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Solve the equation $\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0$
Solve the equation
$$
\lfloor x^2\rfloor-3\lfloor x \rfloor +2=0
$$
where $\lfloor x\rfloor $ denotes floor function.
My Attempt:
Let $x = n+f$, where $n= \lfloor x \rfloor \in \mathbb{Z} $, $f=x-\lfloor x \rfloor = \{x\} $, and $0\leq f<1$. Then using $\lfloor n+f \rfloor = n$ gives us
$$
\begin{align}
\lfloor (n+f)^2 \rfloor -3 \lfloor n+f \rfloor +2 &= 0\\
n^2+\lfloor f^2+2nf \rfloor -3n+2 &= 0
\end{align}
$$
How can I solve the equation from here?
| If $x$ is positive, then it must be at most $4,$ so $\lfloor x \rfloor$ is at most $3.$ If $\lfloor x \rfloor = 0,$ then your equation is $x^2 - 3 x +2 = 0,$ so $(x-1)(x-2) = 0,$ so no solutions. If $\lfloor x \rfloor = 1,$ then if $y$ is the fractional part of $x,$ we have:
$\lfloor (y+1)^2 \rfloor - 1 = 0,$ so $\lfloor (y+1)^2 \rfloor = 1$, so any $y < \sqrt{2} - 1$ works. If $\lfloor x \rfloor = 2,$ then $\lfloor (y+2)^2\rfloor = 4,$ so any $y < \sqrt{5} - 1$ works. You get the picture.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/590388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
} |
Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$ Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$
I know the answer is $2^n - 1$, but how to simplify it?
| let $S= 1+x+x^2+.......+x^{n-1}$ where $x\ne1$
Multiply this equation by $x$ on both sides and get:
$xS= x+x^2+...........+x^{n}$
now subtract the second equality from the first one and you will end up with:
$(1-x)S= 1+x+x^2+.....+x^{n-1}-x-x^2-.......-x^n$
simplify:
$(1-x)S=1-x^n$ which implies that $S=\frac{1-x^n}{1-x} $
when $x=2$; $S=\frac{1-2^n}{1-2}=2^n-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/590733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this inequality $a+b+\sqrt{a^2+b^2}\ge 2(m+n+\sqrt{2mn})$ let $m,n$ is give positive numbers,and such
$$\dfrac{m}{a}+\dfrac{n}{b}=1$$
show that
$$a+b+\sqrt{a^2+b^2}\ge 2(m+n+\sqrt{2mn})$$
use this methods:How to find the minimum of $a+b+\sqrt{a^2+b^2}$
let
$$a=\dfrac{m(x+y)}{x}.b=\dfrac{n(x+y)}{y}$$
then
we have
$$a+b+\sqrt{a^2+b^2}- 2(m+n+\sqrt{2mn})=\cdots\ge 0?$$
so I think this is have AM-GM and Cauchy-Schwarz inequality solve it，Thank you
| let $$a=\dfrac{m(x+y)}{x},b=\dfrac{n(x+y)}{y}$$
then
$$\Longleftrightarrow \dfrac{m(x+y)}{x}+\dfrac{n(x+y)}{y}+(x+y)\sqrt{\dfrac{m^2}{x^2}+\dfrac{n^2}{y^2}}\ge 2(m+n+\sqrt{2mn})$$
$$\Longleftrightarrow (x+y)(my+nx+\sqrt{(my)^2+(nx)^2})\ge 2xy(m+n+\sqrt{2mn})$$
$$\Longleftrightarrow(x+y)\sqrt{m^2y^2+n^2x^2}\ge (x-y)(my-nx)+2xy\sqrt{2mn}$$
note
$$(x+y)^2(m^2y^2+n^2x^2)-((x-y)(my-nx)+2xy\sqrt{2mn})^2=xy(2my-2nx-\sqrt{2mn}(x-y))^2\ge 0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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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 $n^2$.
| First observe that the sum converges (by, say, the root test).
We already know that $\displaystyle R := \sum_{n=0}^\infty \frac{1}{2^n} = 2$.
Let $S$ be the given sum. Then $\displaystyle S = 2S - S = \sum_{n=0}^\infty \frac{2n+1}{2^n}$.
Now use the same trick to compute $\displaystyle T := \sum_{n=0}^\infty \frac{n}{2^n}$: we have $\displaystyle T = 2T - T = \sum_{n=0}^\infty \frac{1}{2^n} = R = 2$. Hence $S = 2T + R = 6$.
One can continue like this to compute $\displaystyle X:= \sum_{n=0}^\infty \frac{n^3}{2^n}$. We have $\displaystyle X = 2X-X = \sum_{n=0}^\infty \frac{3n^2+3n+1}{2^n} = 3S+3T+R = 26$. Sums with larger powers can be computed in the same way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/593996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
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"answer_id": 0
} |
solving a trigonometric identity: $\frac{\sec^2 x + 2 \tan x}{1 + \tan x}=1+\tan x$. $$\dfrac{\sec^2 x + 2 \tan x}{1 + \tan x} = 1 +\tan x$$ I started by making $1/\cos^2 x$ the
$2\tan x/\cos x$ making them but I can't came the the right answer.
| We use the identity $$\sec^2 x = 1 + \tan^2 x$$
This gives us $$\dfrac{\sec^2 x + 2 \tan x}{1 + \tan x} = \dfrac{\overbrace{(1 + \tan^2 x )+ 2\tan x}^{(1 + \tan x)^2}}{1 + \tan x} = \dfrac{(1 + \tan x)^2}{1 + \tan x} = 1 +\tan x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/598264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show $(1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots)^2 = 1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49} + \cdots$ Last month I was calculating $\displaystyle \int_0^\infty \frac{1}{1+x^4}\, dx$ when I stumbled on the surprising identity:
$$\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right) = \frac{\pi}{\sqrt8}$$
and I knew
$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}$$
So if I could find a proof that $$\left(\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right)\right)^2 = \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$ then this could be a new proof that $\zeta(2)=\frac{\pi^2}{6}$. I've thought over this for almost a month and I'm no closer on showing this identity.
Note: Article on the multiplication of conditionally convergent series: http://www.jstor.org/stable/2369519
| Let's have a try.
$$\sum_{n=0}^{+\infty}\frac{(-1)^n}{4n+1}=\int_{0}^{1}\frac{dx}{1+x^4},\qquad S=\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right)=\int_{0}^{1}\frac{1+x^2}{1+x^4}dx,$$
$$ S = \int_{0}^{1}\frac{x+x^{-1}}{x^{-2}+x^2}\frac{dx}{x}=\int_{1}^{+\infty}\frac{z}{(2z^2-1)\sqrt{1-z^2}}\,dz = \int_{0}^{1}\frac{dt}{(2-t^2)\sqrt{1-t^2}},$$
$$ S = \int_{0}^{\pi/2}\frac{d\theta}{2-\sin^2\theta}=\int_{0}^{\pi/2}\frac{d\theta}{1+\cos^2\theta}=\frac{1}{2}\int_{\mathbb{R}}\frac{du}{2+u^2},$$
where in the last integral we used the substitution $\theta=\arctan u$. This gives:
$$ S^2 = \frac{1}{8}\int_{\mathbb{R}^2}\frac{du\,dv}{(1+u^2)(1+v^2)}=\int_{0}^{1}\int_{0}^{+\infty}\frac{1}{(1+z^2)(1+x^2)}dx\,dz$$
On the other hand,
$$\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\int_{0}^{1}\frac{\log y}{y^2-1}dy=\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2y^2)}dx\,dy,$$
where I learned the last equality from the Mike Spivey's note on the Luigi Pace's proof of $\zeta(2)=\frac{\pi^2}{6}$, just here.
By setting $y=\frac{z}{x}$ in the last integral we get $S^2=\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}$, QED.
So it looks like @user17762's proof-by-squaring-the-arctangent-series and Pace's proof can be combined in order to get a very short proof of your claim.
For the sake of exposing a one-line-proof of $\zeta(2)=\frac{\pi^2}{6}$:
$$\zeta(2)=\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy=\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2 y^2}{1+x^2}\right)\right]_{x=0}^{+\infty}dy=\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2 y^2)}dx\,dy=\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{dx\, dz}{(1+x^2)(1+z^2)}=\frac{4}{3}\cdot\frac{\pi}{4}\cdot\frac{\pi}{2}=\frac{\pi^2}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/598995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
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} |
Indefinite Integral Inverse Trigonometric Function $$\int \frac{2}{9w^2+25}dw$$
I already know this will be equal to $\frac{1}{a} \arctan(x/a)$,
but I don't know how to factor out the $9$. I only know how to take out the $2$.
| $$
\begin{align}
& \phantom{={}} 2\int \frac{dw}{9w^2+25} = 2\int\frac{dw}{25(\frac{9}{25} w^2 + 1)} = \frac{2}{25}\int\frac{dw}{\left(\frac35 w\right)^2 + 1} \\[12pt]
& = \frac{2}{3\cdot5} \int \frac{\frac35\, dw}{\left(\frac35 w\right)^2 + 1} = \frac{2}{15} \int \frac{du}{u^2+1}\text{ etc.}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/600714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Prove $\sin^2\theta + \cos^2\theta = 1$ How do you prove the following trigonometric identity: $$ \sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
| Here is a nice proof by squaring the infinite Taylor series of $\sin x$ and $\cos x$.
The proof:
$$\sin x=\frac{x}{1}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\\\\\sin^2x=x^2-x^4\left (\frac{1}{1!3!}+\frac{1}{3!1!}\right )+x^6\left (\frac{1}{1!5!}+\frac{1}{3!3!}+\frac{1}{5!1!}\right )-...\\\\\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\\\\\cos^2x\!=\!1\!-\!x^2\left(\!\frac{1}{0!2!}\!+\!\frac{1}{2!0!}\!\right)\!+\!x^4\left(\!\frac{1}{0!4!}\!+\!\frac{1}{2!2!}\!+\!\frac{1}{4!0!}\!\right)\!-\!x^6\left(\!\frac{1}{0!6!}\!+\!\frac{1}{2!4!}\!+\!\frac{1}{4!2!}\!+\!\frac{1}{6!0!}\!\right)\!+...$$We should have shown that the series for both $\sin x$ and $\cos x$ converge absolutely (since we changed the arrangement), but it's obvious since the absolute value of all terms of $\sin x+\cos x$ add up to $e^x$.$$\sin^2x+\cos^2x=\\=1-x^2\left(\frac{1}{0!2!}-\frac{1}{1!1!}+\frac{1}{2!0!}\right)+x^4\left(\frac{1}{0!4!}-\frac{1}{1!3!}+\frac{1}{2!2!}-\frac{1}{3!1!}+\frac{1}{4!0!}\right)-x^6\left(\frac{1}{0!6!}-\frac{1}{1!5!}+\frac{1}{2!4!}-\frac{1}{3!3!}+\frac{1}{4!2!}-\frac{1}{5!1!}+\frac{1}{6!0!}\right)+...=\\\\=1+\sum_{n=1}^{\infty}(-1)^nx^{2n}\sum_{k=0}^{2n}\frac{(-1)^k\binom{2n}{k}}{(2n)!}$$
Since we can show easily that $\sum_{i=0}^n(-1)^i\binom{n}{i}=0$ by expanding $(1-1)^n$ using Binom's formula. So:$$\sin^2x+\cos^2x=1-0+0-0+...=1$$
I think it's beautiful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/607103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
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"answer_id": 6
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How many terms of the progression $3,6,9,12,\dots$ must be taken to have a sum not less than $2000$? How many terms of the progression $3,6,9,12,\dots$ must be taken to have a sum not less than $2000$?
I tried the calculation, say $n$ terms, as they are in AP so are they want this?
${n\over 2}[2\times 3 (n-1)3]\ge 2000$?
but how to find $n$ then?
| The progression $3,6,9,12, \ldots $ is a sequence $(a_n)_{n \in \mathbb{N}}$, $a_n = 3n$. The sum of this sequence can be written as $$\sum\limits_{i=1}^{n} 3i = 3\cdot\sum_{i=1}^{n}i$$
And we have the sum of a sequence formula that states $\sum\limits_{i=1}^{n}i = \frac{n(n+1)}{2}$. It follows that $$\sum_{i=1}^{n}3i = 3\left(\frac{n(n+1)}{2}\right) = \frac{3n^2+3n}{2}$$.
Now, you want $\frac{3n^2+3n}{2} \geq 2000$. So we'll find the value of $n$ such that $\frac{3n^2+3n}{2} =2000$. Any larger value of $n$ will obviously give you a value greater than $2000$.
$$\frac{3n^2+3n}{2} = 2000$$
$$\Rightarrow 1.5n^2+1.5n=2000$$
$$1.5n^2+1.5n-2000=0$$
Solve the quadratic equation.
$$n = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$n = \frac{-1.5\pm\sqrt{2.25- (-12000)}}{3}$$
$$n = \frac{-1.5\pm 109.5547808}{3}$$
So $n=36.02\ldots$ or $n=-37.02\ldots$. Discard the negative value, as you can't have a series with a negative number of terms. We also need to round up to get an integer term, and we can't round down because we found the minimum $n$ to be $36+\epsilon$, so $36<n$, so $36$ will not produce a satisfactory sum. So $n=37$. We can test it:
$$\frac{3(37^2)+3(37)}{2} = 2109$$
$$\frac{3(36^2)+3(36)}{2} = 1998$$
And that confirms that $37$ is indeed the minimum term to have a satisfactory sum: our progression must have at least $37$ terms. (I.e. our progression will be $3(1),3(2),\ldots,3(37)$, which is $3,6,9,12,\ldots 111$.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Get the last two digits of $16^{100}$ and $17^{100}$ I try to get the last two digits of $16^{100}$ and $17^{100}$
I started with:
$6^{1} = 6$
$6^{2} = 36$
$6^{3} = 216$ means last digit for $6$ is always $6$
That is why the last digit for $16^{100}$ is $6$
And for $17^{100}$ i calculated:
$7^{1} = 7$
$7^{2} = 49$
$7^{3} = 343$
$7^{4} = 2401$
$7^{5} = 16807$
And because $4*25 = 100$ or better said $100mod4 = 0$
The last digit for $17^{100}$ is $1$
But how do i get the penultimate digits? Thanks
| Another way for $16$
We have already found $\displaystyle16^{100}=2^{400}$
$\displaystyle2^{10}=1024\equiv24\pmod{100}\implies 2^{20}\equiv24^2=576\equiv76$
Now, $\displaystyle2^{20a+b}-2^b=2^b(2^{20a}-1)\equiv0\pmod{100}$
if $4$ divides $2^b\iff$ integer $b\ge2$
and $2^{20a}-1$ is divisible by $2^{20}-1$ for any positive integer $a$ and $2^{20}\equiv1\pmod{25}$ by the other answer
So, $\displaystyle2^{20a+b}\equiv2^b\pmod{100}$ if integer $b\ge2,a\ge1$
Setting $\displaystyle b=20\implies2^{20(a+1)}\equiv2^{20}\pmod{100}\equiv76$
Now, set $a+1=20$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/607829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Which is larger :: $y!$ or $x^y$, for numbers $x,y$. This is a generalization of this question :: Which is larger? $20!$ or $2^{40}$?.
No explicit general solution was presented there and I'm just curious :D
Thank-you.
Edit :: I want a most-general solution lfor arbitrary $x$ and $y$; not some specfic cases which can then be solved by direct computation. Below, Ahaan-Rungta shows that the case $x < y$ is the one to be considered.
| $$\frac{20!}{2^{40}}= \frac{20!}{4^{20}} =
\left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right)
\left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)\cdots \left(\frac{20}{4}\right)>\cdots$$
$$ \left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right)
\left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)=\frac{9!}{4^9}=\frac{2835}{2048}>1,$$
so $20!>2^{40}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/609181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is $\lim_{x\to0} \frac{(\cos x + \cos 2x + \dots+ \cos nx - n)}{\sin x^2}$? What is the limit of $$\lim_{x\to0} \frac{\cos x + \cos 2x + \dots+ \cos nx - n}{\sin x^2}$$
| We have
$$\sum_{k=1}^n \cos(kx) = \dfrac{\sin(nx/2) \cos((n+1)x/2)}{\sin(x/2)}$$
Hence, we have
$$\dfrac{\dfrac{\sin(nx/2) \cos((n+1)x/2)}{\sin(x/2)} - n}{\sin(x^2)} = \dfrac{\sin(nx/2) \cos((n+1)x/2) - n\sin(x/2)}{\sin(x^2)\sin(x/2)}$$
Expanding around $0$ gives us
$$\dfrac{\left(nx/2 - \dfrac{(nx/2)^3}{3!} + \mathcal{O}(x^5)\right) \left(1-\dfrac{(n+1)^2x^2/4}{2!} + \mathcal{O}(x^4)\right)-n\left(x/2 - \dfrac{(x/2)^3}{3!} + \mathcal{O}(x^5) \right)}{x^3/2 + \mathcal{O}(x^4)}$$
Now cancel off the terms and conclude the limit.
| {
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"url": "https://math.stackexchange.com/questions/610729",
"timestamp": "2023-03-29T00:00:00",
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Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$ In the following thread
I arrived at the following result
$$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$
Defining
$$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, H_k^{(1)}\equiv H_k $$
But, it was after long evaluations and considering many variations of product of polylogarithm integrals.
I think there is an easier approach to get the solution, any ideas ?
| We are going to evaluate our sum by establishing a system of two relations.
Lets establish the first relation and using the derivative of beta function ( see here) , we have
$$-\int_0^1x^{n-1}\ln^3(1-x)\ dx=\frac{H_n^3}{n}+3\frac{H_nH_n^{(2)}}{n}+2\frac{H_n^{(3)}}{n}$$
divide both sides by $n$ then take the sum with respect to $n$, we get
\begin{align}
R_1&=\sum_{n=1}^\infty\frac{H_n^3}{n^2}+3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=-\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty\frac{x^n}{n}\ dx\\
&=\int_0^1\frac{\ln^4(1-x)}{x}\ dx=\int_0^1\frac{\ln^4x}{1-x}\ dx=24\zeta(5)
\end{align}
Then $$\boxed{R_1=\sum_{n=1}^\infty\frac{H_n^3}{n^2}+3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=24\zeta(5)}$$
To get the second relation, we need to use the sterling number formula ( check here)
$$ \frac{\ln^k(1-x)}{k!}=\sum_{n=k}^\infty(-1)^k \begin{bmatrix} n \\ k \end{bmatrix}\frac{x^n}{n!}$$
letting $k=4$ and using $\displaystyle\begin{bmatrix} n \\ 4 \end{bmatrix}=\frac{1}{3!}(n-1)!\left[\left(H_{n-1}\right)^3-3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)}\right],$ we get $$\frac14\ln^4(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
differentiate both sides with respect to $x$, we get
$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
multiply both sides by $\ln x/x$ then integrate with respect to $x$, we get
\begin{align}
R_2&=\sum_{n=1}^\infty\frac{H_n^3}{n^2}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}\\
&=\int_0^1\frac{\ln^3(1-x)\ln x}{x(1-x)}\ dx=\int_0^1\frac{\ln^3x\ln(1-x)}{x(1-x)}\ dx\\
&=-\sum_{n=1}^\infty H_n\int_0^1x^{n-1}\ln^3x\ dx=6\sum_{n=1}^\infty\frac{H_n}{n^4}=18\zeta(5)-6\zeta(2)\zeta(3)
\end{align}
Then $$\boxed{R_2=\sum_{n=1}^\infty\frac{H_n^3}{n^2}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=18\zeta(5)-6\zeta(2)\zeta(3)}$$
now we are ready to calculate our sum:
$$R_1-R_2=6\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}=6\zeta(5)+6\zeta(2)\zeta(3)$$
or
$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^2}=\zeta(5)+\zeta(2)\zeta(3)$$
And as a bonus:
$$R_1+R_2=2\sum_{n=1}^\infty\frac{H_n^3}{n^2}=42\zeta(5)-6\zeta(2)\zeta(3)-4\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}$$
using the definition of $H_n^{(3)}$ and the partial fraction decomposition, its easy to prove $$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}2\zeta(5)-2\zeta(2)\zeta(3)$$
which follows
$$\sum_{n=1}^\infty\frac{H_n^3}{n^2}=10\zeta(5)+\zeta(2)\zeta(3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/612181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 6,
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Difficult integral involving tangent and square root functions How do I go about integrating this function:
$$\int{\sqrt{\tan{x}}} \,dx$$
| Subbing $x=\arctan{u^2}$, $dx = (2 u)/(1+u^4) du$ produces
$$2 \int du \frac{u^2}{1+u^4}$$
It turns out that
$$1+u^4=(1+\sqrt{2} u+u^2)(1-\sqrt{2} u+u^2)$$
so that we may invoke partial fractions. The result is that the integral becomes
$$\frac1{\sqrt{2}} \int du \left (\frac{u}{1-\sqrt{2} u+u^2} - \frac{u}{1+\sqrt{2} u+u^2} \right )$$
which is equal to
$$\frac1{\sqrt{2}} \int du \left (\frac{u-\frac1{\sqrt{2}}}{\frac12 + \left (u-\frac1{\sqrt{2}}\right )^2} - \frac{u+\frac1{\sqrt{2}}}{\frac12 + \left (u+\frac1{\sqrt{2}}\right )^2} \right )\\+ \frac12 \int du \left (\frac1{\frac12 + \left (u-\frac1{\sqrt{2}}\right )^2}-\frac1{\frac12 + \left (u+\frac1{\sqrt{2}}\right )^2} \right )$$
which evaluates to
$$\frac{1}{2 \sqrt{2}} \log{\left (\frac{1-\sqrt{2} u+u^2}{1+\sqrt{2} u+u^2} \right )} -\frac1{\sqrt{2}} \arctan{\left (\frac1{u^2} \right )}+C$$
Back substitute $u=\sqrt{\tan{x}}$ and you are done. Note that there is some simplification that may be made by observing that
$$\frac1{\tan{x}}=\tan{\left (\frac{\pi}{2}-x \right )}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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two short doubts about the inverse function in a point the function is $F(x,y,z)=(y^2+z^2, z^2+x^2, x^2+y^2)$ the point is (-1,1,-1)
task: find the local inverse of F in that point.
I have already proved that F is actually invertible there. then i solved the system:
$$
\begin{cases}
y^2+z^2=a\\
z^2+x^2=b\\
x^2+y^2=c
\end{cases}
$$
which gave me
$$
x=\pm \sqrt{\frac{c-a+b}{2}}\\
y=\pm \sqrt{\frac{a-b+c}{2}}\\
z=\pm \sqrt{\frac{b-c+a}{2}}\\
$$
The doubt I have is: how should I choose the signs? I thought of looking at the values of x,y,z in the point, which means x,z=negative y=positive
so i picked
$$
x= -\sqrt{\frac{c-a+b}{2}}\\
y= \sqrt{\frac{a-b+c}{2}}\\
z= -\sqrt{\frac{b-c+a}{2}}\\
$$
finding the inverse G:
$$ G(a,b,c)=(-\sqrt{\frac{c-a+b}{2}},\sqrt{\frac{a-b+c}{2}},-\sqrt{\frac{b-c+a}{2}})$$
the second doubt is:
should i leave it written that way or is it better:
$G(y^2+z^2, z^2+x^2, x^2+y^2)=(-x,y,-z)$ ? (with the same doubt about the signs?)
thank you in advance
| The function $F:\>(x,y,z)\mapsto(a,b,c)$ is smooth, and $F(-1,1,-1)=(2,2,2)$. The Jacobian $J_F(x,y,z)=16xyz$ is $\ne 0$ at $(-1,1,-1)$; therefore "by general principles" $F$ maps any sufficiently small neighborhood $U$ of $(-1,1,-1)$ bijectively onto a neighborhood $V$ of $(2,2,2)$. The "local inverse"
$G: \>V\to U$ is again differentiable and has $G(2,2,2)=(-1,1,-1)$.
In your case it was possible to produce via algebraic manipulations explicit expressions for the possible local inverses, and we now have to choose the right candidate. You have already done that and obtained
$$G(a,b,c)=\left(-\sqrt{\frac{c-a+b}{2}},\sqrt{\frac{a-b+c}{2}},-\sqrt{\frac{b-c+a}{2}}\right)\ .\tag{1}$$
This is the explicit expression of $G$ in terms of its inborn variables $a$, $b$, $c$; you could not hope of something better. Note that when $(a,b,c)=(2,2,2)$ all radicands in $(1)$ become $1$, which implies that $G$ is differentiable in the neighborhood of $(2,2,2)$, as guaranteed by the general implicit function theorem.
| {
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Finding the maximum value of a function on an ellipse Let $x$ and $y$ be real numbers such that $x^2 + 9 y^2-4 x+6 y+4=0$. Find the maximum value of $\displaystyle \frac{4x-9y}{2}$.
My solution: the given function represents an ellipse. Rewriting it, we get $\displaystyle (x-2)^2 + 9(y+\frac{1}{3})^2=1$. To find the maximum of $\displaystyle \frac{4x-9y}{2}$, $x$ should be at its maximum and $y$ at its minimum. Solving the equations, I get that $x = 3$ and $\displaystyle y = -\frac{2}{3}$, but the answer i get is wrong. What am I doing wrong?
| This is a classic application of the so-called Lagrange Multiplier. We want to maximise the value of $2x-\tfrac{9}{2}y$ subject to the constraint $x^2 + 9 y^2-4 x+6 y+4=0$. We define
$$\Lambda(x,y,\lambda) = \left(2x-\tfrac{9}{2}y\right)+\lambda\left(x^2 + 9 y^2-4 x+6 y+4\right)$$
The point(s) you are looking for come from solving $\Lambda_x=\Lambda_y=\Lambda_{\lambda}=0$.
Calculating the three partial derivatives gives:
$$(x,y,\lambda) = \left(\frac{6}{5},-\frac{2}{15},\frac{5}{4}\right) \ \ \text{and} \ \ \left(\frac{14}{5},-\frac{8}{15},-\frac{5}{4}\right)$$
The values $(x,y)=(6/5,-2/15)$ correspond to a minimum, where $2x-\tfrac{9}{2}y=3$.
The values $(x,y)=(14/5,-8/15)$ correspond to a maximum, where $2x-\tfrac{9}{2}y=8$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$ I was solving an exercise, so I realized that the one easiest way to do it is using a "weird", but nice identity below. I've tried to found out it on internet but I've founded nothingness, and I wondering how to show easily this identity ? As a matter of the fact, it's a beautiful identity.
$$-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}$$
The one way I think about this I'll let here :
$$\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=\frac{(a-b)(b+c)(c+a)+(b-c)(c+a)(a+b)+(c-a)(c+b)(b+a)}{(a+b)(b+c)(c+a)}$$
Now, I'll take one of the term of RHS
$(a −b)(b +c)(c + a)\\ =(a −b)(b−c + 2c)(a −c + 2c)\\ =(a −b)(b −c)(a −c)+ 2c(a −b)(b−c + a −c + 2c) \\=(a −b)(b −c)(a −c)+ 2c(a −b)(a +b)$
Similarly, We'll do that with the remainder
$(b−c)(a+b)(c + a)+(c − a)(a +b)(b+ c)\\=(a +b)[(b −c)(c + a)+(c −a)(b +c)]\\=(a +b)(bc +ba −c^{2} −ca +cb+c^{2} −ab −ac)\\= (a +b)(2bc − 2ac)\\= −2c(a −b)(a +b)$
and we get :
$$\boxed{\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}}$$
So it's too many work to show this identity. I just want to know if there's a simple way to show that or I don't know. I'm questing that because I didn't find anything on internet.
| I don't know if there is an easier solution but I think that an alternative solution for a,b,c positive real numbers, can be derived using the law of tangents.
http://en.wikipedia.org/wiki/Law_of_tangents
Of course this will not cover all cases since for a,b,c lengths of sides of a triangle we have $a+b>c$, $b+c>a$, $c+a>b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "20",
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Prove that $A + B = C$ I drew the diagram here
I honestly do not see how $A$ and $B$ could possibly equal $C$.
| $\sin A = \dfrac{1}{\sqrt{10}}$, $\cos A = \dfrac{3}{\sqrt{10}}$, $\sin B = \dfrac{1}{\sqrt5}$, $\cos B = \dfrac{2}{\sqrt5}$, $\sin C = \cos C = \dfrac1{\sqrt2}$.
$\sin(A + B)= \sin A \cos B + \cos A \sin B = \dfrac2{5\sqrt{2}} + \dfrac3{5\sqrt{2}} = \dfrac1{\sqrt{2}}= \sin C$
$\cos(A + B) = \cos A \cos B - \sin A \sin B = \dfrac6{5\sqrt{2}} - \dfrac1{5\sqrt{2}} = \dfrac1{\sqrt{2}} = \cos C$
Since their sines and cosines are equal, $C$ and $A + B$ are congruent modulo $2\pi$. Since also $ 0 \lt A \lt B \lt C \lt \dfrac\pi{2}$, $-\dfrac{\pi}2 \lt C - (A + B) \lt \dfrac{\pi}2$, $A + B$ and $C$ must be equal.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$(1+1/x)(1+1/y)(1+1/z) = 3$ Find all possible integer values of $x$, $y$, $z$ given all of them are positive integers. Find all possible integer values of $x$, $y$, $z$ given all of them are positive integers
and
$$(1+1/x)(1+1/y)(1+1/z) = 3.$$
I know
$(x+1)(y+1)(z+1) = 3xyz$ which is no big deal. I can't move forward now.
morever it is given $x$ is less than or equal to $y$ and $y$ is less than or equal to $z$
| Hint: Suppose that $x, y, z \ge 3$. Then $$1 + \frac 1 x \le \frac{4}{3}$$
and likewise for the other two. Then
$$\left(1 + \frac 1 x\right)\left(1 + \frac 1 y\right)\left(1 + \frac 1 z\right) \le \frac{64}{27} < 3$$
So one of the numbers has to be pretty small; now consider cases with $x = 1$ and $x = 2$.
Something else that you know: $3$ is a divisor of the right side, so it's a divisor of the left side. Since $3$ is prime, $3$ has to divide one of the individual terms.
| {
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"timestamp": "2023-03-29T00:00:00",
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remainder when $(x+1)^n$ is divided by $(x-1)^3$, where $n\in \mathbb{N}$ Calculation of remainder when $(x+1)^n$ is divided by $(x-1)^3$, where $n\in \mathbb{N}$
$\bf{My\; Try}::$ Using Division Algorithm:: $p(x) = q(x)\cdot g(x)+r(x)$
Now Let $r(x) = ax^2+bx+c$
So $(x+1)^n=q(x)\cdot (x-1)^3+ax^2+bx+c........................(1)$
Now put $x=1$, we get $\boxed{a+b+c=2^n}$
Now differentitae $(1)$, we get $n(x+1)^{n-1} = q(x)\cdot 3(x-1)^2+(x-1)^3\cdot q^{'}(x)+2ax+b$
again put $x=1$, we get $\boxed{2a+b=n(2)^{n-1}}$
Now again differentitae $(1)$ and then put $x=1$, we get
$\displaystyle \boxed{2a=n(n-1)2^{n-2}\Rightarrow \displaystyle a=\frac{n(n-1)}{2}\cdot 2^{n-2}}$
Similarly we get $\displaystyle b = n(2)^{n-1}-n(n-1)\cdot 2^{n-2}$
Similarly we get $\displaystyle c= 2^{n}+n(n-1)\cdot 2^{n-2}-\frac{n(n-1)}{2}\cdot 2^{n-2}$
So Remainder
$\displaystyle r(x) = \frac{n(n-1)}{2}2^{n-2}x^2+\left\{n(2)^{n-1}-n(n-1) 2^{n-2}\right\}x+2^{n}+n(n-1)\cdot 2^{n-2}-\frac{n(n-1)}{2}2^{n-2}$
is my solution is right , if not then how can i calculate it
or If there is any better method , then plz explain here
Thanks
| Another way : for integer $n\ge0$
$$(x+1)^n=(x-1+2)^n$$
$$\equiv2^n+\binom n12^{n-1}(x-1)+\binom n22^{n-2}(x-1)^2\left(\text{mod}{(x-1)^3}\right)$$
Clearly, the remainder reduces to $(x+1)^n$ for $0\le n\le2$
| {
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Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $ Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $. I provide the answer below.
| The tag says "induction," so step one: prove that
$$\sum_{i=1}^n i^2=\frac {n(n+1)(2n+1)}6$$
by induction.
Step two: Given that
$$f(n)=\sum_{i=1}^n i^2=\frac {n(n+1)(2n+1)}6$$
we have
$$f(2n-1)=\frac {2n(2n-1)(4n-1)}6=\frac {n(2n-1)(4n-1)}3$$
$$4f(n-1)=2\frac {n(n-1)(2n-1)}3=4\sum_{i=1}^{n-1} i^2=\sum_{i=1}^{n-1} (2i)^2$$
and
$$f(2n-1)-4f(n-1)=\sum_{i=1}^n(2i-1)^2=\frac{n(2n-1)(4n-1)}3-2\frac{n(n-1)(2n-1)}3$$
$$=\frac{(n(4n-1)-2n(n-1))(2n-1)}3=\frac {(4n^2-n-2n^2+2n)(2n-1)}3$$
$$=\frac{(2n^2+n)(2n-1)}3=\frac{4n^3-n}3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all integer solutions: $x^4+x^3+x^2+x=y^2$ Find all integer solutions of the following equation:
$$x^4+x^3+x^2+x=y^2$$
| The equation
$$y^2 = x^4 + x^3 + x^2 + x = x(x+1)(x^2+1)\tag{*1}$$
has four trivial solutions over $\mathbb{Z}\times\mathbb{Z}$:
$$(x,y) = (-1,0), (0, 0),(1,\pm 2)$$
To determine whether there are other non-trivial solutions, let's look at an equivalent problem:
$$y^2 = \begin{cases}
z(z + 1)(z^2 + 1), &\text{ for } z = x \ge 2\\
z(z - 1)(z^2 + 1), &\text{ for } z = -x \ge 2\tag{*2}
\end{cases}$$
Notice
$$\gcd(z(z\pm 1),z^2+1) = \gcd( \pm z - 1, z^2 + 1) = \gcd(\pm z - 1,2) = 1 \text{ or } 2$$
Any solution of $(*2)$ must have one of the following two factorizations:
$$\begin{cases}
z(z\pm 1) &= p^2,\\
z^2 + 1 &= q^2,\\
y &= \pm pq;\end{cases}
\quad\text{ or }\quad
\begin{cases}
z(z\pm 1) &= 2p^2,\\
z^2 + 1 &= 2q^2,\\
y &= \pm 2pq.\end{cases}\quad\text{ with } \gcd(p,q) = 1.$$
We can rule out the factorization on the left because for $z \ge 2$, $z^2 + 1$ is never a square. For the factorization on the right, we can rewrite the $x$ part as
$$\left\{\begin{array}{ccrl}
(2 z \pm 1 )^2 &-& 8p^2 &= 1\\
z^2 &-& 2q^2 &= -1
\end{array}\right.\tag{*3}$$
Define $A, B : \mathbb{Z} \to \mathbb{Z}$ by
$$A(k) + B(k) \sqrt{2} = (1+\sqrt{2})^k$$
It is known the positive integer solutions of Diophantine equations like those in $(*3)$
has the form
$$\left\{\begin{array}{ccl}
2 z\pm 1 &=& A(2m)\\
2p &=& B(2m)\\
z &=& A(2n-1)\\
q &=& B(2n-1)\\
\end{array}\right.
\quad\quad\text{ for } m , n \in \mathbb{Z}_{+}.
$$
It is clear $2 z \pm 1 \ge z \implies 2m \ge 2n-1$.
Notice $\frac{A(k+1)}{A(k)} \to 1+\sqrt{2} > 2$ as $k \to \infty$.
For any $n$ large enough such that $A(2n) > 2 A(2n-1) + 1$, there are no way to find
a $m$ to satisfy $(*3)$. A brute force computation shows that this
happens as early as $n \ge 2$.
This leaves us the cases $n = 1$. It is easy to check
for $n = 1$, one can choose $m = 1$ to make $(*3)$ works. However,
$$
m = n = 1 \quad\implies\quad
\left\{\begin{array}{ccl}
2 z\pm 1 &=& A(2) = 3 \\
2p &=& B(2) = 2\\
z &=& A(1) = 1\\
q &=& B(1) = 1\\
\end{array}\right.
\quad\iff\quad
(x,y) = (1,\pm 2)
$$
corresponds to a pair of trivial solutions we have covered before. The conclusion is aside from the four trivial solutions, $(*1)$ doesn't have any non-trivial solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the following equation system using Gaussian Elimination. Solve the following equation system using Gaussian Elimination.
$x_1+2x_2-x_3-x_4+x_5=0$
$x_1+2x_2-2x_4+4x_5=0$
$2x_1+4x_2-2x_3-2x_4+2x_5=0$
$-2x_1-4x_2+4x_3+4x_5=0$
My working so far
Putting the equation system into a coefficient matrix:
$$ \left[
\begin{array}{ccccc|c}
1&2&-1&-1&1&0\\
1&2&0&-2&4&0\\
2&4&-2&-2&2&0\\
-2&-4&4&0&4&0
\end{array}
\right]$$
Using Gaussian Elimination, the matrix reduces to:
$$ \left[
\begin{array}{ccccc|c}
1&2&-1&-1&1&0\\
0&0&1&-1&3&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0
\end{array}
\right]$$
So the equation system is:
$x_1+2x_2-x_3-x_4+x_5=0$
$x_3-x_4+3x_5=0$
This is the point I am up to. I have $2$ equations in $5$ variables, so I will need free variables/parameters. But what do I need to do to get these and have a general solution?
Thanks in advance.
| Note in all combinations the last column will stay zero so we will just drop it.
$$
\begin{pmatrix}
1 & 2 & -1 & -1 & 1 \\
1 & 2 & 0 & -2 & 4 \\
2 & 4 & -2 & -2 & 2 \\
-2 & -4 & 4 & 0 & 4 \\
\end{pmatrix}
\to
\begin{pmatrix}
1 & 2 & -1 & -1 & 1 \\
0 & 0 & 1 & -1 & 3 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & -1 & 3 \\
\end{pmatrix}
\to
\begin{pmatrix}
1 & 2 & 0 & -2 & 4 \\
0 & 0 & 1 & -1 & 3 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}
$$
Which indeed leaves $x_1+2x_2-2x_4+4x_5=0$ and $x_3 - x_4 + 3x_5=0$. Write $x_1,x_3$ in terms of others, to get $x_1 = -2x_2+2x_4-4x_5$ and $x_3 = x_4 - 3x_5$. Now the general solution is
$$
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix}
= \begin{pmatrix} -2x_2+2x_4-4x_5 \\ x_2 \\ x_4 - 3x_5 \\ x_4 \\ x_5 \end{pmatrix}
=
\begin{pmatrix}
-2 & 2 & -4\\
1 & 0 & 0\\
0 & 1 & -3\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}x_2 \\ x_4 \\ x_5 \end{pmatrix}
$$
where $x_2, x_4, x_5$ can take on any real values. In other words any linear combination of the columns of this matrix will be a solution, and no other vector:
$$
\begin{pmatrix}
-2 & 2 & -4\\
1 & 0 & 0\\
0 & 1 & -3\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}$? I am stuck on this one; Its a sum and don't know how to calculate the denominator.
$$\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}$$
| It should be possible to work out the following carefully, but at any rate, using comparisons with integrals, you should be easily able to see that the limit is zero. I could do it in my head.
We are considering
$$\lim_{n\to\infty} \frac{1}{n+1}\sum_{k=1}^n \frac{k}{1 + \sqrt{2} +\sqrt{3} +\cdots+\sqrt{k+1}}.$$
First, note that
$$\sqrt{1}+\sqrt{2}+\cdots+\sqrt{k} \approx \int_1^k s^{1/2} ds \approx \frac{2}{3} k^{3/2}.$$
Next, we then have approximately
$$\frac{2}{3} \cdot \frac{1}{n+1} \sum_{k=1}^n \frac{k}{k^{3/2}} \approx \frac{2}{3n} \int_1^n \frac{1}{k^{1/2}} dk \approx \frac{4}{3n} n^{1/2}.$$
Clearly this goes to zero as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determinant of a special $n\times n$ matrix Compute the determinant of the nun matrix:
$$
\begin{pmatrix}
2 & 1 & \ldots & 1 \\
1 & 2 & \ldots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\ldots & 2
\end{pmatrix}
$$
For $n=2$, I have$$
\begin{pmatrix}
2 & 1 \\
1 & 2
\end{pmatrix}
$$
Then $det = 3$.
For $n=3$, we have
$$
\begin{pmatrix}
2 & 1 & 1\\
1 & 2 & 1\\
1 & 1 & 2 \\
\end{pmatrix}
$$
Then $det = 4$.
For $n=4$ again we have
$$
\begin{pmatrix}
2 & 1 & 1 & 1 \\
1 & 2 & 1 & 1\\
1 & 1 & 2 & 1\\
1 & 1 & 1 & 2
\end{pmatrix}
$$
Then $det = 5$
How can I prove that the determinant of nun matrix is $n+1$.
| Yet another way to do it: note that the matrices in question are of the form $I + J$, where $J$ is the matrix every entry of which is $1$. We have $J^2 = nJ$ by an easy calculation; thus the eigenvalues of $J$ are $0$ and $n$. The eigenspace corresponding to $n$ is the one dimensional subspace spanned by the vector $v = (1, 1, ... , 1)^T$; since $\ker J$ consists of those vectors $w = (w_1, w_2, . . ., w_n)^T$ with $\sum_1^n w_i = 0$, the eigenspace corresponding to $0$ is of dimension $n - 1$; these observations imply the when $J$, being symmetric, is diagonalized one obtains a matrix with $n$ occurring at precisely one place on the diagonal, and zeroes everywhere else. Thus we see that the multiplicity of $0$ as an eigenvalue is $n - 1$; that of $n$ is $1$. Now use the fact that since $Jx = ax \Leftrightarrow (J + I)x = (a + 1)x$ to see that the eigenvalues of $I + J$ are $n + 1$, of multiplicity $1$, and $1$ of multiplicity $n - 1$. Thus $\det (J + I)$, being the product of these eigenvalues, is $n + 1$.
Hope this helps. Happy New Year,
and as always,
Fiat Lux!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Is it possible to prove that $\lim_{x\to 0} {ln(x)x} = 0$ without L'Hospital's rule? Could somebody give me the answer:
Is it possible to prove that $\lim_{x\to 0} {\ln(x)x} = 0$ without L'Hospital's rule?
| One can also use the following power series, which converge for $-1 < x \leq 1$:
$$\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n$$
Then $x \ln x$ becomes
$$x \ln x = (x-1) \ln x + \ln x = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^{n+1} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n = (x-1) + \sum_{n=1}^\infty \left( \frac{(-1)^{n+1}}{n} + \frac{(-1)^{n+2}}{n+1}\right) (x-1)^{n+1} \to -1 + \sum_{n=1}^\infty \left( \frac{(-1)^{n+1+n+1}}{n} + \frac{(-1)^{n+2+n+1}}{n+1} \right) = -1 + \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) = -1+1 = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/631205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Length of chord in circle
*In Figure 3, arc CD is a semicircle. AB is perpendicular to CD, BC = 3, BD = 4. Then the length of AB =
a) 3.25 b) 4.56 c) 3.46 d) 7.00
| HINT :
Note that $\angle CAD=90^\circ$, and so $AC^2+AD^2=CD^2$.
Letting $AB=x$, since
$$AC^2=x^2+3^2, AD^2=x^2+4^2, CD^2=7^2$$
you'll get an equation of $x$ as
$$7^2=9+x^2+x^2+16\iff x^2=12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/631457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares. If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares.
My work:
$3n+1=x^2$
$3n+3=x^2+2$
$3(n+1)=x^2+2$
$(n+1)=\dfrac{x^2+2}{3}$
I have no clue what to do next. Please help!
| This will give the answer : http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares
You need to prove that $n+1=\frac{x^2+2}{3}$ is not of the form $4^k(8m+7)$ for $k,m\in \mathbb{N}$ and its not that difficult to show this.
For Square modulo 8 we know that $x^2 =0,1,4 \mod{8}$, so $x^2 +2 =2,3,6 \mod{8}$.
Since $x^2 +2 =0 \mod{3}$ we obtain that $\frac{x^2 +2}{3} = 1,2,A \mod{8}$ where $A=\frac{2+8a}{3}$, $a$ positive integer. And $A=\psi \mod{8}$ namely, $2=3\psi \mod{8}$ with $\psi\in\{0,1,...,7\}$. The only possible value is $\psi=6$ hence $\frac{x^2 +2}{3} = 1,2,6 \mod{8}$.
If it was of that particular form then:
For $m=0$,$k=0$ it is equal to 7 modulo 8
For $m=0$,$k=1$ it is equal to 4 modulo 8
For $m=0$,$k=2$ it is equal to 0 modulo 8
For $m>0,k>2$ it is equal to 0 modulo 8
So you obtain a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/633651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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} |
Convergence of $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + ...$ How does one use the comparison test to prove that $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$ converges?
Is the following argument valid?
$\quad 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$
$< 1 + 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^3} + \;... $
$= 2 + 1 + \frac{1}{2} + \frac{1}{2^2} + \;... $
$= \frac{2}{1-1/2} = 4 $
So the series is bounded above and since every term is strictly positive, the sequence of partial sums is monotone increasing and hence the series converges.
| Since you only need to prove it converges, and you don't need to find what it converges to, yes, your argument is concise and valid.
Refer to lab bhattacharjee's answer for the actual value it converges to.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Third order ODE initial value problem,solution obeys $y(x) \rightarrow 0 $ as $x \rightarrow \infty$ ??? $y''' + y'' -y' -y=0$
$y(0)=7,y'(0)=-3,y''(0)=\alpha$
Find all values of $\alpha$ for which the solution obeys $y(x) \rightarrow 0 $ as $x
\rightarrow \infty$
Here is my work
I used the cubic characteristic equation $r^3 +r^2 -r -1 = 0$ and find the three roots to be
$-1,-1$and $1$
So my general solution is $y(x) = Ae^x+Be^{-x}+Cxe^{-x}$
and
$y'(x) = Ae^x-Be^{-x}+Ce^{-x}-Cxe^{-x}$
$y''(x) = Ae^x+Be^{-x}+-2Ce^{-x}+Cxe^{-x}$
Use the given initial values, I get
$y(0) = 7 = A+B$
$y'(0) = -3 = A-B+C$
$y''(0)=\alpha = A+B-2C$
What should I do next to determine all the values of $\alpha$ such that the solution obeys $y(x) \rightarrow 0 $ as $x
\rightarrow \infty$ ???
Thanks in advance for the help
| As $x \to \infty$, $e^x \to \infty$, so we must set $A = 0$ in order to guarantee that $y(x) \to 0$ as $x \to \infty$. Then from $y(0) = 7 = A + B$, we see that $A = 0$ implies that $B = 7$.
From $y'(0) = -3 = A - B + C$, we get $-3 = -B + C = -7 + C$, so $C = 4$.
From $y''(0) = \alpha = B - 2C$, we get $\alpha = 7 - 8 = -1$. Then $\alpha = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/635353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that the equation $x^3+7x-14(n^2+1)$ has no integral root for any integer $n$. Show that the equation
$$x^3+7x-14(n^2+1)=0$$
has no integral root for any integer $n$.
My work:
I consider the contraposition that there are integer roots.
Assume that the roots are $\alpha,\beta,\gamma$
We have, $\alpha\beta\gamma=14(n^2+1)$
We also have, $\alpha\beta+\alpha\gamma+\beta\gamma=7$
Since $n^2\ge 1$, we get, $\alpha\beta\gamma\ge 28$
Now, I have no idea! Please help!
| Consider $x^3+7x-14(n^2+1)\equiv0 \mod (7)$, we get $x^3\equiv0 \mod(7)$ and therefore $x\equiv0 \mod (7)$.
Assume $x=7t$, then $(7t)^3+7\cdot 7t-14(n^2+1)=0$. i.e., $7^2t^3+7t-2(n^2+1)=0$. Similar, $7^2t^3+7t-2(n^2+1)\equiv0\mod (7)$, one has $2(n^2+1)\equiv0 \mod (7)$, which is impossible for integer $n$. So the equation has no integral roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/635938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Prove that the limit definition of the exponential function implies its infinite series definition. Here's the problem: Let $x$ be any real number. Show that
$$
\lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \sum_{n=0}^ \infty \frac{x^n}{n!}
$$
I'm sure there are many ways of pulling this off, but there are 3 very important hints to complete the exercise in the desired manner:
*
*Expand the left side as a finite sum using the Binomial Theorem. Call the summation variable $n$.
*Now add into the finite sum extra terms which are $0$ for $n>m$, in order to make it look like an infinite series.
*What happens to the limit on $m$ outside the series?
So far I was able to use Hint 1 to expand the left side:
$$
\lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \lim_{m \to \infty} \sum_{n=0}^m \binom {m}{n} \left( \frac{x}{m} \right)^n
$$
No matter what I do with the binomial coefficients and factorials, I can't figure out what extra terms to add per Hint 2. Any suggestions?
| The answer from robjohn linked in his comment is a must visit. It deals with the definition of $e^{x}$ as $\lim_{n \to \infty}(1 + 1/n)^{nx}$. Another direct approach starting with $(1 + x/n)^{n}$ is given below.
Let $$F_{n}(x) = 1 + x + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}$$ so that $\lim_{n \to \infty}F_{n}(x) = F(x)$ exist and $F(x)$ is given by the convergent power series $$F(x) = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots = \sum_{k = 0}^{\infty}\frac{x^{k}}{k!}$$ First we assume that $x > 0$. We have to establish that $\lim\limits_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n} = F(x)$
If $n > x$ then we have $$\left(1 + \frac{x}{n}\right)^{n} = 1 + x + \frac{n(n - 1)}{2!}\left(\frac{x}{n}\right)^{2} + \cdots + \frac{n!}{n!}\left(\frac{x}{n}\right)^{n}$$ so that $$\left(1 + \frac{x}{n}\right)^{n} < F_{n}(x)$$
Next consider the expression $(1 - (x/n))^{-n}$. By the binomial theorem for negative exponents we have $$\left(1 - \frac{x}{n}\right)^{-n} = 1 + x + \frac{n(n + 1)}{2!}\left(\frac{x}{n}\right)^{2} + \cdots$$ so that $$\left(1 - \frac{x}{n}\right)^{-n} \geq F_{n}(x)$$ It follows that for $0 < x < n$ we have $$\left(1 + \frac{x}{n}\right)^{n} < F_{n}(x) \leq \left(1 - \frac{x}{n}\right)^{-n}\,\,\,\cdots (1)$$
Now note that $\lim_{n \to \infty}(1 + x/n)^{n} = G(x)$ exists and we have $(1 + (x/n))^{n} < G(x)$ for all $x > 0$ and all positive integers $n$. Therefore
$\displaystyle \begin{aligned}\left(1 - \frac{x}{n}\right)^{-n} - \left(1 + \frac{x}{n}\right)^{n} &= \left(1 + \frac{x}{n}\right)^{n}\left\{\left(1 - \frac{x^{2}}{n^{2}}\right)^{-n} - 1\right\}\\
&< G(x)\left\{\left(1 - \frac{x^{2}}{n}\right)^{-1} - 1\right\} = \frac{x^{2}G(x)}{n - x^{2}}\end{aligned}$
so that the expression $(1 - (x/n))^{-n} - (1 + (x/n))^{n}$ tends to $0$ as $n \to \infty$ (using squeeze theorem). It follows that $$\lim\limits_{n \to \infty}\left(1 - \dfrac{x}{n}\right)^{-n} = \lim\limits_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n} = G(x)\,\,\,\cdots (2)$$
Now applying squeeze theorem on equation $(1)$ and using equation $(2)$ above we get $\lim_{n \to \infty}F_{n}(x) = G(x)$ so that $F(x) = G(x)$ for all $x > 0$.
For $x = 0$ the result is trivial. For $x < 0$ we need to use the fact that $F(x + y) = F(x)F(y)$ (this is done by multiplication of series) and thus $F(-x) = 1/F(x)$. Similarly from equation $(2)$ it follows that $G(-x) = 1/G(x)$. Thus the relation $F(x) = G(x)$ holds for every real $x$. Noting the definitions of $F, G$ we see that $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots $$ for all $x \in \mathbb{R}$.
Update: While providing an answer to another question I came up with an alternative proof of $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = \lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^{-n}$$ which is worth mentioning here.
Let us set $$f(n) = \left(1 + \frac{x}{n}\right)^{n}, g(n) = \left(1 - \frac{x}{n}\right)^{-n}$$ and as in the earlier proof we only need to handle the case for $x > 0$. By the use of binomial theorem we can show that the sequence $f(n)$ increases and is bounded by $F(x)$ so that $f(n)$ tends to a limit as $n \to \infty$ and the limit is greater than or equal to $(1 + x)$.
Now we can see that if $0 < x^{2} < n$ then $$\begin{aligned}\frac{f(n)}{g(n)} &= \left(1 - \frac{x^{2}}{n^{2}}\right)^{n}\\
&= 1 - n\cdot\frac{x^{2}}{n^{2}} + \frac{n(n - 1)}{2!}\left(\frac{x^{2}}{n^{2}}\right)^{2} - \cdots + \\
&= 1 - \frac{x^{2}}{n} + \dfrac{\left(1 - \dfrac{1}{n}\right)}{2!}\left(\frac{x^{2}}{n}\right)^{2} - \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\
&= 1 - \phi(n)\end{aligned}$$ where $\phi(n)$ is a finite sum defined by $$\phi(n) = \frac{x^{2}}{n} - \dfrac{\left(1 - \dfrac{1}{n}\right)}{2!}\left(\frac{x^{2}}{n}\right)^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\left(\frac{x^{2}}{n}\right)^{3} - \cdots$$ We can estimate $\phi(n)$ as follows $$\begin{aligned}0 \leq |\phi(n)| &\leq \frac{x^{2}}{n} + \dfrac{\left(1 - \dfrac{1}{n}\right)}{2!}\left(\frac{x^{2}}{n}\right)^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\
&\leq \frac{x^{2}}{n} + \frac{1}{2!}\left(\frac{x^{2}}{n}\right)^{2} + \frac{1}{3!}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\
&\leq \frac{x^{2}}{n} + \frac{1}{2}\left(\frac{x^{2}}{n}\right)^{2} + \frac{1}{2^{2}}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\
&= \frac{x^{2}}{n}\left\{1 + \frac{x^{2}}{2n} + \left(\frac{x^{2}}{2n}\right)^{2} + \cdots\right\}\\
&= \frac{x^{2}}{n}\cdot\dfrac{1 - \left(\dfrac{x^{2}}{2n}\right)^{n}}{1 - \dfrac{x^{2}}{2n}}\\
&\leq \frac{x^{2}}{n}\cdot\dfrac{1}{1 - \dfrac{x^{2}}{2n}}\to 0 \text{ as }n \to \infty\end{aligned}$$ and then we see that $\phi(n) \to 0$ as $n \to \infty$. It follows that $f(n)/g(n) \to 1$ as $n \to \infty$ and therefore $\lim_{n \to \infty}f(n) = \lim_{n \to \infty}g(n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/637255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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pre algebraic factoring with polynomials I really need help solving this particular problem.
$$\frac14x^2y(x-1)^3-\frac54xy(x-1)^2$$
I need help factoring this. It seems like I need to get rid of the fraction but I really just need a little boost.
| Let's see, we have
$$
\frac{1}{4}x^2y(x-1)^3-\frac{5}{4}xy(x-1)^2
$$
Notice the first term has three $x-1$'s and the second term has two of them. So they share two of them, so we can pull out two of them leaving
$$
(x-1)^2\left(\frac{1}{4}x^2y(x-1)^1-\frac{5}{4}xy\right)
$$
Notice, we 'used' two $x-1$ in the first term, leaving only one left and we took out all of them in the second. Anything else? Well, notice both terms have an $x$ and a $y$ in common. So let's take out one of each! This gives us
$$
xy(x-1)^2\left(\frac{1}{4}x^1(x-1)^1-\frac{5}{4}\right)
$$
The first term had only one $y$, which we took out and two $x$'s, taking one out leaves only one left. The second term had only one $x$ and one $y$ so we took both those out leaving nothing there. Are we done? Not quite. Notice the first term and the second term still have a $\frac{1}{4}$ in common, factoring that out yields
$$
\frac{1}{4}xy(x-1)^2\left(x^1(x-1)^1-5\right)
$$
Now all we have to do is clean this up a bit and we get the solution
$$
\frac{1}{4}xy(x-1)^2\left(x(x-1)-5\right)
$$
Some might even distribute the $x$ in $x(x-1)$ to get
$$
\frac{1}{4}xy(x-1)^2\left(x^2-x-5\right)
$$
as their answer because they would say it looks 'nicer'.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/637373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$n^2+3n+2$ is even for any integer $n$ I am trying to prove the following:
If n is an integer, then $n^{2}+3n+2$ is even.
I tried proving this by contradiction, assuming that $n_{2}+3n+2=2k+1$, but I couldn't continue from that point.
Thanks!
| If $n$ is even it can be written as $n=2k$
$$\begin{array}{rcl}n^2 + 3n + 2 &=& (2k)^2 + 3(2k) + 2 \\&=& 2\cdot(2k^2+3k+1)\end{array}$$
If $n$ is odd it can be written as $n=2k+1$
$$\begin{array}{rcl}n^2 + 3n + 2 &=& (2k+1)^2 + 3(2k+1) + 2\\ &=& 4k^2+4k+1+6k+3+2 \\&=& 2\cdot(2k^2+5k+3)\end{array}$$
Hence $n^2 + 3n + 2$ is even for every integer $n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/640023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does $2^{-n} = 5^n \times 10^{-n}$? If we look at the decimal equivilents of $2^{-n}$, we see they resemble $5^n$ with a decimal point in front of them:
$\begin{align}
2^{-1} &= 0.5 \\
2^{-2} &= 0.25 \\
2^{-3} &= 0.125 \\
2^{-4} &= 0.0625 \\
2^{-5} &= 0.03125 \\
...
\end{align}$
It looks like it's as simple as saying $2^{-n} = 5^n \times 10^{-n}$, and when we calculate that out, it's correct:
$\begin{align}
5^1 \times 10^{-1} &= 5 \times 0.1 = 0.5 \\
5^2 \times 10^{-2} &= 25 \times 0.01 = 0.25 \\
5^3 \times 10^{-3} &= 125 \times 0.001 = 0.125 \\
5^4 \times 10^{-4} &= 625 \times 0.0001 = 0.0625 \\
5^5 \times 10^{-5} &= 3125 \times 0.00001 = 0.03125 \\
...
\end{align}$
I calculated this out for $n = [0, 10]$ and it works out, but I have no idea how to prove it fully.
| $\dfrac{1}{2^x}= \dfrac{5^x\cdot2^x}{2^x\cdot10^x}=\dfrac{5^x}{10^x}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\sum_{n=1}^{\infty}\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx$ I want to find the sum:
$$\sum_{n=1}^{\infty}\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx$$
I start with finding the antiderivative of the integrant, which is:
$$\frac{1}{2\sqrt{2}}[\ln(x^2-\sqrt{2}x+1)-\ln(x^2+\sqrt{2}x+1)+2\arctan(\sqrt{2}x-1)+2\arctan(\sqrt{2}x+1)]$$
Then I use the fundamental theorem of calculus to evaluate the integral. It turns out that the result is really ugly and I have no idea how to hanled it. Is there any tricks to tackle this?
| A more manageable way should be the following:
Let us make the change of the variable: $x=\frac{y}{\sqrt{n}}$
$$a_n=\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx=\frac{2}{n^{\frac{3}{2}}}\int_0^1\frac{y^2}{1+\frac{y^4}{n^2}}dy$$
Now, let's expand the integrand:
$$\frac{y^2}{1+\frac{y^4}{n^2}}=\sum_{k=0}^\infty(-1)^k\frac{y^{4k+2}}{n^{2k}}$$ and
$$\int_0^1\frac{y^2}{1+\frac{y^4}{n^2}}dy=\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\frac{1}{n^{2k}}$$ and
$$a_n=\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx=2\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\frac{1}{n^{2k+\frac{3}{2}}}$$
Finally, the sum:
$$S=\sum_{n=1}^\infty a_n=2\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\sum_{n=1}^\infty\frac{1}{n^{2k+\frac{3}{2}}}=$$
$$=2\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\zeta(2k+\frac{3}{2})$$
where $\zeta(s)=\sum_{k=1}^\infty\frac{1}{k^s}$ is the Riemann zeta function.
| {
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Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$ Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$.
I am struggling on this problem very much.
So far I have Basis case = $6\cdot 7^1 - 2\cdot3^1 = 36$ which is divisible by $4$
Assume for a $n$ that $6\cdot 7^n-2\cdot3^n$ is divisible by $4$.
After that I am not sure what to do.
| Let $P(n) = 6\cdot 7^n - 2\cdot 3^n$. If $4| P(1)$ and we assume $P(k)$ to be true for some arbitrary natural number $k$ and we can show $4| P(k+1) - P(k)$ then we have proved that $4|P(n) \ \forall n\in\mathbb{N}$.
$$\begin{align*} P(k+1) - P(k) & = 6\cdot 7^{k+1} - 2\cdot 3^{k+1} - 6\cdot 7^k + 2\cdot 3^k \\ & = 7\cdot 6\cdot 7^k - 3\cdot 2\cdot 3^k - 6 \cdot 7^k + 2\cdot 3^k \\ & = 6\cdot 6 \cdot 7^k - 2 \cdot 2 \cdot 3^k \\ & = 4 \cdot (9\cdot 7^k - 3^k)\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Formula help with this equation I don't know what the answer to this formula is, can someone please help me.
I've tried lots of things but getting no where.
If $x=\dfrac56+\dfrac{15}{18}-\dfrac{10}{12}$, then $(x-1)3=$ ?
| \begin{align}
x&=5/6+15/18-10/12\\
x-1&=5/6+15/18-10/12- 3/3\\
(x-1)3&=(5/6+15/18-10/12- 3/3)3\\
(x-1)3&=3*5/6+3*15/18-3*10/12- 3*3/3\\
(x-1)3&=5/2+15/6-10/4- 3\\
(x-1)3&=5/2+5/2-5/2- 3\\
(x-1)3&=5/2- 3\\
(x-1)3&=5/2- 6/2\\
(x-1)3&=-1/2
\end{align}
| {
"language": "en",
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Magnitude relation of fractions for four real numbers with a condition I would like to know if the following proposition holds or not.
For all $a, b, c, d$ such that $a, b, c, d$ are positive real numbers,
IF $a < b$ and $a - c > b - d \geq 0$ THEN $\frac{c}{a} < \frac{d}{b}$ .
Could you kindly advise?
It is really appreciated if a proof or a counterexample is provided.
Best regards,
| It's false. $a=1$ ; $b=2$ ; $c = 3$ ; $d=5$ is a counterexample.
We have $a<b$, $a-c=-2$ and $b-d = -3$ so $a-c>b-d$.
$\frac{c}{a}=3$ and $\frac{d}{b}=2.5$ so we don't have $\frac{c}{a} <\frac{d}{b}$
EDIT : With your new constraints, I believe this is true.
$a-c>b-d$ so $\frac{a}{c}-1> \frac{b-d}{c}$.
$d>c$ (because $a<b$ and $a-c>b-d$) so $\frac{1}{c} > \frac{1}{d}$ and since now we have $b-d\geq 0$, $\frac{b-d}{c} \geq \frac{b-d}{d} = \frac{b}{d}-1$.
So $\frac{a}{c}-1> \frac{b}{d}-1$ and $\frac{a}{c}> \frac{b}{d}$. So $\frac{c}{a}<\frac{d}{b}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of $(n-k)! \cdot n^k$ as $n$ approaches infinity Is it true that $(n-k)! \cdot n^k$ tends to $n!$ as $n \to \infty$?
I think it is correct but can't think of a satisfying proof.
| Sterling's approximation says $\ln (x!) = x \ln x - x + \frac12\ln(2\pi x) + O(\frac1x)$.
$$\ln \left((n - k)! n^k\right) = k \ln n + \ln \left((n - k)!\right)$$
$$ = k \ln n + (n - k)\ln(n - k) - (n - k) + \frac12 \ln (2 \pi (n - k)) + O(\frac1{n-k})$$
Since $\ln (n - k) = \ln\left(n(1 - \frac{k}n)\right) = \ln n + \ln (1 - \frac{k}n)$
$$ = k \ln n + (n - k)(\ln n + \ln(1 - \frac{k}n)) - n + k + \frac12 \ln (2 \pi (n - k)) + O(\frac1{n-k})$$
$$ = k \ln n + n \ln n - k \ln n + (n - k)\ln(1 - \frac{k}n) - n + k + \frac12 \left(\ln (2 \pi) + \ln n + \ln(1 - \frac{k}n)\right) + O(\frac1{n-k})$$
$$ = n \ln n + (n - k + \frac12)\ln(1 - \frac{k}n) + - n + k \frac12 \left(\ln (2 \pi) + \ln n \right) + O(\frac1{n-k})$$
Since for $x \ll 1$ we have $\ln(1 - x) = -x - O(x^2)$
$$ = n \ln n + (n - k + \frac12)\left(-\frac{k}n - O(\frac{k^2}{n^2})\right) - n + k \frac12\ln(2\pi n) + O(\frac1{n-k})$$
$$ = n \ln n - k + \frac{k^2}n - \frac{k}{2n} - O(\frac{k^2}{n})- n + k +\frac12\ln(2\pi n) + O(\frac1{n-k})$$
$$ = n \ln n - n + \frac12\ln(2\pi n) + O(\frac{k^2}n) $$
$$ = \ln(n!) + O(\frac1n) + O(\frac{k^2}n) = \ln(n!) + O(\frac{k^2}n)$$
So $(n - k)!n^k$ is approximately equal to $n!$ within a factor of about $e^{\frac{k^2}n}$. If you add in more terms of Sterling's approximation you might get a better big-O term.
| {
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Determine for what value of $x$ the series converges $∑_{n=1}^∞ \frac{(3^n+(-2)^n )}{n} (x+1)^n $ Determine for what value of $x$ the series converges
$∑_{n=1}^∞ \frac{(3^n+(-2)^n )}{n} (x+1)^n $
Here is what I got
Using the ratiotest, I got
$D_n =\frac{\frac{(3^{n+1}+(-2)^{n+1} )}{n+1}(x+1)^{n+1}}{\frac {(3^n+(-2)^n )}{n} (x+1)^n}$
$=\frac{n(3^{n+1}+(-2)^{n+1} )}{(n+1)(3^n+(-2)^n)}(x+1) ->3(x+1) <1$ if $x<-2/3$
Is this correct?
I use maple to calculate $lim_{n->\infty} \frac{n(3^{n+1}+(-2)^{n+1} )}{(n+1)(3^n+(-2)^n)}=3$ but I don't know why they got this either.
| First way. The radius of convergence can be found using the ratio test, since
$$
\frac{a_{n+1}}{a_{n}}=\frac{n}{n+1}\cdot \frac{3^{n+1}+(-2)^{n+1}}{3^{n}+(-2)^{n}}=3\cdot
\frac{1}{1+\frac{1}{n}}\cdot \frac{1+\left(-\frac{2}{3}\right)^{n}}{1+\left(-\frac{2}{3}\right)^{n+1}}\,\to\, 3.
$$
Second way. Here you can use the root test as well:
If $\,\,\limsup_{n\to\infty} |a_n|^{1/n}=\ell$, then the radius of convergence is $r=1/\ell$.
Here
$$
\limsup_{n\to\infty}\left|\frac{(3^n+(-2)^n )}{n}\right|^{1/n}=3,
$$
and hence $r=1/3$.
In fact, here the sequence $\left|\frac{(3^n+(-2)^n )}{n}\right|^{1/n}$ converges to $3$ since
$$
\frac{1}{3n}\cdot 3^n=\frac{1}{n} 3^{n-1}=\frac{3^n+(-2)^n }{n}\le \frac{3}{n}\cdot 3^n,
$$
and
$\big(\frac{1}{3n}\cdot 3^n\big)^{1/n}\to 3$ and $\big(\frac{3}{n}\cdot 3^n\big)^{1/n}\to 3$ as well.
| {
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$x^5-x^2+1$ is separable over all fields
Prove that $p(x)=x^5-x^2+1$ is separable over all fields.
When the field is finite or of characteristic zero it is automatically true, since any polynomial is separable. The definition of separability requires to look at the irreducible components, which seems pretty hard.
The only other tool I know of is showing that the derivative is relatively prime to the polynomial, i.e. that $\gcd(x^5-x^2+1,5x^4-2x)=1$ over all fields, and it suffices to show $\gcd(x^5-x^2+1,5x^3-2)=1$, since $0$ is never a root of $p(x)$.
| That idea leads you to the solution.
If the characteristic is $5$ you get $\gcd$ $=1$.
Now assume that characteristic is not $5$. Then you can divide by $5$ (cleaner to multiply by $5$) and compute the gcd.
We get, $\gcd(x^5-x^2+1,5x^3-2)=\gcd(5x^5-5x^2+5,5x^3-2)=\gcd(3x^2-5,5x^3-2)$
If characteristic is $3$ we get the $\gcd=1$. Assume characteristic is not $3$ either.
The $\gcd(3x^2-5,5x^3-2)=\gcd(3x^2-5,15x^3-6)=\gcd(25x-6,3x^2-5)=\gcd(25x-6,75x^2-125)=\gcd(18x-125,25x-6)=\gcd(25\cdot18x-25\cdot125,25x-6)=\gcd(3017,25x-6)$.
We have that $3017=7\cdot 431$. For these primes we get $\gcd\neq1$.
Actually, we get $\gcd=25x-6$. This tells us what the possible multiple root would be.
Let us do the characteristic $7$ case. What this $\gcd$ is telling us is that the solution of $25x-6=0$, i.e. $x=5$, is a root of our polynomial and its derivative.
Divide $x^5-x^2+1$ by $x-5$. We get $x^5-x^2+1=(x-5)^2(x^3+3x^2+5x+2)$.
Using the old definition of separable:
In this case we ($char =7$) we have $x-5$ is an irreducible factor with only one root. We will need to do with the factor $x^3+3x^2+5x+2$, the same study we did for the original polynomial.
Computing the $\gcd$ with its derivative (we proceed as before, multiplying by leading coefficients to ease the division, and taking remainder mod $7$)
$\gcd(x^3+3x^2+5x+2,3x^2+6x+5)=\gcd(3x^3+2x^2x+6,3x^2+6x+5)=\gcd(3x+1,3x^2+6x+5)=\gcd(x+4,0)=x+4$. So $x+4$ is a multiply factor of $x^3+3x^2+5x+2$. To not have to compute any further, notice that because this polynomial is of degree $3$ it should split completely. So, I guess it would be called separable according to the old definition (if I am reading it correctly).
The case of $char=431$ is all yours. But the work is just the same.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int x^3\sqrt{9-x^2} dx$ using a trigonometric substitution The Integral is $$\int x^3\sqrt{9-x^2} dx$$
My method: $$x = 3\sin\theta$$ $$dx = 3\cos\theta d\theta$$
$$\sqrt{9-9\sin^2\theta} = 3\cos\theta$$
$$= \int 27\sin^3\theta (9\cos^2\theta) d\theta$$
$$=243\int (\cos^2\theta - \cos^4\theta)\sin\theta d\theta$$
$$u=\cos\theta$$
$$du=-\sin\theta d\theta$$
$$= -243\int u^2 -u^4 du$$ $$ = -81u^3 + \frac{243u^5}{5}$$ $$= -81(\cos\theta)^3 + \frac{243}{5} (\cos\theta)^5$$ $$\cos\theta = \frac{\sqrt{9-x^2}}{3}$$ $$= -27(\sqrt{9-x^2})^3 +\frac{81}{5}(\sqrt{9-x^2})^5 + C$$
Is this right? I realize that there is an easier substitution but this is how my prof wants it done. I'm confused because Wolfram Alpha gives $-\frac{1}{5}(\sqrt{9-x^2})^3(x^2+6) + C$ as the answer using the more simple substitution.
| Perhaps easier, shorter and/or simpler: by parts
$$u=x^2\;\;,\;\;u'=2x\\v'=x\sqrt{9-x^2}\;\;,\;\;v=-\frac13(9-x^2)^{3/2}$$
so
$$\int x^3\sqrt{9-x^2}\;dx=-\frac13x^2(9-x^2)^{3/2}+\frac23\int x(9-x^2)^{3/2}dx=$$
$$=-\frac13x^2(9-x^2)^{3/2}-\frac2{15}(9-x^2)^{5/2}+C$$
Two things: this answers negatively your last question, and we used twice above the easy-to-check fact that
$$\int f'(x)f(x)^ndx=\frac{f(x)^{n+1}}{n+1}+C\;,\;\;\forall -1\neq n\in\Bbb R$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Transform $\int \frac{x^2}{x - 2}$ to $ \int x +\frac{4}{x-2} + 2 $ I was trying to solve the following integral and I failed so I went to wolframalpha to see the step by step solution, but the following transformation is confusing me.
$\int \frac{x^2}{x - 2} = \int x +\frac{4}{x-2} + 2 $
I am unable to do the transformation by myself. Could you please explain the steps to get from $\int \frac{x^2}{x - 2} $ to $\int x +\frac{4}{x-2} + 2 $ ? The step is labeled as a "long division", but I have no idea what that means.
| You can rewrite the expression in exactly the same way you normally complete the square: $x^2 = (x^2 - 4x + 4) + 4x - 4 = (x-2)^2 + 4x-4$.
Then work with the linear terms in the same way: $4x - 4 = 4(x-2) + 8 - 4 = 4(x-2) + 4$.
Put it together to get $x^2 = (x-2)^2 + 4(x-2) + 4$.
Then you can throw in your denominator of $x-2$ and simplify:
$$\boxed{\frac{x^2}{x-2}} = \frac{(x-2)^2 + 4(x-2) + 4}{x-2} = (x-2 ) + 4 + \frac{4}{x-2} = \boxed{x + \frac{4}{x-2} + 2}$$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this $\prod_{n=2}^{\infty}\left(1-\frac{1}{n^6}\right)$ How find this
$$\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^6}\right)$$
I think we can find this value have closed form
$$\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^{2k}}\right)$$
since
$$1-\dfrac{1}{n^6}=\left(1-\dfrac{1}{n^3}\right)\left(1+\dfrac{1}{n^3}\right)$$
so I think we must find this
$$\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^3}\right)$$
and
$$\prod_{n=2}^{\infty}\left(1+\dfrac{1}{n^3}\right)$$
Thank you
| As Mhenni Benghorbal mentioned, there is a similar problem which has been treated yesterday and I shall use a similar approach to the one he proposed.
$$\prod_{n=2}^{m}\left(1-\dfrac{1}{n^3}\right)=\frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right) \Gamma \left(m-\frac{i
\sqrt{3}}{2}+\frac{3}{2}\right) \Gamma \left(m+\frac{i
\sqrt{3}}{2}+\frac{3}{2}\right)}{3 \pi m^3 \Gamma (m)^2}$$
and
$$\prod_{n=2}^{m}\left(1+\dfrac{1}{n^3}\right)=\frac{(m+1) \cosh \left(\frac{\sqrt{3} \pi }{2}\right) \Gamma \left(m-\frac{i
\sqrt{3}}{2}+\frac{1}{2}\right) \Gamma \left(m+\frac{i
\sqrt{3}}{2}+\frac{1}{2}\right)}{2 \pi \Gamma (m+1)^2}$$ So, the product, from $n=2$ to $n=m$, write $$\frac{(m+1) \cosh ^2\left(\frac{\sqrt{3} \pi }{2}\right) \Gamma \left(m-\frac{i
\sqrt{3}}{2}+\frac{1}{2}\right) \Gamma \left(m-\frac{i
\sqrt{3}}{2}+\frac{3}{2}\right) \Gamma \left(m+\frac{i
\sqrt{3}}{2}+\frac{1}{2}\right) \Gamma \left(m+\frac{i
\sqrt{3}}{2}+\frac{3}{2}\right)}{6 \pi ^2 m^5 \Gamma (m)^4}$$ If $m$ goes to infinity, the limit is then $$\frac{1+\cosh \left(\sqrt{3} \pi \right)}{12 \pi ^2} =\dfrac{\cosh^2\left(\pi\dfrac{\sqrt3}2\right)}{6\pi^2}$$ as shown by Lucian.
| {
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$a+b+c=3, a,b,c>0$, Prove that $a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$ $a+b+c=3, a,b,c>0$, Prove that $$a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$$
My work:
From the given inequality, we can have,
$a^2b^2c^2\ge 27-18(a+b+c)+12(ab+bc+ca)-8abc$
We can also have,$abc\le \bigg(\dfrac{a+b+c}{3}\bigg)^3=1$
So, $0\ge -36+12(ab+bc+ca)$
Again, we can have, $ab+bc+ca\le b(3-b)+\dfrac{1}{b}$
Now, I have to show that, $b(3-b)+\dfrac{1}{b}\le 3$
How can I prove this now? Please help.
| Here's a geometric method, just for variety.
First (as in Yiyuan Lee's solution) we consider cases: how many of the factors on the RHS are negative?
Case 3: All three are negative; then the RHS is negative and we are done.
Case 2: Exactly two are negative; wlog, $3-2a<0$ and $3-2b<0$. This is impossible because $3-2a = a+b+c-2a = -a+b+c$, so $3-2a<0$ means $b+c<a$; similarly $3-2b<0$ means $a+c<b$; adding these inequalities yields $c<0$.
Case 1: Exactly one is negative; same as case 3.
Case 0: All three are positive. Then, as in case 2, we obtain $b+c>a$ and $a+c>b$ and $a+b>c$. Therefore there is a triangle with sides $a$,$b$,$c$. Using a standard formula for the circumradius $R$ and our hypothesis that $a+b+c=3$, we get
$$ R = \frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}
= \frac{abc}{\sqrt{3(3-2a)(3-2b)(3-2c)}} $$
and so the desired inequality is equivalent to $R\ge\frac1{\sqrt3}$. So we want to prove this:
Any triangle with perimeter $3$ has circumradius at least $\frac1{\sqrt3}$.
Equivalently:
Any triangle with circumradius $1$ has perimeter at most $3\sqrt3$.
Equivalently:
Of all triangles with circumradius $1$, the equilateral triangle has the greatest perimeter.
By a standard compactness argument, there exists a triangle with maximum perimeter among those with circumradius $1$; to show that any such triangle is equilateral triangle, apply the following lemma to each vertex.
Lemma. Let $P$ and $Q$ be fixed points on a circle, and let $X$ be a variable point on that circle. If $PX+XQ$ is maximized then $X$ lies on the diameter bisecting the chord $PQ$.
Proof of lemma. Let $XY$ bisect $\angle PXQ$, with $Y$ on $PQ$. Drop perpendiculars from $P$ and $Q$ to $XY$, with feet $P'$ and $Q'$ respectively.
Then
$$ PX+XQ = \frac{PP'+Q'Q}{\sin(\frac12\angle PXQ)}
\le \frac{PY+YQ}{\sin(\frac12\angle PXQ)}
= \frac{PQ}{\sin(\frac12\angle PXQ)}
$$
Note that $\angle PXQ$ does not depend on the position of $X$ (Euclid III:21), only on what side of $PQ$ it falls on*, and that we have equality exactly when $Y$, $P'$, and $Q'$ coincide, that is, $XY$ is perpendicular to $PQ$, as desired.
*It doesn't matter for us which of the two cases is the sharp one, but of course it's when $X$ is on the major arc associated with $PQ$.
| {
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Prove a $n \times n $ matrix has rank 3 I have been examining a problem dealing with finding the rank of a $n \times n $ matrix $M$ as follows:
\begin{bmatrix}
0&1&4&9&16&\cdots &(n-1)^2\\
1&0&1&4&9&\cdots&(n-2)^2\\
4&1&0&1&4&\cdots&(n-3)^2\\
9&4&1&0&1&\cdots&(n-4)^2\\
16&9&4&1&0&\cdots&(n-5)^2\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
(n-1)^2&(n-2)^2&\cdots&(n-4)^2&\cdots&\cdots&0
\end{bmatrix}
That is, the matrix with $k^2$ on its $k^{th}$ super and subdiagonals.
One of my colleagues claims the matrix is of rank 3.
I have tried to factor this matrix somehow, but have not really gotten very far.
Can we develop a proof of the claim that the rank of $M$ is indeed 3 (assume that $n\geq3$).
| When $n\ge4$, let $P=\pmatrix{1&-1\\ &\ddots&\ddots\\ &&\ddots&-1\\ &&&1}^3
=\pmatrix{1&-3&3&-1\\
&\ddots&\ddots&\ddots&\ddots\\
&&\ddots&\ddots&\ddots&-1\\
&&&\ddots&\ddots&3\\
&&&&\ddots&-3\\
&&&&&1}$. Then $P^TMP=\pmatrix{0&1&1\\ 1&-6&1\\ 1&1&0}\oplus 0_{(n-3)\times(n-3)}$. Hence $M$ has rank 3.
Remark.
Knowing that $M$ has rank $3$ allows us to find the eigenvalues of $M$ relatively easily. One can readily verify that $(1-n,3-n,\ldots,n-3,n-1)^T$ is an eigenvector of $M$ corresponding to the eigenvalue $-\frac{n(n^2-1)}{6}$. Since $M$ has zero trace, if we also know $\operatorname{tr}(M^2)$, the other two nonzero eigenvalues can be found. It turns out that
$$
\operatorname{tr}(M^2)=2\sum_{i=1}^{n-1} i(n-i)^4
=\frac1{30}n^2(n^2-1)(2n^2-3)
$$
(see also OEIS A101089) and the other two nonzero eigenvalues of $M$ are
$$
\frac{n(n^2-1)}{12}\pm\frac{n}{2}\sqrt{\frac{(n^2-1)(3n^2-7)}{60}}.
$$
| {
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"url": "https://math.stackexchange.com/questions/676705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How to solve $x^3 - 2x^2 -16x+16=0$? How to solve $x^3 - 2x^2 -16x+16=0$ ?
I tried to factor out $x^2$ but it doesn't work..any hints?
Ok I tried to use rational root theorem and have possible answer $\pm 1, \pm 2, \pm 4, \pm 8 , \pm 16$, but my calculator gives out some decimal numbers...
| To solve the cubic, we first perform a transformation to eliminate the $x^2$ term: by letting $x = y + 2/3$, we then obtain the 'depressed' cubic $$y^3 - \frac{52}{3} y + \frac{128}{27} = 0.$$ Next, recall the triple angle cosine identity: $$4 \cos^3 \theta - 3 \cos \theta = \cos 3\theta.$$ This suggests letting $y = u \cos \theta$ so that $$\begin{align*} 0 &= u^3 \cos^3 \theta - \frac{52}{3}u \cos \theta + \frac{128}{27} \\ &= \frac{u^3}{4} \left( 4 \cos^3 \theta - \frac{208}{3u^2} \cos \theta + \frac{512}{27u^3}\right).\end{align*}$$ Thus if we chose $u = \frac{4 \sqrt{13}}{3}$, it immediately follows that $$\cos 3\theta = - \frac{8}{13\sqrt{13}}.$$ Therefore, the roots are given by $$x = \frac{2}{3} + \frac{4\sqrt{13}}{3} \cos \left( \frac{1}{3} \cos^{-1} \frac{-8}{13 \sqrt{13}} + \frac{2\pi k}{3} \right), \quad k = 0, 1, 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Flux and Gauss theorem I have a problem; There seems to be something wrong with my understanding of gauss theorem.
Let's say $F = [y ; x^2y; y^2z]$. I want to calculate the flux of $F$ going out of $$D = \{1 \le z \le 2 - x^2 - y^2\}$$
Method number 1
LEt's use Gauss theorem.
So $\nabla \cdot F = x^2 + y^2$, and $$\int_{D} x^2 + y^2 = \int_{A} \int_1^{2-x^2-y^2} x^2 + y^2 dz = $$ $$\int_{A} (x^2 + y^2)(1 - x^2 - y^2) dxdy= \int_0^1\int_0^{2\pi} \rho ^3(1-\rho^2)) d\rho d\theta = \frac{\pi}{6}$$
Method number 2
LEt's do the explicit calculations.
So let's write $D$ as $r_{up} = [x ; y ; 2 - x^2 - y^2]$ (for the upper part) and $r_{down} = [x ; y; 1]$
This leads to
$$n_{up} = r_x \wedge r_y = [2x ; 2y ; 1]$$
$$n_{down} = r_x \wedge r_y = [0; 0; -1]$$
(I didn't normalize the normal because in the integral I'm going to write directly $dxdy$ without the norm of $n$)
Then $$\int_{D} F \cdot n_{up} = \int_{D} 2xy + 2x^2y^2 + y^2z = \frac{\pi}{8}$$
and $$\int_{A} F \cdot n_{down} = \int_{A} -y^2 = -\frac{\pi}{4}$$ (where A is the unitary circle)
So the two results are different.
Why is this so?
Thank you in advance
EDIT
I edited $n_{down}$, which I belive must be pointing out of $D$ so I changed it in $(0, 0, -1)$. This does not change the fact that the twto methods yields different results, though
| You have the associated form $$G=ydy\wedge dz+x^2ydz\wedge dx+y^2z dx\wedge dy$$ Then $$dG=({\rm div}\; F)dx\wedge dy\wedge dz=(x^2+y^2)dx\wedge dy\wedge dz$$
Now the flat part of the surface, $S_1$; can be parametrized by $(r\cos t,r\sin t,1)$ with $0\leqslant r \leqslant 1$ and $0\leqslant t\leqslant 2\pi$. The upper cap $S_2$ may be parametrized by $(r\cos t,r\sin t ,2-r^2)$ with the very same domain. We see $dz=0$, so you get
$$\int\limits_{{S_1}} G = \int\limits_0^{2\pi } {\int\limits_0^1 {{r^3}{{\sin }^2}tdrdt} } = \frac{\pi }{4}$$
On the other hand over the upper cap you have $$\eqalign{
& dy = \sin tdr + r\cos tdt \cr
& dx = \cos tdr - r\sin tdt \cr
& dz = - 2rdr \cr} $$
You can calculate $$dx\wedge dy=rdr\wedge dt\\ dy\wedge dz=2r^2\cos tdr\wedge dt\\dz\wedge dx=2r^2\sin tdr\wedge dt$$
Then the integral is $$\eqalign{
& \int_0^1 {\int_0^{2\pi } {\left( {r\sin t\left( {2{r^2}\cos t} \right) + {r^3}{{\cos }^2}t\sin t\left( {2{r^2}\sin t} \right) + {r^2}{{\sin }^2}t\left( {2 - {r^2}} \right)r} \right)dtdr} } = \cr
& \int_0^1 {\int_0^{2\pi } {\left( {2{r^5}{{\cos }^2}t{{\sin }^2}t + {{\sin }^2}t\left( {2 - {r^2}} \right){r^3}} \right)dtdr} } = \cr
& \int_0^{2\pi } {\left( {\frac{2}{6}{{\cos }^2}t{{\sin }^2}t + {{\sin }^2}t\frac{1}{3}} \right)dt} = \frac{2}{6}\frac{\pi }{4} + \frac{\pi }{3} = \frac{\pi }{{12}} + \frac{\pi }{3} =\frac{5\pi}{12}\cr} $$
Then our answers match up $$\frac{5\pi}{12}-\frac{\pi}4=\frac{5\pi}{12}-\frac{3\pi}{12}=\frac{2\pi}{12}=\frac \pi 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/679420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$
I tried AM-GM, CS inequality but all failed.
| Another way.
By Rearrangement
$$\sum_{cyc}\left(\frac{a^3}{bc}-a\right)=\sum_{cyc}\frac{a}{bc}(a^2-bc)\geq\frac{1}{3}\sum_{cyc}\frac{a}{bc}\sum_{cyc}(a^2-bc)=\frac{1}{6}\sum_{cyc}\frac{a}{bc}\sum_{cyc}(a-b)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/679544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to find the minimum value of this function? How to find the minimum value of
$$\frac{x}{3y^2+3z^2+3yz+1}+\frac{y}{3x^2+3z^2+3xz+1}+\frac{z}{3x^2+3y^2+3xy+1}$$,where $x,y,z\geq 0$ and $x+y+z=1$.
It seems to be hard if we use calculus methods. Are there another method? I have no idea.
Thank you.
| To find a minimum of the function I'm going to find a lower bound and show that the lower bound is attained. For that I'm going to use nothing harder than Cauchy-Schwarz inequality.
Let $f(x, y, z)$ denote your function. Using mentioned Cauchy-Schwarz inequality we get:
$$f(x, y, z)\cdot\big(x(3y^2+3z^2+3yz+1) + y(3x^2+3z^2+3xz+1) + z(3x^2+3y^2+3xy+1)\big) \\
\geqslant (x+y+z)^2 = 1.$$
So,
$$f(x, y, z) \geqslant \frac{1}{3(xy^2 + xz^2 + yx^2 + yz^2 + zx^2 + zy^2 + 3xyz) + x+y+z} \\
=\frac{1}{3(x+y+z)(xy+yz+xz) + 1} = \frac{1}{3(xy+yz+xz) + 1}$$
We want to get an upper bound of $xy + yz + xz$ under the condition $x + y + z = 1$. We can do this, for example:
$$\begin{align}
xy + yz + xz &\leqslant x^2 + y^2 + z^2 \\
3(xy + yz + xz) &\leqslant (x + y + z)^2 = 1 \\
xy + yz + xz &\leqslant \frac{1}{3}
\end{align}$$
Finally,
$f(x, y, z) \geqslant \dfrac{1}{3\cdot\frac{1}{3} + 1} = \dfrac{1}{2}.$ So $\dfrac{1}{2}$ is the lower bound.
Since $\displaystyle f\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}$, we can see that we've found a point where the lower bound is attained so this is the minimum of the function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/681153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Proving that $\left(\frac{\pi}{3} \right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k+1)\zeta(2k+2)}{3^{2k+2}}$. I have asked a question or two like this one before and I've tryed to use similar methods to prove this identity(?), but I failed. By using WA it seems that numerically the LHS=RHS
$$
\left(\frac{\pi}{3} \right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k+1)\zeta(2k+2)}{3^{2k+2}}
$$
But a rigorous proof would be better!
The other similar identity is
$$
\left(\frac{\pi}{2}\right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k}}
$$
Thanks in advance.
| The sum
$$\sum_{k=1}^{\infty} \frac{(2 k+1) \zeta(2 k+2)}{3^{2 k+2}} = \sum_{n=1}^{\infty} \frac1{9 n^2}\sum_{k=1}^{\infty} \frac{2 k+1}{(9 n^2)^k}$$
from the definition of the zeta function. The inner sum is a geometrical sum and its derivative. In essence, it is simple to derive the following:
$$\sum_{k=1}^{\infty} (2 k+1) r^k = \frac{r (3-r)}{(1-r)^2}$$
Here, $r=1/(9 n^2)$, so we have the sum
$$\sum_{n=1}^{\infty}\frac1{9 n^2} \frac{27 n^2-1}{(9 n^2-1)^2} = -\sum_{n=1}^{\infty} \frac1{9 n^2} + \sum_{n=1}^{\infty} \frac{9 n^2+1}{(9 n^2-1)^2}$$
The first sum is easy ($-\pi^2/54$), so let's attack the second. This may be accomplished via residues. Without going into details, I state the result:
$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_{k} \operatorname*{Res}_{z=z_k} \pi \, \cot{(\pi z)} \, f(z)$$
Here, $f(z) = (9 z^2+1)/(9 z^2-1)^2$. The poles of $f$ are at $z_{\pm}=\pm 1/3$. The sum of the above residues at these poles are
$$\pi \sum_{\pm} \left [\frac{d}{dz} \frac{\cot{(\pi z)} (9 z^2+1)}{(3 z \pm 1)^2} \right ]_{z=\pm 1/3} = -\frac{4 \pi^2}{27}$$
$$\implies \sum_{n=1}^{\infty} \frac{9 n^2+1}{(9 n^2-1)^2} = \frac{2 \pi^2}{27} - \frac12 $$
Thus,
$$1+2 \sum_{k=1}^{\infty} \frac{(2 k+1) \zeta(2 k+2)}{3^{2 k+2}} = -\frac{\pi^2}{27} + \frac{4 \pi^2}{27} = \frac{\pi^2}{9}$$
as was to be shown.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can 11 be represented by $a^2 - 3b^2$ where a and b are integers? I know the answer is no, just wan't to know how. From a similar question on the site I got that $a^2 - 3b^2$ should always equal a square modulo 3 which 11 is not.
But I don't understand how to get to this conclusion.
Please help
| There are two indefinite binary quadratic forms (well, classes) of discriminant $12.$ These are $x^2 - 3 y^2$ and $3 x^2 - y^2.$ The primes $2,3$ divide $12$ and are given separate consideration; both can be written as $3 x^2 - y^2.$
For any larger (positive) prime with Legendre symbol $(3|p) = 1,$ we have two choices.
If $p \equiv 3 \pmod 4,$ then $(p|3) = - (3|p) = -1,$ so $p \equiv 2 \pmod 3,$ and here $p \equiv 11 \pmod {12}.$ Such a prime is represented by $3 x^2 - y^2.$ So, $$11 = 12 -1, \; \; 23 = 27 - 4, \; \; \; 47 = 48 - 1, \; 59 = 75 - 16,$$ and so on.
If $p \equiv 1 \pmod 4,$ then $(p|3) = (3|p) = 1,$ so $p \equiv 1 \pmod 3,$ and here $p \equiv 1 \pmod {12}.$ Such a prime is represented by $ x^2 - 3y^2.$ So, $$13 = 16 -3, \; \; 37 = 49 - 12, \; \; \; 61 = 64 - 3, \; 73 = 100 - 27,$$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/683135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I cannot find the derivatives of these functions using the product rule? I need help finding the derivative of $y=x^3(3x+7)^2$ at $x=-2$
I tried to simplify the function to $y=x^3 (3x+7)(3x+7)$ and the simplify it into two terms and the derivatives of those terms using the product rule, but that doesn't work.
Since I don't understand how to approach this type of question (with the power on the outside), I couldn't do this one either: $y=(2x+1)^5(3x+2)^4$ at $x=-1$
I don't need help finding the derivative at the specific values of $x$, I just can't seem to get the general case. BTW, I can only use the product rule as my teacher did not teach the quotient and chain rule to us, so she doesn't allow us to use them.
the answers(aka the derivative):
*
*${(3x^2)(3x+7)^2}+{(x^36)(3x+7)}$
*$5(2x+1)^4 (2) (3x+2)^4 + (2x+1)^54(3x+2)^3(3)$
| The easiest solution (by this I mean no fancy theorems needed) is to calculate the product. So we have to way to do so. The first way:
\begin{align}
y&=x^3(3x+7)^2\\&=x^3(9x^2+42x+49), \\
\Longrightarrow y'&=3x^2(9x^2+42x+49)+x^3(18x+42)\\&=45x^4+168x^3+147x^2,
\end{align}
The second way:
\begin{align}
y&=x^3(3x+7)^2\\&=x^3(9x^2+42x+49)\\&=9x^5+42x^4+49x^3, \\\Longrightarrow y'&=45x^4+168x^3+147x^2.
\end{align}
From this point, both ways continue as
\begin{align}
y'&=(x^2)\cdot(45x^2+168x+147)\\
&=(4)\cdot (45\cdot 4-168\cdot 2+147)\\
&=4\cdot(180-336+147)\\
&=4\cdot(327-336)\\
&=-36\end{align}
Note that for the other exercise you can use the following formulas from college algebra,
$$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4,\\
(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+x^5.$$
So,
$$(2x+1)^5=32x^5+80x^4+80x^3+40x^2+10x+1,\\
(3x+2)^4=81x^4+216x^3+216x^2+96x+16.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $C$ if $ 3\sin A + 4\cos B = 6 $ and $ 3\cos A + 4\sin B = 1 $ in a triangle $ABC$
In a triangle $ABC$, it's given that the following two equations are satisfied:
$$ 3\sin A + 4\cos B = 6 $$
$$ 3\cos A + 4\sin B = 1 $$
Source: ISI B-math UGA 2017
We have to find the angle $ C$. Now, it's easy to see that $ \sin C = 0.5 $ (by squaring and adding both the equations). Now we have to decide whether $ C = \pi/6 $ or $5\pi/6$. The solution given in the book goes something like this:
Assume $ C = 5\pi/6$, then even if $B = 0$ and $A = \pi/6$, then the quantity $3\sin A + 4\cos B = 5.5 < 6$ and hence $ C \not= 5\pi/6 $.
But then if we do the same thing by setting $C = \pi/6$, again we face the same problem. So how to find the value of $C$?
EDIT:
Squaring and adding the two equations, we get:
$$ 9(\sin^2 A + \cos^2 A) + 16 (\sin^2 B + \cos^2 B) + 24\sin(A+B) = 37 $$
$$ 24\sin(A+B) = 24\sin C = 12 \implies \sin C = 0.5 $$
| when $C=\dfrac{\pi}{6}, A $ can be large enough,it is different from the case $C=\dfrac{5\pi}{6}$as $A\le \dfrac{\pi}{6}$ that you can't find solution for $A,B$
Now in case $C=\dfrac{\pi}{6}$, it is possible to find solution but you can't use same method to check as $\sin{A} $ can large than $\dfrac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/687156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to move from powers to simple logarithms? I'm following a book that briefly moves from
$$16000 \times 2^{\displaystyle \left (-\frac{x}{24} \right )} = 1600$$
to
$$x = \frac{24 (\log(2) + \log(5))}{\log(2)}$$
adding the comments that
$$\log(1600) = 6\log(2) + 2\log(5) \\
\log(16000) = 7\log(2) + 3\log(5)$$
What general principles are used to achieve this and how to spot possible ways to apply a similar operation in the future?
| In general:
\begin{align*}
\log a^b &= b \log a\\
\log (a \cdot b) &= \log a + \log b
\end{align*}
Specific to the above problem:
$$\log(1600) = \log(2^6 \cdot 5^2) = \log(2^6) + \log (5^2) = 6 \log 2 + 2 \log 5$$
$$\log(16000) = \log(2^7 \cdot 5^3) = 7 \log 2 + 3 \log 5$$
So,
\begin{align*}
16000 \times 2^{-x/24} &= 1600\\
\log 16000 + (-x/24) \log 2 &= \log 1600\\
(-x/24)\log 2 &= -\log 2 - \log 5\\
x &=24 \cdot \frac{\log 2 + \log 5}{\log 2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/693484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplfying $\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$ I am trying to simplify the expression:
$\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$
I tried to square the expression but I can't do that because it is not an equation so I got stuck. Can someone please give me some pointers on how to proceed?
| Set up an equation like:
$$
u = \sqrt{31 -8 \sqrt{15}} + \sqrt{31 + 8 \sqrt{15}}
$$
Square to get it down to:
$$
u^2 = 31 - 8 \sqrt{15} + 31 + 8 \sqrt{15} + 2 \sqrt{31^2 - 8^2 \cdot 15}
= 62 + 2 \cdot 1 = 64
$$
Thus you have $u = \pm 8$. The original expression is clearly positive, so:
$$
\sqrt{31 -8 \sqrt{15}} + \sqrt{31 + 8 \sqrt{15}} = 8
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/696113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to show $\sqrt{3-\sqrt2} \in \mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$ So as title says I wanna show $\sqrt{3-\sqrt2} \in \mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$
So I know that the minimal polynomial of $\mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$ is $x^{4}-6x^{2}+7$ the zeroes of which are $\pm \sqrt{3-\sqrt2}, \pm \sqrt {3+\sqrt 2}$.
So far I have got as far as:
$\sqrt{3-\sqrt2}\sqrt {3+\sqrt 2} = \sqrt7$.
Any tips on how to find $\sqrt7$ in terms of the minimal polynomial of $\mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$ ?
| Since $2,7$ and $2\times 7$ are all nonsquares
in $\mathbb Q$, we see that $K={\mathbb Q}[\sqrt{2},\sqrt{7}]$ has
degree $4$ over $\mathbb Q$. If we put
$\theta=\sqrt{7}+\sqrt{3+\sqrt{2}}$, we have
$$
\theta^2+4=14+\sqrt{2}+2\sqrt{7\times(3+\sqrt{2})}=
\sqrt{2}+2\sqrt{7}\theta \tag{1}
$$
So $\theta$ is a root of the polynomial
$P=X^2-(2\sqrt{7})X+4-\sqrt{2}$. The discriminant of $P$ is
$d=3+\sqrt{2}$. If $d$ were a square in $K$, we would have
$x,y\in{\mathbb Q}[\sqrt{7}]$ such that $(x+y\sqrt{2})^2=3+\sqrt{2}$,
hence $x^2+2y^2=3,2xy=1$. This yields $y=\frac{1}{2x}$,
$x^2+\frac{2}{4x^2}=3$, so $2x^4-6x^2+1=0$ and hence
$x^2=\frac{3\pm\sqrt{7}}{2}$ which can be seen to be impossible in
${\mathbb Q}[\sqrt{7}]$.
We have therefore shown that $d$ is a nonsquare in $K$, so
$[K(\theta):K]=2$. Let $L={\mathbb Q}(\sqrt{7},\theta)$.
By (1) above, $L$ contains $\sqrt{2}=\theta^2-2\sqrt{7}\theta+4$,
so $L$ contains $K(\theta)$. We deduce $L=K(\theta)$ as the other
inclusion is trivial. Thus $[L:{\mathbb Q}]=8$.
Finally, if we had $\sqrt{3+\sqrt{2}}\in M$ where
$M={\mathbb Q}[\sqrt{3-\sqrt{2}}]$, we would have
$\sqrt{7}\in M$ as explained in the OP, so $\theta\in M$
and eventually $M=L$, contradicting $[L:{\mathbb Q}]=8$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if the $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1$ I need to prove that:
If $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1.$
So far I used what's given so I have:
$ax+by=1$ (I wrote the gcd of a and b as a linear combination)
and
$(a+b)u+ab(v)=1$ (I also wrote this as a linear combination)
where do I go from here?
| We're given $\gcd(a,b) = 1$, and from Bézout's identity we have $\gcd(a,b) = 1 \iff \exists x,y: ax + by = 1$ for integer $x$ and $y$.
$1 = ax + by = ax + bx - bx + by = (a + b)x + b(y - x)$, so $\gcd(a+b,b) = 1$. Likewise, $\gcd(a+b,a)=1$ because $(a + b)y + a(x - y) = 1$
$$ \begin{align}
1 &= ((a + b)x + b(y - x))((a + b)y + a(x -y))
\\&= (a+b)(a+b)xy + (a+b)a(x-y) + (a+b)b(y-x)y + ab(y-x)(x-y)
\\&= (a+b)[(a+b)xy + a(x-y) + b(y-x)] + ab[(y-x)(x-y)]
\end{align}$$
So $\gcd(a+b,ab) = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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solve a trigonometric equation $\sqrt{3} \sin(x)-\cos(x)=\sqrt{2}$ $$\sqrt{3}\sin{x} - \cos{x} = \sqrt{2} $$
I think to do :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{2}}$$
but i dont get anything.
Or to divied by $\sqrt{3}$ :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{3}}$$
| A difficulty of this equation is that it contains both $\sin(x)$ and $\cos(x)$. You can use the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$ to express $\sin(x) = \pm \sqrt{1 - \cos^2(x)}$ (or $\cos(x)$ as a function of $\sin(x)$), but it is very unwieldy because it introduces square roots (another difficulty) and you need to distinguish the intervals where $\sin(x)$ is positive or negative. It is simpler if you can recognize another trigonometric identity.
The equation looks a bit like $\cos(u) \sin(x) - \sin(u) \cos(x) = a$ for some values $u$ and $a$ to be determined, only with a multiplicative factor. If you had this, you could apply the sine-difference formula: the equation is equivalent to $\sin(x-u) = a$. Your idea to multiply by a constant was on the right; the other part of the puzzle is this identity which guides you towards a multiplicative constant that helps. The equation constrains $\dfrac{\cos(u)}{\sin(u)} = \dfrac{\sqrt 3}{1}$ which you should recognize as having the solution $u = \frac{\pi}{6}$. Since $\cos(\frac{\pi}{6}) = \frac{\sqrt 3}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$, multiply the original equation by $\frac{1}{2}$ to get
$$ \sin\left(x - \frac{\pi}{6}\right) = \frac{\sqrt 2}{2} $$
Alternatively, to find the multiplicative coefficient, you can remark that $\cos^2(u) + \sin^2(u) = 1$, while here you have $(\sqrt 3)^2 + (1)^2 = 4$. Thus you need to divide the equation by $\sqrt 4$ to get coefficients that are a (cos, sin) pair.
(Here I am essentially deriving the formula shown by Yiyuan Lee from a more common identity.)
Since $\frac{\sqrt 2}{2} = \sin(\frac{\pi}{4})$, the equation is equivalent to
$$
x - \frac{\pi}{6} = \frac{\pi}{4} + 2k\pi
\qquad\text{or}\qquad
x - \frac{\pi}{6} = \pi - \frac{\pi}{4} + 2k\pi
$$
i.e. $x = \frac{5\pi}{12} + 2k\pi$ or $x = \frac{11\pi}{12} + 2k\pi$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof: $ \lfloor \sqrt{ \lfloor x\rfloor } \rfloor = \lfloor\sqrt{x}\rfloor $. I need some help with the following proof:
$ \lfloor \sqrt{ \lfloor x\rfloor } \rfloor = \lfloor\sqrt{x}\rfloor $.
I got:
(1) $[ \sqrt{x} ] \le \sqrt{x} < [\sqrt{x}] + 1 $ (by definition?).
(2) $[ \sqrt{x} ]^2 \le x < ([\sqrt{x}] + 1)^2 $.
(3) $[ \sqrt{x} ]^2 \le [x] \le x < ([\sqrt{x}] + 1)^2$
??
| Note that every nonnegative number $x$ is between the squares of two consecutive nonnegative integers. Say $n^2 \leq x < (n+1)^2$. Then $n \leq \sqrt{x} < n + 1$, so $\lfloor \sqrt{x} \rfloor = n$ in this situation.
So you just have to make sure the left hand side is also $n$. Since $n^2$ is an integer, one has
$n^2 \leq \lfloor x \rfloor \leq x$, so $n^2 \leq \lfloor x \rfloor < (n+1)^2$. So exactly as above one has $\lfloor \sqrt{\lfloor x \rfloor} \rfloor = n$ as well.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
The normal line intersects a curve at two points. What is the other point? The line that is normal to the curve $\displaystyle x^2 + xy - 2y^2 = 0 $ at $\displaystyle (4,4)$ intersects the curve at what other point?
I can not find an example of how to do this equation. Can someone help me out?
| A key to
easily solving this
is lab bhattacharjee's observation that
$x^2+xy-2y^2=(x+2y)(x-y)
$.
Therefore,
the curve is the union
of the lines
$L_1: x+2y = 0$
and
$L_2: x-y = 0$.
Since $(4, 4)$ is on
$L_2$, and not $L_1$,
and the slope of $L_2$
is $1$,
the slope of the normal is $-1$.
The equation of the line through
$(4, 4)$
and slope $-1$ is
$\frac{y-4}{x-4} = -1
$
or
$y-4 = 4-x$
or
$y=-x+ 8$.
This intersects $L_1$
when
$0
=x+2y
= x+2(-x+8)
= -x+16
$
or
$x = 16$.
Then
$y = -x+8
= -8
$,
so the point is
$(16, -8)$.
| {
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"url": "https://math.stackexchange.com/questions/707716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factor $x^4 + 64$ I found a page where this problem was solved but his english is broken so its difficult to understand his explanation. His first step was to divide the constant, 64, by the exponent 4. What is his reasoning behind this step?
http://www.wyzant.com/resources/answers/163/factor_x_4_64
| 1)He added and then subtracted $16x^2$, so the value wouldn't change
$$x^4 + 64 = x^4 + 16x^2 - 16x^2 + 64$$
2)He used the fact that $(A+B)^2=A^2+2AB+B^2$, in your case $A=x^2$ and $B=8$ and the formula was used "backwards" to get $(A+B)^2$ back
$$x^4 + 64 = x^4 + 16x^2 + 64- 16x^2=(x+8)^2-(4x)^2$$
3)$A^2-B^2=(A-B)(A+B)$, where in your case $A=x^2+8$ and $B=4x$, so we have
$$(x^2+8)^2-(4x)^2=(x^2+8-4x)(x+8+4x)$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
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} |
Sum of a simple infinite series
Evaluate:
$$\sum\limits_{n=1}^\infty \frac{n^2}{3^n}.$$
By the ratio test, $\displaystyle\lim_{n\to\infty} \frac{(n+1)^2}{3^{n+1}}\cdot\frac{3^n}{n^2}=1/3,$ which is less than 1, therefore the series is convergent.
Now I am stuck on how to evaluate this series, without the $n^2$ on top, it can be easily calculated by the geometric series formula. Any help would be appreciated.
| Just to expand on David's comment:
\begin{align*}
\frac{1}{1-x} &= 1 + x + x^2 + \cdots + x^n + \cdots \\
\frac{d}{dx} \left( \frac{1}{1-x} \right) =
\frac{1}{(1-x)^2} &= 1 + 2x + 3x^2 + \cdots + nx^{n-1} + \cdots \\
\frac{x}{(1-x)^2} &= x + 2x^2 + 3x^3 + \cdots + nx^{n} + \cdots \\
\frac{d}{dx} \left( \frac{x}{(1-x)^2} \right) =
\frac{1 + x}{(1 - x)^3} &= 1 + 2^2x + 3^2x^2 + \cdots + n^2x^{n-1} + \cdots \\
\frac{x(1 + x)}{(1 - x)^3} &= x + 2^2x^2 + 3^2x^3 + \cdots + n^2x^{n} + \cdots \\
\end{align*}
Now plug in $x = \frac13$.
| {
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Factoring $x^5 + x^4 + x^3 + x^2 + x + 1$ without using $\frac{x^n - 1}{x-1}$? I was at a math team meet today and one of the problems was to factor $x^5 + x^4 + x^3 + x^2 + x + 1$. It also gave the hint that it decomposes into two trinomials and a binomial.
The solution they gave was based on the fact that $\frac{x^6 - 1}{x-1} = x^5 + x^4 + x^3 + x^2 + x + 1$ and from there the solution is pretty straightforward. However, I was not aware of that factorization. The only ones I have really learned are $x^2 - y^2 = (x-y)(x+y)$ and $x^3 \pm y^3 = (x \pm y)(x^2 \mp xy + y^2)$. Is there any other way I could have solved this factorization without using the ones they used?
| Here's how you could have found that identity without having known it before hand.
First, notice that setting $x = -1$ gives $0$. Thus, $(x - (-1)) = x+1$ is a root.
Using polynomial long division, we find that we can reduce it to $(x^4 + x^2 + 1)(x+1)$
Now, what to do with the $x^4 + x^2 + 1$ term? Recall that $x^3 - 1 = (x-1)(x^2 + x + 1)$. If we plug $x = x^2$ into this identity, we find $x^6 - 1 = (x^2 - 1)(x^4 + x^2 + 1)$, so $x^4 + x^2 + 1 = \frac{x^6 - 1}{x^2 - 1}$.
We have now reduced it to $\frac{(x^6 - 1)(x+1)}{(x-1)(x+1)}$ = $\frac{x^6 - 1}{x-1}$ and, as you said, the rest is easy from here using the identities you listed in the OP.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Series Expansion Of An Integral. I want to find the first 6 terms for the series expansion of this integral:
$$\int x^x~dx$$
My idea was to let: $$x^x=e^{x\ln x}$$
From that we have: $$\int e^{x\ln x}~dx$$
The series expansion of $e^x$ is: $$\sum\limits_{n=0}^\infty\frac{x^n}{n!}$$
Then we have:
$$\int e^{x\ln x}~dx=\int\sum\limits_{n=0}^\infty\frac{(x\ln x)^n}{n!}dx$$
I integrated the first 3 terms (since it was the easiest the work with):
$$x+\frac{1}{4}x^2(2\ln x-1)+\frac{1}{54}x^3(9(\ln x)^2-6\ln x+2)+\cdots$$
Is there any other efficient way of getting the first 6 terms of the series expansion of the integral?
| You are in the right approach. Then you have to face $\int x^n(\ln x)^n~dx$ for non-negative integers $n$ :
$\int x^n(\ln x)^n~dx$
$=\int(\ln x)^n~d\left(\dfrac{x^{n+1}}{n+1}\right)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{x^{n+1}}{n+1}d((\ln x)^n)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)x}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^n(\ln x)^{n-1}}{n+1}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{n(\ln x)^{n-1}}{n+1}d\left(\dfrac{x^{n+1}}{n+1}\right)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{x^{n+1}}{n+1}d\left(\dfrac{n(\ln x)^{n-1}}{n+1}\right)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^{n+1}(\ln x)^{n-2}}{(n+1)^2x}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^n(\ln x)^{n-2}}{(n+1)^2}dx$
$=\cdots\cdots$
$\vdots$
$\vdots$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}-\int\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times1)x^n}{(n+1)^n}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}+\dfrac{(-1)^n(n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+1}}+C$
$=\dfrac{(n+1)x^{n+1}(\ln x)^n}{(n+1)^2}-\dfrac{(n+1)nx^{n+1}(\ln x)^{n-1}}{(n+1)^3}+\cdots\cdots+\dfrac{(-1)^{n-1}((n+1)n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^{n+1}}+\dfrac{(-1)^n((n+1)n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+2}}+C$
$=\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}(n+1)!x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+2}}+C$
$=\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$
$\therefore\int\sum\limits_{k=0}^n\dfrac{(x\ln x)}{n!}dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$
Hence $\int x^x~dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Solving for $3^x - 1 = 2^y$ Besides $x=2, y=3$, are there any other solutions?
I know that if there is another solution:
*
*$y$ is odd since $2^y \equiv -1 \pmod 3$
*$x$ is even since $3^x - 1 \equiv 0 \pmod 8$
*$3 | y$ since $-1 \equiv 2^y \pmod 9$
Are there any other solutions? If not, what is the argument for showing that if $3^x > 9$, then $2^y \neq 3^x-1$
Thanks,
-Larry
| You don't need the full strength of Catalan's conjecture here.
Two solutions are found easily. $3^1-1=2^1$ gives $x=y=1$. $3^2-1=2^3$ gives $x=2$ and $y=3$.
To prove that these are the only solutions, assume $x > 2$. As you have noted, $x$ must be even: $x = 2z$. Now we have
$$
3^x-1 = 3^{2z}-1 = (3^z-1)(3^z+1) = 2^y.
$$
It follows that both $3^z-1$ and $3^z+1$ are powers of $2$. But this is impossible, because both numbers are larger than $2$ (since $z > 1$), and they are only two units apart. QED.
PS: I think I've seen this argument somewhere on math.SE.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find any affine transformation that swaps affine lines The task is to find any affine transformation that will swap the following two lines:
$$L_1:(1,1,1) + span((1,0,2))$$
$$L_2:(1,0,1) + span((1,0,-1))$$
From what I understand there is a number of equations I can make:
$$f((1,1,1))=(1,0,1)$$
$$f((1,0,1))=(1,1,1)$$
I am not sure how to create the other two equations that I need.
I also understand that an affine transformation consists of a linear transformation and a translation.
Does this translation have to be $(1,1,1) - (1,0,1)$ or what else could it be.
I'm really confused by the affine transformations.
| \begin{pmatrix} 2\\ 0\\ 0 \end{pmatrix}
\rightarrow
\begin{pmatrix} 2\\ 1\\ 3 \end{pmatrix}.
$$
I will use general approach from "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely" to solve the latter problem. It is shown there, the transformation can be presented as
$$
\vec{A}(\vec{x}) =
(-1)
\frac{
\det
\begin{pmatrix}
0 & (1,0,1)^T & (2,0,0)^T & (1,1,1)^T & (2,1,3)^T \\
\begin{matrix}
x_{1} \vphantom{x_{1}^{(1)}} \\
x_{2} \vphantom{x_{1}^{(1)}} \\
x_{3} \vphantom{x_{1}^{(1)}} \\
\end{matrix} &
%
\begin{matrix}
1 \\
1 \\
1 \\
\end{matrix} &
%
\begin{matrix}
2 \\
1 \\
3 \\
\end{matrix} &
%
\begin{matrix}
1 \\
0 \\
1 \\
\end{matrix} &
%
\begin{matrix}
2 \\
0 \\
0 \\
\end{matrix} \\
%
1 & 1 & 1 & 1 & 1
\end{pmatrix}
}{
\det
\begin{pmatrix}
\begin{matrix}
1 \\
1 \\
1 \\
\end{matrix} &
%
\begin{matrix}
2 \\
1 \\
3 \\
\end{matrix} &
%
\begin{matrix}
1 \\
0 \\
1 \\
\end{matrix} &
%
\begin{matrix}
2 \\
0 \\
0 \\
\end{matrix} \\
%
1 & 1 & 1 & 1
\end{pmatrix}
}.
$$
Doing determinants I get
$$
= \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}
\frac{2 - x_1 + 3 x_2 - x_3}{3} +
\begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}
\frac{x_1 + x_3 - 2}{3} +
\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
\frac{4 - 2 x_1 - 3 x_2 + x_3}{3} +
\begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}
\frac{2 x_1 - x_3 - 1}{3}
$$
or simplified
$$
\vec{A}(\vec{x}) =
\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} x_1 +
\begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} x_2 +
\begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} x_3 +
\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}
$$
or in canonical form
$$
\vec{A}(\vec{x}) =
\begin{pmatrix} 1 & 0 & 0 \\
0 & -1 & 0 \\
1 & 0 & -1 \end{pmatrix}
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} +
\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}.
$$
For more details on the methods used, you can always refer to "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely". The latter contains many problems similar to this one as explained by the authors of the method presented.
| {
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Calculus I: Tangent line at certain point, second derivative of the curve given. How to solve? I have never seen this problem in my homework.. I need help right before final!!
Let $\frac{d^2y}{dx^2}=y''=-x^3$ at every point on a curve. The equation of the tangent line at $(1,1)$ is $y = 3-2x$. Find the equation of the curve.
If someone can help me out with this problem, will greatly appreciate. Thank you.
| Since
$y'' = -x^3, \tag{1}$
the general form of $y'$ is, by integration,
$y' = -\dfrac{1}{4}x^4 + a, \tag{2}$
and so the general form of $y(x)$ must be
$y = -\dfrac{1}{20}x^5 + ax + b; \tag{3}$
since the curve (3) passes through the point $(1, 1)$ we have
$1 = -\dfrac{1}{20} + a + b; \tag{4}$
an since the slope at this point is $y'(1) = -2$ we also have
$-2 = -\dfrac{1}{4} + a \tag{5}$
or
$a = -\dfrac{7}{4}. \tag{6}$
When (6) is used in (4) we find
$b = 1 + \dfrac{1}{20} + \dfrac{7}{4} = \dfrac{14}{5}; \tag{7}$
thus we have
$y(x) = - \dfrac{1}{20}x^5 - \dfrac{7}{4}x + \dfrac{14}{5}. \tag{8}$
A race to the finish with izoec! Some problems only have one answer!
Hope this helps! Cheerio,
and as always,
Fiat Lux!!!
| {
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How prove this $\ln{(x+\sqrt{x^2+1})}<\frac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$ let $0<a<1,x<0$,show that
$$\ln{(x+\sqrt{x^2+1})}<\dfrac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$$
My idea:
$$\Longleftrightarrow \ln{(\sqrt{x^2+1}+x)}<\dfrac{x(a^x-1)\ln{a}}{(a^x+1)\ln{(\sqrt{x^2+1}-x)}}$$
Then following I fell very ugly.Thank you
| If $x<0$, then $x+\sqrt{x^2+1}<x+\sqrt{x^2-2x+1}=x+|x-1|=|x-1|-|x|\leq1$, so
$$\ln(x+\sqrt{x^2+1})<0$$
and $\sqrt{x^2+1}-x>1+0=1$, so
$$\ln(\sqrt{x^2+1}-x)>0$$
Additionally if $0<a<1$, then $a^x-1>0$, $a^x+1>0$ and $\ln a<0$, therefore
$$\frac{x(a^x-1)\ln a}{(a^x+1)\ln(\sqrt{x^2+1}-x)}=\frac{-+-}{++}>0$$
So the inequality holds quite trivially.
| {
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"timestamp": "2023-03-29T00:00:00",
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Remainder of the polynomial A polynomial function $f(x)$ with real coefficients leaves the remainder $15$ when divided by $x-3$, and the remainder $2x+1$ when divided by $(x-1)^2$. Then the remainder when $f(x)$ is divided by $(x-3)(x-1)^2$ is?
What I have thought-The remainder must be of the form $ax^2+bx+c$. Now applying the remainder theorem, I am able to find $2$ equations in $a,b,c$ . eg
Let $f(x)$=$(x-3)$$h(x)$+$15$ ...................................(1)
Also let $f(x)=(x-3)(x-1)^2g(x)+ax^2+bx+c$...........................(2)
Put $x=3$ and using (1) we get $15=9a+3b+c$
Similarly I can get another equation using the other information given. But I am only able to get 2 equations in 3 variables. From where do I get the 3rd equation in $a,b,c$ and hence the remainder?
| Hint $\ f(x) = 2x\!+\!1+ (x\!-\!1)^2 (c + (x-3)g(x))\,$ and $\,15 = f(3) = 7+4c\ $ so $\ c = \,\ldots$
| {
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Show that $89|(2^{44})-1$ Show that $89|(2^{44})-1$
My teacher proved this problem using mod can someone explain the process step by step? Thank you so much!
| With a bit of calculation.
Need to prove $2^{44}\equiv 1\pmod{89}$.
$2^8=256=2\cdot 89+78=3\cdot 89-11$, so $2^8\equiv 78\equiv -11\pmod{89}$ $-$ it is a definition of $\pmod{}$ operation.
Then $2^3\cdot 2^8\equiv 8\cdot 2^8\equiv8\cdot 11\equiv-88 \equiv 1\pmod{89}$ $-$ because $1=1\cdot 89-88$.
So, you have $2^{11}\equiv 1\pmod{89}$ which means $2^{11}=k\cdot 89 +1$, where $k$ is an integer.
You can obtain from it that $(2^{11})^4=m\cdot 89 +1$, where $m$ will be another integer.
| {
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} |
One Diophantine equation I wonder now that the following Diophantine equation:
$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$
have only this formula describing his decision?
$a=-(k^2+2(p+s)k+p^2+ps+s^2)$
$b=2k^2+4(p+s)k+3p^2+3ps+2s^2$
$c=3k^2+4(p+s)k+2p^2+ps+2s^2$
$d=2k^2+4(p+s)k+2p^2+3ps+3s^2$
$k,p,s$ - what some integers.
By your question, I mean what that formula looks like this. Of course I know about the procedure of finding a solution, but I think that the formula would be better.
| My more general method of calculation. Enables us to solve and other factors. Find out whether or when given coefficients solutions and immediately write the formula.
For example, consider the equation:
$4(a^2+b^2+c^2+d^2)=3(a+b+c+d)^2$
Then the solutions are of the form:
$a=-(p^2+4(k+s)p+2k^2+2ks+2s^2)$
$b=p^2+4(k+s)p+6k^2+6ks+2s^2$
$c=p^2+4(k+s)p+2k^2+6ks+6s^2$
$d=3p^2+4(k+s)p+2k^2-2ks+2s^2$
I think the best result by a direct solution of these equations. Brute force search of the law or does not make sense. Spend a lot of effort, but the result does not always work.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find all ordered pairs $(m,n)$ such $\frac{n^3-1}{mn-1}$ is an integer Question:
Find all ordered pairs $(m,n)$ where $m$ and $n$ are positive integers, such that $$\dfrac{n^3-1}{mn-1}$$ is an integer.
I know this problem is very simaler this 1994 IMO,problem 4,But My problem is different.
This IMO problem is :
Find all ordered pairs $(m,n)$ where $m$ and $n$ are positive integers, such that $$\dfrac{n^3+1}{mn-1}$$ is an integer.
this problem solution can see
link
Thank you for you help to solve this
| Note that if $(m, n)$ is a solution then $(mn-1) \mid (n^3-1)$ so $(mn-1) \mid m^3(n^3-1)$. Also $(mn-1)\mid m^3n^3-1$ so $(mn-1) \mid m^3-1$, so $(n, m)$ is also a solution.
We may thus WLOG assume $m \geq n$. If $n=1$ then $\frac{n^3-1}{mn-1}=0$ so $(m, 1)$ is a solution for all positive integers $m>1$. I would reject $(1, 1)$ as that leads to $\frac{0}{0}$.
Otherwise $n \geq 2$. Note that $\frac{n^3-1}{mn-1} \equiv 1 \pmod{n}$ so we may write $\frac{n^3-1}{mn-1}=kn+1$, for some $k \in \mathbb{Z}$.
Now $m, n \in \mathbb{Z}^+$ so $0<\frac{n^3-1}{mn-1}=kn+1$. Since $n \geq 2$, this implies $k \geq 0$.
Since $m \geq n$, we have $kn+1=\frac{n^3-1}{mn-1} \leq \frac{n^3-1}{n^2-1}=\frac{n^2+n+1}{n+1}=n+\frac{1}{n+1}<n+1$. Thus $k<1$.
Therefore $k=0$, so $\frac{n^3-1}{mn-1}=1$, so $n^3-1=mn-1$, so (since $m, n>0$) $m=n^2$.
In conclusion, all positive integer solutions are given by $(m, 1)$ for $m>1$, $(1, n)$ for $n>1$, $(m, m^2)$ for $m>1$, and $(n^2, n)$ for $n>1$.
| {
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If $a+b+c+d = 2$, then $\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}+\frac{c^2}{(c^2+1)^2}+\frac{d^2}{(d^2+1)^2}\le \frac{16}{25}$ If $a+b+c+d = 2$, prove that
$$\dfrac{a^2}{(a^2+1)^2}+\dfrac{b^2}{(b^2+1)^2}+\dfrac{c^2}{(c^2+1)^2}+\dfrac{d^2}{(d^2+1)^2}\le \dfrac{16}{25}$$
Also $a,b,c,d \ge 0$.
| Assume $0 \le a \le b \le c \le d$, with $a+b+c+d=2$,
Then $(48a-4)(a^2+1)^2-125a^2 = (2a-1)^2(12a^3+11a^2+32a - 4) \ge 0$, for $a \ge \frac{1}{8}$;
(See that $32a-4 \ge 0$ and $12a^3+11a^2$ is positive).
That is $\dfrac{a^2}{(a^2+1)^2} \le \dfrac{48a-4}{125}$, and similarly for $b,c,d$ and adding them we have $\sum\limits_{a,b,c,d} \dfrac{a^2}{(a^2+1)^2} \le \sum\limits_{a,b,c,d} \dfrac{48a-4}{125}=\dfrac{80}{125}=\dfrac{16}{25}$.
The case $a < \frac{1}{8}$,
We have $(540x + 108)(x^2+1)^2 - 2197x^2 = (3x-2)^2(60x^3+92x^2+216x+27) \ge 0$, for $x \ge 0$.
Thus $\sum\limits_{x \in \{b,c,d\}} \dfrac{x^2}{(x^2+1)^2} \le \sum\limits_{x \in \{b,c,d\}} \dfrac{540x+108}{2197}=\dfrac{108}{169}-\dfrac{540a}{2197}$
and, $\dfrac{a^2}{(a^2+1)^2} < a^2 < \dfrac{a}{8} < \dfrac{540a}{2197}$.
So, $\sum\limits_{a,b,c,d} \dfrac{a^2}{(a^2+1)^2} \le \dfrac{108}{169} < \dfrac{16}{25}$
Equality occurs iff $a=b=c=d=\dfrac{1}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Linear Transformation from $\alpha$ to $\beta$ T: $R^3$ $\to$ $R^2$
$$[T]_{\beta\alpha} =
\begin{matrix}
2 & 3 & 1 \\
1 & 2 & 1 \\
\end{matrix}
$$
$\alpha$ = {(1, -1, 1), (0, 1, 0), (1, 0, 0)}
$\beta$ = {(3, 2), (2, 1)}
Find: T((x, y, z)) for any x, y, z in $R^3$
My approach:
$[T(v)]_{\beta}$ = $[T]_{\beta\alpha}$$[v]_{\alpha}$
so,
$ z\begin{pmatrix}
1 \\
-1 \\
1 \\
\end{pmatrix} +
(y+z)\begin{pmatrix}
0 \\
1 \\
0 \\
\end{pmatrix}
+(x-z)\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix} = \begin{pmatrix}
x \\
y \\
z \\
\end{pmatrix} $
and
$[(x, y, z)]_{\alpha} = [(z), (y + z), (x - z)]$
then
$[T]_{\beta\alpha}[\begin{pmatrix}x\\y\\z\\\end{pmatrix}]_{\alpha}=\begin{pmatrix}x + 3y + 4z\\x + 2y + 2z\\\end{pmatrix}$
finally,
$(x + 3y + 4z)\begin{pmatrix}3\\2\\\end{pmatrix} + (x + 2y + 2z)\begin{pmatrix}2\\1\\\end{pmatrix} = \begin{pmatrix}5x + 13y + 16z\\3x + 8y + 10z\\\end{pmatrix}$
and
T((x, y, z)) = ((5x + 13y + 16z), (3x + 8y + 10z))
But this does not look like it makes lot of sense, I think I misunderstand something.
Can anyone give me a pointer?
| it is correct in my humble opinion except rhe last step.
finally,
$$(x + 3y + 4z)\begin{pmatrix}2\\1\\\end{pmatrix} + (x + 2y + 2z)\begin{pmatrix}3\\2\\\end{pmatrix} =...$$
instead of
$$(x + 3y + 4z)\begin{pmatrix}3\\2\\\end{pmatrix} + (x + 2y + 2z)\begin{pmatrix}2\\1\\\end{pmatrix} = \begin{pmatrix}5x + 13y + 16z\\3x + 8y + 10z\\\end{pmatrix}$$
| {
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Finding the coefficient of a generating function Given $f(x) = x^4\left(\frac{1-x^6}{1-x}\right)^4 = (x+x^2+x^3+x^4+x^5+x^6)^4$. This is the generating function $f(x)$ of $a_n$, which is the number of ways to get $n$ as the sum of the upper faces of four thrown dice.
How do I calculate a coefficient from said generating function, for example $a_{17}$?
I could of course write it all out, but that would take a lot of time. Is there a quicker way to do it?
I've rewritten the formula as $x^4(1-x^6)^4\left(\frac1{1-x}\right)^4$, giving me to find the coefficient of $x^{17-4}=x^{13}$ of $(1-x^6)^4\left(\frac1{1-x}\right)^4$. I'm, however, stuck there.
| $$f(x) = x^4\left(\frac{1-x^6}{1-x}\right)^4 = (x+x^2+x^3+x^4+x^5+x^6)^4=x^4(1+x+^2+x^3+x^4+x^5)^4=\\
=x^4\left(\sum_{k=0}^{5}x^k\right)^4=x^4\left(\frac{1-x^6}{1-x}\right)^4=x^4(1-x^6)^4(1-x)^{-4}=\\=x^4\sum_{j=0}^{4}(-1)^j\binom{4}{j}x^{6j}\sum_{h=0}^{\infty}(-1)^h\binom{-4}{h}x^h$$
Note that $(-n)!=(-n)(-n-1)(-n-2)\cdots=\infty$, and $\binom{-n}{k}=\frac{(-n)!}{k!(-n-k)!}=(-1)^k\frac{n^{(k)}}{k!}=(-1)^k\frac{(n+k-1)_k}{k!}=(-1)^k\binom{n+k-1}{k}$ so
$$f(x)=x^4\sum_{j=0}^{4}(-1)^j\binom{4}{j}x^{6j}\sum_{h=0}^{\infty}\binom{h+4-1}{h}x^h
$$
Any exponent S of the polynomial will come from the sum $S=4+6j+h \rightarrow h=S-6j-4$ and it coefficient will be
$$c(S)=\sum_{j=0}^{4}(-1)^j\binom{4}{j}\binom{S-6j-1}{S-6j-4}$$
where$\binom{S-6j-1}{3}\neq 0$ iff $S-6j-1\geq 3$ i.e. $j\leq \frac{S-4}{6}$. So
$$c(S)=\sum_{j=0}^{\lfloor\frac{S-4}{6}\rfloor}(-1)^j\binom{4}{j}\binom{S-6j-1}{3};\ S\in [4,24]_{\Bbb N}$$
| {
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How do I integrate $\frac{1}{x^6+1}$ My technique so far was substitution with the intent of getting to a sum of three fractions with squares in their denominators.
$t = x^2 \\
\frac{1}{x^6 + 1} = \frac{1}{t^3+1} = \frac{1}{(t+1)(t^2-t+1)}$
Then I try to reduce this fraction into a sum of two fractions
$\frac{A}{t+1} + \frac{B}{t^2-t+1} = \frac{(At^2-(A-B)t) + A + B}{(t+1)(t^2-t+1)}$
And this is where I reach a dead-end
$\begin{cases}
At^2-(A-B)t &= 0 \\
A + B &= 1
\end{cases}$
Any techniques I'm overlooking?
| Observe that $$x^6 + 1 = (x^2)^3 + 1 = (x^2 + 1)(x^4 - x^2 + 1).$$ To factor the quartic term, suppose there is a factorization of the form $$\begin{align*} x^4 - x^2 + 1 &= (x^2 + ax + 1)(x^2 + bx + 1) \\ &= x^4 + (a+b)x^3 + (ab+2)x^2 + (a+b)x + 1. \end{align*}$$ We thus require $a + b = 0$, and $ab + 2 = -1$, from which we may take without loss of generality that $a = \sqrt{3}$, $b = -\sqrt{3}$. Thus the desired factorization is $$x^6 + 1 = (x^2 + 1)(x^2 + \sqrt{3}x + 1)(x^2 - \sqrt{3}x + 1).$$ Now we suppose there exist variables $A, B, C, D, E, F$ such that $$\frac{1}{x^6+1} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{x^2 + \sqrt{3}x + 1} + \frac{Ex + F}{x^2 - \sqrt{3}x + 1}.$$ Multiplying both sides by $x^6 + 1$ and equating coefficients on the RHS to the LHS (all are zero except the constant coefficient) gives a system of equations in the unknown variables, the solution of which is $$(A,B,C,D,E,F) = \left(0, \tfrac{1}{3}, \tfrac{1}{2\sqrt{3}}, \tfrac{1}{3}, -\tfrac{1}{2\sqrt{3}}, \tfrac{1}{3}\right).$$ This gives us three integrands of the form $$\frac{P(x)}{Q(x)}$$ where $P$ is at most linear, and $Q$ is quadratic, which are easily (but tediously) handled. For instance, $$\int \frac{\sqrt{3}x + 2}{6(x^2+\sqrt{3}x+1)} \, dx = \frac{\sqrt{3}}{12} \int \frac{2x + \sqrt{3}}{x^2 + \sqrt{3}x + 1} \, dx - \frac{1}{12} \int \frac{dx}{x^2 + \sqrt{3}x + 1},$$ and the first integral is simply a logarithm, and the second is an arctangent after completing the square in the denominator: $x^2 + \sqrt{3}x + 1 = (x + \tfrac{\sqrt{3}}{2})^2 + (\tfrac{1}{2})^2$.
| {
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The value of $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}$ is
The value of $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}$ is?
I tried this: $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}=2\binom{50}0\binom{50}1+2\binom{50}1\binom{50}2+\dots+\binom{50}{25}^2$
How can I get that this expression is equal to $\binom{100}{51}$?
| $$(1+x)^{50}=\binom{50}{0}+\binom{50}{1}x+\cdots+\binom{50}{49}x^{49}+\binom{50}{50}x^{50}$$
$$(x+1)^{50}=\binom{50}{0}x^{50}+\binom{50}{1}x^{49}+\cdots+\binom{50}{49}x+\binom{50}{50}$$
Multiplying we observe that our required sum is the coefficient of $x^{49}$ in $(1+x)^{100}$
$$S=\binom{100}{49}=\binom{100}{51}$$
| {
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Hard Olympiad Inequality Let x,y,z be positive real numbers such that $xy+xz+yz=1$. Prove that $$\sqrt{x^3+x}+ \sqrt{y^3+y}+ \sqrt{z^3+z} \geq 2 \cdot \sqrt{x+y+z}$$.
I tried to square expand homogenize then majorize. But I couldn't make it work. Any help would be much appreciated.
| By C-S and Schur we obtain:
$$\sum_{cyc}\sqrt{a^3+a}=\sqrt{\left(\sum_{cyc}\sqrt{a^3+a^2b+a^2c+abc}\right)^2}=$$
$$=\sqrt{\sum_{cyc}\left(a^3+a^2b+a^2c+abc+2\sqrt{(a^2(a+b+c)+abc)(b^2(a+b+c)+abc)}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}(a^3+a^2b+a^2c+abc+2(ab(a+b+c)+abc))}=$$
$$=\sqrt{\sum_{cyc}(a^3+3a^2b+3a^2c+5abc)}=\sqrt{\sum_{cyc}(a^3-a^2b-a^2c+abc+4(a^2b+a^2c+abc))}\geq$$
$$\geq\sqrt{4\sum_{cyc}(a^2b+a^2c+abc)}=\sqrt{4(a+b+c)(ab+ac+bc)}=2\sqrt{a+b+c}.$$
Done!
| {
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Area of triangle A triangle is inscribed in a circle. The vertices of triangle divide the circle into three arcs of length 3, 4 and 5 units, then find the area of triangle.
|
Study the above diagram which describes the problem.
From the calculations of the arc lengths, we have:
$$3 = rA, \qquad4 = rB, \qquad 5 = rC$$
$$3 + 4 + 5 = rA + rB + rC = r(A + B + C)$$
$$r = \frac{12}{A + B + C} = \frac{6}{\pi}$$
We also know that the angles $A,B,C$ are in proportions corresponding to the respective arc length proportions. So,
$$A = \frac{6\pi}{12}, \qquad B = \frac{8\pi}{12}, \qquad C = \frac{10\pi}{12}$$
I believe it is easy to compute the area of the inscribed triangle from here: It is given by
$$\frac{1}{2}r^2\sin A + \frac{1}{2}r^2\sin B + \frac{1}{2}r^2\sin C$$
| {
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If $y = e^{x(y-1)}$ then $y \approx 1-2(x-1)$ when $0Assume $y = e^{x(y-1)}$.
Then $y \approx 1-2(x-1)$ when $0<x-1<<1$
I thought of something like that:
$$ e^{x(y-1)} = e^{-2(x-1)}e^{xy+x-2}=(1-2(x-1)+O(x-1)^2)e^{xy+x-2}$$
But I failed to prove that $e^{xy+x-2} \approx 1$
Any ideas?
Thanks
| This isn't rigorous, but will give you a starting point: since $e^z \approx 1 + z + \dfrac 12 z^2$ you have $y = e^{x(y-1)} \approx 1 +x(y-1) + \dfrac 12 x^2 (y-1)^2$. That is, $$y-1 \approx x(y-1) + \frac 12 x^2 (y-1)^2$$ so that (after canceling the $y-1$) $$1 \approx x + \dfrac 12 x^2 (y-1)$$ and $$y \approx \frac{2}{x^2} (1-x) + 1$$ Since $0 < x-1 << 1$ you can approximate $\dfrac 2{x^2}$ by $2$.
| {
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Why does $\prod_{n=1}^\infty \frac{n^3 + n^2 + n}{n^3 + 1}$ diverge? $\prod_{n=1}^\infty \frac{n^3 + n^2 + n}{n^3 + 1}$ diverges, and I have no idea why? It would seem using L'hop, $\frac{n^3 + n^2 + n}{n^3 + 1}$ goes to 1. So it should end up just being $1\cdot1\cdot1\cdots$, which makes me feel it converges. Is the problem that it always adds that $.000\ldots0001$?
| Notice that for $n\ge1$ you have
$$\frac{n^3+n^2+n}{n^3+1}=
1+\frac{n^2+n-1}{n^3+1} \ge
1+ \frac{n^2-n+1}{n^3+1} =
1+\frac1{n+1} = \frac{n+2}{n+1}.$$
Therefore
$$\prod_{k=1}^n \frac{k^3+k^2+k}{k^3+1} \ge \prod_{k=1}^n \frac{k+2}{k+1} = \frac32 \cdot \frac 43 \cdots \frac{n+2}{n+1} = \frac{n+2}2.$$
And for $n\to\infty$ we have $\lim\limits_{n\to\infty} \frac{n+2}2=\infty$.
| {
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Prove that if b is coprime to 6 then $b^2 \equiv 1 $ (mod 24) Let $\gcd(b,6) = 1$. Prove that $b^2 \equiv 1 $ (mod 24).
Now I have that as $\gcd(b,6) = 1$, we know that $3\nmid b $ and $2\nmid b$ (else the GCD would be 3 or 2 resp.)
So as $2\nmid b$, $b$ must be odd. Hence $b^2$ is also odd.
Then I'm not sure where to take it from here? Maybe I've gone down the wrong path to start with, and should be looking more along the lines of the Chinese Remainder Theorem? I'm not sure.
| Starting with $b$ is odd, so $b = 2k + 1$, and $b^2 - 1 = 4k^2 + 4k = 4k(k+1)$ clearly divisible by $8$. So we need to show that $b^2 - 1$ is divisible by$3$. Since $b$ is not divisible by $3$, $b = 3k + 1$ or $b = 3k +2$. If $b = 3k + 1$, then $b - 1 = 3k$ and $3 | b - 1$, so $3 | b^2 - 1$. And if $b = 3k + 2$, then $b + 1 = 3k + 3$ which is divisible by $3$. So $3 | (b +1 )(b - 1) = b^2 - 1$. So $3 | b^2 - 1$ and since $gcd(3,8) = 1$, $24 | b^2 - 1$.
| {
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Inequality Exercise in Apostol's Calculus I Let p and n denote positive integers. Show that:
$$n^{p} \lt \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^{p}$$
Attempt at Solution
Using the identity $b^{p+1}-a^{p+1} = (b-a)\sum_{k=0}^{p}b^{p-k}a^{k}$, let $b = n+1$ and $a = n$. Then:
$$\frac{(n+1)^{p+1} - n^{p+1}}{p+1} = \frac{\sum_{k=0}^{p}(n+1)^{p-k}n^{k}}{p+1} = \frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1}$$
I'm not sure where to proceed from here. Am I supposed to use a geometric series in here somewhere?
EDIT: I may have figured it out if someone can confirm that $(n+1)^{p-k}n^{k} > n^{p}$, $1 \le k \le p$, $\forall n$
Assuming the above is true, we have:
$$\frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1} \gt \frac{n^{p} + n^{p} + .... + n^{p} + n^{p}}{p+1} = \frac{(p+1)n^{p}}{(p+1)} = n^{p}$$
Similarly:
$$\frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1} \lt \frac{(n+1)^{p} + (n+1)^{p} + .... + (n+1)^{p} + (n+1)^{p}}{p+1} = \frac{(p+1)(n+1)^{p}}{(p+1)} = (n+1)^{p}$$
| Use the mean value theorem. ${}{}{}{}{}{}{}$
| {
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Consider the quadratic equation $ax^2-bx+c=0, a,b,c \in N. $ If the given equation has two distinct real root... Problem :
Consider the quadratic equation $~ax^2-bx+c=0, \quad a,b,c \in N. ~$ If the given equation has two distinct real roots belonging to the interval $~(1,2)~ $ then the minimum possible values of $~a~$ is
$(i) \quad -1 $
$(ii)\qquad 5 $
$(iii)~~\quad 2 $
$(iv)\quad -5 $
$(v) \qquad1 $
My approach :
We know the condition that two roots will between the two numbers viz. $(1,2)$
$(1) \quad f(1) >0 ; \qquad $
$(2)\quad f(2)>0\qquad$
$(3) \quad 1 < \frac{-b}{2a} <2\qquad $
$(4) \quad D \geq 0$
By using the above I got the following :
$(1) \quad a-b+c >0$
$(2)\quad 4a-2b+c >0$
$(3)\quad 1 < \frac{b}{2a} <2$
$(4) \quad b^2-4ac \geq 0$
Please guide further how to get the answer given in above five options. Thanks..
| Let $\alpha,\beta$ be two distinct roots of $ax^2-bx+c=0,$ Where $a,b,c\in \mathbb{N}$
So here $1<\alpha,\beta <2,$ So here $\displaystyle \alpha+\beta = \frac{b}{a}$ and $\displaystyle \alpha \cdot \beta = \frac{c}{a}$
Now Using $\bf{A.M\geq G.M}$ For $0<\alpha, 1-\alpha,\beta, 1-\beta<1$
So $$\frac{1-\alpha+\alpha}{2}\geq \sqrt{\alpha(1-\alpha)}\Rightarrow \alpha (1-\alpha )\leq \frac{1}{4}$$
Similarly $$\beta(1-\beta) \leq \frac{1}{4}$$
So we get $$\alpha \beta (1-\alpha)(1-\beta)< \frac{1}{16}$$
So $$\alpha \beta \left[1-(\alpha +\beta)+\alpha\beta\right]< \frac{1}{16}$$
So $$16c\left(a-b+c\right)<a^2$$
Here $a,b,c\in \mathbb{N}.$ So $\min(c(a-b+c)) = 1$
So we get $a^2>16\Rightarrow a>4$
So we get $\min(a) = 5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/748229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $a^4c+b^4d\ge cd$.
If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $$a^4c+b^4d\ge cd$$
It kind of seems useful to begin with a division of both sides by $cd$:
$$\frac{a^4}{d}+\frac{b^4}{c}\ge1$$
It seems like a simple Cauchy-Schwarz would be of use in this case. However, after some attempts I can't come up with the right solution. It'd be great to hear some ideas. Thanks.
| $\frac{A^4}{d} >= 1-\frac{B^4}{c}$
$\frac{A^4}{d} >= \frac{(c-B^4)}{c}$
$\frac{C\times A^4}{d} >= C-B^4$
$C\times A^4 >= CD -D\times B^4$
$C\times A^4 + D\times B^4 >= CD$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/748413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int_0^\pi \theta^2 \ln^2\big(2\cosh\frac{\theta}{2}\big)d \theta$ Hello I am trying to integrate
$$
I=\int_0^\pi \theta^2 \ln^2\big(2\cosh\frac{\theta}{2}\big)d \theta
$$
which is similar to Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$., however we have hyperbolic functions in this example.
We can possibly write the integrand as
$$
\theta^2 \ln^2\big(2\cosh\frac{\theta}{2}\big)=\theta^2\big(\ln 2+\ln \cosh \frac{\theta}{2}\big)^2=\theta^2\left(\ln^2(2)+\ln^2\cosh\frac{\theta}{2}+2\ln (2)\ln \cosh\big(\frac{\theta}{2}\big)\right).
$$
We can now write
$$
I=\ln^2(2)\int_0^\pi \theta^2d\theta +\int_0^\pi\theta^2 \ln^2 \cosh \frac{\theta}{2}d\theta+2\ln 2 \int_0^\pi\theta^2 \ln \cosh{\frac{\theta}{2}}d\theta.
$$
Change of variables $y=\theta/2$ we obtain
$$
I=\frac{\pi^3\ln^2(2)}{3}+16\ln(2)\int_0^{\pi/2} y^2 \ln \cosh (y) dy+8\int_0^{\pi/2} y^2 \ln^2 \cosh (y) dy.
$$
I am stuck here, thanks.
| $$\int_0^\pi\theta^2\ln^2\left(2\cosh\dfrac{\theta}{2}\right)d\theta$$
$$=\int_0^\pi\theta^2(\ln(e^\frac{\theta}{2}+e^{-\frac{\theta}{2}}))^2~d\theta$$
$$=\int_0^\pi\theta^2(\ln(e^\frac{\theta}{2}~(1+e^{-\theta})))^2~d\theta$$
$$=\int_0^\pi\theta^2\left(\dfrac{\theta}{2}+\ln(1+e^{-\theta})\right)^2~d\theta$$
$$=\int_0^\pi\left(\dfrac{\theta^4}{4}+\theta^3\ln(1+e^{-\theta})+\theta^2(\ln(1+e^{-\theta}))^2\right)d\theta$$
$$=\int_0^\pi\left(\dfrac{\theta^4}{4}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\theta^3e^{-n\theta}}{n}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^{n+k}\theta^2e^{-(n+k)\theta}}{nk}\right)d\theta$$
$$=\left[\dfrac{\theta^5}{20}-\sum\limits_{n=1}^\infty\dfrac{(-1)^n(n^3\theta^3+3n^2\theta^2+6n\theta+6)e^{-n\theta}}{n^5}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^{n+k}((n+k)^2\theta^2+2(n+k)\theta+2)e^{-(n+k)\theta}}{nk(n+k)^3}\right]_0^\pi$$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$$=\dfrac{\pi^5}{20}-\sum\limits_{n=1}^\infty\dfrac{(-1)^n((\pi^3n^3+3\pi^2n^2+6\pi n+6)e^{-\pi n}-6)}{n^5}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^\infty\dfrac{(-1)^{n+k}((\pi^2(n+k)^2+2\pi(n+k)+2)e^{-\pi(n+k)}-2)}{nk(n+k)^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/748956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculation of $\int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2 \cos^2 x}}dx$ Calculate
$$ \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2 \cos^2 x}}dx$$
$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\frac{\pi}{2}-\int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}dx$
Using The formula $\displaystyle \tan^{-1}(x)+\cot^{-1}(x) = \frac{\pi}{2}\Rightarrow \tan^{-1}(x) = \frac{\pi}{2}-\cot^{-1}(x).$
Now Let $\displaystyle J = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos^2 x-\sin^2 x}{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{1}{2}-\frac{\tan^2 x}{2}}dx$
Now How can I solve after that? Help required.
Thanks
| Proposition :
\begin{equation}
\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x}{\cos^2x}}\,dx=\frac{\pi}{2}\left[\arctan\sqrt{2\mu}-\arctan\sqrt{\frac{\mu}{\mu+1}}\right]\quad,\quad\text{for }\,\mu\ge0
\end{equation}
Proof :
Let
\begin{equation}
I(\mu)=\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x}{\cos^2x}}\,dx
\end{equation}
then
\begin{align}
I'(\mu)&=\partial_\mu\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x}{\cos^2x}}\,dx\\
&=\frac{1}{2}\int_0^{\Large\frac{\pi}{4}}\frac{\sqrt{\frac{\mu\cos2x}{\cos^2x}}}{\mu+\frac{\cos2x}{\cos^2x}\mu^2}\,dx\\
&=\frac{1}{2}\int_0^{\Large\frac{\pi}{4}}\frac{\sqrt{\mu(1-2\sin^2x)}}{\mu(1-\sin^2x)+(1-2\sin^2x)\mu^2}\cdot\cos x\,dx\\
&=\frac{1}{2\sqrt{2\mu}}\int_0^{\Large\frac{\pi}{2}}\frac{\sqrt{1-\sin^2\theta}}{\left(1-\frac{1}{2}\sin^2\theta\right)+(1-\sin^2\theta)\mu}\cdot\cos \theta\,d\theta\quad\Rightarrow\quad\sin^2\theta=2\sin^2x\\
&=\frac{1}{\sqrt{2\mu}}\int_0^{\Large\frac{\pi}{2}}\frac{\cos^2\theta}{\sin^2\theta+2(1+\mu)\cos^2\theta}\,d\theta\\
&=\frac{1}{\sqrt{2\mu}}\int_0^{\Large\frac{\pi}{2}}\frac{1}{\tan^2\theta+2(1+\mu)}\,d\theta\\
&=\frac{1}{\sqrt{2\mu}}\int_0^{\infty}\frac{1}{t^2+2(1+\mu)}\cdot\frac{1}{t^2+1}\,dt\quad\Rightarrow\quad t=\tan\theta\\
&=\frac{1}{\sqrt{2\mu}(1+2\mu)}\int_0^{\infty}\left[\frac{1}{t^2+1}-\frac{1}{t^2+2(1+\mu)}\right]\,dt\\
&=\frac{1}{\sqrt{2\mu}(1+2\mu)}\left[\frac{\pi}{2}-\frac{\pi}{2\sqrt{2(1+\mu)}}\right]\\
I(\mu)&=\int\frac{1}{\sqrt{2\mu}(1+2\mu)}\left[\frac{\pi}{2}-\frac{\pi}{2\sqrt{2(1+\mu)}}\right]\,d\mu\\
&=\frac{\pi}{2}\int\left[\frac{1}{\sqrt{2\mu}(1+2\mu)}-\frac{1}{\sqrt{2\mu}(1+2\mu)\sqrt{2(1+\mu)}}\right]\,d\mu\\
\end{align}
where
\begin{align}
\int\frac{1}{\sqrt{2\mu}(1+2\mu)}\,d\mu&=\int\frac{1}{1+y^2}\,dy\qquad\Rightarrow\quad y=\sqrt{2\mu}\\
&=\arctan y+C_1\\
&=\arctan\sqrt{2\mu}+C_1
\end{align}
and
\begin{align}
\int\frac{1}{\sqrt{2\mu}(1+2\mu)\sqrt{2(1+\mu)}}\,d\mu&=\int\frac{1}{(1+y^2)\sqrt{2+y^2}}\,dy\qquad\Rightarrow\quad y=\sqrt{2\mu}\\
\end{align}
Using
\begin{align}
\color{blue}{\int \frac{dx}{(x^2+1)\sqrt{x^2+a}}=\frac{1}{\sqrt{a-1}}\tan^{-1}\left(\frac{x\sqrt{a-1}}{\sqrt{x^2+a}}\right)+C}
\end{align}
It can be derived by using substitution $x=\dfrac{1}{t}$ followed by $z=\sqrt{at^2+1}$.
Hence
\begin{align}
\int\frac{1}{\sqrt{2\mu}(1+2\mu)\sqrt{2(1+\mu)}}\,d\mu&=\arctan\sqrt{\frac{\mu}{\mu+1}}+C_2
\end{align}
then
\begin{equation}
I(\mu)=\frac{\pi}{2}\left[\arctan\sqrt{2\mu}-\arctan\sqrt{\frac{\mu}{\mu+1}}\right]+C
\end{equation}
For $\mu=0$, we have $I(0)=0$ implying $C=0$. Thus
\begin{equation}
I(\mu)=\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x}{\cos^2x}}\,dx=\frac{\pi}{2}\left[\arctan\sqrt{2\mu}-\arctan\sqrt{\frac{\mu}{\mu+1}}\right]\qquad\quad\square
\end{equation}
For $\mu=\frac{1}{2}$, we obtain
\begin{align}
I\left(\frac{1}{2}\right)&=\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\cos2x}{2\cos^2x}}\,dx\\
&=\frac{\pi}{2}\left[\arctan 1-\arctan\left(\frac{1}{\sqrt{3}}\right)\right]\\
&=\frac{\pi}{2}\left[\frac{\pi}{4}-\frac{\pi}{6}\right]\\
&=\frac{\pi^2}{24}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/749493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 0
} |
Solving this equation for $\theta$ $$ 0 = x\cos\theta\cos\phi+y\sin\theta\cos\phi+z\sin\phi$$
Here's what I've tried doing.
$$\begin{align}
x\cos\theta\cos\phi+y\sin\theta\cos\phi & = -z\sin\phi \\
x\cos\theta+y\sin\theta & = -z\tan\phi \\
x\cos\theta+y\sqrt{1-\cos^2\theta} & = -z\tan\phi \\
y\sqrt{1-\cos^2\theta} & = -x\cos\theta-z\tan\phi \\
y^2-y^2\cos^2\theta & = x^2\cos^2\theta-z^2\tan^2\phi+2xz\cos\theta\tan\phi \\
0 & = (x^2+y^2)\cos^2\theta+2xz\tan\phi\cos\theta-y^2-z^2\tan^2\phi
\end{align}$$
$$\begin{align}
\cos\theta & = \frac{-2xz\tan\phi\pm\sqrt{4x^2z^2\tan^2\phi-4(x^2+y^2)(-y^2-z^2\tan^2\phi)}}{2(x^2+y^2)} \\
\cos\theta & = \frac{-2xz\tan\phi\pm2\sqrt{x^2z^2\tan^2\phi+y^2(x^2+y^2)-x^2z^2\tan^2\phi-y^2z^2\tan^2\phi}}{2(x^2+y^2)} \\
\cos\theta & = \frac{-xz\tan\phi\pm y\sqrt{x^2+y^2-z^2\tan^2\phi}}{x^2+y^2} \\
\end{align}$$
However, the $\pm$ is throwing me off. I'm assuming that either the $+$ or $-$ gives a complex solution or something, but I'm not sure what.
| HINT :
$$
x\cos\theta+y\sin\theta=r\cos(\theta-\alpha),
$$
where $r^2=x^2+y^2$ and $\tan\alpha=\dfrac{y}{x}$. You may refer to this link.
$$\\$$
$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/749749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find an integer $x$ satisfying the congruence: $$x \equiv \ 1 \pmod3$$
$$x \equiv \ 2 \pmod5$$
$$x \equiv \ 8 \pmod{11}$$
From the first, I have $x=3k+1$, $x=5j+2$ from the second and $x=11l+8$ from the third.
Subbing the third into the second I get
$11l+8 \equiv 2 \pmod5$
$l \equiv -6\pmod5$
$l \equiv -1\pmod5$
So $l=5m-1$
Subbing back into the third equation I had, I get
$x=11(5m-1)+8=55m-3$
Thus $x \equiv -3\pmod{55}$.
I was only asked to find any integer x which satisfies the system of congruences, so any of x of that form should work; i.e. $x=52$.
I want to know if this is the smallest such $x$ which would work, or could I have approached the question in a different way and arrived at another answer?
If I'd been asked to find the smallest $x$, or all $x$, I'm not sure if this approach would have worked. Is there a way to confirm that this is the smallest such $x$, or that all $x$ of the form $x \equiv -3\pmod{55}$ are the only $x$ which satisfy the system?
| For small moduli, I came up with this method. Make the following table.
\begin{array}{r|ccc}
& \mod 3 & \mod 5 & \mod{11}\\
\hline
55 & 1 & 0 & 0\\
33 & 0 & 3 & 0\\
15 & 0 & 0 & 4\\
\hline
103 & 1 & 3 & 4
\end{array}
Let $N = 3 \times 5 \times 11 = 165$. Then
$55=\dfrac N3,\;
33= \dfrac N5;$ and
$15 = \dfrac N{11}$.
The table consists of those three numbers modulo $3, 5, $ and $11.$ We need for the diagonal, $[1\; 3\; 4],$ to become $[1\; 2\; 8],$
We can convert $3 \pmod 5$ to $2 \pmod 5$ by multiplying $33$ by $-1$.
We can convert $4 \pmod{11}$ to $8 \pmod{11}$ by multiplying $15$ by $2$.
\begin{array}{r|ccc}
& \mod 3 & \mod 5 & \mod{11}\\
\hline
55 & 1 & 0 & 0\\
-33 & 0 & 2 & 0\\
30 & 0 & 0 & 8\\
\hline
52 & 1 & 2 & 8
\end{array}
The number we seek is
$55-33+30 \pmod{3 \times 5 \times 11} = 52 \pmod{165}$
By the CRT, $52$ is the smallest such non negative number.
The smallest such negative number would therefore be $-113$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/749901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
let $a-b=6$ and $b\leq-1$ and $a\leq-2$. Find $A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}$ let $a-b=6$ and $b\leq-1$ and $a\leq-2$. Find $A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}$
My Try $$A=\sqrt{a^2+4a+4}+\sqrt{b^2+2b+1}=$$then I can't
Please just a hint.
| Hint: You need to know that $a\leqslant -2$ iff $a+2\leqslant 0$, and $b\leqslant -1$ iff $b+1\leqslant 0$. And the most important thing: $$ \color{lightblue}{\boxed{\color{white}{\boxed{\color{white}{\overline{\underline{\color{black}{\displaystyle\sqrt{\rm X^2}=|\,\rm X|\,}}}}}}}}$$
I hope this helps.
Best wishes, $\mathcal H$akim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/751579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove the series $\sum_{n = 1}^{ \infty} \frac 1 6 n (\frac 5 6)^{n-1} = 6$
Prove the series $$\sum_{n = 1}^{ \infty} \frac 1 6 n (\frac 5 6)^{n-1} = 6.$$
I've tried various methods for proving the series:
The series is not geometric, but I see that $\frac 1 6 n (\frac 5 6)^{n-1} \rightarrow 0$. Also the series is not telescoping, as far I'm concerned.
What method could solve this ?
| Here's a calculus approach:
We know that:
$$\frac{1}{1 - x} = 1 + x + x^2 + \dots$$
Differentiating both sides,
$$\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + \dots$$
Now, substitute $x = \frac{5}{6}$. Then,
$$\sum_{n = 1}^\infty n\left(\frac{5}{6}\right)^{n-1} = \frac{1}{\left(1 - \frac{5}{6}\right)^2}$$
So we have:
$$\frac{1}{6}\sum_{n = 1}^\infty n\left(\frac{5}{6}\right)^{n-1} = \frac{1}{6\left(1 - \frac{5}{6}\right)^2}\\
= 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/754843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I solve $(x^3-4x^2+5x-6)/(x^2-x-6)=4$ algebraically? How do I solve $\frac {(x^3-4x^2+5x-6)}{(x^2-x-6)=4}$ algebraically?
I tried:
$4(x^2-x-6)=x^3-4x^2+5x-6$
$4x^2-4x-24=x^3-4x^2+5x-6$
$x^3-8x^2+9x+18=0$
I don't know how to solve this algebraically.
| $\dfrac{x^3-4x^2+5x-6}{x^2-x-6}=\dfrac{x^3-4x^2+5x-6}{(x-3)(x+2)}$
Now we are going to use Factor Theorem, to determine one of the factors of the numerator, which I call call $h(x)$.
(Note: Just in case you're not familiar with the Factor Theorem, it says that a polynomial $p(x)$ has a factor $(x-c) \iff p(c)=0$ )
We note that $h(3)=0$ and hence $x-3$ is a factor. Then we apply polynomial long division to obtain the other factors. You should get $x^2-x+2$ as your quotient.
$\dfrac{x^3-4x^2+5x-6}{(x-3)(x+2)}=\dfrac{(x-3)(x^2-x+2)}{(x-3)(x+2)}$
.
Then the $(x-3)$'s will cancel leaving
$\dfrac{(x^2-x+2)}{(x+2)}$
Going back to the original question,
$\dfrac{(x^2-x+2)}{(x+2)}=4 \\
\implies x^2-x+2=4x+8\\
\implies x^2-5x-6=0\\
\implies (x-6)(x+1)=0\\
\implies x=6\text{ or } x=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/755084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the Asymptotic Curves of a Given Surface I have to find the asymptotic curves of the surface given by $$z = a \left( \frac{x}{y} + \frac{y}{x} \right),$$ for constant $a \neq 0$.
I guess that what was meant by that statement is that surface $S$ can be locally parametrized by $$X(u,v) = \left( u, v, a \left( \frac{u}{v} + \frac{u}{v} \right) \right).$$ Do you think that my parametrization is correct (meaning that I read the description of the surface correctly), and do you know of a more convenient parametrization?
Assuming that parametrization, I derived the following ($E$, $F$, $G$, are the coefficients of the first fundamental form; $e$, $f$, $g$ are coefficients of the second fundamental form; $N$ is the normal vector to surface $S$ at a point; these quantities are all functions of local coordinates $(u,v)$):
$$E = 1 + a^2 \left( \frac{1}{v} - \frac{v}{u^2} \right)^2,$$
$$F = -\frac{a^2 (u^2 - v^2)^2}{u^3 v^3},$$
$$G = 1 + a^2 \left( \frac{1}{u} - \frac{u}{v^2} \right)^2.$$
$$N = \frac{1}{\sqrt{E G - F^2}} \left( a \left( \frac{v}{u^2}-\frac{1}{v} \right), a \left( \frac{u}{v^2}-\frac{1}{u} \right), 1 \right).$$
$$X_{u,u} = \left( 0,0, \frac{2 a v}{u^3} \right), X_{u,v} = \left( 0,0, -a \left( \frac{1}{u^2} + \frac{1}{v^2} \right) \right), X_{v,v} = \left( 0, 0, \frac{2 a u}{v^3} \right).$$
$$e = \frac{2 a v}{u^3 \sqrt{E G - F^2}},$$
$$f = - \frac{a (\frac{1}{u^2} + \frac{1}{v^2})}{\sqrt{E G - F^2}},$$
$$g = \frac{2 a u}{v3 \sqrt{E G - F^2}}.$$
Thus, the Gaussian curvature (from these calculations) is:
$$K = -\frac{a^2 u^4 v^4 (u^2 - v^2)^2}{(u^4 v^4 +
a^2 (u^2 - v^2)^2 (u^2 + v^2))^2}.$$
And the mean curvature would be:
$$H = \frac{a u^3 v^3 (u^4 + v^4)}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2
}}.$$
So, the principal curvatures are:
$$k_{\pm} = H \pm \sqrt{H^2 - K} = a u^2 v^2 \frac{u v (u^4 + v^4) \pm \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2}}.$$
In order to find the asymptotic curves, but trying to avoid the differential equation, I was hoping to find the angles $\theta (u,v)$ such that the normal curvature would always be $0$. In other words I was trying:
$0 = k_n = k_{+} \cos{(\theta)}^2 + k_{-} \sin{(\theta)}^2$, and solving for $\theta$.
Assuming sufficient niceness, this calculate would result in: $$(u v (u^4 + v^4) + \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \cos{(\theta)}^2 + (u v (u^4 + v^4) - \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \sin{(\theta)}^2$$
First of all, is this approach (solving for $\theta$ rather than solving the differential equation) valid? If it is, after I find that angle $\theta$, determined by location $(u,v)$ on $S$, what more work do I have to do? How do I find the equations for the asymptotic curves based on this angle?
If this whole method was for naught, how does one solve the differential equation. in this case, of: $$e (u')^2 + 2f u' v' + g (v')^2 = 2a v^4 (u')^2 - 2a u^3 v^3 \left( \frac{1}{u^2} + \frac{1}{v^2} \right)u' v' + 2a u^4 (v')^2 = 0?$$ (Again, assuming sufficient niceness.)
(See: https://math.stackexchange.com/questions/762195/differential-equation-for-the-asymptotic-directions-of-a-given-surface)
Thank you!
|
I believe this surface can be described by two networks of lines. ( Shown in Red and Cyan). Make the substitution $nt$ for $u$ , and separately $nt$ for $t$ ,
so that the new parametrization might look like : $<t,nt,a(1/n + n/1)> $ and/or $<nt,t,a(1/n+ n)>$ these are the asymptotic lines. Then switch the sign of n to pick up the surface below the z axis. I've used $n$ as an integer , but every line in the surface chosen in this way is an asymptotic curve, and a geodesic.
(Added a guess for the solution of the equations based on a study of the surface.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/761350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove the matrix satisfies the equation $A^2 -4A-5I=0$ How to prove that
$$
A=\begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{bmatrix}
$$
satisfies the equation $A^2 -4A-5I=0$?
| If you would like to sped up the calculation (supposing you are doing it on your own) you can use this approach.
Let's take the equation
$$ A^2 - 4A - 5I = 0.$$
And now forget about the matrices. What we have is simple quadratic equation in $x$ written as
$$ x^2 - 4x - 5 = 0,$$
now just find the roots
$$ (x-5)(x+1) = 0.$$
Let's get back to that matrix
$$(A-5I)(A+I) = 0.$$
Simple calculations give
$$
\begin{pmatrix}
-4 & 2 &2\\
2 & -4 &2\\
2 & 2 &-4
\end{pmatrix}\cdot
\begin{pmatrix}
2 & 2 &2\\
2 & 2 &2\\
2 & 2 &2
\end{pmatrix} = 0.
$$
And now it is very simple since you divide by 2
$$
\begin{pmatrix}
-4 & 2 &2\\
2 & -4 &2\\
2 & 2 &-4
\end{pmatrix}\cdot
\begin{pmatrix}
1 & 1 &1\\
1 & 1 &1\\
1 & 1 &1
\end{pmatrix} = 0.
$$
Multiplying those matrices is just adding all the elements in each row of the first matrix. So it is $0$ everywhere. Such tricks might come in handy in more complicated examples.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/762051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How I can refine the proof of $\mathop {\lim }\limits_{(x,y)\, \to \,(3, - 1)} \left( {{x^2} + {y^2} - 4x + 2y} \right) = - \,4$ To prove that $$\mathop {\lim }\limits_{(x,y)\, \to \,(3, - 1)} \left( {{x^2} + {y^2} - 4x + 2y} \right) = - \,4$$ I followed the following process:
Because the hypothesis and the definition of limit, we have:
$\sqrt {{{(x - 3)}^2} + {{(y + 1)}^2}} < \delta \,\,\,;\,\,\,\,\left| {x - 3} \right| < \delta \,\,;\,\,\,\left| {y + 1} \right| < \delta $
Then I have to find an upper bound of:
$\left| {{x^2} + {y^2} - 4x + 2y + 4} \right|$, I complete square like this:
$\begin{array}{c}
\left| {{x^2} + {y^2} - 4x + 2y + 4} \right| = \left| {{x^2} - 6x + 9 + {y^2} + 2y + 1 + 2x - 6} \right|\\
= \left| {{{(x - 3)}^2} + {{(y + 1)}^2} + 2(x - 3)} \right|
\end{array}$
then using the triangle inequality and relate delta epsilon:
$\begin{array}{c}
\left| {{{(x - 3)}^2} + {{(y + 1)}^2} + 2(x - 3)} \right| \le \left| {{{(x - 3)}^2} + {{(y + 1)}^2}} \right| + 2\left| {x - 3} \right|\\
< {\delta ^2} + 2\delta \\
= {(\delta + 1)^2} - 1\\
< {(\delta + 1)^2} = \varepsilon
\end{array}$
but the latter involves that: $\delta = \sqrt \varepsilon-1$
and this is consistent for $\epsilon>1$, and not for $\epsilon$ as small as desired
So how I can refine this demonstration? because every time I find more disconcerting.
| From your last step, you want that: $\delta^2 + 2\delta < \epsilon$, but this means $(\delta + 1)^2 < 1 + \epsilon$, and taking square root $\delta + 1 < \sqrt{1 + \epsilon}$, so $\delta < -1 + \sqrt{1 + \epsilon}$. You can take $\delta = \dfrac{-1 + \sqrt{1 + \epsilon}}{2}$, and it does the job.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/762699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing coefficient of $x^n$ Find the coefficient of $x^n$ in the expansion of $$\left(1 + \frac{x}{1!} + \frac{x^2}{2!}+\cdots +\frac{x^n}{n!} \right)^2$$
How do you even start this problem? Do you use multinomial theorem or binomial theorem?
Could anyone please help? I found this in a textbook of mine. What I feel hard is what to do with the factorials?
| To do this, you need to look at every combination of terms that gives $x^n = x^j \cdot x^{n-j}$. We can do this using the multinomial theorem, or just multiply it out.
\begin{align*}\left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right)^2 &= \left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right)\left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right)\\
&= 1\left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right) + \frac{x}{1!}\left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right) + \ldots\\
&+ \frac{x^n}{n!}\left(1+\frac{x}{1!} + \ldots + \frac{x^n}{n!}\right)\\
&= 1 + 2\frac{x}{1!} + \frac{x^2}{2!} + \left(\frac{x}{1!}\right)^2 + \frac{x^2}{2!} + \ldots
\end{align*}
So the $1$ coefficient is $1$, the $x$ term is $\frac{2}{1!}$, the $x^2$ term is $\frac{1}{2!} + \frac{1}{1!}\frac{1}{1!} + \frac{1}{2!}$, the $x^3$ term is $\frac{1}{3!} + \frac{1}{1!}\frac{1}{2!} + \frac{1}{2!}\frac{1}{1!} + \frac{1}{3!}$. Continuing in this manner, we get that the $x^n$ term is (using the first few terms to predict the pattern):
$$\frac{1}{n!} + \frac{1}{1!}\frac{1}{(n-1)!} + \frac{1}{2!}\frac{1}{(n-2)!} + \cdots + \frac{1}{(n-2)!}\frac{1}{2!} + \frac{1}{(n-1)!}\frac{1}{1!} + \frac{1}{n!}$$
or rewriting as a sum, we could rewrite this as
\begin{align*}\sum_{j=0}^n \frac{1}{j!}\frac{1}{(n-j)!}&= \frac{1}{n!}\sum_{j=0}^n \frac{n!}{j!(n-j)!}\\
&= \frac{1}{n!}\sum_{j=0}^n \left(\begin{array}{c}n\\j\end{array}\right) 1^j 1^{n-j}\\
&= \frac{1}{n!}(1+1)^n\\
&= \frac{1}{n!}2^n
\end{align*}
(the second-to-last equality is the binomial theorem)
(To make sense of this when $j=0$ or $j=n$, we use the convention $0! = 1$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/763047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)$ How to solve this limit
$$
\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)
$$
without using L'Hospital's rule?
| By the formulas
$$
\cos 2x=\cos x ^2-\sin x ^2=1-2\sin x ^2, \sin 2x=2 \sin x \cos x,
$$
we have
$$
1-\cos x=2\sin^2 {\frac{x}{2}}, \sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}.
$$
Then,
$$\begin{align}
\lim_{n\rightarrow \infty} (\frac{1}{\sin x}-\frac{1}{\tan x})
&=\lim_{n\rightarrow \infty} (\frac{1}{\sin x}-\frac{\cos x}{\sin x})\\
&=\lim_{n\rightarrow \infty} \frac{1-\cos x}{\sin x}\\
&=\lim_{n\rightarrow \infty} \frac{2\sin^2 {\frac{x}{2}}}{2 \sin \frac{x}{2} \cos\frac{x}{2}}\\
&=\lim_{n\rightarrow \infty} \frac{\sin {\frac{x}{2}}}{\cos\frac{x}{2}}\\
&=\frac{0}{1}\\
&=0.
\end{align}
$$
P.S. identities among triangle functions are very helpful when one wants to simplify formulas involving them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/764365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 1
} |
Bernoulli numbers identity with binomial coefficient The generating function for the Bernoulli numbers $B_k$ is given by $f(z) = \frac{z}{e^z -1}= \sum_{k=0}^{\infty} \frac{B_k}{k!} z^k$. Applying the identity
$$1 = \frac{e^z -1}{z} \cdot \sum_{k=0}^{\infty} \frac{B_k}{k!} z^k$$ yields $\sum_{n=0}^{k} \binom{k+1}{n} B_n= 0$. Recall that we also have that $f(z)= 1- \frac{z}{2}+ \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!} z^{2k}$ which implies that $$\frac{e^z -1}{z} \cdot \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!} z^{2k}= 1- \frac{e^z -1}{z}(1- \frac{z}{2}).$$ I believe the indices got me all confused, so I keep getting stuck on how to show that this above equality gives us the identity $\sum_{n=0}^{k} \binom{2k+1}{2n} B_{2n}= \frac{1}{2} (2k+1)$ after expanding both sides of the equality. How should I proceed?
| If we don't have to use the equation
$$\frac{e^z -1}{z} \cdot \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!} z^{2k}= 1- \frac{e^z -1}{z}(1- \frac{z}{2}),$$
then we can simply use the already proven
$$\sum_{n=0}^k \binom{k+1}{n}B_n = 0$$
for $k > 0$ and insert $k = 2m$ to obtain
$$\sum_{n=0}^m \binom{2m+1}{2n}B_{2n} = - \sum_{n=0}^{m-1} \binom{2m+1}{2n+1}B_{2n+1},$$
which, since $B_1 = -\frac{1}{2}$ and $B_{2n+1} = 0$ for $n \geqslant 1$ simplifies to
$$\sum_{n=0}^m \binom{2m+1}{2n}B_{2n} = -B_1\binom{2m+1}{1} = \frac{1}{2}(2m+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/764535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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