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Differential of normal distribution Let
$$f(x)=\frac{\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\sigma\sqrt{2\pi}}$$
(Normal distribution curve) Where $\sigma$ is constant. Is my derivative correct and can it be simplified further?
$$\begin{align} f'(x)
&=\frac d{dx}\left(\frac{\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\sigma\sqrt{2\pi}}\right)\\
&=\frac{1}{\sigma\sqrt{2\pi}}\frac d{dx}\left(\exp\left(-\frac{x^2}{2\sigma^2}\right)\right)\\
\end{align}$$
$$\begin{align}
\implies\frac d{dx}\exp\left(-\frac{x^2}{2\sigma^2}\right)&=\frac d{dx}\sum_{n=0}^\infty\frac{\left(-\frac{x^2}{2\sigma^2}\right)^n}{n!}
=\sum_{n=0}^\infty\frac d{dx}\frac{\left(-\frac{x^2}{2\sigma^2}\right)^n}{n!}\\
&=\sum_{n=0}^\infty \frac 1{n!}\frac d{dx}\left(-\frac{x^2}{2\sigma^2}\right)^n=\sum_{n=0}^\infty -\frac 1{n!}\frac{1}{2\sigma^{2n}}\frac d{dx}x^{2n}\\
&=\sum_{n=0}^\infty -\frac 1{n!}\frac{1}{2\sigma^{2n}}2nx^{2n-1}=\sum_{n=0}^\infty -\frac {1}{n!}\frac{2nx^{2n-1}}{2\sigma^{2n}}\\
\end{align}$$
$$\implies f'(x)=\frac{-\sum_{n=0}^\infty \left(\frac {1}{n!}\frac{2nx^{2n-1}}{2\sigma^{2n}}\right)}{\sigma\sqrt{2\pi}}$$
I attempted to simplify the infinite summation to the exponential function but I am failing. How would you simplify this further?
| Here is one approach:
$f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-x^2/2\sigma^2}$. Taking the log of both sides we get:
$ln\left(f(x)\right) = -\frac{x^2}{2 \sigma^2} + ln \left( \frac{1}{\sigma \sqrt{2 \pi}} \right)$.
Let's differentiate both sides to get:
$\frac{f'(x)}{f(x)} = -\frac{x}{\sigma^2}$, implying $f'(x) = -\frac{xf(x)}{\sigma^2}$. Now we can substitute for $f(x)$ to get the final answer:
$f'(x) = -\frac{x}{\sigma^3 \sqrt{2 \pi}} e^{-x^2/2\sigma^2}$.
Algebra is a little less messy than brute-force differentiation, IMHO.
| {
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"url": "https://math.stackexchange.com/questions/461139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Need help with logarithmic differentitation I have the expression
$$y = \sqrt{x^2(x+1)(x+2)}.$$
I have tried looking at videos but I still cannot arrive at the correct answer and don't know how to get there.
By the way, the correct answer is
$$y' = \frac{4x^2+9x+4}{2\sqrt{(x+1)(x+2)}}.$$
Please, help.
| $$y = \sqrt{x^2 (x+1)(x+2)}$$
$$\ln{y} = \ln{x} + \frac12 \ln{(x+1)} + \frac12 \ln{(x+2)}$$
$$\frac{y'}{y} = \frac{1}{x} + \frac12 \frac{1}{x+1} + \frac12 \frac{1}{x+2}$$
$$y' = x \sqrt{(x+1)(x+2)} \left ( \frac{1}{x} + \frac12 \frac{1}{x+1} + \frac12 \frac{1}{x+2}\right) = x \sqrt{(x+1) (x+2)} \left ( \frac{4 x^2+9 x+4}{2 x(x+1)(x+2)}\right)$$
which I believe works out to your answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does this integral have a closed form expression? I have been working with two related integrals. The first one yields a simple expression, but I can't seem to find a simple expression for the second.
Integral 1
$$ \begin{align*}
I_x &= \int_z^{1-y+z} \sqrt{\frac{1}{1-x-y+z} + \frac{1}{x-z}} \, dx\\
&= \int_z^{1-y+z} \sqrt{\frac{1-y}{(1-x-y+z)(x-z)}} \, dx\\
&= \sqrt{1-y} \int_z^{1-y+z} \frac{1}{\sqrt{(1-x-y+z)(x-z)}} \, dx\\
\end{align*} $$
We can substitute $u=\frac{x-z}{1-y}$, so $x=u(1-y)+z$, and $dx=(1-y) \, du$. Then we have:
$$ \begin{align*}
I_x &= \sqrt{1-y} \int_0^1 u^{\frac{1}{2}-1}(1-u)^{\frac{1}{2}-1} \, du\\
&= \sqrt{1-y} \, \operatorname{B} \left( \tfrac{1}{2}, \tfrac{1}{2} \right)\\
&= \sqrt{1-y} \, \pi
\end{align*} $$
So this integral has a simple expression because the substitution led to the exact form of the beta function.
Integral 2
$$ I_z = \int_{\max(0,x+y-1)}^{\min(x,y)} \sqrt{\frac{1}{1-x-y+z} + \frac{1}{x-z} + \frac{1}{y-z} + \frac{1}{z}} \, dz $$
This seems related to the first integral, but I can't seem to simplify it. Even if we assume one of the four "cases" of integration limits ($\int_0^x$, $\int_0^y$, $\int_{x+y-1}^x$, or $\int_{x+y-1}^y$) and try to proceed, it doesn't seem to help.
Am I missing something simple? Is there an obscure integration technique that would help? Or is there just no way to simplify this?
Update (6-26-14): From Jacquet & Szpankowski, 2003, "Markov Types and Minimax Redundancy for Markov Sources" (link), p.10, we have the following, which may (or may not) help here:
$$ \begin{align*}
I_{JS} &= 4 \int_0^1 \frac{1}{\sqrt{(1{-}x)x}} \, dx \int_0^{\min(x,1{-}x)} \frac{1}{(1{-}x{-}y)(x{-}y)} \, dy\\
&= 8 \int_0^{\frac{1}{2}} \frac{\log(1{-}2x) - \log\left(1{-}2\sqrt{(1{-}x)x}\right)}{\sqrt{(1{-}x)x}} \, dx\\
&= 16 \int_0^{\frac{\pi}{4}} \log \left( \frac{\cos(2\theta)}{1{-}\sin(2\theta)} \right) \, d\theta\\
&= 16 \, G\\
\end{align*} $$
where $G \approx 0.915965594$ is Catalan's constant.
| First of all, regarding the limits of integration $a=\max{(0,x+y-1)}$ and $b=\min{(x,y)}$, you do not have to break up your analysis into four separate cases to accommodate the two possible values taken by the min/max functions. These functions can be represented in terms of elementary functions as follows:
$$\max{(a,b)}=\frac{a+b}{2}+\frac12|a-b|=\frac{a+b}{2}+\frac12\sqrt{(a-b)^2},\\
\min{(a,b)}=\frac{a+b}{2}-\frac12|a-b|=\frac{a+b}{2}-\frac12\sqrt{(a-b)^2}.$$
With that in mind, and after making the substitutions $u=2z-x-y$ followed by $w=u+1$, a little algebra reduces the integral to:
$$\begin{align}
\mathcal{I}{\left(x,y\right)}
&=\int_{\max{(0,x+y-1)}}^{\min{(x,y)}}\sqrt{\frac{1}{z}+\frac{1}{z-(x+y-1)}+\frac{1}{x-z}+\frac{1}{y-z}}\,\mathrm{d}z\\
&=\int_{\frac{x+y-1+|1-x-y|}{2}}^{\frac{x+y-|x-y|}{2}}\sqrt{\frac{1}{z}+\frac{1}{z-(x+y-1)}+\frac{1}{x-z}+\frac{1}{y-z}}\,\mathrm{d}z\\
&=\frac12\int_{-1+|1-x-y|}^{-|x-y|}\sqrt{\frac{1}{\frac{u+x+y}{2}}+\frac{1}{\frac{u+x+y}{2}-(x+y-1)}+\frac{1}{x-\frac{u+x+y}{2}}+\frac{1}{y-\frac{u+x+y}{2}}}\,\mathrm{d}u\\
&=\frac12\int_{-1+|1-x-y|}^{-|x-y|}\sqrt{\frac{2}{u+x+y}+\frac{2}{u-x-y+2}+\frac{2}{-u+x-y}+\frac{2}{-u-x+y}}\,\mathrm{d}u\\
&=\frac{\sqrt{2}}{2}\int_{-1+|1-x-y|}^{-|x-y|}\sqrt{\frac{1}{u+x+y}+\frac{1}{u-x-y+2}+\frac{1}{-u+x-y}+\frac{1}{-u-x+y}}\,\mathrm{d}u\\
&=\frac{\sqrt{2}}{2}\int_{|1-x-y|}^{1-|x-y|}\sqrt{\frac{1}{w+x+y-1}+\frac{1}{w-x-y+1}+\frac{1}{1-w+x-y}+\frac{1}{1-w-x+y}}\,\mathrm{d}w\\
&=\frac{\sqrt{2}}{2}\int_{|1-x-y|}^{1-|x-y|}\sqrt{\frac{2w}{w^2-(x+y-1)^2}+\frac{2(1-w)}{(1-w)^2-(x-y)^2}}\,\mathrm{d}w\\
&=\int_{|1-x-y|}^{1-|x-y|}\sqrt{\frac{w}{w^2-|x+y-1|^2}+\frac{(1-w)}{(1-w)^2-|x-y|^2}}\,\mathrm{d}w.
\end{align}$$
At this point I'm not exactly sure how best to proceed. My gut is starting to tell me that this integral is going to wind up being some ghastly thing in terms of elliptic integrals.
| {
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"url": "https://math.stackexchange.com/questions/463127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int_{0}^{\infty}\frac{1}{x}\big (\frac{\sinh ax}{\sinh x}-ae^{-2x}\big )dx$ Some time ago, stumbled out of an integral:
$$\int_{0}^{\infty}\frac{1}x{}\left (\frac{\sinh ax}{\sinh x}-ae^{-2x}\right )dx=\ln\frac{\pi\cos\frac{a\pi}{2}}{\Gamma^2(\frac{a+1}{2})};\left | a \right |<1$$
I have no idea where to start?
| To be specific, let $I(a)$ denote the integral and differentiate the integral with respect to $a$. Then by referring to the contents and notations in my blog posting,
\begin{align*}
I'(a)
&= \int_{0}^{\infty} \left( \frac{x \cosh ax}{\sinh x} - e^{-2x} \right) \, \frac{dx}{x} \\
&= \int_{0}^{\infty} \left( \frac{x e^{-(1-a)x} + x e^{-(1+a)x}}{1 - e^{-2x}} - e^{-2x} \right) \, \frac{dx}{x} \\
&= \int_{0}^{\infty} \left( \frac{1}{2} \frac{x e^{-(1-a)x/2} + x e^{-(1+a)x/2}}{1 - e^{-x}} - e^{-x} \right) \, \frac{dx}{x} \qquad (2x \mapsto x) \\
&= \frac{1}{2} \left\{ \mathrm{ctr} \, \left( \frac{x e^{-(1-a) x/2}}{1 - e^{-x}} \right) + \mathrm{ctr} \, \left( \frac{x e^{-(1+a) x/2}}{1 - e^{-x}} \right) \right\} - \mathrm{ctr} \, (e^{-x}) \\
&= - \frac{1}{2} \left\{ \psi_{0}\left(\frac{1-a}{2}\right) + \psi_{0}\left(\frac{1+a}{2}\right) \right\}.
\end{align*}
Since $I(0) = 0$, we can retrieve $I(a)$ by a simple integration:
$$ I(a) = \int_{0}^{a} I'(t) \, dt = \log\Gamma\left(\frac{1-a}{2} \right) - \log\Gamma\left(\frac{1+a}{2} \right). $$
Now the rest follows by the Euler's reflection formula.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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interesting Integral , alternative solution. Show the following relation:
$$\int_{0}^{\infty} \frac{x^{29}}{(5x^2+49)^{17}} \,\mathrm dx = \frac{14!}{2\cdot 49^2 \cdot 5^{15 }\cdot 16!}.$$
I came across this intgeral on a physics forum and solved by (1) making a substitution (2) finding a recursive formula (with integration by parts).
I would like to see a clever alternative solution to this one, since the only way I would solve such an integral is by finding a recursive formula. But I'm sure there are other ways/methods to do it (maybe more elegant) that I want to discover.
| The given integral is \begin{align}
I=&\int_{0}^{\infty}\frac{x^{29}}{(5x^2+49)^{17}}dx\\
\ =&\frac{1}{2\cdot 5^{15}}\int_{0}^{\infty}\frac{z^{14}}{(z+49)^{17}}dz\quad(\mbox{substituting}\ 5x^2=z)\\
\ =&\frac{1}{2\cdot 5^{15}}\int_{0}^{\pi/2}\frac{(49)^{14}\tan^{28}\theta\cdot2\cdot49\cdot\tan \theta\sec^2\theta}{49^{17}\sec^{34}\theta}dz\quad(\mbox{substituting}\ z=49\tan^2\theta)\\
\ =&\frac{1}{2\cdot 5^{15}\cdot49^2}\cdot2\int_{0}^{\pi/2}\frac{\sin^{29}\theta}{\cos^{29}\theta}\cos^{32}\theta\ d\theta\\
\ =&\frac{1}{2\cdot 5^{15}\cdot 49^2}\cdot\beta\left(15,2\right)\\
\ =&\frac{\Gamma(15)\Gamma(2)}{2\cdot5^{15}\cdot49^2\cdot\Gamma(17)}\\
\ =&\frac{14!}{2\cdot5^{15}\cdot49^2\cdot16!}
\end{align}
Along similar lines of argument, the integral $I(a,b,c,d)$($a,b,c,d$ positive integres) as defined by Hagen von Eitzen becomes(if I have not committed any mistake) $$\frac{1}{2\cdot c^d}\left(\frac{c}{b}\right)^{(n+1)/2}\beta\left(\frac{n+1}{2},d-\frac{n-1}{2}-1\right)\\ =\frac{1}{2\cdot c^{d-\frac{n+1}{2}}\cdot b^{\frac{n+1}{2}}}\cdot\frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(d-1-\frac{n-1}{2}\right)}{(d-1)!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464181",
"timestamp": "2023-03-29T00:00:00",
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Find the limit: $\lim\limits_{x\to1}\dfrac{x^{1/5}-1}{x^{1/3}-1}$ Find the limit of $$\lim_{x\to 1}\frac{x^{1/5}-1}{x^{1/3}-1}$$
How should I approach it? I tried to use L'Hopital's Rule but it's just keep giving me 0/0.
| Hint: $$\frac{x^{\frac{1}{5}}-1}{x^{\frac{1}{3}}-1}=\frac{(x^{\frac{1}{15}})^3-1}{(x^{\frac{1}{15}})^5-1}=\frac{((x^{\frac{1}{15}})-1)(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1)}{((x^{\frac{1}{15}})-1)((x^{\frac{1}{15}})^4+(x^{\frac{1}{15}})^3+(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1)}$$
This is equal to:
$$\frac{(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1}{(x^{\frac{1}{15}})^4+(x^{\frac{1}{15}})^3+(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7) $ The question is to simplify $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$ without using a calculator .
My friend has given me this challenge . I solved it by expanding $$(a+b+c)(-a+b+c)(a-b+c)(a+b-c) = -a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4$$ and then substituting $a,b,c=\sqrt 5 , \sqrt6 , \sqrt7$ respectively to get the answer $104$ .
But I suppose there is a more elegant and easy way to solve this problem .
Can anyone find it ?
| Use $$(a+b)(a-b)=a^2-b^2.$$
With $a=\sqrt6+\sqrt7$, $b=\sqrt5$ you see that the product of first two numbers is $(\sqrt6+\sqrt7)^2-5=8+2\sqrt{42}$. With $a=\sqrt5$, $b=\sqrt7-\sqrt6$ you get that
the product of last two is $5-(\sqrt7-\sqrt6)^2=-8+2\sqrt{42}$. One last application of this rule tells you that the answer is
$$
(2\sqrt{42})^2-8^2=168-64=104.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/465103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Factorise: $2a^4 + a^2b^2 + ab^3 + b^4$ Factorize : $$2a^4 + a^2b^2 + ab^3 + b^4$$
Here is what I did:
$$a^4+b^4+2a^2b^2+a^4-a^2b^2+ab^3+b^4$$
$$(a^2+b^2)^2+a^2(a^2-b^2)+b^3(a+b)$$
$$(a^2+b^2)^2+a^2(a+b)(a-b)+b^3(a+b)$$
$$(a^2+b^2)^2+(a+b)((a^2(a-b)) +b^3)$$
$$(a^2+b^2)^2+(a+b)(a^3-a^2b+b^3)$$
At this point I don't know what to do and am feeling that my direction is wrong. Please help me.
( Wolfram alpha says that the answer is $(a^2-a b+b^2) (2 a^2+2 a b+b^2)$ but how? )
| Hint
We look for a factorization on the form
$$2a^4 + a^2b^2 + ab^3 + b^4=(2a^2+\alpha ab+b^2)(a^2+\beta ab+b^2)$$
We find $\alpha=2,\beta=-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Trisect a quadrilateral into a $9$-grid; the middle has $1/9$ the area Trisect sides of a quadrilateral and connect the points to have nine quadrilaterals, as can be seen in the figure. Prove that the middle quadrilateral area is one ninth of the whole area.
| This is most easily done using vectors. Let the points $A, B, C, D$ be represented by the vectors $a, b, c, d$. The area $[ABCD]$ is equal to $\frac{1}{2}(a-c) \times (b-d) $.
If you are unfamiliar with this, consider triangulation using the origin, and sum up the 4 triangle areas, to get
$$\begin{align} [ABCD] = & \frac{1}{2} a \times b + \frac{1}{2} b \times c + \frac{1}{2} c \times d + \frac{1}{2} d \times a \\ = & (a-c) \times \frac{1}{2} b + (a-c ) \times (-\frac{1}{2} d) \\= & \frac{1}{2}(a-c) \times (b-d) \end{align}$$
It is easy to show that $W= \frac{4a+2b+c+2d} {9}, X = \frac{ 2a+4b+2c+d} { 9},Y = \frac{a+2b+4c+d} { 9} , Z = \frac{ 2a+b+2c+4d} {9} $. Hence the area is
$$
[WXYZ] = \frac{1}{2} ( \frac{3a-3c}{9} ) \times ( \frac{3b-3d}{9} ) = \frac{1}{9} \times \frac{1}{2} (a-c)(b-d) = \frac{1}{9} [ABCD]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of a sequence I need guidance for the following question.
Using the fact that $\sum_1^{\infty}\frac{(-1)^{n+1}}{n}=\log2$, $\sum_1^{\infty}\frac{(-1)^{n}}{n(n+1)}$ equals
$1.$ $1-2\log2$
$2.$ $1+2\log2$
$3.$ $(\log2)^2$
$4.$ $-(\log2)^2$
The given sequence gives us $1-\frac12+\frac13-\frac14+\cdots=log2$, but I am unable to think how this would help me to solve $-\frac12+\frac16-\frac1{12}+\frac1{20}-\cdots$
I wish somebody could help. Thanks in advance!
| $
\begin{array}{l}
\frac{1}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}} \\
\Rightarrow \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{{n\left( {n + 1} \right)}}} = \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}} - \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{{n + 1}}} \\
\Rightarrow - \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{n\left( {n + 1} \right)}}} = \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}} + \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 2} }}{{n + 1}}} \\
\Rightarrow \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{n\left( {n + 1} \right)}}} = - \left( {\sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}} + \sum\limits_{n = 2}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}} } \right) \\
\Rightarrow \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{n\left( {n + 1} \right)}}} = - \left( {\sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}} - 1 + \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}} } \right) \\
\Rightarrow \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{n\left( {n + 1} \right)}}} = - \left( {2\sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}} - 1} \right) \Rightarrow \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{n\left( {n + 1} \right)}}} = 1 - 2\log 2 = 1 - \log 4 \\
\end{array}
$
| {
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"timestamp": "2023-03-29T00:00:00",
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The Laurent series of the digamma function at the negative integers To find the Laurent series of $\psi(z)$ at $z= 0$, I would first find the Taylor series of $\psi(z+1)$ at $z=0$ and then use the functional equation of the digamma function.
Specifically,
$$\begin{align} \psi(z + 1) = \frac{1}{z} + \psi(z) &= \psi(1) + \psi'(1)z + \mathcal{O}(z^{2}) \\ &= -\gamma + \zeta(2) + \mathcal{O}(z^{2}) \\ &= -\gamma + \frac{\pi^{2}}{6}z + \mathcal{O}(z^{2}) \end{align}$$
which implies
$$ \psi(z) = -\frac{1}{z} - \gamma + \frac{\pi^{2}}{6} z + \mathcal{O}(z^{2}).$$
But I'm having trouble finding the Laurent series of $\psi(z)$ at the negative integers.
Since $\psi(z)$ has simple poles at the negative integers with residue $-1$, I know that the first term of the series must be $ \displaystyle\frac{-1}{z+n}$.
But I would like to determine more terms in the series.
EDIT:
The series appears to be $$\begin{align} \psi(z) &= - \frac{1}{z+n} + (H_{n} - \gamma)+ \Big( H_{n}^{(2)} + \zeta(2) \Big) (z+n) + \Big( H_{n}^{(3)} - \zeta(3) \Big) (z+n)^{2} \\ &+ \Big( H_{n}^{(4)} + \zeta(4) \Big) (z+n)^{3} + \Big( H_{n}^{(5)} - \zeta(5) \Big) (z+n)^{4} + \ldots \end{align}$$
SECOND EDIT:
We can find the constant term by evaluating $\lim_{z \to -n} \Big( \psi(z) + \frac{1}{z+n} \Big)$.
Since $$ \begin{align} &\psi(z) + \frac{1}{z+n} \\ &= \psi(z+1) - \frac{1}{z} + \frac{1}{z+n} \\ &= \psi(z+2) - \frac{1}{z+1} - \frac{1}{z} + \frac{1}{z+n} \\ &= \ ... \ = \psi(z+n+1) - \frac{1}{z+n} - \frac{1}{z+n-1} - \ldots - \frac{1}{z+1} - \frac{1}{z} + \frac{1}{z+n}, \end{align}$$
we have $$ \lim_{z \to -n} \Big( \psi(z) + \frac{1}{z+n} \Big) = \psi(1) + 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} = \psi(1) + H_{n} = H_{n} - \gamma.$$
Similarly, we can find the coefficient of the $(z+n)$ term by evaluating $ \lim_{z \to -n} \frac{\psi(z) + \frac{1}{z+n} - H_{n} + \gamma}{z+n}.$
Since $$ \psi_{1}(z) = \psi_{1}(z+n+1) + \frac{1}{(z+n)^{2}} + \frac{1}{(z+n-1)^{2}} + \ldots + \frac{1}{z^2}, $$
we have $$ \lim_{z \to -n} \frac{\psi(z) + \frac{1}{z+n} - H_{n} + \gamma}{z+n} = \lim_{z \to -n} \Big(\psi_{1}(z) - \frac{1}{(z+n)^{2}} \Big) = \psi_{1}(1) + H_{n}^{(2)} = H_{n}^{(2)} + \zeta(2) .$$
And we can find the coefficient of $(z+n)^{2}$ by evaluating $ \lim_{z \to -n} \frac{\psi(z) + \frac{1}{z+n} - H_{n} + \gamma -\big( H_{n}^{(2)} + \zeta(2) \big) (z+n)}{(z+n)^{2}} . $
Since $$\psi_{2}(z) = \psi_{2} (z+n+1) - \frac{2}{(z+n)^{3}} - \frac{2}{(z+n-1)^{3}} - \ldots - \frac{2}{z^{3}},$$
we have $\begin{align} \lim_{z \to -n} \frac{\psi(z) + \frac{1}{z+n} - H_{n} + \gamma -\big( H_{n}^{(2)} + \zeta(2) \big) (z+n)}{(z+n)^{2}} &= \lim_{z \to -n} \frac{\psi_{1}(x) - \frac{1}{(z+n)^{2}} -H_{n}^{(2)} - \zeta(2)}{2(z+n)} \\ &= \lim_{z \to -n} \frac{\psi_{2}(x) + \frac{2}{(z+n)^{3}}}{2} \\ &= \frac{1}{2} \Big( \psi_{2}(1) + 2 H_{n}^{(3)} \Big) \\ &= H_{n}^{(3)} - \zeta(3). \end{align}$
$ $
And so on.
| We will use the Laurent expansion around $x=0$
$$\psi_0(x)=-\frac{1}{x}-\gamma-\sum_{k\geq 1}\zeta(k+1)(-x)^k$$
Look the proof here
Then
$$\tag{1}\psi_0(x+N)=-\frac{1}{x+N}-\gamma-\sum_{k\geq 1}(-1)^k\zeta(k+1)(x+N)^k$$
Now we use that
$$\tag{2} \psi_0(x+N)=\psi_0(x)+\sum_{k=0}^{N-1}\frac{1}{x+k}$$
Now we look at the finite sum
\begin{align}\sum_{k=0}^{N-1}\frac{1}{x+k}&=\sum_{k=0}^{N-1}\frac{1}{k-N+x+N}\\&=\sum_{k=0}^{N-1}\frac{1}{k-N}\frac{1}{1+\frac{x+N}{k-N}}\\ &= \sum_{k=0}^{N-1}\frac{1}{k-N}\sum_{m \geq 0}(-1)^m\left(\frac{x+N}{k-N} \right)^m \\&= \sum_{m \geq 0}(-1)^m(x+N)^m \sum_{k=0}^{N-1}\frac{1}{(k-N)^{m+1}}\\&= -\sum_{m \geq 0} H^{(m+1)}_{N} (x+N)^{m}\end{align}
Hence we have
$$\psi_0(x+N)=\psi_0(x)-\sum_{m \geq 0} H^{(m+1)}_{N} (x+N)^{m}$$
Substitute in (1) to obtain
$$\psi_0(x)-\sum_{m \geq 1} H^{(m+1)}_{N} (x+N)^{m}=-\frac{1}{x+N}-\gamma-\sum_{k\geq 1}(-1)^k\zeta(k+1)(x+N)^k$$
$$\psi_0(x)=-\frac{1}{x+N}-\gamma-\sum_{k\geq 1}(-1)^k\zeta(k+1)(x+N)^k+\sum_{m \geq 0} H^{(m+1)}_{N} (x+N)^{m}$$
$$\psi_0(-x)=\frac{1}{x-N}-\gamma-\sum_{k\geq 1}\zeta(k+1)(x-N)^k+\sum_{m \geq 0}(-1)^m H^{(m+1)}_{N} (x-N)^{m}$$
$$\psi_0(-x)+\gamma=\frac{1}{x-N}+H_N+\sum_{k\geq 1}((-1)^kH^{(k+1)}_{N}-\zeta(k+1))(x-N)^k$$
Proof of (2)
$$\psi_0(x+N)=\psi_0(x)+\sum_{k=0}^{N-1}\frac{1}{x+k}$$
By induction for $N=1$ we get
$$\psi_0(x+1)=\psi_0(x)+\frac{1}{x}$$
Which is true and can be proved using
$$\psi_0(x+1)=-\gamma+\sum_{n\geq 1}\frac{x}{n(n+x)}$$
Now for the inductive step , assume
$$\psi_0(x+N)=\psi_0(x)+\sum_{k=0}^{N-1}\frac{1}{x+k}$$
Then
$$\psi_0(x+N+1)=\psi_0(x+N)+\frac{1}{x+N+1}=\psi_0(x)+\sum_{k=0}^{N}\frac{1}{x+k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
How prove this inequality $\sum\limits_{1\le ilet $n\ge 2,n\in Z$,and $x_{1},x_{2},\cdots,x_{n}\in[0,1]$,
show that
$$\sum_{1\le i<j\le n}ix_{i}x_{j}\le\dfrac{n-1}{3}\sum_{k=1}^{n}kx_{k}$$
This problem is (2013,8.16) chia west compition
my idea:
let
\begin{align*}
&\dfrac{n(2n+1)(n+1)}{6}=n^2+(n-1)^2+\cdots+1\\
&\ge(x_{1}+x_{2}+\cdots+x_{n})^2+(x_{2}+x_{3}+\cdots+x_{n})^2+(x_{3}+\cdots+x_{n})^2+\cdots+(x_{n-1}+x_{n})^2+x^2_{n}\\
&=nx^2_{n}+(n-1)x^2_{n-1}+\cdots+2x^2_{2}+x^2_{1}+2\sum_{1\le i<j\le n}ix_{i}x_{j}\\
&=\sum_{k=1}^{n}kx^2_{k}+2\sum_{1\le i<j\le n}ix_{i}x_{j}
\end{align*}
then
$$\sum_{k=1}^{n}kx^2_{k}\le\dfrac{n(n+1)(2n+1)}{6}-2\sum_{1\le i<j\le n}ix_{i}x_{j}$$
use
Cauchy-Schwarz inequality
$$\sum_{k=1}^{n}kx^2_{k}\sum_{k=1}^{n}k\ge(\sum_{k=1}^{n}(kx_{k})^2\Longrightarrow\sum_{k=1}^{n}kx^2_{k}\ge\dfrac{2}{n(n+1)}\left(\sum_{k=1}^{n}kx_{k}\right)^2$$
then we have
$$\dfrac{2}{n(n+1)}\left(\sum_{k=1}^{n}kx_{k}\right)^2\le\dfrac{n(n+1)(2n+1)}{6}-2\sum_{1\le i<j\le n}ix_{i}x_{j}$$
it suffices to prove that
$$\dfrac{n(n+1)(2n+1)}{6}-\dfrac{2}{n(n+1)}\left(\sum_{k=1}^{n}kx_{k}\right)^2\le\dfrac{n-1}{3}\sum_{k=1}^{n}kx_{k}$$
then I let
$$\sum_{k=1}^{n}kx_{k}=A\in[0,\dfrac{n(n+1)}{2}]$$
then it will prove that
$$\dfrac{n(n+1)(2n+1）}{6}-\dfrac{2}{n(n+1)}A^2\le\dfrac{n-1}{3}A$$
It seem not true, so my idea can't work,and How to prove this inequality? Thank you
| For more convenient notation, let $L_n := \sum_{1 \leq i<j\leq n} i x_i x_j$ and $R_n := \frac{n-1}{3} \sum_{1 \leq i \leq n} i x_i$. We need to prove that $L_n \leq R_n$.
We proceed by induction, the case $n = 1$ being trivial. Suppose that the claim holds for $n$, we will prove it for $n+1$. It will suffice to show that $\Delta L_n \leq \Delta R_n$ where $\Delta L_n := L_{n+1} - L_n$ and $\Delta R_n := R_{n+1} - R_n$.
We have
$$\Delta L_n = \sum_{1 \leq i\leq n} i x_i x_{n+1}$$
and $$\Delta R_n = \frac{1}{3} \sum_{1 \leq i\leq n} i x_i x_{n+1} + \frac{(n+1)n}{3} x_{n+1} = \frac{1}{3} L_n + \frac{2}{3}{n+1 \choose 2} x_{n+1}$$.
Thus, $\Delta R_n \geq \Delta L_n$ is equivalent to ${n+1 \choose 2} x_{n+1} \geq \Delta L_n$. But this is clear because:
$$\Delta L_n = \sum_{1 \leq i\leq n} i x_i x_{n+1} \leq x_{n+1} \sum_{1 \leq i\leq n} i = x_{n+1} {n+1 \choose 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Logarithm integral with golden ratio $\int^1_0 \frac{\log(1+\phi x^2)}{1+x}\, dx$ How to evaluate the following integral
$$ \int^1_0 \frac{\log(1+\phi x^2)}{1+x}\, dx$$
where
$$\phi =\frac{\sqrt{5}+1}{2}$$
Is there a closed form for the integral ?
My first attempt would involve relating the integral to dilogarithms but that seems complicated !
| Let's replace $\phi$ with a positive parameter $a$:
$$
\mathcal{I}(a) = \int_0^1 \frac{\log(1+a x^2)}{1+x} \mathrm{d}x \stackrel{x^2=t}{=} \int_0^1 \frac{\log(1+ a t)}{1+\sqrt{t}} \frac{\mathrm{d}t}{2 \sqrt{t}}
$$
Differentiating under the integral sign:
$$\begin{eqnarray}
\mathcal{I}^\prime(a) &=& \int_0^1 \frac{\sqrt{t}}{2\left(1+a t\right)\left(1+ \sqrt{t}\right)} \mathrm{d}t = \left[\frac{\log \left(1+\sqrt{t}\right)}{1+a}+\frac{\log (1+a t)}{2 a
\left(1+a\right)}-\frac{\arctan\left(\sqrt{a t}\right)}{\sqrt{a}\left(1+a\right)} \right]_0^1 \\
&=& \frac{\log 2}{1+a} + \frac{\log(1+a)}{2 a \left(1+a\right)} - \frac{\arctan(\sqrt{a})}{\sqrt{a} \left(1+a\right)} \\
&=& \frac{\log(1+a)}{2} \left(\frac{1}{a} - \frac{1}{1+a} \right) + \frac{\mathrm{d}}{\mathrm{d}a}\left( \log 2 \cdot \log\left(1+a\right) - \arctan^2\left(\sqrt{a}\right)\right) \\
&=& \frac{\log(1+a)}{2 a} + \frac{\mathrm{d}}{\mathrm{d}a}\left( \log 2 \cdot \log\left(1+a\right) + \frac{\log^2\left(1+a\right)}{4} - \arctan^2\left(\sqrt{a}\right)\right)
\end{eqnarray}
$$
Hence, for $\alpha>0$
$$\begin{eqnarray}
\mathcal{I}\left(\alpha\right) &=& \int_0^\alpha \mathcal{I}^\prime(a) \mathrm{d}a = \left[ \log 2 \cdot \log\left(1+a\right) + \frac{\log^2\left(1+a\right)}{4} - \arctan^2\left(\sqrt{a}\right) \right]_{0}^{\alpha} + \frac{1}{2} \underbrace{\int_0^\alpha \frac{\log(1+a)}{a} \mathrm{d}a }_{-\operatorname{Li}_2\left(-\alpha\right)} \\
&=& \frac{\log^2\left(1+\alpha\right)}{4} + \log(2)\log\left(1+\alpha\right) - \arctan^2\left(\sqrt{\alpha}\right) - \frac{1}{2} \operatorname{Li}_2\left(-\alpha\right)
\end{eqnarray}
$$
Quadrature confirmation:
In[45]:= N[
With[{a = GoldenRatio}, -ArcTan[Sqrt[a]]^2 + Log[2] Log[1 + a] -
1/4 Log[1 + a]^2 - 1/2 PolyLog[2, -a]], 30]
Out[45]= 0.226575874320240735184008669784
In[46]:= NIntegrate[Log[1 + GoldenRatio x^2]/(1 + x), {x, 0, 1},
WorkingPrecision -> 70] - %
Out[46]= 0.*10^-31
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Minima problem? This is a question in my textbook which I can't solve. Any help would be appreciated, thanks.
"A piece of wire 10 metres long is cut into two portions. One piece is bent to form a circle, and the other piece to form a square. Find the circumference of the circle if the sum of areas of the circle and square is to be a minimum. Give your answer in terms of $pi$."
| Given Total length of wire $10$ meter, Now Let $x$ meter cut from it and form a Circle and
remaining $(10-x)$ form a Square.
So Here Radius of Square is $\displaystyle \frac{x}{2\pi}$ and Length of square is $\displaystyle \frac{(10-x)}{4}$
Now Area of Circle $\displaystyle \pi\cdot \frac{x^2}{4\pi^2} = \frac{x^2}{4\pi}$ and Area of Square $\displaystyle \frac{(10-x)^2}{16}$
So Total area $\displaystyle A = \frac{x^2}{4\pi}+\frac{(10-x)^2}{16}.$
Now here we have to minimize $A$
So using $\bf{Cauchy-Schwartz}$ Inequality, We get
$$\displaystyle A = \frac{x^2}{4\pi}+\frac{(10-x)^2}{16}\geq \frac{x+10-x}{4\pi+16} = \frac{10}{4\pi+16}$$
and equality hold when $$\displaystyle \frac{x}{4\pi} = \frac{10-x}{16}\Rightarrow x=\frac{10\pi}{\pi+4}$$
So length of wire to form a circle $\displaystyle x = \frac{10\pi}{\pi+4}$ and
length of wire to form a square $\displaystyle (10-x) = 10-\frac{10\pi}{\pi+4} = \frac{40}{\pi+4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric Identities for $\sin nx$ and $\cos nx $ These are generalizations of simple trigonometric identities for $\sin 2x$ and $\cos 2x$, but in general how can we prove them?
$$\sin nx =\sum_{k=1}^{\left\lceil\frac{n}{2}\right\rceil}(-1)^{k-1}\binom{n}{2k-1}\sin^{2k-1}x\cos^{n-2k+1}x,$$
$$\cos nx =\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}(-1)^{k}\binom{n}{2k}\sin^{2k}x\cos^{n-2k}x,$$
$$\sin nx =\sin x\left(\sum_{k=1}^{\left\lceil\frac n 2\right\rceil}(-1)^{k-1}\binom{n-k}{k-1}(2\cos x)^{n-2k+1}\right),$$
$$\cos nx =\cos x\left((2\cos x)^{n-1}+\sum_{k=1}^{\left\lfloor\frac n 2\right\rfloor}\frac n k(-1)^k\binom{n-k-1}{k-1}(2\cos x)^{n-2k-1}\right).$$
| Have a look at http://en.wikipedia.org/wiki/Chebyshev_polynomials. Once you get the first two identities, you'll easily get the last two. As pbs suggested, you can use the Binomial Theorem. The reason is :
$$\cos(n\theta) + i\sin(n\theta) = e^{in\theta} = (e^{i\theta})^{n} = (\cos(\theta)+i\sin(\theta))^{n}$$
The idea is to expand $(\cos(\theta) + i\sin(\theta))^{n}$ to identify its real and imaginary part. In short, you have :
\begin{eqnarray*}
(\cos(\theta)+i\sin(\theta))^{n} & = & \sum_{k=0}^{n} \begin{pmatrix} n \\ k \end{pmatrix} i^{k} \sin^{k}(\theta) \cos^{n-k}(\theta) \\
& = & \sum_{p=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^{p} \begin{pmatrix} n \\ 2p \end{pmatrix} \sin^{2p}(\theta)\cos^{n-2p}(\theta) - i\sum_{p=0}^{\lfloor \frac{n+1}{2} \rfloor} (-1)^{p} \begin{pmatrix} n \\ 2p-1 \end{pmatrix} \sin^{2p-1}(\theta)\cos^{n-2p+1}(\theta)
\end{eqnarray*}
So, you get the expected identities :
$$\cos(n\theta) = \sum_{p=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^{p} \begin{pmatrix} n \\ 2p \end{pmatrix} \sin^{2p}(\theta)\cos^{n-2p}(\theta)$$
and
$$ \sin(n\theta) = \sum_{p=0}^{\lceil \frac{n}{2} \rceil} (-1)^{p+1} \begin{pmatrix} n \\ 2p-1 \end{pmatrix} \sin^{2p-1}(\theta)\cos^{n-2p+1}(\theta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/476369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Numbers of solutions $e^x=-x^2+2x+5 $ Prove that $e^x=-x^2+2x+5 $ have exactly two solutions.
Is it enoguht that Vertex of the parabola is over $y=e^x$ and arms of it looks down
| This is a little out of the blue, but...
Since $e^x > 0$ for all $x \in \mathbb{R}$, there can only be two points where RH intersects $e^x$ since $e^x = -x^2+2x+5 \implies x^2-2 x+e^x = 5$.
If we rewrite $x^2-2 x+e^x = 5$ as $x^2-2 x+e^x = y$, observe that its minima is at $x^2-2 x+e^x = W(\frac{e}{2}) (W(\frac{e}{2})+2)-1,\, x = 1-W(\frac{e}{2})$
where $W(n)$ denotes the Lambert W function.
As $x$ tends from $1-W(\frac{e}{2}) \to \infty$ or $1-W(\frac{e}{2}) \to -\infty$, $y\to\infty$, thus it is easy to see that $e^x = -x^2+2x+5$ has only two solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the number of solutions I am trying to compare my answer with a friend's and we are both confident in our answers. But the problem is, they are different. So the problem goes:
Suppose I have the equation
$$x+y+z+w = 14$$
where $x,y,z,w \in \mbox{Z}^+$ (i.e. integers that can be zero or positive) such that $x,y,w,z \le 6$.
Friend's approach:
let $X=x+1, Y=y+1, Z=z+1, W=w+1$, so
$$X+Y+Z+W = 10$$
and without the final restriction, applying stars and bars, we get the number of ways :
$$ \left( \begin{array}{cc}
13 \\
3 \end{array} \right)$$
but we need to exclude cases where $1$ of the $4$ has $6$ or more, thus the final answer is:
$$ \left( \begin{array}{cc}
13 \\
3 \end{array} \right)-
\left( \begin{array}{cc}
4 \\
1 \end{array} \right)\left( \begin{array}{cc}
7 \\
3 \end{array} \right)=146$$
My Way:
I incorporated the inclusion-exclusion idea
So using: $$ \left( \begin{array}{cc}
r+M-1 \\
M-1 \\ \end{array} \right)$$ where M represents different kinds of objects and r is the number of such objects,
I first found all of the possible nonnegative solutions with no restrictions:
$$ \left( \begin{array}{cc}
14+4-1 \\
4-1 \\ \end{array} \right) = \left( \begin{array}{cc}
17 \\
3 \\ \end{array} \right)$$
Then I let $x,y,z,w \ge 7$ to be used for the solutions that are not allowed and since there are four variables and their restrictions are all the same, I multiply the following by $C(4,1) = 4$:
$$ 4\left( \begin{array}{cc}
(14-7)+4-1 \\
4-1 \\ \end{array} \right) = 4\left( \begin{array}{cc}
10 \\
3 \\ \end{array} \right)$$
And for the last part, I kept the restriction but now choose two variables of the four. I'll also be multiplying the following by $C(4,2) =6$ since there are six ways of pairing up the variables so:
$$ 6\left( \begin{array}{cc}
(14-7-7)+4-1 \\
4-1 \\ \end{array} \right)= 6\left( \begin{array}{cc}
3 \\
3 \\ \end{array} \right)$$
So by the inclusion-exclusion idea, the number of solutions:
$$\left( \begin{array}{cc}
17 \\
3 \\ \end{array} \right)- 4\left( \begin{array}{cc}
10 \\
3 \\ \end{array} \right) +
6\left( \begin{array}{cc}
3 \\
3 \\ \end{array} \right) = 206$$
Conclusion:
So which one of us is correct? I've asked this problem before and had a member from this site confirm my answer already so I am set on it. But my friend has two confirmations from math/cs majors. Any help is greatly appreciated.
| Your friend is counting the solutions where $ 1 \le w,x,y,z \le 6$; you are counting the solutions where $ 0 \le w,x,y,z \le 6$.
Since the problem statement says $x, y, z, w \in Z^+$, I think your friend has the proper interpretation.
| {
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"url": "https://math.stackexchange.com/questions/479731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\max\{x,y\}=\frac {(x-y)}{2} + \frac {(x+y)}{2}$ Problem: prove $\max\{x,y\}=\frac {(x-y)}{2} + \frac {(x+y)}{2}$
$x$ and $y$ are max elements in two sets.
Here is what i have thought of so far as a concrete problem:
$\max \{4,5\}=d(4,5)= 1$
I know that the $$\begin{align} \max\{4,5\} &= \frac{4-5}{2}+ \frac{4+5}{2} \\
&= \frac{-1}{2}+\frac{1}{2} \\ &= 0\end{align}$$
I am thinking of breaking it down to trichotomy: $x=y, x>y$, and $x<y$.
for $x=y$ what i got is..
$$ \begin{align}
\max\{x,y\} &= \frac{(x-y)}{2}+\frac{(x+y)}{2} \\
&= 0+x \\
&= x \end{align}
$$ (which is the max) but why isn't this 0?
and am i even thinking in the right direction or am i mixing up two different concepts?
| The problem is wrong as $\frac{x+y}2+\frac{x-y}2$ is always equal to $x$
where as max$(x,y)=x$ iff $x\ge y$
In fact max$\displaystyle(x,y)=\frac{x+y}2+\frac{|x-y|}2$
If $x\ge y,$
max$(x,y)=x$ and $|x-y|=x-y\implies \frac{x+y}2+\frac{|x-y|}2=\frac{x+y+(x-y)}2=x$
If $x<y,$
max$(x,y)=y$ and $|x-y|=-(x-y)\implies \frac{x+y}2+\frac{|x-y|}2=\frac{x+y-(x-y)}2=y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/481038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that roots are real I am stuck with this equation, I need to prove the roots are real when $a, b, c \in R$
The equation is $$(a+b-c)x^2+ 2(a+b)x + (a+b+c) = 0$$
If someone could tell me the right way to go about this, so I can attempt it.
Thank you
EDIT: I have made an error in the question. I have now corrected it.
| $$ax^2 +bx^2 + cx^2 + 2ax+2bx + a+b+c = 0$$
or
$$x^2(a+b-c) + 2x(a+b) +a+b+c = 0$$
Let us now find the expression for the discriminant,
$$\Delta = 4(a+b)^2-4(a+b-c)(a+b+c)$$
If we prove that, $\Delta\geq 0$, we done.
$$4(a+b)^2-4(a+b+c)(a+b-c)=$$
$$=4a^2+8ab+4b^2-4(a^2+2ab+b^2-c^2)=$$
$$=4a^2+8ab+4b^2-4a^2-8ab-4b^2+4c^2=$$
$$4c^2\geq 0$$
Hence, indeed
$$\Delta\geq 0$$
and therefore the quadratic equation has real roots.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Solve system equation $\left\{ \begin{array}{l} xy - x - y = 1\\ 4{x^3} - 12{x^2} + 9x = - {y^3} + 6y + 7 \end{array} \right.$ Solve system equation : $$\left\{ \begin{array}{l} xy - x - y = 1\\ 4{x^3} - 12{x^2} + 9x = - {y^3} + 6y + 7 \end{array} \right. ,\quad (x,y\in\mathbb{R}).$$
My solution begin with : Set $z=x-1$ we have : $$\left\{ \begin{array}{l} yz=z+2\\ 4z^3-3z+y^3-6y-6=0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} yz=z+2\\ 4z^3-3y^2z+y^3=0 \end{array} \right.$$
I want to have a difference solution.
| Using the straightforward approach, I find $x=\frac {y+1}{y-1} $ from the first equation and substitute it in the second one $$\frac {{y}^{6}-3\,{y}^{5}-3\,{y}^{4}+11\,{y}^{3}-6\,{y}^{2}+32}{
\left( y-1 \right) ^{3}}=0.
$$ The keypoint is the next step $$y^6-3y^5-3y^4+11y^3-6y^2+32 =(y^2-y+2)(y^2-y-4)^2.$$ The rest is quite simple: $$ \left \{ y=\frac 1 2 \pm \frac {\sqrt{7}i } 2, x= \frac {\frac 1 2 \pm \frac {\sqrt{7}i } 2 +1} {\frac 1 2 \pm \frac {\sqrt{7}i } 2 -1}\right \} $$ or $$\left\{ x=5/4-1/4\,\sqrt {17},y=1/2-1/2\,\sqrt {17} \right\} $$ or $$ \left\{ x=5/4+1/4\,\sqrt {17},y=1/2+1/2\,\sqrt {17} \right\} .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/483835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factoring $a^3-b^3$ I recently got interested in mathematics after having slacked through it in highschool. Therefore I picked up the book "Algebra" by I.M. Gelfand and A.Shen
At problem 113, the reader is asked to factor $a^3-b^3.$
The given solution is:
$$a^3-b^3 = a^3 - a^2b + a^2b -ab^2 + ab^2 -b^3 = a^2(a-b) + ab(a-b) + b^2(a-b) = (a-b)(a^2+ab+b^2)$$
I was wondering how the second equality is derived. From what is it derived, from $a^2-b^2$? I know that the result is the difference of cubes formula, however searching for it on the internet i only get exercises where the formula is already given. Can someone please point me in the right direction?
| The second equality is obtained by first grouping terms in the middle expression and then factoring the grouped terms:
$$\begin{align}
a^3-a^2b+a^2b-ab^2+ab^2-b^3&=(a^3-a^2b)+(a^2b-ab^2)+(ab^2-b^3)\cr
&=a^2(a-b)+ab(a-b)+b^2(a-b)\cr
\end{align}$$
If the OP is wondering where the middle expression came from in the first place, it's kind of a deus ex machina: All you're doing is sticking two $0$'s between $a^3$ and $-b^3$ (the expressions $-a^2b+a^2b$ and $-ab^2+ab^2$ are both obviously equal to $0$), but when you do, lo and behold, the regrouping and factoring work their magic.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
How to find the inverse of 70 (mod 27) The question pertains to decrypting a Hill Cipher, but I am stuck on the part where I find the inverse of $70 \pmod{ 27}$. Does the problem lie in $70$ being larger than $27$?
I've tried Gauss's Method:
$\frac{1}{70} = \frac{1}{16} ^{\times2}_{\times2} = \frac{2}{32} = \frac{2}{5} = \frac{12}{30} = \frac{12}{3} = \frac{132}{33} = \frac{24}{6} = \frac{120}{30} $
And the denominators start repeating so I can never get 1 in the denominator.
And the Euclidean Algorithm
$\ 70 = 2(27) + 16 $
$\ 27 = 1(16) + 11 $
$\ 16 = 1(11) + 5 $
$\ 11 = 1(5) + 6 $
$\ 5 = 1(6) -1 $
Which is also not helpful. I think I'm trying to get + 1 on the last equation for $1 \pmod{ 27}$, but maybe I'm misunderstanding the method.
Am I approaching this incorrectly? I'm new to modular arithmetic.
| We have $70\equiv16\pmod{27}\equiv2^4$
As $(2,27)=1,$
so using Carmichael function $\lambda(27)=\phi(27)=3^{3-1}(3-1)=18$ where $\phi$ is the Totient function
$\implies 2^{18}\equiv1\pmod{27}$
$\implies (2^a)^{-1}\equiv2^{18-a} \pmod{27}$
$\implies (2^4)^{-1}\equiv2^{14} \pmod{27}$
Now, $2^5=32\equiv5\pmod{27}\implies 2^{14}=(2^5)^2\cdot2^4\equiv5^2\cdot16\pmod{27}\equiv(-2)(-11)\equiv22$
We can apply this method to find inverse of any integer $a\pmod m$ where $m$ is an integer provided $(a,m)=1$
In fact, we can find smaller exponent $d(>0)$ than $\lambda(m)$ such that $a^d\equiv1\pmod m$
But,as $2$ is a primitive root of $3^2=9$
using this $,2$ is a primitive root of $3^n$ for integer $n\ge1$
$\implies $ord$_{27}2=\phi(27)=18$
| {
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"timestamp": "2023-03-29T00:00:00",
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Does the derivative of $x^{-1}$ and of $x^3-x$ equal $-\frac{1}{x^{2}}$ and $3x^2-1$? I want to check my answers concerning the derivative of the following functions: $\displaystyle f(x)= \frac{1}{x}$ and of $\displaystyle j(x)= x^3-x$
$$\displaystyle f(x)= \frac{1}{x}$$
$$\begin{align}
f'(x) & = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\
& = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} \\
& = \lim_{h \to 0} \frac{\frac{x}{x(x+h)} - \frac{(x+h)}{x(x+h)}}{h} \\
& = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x^2+hx}}{h} \\
& = \lim_{h \to 0} \frac{\frac{-h}{x^2+hx}}{h} \\
& = \lim_{h \to 0} \frac{-h}{x^2+hx} \times \frac{1}{h} \\
& = \lim_{h \to 0} \frac{-1}{x^2+hx} \\
& = -\frac{1}{x^2}
\end{align}$$
$$j(x) = x^3 - x$$
$$\begin{align}
j'(x) & = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\
& = \lim_{h \to 0} \frac{(x+h)^3 - x^3 - x}{h} \\
& = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x - h - x^3 + x}{h} \\
& = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3 - h}{h} \\
& = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2 - 1)}{h} \\
& = \frac{3x^2 - 1}{1} \\
& = 3x^2 - 1
\end{align}$$
Please, be rude, if you see any error correct me.
| Your title is wrong but your limits in the actual post are correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this is an equilateral triangle in $\Delta ABC$,$AB=c,AC=b,BC=a$and such
$$ab^2\cos{A}=bc^2\cos{B}=ca^2\cos{C}$$
show that
$\Delta ABC$ is an equilateral triangle
this problem I have solution,But not nice, and I think this problem have more nice methods,Thank you everyone.
my solution:
$$ab^2\cdot\dfrac{b^2+c^2-a^2}{2bc}=bc^2\cdot\dfrac{a^2+c^2-b^2}{2ac}=ca^2\cdot\dfrac{a^2+b^2-c^2}{2ab}$$
$$\Longrightarrow a=b=c$$
My other idea:
$$\Longleftrightarrow\dfrac{\sin{B}}{\sin{C}}\cos{A}=\dfrac{\sin{C}}{\sin{A}}\cos{B}=\dfrac{\sin{A}}{\sin{B}}\cos{C}$$
$$\Longleftrightarrow \sin{(2A)}\sin{B}=\sin{(2B)}\sin{C}=\sin{(2C)}\sin{A}$$
then How prove
$$A=B=C$$
and have other nice methods? Thank you
| Hint:divide all the three by abc and then proceed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491292",
"timestamp": "2023-03-29T00:00:00",
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Find min $P$: $P=\frac{1}{(a+b)(b+c)}+\frac{1}{(c+a)(a+b)}+(c+1)(3+a+b)$ Let $a,b,c\geq 0$ and $a+b+c=1$. Know that never have two numbers both zero. Find min $P$:
$$P=\frac{1}{(a+b)(b+c)}+\frac{1}{(c+a)(a+b)}+(c+1)(3+a+b)$$
| The question is tagged integral-inequality. Therefore, I assumed that an integer solution is sought for.
At least one of two integer variables must be unequal to zero
(= greater equal one):
$$a + b \geq 1$$
$$a + c \geq 1$$
$$b + c \geq 1$$
With $a+b+c = 1$ we get:
$$c = 1 - (a+b) \leq 0$$
Because $c \geq 0$, it follows that $c$ must be zero.
Therefore:
$$a \geq 1$$
$$b \geq 1$$
This contradicts $a+b+c=1$.
Thus, there is no integer solution fulfilling the constraint.
A non-integer solution would be $a = 0.5$ and $b = 0.5$.
This leads to $P = 8.0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate limits $ \lim_{x\to+\infty} \frac{3x-1}{x^2+1}$ and $\lim_{x\to-\infty} \frac{3x^3-4}{2x^2+1}$ I want to calculate the following limits
$$\begin{matrix}
\lim_{x\to+\infty} \frac{3x-1}{x^2+1} & \text{(1)} \\
\lim_{x\to-\infty} \frac{3x^3-4}{2x^2+1} & \text{(2)}
\end{matrix}$$
In both cases we have indeterminate forms. Using L'Hôpital's rule on $\text{(1)}$ gives
$$\lim_{x\to+\infty} \frac{3x-1}{x^2+1} = \lim_{x\to+\infty}\frac{3}{2x} = 0$$
Using L'Hôpital's rule on $\text{(2)}$ gives
$$\lim_{x\to-\infty} \frac{3x^3-4}{2x^2+1} = \lim_{x\to-\infty}\frac{9x^2}{4x} = \lim_{x\to-\infty}\frac{18x}{4} = -\infty$$
Is this correct?
| I think we can change the limit $\to0$ to even avoid L'Hosiptal's Rule as follows:
Putting $\frac1x=h$
$$\lim_{x\to\infty}\frac{3x-1}{x^2+1}=\lim_{h\to0}\frac{\frac3h-1}{\frac1{h^2}+1}=\lim_{h\to0}\frac{(3-h)h}{1+h^2}=0$$
$$\lim_{x\to\infty}\frac{3x^3-4}{2x^2+1}=\lim_{h\to0}\frac{\frac3{h^3}-4}{\frac2{h^2}+1}=\lim_{h\to0}\frac{3-4h^3}{h(2+h^2)}=\frac30$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $(a,b)$ if $x^2-bx+a = 0, x^2-ax+b = 0$ both have distinct positive integers roots
If $x^2-bx+a = 0$ and $x^2-ax+b = 0$ both have distinct positive integers roots, then what is $(a,b)$?
My Try:
$$\displaystyle x^2-ax+b = 0\Rightarrow x = \frac{a\pm \sqrt{a^2-4b}}{2}$$
So here $a^2-4b$ is a perfect square.
Similarly
$$\displaystyle x^2-bx+a = 0\Rightarrow x = \frac{b\pm \sqrt{b^2-4a}}{2}$$
So here $b^2-4a$ is a perfect square.
But I did not understand how can I solve after that.
| Suppose $x^2-ax+b$ and $x^2-bx+a$ have positive integer roots. Then
$$x^2-ax+b=(x-c)(x-d)\qquad\text{and}\qquad x^2-bx+a=(x-e)(x-f),$$
for some positive integers $c$, $d$, $e$ and $f$. As the roots are must be distinct, we may assume without loss of generality that $c>d$ and $e>f$. Now compare coefficients.
Hint 1:
It follows that $c+d=a=ef$ and $cd=b=e+f$.
Hint 2:
If $c>d>1$ then $cd>c+d$.
Hint 3:
If $d>1$ and $f>1$ then $cd>c+d=ef>e+f=cd$, a contradiction.
Hint 4:
Without loss of generality we have $f=1$, so $e=c+d$ and $e+1=cd$.
Hint 5:
It follows that $(c-1)(d-1)=2$, so $c=3$ and $d=2$, and hence $e=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can an integer of the form $4n+3$ written as a sum of two squares? Let $u$ be an integer of the form $4n+3$, where $n$ is a positive integer. Can we find integers $a$ and $b$ such that $u = a^2 + b^2$? If not, how to establish this for a fact?
| Lemma 1: $a$ is odd $\Longrightarrow$ $a^2\equiv 1(\operatorname{mod} 4)$.
Proof: $a^2-1=(a-1)(a+1)$. Since $a$ is odd, both $a-1$ and $a+1$ are even, so that $a^2-1$ is divisible by $4$. $\blacksquare$
Lemma 2: $a$ is even $\Longrightarrow$ $a^2\equiv 0(\operatorname{mod} 4)$.
Proof: Trivial. $\blacksquare$
Now, suppose that $u=a^2+b^2$.
(1) If both $a$ and $b$ are even, then $u$ is divisible by four by lemma 2.
(2) If both $a$ and $b$ are odd, then $u\equiv 2(\operatorname{mod}4)$ by lemma 1.
(3) If $a$ is even and $b$ is odd (wlog), then $u\equiv1(\operatorname{mod}4)$ by lemmas 1 and 2.
That is, it is never the case that $u\equiv 3(\operatorname{mod} 4)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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a - b > 0 algebra correction Simple Algebra
$\frac{1}{2}-\frac{1}{5} > 0$
$\frac{1}{2} > \frac{1}{5}$
$5 >2$
Looks correct, but where am I wrong in this,
$\frac{1}{2}-\frac{1}{5} > 0$
$-\frac{1}{5} > -\frac{1}{2}$
$\frac{1}{5} > \frac{1}{2}$
$ 2 > 5$
| $-\frac{1}{5} > -\frac{1}{2}$ implies that 1/2>1/5 not the other way around
| {
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Explicit Formula for a Recurrence Relation for {2, 5, 9, 14, ...} (Chartrand Ex 6.46[b])
Consider the sequence $a_1 = 2, a_2 = 5, a_3 = 9, a_4 = 14,$ etc...
(a) The recurrence relation is: $a_1 = 2$ and $a_n = a_{n - 1} + (n + 1) \; \forall \;n \in [\mathbb{Z \geq 2}]$.
(b) Conjecture an explicit formula for $a_n$. (Proof for conjecture pretermitted here)
I wrote out some $a_n$ to compass to cotton on to an idea or pattern. It seems bootless.
$\begin{array}{cc}
a_2 = 5 & a_3 = 9 & a_4 = 14 & a_5 = 20 & a_6 = 27\\
\hline \\
5 = 2 + (2 + 1) & 9 = 5 + (3 + 1) & 14 = 9 + (4 + 1) & 20 = 14 + (5 + 1) & 27 = 20 + (7 + 1) \\
\end{array}$
The snippy answer only says $a_n = (n^2 + 3n)/2$. Thus, could someone please explicate the (missing) motivation or steps towards this conjecture? How and why would one envision this?
| Let $2\leq n$, then you can write
$$a_{2}-a_{1}=2+1$$
$$a_{3}-a_{2}=3+1$$
$$...$$
$$a_{n}-a_{n-1}=n+1.$$
Now addition of all this equality implies that
$$a_{n}-a_{1}=(2+3+..+n)+(n-1),$$
$$a_{n}=(2+3+..+n)+(n+1)-1,$$
we know that $1+2+..+n=\frac{n^{2}+n}{2}$, so
$$a_{n}=\frac{(n+1)(n+2)}{2}-1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit $ \sqrt{2\sqrt{2\sqrt{2 \cdots}}}$
Find the limit of the sequence $$\left\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots\right\}$$
Another way to write this sequence is $$\left\{2^{\frac{1}{2}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}}2^{\frac{1}{8}},\hspace{5 pt} \dots,\hspace{5 pt}2^{\frac{1}{2^{n+1} - 2}}\right\}$$
So basically we have to find $\lim_{n\to\infty}2^{\frac{1}{2^{n+1} - 2}}$. This equates to $$\lim_{n\to\infty}2^{\frac{1}{2^{\infty+1} - 2}} \Longrightarrow 2^{\frac{1}{2^{\infty}}} \Longrightarrow 1$$
Is this correct? P.S. Finding that $S(n)$ was a pain!
| An alternative proof (of sorts):
If the sequence has a limit, then it will be a fixed point of $f(x)=\sqrt{2x}$, and hence a solution to $x=\sqrt{2x}$. The solutions are clearly $x=0, 2$, and you should be able to show that 0 is out because all of the terms in the sequence are $\gt 1$ and hence the limit is $\geq 1$.
| {
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Find a closed form of $\sum_{i=1}^n i^3$ I'm trying to compute the general formula for $\sum_{i=1}^ni^3$. My math instructor said that we should do this by starting with a grid of $n^2$ squares like so:
$$
\begin{matrix}
1^2 & 2^2 & 3^2 & ... & (n-2)^2 & (n-1)^2 & n^2 \\
2^2 & 3^2 & 4^2 & ... & (n-1)^2 & n^2 & 1^2 \\
3^2 & 4^2 & 5^2 & ... & n^2 & 1^2 & 2^2 \\
\vdots &\vdots & \vdots&\ddots & \vdots & \vdots & \vdots \\
(n-2)^2 & (n-1)^2 & n^2 & ... & (n-5)^2 & (n-4)^2 & (n-3)^2 \\
(n-1)^2 & n^2 & 1^2 & ... & (n-4)^2 & (n-3)^2 & (n-2)^2 \\
n^2 & 1^2 & 2^2 & ... & (n-3)^2 & (n-2)^2 & (n-1)^2 \\
\end{matrix}
$$
And then sum the rows, and then sum all of the sums of rows. How would I then compute the general formula for $\sum_{i=1}^ni^3$?
| Try this link http://mathschallenge.net/library/number/sum_of_cubes It gives full explanation of how to compute it.
| {
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Solve $X^2\equiv-1\pmod{13}$ I've been asked to solve the following equation:
$$X^2 \equiv-1\pmod{13}.$$
I am not sure what to do.
| You can try all possible solutions: $0,1,…,12$, though only half of them need to be tested.
For a more sophisticated solution, note that $13=4+9=2^2+3^2$ and so $3^2 \equiv -2^2 \bmod 13$. Since $2\cdot 7 \equiv 1 \bmod 13$, we conclude that $ -1 \equiv -2^2 \cdot 7^2 \equiv 3^2 \cdot 7^2 \equiv (3\cdot 7)^2 \equiv 8^2 \bmod 13$. The other solution is $13-8=5$.
This is a special case of a fact that plays a key role in Fermat's theorem on primes that are sums of two squares:
A prime $p$ is a sum of two squares iff there is a solution for $x^2\equiv -1 \bmod p$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq 27$ How can I prove $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq 27$, given that $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9$ and $x+y+z=1$.
I've already tried using that: $\frac{1}{x} +\frac{1}{y} +\frac{1}{z}\geq 9$ But I can't seem to manipulate that to prove the above.
| You may also try this :
Apply AM-HM inequality on the set $\{x^2,y^2,z^2\}$ :
$$\frac {x^2+y^2+z^2}{3} \geq \frac {3}{\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}}$$
$$\implies \big(x^2+y^2+z^2\big)\big({\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}}\big) \geq 9$$
$$\implies {\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}} \geq \frac{9}{x^2+y^2+z^2}\,\,\,(♦)$$
Now all that remains is to prove that $$x^2+y^2+z^2 \leq \frac{1}{3} \,\,\,(♣)$$
with the constraint that $$x+y+z=1 $$
I am assuming that you are skilled enough to prove $(♣)$. (You can use Lagrange Multipliers or some other technique.)
Plug $(♣)$ into $(♦)$ to get :
$$ {\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}}\geq \frac{9}{1/3} = 27$$
$$HENCE \,\,\,\, PROVED$$
NOTE : Since $x^2+y^2+z^2$ is in the denominator (♦) , the sign of (♣) flips/reverts , thus proving the desired inequality.
| {
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Three Boolean Algebra Proofs - I just don't get it! I'm having a very difficult time proving the following 3 expressions:
$$\begin{align*}
&x\cdot y\cdot z+x'\cdot z=y\cdot z+x'\cdot z\\
&x\cdot y+y\cdot z+x'\cdot z=x\cdot y+x'\cdot z\\
&(x'\cdot z'+x'\cdot y+x'\cdot z+x\cdot y)'=x\cdot y'
\end{align*}$$
I need to show what law/theorem/postulate is used for each step of the proof - and I don't even know where to start. The last time I did any sort of algebra was at least 7 years ago, and even then it was very basic.
Being thrown into Boolean algebra, only provided a sheet with all the theorems/etc... is proving to be impossible for me. I understand the methodology behind mathematical proofs and boolean simplification, I just don't see what theorems/laws can be used when I look at these....
Any help would be greatly appreciated here!
| Essentially, we manipulate the equation in whatever way possible, trying to make the left-hand side look like the right-hand side. There's no real magic method here, we just keep trying until we succeed.
In the second example, we can do:
$\small
\begin{align*}
x\cdot y+y\cdot z+x'\cdot z &= x\cdot y + 1 \cdot (y\cdot z)+x'\cdot z & \text{identity for } \cdot \\
&= x\cdot y + (x+x') \cdot (y\cdot z)+x'\cdot z & \text{complementation } x+x'=1\\
&=x\cdot y + x\cdot (y\cdot z)+x'\cdot (y\cdot z)+x'\cdot z & \text{distributivity} \\
&=x\cdot y + (x\cdot y)\cdot z+x'\cdot (y\cdot z)+x'\cdot z & \text{associativity} \\
&=(x\cdot y) \cdot 1 + (x\cdot y)\cdot z+x'\cdot (y\cdot z)+x'\cdot z & \text{identity for } \cdot \\
&=(x\cdot y) \cdot (1 + z)+x'\cdot (y\cdot z)+x'\cdot z & \text{associativity} \\
&=(x\cdot y) \cdot (1 + z)+x'\cdot (y\cdot z)+x'\cdot z & \text{distributivity} \\
&=(x\cdot y) \cdot 1+x'\cdot (y\cdot z)+x'\cdot z & \text{annihilator for } + \\
&=x\cdot y+x'\cdot (y\cdot z)+x'\cdot z & \text{identity for } \cdot \\
&=x\cdot y+(x'\cdot z) \cdot y+x'\cdot z & \text{assoc. and comm. of } \cdot \\
&=x\cdot y+(x'\cdot z) \cdot y+(x'\cdot z) \cdot 1 & \text{identity for } \cdot \\
&=x\cdot y+(x'\cdot z) \cdot (y+1) & \text{distributivity} \\
&=x\cdot y+x'\cdot z & \text{annihilator for } +, \\
\end{align*}
$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/509310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How prove this $\frac{1}{2a^2-6a+9}+\frac{1}{2b^2-6b+9}+\frac{1}{2c^2-6c+9}\le\frac{3} {5}\cdots (1)$ let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3}
{5}\cdots (1)$$
I find sometimes,and I find this same problem:
let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$ \frac{1}{5a^2-4a+11}+\frac{1}{5b^2-4b+11}+\frac{1}{5c^2-4c+11}\leq\frac{1}{4} $$
and this problem have some methods,you can see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=223910&start=20
and I like this can_hang2007 methods and Honey_S methods,But for $(1)$ I can't prove it.Thank you
| Let $f(x) = \dfrac{1}{2x^2-6x+9}$. First let us first quickly sketch some properties of $f$ that we'll need. $f(x)=\dfrac{1}{2(x-\frac32)^2+\frac92}$, so its maximum is $f(\frac32)=\frac29$. $f~'(x)=\dfrac{6-4x}{(2x^2-6x+9)^2}$, and $f~''(x)=12\dfrac{2x^2-6x+3}{(2x^2-6x+9)^3}$. Setting $f~''(x)=0$ gives the inflection points $x=\frac32\pm\frac{\sqrt3}2$. $f$ is increasing before $\frac32$, decreasing after $\frac32$, and is concave between the inflection points. Note $f(\frac32\pm\frac{\sqrt3}2)=\frac16$.
If $a$, $b$, and $c$ are all between the inflection points, then by Jensen's inequality
$$f(a)+f(b)+f(c) \le 3f(\dfrac{a+b+c}3) = 3f(1) = \frac35$$
as required.
If at least two of $a$, $b$, or $c$ are outside the inflection points, then
$$f(a)+f(b)+f(c) \le \frac16+\frac16+\frac29=\frac59\lt\frac35$$
If one of $a$, $b$, or $c$ is $\gt\frac32+\frac{\sqrt3}2$, then at least one of $a$, $b$, or $c$ is $\lt\frac32-\frac{\sqrt3}2$, otherwise
$$3=a+b+c\ge 2(\frac32-\frac{\sqrt3}2)+(\frac32+\frac{\sqrt3}2)=\frac92-\frac{\sqrt3}2 \approx 3.7,$$
a contradiction. And so by the previous paragraph, the required result holds.
So the only remaining case is when exactly one of $a$, $b$, or $c$ is $\lt\frac32-\frac{\sqrt3}2$, and the other two are between the inflection points. We need two estimates. First consider the case when the minimum is $\le\frac12$. (Here $\frac12$ is arbitrary, although we need to choose something very close to $\frac12$ for the following argument to work.) Then
$$f(a)+f(b)+f(c) \le f(\frac12) + \frac29 + \frac29 = \frac{2}{13} + \frac49 = \frac{70}{117} \lt \frac35$$
Finally, suppose $a$ is between $\frac12$ and $\frac32-\frac{\sqrt3}2$, and $b$ and $c$ are between the inflection points. By Jensen's inequality, we need only consider the case when $b=c$, by replacing them with their mean. Then $b=(3-a)/2$. So $a\gt\frac12$ implies $b\lt\frac54$, and
$$f(a)+f(b)+f(b) \lt f(\frac32-\frac{\sqrt3}2) + 2f(\frac54) = \frac16 + \frac{16}{37} = \frac{133}{222} \lt \frac35$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$ Question is to check if $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$
we have $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\prod \limits_{n=2}^{\infty}(\frac{n^2-1}{n^2})=\prod \limits_{n=2}^{\infty}\frac{n+1}{n}\frac{n-1}{n}=(\frac{3}{2}.\frac{1}{2})(\frac{4}{3}.\frac{2}{3})(\frac{5}{4}.\frac{3}{4})...$
In above product we have for each term $\frac{a}{b}$ a term $\frac{b}{a}$ except for $\frac{1}{2}$.. So, all other terms gets cancelled and we left with $\frac{1}{2}$.
So, $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\frac{1}{2}$.
I would be thankful if some one can assure that this explanation is correct/wrong??
I am solving this kind of problems for the first time so, it would be helpful if some one can tell if there are any other ways to do this..
Thank you
| Obviously, the product of any positive numbers that less than 1 is still less than one.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
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Probability that the sum of two elements in a set is divisible by 3. Select 2 integers from the set $\{1,2,3,4,5,6,7,8\}$. What is the probability that their sum is divisible by 3? I am assuming this is without replacement although this is not explicitly stated.
My answer: There are $\binom{8}{2}$ ways of selecting two elements. Only 6 of these pairs $(1,2), (4,5), (4,2), (7,8),(3,6), (2,7) $ are divisible by 3. Thus $P=6/\binom{8}{2}$. Is this correct?
| Numbers can be classified as $3n$, $3n+1$, $3n+2$. Now the sum will be divisible by $3$ only if the pair is $(3n,3n)$ or $(3n+1,3n+2)$.
Digits of type: $3n = (3,6)$; type $3n+1= (1,4,7)$; type $3n+2 = (2,5,8)$. So the number of possible would be $1 + 3\cdot 3 = 10$. $\text{Total probability} = 10/28$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/514062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How is $a_n=(1+1/n)^n$ monotonically increasing and bounded by $3$? I was reading about how completeness is required for limits. And I came across this:
the sequence $a_n=(1+1/n)^n$ is monotonically increasing and bounded by 3 and so we expect it to converge, but that it does not converge within $\mathbb{Q}$. More generally it stands to reason that any sequence of real numbers which is increasing and bounded must converge to some real number. This is a consequence of completeness of $IR$
My question is: How is the mentioned sequence monotonically increasing and bounded by $3$ ?
| From Rudin's PMA Theorem $3.31$, by the Binomial theorem,
$$ t_n=\left( 1 + \frac{1}{n} \right)^n $$
$$= 1 + 1 + \underbrace{\frac{1}{2!}\left(1-\frac{1}{n}\right)}_{\text{term } 2} + \underbrace{\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}_{\text{term } 3} + \ldots + \frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{n-1}{n}\right). $$
Since each bracket increases as $n$ increases, each term increases as $n$ increases also, because products and sums of increasing positive functions is also increasing.
Therefore, for all $n\geq 2,$ and all $2\leq k \leq n,$ the $k$-th term of $t_{n+1}$ is greater than the $k$-th term of $t_n.$
So we see that $\left( 1 + \frac{1}{n} \right)^n$ is increasing.
Furthermore, the Binomial expansion above is
$$ \leq 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{n!} = 1 + 1 + \frac{1}{1\cdot 2} + \frac{1}{1\cdot 2\cdot 3} + \frac{1}{1\cdot 2\cdots n} $$
$$ < 1+1+ \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^{n-1}} <3. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots? What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots?
$(x + a)(x + 1991) + 1 = x^2 + (1991 + a)x + (1991a + 1)$
This is of the form $ax^2 + bx + c$. Applying the quadratic formula $\left(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\right)$, we get, that the rooots must equal:
$\frac{-(1991 + a) \pm \sqrt{(1991 + a)^2 - 4(1991a + 1)}}{2} = \frac{-(1991 + a) \pm \sqrt{1991^2 + a^2 + 2\times1991a - 4\times1991a -4}}{2} = \frac{-(1991 + a) \pm \sqrt{(1991 - a)^2 - 2^2}}{2} = \frac{-(1991 + a) \pm \sqrt{(1991 - a - 2)(1991-a+2)}}{2}$
For the formula to yield an integer, the discriminant must be a perfect square or $0$. Clearly, the discriminant will be zero if $a = 1993$ or $a = 1989$. The only thing that remains to be shown is that these are indeed the only possible solutions.
I couldn't think of any way to do so. Are there any alternate solutions, perhaps some that involve more number theory and less algebra?
| (x + a)(x + 1991) + 1 = 0
(x + a)(x + 1991) = -1
if x is integer, then, there are two possible pairs.
I) x+a= 1 and x+1991 = -1
II) x+a = -1, x+1991 = 1.
if we eleminate x in both system euqations, we'll get a = 1989 and a= 1993.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find the $3$ angles of triangle $ABC$ We have a non obtuse triangle $ABC$.
With $$\bf\dfrac{1}{2}\cos(2A)+\sqrt{2}\cos(B)+\sqrt{2}\cos(C)=\dfrac{3}{2}$$
Find the $3$ angles $A,B,C$.
| Given a non obtuse triangle $ABC$,
find the corresponding angles $\alpha,\beta,\gamma$
such that
\begin{align}
\tfrac12\cos2\alpha+\sqrt2\cos\beta+\sqrt2\cos\gamma
&=\tfrac32
\tag{1}\label{1}
\\
\text{or }\quad
\cos^2\alpha+\sqrt2\cos\beta+\sqrt2\cos\gamma
-2&=0
.
\tag{1a}\label{1a}
\end{align}
Using a known identity for triangles
\begin{align}
\cos\alpha+\cos\beta+\cos\gamma
&=1+\frac{r}R
\tag{2}\label{2}
,
\end{align}
where $r$ and $R$ are the radii
of inscribed and circumscribed circles, respectively,
we have
\begin{align}
\sqrt2\cos\alpha+\sqrt2\cos\beta+\sqrt2\cos\gamma
&=\sqrt2\,\left(1+\frac{r}R \right)
\tag{3}\label{3}
.
\end{align}
Let's introduce a parameter
$v=\frac{r}R$. It is also known that
for valid triangles $0<v\le \tfrac12$.
Equations \eqref{1a},\eqref{2} provide
a quadratic equation in $\cos\alpha$
in terms of $v$:
\begin{align}
\cos^2\alpha
-\sqrt2\cos\alpha+\sqrt2(1+v)-2
&=0
\tag{4}\label{4}
.
\end{align}
The two solutions of \eqref{4} are
\begin{align}
s_+&=\tfrac{\sqrt2}2
+\tfrac12\sqrt{10-4\sqrt2\,(v+1)}
,\\
s_-&=\tfrac{\sqrt2}2
-\tfrac12\sqrt{10-4\sqrt2\,(v+1)}
.
\end{align}
Since $s_+>1$ $\forall v\in[0,\tfrac12]$,
the only valid solution is
\begin{align}
\cos\alpha&=\tfrac{\sqrt2}2
-\tfrac12\sqrt{10-4\sqrt2\,(v+1)}
\tag{5}\label{5}
.
\end{align}
The equation \eqref{5} shortens the valid range of $v$
to $[\sqrt2-1.\tfrac12]$,
since corresponding $\cos\alpha<0$
$\forall v\in(0,\sqrt2-1)$.
With the help of
\begin{align}
\cos\gamma&=
\cos(\pi-\alpha-\beta)
=-\cos(\alpha+\beta)
=-\cos\alpha\cos\beta+\sin\alpha\sin\beta
,
\end{align}
\eqref{2} can be transformed to
\begin{align}
\sin\alpha\sin\beta
&=1+v-\cos\alpha-\cos\beta
+\cos\alpha\cos\beta
,\\
\sin^2\alpha\sin^2\beta
&=(1+v-\cos\alpha-\cos\beta+\cos\alpha\cos\beta)^2
,\\
(1-\cos^2\alpha)(1-\cos^2\beta)
&=
(\cos\alpha-1)^2\cos^2\beta
\\
&+2(\cos\alpha-1)(1+v-\cos\alpha)\cos\beta
\\
&+(1+v-\cos\alpha)^2
,\\
\end{align}
and we have a quadratic equation for $\cos\beta$
\begin{align}
\cos^2\beta
-(1+v-\cos\alpha)\cos\beta
-\cos\alpha
+v-\frac{v^2}{2(\cos\alpha-1)}
\tag{6}\label{6}
,
\end{align}
The discriminant of \eqref{6} is
\begin{align}
\Delta&=\tfrac14\,\frac{(\cos\alpha+1)(\cos\alpha^2-2\cos\alpha \,v-1+2v+v^2)}{(\cos\alpha-1)^3}
,
\end{align}
condition $\Delta\ge0$
can be simplified to
\begin{align}
\frac{(\cos\alpha^2-2\cos\alpha \,v-1+2v+v^2)}{(\cos\alpha-1)^3}
&\ge0
,\\
(\cos\alpha^2-2\cos\alpha \,v-1+2v+v^2)
&\le0
,
\end{align}
which we can simplify by substitution
$\cos^2\alpha=2+\sqrt2\cos\alpha-\sqrt2(1+v)$ as
\begin{align}
(\sqrt2-2v)\cos\alpha+1+v^2+(2-\sqrt2)v-\sqrt2
&\le0
,\\
(\sqrt2-2v)
\tfrac{\sqrt2}2
-\tfrac12\sqrt{10-4\sqrt2\,(v+1)}
+1+v^2+(2-\sqrt2)v-\sqrt2
&\le0
\tag{7}\label{7}
.
\end{align}
Considering $v\in[0,\tfrac12]$,
it can be shown that \eqref{7} holds for $v\in[0,\sqrt2-1]$
(rigorous proof is left to anyone interested),
that is,
\begin{align}
D(v)&=0,\quad \text{for }\quad v=\sqrt2-1
,\\
D(v)&<0,\quad \text{for }\quad v>\sqrt2-1
.
\end{align}
Thus, the only valid solution of \eqref{1}
is restricted to $v=\sqrt2-1$,
\begin{align}
\alpha&=\tfrac\pi2,\quad
\beta=\gamma=\tfrac\pi4
.
\end{align}
A 3D illustration - the surfaces
given in implicit form
\begin{align}
\tfrac12\cos2x+\sqrt2\cos y+\sqrt2\cos z
-\tfrac32&=0
,\\
x+y+z-\pi&=0
,\\
x,y,z&\in[0,\tfrac\pi2]
\end{align}
shows only one intersection point, shown in yellow:
| {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Simplifying an inverse trig function? I am trying to figure out how to simplify this expression but I am not quite sure how these inverses work. What sort of approach should I take for this equation?
$$\tan\left(2\cos^{-1}\left(\dfrac{x}{5}\right)\right)$$
| As @Spencer notes, there exists a triangle with hypotenuse 5 and adjacent $x$. Pythagoras tells us that we must therefore have an opposite of $\sqrt{5^2 - x^2} = \sqrt{25-x^2}$
If
$$\cos(\theta) = x/5$$
then
$$\tan(\theta) = \frac{\sqrt{25-x^2}}{x}$$
Given
$$\tan(2\theta) = \frac{2\tan(\theta)}{1 - tan²(\theta)}$$
We now have
$$
\begin{array}{ll}
\tan\left(2\cos^{-1}\left(\frac{x}{5}\right)\right) &= \frac{2\frac{\sqrt{25-x^2}}{x}}{1 - \left(\frac{\sqrt{25-x^2}}{x}\right)^2} \\
&=2\frac{\frac{\sqrt{25-x^2}}{x}}{1 - \frac{25-x^2}{x^2}} \\
&=2x\frac{\sqrt{25-x^2}}{x^2 - (25-x^2)} \\
&=2x\frac{\sqrt{25-x^2}}{2x^2 - 25}
\end{array}
$$
Is that simpler?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the greatest common divisor (gcd) of $f(x) = x^2 + 1$ and $g(x) = x^6 + x^3 + x + 1$ Find the greatest common divisor (gcd) of $f(x) = x^2 + 1$ and $g(x) = x^6 + x^3 + x + 1$.
Since $x^6 + x^3 + x + 1 = (x^2 + 1)(x^4 - x^2 + x + 1)$, $\mathrm{gcd}[f(x),g(x)] = x^2 + 1$.
My question is how could I JUSTIFY that the answer is ACTUAL gcd of $f(x)$ and $g(x)$. thanks
| Clearly $x^{2}+1$ divides both $f$ and $g$, as you have shown. On the other hand, since $g=x^{2}+1$, any polynomial $h$ dividing both $f$ and $g$ must divide $g=x^{2}+1$. Thus, by definition, $x^{2}+1$ is the greatest common divisor.
This example was easy since $g$ divides $f$. In other cases, you could use the Euclidean Algorithm to find the gcd of $f$ and $g$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/521930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Number of irrational roots of the equation $(x-1)(x-2)(3x-2)(3x+1)=21$? The number of irrational roots of the equation $(x-1)(x-2)(3x-2)(3x+1)=21$ is
(A)0
(B)2
(C)3
(d)4
Actually im a 10 class student i don't know any of it,but my elder brother(IIT Coaching) cannot solve them,he told me post these questions on this site someone might know the answers and for now he is not in the town. So can you please help me.
Thank you.
| Hint :
$$ (3x-2)(x-1) = 3x^2-5x+2 $$
and
$$ (3x+1)(x-2) = 3x^2 -5x-2 $$
$$ (x-1)(x-2)(3x-2)(3x+1) = (3x^2-5x-2)(3x^2-5x+2) = 21 $$
Now put
$$ 3x^2-5x = t $$
$$ (t-2)(t+2) = 21 $$
Now, solve
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 2
} |
showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$ I'm having problem with showing that:
$$\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$$
I would need some help in the right direction
| From the Article $240,$ Ex$-5$ of Plane Trigonometry(by Loney),
$$\arctan x+\arctan y=\begin{cases} \arctan\frac{x+y}{1-xy} &\mbox{if } xy<1\\ \pi+\arctan\frac{x+y}{1-xy} & \mbox{if } xy>1\end{cases} $$
$$\implies 2\arctan x=\arctan \frac{2x}{1-x^2}\text{ if }x^2<1$$
Here $\displaystyle x=\frac23$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
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How to find the orthogonal trajectories of the family of all the circles through the points $(1,1)$ and $(-1,-1)$? I'm trying to find the orthogonal trajectories of the family of circles through the points $(1,1)$ and $(-1, -1)$. Now such a family can be given by an equation of the form $$ x^2 + y^2 + 2g(x-y) - 2 = 0, $$ where $g$ is a parameter.
Now upon differentiation with respect to x, we obtain $$ 2x + 2y y^\prime + 2g (1 - y^\prime ) = 0, $$ where $y^\prime$ denotes the derivative of $y$ with respect to $x$. Upon dividing out by $2$, we arrive at $$ x + y y^\prime + g(1 - y^\prime ) = 0, $$ from which we get $$ g = \frac{x + y y^\prime}{y^\prime - 1}. $$ Putting this value of $g$ into the equation of the family of circles, we get $$ x^2 + y^2 +2 \frac{x + y y^\prime}{y^\prime - 1} ( x - y ) - 2 = 0, $$ so $$ (x^2 + y^2 -2 ) (y^\prime - 1 ) + 2 (x + y y^\prime ) (x - y) = 0$$ or $$ (x^2 + y^2 - 2 + 2xy - 2y^2 ) y^\prime + (2x^2 - 2xy - x^2 - y^2 + 2) = 0 $$ or $$ ( x^2 + 2xy - y^2 - 2) y^\prime + (x^2 - 2xy - y^2 + 2) = 0, $$ from which we get $$ y^\prime = - \frac{ x^2 - 2xy - y^2 + 2}{ x^2 + 2xy - y^2 - 2}. $$ Now for the orthogonal trajectories, we get $$ y^\prime = \frac{x^2 + 2xy - y^2 - 2}{x^2 - 2xy - y^2 + 2}. $$ How to solve this differential equation?
| Let $u = y+x$, $v = y-x$, we have
$$\begin{align}
u' = y'+1 = & \frac{2x^2-2y^2}{x^2 - 2xy - y^2+2} = \frac{-2uv}{x^2-2xy-y^2+2}\\
v' = y'-1 = & \frac{4xy-4}{x^2 -2xy- y^2+22} = \frac{u^2 - v^2-4}{x^2-2xy-y^2+2}\\
\end{align}$$
From this, we get
$$u^2 \left( \frac{v^2}{u} + u +\frac{4}{u} \right)' = u^2\left(\frac{v^2}{u}\right)' + (u^2 - 4)u' = 2uvv' + (u^2 - v^2 - 4)u' = 0$$
and hence for some integration constant $K$, one has
$$\frac{v^2}{u} + u +\frac{4}{u} = 2K
\;\;\iff\;\; v^2 + u^2 + 4 = 2K u
\;\;\iff\;\; x^2 + y^2 + 2 = K(x+y)
$$
i.e the orthogonal trajectories is another family of circles.
Update
There is a pure geometric argument why the orthogonal trajectories are circles.
Let $p = (-1,-1), q = (1,1)$. Let $\mathscr{C}$ be a circle centered at $p$ with radius
$|pq| = 2\sqrt{2}$. Consider the inversion $\mu$ with respect to circle $\mathscr{C}$, we have:
$$\mu(p) = \infty\quad\text{ and }\quad \mu(q) = q.$$
This means under $\mu$, the family of circles passing through $p$ and $q$ get mapped to the family of straight lines passing through $q$. The orthogonal trajectories of this family of straight lines is the family of circles centered at $q$.
It is known that inversion is conformal. i.e. preserve angles and hence map orthogonal trajectories to orthogonal trajectories. This means the orthogonal trajectories of the family of circles passing through $p$ and $q$ are the inverse image of $\mu$ of the family of circles centered at $q$.
It is also known that inversion $\mu$ map circles not passing through $q$ to circles.
As a result, the family of orthogonal trajectories we seek are circles themselves.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Ellipse: product of the distance from foci to a tangent is a constant I am supposed to determine what is the result of said product. Given $P(x_0,y_0)$, I need to calculate the distance from the foci of an ellipse to the tangent line that passes through $P$, and then multiply the distances.
In essence it is quite simply. We take:
$$
\frac{x_0}{a^2}x + \frac{y_0}{b^2}y = 1
$$
as the tangent line. Then we simply calculate its distance to each focus $(c,0)$ and $(-c,0)$, using the formula and then, multiplying.
$$
d=\frac{\frac{x_0c}{a^2}±1}{\sqrt{\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}}
$$
$$
\text{Some constant k}=\frac{\frac{x_0^2c^2}{a^4}-1}{\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}
$$
I'm having trouble getting things cancelled here. The constant k is $b^2$, but I can't get to it. Help?
| Here I go:
$$
\frac {\frac{x_0^2c^2}{a^4}-1} {\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}
$$
Simplify
$$
\frac {\frac{x_0^2c^2-a^4}{a^4}} {\frac{x_0^2b^4 + y_0^2a^4}{a^4b^4}}
$$
Then
$$
\frac{(x_0^2c^2-a^4)(a^4b^4)}{a^4(x_0^2b^4 + y_0^2a^4)} \\
$$
Quick cancellation
$$
\frac{(x_0^2c^2-a^4)(b^4)}{x_0^2b^4 + y_0^2a^4}
$$
We have from the original ellipse equation that $y_0^2a^4 = a^4b^2-a^2b^2x_0^2$, so:
$$
\frac{(x_0^2c^2-a^4)(b^4)}{x_0^2b^4 + a^4b^2-a^2b^2x_0^2}
$$
We factor out $b^4$ above, and $b^2$ below
$$
b^2\frac{x_0^2c^2-a^4}{x_0^2b^2 + a^4-a^2x_0^2}
$$
Factorize the $x_0$ below
$$
b^2\frac{x_0^2c^2-a^4}{x_0^2(b^2 - a^2) + a^4}
$$
Using the Pythagorean identity
$$
b^2\frac{x_0^2c^2-a^4}{x_0^2(-c^2) + a^4}
$$
Move around
$$
b^2\frac{x_0^2c^2-a^4}{-(x_0^2c^2 - a^4)}
$$
Nice cancellation
$$
k = -b^2
$$
The $-$ sign shouldn't matter, since the distance formula uses absolute values, and in the end, it would be $±b^2$, right?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Factorise $y^2 -3yz -10z^2$ How do I solve this question? I have looked at the problem several times. However, I cannot find a viable solution. I believe that it is a perfect square trinomial problem.
| $$y^2 - 3 y z - 10 z^2 = \begin{bmatrix} y\\ z\end{bmatrix}^T \begin{bmatrix} 1 & -\frac 32 + t\\ -\frac 32 - t & -10\end{bmatrix} \begin{bmatrix} y\\ z\end{bmatrix}$$
We want to find a $t$ that makes the matrix above rank-$1$. Computing the determinant,
$$\det \begin{bmatrix} 1 & -\frac 32 + t\\ -\frac 32 - t & -10\end{bmatrix} = t^2 - \left(\frac 72\right)^2$$
If $t = \frac 72$, then the matrix is singular and, thus, rank-$1$
$$\begin{bmatrix} 1 & 2\\ -5 & -10\end{bmatrix} = \begin{bmatrix} 1\\ -5\end{bmatrix} \begin{bmatrix} 1\\ 2\end{bmatrix}^T$$
Thus,
$$\begin{array}{rl} y^2 - 3 y z - 10 z^2 &= \begin{bmatrix} y\\ z\end{bmatrix}^T \begin{bmatrix} 1 & 2\\ -5 & -10\end{bmatrix} \begin{bmatrix} y\\ z\end{bmatrix}\\ &= \begin{bmatrix} y\\ z\end{bmatrix}^T \begin{bmatrix} 1\\ -5\end{bmatrix} \begin{bmatrix} 1\\ 2\end{bmatrix}^T \begin{bmatrix} y\\ z\end{bmatrix}\\ &= (y - 5 z) (y + 2 z)\end{array}$$
factoring polynomials quadratic-forms matrices rank-1-matrices
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How to find the center of mass of half a torus? Consider a halved solid torus (half a donut). The radius of the torus are $R_1$ and $R_2$. I need to find its center of mass. The hint they give is that the center of mass of a homogeneous solid object $\Omega \subset \Bbb R^3$ is calculated as
$$\overline{x}=\int_{\Omega}\overline{r}d\overline{r}.$$
I really don't understand this formula, I don't know what $\overline{r}$ means and what is $\Omega$ in this case. I'd appreciate that someone explains what this formula means and how to apply it in this problem. Thanks in advance.
| I found it easiest to use cylindrical coordinates to set up the integrals needed for the center of mass. Before we do so, however, I set my coordinate system up as follows. I have positive $x$ coming out of the screen, positive $y$ going to the right, and positive $z$ up. In cylindrical coordinates $(r,\phi,z)$:
$$x = r \cos{\phi}$$
$$y=r \sin{\phi}$$
where we have the limits defining the region $\Omega$:
$$\phi \in \left [ -\frac{\pi}{2},\frac{\pi}{2} \right]$$
$$z \in [-R_2,R_2]$$
$$r \in \left [ R_1 - \sqrt{R_2^2-z^2}, R_1 + \sqrt{R_2^2-z^2}\right]$$
Also, for an object of constant mass density, the expression for the $x$ component of the center of mass is
$$\bar{x} = \frac{\displaystyle \int_{\Omega} d^3 \vec{x} \, x}{\displaystyle \int_{\Omega} d^3 \vec{x}}$$
(Note that, by symmetry, we have $\bar{y}=0$ and $\bar{z}=0$.) Let's first evaluate the denominator:
$$\begin{align} \int_{\Omega} d^3 \vec{x} &= \int_{-\pi/2}^{\pi/2} d\phi \, \int_{-R_2}^{R_2} dz \, \int_{R_1 - \sqrt{R_2^2-z^2}}^{R_1 + \sqrt{R_2^2-z^2}} dr \, r \\ &= \frac{\pi}{2}\int_{-R_2}^{R_2} dz \left [\left (R_1 + \sqrt{R_2^2-z^2} \right )^2 - \left (R_1 - \sqrt{R_2^2-z^2} \right )^2 \right ]\\ &= 4 \pi R_1 \int_0^{R_2} dz \, \sqrt{R_2^2-z^2}\\ &= \pi^2 R_1 R_2^2 \end{align}$$
Now we evaluate the center of mass:
$$\begin{align}\bar{x} &= \frac{1}{\pi^2 R_1 R_2^2} \int_{-\pi/2}^{\pi/2} d\phi \, \int_{-R_2}^{R_2} dz \, \int_{R_1 - \sqrt{R_2^2-z^2}}^{R_1 + \sqrt{R_2^2-z^2}} dr \, r^2 \cos{\phi} \\ &= \frac{4}{3 \pi^2 R_1 R_2^2} \int_0^{R_2} dz \, \left [\left (R_1 + \sqrt{R_2^2-z^2} \right )^3 - \left (R_1 - \sqrt{R_2^2-z^2} \right )^3 \right ]\\ &= \frac{8}{3 \pi^2 R_1 R_2^2} \int_0^{R_2} dz \,\left [3 R_1^2 \sqrt{R_2^2-z^2} + \left (R_2^2-z^2 \right )^{3/2} \right ] \\ &= \frac{8}{3 \pi^2 R_1 R_2} \left ( \frac{3 \pi}{4} R_1^2 R_2 + \frac{3 \pi}{16} R_2^3 \right )\end{align}$$
Simplifying, I get
$$\bar{x} = \frac{4 R_1^2+R_2^2}{2 \pi R_1}$$
ADDENDUM
As a quick note, in the limits as $R_2 \to 0$, we find that the center of mass becomes
$$\bar{x}=\frac{2}{\pi} R_1$$
which agrees with the center of mass of a uniform wire bent into a semicircle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/531608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Integral $ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ Question
$ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ = ?
we tried to bound it from both sides using $x$ and $(x-1)$, which yield nice estimation ($\frac {24}\pi$) - ($\frac {36}\pi$) but not a precise one.
we also tried to split it to 6 integrals using $\lfloor x\rfloor $ as a different constant each time. $(0\cdot (\cos(...)-\cos(...))+1(\cos(\frac \pi3)-\cos(\frac\pi6))+2\cdot...$ etc
it produced the right result ($\frac {30}\pi$), but the number is not important as achieving the general function.
| Here is another solution using Riemann-Stieltjes integral. Integrating by parts,
\begin{align*}
\int_{0}^{6} \lfloor x \rfloor \sin \left(\frac{6x}{\pi}\right) \, dx
&= \int_{0^{+}}^{6} \lfloor x \rfloor \sin \left(\frac{6x}{\pi}\right) \, dx \\
&= \left[ - \lfloor x \rfloor \frac{\pi}{6} \cos \left(\frac{6x}{\pi}\right) \right]_{0^{+}}^{6} + \int_{0^{+}}^{6} \frac{\pi}{6} \cos \left(\frac{6x}{\pi}\right) \, d\lfloor x \rfloor \\
&= - \pi \cos \left(\frac{36}{\pi}\right) + \sum_{k=1}^{6} \frac{\pi}{6} \cos \left(\frac{6k}{\pi}\right).
\end{align*}
More generally, let $f$ be continuous on $[a, b]$ and $F$ be an anti-derivative of $f$. Then
\begin{align*}
\int_{a}^{b} f(x)\lfloor x \rfloor \, dx
&= \int_{a^{+}}^{b} f(x)\lfloor x \rfloor \, dx \\
&= \left[ F(x)\lfloor x \rfloor \right]_{a^{+}}^{b} - \int_{a^{+}}^{b} F(x) \, d\lfloor x \rfloor \\
&= \lfloor b \rfloor F(b) - \lfloor a \rfloor F(a) - \sum_{a < k \leq b} F(k).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/531962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Logically speaking, why can variables be substituted? Suppose that
$$a^2+a+1=b$$
Suppose also that $a=5/4$. What makes it valid to substitute $5/4$ into the first equation? Is it because equality is transitive?
| The reason why you can make such a substitution is that $a$ and $\dfrac{5}{4}$ are precisely the same, just written differently. Here's a long-winded way of making the substitution you describe.
\begin{align*}
a &= \frac{5}{4} & \\
a^2 &= \left(\frac{5}{4}\right)^2 &\text{squaring both sides}\\
a^2 + a &= \left(\frac{5}{4}\right)^2 + \frac{5}{4} &\text{adding the first two lines}\\
a^2 + a + 1 &= \left(\frac{5}{4}\right)^2 + \frac{5}{4}+1 &\text{adding $1$ to both sides}\\
b &= \left(\frac{5}{4}\right)^2 + \frac{5}{4}+1 &\text{using the fact that $a$ satisfies the given equation}
\end{align*}
The right hand side is precisely what you obtain when you replace $a$ by $\dfrac{5}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/532482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$ Prove that $1<\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.
By using the Mathematical induction. Suppose the statement holds for $n=k$.
Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}=(\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1})+(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$
we know $\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}>1$
What can we do for $(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$?
| You are almost done. Prove that
$$\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}-\frac{1}{k+1}$$ is positive.
To do this it is enough to show that $\frac{1}{3k+2}+\frac{1}{3k+4} \gt \frac{2}{3k+3}$. The left side can be written as $\frac{6k+6}{(3k+2)(3k+4}$. So we want to show that $(3k+3)^2\gt (3k+2)(3k+4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/534117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Matrix multiplication - Express a column as a linear combination Let $A = \begin{bmatrix}
3 & -2 & 7\\
6 & 5 & 4\\
0 & 4 & 9
\end{bmatrix} $ and $B = \begin{bmatrix}
6 & -2 & 4\\
0 & 1 & 3\\
7 & 7 & 5
\end{bmatrix} $
Express the third column matrix of $AB$ as a linear combination of the column matrices of $A$
I don't get this... surely the 3rd column would be an expression of the row matrices of $A$ since the 3rd column of$AB$ would be $
\begin{bmatrix}
3(4) & -2(3) & 7(5)\\
6(4) & 5(3) & 4(5)\\
0(4) & 4(3) & 9(5)
\end{bmatrix} $
As I typed out the question I see my answer...
the 3rd column is $4\begin{bmatrix}
3\\
6\\
0
\end{bmatrix} 3 \begin{bmatrix}
-2\\
5\\
4
\end{bmatrix} 5 \begin{bmatrix}
7\\
4\\
9
\end{bmatrix}$
Is this correct?
| Not quite: we need to add entries. So the third column of matrix $AB$ is given by:
\begin{bmatrix}
3(4) -2(3) + 7(5)\\
6(4) + 5(3) + 4(5)\\
0(4) + 4(3) + 9(5)
\end{bmatrix}
So the third column represented as a linear combination of columns of $A$ is given by:
$$4 \begin{bmatrix}
3\\
6\\
0
\end{bmatrix} + 3 \begin{bmatrix}
-2\\
5\\
4
\end{bmatrix} +5 \begin{bmatrix}
7\\
4\\
9
\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/534873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$? I am trying to show that
$$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$$
This question stems from the underlying homework problem, which asks to show
$$ \frac{\pi}{\sin(\pi z)} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}, $$
to which I am at my wits end. I have a couple of identities on hand, namely
$$ \pi \cot (\pi z) = \frac{1}{z} + \sum_{n \in \mathbb{Z}; n \neq 0} \frac{1}{z - n} + \frac{1}{n} $$
and
$$ \frac{\sin (\pi z)}{\pi} = z \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2} \right) $$
and
$$ \frac{\pi^2}{\sin^2 (\pi z)} = \sum_{n \in \mathbb{Z}} \frac{1}{(z - n)^2} $$
I've tried fooling around with these identities and am getting nowhere. Any hints or suggestions would be greatly appreciated.
| In this answer, it is derived that
$$
\begin{align}
\pi\cot(\pi z)
&=\sum_{k=-\infty}^\infty\frac{1}{z+k}\\
&=\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}\tag{1}
\end{align}
$$
To get the alternating series, note that
$$
\begin{align}
\frac1z+2z\sum_{k=1}^\infty\frac{(-1)^k}{z^2-k^2}
&=\sum_{k=-\infty}^\infty\frac{(-1)^k}{z+k}\\
&=\sum_{k=-\infty}^\infty\frac{2}{z+2k}-\frac1{x+k}\\
&=\sum_{k=-\infty}^\infty\frac{1}{z/2+k}-\frac1{x+k}\\[6pt]
&=\pi\cot(\pi z/2)-\pi\cot(\pi z)\\[7pt]
&=\pi\frac{1+\cos(\pi z)}{\sin(\pi z)}-\pi\frac{\cos(\pi z)}{\sin(\pi z)}\\
&=\frac\pi{\sin(z)}\tag{2}
\end{align}
$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
} |
If $a, b, c$ are integers with $a^2 + b^2 = c^2$, then $a$ and $b$ cannot both be odd If $a, b, c$ are integers with $a^2 + b^2 = c^2$, it's true that $a$ and $b$ cannot both be odd.
But how can we prove it
| We can prove this by contradiction.
$\text{Let }\;\; a = 2m+1, b = 2n + 1 $
So, the LHS is:
$$
\begin{align}
a^2 + b^2 &= 4(m^2+n^2) + 4(m+n) + 2 \\\\
&= 2\left( 2(m^2+n^2) + 2(m+n)+1 \right) \\\\
&= 2\times\text{an odd number}
\end{align}
$$
This can not be a perfect square, as it doesn't have an even power of $2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How prove this$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i+\sum_{j=0}^{n-1}\binom{m-1+j}{j}x^my^j=1$ let $m,n$ be positive numbers,and $x,y>0$ such $x+y=1$,show that
$$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i+\sum_{j=0}^{n-1}\binom{m-1+j}{j}x^jy^m=1$$
My try:
$$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i=\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^n(1-x)^i=\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^n\sum_{k=0}^{i}(-1)^k\binom{i}{k}x^k$$
| Suppose we seek to show that
$$\sum_{q=0}^{m-1} {n-1+q\choose q} x^n (1-x)^q
+ \sum_{q=0}^{n-1} {m-1+q\choose q} x^q (1-x)^m = 1$$
where $n,m\ge 1.$
We will evaluate the second term by a contour integral and show that
is equal to one minus the first term which is the desired result.
Introduce the Iverson bracket
$$[[0\le q\le n-1]]
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z^q}{z^n}
\frac{1}{1-z} \; dz.$$
With this bracket we may extend the sum in $q$ to infinity to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n}
\frac{1}{1-z}
\sum_{q\ge 0} {m-1+q\choose q} z^q x^q (1-x)^m\; dz
\\ = \frac{(1-x)^m}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n}
\frac{1}{1-z}
\sum_{q\ge 0} {m-1+q\choose q} z^q x^q \; dz
\\ = \frac{(1-x)^m}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n}
\frac{1}{1-z}
\frac{1}{(1-xz)^m} \; dz.$$
Now we have three poles here at $z=0$ and $z=1$ and $z=1/x$ and the
residues at these poles sum to zero, so we can evaluate the residue
at zero by computing the negative of the residues at $z=1$ and
$z=1/x.$
Observe that the residue at infinity is zero as can be seen from the
following computation:
$$-\mathrm{Res}_{z=0} \frac{1}{z^2}
z^n \frac{1}{1-1/z}\frac{1}{(1-x/z)^m}
\\ -\mathrm{Res}_{z=0} \frac{1}{z^2}
z^n \frac{z}{z-1}\frac{z^m}{(z-x)^m}
\\ -\mathrm{Res}_{z=0}
z^{n+m-1} \frac{1}{z-1}\frac{1}{(z-x)^m} = 0.$$
Returning to the main thread the residue at $z=1$ as seen from
$$- \frac{(1-x)^m}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n}
\frac{1}{z-1}
\frac{1}{(1-xz)^m} \; dz.$$
is $$-(1-x)^m \frac{1}{(1-x)^m} = -1.$$
For the residue at $z=1/x$ we consider
$$\frac{(1-x)^m}{x^m \times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n}
\frac{1}{1-z}
\frac{1}{(1/x-z)^m} \; dz
\\ = \frac{(-1)^m (1-x)^m}{x^m \times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n}
\frac{1}{1-z}
\frac{1}{(z-1/x)^m} \; dz.$$
and use the following derivative:
$$\frac{1}{(m-1)!}
\left(\frac{1}{z^n} \frac{1}{1-z}\right)^{(m-1)}
\\ = \frac{1}{(m-1)!}
\sum_{q=0}^{m-1} {m-1\choose q}
\frac{(-1)^q (n+q-1)!}{(n-1)! z^{n+q}}
\frac{(m-1-q)!}{(1-z)^{m-q}}
\\ =
\sum_{q=0}^{m-1} \frac{1}{q!}
\frac{(-1)^q (n+q-1)!}{(n-1)! z^{n+q}}
\frac{1}{(1-z)^{m-q}}
\\ =
\sum_{q=0}^{m-1} {n+q-1\choose q}
\frac{(-1)^q}{z^{n+q}}
\frac{1}{(1-z)^{m-q}}.$$
Evaluate this at $z=1/x$ and multiply by the factor in front to get
$$\frac{(-1)^m (1-x)^m}{x^m} \times
\sum_{q=0}^{m-1} {n+q-1\choose q}
(-1)^q x^{n+q}
\frac{1}{(1-1/x)^{m-q}}
\\ = \frac{(-1)^m (1-x)^m}{x^m} \times
\sum_{q=0}^{m-1} {n+q-1\choose q}
(-1)^q x^{n+q}
\frac{x^{m-q}}{(x-1)^{m-q}}
\\ = (-1)^m (1-x)^m \times
\sum_{q=0}^{m-1} {n+q-1\choose q}
(-1)^q x^{n} (-1)^{m-q}
\frac{1}{(1-x)^{m-q}}
\\ =
\sum_{q=0}^{m-1} {n+q-1\choose q}
x^{n} (1-x)^q.$$
This yields for the second sum term the value
$$1 - \sum_{q=0}^{m-1} {n+q-1\choose q}
x^{n} (1-x)^q$$
showing that when we add the first and the second sum by cancellation
the end result is one, as claimed.
This identity generalizes an identity by Gosper to be found at this MSE link.
Remark Fri Jun 9 2017. As written this proof requires $|1/x| \gt \epsilon$ or $1/\epsilon \gt |x|$ for the pole at $1/x$ to be outside the circular contour and for the geometric series to converge. Note however that these are polynomials in $x$ of degree $n+m-1$ and hence this is sufficient to show they agree for all $x.$
| {
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Proving an inequality using the Cauchy-Schwarz inequality Consider four real numbers $a_1, a_2, a_3, a_4$ such that $\sum a_i^3 = 10$. Prove that
$$\sum a_i^4 \geq \sqrt[3]{2500}$$
Applying the Cauchy Schwarz inequality with $a_i^2$ and $a_i$, we get
$$\left(\sum a_i^3\right)^2 \leq \left(\sum a_i^4\right)\left(\sum a_i^2\right)$$
Again, applying the Cauchy Schwarz inequality with $a_i^2$ and $1$, we get:
$$\left(\sum a_i^2\right)^2 \leq 4\left(\sum a_i^4\right)$$
Substituting this into the first inequality, we get:
$$\left(\sum a_i^3\right)^2 \leq 4\left(\sum a_i^4\right)^2$$
Taking the square root of both sides,
$$\left(\sum a_i^3\right) \leq 2\left(\sum a_i^4\right)$$
$$\implies 5 \leq \sum a_i^4$$
But, $\sqrt[3]{2500} = 13.5720881$, which is more than $2$ times $5$. How do I prove the required statement?
| Using the following relations from Cauchy-Schwarz (which has the advantage of being applicable for all reals):
$$\left(\sum a_i^4\right)\left(\sum 1\right)\ge \left(\sum a_i^2\right)^2 \tag{1}$$
$$\left(\sum a_i^4\right)\left(\sum a_i^2\right)\ge \left(\sum a_i^3\right)^2 = 100\tag{2}$$
Noting both LHS and RHS are positive, square both sides of (2) and multiply with (1), then cancel off the positive $(\sum a_i^2)^2 $ from both sides to get:
$$4 \left(\sum a_i^4\right)^3 \ge 100^2 \implies \sum a_i^4 \ge \sqrt[3]{2500}$$
| {
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If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
| Using the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ and denoting $(a+b)$ by $A$, we have (from the assumption)
\begin{align*}
&a^3+b^3+3ab=1\\
&(a+b)(a^2-ab+b^2)+3ab=1\\
&A(A^2-3ab)+3ab=1\\
&(A^3-1)-3abA+3ab=0\\
&(A-1)(A^2+A+1)-3ab(A-1)=0\\
&(A-1)(A^2+A+1-3ab)=0,\\
\end{align*}
where we use the identities $a^2+b^2=(a+b)^2-2ab=A^2-2ab$ and $A^3-1=(A-1)(A^2-A+1)$.
Therefore, $A=1$ or $A^2+A+1-3ab=0$.
In the latter case, replacing $A$ by $a+b$, we can conclude that the only solution is $a=b=-1$ (getting $a^2+(1-b)a+(b^2+b+1)=0$ and using the determinant for real solutions of quadratic equations), which is left to you.
| {
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How can this equation be solved? I have no idea how to solve this equation.
$$x^2y^2+324y^2+64x^2-36xy^2-16x^2y+144xy = 0 $$
Sorry $x,y \in \mathbb{+Z}$
| Consider the equation as a quadratic in $y$, we obtain
$$ y^2 (x^2-36x+324)+y (144x-16x^2) + 64x^2 = 0 $$
First, check that if the coefficient of $y^2$ is 0, then we must have $ x = 18$, which gives $y=8$ (1 solution).
Otherwise, using the quadratic equation, we get that (for $x\neq 18$),
$$y = \frac { 8(x^2-9x \pm 3\sqrt{x^2( 2x - 27)})} {(x-18)^2} $$
Hence, we must have $2x-27 = (2a+1)^2$ for some non-negative $a$. This gives $x = 2a^2 + 2a + 14$. Substituting it into $y$ above, we get
$$ \begin{array} {l l }y & = \frac{ 8 (2a^2 + 2a + 14) ( 2a^2 + 2a + 14 - 9 \pm 3 (2a+1) ) } { (2a^2+2a-4)^2} \\ & = \frac{ 4 (a^2 + a + 7) ( 2a^2 + 2a + 14 - 9 \pm (6a+3) )}{[(a-1)(a+2)]^2} \end{array}$$
Let's first deal with the $+$ case. The numerator actually will cancel out with $(a+2)^2$ in the denominator. Then, using partial fractions, we get
$$ y = 8 + \frac{24}{a-1} + \frac{72}{(a-1)^2}$$
Since $72 = 2^3 \times 3^2$, this means that $(a-1) = -1, 1, 2, 3, 6$.
The $-$ case is similar, and left to you.
| {
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Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it. Can anyone help?
By the way, I've been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$.
Proof : Let
$$f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.$$
We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. Hence, letting $f(0)=1/3$, we know that $f(x)$ is continuous and positive at $x=0$. Hence, since $f(x)\ (0\le x\le {\pi}/2)$ is bounded, there exists a constant $C$ such that $0\lt f(x)\lt C$. Hence, substituting $x={(k\pi)}/{(2n+1)}$ for this, we get
$$0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.$$
Then, the sum of these from $1$ to $n$ satisfies
$$0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.$$
Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.
| Consider the polynomial $S_m(x)$, satisfying $S_m(\sin^2 \theta)=\sin^2(m\theta)$.
These are known as spread polynomials, and may easily be derived from the Chebyshev polynomials $T_m(x)$, via $$1-2S_m(\sin^2(\theta)=1-2\sin^2(m\theta)=\cos(m(2\theta))=T_m(\cos(2\theta))=T_m(1-2\sin^2 \theta)$$ so $1-2S_m(x)=T_m(1-2x)$.
Note that
\begin{align}
&S_{m+1}(\sin^2 \theta)+S_{m-1}(\sin^2 \theta) \\
& =\sin^2(m\theta+\theta)+\sin^2(m\theta-\theta) \\
&=(\sin(m\theta)\cos \theta+\cos(m\theta)\sin \theta)^2+(\sin(m\theta)\cos \theta-\cos(m\theta)\sin \theta)^2 \\
&=2\sin^2(m \theta)\cos^2 \theta+2\cos^2(m \theta) \sin^2(m\theta) \\
&=2(1-\sin^2 \theta)S_m(\sin^2 \theta)+2\sin^2 \theta(1-S_m(\sin^2 \theta))
\end{align}
Thus $S_{m+1}(x)=2(1-2x)S_m(x)-S_{m-1}(x)+2x$.
(We could also have used the more well known recurrence $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$ and derived the recurrence for $S_m$ from there.)
Observe that $\sin^2(\frac{k\pi}{m}), k=0, 1, \ldots, m-1$ are roots of the polynomial equation $S_m(x)=0$. Put $S_m(x)=xP_m(x)$, so that $\sin^2(\frac{k\pi}{m}), k=1, 2, \ldots, m-1$ are roots of the polynomial equation $P_m(x)=0$. The recurrence for $S_m$ gives $$P_{m+1}(x)=2(1-2x)P_m(x)-P_{m-1}(x)+2$$
Now if we write $P_m(x)=a_m+b_mx+x^2Q_m(x)$, it is clear by Vieta's formulas that $$\sum_{k=1}^{m-1}{\frac{1}{\sin^2(\frac{k\pi}{m})}}=\frac{\sum_{k=1}^{m-1}{\prod_{j \not =k}{\sin^2(\frac{j\pi}{m})}}}{\prod_{i=1}^{m-1}{\sin^2(\frac{k\pi}{m})}}=-\frac{b_m}{a_m}$$
We prove by induction on $m$ that $a_m=m^2, b_m=-\frac{(m^2-1)m^2}{3}$.
When $m=1$, we have $S_1(x)=x$ so $P_1(x)=1=(1^2)-\frac{(1^2-1)1^2}{3}x$ so the statement is true for $m=1$.
When $m=2$, we have $S_2(x)=4x(1-x)$ so $P_2(x)=4-4x=2^2-\frac{(2^2-1)2^2}{3}x$ so the statement is true for $m=2$.
Suppose that the statement holds for $m=i-1, i$, where $i \geq 2$. Then
\begin{align}
P_{i+1}(x)&=2(1-2x)P_i(x)-P_{i-1}(x)+2 \\
&=2(1-2x)(a_i+b_ix+x^2Q_i(x))-(a_{i-1}+b_{i-1}x+x^2Q_{i-1}(x))+2 \\
&=(2a_i-a_{i-1}+2)+(2b_i-4a_i-b_{i-1})x+x^2(-4b_i+2Q_i(x)-Q_{i-1}(x))
\end{align}
Thus (after some algebra manipulation)
$$a_{i+1}=2a_i-a_{i-1}+2=(i+1)^2$$
and
\begin{align}
b_{i+1}=2b_i-4a_i-b_{i-1}&=-2\frac{(i^2-1)i^2}{3}-4i^2+\frac{((i-1)^2-1)(i-1)^2}{3} \\
&=-\frac{((i+1)^2-1)(i+1)^2}{3}
\end{align}
We are thus done by induction.
Now,
$$\sum_{k=1}^{m-1}{\frac{1}{\sin^2(\frac{k\pi}{m})}}=-\frac{b_m}{a_m}=-\frac{-\frac{(m^2-1)m^2}{3}}{m^2}=\frac{m^2-1}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$a^2-b^2=bc$ and $b^2-c^2=ac \Rightarrow a^2-c^2=ab$ Some weeks ago our math teacher asked the following question and gave us a week to solve it:
If $a^2-b^2=bc$ and $b^2-c^2=ac ,$ Prove $a^2-c^2=ab$, Where $a,b,c$ are non-zero real numbers.
This seemed really easy at the first, but when i tried to prove it i just failed every time. After a week, I only came up with this idea: Assume our case is true. $a^2-c^2=ab\Rightarrow a^2-b^2+b^2-c^2=ab\Rightarrow bc+ac=ab\Rightarrow\frac{1}{abc}(bc+ac)=\frac{1}{abc}(ab) \Rightarrow$ $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$. Now if we Prove this, our case will be proved as well.
In my opinion this seemed like a really nice question so i wanted to share it with everyone.
| To begin with, note that by adding the two given equations together, we can immediately conclude that $$a^2-c^2=(a+b)c.\tag{$\star$}$$
Now, multiply both sides of $(\star)$ by $a-b,$ giving us $$(a^2-c^2)(a-b)=(a^2-b^2)c\\a^3-a^2b-ac^2+bc^2=bc^2\\a^3-a^2b-ac^2=0\\a(a^2-ab-c^2)=0,$$ so since $a\ne 0,$ we can conclude that $a^2-ab-c^2=0,$ so that $a^2-c^2=ab,$ as desired.
| {
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Solutions to equation involving trigonometry and geometric series $$\cos^2 x + \cos ^3 x +\dots = 1+ \cos x$$
I want to find values of $x$ between $0$ and $ 180$ degrees for which the above equation holds true.
Attempt at a solution: left side is a converging geometric progression, for which $a_1$ is $\cos^2 x$ and $q = \cos x$. Plugging into the known formula for such series yields $\cos^2 x = 1$, which yields $45 $ and $135$ degrees as solutions.
Is this okay?
| As mentioned, the LHS is a geometric series, with first term $ \cos^2 x $ and common ratio $ \cos x $, so it evaluates to $ \dfrac {\cos x}{1 - \cos x} $ and our equation becomes $$ \dfrac {\cos^2 x}{1-\cos x} = 1 + \cos x \implies \cos^2 x = 1 - \cos^2 x \implies \cos^2 x = \dfrac {1}{2}. $$Finally, take the square root of both sides and we have $ \cos x = \pm \dfrac {\sqrt{2}}{2} $ and our answers are $ \boxed {\dfrac{\pi}{4}, \dfrac{3\pi}{4}} $.
Seems I've been beaten by a couple of minutes by ncmathsadist and a few seconds by Oliver!
| {
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Probability task (Find probability that the chosen ball is white.) I have this task in my book:
First box contains $10$ balls, from which $8$ are white. Second box contains $20$ from which $4$ are white. From each box one ball is chosen. Then from previously chosen two balls, one is chosen. Find probability that the chosen ball is white.
The answer is $0.5$. Again I get the different answer:
There are four possible outcomes when two balls are chosen:
$w$ -white, $a$ - for another color
$(a,a),(w,a),(a,w),(w,w)$.
Each outcome has probability:
$\frac{2}{10} \cdot \frac{16}{20}; \frac{8}{10} \cdot \frac{16}{20}; \frac{2}{10} \cdot \frac{4}{20}; \frac{8}{10} \cdot \frac{4}{20};$
In my opinion the probability that the one ball chosen at the end is white is equal to the sum of last three probabilities $\frac{8}{10} \cdot \frac{16}{20} + \frac{2}{10} \cdot \frac{4}{20} + \frac{8}{10} \cdot \frac{4}{20}=\frac{21}{25}$. Am I wrong or there is a mistake in the answer in the book?
| Denote that event that eventually a white ball is drawn by $W$.
Denote that event that a white ball is drawn from the $i$-th box
by $W_{i}$ ($i=1,2$).
Denote that event that not a white ball is drawn from the $i$-th
box by $A_{i}$ ($i=1,2$).
Then $P\left(W\right)=P\left(W|W_{1}\cap W_{2}\right)P\left(W_{1}\cap W_{2}\right)+P\left(W|W_{1}\cap A_{2}\right)P\left(W_{1}\cap A_{2}\right)+P\left(W|A_{1}\cap W_{2}\right)P\left(A_{1}\cap W_{2}\right)+P\left(W|A_{1}\cap A_{2}\right)P\left(A_{1}\cap A_{2}\right)$
Here $P\left(W_{1}\cap W_{2}\right)=P\left(W_{1}\right)\times P\left(W_{2}\right)$,
$P\left(W_{1}\cap A_{2}\right)=P\left(W_{1}\right)\times P\left(A_{2}\right)$..
et cetera because of independency.
This leads to: $P\left(W\right)=1\times\frac{8}{10}\times\frac{4}{20}+\frac{1}{2}\times\frac{8}{10}\times\frac{16}{20}+\frac{1}{2}\times\frac{2}{10}\times\frac{4}{20}+0\times\frac{2}{10}\times\frac{16}{20}=\frac{1}{2}$.
| {
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Proving that $\sqrt{3} +\sqrt{7}$ is rational/irrational I took $\sqrt{3}+\sqrt{7}$ and squared it. This resulted in a new value of $10+2\sqrt{21}$.
Now, we can say that $10$ is rational because we can divide it with $1$ and as for $2\sqrt{21}$, we divide by $2$ and get $\sqrt{21}$.
How do I prove $\sqrt{21}$ to be rational/irrational?
Thanks
| @Doorknob: I think he means that he took the square: $(\sqrt{3} + \sqrt{7})^2 = 10 + 2\sqrt{21}$.
If you want to prove that $\sqrt{21}$ is rational/irrational, do as follows. Suppose $a,b \in \mathbb{N}$ are such that $\frac{a}{b} = \sqrt{21}$ and that $\gcd(|a|,|b|) = 1$. I.e., $\frac{a}{b}$ can not be simplified. Then, we must have that $\frac{a^2}{b^2} = 21$.
Or: $a^2 = b^2\cdot 21$.
Every prime factor of both $a^2$ and $b^2$ must occur an even amount of times in their respective factorisation. We see, however, that in this case, it will not happen, due to the extra factors 3 and 7 on the right hand side of the equation. This yields a contradiction, thus $\sqrt{21}$ is irrational.
| {
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For what $(m, n)$, does $1+x+x^2 +\dots+x^m | 1 + x^n + x^{2n}+\dots+x^{mn}$? For what $(m, n)$, does $1+x+x^2 +\dots+x^m | 1 + x^n + x^{2n}+\dots+x^{mn}$?
Well,
$$\sum_{i = 0}^{m} x^i = \frac{x^{m+1} - 1}{x - 1}$$
and,
$$\sum_{i = 0}^m x^{in} = \frac{x^{n(m+1)} - 1}{x-1}$$
Notice that $x^{m+1} - 1|(x^{m+1})^n - 1$, therefore,
$$x^{n(m+1)} - 1 = q(x^{m+1} - 1)$$
Let
$$\frac{x^{m+1} - 1}{x - 1} = \alpha$$
Therefore,
$$\sum_{i = 0}^{m} x^i = \frac{x^{m+1} - 1}{x - 1} = \alpha$$
and,
$$\sum_{i = 0}^m x^{in} = \frac{x^{n(m+1)} - 1}{x-1} = \frac{q(x^{m+1} - 1)}{x-1} = \alpha q$$
Clearly, $\alpha|\alpha q$. Therefore, all $(m, n)$ such that $m, n \in \mathbb{N}$ should work. But the correct answer, according to the text is that $(m+1)$ and $n$ have to be relatively prime. Where did I go wrong?
| The number $x^{m}-1$ has a unique factor for every number $b \mid m$, say $A_b$.
The number $x^{mn}-1$ has a unique factor for every $b \mid mn$.
The proposed divisor is then the product of $A_b$, where $b\mid m$, except b=1
The proposed quotient is the product of $A_b$, where $b \mid mn$, but not $b \mid n$
So we seek values of $m$, where none of its divisors belong to $n$, and find the two to be co-prime.
| {
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Associativity of a composition $x ∘ y = xy+\sqrt{(x^2-1)(y^2-1)}$ For many hours I had been stucked at this problem. For the following composition
$x ∘ y = xy+\sqrt{(x^2-1)(y^2-1)}$
I have to demonstrate that this composition is associative($(x ∘ y)∘z=x∘(y∘z)$
and that $x ∘ y = xy-\sqrt{(x^2-1)(y^2-1)}$ it's not associative.
| For the second operation, as was pointed out in the comments, it should be easy to just find a counterexample.
For the first one, there's a rather easy way using hyperbolic functions. Assuming that $x,y$ are greater than or equal to $1$, we can find such nonnegative real numbers $a$ and $b$ that $x = \cosh a$ and $y = \cosh b$. Then $\sqrt{x^2-1} = \sinh a$ and $\sqrt{y^2-1} = \sinh b$, according to the well known relations for hyperbolic functions. Then $$x \circ y = \cosh a \cosh b + \sinh a \sinh b = \cosh (a+b),$$ according to another well known relation.
Now to prove associativity see that if $x = \cosh a, y = \cosh b, z = \cosh c$, then
$$
(x\circ y)\circ z = \cosh (a+b) \circ \cosh c = \cosh(a+b+c) =
\cosh a \circ \cosh (b+c) = x \circ (y \circ z).
$$
PS: I assumed the operation is defined on $[1, +\infty)$. I haven't really thought what will happen if we allow numbers less than $1$. Ideally, you should indicate where the operation is defined in the question itself.
UPDATE: Here is another way, without hyperbolic functions. Let us just calculate $(x\circ y)\circ z$ (again, assuming all the variables are $\geq 1$). First, a helpful calculation:
$$
\begin{align}
(x \circ y)^2 - 1 & = \left( xy + \sqrt{(x^2-1)(y^2-1)} \right)^2 - 1 \\
& = 2x^2y^2 - x^2 - y^2 + 2xy\sqrt{(x^2-1)(y^2-1)} \\
& = \left( x\sqrt{y^2-1} + y\sqrt{x^2-1} \right)^2.
\end{align}
$$
Now, moving on to $(x\circ y)\circ z$:
$$
\begin{align}
(x\circ y)\circ z & = (x\circ y)z + \sqrt{((x \circ y)^2-1)(z^2-1)} \\
& = \left(xy + \sqrt{(x^2-1)(y^2-1)}\right)\cdot z + \left( x\sqrt{y^2-1} + y\sqrt{x^2-1} \right)\sqrt{z^2-1} \\
& = xyz + x\sqrt{(y^2-1)(z^2-1)} + y\sqrt{(x^2-1)(z^2-1)} + z\sqrt{(x^2-1)(y^2-1)}.
\end{align}
$$
Now, if you look at $x \circ (y \circ z)$, since the operation is clearly symmetric, it is equal to $(y \circ z) \circ x$. And this one you can get if you substitute $x, y$ and $z$ for each other cyclically in the large formula above. The large formula above is symmetric, so it will remain the same. Therefore $(x\circ y)\circ z = x \circ (y \circ z)$.
Or you can just repeat all the calculations for $x \circ (y \circ z)$ again, makes no matter.
One last word: I do advise to understand the solution with hyperbolic functions too. Relations with hyperbolic functions are even easier to grasp than trigonometric ones: it's just the exponent function wearing a mask. Also, it wasn't just a lucky grouping of terms that led me to the equality $$(x \circ y)^2 - 1 = \left( x\sqrt{y^2-1} + y\sqrt{x^2-1} \right)^2.$$ It was this relation for hyperbolic functions: $\sinh (a + b) = \sinh a \cosh b + \sinh b \cosh a$. So, to sum up: hyperbolic functions are important if you want to really understand the problem. It's almost guaranteed that that is how the author of the problem came up with it.
| {
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How to maximize the function I have a triangle $T=ABC$. I want to calculate $\max (a-b)$, where the the angle $ABC = \beta$, and $|AB|=c$ is fixed (pre-known). My guess is $c\times\cos (\beta)$, but I want to prove it.
Let $A$,$B$,$C$ denote the vertices of $T$, and $|AB|=c$,$|AC|=b$, and $|BC|=a$.
| We can use the cosine formula
$$b^2 = a^2 + c^2 - 2ac\cos \beta \Rightarrow b = \sqrt{a^2 + c^2 - 2ac\cos \beta}\ .$$
Write $f(a) = a-b = a- \sqrt{a^2 + c^2 - 2ac\cos \beta}$, then
$$f'(a) = 1- \frac{a-c\cos\beta}{\sqrt{a^2 + c^2 - 2ac\cos \beta}}\ .$$
As $a^2 + c^2 - 2ac\cos \beta = (a-c\cos\beta)^2 - c^2 \cos^2\beta + c^2 = (a-c\cos\beta)^2 + c^2 \sin^2\beta$, we see that
$$\bigg|\frac{a-c\cos\beta}{\sqrt{a^2 + c^2 - 2ac\cos \beta}} \bigg|\leq 1$$
and $f'(a)>0$ for all $a$. This means $f$ is increasing. So the maximum value of $f$ is never attained (that is the reason you cannot find $a$), but we can still find out the upper bound of $f$ by calculating
$$\lim_{a\to +\infty} f(a) = \lim_{a\to +\infty} \big(a- \sqrt{a^2 + c^2 - 2ac\cos \beta}\big) = c\cos\beta \ \ \text{(How?)}\ .$$
Thus your guess is almost correct: the value $c\cos\beta$ can never be attained, but is the smallest upper bound of $a-b$.
| {
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Express $\cos2\theta$ in terms of $\cos$ and $\sin$ (De Moivre's Theorem) Use De Moivre's to express $\cos2\theta$ in terms of powers of $\sin$ and $\cos$.
What I have is:
$\cos2\theta + i\sin2\theta\\
= (\cos\theta + i \sin\theta)^2\\
= \cos^2\theta + 2 \cos\theta ~i \sin\theta + (i \sin)^2\theta\\
= \cos^2\theta + i(2\cos\theta \sin\theta) - \sin^2\theta\\
= \cos^2\theta - \sin^2\theta + i(2\cos\theta \sin\theta)
$
so $\cos2\theta = \cos^2\theta - \sin^2\theta$
Is this correct?
| Yes, indeed! Since the sine and cosine functions are real-valued functions on the reals, then since $$\cos2\theta+i\sin 2\theta=\cos^2\theta-\sin^2\theta+i(2\sin\theta\cos\theta),$$ we have: $$\cos2\theta=\cos^2\theta-\sin^2\theta\\\sin 2\theta=2\sin\theta\cos\theta$$
| {
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Finding the limit $ \lim_{x\to 0}\frac{(1-3x)^\frac{1}{3} -(1-2x)^\frac{1}{2}}{1-\cos(\pi x)}$ I cannot find this limit:
$$
\lim_{x\to 0}\frac{(1-3x)^\frac{1}{3} -(1-2x)^\frac{1}{2}}{1-\cos(\pi x)}.
$$
Please, help me.
Upd: I need to solve it without L'Hôpital's Rule and Taylor expansion.
| Your limit is
$$\lim_{x\to 0} \frac{(1-3x)^\frac{1}{3}-(1-2x)^{\frac{1}{2} } }{1-\cos\pi x }$$
$$=\lim_{x\to 0} \frac{(1-3x)^\frac{1}{3}-(1-2x)^{\frac{1}{2} } }{1-\cos\pi x }\cdot \frac{1+\cos(\pi x)}{1+\cos(\pi x)}$$
$$=\lim_{x\to 0} \frac{\left((1-3x)^\frac{1}{3}-(1-2x)^{\frac{1}{2} }\right)(1+\cos(\pi x)) }{1-\cos^2\pi x }$$
$$=\lim_{x\to 0} \frac{\left((1-3x)^\frac{1}{3}-(1-2x)^{\frac{1}{2} }\right)(1+\cos(\pi x)) }{\sin^2\pi x }$$
$$=\lim_{x\to 0} \frac{\left((1-3x)^\frac{2}{3}-(1-2x)\right)(1+\cos(\pi x)) }{\left((1-3x)^\frac{1}{3}+(1-2x)^{\frac{1}{2} }\right)\sin^2\pi x } $$
$$=\lim_{x\to 0} \frac{\left((1-3x)^\frac{2}{3}-(1-2x)\right) }{\sin^2\pi x } $$
$$=\lim_{x\to 0} \frac{\left((1-3x)^{2}-(1-2x)^3\right) }{(P(x))\sin^2\pi x } $$
$\lim_{x\to 0} P(x)=3$. Here $P(x)=(1-3x)^{\frac{4}{3}}+(1-3x)^{\frac{1}{3}}(1-2x)+(1-2x)^2$.
Now, $$\lim_{x\to 0} \frac{\left((1-3x)^{2}-(1-2x)^3\right) }{\sin^2\pi x }=-\frac{3}{\pi^2}$$ because
$$=\lim_{x\to 0} \frac{\frac{(1-3x)^{2}-(1-2x)^3}{\pi^2 x^2} }{\frac{\sin^2\pi x}{\pi^2 x^2}}$$
$$=\lim_{x\to 0} \frac{(1-3x)^{2}-(1-2x)^3}{\pi^2 x^2}=\lim_{x\to 0} \frac{8x^3-3x^2}{\pi^2 x^2}=-\frac{3}{\pi^2}$$
Thus your limit is equal to
$$\frac{-3}{\pi^2}\cdot \frac{1}{3}=-\frac{1}{\pi^2}$$
| {
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Find the limit of this function Find
$$\lim_{n\to\infty}\left(1+x^n+\left(\frac {x^2}2\right)^n\right)^{\frac{1}{n}}$$
$(x\ge0)$.
Actually I could not find it when $x\ge1$, but when $0\leq x\lt1$ I found that the limit is $1$. Help me please.
| $$\begin{align} \lim_{n\to\infty} \left(1+x^n+2^{-n}x^{2n}\right)^{1/n} \end{align}$$
Let's consider the roots of the quadratic: $x^n=\frac{-1\pm\sqrt{1-2^{2-n}}}{2^{1-n}}$
Thus, we have
$$\begin{align} &\lim_{n\to\infty} \left(1+x^n+2^{-n}x^{2n}\right)^{1/n} \\
=&\lim_{n\to\infty} (2^{-n})^{1/n}\left(x^n+\frac{1+\sqrt{1-2^{2-n}}}{2^{1-n}}\right)^{1/n}\left(x^n+\frac{1-\sqrt{1-2^{2-n}}}{2^{1-n}}\right)^{1/n} \\
=&\lim_{n\to\infty} \frac{1}{2}\left(x^n+\frac{2}{2^{1-n}}\right)^{1/n}\left(x^n+\frac{0}{2^{1-n}}\right)^{1/n} \\
=&\lim_{n\to\infty} \frac{x}{2}\left(x^n+2^n\right)^{1/n} \\
=&\frac{x}{2}\max(x,2) \end{align}$$
| {
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If 4n+1 and 3n+1 are both perfect sqares, then 56|n. How can I prove this?
Prove that if $n$ is a natural number and $(3n+1)$ & $(4n+1)$ are both perfect squares, then $56$ will divide $n$.
Clearly we have to show that $7$ and $8$ both will divide $n$.
I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd perfect square.
- so we have $4n+1\equiv 1\pmod{8}$; from this $2|n$ so $3n+1$ is a odd perfect square.
- so $3n+1\equiv 1\pmod{8}$ so $8|n$ but I can't show $7|n$. How do I show this?
Thanks for the help.
| For $7 | n$, consider adding $4n+1$ and $3n+1$.
Observe that we get $7n + 2 = x^2 + y^2$, where we can let $4n+1 = x^2$, and $3n+1 = y^2$. Now, we see that mod $7$, the only quadratic residues are $0$, $1$, $2$, $4$. The only way to get two squares summing to 2 mod 7 is either if both are congruent to $1$ mod $7$, or if one is $0$ mod $7$ and one is $2$ mod $7$.
If both are congruent to $1$ mod $7$, we have n is $0$ mod $7$, so we need to show that the other case is not possible.
We now try to multiply $4n+1$ by $3n+1$, to get $12n^2 + 7n + 1$. This mod 7 is $5n^2 + 1$. If one of the squares if $0$ mod 7, it means that the product is also 0 mod 7, meaning that $5n^2 + 1 \equiv 0 $ mod $7$. This means that $5n^2 \equiv 6 $ mod $7$, and hence that $n^2 \equiv 5 $ mod $7$, which is an impossibility due to quadratic residues mod 7.
| {
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Integral $\int_0^\infty\frac{\ln\left(\sqrt{x+1\vphantom{x^0}}-1\right)\,\ln\left(\sqrt{x^{-1}+1}+1\right)}{(x+1)^{3/2}}dx$ Another integral similar to my previous question:
$$\int_0^\infty\frac{\ln\left(\sqrt{x+1\vphantom{x^0}}-1\right)\,\ln\left(\sqrt{x^{-1}+1}+1\right)}{(x+1)^{3/2}}dx$$
Can someone suggest how to evaluate it? Is there a closed form?
| To evaluate
$$
\int^{\infty}_{0}\frac{\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)}{(x+1)^{3/2}}dx.\tag 1
$$
I will evaluate first
$$
I(x)=\int\frac{\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)}{(x+1)^{3/2}}dx\tag 2
$$
and then use limits to find the answer. The evaluation is based on Mathematica program (see Wolfram alpha...etc).
First write
$$
I(x)=-2\int\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)\frac{d}{dx}\left(\frac{1}{\sqrt{x+1}}\right)dx.
$$
Then using integration by parts we find
$$
I(x)=-\frac{2}{\sqrt{x+1}}\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)+
$$
$$
+\int\frac{\log\left(1+\sqrt{1+\frac{1}{x}}\right)}{(1+x)(-1+\sqrt{1+x})}dx+\int\frac{(\sqrt{x^2+x}-x-1)\log\left(-1+\sqrt{1+x}\right)}{x(1+x)^{3/2}}dx=
$$
$$
=-\frac{2}{\sqrt{x+1}}\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)+
$$
$$
+\int\frac{\log\left(1+\sqrt{1+\frac{1}{x}}\right)}{(1+x)(-1+\sqrt{1+x})}dx
+\int\frac{\log\left(-1+\sqrt{1+x}\right)}{(1+x)\sqrt{x}}dx
-\int\frac{\log\left(-1+\sqrt{1+x}\right)}{x\sqrt{1+x}}dx\tag 3
$$
But all three integrals can evaluated with Mathematica and we write for $x>0$
$$
I_1(x)=\int\frac{\log\left(-1+\sqrt{1+x}\right)}{(1+x)\sqrt{x}}dx
-\int\frac{\log\left(-1+\sqrt{1+x}\right)}{x\sqrt{1+x}}dx=
2\pi i\cot^{(-1)}(\sqrt{x})-
$$
$$
-\frac{1}{2}\log\left(-1+\sqrt{1+x}\right)\left(\log 4+\log\left(-1+\sqrt{1+x}\right)-2\log\left(1+\sqrt{1+x}\right)\right)-
$$
$$
-4i\cdot\textrm{Li}\left(2,\frac{1-i\sqrt{x}}{\sqrt{1+x}}\right)+i\cdot\textrm{Li}\left(2,1-\frac{2}{1+i\sqrt{x}}\right)+\textrm{Li}\left(2,\frac{1}{2}\left(1-\sqrt{1+x}\right)\right)+C_1.\tag 4
$$
Also if $y=1/x$, then
$$
I_2(x)=\int\frac{\log\left(1+\sqrt{1+\frac{1}{x}}\right)}{(1+x)(-1+\sqrt{1+x})}dx=-\int\frac{\log\left(1+\sqrt{1+y}\right)}{y+1}dy-\int\frac{\log(1+\sqrt{1+y})}{\sqrt{y(y+1)}}dy=
$$
$$
=-4i\log( x)\arcsin\left(\sqrt{\frac{1}{2}\left(1+\sqrt{1+x^{-1}}\right)}\right)-
$$
$$
-4\arcsin\left(\frac{1}{2}\left(1+\sqrt{1+x^{-1}}\right)\right)^2+\log(x)\log\left(1+\sqrt{1+x^{-1}}\right)-
$$
$$
-2\pi i \log\left(1+\sqrt{1+x^{-1}}\right)-2\log\left(1+\sqrt{1+x^{-1}}\right)\log\left(1+\sqrt{1+x}\right)+
$$
$$
8i\arcsin\left(\sqrt{\frac{1}{2}\left(1+\sqrt{1+x^{-1}}\right)}\right)\log\left(-1+\sqrt{x}+\sqrt{1+x}\right)+
$$
$$
+2\textrm{Li}\left(2,-\sqrt{1+x^{-1}}\right)-4\textrm{Li}\left(2,-\frac{1}{\sqrt{x}}\left(-1+\sqrt{1+x}\right)\right)
$$
Hence
$$
I(x)=I_1(x)+I_2(x)
$$
and we can see easily using Mathematica that
$$
\lim_{x\rightarrow\infty}I(x)=-4G-\left(1-\frac{i}{4}\right)\pi^2+2\pi i \log 2-\frac{1}{2}\log^22
$$
Also
$$
\lim_{x\rightarrow0}I(x)=\left(-\frac{4}{3}+\frac{i}{4}\right)\pi^2+2\pi i\log 2+\frac{1}{2}\log^22
$$
Hence
$$
\int^{\infty}_{0}\frac{\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)}{(x+1)^{3/2}}dx=-4G+\frac{\pi^2}{3}-\log^22
$$
QED
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If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then to prove $x=\sqrt{3}+\sqrt{2}$
If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then we have to prove $x=\sqrt{3}+\sqrt{2}$
The question would have been simple if it asked us to prove the other way round.
We can multiply by $x^3$ and solve the quadratic to get $x^3$ but that would be unnecessarily complicated.Also, as $x^3$ has 2 solutions,I can't see how x can have only 1 value. But the problem seems to claim that x can take 1 value only.Nevertheless,is there any way to get the values of x without resorting to unnecessarily complicated means?
NOTE: This problem is from a textbook of mine.
| $x^3+\frac{1}{x^3}=18\sqrt{3}\Rightarrow x^6-18\sqrt{3}x^3+1=0 $
we asssume $y=x^3$
$y^2-18\sqrt{3}y+1=0\Rightarrow y=\frac{18\sqrt{3}\pm\sqrt{968}}{2}=\frac{18\sqrt{3}\pm22\sqrt{2}}{2}=9\sqrt{3}\pm11\sqrt{2}$
$x^3-(9\sqrt{3}\pm11\sqrt{2})=0$
let $x=a\sqrt{3}+b\sqrt{2}$ now,
$x^3=3\sqrt{3}a^3+2\sqrt{2}b^3+3.3a^2b\sqrt{2}+3.a\sqrt{3}.2b^2=(3a^3+6ab^2)\sqrt{3}+(2b^3+9a^2b)\sqrt{2}$
Now, $3a^3+6ab^2=9$ and $2b^3+9a^2b=11$
i.e., $a^3+2ab^2=3$ and $2b^3+9a^2b=11$.
Now,
for $a^3+2ab^2=3$ one possible case would be $a^3=1,ab^2=1$ (I am not saying this is "the" one)
i.e., $a=1 ,b=\pm 1$
Now,
for $2b^3+9a^2b=11$ one possible case would be $b^3=1,a^2b=1$ (I am not saying this is "the" one)
i.e., $b=1, a=\pm 1$
but then we need both $a^3+2ab^2=3$ and $2b^3+9a^2b=11$ to satisfy at once.
So, only possibility would be $a=1,b=1$ which would imply $x=\sqrt{3}+\sqrt{2}$
Note : I took only $+11$ in $9\sqrt{3}\pm11\sqrt{2}$ , I would leave it to you to do the same for $-11$ case.
Good Luck!
| {
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Ring theory: Ideals being equal Question: Prove directly, without gcd computations, the following equalities of ideals.
(i) $(5, 7) = (1)$ in $\Bbb{Z}$ (of course (1) = $\Bbb Z$).
(ii) $(15, 9) = (3)$ in $Z$.
(iii) $(X^3 −1,X^2 −1) = (X −1) \text{ in } \Bbb{Q} [X]$
Attempted solution:
Try to show that $(1)$ is a subset of $(5,7)$ and that $(5,7)$ is a subset of $(1)$
First thing is $(5,7) = 5x + 7y$ with $x,y$ integers. Then by the closure under addition of the integers $(5,7)$ is a subset of $(1)$ as $(1)$ is the integers
Next to show $(5,7)$ is a subset of $(1)$ I try to say if $x$ belongs to integers, $x=5q + 7p, 1=5q + 7p$, then $x=x . 1= x(5q + 7 p)$ which is $5q' +7p'$, which shows that if $x$ is an integer then it's also in $(5,7)$
| For (i), note that $3 \times 7 - 4 \times 5 = 1$, whence $1 \in (5, 7)$, so that $(5, 7) = Z$;
for (ii), note that $2 \times 15 - 3 \times 9 = 3$, whence $3 \in (15, 9)$, whence $(3) \subset (15, 9)$, and clearly $(15, 9) \subset (3)$, so that $(15, 9) = (3)$;
for (iii), observe that $(x^3 - 1) - x(x^2 -1) = x - 1$, whence $x -1 \in (x^3 -1, x^2 - 1)$, and that $x^3 - 1 = (x - 1)(x^2 + x + 1)$, $x^2 -1 = (x -1)(x + 1)$, whence $x^3 - 1, x^2 - 1 \in (x - 1)$, whence $(x^3 - 1, x^2 - 1) = (x - 1)$.
Call it a wrap, it's in the can, Mr. Hitchcock!
And may The Spirit of Our Lady The Queen of The Sciences, Mathematics, smile upon and shed her blessings on your Thanksgiving Holidays!
Which also means: have some good fun with math this weekend!
Hope this helps, ALL of it! Cheers,
and as always,
Fiat Lux!!!
| {
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Math induction sum of even numbers I need to prove by induction this thing:
$2+4+6+........+2n = n(n+1)$
so, this thing is composed by sum of pair numbers, so its what I do, but I'm stucked.
$2+4+6+\cdots+2n = n(n+1)$
$(2+4+6+\cdots+2n)+(2n+2) = n(n+1) + (2n+2) $
$n(n+1)+(2n+2) = n(n+1)+(2n+2) $
$n^2 + 3n + 2$
$n(n+2+1)+2$
I don't know how to move forward from this.
Thanks.
| Suppose that
$$
2+4+6+\cdots+2n=n(n+1)
$$
add $2n+2$ to both sides:
$$
2+4+6+\cdots+2n+(2n+2)=n(n+1)+(2n+2)
$$
To finish the induction, you want the right side to be $(n+1)(n+2)$. Is it?
$$
n(n+1)+(2n+2)=n^2+n+2n+2=n^2+3n+2=(n+1)(n+2)
$$
| {
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Prove using mathematical induction that $2^{3n}-1$ is divisible by $7$ So, i wanna prove $2^{3n}-1$ is divisible by $7$, so i made this:
$2^{3n}-1 = 7\cdot k$ -> for some $k$ value
$2^{3n+1} = 1+2\cdot1 - 2\cdot1 $
$2^{3n+1} - 1-2\cdot1 + 2\cdot1 $
$2^{3n}\cdot2 - 1-2\cdot1 + 2\cdot1$
$2(2^{3n}-1) -1 +2$
$2\cdot7k+1$ -> made this using the hypothesis.
so, i dont know if its right, or if its wrong, i dont know how to keep going from this, or if its the end.
Thanks.
| If $\displaystyle f(m)=2^{3m}-1$
$\displaystyle f(m+1)=2^{3(m+1)}-1$ and not $2^{3m+1}-1$
So, $2^{3(m+1)}-1=2^3\cdot2^{3m}-1=2^3(2^{3m}-1)-1+2^3$
| {
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Maximizing slope of a secant line Two points on the curve $$ y=\frac{x^3}{1+x^4}$$ have opposite $x$-values, $x$ and $-x$. Find the points making the slope of the line joining them greatest.
Wouldn't the maximum slope of the secant line be with the max/min of the curve?
So $x=3^{1/4}$ and $x=-3^{1/4}$?
| The slope of the line will be
$$slope = \frac{y(x) - y(-x)}{2x} = \frac{2x^3}{2x(1+x^4)} = \frac{x^2}{1+x^4}.$$
Then you take the derivatives of this with respect to $x$ to find the maximum:
$$slope' = \frac{(1+x^4)(2x) - x^2(4x^3)}{(1+x^4)^2} = \frac{2x(1-x^4)}{(1+x^4)^2}.$$
This is zero at $x = 0, -1, and +1$. You'll find through a second derivative test that the answer is $x = 1$ and $x = -1$.
| {
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Help Solving Equation I've been trying to solve this equation for a few hours, but somehow I'm stuck somewhere and I could use some help.
$\frac{1} {x^2} = -\frac {2(x-a)} {a^3} + \frac {1} {a^2}$
I know from the book that there are two solutions (this is the equation of a tangent line to $x^2$). One solution is $x = a$:
$\frac{1} {x^2} - \frac {1} {a^2} = -\frac {2(x-a)} {a^3}$
The second solution is $x = - \frac {a} {2}$, but I don't know how to get to it.
| As lcm$(x^2,a^3,a^2)=x^2a^3$ multiplying either sides by that and rearranging we get
$$a^3-ax^2+2x^2(x-a)=0$$
$$2x^2(x-a)-a(x^2-a^2)=0$$
$$(x-a)\{2x^2-a(x+a)\}=0$$
If $\displaystyle x-a\ne0,2x^2-ax-a^2=0$
Now, $\displaystyle2x^2-ax-a^2=2x^2-2ax+ax-a^2=2x(x-a)+a(x-a)=(x-a)(2x+a)$
But we have already assumed $x-a\ne0$
| {
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probability:picking 3 different numbers from 1-3n. its known that their sum is devided by 3. whats the probability that each of them is devided by 3 we pick 3 different numbers from 1,2,3,...3n , which their sum is devided by 3 without remainder,what is the probability that each of the number is devider by 3 without remainder?
What I tried:
I started with checkis what is the probability that the sum will be devided by 3 without any remainder:
the highest sum of any trio is 3n+(3n-1)+(3n-2)=9n-3
the lowest sum is 1+ 2 + 3 =6
so there are 9n-3-6=9n-9 different summaries avaiable.(9n-9)/3 are devided by Without remainder.
if so the probability 1/3n-3.
the probability of picking 3 number that are devided by 3 from 1...3n is (3n/3 pick 3)=(n pick 3)= (n(n-1)(n-2))/3!
| Let $a$ be the probability that each number is divisible by $3$, and let $b$ be the probability the sum is divisible by $3$. We want $\frac{a}{b}$.
To get a non-zero probability, we need $n\ge 3$. There are $\binom{n}{3}$ ways to choose $3$ distinct numbers divisible by $3$.
Now we need to find the number of ways to choose $3$ numbers so that their sum is divisible by $3$.
The sum can be divisible by $3$ in four different ways. All of the numbers could be divisible by $3$. Or all the numbers could have remainder $1$ on division by $3$. Or all the numbers could have remainder $2$. Or else there could be one of each kind, as in $(3,16,5)$.
For the first three ways, there are in each case $\binom{n}{3}$ choices. For the one of each kind, there are $\binom{n}{1}^3$ choices, for a total of $3\binom{n}{3}+\binom{n}{1}^3$.
Thus the required probability is
$$\frac{\binom{n}{3}}{3\binom{n}{3}+\binom{n}{1}^3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/588035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
logarithms equation Hello I have following problem: solve equation $\log{(x-5)^2}+\log{(x+6)^2}=2$
and I rewrited this equation as
$2\log{(x-5)}+2\log{(x+6)}=\log{100} \implies 2(\log{(x-5)(x+6))=\log{100}} \implies \log{x^2+x-30}=\log10 \implies x^2+x-40=0 $
and I solved this equation, but I obtained only two solutions and there should be four, so I wonder if it is necessary to create $\log{(x-5)^2(x+6)^2}=\log{100}$ or is a simplier way.
| $$\log{(x-5)^2}+\log{(x+6)^2}=2$$
$$\log{((x-5)(x+6))^2}=2$$
$$((x-5)(x+6))^2=10^2=100$$
$$(x-5)(x+6)=\pm10$$Solve from here
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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struggle simplifying $\sqrt{9+\sqrt{5}}$ I need to simplify $\sqrt{9+\sqrt{5}}$
I already do this (proven it) $\sqrt{9-4\sqrt{5}}=2-
\sqrt{5}$
But I couldn't when apply to $\sqrt{9+\sqrt{5}}=\sqrt{9-4\sqrt{5}+5\sqrt{5}}=\sqrt{(2+\sqrt{5})^2+5\sqrt{5}}$
PLEASE help me out
| Let $\sqrt{9+\sqrt{5}}=A+B \sqrt{5}$. Square each side: $9+\sqrt{5} = A^2 + 2AB \sqrt{5} + 5 B^2$. Now we get two equations and two unknowns and solve for $A$ and $B$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/591547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know what to do with the $n^2$.
| Observe that
$$\sum_{n=0}^\infty \frac{n^2}{2^n}=2\sum_{n=0}^\infty \frac{n^2}{2^{n+1}}=2E[X^2]$$
where $X$ is a geometric random variable with support $\mathbb{Z}_{\geq 0}$ and success probability $p=1/2.$ Thus,
$$2E[X^2]=2\left[\text{Var}(X)+(E[X])^2\right]=2\left[\frac{1-p}{p^2}+\left(\frac{1-p}{p}\right)^2\right]=2\frac{(1-p)(2-p)}{p^2}=6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/593996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 8,
"answer_id": 7
} |
Factorization of $ x^2 +xy+5x+m+5 $ I want to factorize $ x^2 +xy+5x+m+5 $ . For what value of m , $ x^2 +xy+5x+m+5 $ can be resolved into linear factors ?
My try :
$ x^2 +xy+5x+m+5 $ = $ x^2 +(5+y)x+(m+5) $
To get the linear factors , we must have the determinant of this eqn is >= 0 .
D = $ (5+y)^2-4(m+5)$ . Then I can not proceed .
| Since now you want to have the discriminant greater than $0$ which in itself is a quadratic polynomial ie $y^2+10y+(5-4m)$ . We see that this will be always greater than $0$ . If its discriminant is always less than $0$ . Because if a quadratic has real roots than it will be positive and negative .
The determinant of $y^2+10y+(5-4m)$ is $ D = 100 -4(5-4m) $ = $80+16m$
Now $D=0$ => $ 80+16m = 0 $ =>$ m = -5 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/595092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Minimum value problem
Find the minimum value of $(x+y)(y+z)$ where $x,y,z$ are positive real numbers satisfying the condition $$xyz(x+y+z)=1$$
Hint?
| Just apply AM-GM.
$(x+y)(y+z)=xy+y^2+xz+yz=y(x+y+z)+zx=\frac y{xyz}+zx=zx+\frac1{zx}\geq 2$
Equality holds if and only if $zx=1$.
Substituting $xz=1$, we have the condition that $y(x+y+\frac{1}{x}) = 1$. As such, $(x+y)(y+ \frac{1}{x}) = xy + \frac{y}{x} + y^2 + 1 = 1 + 1 = 2 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/596080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $4(a^3 + b^3) \ge (a + b)^3$ and $9(a^3 + b^3 + c^3) \ge (a + b + c)^3$
Let $a$, $b$ and $c$ be positive real numbers.
$(\mathrm{i})$ Prove that $4(a^3 + b^3) \ge (a + b)^3$.
$(\mathrm{ii})$Prove that $9(a^3 + b^3 + c^3) \ge (a + b + c)^3.$
For the first one I tried expanding to get $a^3 + b^3 \ge a^2b+ab^2$ but I'm not sure how to prove it.
| (i) By AM-GM, $a^3+a^3+b^3\ge3a^2b$ and $a^3+b^3+b^3\ge3ab^2$.
Adding these inequalities gives $3a^3 + 3b^3 \ge 3a^2b + 3ab^2$, so $4(a^3+b^3)\ge(a+b)^3$.
(ii) Again by AM-GM, $a^3 + b^3 + c^3 \ge 3abc$.
Add two times that to the six cyclic versions of $a^3 + a^3 + b^3 \ge 3a^2b$ to get $8(a^3 + b^3 + c^3)\ge 3a^2b + 3a^2c + 3b^2a+3b^2c+3c^2a+3c^2b+6abc$.
So $9(a^3+b^3+c^3)\ge(a+b+c)^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/596257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Find the limit: $\displaystyle\lim_{x\to \infty} (\sqrt{x^2+x}-\sqrt{x^2-x})$.
Find the following limit: $$ \lim_{x\to \infty} (\sqrt{x^2+x}-\sqrt{x^2-x} )$$
I tried to simplify using conjugation. This gave me the following: $$ \lim_{x\to \infty} \frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} $$
When I plug in the $\infty$, I'm left with $ \frac{\infty}{\infty} $. Did I mess up somewhere, or does the limit not exist?
| For $x\ge 1$ we have
$$
\sqrt{x^2+x}-\sqrt{x^2-x}=\frac{(x^2+x)-(x^2-x)}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}.
$$
It follows that
$$
\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})=\lim_{x\to\infty}\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/597177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Jordan Normal form: problems with finding correct generalized eigenvectors I have been tasked to find the Jordan Normal Form for the matrix $A$ shown below.
\begin{align*}
A = \begin{pmatrix}
2 & 2 & 0 & -1 \\
0 & 0 & 0 & 1 \\
1 & 5 & 2 & -1 \\
0 & -4 & 0 & 4
\end{pmatrix}
\end{align*}
In itself, finding a Jordan Normal form is not that hard, but I'm having some problems with determining the correct coordinate transformation, namely finding the correct generalized eigenvectors.
What I have done so far:
The characteristic polynomial is $\lambda^4 - 8 \lambda^3 + 24 \lambda^2 -32 \lambda + 16 = (\lambda - 2)^4$, so we get eigenvalue $\lambda = 2$ with algebraic multiplicity 4.
The kernel of $(A - 2I)$ is
\begin{align*}
\ker\begin{pmatrix}
0 & 2 & 0 & -1 \\
0 & -2 & 0 & 1 \\
1 & 5 & 0 & -1 \\
0 & -4 & 0 & 2
\end{pmatrix} = \text{L}((0,0,1,0)^T, (-3,1,0,2)^T)
\end{align*}
so our first two eigenvectors are $(0,0,1,0)^T$ and $(-3,1,0,2)^T$, and we are two vectors short.
Following this we determine $\ker((A-2I)^2)$ and we get
\begin{align*}
\ker\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & -4 & 0 & 2 \\
0 & 0 & 0 & 0
\end{pmatrix}
\end{align*}
Now I thought that if we take $v = (0,1,0,2)^T$ and $w = (0,0,0,1)^T$ from $\ker((A-2I)^3)$, then we have $(A - 2I)v \neq 0$ and $(A - 2I)w \neq 0$. Also, the four vectors are linearly independent and generate $\mathbb{R}^4$, so we have a good basis.
However, when using this basis as coordinate transformation, you do not get the Jordan Normal Form $J$ that MATLAB and Wolfram calculate. That is, you do not get a Jordan Normal form at all.
\begin{align*}
J = \begin{pmatrix}
2 & 1 & 0 & 0 \\
0 & 2 & 1 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2
\end{pmatrix}
\end{align*}
My question is thus, where am I making a mistake? Also, I am wondering why is $J$ split into a $3 \times 3$ and a $1 \times 1$ block, is it not supposed to be one big $4 \times 4$ block?
If anyone could provide some hints or point me in the right direction, I'd be forever grateful, thanks!
| JNF of $A$ should be $Q^{-1}AQ$, where $Q$ is the matrix having a Jordan basis as its columns.
The basis which you found is not a Jordan basis, so it is not a disjoint union of Jordan
chains.
Different method of getting JNF:
Theorem.
Let $\lambda$ be an eigenvalue of a matrix $A$ and let $J$ be the JNF of $A$. Then
the number of Jordan blocks of $J$ with eigenvalue $\lambda$ is equal to
$\mbox{nullity} (A − \lambda I)$.
Are you familiar with it?
In your calculations the $\mbox{nullity} (A − 2I)=2$. Thus $J$ will be of the form
$J_{\lambda}^{(k)} \oplus J_{\lambda}^{(l)}$ for some natural numbers $k$,$l$ depending on the characteristic and minimal polynomial of $A$ and such that $k+l=4$, where
Let $J_{\lambda}^{(n)}=\left[ \begin{array}{llll}
\lambda_i & 1 & \; & \; \\
\; & \lambda_i & \ddots & \; \\
\; & \; & \ddots & 1 \\
\; & \; & \; & \lambda_n
\end{array} \right].$
Note that the minimal polynomial of $A$ is $m_A(\lambda)=(\lambda-2)^3$, thus the only option of JNF is $J_{2}^{(3)} \oplus J_{2}^{(1)}=\begin{pmatrix}
2 & 1 & 0 & 0 \\
0 & 2 & 1 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2
\end{pmatrix}.
$
| {
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"url": "https://math.stackexchange.com/questions/599899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Standard form of the equation of a circle involving two points.
For parts (a), (b), and (c), consider the points $\mathrm A=(-1,3)$ and $\mathrm B=(5,3)$ in the $xy$-plane. Distance is measured in meters.
c) Find the standard form of the equation of a circle that has the line segment $\overline{\mathrm{AB}}$ as the diameter. (The circle has $\mathrm A$ and $\mathrm B$ as endpoints of the diameter)
With the following equations I arrived that $(3,0)$ is the center of the line:
\begin{gather*}
X = \frac{(-1 -5)} 2 = -3 \\
Y = \frac{(3 -3)} 2 = 0
\end{gather*}
and then with the following I concluded that.
$$(x-1)^2 + (y-1)^2 = \sqrt{13}$$
Can anyone verify this? I feel unease by my answer.
| The center of the circle is the midpoint of the diameter: $(2, 3).$
The radius has length $3 m$ because the diameter has length $6 m$.
Then the equation for the circle is straightforward:
$$(x-2)^2 + (y-3)^2 = 9 m^2.$$
The $x$ coordinate of the center is $\frac{1}{2} \cdot (5 + (-1)) = 2 m.$
Actually, your $y$ coordinate was off, too: $\frac{1}{2} \cdot (3 + 3) = 3 m.$
You take the average of the coordinates to get the midpoint. This is an addition, not a subtraction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Integrating With Trig Subsitution Can someone please explain how to calculate the integral of
$\frac{\sqrt{1 + x^2}}{x}$ using trig substitution?
| In all the following, I intentionnally forget the constant to add to every primitive, in order to simplify the notations. All equalities with primitives are to be understood "up to a constant".
Using trigonometric functions, and substitution $x=\tan u$ (then $\mathrm{d}x=(1+\tan^2u)\,\mathrm{d}u = \frac{\mathrm{d}u}{cos^2u}$):
$$\int \frac{\sqrt{1+x^2}}{x} \mathrm{d}x=\int \frac{\sqrt{1+\tan^2 u}}{\tan u} \frac{1}{\cos^2u}\mathrm{d}x=\int \frac{1}{|\cos u| \tan u} \frac{1}{\cos^2u} \mathrm{d}u$$
If $\cos u > 0$,
$$\int \frac{\sqrt{1+x^2}}{x} \mathrm{d}x=\int \frac{1}{\sin u\cos^2u} \mathrm{d}u$$
Then
$$\frac{1}{\sin u\cos^2u}=\frac{1}{\cos^2u}\left(\frac{1}{\sin u} - \sin u + \sin u\right) = \frac{1}{\cos^2u}\left( \frac{1-\sin^2 u}{\sin u}+\sin u\right)$$
$$=\frac 1 {\sin u} + \frac{\sin u}{\cos^2 u}$$
And (see here)
$$\int \frac{1}{\sin u} \mathrm{d}u=\log \left|\tan \frac u 2\right|$$
$$\int \frac{\sin u}{\cos^2 u} \mathrm{d}u = \frac 1 {\cos u}$$
So
$$\int \frac{\sqrt{1+x^2}}{x} \mathrm{d}x=\frac 1 {\cos u} +\log \left|\tan \frac u 2\right| = \frac 1 {\cos (\arctan x)} +\log \left|\tan \frac {\arctan x} 2\right| $$
Now $\frac{1}{\cos^2 u}=1+\tan^2 u$ thus $\frac 1{\cos (\arctan x)} = \sqrt{1+x^2}$,
$$\int \frac{\sqrt{1+x^2}}{x} \mathrm{d}x = \sqrt{1+x^2} +\log \left|\tan \frac {\arctan x} 2\right| $$
Also, notice we have always $\cos u = \cos (\arctan x) > 0$ since $\arctan x \in ]-\pi/2, +\pi/2[$. Our assumption above was not too restrictive :-)
There is further simplification, with $\tan \frac{\arctan x}{2}$.
The idea is to write $\tan \frac{\theta}2$ as a function of $\tan \theta$, that is, reversing the usual formula $\tan \theta = \frac{2t}{1-t^2}$ where $t=\tan \frac{ \theta }2$.
That is, we want to solve
$$(1-t^2) \tan \theta -2t = 0$$
$$(\tan \theta) t^2 + 2t - \tan \theta = 0$$
It's a quadratic equation with $\Delta = 4(1+\tan^2\theta$), so the solutions are
$$t = \frac{-1 \pm \sqrt{1+\tan^2 \theta}}{\tan \theta}$$
Notice that $t = \tan \frac{\theta}{2}$ has the same sign than $\tan \theta$ for $\theta \in ]-\pi/2, \pi/2[$, so the sign in $\pm$ is actually a $+$.
$$\tan \frac{\theta}{2}=\frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta}$$
And
$$\tan \frac{\arctan x}{2}=\frac{\sqrt{1+x^2}-1}{x}$$
Therefore, we have
$$\int \frac{\sqrt{1+x^2}}{x} \mathrm{d}x = \sqrt{1+x^2} +\log \left|\frac{\sqrt{1+x^2}-1}{x}\right|=\sqrt{1+x^2} +\log (\sqrt{1+x^2}-1) - \log |x| $$
The answer can be written slightly differently:
$$\sqrt{1+x^2} -\log (\sqrt{1+x^2}+1) + \log |x|$$
It's enough to check that
$$\frac{(\sqrt{1+x^2}+1)(\sqrt{1+x^2}-1)}{|x|^2}=\frac{1+x^2-1}{x^2}=1$$
Thus, taking logarithms,
$$\log (\sqrt{1+x^2}+1) + \log (\sqrt{1+x^2}-1) - 2\log |x| = 0$$
$$\log (\sqrt{1+x^2}-1) - \log |x| = -\log (\sqrt{1+x^2}+1) + \log |x|$$
There is still place for a bit of simplification:
$$\log (\sqrt{1+x^2}+1) - \log |x|=\log \frac{\sqrt{1+x^2}+1}{|x|}=\log \left(\sqrt{1+\frac{1}{|x|^2}}+\frac{1}{|x|}\right)$$
You may recognize $\arg\sinh t = \log (t + \sqrt{1+t^2})$, so
$$\log (\sqrt{1+x^2}+1) - \log |x|=\arg \sinh \frac{1}{|x|}$$
Thus, as a final step, we can write
$$\int \frac{\sqrt{1+x^2}}{x} \mathrm{d}x = \sqrt{1+x^2}-\arg \sinh \frac{1}{|x|}$$
Another solution, using hyperbolic functions, with substitution $x=\sinh u$ (and $\mathrm{d}x=\cosh u \,\mathrm{d}u$), using the fact that $\cosh^2 u - \sinh^2 u = 1$:
$$\int \frac{\sqrt{1+x^2}}{x} \mathrm{d}x = \int \frac{\cosh u}{\sinh u} \cosh u \,\mathrm{d}u= \int \frac{1+\sinh^2u}{\sinh u} \mathrm{d}u$$
$$=\int \left(\frac{1}{\sinh u}+\sinh u\right) \mathrm{d}u$$
And, with the usual $\int \frac 1 {\sinh u} \,\mathrm{d}u = \frac 1 2\log|\tanh \frac u 2|$,
$$\int \frac{\sqrt{1+x^2}}{x} \mathrm{d}x=\frac 1 2\log\left|\tanh \frac u 2\right| + \cosh u = \frac 1 2\log\left|\tanh \frac {\arg \sinh x} 2\right| + \cosh (\arg \sinh x)$$
$$=\sqrt{1+x^2} + \frac 1 2\log\left|\tanh \frac {\arg \sinh x} 2\right|$$
Like with the first solution, it would be possible to simplify further.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this inequality $\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$ let $a,b,c,d,e\in R$,and such
$$a^2+b^2+c^2+d^2+e^2=1$$
find this value
$$A=\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$$
I use computer have this
$$A=\dfrac{2}{\sqrt{10}}$$
then equal holds if we suppose that $a\leq b\leq c\leq d\leq e$, then$$a=\frac{2}{\sqrt{10}},b=-\frac{1}{\sqrt{10}},c=\dfrac{1}{\sqrt{10}},d=-\frac{2}{\sqrt{10}},e=0$$
But I consider sometimes, I want use Cauchy-Schwarz inequality to solve it,
I guess have this follow?
$$(|a-b|+|b-c|+|c-d|+|d-e|+|e-a|)^2\le(\dfrac{1}{10})(a^2+b^2+c^2+d^2+e^2)$$
then I can't. Thank you very much!
| This is a comment that’s too long to fit in the usual format. I prove below
the non-optimal bound $A\leq \sqrt{\frac{5+\sqrt{5}}{10}} \approx 0.8$. Let us denote
$$
x_1=x_6=a, x_2=b, x_3=c, x_4=d, x_5=e
$$
Then, we have
$$
\big(\frac{5+\sqrt{5}}{2}\big)
\bigg(\sum_{k=1}^5 x_k^2\bigg)-
\bigg(\sum_{k=1}^5 (x_k-x_{k+1})^2\bigg)=
\big(\frac{\sqrt{5}+1}{2}\big)
\bigg(x_5+\frac{(\sqrt{5}-1)(x_1+x_4)}{2}\bigg)^2+
\bigg(x_3+x_4-\frac{(\sqrt{5}-1)x_1}{2}\bigg)^2+
\big(\frac{\sqrt{5}-1}{2}\big)
\bigg(x_1+x_3+\frac{(\sqrt{5}+1)x_2}{2}\bigg)^2
$$
We deduce that
$$
\big(\frac{5+\sqrt{5}}{2}\big)
\bigg(\sum_{k=1}^5 x_k^2\bigg) \geq
\bigg(\sum_{k=1}^5 (x_k-x_{k+1})^2\bigg)
$$
So if we put $\varepsilon={\sf min}(|x_{k+1}-x_k|)$, we have
$$
\frac{5+\sqrt{5}}{2} \geq \sum_{k=1}^5 (x_k-x_{k+1})^2 \geq 5\varepsilon^2
$$
and hence
$$
\varepsilon \leq \sqrt{\frac{5+\sqrt{5}}{10}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$
As is the question in the title, I am wishing to find all positive integers $n$ such that $3^n + 5^n$ is divisible by $n^2 - 1$.
I have so far shown both expressions are divisible by $8$ for odd $n\ge 3$ so trivially a solution is $n=3$. I'm not quite sure how to proceed now though. I have conjectured that the only solution is $n=3$ and have tried to prove it but have had little luck. Can anyone point me in the right direction? thanks
| Here are a few necessary conditions on $n$ summarizing my statements from the chat:
*
*$n$ is odd.
*any prime factor $p\mid n^2-1$ is of the form $p \equiv
1,2,4,8 \mod 15$.
*$n \equiv 3,93 \mod 120$
proofs:
1)
It is clear that $3 \nmid 3^n+5^n$ and $5\nmid 3^n+5^n$. Thus $3\mid n$.
Write $n = 3m$, then $$3^n+5^n \equiv 0 \mod n^2-1 \\\Rightarrow (3^{-1}5)^n \equiv -1 \mod 9m^2-1.$$ Suppose $n$ is even. Then $-1$ is a square mod $9m^2-1$, so every odd prime factor $p$ of $9m^2-1$ is $p\equiv 1 \mod 4$. But $9m^2-1 \equiv 3 \mod 4$ as $n$ is even, a contradiction.
2)
The inverse of $3$ is $3^{-1} \equiv 3m^2 \mod 9m^2-1$, so from the identity shown above we get $$(15m^2)^n \equiv -1 \mod 9m^2-1. \\ \Rightarrow (-15) \equiv (15)^{-n+1}m^{-2n} \mod 9m^2-1$$
As $n$ is odd by 1.) we see that for any odd prime $p$ dividing $9m^2-1$ the Legendre-symbol $$1 = \left( \frac{-15}{p} \right) = \left( \frac{-1}{p} \right)\left( \frac{3}{p} \right) \left( \frac{5}{p} \right) = \left( \frac{p}{3} \right)\left( \frac{p}{5} \right)$$
using the Quadratic Reciprocity Theorem. This shows $p \equiv 1,2,4,8 \mod 15$.
3)
By 2) $n^2-1 \equiv 1,2,4,8 \mod 15$ because these residues form a multiplicative subgroup. We already know $3\mid n$ showing $n\equiv 3 \mod 15$.
As $16 \nmid 3^n+5^n$, we know $8 \nmid n^2-1 =(n-1)(n+1)$ and 1) shows $n\equiv 3,5 \mod 8$. Chinese Remainder Theorem gives the result stated in 3).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/612346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 7,
"answer_id": 2
} |
theory of equations finding roots from given polynomial
If the equation $x^4-4x^3+ax^2+bx+1=0$ has four positive roots then $a=\,?$ and $b=\,?$
$\textbf{A.}\,6,-4$
$\textbf{B.}\,-6,4$
$\textbf{C.}\,6,4$
$\textbf{D.}\,-6,-4$
we can replace options and check answers .. are there any other shortcuts we can use
| $$(x-x_1)(x-x_2)(x-x_3)(x-x_4)$$
$$=x^4-x^3\left(\sum_{4\ge i\ge1} x_i\right)+x^2\left(\sum_{4\ge i>j\ge1}x_ix_j\right)-x\left(\sum_{4\ge i>j>k\ge1}x_ix_jx_k\right)+x_1x_2x_3x_4 $$
If $x_i>0,1\le i\le4$ the coefficient of $x^2$ must be $>0$ and that of $x$ must be $<0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/614222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find all postive integer numbers $x,y$,such $x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$ Find all postive integer $x$ and $y$ such that
$x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$
My try: since
$$(x+y)^2-2xy=x^2+y^2$$
I know this well know reslut:
$$xy|(x^2+y^2+1),\Longrightarrow \dfrac{x^2+y^2+1}{xy}=3,x,y\in N$$
and such condition $$(x,y)=(u_{n},u_{n+1})$$
where
$$u_{0}=u_{1}=1,u_{n+2}=3u_{n+1}-u_{n}$$
so I can't,Thank you
| Note that since $(x+y+1)(x+y-1)=(x^2+y^2-1)+2xy$,
$$
x+y-1\mid x^2+y^2-1\iff x+y-1\mid2xy\tag{1}
$$
Since
$$
x+y+1\mid2xy\tag{2}
$$
and $(x+y-1,x+y+1)\mid2$, $(1)$ and $(2)$ imply
$$
(x+y)^2-1\mid4xy\tag{3}
$$
However, because $4xy\le(x+y)^2$, $(3)$ implies $(x+y)^2-1=4xy$, that is,
$$
(x-y)^2=1\tag{4}
$$
Wlog, assume $y=x+1$, then
$$
x+y-1=2x\mid2x(x+1)=2xy\tag{5}
$$
and
$$
x+y+1=2(x+1)\mid2x(x+1)=2xy\tag{6}
$$
Thus, the solutions are $\{(x,y):(x-y)^2=1\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(2,6,3),(3,6,2),(3,2,6),(6,2,3),(6,3,2)}$. Are these the only 10 solutions?
First, none of $x$, $y$ or $z$ can be $1$ ($x$, $y$ and $z$ are positive integers)
If I let $x=2$, then finding all solutions to $1/y+1/z = 1/2$ leads to $(4,4), (3,6)$ and $(6,3)$ which gives me $(x,y,z)$ as $(2,4,4), (2,3,6), (2,6,3)$ but this also means $(4,4,2), (4,2,4), (3,2,6), (3,6,2), (6,2,3), (6,3,2)$ are all valid triples for this equation.
If I let $x=3$, the only different values of $y$ and $z$ are $(3,3)$
How do I prove these are the only ten solutions? (without using any programming)
Known result: If we denote $d(n^2)$ as the number of divisors of $n^2$, then the number of solutions of ${\frac {1} {x} }+{\frac {1} {y} } = {\frac {1} {n} }$ = $d(n^2)$ (For positive $x$, $y$)
For ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$
$z = \frac{xy}{y(x-1)-x}$ where $xy \neq 0$
What happens after that?
Question is: how do we sho there are the only ten solutions? I'm not asking for a solution.
Assuming $x \le y \le z$
$1 \le y \le \frac{xy}{y(x-1)-x}$
$\Longrightarrow 1 \le x \le y \le \frac{2x}{x-1} $
Got the answer. I'll probably call @mathlove's answer. (Any additional answers I'll view later)
Liked @user44197 answer.
| There are 14 possible answers up to order for the 4-term analog of this problem.
2,3,7,42
2,3,8,24
2,3,9,18
2,3,10,15
2,3,12,12
2,4,5,20
2,4,6,12
2,4,8,8
2,5,5,10
2,6,6,6
3,3,4,12
3,3,6,6
3,4,4,6
4,4,4,4
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$ Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$
how to do it? I think it should be $3^6$ since $(3x^2)^6=3^6x^8$. (this is false)
Is this true?
| Notice
$$\frac{1}{(1+x)^2} = 1 -2x+3x^2-4x^3+5x^4-6x^5+7x^6-8x^7 + 9x^8 + O(x^9)$$
So the coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$ is the same as the
one in
$$\left[\frac{1}{(1+x)^2} + \big(8x^7 - 9x^8\big)\right]^6
= \frac{1}{(1+x)^{12}} + \binom{6}{1}\frac{8x^7-9x^8}{(1+x)^{10}} + O(x^9)
$$
Above equality is true because when we expand LHS by binomial theorem, terms containing factor $(8x^7-9x^8)^k$
is of the order $o(x^{13})$ for $k > 1$. Since
$$\frac{1}{(1+x)^{\alpha}} = \sum_{n=0}^{\infty} (-1)^n \binom{\alpha+n-1}{n}x^n$$
The coefficient we want is
$$(-1)^8\binom{12+8-1}{8} + 6\times\left[8 \times (-1)^1\binom{10+1-1}{1} - 9\right]\\
= \binom{19}{8} - 6\times 89 = 75582 - 534 = 75048 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Density of the sum of $n$ uniform(0,1) distributed random variables I am working on the following problem:
Let $X_1, X_2, \ldots, X_n, \ldots$ be iid. random variables, each of Uniform$(0,1)$ distribution. Denote by $f_n(x)$ the density of the random variable $S_n := \sum_{k = 1}^n X_k$. Then
\begin{align*}
f_n(x) = \frac{1}{(n-1)!} \sum_{k = 0}^{[x]} (-1)^k \binom{n}{k} (x-k)^{n-1},
\end{align*}
where $[x]$ denotes the floor function.
I tried this, but I think something went wrong:
Prove by induction. For $n = 1$
\begin{align*}
f_1(x) = \frac{1}{(1-1)!} \sum_{k = 0}^{[x]} (-1)^k \binom{1}{k} (x-k)^{1-1} &= \sum_{k = 0}^{[x]} (-1)^k \binom{1}{k}
=\begin{cases}
1, & 0 \le x < 1 \\
0, & \text{otherwise}
\end{cases}
\end{align*}
Furthermore
\begin{align*}
f_{n+1}(x) &= \mathbb P(S_{n+1} = x) = \mathbb P(S_n + X_{n+1} = x) \\
&= \sum_{m = 0}^{\infty} \mathbb P(S_n + X_{n+1} = x \mid X_{n+1} = m) \mathbb P(X_{n+1} = m) \\
&= \sum_{m = 0}^{\infty} \mathbb P(S_n + m = x) \cdot 1_{[0,1]}(m)
= \mathbb P(S_n = x) + \mathbb P(S_n = x-1) \\
&= \frac{1}{(n-1)!} \sum_{k = 0}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \frac{1}{(n-1)!} \sum_{k = 0}^{[x-1]} (-1)^k\binom{n}{k}(x-1-k)^{n-1} \\
&= \frac{1}{(n-1)!}\left(\sum_{k = 0}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \sum_{k = 0}^{[x]-1} (-1)^k\binom{n}{k}(x-1-k)^{n-1}\right) \\
&= \frac{1}{(n-1)!}\left(x^{n-1}+\sum_{k = 1}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \sum_{k = 1}^{[x]} (-1)^{k-1}\binom{n}{k-1}(x-1-(k-1))^{n-1}\right) \\
&= \frac{1}{(n-1)!}\left(x^{n-1}+\sum_{k = 1}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} - (-1)^{k}\binom{n}{k-1}(x-k)^{n-1}\right).
\end{align*}
How can I solve this?
Edit:
\begin{align*}
f_{n+1}(x) &= (f_n * f_{X_{n+1}})(x)
= \int_{-\infty}^\infty f_{n}(y) f_{X_{n+1}}(x-y) \, dy \\
&= \int_{-\infty}^\infty \frac{1}{(n-1)!} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \cdot 1_{(0,1)}(x-y) \, dy \\
&= \frac{1}{(n-1)!} \int_{-\infty}^{\infty} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \cdot 1_{(x-1, x)}(y)\, dy \\
&= \frac{1}{(n-1)!} \int_{x-1}^{x} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \, dy
\end{align*}
I want to swap the sum and the integral, but the sum depends on $y$.
| $\int f_{n}\left(y\right)f_{1}\left(x-y\right)dy=\int\frac{1}{\left(n-1\right)!}\sum_{k=0}^{\left\lfloor y\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\left(y-k\right)^{n-1}1_{\left(0,1\right)}\left(x-y\right)dy$
$=\frac{1}{\left(n-1\right)!}\sum_{k=0}^{\left\lfloor x-1\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\int_{x-1}^{\left\lfloor x\right\rfloor }\left(y-k\right)^{n-1}dy+\frac{1}{\left(n-1\right)!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\int_{\left\lfloor x\right\rfloor }^{x}\left(y-k\right)^{n-1}dy$
$=\frac{1}{n!}\sum_{k=0}^{\left\lfloor x-1\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\left[\left(\left\lfloor x\right\rfloor -k\right)^{n}-\left(x-1-k\right)^{n}\right]+\frac{1}{n!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\left[\left(x-k\right)^{n}-\left(\left\lfloor x\right\rfloor -k\right)^{n}\right]$
$=\frac{1}{n!}\sum_{k=0}^{\left\lfloor x-1\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\left[\left(x-k\right)^{n}-\left(x-1-k\right)^{n}\right]+\left(-1\right)^{\left\lfloor x\right\rfloor }\binom{n}{\left\lfloor x\right\rfloor }\left(x-\left\lfloor x\right\rfloor \right)^{n}$
$=\frac{1}{n!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\left(x-k\right)^{n}+\frac{1}{n!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\binom{n}{k-1}\left(x-k\right)^{n}$
(here $\binom{n}{-1}:=0$)
$=\frac{1}{n!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\left[\binom{n}{k}+\binom{n}{k-1}\right]\left(x-k\right)^{n}$
$=\frac{1}{n!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\binom{n+1}{k}\left(x-k\right)^{n}$
pfffffff.....In my language: een echte 'rotsom'.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $a,b,c\in(0;+\infty)$, prove that $\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$. If $a,b,c\in(0;+\infty)$ and $$\frac{c}{1+a+b}+\frac{a}{1+b+c}+\frac{b}{1+c+a}\ge\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ca}{1+c+a}$$Prove that $$\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$
I know that $$a^2+b^2+c^2\ge\frac{(a+b+c)^2}{3}\ge ab+bc+ac$$
So we could prove that $$a+b+c+3\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$
I.e. $$2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\le a+b+c+3$$
By using AM-GM we see that we could prove that: $$2(a+b+c)\le a+b+c+3\Rightarrow a+b+c\le3$$
So this didn't work.
| Here are some thought I have, but no proof (yet):
Both the condition and the inequality to be proven have equality when $(a,b,c)=(x,x,x)$ and when $(a,b,c)=(1,1,4)$ (or permutations). We know that the inequalities of means only have equality when the term we apply them to are equal. Thus, we can never use AM-GM on $a$, $b$ and $c$, because $(1,1,4)$ wouldn't give equality and thus the inequality is to lossy.
If we apply AM-GM on $f(a,b,c)$, $g(a,b,c)$ and $h(a,b,c)$, we want $f(1,1,4)=g(1,1,4)=h(1,1,4)$, $f(1,4,1)=g(1,4,1)=h(1,4,1)$ and the same for $(1,1,4)$. I think it is not easy to find functions that satisfy this (apart from $f=g=h$, but then, you won't get any information from the inequality).
We can thus only use the given inequality and use rewriting of terms. Maybe we can also use Cauchy-Schwarz, because it has not-so-trivial cases for equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$3$ never divides $n^2+1$ Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.
Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.
If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.
And that is where I am stuck. I try to plug in numbers for $n$ but I want a more general form of showing that $3$ can't divide $n^2+1$ when $n$ is even.
| If $3$ divides $n^2+1$ then it must have solution modulo $3$. But clearly
$0^2+1\equiv 1 \pmod 3$
$1^2+1 \equiv 2 \pmod 3$
$2^2+1 \equiv 5 \equiv 2 \pmod 3$
Otherwise put $n=3k,3k+1,3k+2$ and see that $3$ never divides it
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 2
} |
proving the inequality $\triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$ If $\triangle$ be the area of $\triangle ABC$ with side lengths $a,b,c$. Then show that $\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$
and also show that equality hold if $a=b=c$.
$\bf{My\; Try}::$ Here we have to prove $4\triangle\leq \sqrt{(a+b+c)\cdot abc}$
Using the formula $$\triangle = \sqrt{s(s-a)(s-b)(s-c)},$$ where $$2s=(a+b+c)$$
So $$4\triangle = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{(a+b+c)\cdot(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$
Now using $\bf{A.M\geq G.M}$ for $(b+c-a)\;,(c+a-b)\;,(a+b-c)>0$
$$\displaystyle \frac{(b+c-a)+(c+a-b)+(a+b-c)}{3}\geq \sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$
So we get $\displaystyle (a+b+c)\geq 3\sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$
But I did not understand how can I prove above inequality
help Required
Thanks
| We know
$$S=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4}.$$
So, all we need is to prove the following :
$$(-a+b+c)(a-b+c)(a+b-c)\le abc.$$
This is a well known inequality. Proof is here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Prove that if $10|A$ then $100|A$. For : $A=a^2+ab+b^2$ with $a,b \in \mathbb{N}$.
We known that $10|A$ prove that $100|A$.
| We show that $4 \mid A$ and $25 \mid A$.
Suppose $a^2+ab+b^2 \equiv 0 \bmod 2$. As $a \equiv 1 \equiv b$ is not a solution, at least one of them is $\equiv 0$, forcing the other to be $\equiv 0$. So $2\mid a,b$, showing $4\mid a^2+ab+b^2 = A$.
Suppose $a^2+ab+b^2 \equiv 0 \bmod 5$. Then $(a+3b)^2 \equiv a^2+ab+4b^2\equiv 3b^2 \bmod 5$. But $3$ is not a quadratic residue $\bmod 5$, hence $b\equiv 0$ showing $a\equiv 0$. Thus $25 \mid a^2+ab+b^2=A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/622243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
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