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Highschool Exam Question About Cube Factoring Given;
$ a^3 - 3ab^2 = 10 $ and $ b^3 - 3ba^2 = 5$
What is the value of $ a^2 + b^2 $ ?
|
\begin{align}
10 &= a^3 - 3ab^2 \\
5 &= b^3 - 3a^2b \\
\hline
100 &= a^6 -6a^4b^2 + 9a^2b^4 \\
25 &= b^6 - 6a^2b^4 + 9a^4b^2 \\
\hline
125 &= a^6 + 3a^4b^2 + 3a^2b^4 + b^6 \\
5^3 &= (a^2 + b^2)^3 \\
5 &= a^2 + b^2
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1775275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Using Finite Differences and Integration to prove result If $f(x)$ is a polynomial in $x$ of third degree and:
$$u_{-1}=\int_{-3}^{-1}f(x)dx\ ;\ u_{0}=\int_{-1}^{1}f(x)dx\ ; u_{1}=\int_{1}^{3}f(x)dx$$
then show that
$$f(0) = \frac{1}{2}\Bigg(u_0-\frac{\Delta^2u_{-1}}{12}\Bigg)$$
I attempted this question by assuming that the function $f(x)$ is of the following form:
$$f(x) = a + bx + cx^3$$
In this case I obtained the following values and was able to arrive at the correct solution:
$$
\begin{align}
u_{-1}&=\int_{-3}^{-1}f(x)dx\\&=\int_{-3}^{-1}(a+bx+cx^3)dx\\&=\Bigg[ax+b\frac{x^2}{2} + c\frac{x^4}{4}\Bigg]_{-3}^{-1}=2a - 4b - 20c
\end{align}
$$
$$
\begin{align}
u_0 = \int_{-1}^{1}f(x)dx = \int_{-1}^{1}(a+bx+cx^3)dx=2a
\end{align}
$$
$$
\begin{align}
u_{-1}&=\int_{1}^{3}f(x)dx\\&=\int_{1}^{3}(a+bx+cx^2)dx\\&=\Bigg[ax+b\frac{x^2}{2} + c\frac{x^4}{4}\Bigg]_{1}^{3}=2a + 4b + 20c
\end{align}
$$
Now,
$$
\begin{align}
RHS\ =\ \frac{1}{2}\Bigg[u_0-\frac{\Delta^2u_{-1}}{12}\Bigg]&=\frac{1}{2}\Bigg[u_0-\frac{1}{24}(u_{-1}-2u_{0}+u_{1})\Bigg]\\&=\frac{1}{2}\big[2a-\frac{1}{24}(4a-4a)\big] \\&= a = f(0) = LHS
\end{align}
$$
However, I was unable to prove in case I assumed that $f(x)$ was of the form:
$$
f(x) = a + bx + cx^2 + dx^3
$$
Will my proof be sufficient for the given question or is it necessary to also show for the latter form?
|
This is a straightforward power-rule calculation:
$$
u_{-1} = \int _{-3}^{1} f(x)\,dx = 2 a-4 b+\frac{26 c}{3}-20 d;
$$
$$
u_0 = \int _{-1}^1 f(x)\,dx = 2 a+\frac{2 c}{3};
$$
$$
u_1 = \int _1^3 f(x)\,dx = 2 a+4 b+\frac{26 c}{3}+20 d
$$With your notation,
$$
\Delta^2 u_{-1} = \frac{1}{2}\left(u_{-1}-2u_0+u_1\right) = 8c;
$$so we have
$$
\frac{1}{2}\left(2a+\frac{2}{3}c - \frac{1}{12}\cdot 8c\right) = a = f(0),
$$as was to be shown.
|
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|
Factor $6x^2β β7xβ5=0$ I'm trying to factor
$$6x^2β β7xβ5=0$$
but I have no clue about how to do it. I would be able to factor this:
$$x^2-14x+40=0$$
$$a+b=-14$$
$$ab=40$$
But $6x^2β β7xβ5=0$ looks like it's not following the rules because of the coefficient of $x$. Any hints?
|
Here's an alternative (but you should just learn the quadratic equation)
$6x^2 - 7x - 5= 6(x + a)(x + b)$
So $ab = -5/6$ and $a + b = -7/6$.
$a = \pm 1, \pm 5, \pm 1/2, \pm 5/2, \pm 1/3, \pm 5/3, \pm 1/6, \pm 5/6$
$b = \mp 5/6, \mp 1/6, \mp 5/3, \mp 1/3, \mp 5/2, \mp 1/2, \mp 5, \mp 1$.
So $a + b = \pm 2/3, \pm 29/6, \mp 7/6, \pm 13/6, etc....$
So $a = 1/2$ and $b= -5/3$ works.
$6(x + 1/2)(x - 5/3) = (2x + 1)(3x - 5) = 6x^2 - 7x -5$.
|
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|
Evaluate $\int_0^{2\pi}\frac{\sin^2(x)}{a + b\cos(x)}\ dx$ using a suitable contour I need to find a good contour for $\int_0^{2\pi}\frac{\sin^2(x)}{a + b\cos(x)}\ dx$ but I don't know which one to choose. Both a semicircular, and rectangular contour look ugly for this.
I've been looking at a semicircular contour of radius $2\pi$, but then I have the problem that I don't know whether the singularity is inside or outside the closed region.
If it helps, the answer is $\frac{2\pi}{b^2}\left[a - \sqrt{a^2 - b^2}\right]$
|
Note that the integral diverges for $a\le b$. Therefore, we assume throughout the development that $a>b$.
We can simplify the task by rewriting the integrand as
$$\begin{align}
\frac{\sin^2(x)}{a+b\cos(x)}&=\frac{a}{b^2}-\frac{1}{b}\cos(x)-\left(\frac{a^2-b^2}{b^2}\right)\frac{1}{a+b\cos(x)}
\end{align}$$
Then, the integral of interest reduces to
$$\int_0^{2\pi}\frac{\sin^2(x)}{a+b\cos(x)}\,dx=\frac{2\pi a}{b^2}-\frac{a^2-b^2}{b^2}\int_0^{2\pi}\frac{1}{a+b\cos(x)}\,dx \tag 1$$
We enforce the substitution $z=e^{i x}$ in the integral on the right-hand side of $(1)$ and obtain
$$\begin{align}
\int_0^{2\pi}\frac{1}{a+b\cos(x)}\,dx& =\oint_{|z|=1}\frac{1}{a+b\left(\frac{z+z^{-1}}{2}\right)}\frac{1}{iz}\,dz\\\\
&=\frac2{ib}\oint_{|z|=1}\frac{1}{(z+(a/b)-\sqrt{(a/b)^2-1})(z+(a/b)+\sqrt{(a/b)^2-1})}\,dz\\\\
&=2\pi i \frac2{ib} \frac{1}{2\sqrt{(a/b)^2-1}}\\\\
&=\frac{2\pi}{\sqrt{a^2-b^2}}
\end{align}$$
Putting it all together, the integral of interest is
$$\bbox[5px,border:2px solid #C0A000]{\int_0^{2\pi}\frac{\sin^2(x)}{a+b\cos(x)}\,dx=\frac{2\pi}{b^2}\left(a-\sqrt{a^2-b^2}\right)}$$
as was to be shown!
|
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|
Show that, $(s+1)\gamma-\int_0^1\left(\frac{1}{\ln(x)}+\frac{1}{1-x}\right)\sum_{n=0}^{s}x^ndx=\sum_{i=1}^{s}H_i-\ln(s+1)!$ harmonic numbers
$H_n=\sum_{i=1}^{n}\frac{1}{i}$
Euler's constant
$\gamma=\lim_{n\to \infty} [H_n-\ln(n)]$
Factorial
$n!=n(n-1)(n-2)\cdots2\cdot1$; valid for all non-negative integers
Show that,
$$(s+1)\gamma-\int_0^1\left(\frac{1}{\ln(x)}+\frac{1}{1-x}\right)\sum_{n=0}^{s}x^ndx=\sum_{i=1}^{s}H_i-\ln(s+1)!$$
Inspiring from the integral of Euler's constant; see Wikipedia
$$\int_0^1\left(\frac{1}{\ln(x)}+\frac{1}{1-x}\right)=\gamma$$
Mathematical experimental
We did trial and error using wolfram integrator and observe the numerical values of the integral and we was able to come up with a closed form for it.
Unfortunately we are unable to provide a proof for it, can anyone provide us a prove of it?
|
Here is my two pence worth and is not a complete answer but paves the way I think.
Theorem [Euler 1731]
The limit
$$\gamma = \lim_{n \rightarrow \infty}(H_n-\log n)$$
Is given by the convergent series
$$\gamma = \sum_{n=2}^{\infty}(-1)^{n}\frac{\zeta(n)}{n}$$
We can evaluate this formula using Mercator's expansion
$$\log(1+x) = \sum_{k=1}^{\infty}(-1)^{k}\frac{x^{k}}{k}$$
and in doing so we find
\begin{align}
\log 2 &= 1-\frac{1}{2}\left(\frac{1}{1}\right)^{2}+\frac{1}{3}\left(\frac{1}{1}\right)^{3}-\ldots\\
\log \frac{3}{2} &= \frac{1}{2}-\frac{1}{2}\left(\frac{1}{2}\right)^{2}+\frac{1}{3}\left(\frac{1}{2}\right)^{3}-\ldots\\
\log \frac{4}{3} &= \frac{1}{3}-\frac{1}{2}\left(\frac{1}{3}\right)^{2}+\frac{1}{3}\left(\frac{1}{3}\right)^{3}-\ldots\\
\log \frac{5}{4} &= \frac{1}{4}-\frac{1}{2}\left(\frac{1}{4}\right)^{2}+\frac{1}{3}\left(\frac{1}{4}\right)^{3}-\ldots\\
\end{align}
summing columns of the first $n$ terms gives
$$\log(n+1) = H_n-\frac{1}{2}H_{n, 2}+\frac{1}{3}H_{n, 3}-\ldots$$
or
$$H_n-\log(n+1) = \frac{1}{2}H_{n, 2}-\frac{1}{3}H_{n, 3}+\ldots$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1779314",
"timestamp": "2023-03-29T00:00:00",
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|
Calculate $ \lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx}$ My attempt:
\begin{align*}
\lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx} &= (\frac{\ln1}{0}) \text{ (we apply L'Hopital's rule)} \\
&= \lim_{x \to0}\frac{\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{x^n+x^{n-1}+\dots+1}}{n} \\
&= \lim_{x \to0}\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{n(1+x+\dots+x^n)} \\
&= \frac{1}{n}.
\end{align*}
Are my steps correct? Thanks.
|
It is well-known that $\ln (1+z)= z+o(z)$ for small $z$, so plugging in $z=x+x^2+\cdots+x^n$ yields
$$\frac{\ln(1+x+\cdots+x^n)}{x}=\frac{x+\cdots+x^n+o(x+\cdots+x^n)}{x}=\frac{x+o(x)}{x}\to 1$$
as $x\to 0$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Complex number and polar coordinates True or false: the polar coordinates of $-1-i$ are $-\sqrt{2}\operatorname{cis}\frac{\pi}{4}$
In my opinion it's true:
$\tan\theta=\frac{-1}{-1}=1\Rightarrow \theta=\frac{5\pi}{4}$, $r=\sqrt{(-1)^2+(-1)^2}=\sqrt{2}$
Therefore we get: $z=\sqrt{2}\operatorname{cis}\frac{5\pi}{4}=\sqrt{2}(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4})=\sqrt{2}(-\cos\frac{\pi}{4}-i\sin\frac{\pi}{4})=-\sqrt{2}\operatorname{cis}\frac{\pi}{4}$
that means it's true. But it was written in the algebra book that because of $-2$, it's not polar coordinate. Is the proof right?
|
No, it's false. The $r$ part must be positive and is computed by
$$
r=\sqrt{(-1-i)(-1+i)}=\sqrt{1+1}=\sqrt{2}
$$
This already answers the true/false question. If you want to find the $\theta$ part, you need to find $\theta$ such that
$$
\cos\theta=-\frac{1}{\sqrt{2}},\quad\sin\theta=-\frac{1}{\sqrt{2}}
$$
and this is clearly
$$
\theta=\pi+\frac{\pi}{4}=\frac{5\pi}{4}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1784148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Deadly integral $\int_0^1\frac{x^{2}+x+1}{x^{4}+x^{3}+x^{2}+x+1}dx$. How to solve this question
$$\int\limits_0^1\frac{x^{2}+x+1}{x^{4}+x^{3}+x^{2}+x+1}dx$$
. Please help me in solving this short way
my approach is in the answer
Is it correct and can it be solved in a shorter way ?
|
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\iff}{\Leftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
&\color{#f00}{\int_{0}^{1}{x^{2} + x + 1 \over x^{4} + x^{3} + x^{2} + x + 1}
\,\dd x} =
\int_{0}^{1}{\pars{x^{2} + x + 1}\pars{x - 1} \over
\pars{x^{4} + x^{3} + x^{2} + x + 1}\pars{x - 1}}\,\dd x =
\int_{0}^{1}{1 - x^{3} \over 1 - x^{5}}\,\dd x
\\[3mm] \stackrel{x^{5}\ \to x}{=} &
\int_{0}^{1}{1 - x^{3/5} \over 1 - x}\,{1 \over 5}\,x^{-4/5}\,\dd x =
{1 \over 5}\bracks{%
\int_{0}^{1}{1 - x^{-1/5} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{-4/5} \over 1 - x}\,\dd x}
\end{align}
However,
$\ds{\Psi\pars{z} + \gamma = \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t\,,\ \Re\pars{z} > 0}$. $\Psi$ and $\gamma$ are the digamma function and the
Euler-Mascheroni constant, respectively. See, for example, Abramowitz $\&$ Stegun Table. Then,
\begin{align}
&\color{#f00}{\int_{0}^{1}{x^{2} + x + 1 \over x^{4} + x^{3} + x^{2} + x + 1}
\,\dd x} =
{1 \over 5}\braces{\bracks{\Psi\pars{-\,{1 \over 5} + 1} + \gamma} -
\bracks{\Psi\pars{-\,{4 \over 5} + 1} + \gamma}}
\\[3mm] = &
{1 \over 5}\bracks{\Psi\pars{4 \over 5} - \Psi\pars{1 \over 5}} =
{1 \over 5}\,\pi\cot\pars{\pi\,{1 \over 5}}
\end{align}
In the last step we used the PolyGamma's Euler identity.
\begin{align}
&\color{#f00}{\int_{0}^{1}{x^{2} + x + 1 \over x^{4} + x^{3} + x^{2} + x + 1}
\,\dd x} =
{1 \over 5}\,\pi\cot\pars{\pi \over 5} =
{1 \over 5}\,\pi\root{1 + {2 \over \root{5}}}
\\[3mm] = &
\color{#f00}{{1 \over 25}\,\pi\root{10\root{5} + 25}} \approx 0.8648
\end{align}
|
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|
Prove $\frac{2\cos x}{\cos 2x + 1 }= \sec x$ Prove that $\dfrac{2\cos x}{\cos 2x + 1 }= \sec x$.
So far I have:
$\dfrac{2\cos x}{\cos 2x + 1 }= \dfrac 1 {\cos x}$
Where do I go from here?
|
It seems to be a widespread practice for students to begin attempted proofs of trigonometric identities by writing things like this:
$$\frac{2\cos x}{\cos 2x + 1 }= \sec x$$
$$\frac{2\cos x}{\cos 2x + 1 }= \frac 1 {\cos x}$$
That's ok in scratchwork, but the finished proof should go like this:
$$
\sec x = \frac 1 {\cos x} = \cdots \cdots \cdots = \frac{2\cos x}{\cos2x + 1}
$$
or like this:
$$
\frac{2\cos x}{\cos2x + 1} = \cdots \cdots \cdots = \frac 1 {\cos x} = \sec x
$$
and of course to finish it you need to figure out what goes where all those dots are.
In other words, you put $\text{β}=\text{''}$ between things that you already know are equal.
Seeing $\cos(2x)$, you should recall the double-angle formula that says $\cos(2x) = \cos^2 x - \sin^2 x$. Then you have
\begin{align}
\frac{2\cos x}{\cos(2x)+1} & = \frac{2\cos x}{\cos^2 x - \sin^2 x + 1} \\[10pt]
& = \frac{2\cos x}{\cos^2 x + \cos^2 x} & & \text{since }-\sin^2x+1 = \cos^2 x \\[10pt]
& = \frac{2\cos x}{2\cos^2 x} \\[10pt]
& = \frac 1 {\cos x} \\[10pt]
& = \sec x.
\end{align}
|
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|
Laurent Series of $1/\tan z$ How can we find the Laurent series of the function $$f(z)=\frac{1}{\tan z }$$ around 0.
Thank you very much.
|
I read your comment just now (looking for the general form), but I'll leave the answer below: a way to get the series term by term.
Well there' a formula for the coefficients (see wiki), but I suppose you're looking for another way to find the series expansion of $\cot z$.
Dividing the series for $\sin z$ term by term by $z$ gives:
$$\frac{\sin z}{z} = 1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots$$
Now after recognizing the sum of a geometric series, you get the expansion of $\tfrac{z}{\sin z}$:
$$\begin{array}{rl}
\displaystyle \frac{z}{\sin z}
& \displaystyle = \frac{1}{1-\left( \frac{z^2}{3!}-\frac{z^4}{5!}-\cdots \right)} \\[8pt]
& \displaystyle = 1 + \left( \frac{z^2}{3!}-\frac{z^4}{5!}-\cdots \right) + \left( \frac{z^2}{3!}-\frac{z^4}{5!}-\cdots \right)^2 + \cdots \\[8pt]
& \displaystyle = 1 + \frac{1}{6}z^2 + \frac{7}{360}z^4 + \cdots
\end{array}$$
Now you can multiply this series with the one for $\cos z$:
$$\cos z = 1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots$$
To get:
$$\begin{array}{rl}
\displaystyle \frac{z \cos z}{\sin z}
& \displaystyle = \left( 1-\frac{z^2}{2}+\frac{z^4}{24}-\cdots\right)\left( 1 + \frac{1}{6}z^2 + \frac{7}{360}z^4 + \cdots\right) \\[8pt]
& \displaystyle = 1-\frac{z^2}{3}-\frac{z^4}{45}-\frac{2z^6}{945}+ \cdots
\end{array}$$
And now you only have to divide by $z$ to get:
$$\cot z = \frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}-\frac{2z^5}{945}+ \cdots$$
|
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|
Evaluate $\int{\sqrt{a^2 - x^2}}dx$ I'm trying to solve the following integral, but seems these 2 methods led to different answers. I think one of the methods must be incorrect. But why doesn't one of them work?
Evaluate $\int{\sqrt{a^2 - x^2}}\ dx$
My friend evaluated this way:
First let $x=a\cos{\theta}$, so
$a^2-x^2=a^2(1-\cos^2\theta)=a^2\sin^2{\theta}$ $$ \int{\sqrt{a^2 -
x^2}}\ dx = -\int{a\sin{\theta}}\ d(a\cos{\theta}) =
-\int{a^2\sin^2{\theta}}\ d\theta=-a^2\int{\frac{1-\cos{2\theta}}{2}}\ d\theta
$$ $$ = -\frac{a^2}{2}\int(1-\cos{2\theta})\ d\theta =
-\frac{a^2}{2}\left(\theta - \frac{\sin{2\theta}}{2}\right) $$ $$ = \frac{a^2}{2}\left(-\cos^{-1}{\frac{x}{a}}+\frac{{x}\sqrt{a^2-x^2}}{a^2}\right)
$$
However I've done this way:
First let $x=a\sin{\theta}$, so
$a^2-x^2=a^2(1-\sin^2\theta)=a^2\cos^2{\theta}$ $$ \int{\sqrt{a^2 -
x^2}}\ dx = \int{a\cos{\theta}}\ d(a\sin{\theta}) =
\int{a^2\cos^2{\theta}}\ d\theta=a^2\int{\frac{1+\cos{2\theta}}{2}}\ d\theta
$$ $$ = \frac{a^2}{2}\int(1+\cos{2\theta})\ d\theta =
\frac{a^2}{2}(\theta + \frac{\sin{2\theta}}{2}) $$ $$=
\frac{a^2}{2}\left(\sin^{-1}{\frac{x}{a}}+\frac{{x}\sqrt{a^2-x^2}}{a^2}\right) $$
|
$\cos(\pi/2-x)=\sin(x)$ and vice versa. If you haven't seen this before, the geometric explanation is that the sine of one of the acute angles in a right triangle is the cosine of the other.
Therefore $-\cos^{-1}(x)$ and $\sin^{-1}(x)$ are the same up to a constant. Since indefinite integrals are only defined up to a constant (and the factor on the outside is a constant), your two solutions are consistent with each other.
|
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|
Prove Why $B^2 = A$ exists?
Define
$$A =
\begin{pmatrix}
8 & β4 & 3/2 & 2 & β11/4 & β4 & β4 & 1 \\
2 & 2 & 1 & 0 & 1 & 0 & 0 & 0 \\
β9 & 8 & 1/2 & β4 & 31/4 & 8 & 8 & β2 \\
4 & β6 & 2 & 5 & β7 & β6 & β6 & 0 \\
β2 & 0 & β1 & 0 & 1/2 & 0 & 0 & 0 \\
β1 & 0 & β1/2 & 0 & β3/4 & 3 & 1 & 0 \\
1 & 0 & 1/2 & 0 & 3/4 & β1 & 1 & 0 \\
2 & 0 & 1 & 0 & 0 & 0 & 0 & 5 \\
\end{pmatrix} \in M_8(\mathbb R)$$
Justify why there is a matrix $B$ such that $B^2 = A$.
I think it has something to do with the Jordan normal form so I found it.
\begin{pmatrix}
2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 2 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 5 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 5 \\
\end{pmatrix}
(I know the ones are under the diagonal, but I think it doesn't make a difference.)
I have no idea what should I do next, a help will be appreciated!
Thanks!
|
By putting $A$ into Jordan normal form you have found a non-singular matrix $P$ such that
$$
A = P^{-1}CP
$$
where $C$ is your matrix shown above. It is easy to find a matrix sqare root of $C$, for example, in the upper left start with $\pmatrix{\sqrt{2} &0\\\frac14\sqrt{2}& \sqrt{2}}$
Then $$(P^{-1}DP)^2 = P^{-1}D(PP^{-1})DP = P^{-1}D^2P=P^{-1}CP=A
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing $\lim_{x\to-\frac\pi2}\frac{e^{\tan x}}{\cos^2x}$ how can i compute: $\lim_{x\to-\frac\pi2}\frac{e^{\tan x}}{\cos^2x}$?
i tried l'hopital's rule but it's like a loop.
also if it can be done without that rule i'd like to know how.
Thanks.
|
Hint.
Let $x \to -\dfrac \pi2^-$, then
$$
\begin{align}
\tan x &=\frac{-\cos (x+\frac{\pi}{2})}{\sin (x+\frac{\pi}{2})} =\frac{-1}{x+\frac{\pi }{2}}+O\left(x+\frac{\pi }{2}\right)
\\\\\cos^2 x&=\sin^2 (x+\frac{\pi}{2})\sim\left(x+\frac{\pi}{2}\right)^2
\end{align}
$$ giving, as $x \to -\dfrac \pi2^-$,
$$
\frac{e^{\tan x}}{\cos^2x} \sim \frac{e^{\large -\frac{1}{x+\frac{\pi }{2}}}}{\left(x+\frac{\pi}{2}\right)^2} \to \infty\, \left(=\text{"}\frac{e^{+\infty}}{0^+}\text{"} \right).
$$
Let $x \to -\dfrac \pi2^+$, then
$$
\begin{align}
\tan x &=\frac{-\cos (x+\frac{\pi}{2})}{\sin (x+\frac{\pi}{2})} =\frac{-1}{x+\frac{\pi }{2}}+O\left(x+\frac{\pi }{2}\right)
\\\\\cos^2 x&=\sin^2 (x+\frac{\pi}{2})\sim\left(x+\frac{\pi}{2}\right)^2
\end{align}
$$ giving, as $x \to -\dfrac \pi2^+$,
$$
\frac{e^{\tan x}}{\cos^2x} \sim \frac{e^{\large -\frac{1}{x+\frac{\pi }{2}}}}{\left(x+\frac{\pi}{2}\right)^2} \to 0\, \left(=\text{"}\frac{e^{-\infty}}{0^+}\text{"} \right).
$$
|
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|
sum of all Distinct solution of the equation $ \sqrt{3}\sec x+\csc x+2(\tan x-\cot x) = 0\;,$
The sum of all Distinct solution of the equation $\displaystyle \sqrt{3}\sec x+\csc x+2(\tan x-\cot x) = 0\;,$
Where $x\in (-\pi,\pi)$ and $\displaystyle x\neq 0,\neq \frac{\pi}{2}.$
$\bf{My\; Try::}$ We can write equation as $$\frac{\sqrt{3}\sin x+\cos x}{\sin x\cos x}-2\frac{\cos 2x}{\sin x\cos x} = 0$$
So we get $$2\frac{\cos \left(x+\frac{\pi}{3}\right)}{\sin x\cos x}-2\frac{\cos 2x}{\sin x\cos x}=0$$
So we get $$\cos (2x) = \cos\left(x+\frac{\pi}{3}\right)$$
So we Get $\displaystyle 2x=2n\pi\pm (x+\frac{\pi}{3})\;,n\in \mathbb{Z}$
So we get $\displaystyle x= 2n\pi+\frac{\pi}{3}$ or $\displaystyle x = \frac{2n\pi-\frac{\pi}{3}}{3}\;,$ Where $n\in \mathbb{Z}$
So we get $$x=\frac{\pi}{3} = \frac{3\pi}{9}\;\;,-\frac{5\pi}{9}\;\;,-\frac{7\pi}{9}$$
So Sum of distinct Roots is $\displaystyle = -\pi$
But it is wrong, Where i have done wrong, Help me
Thanks
|
Observe that:$$\sqrt{3}\sin x+\cos x = 2\cdot \left(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x\right)$$ $$ = 2\cdot \left(\sin\frac{\pi}{3}\sin x+cos\frac{\pi}{3}\cos x\right)$$ $$=2\cos(x\color{red}{-}\frac{\pi}{3})$$
But instead of $2\cos(x\color{red}{-}\frac{\pi}{3})$, you have written $2\cos(x\color{red}{+}\frac{\pi}{3})$ in the following step and proceeded accordingly:
So we get $$2\frac{\cos \left(x+\frac{\pi}{3}\right)}{\sin x\cos x}-2\frac{\cos 2x}{\sin x\cos x}=0$$
|
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|
Check convergence of $\sum^{\infty}_{n=1} \frac{2^{n} +n ^{2}}{3^{n} +n ^{3}}$ Zoomed version: $$\sum^{\infty}_{n=1} \frac{2^{n} +n ^{2}}{3^{n} +n ^{3}}$$
So, I've seen similar example at Convergence or divergence of $\sum \frac{3^n + n^2}{2^n + n^3}$
And I liked that answer :
$$3^n+n^2\sim_03^n,\quad2^n+n^3\sim_\infty2^n,\enspace\text{hence}\quad \frac{3^n+n^2}{2^n+n^3}\sim_\infty\Bigr(\frac32\Bigl)^n,$$
which doesn't even tend to $0$.
Could you explain why does $3^n+n^2\sim_03^n$ and $2^n+n^3\sim_\infty2^n$ when we have only $n \to \infty$ ?
|
To begin, notice that for all $n\geq 1$ one has $2^n\geq n^2$
$$\sum\limits_{n=1}^\infty \frac{2^n+n^2}{3^n+n^3}\leq \sum\limits_{n=1}^\infty\frac{2^n+2^n}{3^n+n^3}\leq 2\sum\limits_{n=1}^\infty \frac{2^n}{3^n}= 4$$
Since all terms of the original series are non-negative and the series is bounded above, it must converge by the dominated convergence theorem.
Note: the first inequality is true since $n^2\leq 2^n$.
The second inequality is true since $n^3\geq 0$, we are dividing by a smaller denominator at each step thereby making each number larger.
The final equality is true by properties of geometric series.
A more relaxed approach is to simply note that $2^n+n^2$ "acts like" $2^n$, and similarly $3^n+n^3$ "acts like" $3^n$, to note that $\sum\limits_{n=1}^\infty \frac{2^n+n^2}{3^n+n^3}$ "acts like" $\sum\limits_{n=1}^\infty \frac{2^n}{3^n}$
To use this in action, we may use instead of direct comparison as above, the ratio test.
If $\lim\limits_{n\to\infty}|\frac{a_{n+1}}{a_{n}}|<1$ then $\sum\limits_{n=1}^\infty a_n$ converges.
Here, we look at $\frac{a_{n+1}}{a_n} = \frac{2^{n+1}+(n+1)^2}{3^{n+1}+(n+1)^3}\cdot \frac{3^n+n^3}{2^n+n^2} = \frac{2\cdot 6^n + 2^{n+1}n^3+3^n(n+1)^2+n^3(n+1)^2}{3\cdot 6^n + 2^n(n+1)^3+3^{n+1}n^2+(n+1)^3n^2}$
By dividing top and bottom by $6^n$ and taking the limit as $n\to\infty$ we see that $\frac{a_{n+1}}{a_n}\to \frac{2}{3}$ since every term but the first for each of the numerator and denominator approach zero.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Quaternary numeral system: fractions I have a question related to the expression of a real number in base 4. Consider the table here: it is clear to me how all columns of the table are obtained except the fourth one: how do they get the positional representation in quaternary base?
|
The non-repeating ones like $\frac{1}{2},\frac{1}{4}$ are obvious. For the others we repeatedly use $\frac{1}{1-x}=1+x+x^2+x^3+\dots$.
We have $\frac{1}{3}=\frac{1}{4}\frac{1}{1-\frac{1}{4}}=\frac{1}{4}(1+\frac{1}{4}+\left(\frac{1}{4}\right)^2+\dots=\frac{1}{4}+\left(\frac{1}{4}\right)^2+\left(\frac{1}{4}\right)^2+\dots=0.111\dots$.
Hence $\frac{1}{12}=0.011111\dots$ and $\frac{1}{6}=0.02222\dots$
We have $\frac{1}{15}=\frac{1}{16}\frac{1}{1-\frac{1}{16}}=\frac{1}{16}+\left(\frac{1}{16}\right)^2+\left(\frac{1}{16}\right)^3+\dots=0.010101\dots$ Hence $\frac{1}{5}=0.030303\dots$.
We have $\frac{1}{7}=\frac{9}{63}=\frac{9}{64}\frac{1}{1-\frac{1}{64}}=\frac{9}{64}\left(1+\frac{1}{64}+\left(\frac{1}{64}\right)^2+\dots\right)=0.021021021\dots$. Hence $\frac{2}{7}=0.102102102\dots$ and so $\frac{1}{14}=0.0102102102\dots$.
and so on.
|
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|
A cube and a sphere have equal volume. What is the ratio of their surface areas? The answer is supposed to be $$ \sqrt[3]{6} : \sqrt[3]{\pi} $$
Since $$ \ a^3 = \frac{4}{3} \pi r^3 $$
I have expressed it as: $$ \ a = \sqrt[3]{ \frac{4}{3} \pi r^3} $$
and,
$$ \ 6 \left( \sqrt[3]{ \frac{4}{3} \pi r^3 } \right) ^2 : 4 \pi r^2 $$
But I am not really sure how to arrive at the desired result. I have tried to simplify it, but apparently I am missing some step in the process and come to a result that is far from the correct one.
Could you please help? Thank you.
|
Unit volume assumed:
$$
1 = \frac{4}{3}\pi r^3 = a^3
$$
So we have
$$
r = \sqrt[3]{\frac{3}{4\pi}} \\
\quad a = 1
$$
So the surface ratio is
$$
A_s = 4 \pi r^2
= 4 \pi \left( \frac{3}{4\pi} \right)^{2/3}
= \sqrt[3]{4\pi \cdot 9}
=\sqrt[3]{\pi} \, 6^{2/3} \\
A_c = 6
$$
So we get
$$
A_s : A_c =
\sqrt[3]{\pi} : \sqrt[3]{6}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1798726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Check for convergence $$\sum_{n = 2}^\infty (-1)^n \sin\left( \frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)$$
I tried to use Maclaurin series, but failed to evaluate little-o.
|
Write $\sin x=x-x^3/6+x^3\varepsilon(x)$, where $\varepsilon$ is such that $\lim_{x\to 0}\varepsilon(x)=0$. Consequently,
$$ (-1)^n \sin\left( \frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)=(-1)^n\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)-\frac{(-1)^n}6\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)^3\\
+\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)^3\delta_n,$$
where the sequence $\left(\delta_n\right)_{n\geqslant 1}$ converges to $0$ as $n$ goes to infinity. Since
$$\left|\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n (\ln n)^2}\right)^3\right|\leqslant cn^{-3/2}$$
for some $c$ independent of $n$ and the series $\sum_{n\geqslant 2}1/(n(\ln n)^2)$ is convergent, the problem reduces to the simpler one: determine the convergence of the series $\sum_{n\geqslant 2}(-1)^n\sin(3n)/\sqrt n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a, b, c >0$ prove that $ [(1+a)(1+b)(1+c)]^7 > 7^7a^4b^4c^4 $. I solved it using AM, GM inequalities and reached to $[(1+a)(1+b)(1+c)]^7 > 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.
|
We only need to prove that for positive $x$ we have $(1+x)^7\gt 7^{7/3}x^4$.
For positive $x$, let
$$f(x)=\frac{(1+x)^7}{x^4}.$$
By using $f'(x)$ we find that $f(x)$ reaches a minimum at $x=4/3$. The minimum value of $f(x)$ is
$$\frac{7^7}{3^34^4,}$$
which is greater than $7^{7/3}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1802976",
"timestamp": "2023-03-29T00:00:00",
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|
Solve $636^{369}\equiv x\pmod{126}$
Solve
$$636^{369}\equiv x\pmod{126}$$
My attempt:
$$126=2\times 3^2 \times 7$$
$$\varphi(126)=\varphi(2)\times \varphi(3^2)\times \varphi(7)=36$$
$$\color{gray}{636=6\pmod{126}}$$
$$6^{369}\equiv x \pmod{126}$$
$$2^{369}3^{369}\equiv x \pmod{126}$$
I am stuck here
|
$636 \equiv 06 \equiv 6 $ (mod 7)
$\varphi(7)=6$
$369 \equiv 09 \equiv 3$ (mod 6)
So
$636^{369}\equiv 6^3 \equiv 36 \times 6 \equiv 6$ (mod 7)
(Note that $gcd(636,7)=1$)
Also, $636\equiv06 \equiv 6$ (mod 9)
and $6\times6=36\equiv 0$ (mod 9)
So
$636^n\equiv 0$ (mod 9) for $n>1$
Therefore, we have $636^{369}=6+7k\equiv 0$ (mod 18)
Hence $k \equiv-\frac{6}{7}\equiv-\frac{42}{7}\equiv-6\equiv12$ (mod 18)
So, finally, $636^{369} =6+7(12+18k')\equiv6+84\equiv90$ (mod 126)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the following equation for $x$,$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$ I am not great at transposition and wolfram alpha confused me so I would like to see the steps in solving for x.
$$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$$
Wolfram alpha gave me two incompatible answers
$$x=\pm\sqrt{n}, \ \ \ \ \ \ \ \ \sqrt{n}\not=0$$
$$x=\pm\frac{\sqrt{n}}{\sqrt{3}}, \ \ \ \ \ \ \ \ \sqrt{n}\not=0$$
The first answer is wrong the second is correct.
Can anyone explain why the first solution is wrong and how to derive the second?
|
Another way to look at the problem is to rewrite (assuming $x\neq 0$) $$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)\implies 4 x^2 \left(\frac{\left(n-x^2\right)^2}{4 x^2}-\frac{1}{2}
\left(n-x^2\right)\right)=0$$ Expand and simplify to get $$3 x^4-4n x^2+n^2=0$$ Now, using $y=x^2$, the equation is just the quadratic in $y$ $$3y^2-4ny+n^2=0$$ the roots of which being $y=n$ and $y=\frac n 3$.
So $x=\pm \sqrt n$ and $x=\pm \sqrt{\frac n3}$ are the roots of the original equation.
|
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|
Prove $(x+y+z) \cdot \left( \frac1x +\frac1y +\frac1z\right) \geqslant 9 + \frac{4(x-y)^2}{xy+yz+zx}$ $x,y,z >0$, prove
$$(x+y+z) \cdot \left( \frac1x +\frac1y +\frac1z\right) \geqslant 9 + \frac{4(x-y)^2}{xy+yz+zx}$$
The term $\frac{4(x-y)^2}{xy+yz+zx}$ made this inequality tougher. It remains me of this inequality. I think one can prove it by expanding everything, but is there any elegant solution for this problem ?
|
There is a very nice solution only using SOS, I was found it!
We need to prove:$$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) - 9 \geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$(\frac{x}{y}+\frac{y}{x}-2) + (\frac{y}{z}+\frac{z}{x}-2)+(\frac{z}{x}+\frac{x}{z}-2) \geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$\frac{(x-y)^2}{xy}+\frac{(x-y)^2}{yz+zx}+(\frac{(y-z)^2}{yz}+\frac{(z-x)^2}{zx} -\frac{(x-y)^2}{yz+zx})\geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$(x-y)^2(\frac{1}{xy}+\frac{1}{yz+zx})+\frac{(xz+yz-2xy)^2}{xyz(x+y)} \geq \frac{4(x-y)^2}{xy+yz+zx}$$
or $$(x-y)^2(\frac{1}{xy}+\frac{1}{yz+zx}-\frac{4}{xy+yz+zx})+\frac{(xz+yz-2xy)^2}{xyz(x+y)} \geq 0$$
or $$\frac{(x-y)^2(xz+yz-xy)^2}{xyz(x+y)(xy+yz+zx)}+\frac{(xz+yz-2xy)^2}{xyz(x+y)} \geq 0$$
Which is obvious!
Equality holds when $x=y=z$
|
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|
Problem based on sum of reciprocal of $n^{th}$ roots of unity
Let $1,x_{1},x_{2},x_{3},\ldots,x_{n-1}$ be the $\bf{n^{th}}$ roots of unity. Find: $$\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+......+\frac{1}{1-x_{n-1}}$$
$\bf{My\; Try::}$ Given $x=(1)^{\frac{1}{n}}\Rightarrow x^n=1\Rightarrow x^n-1=0$
Now Put $\displaystyle y = \frac{1}{1-x}\Rightarrow 1-x=\frac{1}{y}\Rightarrow x=\frac{y-1}{y}$
Put that value into $\displaystyle x^n-1=0\;,$ We get $\displaystyle \left(\frac{y-1}{y}\right)^n-1=0$
So we get $(y-1)^n-y^n=0\Rightarrow \displaystyle \left\{y^n-\binom{n}{1}y^{n-1}+\binom{n}{2}y^2-......\right\}-y^n=0$
So $\displaystyle \binom{n}{1}y^{n-1}-\binom{n}{2}y^{n-1}+...+(-1)^n=0$ has roots $y=y_{1}\;,y_{2}\;,y_{3}\;,.......,y_{n-1}$
So $$y_{1}+y_{2}+y_{3}+.....+y_{n-1} = \binom{n}{2} =\frac{n(n-1)}{2}\;,$$ Where $\displaystyle y_{i} = \frac{1}{1-x_{i}}\;\forall i\in \left\{1,2,3,4,5,.....,n-1\right\}$
My Question is can we solve it any less complex way, If yes Then plz explain here, Thanks
|
Let $f(x) = \frac{(x^n-1)}{(x-1)} = x^{n-1}+x^{n-2}+...+x+1$
Then $\frac{f'(x)}{f(x)} = \frac{1}{x-x_1} + \frac{1}{x-x_2}+ ... + \frac{1}{x-x_{n-1}}$
So what you want is just $\frac{f'(1)}{f(1)}$, which is easy to compute as $\frac{n-1}{2}$.
|
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|
Evaluating a strange integral... I was given this integral by a friend and have been unable to evaluate it. He said it is easily possible to do it by hand (no calculator or computational aids necessary)
$$\int_{0}^{1} \mathrm{frac}\left(\frac{1}{x}\right)\cdot\mathrm{frac}\left(\frac{1}{1-x}\right)dx $$
frac is a function that returns the fractional part of x. I attempted to "simplify" this to x-floor(x), but was still unable to get anywhere.
He also gave me a hint that it involves the euler mascheroni constant. I looked at the integral representation of it but was unable to make any progress
|
Making the variable change, $u = 1-x,$ we have
$$\int_{1/2}^1 \left\{\frac{1}{x} \right\}\left\{\frac{1}{1-x} \right\} \, dx = \int_{0}^{1/2} \left\{\frac{1}{1-u} \right\}\left\{\frac{1}{u} \right\} \, du. $$
Hence,
$$\int_0^1 \left\{\frac{1}{x} \right\}\left\{\frac{1}{1-x} \right\} \, dx = 2\int_0^{1/2} \left\{\frac{1}{x} \right\}\left\{\frac{1}{1-x} \right\} \, dx \\ = 2\lim_{n \to \infty} \sum_{k =2}^n \int_{1/(k+1)}^{1/k} \left\{\frac{1}{x} \right\}\left\{\frac{1}{1-x} \right\} \, dx. $$
For $1/(k+1) < x < 1/k$ we have $k < 1/x < k+1$ and $1 + 1/k < 1/(1-x) < 1 + 1/(k-1)$.
Hence, for $k \geqslant 2$,
$$\left\{\frac{1}{x} \right\} = \frac{1}{x} - k, \\ \left\{\frac{1}{1-x} \right\} = \frac{1}{1-x} - 1,$$
and
$$\int_0^1 \left\{\frac{1}{x} \right\}\left\{\frac{1}{1-x} \right\} \, dx = 2\lim_{n \to \infty} \sum_{k =2}^n \int_{1/(k+1)}^{1/k} \left(\frac{1}{x}-k \right) \left(\frac{1}{1-x} -1\right) \, dx.$$
Proceeding,
$$\begin{align}\int_{1/(k+1)}^{1/k} \left(\frac{1}{x}-k \right) \left(\frac{1}{1-x} -1\right) \, dx &= \int_{1/(k+1)}^{1/k} \left(\frac{1-kx}{1-x} \right) \, dx \\ &=\int_{1/(k+1)}^{1/k} \left(\frac{(1-k) + k(1-x)}{1-x} \right) \, dx \\ &= [(k-1) \log(1-x) +kx]_{1/(k+1)}^{1/k} \\ &= (k-1) \log \frac{k-1}{k} - (k-1)\log \frac{k}{k+1} +\frac{1}{k+1} \\ &= k\log \frac{k+1}{k} - (k-1) \log \frac{k}{k-1} - [\log(k+1) - \log k] + \frac{1}{k+1}\end{align}$$
Summing, we get
$$\begin{align}\sum_{k =2}^n \int_{1/(k+1)}^{1/k} \left(\frac{1}{x}-k \right) \left(\frac{1}{1-x} -1\right) \, dx &= \sum_{k=2}^n \frac{1}{k+1} -\log(n+1) + n \log \frac{n+1}{n} \\ &= -\frac{3}{2} + \sum_{k=1}^{n+1} \frac{1}{k} - \log(n+1) + \log \left[\left( 1 + \frac{1}{n} \right)^n \right]\end{align}.$$
Taking the limit as $n \to \infty$ we find
$$\lim_{n \to \infty}\sum_{k =2}^n \int_{1/(k+1)}^{1/k} \left(\frac{1}{x}-k \right) \left(\frac{1}{1-x} -1\right) \, dx = -\frac{1}{2} + \gamma ,$$
and
$$\int_0^1 \left\{\frac{1}{x} \right\}\left\{\frac{1}{1-x} \right\} \, dx = -1 + 2\gamma.$$
|
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|
Evaluate the limit $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$
Evaluate the limit $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$, without using a Riemann sum
$\bf{My\; Try:}$ Using the graph of $\displaystyle f(x) = \frac{1}{\sqrt{x}}\;,$ we get
$$\int_{1}^{n+1}\frac{1}{\sqrt{x}}dx <\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<1+\int_{1}^{n}\frac{1}{\sqrt{x}}dx$$
So we get $$2\sqrt{n+1}-2<\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<2\sqrt{n}-1$$
So $$\lim_{n\rightarrow \infty}\frac{2\sqrt{n+1}-2}{\sqrt{n}}=2<\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<\lim_{n\rightarrow \infty}\frac{2\sqrt{n}-1}{\sqrt{n}} = 2$$
So using the Sandwich Theorem, we get $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)=2$
My question is can we solve it by using any other method? If yes then please explain it here.
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Notice that:
$$\frac1{\sqrt n}\sum_{k=1}^n\frac1{\sqrt k}=\frac1n\sum_{k=1}^n(k/n)^{-1/2}$$
As $n\to\infty$, we get a Riemann sum:
$$\lim_{n\to\infty}\frac1n\sum_{k=1}^n(k/n)^{-1/2}=\int_0^1x^{-1/2}~\mathrm dx=2x^{1/2}\bigg|_0^1=2$$
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"timestamp": "2023-03-29T00:00:00",
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|
Limit with root and fraction I have this limit:
$$\lim_{x\to 0^+} \frac{\sqrt{1+2x} - e^x}{x\arctan{x}}$$
And I try this
\begin{align}\lim_{x\to 0^+} \frac{\sqrt{1+2x} - e^x}{x\arctan{x}} & = \lim_{x\to 0^+} \frac{\sqrt{1+2x} - e^x}{x\arctan{x}}\frac{\sqrt{1+2x} + e^x}{\sqrt{1+2x} + e^x} \\ & \stackrel{(*)}= \lim_{x\to 0^+} \frac{1+2x - e^x}{x^2(\sqrt{1+2x}+e^x)} \end{align}
$(*)$: $\arctan x\simeq x$
But now I'm blocked... Any advice?
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Considering $$A=\frac{\sqrt{1+2x} - e^x}{x\,\tan^{-1}(x)}$$ there are two possibilities.
The first one is to use L'HΓ΄pital's rule just as Workaholic commented, setting $$u=\sqrt{1+2x} - e^x\qquad , \qquad v=x\,\tan^{-1}(x)$$ and you will probaly need to apply it more than once.
The second one would involve Taylor expansions $$\sqrt{1+2x}=1+x-\frac{x^2}{2}+\frac{x^3}{2}+O\left(x^4\right)$$ $$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)$$ $$\tan^{-1}(x)=x-\frac{x^3}{3}+O\left(x^4\right)$$ which make $$A=\frac{-x^2+\frac{x^3}{3}+O\left(x^4\right)}{x^2-\frac{x^4}{3}+O\left(x^5\right)}$$ Now, long division to get $$A=-1+\frac{x}{3}+O\left(x^2\right)$$ which shows the limit and also how it is approached.
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|
Showing $\int_{0}^{1}{1-x^2\over x^2}\ln\left({(1+x^2)^2\over 1-x^2}\right)dx=2$ Showing
$$I=\int_{0}^{1}{1-x^2\over x^2}\ln\left({(1+x^2)^2\over 1-x^2}\right)dx=2\tag1$$
$$I=\int_{0}^{\infty}{1-x^2\over x^2}\ln\left({1+x^2\over 1-x^2}\right)+\int_{0}^{\infty}{1-x^2\over x^2}\ln(1+x^2)dx\tag2$$
Let $$J=\int_{0}^{\infty}{1-x^2\over x^2}\ln\left({1+x^2\over 1-x^2}\right)dx\tag3$$
Apply integration by part to (3)
$u=\ln\left({1+x^2\over 1-x^2}\right)\rightarrow du={4x\over (1-x^2)^2}dx$
$dv=1-x^{-2}\rightarrow v=x-x^{-1}$
$$J=\left.-{x+1\over x}\ln\left({1+x^2\over 1-x^2}\right)\right|_{0}^{1}+\int_{0}^{1}{4\over (1+x)(1-x)^2}dx\tag4$$
$$J=\left.-{x+1\over x}\ln\left({1+x^2\over 1-x^2}\right)\right|_{0}^{1}+\int_{0}^{1}\left({1\over 1+x}+{1\over 1-x}+{2\over (1-x)^2}\right)dx\tag5$$
$$J=\left.-{x+1\over x}\ln\left({1+x^2\over 1-x^2}\right)\right|_{0}^{1}+\left.\ln\left({1+x\over 1-x}\right)-{2\over 1-x}\right|_{0}^{1}\tag6$$
I am going to stop here. Evaluate these limits is not valid because of ${1\over 0}$. I have try substitution it looks more messier and complicated than this. I need some help, thank.
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The Taylor series approach is quite straightforward, too. We have:
$$ 2\log(1+x^2)-\log(1-x^2)=\sum_{n\geq 1}\frac{1-2(-1)^n}{n}x^{2n} \tag{1}$$
so by multiplying the RHS by $\frac{1}{x^2}-1$ we get:
$$ \frac{1-x^2}{x^2}\left(2\log(1+x^2)-\log(1-x^2)\right)=3+\sum_{n\geq 1}\frac{-1+(2+4n)(-1)^n}{n(n+1)}x^{2n} \tag{2}$$
and by integrating over $(0,1)$ the original integral turns into:
$$ 3-\sum_{n\geq 1}\frac{1}{n(n+1)(2n+1)}+2\sum_{n\geq 1}\frac{(-1)^n}{n(n+1)}=3-(3-4\log 2)+(2-4\log 2)\tag{3}$$
i.e. $\color{red}{\large 2}$, by partial fraction decomposition. We may also notice that:
$$ -\sum_{n\geq 1}\frac{1}{n(n+1)(2n+1)}+\sum_{n\geq 1}\frac{2(-1)^n}{n(n+1)} \\= -1+\sum_{n\geq 1}\left(\frac{1}{n(2n+1)}-\frac{1}{(n+1)(2n+1)}-\frac{1}{n(n+1)(2n+1)}\right)=-1+\sum_{n\geq 1}\color{red}{0}\tag{4}$$
and avoid partial fraction decomposition at all!
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|
Equation involving Wilson's theorem Find all primes $p$ such that
$$(p-1)!=p^k-1.$$
Where $k$ is a natural number.
|
this is a pretty known one:
for $p=2$, we have $2^k-1=1$ which gives $k=1$.
for $p=3$, we have $3^k-1=2$ which also gives $k=1$.
for $p=5$, we have $5^k-1=4!=24$ which gives $k=2$.
For $p>5$:
note that $p-1|(p-2)!$ for every prime $p>5$. Why?
Because $p-1=2\frac{p-1}{2}$ and given that $2$, and $\frac{p-1}{2}$ are smaller than $p-2$, not equal to each other and that the second one is integer (i.e. $p$ is odd), we can find both of them in $(p-2)!$ expansion.
If $2 \geq p-2$ then $p \leq 4$ so $p=2$ or $3$, contradiction. Similarly, if $\frac{p-1}{2}=p-2$ then $p=3$ and when $\frac{p-1}{2}=p-2$ then $p=5$, both contradicting $p>5$. Also,$p$ is even since $p>5>2$.
Therefore, $(p-1)^2|(p-1)!$ so $(p-1)^2|p^k-1=(p-1)(1+p+ \ldots +p^{k-1})$.
Therefore, $p-1|1+p+ \ldots +p^{k-1} \equiv 1+1+ \ldots +1 \equiv k \pmod{p-1}$.
Therefore, $p-1|k$.
This gives $p^{p-1}-1| \leq p^k-1=(p-1)!$ which after some routine gives $p=2$.
Indeed, after substitution, $(2,1),(3,1),(5,2)$ are the only solution pairs.
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|
Integrate $\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx$ $\displaystyle\int \dfrac {x^{2}} {\sqrt {x^{2}-16}}dx$
Effort 1:
Let be $x=4\sec u$
$dx=4.\sin u.\sec^2u.du$
Then integral;
$\displaystyle\int \dfrac {\sec^2u \; .4.\sin u.\sec^2u.du} {\sqrt {16\sec^2u-16}}=\displaystyle\int \sec^3.du$
After I didn't nothing.
Effort 2:
$\displaystyle\int \dfrac {x^{2}dx} {\sqrt {x^{2}-16}}$
Let's doing integral by parts;
$du=\dfrac{x}{\sqrt{x^2-16}}$
$v=x$
$\displaystyle\int \dfrac {x^{2}dx} {\sqrt {x^{2}-16}}=x.\sqrt{x^2-16}-\displaystyle\int\sqrt{x^2-16}dx$
We have $\quad \displaystyle\int\sqrt{x^2-16}dx$
let be $\quad x=4\sec j$
$dx=4\dfrac{\sin j}{\cos^2 j}dj$
and;
$\displaystyle\int\sqrt{x^2-16}\;dx=16.\displaystyle\int\dfrac{\sin j}{\cos j}\dfrac{\sin j}{\cos^2 j}dj=16.\displaystyle\int \sec j.tan^2j.dj$
After I didn't nothing.
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Take $x=4\sec\left(u\right)
$. We get $$I=\int\frac{x^{2}}{\sqrt{x^{2}-16}}dx=16\int\sec^{3}\left(u\right)du
$$ and now for the reduction formula for powers of $\sec\left(u\right)
$ we have $$I=8\tan\left(u\right)\sec\left(u\right)+8\int\sec\left(u\right)du
$$ $$=8\tan\left(u\right)\sec\left(u\right)+8\log\left(\tan\left(u\right)+\sec\left(u\right)\right)
$$ and now substitute back.
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|
Solutions to a quadratic diophantine equation $x^2 + xy + y^2 = 3r^2$. Let $k,i,r \in\Bbb Z$, $r$ constant. How to compute the number of solutions to $3(k^2+ki+i^2)=r^2$, perhaps by generating all of them?
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There are infinitely many solutions to this Diophantine equation.
I change your variables. We have the Diophantine equation$$3(a^2+ab+b^2)=c^2$$Let's assume $c$ is not constant and find all possible solutions! $c=0$ gives $a=b=0$. W.L.O.G. suppose $c>0$. It's easy to see that $c=3d$ for some positive integer $d$. Equation becomes$$a^2+ab+b^2=3d^2$$For $a=b$ we get $a=b=\pm d$. W.L.O.G. suppose that $a>b$. Notice that$$a^2+ab+b^2 \equiv 0 \pmod 3$$It follows that $a^3 \equiv b^3 \pmod 3$ and $a \equiv b \pmod 3$. Let $a-b=3e$ for some positive integer $e$. Equation turns into$$b^2+3be+3e^2=d^2$$In order to get integer values for $b$ discriminant of this quadratic must be a perfect square, that is$$\Delta_{b}=-3e^2+4d^2=f^2$$For some integer $f$.
$f=0$ gives $\sqrt{3}=2\frac{d}{e}$, which is impossible. W.L.O.G. suppose that $f>0$. Therefore, we need to solve$$f^2+3e^2=4d^2$$In positive integers. Now this is of the form of extended Pythagorean equation $Ax^2+By^2=Cz^2$, which is widely studied. Even there are some questions here and here, which exactly discuss your particular case!
Once you find parametric forms of $f$, $e$ and $d$, you can substitute backwards and find parametric form of your original variables $a$, $b$ and $c$.
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|
Finding solution of the irrational equation Given equation $\sqrt{x + 3 - 2\sqrt{x + 2}} + \sqrt{x + 27 - 10\sqrt{x + 2}} = 4$, find its solution(s).
At first, finding the domain of the function. Noting that $\sqrt{x + 2} \geq 0 \implies x \geq -2$. Then, solving inequalities for $x + 3 - 2\sqrt{x + 2} \geq 0$ and $x + 27 - 10\sqrt{x + 2} \geq 0$ we get that their domain $D = \mathbb{R}$, because quadratics which arise from those inequalities have exactly one root - $S_1 = \{-1\}$ and $S_2 = \{23\}$. Because all expressions are greater than $0$, I am omitting absolute values.
Solving the equation via substitution $t = \sqrt{x + 2}$ we get:
$$\sqrt{x + 3 - 2t} + \sqrt{x + 27 - 10t} = 4$$
$$x + 3 - 2t = 16 - 8\sqrt{x + 27 - 10t} + x + 27 - 10t$$
$$t^2 - 10t + 25 = x + 27 - 10t$$
$$t^2 = x + 2$$
$$x + 2 = x + 2 \implies S = [-2, \infty).$$
But according to my textbook (which I use as a preparation for entrance exams) the solution must be $S = [-1, 23]$, where -1 and 23 are, obviously not coincidently, solutions to the quadratics, which we obtain while solving expressions under square roots. Where do I have a mistake, then?
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Put $t=\sqrt{x+2}$, so we require $t\ge0$. Then $\sqrt{x+3-2t}=\sqrt{t^2+1-2t}=\sqrt{(t-1)^2}=|t-1|$. Similarly, $\sqrt{x+27-10t}=\sqrt{(t-5)^2}=|t-5|$. So we have $|t-1|+|t-5|=4$. That holds for $t\in[1,5]$.
So we must have $1\le\sqrt{x+2}\le5$ and hence $1\le x+2\le25$, so $-1\le x\le 23$.
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|
Sum of infinite series with Irrational terms
The sum of the series $\displaystyle 1+\frac{\sqrt{2}-1}{2\sqrt{2}}+\frac{3-2\sqrt{2}}{12}+\frac{5\sqrt{2}-7}{24\sqrt{2}}+\frac{17-12\sqrt{2}}{80}+.....\infty$
$\bf{My\; Try::}$Here $3-2\sqrt{2} = (\sqrt{2}-1)^2$ and $5\sqrt{2}-7 = (\sqrt{2}-1)^3$
similaryly $17-12\sqrt{2} = (\sqrt{2}-1)^4\;,$ Now let $(\sqrt{2}-1)=x>0$
So our series convert into $$1+\frac{x^2}{2\sqrt{2}}+\frac{x^4}{12}+\frac{x^6}{24\sqrt{2}}+\frac{x^8}{80}+........$$
Now how can i solve after that, Help required, Thanks
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From the suggestion by @Winther, express the series as
$$S = 1+\sum_{n=1}^{\infty} a_n x^n $$
where $a_n = \frac1{n (n+1)}$ and $x=1-\frac1{\sqrt{2}} $. Then
$$\begin{align}S &= 1+\sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{n+1}\\ &= 1-\log{(1-x)} - \frac1x\left (-\log{(1-x)}-x \right )\\ &= 2-\left (1-\frac1x \right ) \log{(1-x)}\end{align}$$
Now
$$1-\frac1x = 1-\frac{\sqrt{2}}{\sqrt{2}-1} = -\frac1{\sqrt{2}-1}$$
Thus
$$S=2 - \frac12 (\sqrt{2}+1) \log{2} $$
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Find $3^{333} + 7^{777}\pmod{ 50}$ As title say, I need to find remainder of these to numbers. I know that here is plenty of similar questions, but non of these gives me right explanation. I always get stuck at some point (mostly right at the beginning) and don't have idea how to start.
Thanks in advance.
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I always start by writing a couple of powers, in this case modulo $50$:
$$\begin{align}3^1\equiv 3&\mod 50\\
3^2\equiv 9&\mod 50\\
3^3\equiv 27&\mod 50\\
3^4=81\equiv 31&\mod 50\\
3^5\equiv 93\equiv 43&\mod 50\\
3^6\equiv 129\equiv 29&\mod 50\\
3^7\equiv 87\equiv 37&\mod 50\\
3^8\equiv 111\equiv 11&\mod 50\\
3^9\equiv 33&\mod 50\\
3^{10}\equiv 99\equiv-1&\mod 50\\
\end{align}$$
Oh look, a pattern! If $3^{10}\equiv -1\pmod {50}$, this means that $3^{20}\equiv 1\pmod {50}$,
so from here, I know that $3^{333}=(3^{20})^{16}\cdot 3^{10}\cdot 3^3\equiv 1\cdot(-1)\cdot 27=-27\equiv 23\pmod {50}$.
For the $7$, it's even easier, since $7^2\equiv -1\pmod {50}$, meaning that $7^{777}=(7^2)^{388}\cdot 7\equiv (-1)^{388}\cdot 7=7\pmod {50}$
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|
Cauchy sequence $x_n=\sqrt{a+x_{n-1}}$ I have to show that this sequence
$$
x_n=\sqrt{a+x_{n-1}} \hbox{ with } x_1=\sqrt{a}
$$
is a Cauchy sequence for every $a>0$. I have done the following calculations:
$$
\left| x_{n+2}-x_{n+1} \right|=\left| \sqrt{a+x_{n+1}}-x_{n+1}\right|=\left|\frac{a+x_{n+1}-x_{n+1}^2}{\sqrt{a+x_{n+1}}+x_{n+1}} \right|= \left|\frac{x_{n+1}-x_n}{\sqrt{a+x_{n+1}}+x_{n+1}}\right|
$$
I don't come up with a boundary for the denominator so that
$$
\left|\frac{x_{n+1}-x_n}{\sqrt{a+x_{n+1}}+x_{n+1}}\right|<k\cdot\left|x_{n+1}-x_n\right| \hbox{ with } k<1
$$
Any hint? I know I can use induction to easily prove that the sequence is convergent, but I'd like to prove it's Cauchy without using convergence. Thank you in advance.
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Claim. The given sequence is bounded from above by $$ \frac{1 + \sqrt{4a + 1}}{2} = \alpha $$
Proof. We proceed by induction. For $ k = 1 $ we have that
$$ \sqrt{a} = \frac{2\sqrt{a}}{2} = \frac{\sqrt{4a}}{2} < \frac{1 + \sqrt{4a + 1}}{2} $$
Now, assuming that the statement is true for $ k = n $, we have
$$ x_{n+1} = \sqrt{a + x_n} < \sqrt{ \frac{1 + 2a + \sqrt{4a + 1}}{2} } $$
Observe that
$$ \left( \frac{1 + \sqrt{4a + 1}}{2} \right)^2 = \frac{4a + 2 + 2\sqrt{4a + 1}}{4} = \frac{1 + 2a + \sqrt{4a + 1}}{2} $$
which establishes the bound for $ x_{n+1} $.
Now, we prove that the sequence is monotone increasing. For this, the following inequality is crucial:
Claim. For $ 0 \leq x < \alpha $, we have that $ x < \sqrt{a+x} $.
Proof. The given inequality is true precisely when $ x^2 - x - a < 0 $. The inequality factors as $ (x - \alpha)(x - \alpha') < 0 $ where $ \alpha' < 0 $, therefore the second factor is always positive when $ x \geq 0 $, and the first factor is negative precisely when $ x < \alpha $.
Now, since we have $ 0 \leq x_k < \alpha $ for all $ k $, it follows by this claim that
$$ x_{k+1} = \sqrt{a + x_k} > x_k $$
Since the sequence $ x_k $ is monotone increasing and bounded from above, by the monotone convergence theorem it is convergent; and thus Cauchy.
|
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|
Find the order of the pole How would you go about finding the order of the pole at $z=0$ of the following function?
$$f(z)=\frac{1}{(2\cos(z)-2+z^2)^2}$$
I feel like you might need to rewrite $\cos(z)$ as a Maclaurin series but I'm not entirely sure what you'd do next.
Also, once you think you've worked out the order of the pole, is there any way you can check that it's correct? Like some kind of test maybe.
Any help would be greatly appreciated!
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Consider the expansion of $f(z)=\cos(z)$ at $z=0$ namely
$$\cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}+\mathcal{O}(z^6)$$
We can put this into the expression,
\begin{align*}
\frac{1}{(2\cos(z)-2+z^2)^2}&=\frac{1}{\left(2\left(1-\frac{z^2}{2}+\frac{z^4}{4!}+\mathcal{O}(z^6)\right)-2+z^2\right)^2}\\
&=\frac{1}{\left(2\left(-\frac{z^2}{2}+\frac{z^4}{4!}+\mathcal{O}(z^6)\right)+z^2\right)^2}\\
&=\frac{1}{\left(-z^2+z^2+2\left(\frac{z^4}{4!}-\frac{z^6}{6!}+\mathcal{O}(z^8)\right)\right)^2}\\
&=\frac{1}{2(z^4)^2\left(\frac{1}{4!}-\frac{z^2}{6!}+\frac{z^4}{8!}+\mathcal{O}(z^6)\right)^2}\\
&=\frac{1}{2z^8(1/4!+\beta)^2}. \tag{*}
\end{align*}
Here,
$$\beta=-\frac{z^2}{6!}+\frac{z^4}{8!}+\mathcal{O}(z^6)$$
which is identically zero when $z=0$. Thus, from ($*$) we can see that our function has a $1/z^8$ term only, which is a pole of order $8$.
|
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How to solve this homogeneous recurrence relation of 2nd order? I have this homogeneous recurrence relation: $x_n = 3x_{n-1} + 2x_{n-2}$ for $n \geq 2$ and $x_0 = 0$, $x_1 = 1$. I form the characteristic polynomial: $r^2 - 3r -2 = 0$ which gives the roots $r = \frac{3}{2} - \frac{\sqrt{17}}{2}$, $\frac{3}{2} + \frac{\sqrt{17}}{2}$. Now when I pluck these values into the function $a_n = \alpha (\frac{3}{2} - \frac{\sqrt{17}}{2})^n + \beta (\frac{3}{2} + \frac{\sqrt{17}}{2})^n$ and solve it for $\alpha$ and $\beta$ I get enormous fractions which make my head hurt. Is there an easier and neater way of solving this or am I doing something wrong?
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Name the roots $r$ and $s$ and solve for the initial conditions
$$ar^0+bs^0=0=a+b,\\ar^1+bs^1=1=ar+bs.$$
Then
$$a=-b=\frac1{r-s}.$$
The difference between the roots being $\sqrt{17}$, the "enormous fraction" is
$$x_n=\frac{\left(\frac{3+\sqrt{17}}2\right)^n-\left(\frac{3-\sqrt{17}}2\right)^n}{\sqrt{17}}=\frac{(3+\sqrt{17})^n-(3-\sqrt{17})^n}{2^n\sqrt{17}}.$$
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|
How to apply Fubini's theorem? I was asked to show the equality of these integrals
$$\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\log(4+\sin x)dydx
=\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\log(4+\sin x)dxdy\tag{1}$$
Which can be answered by using Fubini's theorem but in order to use Fubini here is what I did
Let $f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\log(4+\sin x)$. Now
\begin{align*}
|f(x,y)|&= \biggl|\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\log(4+\sin x)\biggl|\\
&\leq \log5\biggl|\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\biggl|\\
&\leq\log 5\biggl(\frac{x^2+y^2}{(x^2+y^2)^{3/2}}\biggl)=\frac{\log5}{(x^2+y^2)^{1/2}}:=g(x,y)
\end{align*}
To apply Tonelli Theorem on a non negative function
$$\int_0^1\int_0^1\frac{\log5}{(x^2+y^2)^{1/2}}dxdy\tag{2}$$
This is the part I get stuck. If the above integral in $(2)$ is finite, then we can apply Fubini and $(1)$ follows immediately. Is my choice of $g(x,y)$ fine? If so, how can I evaluate $(2)$. But if not, what is the best choice to bound $|f(x,y)|$?
|
Hint: change to polar coordinates.
|
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|
Pell equation in ${\mathbb Q}(x)$ Is it known whether the equation $A^2-(x^2+3)B^2=1$ has a solution $A,B\in{\mathbb Q}(x)$ with $B\neq 0$ ?
My thoughts : I think that there is no solution, as the fundamental solution of $A^2-(x^2+3)B^2=1$ for $x\in {\mathbb Z}$ seems to vary uncontrollably.
|
$$ \left( 2x^2 + 3 \right)^2 - \left( x^2 + 3 \right) \left( 2x \right)^2 = 9 $$
$$ \left( \frac{2}{3}x^2 + 1 \right)^2 - \left( x^2 + 3 \right) \left( \frac{2}{3}x \right)^2 = 1 $$
From $ \color{red}{\mbox{Jyrki's}} $ comment, material I had not realized applied here: we get a matrix that takes a solution ( as a column vector) to the next one, in matrix form
$$
M =
\left(
\begin{array}{cc}
\frac{2}{3} x^2 + 1 & \frac{2}{3} x^3 + 2x \\
\frac{2}{3} x & \frac{2}{3} x^2 + 1
\end{array}
\right)
$$
In turn, we have solutions $(A_n, B_n)$ with $A_0 = 1, A_1 = \frac{2}{3} x^2 + 1,$ then $B_0 = 0, B_1 = \frac{2}{3} x,$ finally from Cayley-Hamilton
$$ A_{n+2} = \left(\frac{4}{3} x^2 + 2 \right) A_{n+1} - A_n $$ and
$$ B_{n+2} = \left(\frac{4}{3} x^2 + 2 \right) B_{n+1} - B_n. $$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
$$ \left( \frac{8}{163}x^2 + \frac{8}{163}x + \frac{165}{163} \right)^2 - \left( x^2 + x + 41 \right) \left( \frac{8}{163}x + \frac{4}{163} \right)^2 = 1 $$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
$$ \left( x^2 + \frac{k}{2} \right)^2 - \left( x^2 + k \right) \left( x \right)^2 = \frac{k^2}{4} $$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
$$ \left( x^2 + \frac{4k+1}{8} \right)^2 - \left( x^2 + x + k \right) \left( x + \frac{1}{2} \right)^2 = \frac{(4k-1)^2}{64} $$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
For these principal forms, we can always adjust by multiplying the appropriate terms by a rational constant to get $1$ on the right hand side, either $2/k$ or $8/(4k-1).$
|
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|
Is there an elegant way to evaluate $ I={ \int \sqrt[8]{\frac{x+1}{x}} \ \mathrm{d}x}$? Is there an elegant way to evaluate the following integral?
$$ I={ \int \sqrt[8]{\dfrac{x+1}{x}} \ \mathrm{d}x}$$
This seems to me a very lengthy question, yet it was given in my weekly worksheet, so there must be an elegant solution.
Any help will be appreciated.
|
Just to rework @H. R.'s solution by hand. First off setting
$$t=\frac{x+1}x\ge0$$
was definitely a good step. Solving for $x$,
$$x=\frac1{t-1}$$
Note that this implies that either $0\le t<1$ or $t>1$, a property which we will use later to simplify the result. Then
$$\int\sqrt[8]{\frac{x+1}x}dx=-\int\frac{t^{1/8}}{(t-1)^2}dt=-8\int\frac{u^8}{(u^8-1)^2}du$$
Where we have let $t=u^8$, just as @H. R. Further, observe that
$$\frac d{du}\left(\frac u{u^8-1}\right)=\frac{-7u^8-1}{(u^8-1)^2}$$
So that
$$\int\sqrt[8]{\frac{x+1}x}dx=\int\left[\frac d{du}\left(\frac u{u^8-1}\right)-\frac1{u^8-1}\right]du=\frac u{u^8-1}-\int\frac{du}{u^8-1}=\frac u{u^8-1}-J$$
Then partial fractions is easy:
$$\frac1{u^8-1}=\frac1{\prod_{k=0}^7(u-\omega_k)}=\sum_{k=0}^7\frac{A_k}{u-\omega_k}$$
Where $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$ and $\theta_k=\frac{\pi k}4$. Evaluating
$$\lim_{u\rightarrow\omega_k}\frac{u-\omega_k}{u^8-1}=\lim_{u\rightarrow\omega_k}\frac1{8u^7}=\frac1{8\omega_k^7}=\frac{\omega_k}8=\lim_{u\rightarrow\omega_k}\sum_{j=0}^7A_j\frac{u-\omega_k}{u-\omega_j}=\sum_{j=0}^7A_j\delta_{jk}=A_k$$
So
$$\begin{align}J&=\frac18\sum_{k=0}^7\int\frac{\omega_k}{u-\omega_k}du=\frac18\int\left[\frac1{u-1}-\frac1{u+1}+\sum_{k=1}^3\frac{2u\cos\theta_k-2}{u^2-2u\cos\theta_k+1}\right]du\\
&=\frac18\int\left[\frac1{u-1}-\frac1{u+1}+2\sum_{k=1}^3\frac{(u-\cos\theta_k)\cos\theta_k-\sin^2\theta_k}{(u-\cos\theta_k)^2+\sin^2\theta_k}\right]du\\
&=\frac18\left[\ln\left|\frac{u-1}{u+1}\right|+\sum_{k=1}^3\left(\cos\theta_k\ln\left(u^2-2u\cos\theta_k+1\right)-2\sin\theta_k\tan^{-1}\left(\frac{u-\cos\theta_k}{\sin\theta_k}\right)\right)\right]+C\end{align}$$
We may apply the values $\cos\theta_1=-\cos\theta_3=\sin\theta_1=\sin\theta_3=\frac1{\sqrt2}$, $\cos\theta_2=0$, and $\sin\theta_2=0$ to arrive at
$$\begin{align}J&=\frac18\left[\ln\left|\frac{u-1}{u+1}\right|+\frac1{\sqrt2}\ln\left(\frac{u^2-\sqrt2u+1}{u^2+\sqrt2u+1}\right)\right.\\
&\left.-\sqrt2\tan^{-1}\left(\sqrt2u-1\right)-\sqrt2\tan^{-1}\left(\sqrt2u+1\right)-2\tan^{-1}u\right]+C\end{align}$$
Now we can add two of those arctangents because $u$ can't take on values that make the new denominator $0$ nor change the sign of the numerator when the denominator is negative:
$$\begin{align}J&=\frac18\left[\ln\left|\frac{u-1}{u+1}\right|+\frac1{\sqrt2}\ln\left(\frac{u^2-\sqrt2u+1}{u^2+\sqrt2u+1}\right)-\sqrt2\tan^{-1}\left(\frac{\sqrt2u}{1-u^2}\right)-2\tan^{-1}u\right]+C\end{align}$$
We now have the solution
$$\begin{align}\int\sqrt[8]{\frac{x+1}x}dx&=\frac{\sqrt[8]{\frac{x+1}x}}{\frac{x+1}x-1}-\frac18\left[\ln\left|\frac{\sqrt[8]{\frac{x+1}x}-1}{\sqrt[8]{\frac{x+1}x}+1}\right|+\frac1{\sqrt2}\ln\left(\frac{\sqrt[4]{\frac{x+1}x}-\sqrt2\sqrt[8]{\frac{x+1}x}+1}{\sqrt[4]{\frac{x+1}x}+\sqrt2\sqrt[8]{\frac{x+1}x}+1}\right)\right.\\
&\left.-\sqrt2\tan^{-1}\left(\frac{\sqrt2\sqrt[8]{\frac{x+1}x}}{1-\sqrt[4]{\frac{x+1}x}}\right)-2\tan^{-1}\sqrt[8]{\frac{x+1}x}\right]+C\end{align}$$
Which matches the Mathematica solution.
|
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|
Prove that $ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $ Prove:
$$ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $$
Hypothesis:
$$ F(x) = 1+2q+3q^2+...+xq^{x-1} = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} $$
Proof:
$$ P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} + (x+1)q^x = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2} $$
$$ P2 | \frac{1-(x+1)q^x+xq^{x+1}+[(x+1)(1-q)^2]q^x}{(1-q)^2} = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2} $$
$$ P3| \frac{x\color{red}{q^{x+1}}+[-(x+1)]\color{red}{q^x}+1+[(x+1)(1-q)^2]\color{red}{q^x}}{(1-q)^2} = \frac{x\color{red}{q^{x+2}}-(x+2)\color{red}{q^{x+1}}+1}{(1-q)^2} | $$
Here I just reorganize both sides of the equation, so LHS is explicity an expression with a degree of x+1, while the degree of RHS is x+2. Both LHS' $\color{red}{q^x}$ are added next.
$$P4| \frac{xq^{x+1}+[-(x+1)+(x+1)(<1^2q^0+\binom{2}{1}1q-1^0q^2>)]q^x+1}{(1-q)^2}=\frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2} $$
$$P5 | \frac{xq^{x+1}+[2xq-xq^2+2q-q^2]q^x+1}{(1-q)^2} = \frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2} $$
I get stuck at this point. I don't know if i'm approaching the problem the right way. So, any help would be appreciated.
Thanks in advance.
|
Ok, so, I had some mistakes in my first approach to the problem, specially with signs, but I already proved it, so, here it is.
Hypothesis :
$$F(x) = 1+2q+3q^2+...+xq^{x+1}=\frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2}$$
Proof:
$P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2}+(x+1)q^x = \frac{1-(x+2)q^{x+1}+(x+1)q^{x+2}}{(1-q)^2}$
$P2 | \frac{1-(x+1)q^x+xq^{x+1}+[(x+1)(1-q^2)]q^x}{(1-q)^2} = \frac{1-(x+2)q^{x+1}+(x+1)q^{x+2}}{(1-q)^2} $
$P3 | \frac{xq^{x+1}+[-(x+1)+(x+1)(1^2-\binom{2}{1}*1*q+q^2)]q^x+1}{(1-q)^2} = \frac{(x+1)q^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2} $
$ P4 | \frac{x\color{red}{q^{x+1}}+[xq^2-2xq+q^2-2q]\color{red}{q^x}+1}{(1-q^2)} = \frac{(x+1)\color{red}{q^{x+2}}-(x+2)\color{red}{q^{x+1}}+1}{(1-q)^2}$
$ P5 | \frac{xq^{x+1} + xq^{x+2} - 2xq^{x+1} + q^{x+2} - 2q^{x+1}+1}{(1-q)^2} = \frac{xq^{x+2}+q^{x+2}-(xq^{x+1}+2q^{x+1})+1}{(1-q)^2} $
$ β΄ | \frac{xq^{x+2}-xq^{x+1}+q^{x+2}-2q^{x+1}+1}{(1-q)^2} = \frac{xq^{x+2}-xq^{x+1}+q^{x+2}-2q^{x+1}+1}{(1-q)^2} $
Quod erat demonstrandum
|
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|
Prove that the square of an integer $a$ is of the form $a^2=3k$, or $a^2=3k+1$, where $k\in \mathbb{Z} $ Here's my attempt to prove this. I'm not sure about it, but i hope i'm correct.
Let $a=Ξ», Ξ»\in \mathbb{Z}$. Then $a^2=Ξ»^2=3\frac{Ξ»^2}{3}$.
When $Ξ»^2$ is divided by 3 there are three possible remainders $0,1,2$. So
$$ Ξ»^2=3k \lor Ξ»^2=3k+1 \lor Ξ»^2=3k+2 $$
*
*$ Ξ»^2=3k \rightarrow a^2=3\frac{3k}{3}=3k $
*$ Ξ»^2=3k+1 \rightarrow a^2=3\frac{3k+1}{3}=3k+1 $
*$ Ξ»^2=3k+2 \rightarrow a^2=3\frac{3k+2}{3}=3k+2=3ΞΌ, ΞΌ\in \mathbb{Z} $
Therefore $$a^2=
\begin{cases}
3k \\
3k+1
\end{cases}
$$
where $k$ is an integer.
Can someone verify this? Any help will be appreciated. Thanks in advance!
|
By the division algorithm
\begin{equation}
x=3q+r, \quad r \in \{0, 1, 2\}
\end{equation}
writing
\begin{align}
x^{2} &= 9q^{2}+r^{2} +6qr \\
&= 3(3q^{2}+2qr)+r^{2}
\end{align}
The result follows if for a given $x$, $r=0, 1$. If $r=2$ then clearly $r^{2}=4=1+3$ and thus
\begin{align}
x^{2} &=3 \lambda + 1 + 3 \\
&= 3(\lambda + 1) +1
\end{align}
(I should make it clear that it is obvious why $\lambda$ is an integer)
Q.E.D
|
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|
Finding an invertible matrix.
I want to find an invertible matrix $P$ where $P^tAP$ is a diagonal matrix.
$$A=\begin{pmatrix} 1 & 2 & 1 \\ 2 & 0 & 2 \\ 1 & 2 & 1 \end{pmatrix} $$
I have calculated the eigenvalues of $A$: $0,-2,4$ so the diagonal of $A$ should be $$D=\begin{pmatrix} -2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$
How can I calculate a $P$ which verifies $D=P^tAP$?
I thought I could find $P$ doing
$$P=\begin{pmatrix} a & b & c \\ d & e & f \\ g & h&i \end{pmatrix}$$
and
multiplying $P$ and $P^t$ with $A$ but it is a long process and I am pretty sure there should be a better method to find $P$.
|
The characteristic polynomial of $A$ is $q(x)=x^3-2x^2-8x$, hence the spectrum is $\{-2,0,4\}$.
The nullspace of $A+2I$ is generated by $(1,-2,1)$, the nullspace of $A$ is generated by $(1,0,-1)$ and the nullspace of $A-4I$ is generated by $(1,1,1)$, hence with your $D$
$$ P = \begin{pmatrix}\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{2}} \\ -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}}& 0 \\ \frac{1}{\sqrt{6}} &\frac{1}{\sqrt{3}} &-\frac{1}{\sqrt{2}} \end{pmatrix}$$
and $P^{-1}=P^T$.
|
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|
Why does an argument similiar to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$ show that $2+4+8+...=-2$ See how to prove $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$
$x=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
$2x=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
Then:
$x=1$
Now I use the same argument to prove $2+4+8+...=-2$
$x=2+4+8+...$
$2x=4+8+16+...$
Then:
$x=-2$
|
Hint the series converges only if $|r|<1$ so your second proof is wrong as $|r|=2$
|
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|
Evaluate $\cos \frac{\pi}{7} \cos \frac{2\pi}{7}\cos \frac{4\pi}{7}$ Evaluate $$\cos \frac{\pi}{7} \cos \frac{2\pi}{7}\cos \frac{4\pi}{7}.$$
The first thing i noticed was that $$\cos \frac{\pi}{7}=\frac{\zeta_{14}+\zeta_{14}^{-1}}{2},$$ where $\zeta_{14}=e^{2\pi i/14}$ is the 14-th root of unity.
Substituting this into the expression and simplifying, the expression is then transformed to $$\frac{1}{4}Re (\zeta_{14}+\zeta_{14}^3+\zeta_{14}^5+\zeta_{14}^7).$$
But then i have no idea how to get the real part of that thing. I only observed that $\zeta_{14}^7=-1$ and all the terms are primitive 14-th roots of unity. Please helps. Thanks!
|
Multiply by $8\sin\tfrac{\pi}{7}$ and simplify using $2\sin x \cos x = \sin 2x$ three times:
$$8\sin\tfrac{\pi}{7} \cos \tfrac{\pi}{7} \cos \tfrac{2\pi}{7}\cos \tfrac{4\pi}{7}
= 4 \sin\tfrac{2\pi}{7} \cos \tfrac{2\pi}{7}\cos \tfrac{4\pi}{7}
= 2 \sin\tfrac{4\pi}{7} \cos \tfrac{4\pi}{7}
= \sin\tfrac{8\pi}{7}$$
But you also have:
$$\sin\tfrac{8\pi}{7} = \sin\left( \pi - \tfrac{8\pi}{7} \right) = -\sin\tfrac{\pi}{7}$$
Now divide again by $8\sin\tfrac{\pi}{7}$.
|
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|
Number of solutions in non-negative integers question (Stars and bars) Q How many solutions are there in non-negative integers $a, b
, c, d$ to the equation:
$$ a + b + c + d = 79 $$
with the restrictions that $a \geq 10$, $b \leq 40$ and $20 \leq c \leq 30?$
If anyone could point me to any notes regarding the topic that would be greatly appreciated.
|
Using generating functions, we want to find the coefficient of $x^{79}$ in the product
$\displaystyle(x^{10}+x^{11}+x^{12}+\cdots)(1+x+\cdots+x^{40})(x^{20}+x^{21}+\cdots+x^{30})(1+x+x^2+\cdots)$
$\displaystyle=\left(\frac{x^{10}}{1-x}\right)\left(\frac{1-x^{41}}{1-x}\right)\left( x^{20}\cdot\frac{1-x^{11}}{1-x}\right)\left(\frac{1}{1-x}\right)=x^{30}\cdot\frac{(1-x^{41})(1-x^{11})}{(1-x)^4}$.
Therefore we need to find the coefficient of $x^{49}$ in
$\displaystyle\left(1-x^{11}-x^{41}+x^{52}\right)\sum_{n=0}^{\infty}\binom{n+3}{3}x^n$, which is given by$\displaystyle\binom{52}{3}-\binom{41}{3}-\binom{11}{3}$
|
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|
Difficulty in finding appropriate $\delta$ I'm trying to prove that for every $\epsilon > 0$ there exists $\delta = $___
s.t for every $0 < \lvert\lvert(x,y) - (1, 1)\rvert\rvert < \delta$ :
$\lvert x^2y\rvert+\lvert y \rvert + 1 < \epsilon$
Can you help me find such $\delta$?
This is what I got so far:
$\lvert x^2y\rvert+\lvert y \rvert + 1 = \lvert y \rvert ( \lvert x \rvert ^2 + 1) + 1 \leq (\delta + 1)( \lvert x \rvert ^2 + 1) +1$
($\lvert y-1 \rvert = \sqrt{(y-1)^2} \leq \sqrt{(x-1)^2 + (y-1)^2} = \lvert\lvert(x,y) - (1, 1)\rvert\rvert < \delta$)
($\lvert y-1 \rvert < \delta \implies \lvert y \rvert < \delta + 1$)
|
I'm going to assume this is related to this question Is $f$ continuous at $(0, 1)$ , $(1,1)$?.
The similarities are too great.
So that question was to show
$f(x,y) =
\left\{
\begin{array}{ll}
x^2y & \mbox{if } x \in \mathbb{Q} \\
y & \mbox{if } x \notin \mathbb{Q}
\end{array}
\right.$
show that $f$ is continuous at $(1,1)$
So you want to show for any $\epsilon > 0$ there is a $\delta$ such that
$||(x,y) - (1,1)|| < \delta \implies |f(x,y) - 1| < \delta$.
==== argh! here is a much better answer =====
Let $0 < 1/n < \epsilon/4$
Let $\delta = 1/n$.
If $|(x,y) - (1,1)| < 1/n$ then $(n-1)/n< y < (n+1)/n \implies |y -1| < 1/n < \epsilon$.
Likewise $(n-1)/n< x < (n+1)/n$ so $[(n-1)/n]^3 < x^2y < [(n+1)/n]^3$ so
$1 - 3/n + 3/n^2 - 1/n^3 = 1- 1/n[3-3/n+1/n^2] < x^2y < 1 + 3/n + 3/n^2 + 1/n^3= 1 + 1/n[3+3/n+1/n^2]$
$1 - 4/n < 1- 1/n[3-3/n+1/n^2] < x^2y < 1 + 3/n + 3/n^2 + 1/n^3= 1 + 1/n[3+3/n+1/n^2] < 1 + 4/n$
So $|x^2y - 1| < 4/n < \epsilon$.
So as $f(x,y) = x^2y$ or $f(x,y) = y$. $|f(x,y) - 1| < \epsilon$
===== so here is my old inelegant solution to see how convoluted my mind can get...=====
====
What your wrote ($\lvert x^2y\rvert+\lvert y \rvert + 1 < \epsilon$) doesn't work as Soke pointed out, as your expression is greater than 1.
What you want is $|f(x,y) - 1| \le \max(|x^2y - 1|,|y-1|) \le \epsilon$ so ...
$\sqrt{(1 - x)^2 + (1- y)^2} < \delta \implies$
$|1-x| < \delta$ and $|1 - y| < \delta$ so if $\delta \le \epsilon$ we have $|y - 1| < \epsilon$.
$|1 -x| < \delta \implies |x| < 1 \pm \delta$ and $|1 - y| < \delta \implies |y| < 1 \pm \delta$ so $x^2y < 1 + 3\delta + 3\delta^2 + \delta^3$.
So for any $\epsilon > 0$ chose $\delta$ such that $3\delta + 3\delta^2 + \delta^3 \le \epsilon$. We don't actually have to solve for $\delta$-- if $\delta < \epsilon/4$ will be good enough for small enough epsilon.
$|(x,y) - (1,1)| < \epsilon/4$
$|y-1| < \epsilon/4 < \epsilon$
$|x^2y - 1| < (1+ \epsilon/4)^3 - 1 = 3\epsilon/4 + 3\epsilon^2/4 + \epsilon^3 < \epsilon$ for small enough epsilon.
So $|f(x,y) - 1| = \{|y-1|, |x^2y-1|\} < \epsilon$.
|
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|
Bounds for $\sum_{n\geq 2}\frac{n+2}{(n^2-1)(n+3)}$ Use the integral test to prove this inequaliy
I calculated the integral $\int_{2}^{\infty}\frac{2+x}{(x^2-1)(x+3)}dx$
How can I use the integral test to show that $0.45 < \sum_{n=2}^{\infty}\frac{2+n}{(n^2-1)(n+3)} < 0.75$ ?
I know I can use the integral test but I don't quiet understand how this helps
|
Why do you need the integral test? The series can be computed through partial fraction decomposition:
$$\frac{n+2}{(n-1)(n+1)(n+3)}=\frac{3}{8}\cdot\frac{1}{n-1}-\frac{1}{4}\cdot\frac{1}{n+1}-\frac{1}{8}\cdot\frac{1}{n+3}\tag{1}$$
together with:
$$ \sum_{n\geq 2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right) = \frac{3}{2},$$
$$ \sum_{n\geq 2}\left(\frac{1}{n-1}-\frac{1}{n+3}\right) = \frac{25}{12}\tag{2}$$
lead to:
$$\sum_{n\geq 2}\frac{n+2}{(n-1)(n+1)(n+3)}=\color{red}{\frac{61}{96}}=0.63541666\ldots\tag{3}$$
We may notice that the value of the integral $\int_{2}^{+\infty}\frac{x+2}{(x^2-1)(x+3)}\,dx = \frac{\log(45)}{8} = 0.47583\ldots$ is just a weak lower bound and the value of the integral $\int_{3/2}^{+\infty}\frac{x+2}{(x^2-1)(x+3)}\,dx = \frac{\log(15)}{4} = 0.67701\ldots $ is just a slightly better upper bound.
|
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|
How to find range of $\frac{\sqrt{1+2x^2}}{1+x^2}$? How to find range of $$\frac{\sqrt{1+2x^2}}{1+x^2}$$ ?
I tried put it equal to $y$ and squaring but I'm getting $4$th degree equation.
|
You indeed get a fourth degree equation, but $x$ only appears with even exponent:
$$
y=\frac{\sqrt{1+2x^2}}{1+x^2}
$$
means that $y>0$ and that
$$
(1+x^2)^2y^2=1+2x^2
$$
Expanding and reordering gives
$$
y^2x^4+2(y^2-1)x^2+y^2-1=0
$$
and the usual quadratic formula provides the value for $x^2$; it's common to advise setting $z=x^2$ and solving $y^2z^2+2(y^2-1)z+y^2-1=0$:
$$
z=\frac{1-y^2\pm\sqrt{1-y^2}}{y^2}
$$
Note that the discriminant should be nonnegative, thus $y^2\le1$. In this case one of the roots of the quadratic is negative and the other one is positive (use Descartes' rule of signs). Hence, for $y^2\le1$, the equation
$$
x^2=\frac{1-y^2+\sqrt{1-y^2}}{y^2}
$$
has two solutions (except for $y=1$).
Now the conditions
$$
\begin{cases}
y>0 \\[4px]
y^2\le1
\end{cases}
$$
give you the range: $y\in(0,1]$.
|
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|
Determine matrix of linear map Linear map is given through:
$\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $
$\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$
Determine matrix $A$ linear map.
Here I have solution but I dont understand how to get it.
$A=\begin{pmatrix} -3 & -3 \\ -2 & 4 \end{pmatrix} $
|
Denote
$$A=\begin{pmatrix} a & b \\ c & d\end{pmatrix} $$
and solve the system of $4$ equations
$$A\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $$
$$A\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$$
for the unknowns $a,b,c$ and $d$.
|
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|
Finding the action of T on a general polynomial given a basis I am given a question as follows:
Suppose $T: P_{2} \rightarrow M_{2,2}$ is a linear transformation whose action on a basis for $P_{2}$ is
$$T(2x^2+2x+2)=\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix} \\T(4x^2+2x+2)=\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix} \\T(1)=\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix}.$$
Find the action of $T$ on a general polynomial with $a,b,c$ as constants.
Assuming that $T$ is a linear transformation, then for any vectors $\vec{v_{1}}, ..., \vec{v_{n}}$ and any scalars $a_{1}, ..., a_{n}$, $T(a_{1}\vec{v_{1}}+...+a_{n}\vec{v_{n}})=a_{1}T(\vec{v_{1}})+...+a_{n}T(\vec{v_{n}})$.
The answer is supposed to be in matrix form.
What I have tried doing so far is taking the basis polynomials and putting them into a matrix, which I then multiplied by a constant vector consisting of $a,b,c$:
$$\begin{bmatrix} 2&4&0 \\ 2&2&0 \\ 2&2&1\end{bmatrix} \begin{bmatrix} a \\ b \\c \end{bmatrix}= \begin{bmatrix} 2a+4b \\ 2a+2b \\ 2a+2b+c \end{bmatrix}$$
Given what I have done so far, I am not sure if this is the right approach, and I am not sure where to proceed from here. Any help would be appreciated!
|
First determine the matrix representation relative to the basis, $T(1), T(x),T(x^2)$.$$T(x^2)=\frac{T(4x^2+2x+2)-T(2x^2+2x+2)}{2}\\=(\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix}-\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix})/2 =\begin{bmatrix}1&1 \\ -2&1 \end{bmatrix}$$
$$T(x)=\frac{T(2x^2+2x+2)-2T(x^2)-2T(1)}{2}=(\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix}-2\begin{bmatrix}1&1 \\ -2&1 \end{bmatrix}-2\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix})/2=\begin{bmatrix} -1&0 \\ 1&2 \end{bmatrix}$$
$$T(1)=\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix}$$Therefore,$$T(ax^2+bx+c)=aT(x^2)+bT(x)+cT(1)=\begin{bmatrix} a-b+c&a+c \\ -2a+b+2c&a+2b+c \end{bmatrix}$$
|
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|
If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$then.. If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$ then find the value of $f(0) + f'(0) + f''(0)$.
I tried differentiating the given. But it is getting too long and complicated. So there must be a way to simplify $f(x)$. What is it?
|
We can try to simplify the fraction using trigonometric formulae :
$$\cos (x) + 5\cos (3x) + \cos (5x)=\cos(x) (2 \cos(2 x)-1) (2 \cos(2 x)+5)$$
and $$\cos (6x) + 6\cos(4x) + 15\cos(2x) + 10=32 \cos^6(x)$$
So $$f(x)=\frac{(2 \cos(2 x)-1) (2 \cos(2 x)+5)}{32\cos^5(x)}=\frac{4\cos^2(2x)+8\cos(2x)-5}{32\cos^5(x)}$$
$$f'(x)=\frac{1}{32}(5 (12 \cos^2(2 x)-5) \tan(x) \sec^5(x)-48 \sin(2 x) \cos(2 x) \sec^5(x))\\=\frac{1}{32}\sec^5(x)(5 (12 \cos^2(2 x)-5) \tan(x)-48 \sin(2 x) \cos(2 x) )$$
But the calculation of $f''(x)$ is still pretty tedious...
|
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|
Minimum value of $4a+b$ Let $ax^2+bx+8=0$ be an equation which has no distinct real roots then what is the least value of $4a+b$ where $a,b\in \Bbb R$.
My Try:
I differentiated the given function to get $f'(x)=2ax+b$ now if we put $x=2$ in this we get the required value . Now it's given that equation has no distinct roots so it can be a perfect square thus $b^2\leq 32a$. But now I don't know how to continue then. Also note that it's given that a,bare real so we cannot apply AM-GM inequality as they maybe negative. Thanks.
|
This is the solution of DeepSea, just a bit more detailed. Since the constraint given in the question has to do with the roots of the polynomial, try solving it by completing the square (assuming $a > 0$)
$$
ax^2 + bx + 8 = 0 \iff x^2 + \frac ba x + \frac 8a = 0 \iff \left(x + \frac{b}{2a}\right)^2 + \frac 8a - \frac{b^2}{4a^2} = 0,
$$
and conclude that
$$
x_{1,2} = -\frac{b}{2a} \pm \sqrt{\frac{b^2}{4a^2} - \frac 8a} = \frac{-b \pm \sqrt{b^2 - 32a}}{2a}.
$$
Note that if $b^2 - 32a = 0$ we get a double root, and if $b^2 - 32a < 0$ we get no real roots. Therefore for the polynomial to have no distinct real roots, we wish that $b^2 - 32a \leq 0$ or equivalently
$$
32a \geq b^2 \iff 4a \geq \frac{b^2}{8}.
$$
Adding $b$ to each side we get that
$$
4a + b \geq \frac{b^2}{8} + b = \frac{(b+4)^2 - 16}{8},
$$
where we once again completed the square to get the final expression. Realize that the least value of $4a + b$ is the minimal value of the right hand side, with $b \in \mathbb{R}$. Since $(b+4)^2 \geq 0$ for all values of $b$ conclude that the minimal value is $-16/8 = -2$.
|
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|
Prove $\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0$ without $\varepsilon - \delta$. Unlike Multivariable Delta Epsilon Proof $\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}$ --- looking for a hint I would like to avoid the $\varepsilon - \delta$ criterium.
Prove $$\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0 \,.$$
Approaching this limit from $y=0$, $x=0$, $y=x$, $y=x^2$ etcetera all yields 0 as value, so my proposal is that this limit is indeed 0.
I have been able to solve most similar limits so far by finding some convergent upper bound for the absolute limit, but with this one the difference between the numerator and the denominator is so small I can't find anything to fit inbetween. For example, $(x,y)\to(0,0)$,
$$ \left| \frac{x^2 y}{x^2 + y^2} \right| \le \left| \frac{(x^2 + y^2)y}{x^2 + y^2} \right| \to 0 \,. $$
Also, Continuity of $\frac{x^3y^2}{x^4+y^4}$ at $(0,0)$? and Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous. contain some helpful hints.
|
In polar coordinates the expression is
$$r\frac{\cos^2t\sin^3t}{\cos^4 t + \sin^4 t}.$$
The denominator in this fraction has a positive minimum; thus the fraction is a bounded function of $t,$ and the $r$ in front guarantees a limit of $0.$
|
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|
Find the value of $h$ if $x^2 + y^2 = h$
Consider equation $x^2 + y^2 = h$ that touches the line $y=3x+2$ at some point $P$. Find the value of $h$
I know that $x^2 + y^2 = h$ is a circle with radius $\sqrt{h}$.
Also, since $y = 3x + 2 $ is a tangent, we know that the slope of the radius perpendicular to the tangent is $M_{OP}= -\frac{1}{3}$. I'm not sure how I can determine the value of $h$ though?
|
You're almost there! We want to find the equation of line $OP$. We know its slope, and (since it passes through the origin) we know its $y$-intercept, so its equation is:
$$
y = \tfrac{-1}{3}x
$$
We can now find the intersection point of the radius and tangent by solving the system of equations. Equating, we obtain:
\begin{align*}
\tfrac{-1}{3}x = y = 3x + 2
&\implies -x = 9x + 6 \\
&\implies -10x = 6 \\
&\implies x = \tfrac{-3}{5} \\
&\implies y = \tfrac{-1}{3} \cdot \tfrac{-3}{5} = \tfrac{1}{5}
\end{align*}
Thus, we conclude that:
$$
h = x^2 + y^2 = (\tfrac{-3}{5})^2 + (\tfrac{1}{5})^2 = \tfrac{9 + 1}{25} = \boxed{\tfrac{2}{5}}
$$
|
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|
Find a Jordan basis for the endomorphism $g:M_2(R)\longrightarrow M_2(R)$ such that...
Find a Jordan basis for the endomorphism $g:M_2(R)\longrightarrow M_2(R)$ such that
$M(g,B) = \begin{pmatrix} 2&0&3&0\\ 1&2&0&3\\0&0&2&0\\ 0&0&1&2 \end{pmatrix}$, where $B=(\begin{pmatrix} 0&1\\ 1&1 \end{pmatrix},\begin{pmatrix} 1&0\\ 1&1 \end{pmatrix},\begin{pmatrix} 1&1\\ 0&1 \end{pmatrix},\begin{pmatrix} 1&1\\ 1&0 \end{pmatrix})$
I know how to calculate a Jordan basis. However, I am stuck at this problem, I really don't know how to start.
Thank you
|
Let $\alpha=\{E_1,E_2,E_3,E_4\}$ where
\begin{align*}
E_1 &=
\left[\begin{array}{rr}
0 & 1 \\
1 & 1
\end{array}\right]
&
E_2 &=
\left[\begin{array}{rr}
1 & 0 \\
1 & 1
\end{array}\right]
&
E_3 &=
\left[\begin{array}{rr}
1 & 1 \\
0 & 1
\end{array}\right]
&
E_4 &=
\left[\begin{array}{rr}
1 & 1 \\
1 & 0
\end{array}\right]
\end{align*}
You are given that
$$
[g]_\alpha^\alpha=
\left[\begin{array}{rrrr}
2 & 0 & 3 & 0 \\
1 & 2 & 0 & 3 \\
0 & 0 & 2 & 0 \\
0 & 0 & 1 & 2
\end{array}\right]
$$ and you wish to find a basis $\beta=\{F_1,F_2,F_3,F_4\}$ such that $[g]_\beta^\beta$ is in Jordan form.
You state that you know how to find Jordan bases so I'll assume that you can show that $[g]_\alpha^\alpha=PJP^{-1}$ where
\begin{align*}
P &=
\left[\begin{array}{rrrr}
0 & 3 & 0 & 1 \\
6 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & -\frac{1}{3}
\end{array}\right]
&
J &=
\left[\begin{array}{rrrr}
2 & 1 & 0 & 0 \\
0 & 2 & 1 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2
\end{array}\right]
\end{align*}
We can interpret $P$ as the change of basis matrix from $\alpha$ to $\beta$. This means
$$
[g]_\alpha^\alpha = \underbrace{[\DeclareMathOperator{id}{id}\id]_\beta^\alpha}_{=P}\underbrace{[g]_\beta^\beta}_{=J}\underbrace{[\id]_\alpha^\beta}_{=P^{-1}}
$$
Since
$$
P^{-1}=
\left[\begin{array}{rrrr}
0 & \frac{1}{6} & 0 & 0 \\
\frac{1}{6} & 0 & 0 & \frac{1}{2} \\
0 & 0 & 1 & 0 \\
\frac{1}{2} & 0 & 0 & -\frac{3}{2}
\end{array}\right]
$$
it follows that
\begin{array}{rcrcrcrcrcrcrcrcrc}
F_1 &=& 0\cdot E_1 &+& \frac{1}{6}\cdot E_2 &+& 0\cdot E_3 &+& \frac{1}{2}\cdot E_4\\
F_2 &=& \frac{1}{6}\cdot E_1 &+& 0\cdot E_2 &+& 0\cdot E_3 &+& 0\cdot E_4\\
F_3 &=& 0\cdot E_1 &+& 0\cdot E_2 &+& 1\cdot E_3 &+& 0\cdot E_4\\
F_4 &=& 0\cdot E_1 &+& \frac{1}{2}\cdot E_2 &+& 0\cdot E_3 &-& \frac{3}{2}\cdot E_4\\
\end{array}
Your desired basis is thus
\begin{align*}
F_1 &=
\left[\begin{array}{rr}
\frac{2}{3} & \frac{1}{2} \\
\frac{2}{3} & \frac{1}{6}
\end{array}\right]
&
F_2 &=
\left[\begin{array}{rr}
0 & \frac{1}{6} \\
\frac{1}{6} & \frac{1}{6}
\end{array}\right]
&
F_3 &=
\left[\begin{array}{rr}
1 & 1 \\
0 & 1
\end{array}\right]
&
F_4 &=
\left[\begin{array}{rr}
-1 & -\frac{3}{2} \\
-1 & \frac{1}{2}
\end{array}\right]
\end{align*}
|
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|
Finding arc length of the curve $6xy=x^4+3$ from $x=1$ to $x=2$ Looking at this as a graph of a function of $y$ is more convenient
$$
y=\frac{x^4+3}{6x}\Rightarrow \frac{dy}{dx}=\frac{x^3-1}{2x^2}\Rightarrow \left( \frac{dy}{dx} \right)^2=\frac{x^6-2x^3+1}{4x^4}
$$
And the integral for arc length:
$$
\int_1^2 \sqrt{1+\frac{x^6-2x^3+1}{4x^4}} \, \mathrm dx=\int_1^2 \frac{1}{2x^2}
\sqrt{4x^4+x^6-2x^3+1} \, \mathrm dx
$$
Which I am not sure how to evaluate. Substitutions don't seem fruitful and integration by parts looks like it will be messy. Did I set things up wrong? Is there an easy way to evaluate this integral?
edit: As Dr. MV pointed out, I made a differentiation error which he corrects below.
|
HINT:
$$1+y'^2(x)=\frac14\left(x^2+\frac{1}{x^2}\right)^2$$
|
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|
Differential equation $\left(x^2+xy\right)y'=x\sqrt{x^2-y^2}+xy+y^2$ I am not sure which type of differential equation this falls into:
$$\left(x^2+xy\right)y'=x\sqrt{x^2-y^2}+xy+y^2$$ any hints? P.S. I first tried reornazing it so I have $y'$ alone, and hoping that I would get a homogeneous equation, but no such luck.
|
Divide $x^2$ from both sides and then let $z=\dfrac{y}x$.
Note that $y'=z+xz'$.
The new equation becomes $(1+z)y'=\sqrt{1-z^2}+z+z^2$.
$y'=\sqrt{\dfrac{1-z}{1+z}}+z$
$x\dfrac{\mathrm dz}{\mathrm dx}=\sqrt{\dfrac{1-z}{1+z}}$
$\dfrac{\mathrm dx}{x}=\sqrt{\dfrac{1+z}{1-z}}\ \mathrm dz$
$\ln(x)=\sqrt{\dfrac1{1-z}}\left((z-1)\sqrt{z+1}+2\sqrt{1-z}\sin^{-1}\left(\dfrac{\sqrt{1+z}}{\sqrt2}\right)\right)+C$
And... it gets too complicated.
|
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|
Prove inequality $\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+d^3}{2}}+\sqrt[3]{\frac{d^3+a^3}{2}} \le 2(a+b+c+d)-4$
Let $a,b,c,d$ positive real numbers, such that $$\frac1a+\frac1b+\frac1c+\frac1d=4.$$
Prove inequality
$$\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+d^3}{2}}+\sqrt[3]{\frac{d^3+a^3}{2}} \le 2(a+b+c+d)-4$$
My work so far:
$$1=\frac{4}{\frac1a+\frac1b+\frac1c+\frac1d}\le \sqrt[4]{abcd}\le\frac{a+b+c+d}{4}$$
|
$\sum\limits_{cyc}\left(2a-\sqrt[3]{\frac{a^3+b^3}{2}}\right)\geq\sum\limits_{cyc}\left(a+b-\frac{a^2+b^2}{a+b}\right)=2\sum\limits_{cyc}\frac{ab}{a+b}=2\sum\limits_{cyc}\frac{1}{\frac{1}{a}+\frac{1}{b}}\geq\frac{32}{\sum\limits_{cyc}\left(\frac{1}{a}+\frac{1}{b}\right)}=4$
|
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|
Minimum value of $\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$ Recently I was solving one question, in which I was solving for the smallest value of this expression
$$f(\theta)=\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$$
My first attempt:
$$\begin{align}
f(\theta) &=3+2\sin^2\theta-6\sin\theta \cos\theta \\
&=3(1-2\sin\theta \cos\theta)+2\sin^2\theta \\
&=3(\sin\theta-\cos\theta)^2 + 2\sin^2\theta
\end{align}$$
Hence the minimum value of $f(\theta)=2\sin^2\theta$ when $\theta=\pi/4$
hence minimum value of $f(\theta)=1$.
But then again I tried to do question differently by making substitutions in order to change the whole $f(\theta)$ in the form of $\cos x+\sin x$
Then $f(\theta)$ came out to be
$$f(\theta)= 4-(\cos(2\theta)+3\sin(2\theta))$$
The minimum value of this expression is surely $$(f(\theta))_{min}=4-\sqrt{10}$$
Can anybody explain me algebraically why my first method gave the wrong result?
|
hint: Alternatively, using Lagrange Multiplier we have:
$x = \cos \theta, y = \sin \theta\implies f(x,y) = x^2-6xy+3y^2+2, x^2+y^2 = 1\implies f(x,y) = (x^2+y^2)+2y^2+2 - 6xy = 3-6xy+2y^2 \implies f_x = -6y= 2\lambda x, f_y = 4y-6x= 2\lambda y\implies4\lambda y - 6\lambda x = 2\lambda^2y\implies 4\lambda y+18y=2\lambda^2y\implies y(\lambda^2-2\lambda-9) = 0\implies y = 0 $ or $\lambda = 1\pm \sqrt{10}$ . Can you continue at this point?
|
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|
Solve the congruence $6x+15y \equiv 9 \pmod {18}$ Solve the congruence $6x+15y \equiv 9\pmod {18}$
Approach:
$(6,18)=6$, so $$15y \equiv 9\pmod 6$$
$$15y \equiv 3\pmod 6$$
So the equation will have $(15,6)$ solutions. Now we divide by 3
$$5y \equiv 1\pmod 2$$.
Solving the Diophantine equation we get $y \equiv1\pmod 2 $, so $y=1+2m$
$$6x \equiv 9-15y\pmod {18}$$
$$6x \equiv 9-15(1+2m)\pmod {18}$$
$$6x \equiv -6-30m\pmod {18}$$
Divide by 6
$$x \equiv -1-5m\pmod 3$$
The right solution is $x-m \equiv 2\pmod 3$. I have $x+5m \equiv 2\pmod 3$
$$x-m+6m \equiv 2\pmod 3$$ $$x-m \equiv 2\pmod 3$$
Wolfram alpha says $y=1+2t$ and $x=t+3d+2$
|
Wolfram answer:
$y = 2c + 1$
$x = c + 3d + 2$
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Can someone suggest a way to simplify $x_1(y_1 - x^Ty) + x_1(w_1 - x^Tw)^2 - x_1^2(w_1 - x^Tw)^2 + x_1x_2(w_1 - x^Tw)^2$ Let $x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$, $y = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}$, $w =\begin{bmatrix} w_1 \\ w_2 \end{bmatrix}$
I have the following vector:
$V = \begin{bmatrix} x_1(y_1 - x^Ty) + x_1(w_1 - x^Tw)^2 - x_1^2(w_1 -
x^Tw)^2 + x_1x_2(w_1 - x^Tw)^2 \\ x_2(y_2 - x^Ty) + x_2(w_2 - x^Tw)^2
- x_2^2(w_2 - x^Tw)^2 + x_2x_1(w_2 - x^Tw)^2 \end{bmatrix}$
As you can see, the components has some interesting structures
I wish to put them into a much more compact form but I am not sure how to proceed
But I can do it for the first "part" in the above vector, consider $$V^* = \begin{bmatrix} x_1(y_1 - x^Ty)\\ x_2(y_2 - x^Ty)\end{bmatrix}$$
I can put this into $V^* = \text{Diag}(x)[y - x^Ty \mathbf{1}]$
Can something similar be done to $V$ itself?
|
$$B \begin{array}[t]{l} = \begin{bmatrix} x_1(w_1 - \mathbf x^T\mathbf w)^2 - x_1^2(w_1 -
\mathbf x^T\mathbf w)^2 + x_1x_2(w_1 - \mathbf x^T\mathbf w)^2 \\ x_2(w_2 - \mathbf x^T\mathbf w)^2
- x_2^2(w_2 - \mathbf x^T\mathbf w)^2 + x_2x_1(w_2 - \mathbf x^T\mathbf w)^2 \end{bmatrix}
\\[3ex]
=\begin{bmatrix} (w_1 - \mathbf x^T \mathbf w)^2 & 0 \\ 0 & (w_2 - \mathbf x^T\mathbf w)^2 \end{bmatrix}\cdot\begin{bmatrix} x_1- x_1^2+x_1x_2 \\ x_2-x_2^2 + x_1x_2\end{bmatrix}
\\[3ex]
= \left[\operatorname{diag}(\mathbf w-\mathbf x^T \mathbf w\cdot \mathbf 1)\right]^2\cdot\begin{bmatrix} x_1 & 0\\0&x_2 \end{bmatrix}\cdot \begin{bmatrix} 1-x_1+x_2\\
1-x_2+x_1 \end{bmatrix}
\\[3ex]
=\left[\operatorname{diag}(\mathbf w-\mathbf x^T\mathbf w\cdot \mathbf 1)\right]^2\cdot\operatorname{diag}(\mathbf x)\cdot
\begin{bmatrix} 1-x_1+x_2\\
1-x_2+x_1 \end{bmatrix}
\\[3ex]
=\left[\operatorname{diag}(\mathbf w-\mathbf x^T\mathbf w\cdot \mathbf 1)\right]^2\cdot\operatorname{diag}(\mathbf x)\cdot\left(\mathbf 1-\mathbf x + \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\cdot\mathbf x \right),
\end{array}$$
where $\mathbf 1 = \begin{bmatrix} 1\\1 \end{bmatrix}. $ Is this compact enough? The final answer will be $V = V^* + B$.
|
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|
Smallest number whose $\sin(x)$ in radian and degrees is equal Question:
What is the smallest positive real number $x$ with the property that the sine of $x$ degrees is equal to the sine of $x$ radians?
My try: 0. But zero isn't a positive number. How do I even begin to solve it? I tried taking inverse on both sides of $\sin \theta = \sin x$, but that didn't help.
|
One degree is $\frac{\pi}{180}$ radians, so what we want here is
$$\sin(x)=\sin(\frac{\pi x}{180})$$
And so
$$x = \frac{\pi x}{180}+2\pi k, \qquad \text{or} \qquad x = \frac{-\pi x}{180}+\pi (2k+1)$$
Where $k$ is any integer (since $\sin(a) = \sin(b)$ iff $a-b$ is an integer multiple of $2\pi$, or if $a+b$ is an odd multiple of $\pi$). Solving for $x$,
$$x(1 - \frac{\pi}{180}) = 2\pi k, \qquad \text{or} \qquad x(1 + \frac{\pi}{180}) = \pi (2k+1)$$
And now we just need to find the $k$ that gives us the smallest positive solution. Since the number on the left hand side is positive (in both cases), we choose $k = 1$ for the left, and $k = 0$ for the right. Then
$$x = \frac{2\pi}{1-\frac{\pi}{180}}, \qquad \text{or} \qquad x = \frac{\pi}{1+\frac{\pi}{180}}$$
The right hand solution is smaller, so that's our answer.
|
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|
How to find $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$ How to find ?$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$$
I tried using the substitution $x^2=z$.But that did not help much.
|
You can apply $u=x^2$ :
$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx=\frac{1}{2}\int \frac{x^2-1}{x^4\sqrt{2x^4-2x^2+1}} 2xdx=\frac{1}{2}\int \frac{u-1}{u^2\sqrt{2u^2-2u+1}} du$$
Now note that :
$$\int \frac{u-1}{u^2\sqrt{2u^2-2u+1}} du=\int \frac{u-1}{u^2\sqrt{2(u^2-u)+1}} du$$
we can substitute $v=\frac{1}{u}$ :
$$\int \frac{u-1}{u^2\sqrt{2(u^2-u)+1}} du=\int \frac{1-u}{\sqrt{2(u^2-u)+1}} \frac{-du}{u^2}=\int \frac{1-\frac{1}{v}}{\sqrt{2(\frac{1}{v^2}-\frac{1}{v})+1}} dv\\=\int \frac{\frac{v-1}{v}}{\sqrt{\frac{2-2v+v^2}{v^2}}}dv=\int \frac{v-1}{\sqrt{v^2-2v+2}}dv$$
Substitute $y=v^2-2v+2$ and you should be able to finish.
|
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|
How do I evaluate this integral using cauchy's residue theorem. $$\int_0^{2\pi} \dfrac{\cos 2 \theta}{1+\sin^2 \theta}d\theta$$
$$=\dfrac{-2}{i}\oint_{|z|=1} \dfrac{z^4+1}{z(z^4-4z^2-2z+1)}dz $$
I am stuck on how to use Cauchy's residue theorem since the bottom does not factor nicely. I know $z=-1$, $z=0.3111$ and $z=0$ lie within the contour but not sure where to go from here.
|
Using trig. identities and eulers formula and long division:
$$
\int_0^{2 \pi} \dfrac{ \cos 2 \theta}{1+\sin^2 \theta} d \theta=\int_0^{2 \pi} \dfrac{ \cos 2 \theta}{1+\frac{1}{2}(1-\cos 2\theta)}$$
$$=2\int_0^{2 \pi} \dfrac{ \cos 2 \theta}{3-\cos 2\theta}$$
$$= 2 \int_0^{2 \pi} - d\theta + 6\int_0^{2\pi}\dfrac{d \theta}{3- \cos 2\theta}$$
$$-4\pi +\dfrac{12}{i}\oint_{|z|=1}\dfrac{zdz}{(-z^4+6z^2-1)}$$
$$-4 \pi +\dfrac{12}{i}\oint_{|z|=1}\dfrac{zdz}{(z-1-\sqrt{2})(z+1+\sqrt{2})(z-\sqrt{2}+1)(z+\sqrt{2}-1)}$$
$$=-4 \pi +24\pi Res(f;\sqrt{2}-1)+24 \pi Res(f;-\sqrt{2}+1) $$
$$-4\pi + \frac{3}{2}\sqrt{2}\pi+\frac{3}{2} \sqrt{2} \pi $$
$$\pi(3\sqrt{2}-4)$$
This what I was looking for. I'm not sure if this is correct or not but it did match one of the answers.
|
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|
How to find limit $\lim \limits_{ x\rightarrow +\infty }{ \tan { { \left( \frac { \pi x }{ 2x+1 } \right) }^{ 1/x } } } $ I need to find this limit $\lim\limits_{ x\rightarrow +\infty }{ \tan { { \left( \frac { \pi x }{ 2x+1 } \right) }^{ 1/x } } } $.
Give a hint please.Thanks
|
Hints:
$$\frac{\pi x}{2x+1} = \frac{\pi}{2}-\frac{\pi}{4x+2}\tag{1}$$
$$\tan\left(\frac{\pi x}{2x+1}\right) = \cot\left(\frac{\pi}{4x+2}\right)\tag{2}$$
$$\frac{1}{x}\,\log\cot\left(\frac{\pi}{4x+2}\right)=\frac{\log\frac{4}{\pi}+\log x}{x}+O\left(\frac{1}{x^2}\right)\text{ as }x\to+\infty.\tag{3}$$
|
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How to find whether the sum of a geometric series is prime or divisible by 3, 13 or 125? Let $A = 1+5+5^2+\dots+5^{99}$, then $A$ is:
*
*A prime number
*not divisible by 3
*divisible by 13
*divisible by 125
I know this is a sum of a Geometric Progression, so $ A = (5^{100}-1)/4$ but I cannot find $5^{100}$ So I thought of finding a pattern between the progression starting from 1. I get 1,6,31,131,... but cannot seem to be finding any pattern. May someone Help?
|
I will assume that you mean $1+5+5^2+\cdots +5^{99}$. Let's deal with the choices one at a time.
Of course 4.) is false, all the terms except one are multiples of $5$. So the sum cannot be a multiple of $5$.
1.) is clearly false, we have an even number of odd numbers. The sum is therefore even, and thus not prime.
2.) Group in groups of $2$. Note that $1+5$ is a multiple of $3$, and so is $5^2+5^3$, which is $5^2(1+5)$. The same is true of $5^4+5^5=5^4(1+5)$, and so on. So our whole sum is a multiple of $3$.
3.) Note that $1+5+5^2+5^3$ is a multiple of $13$, and therefore so is $5^4+5^5+5^6+5^7$ and so on. (We are grouping in groups of $4$.)
|
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Find $f'(c)$ using the derivative definition for $f(x)=\frac{1}{x^2}$ So $f:\mathbb R \backslash\{0\}\rightarrow\mathbb R$ by $f(x)=\frac{1}{x^2}$
I'm need to use the definition of the derivative to find $f'(c)$ for $\frac{1}{x^2}$. If I use standard differentiation techniques I get:
$-\frac{2}{c^3}$
However, using the following definition:
$f'(c) = \frac{f(x)-f(c)}{x-c}$
I end up with $\frac{c^2-x^2}{x^2c^2(x-c)}$
What am I doing wrong?
|
set $x-c=h$ then $$f^{ \prime }\left( c \right) =\lim _{ h\rightarrow 0 }{ \frac { f\left( c+h \right) -f\left( c \right) }{ h } } =\lim _{ h\rightarrow 0 }{ \frac { \frac { 1 }{ { \left( c+h \right) }^{ 2 } } -\frac { 1 }{ { c }^{ 2 } } }{ h } } =$$ $$ \\ \\ =\lim _{ h\rightarrow 0 }{ \frac { { c }^{ 2 }-{ \left( c+h \right) }^{ 2 } }{ { \left( c+h \right) }^{ 2 }{ c }^{ 2 }h } } =\lim _{ h\rightarrow 0 }{ \frac { -h\left( 2c+h \right) }{ { \left( c+h \right) }^{ 2 }{ c }^{ 2 }h } } =\frac { -2 }{ { c }^{ 3 } } $$
|
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Matrix equation $A^2+A=I$ when $\det(A) = 1$ I have to solve the following problem:
find the matrix $A \in M_{n \times n}(\mathbb{R})$ such that:
$$A^2+A=I$$ and $\det(A)=1$.
How many of these matrices can be found when $n$ is given?
Thanks in advance.
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Consider the Jordan Canonical Form for $A$; that is, $A = PJP^{-1}$ for some invertible $P$ and block diagonal matrix $J$ whose blocks are either diagonal or Jordan (same entry on the diagonal, have $1$s on the the diagonal above the main diagonal, and $0$s elsewhere).
Then, the equation reduces to $J^2 + J = I$. Looking at the diagonal entries on both sides, this reduces to solving $r^2 + r = 1$, which has solutions $r = \frac{-1 \pm \sqrt{5}}{2}$. In other words, $J$ has diagonal entries are $\frac{-1 \pm \sqrt{5}}{2}$. However, we want $|A| = 1$, which is equivalent to $|J| = 1$. This is only possible if $n$ is even and $J$ has an equal even number of $\frac{-1 + \sqrt{5}}{2}$ and $\frac{-1 - \sqrt{5}}{2}$ on its diagonal, due to $(\frac{-1 + \sqrt{5}}{2})(\frac{-1 -\sqrt{5}}{2}) = -1$ and no positive integral power of $\frac{-1 \pm \sqrt{5}}{2}$ equaling $\pm 1$.
As a summary, there are no solutions when $n$ is not a multiple of $4$.
Otherwise, when $n$ is not a multiple of $4$, there are infinitely many solutions. To illustrate this, let $J$ be diagonal with an equal and even number of $\frac{-1 + \sqrt{5}}{2}$ and $\frac{-1 - \sqrt{5}}{2}$ on its diagonal. Letting $P$ vary (as there are infinitely many non-invertible matrices that do not commute with $J$) will give infinitely many possibilities for $A$ when $n$ is a multiple of $4$.
|
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solve $x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$ Help with this excercise.. :)
$$x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$$
the book says it is an exact differential equation, but how?
$$x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$$
$$x(x^2+y^2)^{-1/2}dx+y(x^2+2y^2)dy=0$$
$M=x(x^2+y^2)^{-1/2}$
$N=y(x^2+2y^2)$
$$\frac{\partial M}{\partial y}=-\frac{xy}{(x^2+y^2)^{3/2}}$$
$$\frac{\partial N}{\partial x}=2xy$$
I cant find the integrating factor,, :(
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Obviously $x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$ isn't an exact differential equation.
So, there is a typo. Moreover the equation is certainly simple. The simplest guess is :
$$x(x^2+y^2)+yy\prime(x^2+2y^2)=0$$
$$x(x^2+y^2)dx+y(x^2+2y^2)dy=0$$
If it is the exact differential of $F(x,y)$ then :
$\begin{cases}
\frac{\partial F}{\partial x}=x(x^2+y^2) \quad\to\quad F=\frac{x^4}{4}+\frac{x^2y^2}{2}+f(y)\\
\frac{\partial F}{\partial y}=y(x^2+2y^2) \quad\to\quad F=\frac{x^2y^2}{2}+2\frac{y^4}{4}+g(x)
\end{cases}$
which implies $f(y)=\frac{y^4}{2}$ and $g(x)=\frac{x^4}{4}$
$F=\frac{x^4}{4}+\frac{x^2y^2}{2}+\frac{y^4}{2}=\text{constant}\quad$ because $dF=0$
$$x^4+2x^2y^2+2y^4=c$$
Solving it for $y$ leads to :
$$y(x)=\pm\sqrt{\frac{-x^2\pm \sqrt{2c-x^4}}{2}}$$
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Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Attempt:
$$\gcd(4n-5,60-12n)=(4n-5,-8n+55)=(4n-5,-4n+50)=(4n-5,45)$$
$$n=1: (4-5,45)=1\quad \checkmark\\
n=2: (3,45)=3\quad \times\\
n=3: (7,45)=1\quad \checkmark\\
n=4: (11,45)=1\quad \checkmark\\
n=5: (15,45)=15\quad \times\\
n=6: (19,45)=1\quad \checkmark\\
n=7: (23,45)=1\quad \checkmark\\
n=8: (27,45)=9\quad \times\\
\vdots$$
So the answer is that it can't be reduced for $n=1,3,4,6,7,..$ i.e
$$\bigg\{n\bigg|n\notin \begin{cases}a_1=2\\a_n=a_{n-1}+3\end{cases}\bigg\}$$
I want to verify that my solution is correct
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If $d$ is the common divisor of $4n-5$ and $60-12n$, then it should also divide $3(4n-5)+(60-12n) =45$. Thus when numerator and denominator are not divisible by 3 and 5, the fraction is irreducible. Hence $n \not \equiv 2 \pmod 3$ and $n \not \equiv 0 \pmod 5$.
|
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The shortest distance from surface to a point Many have asked the question about finding the shortest distance from a point to a plane. I have checked those questions and answers and haven't found what I am looking for. Might still have missed some question though.
I have a problem where I am to calculate the shortest distance from the surface $z = x^2 + 3y^2$ to the point $P = (5, 0, 1)$
I want to use the Langrange multiplier for this.
This is what I have done:
I am to minimize $$(x-5)^2 + y^2 + (z - 1)^2$$
s.t.
$$x^2 + 3y^2 - z = 0$$
So the Lagrange function is
$$L(x, y, z, \lambda) = (x-5)^2 + y^2 (z-1)^2 + \lambda(x^2 + 3y^2 - z)$$
I have taken the gradient of $L$ and am looking for where each partial derivative equals to zero.
And I am stuck. Is $\lambda = 1/3$? It seems like that, otherwise $y=0$. But if $y\neq 0$ and lambda is $1/3$ then I get that $y=0$ anyway :/
|
Set
$$f(x,y,z)=(x-5)^2+y^2+(z-1)^2$$
$$g(x,y,z)=z-x^2-3y^2=0$$
By application of Lagrange method, we have
$$\nabla f=\lambda \nabla g$$
$$(2x-10\,,\,2y\,,\,2z-2)=\lambda(-2x\,,\,-6y\,,\,1)$$
therefore
$$\left\{ \begin{align}
& y=0\,\,\,\,\quad\Rightarrow \,\,\,z={{x}^{2}} \\
& \lambda =-\frac{1}{3}\,\,\,\Rightarrow \,\,\,x=\frac {15} 2\,\,\,\,,\,\,z=\frac{5}{6} \\
\end{align} \right.$$
Case 1
$y=0\quad,\quad z=x^2$
$$f(x,0,x^2)=(x-5)^2+(x^2-1)^2=x^4-x^2-10x+26$$
Now you should minimize $f(x,0,x^2)$
Case 2
$x=\frac {15} 2\,\,\,\,,\,\,z=\frac{5}{6} $
$$g\left(\frac {15} 2,\,y\,,\frac{5}{6}\right)=\frac{5}{6}-\frac{225}{4}-3y^2=0$$
therefore
$$y=\pm \frac{\sqrt{185}}{3}$$
Now you should calculate $f\left( \frac{15}{2},\frac{\sqrt{185}}{3},\frac{5}{6} \right)$ and $f\left( \frac{15}{2},-\frac{\sqrt{185}}{3},\frac{5}{6} \right)$
|
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Proving that $1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } $ by induction
Prove that
$$1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 }. $$
I can get to $1/3(k+1)(k+2) + (k+1)(k+2)$ but then finishing off and rearranging the problem to $1/3(k+2)(k+3)$ is where I always get stuck.
|
Your proposition is not true.
try $n = 2$
$1\cdot 2 + 2\cdot3 \ne \frac 13 3\cdot4$
Which is why you are having a tough time proving this by induction.
Induction is usefull to prove things, but it doesn't always have a whole lot of insight why things are the way they are.
This is probably what you should have:
$\sum_\limits{k=1}^{n} (k)(k+1) = \frac 13 n(n+1)(n+2)$
base case:
$n = 1\\
1\cdot2 = \frac13 1\cdot2\cdot 3$
Inductive hypothesis:
$\sum_\limits{k=1}^{n} (k)(k+1) = \frac 13 n(n+1)(n+2)$
Show that
$\sum_\limits{k=1}^{n+1} (k)(k+1) = \frac 13 (n+1)(n+2)(n+3)$
based on the inductive hypothesis.
$\sum_\limits{k=1}^{n+1} (k)(k+1)\\
\sum_\limits{k=1}^{n} (k)(k+1) + (n+1)(n+2)\\
\frac 13 n(n+1)(n+2) + (n+1)(n+2)\\
(n+1)(n+2) (\frac 13 n + 1)\\
\frac 13(n+1)(n+2)(n + 3)$
QED
|
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|
How to get a whole number from $y = \frac{1}{x + 2}$ How to come up with a whole number for y.
I keep coming up with fractions from $y = \frac{1}{x +2}$
I've tried numerous numbers, as in, $ 1, 2, 3 ,4 , 5, -1, -2, -3, -4, -5$.
For example, $y = \frac{1}{1 + 2} = \frac{1}{3}$.
It's suppose to make two separate identical curves on a Cartesian plane, and one is positive and negative, the other is in the negatives. When the $x$ values are positive, the $y$ value is less then $1$. When the $x$ value is negative the $y$ value goes greater then $1$.
|
Because the numerator is $1$ in this case, using whole numbers for $x$ the only way that the fraction $\frac{1}{x+2}$ can be a whole number is when the denominator is either $1$ or $-1$. Knowing this, the only solutions are when $x$ is either $-1$ or $-3$. This gives us $y$ values of $1$ and $-1$ respectively.
If you just want $y$ to be an integer, but you don't need $x$ to be an integer, you could first solve your function for $x$ like this:
$$
y = \frac{1}{x+2} \qquad\implies\qquad x = \frac{1-2y}{y}
$$
Now supposing you want the $y$-value of your original function to be $y=5$, you just evaluate this new form at $y=5$ and find $x$, so
$$
x = \frac{1-2y}{y} = \frac{1-2(5)}{(5)} = -\frac{9}{5}\;.
$$
So the $x$-value of $-9/5$ evaluates to $5$ when used in your original function.
|
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|
Prove/disprove : $\left(1+\frac{(-1)^{n}}{\sqrt{n}} \right)^{-1}=\left(1-\frac{(-1)^{n}}{\sqrt{n}}+o\left(\frac{1}{n}\right) \right)$ Why we have that
$$\begin{align*}
\frac{(-1)^n}{\sqrt{n}}\left(1+\frac{(-1)^n}{\sqrt{n}} \right)^{-1} & =\frac{(-1)^n}{\sqrt{n}}\left(1-\frac{(-1)^n}{\sqrt{n}}+O\left(\frac{1}n\right) \right) \tag{$E_1$} \\
\end{align*}$$
insted of $$\left(1-\frac{(-1)^n}{\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right) \right) \tag{$E_2$} ?$$
Is that because we have $\frac{1}{\sqrt{n}}=o\left(\frac{1}{n}\right)$? But this is wrong!
*
*Could someone explain to me why we have $E_1$ instead of $E_2$
|
By Taylor expansion,
\begin{align*}
\frac{1}{1+\frac{(-1)^n}{\sqrt{n}}} &= 1+ \frac{(-1)^n}{\sqrt{n}} + \frac{1}{n} + o\left(\frac{1}{n}\right)\\
&=1+ \frac{(-1)^n}{\sqrt{n}} + O\left(\frac{1}{n}\right).
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing that $2^6$ divides $3^{2264}-3^{104}$
Show that $3^{2264}-3^{104}$ is divisible by $2^6$.
My attempt: Let $n=2263$. Since $a^{\phi(n)}\equiv 1 \pmod n$ and
$$\phi(n)=(31-1)(73-1)=2264 -104$$
we conclude that $3^{2264}-3^{104}$ is divisible by $2263$.
I have no idea how to show divisibility by $2^6$.
|
A variant with some details:
By *Euler's theorem, $\;3^{\varphi(2^6)}\equiv 3^{32}\equiv=1\mod 2^6$. So
$$3^{2264}-3{104}\equiv3^{2264\bmod\varphi(2^6)}-3^{104\bmod\varphi(2^6)}\equiv 3^{24}-3^{8}\equiv3^8(3^{16}-1)\mod 2^6,$$
so it is enough to prove $3$ has order $16$ mod $2^6=64$.
Let's use the fast exponentiation algorithm mod $64$:
$$\begin{array}{r| l}
k & 3^k \\\hline
1&3\\2&9\\4&81\equiv 17\\8&289 \equiv-31\\16&961\equiv\color{red}{1}
\end{array}$$
A nicer and faster computation (after @Joffan's suggestion):
As $x\equiv a\mod 2^k\implies x^2\equiv a^2\mod 2^{k+1}$, we have
$$ 3 \equiv -1\mod 2^2\implies3^2 \equiv 1\mod2^3,$$
whence $3^{16}\equiv 1\mod 2^6$.
|
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|
Is it possible that $a^2+b^2+c^2 = d^2+e^2+f^2$?
Let $a,b,c$ be nonnegative integers such that $a \leq b \leq c, 2b \neq a+c$ and $\frac{a+b+c}{3}$ is an integer. Is it possible to find three nonnegative integers $d,e,$ and $f$ such that $d \leq e \leq f, f \neq c,$ and such that $a^2+b^2+c^2 = d^2+e^2+f^2$?
I have tried it for examples such as $a=1,b=2,c=6$, we have $1^2+2^2+6^2 = 41 = 0^2+4^2+5^2$ and it always seems to work. I am wondering how to use the fact that $\frac{a+b+c}{3}$ is an integer and wondering if it may be related to the AM-GM inequality. How should I use this information to solve the question?
|
Short answer (explained in the comments above):
$$ (4x)^2+(4x^2-1)^2 = (4x^2+1)^2 $$
can be "interlaced" with
$$ (4x^2+1)^2 + (8x^4+4x^2)^2 = (8x^4+4x^2+1)^2 $$
to get:
$$ (4x)^2 + (4x^2-1)^2 + (8x^4+4x^2)^2 = 0^2 + 0^2 + (8x^4+4x^2+1)^2 $$
or:
$$ (12x)^2 + (12x^2-3)^2 + (24x^4+12x^2)^2 = 0^2 + 0^2 + (24x^4+12x^2+3)^2 $$
i.e. an infinite number of solutions.
Easier approach: we may take three positive integers $N=4u+1$, $M=4v-1$, $ N+M=4(u+v)$
and write them as a difference of two squares,
$$N=a^2-d^2,\quad M=b^2-e^2,\quad N+M=f^2-c^2.$$
We get $a^2+b^2+c^2=d^2+e^2+f^2$ with $f> c$ and we just need very few work to ensure the other constraints.
|
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|
Show that $u_1^3+u_2^3+\cdots+u_n^3$ is a multiple of $u_1+u_2+\cdots+u_n$
*
*Let $k$ be a positive integer.
*Define $u_0 = 0\,,\ u_1 = 1\ $ and $\ u_n = k\,u_{n-1}\ -\ u_{n-2}\,,\
n \geq 2$.
*Show that for each integer $n$, the number
$u_{1}^{3} + u_{2}^{3} + \cdots + u_{n}^{3}\ $ is a multiple of
$\ u_{1} + u_{2} + \cdots + u_{n}$.
Computing a few terms I found
\begin{align*}u_0 &= 0\\u_1 &= 1\\u_2 &= k\\u_3 &= k^2-1\\u_4 &= k(k^2-1)-k = k^3-2k\\u_5 &= k(k^3-2k)-(k^2-1) = k^4-3k^2+1\\u_6 &= k(k^4-3k^2+1)-(k^3-2k) = k^5-4k^3+3k.\end{align*}
I am not sure how we can use this to solve the question, but I think it may help. Cubing these expressions seems very computational so there must be an easier way.
|
fiddling with small $k.$ Take $x = \Sigma u_j, \; \; y = \Sigma u_j^3.$
CONCLUSION: for $k \geq 2,$
$$ \color{blue}{ y = \frac{x^2 ((k-2) x + 3)}{k+1},} $$
while $\color{blue}{x \equiv 0,1 \pmod {k+1}}.$
When $k=2,$ $$ y = x^2. $$ This comes up pretty often, the sum of the consecutive cubes (starting with $1$) is the square of the sum of the consecutive numbers.
When $k=3,$ $$ y = \frac{x^2 (x + 3)}{4}. $$
For this one, you need to know that $x \equiv 0,1 \pmod 4.$
This already suggests that $k=4$ gives $y = a x^4 + b x^3 + c x^2 + d x,$ with rational coefficients, and some restrictions on $x$ that make $y$ an integer. If true, the coefficients can be found by taking four $x$ points, then making and inverting a certain four by four rational matrix. NOPE, not that much effort required. Stays cubic, a three by three matrix would have been enough...
Not quite what I expected: for $k=4,$
$$ y = \frac{x^2 (2 x + 3)}{5}, $$
while $x \equiv 0,1 \pmod 5.$
For $k=5,$
$$ y = \frac{x^2 (3 x + 3)}{6}, $$
while $x \equiv 0,1 \pmod 6.$
For $k=6,$
$$ y = \frac{x^2 (4 x + 3)}{7}, $$
while $x \equiv 0,1 \pmod 7.$
Apparently, for $k \geq 2,$
$$ \color{red}{ y = \frac{x^2 ((k-2) x + 3)}{k+1},} $$
while $x \equiv 0,1 \pmod {k+1}.$
Recall $x = \Sigma u_j, \; \; y = \Sigma u_j^3.$
|
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|
Evaluation of $\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$ Evaluate the following limit:
$$L=\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$$
Using $\ln(1+x)=x-x^2/2+x^3/3-\cdots$
I got $(1+x)^{1/x}=e^{1-x/2+x^2/3-\cdots}$
Could some tell me how to proceed further?
|
It's a well-known fact that $$\log(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3+\underset{x\to 0}{o}(x^3)$$
Hence we get $$\dfrac 1x\log(1+x)=1+u(x)$$ where $$u(x)=-\dfrac x2+\dfrac{x^2}3+\underset{x\to 0}{o}(x^2)$$
Notice that $$\lim_{x\to 0} u(x)=0$$
We can write $$(1+x)^{1/x}=ee^{u(x)}$$
But we know that $$e^u=1+u+\dfrac{u^2}2+\underset{u\to 0}o(u^2)$$
Now using the identity $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$$
we quickly find, since we only need the terms whose degrees are less than three (the other terms are all some $\underset{x\to 0}o(x^2)$), that
$$(u(x))^2=\left(-\dfrac x2+\dfrac{x^2}3+\underset{x\to 0}{o}(x^2)\right)^2=\dfrac{x^2}4+\underset{x\to 0}o(x^2)$$
Hence, since $\displaystyle\lim_{x\to 0}u(x)=0$, we see than an $\underset{u\to 0}o(u^2)$ function is an $o(x^2)$ when $x$ tends to $0$.
This leads to
$$\begin{align*}(1+x)^{1/x} & = e\left(1+\left(-\dfrac x2+\dfrac{x^2}3+\underset{x\to 0}{o}(x^2)\right)+\dfrac 12\left(\dfrac{x^2}4+\underset{x\to 0}o(x^2)\right)+\underset{x\to 0}{o}(x^2)\right)\\ & = e-\dfrac{ex}2+\dfrac{11e}{24}x^2+\underset{x\to 0}o(x^2)\end{align*}$$
And so :
$$\dfrac{(1+x)^{1/x}-e+\dfrac{ex}2}{x^2}=\dfrac{11e}{24}+\underset{x\to 0}o(1)$$
which means exactly that $$\lim_{x\to 0}\dfrac{(1+x)^{1/x}-e+\dfrac{ex}2}{x^2}=\dfrac{11e}{24}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\lim\limits_{x\to\infty} f(x)$ exists if $f'(x) = \frac{1}{x^2+f(x)^2}$ and $f(1)=1$
Let $f(x)$ be a real differentiable function defined for $x\geq 1$ such that $f(1)=1$ and $f'(x)=\dfrac{1}{x^2+f(x)^2}.$ Show that $$\lim_{x\to \infty}f(x)$$ exists and is less than $1+\frac{\pi}{4}$
I have no idea how to tackle this question... Help!
|
Notice that $f'(x)$ is always positive, so $f$ is increasing. Hence for $x > 1$ we have $f(x) > 1$. Consequently, for $x > 1$, we have $f'(x) < \frac{1}{x^2 + 1}$.
We also have that $f'(x)$ is continuous for $x > 1$ since it's the quotient of continuous functions and the denominator is not zero. (The numerator is the continuous function g(x) = 1, and the decnominator is the sum of continuous functions $h(x) = x^2$ and $f(x)$.) Thus, we can apply the fundamental theorem of calculus, which tells us
$$
\int_1^t f'(x)dx = f(t) - f(1).
$$
for any $t > 1$.
For any $t > 1$, we then have
$$
\begin{align*}
f(t)
&= f(1) + f(t) - f(1) \\
&= f(1) + \int_1^t f'(x)dx \\
&< f(1) + \int_1^t \frac{1}{x^2 + 1}dx \\
&= f(1) + \left.\tan^{-1}(x)\right\rvert_{x=1}^{x=t} \\
&< 1 + \frac{\pi}{4}.
\end{align*}
$$
Since $f(t)$ is increasing and is bounded above, the limit exists.
Then, taking the limit as $t$ approaches infinity, we have
\begin{align*}
\lim_{t \to \infty} f(t)
&= f(1) + \lim_{t \to \infty} \int_1^t f'(x)dx \\
&< f(1) + \lim_{t \to \infty} \int_1^t \frac{1}{x^2 + 1}dx \\
&= 1 + \frac{\pi}{4}
\end{align*}
as desired.
|
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Proof about Pythagorean triples Show that if $(x,y,z)$ is a Pythagorean triple, then $10\mid xyz$
Proof
First, if $x$, $y$, $z$ are all odd, then so are $x^2$, $y^2$, $z^2$, so $x^2+y^2$ is even, which means that $x^2+y^2 \neq z^2 $. Hence, at least one of $x$, $y$, $z$ is even, so $2\mid xyz$ (clear).
Next, for any $n \in Z$, if $5$ doesn't divide $n$, then $(n^2)^2=n^4 \equiv 1\pmod{5}$ (as you can check or quote Euler's theorem), and therefore $n^2 \equiv \pm 1\pmod{5}$. Now, if $5$ doesn't divide $xy$, then $x^2 \equiv \pm 1\pmod{5}$ and $y^2 \equiv \pm 1\pmod{5}$, so
$$x^2+y^2 \equiv -2,0,2 \pmod{5}$$
Therefore if $x^2+y^2=z^2$ and $5$ doesn't divide $xy$ then $z^2=x^2+y^2 \equiv -2,0,2\pmod{5}$, so $5\mid z^2$ (why? it looks weird to me because $z^2$ can be also congruent to $2$ and $-2$) and hence $5\mid z$ (otherwise $z^2 \equiv \pm 1\pmod{5}$). It follows that $x^2+y^2=z^2$, then either $5\mid xy$ or $5\mid z$ (how does this follow? if $5$ doesn't divide $z$ then how can $5$ divide $xy$), so in any case $5\mid xyz$
Finally, we can conclude that if $x^2+y^2=z^2$ then $2\mid xyz$ and $5\mid xyz$, so $10\mid xyz$ (intuitively it looks right but I can't prove it!)
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The general solution of the above Diophantine Equation is: $x = m^2 - n^2, y = 2mn, z = m^2+n^2$. Its better if you start at this point. Then your analysis yields the followings: $m^2 = \pm 1 \pmod 5, n^2 = \pm 1 \pmod 5$. Since $5 \nmid xy\implies 5 \nmid x, 5 \nmid y\implies m^2 = 1\pmod 5, n^2 = -1\pmod 5$ or $m^2 = -1\pmod 5, n^2 = 1\pmod 5$ for otherwise $x = m^2 - n^2 = \pm1 - \pm1 = 0\pmod 5$, contradicting the assumption that $5 \nmid x$. Thus: $m^4 + 2m^2n^2 + n^4 = 1 + 2(-1) + 1 = 0 \pmod 5\implies 5 \mid (m^2+n^2)^2 = z^2$.
|
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nature of the series $\sum (-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \right)}$ I would like to study the nature of the following serie:
$$\sum_{n\geq 0}\ (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)} $$
we can use simply this question :
Show : $(-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \right)}=\tfrac{(-1)^{n}}{n}+\mathcal{O}\left(\tfrac{\ln(n)}{n^{2}} \right)$
$$\sum_{n\geq 1} (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)}=\sum_{n\geq 1}\dfrac{(-1)^{n}}{n}+\sum_{n\geq 1} \mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right) $$
Since :
*
*Alternating harmonic series $\sum_{n\geq 1}\dfrac{(-1)^{n}}{n}$ is convergent
*$\sum_{n\geq 1} \mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right) $ is convergent since $\left| \mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)\right|\leq \dfrac{\ln(n)}{n^{2}}$
thus $\sum_{n\geq 0}\ (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)} $ is convergent
My question :
*
*Why we can't take expansion up to $2$ $\frac{1}{1-z}=1+z+O(z^2)$ instead of $\frac{1}{1-z}=1+z+z^2+O(z^3)$
\begin{align}
\tan\left(\frac{\pi}{4}+\frac1n\right)&=\left(1+\tan(1/n)\right)\left(1-\tan(1/n)\right)^{-1}\\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right) \\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)+\mathcal{O}\left(\left(\frac{-1}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)^{2} \right) \right) \\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{2}}\right) \right) \\
&=1-\frac{1}{n^2}+\mathcal{O}\left( \frac{1}{n^{2}}\right)
\end{align}
then :
\begin{align}
e^{-\tan\left(\frac{\pi}{4}+\frac1n\right)\log(n)}&=e^{\left(-1-\frac{1}{n^2}+\mathcal{O}\left( \frac{1}{n^{2}}\right)\right)\log(n)}\\\\
&=\frac1n\,e^{\left(-\frac{\log(n)}n^{2}+O\left(\frac{\log(n)}{n^2}\right)\right)}\\\\
&=\frac1n\,\left(1-\frac{\log(n)}{n^{2}}+O\left(\frac{\log(n)}{n^2}\right)+O\left(\frac{\log^{2}(n)}{n^4}\right) \right)\\\\
&=\frac1n +O\left(\frac{\log(n)}{n^3}\right)
\end{align}
$$(-1)^{n}e^{-\tan\left(\frac{\pi}{4}+\frac1n\right)\log(n)}=(-1)^{n}\frac1n +O\left(\frac{(-1)^{n}\log(n)}{n^3}\right) $$
and since $\sum O\left(\frac{\log(n)}{n^3}\right)$ and $(-1)^{n}\frac1n$ are convergent
i prove that the series is conevrgent only by using expansion up to 2 why we can't take it
|
You made a mistake in
\begin{align}
\tan\left(\frac{\pi}{4}+\frac1n\right)&=\left(1+\tan(1/n)\right)\left(1-\tan(1/n)\right)^{-1}\\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right) \\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)+\mathcal{O}\left(\left(\frac{-1}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)^{2} \right) \right) \\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{2}}\right) \right) \\
&=1-\frac{1}{n^2}+\mathcal{O}\left( \frac{1}{n^{2}}\right)
\end{align}
It should be
\begin{align}
\tan\left(\frac{\pi}{4}+\frac1n\right)&=\left(1+\tan(1/n)\right)\left(1-\tan(1/n)\right)^{-1}\\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1-\left(\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\right)^{-1} \\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1+\left(\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)+\mathcal{O}\left(\left(\frac{1}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)^{2} \right) \right) \\
&= \left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{3}}\right)\right)\left(1+\frac{1}{n}+\mathcal{O}\left( \frac{1}{n^{2}}\right) \right) \\
&=1+\frac{2}{n}+\mathcal{O}\left( \frac{1}{n^{2}}\right)
\end{align}
That is, you effectively wrote
\begin{align}
{1+x\over1-x}&=(1+x)(1-x)^{-1}\\
&=(1+x)(1-x+x^2+\cdots)
\end{align}
instead of
\begin{align}
{1+x\over1-x}&=(1+x)(1-x)^{-1}\\
&=(1+x)(1+x+x^2+\cdots)
\end{align}
(It looks like you also omitted the exponent $-1$ when changing from $(1-\tan(1/n))$ to its Taylor expansion, but I think that was just a typo.)
|
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|
Compute $1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \cdots + n \cdot \frac {1}{2^n} + \cdots $ I have tried to compute the first few terms to try to find a pattern but I got
$$\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}$$
but I still don't see any obvious pattern(s). I also tried to look for a pattern in the question, but I cannot see any pattern (possibly because I'm overthinking it?) Please help me with this problem.
|
Summation by parts gives:
$$ \sum_{n=1}^{N}\frac{n}{2^n} = N\left(1-\frac{1}{2^N}\right)-\sum_{n=1}^{N-1}\left(1-\frac{1}{2^k}\right)=1-\frac{N}{2^N}+\left(1-\frac{1}{2^{N-1}}\right)=\color{red}{2-\frac{N+2}{2^N}}. $$
|
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|
Solve $\sec (x) + \tan (x) = 4$ $$\sec{x}+\tan{x}=4$$
Find $x$ for $0<x<2\pi$.
Eventually I get $$\cos x=\frac{8}{17}$$
$$x=61.9^{\circ}$$
The answer I obtained is the only answer, another respective value of $x$ in $4$-th quadrant does not solve the equation, how does this happen? I have been facing the same problem every time I solved this kind of trigonometric equation.
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$$\tan x +\sec x =\frac{1+ \sin x}{ \cos x}$$
$$=\frac{(\cos\frac{x}{2}+ \sin \frac{x}{2})^2}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}$$
$$=\tan \left(\frac{\pi}{4}+\frac{x}{2} \right)$$
$$\frac{\pi}{4}+\frac{x}{2} =n\pi +tan^{-1}4$$
$$ x =2 n\pi +2 tan^{-1}4 -\frac{\pi}{2}$$
For solution to be in $[0,2\pi]$
$$ n =0 $$
$x=1.080 $ or $ 61.9^{\circ}$
You can also see from the graph that in $[0,2\pi]$ there is only solution lying .
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|
Can someone explain this part of the definition of a linear combination of column vectors to me? This is what I need explained: "The system Ax=b is consistent iff b can be expressed as a linear combination, where the coefficients of the linear combination are a solution of the system."
I thought a linear combination of column vectors was something like
$$x\begin{pmatrix}1\\2\end{pmatrix}+y\begin{pmatrix}3\\4\end{pmatrix}$$
where you have to have coefficients AND variables and multiple matrices that are added together, but a system with a single solution would have just coefficients. And any solution would have just one column matrix, not a few added together.
Also, what is a situation that would cause b to NOT be able to be expressed as a linear combination? Would that just be when the system has no solution?
Am I reading too much into this, is it just saying that the system is consistent if it has a solution like
$$t\begin{pmatrix}3\\4\end{pmatrix}\text{, or just }\begin{pmatrix}2\\1\end{pmatrix}?$$
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$$x\begin{pmatrix}1\\2\end{pmatrix}+y\begin{pmatrix}3\\4\end{pmatrix}=\underbrace{\begin{pmatrix}1&3\\2&4\end{pmatrix}}_A\underbrace{\begin{pmatrix}x\\y\end{pmatrix}}_{\bf x}$$
If we change this slightly to $$x\begin{pmatrix}1\\2\end{pmatrix}+y\begin{pmatrix}2\\4\end{pmatrix}=\underbrace{\begin{pmatrix}1&2\\2&4\end{pmatrix}}_A\underbrace{\begin{pmatrix}x\\y\end{pmatrix}}_{\bf x}$$
then the result will never be $\begin{pmatrix}1\\1\end{pmatrix}$, for any values of $x,y$.
|
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|
Inner Product Examples, what is the points? Example:
For $ -\pi<x<\pi$,
$$x =-2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sin(nx)$$
and
$$x^3 =-2 \sum_{n=1}^{\infty} \left( \frac{\pi^2}{n}-\frac{6}{n^3} \right)(-1)^n \sin(nx)$$
by using inner products of these two functions, the value of
$$\sum_{n=1}^{\infty} \frac{1}{n^4}$$
is equal to $\frac {\pi^4}{90} $ .
My question is what is the point in this example that the author gives
the solution without any detail? any users can tips me how this value
is reached?
|
HINT:
$$\frac 1{\pi}\int_{-\pi}^\pi x^4dx=4\sum_{n=1}^\infty \frac{\pi^2}{n^2}-\frac 6{n^4} $$
FURTHER EXPLANATION:
You have two functions $f(x)=x$ and $g(x)=x^3$. The coefficients of the expansion of $f$ in the basis $\sin(nx)$ are of the form $$a_n=-2\times \frac{(-1)^n}{n}$$
and the coefficients for $g(x)$ are: $$b_n=-2\times\left(\frac{\pi^2}{n^2}-\frac 6{n^3}\right)\times (-1)^n$$
Then the inner product is given by:
$$\frac 1\pi\int_{-\pi}^{\pi}f(x)g(x)=\sum_{n=1}^\infty a_n b_n$$
Replacing $f,g,a_n$ and $b_n$, using the fact that $\sum 1/n^2=\pi^2/6$ and manipulating, gives you the answer.
SOLUTION:
$$\frac 1{\pi}\int_{-\pi}^\pi x^4dx=4\sum_{n=1}^\infty \left(-2\times \frac{(-1)^n}{n}\right)\times(-2)\left(\left(\frac{\pi^2}{n^2}-\frac 6{n^3}\right)\times (-1)^n\right) $$
$$\frac 1{\pi}\int_{-\pi}^\pi x^4dx=4\sum_{n=1}^\infty \frac{\pi^2}{n^2}-\frac 6{n^4} $$
$$\frac{2\pi^5}{5\pi}=4\pi^2\times\frac{\pi^2}6-24\sum_{n=1}^\infty\frac 1{n^4}$$
$$24\sum_{n=1}^\infty\frac 1{n^4}=\pi^4\left( \frac 23 - \frac 25\right) $$
$$\sum_{n=1}^\infty\frac 1{n^4}=\frac{\pi^4}{90}$$
|
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|
Prove using induction the following equation is true. If $$(1-x^2)\frac{dy}{dx} - xy - 1 = 0$$
Using induction prove the following for any positive integer n$$(1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+3)x\frac{d^{n+1}y}{dx^{n+1}} - (n+1)^2\frac{d^ny}{dx^n} = 0$$
I know Leibtniz can be used to solve it easier but I need the proof to use induction.
|
$(1βx^2)\frac{dy}{dx}βxyβ1=0$
differentiate
$-2x \frac{dy}{dx} + (1+x^2)\frac{d^2y}{dx^2}βy - x\frac{dy}{dx}=0\\
(1+x^2)\frac{d^2y}{dx^2} -3x \frac{dy}{dx} - y=0$
This covers the base case $n=0$
Suppose,
$(1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+3)x\frac{d^{n+1}y}{dx^{n+1}} - (n+1)^2\frac{d^ny}{dx^n} = 0$
(this is the inductive hypothesis)
$\frac {d}{dx} \big((1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+3)x\frac{d^{n+1}y}{dx^{n+1}} - (n+1)^2\frac{d^ny}{dx^n} = 0\big)\\
-2x\frac{d^{n+2}y}{dx^{n+2}} + (1-x^2)\frac{d^{n+3}y}{dx^{n+3}} - (2n+3)\frac{d^{n+1}y}{dx^{n+1}} - (2n+3)x\frac{d^{n+2}y}{dx^{n+2}} - (n+1)^2\frac{d^{n+1}y}{dx^{n+1}} = 0\\
(1-x^2)\frac{d^{n+3}y}{dx^{n+3}} - (2n+5)x\frac{d^{n+2}y}{dx^{n+2}} - (n^2 + 4n +4)\frac{d^{n+1}y}{dx^{n+1}} = 0\\
(1-x^2)\frac{d^{n+3}y}{dx^{n+3}} - (2(n+1)+3)x\frac{d^{n+2}y}{dx^{n+2}} - (n+2)^2\frac{d^{n+1}y}{dx^{n+1}} = 0\\$
QED
|
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|
Integrating $\displaystyle\int \frac{1+x^2}{1+x^4}dx$ I am trying to integrate this function, which I got while solving $\int\frac{1}{\sin^4( x) + \cos^4 (x)}$:
$$\int \frac{1+x^2}{1+x^4}\mathrm dx$$
I think to factorise the denominator, and use partial fractions. But I cant seem to find roots of denominator. I also am unable to think substitution.
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$$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$
$$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$
$$\sin^4 x + \cos ^4 x =\frac{1}{4}\left[(1- \cos2x)^2 +(1+ \cos2x)^2 \right]
=\frac{1}{2}((1+ \cos^2 2x)$$
$$\implies 2 \int \frac{dx}{1+ \cos^2 2x}$$
$$= 2 \int \frac{\sec^2 2x}{\sec^2 2x+ 1}{dx}$$
$$t= \tan2x$$
$$\implies 2 \int \frac{dt}{2 + t^2}$$
$$\frac{1}{\sqrt2} \tan^{-1} \left(\frac{t}{\sqrt2}\right)$$
|
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|
Evaluation of $\int^{\pi/2}_{0} \frac{x \tan(x)}{\sec(x)+\tan(x)}dx$ Evaluate the given integral:
$$\int^{\pi/2}_0 \frac{x \tan(x)}{\sec(x)+\tan(x)}dx$$
I multiplied and divided by $\sec(x)+\tan(x)$ to get denominator as $1$ but In calculation of integral, $x$ is creating problem. Is there any way to eliminate $x$ here, like it would have been eliminated if upper limit was $\pi$?
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Hint: Try the sub $x \mapsto \frac{\pi}{2} - x$ to get $$\int_0^{\pi/2} \frac{(\pi/2 - x) \cot x}{\csc x + \cot x} \, \mathrm{d}x = \int_0^{\pi/2} \frac{\pi/2 - x}{1 + \sec x} \, \mathrm{d}x$$
Then the rest is do-able using $$\frac{1}{\sec x + 1} = \frac{\cos x}{\cos x + 1} = \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}} = \frac{1}{2} - \frac{\tan^2 \frac{x}{2}}{2}$$
And so $$ \int \frac{\mathrm{d}x}{1 + \sec x} = \frac{1}{2} \int 1 - \left(-1+\sec^2 \frac{x}{2}\right) \, \mathrm{d}x = x - \tan \frac{x}{2} + \mathrm{C}$$
|
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|
How to show $\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$
How does one show that $$\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$$ for each nonnegative integer $n$?
I tried using the Snake oil technique but I guess I am applying it incorrectly. With the snake oil technique we have $$F(x)= \sum_{n=0}^{\infty}\left\{\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}\right\}x^{n}.$$ I think I have to interchage the summation and do something. But I am not quite comfortable in interchanging the summation. Like after interchaging the summation will $$F(x)=\sum_{k=0}^{n}\sum_{n=0}^{\infty}\binom{n+k}{k}\frac{1}{2^k}x^{n}?$$ Even if I continue with this I am unable to get the correct answer.
*
*How does one prove this using the Snake oil technique?
*A combinatorial proof is also welcome, as are other kinds of proofs.
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A proof by induction is possible, if a bit messy. For $n\in\Bbb N$ let $$s_n=\sum_{k=0}^n\binom{n+k}k\frac1{2^k}\;.$$ Clearly $s_0=1=2^0$. Suppose that $s_n=2^n$ for some $n\in\Bbb N$. Then
$$\begin{align*}
s_{n+1}&=\sum_{k=0}^{n+1}\binom{n+1+k}k\frac1{2^k}\\\\
&=\binom{2n+2}{n+1}\frac1{2^{n+1}}+\sum_{k=0}^n\left(\binom{n+k}k+\binom{n+k}{k-1}\right)\frac1{2^k}\\\\
&=\binom{2n+2}{n+1}\frac1{2^{n+1}}+\sum_{k=0}^n\binom{n+k}k\frac1{2^k}+\sum_{k=0}^n\binom{n+k}{k-1}\frac1{2^k}\\\\
&=\binom{2n+2}{n+1}\frac1{2^{n+1}}+\sum_{k=0}^n\binom{n+k}k\frac1{2^k}+\sum_{k=1}^n\binom{n+k}{k-1}\frac1{2^k}\\\\
&=\binom{2n+2}{n+1}\frac1{2^{n+1}}+s_n+\sum_{k=0}^{n-1}\binom{n+1+k}k\frac1{2^{k+1}}\\\\
&=s_n+\binom{2n+2}{n+1}\frac1{2^{n+1}}+\frac12\sum_{k=0}^{n-1}\binom{n+1+k}k\frac1{2^k}\\\\
&=2^n+\left(\binom{2n+1}{n+1}+\binom{2n+1}n\right)\frac1{2^{n+1}}+\frac12\sum_{k=0}^{n-1}\binom{n+1+k}k\frac1{2^k}\\\\
&=2^n+\binom{2n+1}{n+1}\frac1{2^{n+1}}+\frac12\sum_{k=0}^n\binom{n+1+k}k\frac1{2^k}\\\\
&\overset{(*)}=2^n+\frac12\binom{2n+2}{n+1}\frac1{2^{n+1}}+\frac12\sum_{k=0}^n\binom{n+1+k}k\frac1{2^k}\\\\
&=2^n+\frac12\sum_{k=0}^{n+1}\binom{n+1+k}k\frac1{2^k}\\\\
&=2^n+\frac12s_{n+1}\;,
\end{align*}$$
where the step $(*)$ follows from the fact that
$$\binom{2n+2}{n+1}=\binom{2n+1}{n+1}+\binom{2n+1}n=2\binom{2n+1}{n+1}\;.$$
Thus, $\frac12s_{n+1}=2^n$, and $s_{n+1}=2^{n+1}$, as desired.
Added: I just came up with a combinatorial argument as well. Flip a fair coin until either $n+1$ heads or $n+1$ tails have appeared. Let $k$ be the number of times the other face of the coin has appeared; clearly $0\le k\le n$. The last flip must result in the $(n+1)$-st instance of the majority face, but the other $n$ instances of that face and $k$ of the other can appear in any order.
Now imagine that after achieving the desired outcome we continue to flip the coin until weβve flipped it $2n+1$ times. There are altogether
$$\binom{n+k}k2^{(2n+1)-(n+k)}=\binom{n+k}k2^{n+1-k}$$
sequences of $2n+1$ flips that decide the outcome at the $(n+k+1)$-st toss, so
$$\sum_{k=0}^n\binom{n+k}k2^{n+1-k}=2^{2n+1}\;,$$
and
$$\sum_{k=0}^n\binom{n+k}k\frac1{2^k}=2^n\;.$$
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|
To find area enclosed within curve $x^4+y^4=x^2+y^2$ Find area enclosed by the curve
$x^4+y^4=x^2+y^2$
My attempt:
How to describe $y$ as function of $x$.
Only after that we can integrate.
Or can this curve be described parametrically.
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$$y^4-y^2+x^4-x^2=0$$
$$r^2(\cos^4(\theta)+\sin^4(\theta))=1$$
$$r^2=\frac 1 {\cos^4(\theta)+\sin^4(\theta)}$$
$$A/4=\frac 12\int_0^{\pi/2}\frac 1 {\cos^4(\theta)+\sin^4(\theta)}d\theta$$
$$A/4=\frac 12\int_0^{\pi/2} \frac 1{\cos^4 \theta}\frac{1}{1+\tan^4\theta}d\theta $$
Let $s=\tan(\theta)$ then $$A/4=\frac 12\int_0^{\infty}\frac{s^2+1}{s^4+1}ds$$
Decomposing the rational function:
$$A/4=\frac 14\int_0^{\infty}\frac1{s^2+\sqrt 2 s +1}-\frac1{-s^2+\sqrt 2 s -1}ds$$
$$A/4=\frac 14\int_0^{\infty}\frac1{(s+1/\sqrt2)^2+1/2}-\frac1{-(s+1/\sqrt2)^2-1/2}ds$$
And keeping in mind that $$\int \frac{1}{a^2+(x+b)^2}dx=\frac 1a\tan^{-1}\left(\frac{x+b}a\right)$$
You find that $$\frac A 4=\frac {\pi}{2\sqrt 2}$$
EDIT: Another, more interesting way of solving the integral: $$\frac{1}{\cos^4(\theta)+\sin^4(\theta)}=\frac{1}{1-2\sin^2(\theta)+2\sin^4(\theta)}=\frac{1}{1-\sin^2(2\theta)/2}=\frac{1}{1-\dfrac{1-\cos^2(2\theta)}{2}}=\frac{1}{\dfrac{2-1+\cos^2(2\theta)}{2}}=\frac{2}{1+\cos^2(\theta)}=\frac{1+\tan^2(2\theta)}{1+\dfrac{\tan^2(2\theta)}2}$$
If $u=\dfrac{\tan(2\theta)}{\sqrt2}$ then $du=\sqrt2(1+\tan^2(2\theta)) d\theta$, so that the integral becomes:$$\frac 1{2\sqrt2}\int \frac{du}{1+u^2}=\frac 1{2\sqrt2}\tan^{-1}(u)=\frac 1{2\sqrt2}\tan^{-1}\left(\dfrac{\tan(2\theta)}{\sqrt2}\right)$$
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Convergence/Divergence of $\sum_{n=1}^{\infty}\frac{n+n^2+\cdots+n^n}{n^{n+2}}$
$$\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}$$
$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \sum_{n=1}^{\infty}\frac{n^n+n^n+\cdots+n^n}{n^{n+2}}=\sum_{n=1}^\infty \frac{n^{n+1}}{n^{n+2}}=\sum_{n=1}^\infty \frac{1}{n}$$
But is still does not help to conclude about converges/diverges
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If $n>1$ we have $n+n^2+\dots + n^n=\frac{n^{n+1}-1}{n-1}-1\leq\frac{n^{n+1}}{n-1}\leq 2n^{n}$
So $\frac{n+n^2+\dots + n^n}{n^{n+2}}\leq n^{-2}$
Therefore $\sum_{n=2}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq\sum_{n=2}^\infty n^{-2}<\infty$
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Formula for $1^k+2^k+3^k...n^k$ for $n,k \in \mathbb{N}$ So I've been looking for a formula where I can input the parameter $k$ and it will give me a formula for $1^k+2^k+3^k...+ n^k$ with $n,k \in \mathbb{N}$. The result is always a polynomial with $k+1$ as highest power. I've taken the time to calculate the polynomes for $k=1$ to $k=10$ by hand and using the interpolation feature of Wolfram Alpha. Here are the results (I'll only show the coefficients for the sake of clarity. the coefficients are always from $n^{k+1}$ to $n^1$. the constant is always $0$. So $\frac{1}{2},-\frac{1}{2}$ becomes $\frac{1}{2}n^2-\frac{1}{2}n$):
*
*$k=1$ : $\frac{1}{2},-\frac{1}{2}$
*$k=2$ : $\frac{1}{3},\frac{1}{2},\frac{1}{6}$
*$k=3$ : $\frac{1}{4},\frac{1}{2},\frac{1}{4},0$
*$k=4$ : $\frac{1}{5},\frac{1}{2},\frac{1}{3},0,-\frac{1}{30}$
*$k=5$ : $\frac{1}{6},\frac{1}{2},\frac{5}{12},0,-\frac{1}{12},0$
*$k=6$ : $\frac{1}{7},\frac{1}{2},\frac{1}{2},0,-\frac{1}{6},0,\frac{1}{42}$
*$k=7$ : $\frac{1}{8},\frac{1}{2},\frac{7}{12},0,-\frac{7}{24},0,\frac{1}{12},0$
*$k=8$ : $\frac{1}{9},\frac{1}{2},\frac{2}{3},0,-\frac{7}{15},0,\frac{2}{9},0,-\frac{1}{30}$
*$k=9$ : $\frac{1}{10},\frac{1}{2},\frac{3}{4},0,-\frac{7}{10},0,\frac{1}{2},0,-\frac{3}{20},0$
*$k=10$ : $\frac{1}{11},\frac{1}{2},\frac{5}{9},0,1,0,1,0,-\frac{1}{2},0,\frac{5}{66}$
There are a few things i notice: Firstly, the coefficient of the highest power seems to be $\frac{1}{k+1}$. Secondly, the coefficient of the second highest power seems to be $\frac{1}{2}$ with the exception of $k=1$. Thirdly, all coefficients of the fourth, sixth, eight highest power and so on seem to be $0$.
What is the formula that will output the coefficients for any value of $k$?
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I'll give you a derivation which I don't think is all that known. You just need to know four things:
(1) A version of the umbral Taylor series
Suppose that we have a sequence:
$$a_0,a_1,...a_k$$
And we want to find the function of $n$ that defines $a_n$.
To do this we start by letting $a_{n+1}-a_n=\Delta a_n$ and we call this operation on $a_n$ the forward difference. Then given $\Delta a_n$ we can find $a_n$. Sum both sides of the equation from $n=0$ to $x-1$, and note that we have a telescoping series:
$$\sum_{n=0}^{x-1} \Delta a_n=\sum_{n=0}^{x-1} (a_{n+1}-a_n)=a_{x}-a_{0}$$
Hence $a_n=a_0+\sum_{i=0}^{n-1} \Delta a_i$. Also $\Delta a_n=\Delta (0)+\sum_{i=0}^{n-1} \Delta^2 a_n$...and so forth. Using this we must have if the series converges:
$$a_n=a_0+\Delta (0) \sum_{x_0=0}^{n-1} 1+\Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} 1+\Delta \Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} \sum_{x_2=0}^{x_1-1} 1+\cdots$$
Where $\Delta^i (0)$ denotes the first term ($n=0$) of the $i$ th difference sequence of $a_n$.
Through a combinational argument, If we take $\Delta^0 (0)=a_0$ and ${n \choose 0}=1$ we may get:
$$a_n=\sum_{i=0}^{\infty} \Delta^i(0) {n \choose i}$$
(2) Hockey stick identity Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$
(3) The forward difference of a polynomial of degree $k$ is of degree $k-1$ (follows from binomial theorem) hence $k+1$ forward differences survey results in $0$ as $k$ forward differences must result in a constant sequence.
(4) A polynomial of degree $k$ can be uniquely defined by the sequence $f(0),f(1),f(2),...f(k)$.
Using (1), (2), (3), and (4) you may come up with the formula:
$$\sum_{s=1}^{n} s^k=\sum_{s=1}^{k} b_s{n+1\choose s+1}$$
By considering how to represent $a_s=s^k$ as a sum of multiples of binomials.
$$s^k=0+b_1{ s \choose 1}+b_2{s \choose 2}+\cdots+ b_k{s \choose k}+0+0+\cdots=\sum_{i=1}^{k} b_i{s \choose i}$$.
$$\sum_{s=1}^{n}\sum_{i=1}^{k} b_i{s \choose i}=\sum_{s=1}^{n}b_1{ s \choose 1}+\sum_{s=1}^{n}b_2{s \choose 2}+\cdots +\sum_{s=1}^{n}b_k{s \choose k}$$.
Taking the constants $b_1,b_2,...$ out of the sums and utilizing the hockey stick identity we get our desired result.
Now see Function $f: \mathbb{Z} \to \mathbb{Z}^n$ related to $\sum_{k=1}^{x} k^n$.
We have that:
$$b_s=s!S(k,s)$$
Where ! denotes the factorial and $S( , )$ denotes the stirling numbers of the second kind.
Hence we have that:
$$\sum_{s=1}^{n} s^k=\sum_{s=1}^{k} s!S(k,s){n+1 \choose s+1}$$
Here are some examples.
$k=1$
$$\sum_{s=1}^{n} s^1=1{n+1 \choose 2}$$
$$=1\frac{(n+1)(n)}{2!}$$
$k=2$
$$\sum_{s=1}^{n} s^2=1{n+1 \choose 2}+2{n+1 \choose 3}$$
$$=1\frac{(n+1)(n)}{2!}+2\frac{(n+1)(n)(n-1)}{3!}$$
$k=3$:
$$\sum_{s=1}^{n} s^3=1{n+1 \choose 2}+6{n+1 \choose 3}+6{n+1 \choose 4}$$
$$=1\frac{(n+1)(n)}{2!}+6\frac{(n+1)(n)(n-1)}{3!}+6\frac{(n+1)(n)(n-1)(n-2)}{4!}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1878810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
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|
How many solutions are there to $x_1 + x_2 + ... + x_5 = 21$? How many solutions are there to the equation
$x_1 + x_2 + x_3 + x_4 + x_5 = 21$
where $x_1, i = 1,2,3,4,5$ is a nonnegative integer such that $0 β€ x_1 β€ 3, 1 β€ x_2 < 4$, and $x_3 β₯ 15$?
I have correctly completed the previous parts of the question but am having trouble with this particular restriction.
The total number of solutions to the equation is $C(25,21) = 12650$.
I have tried to do each restriction individually:
For $0 β€ x_1 β€ 3$, I solved this by subtracting the number of solutions when $x_1 β₯ 4$ from the total number of solutions. This is $C(25,21) - C(21,17) = 5985$.
For $1 β€ x_2 < 4$, I have $C(24,20) - C(21,17) = 5985$.
For $x_3 β₯ 15$, I have $C(10,6) = 210$.
However, the restriction is when these are all together. I am not sure how to proceed. The answer is $106$.
|
An approach using generating functions:
In order to count up the number of solutions that sum to $n$, we can look at the coefficient of $x^n$ in the generating function $f(x)$:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=0}^{3}x^n\right)\left(\sum_{n=1}^{3}x^n\right)\left(\sum_{n=15}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}x^n\right)$$
The $[x^n]$ operator refers to the coefficient of $x^n$.
Factor out $x$'s so you can shift the indices down to $n=0$ for each sum:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=0}^{3}x^n\right)x\left(\sum_{n=0}^{2}x^n\right)x^{15}\left(\sum_{n=0}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}x^n\right)\left(\sum_{n=0}^{\infty}x^n\right)$$
Use the identities $\displaystyle \sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$ and $\displaystyle \sum_{n=0}^{k}x^n = \frac{1-x^{k+1}}{1-x}$ to convert each piece:
$$[x^{n}]f(x) = [x^{n}]\left(\frac{1-x^4}{1-x}\right)x\left(\frac{1-x^3}{1-x}\right)x^{15}\left(\frac{1}{1-x}\right)^3$$
Simplify:
$$[x^{n}]f(x) = [x^{n}]\left(\frac{x^{23}-x^{20}-x^{19}+x^{16}}{(1-x)^5}\right)$$
Factor out $\dfrac{1}{(1-x)^5}$ and use the identity $\displaystyle \frac{1}{(1-x)^{m+1}} = \sum_{n=0}^{\infty}\binom{n+m}{m}x^n$:
$$[x^{n}]f(x) = [x^{n}]\left(x^{23}-x^{20}-x^{19}+x^{16}\right)\sum_{n=0}^{\infty}\binom{n+4}{4}x^n$$
Distribute:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=0}^{\infty}\binom{n+4}{4}x^{n+23}-\sum_{n=0}^{\infty}\binom{n+4}{4}x^{n+20}-\sum_{n=0}^{\infty}\binom{n+4}{4}x^{n+19}+\sum_{n=0}^{\infty}\binom{n+4}{4}x^{n+16}\right)$$
Shift indices:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=23}^{\infty}\binom{n-19}{4}x^{n}-\sum_{n=20}^{\infty}\binom{n-16}{4}x^{n}-\sum_{n=19}^{\infty}\binom{n-15}{4}x^{n}+\sum_{n=16}^{\infty}\binom{n-12}{4}x^{n}\right)$$
If we look at the lower bounds of the indices, the smallest $x^n$ that we could get the coefficient for is $x^{16}$. This is because $16$ is the smallest possible value of $n$ under the $x_i$ constraints, which occurs when $x_1 = x_4 = x_5 = 0$, $x_2=1$, and $x_3=15$. The coefficients for all $n<16$ are simply $0$.
Now we can invoke the coefficient operator:
$$[x^n]f(x) = \binom{n-19}{4}-\binom{n-16}{4}-\binom{n-15}{4}+\binom{n-12}{4}$$
Set $n=21$ to get the number of solutions to the original problem:
$$[x^{21}]f(x) = 106$$
Side note: If you use $n<19$ in the expression above, you'll start to see negative binomial coefficients. For those, you need to use the identity $\binom{-n}{k} = (-1)^k\binom{n+k-1}{k}$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Integral $\int_0^1 \frac{\ln (2-x)}{2-x^2} \, dx$ Here is an integral that I am trying to solve for quite some time. Find a closed form for the integral:
$$\mathcal{J}=\int_0^1 \frac{\log (2-x)}{2-x^2} \, {\rm d}x$$
Here is what I've done.
\begin{align*}
\int_{0}^{1}\frac{\log (2-x)}{2-x^2} \, {\rm d}x &= \int_{0}^{1} \frac{\log (2-x)}{\left ( \sqrt{2}-x \right )\left ( \sqrt{2}+x \right )} \, {\rm d}x\\
&= \int_{0}^{1}\left [ \frac{\log (2-x)}{2\sqrt{2}(x+\sqrt{2})} + \frac{\log (2-x)}{2\sqrt{2} \left ( \sqrt{2}-x \right )} \right ] \, {\rm d}x \\
&=\frac{1}{2\sqrt{2}} \left [ \int_{0}^{1} \frac{\log (2-x)}{\sqrt{2}+x}\, {\rm d}x + \int_{0}^{1} \frac{\log (2-x)}{\sqrt{2}-x} \, {\rm d}x \right ]
\end{align*}
Now W|A is able to evaluate the last integrals. Indeed they boil down to:
*
*$\displaystyle \int_{0}^{1}\frac{\log(2-x)}{\sqrt{2}+x} \, {\rm d}x = {\rm Li}_2 \left ( 1-\frac{1}{\sqrt{2}} \right ) -{\rm Li}_2 \left ( 2-\sqrt{2} \right ) - \log 2 \log \left ( \sqrt{2}-1 \right )$
*$\displaystyle \int_{0}^{1}\frac{\log(2-x)}{\sqrt{2}-x} \, {\rm d}x = -{\rm Li}_2 \left ( 1+\frac{1}{\sqrt{2}} \right ) +{\rm Li}_2 \left ( 2+\sqrt{2} \right ) + i \pi \log 2 + \log 2 \log \left ( \sqrt{2}+1 \right )$
However I cannot simplify the dilogs here. Before this approach I had done the following:
\begin{align*}
\int_{0}^{1} \frac{\log(2-x)}{2-x^2} &=\frac{1}{2}\int_{0}^{1} \frac{\log(2-x)}{1- \frac{x^2}{2}} \, {\rm d}x \\
&=\frac{1}{2}\int_{0}^{1} \frac{\log(2-x)}{1- \left ( \frac{x}{\sqrt{2}} \right )^2} \, {\rm d}x \\
&= \frac{1}{2} \int_{0}^{1} \log(2-x) \sum_{n=0}^{\infty} \left ( \frac{x}{\sqrt{2}} \right )^{2n} \, {\rm d}x\\
&=\frac{1}{2} \int_{0}^{1} \log(2-x) \sum_{n=0}^{\infty} \frac{x^{2n}}{2^n} \, {\rm d}x \\
&= \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n} \int_{0}^{1} x^{2n} \log(2-x) \, {\rm d}x \\
&=??
\end{align*}
In this approach I am unable to evaluate the integral $\displaystyle \int_0^1 x^{2n} \log (2-x) \, {\rm d}x$.
As a side note if we replace that $2-x$ with $1-x$ then we have that:
\begin{align*}
\int_{0}^{1}\frac{\ln (1-x)}{2-x^2} \, {\rm d}x &=\frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^n} \int_{0}^{1}x^{2n} \ln (1-x) \, {\rm d}x \\
&\overset{(*)}{=} -\frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n} \frac{\mathcal{H}_{2n+1}}{2n+1}\\
&=-\frac{1}{2}\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)}
\end{align*}
$(*)$ since $\displaystyle \int_{0}^{1}x^n \ln (1-x) \, {\rm d}x = -\frac{\mathcal{H}_{n+1}}{n+1}$. The series is quite easy to calculate although it shall still contain polylogs.
Questions:
*
*Can we simplify those polylogs in the first attempt I have made?
*Do you see another way of evaluating the original integral?
|
Here's a way to calculate the integral that circumvents the use of polylogarithms entirely.
Let $I$ denote the value of the integral,
$$I:=\int_{0}^{1}\frac{\ln{\left(2-x\right)}}{2-x^{2}}\,\mathrm{d}x.$$
Using a clever choice of substitution, we find
$$\begin{align}
I
&=\int_{0}^{1}\frac{\ln{\left(2-x\right)}}{2-x^{2}}\,\mathrm{d}x\\
&=\int_{0}^{1}\frac{\ln{\left(\frac{2}{2-y}\right)}}{\left(2-y^{2}\right)}\,\mathrm{d}y;~~~\small{\left[x=\frac{2\left(1-y\right)}{2-y}\right]}\\
&=\int_{0}^{1}\frac{\ln{\left(2\right)}}{2-y^{2}}\,\mathrm{d}y-\int_{0}^{1}\frac{\ln{\left(2-y\right)}}{2-y^{2}}\,\mathrm{d}y\\
&=\ln{\left(2\right)}\int_{0}^{1}\frac{\mathrm{d}y}{2-y^{2}}-I.\\
\end{align}$$
Then,
$$\begin{align}
I
&=\frac12\ln{\left(2\right)}\int_{0}^{1}\frac{\mathrm{d}y}{2-y^{2}}\\
&=\frac{1}{2\sqrt{2}}\ln{\left(2\right)}\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\mathrm{d}t}{1-t^{2}};~~~\small{\left[y=\sqrt{2}\,t\right]}\\
&=\frac{1}{2\sqrt{2}}\ln{\left(2\right)}\operatorname{arctanh}{\left(\frac{1}{\sqrt{2}}\right)}.\\
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1880491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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|
Can this systems of equations be solved? There is a system of equations as follow:
$$\left[\begin{matrix}x\\ y\\ \end{matrix} \right] = \left[\begin{matrix}a& b\\b & c\\\end{matrix} \right]\left[\begin{matrix}\frac{1}{x}\\ \frac{1}{y}\\ \end{matrix} \right]$$
Given $ \left[\begin{matrix}a& b\\b & c\\\end{matrix} \right]$, can this system of equations be solved for $x$ and $y$?
And how to solve the extended case?
$$\left[\begin{matrix}x_1\\ x_2\\ \vdots \\x_n \end{matrix} \right] =
\left[\begin{matrix}a_{11}& a_{12} & \dots &a_{1n}\\a_{12} & a_{22} & \dots &a_{2n}\\ \vdots & \vdots & &\vdots \\a_{1n} & a_{2n} & \dots &a_{nn}\end{matrix} \right]\left[\begin{matrix}\frac{1}{x_1}\\ \frac{1}{x_2}\\ \vdots \\ \frac{1}{x_n}\\ \end{matrix} \right]$$
|
This is basically asking for $\frac{a}{x}+\frac{b}{y}=x$ and $\frac{b}{x}+\frac{c}{y}=y$. From the first equation, $y=\frac{xb}{x^2-a}$. You can substitute this into the second equation to get: $$\frac{b}{x}+\frac{c(x^2-a)}{bx}=\frac{xb}{x^2-a}$$
This should be translatable into a polynomial. However, the smallest polynomial that you can multiply this by to remove all denominators is $bx(x^2-a)=bx^3-abx$. Doing this gives $$b^2(x^2-a)+c(x^2-a)^2=b^2x^2.$$
This results in a quartic polynomial, which is always solvable, though tedious. There's probably a shortcut that I'm missing, but at least this is a solvable problem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1881765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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|
Decide if the series converges and prove it using comparison test: $\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}$
Decide if the series converges and prove it using comparison test:
$\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}$
$$\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}< \frac{k^{2}+k}{k^{4}+k^{3}} < \frac{k^{2}}{k^{4}} \leq \frac{1}{k^{2}}$$
We know (from our readings) that $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$ is a converging series.
Thus the complete series will converge.
Did I do everything correctly?
|
IMHO this could be written more precisely. For example the second inequality does not look right. I would suggest the following.
$$\frac{3k^2+k+1}{k^4+k^3+4}=\frac{k^2}{k^4} \cdot \frac{3+1/k +1/k^2}{1+1/k} \frac{1}{k^2} \cdot \frac{3 + 1+ 1}{1+1}. $$
Now one can conclude that each of the entries of the sum is smaller or equal to $\frac{5}{2k^2}$. Therefore, according to your readings, the sum converges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1882141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Calculate $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$
Calculate $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$
\begin{align}\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}&=3\sum_{k=2}^{\infty}\frac{1}{5^{k-1}}\\&=3\sum_{k=2}^{\infty}\frac{1}{5^{k}}\cdot\frac{1}{5^{-1}}\\&=3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5\\&= 15\left( \sum_{k=0}^{\infty}\left( \frac{1}{5} \right )^{k}-2 \right )\\&=15\left( \frac{1}{1-\frac{1}{5}}-2 \right )\\&=-\frac{45}{4}\end{align}
Did I do it correctly?
|
Your summation should be
$$15\times \sum \limits^{\infty }_{k=2}\left( \frac{1}{5} \right) ^{k}=15\times \frac{\left( \frac{1}{5} \right) }{1-\frac{1}{5} } ^{2}=15\times \frac{1}{20} =\frac{3}{4} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1882243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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|
How to solve $\int_0^{2\pi} \frac{dt}{3+\sin(t)}$? Got stuck with the integral.
I rewrote it as $$I=\int_0^{2\pi} \frac{2idt}{6i+e^{it}-e^{-it}},$$
then I took $z:=e^{it}\implies dz=ie^{it}dt\implies dt = -\frac{idz}{z}$, so I'd get:
$$\int_\gamma \frac{2dz}{6i+z-\frac{1}{z}}=\int_\gamma \frac{2zdz}{6iz+z^2-1}.$$
It can be observed that $-i(2\sqrt{2}\pm 3)$ are the singular points of the integrand.
Then the residues are $\pm \frac{i}{4\sqrt{2}}$. And the integral would become:
$$ I=0 $$
as per the Cauchy Residue Theorem.
Where am I mistaken? Appreciate your input.
|
A real-analytic way is to break the integration range into sub-intervals with length $\frac{\pi}{2}$ to get
$$ I = \int_{0}^{\pi/2}\left(\frac{1}{3+\sin z}+\frac{1}{3+\cos z}+\frac{1}{3-\sin z}+\frac{1}{3-\cos(z)}\right)\,dz \tag{1}$$
from which:
$$ I = \int_{0}^{\pi/2}\frac{12}{9-\cos^2 z}\,dz =\int_{0}^{+\infty}\frac{12\,dt}{9(1+t^2)-1}\tag{2}$$
through the substitution $z=\arctan t$. The last integral is straightforward to compute:
$$ I = \int_{0}^{+\infty}\frac{12\,dt}{9t^2+8}=\frac{2\sqrt{2}}{3}\int_{0}^{+\infty}\frac{12\,du}{8u^2+8}=\frac{2\sqrt{2}}{3}\cdot\frac{12}{8}\cdot\frac{\pi}{2}=\color{red}{\frac{\pi}{\sqrt{2}}}.\tag{3}$$
An alternative way may be the following. We have:
$$ I = \sum_{n\geq 0}\frac{1}{3^{n+1}}\int_{0}^{2\pi}\sin^n(x)\,dx = \sum_{n\geq 0}\frac{1}{3^{2n+1}}\int_{0}^{2\pi}\sin^{2n}(x)\,dx\tag{4} $$
so:
$$ I = \frac{2\pi}{3} \sum_{n\geq 0}\frac{\binom{2n}{n}}{36^n}=\frac{2\pi}{3\sqrt{1-\frac{4}{36}}}=\color{red}{\frac{\pi}{\sqrt{2}}}\tag{5}$$
also follows from a well-known generating function.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1883026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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|
Convex Quadrilaterals Let $n>4$.
In how many ways can we choose $4$ vertices of a convex $n$-gon so as to form a convex quadrilateral, such that at least $2$ sides of the quadrilateral are sides of the $n$-gon?
Explain your answer, which should be expressed in terms of $n$.
|
We need to count the number of ways to choose 4 vertices so that at least two pairs of vertices are adjacent.
There are $\dbinom{n-3}{4}-\dbinom{n-5}{2}$ ways to choose them so that none are adjacent,
and there are $n\dbinom{n-5}{2}$ ways to choose them so that exactly two are adjacent;
so this gives a total of $\displaystyle\binom{n}{4}-\binom{n-3}{4}-(n-1)\binom{n-5}{2}=\frac{n(3n-13)}{2}$ ways to choose the vertices.
Alternate solution:
1) There are $n$ ways to choose all 4 vertices adjacent
2) There are $n(n-5)$ ways to choose the vertices so that exactly 3 are
adjacent.
3) There are $\displaystyle\frac{n(n-5)}{2}$ ways to choose the vertices so that there are 2 separate pairs which are adjacent.
This gives a total of $\displaystyle n+n(n-5)+\frac{n(n-5)}{2}=\frac{n(3n-13)}{2}$ possibilities.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1883638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
What is the expected number of red apples left when all the green apples are picked? We have 4 green apples and 60 red apples. Each time we pick one out without replacement. Then what is the expected number of red apples left when all 4 green apples are picked?
|
Suppose we have $G$ green apples and $R$ red apples and ask about the
number of red apples left when all green apples have been picked.
The probability of the last green apple being picked as apple number
$q$ is
$${R+G\choose R}^{-1} {q-1\choose G-1}.$$
We now verify that this is indeed a proability. We use the Egorychev
method to evaluate the sum
$${R+G\choose R}^{-1} \sum_{q=G}^{R+G} {q-1\choose G-1}.$$
Introducing
$${q-1\choose G-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^G} (1+z)^{q-1} \; dz$$
we obtain for the sum of the probabilities
$${R+G\choose R}^{-1}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^G} \frac{1}{1+z}
\sum_{q=G}^{R+G} (1+z)^q \; dz
\\ = {R+G\choose R}^{-1}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^G} (1+z)^{G-1}
\sum_{q=0}^{R} (1+z)^q \; dz
\\ = {R+G\choose R}^{-1}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^G} (1+z)^{G-1}
\frac{(1+z)^{R+1}-1}{1+z-1}
\; dz
\\ = {R+G\choose R}^{-1}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{G+1}} (1+z)^{G-1}
((1+z)^{R+1}-1)
\; dz$$
This has two pieces namely $[z^G] (1+z)^{R+G}$ and $[z^G] (1+z)^{G-1}$
and we get
$${R+G\choose R}^{-1} {R+G\choose G} = 1.$$
So we indeed have a probability here.
Continuing with the expectation we find
$${R+G\choose R}^{-1} \sum_{q=G}^{R+G} {q-1\choose G-1}
(R+G-q)
\\ = R+G
- {R+G\choose R}^{-1} \sum_{q=G}^{R+G} q {q-1\choose G-1}
\\ = R+G
- G {R+G\choose R}^{-1} \sum_{q=G}^{R+G} \frac{q}{G} {q-1\choose G-1}
\\ = R+G
- G {R+G\choose R}^{-1} \sum_{q=G}^{R+G} {q\choose G}.$$
Introducing
$${q\choose G} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{G+1}} (1+z)^{q} \; dz$$
we obtain for the sum component of the expectation
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{G+1}}
\sum_{q=G}^{R+G} (1+z)^q
\; dz
\\ \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{G+1}} (1+z)^G
\sum_{q=0}^{R} (1+z)^q
\; dz
\\ \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{G+1}} (1+z)^G
\frac{(1+z)^{R+1}-1}{1+z-1}
\; dz
\\ \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{G+2}} (1+z)^G
((1+z)^{R+1}-1)
\; dz.$$
We have the two pieces $[z^{G+1}] (1+z)^{G+R+1}$ and $[z^{G+1}]
(1+z)^G$ and obtain
$${R+G+1\choose G+1}.$$
Returning to the expectation we have
$$R+G - G {R+G\choose R}^{-1} {R+G+1\choose G+1}
\\ = R+G - G
\frac{(R+G+1)! \times R! \times G!}{(R+G)! \times (G+1)! \times R!}.$$
This simplifies to
$$\bbox[5px,border:2px solid #00A000]{
R+G - G \frac{R+G+1}{G+1}
= R - G \frac{R}{G+1}
= \frac{R}{G+1}.}$$
In particular for $4$ green apples and $60$ red apples we get
$$64- 4\frac{65}{5} = 64 - 4\times 13
= 12 = \frac{60}{5}.$$
This formula may be verified with the following Maple code (warning --
total enumeration, use on small values for the number of apples)
with(combinat);
X :=
proc(G, R)
option remember;
local perm, src, pos, seen, res;
src :=
[seq(GG, q=1..G), seq(RR, q=1..R)];
res := 0;
for perm in permute(src) do
seen := 0;
for pos to R+G do
if perm[pos] = GG then
seen := seen + 1;
if seen = G then
break;
fi;
fi;
od;
res := res + (R+G)-pos;
od;
res/binomial(R+G, R);
end;
Y :=(G, R) -> R/(G+1);
|
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"url": "https://math.stackexchange.com/questions/1883733",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.