13\\
& = 59
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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At which values of the parameter $k$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0$?
Problem:
At which values of the parameter $k$, there is no solution to the inequality $$(k+1)x^2-2kx+2k+2<0.$$
The solution in my textbook is as follows:
$a=k+1, D=4\left[k^2-2(k+1)^2\right]$
$\begin{cases} k+1>0 \\k^2-2(k+1)^2<0 \end{cases} \Longrightarrow \begin{cases} k>-1 \\ -k^2-2k-2 <0 \end{cases} \Longrightarrow -1<k<+\infty.$
Answer: That is, there is no solution to the inequality of $ (k + 1) x ^ 2-2kx + 2k + 2 <0$ for the $ k $ `s that satisfy the $ -1 <k <+ \infty $ condition.
Firstly, I understand the question as follows:
At which values of the parameter $k$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0$ , for all $x\in\mathbb{R}.$
The last sentence is logically equivalent to :
At which values of the parameter $k$, the inequality $(k+1)x^2-2kx+2k+2\geq 0$ holds on for all $x\in\mathbb{R}.$
If I understand the question correctly, here is my solution:
It is obvious that, for $k=-1$ is not a solution.
$\color{black}{\large\text{Case} \thinspace 1:}$ $k+1>0$
We have,
$$x^2-\dfrac{2k}{k+1}x+\dfrac{2k+2}{k+1}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2-\dfrac{k^2}{(k+1)^2}+\dfrac{2k+2}{k+1}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2 + \dfrac{2k+2}{k+1}-\dfrac{k^2}{(k+1)^2}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac {2(k+1)^2-k^2}{(k+1)^2} \geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\geq 0, ∀ x\in\mathbb {R}$$
Then, applying $x=\dfrac{k}{k+1}$ we get, $\dfrac{k^2+4k+2}{(k+1)^2}\geq 0$. We have,
$$\begin{cases} k+1>0 \\ \dfrac{k^2+4k+2}{(k+1)^2}\geq 0 \end{cases} \Longrightarrow k\geq -2+\sqrt2.$$
$\color{black}{\large\text{Case} \thinspace 2:}$ $k+1<0$
We have,
$$x^2-\dfrac{2k}{k+1}x+\dfrac{2k+2}{k+1}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2-\dfrac{k^2}{(k+1)^2}+\dfrac{2k+2}{k+1}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2 + \dfrac{2k+2}{k+1}-\dfrac{k^2}{(k+1)^2}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac {2(k+1)^2-k^2}{(k+1)^2} \leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\leq 0, ∀ x\in\mathbb {R}$$
For sufficiently large $ x $ `s, we have $\left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\geq 0$, which gives a contradiction.
Finally we deduce that, for all $x\in\mathbb{R}$ satisfying the condition $k\in\mathbb[\sqrt 2-2; +\infty)$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0.$
It does not match the solution in my book. Probably, maybe I misunderstand the question or my solution is wrong. Or the book says it wrong.
Did I get the question right? If so, is my solution correct?
Thank you very much.
| Here is an alternative, perhaps simpler, approach. Rewrite the given inequality $(k+1)x^2-2kx+2k+2<0$ as
$$k< -\frac{x^2+2}{x^2-2x+2}$$
In order for the inequality not to hold, the values of $k$ has to be great than or equal to the maximum value of the RHS, which can be obtained by evaluating its derivative and setting it zero, i.e.
$$\frac{2(x^2-2)}{(x^2-2x+2)^2}=0\implies x=\pm\sqrt2$$
It is straightforward to verify that the maximum $-2+\sqrt2$ occurs at $x = -\sqrt2$. Thus,
$$k\ge-2+\sqrt2$$
| {
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"url": "https://math.stackexchange.com/questions/3573278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$ Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$
I don't really know how to approach this. I was thinking doing something like squaring or cubing $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$ would help, but it doesn't really work out...Any help???
| $$
\begin{aligned}
a+b+c-3\sqrt[3]{abc}&=\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}}-\sqrt[3]{ab}-\sqrt[3]{ac}-\sqrt[3]{bc}\right)\\
\\
&=0
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is ${(-1)}^{\frac{1}{6}}=i$? I just thought of this and tried to simplify. ${(-1)}^\frac{1}{6}$.
$${(-1)}^\frac{1}{6}={[{(-1)}^\frac{1}{3}]}^{\frac{1}{2}}$$
Since the cube root of $-1$ is $-1$,
$${(-1)}^{\frac{1}{6}}=\sqrt{-1}\to{(-1)}^{\frac{1}{6}}=i$$
Is this a correct conclusion? If so, does this work to every roots with even index, where the radicand is negative?
| If $z = (-1)^\frac 16$ then $z^6 = -1$ or $z^{6} + 1 = 0$
This we can factor!
$(z^{2} + 1)(z^4 - z^2 + 1) = 0$
From the left hand factor, $z = \pm i$ are indeed solutions.
The right hand factor is a little trickier.
$z^4 - z^2 + 1 = (z^2 + \sqrt 3 z + 1)(z^2 - \sqrt 3z + 1)$
$z = \pm\frac {\sqrt {3}}{2} \pm \frac {1}{2} i$
There are 6 solutions.
These are called the roots of unity.
There are other methods to find these roots.
If you know De Moivre's law you can apply it.
$z^n = |z|^n(\cos \theta + i\sin \theta)^n = |z|^n(\cos n\theta + i\sin n\theta)$
Using this approach:
$(-1)^\frac 16 = (\cos \pi + i\sin \pi)^\frac 16 = \cos \frac {\pi}{6} + i\sin \frac {\pi}6$
$\cos \frac {(2k-1)\pi}{6} + i\sin \frac {(2k-1)\pi}6$ are also roots.
And by Euler formula
$e^{\pi i} = -1\\
e^{\frac {\pi}{6} i} = (-1)^\frac 16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3575653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Confusion concerning logarithmic differentiation The problem is stated as follows:
Find the equation of the line perpendicular to $f(x)=\pi^2+2^x+x^2+x^{1/x}$ at $x=1$.
My approach was this:
Notice that $f(1)=4+\pi^2$ so I am interested in the point $P(1, 4+\pi^2)$
Then taking the derivative
\begin{align}
\frac{df}{dx}&=\frac{d}{dx}\pi^2+\frac{d}{dx}2^x+\frac{d}{dx}x^2+\frac{d}{dx}x^{1/x}\\
&=2^x\ln{2}+2x+\frac{d}{dx}e^{1/x\ln{x}}\\
&=2^x\ln{2}+2x+x^{1/x}\cdot\left(-\frac{1}{x^2}\ln{x}+\frac{1}{x^2}\right)\\
\end{align}
Now finding $f'(1)$:
\begin{align}
f'(1)&=2\ln{2}+2+1\cdot\left(-\frac{1}{1^2}\ln{1}+\frac{1}{1^2}\right)\\
&=2\ln{2}+3
\end{align}
Now taking the opposite reciprocal gives
\begin{align}
-\frac{1}{2\ln{2}+3}
\end{align}
Now using point $P$ we get the following equation
\begin{align}
y-(4+\pi^2)&=-\frac{1}{2\ln{2}+3}\cdot(x-1)\\
y&=-\frac{(x-1)}{2\ln{2}+3}+4+\pi^2
\end{align}
I am quite certain that this is the correct answer. However, when I try a different method, namely logarithmic differentiation I get a different derivative:
Taking the natural log of both sides:
\begin{align}
\ln{f(x)}=\ln\pi^2+\ln{2^x}+\ln{x^2}+\ln{x^{1/x}}\\
\end{align}
And differentiating both sides:
\begin{align}
\frac{f'(x)}{f(x)}&=\frac{d}{dx}\ln{\pi^2}+\frac{d}{dx}x\cdot\ln{2}+\frac{d}{dx}2\ln{x}+\frac{d}{dx}\frac{1}{x}\ln{x}\\
f'(x)&=\left(\ln{2}+\frac{2}{x}+\frac{1}{x^2}-\frac{1}{x^2}\ln{x}\right)\cdot(\pi^2+2^x+x^2+x^{1/x})
\end{align}
Clearly something has gone very wrong here. Is there a flaw in my original answer or am I misunderstanding the logarithmic differentiation? Any help is appreciated, thanks!
| $$f(x)=\pi^2+2^x+x^2+x^{1/x}$$
$$\ln |f(x)|=\ln |\pi^2+2^x+x^2+x^{1/x}|$$
You don't have a product of factors on RHS.This line is not correct:
$$\ln{f(x)}\ne \ln\pi^2+\ln{2^x}+\ln{x^2}+\ln{x^{1/x}}\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578297",
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The Arts and Crafts of Problem Solving Problem 2.2.20 Let $n$ be a positive integer.
We can find integers $a,b$ such that $1 \leq a < b \leq n$, $a+b > n$ and $a,b$ are relatively prime.
Then we find $\frac{1}{ab}$.
Then we sum all of the found fractions.
For example if $n=6$, the ordered pairs are
$
(1,6),(5,6),(2,5),(3,5),(4,5),(3,4)
$
and the corresponding sum is
$
\frac{1}{6} + \frac{1}{30} + \frac{1}{10} + \frac{1}{15} + \frac{1}{20} + \frac{1}{12} = \frac{1}{2}
$
Investigate what happens with the other values of $n$, and conjecture something.
What I have found so far is that the sum always stays the same but I am having issues proving that it is true.
$\textbf{Partial Solution}$
If we start experimenting with more values of $n$, such as $n = 2$. Then the ordered pairs are
(1,2)
Then the corresponding sum is
$
\frac{1}{2} = \frac{1}{2}
$
$n = 5$. Then the ordered pairs are
(1,5),(2,5),(3,5),(4,5),(3,4)
Then the corresponding sum is
$
\frac{1}{5} + \frac{1}{10} + \frac{1}{15} + \frac{1}{20} + \frac{1}{12} = \frac{1}{2}
$
From this we can assume that the sum stays the same for all $n$
in the positive integers
| (My deleted solution shows more details.)
Hint: Induction.
Consider going from $n $ to $ n+1 $. (Use $n = 6$ to help think it through).
Step 1. Completely categorize all the pairs that you lose.
Step 2. Completely categorize all the pairs that you gain.
Step 3. Show that the sum from terms that you lose is equal to the sum from terms that you gain.
Step 4: Hence, the sum is a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3581239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the sum of the following infinite series? $$
\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} + \cdots
$$
So basically I separated it into two series
where:
one of them is $\left(\frac{1}{3}\right)^n$
And I use geometric series formula to find that this series equals $\frac{1}{2}$.
But I can't figure out the series of the other one.
Apparently the answer for the series combined is: $\frac{5}{8}$
What is the other series?
| $$9\left(\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729}+\cdots\right)=3+2+\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \cdots$$
so that
$$9S=5+S.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Parametrization Of A Curve - Intersection $x^2+y^2+z^2=1$ And $x+y=1$
Find parametrization of the curve given by the intersection of the unit sphere $x^2+y^2+z^2=1$ and $x+y=1$
So look on the intersection:
$$x^2+(1-x)^2+z^2=1\iff 2x^2-2x+z^2=0\iff 4(x-\frac{1}{2})^2+2z^2=1$$
Plug in just $x=\cos t,y=\sin t$ would not be useful?
we need to chose functions of $\cos t$ and $\sin t$ that will be based on $\cos^2x+\sin^2x=1$? or $x=\frac{1}{2}\cos t+\frac{1}{2}$ and $z=\frac{1}{\sqrt{2}} \sin t$?
| This yields the same result but gives insight on the intersection shape.
Define $u=\frac{1}{\sqrt{2}}(x+y)$ and $v=\frac{1}{\sqrt{2}}(x-y)$. We get $u^{2}+v^{2}+z^{2}=1$, $u=\frac{1}{\sqrt{2}}$, and consequently $v^{2}+z^{2}=\frac{1}{2}$.
This is a circle on the $yz$ plane, centered on origin with radius $\frac{1}{\sqrt{2}}$, translated $\frac{1}{\sqrt{2}}\hat{i}$ and rotated $\frac{\pi}{4}\hat{k}$.
The parametrization is $u=\frac{1}{\sqrt{2}}$, $v=\frac{1}{\sqrt{2}}\cos{t}$, $z=\frac{1}{\sqrt{2}}\sin{t}$ or $x=\frac{1}{2}(1+\cos{t})$, $y=\frac{1}{2}(1-\cos{t})$, $z=\frac{1}{\sqrt{2}}\sin{t}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$
Question: If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$.
My approach: Since $$\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b} \\ \implies \left(\frac{\sin^4x}{a}+\frac{\cos^4x}{b}\right)^2=\frac{1}{(a+b)^2} \\ \implies \frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}-\sin^2x\cos^2x\left(\frac{\sin^2x}{a}-\frac{\cos^2x}{b}\right)^2=\frac{1}{(a+b)^2}.$$
Therefore, if we can prove that $$\frac{\sin^2x}{a}-\frac{\cos^2x}{b}=0,$$ then we are done. But, I am not able to prove the same.
| Hint
Set $\sin^2x=s\iff\cos^2x=1-s$ in the given condition to form a quadratic equation in $s$
Solve to find $s=\sin^2x=\dfrac a{a+b}$
$\cos^2x=?$
| {
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"source": "stackexchange",
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The equation $x^4-2x^3-3x^2+4x-1=0$ has four distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1The equation $x^4-2x^3-3x^2+4x-1=0$ has four distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1<x_2<x_3<x_4$ and product of two roots is unity, then:
$Q-1$: Find $x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4$
$Q-2$: Find $x_2^3+x_4^3$
My attempt is as follows:-
$A-1$ : First I tried to find any trivial root, but was not able to find any. After that I tried following:-
$$x_1\cdot x_2+x_1\cdot x_3+x_1\cdot x_4+x_2\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_1\cdot x_4-x_2\cdot x_3$$
$$x_1\cdot x_2\cdot x_3\cdot x_4=-1$$
$$x_1\cdot x_4=\dfrac{-1}{x_2\cdot x_3}$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_1\cdot x_4-x_2\cdot x_3$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_2\cdot x_3+\dfrac{1}{x_2\cdot x_3}$$
But from here I was not able to proceed as I was not able to calculate value of $x_2\cdot x_3$
$A-2$ : $(x_2+x_4)(x_2^2+x_4^2-x_2\cdot x_4)$
Now here I was not getting any idea for how to proceed.
Please help me in this.
| Hint The product of two roots is $-1$ and the product of the other two roots is $1$.
Therefore
$$x^4-2x^3-3x^2+4x-1=(x^2+ax+1)(x^2+bx-1)$$
Oppening the brackets gives
$$a+b=-2\\
ab=-3 \\
b-a=4$$
which is trivial to solve.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to simplify series in a fraction Given
$$\\ A =\frac{1}{1 \cdot 2} + \frac{1}{3 \cdot 4} + \frac{1}{5 \cdot 6} + \ ... \ + \frac{1}{1997 \cdot 1998} \\ B =\frac{1}{1000 \cdot 1998} + \frac{1}{1001 \cdot 1997} + ... + \frac{1}{1998 \cdot 1000}$$
Simplify $\frac{A}{B}$
First of all, I broke down each fraction in A term to partial fractions:
$\frac{1}{1 \cdot 2} + \frac{1}{3 \cdot 4} + \frac{1}{5 \cdot 6} + \ ... \ + \frac{1}{1997 \cdot 1998} \\ = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \ ... \ + \frac{1}{1997} - \frac{1}{1998}$
At this point, I noticed that the series of the partial fractions can be formed in Catalan Identity
Catalan Identity: $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ ... \ + \frac{1}{2n - 1} - \frac{1}{2n} = \frac{1}{n + 1} + \frac{1}{n + 2} + \ ... \ + \frac{1}{2n}$
Then I applied this identity to the series:
$A = \frac{1}{1000} + \frac{1}{1001} + \ ... \ + \frac{1}{1998}$
Subtitute it to A, I get:
$\frac{A}{B} = \frac{\frac{1}{1000} + \frac{1}{1001} + \ ... \ + \frac{1}{1998}}{\frac{1}{1000 \cdot 1998} + \frac{1}{1001 \cdot 1997} + ... + \frac{1}{1998 \cdot 1000}}$
That's how far I got. I'm stucked at this point. Can anybody show me how to finish this problem?
| For any $a \neq 0$ and $b \neq 0$, you have
$$\frac{a + b}{ab} = \frac{1}{a} + \frac{1}{b} \tag{1}\label{eq1A}$$
Note the sum of each of the $2$ factors in the denominators of the summation in $B$ is the same, e.g., $2998 = 1000 + 1998 = 1001 + 1997 = \ldots$. Using this, along with \eqref{eq1A}, you have
$$\begin{equation}\begin{aligned}
2998\sum_{i=0}^{998}\frac{1}{(1000 + i)(1998 - i)} & = \sum_{i=0}^{998}\left(\frac{1}{1000 + i} + \frac{1}{1998 - i}\right) \\
& = \sum_{i=0}^{998}\left(\frac{1}{1000 + i}\right) + \sum_{i=0}^{998}\left(\frac{1}{1998 - i}\right) \\
& = 2\sum_{i=0}^{998}\left(\frac{1}{1000 + i}\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
The last line above comes from the second summation being of the same values as the first one, just in the opposite order. You now get
$$\sum_{i=0}^{998}\frac{1}{(1000 + i)(1998 - i)} = \frac{1}{1499}\sum_{i=0}^{998}\left(\frac{1}{1000 + i}\right) \tag{3}\label{eq3A}$$
Using your equation and \eqref{eq3A}, you thus have
$$\begin{equation}\begin{aligned}
\frac{A}{B} & = \frac{\frac{1}{1000} + \frac{1}{1001} + \ ... \ + \frac{1}{1998}}{\frac{1}{1000 \cdot 1998} + \frac{1}{1001 \cdot 1997} + ... + \frac{1}{1998 \cdot 1000}} \\
& = \frac{\sum_{i=0}^{998}\left(\frac{1}{1000 + i}\right)}{\frac{1}{1499}\sum_{i=0}^{998}\left(\frac{1}{1000 + i}\right)} \\
& = \frac{1}{\frac{1}{1499}} \\
& = 1499
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
| {
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Calculate the limit (verifying my answer). I have to calculate
$$\lim_{x\rightarrow 1^+} \frac{\sin(x^3-1)\cos(\frac{1}{1-x})}{\sqrt{x-1}}$$
Making the substitution $x-1=y$ and doing the math, I get that,
$$=\lim_{y\rightarrow 0^+} \frac{\sin(y(y^2+3y+3))}{y(y^2+3y+3)}\cdot\cos\Big(\dfrac{1}{y}\Big)\cdot\sqrt{y}(y^2+3y+3)$$
Since the first fraction goes to $0$. I have to worry with the $\cos(1/y)$, but I realized that $\cos(x)$ is bounded above and below, then, this kind of function times something that goes to $1$ results in $0$ (I studied this theorem). Since $\sqrt{y}$ goes to $0$. Then, the asked limit is $0$. Is that correct?
| hint
$$x^3-1=(x-1)(x^2+x+1)$$
hence
$$\sin(x^3-1)\sim 3(x-1) \;\;(x\to 1^+)$$
and
$$\frac{\sin(x^3-1)}{\sqrt{x-1}}\cos(\frac{1}{x-1} )\sim 3\sqrt{x-1}\cos(\frac{1}{x-1}) \;\;(x\to 1^+)$$
but
$$|\sqrt{x-1}\cos(\frac{1}{x-1})|\le \sqrt{x-1}$$
thus the limit is zero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cauchy-Schwarz inequality for $a_1^4 + a_2^4 + \cdots + a_n^4 \geqslant n$
Let $a_1+a_2,...,a_n \in \mathbb{R}.$ Show that if $a_1+a_2+...+a_n=n$, then $$a_1^4+a_2^4+...+a_n^4 \geqslant n.$$
The proposed solution for this was the following:
Using the Cauchy-Schwarz inequality twice we get
$a_1^4+a_2^4+...+a_n^4 \geqslant \frac{(a_1^2+a_2^2+...+a_n^2)^2}{n} \geqslant \frac{\frac{((a_1+a_2+...+a_n)^2)^2}{n}}{n} = \frac{(\frac{n^2}{n})^2}{n} = n$
I can see that we can deduce this straight from the definition, but where on earth does the denominator $n$ come for $a_1^4+a_2^4+...+a_n^4 \geqslant \frac{(a_1^2+a_2^2+...+a_n^2)^2}{n}$.
From Cauchy-Schwarz we can come up with $a_1^4+a_2^4+...+a_n^4 \geqslant (a_1^2+a_2^2+...+a_n^2)^2$, but I don't see where the denominator comes from. Could someone enlighten me?
| Also we can use Jensen. Let $f:\mathbb R \to \mathbb R, x\mapsto x^4$. Then $f$ is convex and thus by Jensen,
$$\frac{a_1^4+\dots+a_n^4}n=\frac{f(a_1)+\dots+f(a_n)}n\geq f\left(\frac{a_1+\dots+a_n}n\right)=\frac{(a_1+\dots+a_n)^4}{n^4}=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the Taylor polynomial for a function $f(x) = (1+x^2)^{1/3}$ I have to find the Taylor polynomial $T_2(x)$for a function $f$ of grade $2$ around $x_0 = 0$ given by
$$
f(x) = (1+x^2)^{\frac{1}{3}} \ \text{for} \ x \in \mathbb{R}
$$
I am very new to Taylor polynomials so I think what I have done is correct but I am just unsure whether or not they should give something 'pretty'. I know that they are approximations so they probably shouldn't but are you able to tell me whether or not my calculations are correct as I need this to be correct to be able to do some other questions which relies on me having the Taylor polynomial correct.
I have done the following:
\begin{align*}
T_2(x) & = \sum_{n=0}^2 \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n = f(x_0) + f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2 \\
& = f(0) + f'(0)(x-0) + \frac{f''(0)}{2!}(x-0)^2 = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2
\end{align*}
Thus
$$
f(0) = (1+0^2)^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1
$$
and
$$
f'(0) = \frac{2}{3} \cdot \frac{1}{(1+0^2)^{\frac{2}{3}}} \cdot 0 = 0
$$
and lastly
\begin{align*}
f''(x) & = (f'(x))' = \bigg( \frac{2}{3} \cdot \frac{1}{(1+x^2)^{\frac{2}{3}}} \cdot x \bigg)' = \frac{2}{3} \cdot \bigg(\frac{1}{(1+x^2)^{\frac{2}{3}}} \cdot x \bigg)' \\
& = \frac{2}{3} \cdot \Bigg( \bigg(\frac{1}{(1+x^2)^{\frac{2}{3}}} \bigg)' \cdot x + \frac{1}{(1+x^2)^{\frac{2}{3}}} \cdot x' \Bigg) \\
& = \frac{2}{3} \cdot \Bigg( - \frac{1}{((1+x^2)^\frac{2}{3})^2} \cdot \frac{2}{3} \cdot (1+x^2)^{\frac{2}{3}-1} \cdot 2x \cdot x + \frac{1}{(1+x^2)^{\frac{2}{3}}} \cdot 1 \Bigg)\\
& = \frac{2}{3} \cdot \Bigg( - \frac{1}{(1+x^2)^\frac{4}{3}} \cdot \frac{4}{3} \cdot \frac{1}{(1+x^2)^\frac{1}{3}} \cdot x^2 + \frac{1}{(1+x^2)^{\frac{2}{3}}} \Bigg) \\
& = \frac{2}{3} \cdot \Bigg( - \frac{4x^2}{3(1+x^2)^\frac{5}{3}} + \frac{1}{(1+x^2)^{\frac{2}{3}}} \Bigg) = \frac{2}{3} \cdot \Bigg (- \frac{4x^2}{3(1+x^2)^\frac{5}{3}} + \frac{3(1+x^2)}{3(1+x^2)^{\frac{5}{3}}} \Bigg) \\
& = \frac{2}{3} \cdot \Bigg ( \frac{x^2+3}{3(1+x^2)^{\frac{5}{3}}} \Bigg) = \frac{2(x^2+3)}{9(1+x^2)^{\frac{5}{3}}}
\end{align*}
which if we evaluate in $x_0 = 0$ gives
$$
f''(0) = \frac{6}{9^{\frac{5}{3}}}
$$
Thus
$$
T_2(x) = 1 + \frac{6}{2! \cdot 9^{\frac{5}{3}}}\cdot x^2
$$
I know that this is a lot of calculations but I hope some of you still wants to help me.
Thank you very much.
Regards
Mathias
| sorry your evaluation is wrong ist is NOT 9^(5/3) but just 9 so f''(0)=2/3
you have much less work, wenn you set x=0 in your first equation. so you have not to evaluate the derivative in (...)'*x
| {
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How can I show that the all of the roots of the polynomial $f = 2x^4-12x^2+2$ are real. I am given the polynomial
$$f = 2x^4-12x^2+2$$
and I have to show that all of the roots of this polynomial are real. I have no idea how to approach this. I tried plugging in the rational roots given by the rational root theorem, but none of them ($\pm1, \pm 2$) turned out to actually be roots, so I'm kind of lost.
| Just find them all.
By quadratic equation
$x^2 = \frac {12\pm \sqrt{144- 16}}{4}$.
So the four solutions are
$\sqrt{\frac {12+ \sqrt{144- 16}}{4}}, -\sqrt{\frac {12+ \sqrt{144- 16}}{4}},\sqrt{\frac {12- \sqrt{144- 16}}{4}}, -\sqrt{\frac {12- \sqrt{144- 16}}{4}} $
Now as $144 > 164$ then $\sqrt{144-16}$ is real. And as $\sqrt{144-16} < \sqrt {144} = 12$ then $12 - \sqrt{144-16} > 0$ so $\sqrt{\frac {12- \sqrt{144- 16}}{4}}$ is real and all the roots are real.
| {
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Pairing 8 numbers, counting probability with or without permutation This is a problem posted in this website yesterday, although it has been closed, this is quite an interesting problem.
Numbers $1,2,...,8$ are divided into pairs. Then the lower number in each pair is eliminated, leaving $4$ numbers. This is repeated, dividing the $4$ remaining numbers into pairs and eliminating the lower number in each pair, leaving $2$ numbers. What is the probability that $4$ is one of the two remaining numbers?
my attempt:
Alternative 1:
$4$ can survive twice if none of $1,2,3,4$ are paired with $5,6,7,8$ on both repetitions. Number of ways to divide $1,2,3,4$ into pairs is $3$, so is the number of ways to divide $5,6,7,8$ into pairs. Number of ways to divide $1,2,...,8$ into pairs is $\binom{8}{2\ 2\ 2\ 2}\frac{1}{4!}=105$. Thus the probability is $\frac{3\times 3}{105}=\frac{3}{35}$
Then $4$ needs to be paired with the other survivor from $1,2,3,4$. Probability is $\frac{1}{3}$. Therefore the probability of $4$ surviving is $\frac{3}{35}\times\frac{1}{3}=\frac{1}{35}$
Alternative 2:
We do tournament with $8$ slots. In order to have $4$ surviving to semifinal, we need to have $1,2,3,4$ in the first $4$ slots or the last $4$ slots ($2$ ways). Total number of choosing $4$ numbers from $1,2,...,8$ for the first $4$ slots is $\binom{8}{4}$. So the probability is $\frac{2}{\binom{8}{4}}=\frac{1}{35}$
Could You guys check my solutions and is there any other solutions? Thanks.
| Your solutions look good.
A perhaps somewhat simpler solution: In the knockout tournament, $4$ survives the first two rounds exactly if the other three numbers in its bracket are $1,2,3$. That’s one out of $\binom73=35$ possible choices for these three numbers.
| {
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Integrate $\int \frac{16x+16}{(x^2+2x+8)^9}dx$ using u substitution. Section 5.2
Can somebody verify this solution for me?
Integrate $\int \frac{16x+16}{(x^2+2x+8)^9}dx$ using u substitution.
Let $u=x^2+2x+8$. Then $\frac{du}{dx}=2x+2$ and so $\frac{du}{2x+2}=dx$. Thus we have:
$\int \frac{16x+16}{(x^2+2x+8)^9}dx$
$= \int \frac{16x+16}{u^9}\frac{du}{2x+2}$
$= \int \frac{8(2x+2)}{u^9}\frac{du}{2x+2}$
$= \int \frac{8}{u^9}du$
$= 8 \int u^{-9} du$
$= 8 \frac{u^{-8}}{-8}+C$
$= 8 \frac{(x^2+2x+8)^{-8}}{-8}+C$
$= -(x^2+2x+8)^{-8}+C$
| Your solution is correct. You can also use the fact that
$$\int f'(x)\,f(x)^{-9}\,dx=\frac{f(x)^{-8}}{-8}+C$$
and thus
$$\int\frac{16x+18}{(x^2+2x+8)^9}dx=8\int\overbrace{(2x+2)}^{=(x^2+2x+8)'}(x^2+2x+8)^{-9}dx=$$
$$=8\frac{(x^2+2x+8)^{-8}}{-8}+C=-\frac1{(x^2+2x+8)^8}+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Floor function repeated addition I've come across this problem: $$\left\lfloor r + \frac{19}{100}\right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100}\right\rfloor + \dots + \left\lfloor r + \frac{91}{100}\right\rfloor = 546$$ $$\text{Find} \
\lfloor 100r\rfloor. \textit{(Source: AIME)}$$
Here is my work:
$\lfloor r + \frac{19}{100}\rfloor = r + \frac{19}{100} - \{r + \frac{19}{100}\}$, so the figure can be restated as $\{r + \frac{19}{100}\} = r + \frac{19}{100} + a - 546$, where $a = \lfloor r + \frac{20}{100}\rfloor + \lfloor r + \frac{21}{100}\rfloor + \dots + \lfloor r + \frac{91}{100}\rfloor$.
Because $\{r + \frac{19}{100}\}$ is the fractional part, $0 \le r + \frac{19}{100} + a - 546 < 1$, so after some more maniuplation, $545 + \frac{81}{100} \le r + a < 546 + \frac{81}{100}$. $a$ is an integer, so the fractional part of $r$ must be $\frac{81}{100}$.
$r = \lfloor r\rfloor + \{r\}$, so $\lfloor r + \frac{19}{100}\rfloor = \lfloor \lfloor r\rfloor + \frac{81}{100} + \frac{19}{100}\rfloor$ = $\lfloor \lfloor r\rfloor + 1\rfloor$. Because both of the terms inside that floor function are integers, it must equal $\lfloor r\rfloor + 1$. This same reasoning can be applied to each of the individual floor functions of the given figure's LHS, and they each turn out to be $\lfloor r\rfloor + 1$.
Therefore, $73 \lfloor r\rfloor + 73 = 546$, so $73\lfloor r\rfloor = 473$. However, this cannot be true unless there is no answer (which I assume is not the case), or unless I did something wrong in my process, because then $\lfloor r\rfloor$ is not an integer.
If you do see the solution, it would be really nice if you did not give the answer in your response! Instead, maybe some helpful hints or partial solutions would be preferred.
| Don't worry about the fractional part. And don't worry about $[r]$. Find the precise value of $k$ where $[r + \frac k{100}]\ne [r+\frac {k+1}{100}]$. That is where $r + \frac {k}{100} < m \le r + \frac {k+1}{100}$ for some integer $m$.
====== my answer below ====
Well, what jumps at me is $0 < \frac k{100} < 1$ so all the $[r +\frac {k}{100}]$ are either one integer, call it $n$ or the next, $n+1$.
.
So if $b$ of them equal $n+1$ and $(73 -b)$ of them equal $n$ we have $(73-b)n + b(n+1) = 73n + b =546$ where $0\le b < 73$.
.
So as $546\equiv 35 \pmod {73}$ and $546= 7*73 + 35$ so $b=35$ and $n=7$.
.
So we have (I'll have to be careful not to do a fencepost error... $91-35=56$ so....) $[r+\frac{19}{100}],....,[r+\frac{56}{100}] = 7$ and $[r+\frac {57}{100}],...,[r+\frac {91}{100}] = 8$.
.
So $r + \frac {56}{100} < 8\le r+\frac {57}{100}$ so
.
$100r + 56 < 800 \le 100r + 57$
.
$100r < 744 \le 100r + 1$
.
And $100r -1 < 743 \le 100r$ so $743 \le 100r < 744$
.
So $[100r] = 743$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Using number bases to prove combinatorics problem
Rewriting the grid in base 8,
I claim that at this point, it is essentially proven. If I choose any set of numbers as described in the question, the numbers I select must contain exactly one every last digit(0-7), and exactly one of every second digit 0-7. Note that last digit of 0 would imply that it is no longer in the same residue since the numbers start from 1, so the number chosen that last digit of 0 bears an extra value of 8 rather than 0.
This means the sum of the numbers chosen are,
$$S=(70+60+50+40+30+20+10+1+2+3+4+5+6+7+10)_8$$
$$S=(8(7+6+5+4+3+2+1) + 1+2+3+4+5+6+7+8)_{10}$$
$$S=260_{10} \ \blacksquare$$
(Please confirm the validity of my proof, it's very different from the given answer to the question)
| Each numbercan be represented as $8r+c$ where $c$ is the column number and $r$ is the row number.(starting from 0) One number must be selected from every row, this means that every distinct $r$ and distinct $c$ must be selected.
$$\frac{8 \cdot 9}{2} + \frac {8 \cdot(7\cdot 8)}{2}=36+224=260$$
So your proof is valid.
| {
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solve for real x and y: $x(x^2-3y^2)=2, y(3x^2-y^2)=11$ Solve for real x and y:
$x(x^2-3y^2)=2$
$y(3x^2-y^2)=11$
My attempt: I got $(x-y)^3=13$ but this doesn't always hold, I got a solution $(2,1)$.
How to proceed?
| The system can be written as
\begin{align*}
x^3-3xy^2&=2\\
3x^2y-y^3&=11
\end{align*}
Then
\begin{align*}
x^3-3xy^2+(3x^2y-y^3)i&=x^3+3ix^2y-3xy^2-iy^3=2+11i\\
(x+iy)^3&=2+11i\quad (\star)
\end{align*}
but
$$2+11i=8+12i-6-i=2^3+3(2^2)(i)+3(2)(i^2)+i^3=(2+i)^3$$
Now, there are three cubic roots of $2+11i$: $2+i,\;(2+i)\omega\;\text{and}\;(2+i)\omega^2$ where $\omega=\frac{-1+i\sqrt 3}{2}$.
From equation $(\star)$ we get
a) $x+iy=2+i\qquad \implies\qquad (x,y)=(2,1)$
b) $x+iy=(2+i)\omega=(2+i)\left(\frac{-1+i\sqrt 3}2\right)=-1-\frac{\sqrt 3}2+\left(-\frac12+\sqrt 3\right)i$, then $(x,y)=\left(-1-\frac{\sqrt 3}2,\;-\frac12+\sqrt 3\right)$
c) $x+iy=(2+i)\omega^2=(2+i)\left(\frac{-1-i\sqrt 3}2\right)=-1+\frac{\sqrt 3}2+\left(-\frac12-\sqrt 3\right)i$, then $(x,y)=\left(-1+\frac{\sqrt 3}2,\;-\frac12-\sqrt 3\right)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction: $\dfrac{d^{2n}}{dx^{2n}}(x^2-1)^n = (2n)!$ Let $P_n$ be the statement that $\dfrac{d^{2n}}{dx^{2n}}(x^2-1)^n = (2n)!$
Base case: n = 0, $\dfrac{d^0}{dx^0}(x^2-1)^0 = 1 = 0!$
Assume $P_m = \dfrac{d^m}{dx^m}(x^2-1)^m = m!$ is true.
Prove $P_{m+1} = \dfrac{d^{2(m+1)}}{dx^{2(m+1)}}(x^2-1)^{m+1} = [2(m+1)]!$
$\dfrac{d^{2(m+1)}}{dx^{2(m+1)}}(x^2-1)^{m+1}$
= $\dfrac{d^{2m}}{dx^{2m}}\left(\dfrac{d^2}{dx^2}(x^2-1)^{m+1}\right)$
= $\dfrac{d^{2m}}{dx^{2m}}\left(2x(m)(m+1)(x^2-1)^{m-1}\right)$
= $[\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m][2x(m)(m+1)(x^2-1)^{-1}]$
From the inductive hypothesis,
= $(2m)! [2x(m)(m+1)(x^2-1)^{-1}]$
I got stuck here, and not sure if I have done correctly thus far? I did not know how to get to $[2(m+1)]!$. Please advise. Thank you.
| Assume:
$$P_m = \dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m = (2m)!$$
Then:
$$P_{m+1}=\dfrac{d^{2m}}{dx^{2m}}\left(\dfrac{d^2}{dx^2}(x^2-1)^{m+1}\right)=
\dfrac{d^{2m}}{dx^{2m}}\left(\dfrac{d}{dx}\left[2x(m+1)(x^2-1)^m\right]\right)=\\
2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((x^2-1)^m+2\overbrace{x^2}^{x^2-1+1}m(x^2-1)^{m-1}\right)=\\
\color{blue}{2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((x^2-1)^m+2m(x^2-1+1)(x^2-1)^{m-1}\right)}=\\
\color{blue}{2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((x^2-1)^m+2m(x^2-1)^{m}+2m(x^2-1)^{m-1}\right)}=\\
2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((2m+1)(x^2-1)^m+2m(x^2-1)^{m-1}\right)=\\
2(m+1)(2m+1)\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m+\overbrace{4m(m+1)\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^{m-1}}^{0}=\\
\color{blue}{2(m+1)(2m+1)\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m+\overbrace{4m(m+1)\dfrac{d^{2m}}{dx^{2m}}P_{2m-2}}^0}=\\
(2m+2)(2m+1)(2m)!=(2(m+1))!$$
$\color{blue}{\text{where $P_{2m-2}$ is a polynomial of degree $2m-2$, whose $2m$-th order derivative is zero.}}$
| {
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How to prove $x^2 + y^2 + z^2\geq xy + xz + yz$ Result: Let $, , ∈ ℝ$. Then we have $^2 + ^2 + ^2 ≥ + + $
Need some help proving this, just a few steps with work.
Was thinking you start with $x^2+y^2+z^2−xy−xz−yz$
then factor?
can anyone show me how to solve this?
$x^2+y^2+z^2≥xy+yz+zx⇔ $
$⇔2(x^2+y^2+z^2)≥2(xy+yz+zx)⇔$
$⇔x^2−2xy+y^2+y^2−2yz+z^2+z^2−2xz+x^2≥0⇔$
$⇔(x−y)^2+(y−z)^2+(z−x)^2≥0$
Should I do anything else? Or is this right?
| $$^2 + ^2 + ^2 ≥ + + $$
$$2x^2+2y^2+2z^2 ≥ 2xy+2xz+2yz$$
$$(x^2-2xy+y^2)+(x^2-2xz+z^2)+(y^2-2yz+z^2) ≥ 0$$
$$(x-y)^2+(x-z)^2+
(y-z)^2≥0$$
| {
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Chess tournament combinatorics There are $2n$ players in a chess tournament. The first round consists of pairing the players to
participate in $n$ matches with every player playing one match. In terms of $n,$ how many ways can
this pairing take place?
The first pair of people can be chosen from $\binom{2n}{2}.$
Since we chose the first pair, the next pair will be chosen from an amount of $\binom{2n-2}{2}.$
This pattern continues till $\binom{2}{2}$ which is for the $n$th pair.
Hence, the pairings will be $$\binom{2n}{2}\cdot\binom{2n-2}{2}\cdot...\cdot\binom{2}{2}.$$
Dividing by $n!$ gets the ways $2n$ people can be chosen to play $n$ games.
How do I simplify this?
| After solving the problem (which you did correctly), you could have simplified by observing that
$$ \binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1),$$
$$ \binom{2n-2}{2} = \frac{(2n-2)(2n-1)}{2} = (n-1)(2n-1),$$
and in general
$$ \binom{2k}{2} = \frac{2k(2k-1)}{2} = k(2k-1).$$
So the numerator you computed is
$$ n(2n-1)\cdot(n-1)(2n-3)\cdot(n-2)(2n-5)\cdots3(5)\cdot2(3)\cdot1(1).$$
In the denominator you have $n!$ canceling the terms
$n,$ $n-1,$ $n-2,\ldots,$ $3,2,1$ in the numerator, leaving
$$ (2n-1)\cdot(2n-3)\cdot(2n-5)\cdots(5)\cdot(3)\cdot(1),$$
which is the same formula you have seen derived in another way already.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3607398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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An algebraic way to do this question? Find minimum and maximum values of $|z_1+iz|$ where $|z-i|\leq5,\:\:z_1=5+3i$.
Let $z$ be a complex number such that$|z-i|\leq5$, and let $z_1=5+3i$.
Find the minimum and maximum values of $|z_1+iz|$.
The geometric way to do this is easy, just draw a circle of radius $5$ centered at $(0,1)$ and find the minimum and maximum distances from there. But is there a way to do this purely algebraically?
My attempt:
Let $z=a+ib$
$\sqrt{a^2+(b-1)^2}\leq5$
$a^2+b^2-2b+1\leq25\qquad[1]$
Now, $|z_1+iz|=\sqrt{(5-b)^2+(3+a)^2}$
Adding $6a-8b+33$ to $[1]$, we get $|z_1+iz|^2\leq58+6a-8b$
I don't know where to go from here. Please help.
| For a purely algebraic solution:
We have
$f(z) = |z_1+iz|^2 = (5-b)^2 + (3+a)^2 = 58 +6a - 8b = 50 + 6a - 8(b-1)$
It is clear that to maximise $f(z)$ subject to the constraint $a^2 + (b-1)^2 \le 25$, we must have $a^2 + (b-1)^2 = 25$, otherwise if $a^2 + (b-1)^2 < 25$ we could increase $a$ and/or decrease $b$ and so increase $f(z)$. So let $a=5 \sin \theta$ and $b-1 = 5 \cos \theta$. Then
$f(z) = 50 + 30 \sin \theta - 40 \cos \theta \\
\Rightarrow \frac {df}{d \theta} = 30 \cos \theta + 40 \sin \theta$
So $f$ has maximum and minimum values when
$30 \cos \theta + 40 \sin \theta = 0
\\ \Rightarrow \tan \theta = -\frac{3}{4}
\\ \Rightarrow (\sin \theta, \cos \theta) = (\frac 3 5, - \frac 4 5) \text{ or } (- \frac 3 5, \frac 4 5)$
To maximise $f(z)$ we take the first pair of values, so
$f(z)_{max} = 50 + \frac {90} 5 + \frac {160} 5 = 100 \\
\Rightarrow |z_1+iz| = 10$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
On the reducibility of $x^4+ a$ over the rationals Let $a \in \mathbb{Z^{+}}$. Show that $x^4+a$ is reducible over $\mathbb{Q}$ if and only if $a=4b^4$ for some integer $b$.
My idea for one implication was to assume reducibility and write $x^4+a = (x^2+\alpha_1x + \beta_1)(x^2+\alpha_2x + \beta_2)$, leading to the system of equations: $\alpha_1 + \alpha_2 = 0$, $\beta_1+\beta_2+\alpha_1 \alpha_2 = 0$, $\alpha_2 \beta_1 + \alpha_1 \beta_2 = 0$, $\beta_1 \beta_2 = a$. But I wasn't able to get much farther than this. This is a practice problem for an exam, any help is appreciated.
| Another solution is to first factor out into linear factors over $\mathbb{C}$, and multiply conjugate factors to obtain the real quadratic factors:
$$X^4 + a = \left(X^2 + ia^{1/2}\right)\left(X^2 - ia^{1/2}\right) = \color{red}{\left(X + \frac{1-i}{\sqrt{2}}a^{1/4}\right)}\color{blue}{\left(X - \frac{1-i}{\sqrt{2}}a^{1/4}\right)}\color{red}{\left(X + \frac{1+i}{\sqrt{2}}a^{1/4}\right)}\color{blue}{\left(X - \frac{1+i}{\sqrt{2}}a^{1/4}\right)}$$
$$ = \color{red}{\left(X^2 + \sqrt{2}a^{1/4}X + a^{1/2}\right)}\color{blue}{\left(X^2 - \sqrt{2}a^{1/4}X + a^{1/2}\right)}$$
where we used $\sqrt{i} = \pm \frac{1+i}{\sqrt{2}}$ and $\sqrt{-i} = \pm\frac{1-i}{\sqrt{2}}$.
(you can also check the factorization directly without using complex numbers).
From this expression it is clear that we need $\sqrt{2}a^{1/4} = 2r$ for some $r \in \mathbb{Q}$ which implies $a = 4r^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find the polynomial of integral coefficient with minimum degree and root $z+z^3+z^9$.
Let $z$ be a 13th root of unity $(z\neq 1)$. Find the polynomial of integral coefficient with minimum degree and root $z+z^3+z^9$.
My idea:
since $z$ such
$$0=z^{12}+z^{11}+\cdots+z+1=\prod_{k=1}^{12}(z-z_{k})$$ maybe use
$$\sum_{k=1}^{n}z_{k}=-1,\sum_{i<j}z_{i}z_{j}=1,\sum_{i<j<k}z_{i}z_{j}z_{k}=-1,\cdots$$
then we must find
$$\sum_{k=1}^{12}(z_{k}+z^3_{k}+z^9_{k})=?$$
$$\sum_{k<i}(z_{k}+z^3_{k}+z^9_{k})(z_{i}+z^3_{i}+z^9_{i})$$
$$\cdots$$
I think it's too complicated, and it seems like an interesting question, how to think correctly
| The hint.
Let $z+z^3+z^9=a$, $z^2+z^5+z^6=b$, $z^4+z^{10}+z^{12}=c$ and $z^7+z^8+z^{11}=d$.
Now, show that:
$$a+b+c+d=-1,$$
$$ac=3+b+d,$$
$$bd=3+a+c,$$
$$bc=a+b+c$$ and get the polynomial:
The second equation gives $$ac=2-a-c$$ or $$c=\frac{2-a}{a+1}.$$
From the fourth equation we obtain:
$$b(c-1)=a+c,$$ which gives
$$b=\frac{a^2+2}{1-2a}.$$
Also, $$d=-1-a-b-c=\frac{a^3+3a-5}{(a+1)(1-2a)}.$$
Thus, from the third equation we obtain:
$$\frac{a^2+2}{2a-1}\cdot\frac{a^3+3a-5}{(a+1)(2a-1)}=3+a+\frac{2-a}{a+1}$$ or
$$(a-5)(a^4+a^3+2a^2-4a+3)=0$$ or
$$a^4+a^3+2a^2-4a+3=0.$$
For example, $$ac=3+b+d$$ because
$$ac=(z+z^3+z^9)(z^4+z^{10}+z^{12})=$$
$$=z^5+z^{11}+1+z^7+1+z^2+1+z^6+z^8=$$
$$=3+z^2+z^5+z^6+z^7+z^8+z^{11}=3+b+d.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Recursive formula, eigenvalue problem A sequence $ \{X_i \}_{i \geq 0} $ is defined recursive by $ X_{n+1} = 3 X_{n} - 2X_{n-1}, \qquad n \geq1. $
and $ X_0 = \begin{pmatrix}
1 & 0 \newline
1 & 1
\end{pmatrix},X_1 = \begin{pmatrix}
1 & 1 \newline
1 & 1
\end{pmatrix}. $ Find an explicit formula for $ X_n $.
My attempt;
We can write $ = \begin{pmatrix}
X_{n+1} \newline X_{n} \end{pmatrix} = \begin{pmatrix} 3 & -2 \newline 1 & 0 \end{pmatrix} \begin{pmatrix} X_{n} \newline X_{n-1} \end{pmatrix} $ for $ n \geq 1. $ Diagonalizing the matrix yields
$\begin{pmatrix} X_{n+1} \newline X_{n} \end{pmatrix} = \begin{pmatrix} 2 & 1 \newline 1 & 1 \end{pmatrix} \begin{pmatrix} 2^n & 0 \newline 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \newline -1 & 2 \end{pmatrix} \begin{pmatrix} X_{n} \newline X_{n-1} \end{pmatrix}$
This is and old exam; now the answer concludes that $X_n $ can be written as $ X_n = A + 2^n B $, for some matrices A and B, which can be found by plugging in $ X_1 $ and $ X_0.$ My question is how did he draw this conclusion, is it only from the eigenvalues? How would one approach questions on this form if the matrix couldnt be diagonalized?
| We may prove the claim $X_n = A + 2^n B$ by induction. The inductive step is easy: for $n\geq 1$,
$$X_{n+1} = 3 X_{n} - 2X_{n-1}=3(A + 2^n B)-2(A + 2^{n-1} B)=A + 2^{n+1} B.$$
It remains to find $A$ and $B$ by considering the cases $n=0$ and $n=1$:
$$\begin{cases}
X_0=A + B\\
X_1=A + 2 B
\end{cases}\implies \begin{cases}
B=X_1-X_0\\
A=X_0-B=2X_0-X_1
\end{cases}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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transformation of functions. Let $f: A \rightarrow K, f(x)=\sqrt{2 x+c}-1,$ where $A$ is the maximal domain of $f$ and $c$ is a real number.
i. For what value(s) of $c$ does $f(x) = f^{-1}(x)$ have no real roots?
ii. For what value(s) of $c$ does $f(x) = f^{-1}(x)$ have exactly one real solution?
My answers:
i)
\begin{aligned}
&\sqrt{2 x+c}-1=x\\
&2 x+c=(x+1)^{2}\\
&x^{2}+2 x+1=2 x+c\\
&\begin{array}{rl}
& x^{2}+1-c=0 \\
a=1 & b=0 \quad c=(1-c)
\end{array}
\end{aligned}
.
.
\begin{array}{c}
D<0, b^{2}-4 a c<0 \\
0^{2}-4(1)(1-c)<0 \\
-4+4 c<0 \\
-1+c<0 \\
\therefore c<1
\end{array}
ii)
\begin{aligned}
&\begin{array}{l}
x^{2}+1-c=0 \\
b^{2}-4 a c=0 \\
0^{2}-4(1)(1-c)=0
\end{array}\\
&-4(1-c)=0\\
&\begin{array}{l}
1-c=0 \\
c=1
\end{array}
\end{aligned}
Answer is $c=1$ or $c>2$
How to do I know that $C>2$ without a calculator
| You first need to find $f^{-1}(x)$, whatever the value of $c$ is. First note that the range of $f$ is $[-1,\infty)$ (since $\sqrt{2x+c}$ may attain any nonnegative value).
Here's the easy way to find $f^{-1}(x)$: from $x$ to $f(x)$ we
*
*multiply $x$ by $2$;
*add $c$
*take the square root
*subtract one
The opposite process is
*
*add one
*square
*subtract c
*divide by $2$
So $f^{-1}(x)=\frac{(x-1)^2-c}{2}$, defined for $x\geq -1$
So the equation "$f(x)=f^{-1}(x)$" actually becomes
$$\sqrt{2x+c}-1=\frac{(x-1)^2-c}{2}$$
which is what you should try to solve.
Here's an equivalent way, which might be easier: Note that "$f(x)=f^{-1}(x)$" is equivalent ot $x=f^{-1}(f^{-1}(x))$,
$$x=\frac{(\frac{(x-1)^2-c}{2}-1)^2-c}{2}$$
As a further alternative, which avoid computing $f^{-1}(x)$, you can note that "$f(x)=f^{-1}(x)$" is equivalent to $f(f(x))=x$, which becomes
$$\sqrt{2(\left(\sqrt{2x+c}-1)+c\right)}-1=x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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General formula for $f(n)$
Let for $n\geq 3, C_n$ denote the $(2n) \times (2n)$ matrix such that all entries along the diagonal are $2$, all entries along the sub- and super-diagonal are $1$, all entries along the antidiagonal are $1$, all entries along the diagonals directly above and below the antidiagonal are $2$, and all other entries are zero.
Let $f(n) : \mathbb{N}\to\mathbb{N}, f(n) = \det(C_n)$ for $n\geq 3.$ Prove that $$f(n) = \begin{cases}0,&\text{if }n = 3k+2,\ k\in\mathbb{N}\\
3^n,& \text{otherwise}\end{cases}.$$
I'm not sure how to go about doing this. I tried cofactor expansion along the first column, but I couldn't make much progress. I can't seem to find a recursive relationship. So I just tried converting $C_n$ to an upper triangular matrix using row operations. This results in a matrix satisfying certain patterns, but I can't seem to find a way to prove why reducing the matrix always produces these patterns (I can prove that the $k$th diagonal entry of the resulting upper triangular matrix is $\frac{k+1}k,$ where $1\leq k\leq n$ but I can't deal with the other $n$ diagonal entries well).
| $\newenvironment{vsmatrix}{\left|\begin{smallmatrix}}{\end{smallmatrix}\right|}\def\mycolor#1{{\color{blue}#1}}$Denote by $D_n$ the determinant of the $2n × 2n$ matrix to be solved with $2$ replaced by $a$ and $1$ replaced by $b$. In the following calculation, rows to which Laplace's formula for determinants are applied are colored blue. If a determinant has no colored entry, it means the determinant after the next equal sign is derived by applying elementary transformations.$$
D_1 = \begin{vmatrix}a & b \\ b & a\end{vmatrix} = a^2 - b^2,\ D_2 = \begin{vmatrix}
a & b & a & b\\
b & a & b & a\\
a & b & a & b\\
b & a & b & a
\end{vmatrix} = 0,
$$\begin{align*}
D_3 &= \begin{vmatrix}
a & b &&& a & b\\
b & a & b & a & b & a\\
& b & a & b & a &\\
& a & b & a & b &\\
a & b & a & b & a & b\\
b & a &&& b & a
\end{vmatrix}
= \begin{vmatrix}
a & b &&& a & b\\
b & 0 & 0 & 0 & 0 & a\\
& b & a & b & a &\\
& a & b & a & b &\\
a & 0 & 0 & 0 & 0 & b\\
b & a &&& b & a
\end{vmatrix}
= \begin{vmatrix}
a & b &&& a & b\\
b & 0 & 0 & 0 & 0 & a\\
& \mycolor{0} & \mycolor{a} & \mycolor{b} & \mycolor{0} &\\
& \mycolor{0} & \mycolor{b} & \mycolor{a} & \mycolor{0} &\\
a & 0 & 0 & 0 & 0 & b\\
b & a &&& b & a
\end{vmatrix}\\
&= (a^2 - b^2) \begin{vmatrix}
a & b & a & b\\
b & 0 & 0 & a\\
a & 0 & 0 & b\\
b & a & b & a
\end{vmatrix}
= (a^2 - b^2) \begin{vmatrix}
0 & b & a & 0\\
\mycolor{b} & \mycolor{0} & \mycolor{0} & \mycolor{a}\\
\mycolor{a} & \mycolor{0} & \mycolor{0} & \mycolor{b}\\
0 & a & b & 0
\end{vmatrix}\\
&= -(a^2 - b^2)^2 \begin{vmatrix}b & a \\ a & b\end{vmatrix} = (a^2 - b^2)^3.
\end{align*}
For any $n \geqslant 4$,\begin{align*}
D_n &= \begin{vsmatrix}
⋱ & ⋱ &&&&& ⋰ & ⋰\\
⋱ & a & b &&& a & b & ⋰\\
& b & a & b & a & b & a &\\
&& b & a & b & a &&\\
&& a & b & a & b &&\\
& a & b & a & b & a & b &\\
⋰ & b & a &&& b & a & ⋱\\
⋰ & ⋰ &&&&& ⋱ & ⋱
\end{vsmatrix}_{2n × 2n}\mskip-18mu
= \begin{vsmatrix}
⋱ & ⋱ &&&&& ⋰ & ⋰\\
⋱ & a & b &&& a & b & ⋰\\
& b & 0 & 0 & 0 & 0 & a &\\
&& b & a & b & a &&\\
&& a & b & a & b &&\\
& a & 0 & 0 & 0 & 0 & b &\\
⋰ & b & a &&& b & a & ⋱\\
⋰ & ⋰ &&&&& ⋱ & ⋱
\end{vsmatrix}_{2n × 2n}\mskip-18mu
= \begin{vsmatrix}
⋱ & ⋱ &&&&& ⋰ & ⋰\\
⋱ & a & b &&& a & b & ⋰\\
& b & 0 & 0 & 0 & 0 & a &\\
&& \mycolor{0} & \mycolor{a} & \mycolor{b} & \mycolor{0} &&\\
&& \mycolor{0} & \mycolor{b} & \mycolor{a} & \mycolor{0} &&\\
& a & 0 & 0 & 0 & 0 & b &\\
⋰ & b & a &&& b & a & ⋱\\
⋰ & ⋰ &&&&& ⋱ & ⋱
\end{vsmatrix}_{2n × 2n}\\
&= (a^2 - b^2) \begin{vsmatrix}
⋱ & ⋱ &&&&& ⋰ & ⋰\\
⋱ & a & b &&& a & b & ⋰\\
& b & a & b & a & b & a &\\
&& b & 0 & 0 & a &&\\
&& a & 0 & 0 & b &&\\
& a & b & a & b & a & b &\\
⋰ & b & a &&& b & a & ⋱\\
⋰ & ⋰ &&&&& ⋱ & ⋱
\end{vsmatrix}_{2(n - 1) × 2(n - 1)}\mskip-54mu
= (a^2 - b^2) \begin{vsmatrix}
⋱ & ⋱ &&&&& ⋰ & ⋰\\
⋱ & a & b &&& a & b & ⋰\\
& b & 0 & b & a & 0 & a &\\
&& \mycolor{b} & \mycolor{0} & \mycolor{0} & \mycolor{a} &&\\
&& \mycolor{a} & \mycolor{0} & \mycolor{0} & \mycolor{b} &&\\
& a & 0 & a & b & 0 & b &\\
⋰ & b & a &&& b & a & ⋱\\
⋰ & ⋰ &&&&& ⋱ & ⋱
\end{vsmatrix}_{2(n - 1) × 2(n - 1)}\\
&= -(a^2 - b^2)^2 \begin{vsmatrix}
⋱ & ⋱ &&&&& ⋰ & ⋰\\
⋱ & a & b &&& a & b & ⋰\\
& b & a &&& b & a &\\
&& b & b & a & a &&\\
&& a & a & b & b &&\\
& a & b &&& a & b &\\
⋰ & b & a &&& b & a & ⋱\\
⋰ & ⋰ &&&&& ⋱ & ⋱
\end{vsmatrix}_{2(n - 2) × 2(n - 2)}\mskip-54mu
= -(a^2 - b^2)^2 \begin{vsmatrix}
⋱ & ⋱ &&&&& ⋰ & ⋰\\
⋱ & a & b &&& a & b & ⋰\\
& b & a &&& b & a &\\
&& \mycolor{0} & \mycolor{b} & \mycolor{a} & \mycolor{0} &&\\
&& \mycolor{0} & \mycolor{a} & \mycolor{b} & \mycolor{0} &&\\
& a & b &&& a & b &\\
⋰ & b & a &&& b & a & ⋱\\
⋰ & ⋰ &&&&& ⋱ & ⋱
\end{vsmatrix}_{2(n - 2) × 2(n - 2)}\\
&= (a^2 - b^2)^3 \begin{vsmatrix}
⋱ & ⋱ &&&&& ⋰ & ⋰\\
⋱ & a & b &&& a & b & ⋰\\
& b & a & b & a & b & a &\\
&& b & a & b & a &&\\
&& a & b & a & b &&\\
& a & b & a & b & a & b &\\
⋰ & b & a &&& b & a & ⋱\\
⋰ & ⋰ &&&&& ⋱ & ⋱
\end{vsmatrix}_{2(n - 3) × 2(n - 3)}\mskip-54mu
= (a^2 - b^2)^3 D_{n - 3}.
\end{align*}
Therefore,$$
D_n = \begin{cases}
(a^2 - b^2)^n; & n \not\equiv 2 \pmod{3}\\
0; & n \equiv 2 \pmod{3}
\end{cases}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve the following expression I have the following expression: $2^{n-3}+\sum_{k=1}^{n-3} 2^{k-1}(n-k+1)^2$ and I have no idea how to solve it. I tried plugging it in WolframAlpha to get an idea but it gives me $-n^2 - 4 n + 7 \cdot 2^{n - 1} - 6$ and I have even less of an idea how I would get there.
I don't need the complete solution done. I just would like pointers on how to solve it.
| Expand:
$$E=2^{n-3}+\sum_{k=1}^{n-3} 2^{k-1}(n-(k-1))^2=\\
2^{n-3}+\color{red}{n^2\sum_{k=1}^{n-3} 2^{k-1}}-2n\color{green}{\sum_{k=1}^{n-3}(k-1)2^{k-1}}+\color{blue}{\sum_{k=1}^{n-3}(k-1)^22^{k-1}}$$
One:
$$\color{red}{n^2\sum_{k=1}^{n-3} 2^{k-1}}=n^2\cdot \frac{2^{n-3}-1}{2-1}=\color{red}{n^2\cdot 2^{n-3}-n^2}$$
Two:
$$f(x)=\sum_{k=1}^{n-3} x^{k-1}=\frac{x^{n-3}-1}{x-1}\\
f'(x)=\sum_{k=1}^{n-3} (k-1)x^{k-2}=\frac{(n-3)x^{n-4}(x-1)-x^{n-3}+1}{(x-1)^2}\\
xf'(x)=\sum_{k=1}^{n-3} (k-1)x^{k-1}=\frac{(n-4)x^{n-2}-(n-3)x^{n-3}+x}{(x-1)^2}\\
2f'(2)=\color{green}{\sum_{k=1}^{n-3} (k-1)2^{k-1}}=(n-4)2^{n-2}-(n-3)2^{n-3}+2=\\
\color{green}{n\cdot 2^{n-3}-5\cdot 2^{n-3}+2}$$
Three:
$$f(x)=\sum_{k=1}^{n-3} (k-1)x^{k-1}=\frac{(n-4)x^{n-2}-(n-3)x^{n-3}+x}{(x-1)^2}\\
f'(x)=\sum_{k=1}^{n-3} (k-1)^2x^{k-2}=\cdots\\
xf'(x)=\sum_{k=1}^{n-3} (k-1)^2x^{k-1}=\cdots\\
2f'(2)=\color{blue}{\sum_{k=1}^{n-3} (k-1)^22^{k-1}}=\cdots=\\
\color{blue}{n^2\cdot 2^{n-3}-5n\cdot 2^{n-2}+27\cdot 2^{n-3}-6}$$
If all plugged in, it results in the stated answer.
Wolfram answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is: If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:
$A)$ $\le3$ , $B)$ $\ge3$, $(C)$ $\ge2$, $(D)$ $\le2$
My attempt is as follows:-
$$\dfrac{1}{2}\left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right)$$
Let $y=\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}$
$$A.M\ge H.M$$
$$\dfrac{\dfrac{s-a}{a}+\dfrac{s-b}{b}+\dfrac{s-c}{c}}{3}\ge \dfrac{3}{y}$$
$$\dfrac{\dfrac{s\cdot (ab+bc+ca)}{abc}-3}{3}\ge\dfrac{3}{y}$$
$$y\ge\dfrac{9}{\dfrac{(a+b+c)(ab+bc+ca)}{2abc}-3}\tag{1}$$
Let $z=\dfrac{(a+b+c)(ab+bc+ca)}{abc}$
Equation $(1)$ will give us $y_{min}$, so for that we need to find maximum value of $z$
But unfortunately I was able to find minimum value of $z$ in the following way
$$\dfrac{a+b+c}{3}\ge \dfrac{3abc}{ab+bc+ca}$$
$$\dfrac{(a+b+c)(ab+bc+ca)}{abc}\ge 9$$
But nevertheless, I tried plugging this minimum value of $z$ into equation $(1)$ and I got $y\ge 6$ and as the original expression was $\dfrac{y}{2}$, so $\dfrac{y}{2}\ge 3$ and surprisingly this answer is correct. What am I missing here?
| I don't see what's wrong with your answer, but it is a long method that gives an incorrect answer, so I'd suggest another way, such as the following. By scaling we can assume that $a + b + c = 1$, in which case we want to find a lower bound for
$${a \over 1- 2a} + {b \over 1 - 2b} + {c \over 1 - 2c} $$
$$-{3 \over 2} + {1 \over 2}\bigg({1 \over 1- 2a} + {1 \over 1 - 2b} + {1 \over 1 - 2c}\bigg)$$
Apply the AM-HM inequality in the expression in parentheses and the above is greater than or equal to
$$-{3 \over 2} + {9 \over 2}\bigg({ 1- 2a} + { 1 - 2b} + { 1 - 2c}\bigg)^{-1}$$
Since $a + b + c = 1$ this is exactly equal to $3$, which is achieved when $a = b = c = {1 \over 3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Verify triginometric result of cubic equation $x^3-x^2-p^2x+p^2=0$ Consider the following cubic function,
$$f(x):=(x+p)(x-p)(x-1)=x^3-x^2-p^2x+p^2$$
where $p\in(0,1)$ is a fixed parameter.
Then the sum of the absolute value of the three roots is
$$S_1:=1+2p$$
On the other hand, if I use the trigonometric root formula to calculate the absolute sum, I obtained
$$ S_2:=\frac{4}{3} \sqrt{1+3p^2}\, \cos \Bigg(\frac{1}{3}\arccos\Bigg(\frac{1-9p^2}{(1+3p^2)^{\frac{3}{2}}}\Bigg)-\frac{\pi}{3}\Bigg)+\frac{1}{3}$$
Then, obviously, the $\textbf{following is true}$ $$S_1=S_2 \quad \forall p\in(0,1)$$
$\textbf{Question:}$ Is there a way to verify $S_1=S_2$ by direct calculation?
(I've tried with the Taylor series of the $\arccos(x)$ but I've not succeded.)
| Let
$$ \theta = \arccos \left(\frac{1-9p^2}{(1+3p^2)^{\frac{3}{2}}} \right)-\pi$$
Then,
$$ S_2=\frac{4}{3} \sqrt{1+3p^2}\, \cos \frac{\theta}3+\frac{1}{3}\tag1$$
and
$$\cos\theta =-\frac{1-9p^2}{(1+3p^2)^{\frac{3}{2}}}=4\cos^3\frac{\theta}3 - 3\cos\frac{\theta}3$$
Factorize
$$\left(2\cos\frac{\theta}3 - \frac{1+3p}{\sqrt{1+3p^2}} \right)
\left(2\cos^2\frac{\theta}3 + \frac{1+3p}{\sqrt{1+3p^2}} \cos\frac{\theta}3
+\frac{3p-1}{1+3p^2}\right) =0 $$
to get
$$\cos\frac{\theta}3 = \frac{1+3p}{2\sqrt{1+3p^2}}$$
Plug into (1) to obtain
$$ S_2
=\frac{4}{3} \sqrt{1+3p^2}\cdot \frac{1+3p}{2\sqrt{1+3p^2}}+\frac{1}{3}
=1+2p=S_1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3628351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find locus of $S$ denoting set of complex numbers $\frac{z+1}{z-3}$, where $z$ varies over set of $|z|=1$.
Question: Let $S$ denote the set of all complex numbers of the form $\frac{z+1}{z-3}$, where $z$ varies over the set of all complex numbers with $|z|=1.$ Find the locus of the points in set $S$.
My approach:
Let $z=x+iy$, with $x,y\in\mathbb{R}$ such that $|z|=1\implies |z|^2=1$. This implies that we must have $x^2+y^2=1$. Now $z+1=(x+1)+iy$ and $z-3=(x-3)+iy$.
Thus
\begin{align*}
\frac{z+1}{z-3}&=\frac{(z+1)(\overline{z-3})}{|z-3|^2}\\
&=\frac{(z+1)(\overline{z}-3)}{|z-3|^2}\\
&=\frac{x^2-2x-3+y^2-4iy}{(x-3)^2+y^2}\\
&=\frac{-2x-2-4iy}{10-6x}\\
&=\frac{-x-1-2iy}{5-3x}.
\end{align*}
Thus we have
$$
\Re\left(\frac{z+1}{z-3}\right)=\frac{x+1}{3x-5}
$$
and
$$
\Im\left(\frac{z+1}{z-3}\right)=\frac{2y}{3x-5},$$ and our task is to find a relationship between these two given that $|z|=1$.
For our ease let us have $\alpha=\frac{z+1}{z-3}\,\, \forall z$ satisfying $|z|=1$.
Now we obtain two useful information using the triangle inequality. We have
$$
|z+1|\le |z|+1=2
$$
and
$$
|z-3|\le |z|+3=4.
$$
From here we can conclude that $$|z+1|^2=(x+1)^2+y^2=2+2x\le 4 \iff 1+x\le 2\text{ and } |z-3|^2=5-3x\le 16.$$
Thus we have $$0\le 1+x\le 2 \text{ and } 0\le 5-3x\le 16\implies -1\le x\le 1.$$
Observe that even from $x^2+y^2=1$, we can directly conclude that $-1\le x,y\le 1$.
But this doesn't help much to find a relationship between $\Re(\alpha)$ and $\Im(\alpha)$.
How to proceed?
| Let $w=\frac{z+1}{z-3}\implies z=\frac{3w+1}{w-1}$. Then, the given $|z|^2=1$ leads to
$$\frac{3w+1}{w-1}\frac{3\bar w+1}{\bar w-1}=1\implies |w|^2+\frac12(w+\bar w)=0$$
which is just
$$|w+\frac12|^2=\frac1{2^2}$$
Thus, the locus is a circle with center at $-\frac12$ and radius $\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Calculus A Level Line Tangent to Circle How can you find values of $k$ such that $y = kx + 1$ is tangent to the circle $(y-1)^2 + (x-5)^2 = 9 $?
I first rewrote the circle equation in terms of y:
$$ (y-1)^2 = -(x-5)^2 + 9 \\y-1 = \pm\sqrt{-x^2+10x -16} \\ y = \pm\sqrt{-x^2+10x -16} + 1 $$
Then I took two derivatives, one for the positive equation and one for the negative equation:
$$y\prime_{_{+}} = \frac{1}{2}(-x^2+10x-16)^\frac{-1}{2}(-2x+10) \\=-(x-5) (-x^2+10x-16)^\frac{-1}{2}\\y\prime_{_{-}} = -\frac{1}{2}(-x^2+10x-16)^\frac{-1}{2}(-2x+10) \\=(x-5) (-x^2+10x-16)^\frac{-1}{2}$$
So if I look at just the positive semicircle, I have a line $y = kx + 1$ which intersects $y = \sqrt{-x^2+10x-16} + 1 $ at one spot and tangent at that spot $ k = -(x-5)(-x^2+10x-16)^\frac{-1}{2} $.
How do I actually find the value of $k$?
| If the line is a tangent to the circle, it means that the line and circle have one point in common.
Substituting $y = kx + 1$ into the equation of the circle gives $9 = (kx)^2 + (x-5)^2$, which simplifies to $0 = (k^2+1)x^2 - 10x + 16$.
The discriminant of the quadratic must be zero:
$\Delta = (-10)^2 -4(k^2+1)(16) \rightarrow k = \pm \frac{3}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3632373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that sum of matrices equals zero
The matrix $A$ has size $3 \times 3$ and we know that for any column
vector $v\in \mathbb{R}^{3}$ the vectors $Av$ and $v$ are orthogonal. Prove that
$A^{T} + A = 0$, where $A^{T}$ is the transposed matrix $A$.
So if
$$A = \begin{pmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{pmatrix}$$
and
$$v = \begin{pmatrix}
x\\
y\\
z
\end{pmatrix}$$
orthogonality of $Av$ and $v$ brought me to the equation
$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) + a_{11} \cdot x^{2} + a_{22} \cdot y^{22} + a_{33} \cdot z^{2} = 0$
and the matrix $A^{T} + A$ equals
$$A^{T} + A = \begin{pmatrix}
2a_{11} & a_{12} + a_{21} & a_{13} + a_{31}\\
a_{12} + a_{21} & 2a_{22} & a_{23} + a_{32}\\
a_{13} + a_{31} & a_{23} + a_{32} & 2a_{33}
\end{pmatrix}$$
However I don't see how then prove that $A^{T} + A = 0$.
| You know that $A.(1,0,0)$ and $(1,0,0)$ are orthogonal. Since $A.(1,0,0)=(a_{11},a_{21},a_{31})$, this means that $a_{11}=0$. By a similar argument, $a_{22}=a_{33}=0$.
So,$$A=\begin{bmatrix}0&a_{12}&a_{13}\\a_{21}&0&a_{23}\\a_{31}&a_{32}&0\end{bmatrix}.$$And so $A.(1,1,0)=(a_{12},a_{21},a_{31}+a_{32})$. But this vector is orthogonal to $(1,1,0)$, which means that $a_{12}=-a_{21}$. By a similar argument, $a_{13}=-a_{31}$ and $a_{23}=-a_{32}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\sec x + \csc x =p$ has four distinct solutions between $(0,2\pi)$, then which if the following is incorrect?
a) $p^2-8>0$
b) $p=\sqrt 2$
c) $p=-\sqrt 2$
d) $p=0$
My attempt
$$\frac{\sec x +\csc x}{2} \ge \sqrt {\sec x \csc x}$$
$$\frac{\sin x +\cos x}{\sin x \cos x }\ge 2\sqrt {\frac{1}{\sin x \cos x}}$$
$$\sin x +\cos x \ge 2\sqrt {\sin x \cos x}$$
$$(\sqrt {\sin x}-\sqrt {\cos x})^2\ge 0$$
$$\sin x \ge \cos x$$
I realise that some of that squaring might have removed or added some roots, but I don’t know what else to do
From this result, the interval for $x$ is $[\frac{\pi}{4}, \pi]\cup [\frac{5\pi}{4}, \frac{3\pi}{2}]$
Don’t know what do next. Can I get some insight?
Another attempt
$$\sin x +\cos x =p \sin x \cos x$$
$$1+2\sin x\cos x =\frac 14 p^2 4\sin^2x \cos ^2x$$
$$1+\sin 2x =\frac 14 p\sin^2 2x$$
$$p\sin^22x-4\sin 2x -4=0$$
| The derivative of $p$ is equal to $\dfrac{\sin(x)}{\cos^2(x^2)}-\dfrac{\cos(x)}{\sin^2(x)}$ so$p$ has a minimum at $x=\dfrac{\pi}{4}$ and a maximum at $x=\dfrac{5\pi}{4}$. Consequently $p$ has four values in $[0,2\pi]$ for $\sec x + \csc x \gt2\sqrt2$ and for
$\sec x + \csc x \lt-2\sqrt2$.
Thus only $p^2-8\gt0$ is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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When does $\sqrt{x+\sqrt{x+1+\sqrt{x+2+...}}}=0$? Consider the function $f$ defined as the limit of the functions
$$f_0(x)=\sqrt{x}$$
$$f_1(x)=\sqrt{x+\sqrt{x+1}}$$
$$f_2(x)=\sqrt{x+\sqrt{x+1+\sqrt{x+2}}}$$
$$...$$
so that $f(x)$ is defined iff $f_n(x)$ is defined for some $n$. The unique root $x_0$ of the function $f$ satisfies $f(x_0)=0$, and it can be alternatively expressed as the limit of the roots of $f_0, f_1, f_2, ...$. See the graphs below:
Can anyone find an expression equal to this limit? I realize that the chances of something nice and closed-form are slim - can we find a series, integral, or even nested radical representation of the real root of $f(x)$?
| $\color{brown}{\textbf{The task standing.}}$
Let $-r$ is the root of $f(x).$
Since $f(x)$ is increasing function and
$$f(-r)=0,\quad f(-1) > f_2(-1) = 0,$$
then $r>1.$
At the same time,
\begin{cases}
f(x+1) = f^2(x)-x\\
f(-r)=0\\
f(-r+1) = r\\
f(-r+2) = r^2+r-1\\
f(-r+3)= (r^2+r-1)^2 + r-2 = (r-1)(r^4+3r^2+2r+1)\dots.\tag1
\end{cases}
Besides,
\begin{cases}
f'(x+1) = 2f(x)f'(x)-1\\
f'(-r) = \infty\\
f'(-r+2) = 2rf'(-r+1)-1\\
f'(-r+3) = 2(r^2-r+1)f'(-r+2)-1\dots\tag2
\end{cases}
Conditions $(1),(2)$ allow to define unknown parameters in the various parametric representations of $f(x).$
$\color{brown}{\textbf{Inverse polynomial model.}}$
Let the inverse function is $f^{-1}(x)=g(y),$ then
\begin{cases}
g(0) = -r\\
g'(0)=0\\
g(r) = -r+1\\
g(r^2+r-1) = -r+2\\
g((r-1)(r^4+3r^2+2r+1)) = -r+3,\tag3
\end{cases}
and the coefficients of the cubic approximating polynomial in the form of
$$g(y)=-r+qy^2+py^3\tag4$$
can be obtained from the algebraic system
\begin{cases}
-r+1 = -r + pr^2 + qr^3\\
-r+2 = -r + p(r^2+r-1)^2 + q(r^2+r-1)^3\\
-r+3 = -r + p(r-1)^2(r^4+3r^2+2r+1)^2 + q(r-1)^3(r^4+3r^2+2r+1)^3,
\end{cases}
with the single positive solution
$$p\approx 0.0622998,\quad q\approx 0.606587,\quad r\approx1.21088$$
(see also Wolfram Alpha results).
Therefore, the cubic model estimation of root of $f(x)$ is $\color{brown}{\mathbf{-1.21088}}.$
$\color{brown}{\textbf{Power model.}}$
Taking in account conditions $(1),$ can be used the explicit power model
$$f(x) = r(x+r)^p,\tag5$$
where
\begin{cases}
r^2+r-1 = r\cdot 2^p\\
(r-1)(r^3+3r^2+2r+1) = r\cdot 3^p,
\end{cases}
The power model estimation of root of $f(x)$ is $\color{brown}{\mathbf{-1.21168}}.$
Plot of the inverse functions for the considered models see below.
Using of detalized models can improve approximations accuracy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635666",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "25",
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If $a,b,c>0$ prove that $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge ab+bc+ca$
If $a,b,c>0$ prove that $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge ab+bc+ca$
Simplifying yields
$a^4c+b^4a+c^4b \ge ab^2c^2+a^2b^2c+a^2bc^2$
Which readily follows from muirhead. I read some where that all muirhead 'like' inequalities can be proven with AM GM HM basic inequalities. I tried to prove it using AM GM,but failed. Maybe a clever substitution can clear the clouds?? Is this even possible to do it?If yes, would you share it?
| Hint:
$$\frac{a^3}{b} + ab \geq 2a^2 \\ \frac{b^3}{c} + bc \geq 2b^2
\\ \frac{c^3}{a} + ac \geq 2c^2 \\
$$
Edit: Basically it's AM-GM , note : $$\frac{a^3}{b}+ab \ge2 \sqrt{\frac{a^3}{b}\cdot ab}=2a^2$$
You can take it from here.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A question about meaning of a notation When we have a sum of the form
$$ \sum_{cyc} \dfrac{ (a+b)(a+c) - bc }{(b-c)(b^3-c^3)} $$
Does this mean:
$$ \dfrac{ (a+b)(a+c) - bc }{(b-c)(b^3-c^3)} + \dfrac{ (b+c)(b+a) - ac }{(a-c)(a^3-c^3)} + \dfrac{ (c+b)(c+a) - ba }{(b-a)(b^3-a^3)} $$
?
| It means that you take the sum over the cycles based on $(a \ b \ c)$ which are
$$\begin{cases}
(a \ b \ c) \to (a \ b \ c)\\
(a \ b \ c) \to (b \ c \ a)\\
(a \ b \ c) \to (c \ a \ b)
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3637088",
"timestamp": "2023-03-29T00:00:00",
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If $x=y^2+z^2$, $y=z^2+x^2$, and $z=x^2+y^2$, prove $\frac{x}{x+1} + \frac{y}{y+1} + \frac{z}{z+1} = 1$ If $x=y^2+z^2$, $y=z^2+x^2$, and $z=x^2+y^2$,
prove
$$\frac{x}{x+1} + \frac{y}{y+1} + \frac{z}{z+1} = 1$$
| First of all there is an evident solution $x=y=z=0$ for which the second identity is... false.
We will assume $x,y,z \neq 0$ in the following.
A first remark is that, evidently $x,y,z >0$
Let us call (1),(2),(3) your equations.
Make the difference (1)-(2):
$$x-y=y^2-x^2 \ \iff x-y=-(x-y)(x+y)$$
Let us assume for a while that $x-y=0$ : in this case, we would deduce that $x+y=-1$ which is impossible due to the positivity of y o$x $ and $y$.
Therefore, $x=y$.
For the same reason, using difference (1)-(3), we get $x=z$.
Replacing $y$ and $z$ by $x$ in any of the 3 equations, we get $x=2x^2$. As $x \ne 0$, we deduce that $x=\tfrac12$ and globally :
$$x=y=z=\tfrac12$$
We can check that with these values, the targetted relationship is indeed fullfilled.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{1}{a_1} + \frac{2}{a_1+a_2} + \frac{3}{a_1+a_2+a_3}<2(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}) $ Prove that $$\frac{1}{a_1} + \frac{2}{a_1+a_2} + \frac{3}{a_1+a_2+a_3}<2(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}) $$ where $a_1, a_2, a_3 >0$.
From AM-HM I got that $\frac{2}{a_1+a_2}\le \frac{1}{2}(\frac{1}{a_1}+\frac{1}{a_2})$ and $\frac{3}{a_1+a_2+a_3}\le \frac{1}{3}(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3})$, but adding these is not enough.
| Adding the inequalities given by you:
$$\dfrac{1}{a_1}+\dfrac{2}{a_1+a_2}+\dfrac{3}{a_1+a_2+a_3}<\dfrac{1+\frac{1}{2}+\frac{1}{3}}{a_1}+\dfrac{\frac{1}{2}+\frac{1}{3}}{a_2}+\dfrac{\frac{1}{3}}{a_3}$$
$$\dfrac{1+\frac{1}{2}+\frac{1}{3}}{a_1}+\dfrac{\frac{1}{2}+\frac{1}{3}}{a_2}+\dfrac{\frac{1}{3}}{a_3}<\dfrac{2}{a_1}+\dfrac{2}{a_2}+\dfrac{2}{a_3}$$
$\therefore \dfrac{1}{a_1}+\dfrac{2}{a_1+a_2}+\dfrac{3}{a_1+a_2+a_3}<\dfrac{2}{a_1}+\dfrac{2}{a_2}+\dfrac{2}{a_3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the first four non-zero terms of the power series for $f(x) = \sin x \cos x$ The given function is
$$f(x)= \sin (x) \cos (x)$$
Now first we rewrite the function, knowing that $ \sin x \cos x=\frac{1}{2}(2\sin(x)\cos(x))$ we can now rewrite our given into:
$$\frac{1}{2}\sin(2x)$$
Then if we see the maclaurin series of $\sin(x)$ is:
$$\sin(x)=x -\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+....$$
Then manipulating this series into our given results in the series being:
$$\frac{1}{2}\left[2x+\frac{(2x)^{3}}{3!}-\frac{(2x)^{5}}{5!}-\frac{(2x)^{7}}{7!} \right]$$
$$f(x)=x-\frac{2}{3}x^3+\frac{2}{15}x^5-\frac{4}{315}x^7+...$$
Am I correct in my calculations?
| $$\sin(x)=x -\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+....$$
You forgot the factorial at the denominator and the power of $2$:
$$\sin(2x)=2x -\frac{2^3x^3}{3!}+\frac{2^5x^5}{5!}-\frac{2^7x^7}{7!}+....$$
$$\frac 12\sin(2x)=x -\frac{2^2x^3}{3!}+\frac{2^4x^5}{5!}-\frac{2^6x^7}{7!}+....$$
$$\frac 12\sin(2x)=x -\frac{2x^3}{3}+\frac{2x^5}{15}-\frac{4x^7}{315}+....$$
| {
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$\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq 2 $ Question -
Suppose that $a, b, c, d$ are four positive real numbers with sum 4.
Prove that
$$
\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq 2
$$
my doubt -
Solution.- According to AM-GM, we deduce that
$$
\begin{aligned}
\frac{a}{1+b^{2} c} &=a-\frac{a b^{2} c}{1+b^{2} c} \geq a-\frac{a b^{2} c}{2 b \sqrt{c}}=a-\frac{a b \sqrt{c}}{2} \\
&=a-\frac{b \sqrt{a \cdot a c}}{2} \geq a-\frac{b(a+a c)}{4}
\end{aligned}
$$
According to this estimation,
$$
\sum_{c y c} \frac{a}{1+b^{2} c} \geq \sum_{c y c} a-\frac{1}{4} \sum_{c y c} a b-\frac{1}{4} \sum_{c y c} a b c
$$
By AM-GM inequality again, it's easy to refer that
$$
\sum_{c y c} a b \leq \frac{1}{4}\left(\sum_{c y c} a\right)^{2}=4 \quad ; \quad \sum_{c y c} a b c \leq \frac{1}{16}\left(\sum_{c y c} a\right)^{3}=4
$$
now i did not understand how they got to this both results using am - gm
$$
\sum_{c y c} a b \leq \frac{1}{4}\left(\sum_{c y c} a\right)^{2}=4 \quad ; \quad \sum_{c y c} a b c \leq \frac{1}{16}\left(\sum_{c y c} a\right)^{3}=4
$$
i think i am missing something easy...
any hints ??
thankyou
| $\sum_{cyc} ab = (a+c)(b+d) \leq \left(\frac{ (a+b+c+d)}{2} \right)^2$
This happens to work for 4 variables, but not other cases.
In Pham's book, example 1.1.4 is:
$16 (abc+bcd+cda+dab) = 16ab(c+d)+16cd(a+b) $
$\leq 4(a+b)^2(c+d) + 4(c+d)^2 (a+b)$
$= 4 (a+b+c+d)(a+b)(c+d)$
$\leq (a+b+c+d)^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
what is the value of $\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\binom{n}{10}+\binom{n}{13}+\dots$ what is the value of $$\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\binom{n}{10}+\binom{n}{13}+\dots$$ in the form of number,
cos, sin
attempts : I can calculate the value of $$\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\binom{n}{9}+\binom{n}{12}+\dots=\frac{1}{3}\left(2^n+2\cos \frac{n\pi}{3}\right)$$ by use primitive $3^\text{rd}$ root of the unity
but this problem i cant solve it.
| Let $\omega=\exp(2\pi i/3)$. Then
$$\frac{1+\omega^k+\omega^{2k}}{3}=
\begin{cases}
1 &\text{if $3\mid k$}\\
0 &\text{otherwise}
\end{cases}$$
So
\begin{align}
\sum_{k=0}^\infty \binom{n}{3k+1}
&=\sum_{k=0}^\infty \binom{n}{k+1}\frac{1+\omega^k+\omega^{2k}}{3} \\
&=\sum_{k=1}^\infty \binom{n}{k}\frac{1+\omega^{k-1}+\omega^{2(k-1)}}{3} \\
&=\frac{1}{3}\sum_{k=1}^\infty \binom{n}{k} + \frac{1}{3\omega}\sum_{k=1}^\infty \binom{n}{k}\omega^k + \frac{1}{3\omega^2}\sum_{k=1}^\infty \binom{n}{k} \omega^{2k} \\
&=\frac{1}{3}(2^n-1) + \frac{1}{3\omega}((1+\omega)^n-1) + \frac{1}{3\omega^2}((1+\omega^2)^n-1)\\
&=\frac{1}{3}(2^n-1) + \frac{\omega^2}{3}((1+\omega)^n-1) + \frac{\omega}{3}((1+\omega^2)^n-1)\\
&=\frac{2^n + \omega^2(1+\omega)^n + \omega(1+\omega^2)^n}{3}
\end{align}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Sum $\sum_{n=1}^{\infty} \frac{1}{n}-\frac{1}{n+2}$ I have to first prove whether the following series is convergent, and then find its limit if it exists:
$\sum_{n=1}^{\infty} \frac{1}{n}-\frac{1}{n+2}$
So, I have proved that it is convergent, but I'm having trouble in finding the limit. Can someone help me out?
| The ans is $\frac{3}{2}$. Compute the sequence of partial sum and you'll figure it out.
Spoiler :
$S_n = (1+\frac{1}{2}+...+\frac{1}{n})-(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n+2})$
$= (1+\frac{1}{2}+...+\frac{1}{n})+(1+\frac{1}{2})-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n+2})$
$=(\gamma_{n} + ln {n}) + \frac{3}{2} - (\gamma_{n+2} + ln (n+2))$
$=(\gamma_n - \gamma_{n+2}) +ln {\frac{n}{n+2}}+ \frac{3}{2}$
Therefore $\lim_{n\to \infty} S_n = \frac{3}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Fermat's little theorem and modular arithmetic Below I have the answer to the question solve: $23^{119} \ \mathrm{mod} \ 5$.
We have $119 = 29\cdot4 + 3$.
$23^{119} \ \mathrm{mod} \ 5
\\= (23^4)^{29}\cdot 23^3 \ \mathrm{mod} \ 5
\\
= 1 \cdot 23^3 \ \mathrm{mod} \ 5
\\
= 23^3 \ \mathrm{mod} \ 5
\\
= 3^3 \ \mathrm{mod} \ 5
\\
= 27 \ \mathrm{mod} \ 5
\\
= 2 \ \mathrm{mod} \ 5.$
I can follow the problem up to the point when we say that $23^3\ \mathrm{mod} \ 5 = 3^3 \ \mathrm{mod} \ 5$. I'm able to solve the problem by manually cubing $27$ but I don't understand how to make the logical jump from $23^3 \ \mathrm{mod} \ 5 = 3^3 \ \mathrm{mod} \ 5$.
| $23 = 3 + 5(4)$, so \begin{align*}
23 &\cong 3 \pmod{5} \\
23^2 \cong 23 \cdot 23 &\cong 3 \cdot 3 \cong 3^2 \pmod{5} \\
&\vdots \\
23^{k} \cong 23^{k-1} \cdot 23 &\cong 3^{k-1} \cdot 3 \cong 3^{k} \pmod{5},
\end{align*}
for any integer $k \geq 1$. (Actually true for any integer $k$.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given $ I_n = \int_0^1 \frac{(x^2 + x + 1)^n - x}{x^2 + 1} dx$ show that we have $I_{4n+1} \in \mathbb{Q}$, for any $n \in \mathbb{N}$. Consider the integral:
$$I_n = \int_0^1 \frac{(x^2+x+1)^n - x}{x^2 + 1} dx$$
I have to show
$$I_{4n+1} \in \mathbb{Q}$$
for any $n \in \mathbb{N}$. I wasn't able to see any "direct" way of doing this, so I thought about induction. However, I couldn't find a way of expressing $I_{4n+1}$ in terms of $I_{4n-3}$ in order to use the induction step, so maybe induction is not the way either.
So how should I approach this?
| Note that
$$(x^2+x+1)^n=\big((x^2+1)+x\big)^n=\sum_{k=0}^n\binom{n}k(x^2+1)^kx^{n-k}\;,$$
so the integrand is
$$\frac{\sum_{k=0}^n\binom{n}k(x^2+1)^kx^{n-k}-x}{x^2+1}=\frac{x^n-x}{x^2+1}+\sum_{k=1}^n\binom{n}k(x^2+1)^{k-1}x^{n-k}\;.$$
The summation is a polynomial with integer coefficients, so its integral from $0$ to $1$ is rational, and we need only worry about the first term. When the exponent is $4n+1$ rather than $n$, it becomes
$$\begin{align*}
\frac{x(x^{4n}-1)}{x^2+1}&=\frac{x}{x^2+1}\big((x^4)^n-1^n\big)\\
&=\frac{x}{x^2+1}(x^4-1)\sum_{k=0}^{n-1}x^{4k}\\
&=\frac{x}{x^2+1}(x^2+1)(x^2-1)\sum_{k=0}^{n-1}x^{4k}\\
&=x(x^2-1)\sum_{k=0}^{n-1}x^{4k}\;,
\end{align*}$$
another polynomial with integer coefficients.
| {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Prove $\sum_{n\geq1}\frac{2^n (1-\cos\frac{x}{2^n})^2}{\sin\frac{x}{2^{n-1}}}=\tan\frac{x}{2}-\frac{x}{2}$ How to prove for $|x|<\pi$:
*
*$\sum_{n\geq1}\frac{2(1-\cos(\frac{x}{2^n}))}{\sin(\frac{x}{2^{n-1}})}=\tan(\frac{x}{2})$
*$\sum_{n\geq1}\frac{2^n (1-\cos(\frac{x}{2^n}))^2}{\sin(\frac{x}{2^{n-1}})}=\tan(\frac{x}{2})-\frac{x}{2}$
Any help will be appreciated.
|
Here is a geometric proof for
$$\sum_{n\geq1}\frac{2(1-\cos\frac{x}{2^n})}{\sin\frac{x}{2^{n-1}}}=\tan\frac{x}{2}$$
Let $OA_1,\>OA_2,\>OA_3 …$ successively bisect the vextex angle $\angle O=\frac x2$ of the right triangle $OTA$, which leads to $\frac{AA_1}{TA_1} = \frac{OA}{OT}$, or,
$$AA_1= OA\cdot TA_1 = \sec\frac x2\tan\frac x4 = \frac{2\sin^2\frac x4}{\cos\frac x2\sin\frac x2} = \frac{2(1-\cos\frac x2)}{\sin x}$$
and, likewise,
$$A_1A_2 = \frac{2(1-\cos\frac x4)}{\sin \frac x2},\>\>\>\>\>
A_2A_3 = \frac{2(1-\cos\frac x8)}{\sin \frac x4},\>\>\>\>\>A_3A_4 = ...$$
From the diagram, we have
$$\tan \frac x2 = AT = AA_1 + A_1A_2 + A_2A_3 + … = \sum_{n\geq1}\frac{2(1-\cos\frac{x}{2^n})}{\sin\frac{x}{2^{n-1}}}$$
| {
"language": "en",
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Proof of $ \nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)} - k^2 \mathbf{A} = \mathbf{0}$ I have this exercise and I don't know how to do it.
These are the instructions:
Show that if the vector A satisfy the equation:
1.$ \nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)} - k^2 \mathbf{A} = \mathbf{0}$
It will also satisfy:
2. $ \nabla^2 \mathbf{A} + k^2 \mathbf{A} = 0$
and
3. $ \nabla \cdot \mathbf{A} = 0$
Notice that $k$ is a constant.
Hint: apply $ \nabla \cdot $ to the first equation.
So I tried to apply the $ \nabla \cdot $ to the first equation and I had this:
$ \nabla\cdot \mathbf{(} \nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)}- k^2 \mathbf{A} \mathbf{)}$
$\delta_{ij}$ $\partial_i \mathbf{(} \nabla \times \mathbf{(} \nabla \times \mathbf{A} \mathbf{)}- k^2 \mathbf{A} \mathbf{)}_j$
$\partial_i \epsilon_{ijk} \partial_j (\nabla \times \mathbf{A})_k - \partial_ik^2 \mathbf{A}_i$
$\epsilon_{ijk} \epsilon_{klm}\partial_i \partial_j \partial_l \mathbf{A}_m - \partial_ik^2 \mathbf{A}_i$
$(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})$$(\partial_i \partial_j \partial_l \mathbf{A}_m) - \partial_ik^2 \mathbf{A}_i$
$(\delta_{il}\delta_{jm} \partial_i \partial_j \partial_l \mathbf{A}_m-\delta_{im}\delta_{jl}\partial_i \partial_j \partial_l \mathbf{A}_m) - \partial_ik^2 \mathbf{A}_i$
$(\partial_i \partial_j \partial_i \mathbf{A}_j - \partial_i \partial_j \partial_j \mathbf{A}_i) - \partial_ik^2 \mathbf{A}_i$
So from here I just don't know how to change the index notation to vector notation, also, I can't see how this satisfies the equations 2 and 3. So please let me know if I made a mistake in any of the previous steps and help me telling me what I should do next.
Thank you very much.
| For the first one, it holds for all vector $\mathbf{X}$ that
$$\nabla\cdot (\nabla \times \mathbf{X})=0$$
Applying this to your equation, we get
$$0=\nabla\cdot 0=\nabla\cdot[\nabla \times(\nabla \times \mathbf{A})]-k^2\nabla \cdot\mathbf{A}=0-k^2\nabla \cdot\mathbf{A},$$
so if $k\neq 0$ then you have that $\nabla\cdot \mathbf{A}=0$.
Now, another well-known identity is that
$$\nabla \times (\nabla\times \mathbf{X})=\nabla(\nabla\cdot \mathbf{X})-\nabla^2\mathbf{X}.$$
For our case, and provided that $\nabla\cdot \mathbf{A}=0$, then
$$0=\nabla \times (\nabla\times \mathbf{A})-k^2\mathbf{A}=\nabla(\nabla\cdot \mathbf{A})-\nabla^2\mathbf{A}-k^2\mathbf{A}=0-\nabla^2\mathbf{A}-k^2\mathbf{A}$$
and you get that
$$\nabla^2\mathbf{A}+k^2\mathbf{A}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653050",
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"source": "stackexchange",
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Establishing $\frac{ \sin mx}{\sin x}=(-4)^{(m-1)/2}\prod_{1\leq j\leq(m-1)/2}\left(\sin^2x-\sin^2\frac{2\pi j}{m}\right) $ for odd $m$
Let $m$ be an odd positive integer. Prove that
$$ \dfrac{ \sin (mx) }{\sin x } = (-4)^{\frac{m-1}{2}} \prod_{1 \leq j
\leq \frac{(m-1)}{2} } \left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi
j }{m } \right) \right) $$
Atempt to the proof
My idea is to use induction on $m$. The base case is $m=3$ and we obtain
$$ \dfrac{ \sin (3x) }{\sin x } = (-4) ( \sin^2 x - \sin^2 (2 \pi /3 ) ) $$
and this holds if one uses the well known $\sin (3x) = 3 \sin x - 4 \sin^3 x $ identity.
Now, if we assume the result is true for $m = 2k-1$, then we prove it holds for $m=2k+1$. We have
$$ \dfrac{ \sin (2k + 1) x }{\sin x } = \dfrac{ \sin [(2k-1 + 2 )x] }{\sin x } = \dfrac{ \sin[(2k-1)x ] \cos (2x) }{\sin x } + \dfrac{ \cos [(2k-1) x ] \sin 2x }{\sin x } $$
And this is equivalent to
$$ cos(2x) \cdot (-4)^{k-1} \prod_{1 \leq j
\leq k-1 }\left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi
j }{m } \right) \right) + 2 \cos [(2k-1) x ] \cos x $$
Here I dont see any way to simplify it further. Am I on the right track?
| Too long for a comment: Look up Chebyshev polynomial of second kind. They are literally what you are dealing with:
$$U_n(\cos x) = \dfrac{\sin((n+1)x)}{\sin x}.$$
Your attempt at induction basically reduces it to an equivalent problem that uses the first kind of Chebyshev polynomials, so I would not be fixated on inductive approach.
| {
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"source": "stackexchange",
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Reference request - identity in central factorial numbers Knuth (in arXiv:math/9207222 [math.CA], page 10) gives an odd polynomial identity like
$$n^{2m-1} = \sum_{k=1}^{m} (2k-1)! T(2m,2k) \binom{n+k-1}{2k-1},$$
where $T(m,k)$ is cental factorial numbers. However, the following identity holds as well
$$(2k-1)! T(2m,2k) = \frac{1}{k} \sum_{j=0}^{k} (-1)^j \binom{2k}{j} (k-j)^{2m} \quad \quad (1.0)$$
Or by symmetry of binomial coefficients
$$(2k-1)! T(2m,2k) = \frac{1}{k} \sum_{j=0}^{k} (-1)^{k-j} \binom{2k}{k-j} j^{2m} \quad \quad (1.1)$$
The question: Is there any reference to identities (1.0), (1.1) ?
Alternative question: Prove the identities (1.0), (1.1)
PS It might seem that (1.0) is got from stirling numbers of second kind
PSS Bounty started
| Some references: We find in
*
*Combinatorial Identities by J. Riordan (1963), ch. 6.5 the formula (24):
\begin{align*}
k!T(n,k)=\sum_{j=0}\binom{k}{j}(-1)^j\left(\frac{1}{2}k-j\right)^n\tag{24}
\end{align*}
*The divided central differences of zero by L. Carlitz and J. Riordan (1961) the formula (10a):
\begin{align*}
K_{rs} = \frac{1}{(2s)!}\sum_{t=0}^{2s}(-1)^t\binom{2s}{t}(s-t)^{2r+2}\tag{10a}
\end{align*}
*Interpolation by J.F. Steffenson (1927) section 58:
The development of $x^r$ in central factorials
\begin{align*}
x^r=\sum_0^r x^{[\nu]}\frac{\delta^{\nu}0^r}{\nu!}
\end{align*}
leads to central differences of nothing, that is
\begin{align*}
\delta^m0^r=\sum_0^m(-1)^{\nu}\binom{m}{\nu}\left(\frac{m}{2}-\nu\right)^r
\end{align*}
Comment: The meaning of the left hand side $\delta^m0^r$ is given in the derivation below.
Here I'd like to show the derivation of (24) above following J. Riordan. It is based upon three ingredients: operators, a recurrence relation and Newton's formula.
Operators: We recall the shift operator $E^a$ and the difference operator $\Delta$:
\begin{align*}
E^af(x)&=f(x+a)\\
\Delta f(x)&=f(x+1)-f(x)
\end{align*}
and introduce the central difference operator $\delta$:
\begin{align*}
\delta f(x)=f\left(x+\frac{1}{2}\right)-f\left(x-\frac{1}{2}\right)
\end{align*}
We can write the $\delta$ operator using shift and difference operator as:
\begin{align*}
\delta f(x)&= \left(E^{\frac{1}{2}} - E^{-\frac{1}{2}}\right)f(x)\tag{1}\\
&=\Delta E^{\frac{1}{2}}f(x)=E^{\frac{1}{2}}\Delta f(x)\tag{2}\\
\end{align*}
We obtain from (1) by successive application of $\delta$
\begin{align*}
\delta^kf(x)&=\left(E^{\frac{1}{2}}-E^{-\frac{1}{2}}\right)^kf(x)\\
&=\sum_{j=0}^k\binom{k}{j}(-1)^jE^{-\frac{j}{2}}E^{\frac{k-j}{2}}f(x)\\
&=\sum_{j=0}^k\binom{k}{j}(-1)^jf\left(x-j+\frac{k}{2}\right)\tag{3}
\end{align*}
Note that (3) has already the shape of (24). So, this step looks promising and it's interesting to see how J. Riordan continues. The next step is to introduce the main actor.
Central factorials: We denote with $x^{[n]}$ the central factorial defined as
\begin{align*}
x^{[n]}&=x\left(x+\frac{n}{2}-1\right)^{\underline{n-1}}\\
&=x\left(x+\frac{n}{2}-1\right)\left(x+\frac{n}{2}-2\right)\cdots\left(x+\frac{n}{2}-n+1\right)
\end{align*}
where we use Don Knuths notation for falling factorials $x^{\underline{n}}=x(x-1)\cdots(x-n+1)$.
The central factorials obey an important recurrence relation which is the key for all what follows. We have according to (2):
\begin{align*}
\delta x^{[n]}&=\Delta E^{-\frac{1}{2}}x^{[n]}\\
&=\Delta \left(x-\frac{1}{2}\right)^{[n]}\\
&=\Delta\left(x-\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{3}{2}\right)^{\underline{n-1}}\\
&=\left(x+\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{1}{2}\right)^{\underline{n-1}}-\left(x-\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{3}{2}\right)^{\underline{n-1}}\\
&=\left(x+\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{3}{2}\right)^{\underline{n-2}}\\
&\qquad-\left(x-\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{3}{2}\right)^{\underline{n-2}}\left(x+\frac{n}{2}-\frac{3}{2}-n+2\right)\\
&=nx^{[n-1]}\tag{4}
\end{align*}
The recurrence relation reminds us on $\frac{d}{dx}x^n=nx^{n-1}$ and we will use it accordingly.
Newton's formula:
We consider a series expansion of $f(x)$ in terms of $x^{[n]}$ and apply the central difference operator $\delta$:
\begin{align*}
f(x)&=\sum_{n\geq 0}a_nx^{[n]}\\
\delta^j f(x)&=\sum_{n\geq 0}a_n\delta^j x^{[n]}=\sum_{n\geq 0}a_n n^{\underline{j}}x^{[n-j]}\tag{5}\\
\delta^j f(0)&=\sum_{n\geq 0}a_n n^{\underline{j}}\delta_{n,j}=a_jj!\tag{6}
\end{align*}
Comment:
*
*In (5) we apply the central difference operator $\delta$ $j$ times and use the recurrence relation (4).
*In (6) we evaluate $f(x)$ at $x=0$ using the Kronecker delta symbol.
We obtain from (6) the following representation of Newton's formula:
\begin{align*}
f(x)&=\sum_{n\geq 0}\frac{x^{[j]}}{j!}\delta^jf(0)\tag{7}
\end{align*}
The final step is setting $f(x)=x^n$ in (7), denoting the coefficients $a_k$ with $T(n,k)$ and obtain with (6)
\begin{align*}
x^n&=\sum_{k=0}^nT(n,k)x^{[k]}\\
\delta^k0^n&=T(n,k)k!\tag{8}\\
\end{align*}
and using (3) we obtain from (8):
\begin{align*}
k!T(n,k)=\sum_{j=0}^k\binom{k}{j}(-1)^jf\left(\frac{1}{2}k-j\right)^n
\end{align*}
which is formula (24) and the claim follows.
| {
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Convergence of $ \begin{array}{l}\sum _{n=0}^{\infty }\frac{\left(2k^2+2k+1\right)}{k^4+2k^3+k^2}\end{array}$ Prove that :$ \begin{array}{l}\sum _{n=0}^{\infty }\frac{\left(2k^2+2k+1\right)}{k^4+2k^3+k^2}\end{array}$ where $k=2n+1$ converges. Also find the value it converges to.
| Hint:
Use partial fraction decomposition.
$$\sum_{n=0}^{\infty} \frac{1}{k^2} +\frac{1}{(k+1)^2} = \sum_{n=1}^{\infty} \frac{1}{n^2} = ?$$
| {
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How to simplify this differentiated equation further? The question I have is sqrt(2x-1) / 2e^3x and I am asked to differentiate and simplify.
This is my working below:
I am unsure of how to get to simplify this to get the final answer in the textbook shown below:
What do I have to do?
| Starting from your result, since $\frac{e^{3x}}{3^{6x}} = e^{-3x}$, $(2x - 1)^{\frac{1}{2}} = (2x - 1)^{-\frac{1}{2}}(2x - 1)$ and $(2x - 1)^{-\frac{1}{2}} = \frac{1}{\sqrt{2x - 1}}$,
$$\begin{equation}\begin{aligned}
& \frac{2e^{3x}\left((2x - 1)^{-\frac{1}{2}} - 3(2x - 1)^{\frac{1}{2}}\right)}{4e^{6x}} \\
& = \frac{e^{-3x}\left((2x - 1)^{-\frac{1}{2}} - 3(2x - 1)^{-\frac{1}{2}}(2x - 1)\right)}{2} \\
& = \frac{e^{-3x}(2x - 1)^{-\frac{1}{2}}\left(1 - 3(2x - 1)\right)}{2} \\
& = \frac{e^{-3x}\left(1 - 6x + 3)\right)}{2\sqrt{2x - 1}} \\
& = \frac{e^{-3x}\left(2\left(2 - 3x\right)\right)}{2\sqrt{2x - 1}} \\
& = \frac{\left(2 -3x\right)e^{-3x}}{\sqrt{2x - 1}}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This matches the final result shown for $g'(x)$ in $242$ in your image.
| {
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Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$ Given that $0 < a , b , c < 1$.
Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$.
I tried using modified C.S. and brute-force. But , it demands a lot of calculation. So , I want some better solution than this. Thank you.
| The RHS is
$$\sum_{n=0}^\infty(a^n+b^n+c^n).$$
The LHS is
$$\sum_{n=0}^\infty(\sqrt{a^nb^n}+\sqrt{a^nc^n}+\sqrt{b^nc^n}).$$
For each $n$,
$$a^n+b^n+c^n=\frac{a^n+b^n}2+\frac{a^n+c^n}2+\frac{b^n+c^n}2
\ge\sqrt{a^nb^n}+\sqrt{a^nc^n}+\sqrt{b^nc^n}$$
by AM/GM
| {
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"question_score": "4",
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} |
Diophantine equation $1+5^k=2^y+2^z\cdot 5^t$ Find all positive integers solutions of the equation $1+5^k=2^y+2^z\cdot 5^t$.
My idea is, let $y,z\geq 2$. Then $4\mid RHS$. But $LHS\equiv 2(mod4)$ which is contradiction.
Let $y=1$. Then the equation becomes $1+5^k=2+2^z\cdot 5^t\implies 5^k-2^z\cdot5^t=1$ which is false by $(mod5)$.
And I don't know how to continue this solution if $z=1$. I know that $(k,y,z,t)=(2,4,1,1)$ is a solution and I'm trying to prove $k\leq2$. Any hint on this problem?
| In case of $z = 1$ I will prove that $k$ and $t$ cannot be both greater than $1$ (the remaining cases of $k = 1$ and $t = 1$ can be finished easily).
In this case we get that $25$ divides $2^{y} - 1$. For that to be true $y$ has to be divisible by $4$ as the order of $2$ mod $5$ is $4$. Therefore $y = 4x$ and then:
$$25 \mid 2^{y} - 1 = 2^{4x} - 1 = 16^{x} - 1$$
By Lifting the Exponent Lemma we get that:
$$2 \le v_{5}(16^{x} - 1) = v_{5}(16 - 1) + v_5(x) = 1 + v_5(x)$$
Where $v_{p}(x)$ is the highest power of $p$ that divides $x$. So we get that $v_{5}(x) \ge 1$ which is equivalent to $5 \mid x$. Therefore $x = 5w$ and we get:
$$5^{k} - 2 \cdot 5^{t} = 2^{y} - 1 = 16^{x} - 1 = 16^{5w} - 1 = 1048576^{w} - 1 = (1048576 - 1)(...) = 1048575(...)$$
$104857 = 3 \cdot 5^2 \cdot 11 \cdot 31 \cdot 41$ so we get that all of these five primes divide:
$$5^{k} - 2 \cdot 5^{t} = 5^{t}(5^{k - t} - 2)$$
And therefore $3$, $11$, $31$ and $41$ must divide $5^{k - t} - 2$. $3$ and $41$ are of no use but in fact $11$ and $31$ give us the contradiction as no power of five is $2$ mod these primes. $11$ is smaller but $31$ is easier to check by hand as $5^{3} = 125 = 4 \cdot 32 + 1$ so the order of $5$ mod $31$ is just three and we have to eliminate just the first two powers.
| {
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Field extension of degree three It's well known that
$$\mathbb{Q}(\sqrt{2},\sqrt{5}) = \mathbb{Q}(\sqrt{2} + \sqrt{5})$$
This property is also true with the cubic root (for a great general theorem), but I want to prove this via an elementary proof. Anyone know how to prove that
$$[\mathbb{Q}(\sqrt[3]{2} + \sqrt[3]{5}) : \mathbb{Q}] = 9$$ ?
The minimal polynomial must be extremely large and very difficult to be computed.
| Let $\alpha=\sqrt[3]{3}$ and $\beta=\sqrt[3]{5}$.
Write $(\alpha+\beta)^i$ for $i=0, \ldots, 8$
in the basis
$\{1, \alpha, \alpha^2, \beta, \alpha \beta, \alpha^2 \beta, \beta^2, \alpha \beta^2, \alpha \beta^2, \alpha^2 \beta^2\}$:
$$
\begin{pmatrix}
1 \\ \alpha+\beta \\ (\alpha+\beta)^2 \\
(\alpha+\beta)^3 \\ (\alpha+\beta)^4 \\ (\alpha+\beta)^5 \\
(\alpha+\beta)^6 \\ (\alpha+\beta)^7 \\ (\alpha+\beta)^8
\end{pmatrix}
=\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 2 & 0 & 1 & 0 & 0 \\
8 & 0 & 0 & 0 & 0 & 3 & 0 & 3 & 0 \\
0 & 23 & 0 & 17 & 0 & 0 & 0 & 0 & 6 \\
0 & 0 & 53 & 0 & 40 & 0 & 35 & 0 & 0 \\
334 & 0 & 0 & 0 & 0 & 93 & 0 & 75 & 0 \\
0 & 709 & 0 & 613 & 0 & 0 & 0 & 0 & 168 \\
0 & 0 & 1549 & 0 & 1322 & 0 & 1117 & 0 & 0 \\
\end{pmatrix}
\begin{pmatrix}
1 \\ \alpha \\ \alpha^2 \\ \beta \\ \alpha\beta \\ \alpha^2\beta \\ \beta^2 \\ \alpha \beta^2 \\ \alpha^2 \beta^2
\end{pmatrix}
$$
Now prove that the matrix is non-singular (for instance, see here)
| {
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Counter-intuitive inequality : $a^{\frac{a^2}{a^2+b^2}}b^{\frac{b^2}{a^2+b^2}}+a^{\frac{a}{a+b}}b^{\frac{b}{a+b}}\leq \frac{2(a^2+b^2)}{a+b}$ I'm studing the following inequality :
Let $a,b>0$ then we have :$$a^{\frac{a^2}{a^2+b^2}}b^{\frac{b^2}{a^2+b^2}}+a^{\frac{a}{a+b}}b^{\frac{b}{a+b}}\leq \frac{2(a^2+b^2)}{a+b}$$
It's related to the function :
Let $x>0$ then we have :$$x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}\leq \frac{2(x^2+1)}{x+1}$$
Where $x=\frac{a}{b}$
It (RHS-LHS) increases slowly ,equivalently to the logarithmic function so I try to find a refinement in this sense :
$\exists\,x>0$ such that :
$$x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}+\ln(x)\leq \frac{2(x^2+1)}{x+1}+C$$ Where $C$ is a constant .
To prove it I have tried to use power series without success.
I was thinking of asymptotic evaluation.
Finally we have :
$$\lim_{x \to +\infty} x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}+\ln(x)-\frac{2(x^2+1)}{x+1}=2$$
My question :
How to prove the initial inequality ?
So if you have an idea to prove it or hints(I prefer it).
Thanks a lot for all your contributions .
P.S. We don't have :$$\lim_{x \to +\infty} x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}-\frac{2(x^2+1)}{x+1}=0$$
| By Bernoulli we obtain:
$$a^{\frac{a^2}{a^2+b^2}}=(1+a-1)^{\frac{a^2}{a^2+b^2}}\leq1+\frac{(a-1)a^2}{a^2+b^2}=\frac{a^3+b^2}{a^2+b^2}.$$
Similarly, $$b^{\frac{b^2}{a^2+b^2}}\leq\frac{b^3+a^2}{a^2+b^2},$$
$$a^{\frac{a}{a+b}}\leq\frac{a^2+b}{a+b}$$ and $$b^{\frac{b}{a+b}}\leq\frac{b^2+a}{a+b}.$$
Id est, it's enough to prove that
$$\frac{(a^3+b^2)(b^3+a^2)}{(a^2+b^2)^2}+\frac{(a^2+b)(b^2+a)}{(a+b)^2}\leq\frac{2(a^2+b^2)}{a+b}$$
or
$$ab(a^4+a^3b+4a^2b^2+ab^3+b^4)-ab(3a^3+5a^2b+5ab^2+3b^3)+a^4+a^3b+4a^2b^2+ab^3+b^4\geq0$$ and after using AM-GM it's enough to prove that
$$4ab(a^4+a^3b+4a^2b^2+ab^3+b^4)^2\geq a^2b^2(3a^3+5a^2b+5ab^2+3b^3)^2.$$
Let $a^2+b^2=2uab.$
Thus, $u\geq1$ and we need to prove that
$$4(a^4+a^3b+4a^2b^2+ab^3+b^4)^2\geq ab(a+b)^2(3a^2+2ab+3b^2)^2$$ or
$$4(4u^2-2+2u+4)^2\geq (2u+2)(6u+2)^2$$ or $$(u-1)(8u^3+7u^2+2u-1)\geq0,$$ which is obvious.
Also, we have:
$$a^{\frac{a^2}{a^2+b^2}}b^{\frac{b^2}{a^2+b^2}}+a^{\frac{a}{a+b}}b^{\frac{b}{a+b}}\leq a+b\leq \frac{2(a^2+b^2)}{a+b}.$$
| {
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If rational $a,b,c$ are such that $a\sqrt[3]{3}+b\sqrt[3]{4}+c\sqrt[3]{5}=0$, then $a=b=c=0$ Problem:
Let $a,b,c \in\mathbb{Q}$ such that $a \sqrt[3]{3}+b \sqrt[3]{4}+c\sqrt[3]{5} = 0$.
Show that $a=b=c=0$.
Case if $a,b,c \in\mathbb{Z}$ is easy, I solved, but I don't have any idea how can I solve the case when $a,b,c \in\mathbb{Q}$. Maybe with polynomial $f = a \sqrt[3]{3}X^3+b \sqrt[3]{4}X^2+c \sqrt[3]{5}X$ and see $f(1) = 0$, but idk.
Any ideas?
| Assume one of these three is not $0$. Let's say $c$.
So $a=0, b=0,$ and $c= \frac{p}{q},$ where $p,q \in \mathbb{Z}$
Our equation $\implies \frac{p}{q}\sqrt[3]{5}=0 \implies \frac{p}{q} = 0$
Since $q \neq 0 \implies p=0$
But then $c=\frac{0}{q} =0.$
(Is there something wrong with what I did?)
| {
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Prove $\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}.$ Let $a,\,b,\,c$ are non-negative such that $ab+bc+ca>0.$ Prove that
$$\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}. \quad (1)$$
Note. Because
$$\sum \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)} = 2.$$
So, from $(1)$ we get know inequality of Darij Grinberg
$$\frac{a(b+c)}{b^2+bc+c^2}+\frac{b(c+a)}{c^2+ca+a^2}+\frac{c(a+b)}{a^2+ab+b^2} \geqslant 2.$$
My proof use sum of squares method.
| BW helps!
We need to prove that:
$$3(b+c)a^3-(5b^2+2bc+5c^2)a^2+(b+c)(b^2+bc+c^2)a+(b-c)^2(b^2+bc+c^2)\geq0.$$
Thus, since our inequality is symmetric respect to $b$ and $c$, we have two cases only:
*
*$a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, $u\geq0$, $v\geq0$ and we need to prove that:
$$3(u^2+v^2)a^2+(u+v)(4u^2-5uv+4v^2)a+(u-v)^2(u^2+uv+v^2)\geq0,$$ which is obvious;
*$b=\min\{a,b,c\}$, $a=b+u$ and $c=b+v$.
Thus, we need to prove that:
$$3(2u^2-2uv+v^2)b^2+(6u^3-3u^2v-5uv^2+4v^2)b+v(3u+v)(u-v)^2\geq0,$$ which is obvious again.
| {
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Show $\int_{0}^{1}\frac {x^2\ln x }{{(1-x^2)}{(1+x^4)}}dx=\frac{-π^2}{16(2+\sqrt{2})}$ Question:
Prove that $$\int_{0}^{1}\frac {x^2\ln x}{{(1-x^2)}{(1+x^4)}}dx=\frac{-π^2}{16(2+\sqrt{2})}$$
Using partial fraction,we get
\begin{align}
&\int_{0}^{1}\frac {x^2\ln x}{{(1-x^2)}{(1+x^4)}}dx\\
= &\frac{1}{4}\int_{0}^{1}\frac{\ln x}{1-x}dx+\frac{1}{4}\int_{0}^{1}\frac{\ln x}{1+x}dx+\frac{1}{2}\int_{0}^{1}\frac{(x^2-1)\ln x}{1+x^4}dx
\end{align}
I got first integral as
$\frac{-\pi^2}{6}$ and second integral as $\frac{-\pi^2}{12}$
$$\int_{0}^{1}\frac {x^2\ln x}{{(1-x^2)}{(1+x^4)}}dx=\frac{-\pi^2}{16}+\frac{1}{2}\int_{0}^{1}\frac{(x^2-1)\ln x}{1+x^4}dx$$
I got stuck with third integral, which seems difficult to evaluate.
${}{}{}$
A note is also written saying that:-
The reader should evaluate the family of integrals ${I_{n}=\int_{0}^{1}\frac {x^{2n}\ln x}{{(1-x^2)}{(1+x^4)^n}}dx{,n} \in N}$
The computation of the first few special values indicates an interesting arithmetic structure of the answer.
How to evaluate integral for $n$?
| Rewrite the third integral as
$$I=\int_{0}^{1}\frac{(x^2-1)\ln x}{1+x^4}dx
= -\int_{0}^{\infty}\frac{\ln x}{1+x^4}dx
$$
Use the general result
$$J(a)= \int_{0}^{\infty}\frac{x^{a-1}}{1+x^m}dx=\frac{\pi}m\csc\frac{\pi a}m
$$
and let $m=4$ to obtain
$$I= - J’(1)= \frac{\pi^2}{8\sqrt2} $$
| {
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The leg of an isosceles triangle; given base and area
The area of an isosceles triangle $\triangle ABC$ with base $AB=2c$ is $S.$ Find the leg of the triangle.
Let $AC=BC$ and $CH$ and $BP$ be the altitudes through $C$ and $B$, respectively.
I am not sure how to approach the problem. The area of the triangle is $S_{\triangle ABC}=\dfrac{AB.CH}{2}=\dfrac{AC.BP}{2}.$ What next?
| Remember the definition of a right angle and the perpendicular bisector is that the perpendicular bisector of $AB$ is all the points $x$ so that $Ax = Bx$ and a right angle with vertex $M$ is one so that if $A,M,B$ are colinear with $AM = BM$ and $M$ between $A$ and $B$ then $\angle AMC$ is a right angle if and only if $AC = BC$.
So as the legs, $AC$ and $BC$ of an isosceles triangle are equal, then $C$ is on the perpendicular bisector of $AB$ and if $M$ is the midpoint of $AB$ then $\triangle AMC$ and $\triangle BMC$ are two congruent right triangles.
Use the Pythogorean Theorem to find $AC = BC$ via
$AC^2 = AM^2 + MC^2$
What is $AM$?
$AB = 2c$ so $AM = \frac 12 AB = \frac 12 2c = c$.
What is $MC$?
$MC$ is the height of $\triangle ABC$. The area of a triangle is $\frac 12 base*height = \frac 12 AB*MC$. The area of $\triangle ABC$ is $S$ and $AB =2c$ so $S = c*MC$. So $MC = \frac Sc$.
So what is $AC$?
$AC^2 = c^2 + (\frac Sc)^2$ so $AC =\sqrt{c^2 + (\frac Sc)^2}$.
| {
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A right triangle has legs $a$ and $b$ and hypotenuse $c.$ Find the largest possible value of $\frac{a + b}{c}$.
A right triangle has legs $a$ and $b$ and hypotenuse $c.$ Find the largest possible value of
$$\frac{a + b}{c}.$$
I used the QM-AM inequality. For a set of numbers $\{a_1, a_2, \dots, a_n\}$ this inequality is,
$$\sqrt{\frac{a_1^2 + a_2^2+\dots+a_n^2}{n}}\geq\frac{a_1 + a_2 + \dots + a_n}{n}.$$
Since $(a,b,c)$ form a right triangle with $c$ being the hypotenuse, we have $a^2 + b^2 = c^2$. Using the QM-AM inequality on $a, b, $ and $c$, we have,
$$\frac{a+b+c}{3}\leq\sqrt{\frac{a^2+b^2+c^2}{3}}.$$
Multiplying both sides by $9$ and plugging in $a^2 + b^2 =c^2$, we have
$$a+b+c\leq \sqrt{6c^2}.$$
Taking the $c$ out of the square root, we have,
$$a+b+c\leq c\sqrt6.$$
Subtracting $c$ from both sides, we have,
$$a+b\leq c\sqrt{6}-c.$$
Factoring $c$ out, we have $$a+b\leq(1-\sqrt{6})\cdot c.$$ I don't know where to go from here. I know for a fact that $1-\sqrt{6}$ isn't the answer, and this is probably because $a, b, c$ have to be positive, but I don't know how to solve for it. Any help? Thanks!
| A suggestion, if you may.
In terms of one of the acute angles $\theta$ in that triangle the value to maximize is:
$\sin \theta + \cos \theta =\sqrt{2} \cos \left(\theta - \dfrac{\pi}4 \right)$
https://math.stackexchange.com/a/3646439/719444
And for $0 \le \theta \le \dfrac{\pi}2$, there is a maximum at $\theta=\dfrac{\pi}4$ which gives the same answer in general.
| {
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Does it follow for $x \ge 785$, that Gautschi's Inequality implies that $\frac{\Gamma(2x + 3 - \frac{1.25006}{\ln n})}{\Gamma(2x+1)} > x^2$ Does it follow for $x \ge 785$, that Gautschi's Inequality implies that $\frac{\Gamma(2x + 3 - \frac{1.25006}{\ln n})}{\Gamma(2x+1)} > x^2$
Here's my reasoning. Please let me know if I made any mistakes or made any jumps in my logic.
(1) From Gautschi's Inequality, from any real $z$ and any real $s$ where $0 < s < 1$, it follows that:
$$z^s > \frac{\Gamma(z+s)}{\Gamma(z)} > (z)(z+1)^{s-1}$$
(2) Setting $z = 2x+2$ gives us:
$$(2x+2)^s > \frac{\Gamma(2x+2+s)}{\Gamma(2x+2)} > (2x+2)(2x+3)^{s-1}$$
(3) Multiplying $2x+1$ to both sides:
$$(2x+1)(2x+2)^s > \frac{\Gamma(2x+2+s)}{\Gamma(2x+1)} > (4x^2+6x+2)(2x+3)^{s-1}$$
(4) Since $\dfrac{1.25506}{\ln x} < 1$ for $x \ge 4$, setting $s = 1 - \dfrac{1.25506}{\ln x}$ gives us:
$$\frac{\Gamma(2x+3-\frac{1.25506}{\ln x})}{\Gamma(2x+1)} > (4x^2+6x+2)(2x+3)^{-\frac{1.25506}{\ln x}} = \frac{4x^2+6x+2}{(2x+3)^{\frac{1.25506}{\ln x}}}$$
(5) Since for $x \ge 785$ (see here for details), $(2x+3)^{\frac{1.25506}{\ln x}} < 4$, it follows that for $x \ge 785$:
$$\frac{\Gamma(2x+3-\frac{1.25506}{\ln x})}{\Gamma(2x+1)} > x^2$$
| I have not found any mistakes.
In the following, let us prove, with the help of WolframAlpha, that if $x \ge 785$, then $(2x+3)^{\frac{1.25506}{\ln x}} < 4$ since it seems that you forgot to show the details.
The inequality is equivalent to $f(x)\gt 0$ where
$$ f(x)=(\ln 4)(\ln x)-1.25506\ln(2x+3)$$
with
$$f'(x)=\frac{(\ln(16)-2.51012)x+3\ln 4}{x(2x+3)}\gt 0$$
since $\ln(16)-2.51012\gt 0\ $ (see here).
Since $f(x)$ is increasing with
$$f(785)=(\ln 4)(\ln 785)-1.25506\ln(1573)\gt 0$$
(see here), it follows that $f(x)\gt 0$ for $x\ge 785$.
| {
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What is a method for solving principal value integral of $\frac{1}{\pi}\int_{-B}^{B} \frac{x \sqrt{B^2-x^2}}{x-y}\mathrm{d} x$? Question:
I am trying to solve a principal value integral involving a square root. Using Mathematica I can get an answer but I would like to know a general approach to obtain them by hand. To be clear I am interested in a clear explanation of the method not just the solution.
The principal value integral is of the form:
$$ I_{B}\left(y\right)=\frac{1}{\pi} \int_0^B J\left(x\right)\left(\frac{\mathcal{P}}{x+y}+\frac{\mathcal{P}}{x-y}\right)\mathrm{d}x,$$
where $\mathcal{P}$ denotes a principal value integral and $0\le y\le B$ (we only consider real numbers). I would be interested in a method of solution for $J\left(x\right)$ given by:
$$J\left(x\right)=x\sqrt{B^2-x^2}.$$
Using Mathematica I have ascertained that $I\left(y\right)=\frac{B^2}{2}-y^2$ and since $J\left(x\right)$ is odd we have:
$$ I_{B}\left(y\right)=\frac{1}{\pi} \int_{-B}^B J\left(x\right)\frac{\mathcal{P}}{x-y}\mathrm{d}x,$$
so I wonder if it could be solved with something akin to contour integration but I am at a loss as to how to proceed further.
Context:
The integral is required to map spectral densities of Hamiltonians used in Open Quantum Systems.
Bonus:
I will accept any answer that provides a step-by-step (analytic) method of solution for $I$ but I would also be interested in methods of solution for a couple of other integrals. I can post these as a separate question if people prefer.
Firstly:
$$ L_{\left(C,D\right)}\left(y\right)=\frac{1}{\pi} \int_{C}^{D} J_2\left(x\right)\frac{\mathcal{P}}{x-y}\mathrm{d}x,$$
with $0\le C<y<D$, all positive real numbers and:
$$J_2\left(x\right)=\sqrt{\left(D-x\right)\left(x-C\right)}.$$
Unfortunately, I have been unable to compute $L$ with Mathematica, but I have used alternative methods to ascertain $L_{\left(C,D\right)}\left(y\right)=\frac{C+D}{2}-y$.
And also:
$$ K_{\left(A,B\right)}\left(y\right)=\frac{1}{\pi} \int_{A}^{B} J_1\left(x\right)\left(\frac{\mathcal{P}}{x+y}+\frac{\mathcal{P}}{x-y}\right)\mathrm{d}x,$$
with $0\le A<y<B$ all positive real numbers and:
$$J_1\left(x\right)=\sqrt{\left(B^2-x^2\right)\left(x^2-A^2\right)}.$$
Since:
$$ \frac{1}{\pi} \int_{A}^{B} J_1\left(x\right)\left(\frac{1}{x+y}+\frac{1}{x-y}\right)\mathrm{d}x=\frac{1}{\pi} \int_{A}^{B} J_1\left(x\right)\frac{2x}{x^2-y^2}\mathrm{d}x=\frac{1}{\pi} \int_{A^2}^{B^2} J_1\left(\sqrt{w}\right)\frac{1}{w-y^2}\mathrm{d}x=L\left(y^2\right)=\frac{A^2+B^2}{2}-y^2,$$
where we have used the substitution $w=x^2$ and defined $C=A^2$ and $D=B^2$.
It is probably also worth noting that:
$$ L_{\left(C,D\right)}\left(y\right)=\frac{1}{\pi} \int_{0}^{D-C} \sqrt{w}\sqrt{D-C-w}\frac{\mathcal{P}}{w-(y-C)}\mathrm{d}w,$$
which can be found using the substitution $w=x-C$.
| The evaluation of the Cauchy principal value integral via contour integration is relatively straightforward. To begin, consider the contour integral
$$\oint_C dz \, \frac{z \sqrt{z^2-B^2}}{z-y} $$
where $C$ is the following contour:
There are semicircular detours of radius $\epsilon$ around the branch points at $z=\pm B$ and the pole at $z=y$. Also, the large circle has a radius $R$. The pieces of the contour integral as labeled in the figure are as follows. (Yes, there are a lot of pieces, but as you will see, most will vanish or cancel.)
$$\int_{AB} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{-R}^{-B-\epsilon} dx \frac{x \sqrt{x^2-B^2}}{x-y}$$
$$\int_{BC} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \, \frac{(-B+\epsilon e^{i \phi})\sqrt{(-B+\epsilon e^{i \phi})^2-B^2}}{-B+\epsilon e^{i \phi}-y} $$
$$\int_{CD} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{-B+\epsilon}^{y-\epsilon} dx \frac{x \sqrt{x^2-B^2}}{x-y} = \int_{-B+\epsilon}^{y-\epsilon} dx \frac{x i \sqrt{B^2-x^2}}{x-y}$$
$$\int_{DE} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \, \frac{(y+\epsilon e^{i \phi})\sqrt{(y+\epsilon e^{i \phi})^2-B^2}}{\epsilon e^{i \phi}} \\ = i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \, \frac{(y+\epsilon e^{i \phi})i \sqrt{B^2-(y+\epsilon e^{i \phi})^2}}{\epsilon e^{i \phi}}$$
$$\int_{EF} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{y+\epsilon}^{B-\epsilon} dx \frac{x \sqrt{x^2-B^2}}{x-y} = \int_{y+\epsilon}^{B-\epsilon} dx \frac{x i \sqrt{B^2-x^2}}{x-y}$$
$$\int_{FG} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \, \frac{(B+\epsilon e^{i \phi}) \sqrt{(B+\epsilon e^{i \phi})^2-B^2}}{B+\epsilon e^{i \phi}-y} $$
$$\int_{GH} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{B-\epsilon}^{y+\epsilon} dx \frac{x \sqrt{x^2-B^2}}{x-y} = \int_{B-\epsilon}^{y+\epsilon} dx \frac{x (-i) \sqrt{B^2-x^2}}{x-y}$$
$$\int_{HI} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i \epsilon \int_{0}^{-\pi} d\phi \, e^{i \phi} \, \frac{(y+\epsilon e^{i \phi})\sqrt{(y+\epsilon e^{i \phi})^2-B^2}}{\epsilon e^{i \phi}} \\ = i \epsilon \int_{0}^{-\pi} d\phi \, e^{i \phi} \, \frac{(y+\epsilon e^{i \phi})(-i) \sqrt{B^2-(y+\epsilon e^{i \phi})^2}}{\epsilon e^{i \phi}} $$
$$\int_{IJ} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{y-\epsilon}^{-B+\epsilon} dx \frac{x \sqrt{x^2-B^2}}{x-y} = \int_{y-\epsilon}^{-B+\epsilon} dx \frac{x (-i) \sqrt{B^2-x^2}}{x-y}$$
$$\int_{JK} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i \epsilon \int_{0}^{-\pi} d\phi \, e^{i \phi} \, \frac{(-B+\epsilon e^{i \phi})\sqrt{(-B+\epsilon e^{i \phi})^2-B^2}}{-B+\epsilon e^{i \phi}-y} $$
$$\int_{KL} dz \frac{z \sqrt{z^2-B^2}}{z-y} = \int_{-B-\epsilon}^{-R} dx \frac{x \sqrt{x^2-B^2}}{x-y}$$
$$\int_{LA} dz \frac{z \sqrt{z^2-B^2}}{z-y} = i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{R e^{i \theta} \sqrt{R^2 e^{i 2 \theta}-B^2}}{R e^{i \theta}-y} $$
Note that, on the branch above the real axis, $-1=e^{i \pi}$ and on the branch below the real axis, $-1=e^{-i \pi}$. Thus, the sign of $i$ in front of the square root when $|x| \lt B$ is positive above the real axis and negative below the real axis.
First, note that the integrals over $AB$ and $KL$ cancel because the square root does not introduce any phase change there.
Second, as $\epsilon \to 0$, the integrals over $BC$, $FG$, and $JK$ vanish.
Third, as $\epsilon \to 0$, the integrals over $DE$ and $HI$ cancel. In this case, the phase difference over the branch cut due to the square root turns what is normally a constructive interference (i.e., the contributions usually add) into a destructive interference (i.e., they cancel.)
Fourth, as $\epsilon \to 0$, the integrals over $CD$ and $EF$ combine to form
$$i PV \int_{-B}^B dx \frac{x \sqrt{B^2-x^2}}{x-y}$$
and the integrals over $GH$ and $IJ$ combine to form
$$-i PV \int_{B}^{-B} dx \frac{x \sqrt{B^2-x^2}}{x-y}$$
so together, the contribution to the contour integral over these four intervals is
$$i 2 PV \int_{-B}^B dx \frac{x \sqrt{B^2-x^2}}{x-y}$$
Fifth, the final contribution to the contour integral is the integral over $LA$ in the limit as $R \to \infty$. That limit is evaluated as follows:
$$\begin{align} \int_{LA} dz \frac{z \sqrt{z^2-B^2}}{z-y} &= i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{R e^{i \theta} \sqrt{R^2 e^{i 2 \theta}-B^2}}{R e^{i \theta}-y} \\ &= i R^2 \int_{-\pi}^{\pi} d\theta \, e^{i 2 \theta} \left (1-\frac{B^2}{R^2 e^{i 2 \theta}} \right )^{1/2} \left (1-\frac{y}{R e^{i \theta}} \right )^{-1} \\ &= i R^2 \int_{-\pi}^{\pi} d\theta \, e^{i 2 \theta} \left [1+y \frac{1}{R e^{i \theta}} -\left (\frac{B^2}{2} - y^2 \right ) \frac1{R^2 e^{i 2 \theta}} + O \left ( \frac1{R^3} \right ) \right ]\end{align}$$
After integration, the first two contributions inside the brackets vanish. Further, as $R \to \infty$, all terms $O(1/R^3)$ will also vanish. This simply leaves the $1/R^2$ term in the integrand, and we may finally write an expression for the contour integral:
$$\oint_C dz \, \frac{z \sqrt{z^2-B^2}}{z-y} = i 2 PV \int_{-B}^B dx \frac{x \sqrt{B^2-x^2}}{x-y} - i 2 \pi \left (\frac{B^2}{2} - y^2 \right )$$
By Cauchy's theorem, the contour integral is equal to zero. Therefore, when $y \in (-B,B)$
$$\frac1{\pi} PV \int_{-B}^B dx \frac{x \sqrt{B^2-x^2}}{x-y} = \frac{B^2}{2} - y^2$$
as asserted by the OP.
ADDENDUM
The case $|y| \gt B$ is a straightforward application of the residue theorem and the result is
$$\frac1{\pi} \int_{-B}^B dx \frac{x \sqrt{B^2-x^2}}{x-y} = \sqrt{y^2-B^2} \left (y - \sqrt{y^2-B^2} \right ) - \frac{B^2}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Express $ \operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right) $ in terms of $m$ and $n$
Let $m$ be a positive integer with $\operatorname{gcd}(m, n)=1 .$ Express
$
\operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right)
$
in terms of $m$ and $n$
My work -
let $d=\operatorname{gcd}(5^m +7^m,5^n +7^n)$ then
$5^{2m} \equiv 7^{2m}$ mod(d)
$5^{2n} \equiv 7^{2n}$ mod(d)
and obviously $gcd(5,d)=gcd(7,d)=1$ so,
$5^{gcd(2m,2n)} \equiv 7^{gcd(2m,2n)}$ (mod d)
$5^2 \equiv 7^2$ (mod d)
$d= 1,2,3,4,6,8,12,24$
now i find values of d how to express this in terms of $m$ and $n$ ???
| When $m$ and $n$ are both odd and coprime and $a,b$ are coprime, $gcd(a^m + b^m, a^n + b^n) = a + b$.
As stated by @Geoffrey in the comments, $a+b \mid a^k + b^k$ for odd $k$:
$$a^m+b^m = (a + b)\underbrace{(a^{m-1} - a^{m-2}b + a^{m-3}b^{2} - \dots + b^{m-1})}_{m\space terms}$$
$$a^n+b^n = (a + b)\underbrace{(a^{n-1} - a^{n-2}b + a^{n-3}b^{2} - \dots + b^{n-1})}_{n\space terms}$$
The rightmost polynomials in both equations have no common factor since $m$ and $n$ have no common factor.
If $m,n$ are both even, then they cannot be coprime. If one of $m, n$ is even, then...?
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Sum of two cubes equal to prime square If $a,b\in \mathbb{N}$ find all primes $p$ such that $a^3+b^3=p^2$
My approach-
$a^3+b^3=(a+b)(a^2-ab+b^2)=p^2$ suppose $a+b=x$ and $a^2-ab+b^2=y$ then there are two cases- $(x,y)=(p^2,1),(p,p)$
Now I am struggling for case 02 where $(x,y)=(p,p)$
| In the case that $a+b=a^2-ab+b^2$, we have $a^2+b^2=ab+a+b$, so $a^2+b^2=(a+1)b+a$. Without loss of generality (since the formula is symmetric in $a,b$) that $a\leq b$. Clearly, since $a+b$ is a prime, and $a,b=1$ isn't a solution, $a\neq b$. And, if $b>a+1$, then $a^2\geq a$ and $b^2>(a+1)b$, so $a^2+b^2>(a+1)b+a$, so there are no solutions.
Hence, any solutions to this case will be of the form $a, a+1$, for some natural $a$.
Now, we make that substitution to get $2a^2+2a+1=a^2+a+2a+1$, so $a^2=a$, so $a=0,1$. So, the only solution is $a=1,b=2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the volume of the solid determined by $S_1$ and $S_2$ I want to calculate the volume of the solid determined by this tho surfaces:
$$S_1=\{(x,y,z)\in\mathbb{R}:x^2+y^2+z^2=R^2\}$$
$$S_2=\{(x,y,z)\in\mathbb{R}:x^2+y^2=Rx\}$$
The solid is the intersection of a sphere of radius $R$ ($S_1$) and a cylinder of diameter $R$ (centered in $(R/2,0,0)$)($S_2$)
I guess i must change to spherical or cylindrical coordinates, and that's what i have problems with. I'm stucked in finding the new values of the variables. Also, which coordinate system will work better for this problem? Spherical or cylindrical? I will thank any help.
| I use cylindrical coordinate to obtain the answer.
For $S_2$, if $x^2+y^2=Rx$ a, then we have $r^2=Rr\cos \theta$, hence $r=R\cos \theta$.
The intersection region involves the first and fourth quadrant.
Hence, we want to evaluate
\begin{align}
\int_{-\frac{\pi}2}^\frac{\pi}2 \int_0^{R\cos \theta} \int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}} r\, dz \,dr \, d\theta
\end{align}
By using symmetry, we can simplify the expression to
\begin{align}
&4\int_{0}^\frac{\pi}2 \int_0^{R\cos \theta} \int_{0}^{\sqrt{R^2-r^2}} r\, dz \,dr \, d\theta \\
&= 4\int_{0}^\frac{\pi}2 \int_0^{R\cos \theta} r\sqrt{R^2-r^2} \,dr \, d\theta \\
&=-2\int_{0}^\frac{\pi}2 \int_0^{R\cos \theta} (-2r)\sqrt{R^2-r^2} \,dr \, d\theta \\
&=-\frac43\int_{0}^\frac{\pi}2 \left[(R^2-r^2)^\frac32 \right]_0^{R\cos \theta} \, d\theta \\
&= - \frac43 \int_0^\frac{\pi}2 (R^3\sin^3 \theta - R^3) \, d\theta \\
&= \frac{4}{3}R^3 \int_0^\frac{\pi}2 (1-\sin^3 \theta) \, d\theta \\
&=\frac{4}{3}R^3 \int_0^\frac{\pi}2 (1-\sin \theta(1-\cos^2\theta)) \, d\theta \\
&= \frac{4}{3}R^3 \int_0^\frac{\pi}2 (1-\sin \theta- (-\sin \theta)\cos^2\theta) \, d\theta \\
&= \frac43 R^3\left[ \theta +\cos \theta- \frac{\cos^3 \theta}{3}\right]_0^\frac{\pi}2 \\
&= \frac43 R^3\left[\frac{\pi}2-1+\frac13 \right] \\
&= \frac{2(3\pi-4)}9 R^3
\end{align}
Remark: In the event that $R$ is not specified to be a nonnegative quantity such as radius or diameter, that is if $R$ can take negative value, by symmetry, the answer is $\frac{2(3\pi-4)}9 |R|^3$
| {
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"source": "stackexchange",
"question_score": "1",
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Number of ordered Pairs satisfying $4^m-3^n=1$ Find the Number of ordered Pairs $(m,n)$ of positive integers satisfying $4^m-3^n=1$
Mt try:
Trivially $m=n=1$ satisfies
Let $m \gt 1$
$$4^m-3^n=(1+3)^m-3^n=1$$
$\implies$
$$3\binom{m}{1}+3^2\binom{m}{2}+3^3\binom{m}{3}+\cdots+3^m=3^n$$
Now since LHS is not a power of $3$ and RHS is, this is possible only when $m=1$
Hence the only ordered pair is $(1,1)$
is this the right way?
| Let $(m,n)$ be a pair with $n>1$ such that $4^m-3^n=1$. Looking at both sides modulo $4$, we see that $n$ must be odd, so of the form $n=2k+1$. Thus we now have $4^m-3\cdot 9^k=1$.
Now looking at the equation modulo $9$, we see that $m$ must be a multiple of $3$, so of the form $m=3l$. Thus we have $64^l-3\cdot 9^k=1$.
However, comparing both sides modulo $7$, we now must have $1-3\cdot 2^k\equiv 1\pmod 7$, or $2^k\equiv 0\pmod 7$. This is clearly impossible, hence $(1,1)$ in fact is the only solution.
PS: I use $n>1$, or, equivalently, $k>0$, to say $3\cdot 9^k\equiv0\pmod9$.
| {
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"source": "stackexchange",
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Find $\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min$ I'm trying to solve
\begin{align*}
&\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min \\
&x + y + z = 1 \\
&x, y, z > 0,
\end{align*}
using only inequalities.
How can i solve it? I used am-gm, tried to "break" terms, but got only after all $(xy + yz + xz) = \frac{1}{2}$
| You can use Cauchy-Schwarz:
$$\sum\dfrac{x^2}{y}\sum y\geq(x+y+z)^2$$
or an even simpler one is:
$$\sum\dfrac{x^2}{y}-\sum y = \sum\left(\dfrac{x^2}{y} - 2x+y\right)=\sum\dfrac{(x-y)^2}{y}\geq 0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the dimension and base of the following vectors' sum and intersection? I have 2 Vector Subspaces of $\mathbb{R}^3$, namely $U = \operatorname{Span}(\begin{pmatrix} 2 \\ 5 \\ 9\end{pmatrix}, \begin{pmatrix} 0 \\ -1 \\ -3\end{pmatrix})$ and $W = \operatorname{Span}(\begin{pmatrix} -3 \\ 1 \\ 6\end{pmatrix}, \begin{pmatrix} 5 \\ 3 \\ 0\end{pmatrix})$. Now I want to find the dimension and base of $U + V$ and of $U \cap W$. This is what I have done so far:
Since the vectors of $U$ and $W$ are lineary independent, they have dimension $\dim(U) = \dim(W) = 2$. I can see that $U+W = \operatorname{Span}(\begin{pmatrix} -1 \\ 6 \\ 15\end{pmatrix}, \begin{pmatrix} 5 \\ 2 \\ -3 \end{pmatrix})$ and it is also lineary independent, so $\dim(U+W) = 2$.
This means that $\dim(U\cap W) = \dim(U) + \dim(W) - \dim(U+W) = 2 + 2 - 2 = 2$.
Now, going further, I want to find the base of $U\cap W$, so I construct the following matrix and reduce it:
$$\begin{pmatrix} 2 & 0 & 3 & -5 \\ 5 & -1 & -1 & -3 \\ 9 & -3 & -6 & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{pmatrix}$$
So, this means that the two vector spaces are intersecting exactly when $$\lambda \begin{pmatrix} 2 \\ 5 \\ 9\end{pmatrix} + \lambda \begin{pmatrix} 0 \\ -1 \\ -3\end{pmatrix} = \lambda \begin{pmatrix} -1 \\ 6 \\ 15\end{pmatrix} + \lambda \begin{pmatrix} 5 \\ 2 \\ -3 \end{pmatrix}$$
So, a base of $U\cap W$ would be $\begin{pmatrix} 2 \\ 4 \\ 6\end{pmatrix}$
Now, $U \cap W $ has dimension 1. But earlier I found that it should be 2. Where is the error? Do I make some error when I calculate $U + W$? Can someone point me in the right direction?
| You only formed the sum of the chosen basis elements of $U$ and $W$ instead of all vectors therein.
Your calculation for the intersection is correct, hence $\dim(U+V)=3$, i.e. $U+V=\Bbb R^3$, that is, every vector can be expressed as a sum $u+v$ with $u\in U$ and $v\in V$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can this function be defined in a way to make it continuous at $x=0$? We have $$f=\frac{x}{\vert x-1 \vert - \vert x +1 \vert}$$
If we want to "define" this function to be continuous at $x=0$, it's limit at $0$ must equal $f(0)$. So we should find this limit and assign it to be equal to $f(0)$, then the function is continuous at $0$. Since we are looking at the function when $x\to 0$, $x\neq 0$. Lets divide both sides by $x$.
$$f=\frac{x}{\vert x-1 \vert - \vert x +1 \vert}=\frac{1}{\frac{\vert x-1 \vert}{x}-\frac{\vert x+1\vert}{x}}=\frac{1}{\vert 1-\frac{1}{x}\vert - \vert 1+ \frac{1}{x}\vert }$$
We can use $\lim \phi(x)^{-1}=\frac{1}{\lim \phi(x)}$ here ( the limit $\neq$ 0, by hypothesis ). The inverse of the limit of $\phi(x)=\vert 1 - \frac{1}{x} \vert-\vert 1+\frac{1}{x}\vert$, when $x\to 0$. If $x<1$, we have that $$\frac{1}{x}>1\implies0>1-\frac{1}{x}\implies \Bigg\vert 1-\frac{1}{x}\Bigg\vert=-\Big(1-\frac{1}{x}\Big)$$
Now if $x>0$, we have that $$\Bigg\vert 1 - \frac{1}{x} \Bigg\vert-\Bigg\vert 1+\frac{1}{x}\Bigg\vert=-2$$
and if $x<0$, then $$\Bigg\vert 1 - \frac{1}{x} \Bigg\vert-\Bigg\vert 1+\frac{1}{x}\Bigg\vert=1-\frac{1}{x}-1-\frac{1}{x}=\frac{(-2)}{x}$$
The limit of $f$ when $x\to 0$ appears to be $\frac{-1}{2}$. Could anyone tell me what errors I made in the limit finding process?
| Another way could be:
Multiply by $\frac{|x-1|+|x+1|}{|x-1|+|x+1|}$, in this way you have at the denominator:
$$(|x-1|-|x+1|)(|x-1|+|x+1|)=(x-1)^2-(x+1)^2=-4x$$
and at the nominator:
$$x(|x-1|+|x+1|)$$
you can simplify the $x$ and it remains:
$$\frac{|x-1|+|x+1|}{-4}$$
and the limit for the numerator is $2$.
| {
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"url": "https://math.stackexchange.com/questions/3700549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Perimeter of an ellipse solution and elliptic integrals. I developed a formula for finding the length of a curve; let's call it $L(x)$:
$$L(x)=\int_{0}^x ((\frac{dy}{dx})^2+1)^\frac{1}{2}dx$$
Applying this to an ellipse of equation
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
and therefore $\frac {dy}{dx}={-xb\over a\sqrt{a^2-x^2}}$
so $((\frac{dy}{dx})^2+1)={(b^2-a^2)x^2+a^4\over a^2({a^2-x^2})}$
This means a quarter of an ellipse's perimeter is
$$\int_{0}^a\sqrt{(b^2-a^2)x^2+a^4\over a^2({a^2-x^2})}dx$$
which is
$$\int_{0}^a\frac{1}{a}\sqrt{(b^2-a^2)x^2+a^4\over ({a^2-x^2})}dx $$
This means the perimeter of an ellipse is equal to
$$\frac{4}{a}\int_{0}^a\sqrt{(b^2-a^2)x^2+a^4\over ({a^2-x^2})}dx $$
Can anyone integrate this for me please? I don't have the necessary skills yet. I've been told that to integrate this I'd need elliptic integrals, but I don't understand what they are. If you do, could you please explain what they are and how they work? Thanks a lot.
| Your integral is correct and there're alternative representations for the arclength:
\begin{align}
(x,y) &= (a\sin \theta,b\cos \theta) \\
k &= \sqrt{1-\frac{b^2}{a^2}} \\
k' &= \frac{b}{a} \\
s &= \int_0^\theta \sqrt{a^2\cos^2 \phi+b^2\sin^2 \phi} \, d\phi \\
&= a\int_0^{\frac{x}{a}} \sqrt{\frac{1-k^2 t^2}{1-t^2}} \, dt \\
&= b\int_{\frac{y}{b}}^1 \sqrt{\frac{k'^2+k^2 t^2}{1-t^2}} \, dt \\
&= a\int_{b\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}}}^1
\frac{t^2 dt}{\sqrt{(1-t^2)(t^2-k'^2)}} \\
&= aE(\theta,k)
\end{align}
where $E(.,.)$ is called incomplete elliptic integral of the second kind.
May compare these with Jacobi elliptic functions:
\begin{align}
(x,y) &= (a\operatorname{sn} u,b\operatorname{cn} u) \\
u &= \int_0^{\operatorname{sn} u} \frac{1}{\sqrt{(1-t^2)(1-k^2 t^2)}} \, dt \\
&= \int_{\operatorname{cn} u}^1 \frac{1}{\sqrt{(1-t^2)(k'^2+k^2 t^2)}} \, dt \\
&= \int_{\operatorname{dn} u}^1 \frac{dt}{\sqrt{(1-t^2)(t^2-k'^2)}} \\
&= F(\operatorname{sn u},k)
\end{align}
where $F(.,.)$ is called incomplete elliptic integral of the first kind.
See the link here in other ways for calculating the perimeter.
Further points to be noticed
*
*$k$ is elliptic modulus of elliptic integrals/functions which also equals to the eccentricity of the ellipse.
*The integrals are the arclength measured between the fixed point $(0,b)$ and an arbitrary point $(x,y)$ on the ellipse within one quadrant.
*A quarter of perimeter is between $(x,y)=(0,b)$ and $(x,y)=(a,0)$.
*The perimeter is $$4aE\left( \dfrac{\pi}{2}, k \right)=4aE(k)$$ where $E(k)$ is a complete elliptic integral of the second kind.
*We can derive the above integrals by change of variables
*
*$t=\dfrac{x}{a}=\sin \phi=\operatorname{sn} (u,k)$
*$t=\dfrac{y}{b}=\cos \phi=\operatorname{cn} (u,k)$
*$t=b\sqrt{\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}}
=\sqrt{1-k^2\sin^2 \phi}=\operatorname{dn} (u,k)$
*If $p$ is the perpendicular distance of a tangent, at $(x,y)$ on the ellipse, from the origin
$$p=\frac{1}{\sqrt{\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}}}$$
we have arclength element $$ds=\frac{ab\, d\phi}{p}$$ and curvature at $(x,y)$ $$\kappa=\frac{p^3}{a^2 b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3702212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How Can I prove $\sum_{n=1}^{\infty}{\frac{n^4}{5^n}}=\frac{285}{128}$ Question:Prove that $ \sum_{n=1}^{\infty}{\frac{n^4}{5^n}}=\frac{285}{128}$
While doing questions on series and products,i got stuck in this question.I was not able to figure out any way how to prove this one.Convergence tests shows that this series converges but i don't know how to find its value.
| Equivalently, we want to evaluate $\left.\sum_nn^4x^n\right|_{x=1/5}$ with $\sum_n:=\sum_{n=\color{blue}{0}}^\infty$. By the binomial theorem, if $|x|<1$ then$$\begin{align}(1-x)^{-1}&=\sum_nx^n,\,\\(1-x)^{-2}&=\sum_n(n+1)x^n,\,\\(1-x)^{-3}&=\sum_n\tfrac12(n+1)(n+2)x^n,\,\\(1-x)^{-4}&=\sum_n\tfrac16(n+1)(n+2)(n+3)x^n,\,\\(1-x)^{-5}&=\sum_n\tfrac{1}{24}(n+1)(n+2)(n+3)(n+4)x^n.\end{align}$$We'll find a linear combination by first getting the $n^4$ coefficient right, then $n^3$ etc. Since$$\begin{align}n^4&=24\cdot\tfrac{1}{24}(n+1)(n+2)(n+3)(n+4)\\&-60\cdot\tfrac16(n+1)(n+2)(n+3)\\&+50\cdot\tfrac12(n+1)(n+2)\\&-15\cdot(n+1)\\&+1,\end{align}$$we have$$\begin{align}\sum_nn^4x^n&=24(1-x)^{-5}-60(1-x)^{-4}\\&+50(1-x)^{-3}-15(1-x)^{-2}+(1-x)^{-1}.\end{align}$$Substituting $x=\tfrac15$ gives the desired result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Closed form of $\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx$
Can a closed form solution for the following integral be found:
$$\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx\,?$$
I have tried all the standard tricks such as integration by parts, various substitutions, and parametric differentiation (Feynman's trick), but all to no avail.
An attempt is letting
$$f(t):=\int_0^\infty\,\arctan^2\left(\frac{2tx}{1+x^2}\right)\,\text{d}x\,.$$
Therefore,
$$f'(t)=\int_0^\infty\,\frac{8x^2(x^2+1)}{\big(x^4+2(2t^2+1)x^2+1\big)^2}\,\left(1+x^2-4tx\arctan\left(\frac{2tx}{1+x^2}\right)^{\vphantom{a^2}}\right)\,\text{d}x\,.$$
This doesn't seem to go anywhere. Help!
| $$I=\int_0^\infty \arctan^2 \left (\frac{2x}{x^2 + 1} \right ) dx\overset{IBP}=4\int_0^\infty \frac{x(x^2-1)\arctan\left(\frac{2x}{x^2+1}\right)}{x^4+6x^2+1}dx$$
We have that: $$4\int\frac{x(x^2-1)}{x^4+6x^2+1}dx=(\sqrt 2 +1)\ln(x^2+(\sqrt 2+1)^2)-(\sqrt 2-1)\ln(x^2+(\sqrt 2-1)^2)$$
$$\frac{d}{dx}\arctan\left(\frac{2x}{x^2+1}\right)=\frac12\left(\frac{\sqrt 2+1}{x^2+(\sqrt 2+1)^2}-\frac{\sqrt 2-1}{x^2+(\sqrt 2-1)^2}\right)$$
Thus integrating by parts again and simplifying we obtain:
$$I=\int_0^\infty \frac{(\sqrt 2+1)^2 \ln(x^2+(\sqrt 2+1)^2)}{x^2+(\sqrt 2+1)^2}dx+\int_0^\infty \frac{(\sqrt 2-1)^2 \ln(x^2+(\sqrt 2-1)^2)}{x^2+(\sqrt 2-1)^2}dx$$
$$-\int_0^\infty \frac{\ln(x^2+(\sqrt 2-1)^2)}{x^2+(\sqrt 2+1)^2}dx-\int_0^\infty \frac{\ln(x^2+(\sqrt 2+1)^2)}{x^2+(\sqrt 2-1)^2)}dx$$
From here we have the following result:
$$\int_0^\infty \frac{\ln(x^2+a^2)}{x^2+b^2}dx=\frac{\pi}{b}\ln(a+b), \ a,b>0$$
So using this result and with some algebra everything simplifies to:
$$\boxed{\int_0^\infty \arctan^2 \left (\frac{2x}{x^2 + 1} \right ) dx=2\pi \ln(1+\sqrt 2)-\sqrt 2\pi \ln 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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I have to find the marginal pmf's $f_X$, $f_Y$ and $f_{X+Y}$ of $X+Y$. I have to find $D > 0$ such that $f(x, y)$ = $D$($\frac{1}{x+y−1}$$+$ $\frac{1}{x+y+1}$ $−$ $\frac{2}{x+y}$) is the joint pmf $f_{X,Y} (x, y)$ of a random vector $(X,Y)$ in
{$1,2,..$}$^2$. And then I have to find the marginal pmf's $f_X$, $f_Y$ and $f_{X+Y}$ of $X+Y$.
My attempt: First I tried to find $D$
$\sum_{x=1}^{\infty}$$\sum_{y=1}^{\infty}$ $D$($\frac{1}{x+y−1}$$+$ $\frac{1}{x+y+1}$ $−$ $\frac{2}{x+y}$) = $1$
D$\sum_{x=1}^{\infty}$ $\sum_{y=1}^{\infty}$($\frac{2}{(x+y−1)(x+y+1)(x+y)})$ = $1$
$D$$\sum_{x=1}^{\infty}$ $\frac{1}{x^2+x}$= $1$
$D$ = $1$
Then:
$f_X(x)$ = $\sum_{y=1}^{\infty}$$D$($\frac{2}{(x+y−1)(x+y+1)(x+y)})$
$f_X(x)$ = $\sum_{y=1}^{\infty}$$\frac{2}{(x+y−1)(x+y+1)(x+y)}$
$f_X(x)$ = $\frac{1}{x^2+x}$
Then:
$f_Y(y)$ = $\sum_{x=1}^{\infty}$$D$($\frac{2}{(x+y−1)(x+y+1)(x+y)})$
$f_Y(y)$ = $\sum_{x=1}^{\infty}$$\frac{2}{(x+y−1)(x+y+1)(x+y)}$
$f_Y(y)$ = $\frac{1}{y^2+y}$
For $f_{X+Y}$ I was thinking of the following:
Set $U=X$ and $V=X+Y$, then use the change of variables theorem, note that:
$\begin{pmatrix}
\frac{\partial U}{\partial X} & \frac{\partial U}{\partial Y} \\
\frac{\partial V}{\partial X} & \frac{\partial V}{\partial Y}
\end{pmatrix}$
= $\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix}$ has determinant one. Hence,
$f_{U,V}$ = $f_{X,Y}(x(u,v),y(u,v))$ = $\frac{2}{(x+y−1)(x+y+1)(x+y)}$ = $\frac{2}{(u+v-u-1)(u+v-u+1)(u+v-u)}$ = $\frac{2}{(v-1)(v+1)v}$ by the change of variables
Now we get via $f_V$
$f_V(v)$ = $\sum_{v=2}^{\infty}$ $\frac{2}{(v-1)(v+1)v}$ = $\frac{1}{2}$
Now I have to calculate the conditional pmf's $f_{X|Y}(x|y)$ and $f_{Y|X}(y|x)$
I know that we have to use:
$f_{X|Y}(x|y)$ = $P(X=x|Y=y)$ = $\frac{P(X=x,Y=y)}{P(Y=Y)}$
I am not sure if the above answers are correct.
So any help would be grateful. Thanks in advance.
| You are correct concerning $D$, $f_X$ and $f_Y$.
For finding $f_{X+Y}$ realize that for $n=2,3,\dots$ we have:
$P\left(X+Y=n\right)=\sum_{x+y=n}\left[\frac{1}{x+y-1}+\frac{1}{x+y+1}-\frac{2}{x+y}\right]=\left(n-1\right)\left[\frac{1}{n-1}+\frac{1}{n+1}-\frac{2}{n}\right]$
| {
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"url": "https://math.stackexchange.com/questions/3705536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A problem posed by Ramanujan involving $\sum e^{-5\pi n^2}$ While going through the list of problems posed by Ramanujan in Journal of Indian Mathematical Society I came across this problem involving theta functions:
Prove that $$\frac{1}{2}+\sum_{n=1}^{\infty} e^{-\pi n^2x}\cos(\pi n^2\sqrt{1-x^2})=\frac{\sqrt{2}+\sqrt{1+x}}{\sqrt{1-x}}\sum_{n=1}^{\infty}e^{-\pi n^2x}\sin(\pi n^2\sqrt{1-x^2})$$ and deduce the following:
*
*${\displaystyle \frac{1}{2}+\sum_{n=1}^{\infty} e^{-\pi n^2}=\sqrt{5\sqrt{5}-10}\left(\frac{1}{2}+\sum_{n=1}^{\infty} e^{-5\pi n^2}\right)} $
*${\displaystyle \sum_{n=1}^{\infty} e^{-\pi n^2}\left(\pi n^2-\frac{1}{4}\right)=\frac{1}{8}} $
The sums in above problem are clearly based on theta functions and we use a simplified notation here to define them. If $\tau$ is any complex number with positive imaginary part then we define $$\vartheta(\tau) =\sum_{n\in\mathbb {Z}} e^{\pi i\tau n^2}$$ and one of the key properties of theta function defined above is $$\vartheta(\tau) =(-i\tau) ^{-1/2}\vartheta(-1/\tau)$$ Ramanujan's first formula probably assumes that $x\in(0,1)$ and hence one can write $x=\cos t$ with $t\in(0,\pi/2)$ and we can consider the complex number $\tau=\sin t +i\cos t$ which clearly has positive imaginary part. The choice of $\tau$ in this manner is done because it gives us $$(-i\tau) ^{-1/2}=\cos(t/2) +i\sin(t/2)=\sqrt{\frac{1+x}{2}}+i\sqrt{\frac{1-x}{2}}$$ and $$-1/\tau=-\sin t+i\cos t=-\sqrt{1-x^2}+ix$$ Using this value of $\tau$ in the transformation formula for theta functions we get $$1+2A+2iB=\frac{\sqrt{1+x}+i\sqrt{1-x}}{\sqrt{2}}(1+2A-2iB)$$ where $$A=\sum_{n=1}^{\infty}e^{-\pi n^2x}\cos(\pi n^2\sqrt{1-x^2}),B=\sum_{n=1}^{\infty} e^{-\pi n^2x}\sin(\pi n^2\sqrt{1-x^2})$$ and equating real parts we get $$1+2A=(1+2A)\sqrt {\frac{1+x}{2}}+2B\sqrt{\frac{1-x}{2}}$$ or $$\frac{1}{2}+A=\frac{\sqrt{2}+\sqrt{1+x}}{\sqrt{1-x}}B$$ In this manner the key formula of Ramanujan is established.
Out of the next two corollaries I was able to prove the second one easily by dividing the main formula by $\sqrt{1-x^2}$ and then taking limits as $x\to 1^{-}$. The first one dealing with $\sum e^{-5\pi n^2}$ was really looking difficult to obtain.
My question is
How to obtain the first corollary dealing with $\sum e^{-5\pi n^2}$ from the main formula of Ramanujan?
Since the formula appears to be using $x\in(0,1)$ I don't see a way to put $x=5$. Even if one does that both sides will contain the sums involving $\sum e^{-5\pi n^2}$ and it appears rather mysterious to obtain a link between $\sum e^{-\pi n^2}$ and $\sum e^{-5\pi n^2}$.
The link between these two sums can be obtained using a modular equation of degree 5, but the calculations involved are tedious (for this technique in action see this answer which evaluates $\sum_{n\in\mathbb {Z}} e^{-3\pi n^2}$). I was therefore hoping for some easier approach as indicated by Ramanujan. Maybe I am mising something obvious here.
| Finally I have managed to prove the identity in question. It appears that the main formula (proved in question) as well as its first corollary are both derived from the transformation formula for theta functions rather than being derived from each other. This means that the issue at hand is not really the corollary of the main result as I was expecting.
Let us write $$a=\vartheta(i), b=\vartheta(5i),c=\vartheta(i/5)\tag{1}$$ and we have to prove that $$a=b\sqrt {5\sqrt{5}-10}\tag{2}$$ This is done in two steps and the first one out of these two is obvious. Putting $\tau=5i$ in the transformation formula for theta functions (see the question) we get $$c=b\sqrt{5}\tag{3}$$ In order to prove $(2)$ we need another relation between $a, b$ and $c$ and use it together with $(3)$.
This is the second step where we put $\tau=i+2$ so that $$(-i\tau) ^{-1/2}=\frac{\sqrt{1+2i}}{\sqrt{5}}=\sqrt{\frac{\sqrt{5}+1}{10}}+i\sqrt{\frac{\sqrt{5}-1}{10}}=p+iq\text{ (say)} \tag{4}$$ and $$-\frac{1}{\tau}=\frac{i-2}{5}$$ Using these values in the transformation formula for theta function (and also noting that $\vartheta(\tau +2)=\vartheta(\tau)$) we get $$a=(p+iq)\left\{1+2\sum_{n=1}^{\infty} e^{-\pi n^2/5}\left(\cos\frac{2\pi n^2}{5}-i\sin\frac{2\pi n^2}{5}\right)\right\}$$ Note that the left hand side is purely real and hence equating real parts we get $$a=p\left(1 +2\sum_{n=1}^{\infty} e^{-\pi n^2/5}\cos\frac{2\pi n^2}{5}\right)+2q\sum_{n=1}^{\infty}e^{-\pi n^2/5}\sin\frac{2\pi n^2}{5}$$ and equating imaginary parts we get $$ 2p\sum_{n=1}^{\infty} e^{-\pi n^2/5}\sin\frac{2\pi n^2}{5}=q\left(1+2\sum_{n=1}^{\infty} e^{-\pi n^2/5}\cos\frac{2\pi n^2}{5}\right)$$ Combining these equations we have $$a=\frac{p^2+q^2}{p}\left(1+2\sum_{n=1}^{\infty}e^{-\pi n^2/5}\cos\frac{2\pi n^2}{5}\right)$$ And now we have the magic happening here. If $5\mid n$ then the cosine term equals $1$ otherwise it equals $\cos(2\pi/5)$. We can thus rewrite the above equation as $$a=\frac{p^2+q^2}{p}\left(1+2\cos\frac{2\pi}{5}\sum_{n>0,5\nmid n} e^{-\pi n^2/5}+2\sum_{n=1}^{\infty} e^{-5\pi n^2}\right)$$ and this can be further rewritten as $$a=\frac{p^2+q^2}{p}\left\{1+2\cos\frac{2\pi}{5}\sum_{n=1}^{\infty} e^{-\pi n^2/5}+2\left(1-\cos\frac{2\pi}{5}\right)\sum_{n=1}^{\infty}e^{-5\pi n^2}\right\}$$ Finally this means that
\begin{align} a&=\frac{p^2+q^2}{p}\left(1+(c-1)\cos\frac{2\pi}{5}+2(b-1)\sin^2\frac{\pi}{5}\right)\notag\\
&=b\cdot\frac{p^2+q^2}{p}\left(\sqrt{5}\cos\frac{2\pi}{5}+1-\cos\frac{2\pi}{5}\right)\text{ (using (3))}\notag\\
&=\frac{b} {p\sqrt{5}}\left(1+\frac{(\sqrt{5}-1)^2}{4}\right)\text{ (using (4))}\notag\\
&=\frac{b}{p}\cdot\frac{\sqrt{5}-1}{2}\notag\\
&=b\sqrt{\frac{5(\sqrt{5}-1)(3-\sqrt{5})}{4}}\notag\\
&=b\sqrt{5(\sqrt{5}-2)}\notag
\end{align}
I think this is almost what Ramanujan had in his mind when he posed the problem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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How to find $a, b$ and $c$? I was given the following two equations and told to find $a, b$ and $c$:
$$\left\{\begin{array}{c}
a+b-c=1\\
a^2+b^2-c^2=-1\end{array}\right.$$
I tried to form a matrix to solve it and I made this:
$$\left[\begin{array}{ccc|c}
1 & 1 & -1 & 1\\
a & b & -c & -1
\end{array}
\right]$$
Then I performed elimination and got this:
$$\left[\begin{array}{ccc|c}
1&0&\frac{-b+c}{-a+b}&\frac{b+1}{-a+b}\\
0&1&\frac{a-c}{-a+b}&-\frac{a+1}{-a+b}
\end{array}\right]$$
After this I tried to find a specific solution when $c=0$ and a made this:
$$\left[\begin{matrix}
\frac{1\pm i\sqrt 3}{2}\\
\frac{1\mp i\sqrt 3}{2}\\
0
\end{matrix}\right]$$
Then, to find all the solutions I tried to find the null space while making $c=1$, then I found that all values are of the form:
$$\left[\begin{matrix}
\frac{1\pm i\sqrt 3}{2}\\
\frac{1\mp i\sqrt 3}{2}\\
0
\end{matrix}\right]
%
+C\left[\begin{matrix}
0/1\\
1/0\\
1
\end{matrix}\right]$$
Note: ‘/’ denotes $“or”$ and $C$ can be any integer
But when tried to plug this back into my question taking $C=1$ it didn’t work, but when $C=0$, it works, please help me.
| There are only $2$ equations for $3$ variables so you can try to find two of them in function of the third one.
Since $a,b$ play a symmetrical role, we can choose between $b$ or $c$ as free variable. The choice of $c$ leads to solve quadratic equations, so we will choose $b$ instead.
First let $b=1$ then the system becomes $\begin{cases}a=c\\a^2-c^2=-2\iff 0=-2\end{cases}\quad$ the system has no solution.
So now we can exploit $a^2-c^2=\underbrace{(a-c)}_{1-b}(a+c)=-1-b^2\quad$ and solve
$\begin{cases}a=\frac b{b-1}\\c=\frac{b^2-b+1}{b-1}\end{cases}$
| {
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"url": "https://math.stackexchange.com/questions/3709466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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