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Find the maximum value of the expression $ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ Find the maximum value of the expression $ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ where $x, y, z$ are real numbers satisfying the condition $x + y + z = 1 $.
My Attempt
$$
\begin{align}
\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}&=\frac{1}{x+\frac{1}{x}}+\frac{1}{y+\frac{1}{y}}+\frac{1}{z+\frac{1}{z}}\\&\geq \frac{(1+1+1)^2}{x+\frac{1}{x}+y+\frac{1}{y} + z+\frac{1}{z}}\\&=\frac{9}{x+y+z +(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}
\end{align}
$$
From AM-HM
$$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 3^2 $$
$$\implies (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 9 $$
$$\implies (x+y+z)+ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 10$$
From this point i am not being able o proceed as inequality signs are getting mixed. Can anyone help out? Thanks in advance.
|
Ok rugi I will solve it using derivatives for your question consider x=y=z because for any condition like x+y+z=a we need to consider x=y=z to get maxima .So answer is 9÷10=0.9.This is for a <3.And when the degree in the numerator is less than the degree in denominator in which the value is to be found out
|
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"timestamp": "2023-03-29T00:00:00",
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|
Conjecture regarding integrals of the form $\int_0^\infty \frac{(\log{x})^n}{1+x^2}\,\mathrm{d}x$. I have been playing around a bit with integrals of the form $$I(n)=\int_0^\infty \frac{(\log{x})^n}{1+x^2}\,\mathrm{d}x,\,\,n\in\mathbb{Z}^+,$$ and I am trying to obtain a closed form solution for $I(n).$ I believe the special cases $I(1)$ and $I(2)$ are somewhat well-known, but I will go over them. When $n=1,$ we have $$I(1)=\int_0^\infty \frac{\log{x}}{1+x^2}\,\mathrm{d}x=\int_0^1 \frac{\log{x}}{1+x^2}\,\mathrm{d}x+\int_1^\infty \frac{\log{x}}{1+x^2}\,\mathrm{d}x.$$ This can be easily shown to be zero by performing the substitution $x=1/y,$ which will yield $$\int_0^1 \frac{\log{x}}{1+x^2}\,\mathrm{d}x=-\int_1^\infty \frac{\log{x}}{1+x^2}\,\mathrm{d}x.$$ Thus, $I(1)=0.$ Clearly, this can be generalized to all odd integers, and $I(2n+1)=0.$ In the case of $n=2$, first observe through the same substitution as above that $$\int_0^1 \frac{(\log{x})^2}{1+x^2}\,\mathrm{d}x=\int_1^\infty \frac{(\log{x})^2}{1+x^2}\,\mathrm{d}x.$$ This implies that $$I(2)=2\int_0^1 \frac{(\log{x})^2}{1+x^2}\,\mathrm{d}x,$$ which is an easier integral to work with. Performing the substitution $x=e^{-y}$ yields $$I(2)=2\int_0^\infty y^2\left(e^{-y}-e^{-3y}+e^{-5y}-\cdot\cdot\cdot\right)\,\mathrm{d}y.$$ Using the identity $$\int_0^\infty x^2 e^{-ax}=\frac{2}{a^3},$$ we obtain $$I(2)=4\left(\frac{1}{1^3}-\frac{1}{3^3}+\frac{1}{5^3}-\cdot\cdot\cdot\right)=4\cdot\frac{\pi^3}{32}=\frac{\pi^3}{8}.$$
I'm not sure how this infinite series is evaluated, but I found this result in a book. I used Mathematica to check a few more values, and I found that $I(4)=5\pi/32,\,I(6)=61\pi/128,\,$ and $I(8)=1385\pi/512.$ Clearly the pattern is $$I(2n)=A_{2n}\left(\frac{\pi}{2}\right)^{2n+1},$$ where $A_{2n}$ is some constant. It turns out that these constants are the Euler numbers, which are the coefficients $E_k$ corresponding to the series $$\operatorname{sech}x=\sum_{k=0}^\infty\frac{E_k}{k!}x^k.$$
All Euler numbers corresponding to odd $n$ are zero, and the first few even Euler numbers are $E_0=1,\, E_2=-1,\, E_4=5, \,E_6=-61,\,$ and $E_8=1385.$ Thus, I have conjectured that $$I(2n)=(-1)^n E_{2n} \left(\frac{\pi}{2}\right)^{2n+1}$$ for all $n\in\mathbb{Z}^+.$ I suppose this could be extended to $n\in\mathbb{Z}_{\geq 0}$ thusly:
$$I(n)=i^n E_n \left(\frac{\pi}{2}\right)^{n+1},$$ since $E_n=0$ for odd $n.$ So the question, of course, is how to prove this. I tried generalizing the method I used for $I(2),$ and I found that $$\int_0^\infty x^n e^{-ax}=\frac{n!}{a^{n+1}},\,\,n\in\mathbb{Z}_{\geq 0}.$$ Using this, I obtained $$I(2n)=n!\left(\frac{1}{1^{n+1}}-\frac{1}{3^{n+1}}+\frac{1}{5^{n+1}}-\cdot\cdot\cdot\right)=n!\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^{n+1}}.$$ Mathematica wasn't able to evaluate this sum, even for the case of $n=2.$ It gives some expression involving multiple zeta functions, with which I have no experience. Even if we can't prove this, I would be interested to know why the Euler numbers might appear here. Any help would be greatly appreciated.
Edit:
As Claude Leibovici helped point out, there final series expression should be $$I(2n)=2(2n)!\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}=\frac{(2n)!}{2^{4n+1}}\left[\zeta\left(2n+1, \frac{1}{4}\right)-\zeta\left(2n+1, \frac{3}{4}\right)\right].$$
|
\begin{align*}J_{n}&=\int_0^\infty \frac{\ln^{2n} x}{1+x^2}dx\\
J_0&=\int_0^1 \frac{1}{1+x^2}dx=\frac{\pi}{2}\\
K_{n}&=\int_0^\infty\int_0^\infty\frac{\ln^{2n}(xy)}{(1+x^2)(1+y^2)}dxdy\\
&=\sum_{k=0}^{n}\binom{2n}{2k}J_kJ_{n-k}\\
J_n=&\overset{u(x)=yx}=\int_0^\infty\int_0^\infty\frac{y\ln^{2n}u}{(y^2+u^2)(1+y^2)}dudy \\
&=\frac{1}{2}\int_0^\infty \frac{\ln^{2n} u}{1-u^2}\left[\ln\left(\frac{y^2+u^2}{y^2+1}\right)\right]_0^\infty du\\
&=-\int_0^\infty \frac{\ln^{2n+1} u}{1-u^2}du\\
&=-2\int_0^1 \frac{\ln^{2n+1} u}{1-u^2}du\\
&=\underbrace{\int_0^1 \frac{2u\ln^{2n+1} u}{1-u^2}du}_{z=u^2}-2\int_0^1 \frac{\ln^{2n+1} u}{1-u}du\\
&=\left(\frac{1}{2^{2n+1}}-2\right)\int_0^1 \frac{\ln^{2n+1} u}{1-u}du\\
&=2(2n+1)!\left(1-\frac{1}{2^{2n+2}}\right)\zeta(2n+2)\\
\end{align*}
Therefore,
\begin{align}\boxed{J_n=\frac{1}{\pi}\left(2(2n+1)!\left(1-\frac{1}{2^{2n+2}}\right)\zeta(2n+2)-\sum_{k=1}^{n-1}\binom{2n}{2k}J_kJ_{n-k}\right)}\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that
$x^5 + \frac{1}{x^5} = 1$ as well.
|
Notice that: $$x + \frac{1}{x} = 1 \;\implies\; x^2 - x + 1 = 0 \;\implies\; x^3 + 1 = (x+1)(x^2-x+1) = 0 \;\implies\; x^3=-1$$
Then $x^6=1$ so: $$x^5 + \frac{1}{x^5} = \frac{x^6}{x} + \frac{x}{x^6} = \frac{1}{x} + x = 1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$ I've tried this but didn't get the answer :
Let $S=2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$
Using $n^3-(n-1)^3 = 3n^2-3n+1$,
\begin{align}
S &= 3(2014^2)-3(2014)+1+3(2012^2)-3(2012)+1+\ldots+3(2^2)-3(2)+1
\\&= 3\left ( 2014(2013)+2012(2011)+2010(2009)+ \ldots+2(1) \right ) + 1(1007)
\\&= 3\left ( \sum_{n=1}^{1007}2n(2n-1) \right )+1007\\
=& \left ( \sum_{n=1}^{1007}4n^2-\sum_{n=1}^{1007}2n \right )+1007
\\=&\frac{12(1007)(1008)(2015)}{6}-\frac{2(1007)(1008)(3)}{2}+1007
\end{align}
This is divisible by $1007$ but not by $1007^2$ which is the correct answer. Where have I gone wrong ?
|
$$\begin{align}
\sum_{n=1}^{2m}(-1)^n n^3&=\sum_{n=1}^m (2n)^3-(2n-1)^3\\
&=\sum_{n=1}^m 12n^2-6n+1\\
&=\sum_{n=1}^m 24 \binom n2+6\binom n1+1\\
&=24\binom {m+1}3+6\binom {m+1}2+m\\
&=m\; \big[\;4(m+1)(m-1)+3(m+1)+1\;\big]\\
&=m^2(4m+3)\\
\end{align}$$
which is divisible by $m^2$.
Putting $m=1007$ gives the answer required.
|
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|
Olympic problem - Probability A jar contains 3 cookies filled with chocolate, 4 cookies filled with strawberry and 3 cookies filled with vainilla. What is the probability of choosing 3 cookies with the same filling from the jar?
EDIT: I've tried this:
What I want to find is:
$$\frac{Favorable\ outcomes}{Possible\ outcomes}$$
Since I only care about choosing three cookies, I think the possible outcomes are $\binom{10}{3}$.
Now, the favorable outcomes should be: 6.
Only one option to get the three chocolate filled cookies, same with the vanilla filling. I have $\binom{4}{3}=4$ ways to choose three cockies from the strawberry cookies.
Now, the probability I want should be:
$$\frac{1+1+\binom{4}{3}}{\binom{10}{3}}=\frac{6}{120}=\frac{1}{20}.$$
But I'm not sure if I'm correct.
|
Look at the problem as a choice tree and ask yourself: how many branches do I have that satisfy the conditions above?
In this case you have a tree with depth three (3). How many branch ends are there to satisfy your condition, and what is the probability of getting to each of them?
P(3 vanilla) = $\frac{3}{10} * \frac{2}{9} * \frac{1}{8} = \frac{6}{720}$
P(3 straberry) = $\frac{4}{10} * \frac{3}{9} * \frac{2}{8} = \frac{24}{720}$
P(3 chocolate) = $\frac{3}{10} * \frac{2}{9} * \frac{1}{8} = \frac{6}{720}$
Sum it up and you get $\frac{36}{720} = \frac{1}{20}$
|
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|
How would you find the trigonometric roots of a cubic?
Question: How would you find the roots of the cubic$$x^3+x^2-10x-8=0\tag{1}$$
I'm not too sure where to begin. I'm thinking of somehow, implementing $\cos 3\theta=4\cos^3\theta-3\cos\theta$.
I've tried substituting $x$ with $t+t^{-1}$, but didn't get anywhere, and using Vieta's trigonometric solution formula, I got
$$x_1=\frac {2\sqrt{31}}3\cdot\cos\left(\frac {\arccos \frac {2}{\sqrt{31}}}3\right)-\frac 13\\x_2=\frac {2\sqrt{31}}3\cdot\cos\left(\frac {2\pi+\arccos\frac {2}{\sqrt{31}}}3\right)-\frac 13\\x_3=\frac {2\sqrt{31}}{3}\cdot\cos\left(\frac {4\pi+\arccos\frac 2{\sqrt{31}}}3\right)-\frac 13$$
But that's not the form I want. I'm looking for a form of $2\left(\cos\frac {\text{something}}{31}+\cos\frac {\text{something}}{31}+\cos\frac {\text{something}}{31}\right)$.
I've spent so much time, that I'm practically burnt out. -.-
|
If you wish to express as roots of cubics the sums of trigonometric functions with arguments that use a prime of form $p=6\color{blue}n+1$, then the number of addends is $\color{blue}n$. Thus, the reason you couldn't find $p=31$ was that you should have used $\color{blue}5$ addends.
$p=31$
A root of
$$x^3+x^2-(2\times\color{blue}5)x-8=0\tag1$$
(note the $5$) is given by
$$x_1 = 2\left(\cos\tfrac {2\pi}{31}+\cos\tfrac {4\pi}{31}+\cos\tfrac {8\pi}{31}+\cos\tfrac {16\pi}{31}+\cos\tfrac {32\pi}{31}\right) =3.083872\dots$$
More succinctly, let $\displaystyle\beta=\frac{2\pi}{31}$, then,
$$x_1=2\sum_{k=1}^5\cos\big(2^k\times\beta\big)=3.083872\dots\\x_2=2\sum_{k=1}^5\cos\big(2^k\times3\beta\big)=-0.786802\dots\\x_3=2\sum_{k=1}^5\cos\big(2^k\times5\beta\big)=-3.29707\dots$$
$p=43$
Similarly, given the cubic,
$$x^3+x^2-(2\times\color{blue}7)x+8=0\tag2$$
Let $\displaystyle\gamma=\frac{2\pi}{43}$, then the roots are,
$$x_1=2\sum_{k=1}^7\cos\big(2^k\times\gamma\big)=2.88824\dots\\x_2=2\sum_{k=1}^7\cos\big(2^k\times3\gamma\big)=0.615072\dots\\x_3=2\sum_{k=1}^7\cos\big(2^k\times7\gamma\big)=-4.50331\dots$$
though not all primes $p=6n+1$ will have such neat cubic roots. The family $p=31,43,109,\dots$ is discussed in this post.
P.S. See also mercio's general answer here which uses cosets.
|
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|
How to solve $\sin(x) + 2\sqrt{2}\cos x =3$ How to solve $\sin(x) + 2\sqrt{2}\cos x=3$ ?
What is general method for doing these kind of questions?
Thanks
|
$$\sin x+2\sqrt { 2 } \cos { x } =3\\ 2\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } +2\sqrt { 2 } \left( \cos ^{ 2 }{ \frac { x }{ 2 } -\sin ^{ 2 }{ \frac { x }{ 2 } } } \right) =3\sin ^{ 2 }{ \frac { x }{ 2 } +3\cos ^{ 2 }{ \frac { x }{ 2 } } } } } \\ \left( 3+2\sqrt { 2 } \right) \sin ^{ 2 }{ \frac { x }{ 2 } } -2\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } } } +\left( 3-2\sqrt { 2 } \right) \cos ^{ 2 }{ \frac { x }{ 2 } } =0\\ \left( 3+2\sqrt { 2 } \right) \tan ^{ 2 }{ \frac { x }{ 2 } } -2\tan { \frac { x }{ 2 } } +\left( 3-2\sqrt { 2 } \right) =0\\ \tan { \frac { x }{ 2 } =\frac { 2\pm \sqrt { 3 } }{ 2\left( 3+2\sqrt { 2 } \right) } } \\ \\ \\ \\ $$
Can you take here?
|
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|
Mathemathic induction proof I need to prove that $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}=\frac{n}{n+1}$$
Here is what I tried:
\begin{align}
& \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}+\frac{1}{n+1(n+2)} \\[10pt]
= {} & \frac{n}{n+1}+\frac{1}{n+1(n+2)} = \frac{(n+1)(n+2)+(n+1)}{n^2(n+2)}
\end{align}
I am stuck from here, I feel there must be someway to simplify the last part to get to
$$\frac{n+1}{n+2}$$
|
Hint: write
$$\frac1{k(k+1)}=\frac1k-\frac1{k+1}$$
and now do the telescopic sum carefully.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How would you go about find the values of a,b and c such that they f(0) = f'(0) = f''(0) = 1 ? The function is:
\begin{align}
f( x) &= \frac{a+bx}{1+cx} \\\\
\end{align}
The question also tells us that the function and both it's first and second derivatives are equal to the corresponding values of:
$$\ e^{x}$$
Would finding the values of a,b and c simply be a trial and error process?
|
Given $f(0) = f'(0) = f''(0) = 1$ with $$f(x) = \frac{a+bx}{1+cx}$$ quickly leads to $a=1$ and $$f(x) = \frac{1+bx}{1+cx} = \frac{b}{c} - \frac{b-c}{c} \, \frac{1}{1+cx}.$$
Taking derivatives provides:
\begin{align}
f'(x) &= \frac{b-c}{(1+cx)^{2}} \\
f''(x) &= - \frac{2c (b-c)}{(1+cx)^{3}}.
\end{align}
Using the remaining conditions leads to $1 = b-c$ and $1 = -2c(b-c) = -2c$ which is $b = \frac{1}{2}$ and $c = - \frac{1}{2}$. The function then becomes
$$f(x) = \frac{2+x}{2-x}$$
|
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|
Are two sequences equal if the sums and sums of squares are equal? Are two sequences $(x_i)_{i=1,\ldots,n}$ and $(y_i)_{i=1,\ldots,n}$ equal if $\sum_{i=1}^nx_i=\sum_{i=1}^ny_i$ and $\sum_{i=1}^nx_i^2=\sum_{i=1}^ny_i^2$?
|
Something interesting. On all $3 \times 3$ magic squares, the sums of the top and bottom rows is the same (of course) but, also, the sums of the squares of the top and bottom rows is the same. For example
\begin{array}{|c|c|c|}
\hline
8 & 3 & 4 \\
\hline
1 & 5 & 9 \\
\hline
6 & 7 & 2 \\
\hline
\end{array}
$8 + 3 + 4 = 6 + 7 + 2 = 15$
$8^2 + 3^2 + 4^2 = 6^2 + 7^2 + 2^2 = 89$
Also
$8 + 1 + 6 = 4 + 9 + 2 = 15$
$8^2 + 1^2 + 6^2 = 4^2 + 9^2 + 2^2 = 101$
This seems to work for some families of larger order magic squares, but I haven't been able to prove it.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding an unknown coefficient of a polynomial given a factor Q:Find the value of $a$ given that $x^2+1$ is a factor of $x^4-3x^3+3x^2+ax+2$
No idea where to start, I was going to use the factor theorem but it didn't work out.
Question from year 10 Cambridge maths textbook
|
Divide your polynomial by $x^{2} + 1$ and set the remainder equal to zero.
Or, since you are only interested in the remainder, set $x^{2} = -1$ in the polynomial to get that the remainder is
$$
(x^2)^{2}-3 x x^2+3 x^2+ax+2
=
(-1)^{2} + 3 x - 3 + a x + 2
=
(a + 3) x.
$$
So $x^{2} + 1$ divides your polynomial iff $a = -3$.
|
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|
Prove the following Inequality. Given that $\alpha,\beta$ and $\gamma\in (0,\pi)$ and $\alpha+\gamma+\beta=\pi$, show that $$\cos\alpha+\cos\gamma+\sin\beta\leq\frac{3\sqrt3}{2}.$$
Now I am aware of the following Inequality: $\cos\alpha+\cos\gamma+\cos\beta\leq\frac{3}{2}$, but notice that the term $\cos\beta$ has been replaced with $\sin\beta$. I also tried substituting $\sin\beta=\sin(\pi-(\alpha+\gamma))=\sin(\alpha+\gamma)$ and expanding the resulting expression but I was unable to deduce anything meaningful.
|
Because if $\cos\frac{\beta}{2}=x$ by AM-GM we obtain:
$$\cos\alpha+\cos\gamma+\sin\beta=2\sin\frac{\beta}{2}\cos\frac{\alpha-\gamma}{2}+\sin\beta\leq2\sin\frac{\beta}{2}+\sin\beta=$$
$$=2\sqrt{1-x^2}(1+x)=2\sqrt{(1-x)(1+x)^3}=2\sqrt{27(1-x)\left(\frac{1}{3}+\frac{x}{3}\right)^3}\leq$$
$$\leq2\sqrt{27\left(\frac{1-x+3\left(\frac{1}{3}+\frac{x}{3}\right)}{4}\right)^4}=\frac{3\sqrt3}{2}$$
|
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|
Let $\alpha$, $\beta$, and $\gamma$ be acute angles such that $\alpha$ + $\beta$ = $\gamma$. Show that Let $\alpha$, $\beta$, and $\gamma$ be acute angles such that $\alpha$ + $\beta$ = $\gamma$. Show that
cos$\alpha$ + cos$\beta$ + cos$\gamma$ - 1 $\geq$ $2\sqrt{\cos\alpha\times\cos\beta\times\cos\gamma}$.
Here's what I've tried. I know that $0 < \alpha, \beta, \gamma < \pi/2$ since they are acute angles. So I substituted $\alpha+\beta$ for $\gamma$
$\cos\alpha$ + $\cos\beta$ + $\cos(\alpha+\beta)$- 1 $\geq$ $2\sqrt{\cos\alpha\cos\beta\cos(\alpha+\beta)}$.
$\cos\alpha + \cos\beta + \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)- 1 \geq 2\sqrt{\frac{1}{2}[\cos(\alpha-\beta)+\cos(\alpha+\beta)]\cos(\alpha+\beta)}.$
$\cos\alpha + \cos\beta + \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)- 1 \geq 2\sqrt{\frac{1}{2}[\cos(\alpha-\beta)\cos(\alpha+\beta)+\cos^{2}(\alpha+\beta)]}$
$\cos\alpha + \cos\beta + \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)- 1 \geq 2\sqrt{\frac{1}{4}[\cos(-2\beta)+\cos(2\alpha)]+\cos^{2}(\alpha+\beta)}$
I got stuck from there and I'm not sure where to go from here. Any help would be appreciated.
|
Using that
$$
\cos\alpha\cos\beta-\cos\gamma = \sin\alpha\sin\beta
= 4 \sin\frac\alpha2 \cos\frac\alpha2 \sin\frac\beta2 \cos\frac\beta2 \ge 0
$$
and
$$
(1-\cos\alpha)(1-\cos\beta) = 4 \sin^2\frac\alpha2 \sin^2\frac\beta2,
$$
a possible proof:
$$
\cos\gamma - 2\sqrt{\cos\alpha\cos\beta}\cdot\sqrt{\cos\gamma}+
\cos\alpha\cos\beta
\stackrel{?}{\ge}
(1-\cos\alpha)(1-\cos\beta)
$$
$$
\Big(\underbrace{\sqrt{\cos\alpha\cos\beta}-\sqrt{\cos\gamma}}_{\ge0}\Big)^2
\stackrel{?}{\ge}
4\sin^2\frac\alpha2 \sin^2\frac\beta2
$$
$$
\sqrt{\cos\alpha\cos\beta}-\sqrt{\cos\gamma}
\stackrel{?}{\ge}
2 \sin\frac\alpha2 \sin\frac\beta2
$$
$$
\cos\alpha\cos\beta
\stackrel{?}{\ge}
\Big(\sqrt{\cos\gamma} + 2 \sin\frac\alpha2 \sin\frac\beta2\Big)^2
$$
$$
\cos\alpha\cos\beta-\cos\gamma
\stackrel{?}{\ge}
4 \sin\frac\alpha2 \sin\frac\beta2 \sqrt{\cos\gamma}
+ 4 \sin^2\frac\alpha2 \sin^2\frac\beta2
$$
$$
4 \sin\frac\alpha2 \cos\frac\alpha2 \sin\frac\beta2 \cos\frac\beta2 \stackrel{?}{\ge}
4 \sin\frac\alpha2 \sin\frac\beta2 \sqrt{\cos\gamma}
+ 4 \sin^2\frac\alpha2 \sin^2\frac\beta2
$$
$$
\cos\frac\alpha2 \cos\frac\beta2
- \sin\frac\alpha2 \sin\frac\beta2
\stackrel{?}{\ge}
\sqrt{\cos\gamma}
$$
$$
\cos\frac\gamma2 \stackrel{?}{\ge}
\sqrt{\cos\gamma}
$$
$$
\sqrt{\frac{1+\cos\gamma}2} \stackrel{?}{\ge}
\sqrt{\cos\gamma}
$$
$$
1 \ge \cos\gamma.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1986571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
Prove that $\lim ( \sqrt{n^2+n}-n) = \frac{1}{2}$ Here's what I have so far:
Given $\epsilon > 0$, we want to find N such that $\sqrt{n^2+n}-n < \epsilon$ for all $n>N$. And so:
$( \sqrt{n^2+n}-n-\frac{1}{2}) \cdot \frac{\sqrt{n^2+n}-(n+\frac{1}{2})}{\sqrt{n^2+n}-(n+\frac{1}{2})}$
$= \frac{(n^2+n)-(n+\frac{1}{2})}{\sqrt{n^2+n} + (n+\frac{1}{2})}$
And I'm not sure how to go on from here. Help would be appreciated.
|
Hint:
\begin{align}
\sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n}
\end{align}
and
\begin{align}
\left|\frac{n}{\sqrt{n^2+n}+n} -\frac{1}{2}\right| = \left|\frac{n-\sqrt{n^2+n}}{2\sqrt{n^2+n}+2n}\right| = \frac{n}{2(\sqrt{n^2+n}+n)^2} \leq \frac{n}{8n^2}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1988705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Find the power series of $f(x)=\frac{1}{x^2+x+1}$ I want to find the power series of $$f(x)=\frac{1}{x^2+x+1}$$
How can I prove the following?
$$f(x)=\frac{2}{\sqrt{3}} \sum_{n=0}^{\infty} \mathrm{sin}\frac{2\pi(n+1)}{3} x^n \,\,\,\, |x|<1$$
In particular I would like to know how to proceed in this case. The polinomial $x^2+x+1$ has no roots so here I cannot use partial fraction decomposition: what method should I use?
|
The polynomial $x^2+x+1=\Phi_3(x)$ has no real roots, but it vanishes at $x=e^{\pm\frac{2\pi i}{3}}$. In particular, by setting $\omega=\exp\left(\frac{2\pi i}{3}\right)$ and $\overline{\omega}=\omega^2=\exp\left(\frac{4\pi i}{3}\right)$,
$$ \frac{1}{x^2+x+1} = \frac{1}{(x-\omega)(x-\omega^2)} = \frac{i\omega^2}{\sqrt{3}}\cdot\frac{1}{1-\omega^2 x}-\frac{i\omega}{\sqrt{3}}\cdot\frac{1}{1-\omega x} $$
where the RHS, expanded as the difference between two geometric series, equals
$$ \frac{i}{\sqrt{3}}\sum_{n\geq 0}\left(\omega^{2n+2}-\omega^{n+1}\right)x^n =\frac{2}{\sqrt{3}}\sum_{n\geq 0}\sin\left(\frac{2\pi(n+1)}{3}\right)x^n$$
as wanted. That clearly simplifies, since
$$ \frac{1}{1+x+x^2}=\frac{1-x}{1-x^3} = \sum_{m\geq 0}\left(x^{3m}-x^{3m+1}\right),$$
too, and the Taylor series at the origin is unique.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1990704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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|
Finding maximum value from a statement/equation The statement has values of $x$ and $y$ as positive integers: $$\sqrt{x} - \sqrt{11} = \sqrt{y}$$
I have to find the maximum possible values of $\frac{x}{y}$, this what I have done so far:
$$x = (\sqrt{y} + \sqrt{11})^2$$
$$y = (\sqrt{x} - \sqrt{11})^2$$
therefore: $$\frac{x}{y} = \frac{(\sqrt{y} + \sqrt{11})^2}{(\sqrt{x} - \sqrt{11})^2}$$
...
$$\frac{x}{y} = \frac{y + 11 + 2\sqrt{y}\sqrt{11}}{x + 11 - 2\sqrt{x}\sqrt{11}} = \frac{y+\sqrt{y}}{x-\sqrt{x}} + 1$$
and from here I'm not sure what to do...
(Again better title suggestions are also welcome)
|
Dividing by $\sqrt{y}$ we get
$$\sqrt{\frac{x}{y}}=\sqrt{\frac{11}{y}}+1$$
and by squaring
$$\frac{x}{y}=\frac{11}{y}+2\sqrt{\frac{11}{y}}+1$$
So $\sqrt{\frac{11}{y}}$ should be rational and y is integer then for maximum we take y=11. so you get
$\frac{x}{y}=4$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1991477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
Prove $\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2$ I stumbled upon the interesting definite integral
\begin{equation}
\int\limits_0^1 \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2
\end{equation}
Here is my proof of this result.
Let $u=\sin^{-1}(x)$ then integrate by parts,
\begin{align}
\int \frac{\sin^{-1}(x)}{x} dx &= \int u \cot(u) du \\
&= u \ln\sin(u) - \int \ln\sin(u) du
\tag{1}
\label{eq:20161030-1}
\end{align}
\begin{align}
\int \ln\sin(u) du &= \int \ln\left(\frac{\mathrm{e}^{iu} - \mathrm{e}^{-iu}}{i2} \right) du \\
&= \int \ln\left(\mathrm{e}^{iu} - \mathrm{e}^{-iu} \right) du \,- \int \ln(i2) du \\
&= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \int \ln\mathrm{e}^{iu} du \,-\, u\ln(i2) \\
&= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \frac{i}{2}u^{2} -u\ln2 \,-\, ui\frac{\pi}{2}
\tag{2}
\label{eq:20161030-2}
\end{align}
To evaluate the integral above, let $y=\mathrm{e}^{-i2u}$
\begin{equation}
\int \ln\left(1 - \mathrm{e}^{-i2u} \right) du = \frac{i}{2} \int \frac{\ln(1-y)}{y} dy
= -\frac{i}{2} \operatorname{Li}_{2}(y) = -\frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2u}
\tag{3}
\label{eq:20161030-3}
\end{equation}
Now we substitute equation \eqref{eq:20161030-3} into equation \eqref{eq:20161030-2}, then substitute that result into equation \eqref{eq:20161030-1}, switch variables back to (x), and apply limits,
\begin{align}
\int\limits_{0}^{1} \frac{\sin^{-1}(x)}{x} dx
&= \sin^{-1}(x)\ln(x) + \sin^{-1}(x)\left(\ln2 + i\frac{\pi}{2}\right) \\
&- \frac{i}{2}[\sin^{-1}(x)]^{2} + \frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2\sin^{-1}(x)} \Big|_0^1 \\
&= \frac{\pi}{2}\ln2
\end{align}
I would be interested in seeing other solutions.
|
Integration by parts reduces the integral to,
$$\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^2}} dx$$
And the substitution $x=\sin u$ reduces the integral to,
$$I=\int_{0}^{\frac{\pi}{2}} \ln (\sin u) du$$
And the substitution $v=\frac{\pi}{2}-x$ reduces the integral to,
$$I=\int_{0}^{\frac{\pi}{2}} \ln (\cos v) dv$$
$$I=\int_{0}^{\frac{\pi}{2}} \ln (\cos u) du$$
Now adding the integrals and noting properties of logarithms we have,
$$2I=\int_{0}^{\frac{\pi}{2}} \left( \ln (2 \sin x \cos x)-\ln 2\right) dx$$
Double angle,
$$2I=\int_{0}^{\frac{\pi}{2}} \ln (\sin 2x) dx -\frac{\pi}{2} \ln 2$$
The substitution $s=2x$ gives
$$2I=\frac{1}{2}\int_{0}^{\pi} \ln (\sin s) ds -\frac{\pi}{2} \ln 2$$
But
$$\int_{0}^{\pi} \ln (\sin s) ds=2I$$
Follows from the substitution $w=\frac{\pi}{2}-s$ and the evenness of the function $f(w)=\ln (\cos w)$:
$$\int_{0}^{\pi} \ln (\sin s) ds$$
$$=-\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \ln (\cos w) dw$$
$$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln (\cos w)dw $$
$$=2 \int_{0}^{\frac{\pi}{2}} \ln (\cos w) dw=2I$$
So,
$$2I=I-\frac{\pi}{2}\ln 2$$
$$I=-\frac{\pi}{2}\ln 2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1992462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 1
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|
Show a chain is a composition series for a matrix ring
Consider the ring
$$
R =
\begin{pmatrix}
\mathbb{Q} & 0 \\
\mathbb{R} & \mathbb{R}
\end{pmatrix}
$$
as a left $R$-module over itself. I want to show that the following is a composition series
$$
0=\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}\subset
\begin{pmatrix}
0 & 0 \\
\mathbb{R} & 0
\end{pmatrix}\subset
\begin{pmatrix}
0 & 0 \\
\mathbb{R} & \mathbb{R}
\end{pmatrix}\subset
\begin{pmatrix}
\mathbb{Q} & 0 \\
\mathbb{R} & \mathbb{R}
\end{pmatrix}=R
$$
for $R$.
So I consider the quotients of the terms, for example $ \begin{pmatrix}
\mathbb{Q} & 0 \\
\mathbb{R} & \mathbb{R}
\end{pmatrix} / \begin{pmatrix}
0 & 0 \\
\mathbb{R} & \mathbb{R}
\end{pmatrix}$. I see that this is isomorphic to $\mathbb{Q}$. Can I use this to show that the quotient is simple?
Similarly how can I do the same thing for the quotient $\begin{pmatrix}
0 & 0 \\
\mathbb{R} & 0
\end{pmatrix} / \begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}$ ?
|
$ \begin{pmatrix}
\mathbb{Q} & 0 \\
\mathbb{R} & \mathbb{R}
\end{pmatrix} / \begin{pmatrix}
0 & 0 \\
\mathbb{R} & \mathbb{R}
\end{pmatrix}$. I see that this is isomorphic to $\mathbb{Q}$. Can I use this to show that the quotient is simple?
That alone is probably not a transparent enough explanation of why it is simple. You can, if you like, start with the fact it is isomorphic to $\begin{pmatrix}
\mathbb Q & 0 \\
0 & 0
\end{pmatrix}$ as a left $R$ module, and note that for any two nonzero elements $\begin{pmatrix}x & 0\\0&0\end{pmatrix}$ and $\begin{pmatrix}y & 0\\0&0\end{pmatrix}$, you can find $\begin{pmatrix}q & 0\\r&s\end{pmatrix}$ such that $\begin{pmatrix}q & 0\\r&s\end{pmatrix}\begin{pmatrix}x & 0\\0&0\end{pmatrix}=\begin{pmatrix}y & 0\\0&0\end{pmatrix}$. (When a ring acts transitively on the nonzero elements of a module, that means it is simple.)
Similar logic holds for $\begin{pmatrix}
0 & 0 \\
\mathbb{R} & \mathbb R
\end{pmatrix} / \begin{pmatrix}
0 & 0 \\
\mathbb R & 0
\end{pmatrix}\cong\begin{pmatrix}
0 & 0 \\
0&\mathbb R
\end{pmatrix}$ and $\begin{pmatrix}
0 & 0 \\
\mathbb{R} & 0
\end{pmatrix} / \begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}\cong \begin{pmatrix}
0 & 0 \\
\mathbb R & 0
\end{pmatrix}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1993621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Stuck on showing $\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$ converges I am trying to show if this series converges or diverges and I know it converges since for very large values of n,
$$\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$$ becomes $$\sum_{n=0}^\infty \frac{(-1)^n1}{n}$$
which is convergent from the alternating series test since lim 1/n = 0 and 1/n is a decreasing sequence. I am going back and forth using the limit comparison test and alternating series test and can't seem to come up with anything. So right now i have $$\frac{(-1)^n(n^2+3n-7)}{n^3+1} \lt \frac{(-1)^n(n^2+3n-7)}{n^3}$$
and whenever I try to show the right handed side to be convergent I come back to the same problem as with the left hand side.
|
We can write
$$\frac{n^2+3n-7}{n^3+1}= \frac{n^2-n +1}{n^3+1} + \frac{4(n +1)}{n^3+1} - \frac{12}{n^3+1}= \\
= \frac{1}{n+1} + \frac{4}{n^2-n +1} - \frac{12}{n^3+1}.
$$
Now, $\sum\limits_{n=0}^{\infty}{ \frac{(-1)^n}{n+1}}$ converges conditionally, however, $4\sum\limits_{n=0}^{\infty}{ \frac{(-1)^n}{n^2-n +1}}$ and $12\sum\limits_{n=0}^{\infty}{ \frac{(-1)^n}{n^3+1}}$ are absolutely convergent.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1993724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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|
Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothing.
|
Use:
*
*$$\log_2(3)=\frac{\ln(3)}{\ln(2)}$$
*$$\log_3(6)=\frac{\ln(6)}{\ln(3)}=\frac{\ln(2\cdot3)}{\ln(3)}=\frac{\ln(2)+\ln(3)}{\ln(3)}=1+\frac{\ln(2)}{\ln(3)}=1+\frac{1}{\log_2(3)}$$
So, you need to prove that:
$$\log_2(3)<\log_3(6)\space\space\space\Longleftrightarrow\space\space\space\log_2(3)<1+\frac{1}{\log_2(3)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1994320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
}
|
How to determine if a function is increasing. How would I determine whether a function is increasing, decreasing or neither without using calculus? Like whether x^0.5 is increasing in interval [0, infinity)
Just curious
Thanks
|
There are many non-calculus techniques that can be applied to show a function is increasing. One approach is to show for $k > 0$ that $f(x + k)$ is larger than $f(x)$.
Here’s a simple example.
Suppose that $f(x) = mx + b$
Now take $k > 0$ and compare $f(x)$ to $f(x+k)$.
$$f(x+k) - f(x) = mk$$
Since $k > 0$ we know that $mk$ is positive iff $m$ is positive, hence $f(x)$ is increasing if $m$ is positive.
Another example
Take $f(x) = x^3$
$$f(x+k) = (x+k)^3 = x^3 + k^3 + 3k^2x + 3x^2k$$
$$f(x+k) - f(x) = 3kx^2 + 3k^2x + k^3$$
For $x > 0$ all terms are positive, hence the function is increasing.
At $x = 0$ the difference is $k^3$ which is also positive by definition.
For $x < 0$ we can take $k$ arbitrarily small (but positive) to show that $f(x)$ is still increasing.
Let $k < |x|$:
$$f(x+k) - f(x) = 3kx^2 + 2k^2x + k^3$$
$$f(x+k) - f(x) = k(3x^2 +3kx) + k^3 > k(3x^2 + 3kx) > k(3x^2 + 3|x|x)$$
Since we said $x$ is negative we have
$$f(x+k) - f(x) = k(3x^2 - 3x^2) = 0$$
The difference is positive, hence the function is increasing.
Now you can try this method with $\sqrt{x}$
$\sqrt{x}$
$$f(x+k) = \sqrt{x+k}$$
$$f(x+k) - f(x) = \sqrt{x+k} - \sqrt{x}$$
To solve it, lets' multiply and divide by
$$\frac{\sqrt{x+k} + \sqrt{x}}{\sqrt{x+k} + \sqrt{x}}$$
So we have
$$(\sqrt{x+k} - \sqrt{x})\cdot \frac{\sqrt{x+k} + \sqrt{x}}{\sqrt{x+k} + \sqrt{x}} = \frac{k}{\sqrt{x+k} + \sqrt{x}}$$
Since $k> 0$ and since the square root admits only positive real numbers (in $\mathbb{R}$), and since the denominator is a sum of positive terms, you have that
$$ \frac{k}{\sqrt{x+k} + \sqrt{x}}$$
Is always positive, hence the function $\sqrt{x}$ is increasing.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1996569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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|
Prove $(1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)$ How one can prove the following.
Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds:
$(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$
WLOG one can assume that $0\le a\le b\le c$.
It is not difficult to prove the statement when $0\le a\le b\le c\le 1$ or $1\le a\le b\le c$.
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $$(1+a+b)(1+a+c)(1+b+c)\geq9(ab+ac+bc)$$ or
$$1+2(a+b+c)+\sum_{cyc}(a^2+3ab)+\prod\limits_{cyc}(a+b)\geq9(ab+ac+bc)$$ or
$$1+6u+9u^2+3v^2+9uv^2-w^3\geq27v^2,$$ which is a linear inequality of $v^2$, which says that it remains to prove our inequality for the extremal value of $v^2$.
$a$, $b$ and $c$ are three non-negative roots of the equation
$(x-a)(x-b)(x-c)=0$ or $3v^2x=-x^3+3ux^2+w^3$.
Thus, a line $y=3v^2x$ and a graph of $y=-x^3+3ux^2+w^3$ have three common points.
Hence, $v^2$ gets an extremal value, when a line $y=3v^2x$ is a tangent line to the graph of $y=-x^3+3ux^2+w^3$, which happens for equality case of two variables.
Let $b=a$.
Hence, we need to prove that
$$(2a+1)c^2+2(2a^2-6a+1)c+2a^3-4a^2+4a+1\geq0,$$ for which it's enough to probe that
$$(2a^2-6a+1)^2-(2a+1)(2a^3-4a^2+4a+1)\leq0,$$ which is $a(a-1)^2\geq0$.
Done!
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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|
Calculating limit of $\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$ As the title says we want to calculate:
$$\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$$
By multiplying nominator and denominator in their conjugates
$=\lim_{x\to\infty}\dfrac{(\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4})(x+1+x+2\sqrt{x(x+1)}-4(x+2))}{(\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x})(x+2+x-4+2\sqrt{(x+2)(x-4)})-4x)}$
$=\lim_{x\to\infty}\dfrac{(\sqrt{x+2}+2\sqrt{x}+\sqrt{x-4})(-2x-7+2\sqrt{x^2+x})}{(\sqrt{x+1}+2\sqrt{x+2}+\sqrt{x})(-2x-2+2\sqrt{x^2-2x-8})}$
I think now we can take $$2x\approx2\sqrt{x^2+x}\approx2\sqrt{x^2-2x-8}\\[2ex]
\sqrt{x}\approx\sqrt{x+1}\approx\sqrt{x+2}\approx\sqrt{x-4}$$ as $x$ goes to infinity. Hence the limit of above fraction would be $\dfrac{7}{2}$, but wolframalpha gives me $\dfrac{3}{2}$ as the limit of the above fraction.
What am I doing wrong?
|
Divide top and bottom by $\sqrt{x}$
$$ \lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}} = \lim_{x\to\infty}\dfrac{\sqrt{1+\frac1{x}}-2\sqrt{1+\frac{2}{x}}+\sqrt{1}}{\sqrt{1+\frac{2}{x}}-2\sqrt{1}+\sqrt{1-\frac{4}{x}}}$$
$$ \to \frac{0}{0}$$
so use L'Hopitals rule differentiate top and bottom
$$\lim_{x\to\infty}\dfrac{-\frac1{x^2} \frac1{2} \frac1{\sqrt{1 + \frac1{x} }}+ \frac{2}{x^2}\frac1{\sqrt{1 + \frac{2}{x}}}}{ - \frac1{x^2} \frac1{\sqrt{1 + \frac{2}{x}}} + \frac{2}{x^2} \frac1{\sqrt{1 - \frac{4}{x}}}} $$
Multiply top and bottom by $x^2$ and take the limit
$$ \dfrac{-\frac1{2} + 2}{-1 + 2} = \frac{3}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1998813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Probability: $a$ chosen randomly from $\{1,2,\dots,n\}$, find $P(a^2=1 \mod 10)$.
A number $a$ is chosen randomly from the set $\{1, 2,\dots , n\}$. Find
*
*The probability $p_n$ that $a^2=1 \mod 10 $.
*$\displaystyle \lim_{n\to\infty} p_n$.
I found that in $n$ numbers $\{1,2,3,\dots, n\}$ there are $(n-1)/10$ numbers that number $10=1$, $(11,21,31,41,51, \dots)$
Hope for any help. Thanks!
|
Note that if $a^2 \equiv 1 \pmod{10}$ then we must have that $a^2 \equiv 1 \pmod{5}$ and $a^2 \equiv 1 \pmod{2}$. As both $5$ and $2$ are primes we must have that $a \equiv \pm 1 \pmod 5$ and $a \equiv 1 \pmod 2$. Combining these two with the Chinese Remainder Theorem we have that:
$$a \equiv \pm 1 \pmod{10}$$
So this proves that the only solutions to $a^2 \equiv 1 \pmod{10}$ are the numbers of the type $a=10k + 1$ and $a=10k-1$. To calculate that number you can use String's formula in the comments.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
find the range of the $b$, if $a^2+b^2+c^2=21~~~~2b=a+c$ Let $\Delta ABC$ such $|AC|=b,|BC|=a,|AB|=c$,and such $$\begin{cases}
2b=a+c\\
a^2+b^2+c^2=21
\end{cases}$$
find the range of the $b$
since
$$(a+c)^2-2ac+b^2=21$$
so we have
$$2ac=5b^2-21$$
then $a,c$ is a equation
$$x^2-2bx+\dfrac{5b^2-21}{2}=0$$ roots.so we have
$$\Delta =4b^2-2(5b^2-21)=42-6b^2\ge 0\Longrightarrow -\sqrt{7}\le b\le \sqrt{7}$$
the book aswer: $\sqrt{6}<b\le\sqrt{7}$
|
Upper bound
Since $a+c=2b$ we have that $a+b+c=3b.$ Now, squaring we get
$$9b^2=(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc.$$ Using that $a^2+b^2+c^2=21$ and $ab+ac+bc\le a^2+b^2+c^2=21$ we get $b^2\le 7.$ So, we have that upper bound $b\le \sqrt{7}.$
|
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|
What is this integral $\int {\sqrt{\frac{x+1}{x}}}\:dx$? I tried a lot of different approaches but found this problem very hard.
So can you help me with this integral?
$$\int {\sqrt{\frac{x+1}{x}}}\:dx$$
Thanks.
|
I have finally understood what Surb meant in his answer and I'm posting it here for the benefit of future readers. I can't post it as a comment due to space restrictions.
$\int{\sqrt{\frac{x+1}{x}}}$
Let u = $\sqrt{\frac{x+1}{x}}$ and v = $\frac{x+1}{x} = 1 + \frac{1}{x}$
$\therefore u = \sqrt{v}$
$\frac{dv}{dx} = \frac{-1}{x^{2}}$
$\frac{du}{dx} = \frac{du}{dv} \frac{dv}{dx} = \frac{1}{2 \sqrt{v}} \frac{-1}{x^{2}} = \frac{-1}{2ux^{2}}$
$u^{2} = 1 + \frac{1}{x} \therefore u^{2} - 1 = \frac{1}{x} \therefore x^{2} = \frac{1}{( u^{2} - 1)^{2}}$
$\therefore \frac{du}{dx} = \frac{- ( u^{2} - 1)^{2}}{2u}$
$\therefore dx = \frac{-2u}{ ( u^{2} - 1)^{2}} du$
$\therefore \int{\sqrt{\frac{x+1}{x}}}dx = \int{ \frac{-2u^{2}}{ ( u^{2} - 1)^{2}}}du = -2 \int{ \frac{u^{2}}{ ( u^{2} - 1)^{2}}}du = -2 \int{ u \times \frac{u}{ ( u^{2} - 1)^{2}}}du$
Using rule $\int{a b'} = ab - \int{b a'}$ where a = u and b' = $\frac{u}{ ( u^{2} - 1)^{2}}$
Let v = $u^{2} - 1$ then
b = $\int{b'} = \int{ \frac{u}{ ( u^{2} - 1)^{2}}}du = \frac{1}{2} \int{\frac{1}{v^{2}}}dv = \frac{-1}{2(u^{2} - 1)}$
$\int{a b'} = u \times \frac{-1}{2(u^{2} - 1)} - \int{ \frac{-1}{2(u^{2} - 1)}}du = \frac{-u}{2(u^{2} - 1)} + \frac{1}{2} \int{\frac{1}{u^{2} - 1}}du$
$\frac{1}{u^{2} - 1} = \frac{1}{u+1} \frac{1}{u-1} = \frac{1}{2(u-1)} - \frac{1}{2(u+1)}$
$\therefore \int{\frac{1}{u^{2} - 1}}du = \int{\frac{1}{2(u-1)} - \frac{1}{2(u+1)}}du = \frac{log(u-1) - log(u+1)}{2}$
$\therefore \int{a b'} = \frac{-u}{2(u^{2} - 1)} + \frac{log(u-1) - log(u+1)}{4}$
$\therefore \int{\sqrt{\frac{x+1}{x}}}dx = \frac{u}{u^{2} - 1} + \frac{log(u+1) - log(u-1)}{2} = x \sqrt{\frac{x+1}{x}} + \frac{log( \sqrt{\frac{x+1}{x}}+1 ) - log( \sqrt{\frac{x+1}{x}} - 1)}{2}$
Drumroll please. And thanks Surb for the hint. You are awesome. You really helped Jupiter.
|
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|
Show that exactly half of $1^{\frac{p-1}{2}}, 2^{\frac{p-1}{2}}, \dots, (p-1)^{\frac{p-1}{2}}$ are congruent to 1 modulo $p$
Let $p$ be an odd prime number. Look at the numbers in the set
\begin{align*}
S \in \{1^{\frac{p-1}{2}}, 2^{\frac{p-1}{2}}, \dots, (p-1)^{\frac{p-1}{2}}\}
\end{align*}
Show that exactly half of these numbers are congruent to 1 modulo $p$.
I define two polynomials
\begin{align*}
&f(x) = x^{\frac{p-1}{2}} - 1 \\
&g(x) = x^{\frac{p-1}{2}} + 1
\end{align*}
According to Lagranges theorem, the congruences
\begin{align*}
&f(x) \equiv 0 \pmod{p} \\
&g(x) \equiv 0 \pmod{p}
\end{align*}
or
\begin{align*}
&x^{\frac{p-1}{2}} \equiv 1 \pmod{p} \\
&x^{\frac{p-1}{2}} \equiv -1 \pmod{p}
\end{align*}
will have maximum $\frac{p-1}{2}$ slutions each. Thus, we can say that maximum half of the numbers in $S$ will be congruent to either 1 or -1 modulo $p$. How can I show that exactly half of the numbers are congruent to 1 or -1 modulo $p$?
|
$x^p-1=0$ has $p-1$ solutions.
$$x^p-1=(x^{\frac{p-1}{2}}-1)(x^{\frac{p-1}{2}}+1)$$ and since each of
$x^{\frac{p-1}{2}}-1$ and $x^{\frac{p-1}{2}}+1$ has at most $\frac{p-1}{2}$ many solutions both must have exactly $\frac{p-1}{2}$ many solutions.
|
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|
Jensen's inequality application I tried finding the maximum value of the expression
$$\frac{2a}{3a+b}{\sqrt{3a+b}}+\frac{2b}{3b+c}{\sqrt{3b+c}}+\frac{2c}{3c+a}{\sqrt{3c+a}}$$
where a,b,c are positive real numbers.
I used Jensen's inequality for concave function: $$f(x)=\sqrt{x}$$
The expression is homogeneous, so WLOG I assume:
$$\frac{2a}{3a+b}+\frac{2b}{3b+c}+\frac{2c}{3c+a}=1$$
Therefore: $$\frac{2a}{3a+b}{\sqrt{3a+b}}+\frac{2b}{3b+c}{\sqrt{3b+c}}+\frac{2c}{3c+a}{\sqrt{3c+a}}\le\sqrt{\frac{2a}{3a+b}(3a+b)+\frac{2b}{3b+c}(3b+c)+\frac{2c}{3c+a}(3c+a)}=\sqrt{2a+2b+2c}$$
And that is not true. Is there someone nice who could tell me why is this solution wrong?
|
The maximum does not exist. Try $a=b=c\rightarrow+\infty$.
Your reasoning is true and gives
$\sum\limits_{cyc}\frac{2a}{3a+b}\sqrt{3a+b}\leq\sqrt{2a+2b+2c}=\sqrt{(2a+2b+2c)\sum\limits_{cyc}\frac{2a}{3a+b}}$,
which gives again that the maximum does not exist.
|
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|
Find the coefficient of $x^{16}$ in expansion of $(x^2+5x+6)^4(x^3+6x^2+11x+6)^3$.
Find the coefficient of $x^{16}$ in expansion of $(x^2+5x+6)^4(x^3+6x^2+11x+6)^3$.
I simplified the expression and it turned out to be $(x+3)^7(x+2)^7(x+1)^3$. Now I'm stuck. I only know how to deal with 2 terms but there are 3 over here. I would appreciate a hint. Thanks
|
Considering $f(x) = (x+3)^7(x+2)^7(x+1)^3$
In expanded form, the sum of roots of this polynomial would be
$$S = \frac{-b}{a} $$
Here , $b$ would be the coefficient of $x^{16}$, and $a$ the coefficient of $x^{17}$[This is a standard result for any polynomial, that sum of roots is negative of coefficient of second highest power divided by coefficient of highest power (Can be 0)]
Now, $a = 1$ (Easy to check)
Hence, Coefficient of $x^{16} = -S = -((-3)*7+(-2)*7 +(-1)*3) = 38 $
Approach 2
You have $f(x) = (x+3)^7(x+2)^7(x+1)^3$
You can obtain $x^{16}$ by the following cases:
1) $x^7$ from $(x+3)^7$ , $x^7$ from $(x+2)^7$ , $x^2$ from $(x+1)^3$
2) $x^7$ from $(x+3)^7$ , $x^6$ from $(x+2)^7$ , $x^3$ from $(x+1)^3$
3) $x^6$ from $(x+3)^7$ , $x^7$ from $(x+2)^7$ , $x^3$ from $(x+1)^3$
From
1) 7C7*7C7*3C2 ,
2) 7C6 *(3) *7C7*3C3 ,
3) 7C7*7C6 *(2) *3C3
Adding these , we get Coefficient $= 38$
|
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|
Limit square roots of polynomials I am trying to find $\lim \limits_{n \to \infty} {\sqrt{n^3+1}-n\sqrt{n} \over \sqrt{n^2+1}-n}$. I rewrite the fraction as
$${(\sqrt{n^3+1}-n\sqrt{n})(\sqrt{n^3+1}+n\sqrt{n}) \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})} = {1 \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})}$$
I notice $$\sqrt{n^2+1}-n > 0$$
so the denominator is growing to infinity while the whole limit is $0$. In the material I'm covering, the properties of polynomials haven't been discussed (but the properties of the square root have been). Is there a more basic way to conclude about the denominator going to infinity?
|
One may proceed the following way, as $n \to \infty$,
$$
\begin{align}
{1 \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})}&={\sqrt{n^2+1}+n \over ((n^2+1)-n^2)\cdot(\sqrt{n^3+1}+n\sqrt{n})}
\\\\&={\sqrt{n^2+1}+n \over 1\cdot(\sqrt{n^3+1}+n\sqrt{n})}
\\\\&=\frac1{n^{1/2}}{\sqrt{1+1/n^2}+1 \over (\sqrt{1+1/n^3}+1)}
\\\\&\sim \frac1{n^{1/2}}
\end{align}
$$ then one may conclude easily.
|
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|
Find minimum of the $a+c+e$ if $a+b+c+d+e=100$.
Let $a\ge b\ge c \ge d \ge e \ge0$ and $a+b+c+d+e=100$. Find minimum of
$$a+c+e$$
My work so far:
$100=a+b+c+d+e\le 5a \Rightarrow a\ge20$
I think that $a+c+e \ge 50$ (if $a=40,b=40,c=10,d=10,e=0$ or $a=b=49, c=d=e=1$).
|
$$100=a+b+c+d+e\le a+a+c+c+e=2a+2c+2e-e.$$ Thus $$50\le 50+\frac e2\le a+c+e.$$ So, the minimum of $a+c+e$ is $50$ and is achieved if $e=0.$
|
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|
Solving simultaneous equations involving a quadratic I have the question
Solve the simultaneous equation pair
$$x^2 + y^2 = 25\tag1$$
$$2x - y = 5\tag2$$
I have found the value of $y$ from the second equation which is $2x-5$ and substituted this into the first equations $y$ value.
I get $x^2 + (2x -5)^2 = 25$
When I expand the brackets I get the equation
$$x^2+4x^2-20 =0\tag3$$
However, when I checked the solutions the equation should simplify to $x^2 - 4x = 0$ and I do not understand how this is achieved.
|
$$x^2 + y^2 = 25 $$
$$2x - y = 5 \Leftrightarrow y=2x-5$$
$$ x^2 + (2x-5)^2 = 25 $$
$$ x^2 +4x^2 -20x +25 = 25 $$
$$ 5x^2 -20x=0 $$
|
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|
Use mathematical induction to prove an assertion The assertion: $n^3 + 5n$ is divisible by $6$
I have completed the basis step $(n=1)$ and the first part of the induction step $(n=k)$, but I am stuck on the second part $(n=k+1)$.
This is what I have so far:
For $n=k$: $k^3 + 5k = 6t$
For $n= k+1$:
$(k+1)^3 + 5(k+1)$
$= k^3 + 3k^2 + 8k + 6$
$= (k^3 + 5k) + 3k + 3k^2 + 6$
$= 3k^2 + 3k + 6t + 6$
$= ???$
I cannot pull out a $6$ because that would leave halves that cannot be counted as integers. How should I proceed?
|
You can complete your work this way:
$$
(k+1)^3 + 5(k+1)=(k^3 + 5k) + 3k + 3k^2 + 6=6t+3k+3k^2+6=6 \left(t+1+\frac{k(k+1)}{2} \right)
$$
and note that $k$ or $k+1$ is even.
|
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|
Inequality on Quadratic Equation Coefficients
Let $a \in \mathbb{Z}_{>0}$ and $b, c \in \mathbb{Z}$, such that $$ax^2 + bx + c = 0$$ has two different solutions in the interval $(0, \frac{1}{2}]$. Prove that $a \geq 6$.
My work consists of just some observations:
*
*if the roots are $\alpha$ and $\beta$, then $\alpha + \beta = \frac{-b}{a} \in (0, 1)$
*from the quadratic equation, $\alpha - \beta = \frac{\sqrt{b^2 - 4ac}}{a}$ (WLOG assuming $\alpha > \beta$) which gives $\frac{\sqrt{b^2 - 4ac}}{a} \in (0, \frac{1}{2}) \implies \frac{a^2}{4} > b^2 - 4ac$
*graphically, for the roots to be close together, the curve has to be steep enough which is affected primarily by $a$ (not rigorous at all but intuitive to me). A further thought that occurred to me was that this question might be generalizable, for different intervals and bounds on $a$.
How do I solve this?
|
\begin{align}
&f(x):=ax^2+bx+c\\\\
&f(0)>0\quad\rightarrow\quad c>0\tag1\\
&f\left(\frac12\right)>0\quad\rightarrow\quad a+2b+4c>0\tag2\\
&0<-\frac{b}{2a}<\frac12\quad\rightarrow\quad b^2<a^2\tag3\\
&b^2-4ac>0\quad\rightarrow\quad a<\frac{b^2}{4c}\tag4\\\\
&\text{As $a$ is maximuzed when $c$ is 1, let's set $c=1$. Then, from (3) and (4),}\\
&a<\frac{b^2}4<\frac{a^2}4\\
&\text{As $a$ and $b$ are integers,}\\
&a+2\le\frac{b^2}4+1\le\frac{a^2}4\\
&a+2\le\frac{a^2}4\quad\rightarrow\quad a^2-4a-8\ge0\\
&a\ge2+2\sqrt3\approx 5.46\\
\therefore\space&a\ge6
\end{align}
You could generalize this, for example, by replacing $\frac12$ into $k$.
Also, I didn't use the condition (2), but it seems that it is not required to use it as we are only interested in the range of $a$ here.
|
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|
factoring $(x^6 - y^6)$: what is going on here? I apologize if this question already exists, but it was quite difficult to word.
In my math class today, we learned how to factor a difference of two perfect cubes. One of our practice questions was:
factor $x^6 - 64$
almost everyone other than me (including my teacher, as he was rushing) ended up with:
$(x^2 - 4)(x^4 + 4x^2 + 16)$
I pointed out two things:
1:
$(x^6 - 64)$ was also a difference of two perfect squares, and we could factor it into this:
$(x^3 + 8)(x^3 - 8)$
Which could then be factored further into:
$(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$
and 2:
the $(x^2 - 4)$ was also a difference of two perfect squares, and we could factor further:
$(x + 2)(x - 2)(x^4 + 4x^2 + 16)$
Now for my question:
How are $(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$ and $(x + 2)(x - 2)(x^4 + 4x^2 + 16)$ equivalent?
The teacher didn't seem to know off the top of his head, and I can't figure it out after trying for a half an hour. I've heard that you can factor the sum of two perfect squares with imaginary numbers, so maybe i can do something there to help explain this?
edit: fixed instances of "- 16" to "+ 16", and "4x" to "4x^2" (thanks for pointing that out)
|
A complete factorization of $x^6 - y^6$ can be performed as follows: $$\begin{align*}
x^6 - y^6 &= (x^3)^2 - (y^3)^2 \\
&= (x^3 - y^3)(x^3 + y^3) \\
&= (x-y)(x^2 + xy + y^2)(x+y)(x^2 - xy + y^2). \end{align*}$$
The quadratic factors are irreducible in the ring of polynomials with rational coefficients $\mathbb Q[x]$ (more informally, they don't factor "nicely").
We can recombine these factors in carefully chosen ways to get expressions that might have resulted from an incomplete factorization; e.g., if we let $a = x^2 + y^2$ and $b = xy$, then
$$\begin{align*} (x^2 + xy + y^2)(x^2 - xy + y^2) &= (a+b)(a-b) \\
&= a^2 - b^2 \\
&= (x^2+y^2)^2 - (xy)^2 \\
&= x^4 + 2x^2 y^2 + y^4 - x^2 y^2 \\
&= x^4 + x^2 y^2 + y^4. \end{align*}$$
And if we combine $(x-y)(x+y)$ we get $x^2 - y^2$; thus we recover the incomplete factorization of $x^6 - y^6$ as a difference of cubes $$(x^2)^3 - (y^2)^3 = (x^2 - y^2)(x^4 + x^2 y^2 + y^4).$$ There are of course other ways to combine these factors, each leading to a different expression, but all are algebraically equivalent.
|
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|
Show that a given equation has no solutions in integers Prove that there are no integers $x$ and $y$ such that
$$x^5-x^4y-13x^3y^2+13x^2y^3+36xy^4-36y^5=77.$$
Firstly setting $x=y$ gives LHS=0. This suggests that $x-y$ is a factor of it, factoring gives
\begin{align*}
(x-y)(x^4-13x^2y^2+36y^4) &=(x-y)(x^2-9y^2)(x^2-4y^2)\\ &=(x-3y)(x-2y)(x-y)(x+2y)(x+3y)=77.
\end{align*}
Im not sure how to proceed.
|
By my comment above, we have that $(x-y)=\pm 1, \pm 7, \pm 11$ or $\pm 77$. Similarly, $(x-2y)=\pm 1, \pm 7, \pm 11$ or $\pm 77$.
Case I: Assume $x-y=1=x-2y$, then $y=0$ and $x=1$, but then $(x-3y)(x-2y)(x-y)(x+2y)(x+3y)=1\neq 77$.
Case II: Assume $x-y=-1$ and $x-2y=1$, then $y=-2$ and $x=-3$, but then $x+3y=-9$ which is not a divisor of $77$.
Case III: Assume $x-y=7$ and $x-2y=1$, then $y=6$ and $x=13$, but then $x+2y=25$ is not a divisor of $77$.
And so on. In principle you can proceed in this fashion and exclude all $64$ cases. You can remove many cases by making some more observations.
You can also work in a reverse way. Notice that $77=1\cdot 77$ or $ 7\cdot 11$. (or $-1\cdot -77$ or $-7\cdot -11$). Thus either one factor is $77$ and all others are $1$. Or one factor is $7$, one other factor is $11$ and all others are $1$ (up to signs). I admit that this seems like a brute force method, but it is doable and will work.
|
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|
$\lim_\limits{n\to \infty}\ (\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})$? what is the value given to this limit?
$$\lim_{n\to \infty}\ \bigg(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + \cdots + \frac{1}{n+1}\bigg)$$
Is it simply 0 because each term tends to 0 and you are just summing up zeros?
|
Alternatively,
\begin{align*}
1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n} &=
\ln n+\gamma+\frac{1}{2n}+O\left( \frac{1}{n^2} \right) \\
\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n} &=
\left[ \ln (2n)+\gamma+\frac{1}{2(2n)} \right]-
\left( \ln n+\gamma+\frac{1}{2n} \right)+
O\left( \frac{1}{n^2} \right) \\
&= \ln 2-\frac{1}{4n}+O\left( \frac{1}{n^2} \right)
\end{align*}
|
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|
Question about solution to Putnam 1995 A2. To provide context Putnam 1995 A2 states:
For what pairs of integers $(a,b)$ of positive real numbers does the improper integral
$$ \int_{b}^{\infty} \left( \sqrt{ \sqrt{x+a \vphantom{b}} -\sqrt{x}} - \sqrt{\sqrt{x} - \sqrt{x-b}} \right) $$
converge?
The solution then proceeds to make the following claims:
$$(1+x)^{\frac{1}{2}} = 1 + x/2 + O(x^2)$$
So:
$$\sqrt{x-a} - \sqrt{x} = x^{1/2}(\sqrt{1 + a/x} - 1)$$
Subbing in the expansion we have:
$$x^{1/2}(1 + a/(2x) + O(x^{-2}))$$
Here I didn't agree, I thought it would be:
$$x^{1/2}(1 + a/(2x) + O(x^{-2}) - 1)$$
which becomes:
$$x^{1/2}(a/(2x) + O(x^{-2}))$$
Furthermore, after taking another square root they get:
$$x^{1/4}\left(a/(4x) + O(x^{-2})\right)$$
I would imagine it should be:
$$x^{1/4}\sqrt{a/(2x) + O(x^{-2})}$$
Can someone explain what I'm doing wrong or not understanding?
|
I do not understand your change in notation; I will stick by the original. In any case, it is
$$\int_b^{\infty} dx \, \left (\sqrt{\sqrt{x+a}-\sqrt{x}} -\sqrt{\sqrt{x}-\sqrt{x-b}}\right ) $$
Note that we should have $a$ and $b \gt 0$ so that we have real quantities throughout the integration region.
We need to worry about the convergence at $\infty$. We want to find the conditions under which the iintegrand falls off faster than $1/x$ as $x \to \infty$.
The way I solved this is to keep using the difference of two squares. In this respect, everything is exact and we need not appeal to crude approximations until the very end. For example, we may rewrite the integrand $f(x)$ as
$$\begin{align} f(x) &= \left (\sqrt{\sqrt{x+a}-\sqrt{x}} -\sqrt{\sqrt{x}-\sqrt{x-b}}\right )\\ &= \frac{\sqrt{a}}{\sqrt{\sqrt{x+a}+\sqrt{x}}} - \frac{\sqrt{b}}{\sqrt{\sqrt{x}+\sqrt{x-b}}} \\ &= \frac{\sqrt{a}\sqrt{\sqrt{x}+\sqrt{x-b}} - \sqrt{b} \sqrt{\sqrt{x+a}+\sqrt{x}} }{\sqrt{\left (\sqrt{x+a}+\sqrt{x} \right )\left (\sqrt{x}+\sqrt{x-b} \right )}} \\ &= \frac{(a-b)\sqrt{x} + a \sqrt{x-b}-b \sqrt{x+a}}{\left (\sqrt{a}\sqrt{\sqrt{x}+\sqrt{x-b}}+\sqrt{b} \sqrt{\sqrt{x+a}+\sqrt{x}} \right ) \sqrt{\left (\sqrt{x+a}+\sqrt{x} \right )\left (\sqrt{x}+\sqrt{x-b} \right )}}\end{align} $$
Now, let's examine this integrand. Note that
$$a \sqrt{x-b}-b \sqrt{x+a} = \frac{(a^2-b^2) x - ab (a+b)}{a \sqrt{x-b}+b \sqrt{x+a}} $$
No matter what we do, it seems that the best we have is that the numerator behaves as $O(x)$ and the denominator as $O(x^{5/4})$ as $x \to \infty$, which leads to a nonconvergent integral. Nevertheless, if $a=b$, then the pieces dependent on $x$ in the numerator vanish, and the denominator still behaves as $O(x^{5/4})$ as $x \to \infty$. Thus, when $a=b$ we have a convergent integral.
|
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|
Solve a congruence $1978^{20}\equiv x\pmod{125}$ I have checked the solution ($x=26$).
Solving modulo $5$ gives
$$1978^{20}\equiv 1978^{2\cdot 10}\equiv 1\pmod{5}$$
Solving modulo $25$ also gives
$$1978^{20}\equiv 1\pmod{5}$$
How to evaluate the remainder $x$?
|
Note that, modulo $125$, $11^2=121\equiv-4$ and $2^7=128\equiv3$.
Therefore, $1978^{20}\equiv(-22)^{20}\equiv 2^{20}11^{20}\equiv 2^{20}(-4)^{10}\equiv 2^{40}\equiv2^{35}2^5\equiv3^52^5\equiv6^5\equiv26$.
|
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|
Find the $n$ derivative of $y= e^{2x}\sin^2 x$ We have
\begin{align*}
y&= e^{2x}\sin^2 x\\
&= e^{2x}\left(\frac{1-\cos 2x}{2}\right)\\
&= \frac{e^{2x}}{2} - \frac{e^{2x}\cos 2x}{2}
\end{align*}
Then
\begin{align*}
y^{(n)} &= \left(\frac{e^{2x}}{2}\right)^{(n)} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}\\
&= 2^{n-1}e^{2x} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}
\end{align*}
I don't know how to proceed with the rightmost term. So far I've been applying the Leibniz Rule whenever I've had to find the $n$ derivative of a function of the form $f(x)g(x)$, because is clear that either $f(x)$ or $g(x)$ has a derivative a $k$ derivative ($1<k<n$) equal to zero, which simplifies the expression nicely.
But here
$$\frac{e^{2x}\cos 2x}{2}$$
both functions are infinitely differentiable on $\mathbb{R}$ which makes things a bit different.
My only attempt was to write its n derivative in this form
\begin{align*}
&=\frac{1}{2}\sum_{k=0}^n{n \choose k} \big(2^k e^{2x}\big)\bigg(2^{n-k}\cos \left[2x + \frac{\pi(n-k)}{2}\right]\bigg)\\
&=\sum_{k=0}^n {n \choose k} 2^{n-1}e^{2x}\cos \left[2x + \frac{\pi(n-k)}{2}\right]
\end{align*}
So
\begin{align*}
y^{(n)} &= 2^{n-1}e^{2x} -\sum_{k=0}^n{n \choose k} 2^{n-1}e^{2x}\cos \left[2x + \frac{\pi(n-k)}{2}\right]\\
&= 2^{n-1}e^{2x}\left(1 -\sum_{k=0}^n{n \choose k} \cos \left[2x + \frac{\pi(n-k)}{2}\right]\right)
\end{align*}
But the textbook's answer is
$$2^{n-1}e^{2x}\left(1 -2^{n/2}\cos \left[2x + \frac{\pi n}{4}\right]\right)$$
For some reason I have the feeling that a little of modular arithmetic has to be applied on $\frac{\pi(n-k)}{2}$
|
Taking into consideration just the second part, let's call
$$p(x)=e^{2x}\cos 2x$$
then
\begin{align}
&p^{(1)}(x)=2e^{2x}\cos 2x-2e^{2x}\sin 2x\\
&p^{(1)}(x)=2e^{2x}(\cos 2x-\sin 2x)=2e^{2x}\sqrt{2}\left(\frac{\sqrt{2}}{2}\cos 2x-\frac{\sqrt{2}}{2}\sin 2x\right)\\
&p^{(1)}(x)=2^{3/2}e^{2x}\left(\cos(\pi/4)\cos 2x-\sin(\pi/4)\sin 2x\right)\\
&p^{(1)}(x)=2^{3/2}e^{2x}\cos\left( 2x+\pi/4 \right)
\end{align}
and following the same calculation that we just did we can conclude that
$$p^{(2)}(x)=2^{3/2}2^{3/2}e^{2x}\cos\left( 2x+\pi/4+\pi/4 \right)$$
Now we can see the pattern. So we can prove by finite induction that
$$p^{(n)}(x)=2^{3n/2}e^{2x}\cos\left( 2x+n\pi/4\right)$$
and finaly:
$$\left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}=2^{3n/2-1}e^{2x}\cos\left( 2x+n\pi/4\right)$$
|
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|
Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!
|
$$
x^2 + 2x + 2 = (x+1)^2 + 1
$$
and, with $\ s = x+1 \ $ you have
$$
\frac 1{x^2 + 2x + 2} = \frac 1{(x+1)^2 + 1} = \frac 1{s^2 + 1} = \text{D} ( \arctan s ).
$$
Now, just derive the Taylor expansion of $\ \arctan s \ $ and subsitute $\ s = x +1 $
|
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|
Showing $\det(aM+bN)-\det(aM)-\det(bN)=ab\left[\det(M+N)-\det(M)-\det(N)\right]$
Why is $\det(aM+bN)-\det(aM)-\det(bN)=ab\left[\det(M+N)-\det(M)-\det(N)\right]$
$M,N$ are $2\times 2$ matrices, $a,b\in K$ for some field $K$. If I write $M=(m_{ij})_{i,j}$ for $\ 1\le i,j\le2$ then it is true, but a a long computation. Is there a shorter way to verify it
|
Just applying properties of determinants, we have:
$$\begin{vmatrix}aa_{11}+bb_{11} & aa_{12}+bb_{12}\\ aa_{21}+bb_{21}& aa_{22}+bb_{22}\end{vmatrix}- \begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}- \begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}$$
$$=\begin{vmatrix}aa_{11} & aa_{12}+bb_{12}\\ aa_{21}& aa_{22}+bb_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & aa_{12}+bb_{12}\\ bb_{21}& aa_{22}+bb_{22}\end{vmatrix}- \begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}- \begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}$$
$$=\begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}+\begin{vmatrix}aa_{11} & bb_{12}\\ aa_{21}& bb_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & aa_{12}\\ bb_{21}& aa_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}- \begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}- \begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}$$
$$=\begin{vmatrix}aa_{11} & bb_{12}\\ aa_{21}& ba_{22}\end{vmatrix}+\begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& ba_{22}\end{vmatrix}$$
$$=ab\left(\begin{vmatrix}a_{11} & b_{12}\\ a_{21}& b_{22}\end{vmatrix}+\begin{vmatrix}b_{11} & a_{12}\\ b_{21}& a_{22}\end{vmatrix}\right)$$
$$=ab\left(\begin{vmatrix}a_{11}+b_{11} & a_{12}+b_{12}\\ a_{21}+b_{21}& a_{22}+b_{22}\end{vmatrix}- \begin{vmatrix}a_{11} & a_{12}\\ a_{21}& a_{22}\end{vmatrix}- \begin{vmatrix}b_{11} & b_{12}\\ b_{21}& b_{22}\end{vmatrix}\right)$$
|
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|
How many positive integers less than 1,000,000 have the sum of their digits equal to 19? How many positive integers less than $1,000,000$ have the sum of their digits equal to $19$ ?
I tried to answer it by using stars and bars combinatorics method.
The question says that sum of $6$ digit numbers { considering the number as $abcdef$ } is 19.
Hence, I can write it as $a + b + c + d + e + f = 19$ ,but this assumes that $a,b,c,d,e,f >= 0$ .
But, here each digit lies between 0 and 9. So, how stars and bars method will be modified according to this domain ?
|
If we treat each positive integer with fewer than six digits as a six-digit string with leading zeros, we seek the number of solutions of the equation
$$a + b + c + d + e + f = 19 \tag{1}$$
in the non-negative integers subject to the restrictions that $a, b, c, d, e, f \leq 9$. A particular solution of equation 1 corresponds to the placement of five addition signs in a row of $19$ ones. For instance,
$$1 1 1 + 1 1 + 1 1 1 1 + 1 1 1 1 1 1 + + 1 1 1 1$$
corresponds to the solution $a = 3$, $b = 2$, $c = 4$, $d = 6$, $e = 0$, and $f = 4$. Hence, if there were no such restrictions, the number of solutions would be equal to the number of ways we can insert five addition signs into a row of $19$ ones, which is
$$\binom{19 + 5}{5} = \binom{24}{5}$$
since we must choose which five of the $24$ symbols (five addition signs and nineteen ones) are addition signs.
However, we have counted solutions in which one or more of the variables exceeds $9$. There can only be one such variable since $2 \cdot 10 = 20 > 19$. We must exclude solutions in which one of the variables exceeds $9$.
We count the number of solutions in which $a > 9$. Since $a$ is an integer, then $a \geq 10$. Hence, $a' = a - 10$ is a non-negative integer. Substituting $a' + 10$ for $a$ in equation 1 yields
\begin{align*}
a + b + c + d + e + f & = 19\\
a' + 10 + b + c + d + e + f & = 19\\
a' + b + c + d + e + f & = 9 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers. Consequently, the number of solutions of equation 1 in which $a > 9$ is
$$\binom{9 + 5}{5} = \binom{14}{5}$$
By symmetry, there are
$$\binom{14}{5}$$
solutions for each of the six variables that could exceed $9$. Hence, the number of positive integers less than $1,000,000$ with digit sum $19$ is
$$\binom{24}{5} - \binom{6}{1}\binom{14}{5}$$
|
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|
point of symmetry of the graph of a function the graph of the function $$f(x)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\cdots+\frac{2^{n-1}x^{2^{n-1}-1}}{1+x^{2^{n-1}}}+\cdots$$ where $x \in (-1,1)$ is symmetric about the point---
I tried to integrate the function and could not proceed to find centre of symmetry.I tried it in desmos and i got the following curve.Please help me out.Thanks
|
Surprisingly $f(x)$ is a very simple function as shown below. To get it, we start from :
$F_0(x)=(1+x)$
$F_1(x)=(1+x)(1+x^2)=(1+x)+x^2(1+x)=1+x+x^2+x^3$
$\begin{cases}
F_2(x)=(1+x)(1+x^2)(1+x^4)=(1+x^2+x^3)+x^4(1+x^2+x^3) \\
F_2(x)=1+x^2+x^3+x^4+x^5+x^6+x^7
\end{cases}$
and so on ...
$\begin{cases}
F_n(x)=(1+x)(1+x^2)...(1+x^{(2^n)}) \\
F_n(x)=1+x+x^2+...+x^{(2^{n+1}-1)}
\end{cases}$
$$F_n(x)=\displaystyle\prod_{k=0}^{n}(1+x^{(2^k)})=\displaystyle\sum_{k=0}^{2^{n+1}-1}x^k=\frac{1-x^{(2^{n+1})}}{1-x}$$
$$\displaystyle\prod_{k=0}^{\infty}(1+x^{(2^k)})=\frac{1}{1-x}$$
$$\ln\left(\displaystyle\prod_{k=0}^{\infty}(1+x^{(2^k)})\right) =\displaystyle\sum_{k=0}^{\infty}\ln(1+x^{(2^k)})=-\ln(1-x)$$
The differentiation leads to :
$$\displaystyle\sum_{k=0}^{\infty}\frac{2^kx^{(2^k)-1} }{1+x^{(2^k)}}=\frac{1}{1-x}$$
We recognize $f(x)$ :
$$f(x)=\frac{1}{1-x}$$
All above is valid for $-1<x<1$. So, $f(x)$ is an arc of hyperbola.
Obviously, the symmetry is with respect to the straight line $y=1-x$ with center $(x=0\:,\:y=1)$.
This is in good agreement with your own graph (magnified) :
|
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|
Find $\int_{|z|=3} {1 \over P(z)}dz$ where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$
Evaluate
$$\int_{|z|=3} {1 \over P(z)}dz$$
where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$.
I proved that $P(z)\neq 0$ for all $z$ outside of the ball of radius 3, except for z=9, so all the poles of ${1 \over P(z)}$ except $z=9$ are inside the ball.
$$P(z)=(z-9)(2z^4+2z^3+z^2-2z+2)=(z-9)(z-a_1)(z-\bar a_1)(z-a_2)(z-\bar a_2).$$
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The roots of $P(z)= 2z^4 + 2z^3+ z^2-2z+2 $ are $a_1 = -1-i$, $a_2 = \frac{-a_1}{2}$ , $\overline{a_1}$ and $\overline{a_2}$. Now by the residue Theorem, you get,
\begin{equation*}
\int_{|z|=3} \frac{dz}{P(z)} = 2i\pi \left( \frac{1}{(a_1 -9)P'(a_1)} + \frac{1}{(\overline{a_1}-9)P'(\overline{a_1})} + \frac{1}{(a_2-9)P'(a_2)} + \frac{1}{(\overline{a_2}-9)P'(\overline{a_2})} \right)
\end{equation*}
Since $P'$ has real coefficients, this sum is easier to compute....
|
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|
Find an $v,w \in \mathbb{C}$ such that $(1+2i)v=1-i$ and $w^{2}-i=0$ Task from an old exam:
Find an $v,w \in \mathbb{C}$ such that $(1+2i)v=1-i$ and $w^{2}-i=0$
and write the solutions in this form: $a+bi$
I'm currently trying to solve this, probably written several pages and always failed calculating it with success so I came here. I think there is some info missing for this task, about definition but anyway maybe you can give me some ideas of how this could be solved?
|
We obtain from $(1+2i)v=1-i$
\begin{align*}
v&=\frac{1-i}{1+2i}\\
&=\frac{(1-i)(1-2i)}{(1+2i)(1-2i)}\\
&=\frac{-1-3i}{5}\\
&=-\frac{1}{5}-\frac{3}{5}i
\end{align*}
We recall
\begin{align*}
w^2&=a+ib\\
&=r\left(\cos\phi+i\sin \phi\right)\\
w_{1,2}&=\pm\sqrt{r}\left(\cos\frac{\phi}{2}+i\sin \frac{\phi}{2}\right)\\
\end{align*}
We obtain from $w^2-i=0$
\begin{align*}
w^2&=i\\
&=1\left(\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}\right)\\
\\
w_{1,2}&=\pm\sqrt{1}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)\\
&=\pm\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)
\end{align*}
|
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|
Given $a^2+b^2=2$ prove $a+b\le2$
*
*Given $a^2+b^2=2$ prove $a+b\le2$
*Given $a+b=2$ prove $a^4+b^4\ge2$
I was trying to prove these using the fact that we know $a^2+b^2\ge2ab$ but not sure where to start.
|
Courtsey : $\rightarrow$ Cauchy Schwarz
$$(a+b)^2\le(1+1)(a^2+b^2)\\\implies a+b\le |a+b|\le 2$$
$$$$Now by the fact that $a+b=2$ and $a^2+b^2\ge 2ab$, we deduce that $2(a^2+b^2)\ge 4\implies a^2+b^2\ge 2$ $$(a^2+b^2)^2\leq2(a^4+b^4)\\ \implies a^4+b^4\ge 2$$
BINGO!
|
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|
total number of positive and negative roots of the equation $ax^3+bx^2+cx+d=0$ Suppose $a,b,c,d$ are non zero real numbers and $ab>0,$ and
$\displaystyle \int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx = \int^{2}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0$
Then total number of positive and negative roots of the equation $ax^3+bx^2+cx+d=0.$
what i have try $\displaystyle \int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx = \int^{2}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0$
$\displaystyle \int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=\int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx +\int^{2}_{1}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0$
Am I not able to go further. Could someone help me with this? Thanks
|
Observe that all the functions involved in the integrand are continuous, furthermore the function $(1+e^{x^2})>0$ for all $x$.
From the first condition
$$\int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0.$$
This means the polynomial $ax^3+bx^2+cx+d$ will have at least one root in the interval $[0,1]$. So at least one positive root.
From the last step you get
$$\int^{2}_{1}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0.$$
This means the polynomial $ax^3+bx^2+cx+d$ will have at least one root in the interval $[1,2]$. So at least two positive root.
By Viete's relations: if $\alpha, \beta, \gamma$ are the roots, then
$$\alpha+\beta+\gamma = -\frac{b}{a}.$$
Since $ab>0$, both $a$ and $b$ are of the same sign. This means at least one root of the cubic must be negative (because we have already established that two roots in $[0,1]$ and $[1,2]$.
|
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|
What's wrong with my approach - reciprocal integral I want to integrate $\int \frac{1}{2x}dx$
The way I would solve it: $$\frac{1}{a}f(ax + b)$$ where $$f(x) = \ln x$$
My answer $\frac{1}{2}\ln(2x) + k$
The correct way seems to be to factor out $\frac{1}{2}$ from the initial integral, leaving $\int \frac{1}{x}$. The correct answer $\frac{1}{2}\ln(x) + k$
Where did I go wrong?
|
You are not wrong; however, your answer is just not simplified.
To check that your answer is correct, differentiate both your answer and the "correct" answer:
$$\frac{d}{dx} (\frac{1}{2} ln(2x) + k) = \frac{1}{2x}$$
$$\frac{d}{dx} (\frac{1}{2} ln(x) + k) = \frac{1}{2x}$$
To simplify:
$\frac{1}{2} ln(2x) + k = \frac{1}{2} (ln(x) + ln(2)) + k$
$=> \frac{1}{2} ln(x) + \frac{1}{2} ln(2) + k = \frac{1}{2} ln(x) + (\frac{1}{2} ln(2) + k)$
$(\frac{1}{2} ln(2) + k)$ is another constant, so let's just call that C.
$=> \frac{1}{2} ln(2x) + k = \frac{1}{2} ln(x) + C$
C and k are just constants and do not affect the rate of change of the function; therefore, it doesn't matter what constant you add at the end of your equation as long as it is indeed constant.
|
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|
Convergence of $\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$ I am checking for convergence of
$$\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$$
First, we notice
$${n+1 \over n-1}= {1+{2 \over n-1}}$$
Then, we use $\ln(1+x) \le x$ and get
$$\ln\left({1 + {2 \over n-1}}\right) \le {2 \over n-1}$$
$${1 \over \sqrt{n}}\ln{n+1 \over n-1} \le {1 \over \sqrt{n}}{2 \over n-1}$$
We now use $n-1 > {n \over 2}$
$${1 \over \sqrt{n}}\ln{n+1 \over n-1} \le {1 \over \sqrt{n}}{2 \over n-1} < {1 \over \sqrt{n}}{2 \over {n \over 2}} = 4{1 \over n\sqrt{n}}$$
$\sum_{n=2}^{\infty}4{1 \over n\sqrt{n}}$ converges and so does $\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$. Is this correct? Alternatively, is there a quicker way?
|
You are right. Alternatively you can use the asymptotic comparison test by noting that
$${1 \over \sqrt{n}}\ln\left({n+1 \over n-1}\right)=
{1 \over \sqrt{n}}\ln\left({1+{2 \over n-1}} \right)\sim \frac{2}{n\sqrt{n}}=\frac{2}{n^{3/2}}$$
where we used that $\ln(1+x)\sim x$ when $x\to 0^+$.
It is similar to your approach but you do not deal with inequalities.
|
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|
Evaluate the integral $\int_0^1 e^{-x^{4}}(1-x^{4}) dx $ How to find the definite integral of $$\int_0^1 e^{-x^{4}}(1-x^{4})dx $$ I tried solving this by using integration by parts and then by substitution but want able to solve this by either of those methods .
|
Let $y=x^{4}$
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} dx = \frac{1}{4} \int\limits_{0}^{1} \mathrm{e}^{-y} y^{-3/4} dy
= \frac{1}{4} \gamma\left(\frac{1}{4},1 \right)
\end{equation}
Using the same substitution, we also have
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} x^{4} dx = \frac{1}{4} \int\limits_{0}^{1} \mathrm{e}^{-y} y^{1/4} dy
= \frac{1}{4} \gamma\left(\frac{5}{4},1 \right)
\end{equation}
Thus we obtain
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} (1-x^{4}) dx
= \frac{1}{4} \Big[ \gamma\left(\frac{1}{4},1 \right) - \gamma\left(\frac{5}{4},1 \right) \Big]
\approx 0.7256
\end{equation}
We have used the lower incomplete gamma function:
\begin{equation}
\gamma(s,z) = \int\limits_{0}^{z} \mathrm{e}^{-x} x^{s-1} dx
\end{equation}
|
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|
How to show $\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+...\right)^2 = 1 +\frac{2}{3}x+\frac{7}{36}x^2+\frac{1}{30}x^3+...$? I tried to find the right handside of the equation by manipulating the series but I failed at getting the right handside of it.
$$\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+...\right)^2 = 1 +\frac{2}{3}x+\frac{7}{36}x^2+\frac{1}{30}x^3+...$$
Closed form of the left handside in the parantheses would be
$$\sum_{n=0}^{\infty}\frac{2}{n!(n+2)!}x^n$$
Any hint would be appreciated.
|
We first calculate the Cauchy product/product of power series
\begin{align*}
\left(\sum_{n=0}^\infty \frac{2}{n!(n+2)!}\cdot x^n\right)^2 &= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{2}{k!(k+2)!}\cdot \frac{2}{(n-k)!(n-k+2)!} \right)\cdot x^n\\
&= \sum_{n=0}^\infty \left(\frac{4}{(n+2)!^2}\cdot \sum_{k=0}^n \binom {n+2}{k} \cdot \binom{n+2}{k+2}\right)\cdot x^n.
\end{align*}
By the Vandermonde convolution identity we have
$$
\sum_{k=0}^n\binom{n+2}{k}\cdot\binom{n+2}{k+2} = \sum_{k=0}^n \binom{n+2}{k} \cdot \binom{n+2}{n-k} = \binom{2n+4}{n}.
$$
Thus, we have
$$
\left(\sum_{n=0}^\infty \frac{2}{n!(n+2)!}\cdot x^n\right)^2 = \sum_{n=0}^\infty \frac{4}{(n+2)!^2}\cdot \binom{2n+4}{n}\cdot x^n.
$$
|
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|
What is wrong with the "proof" for $\ln(2) =\frac{1}{2}\ln(2)$? I have got a question which is as follows:
Is $\ln(2)=\frac{1}{2}\ln(2)$??
The following argument seems suggesting that the answer is yes:
We have the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
which has a mathematically determined value $\ln(2)=0.693$.
Now, let's do some rearrangement:
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......$$
$$
(1-\frac{1}{2})-\frac{1}{4}+(\frac{1}{3}-\frac{1}{6})-\frac{1}{8}+(\frac{1}{5}-\frac{1}{10})-\frac{1}{12}.......$$
$$\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}......$$
$$\frac{1}{2}(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......)$$
$$\frac{1}{2}\ln(2).$$
I know that mathematics can't be wrong, and I have done something wrong. But here is my question: where does the argument above go wrong?
|
You may not rearrange the series if it's not absolutely convergent. Otherwise you can get results like these, and like $\sum_{n=0}^\infty (-1)^k = \frac{1}{2}$, which both are false (the last one being the famous Grandi's series, https://en.wikipedia.org/wiki/Grandi%27s_series). All of this is in the Riemann series theorem (https://en.wikipedia.org/wiki/Riemann_series_theorem). Basically, that's the reason why your result is incorrect, and I hope this helps you!
|
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|
Condition on a,b and c satisfying an equation(TIFR GS 2017) Let $a,b,c$ be positive real numbers satisfying $$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=16,$$ then $a+b+c=3$.
I thought about the application of the AM-GM-HM inequality, but in vain. I also thought about splitting 16 into factors and comparing but went nowhere. Any ideas. Thanks beforehand.
|
By the AM-HM inequality $$\frac{1 + a + b + c}{4} \geq \frac{4}{\frac{1}{1} + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$ and equality holds if and only if $1 = a = b = c$. Since the problem statement tells us that equality holds we find $a + b + c = 3$.
|
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|
Find the solution for $x$ with $0 \le x \lt 13$ so $13 \mid 3x^2+9x+7 $. Find the solution for $x$ with $0 \le x \lt 13$ so $13 \mid 3x^2+9x+7 $.
I got this question in my discrete mathematics class.
I don't really get the idea how 13 can divide $3x^2+9x+7 $
All of my friend told me to try all the number between 0 and 12, but I know that's not how to solve this question.
Can anyone help me?
Thanks.
|
Let's work in $\mod 13$. We want $3x^2+9x+7 \equiv 0 \pmod {13}$. Now $7 \equiv -6$ so you have $3x^2+9x-6 \equiv 0 \pmod {13}$. Because $\gcd(3, 13) = 1$, we can "divide both sides" by $3$ and get $x^2+3x-2 \equiv 0 \pmod {13}$
Notice that $-2 \equiv -28$, so $x^2+3x-28 \equiv 0 \pmod {13}$ which you can factor to $(x-4)(x+7)\equiv 0 \pmod {13}$. Because $13$ is prime, we have $x \equiv 4$ or $x \equiv -7 \equiv 6$.
|
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|
Is $g(u)= \frac{E [ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} ] }{E [ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} ]}$ decreasing in $u$ Let $X$ be a positive random variable, let us define a function
\begin{align}
g(u,a)= \frac{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} \right] }{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} \right]}.
\end{align}
Question: Can we show that the above integral is monotonically decreasing in $u$ ( for $u>0$ ) for all $a > 1$.
Note that $X$ here represents the variance of standard normal. That is we consider the variance to be a random variable.
I can show that $g(u,a)$ is bounded by $1$ and continuous but can not establish that it is decreasing. Also, note that the function $g(u,a)$ is symmetric around $u=0$.
What I tried:
I was able to show that for $p,q\ge 1$ and $\frac{1}{q}+\frac{1}{p}=1$ and $a^2 \ge \frac{1}{p}$ we have
\begin{align}
g(u,a) \le \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}},
\end{align}
where $\beta=\sqrt{\frac{q(a^2-\frac{1}{p})}{a^2}}$.
Proof:
By using Holder's inequality
\begin{align}
E \left[ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} \right] &=E \left[ \frac{1}{\sqrt{X}} e^{-\frac{ (a^2-\frac{1}{p})u^2}{2X}} e^{-\frac{ \frac{1}{p}u^2}{2X}} \right] \\
&\le E^\frac{1}{q} \left[ \frac{1}{\sqrt{X}} e^{-\frac{ q(a^2-\frac{1}{p})u^2}{2X}} \right] E^\frac{1}{p} \left[ \frac{1}{\sqrt{X}} e^{-\frac{ u^2}{2X}} \right].
\end{align}
Therefore,
\begin{align}
g(u,a) \le \left( \frac{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{q(a^2-\frac{1}{p})u^2}{2X}} \right] }{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} \right]} \right) ^\frac{1}{q}
&= \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}},
\end{align}
and
\begin{align}
g(u,a) \le \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}},
\end{align}
where $\beta=\sqrt{\frac{q(a^2-\frac{1}{p})}{a^2}}$.
Thank you. Looking forward to seeing your approaches.
|
If by "monotonically decreasing" you mean decreasing for $u \in (-\infty, \infty)$, then the answer is trivially no: $g$ is an even function in $u$ on the reals, thus it is either constant or it has a local extremum. And we can already tell intuitively that $g$ is not constant.
|
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|
2-form integration example I encountered the following example and I'm not getting $8\pi$ as the answer. It seems to me that
$\int_{D} (x^2+y^2) \mathrm{dx}\wedge \mathrm{dy} = \int\int r^3 \mathrm{dr}\mathrm{d\theta}$
with $0<r<1$ and $0 < \theta < 2\pi$. This gives $\frac{\pi}{2}$. Am I missing something?
|
You are right, the correct value is $\frac{\pi}{2}$, not $8\pi$. The wrong value is however the only error, the argument for transforming the integral into
$$\int_0^{2\pi} \int_0^1 r^3\,dr\,d\theta$$
is correct.
Let's go into "just stick to the definitions and plug things in" mode to verify the value of $\frac{\pi}{2}$. We can parametrise the curve $C$ by $\gamma \colon [-\pi,\pi] \to \mathbb{R}^3$,
$$\gamma(t) = (\cos t, \sin t, \cos t + 1).$$
Then doggedly compute
\begin{align}
\int_C \mathbf{F}\cdot ds &= \int_{-\pi}^{\pi} \langle \mathbf{F}(\gamma(t)), \gamma'(t)\rangle\,dt \\
&= \int_{-\pi}^{\pi} (-\cos^2 t\sin t)\cdot (-\sin t) + (\cos t\sin^2 t)\cdot (\cos t) + (\cos t + 1)^3\cdot(-\sin t)\,dt \\
&= \int_{-\pi}^{\pi} 2\cos^2 t \sin^2 t - (\cos t + 1)^3\sin t\,dt \\
&= \frac{1}{2} \int_{-\pi}^{\pi} (2\sin t \cos t)^2\,dt - \int_{-\pi}^{\pi} (\cos t + 1)^3\sin t\,dt \\
&= \frac{1}{2} \int_{-\pi}^{\pi} \sin^2 (2t)\,dt \tag{1} \\
&= \frac{1}{4} \int_{-2\pi}^{2\pi} \sin^2 u\,du \\
&= \frac{1}{8} \int_{-2\pi}^{2\pi} 1 - \cos (2u)\,du \tag{2} \\
&= \frac{4\pi}{8} -\frac{1}{16}\sin (2u)\biggr\rvert_{-2\pi}^{2\pi} \\
&= \frac{\pi}{2}.
\end{align}
In line $(1)$, we have used the double-angle formula $\sin (2t) = 2\sin t\cos t$, and the fact that $(\cos t + 1)^3\sin t$ is an odd function, so its integral over a symmetric interval - in this case $[-\pi,\pi]$ - is $0$. In line $(2)$ we use the double-angle formula $\cos (2u) = \cos^2 u - \sin^2 u$ and $\cos^2 u + \sin^2 u = 1$ to rewrite $\sin^2 u = \frac{1}{2}(1-\cos (2u))$.
|
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|
Inverse laplace 2/((s-1)^2+1)^2
Find inverse laplace transform of $$\frac{2}{((s-1)^2+1)^2}$$
I tried to decompose the fraction using
$$\because (s-1)^2+1=s^2-2s+2$$
$$\rightarrow \frac{2}{((s-1)^2+1)^2}=\frac{As+B}{s^2-2s+2}+\frac{Cs+D}{(s^2-2s+2)^2}$$
yet I get D=2, which leads me back to the same exact equation, any help?
(I have also tried the $\frac{-d}{ds}(L{\{sint\}})$ propertie, but failed since there's no variable s in the numerator)
|
While there is a defined inverse Laplace transform, the integral is often difficult. Usually, it is easier to take the transforms we know and noodle with them until we find a fit.
$L\{e^t\cos t\} = \frac {s-1}{(s-1)^2+1}\\
L\{e^t\sin t\} = \frac 1{(s-1)^2+1} = \frac {s^2-2s +2}{((s-1)^2+1)^2}\\
L\{te^t\cos t\} = -\frac {d}{ds} \frac {(s-1)}{(s-1)^2+1} = \frac {s^2 - 2s}{((s-1)^2+1)^2}\\
L\{te^t\sin t\} = -\frac {d}{ds} \frac 1{(s-1)^2+1} = \frac {2s-2}{((s-1)^2+1)^2}$
Some combination of $(s^2 - 2s + 2), (s^2-2s),(2s-2) = 2$
$e^t \sin t - te^t \cos t$
|
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|
what is the value of $ x$? If $\log_{2}3^4\cdot\log_{3}4^5\cdot\log_{4}5^6\cdot....\log_{63} {64}^{65}=x!$, what is the value of $ x$?
I've tried
$$\log_2 3^4\cdot\log_3 4^5\cdot\log_4 5^6\cdot.... \log_{63} 64^{65}$$
$$=\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\log 64}{\log 63}$$
don't know how to solve futher steps, please help.
Thanks
|
You get $$4\cdot 5\cdot 6\cdots 65\cdot {\log 64\over \log 2}\\=1\cdot2\cdot3\cdot4\cdot5\cdots65=x!$$
Hence $x=65$
|
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How to find the value of an unknown exponent? E.g. I have the question:
$$2^{4x+1} = 128$$
I solved this by knowing that $128 = 2^7$ and therefore $x$ must equal $1.5$.
However, is there a way of solving this without knowing that $128 = 2^7$?
|
... and without logarithms or knowing any powers of $2$ other than the most trivial one ... \begin{align*}
2^{4x+1} &= 128 \\
2^{4x+1-1} = 2^{4x} &= 128/2 = 64 \\
2^{4x-1} &= 32 \\
2^{4x-2} &= 16 \\
2^{4x-3} &= 8 \\
2^{4x-4} &= 4 \\
2^{4x-5} &= 2 \\
2^{4x-6} &= 1 = 2^0 \text{,} \\
\end{align*}
so $4x-6 = 0$ and $x = 6/4 = 3/2$.
|
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|
Solve $\sqrt{3}^n=3^3$ I got answer but i could not understand, can anyone explain me ?
$\sqrt{3}^n=3^3$
$\implies{3}^{n/2}=3^3$
$\implies \frac{n}{2}=3\implies n=6$
How we get ${3}^{n/2}$ in the second line and how we get $\frac{n}{2}$ in third line.
|
$ {\sqrt{3}}^{n} = 3^3$
$\Rightarrow$ $({\sqrt{3}}) ^{n} = 3^3$
$\Rightarrow$ $({3}^{1/2}) ^{n} = 3^3$
$\Rightarrow$ $({3}) ^{{(1/2)}.{n}} = 3^3$
$\Rightarrow$ $({3}) ^{n/2} = 3^3$
$\Rightarrow$ $n/2 = 3$
$\Rightarrow$ $n = 6$
|
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|
Evaluating $\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$ The following is a problem from an older exam which the instructor didn't provide solutions to.
Evaluate $$\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$$
using only real-analytic techniques.
My work: this boils down to factoring $x^4 - x^3 + x^2 - x+ 1$ which I am unable to do by hand.
|
Hint
$$x^4 - x^3 + x^2 - x+ 1=\left(x^2-\frac{\sqrt{5}+1}{2}x+1\right)\left(x^2+\frac{\sqrt{5}-1}{2}x+1\right)$$
Now apply the partial fraction decomposition method.
Edit
\begin{align*}x^4 - x^3 + x^2 - x+ 1&=x^2\left(x^2-x+1-\frac{1}{x}+\frac{1}{x^2}\right)\\
& =x^2\left((x^2+\frac{1}{x^2})-(x+\frac{1}{x})+1\right)\\
&=x^2\left((x+\frac{1}{x})^2-(x+\frac{1}{x})-1\right)\\
&=x^2\left(x+\frac{1}{x}-\frac{1+\sqrt{5}}{2}\right)\left(x+\frac{1}{x}-\frac{1-\sqrt{5}}{2}\right)\\
&=\left(x^2-\frac{\sqrt{5}+1}{2}x+1\right)\left(x^2+\frac{\sqrt{5}-1}{2}x+1\right)
\end{align*}
|
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|
Finding the linear combination of a vector that is in a span So say we have
Span S =
$
\begin{Bmatrix}
\begin{bmatrix}
1 \\
0 \\
1 \\
\end{bmatrix}
\begin{bmatrix}
-1 \\
1 \\
1 \\
\end{bmatrix}
\begin{bmatrix}
1 \\
1 \\
3 \\
\end{bmatrix}
\end{Bmatrix}
$
I know that
$
\begin{bmatrix}
-1 \\
4 \\
7 \\
\end{bmatrix}
$
is in the span because the reduced row echelon form of [A v] is consistent. Now
how do I find what exactly the linear combination that makes the vector?
|
$$\begin{pmatrix}-1\\4\\7\end{pmatrix}=3\begin{pmatrix}1\\0\\1\end{pmatrix}+4\begin{pmatrix}-1\\1\\1\end{pmatrix}.$$
$$\begin{pmatrix}-1\\4\\7\end{pmatrix}=-5\begin{pmatrix}1\\0\\1\end{pmatrix}+4\begin{pmatrix}1\\1\\3\end{pmatrix}.$$
|
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|
Asymptotic of integral I have this definite integral:
$$ \int_{0}^{\frac \pi 2} \left( 1 + t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta , t\in (0,1)$$
where $n$ is an integer. I need to find the asymptotic as a function on $n$.
I suspect it should be $O(\frac{1}{\sqrt{n}})$ but wasn't able to complete the calculation.
Any ideas?
|
*
*A behaviour as $n \to \infty$. Laplace's method ($3.$, p. $2$)
may be applied here, $$ \begin{align} \int_a^bf(x)e^{-\lambda
g(x)}dx\sim f(a)e^{-\lambda g(a)}\sqrt{\frac{\pi}{2\lambda
g''(a)}},\qquad \lambda \to \infty, \tag1 \end{align} $$ with
$g'(a)=0$, $g''(a)>0$, $f(a)\neq 0$.
One may write $$ \begin{align} \int_{0}^{\large \frac \pi 2} \left( 1
+ t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta&=\int_{0}^{\large \frac \pi 2} \left( 1 + t^2 \tan^2\theta
\right )^{-\frac{n-1}{2}} d \theta \\\\&=\int_{0}^{\infty}
\frac{1}{1+x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx
\\\\&=\int_{0}^{\infty} \frac{1}{1+x^2}e^{-\large \frac{n-1}{2}\cdot
\ln(1 + t^2 x^2)} dx \end{align} $$ here $ \lambda =\frac{n-1}{2}$,
$g(x)=\ln(1 + t^2 x^2)$, $g(0)=0$, $g'(0)=0$, $g''(0)=2t^2>0$,
$f(0)=1\neq 0$, we then get
$$ \int_{0}^{\large \frac \pi 2} \left( 1 + t^2 \cot^2\theta \right )^{-\large\frac{n-1}{2}} d \theta \sim
\frac1t\sqrt{\frac{\pi}{2}}\cdot
\frac1{\sqrt{n}}=\mathcal{O}\left(\frac1{\sqrt{n}}\right),\qquad n
\to \infty. \tag2 $$
*A closed form. One may evaluate $$ \int_{0}^{\large \frac \pi 2}
\left( a +1+ \cot^2\theta \right )^{-\frac{1}{2}} d
\theta=\frac{\arctan \sqrt{a}}{\sqrt{a}},\qquad a>0 $$ then
differentiating $n$ times with respect to $a$ setting
$t=\dfrac1{\sqrt{a+1}}$, gives
$$ \int_{0}^{\large \frac \pi 2} \left(1+t^2 \cot^2\theta \right )^{-\frac{2n+1}{2}} d \theta=\frac{(-1)^n}{t^{2n+1}}
\frac{(2n)!}{2^{2n-1}n!}\left[\frac{d^n}{da^n}\left(\frac{\arctan
\sqrt{a}}{\sqrt{a}}\right)\right]_{t=\frac1{\sqrt{a+1}}}.\tag3 $$
Similarly, from the evaluation $$ \int_{0}^{\large \frac \pi 2}
\left(a+1+\cot^2\theta\right)^{-1} d
\theta=\dfrac1{a+1+\sqrt{a+1}},\qquad a>0 $$ then differentiating $n$
times with respect to $a$ setting $t=\dfrac1{\sqrt{a+1}}$ we get
$$ \int_{0}^{\large \frac \pi 2} \left(1+t^2 \cot^2\theta \right )^{\large-\frac{2n+2}2} d \theta=\frac{(-1)^n}{t^{2n+2}}
\frac{1}{n!}\left[\frac{d^n}{da^n}\left(\dfrac1{a+1+\sqrt{a+1}}\right)\right]_{t=\frac1{\sqrt{a+1}}}.
\tag4 $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int_{-\infty}^{\infty}\frac{x^2 \cos(x)}{(x^2+1)^2}dx$ by method of residues. Im trying to solve $$\int_{-\infty}^{\infty}\frac{x^2 \cos(x)}{(x^2+1)^2}dx$$ using the method of residues. This function has two simple poles at $x=\pm i$ and so $$\int_{-\infty}^{\infty}\frac{x^2 \cos(x)}{(x^2+1)^2}dx=2\pi i\text{Res}_{z=i }\frac{z^2 \cos(z)}{(z^2+1)^2}=2\pi i \lim_{z\to i} \frac d{dz}\Big[ (x-i)\frac{z^2 \cos(z)}{(z^2+1)^2}\Big]=2\pi i\lim_{z\to i}\frac d{dz}\Big[\frac{z^2\cos(z)}{(z+i)^2} \Big]=2\pi i \lim_{z\to i} \Big[ \frac{(z+i)^2(2z\cos z-z^2 \sin z)-2z^2(z+i)\cos z}{(z+i)^4} \Big]=2\pi i(-\frac{ei}{4})=\frac{\pi e}{4}$$
However wolfram alpha says that $$\int_{-\infty}^{\infty}\frac{x^2 \cos(x)}{(x^2+1)^2}dx=0$$
I'm almost certain that the residue calculated above is correct, so why am I not able to apply the method here?
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Let $C$ be a closed contour comprised of $(i)$ the real-line segment from $-R$ to $R$ and $(ii)$ the semi-circular arc of radius $R$, centered at the origin, and in the upper-half plane. Then, we have for $R>1$
$$\begin{align}
\oint_C \frac{z^2e^{iz}}{(z^2+1)^2}\,dz&=2\pi i \text{Res}\left(\frac{z^2e^{iz}}{(z^2+1)^2},z=i\right)\\\\
&=2\pi i \lim_{z\to i}\frac{d}{dz}\left(\frac{z^2e^{iz}}{(z+i)^2}\right)\\\\
&=2\pi i \lim_{z\to i}\left(\frac{2ze^{iz}+iz^2e^{iz}}{(z+i)^2}-\frac{2z^2e^{iz}}{(z+i)^3}\right)\\\\
&=2\pi i \left(-ie^{-1}/4 +ie^{-1}/4\right)\\\\
&=0 \tag 1
\end{align}$$
We also have
$$\begin{align}
\oint_C \frac{z^2e^{iz}}{(z^2+1)^2}\,dz&=\int_{-R}^R\frac{x^2e^{ix}}{(x^2+1)^2}\,dx+\int_0^\pi \frac{R^2e^{i2\phi}e^{iRe^{i\phi}}}{(R^2e^{i2\phi}+1)^2}\,iRe^{i\phi}\,d\phi \tag 2
\end{align}$$
As $R\to \infty$, the second integral on the right-hand side of $(2)$ approaches zero. Therefore, putting everything together we find that
$$\int_{-\infty}^\infty\frac{x^2e^{ix}}{(x^2+1)^2}\,dx=0$$
and therefore since $e^{ix}=\cos(x)+i\sin(x)$
$$\int_{-\infty}^\infty\frac{x^2\cos(x)}{(x^2+1)^2}\,dx=0$$
|
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"url": "https://math.stackexchange.com/questions/2064290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
An arctan exponential integral Prove the following for $\Re(z) >0$
$$ \log(\Gamma(z)) = 2\int_{0}^{\infty} \tan^{-1} \left(\dfrac{t}{z}\right)\dfrac{\mathrm{d}t}{e^{2\pi t} - 1} + \dfrac{\log(2z)}{2} + \left( z - \dfrac{1}{2} \right)\log (z) -z $$
I found this identity on Wolfram Functions here. I can't see where to start solving this problem.
I'm looking for elementary solution, I haven't learned about Complex Analysis or Abel Plana formula yet.
Please help.
Thanks.
|
Proposition : $$\int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x = \pi \left[\dfrac{1}{2} \log (2a\pi ) + a(\log a - 1) - \log(\Gamma(a+1)) \right]$$
Proof : Let $ \displaystyle \text{I} (a) = \int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x$
$\displaystyle = -\sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{e^{-2ar \pi x}}{1+x^2} \mathrm{d}x $
$\displaystyle = -\sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \int_{0}^{\infty} e^{-x(2ar \pi + y)} \sin y \ \mathrm{d}y \ \mathrm{d}x $
$\displaystyle = -\sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{\sin y}{2ar\pi + y} \mathrm{d}y $
$\displaystyle = - \sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{\sin x}{2ar\pi + x} \mathrm{d}x$
Substitute $\displaystyle x \mapsto 2 r \pi x$
$\displaystyle \implies \text{I}(a) = - \sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{\sin 2 r \pi x}{x + a} \mathrm{d}x $
$\displaystyle = -\int_{0}^{\infty} \dfrac{1}{x+a} \sum_{r=1}^{\infty} \dfrac{\sin(2 r \pi x)}{r} \mathrm{d}x $
$\displaystyle = -\pi \int_{0}^{\infty} \dfrac{1}{x+a} \left(\dfrac{1}{2} - \{x\} \right) \mathrm{d}x \quad \left( \because \sum_{r=1}^{\infty} \dfrac{\sin (2 r \pi x)}{r} = \dfrac{\pi}{2} - \pi \{x\} \right) $
$\displaystyle = -\pi \left[ \int_{0}^{\infty} \dfrac{\mathrm{d}x}{2(x+a)} - \int_{0}^{\infty} \dfrac{\{x\}}{x+a} \mathrm{d}x \right] $
$\displaystyle = -\pi \lim_{n \to \infty} \left[ \dfrac{1}{2} \log \left(\dfrac{a+n}{a}\right) - \text{A}(n) \right]$
where $\displaystyle \text{A}(n) = \int_{0}^{n} \dfrac{\{x\} }{x+a} \mathrm{d}x$
Now,
$\displaystyle \text{A}(n) = \int_{0}^{n} \dfrac{\{x\} }{x+a} \mathrm{d}x $
$\displaystyle = \sum_{k=0}^{\infty} \int_{k}^{k+1} \dfrac{x-k}{x+a} \mathrm{d}x $
$\displaystyle = \sum_{k=0}^{\infty} \left[1 - (k+a)\log \left(\dfrac{k+a+1}{k+a}\right) \right] $
$\displaystyle = n - \sum_{k=0}^{\infty} \left[ (k+a+1) \log (k+a+1) - (k+a) \log (k+a) - \log (k+a+1) \right] $
$\displaystyle = n + a\log a - (a+n) \log (a+n) +\log(a \cdot (a+1) \cdot \ldots \cdot (a+n)) - \log a $
$\displaystyle \implies \text{I} (a) = -\pi \lim_{n \to \infty} \left[ \dfrac{1}{2} \log \left(\dfrac{a+n}{a}\right) - n - a\log a + (a+n) \log (a+n) - \log(a \cdot (a+1) \cdot \ldots \cdot (a+n)) + \log a \right] $
Note that $\displaystyle \lim_{n \to \infty} \dfrac{ n! n^t}{t \cdot (t+1) \cdot \ldots \cdot (t+n)} = \Gamma(t) $
$\displaystyle \implies \text{I} (a) = -\pi \lim_{n \to \infty} \left[ \dfrac{1}{2} \log \left(\dfrac{a+n}{a}\right) - n - a\log a + (a+n) \log (a+n) - a\log(n) - \log (n!) + \log(\Gamma(a)) + \log a \right] $
Simplifying using Stirling's Approximation and $\displaystyle \lim_{n \to \infty } n \log \left(1 + \dfrac{a}{n} \right) = a $, we have,
$\displaystyle \text{I} (a) = -\pi \left[\log(\Gamma(a+1)) - \dfrac{1}{2} \log(2a \pi) - a (\log(a) -1) \right] $
$\displaystyle = \pi \left[ \dfrac{1}{2} \log(2a \pi) + a (\log(a) -1) - \log(\Gamma(a+1)) \right] \quad \square $
Now,
Applying Integration By Parts on the proposition and simplifying, we get,
$ \displaystyle \int_{0}^{\infty} \tan^{-1} \left(\dfrac{t}{a}\right)\dfrac{\mathrm{d}t}{e^{2\pi t} - 1} = \dfrac{1}{2} \left[ \log(\Gamma(a)) - \dfrac{\log(2a)}{2} - \left( a - \dfrac{1}{2} \right)\log (a) +a\right] $
$$\therefore \log(\Gamma(a)) = 2\int_{0}^{\infty} \tan^{-1} \left(\dfrac{t}{a}\right)\dfrac{\mathrm{d}t}{e^{2\pi t} - 1} + \dfrac{\log(2a)}{2} + \left( a - \dfrac{1}{2} \right)\log (a) -a \quad \square $$
|
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|
Integral $ \int_{0}^{\infty} \ln x\left[\ln \left( \frac{x+1}{2} \right) - \frac{1}{x+1} - \psi \left( \frac{x+1}{2} \right) \right] \mathrm{d}x $
Prove That : $$ \int_{0}^{\infty} \ln x\left[\ln \left( \dfrac{x+1}{2} \right) - \dfrac{1}{x+1} - \psi \left( \dfrac{x+1}{2} \right) \right] \mathrm{d}x = \dfrac{\ln^2 2}{2}+\ln2\cdot\ln\pi-1 $$
where $\psi(z)$ denotes the Digamma Function.
This integral arose from my attempt to find an alternate solution to Problem 5, i.e,
$$ {\large\int}_0^\infty\frac{\ln\left(x^2+1\right)\,\arctan x}{e^{\pi x}-1}dx=\frac{\ln^22}2+\ln2\cdot\ln\pi-1 $$
Here's my try : We have the identity,
$$ \displaystyle \int_0^\infty \frac{\ln y}{(y+a)^2 + b^2}\,\mathrm{d}y \; = \; \tfrac{1}{2b}\,\tan^{-1}\tfrac{b}{a}\,\ln(a^2+b^2) $$
Since substituting $y \mapsto \dfrac{a^2+b^2}{y}$ gives,
$$\displaystyle \int_0^\infty \frac{\ln y}{(y+a)^2+b^2} \, \mathrm{d}y =\ln(a^2+b^2)\int_0^\infty \frac{dy}{y^2+2ay+a^2+b^2} - \int_0^\infty \frac{\ln y}{(y+a)^2+b^2} \, \mathrm{d}y $$
$$ \implies \displaystyle \int_0^\infty \frac{\ln y}{(y+a)^2 + b^2}\,\mathrm{d}y \; = \; \tfrac{1}{2b}\,\tan^{-1}\tfrac{b}{a}\,\ln(a^2+b^2) $$
Putting $a=1$ and $b=x$, we have,
$ \displaystyle \int_0^\infty \frac{2x \ln y}{(y+1)^2 + x^2}\,\mathrm{d}y \; = \; \,\,\ln(1+x^2) \ \tan^{-1}x \tag{1} $
Now, we have to prove,
$$\displaystyle {\large\int}_0^\infty\frac{\ln\left(x^2+1\right)\,\tan^{-1}x}{e^{\pi x}-1} \mathrm{d}x=\frac{\ln^22}2+\ln2\cdot\ln\pi-1$$
Let,
$$\displaystyle \text{I} = {\large\int}_0^\infty\frac{\ln\left(x^2+1\right)\,\tan^{-1}x}{e^{\pi x}-1} \mathrm{d}x $$
$$\displaystyle = \int_{0}^{\infty} \int_{0}^{\infty} \frac{2x \ln y}{[(y+1)^2 + x^2][e^{\pi x} - 1]} \mathrm{d}x \ \mathrm{d}y \quad (\text{From 1}) \tag{2}$$
The inner integral is of the form,
$$ \displaystyle \text{J} = \int_{0}^{\infty} \dfrac{x}{(x^2+a^2)(e^{\pi x} - 1)} \ \mathrm{d}x \ ; \ a = (y+1)$$
I have proved here that,
$\displaystyle \int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x = \pi \left[\dfrac{1}{2} \log (2a\pi ) + a(\log a - 1) - \log(\Gamma(a+1)) \right] \tag{3}$
Differentiating both sides w.r.t. $a$, substituting $ a \mapsto \frac{a}{2} $ and $ x \mapsto \frac{x}{a} $, we get,
$\displaystyle \int_{0}^{\infty} \dfrac{x}{(x^2+a^2)(e^{\pi x} - 1)} \ \mathrm{d}x = \dfrac{1}{2} \left[ \dfrac{1}{a} + \ln \left( \dfrac{a}{2} \right) - \psi \left( \dfrac{a}{2} + 1 \right) \right] \tag{4}$
Putting $(4)$ in $(2)$, we have,
$$ \displaystyle \text{I} = \int_{0}^{\infty} \left[ \dfrac{\ln y}{y+1} + \ln y \ln \left( \dfrac{y+1}{2} \right) - \ln y \ \psi \left( \dfrac{y+1}{2} + 1 \right) \right] \mathrm{d}y $$
$ = \displaystyle \int_{0}^{\infty} \ln x\left[\ln \left( \dfrac{x+1}{2} \right) - \dfrac{1}{x+1} - \psi \left( \dfrac{x+1}{2} \right) \right] \mathrm{d}x \tag{*}$
Since the original question has already been proved in the link, so $(*)$ must be equal to the stated closed form. It also matches numerically.
I'm looking for some method to evaluate $(*)$ independent of Problem 5.
Any help will be greatly appreciated.
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Two Auxiliary Identities
First, we shall establish two simple identities.
$\textbf{Identity }(*)$
$$\int^1_0\left[\frac{1}{\log{x}}+\frac{1}{1-x}\right]x^{s-1}\ dx=\log{s}-\psi_0(s)\tag{*}$$
$\text{Proof Outline: }$ Differentiate with respect to $s$, recognise the integral representation of the trigamma function, then integrate back.
$\textbf{Identity }(**)$ $$\int^\infty_0e^{-ax}\log{x}\
dx=-\frac{\gamma+\log{a}}{a}\tag{**}$$
$\text{Proof Outline: }$ Differentiate the result $\int^\infty_0x^{s-1}e^{-ax}\ dx=a^{-s}\Gamma(s)$ with respect to $s$ and set $s=1$.
The Integral In Question
Applying these two identities and switching the order of integration gives us
\begin{align}
I
&:=\int^\infty_0\log{x}\left[\log\left(\frac{x+1}{2}\right)-\psi_0\left(\frac{x+1}{2}\right)-\frac{1}{x+1}\right]\ dx\\
&=\int^\infty_0\log{x}\int^1_0\left[\frac{1}{\log{t}}+\frac{1}{1-t}-\frac{1}{2}\right]t^{\frac{x-1}{2}}\ dt\ dx\\
&=\int^1_0\frac{1}{\sqrt{t}}\left[\frac{1}{\log{t}}+\frac{1}{1-t}-\frac{1}{2}\right]\int^\infty_0 \exp\left[-\left(-\frac{\log{t}}{2}\right)x\right]\log{x}\ dx\ dt\\
&=2\int^1_0\frac{\gamma+\log\left(-\tfrac{\log{t}}{2}\right)}{\sqrt{t}\log{t}}\left[\frac{1}{\log{t}}+\frac{1}{1-t}-\frac{1}{2}\right]\ dt\\
&=2\int^\infty_0\frac{e^{-u}(\gamma+\log{u})}{u}\left[\frac{1}{2u}-\frac{1}{1-e^{-2u}}+\frac{1}{2}\right]\ du
\end{align}
Let us define
$$I(s)=2\int^\infty_0u^{s-1}e^{-u}(\gamma+\log{u})\left[\frac{1}{2}+\frac{1}{2u}-\frac{1}{1-e^{-2u}}\right]\ du$$
so that $I=I(0)$. We have
\begin{align}
I(s)
&=\color{red}{\gamma\Gamma(s)+\gamma\Gamma(s-1)}+\color{blue}{\Gamma'(s)+\Gamma'(s-1)}-2\sum^\infty_{n=0}\int^\infty_0 u^{s-1}e^{-(2n+1)u}(\gamma+\log{u})\ du\\
&=\color{red}{\frac{\gamma\Gamma(1+s)}{s}\frac{s}{s-1}}\color{blue}{-\frac{s\Gamma(s)\psi_0(s)}{1-s}-\frac{\Gamma(s)}{(1-s)^2}}-2\sum^\infty_{n=0}\frac{\Gamma(s)\left(\gamma+\psi_0(s)-\log(2n+1)\right)}{(2n+1)^s}\\
&=\color{red}{\frac{\gamma\Gamma(1+s)}{s-1}}\color{blue}{-\frac{s\Gamma(s)\psi_0(s)}{1-s}-\frac{\Gamma(s)}{(1-s)^2}}\color{green}{-2(1-2^{-s})\Gamma(s)\zeta(s)(\gamma+\psi_0(s))}\\
&\color{purple}{\ \ \ \ -2\Gamma(s)\frac{d}{ds}(1-2^{-s})\zeta(s)}\\
\end{align}
With the expansions
\begin{align}
\Gamma(s)&\sim_0\frac{1}{s}-\gamma+\mathcal{O}(s)\\
\gamma+\psi_0(s)&\sim_0 -\frac{1}{s}+\mathcal{O}(s)\\
(1-2^{-s})\zeta(s)&\sim_0\left(s\log{2}-\frac{s^2\log^2{2}}{2}+\mathcal{O}(s^3)\right)\left(-\frac{1}{2}-\frac{s\log(2\pi)}{2}+\mathcal{O}(s^2)\right)\\
&\sim_0 -\frac{s\log{2}}{2}-\left(\frac{\log^2{2}}{4}+\frac{\log 2\log{\pi}}{2}\right)s^2+\mathcal{O}(s^3)
\end{align}
we obtain
\begin{align}
I(s)
&\sim_0\color{red}{-\gamma}\color{blue}{-1+\gamma}\color{green}{-\frac{\log{2}}{s}+\gamma\log{2}-\frac{\log^2{2}}{2}-\log{2}\log\pi}+\color{purple}{\frac{\log{2}}{s}-\gamma\log{2}}\\
&\color{purple}{\ \ \ \ \ \ \ +\log^2{2}+2\log{2}\log\pi}+\mathcal{O}(s)\\
&\sim_0 -1+\log{2}\log{\pi}+\frac{\log^2{2}}{2}+\mathcal{O}(s)
\end{align}
as was to be shown.
Note:
One can obtain the values of the derivatives of the zeta function at $0$ by differentiating Riemann's functional equation and using the Laurent series of the zeta function at $1$.
|
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|
Prob. 4, Chap. 3 in Baby Rudin: How to show that these are the limit superior and the limit inferior? Here's Prob. 4, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Find the upper and lower limit of the sequence $\left\{ s_n \right\}$ defined by $$s_1 = 0; \ s_{2m} = \frac{s_{2m-1}}{2}; \ s_{2m+1} = \frac{1}{2}+ s_{2m}.$$
My effort:
We note that $s_1 = 0$, $s_2 = 0$, $s_3 = \frac{1}{2}$, $s_4 = \frac{1}{4}$, $s_5 = \frac{1}{2} + \frac{1}{4}$, $s_6 = \frac{1}{4} + \frac{1}{8}$, $s_7 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8}$, $s_8 = \frac{1}{4} + \frac{1}{8} + \frac{1}{16}$, $s_9 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}$, $s_{10} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32}$, $s_{11} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32}$, $s_{12} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64}$, and continuing in this way we obtain
$$s_{2m} = \frac{1}{2}- \frac{1}{2^m} \ \mbox{ and } \ s_{2m+1} = 1- \frac{1}{2^m} $$ for $m = 0, 1, 2, 3, \ldots$. So $$\lim_{m \to \infty} s_{2m} = \frac{1}{2} \ \mbox{ and } \ \lim_{m \to \infty} s_{2m+1} = 1.$$
Am I right?
Using the above, how to rigorously prove that $$\lim\inf_{n \to \infty} s_n = \frac{1}{2} \ \mbox{ and } \ \lim\sup_{n\to\infty} s_n = 1?$$
|
Suppose there exists a sub sequence of $\{s_n\}$ converges to a point $s$. That is $\{s_{n_k} \}\to s$. Define $S_1$ as the set of terms with odd indices. That is, $n_k$ is odd and similarly $S_2$ for even terms. Note that either $S_1$ or $S_2$ has infinite elements, but not both. Hence, $s=1$ or $s=1/2$. Thus for any limit $s$ of sub sequence we have $1/2\leq s\leq 1$ and the result follows.
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|
Equilateral triangle trisected In an equilateral triangle $ABC$ the side $BC$ is trisected by points $D$ and $E$. Prove $9|AD|^2 = 7|AB|^2$.
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Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = $\frac{BC}{2}$ = $\frac{a}{2}$
And using phythagorous theorem,
AE = $\sqrt{AB^2 - BE^2}$
AE = $\frac{\sqrt{3}a}{2}$
Given that, BD = $\frac{1}{3}$ BC
∴ BD = $\frac{a}{3}$
DE = BE - BD = $\frac a{2} - \frac a{3} = \frac a{6}$
Applying Pythagoras theorem in ΔADE, we obtain
$AD^2 = AE^2 + DE^2$
= $\left(\frac{\sqrt{3}a}{2}\right)^2 + \left(\frac{a}{6}\right)^2$
=$\left(\frac{3a^2}{4}\right) + \left(\frac{a^2}{36}\right)$
$AD^2 = \frac{28}{36} a^2$
$9 AD^2 = 7 AB^2$
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|
Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
Let $a$ and $b$ be integers. Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
I saw that $a^3-b^3 = (a-b)(a^2+ab+b^2)$ and $(a^2+ab+b^2) = (a+b)^2-ab$. How can we use the fact that $10 \mid (a^2+ab+b^2)$ to solve this question?
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Clearly, $a^2+ab+b^2$ has to be even $\implies a,b$ both must be even(why?)
WLOG $a=2A,b=2B$
Now as $10|(a^2+ab+b^2),5|(a^2+ab+b^2)$
$$a^2+ab+b^2=(A+2B)^2+3A^2\implies(A+2B)^2\equiv-3A^2\pmod5$$
Again, for any integer $r,r\equiv0,\pm1,\pm2\pmod5\implies r^2\equiv0,\pm1$
$\implies(A+2B)^2\equiv0,\pm3\pmod5$ which is only possible if $5|(A+2B)$
$\implies5|B^2\implies5|B$ and $\implies5|(A+2B)\implies5|A$
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Prove A New Method For Finding Primes. I was recently messing around with harmonic numbers - that is
$H_x = \sum^x_{k=1}\frac{1}{k}$ - and I was thinking what would happen if you applied Gauss' trick to the sum, which is to add up the first and last term, then the second and second-to-last term and so on. This is usually used in adding natural numbers, but when I applied it to the sum I noticed a recurring numerator.
$$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\to\frac{7}{6}+\frac{7}{10}+\frac{7}{12}$$
I also noticed there were some cases where there wasn't, like $H_{8}$
$$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\to\frac{9}{8}+\frac{9}{14}+\frac{\textbf{1}}{2}+\frac{9}{20}$$
From this I guessed that if $H_x$ with Gauss' trick applied has a constant numerator, then $x+1$ is prime. I tried proving this, but I'm not sure if I'm right.
*
*Gauss Trick
*Harmonic Numbers
|
If you consider the $n$-th harmonic number, you add the terms $\frac{1}{k}$ and $\frac{1}{n - k + 1}$. Their sum is
$$\frac{1}{k} + \frac{1}{n - k + 1} = \frac{n - k + 1}{k(n - k + 1)} + \frac{k}{k(n - k + 1)} = \frac{n + 1}{k(n - k + 1)}$$
For $k = 1$ we get $\frac{n + 1}{n}$, which is a fraction in reduced form, since $\gcd(n + 1, n) = 1$. This means that the only possible value for the common numerator is $n + 1$. But this will be the case if and only if $\gcd(n + 1, k) = 1$ and $\gcd(n + 1, n - k + 1) = 1$. For the latter we have
$$\gcd(n, n - k + 1) = \gcd(n + 1, n + 1 - k) = \gcd(n + 1, k).$$
This means that all fractions will be in reduced form if and only if $\gcd(n + 1, k) = 1$ for all $1 \le k \le \lfloor \frac{n}{2}\rfloor$, which is equivalent to $n + 1$ being prime.
|
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|
Prove that $\sum\limits_{cyc}\left(\frac{a+b}{a+b+c}\right)^2\geq\frac{16}{9}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\left(\frac{a+b}{a+b+c}\right)^2+\left(\frac{b+c}{b+c+d}\right)^2+\left(\frac{c+d}{c+d+a}\right)^2+\left(\frac{d+a}{d+a+b}\right)^2\geq\frac{16}{9}$$
I tried C-S and more, but without success.
I am looking for an human proof, which we can use during competition.
|
HINT
Using substitutions $x =a+b+c, y = b+c+d, z = c+d+a, t = a+b+d$ we get $$(\frac{a+b}{a+b+c})^2=\frac 1 9(2 + \frac t x - \frac y x -\frac z x)^2$$
Similar for the other terms. Eliminate the squares then use the fact that $r + \frac 1 r \ge 2, r \gt 0$.
|
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|
Simplifying $9^{3/4}$, I get $3\sqrt[4]{9}$, but that's not the answer. Why? I am trying to simplify:
$9^\frac{3}{4}$
So this is what I did:
$9^\frac{3}{4} = \sqrt[4]{9^3}$
$\sqrt[4]{3*3*3*3*3*3}$
$3\sqrt[4]{3*3}$
$3\sqrt[4]{9}$
$3\sqrt[4]{3^2}$
I don't see how I can simplify this even more, however the answer I provided is incorrect. How can I simplify this even more?
|
Simplify it this way:
$
9^\frac{3}{4} \\
= (3^2)^\frac{3}{4}\\
= 3^\frac{2\times 3}{4}\\
= 3^\frac{3}{2}\\
= 3^{\left(1 + \frac{1}{2}\right)}\\
= 3 \times 3^\frac{1}{2}\\
= 3 \sqrt{3}
$
|
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|
Volume of sphere-cube intersection I am looking for a formula for the volume $V_i(r,L)$ of the intersection between a sphere of radius $r$ and a cube of edge $L$ with the same center.
For $r<\frac L 2$, $V_i$ is simply equal to the volume of the sphere:
$$V_i(r,L) = \frac 4 3 \pi r^3 \ \ \ \ r<\frac L 2$$
For $r>\frac{\sqrt 3} 2 L $, it is equal to the volume of the cube:
$$V_i(r,L) = L^3 \ \ \ \ r>\frac {\sqrt 3} 2 L$$
For $\frac L 2 < r < \frac{\sqrt 2} 2 L$, it is equal to the volume of the sphere minus six times the volume of a spherical cap. The result is
$$V_i(r,L) = \frac 4 3 \pi r^3 -\frac \pi 4(2r-L)^2(4r+L) \ \ \ \ \frac L 2 < r < \frac{\sqrt 2} 2 L$$
However, for $\frac{\sqrt 2} 2 L < r < \frac{\sqrt 3} 2 L$ I have not been able to think of an easy way to calculate $V_i(r,L)$.
|
As John Hughes notes, you can fix the side of the cube or radius of the sphere and work in terms of the ratio. The strategy here is to fix the cube's edge length to be $2$, calculate the area of plane slices parallel to one face of the cube, then apply Cavalieri's theorem.
Theorem: The intersection of the square $[-1, 1]^{2}$ with the disk of radius $\rho$ centered at the origin has area
$$
A(\rho) = \begin{cases}
\pi \rho^{2} & 0 \leq \rho \leq 1, \\
\pi \rho^{2} - 4\rho^{2} \arccos(1/\rho) + 4\sqrt{\rho^{2} - 1} & 1 < \rho < \sqrt{2}, \\
4 & \sqrt{2} \leq \rho.
\end{cases}
$$
The intersection of the cube $[-1, 1]^{3}$ with the ball of radius $r$ centered at the origin has volume
$$
V(r) = \int_{-\min(1, r)}^{\min(1, r)} A(\sqrt{r^{2} - x^{2}})\, dx
= 2\int_{0}^{\min(1, r)} A(\sqrt{r^{2} - x^{2}})\, dx.
$$
Trivial Cases/Sanity Checks:
*
*If $0 \leq r \leq 1$, then $0 \leq \sqrt{r^{2} - x^{2}} \leq 1$ for $|x| \leq r$, so $A(\sqrt{r^{2} - x^{2}}) = \pi(r^{2} - x^{2})$ and $V(r) = \frac{4}{3}\pi r^{3}$.
*If $\sqrt{3} \leq r$, then $\sqrt{2} \leq \sqrt{r^{2} - x^{2}}$ for $|x| \leq 1$, so $A(\sqrt{r^{2} - x^{2}}) = 4$ and $V(r) = 8$.
Interesting Cases (edited to fix mishap with limits of integration):
*
*If $1 < r \leq \sqrt{2}$ (dark gray), then
*
*$\sqrt{r^{2} - x^{2}} \leq 1$ for $\sqrt{r^{2} - 1} \leq |x| \leq 1$,
*$1 \leq \sqrt{r^{2} - x^{2}} \leq \sqrt{2}$ for $|x| \leq \sqrt{r^{2} - 1}$,
so the volume is
\begin{align*}
V(r) &= 2\int_{0}^{1} A(\sqrt{r^{2} - x^{2}})\, dx \\
&= 8\int_{0}^{\sqrt{r^{2} - 1}} \bigl[-(r^{2} - x^{2})\arccos\left(\frac{1}{\sqrt{r^{2} - x^{2}}}\right) + \sqrt{r^{2} - 1 - x^{2}}\bigr]\, dx \\
&\quad + 2\pi\int_{\sqrt{r^{2}-1}}^{1} (r^{2} - x^{2})\, dx.
\end{align*}
(The second integral evaluates to $\frac{2}{3}\pi\bigl[(3r^{2} - 1) - (2r^{2} + 1)\sqrt{r^{2} - 1}\bigr]$, of course. I haven't verified explicitly that $V(r)$ evaluates to $\frac{4}{3}\pi r^{3} - 2\pi(r - 1)^{2}(2r + 1)$; the integral is elementary, but integrating by parts and making the obvious trig substitution gives a rational function of sine that looks tedious to integrate.)
*If $\sqrt{2} < r < \sqrt{3}$ (light gray), then
*
*$1 \leq \sqrt{r^{2} - x^{2}} \leq \sqrt{2}$ for $\sqrt{r^{2} - 2} \leq |x| \leq 1$,
*$\sqrt{2} \leq \sqrt{r^{2} - x^{2}} \leq \sqrt{3}$ for $|x| \leq \sqrt{r^{2} - 2}$,
so the volume is
\begin{align*}
V(r) &= 8\sqrt{r^{2} - 2} \\
&\quad+ 8\int_{\sqrt{r^{2}-2}}^{1} \bigl[-(r^{2} - x^{2})\arccos\left(\frac{1}{\sqrt{r^{2} - x^{2}}}\right) + \sqrt{r^{2} - 1 - x^{2}}\bigr]\, dx.
\end{align*}
Again, this integral is elementary, and can be evaluated exactly (in principle).
|
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|
Finding necessary and sufficient condition for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$ My question is that:
Find the necessary and sufficient condition for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$.
I know that answer is $k=-3$, as I know that $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.
But I can't prove that $k=-3$ is also a necessary condition. A mathematical proof is needed.
Thanks.
|
A low key answer is: We have $x^3+y^3+z^3+kxyz = (3+k)xyz + (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. Thus this means $(3+k)xyz$ must be divisible to $x+y+z$ for all $x,y,z$ integers. Put $x = y = 1$ we have $(3+k)z$ is divisible to $z+2$, thus you can write $(3+k)z = n(z+2)$, for all $z$ integer. Thus $n = \dfrac{(k+3)z}{z+2} = k+3 - \dfrac{2k+6}{z+2} \in \mathbb{Z}$ for all $z$ integers. Thus $2k+6 = 0 \implies k = -3$ as claimed.
|
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|
Let $P(x)=4x^2+6x+4$ and $Q(x) =4y^2-12y+25$. If $x,y$ satisfy the equation $P(x)Q(y)=28$, then find the value of $11y-26x$.
Let $P(x)=4x^2+6x+4$ and $Q(y) =4y^2-12y+25$. If $x,y$ satisfy the equation $P(x)Q(y)=28$, then find the value of $11y-26x$.
I know this question has been asked earlier here but the answer given states "Note that we have...". I want to know how we can arrive at that statement mathematically instead of trial and error method, because I wasn't able to get that while trying to solve it.
Any other way of solving this question would also be helpful.
|
It can be done by completing the square method,
$P(x)=4x^2+6x+4$
Let P(x) = 0.
Then,
$4x^2+6x+4 = 0$
First divide equation by 4,
So we have,
$x^2+\frac64x+1 = 0$
$x^2+\frac32x = -1$
The term of x is positive so compare it with,
$a^2 + 2ab + b^2 = (a + b)^2$
So, a = x,
And 2ab = $\frac32x$
2b = $\frac32$
b = $\frac34$
Then $b^2 = \frac{9}{16}$
So above equation become,
$x^2+\frac32x+\frac9{16}= -1+\frac{9}{16}$
$(x+\frac34)^2 = \frac{-7}{16}$
$(x+\frac34)^2 + \frac{7}{16} = 0$
In Q(x) also divide the equation by 4 and then compare with
$a^2 - 2ab + b^2 = (a - b)^2$ as term containing x is negative.
|
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|
finding value of $ \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$ finding value of $\displaystyle \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$
Substituting $x=\tan^2 \theta\;, dx = 2\tan \theta \sec^2 \theta$
integral is $=\displaystyle \int \frac{2\tan \theta \sec^2 \theta }{\tan^6 \theta \cdot \sec^3 \theta}d\theta= 2\int\frac{\cos^6 \theta}{\sin^5 \theta}d\theta $
wan,t be able to proceed after, could some help me
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A big partial fraction decomposition!
We set $u =\cos x$ to get $$I = -\int \frac{\cos ^6 x}{(\cos ^2x-1)^3} \sin x dx = \int \frac{u^6}{(u^2-1)^3} du =\int (\frac{3u^4-3u^2+1}{(u^2-1)^3} + 1) du = I_1 + I_2$$
For $I_1$, perform the partial fraction decomposition $$I_1 = \int \frac{3u^4-3u^2+1}{(u-1)^3(u+1)^3} du =-\int \frac{15}{16(u+1)} du +\int \frac{9}{16(u+1)^2} du -\int \frac{1}{8(u+1)^3} du +\int \frac{15}{16(u-1)} du + \int \frac{9}{16(u-1)^2} du +\int \frac{1}{8(u-1)^3} du $$ The integration of $I_2$ is however very easy. Hope you can take it from here.
|
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|
Prove that $A = \dfrac{f^2(1)+f^2(-1)}{2}$ is a composite number
Given $f(x) = a^{2016}x^2+bx+a^{2016}c-1$ where $a,b,c \in \mathbb{Z}$, suppose that the equation $f(x) = -2$ has two positive integer solutions. Prove that $A = \dfrac{f(1)^2+f(-1)^2}{2}$ is a composite number.
Let $g(x) = f(x)-2 = a^{2016}x^2+bx+a^{2016}c-3$, and let $-r,-s$ be the two positive integer roots of $g(x)$. Then by Vieta's Formula, $r+s = \dfrac{b}{a^{2016}}$ and $rs = \dfrac{a^{2016}c-3}{a^{2016}}$. Thus $a = \pm 1$ and so $f(x) = x^2+bx+c-1$ and so $f(1) = b+c$ and $f(-1) = c-b$. Then $$A = \dfrac{[(b+c)^2+b(b+c)+c-1]+[(c-b)^2+b(c-b)+c-1]}{2} = b^2+bc+c^2+c-1.$$ How do we prove this is composite?
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The claim is not true: taking $a=1, \ b=2, \ c=3$, the equation $f(x) - 2 = 0$ is $x^2 + 2x = 0$, having both roots integers ($0$ and $-2$), but $A = \dfrac {5^2 + 1^2} 2 = 13$, clearly not a composite number.
|
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|
For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$? For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$?
1.$1$
2.$2$
3.$3$
4.$4$
5.$5$
My attempt:It's clear that it is true for $n=2$.Because:
$(a^3-1)(a^2-1) \ge 0$
Is true because $a^3-1$ and $a^2-1$ are both negative or both positive.But there is a problem:How can we make sure it doesn't hold for any other $n$?
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Clearly $\;n=5\;$ is out of question, as $\;a^3\le1\;$ isn't true for all $\;a>0\;$.
Something similar with $\;n=4\;$ , as then
$$a^5+1\ge a^3+a^4\iff (a+1)(a^4-a^3+a^2-a+1)\ge a^3(a+1)$$
which is false for values close to $\;1\;$ from the right, for example $\;a=1.1\;$ (check this).
With $\;n=3\;$ we'd get
$$a^5+1\ge 2a^3\;,\;\;\text{ false, again for}\;\;a=1.1$$
Finally, for $\;n=1\;$ we'd get
$$a^5+1\ge a^3+a\;\;,\;\;\text{ false for}\;\;a=0.9 $$
|
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|
Is probability that $x<2$ equal to the probability that $x^2<4$ given $1Assuming $x$ is a real number uniformly distributed over the interval $(1,3).$
so $x^2$ is also uniformly distributed over the interval $(1,9)${As for every $x=a\in (1,3) $ there exists $x^2=a^2\in (1,9)$}.
Probability that $x<2$ would be $\frac{1}{2}$ as $x$ can be in $(1,2)$ where sample set of $x$ is $(1,3)$, while probability that $x^2<4$ is $\frac{3}{8}$ as $x^2$ can be in $(1,4)$ where sample set of $x^2$ is $(1,9).$
So why is the probability that $x<2$ different from $x^2<4$ if both are identical?
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If $X$ is uniformly distributed on $[1,3]$, then $P(X^2\le a^2)=P(X\le a)=\frac{a-1}2$. Therefore, $P(X^2\le a)=\frac{\sqrt{a}-1}2$. Thus, the PDF of $X^2$ is $\frac1{4\sqrt{a}}$.
That is, $X^2$ is not uniformly distributed on $[1,9]$.
|
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|
Find all natural numbers $k$ for which there exist natural numbers $n,m$ such that $m(m+k) = n(n+1)$
Find all natural numbers $k$ for which there exist natural numbers $n,m$ such that $m(m+k) = n(n+1)$.
Rearranging the given equation gives $k = \dfrac{n(n+1)}{m}-m$. Thus, $m \mid n(n+1)$ and so we have cases.
Case 1: $n = am$ for some $a \in \mathbb{Z}^+$
In this case we have $k = a(am+1)-m$, which is positive for all $a,m$.
Case 2: $n+1= am$ for some $a \in \mathbb{Z}^+$
In this case we have $k = a(am-1)-m$ and we need $k$ to be positive and so $a(am-1)-m \geq 1$, or $am \geq m+1$. This is true if and only if $a \geq 2$.
I didn't see how to solve the other case where both are not multiples of $m$.
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For each positive integer $N$, let $\tau(N)$ denote the number of positive integers $d$ that divides $N$. For a given positive integer $u$, we also denote by $\bar{\tau}_u(N)$ the number of odd positive integers $d<u$ that divide $N$. If $N=2^r\tilde{N}$, where $r\in\mathbb{Z}_{\geq 0}$ and $\tilde{N}\in\mathbb{Z}$ is odd, then write $\tilde{\tau}_u(N)$ the number of positive integers $d<u$ of the form $2^r\tilde{d}$, where $\tilde{d}$ is a divisor of $\tilde{N}$. Obviously, $\bar{\tau}_1(N)=\tilde{\tau}_1(N)=0$ and $\bar{\tau}_u(N)\geq 1$ for $u>1$.
We claim that, for a given positive integer $k>1$, the number of solutions $(m,n)\in\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}$ to the Diophantine equation
$$m(m+k)=n(n+1)$$
is
$$s_k:=\left\{
\begin{array}{ll}
\bar{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)+\tilde{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)&\text{if }k\text{ is odd}\,,
\\
\dfrac{1}{2}\,\tau(k^2-1)-1&\text{if }k\text{ is even}\,.
\end{array}\right.$$
Moreover, we set $s_1:=\infty$. In particular, $s_k=0$ if and only if $k=2$ or $k=3$. For an even integer $k>3$, there exists a unique solution if and only if $(k-1,k+1)$ is a pair of twin primes. For an odd integer $k>3$, there exists a unique solution if and only if $k=7$ or $k=2F-1$, where $F$ is a Fermat prime (i.e., $F$ is a prime such that $F=2^{2^{l}}+1$ for some nonnegative integer $l$).
When $k=1$, there are $s_1=\infty$ solutions to the case $k=1$ given by $m=n$. Suppose from now on that $k>1$. Clearly, we hae $m<n$. Furthermore,
$$(2n+1)^2< (2n+1)^2+(k^2-1)=4n(n+1)+k^2=4m(m+k)+k^2=(2m+k)^2\,.$$
That is, $2n+2\leq 2m+k$. If $m=n-t$, then we have
$$1\leq t \leq \frac{k}{2}-1\,.$$
(At this stage, it is clear that, for a solution to exist, we must have $\dfrac{k}{2}-1\geq 1$, or $k\geq 4$.)
We then rewrite $m(m+k)=n(n+1)$ as
$$n(k-2t-1)=t(k-t)\,.$$
If $k$ is odd, we have
$$2n\,\left(\frac{k-1}{2}-t\right)=t(k-t)\,.$$
That is, since $t\equiv \frac{k-1}{2}\pmod{\frac{k-1}{2}-t}$, we have
$$\dfrac{k-1}{2}-t \,\mid\,\dfrac{k^2-1}{4}\,.$$
Indeed, for any divisor $d\in\mathbb{Z}_{>0}$ of $\dfrac{k^2-1}{4}$ such that $d<\dfrac{k-1}{2}$, we take $t:=\dfrac{k-1}{2}-d$, $$n:=\frac{\left(\frac{k-1}{2}-d\right)\left(\frac{k+1}{2}+d\right)}{2d}\text{ and }m:=n-t=n-\frac{k-1}{2}+d$$
to obtain a solution. Now, note that $n$ defined in this way is a positive integer if and only if $d$ is odd, or $d$ is a maximally even divisor of $\dfrac{n^2-1}{4}$ (that is, $d$ is an even divisor such that $2d$ is not a divisor). There are $s_k=\bar{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)+\tilde{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)$ possible values of $d$.
If $k$ is even, we have
$$4n\,(k-2t-1)=(2t)\,(2k-2t)\,.$$
Note that $2t\equiv k-1\pmod{k-2t-1}$, so
$$k-2t-1\,\mid\,k^2-1\,.$$
Indeed, let $d\in\mathbb{Z}_{>0}$ be a divisor of $k^2-1$ such that $d<k-1$. We obtain a solution by setting $t:=\dfrac{k-1-d}{2}$,
$$n:=\frac{(k-1-d)(k+1+d)}{4d}\,,\text{ and }m:=n-t=n-\frac{k-1-d}{2}\,.$$
There are $s_k=\frac{1}{2}\,\tau(k^2-1)-1$ values of $d$.
The first odd value of $k>1$ with $s_k>1$ is $k=11$, where we have $s_{11}=3$ solutions $$(m,n)=(1,3)\,,\,\,(m,n)=(3,6)\,,\text{ and }(m,n)=(10,14)\,.$$ The first even value of $k>1$ with $s_k>1$ is $k=8$, where we have $s_8=2$ solutions $$(m,n)=(2,4)\text{ and }(m,n)=(12,15)\,.$$ In both odd and even cases, quasi's solutions correspond to $d=1$.
|
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|
Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$. Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$.
My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
$a=2^{x_{1}}3^{y_{1}}5^{z_{1}}7^{w_{1}}11^{t_{1}}$
$b=2^{x_{2}}3^{y_{2}}5^{z_{2}}7^{w_{2}}11^{t_{2}}$
$c=2^{x_{3}}3^{y_{3}}5^{z_{3}}7^{w_{3}}11^{t_{3}}$
where $x_{1}+x_{2}+x_{3}=1$; $x_{1},x_{2},x_{3}$ being non-negative integers.
Number of solutions of this equation clearly $3$ and similarly for other variables as well.
So,number of solutions $=3^5$.
But answer given is $40$ .
Am I misinterpreting the question
|
Since we are looking for a set, $\{2310,1,1\}$ is not an answer - that is, not all the prime factors are gathered in the same factor.
Then, to avoid double-counting, we can say, for example, that $11$ always divides $a$ and the largest prime factor $q$ that does not divide $a$, divides $b$.
So for $q=7$, we can distribute the smaller primes arbitrarily among the three factors, giving $3^3$ options. Similarly for $q=\{5,3,2\}$ there are $\{3^2, 3^1, 3^0\}$ options giving a total of $27+9+3+1=\fbox{40}$ options altogether.
|
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|
When is the fraction $\frac{n+7}{2n+7}$ the square of a rational number?
Find all positive integers $n$ such that the fraction $\dfrac{n+7}{2n+7}$ is the square of a rational number.
The answer says $n = 9,57,477$, but how do we prove this? I wrote $\dfrac{n+7}{2n+7} = \dfrac{a^2}{b^2}$ for some $a,b \in \mathbb{Z^+}$ and $\gcd(a,b) = 1$. We know that $\gcd(n+7,2n+7) = \gcd(n+7,7) = 1,7$ and so $n+7$ and $2n+7$ must both be perfect squares or their greatest common divisor is $7$. How do we find all values for which this is the case?
|
There are more solutions (n<1000000):
$$
\begin{array}{c}
9 \\
57 \\
168 \\
477 \\
2109 \\
5880 \\
16377 \\
71817 \\
199920 \\
556509 \\
\end{array}
$$
so you just plug these values of $n$ and check otherwise the question does not make sense. Also One solution in your list is missing for n<500.
Added after observation of ross-millikan:
$n+7$ and $2n+7$ can have common multiplier, is so than $2*(n+7)-(2n+7)$ is also devisible by the same number which is clearly $7$, so the theorem is that if numerator and denominator have a common multiplier it can only be $7$. Thus we have two possibilities
a) $n+7=a^2$ and $2n+7=b^2$
b) $n+7=7 a^2$ and $2n+7=7 b^2$
for b) n should be divisible by $7$
|
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|
Integer solutions to nonlinear system of equations $(x+1)^2+y^2 = (x+2)^2+z^2$ and $(x+2)^2+z^2 = (x+3)^2+w^2$
Do there exist integers $x,y,z,w$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2?\end{align*}
I was thinking about trying to show by contradiction that no such integers exist. The first equation gives $y^2 = 2x+3+z^2$ while the second gives $z^2 = 2x+5+w^2$. How can we find a contradiction from here?
|
The differences between consecutive squares are consecutive odd numbers. So we have:
$y^2-z^2=$ odd number
$z^2-w^2=$ next odd number
Of course if these were reversed (if the word "next" were moved one equation up) then the solution would be simple; any three consecutive numbers would work. As it is, it's trickier, but certainly not impossible.
Phrasing the equations as I have, a little thought will reveal that any sequence of consecutive odd numbers which can be partitioned such that the smaller numbers add up to an odd number that is $2$ greater than the sum of the larger numbers of the partition, will provide a solution.
What's more, any solution to the original equation will exhibit the characteristic described in the preceding paragraph; the two questions are equivalent.
Borrowing from another existing answer, let's check this solution against my statement above:
$$31^2 + 12^2 = 32^2 + 9^2 = 33^2 + 4^2$$
The odd numbers which comprise the difference between $4^2$ and $9^2$ are $$(4+5), (5+6),(6+7),(7+8),(8+9)$$ or in other terms: $$9,11,13,15,17$$
The odd numbers which comprise the difference between $9^2$ and $12^2$ are $$19,21,23$$
The first five numbers of course add up to $13\times5=65$, and the latter three numbers add up to $21\times3=63$, which illustrates my earlier conclusion.
And, of course, $65$ and $63$ are the differences between $(x+2)^2$ and $(x+3)^2$, and $(x+1)^2$ and $(x+2)^2$, respectively, which tells us that $x$ will equal half of the smaller of these two odd numbers, after the extra $3$ units are subtracted: $$\frac {63-3} 2=30$$ (This is because the smaller number, $63$, equals $(x+1)+(x+2)$.)
|
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|
Understanding the proof of $\displaystyle\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$ I need help understanding this:
Prove that $$\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$$
$$(\sqrt n)^\frac{1}{n}\geq
(\sqrt1)^\frac{1}{n}=1, \forall n\in\Bbb N$$
By binomial theorem we get for $n\geq 2$
$$n=((\sqrt n)^\frac{1}{n})^n=[1+((\sqrt n)^\frac{1}{n}-1)]^n=\sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k$$
$$\geq 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2=1+\frac{n(n-1)}{2}(\sqrt n^\frac{1}{n}-1)^2$$
$$\Rightarrow (\sqrt n^\frac{1}{n}-1)^2\leq \frac{2}{n}$$
$$\Rightarrow \sqrt n^\frac{1}{n}\leq 1+\frac{\sqrt 2}{\sqrt n}$$
$1+\frac{\sqrt 2}{\sqrt n}$ approaches $1$ for $n \rightarrow \infty$ so by the sandwich theorem we get $\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$
I don't understand this part:
$$
\sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k\geq 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2.$$ Could someone explain how they got that on the RHS?
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Since $n^{1/n}\ge 1$, then all of the terms in the binomial expansion
$$\left(1+(\sqrt[n]{n}-1)\right)^n=\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k$$
are non-negative. Therefore, we have for any $m\le n$
$$\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k\ge \sum_{k=0}^m\binom{n}{k}(\sqrt[n]{n}-1)^k$$
Taking $m=2$ and $n\ge 2$ yields
$$\begin{align}
\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k&\ge \sum_{k=0}^2\binom{n}{k}(\sqrt[n]{n}-1)^k\\\\
&=1+n(\sqrt[n]{n}-1)+\frac{n(n-1)}{2}(\sqrt[n]{n}-1)^2\\\\
&\ge 1+\frac{n(n-1)}{2}(\sqrt[n]{n}-1)^2
\end{align}$$
where the last inequality is true since $n(\sqrt[n]{n}-1)\ge 0$.
|
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|
roots of quadratic polynomials are positive integer numbers Find positive integer numbers $a,b,c$ such that all the roots of these quadratic polynomials are positive integer numbers
1) $x^2−2ax+b=0$
2) $x^2−2bx+c=0$
3) $x^2−2cx+a=0$
Sorry but I don't know what to do :(( please help me :((
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The only solution in positive integers is $a=b=c=1$.
For the equations to have integer roots, their discriminants must be perfect squares. Taking the first equation for example, this means $a^2-b$ must be a perfect square. But $a^2-b \lt a^2$ thus it must be less than or equal the next smaller perfect square below $a^2$ which is $(a-1)^2$. Therefore:
$$a^2-b \le (a-1)^2 = a^2 - 2a +1 \quad \iff \quad b + 1 \ge 2a$$
Adding the $3$ similar inequalities together gives:
$$a+b+c+3 \ge 2(a+b+c) \quad \iff \quad a+b+c \le 3$$
The latter has $a=b=c=1$ as the unique solution in positive integers, in which case all $3$ equations reduce to $x^2-2x+1=(x-1)^2=0$ with the positive integer $1$ as a double root.
|
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|
Function is small $o$ of $x^2$ I have to solve the following exercise:
Give $a$, $b$, $c \in \mathbb R$ such that
$$\frac{1}{1-\cos x} = \frac{a}{x^2} + b +cx^2 + o(x^2)$$
for $x\to 0$.
Here's my attempt:
I know that $$\cos x = \frac{1}{2}\left( e^{ix} + e^{-ix}\right) = \frac{1}{2} \sum_{n=0}^\infty \frac{(ix)^{2n}}{(2n)!} = \frac{1}{2}\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!} = \frac{1}{2}\left(1-\frac{x^2}{2}+o(x^2)\right)$$
for $x\to 0$.
The question now is how to resolve $\frac{1}{1-\cos x}$. This much looks like an application of the geometric series here. I assume I can use it since $\cos x < 1$ for all sufficiently small $x \neq 0$. I wouldn't know how to solve this exercise otherwise. Proceeding, it follows that
$$\frac{1}{1-\cos x} = \frac{1}{1-\left(\frac{1}{2} - \frac{x^2}{4} + o(x^2)\right)} = \sum_{n=0}^\infty\left(\frac{1}{2} - \frac{1}{4}x^2+o(x^2)\right)^n = 1+ \frac{1}{2}-\frac{1}{4}x^2+o(x^2).$$
Choosing $a=0$, $b = 1,5$ and $c=-\frac{1}{4}$ establishes the case.
|
Your expansion of $\cos(x)$ at $0$ is obviously wrong, since it implies $\lim_{x\to 0}\cos(x)=\frac 12$.
You needn't use the power series for $e^{ix}$. I'd suggest that you rather use Taylor expansion at order $6$ for $\cos$, since the derivatives at $0$ are very simple. This method yields $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+o(x^6)$$
Hence $$\begin{align}
\frac{1}{1-\cos x}
&=\frac{1}{1-\left(1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+o(x^6) \right)}\\
&=\frac{2}{x^2}\frac{1}{1-\left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)}\\
&=\frac{2}{x^2} \left( 1+\left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)+ \left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)^2+ o\left(\left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)^2\right)\right)\\
&=\frac{2}{x^2}\left( 1+\frac{x^2}{12}-\frac{x^4}{360}+\frac{x^4}{144}+o\left(x^4\right)\right)\\
&=\frac{2}{x^2} + \frac 16 + \frac{x^2}{120}+o(x^2)
\end{align}$$
|
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|
About an integration technique My professor wrote this :
$\displaystyle \int x^2\log(\sqrt{1-x^2}) \,\mathrm{d}x = \frac{1}{3}\int \log(\sqrt{1-x^2}) \, \mathrm dx^3$
Can someone explain me what is going on with the integration variable ?
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$
\int x^2 \log \sqrt{1 - x^2} \ dx \\
= \int x^2 \log \sqrt{1 - x^2} \ \frac{dx}{dx^3} \ dx^3 \\
= \int x^2 \log \sqrt{1 - x^2} \ \frac{1}{3x^2} \ dx^3 \\
= \frac13 \int \log \sqrt{1 - x^2} \ dx^3
$
|
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|
Proof of binomial formula to extract coefficients of a generating function A generating function ends up rearranged into a form:
\begin{align}
(1+x+x^2+\dotsb)^n&=\frac{1}{(1-x)^n}\\[6px]
&= 1+\binom{1+n-1}{1}x+\binom{2+n-1}{2}x^2+\binom{3+n-1}{3}x^3\\
&\phantom{=\;1}+\dots+\binom{r+n-1}{r}x^r+\dotsb
\end{align}
used to extract coefficients.
I can't find a proof (some construction akin to Pascal's triangle, for example) of these coefficients working for all cases.
Probably, it is just a matter of knowing the term to include in the online search, and if this is the case, I'll be happy to delete the question.
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The statement is obviously true for $n=1$; suppose it is for $n$; then we can see that
\begin{align}
(1+x+\dotsb)^{n+1}=\frac{1}{n}D\frac{1}{(1-x)^n}=
\frac{1}{n}\sum_{k\ge1}k\binom{k+n-1}{k}x^{k-1}=
\sum_{k\ge0}\frac{k+1}{n}\binom{k+n}{k+1}x^k
\end{align}
and it's just a matter of proving that
$$
\frac{k+1}{n}\binom{k+n}{k+1}=\binom{k+n}{k}
$$
Indeed,
$$
\frac{k+1}{n}\binom{k+n}{k+1}=
\frac{k+1}{n}\frac{(k+n)(k+n-1)\dotsm(k+n-(k+1)+1)}{(k+1)!}=
\binom{k+n}{k}
$$
|
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|
Differentiating $y=x^6e^{-4x^3}$ Can some one explain, how to solve this derivative.
I'm total beginner.
Would be preferable if some one explain step-by-step.
$$y=x^6e^{-4x^3}$$
|
Breaking everything, let me start by remembering a couple of rules, if $f, g$ are functions:
$$(fg)' = f'g + fg' \\ [f(g)]' = f'(g)\cdot g$$
those are just the rules of the derivatives of the product and of the composition.
Let $f(x) = x^6$ and $g = e^{-4x^3}$. What you have is
$$y = fg$$
Therefore differentiating it gives
$$y' = (fg)' = f'g + fg' = (x^6)'\cdot e^{-4x^3} + x^6\cdot (e^{-4x^3})'$$
Let us now write $h(x) = e^x$, $j(x) = -4x^3$. Notice that the rightmost part is
$$x^6\cdot (e^{-4x^3})' = x^6\cdot [h(j(x))]'$$
Using the chain rule,
$$x^6\cdot [h(j(x))]' = x^6\cdot j(x)'\cdot h'(j(x)) = x^6 \cdot (-4x^3)' \cdot h'(-4x^3)$$
Now notice that $(e^x)' = e^x$ and therefore $h' = h$:
$$x^6 \cdot (-4x^3)' \cdot h'(-4x^3) = x^6 \cdot (-4x^3)' \cdot e^{-4x^3}$$
Assuming you know the rule $(ax^k)' = (kax^{k-1})$, we can use it to finish everything. We had $y' = (x^6)'\cdot e^{-4x^3} + x^6\cdot (e^{-4x^3})'$. We substitute the rightmost part by what we got:
$$y' = (x^6)'\cdot e^{-4x^3} + x^6\cdot (-4x^3)' \cdot e^{-4x^3}$$
and apply the rule I just told you about: $(x^6)' = 6x^5, (-4x^3)' = -12x^2$ giving
$$y' = 6x^5\cdot e^{-4x^3} + x^6\cdot -12x^2 \cdot e^{-4x^3} = 6x^5\cdot e^{-4x^3} - 12x^8\cdot e^{-4x^3}$$
|
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|
Show that a limit of a sequence is zero Let $(a_n)$ positive sequence where $$\dfrac{n-1}{n} \leq \dfrac{a_{n+1}}{a_n} \leq \dfrac {n}{n+1}$$
Prove that: $\lim_{n\to\infty} a_n = 0$.
I was stuck here:
I: $$\lim_{n\to\infty} \dfrac{n-1}{n} = \lim_{n\to\infty}1-\dfrac{1}{n} = 1$$
II: $$\lim_{n\to\infty} \dfrac {n}{n+1} = \lim_{n\to\infty} \dfrac{1}{1+ \dfrac{1}{n}} = 1$$
Then: $$\lim_{n\to\infty} \dfrac{a_{n+1}}{a_n} = 1$$
How do I continue from here?
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Just observe that $$a_{n+1}\le \frac{n}{n+1}a_n\le \frac{n}{n+1}\cdot\frac{n-1}{n}a_{n-1}=\frac{n-1}{n+1}a_{n-1}\le\cdots\le \frac{1}{n}a_1\\a_{n+1}\ge \frac{n-1}{n}a_n\ge \frac{n-1}{n}\cdot\frac{n-2}{n-1}a_{n-1}=\frac{n-2}{n}a_{n-1}\ge\cdots\ge \frac{1}{n}a_2\\\implies \lim\sup_n a_n\le 0\le \lim\inf_n a_n\\\implies \lim_n a_n=0$$ Actually the lower bound on $a_n/a_{n-1}$ is redundant as $a_n$ is a positive sequence.
|
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|
Problem with the area calculation I have to calculate the area of {$x^2+y^2+z^2\le1$}$\cap${$\frac{1}{2}\le z \le1$}.
So i try with the follow integral:
$\int_0^{2\pi}\int_0^{\sqrt{\frac{3}{4}}}\frac{2\rho}{\sqrt{4-\rho^2}}d\rho d\theta$
I got $z$ in function of $x$ and $y$
$z=\sqrt{1-x^2-y^2}$
Then i used the formula:
$\int_a^b\int_c^d \sqrt{1+[\frac{d}{dx}f(x,y)]^2+[\frac{d}{dy}f(x,y)]^2}$
by obtainig the first integral replacing with poolar coordinates, and considering $\frac{1}{2}\le z \le1$ I got $0\le\rho\le\sqrt{\frac{3}{4}}$.
But there is an error, that i can't undertand. where is?
|
Using spherical coordinates,
$$
x = \rho \cos \theta \sin \varphi
\\
y = \rho \sin \theta \sin \varphi
\\
z = \rho \cos \varphi\mbox{,}
$$
with $\rho \in (0 , \infty)$, $\theta \in (0 , 2 \pi)$ and $\varphi \in (0 , \pi)$, you can obtain, by the inequalities in
$$
A = \left\{(x , y , z) \in {\mathbb{R}}^3 : x^2 + y^2 + z^2 \leq 1, \frac{1}{2} \leq z \leq 1\right\}\mbox{,}
$$
that $\rho \in \left[\frac{1}{2 \cos \varphi} , 1\right]$ and $\varphi \in \left(0 , \frac{\pi}{3}\right)$, so
$$
V(A) = \int_{\theta = 0}^{2 \pi} \left(\int_{\varphi = 0}^{\frac{\pi}{3}} \left(\int_{\rho = \frac{1}{2 \cos \varphi}}^1 {\rho}^2 \sin \varphi d \rho\right) d \varphi\right) d \theta = 2 \pi \int_{\varphi = 0}^{\frac{\pi}{3}} \left(\int_{\rho = \frac{1}{2 \cos \varphi}}^1 {\rho}^2 \sin \varphi d \rho\right) d \varphi =
$$
$$
= 2 \pi \int_0^{\frac{\pi}{3}} \frac{\sin \varphi}{3} \left(1 - \frac{1}{8 {\cos}^3 \varphi}\right) d \varphi = \frac{2 \pi}{3} \int_0^{\frac{\pi}{3}} \left(\sin \varphi - \frac{\sin \varphi}{8 {\cos}^3 \varphi}\right) d \varphi = \frac{5 \pi}{24}\mbox{.}
$$
|
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|
Write a product of two numbers in non decimal base If $A=(b-1) (b-1) (b-1)$ and $B=(b-1) (b-1)$ are written in base $b$, what is $A\times B$ in base $b$?
I've tried developing $A \times B$ in decimal base but couldn't get this product back in base $b$.
Beside this, is there a general procedure to compute product in non decimal base.
Thanks in advance for your help,
|
Let throw away the usual meanings of $8,9,10$ and replace them with $10$ means $b$, $9$ means simply $b -1$ and $8$ means $b-2$. We assume $b \ge 3$ and so if we were dealing with base $7$ or base $32$ then $9$ and $8$ means "six" and "five" in base $7$, and $9$ and $8$ means "thirty-one" and "thirty" in base $32$.
Okay. So
$A = (b-1)(b-1)(b-1) = 999 = b^3 -1 = 1000 - 1$
$B = (b-1)(b-1) = 99 = b^2 - 1 = 100 - 1$
So $AB = (b^3 -1)(b^2 - 1) = (1000 -1)(100 - 1)$
$100000 - 1000 - 100 + 1 = $
$99000 - 100 + 1 = $
$98900 + 1 = 98901 = (b-1)(b-2)(b-1)01$
This will be true for all $b \ge 3$
So $b=3: 21201$
$b=4: 32301$
$b=5: 43401$.....
$b=9: 87801$
$b=10: 98901$
$b=11: A9A01$.....
$b =16: FEF01$, etc.
|
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|
Decomposition of this partial function I came across this
$$\int \frac{dx}{x(x^2+1)^2}$$
in "Method of partial functions" in my Calculus I book.
The thing is that, however, decomposing $\frac{1}{x(x^2+1)^2}$ would take long in this way:
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
So is there a quicker or a more practical way that I can use?
|
Let $\displaystyle \mathcal{I} = \int\frac{1}{x(x^2+1)^2}dx$
substitute $\displaystyle x = \frac{1}{t}$ and $\displaystyle dx = -\frac{1}{t^2}$
So $\displaystyle \mathcal{I} = -\int\frac{t^5}{(1+t^2)^2}\cdot \frac{1}{t^2}dt = -\frac{1}{2}\int\frac{t^2\cdot 2t }{(1+t^2)^2}dt$
put $1+t^2=u\;,$ then $2tdt = du$
So $\displaystyle \mathcal{I} = -\int\frac{(u-1)}{u^2}du = +\int u^{-2}du+\int \frac{1}{u}du$
|
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|
Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$ Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2}{\sin(x+h)}-\frac{x^2}{\sin x}}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2\sin x-x^2\sin(x+h)}{\sin(x+h)\sin x}}{h}$$
$$f'(x)=\lim_{h\to0}\frac{(x+h)^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)}$$
$$f'(x)=\lim_{h\to0}\frac{x^2}{\sin x}\frac{(1+\frac{h}{x})^2\sin x-\sin(x+h)}{h\sin(x+h)}$$
I am stuck here.
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Note that $$\lim _{ h\to 0 } \frac { 2\sin ^{ 2 }{ \frac { h }{ 2 } } }{ h } =\lim _{ h\to 0 } \frac { \sin { \frac { h }{ 2 } \cdot \sin { \frac { h }{ 2 } } } }{ \frac { h }{ 2 } } =\lim _{ h\to 0 } \frac { \sin { \frac { h }{ 2 } } }{ \frac { h }{ 2 } } \sin { \frac { h }{ 2 } } =0$$
so
$$
\begin{align}
f'(x)&=\lim _{ h\to 0 } \frac { (x+h)^{ 2 }\sin x-x^{ 2 }\sin (x+h) }{ h\sin x\sin (x+h) }\\&=\lim _{ h\to 0 } \frac { { x }^{ 2 }\sin x+2xh\sin { x+{ h }^{ 2 }\sin { x } } -x^{ 2 }\sin x\cos { h } -{ x }^{ 2 }\sin { h\cos { x } } }{ h\sin x\sin (x+h) }\\ &=\lim _{ h\to 0 } \frac { { x }^{ 2 }\sin x\cdot \left( 1-\cos { h } \right) +h\sin { x\cdot \left( 2x+h \right) } -{ x }^{ 2 }\sin { h\cos { x } } }{ h\sin x\sin (x+h) }\\&=\lim _{ h\to 0 } \frac { { x }^{ 2 }\sin x\cdot \frac { 2\sin ^{ 2 }{ \frac { h }{ 2 } } }{ h } +\sin { x\cdot \left( 2x+h \right) } -{ x }^{ 2 }\frac { \sin { h } }{ h } \cdot \cos { x } }{ \sin x\sin (x+h) }\\& =\frac { 2x\sin { x-{ x }^{ 2 }\cos { x } } }{ \sin ^{ 2 }{ x } }
\end{align}$$
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"timestamp": "2023-03-29T00:00:00",
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Compute $5^{10,000}\equiv \mod 52$ I'm unsure how to solve questions like these, we are given an example below but I'm still confused about how they got from $5^{100\times10+0}$ to $5^{0}$.
ie) Compute $5^{1000} \bmod 77$
$$
5^{1000}=5^{(166⋅6+4)}=5^4=25^2=4^2=16=2 \pmod 7\\
5^{1000}=5^{(100⋅10+0)}=5^0=1 \pmod{11}\\
5^{1000}=2⋅2⋅11+1⋅8⋅7=44+56=100=23 \pmod{77}
$$
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What you have to understand from the examples is that if $n$ is relatively prime to $m$, then there exists a exponent $k$ such that $n^k\equiv1\mod m$.
One exponent that works every time is $m-1$, but there may be others (divisors of $m-1$). For example, $5^6\equiv1\mod 7$, so
$$5{(166.6+4}\equiv {(5^6)}^{166}.5^4\equiv 1^{166}.5^4\equiv5^4\mod7$$
So now you have to find the first power of $5$ equivalent to $1$ modulo $52$, and use the same principle.
Hint : you should find $5^1000\equiv1\mod52$ ;-)
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"timestamp": "2023-03-29T00:00:00",
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How to solve this equation? $a^2 \cdot \arcsin(\frac 4a)+4 \cdot\sqrt {a^2-16}=40$ Please, help me solve this equation:
$$a^2 \cdot \arcsin \left(\frac 4a \right)+4 \cdot\sqrt {a^2-16}=40$$
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There may not exist an explicit solution for $a$ in terms of elementary functions.
However, you can apply the Newton-Raphson Method (A numerical method) using a spreadsheet, or with more sophisticated software such as MATLAB:
$$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)} \tag{1}$$
You have your function:
$$f(a)=a^2 \cdot \arcsin \left(\frac 4a \right)+4 \cdot\sqrt {a^2-16}-40$$
Differentiating it:
$$f'(a)=\frac{4a}{\sqrt{a^2-16}}-\frac{4}{\sqrt{1-\frac{16}{a^2}}}+2a\arcsin\left(\frac{4}{a}\right)$$
Substituting into $(1)$:
$$a_{n+1}=a_n-\frac{a_n^2 \cdot \arcsin \left(\frac {4}{a_n} \right)+4 \cdot\sqrt {a_n^2-16}-40}{\frac{4a_n}{\sqrt{a_n^2-16}}-\frac{4}{\sqrt{1-\frac{16}{a_n^2}}}+2a_n\arcsin\left(\frac{4}{a_n}\right)} \tag{2}$$
Let us start with an initial value $a_0=11$. You can choose these values by simply 'guessing' a value close to the actual solution.
Now, substitute the value of $a_0$ for $a_n$, and find the value of $a_1$ using equation $(2)$. Continue this iteration until you get closer and closer to the value of $a$.
I get a value of $a\approx 5.52978022$.
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"timestamp": "2023-03-29T00:00:00",
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Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a convex function, so I can use jensen's inequality, but I can't come up with one which works.. Has anyone an idea?
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We need to prove that
$$\sum\limits_{cyc}\left(\frac{a}{a+3b+3c}-\frac{1}{7}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{2a-b-c}{a+3b+3c}\geq0$$ or
$$\sum\limits_{cyc}\frac{a-b-(c-a)}{a+3b+3c}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{1}{a+3b+3c}-\frac{1}{b+3c+3a}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2}{(a+3b+3c)(b+3c+3a)}\geq0.$$
Done!
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"timestamp": "2023-03-29T00:00:00",
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What's the derivative of: $ \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$? let $ y=\displaystyle \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$, i'm really interesting to know how do I find :
$\displaystyle \frac{dy}{dx}$ ?.
Note: I have used the definition of derivative of the square root function but i don't succed .
Thank you for any help
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Clarification to solve the question
The question posed is calcule Derivative $\sqrt{x+\sqrt{x+\sqrt{x+....}}}$
we put $y=\sqrt{x+\sqrt{x+\sqrt{x+....}}} $
Before calculating the derivative we simplify the expression $y$
$y^{2}=x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}=x+y$
$y^{2}-y-x=0$
We solve the equation $y^{2}-y-x=0$
The solution to this equation is $y=\frac{-b+\sqrt{\bigtriangleup }}{2a}=\frac{1+\sqrt{1+4x}}{2}$
After simplifying the relationship y we calculate $\frac{dy}{dx}$
we've got $\frac{dy}{dx}=\frac{1}{2}.\frac{4}{2\sqrt{1+4x}}=\frac{1}{\sqrt{1+4x}}$
In the general case, if
$y=\sqrt{f(x)+\sqrt{f(x)+\sqrt{f(x)+...}}}$
$\Rightarrow y^{2}=f(x)+y$
$y^{2}-y-f(x)=0$
$y=\frac{-b+\sqrt{\bigtriangleup }}{2a}=\frac{1+\sqrt{1+4f(x)}}{2}$
$\frac{dy}{dx}=\frac{1}{2}\frac{4\acute{f}(x)}{2\sqrt{1+4f(x)}}=\frac{\acute{f}(x)}{\sqrt{1+4f(x)}}$
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{
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"url": "https://math.stackexchange.com/questions/2101745",
"timestamp": "2023-03-29T00:00:00",
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Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers? $F =$ The probability that the dice land on different numbers
$F = {(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)}$
$F = \frac{30}{36} = \frac{5}{6}$
$E =$ The event that at least one lands on 6
$E = {(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}$
$E = \frac{11}{36}$
What is the probability of E, when given that F has occurred?
$P = \frac{P(EF)}{P(F)}$
$P = \frac{1-P(EF^c)}{P(F)}$
$P = \frac{\frac{11}{36}}{\frac{5}{6}}$
$P = \frac{11}{30}$
This answer was incorrect it is $\frac{1}{3}$ I think there is an issue in identifying $P(EF)$, what should that be and why? Identifying F is easy because that is what is given to have occurred.
Added: Is $EF$ equivalent to $E \cap F$?
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You have made a mistake. In event E, $(6,6)$ is not an option, since it is given that dices are to land on different numbers. So you have only 10 such. So the probability is, going by your method, ${10\over{36}}\over{5\over6}$
This is $1\over3$.
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"url": "https://math.stackexchange.com/questions/2102523",
"timestamp": "2023-03-29T00:00:00",
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Fractions in Questions and Answers
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