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The Decision of three methods of the solutions $dx/P=dy/Q=dz/R$ Question:
(A) $$\frac{adx}{(b-c)yz}= \frac{bdy}{(c-a)xz}=\frac{cdz}{(a-b)xy}$$
(B) $$\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{1-z^2}$$
These are simultaneous diff eq. of the first order and the first degree in three variables
And I know that three methods of solution of $dx/P=dy/Q=dz/R$
First method:
Second method:
Third method:
I know them. But I cannot decide which method I need to use for above two questionss. I dont want to solve these two questions. Only I want to learn which methods I use respectively? Please give me suggestion or hint? Thank a lot.
| A.
$$\frac{a dx}{(b -c ) yz} = \frac{b dy}{(c -a ) xz} = \frac{c dx}{(a -b ) xy}$$
Now multiply the numerator and denominator first fraction by $x$ for second fraction by $y$ and third fraction by $z$ and add all of them and get it
$$\frac{ax dx}{(b-c)xyz} = \frac{by dy}{(c-a)xz} = \frac{cz dz}{(a-b)xy} = \frac{ax dx + by dy + cz dz}{(b-c+c-a + a - b) xyz} $$
So you shall get $ax dx + by dy + cz dz = 0$ from the last fraction. One solution $a\frac{x^2}{2} + b\frac{y^2}{2} + c\frac{z^2}{2} =\text{constant}$ will comes from here.
For other solution take any two of the fractions such as
$$\frac{adx}{(b-c)yz} = \frac{bdy}{(c - a)xz}$$
and see one term will be cancelled from the denominator.
So you shall get
$a(c-a)x dx = b(b-c)y dy$
i.e. $a(c-a)\frac{x^2}{2} - b(b-c)\frac{y^2}{2}$ = constant.
I hope one shall get it easily.
B.
$$\frac{dx}{xz - y} = \frac{dy}{yz - x} = \frac{dz}{1 - z^2}$$
Do it as follows.
$$\frac{dz}{1 - z^2} = \frac{dx + dy + 0 dz}{(x + y)z - (x+y) + 0. (1 - z^2)}$$
Simplify it and get one solution from the relation $\frac{dz}{1 - z^2} = \frac{d(x+y)}{x+y}$.
For he other solution take $dx - dy + 0dz$ instead of $dx + dy + 0 dz$.
I hope everybody will be satisfied.
| {
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"timestamp": "2023-03-29T00:00:00",
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integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^3$ (1) Total no. of integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^2$
(2) Total no. of integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^3$
(3) Total no. of integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^4$
where $(x,y)\in \mathbb{N}.$
$\underline{\bf{My\;Try}}::$ for (1) one:: If $x\geq 4.$ Then $(1!+2!+3!+......+x!)$ conatin last digit is $ 3$
and we know that square of any integer does not have last digit is $3$.
So $x<4$, So we will check for $x = 1,2,3$.
If $x=1$, Then $y^2=1\Rightarrow y = 1$. So integer ordered pairs $(1,1)$.
If $x=2$, Then $y^2=3\Rightarrow y = \sqrt{3}$. So no integer ordered pairs.
If $x=3$, Then $y^2=9\Rightarrow y = 3$. So integer ordered pairs $(3,3)$.
So for (1) one integer ordered pair,s is $(x,y) = (1,1)\;\;,(3,3)$
But I did not understand how can i solve (2) and (3) one,
Help Required
Thanks
| For Q1, you have already noticed that for $x = 4$, the LHS ends in $3$, and for $x > 4$, the terms that are added have last digit $0$. As no square ends in $3$, we do not have any solution for $x \ge 4$.
For Q2, note that if $x \ge 9$, the LHS is divisible by $3$, so if this is the cube of an integer, that integer must also be divisible by $3$. In fact the power must be divisible by $27$. But $1! + 2! + ... + 8! \equiv 9 \pmod {27}$, hence there are no solutions for $x \ge 9$.
For Q3, you could notice that a fourth power is also a square and hence apply the answer to Q1, or extend the argument of Q2 (which in fact holds true for any larger power).
| {
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real analysis The e number How can show
$$\frac{{{{\left( {1 + \frac{1}{n}} \right)}^{{n^2}}}}}{{{{\left( {1 + \frac{1}{{n + 1}}} \right)}^{{{(n + 1)}^2}}}}} < \frac{1}{{{{\left( {1 + \frac{1}{n}} \right)}^n}}}$$
| Using Bernoulli's Inequality,
$$
\begin{align}
\frac{\left(1+\frac1{n+1}\right)^{(n+1)^2}}{\left(1+\frac1n\right)^{n^2}}
&=\left(\frac{n(n+2)}{(n+1)^2}\right)^{(n+1)^2}\left(1+\frac1n\right)^{2n+1}\\
&=\left(1-\frac1{(n+1)^2}\right)^{(n+1)^2}\left(1+\frac1n\right)^{n+1}\left(1+\frac1n\right)^n\\
&\ge\left(1-\frac1{n+1}\right)^{n +1}\left(1+\frac1n\right)^{n+1}\left(1+\frac1n\right)^n\\
&=\left(1+\frac1n\right)^n
\end{align}
$$
Second approach:
Using this result
$$
\left(1+\frac1n\right)^n\le\left(1+\frac1{n+1}\right)^{n+1}
$$
Raise both sides to the $n+1^\text{st}$ power
$$
\left(1+\frac1n\right)^{n(n+1)}\le\left(1+\frac1{n+1}\right)^{(n+1)^2}
$$
Divide both sides by $\left(1+\frac1n\right)^n\left(1+\frac1{n+1}\right)^{(n+1)^2}$
$$
\frac{\left(1+\frac1n\right)^{n^2}}{\left(1+\frac1{n+1}\right)^{(n+1)^2}}\le\frac1{\left(1+\frac1n\right)^n}
$$
| {
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system of equations $\sqrt{x}+y = 11$ and $x+\sqrt{y} = 7$. If $x,y\in \mathbb{R}$ and $\sqrt{x}+y = 11\;$ and $x+\sqrt{y} = 7$. Then $(x,y) = $
$\underline{\bf{My\;\; Try::}}$ Let $x=a^2$ and $y=b^2$, Then equation is $a+b^2 = 11$ and $a^2+b = 7$.
$(a+b)+(a+b)^2-2ab = 18$ and Now Let $a+b=S$ and $ab=P$, we get $S+S^2-P=18$
Now I did not understand how can I solved it.
Help required
Thanks
| You were on the right track, all you need to do is subtract rather than add.
Doing that will give you $(a+b^2)-(a^2+b)=4$
Which can become
$(b^2-a^2)-(b-a)=4=(b-a)(b+a)-(b-a)=(b-a)(b+a-1)$
Since the only way to can obtain 4 by multiplying is either $2*2$ or $4*1$, either
$b-a=2$ and $b+a-1=2$
or
$b-a=1$ and $b+a-1=4$
if $b-a=b+a-1=2$, then $a=1/2$ and $b=5/2$, which means that $(x,y)=(1/4,25/4)$ and that does not satisfy that initial conditions
if $b-a=1$ and $b+a-1=4\implies b+a=5$
then one can easily determine that $b=3$ and $a=2$, which means that $(x,y)=(4,9)$, which satisfies the initial conditions.
| {
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Integral $ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ Question
$ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ = ?
we tried to bound it from both sides using $x$ and $(x-1)$, which yield nice estimation ($\frac {24}\pi$) - ($\frac {36}\pi$) but not a precise one.
we also tried to split it to 6 integrals using $\lfloor x\rfloor $ as a different constant each time. $(0\cdot (\cos(...)-\cos(...))+1(\cos(\frac \pi3)-\cos(\frac\pi6))+2\cdot...$ etc
it produced the right result ($\frac {30}\pi$), but the number is not important as achieving the general function.
| x is real number, m is integer, and $\mathbb{Z}$ is the set of integers (positive, negative, and zero).
$$\lfloor x \rfloor=\max\, \{m\in\mathbb{Z}\mid m\le x\}$$
So
$$\begin{alignat*}{1}
\int_{0}^{6}\lfloor x\rfloor\sin(\frac{6x}{\pi})\ \mathrm{d}x= & \int_{0}^{1}0.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{1}^{2}1.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{2}^{3}2.\sin(\frac{6x}{\pi})\ \mathrm{d}x\\
& +\int_{3}^{4}3.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{4}^{5}4.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{5}^{6}5.\sin(\frac{6x}{\pi})\ \mathrm{d}x\\
= & \frac{1}{6}\pi(\cos(\frac{6}{\pi})-\cos(\frac{12}{\pi}))+\frac{1}{3}\pi(\cos(\frac{12}{\pi})-\cos(\frac{18}{\pi}))\\
& +\frac{1}{2}\pi(\cos(\frac{18}{\pi})-\cos(\frac{24}{\pi}))+\frac{2}{3}\pi(\cos(\frac{24}{\pi})-\cos(\frac{30}{\pi}))\\
& +\frac{5}{6}\pi(\cos(\frac{30}{\pi})-\cos(\frac{36}{\pi}))\\
= & \frac{1}{6}\pi\left(\cos(\frac{6}{\pi})+\cos(\frac{12}{\pi})+\cos(\frac{18}{\pi})+\cos(\frac{24}{\pi})+\cos(\frac{30}{\pi})-5\cos(\frac{36}{\pi})\right)
\end{alignat*}$$
| {
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Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$ Prove that $1<\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.
By using the Mathematical induction. Suppose the statement holds for $n=k$.
Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}=(\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1})+(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$
we know $\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}>1$
What can we do for $(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$?
| By cauchy-schwarz inequality
$$\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{3n+1}\right)(n+1+n+2+\cdots+3n+1)>(1+1+\cdots+1)^2=(2n+1)^2$$
note$$ (n+1+n+2+\cdots+3n+1)=\dfrac{(n+1+3n+1)(2n+1)}{2}=(2n+1)^2$$
| {
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If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
| Hint: \begin{align} x^3+y^3+z^3-3xyz& =(x+y+z)(x^2+y^2+z^2-xy-xz-yz) \\ &=(x+y+z)\left(\frac{(x-y)^2+(x-z)^2+(y-z)^2}{2}\right) \end{align}
Solution:
$$0=a^3+b^3+(-1)^3-3(a)(b)(-1)=(a+b-1)\left(\frac{(a-b)^2+(a+1)^2+(b+1)^2}{2}\right)$$
so $a+b=1$ or $a=b=-1$. The latter gives $a+b=-2$.
| {
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Prove that for $\cos (\alpha ) = \frac{1}{3}$, $\alpha < \frac{\pi}{2} - \frac{1}{3}$ I have the following question in a mock exam:
$\beta = \frac{\pi}{2} - \alpha$, show that $\beta > \frac{1}{3}$
From the earlier part of the question we know that $\alpha$ is an angle between two vectors (so it's between $0$ and $\pi$) and that $\cos(\alpha) = \frac{1}{3}$ (so it's really between $0$ and $\frac{\pi}{2}$).
Using this information I simplified the inequality to:
$\cos (\alpha ) = \frac{1}{3}$, $\alpha < \frac{\pi}{2} - \frac{1}{3}$
Now I believe the answer will have something to do with the monotonicity of cos between 0 and $\frac{\pi}{2}$, but I don't know how to show this. Thanks for your help.
| You mean that: if $0<\alpha<\frac{\pi}{2}$ and $\cos\alpha=\frac{1}{3}$, then $\alpha<\frac{\pi}{2}-\frac{1}{3}$?
Let $F(x)=\cos x-\frac{1}{3}$, then $F(0)=\frac{2}{3}$ and $F(\frac{\pi}{2}-\frac{1}{2})=\sin\frac{1}{3}-\frac{1}{3}<0$ since $\sin x<x (0<x<\frac{\pi}{2})$.
This shows that $\alpha<\frac{\pi}{2}-\frac{1}{3}$.
| {
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Show $g(z+1) = zg(z)$ This is for homework, and I am in need of a hint. Given the product
$$ g(z) = \prod_{k=1}^{\infty} \frac{k}{z+k}\left( 1 + \frac{1}{k} \right)^z, $$
I am trying to show that $g(z+1) = zg(z)$. Here is what I have so far.
I tried to get a better sense of $g$, and wrote
$$ g(z) = \frac{1}{z+1}\left( 1 + \frac{1}{1} \right)^z\frac{2}{z+2}\left( 1 + \frac{1}{2} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{3} \right)^z\dotsb. $$
Then, after some manipulation, I found that
\begin{align*}
g(z+1) &= \frac{1}{(z+1)+1}\left( 1 + \frac{1}{1} \right)^{z+1}\frac{2}{(z+1)+2}\left( 1 + \frac{1}{2} \right)^{z+1}\frac{3}{(z+1)+3}\left( 1 + \frac{1}{3} \right)^{z+1}\dotsb \\
&= \frac{2}{z+2}\left( 1 + \frac{1}{1} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{2} \right)^z\frac{4}{z+4}\left( 1 + \frac{1}{3} \right)^z\dotsb.
\end{align*}
Then I tried to calculate $zg(z)$ and see if they match. I found
\begin{align*}
zg(z) &= \frac{z}{z+1}\left( 1 + \frac{1}{1} \right)^z\frac{2}{z+2}\left( 1 + \frac{1}{2} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{3} \right)^z\dotsb \\
&= \frac{z}{z+1}\frac{2}{z+2}\left( 1 + \frac{1}{1} \right)^z\frac{3}{z+3}\left( 1 + \frac{1}{2} \right)^z\frac{4}{z+4}\left( 1 + \frac{1}{3} \right)^z\dotsb,
\end{align*}
where I am doing a rearrangement in the second line. However, now it looks like $g(z+1) = zg(z)$ if and only if $\frac{z}{z+1} = 1$. What am I doing wrong here?
| Notice that $g(z+1)$ is the product for $g(z)$ "shifted" by one: that is,
$$ g(z+1) = \prod_{k=2}^\infty \frac{k}{z+k}\left(1+\frac{1}{k-1}\right)^z.$$
We can try to rewrite this in terms of the product $\prod\frac{1}{z+k}(1+\frac{1}{k})^z$:
$$\prod_{k=2}^\infty \frac{k}{z+k}\left(1+\frac{1}{k-1}\right)^z = \frac{z+1}{1}\prod_{k=1}^\infty \frac{k}{z+k}\left(1+\frac{1}{k}\right)^z$$
where I have multiplied and divided the expression by $\frac{1}{z+1}$ to obtain the right-hand side. Rewriting, $g(z+1) = (z+1)g(z)$.
| {
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Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it. Can anyone help?
By the way, I've been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$.
Proof : Let
$$f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.$$
We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. Hence, letting $f(0)=1/3$, we know that $f(x)$ is continuous and positive at $x=0$. Hence, since $f(x)\ (0\le x\le {\pi}/2)$ is bounded, there exists a constant $C$ such that $0\lt f(x)\lt C$. Hence, substituting $x={(k\pi)}/{(2n+1)}$ for this, we get
$$0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.$$
Then, the sum of these from $1$ to $n$ satisfies
$$0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.$$
Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.
| Note that
$$\frac{1}{\sin^2(x)}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}$$
using this identity we can write
$$\begin{align}\sum_{k=0}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}&=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(\frac{x+k\pi}{m}+n\pi)^2}\\
&=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{\frac{(x+k\pi+mn\pi)^2}{m^2}}\\
&=m^2\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(x+k\pi+mn\pi)^2}\\
&=m^2\sum_{n\in\mathbb{Z}}\sum_{k=0}^{m-1}\frac{1}{(x+(k+mn)\pi)^2}=\frac{m^2}{\sin^2(x)}\end{align}$$
and
$$\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}=\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}$$
Hence,
$$\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{k\pi}{m})}=\lim_{x\to0}\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}=\frac{m^2-1}{3}.$$
| {
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perfect squares possible? If we let a, b, c, d, and x be integers is it possible that $$x^2+a^2 = (x+1)^2 + b^2 = (x+2)^2 + c^2 = (x+3)^2 + d^2$$
My initial thought is no way! I tried expanding and simplifying, getting $$a^2 = 2x+1 + b^2 = 4x+4 + c^2 = 6x+9 + d^2$$.
It seems that the difference between these perfect squares is impossible - but why? Consecutive perfect squares always differ by some $2n+1$ term, and each term must continue to increase by an odd amount...
Any ideas in the right direction are greatly appreciated!
| I get it. Mod 4 suffices. Well, mod 8 anyway. One of $x,x+1,x+2,x+3$ is divisible by 4. If its matching partner out of $a,b,c,d$ is even, the common sum is divisible by 4, and we actually cannot have any odd numbers involved, which cannot occur with consecutive $x,x+1,x+2,x+3.$ Required detail: the sum of two odd squares is $2 \pmod 4.$
So, the partner is odd, the sum is $1 \pmod 8.$ One of the other $x+j$ numbers is $2 \pmod 4,$ so that sum must be $5 \pmod 8.$ So, actually, impossible.
Done.
| {
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What is the last digit of $\operatorname{lcm}(3^{2003}-1,3^{2003}+1)$? What is the last digit of $\operatorname{lcm}(3^{2003}-1,3^{2003}+1)$?
I am able to find out that LCM is $\dfrac{3^{4006}-1}2$. Since $3^{4006}$ has last digit as $8$, now second last digit can be anything from $0-9$. Based on that second last digit, my answer will vary. Please help how to go further?
| The two numbers have GCD equal to $2$. So, you're basically asking for the last digit of $(3^{2003}-1)(3^{2003}+1)/2 = (3^{4006}-1)/2$.
By Fermat's theorem, $3^{10 \cdot 4} \equiv 1 \pmod{100}$, since: $$\phi(100) = \phi(5^2 \cdot 2^2) = 5 \cdot 2 \cdot (5-1) \cdot (2-1) = 40 .$$
Thus, the two last digits of $(3^{4006}-1)$ are the same as those of $3^{6}-1 = 728$. Hence, the last digit of $(3^{4006}-1)/2$ is $4$, since you have no carry from the tens.
| {
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Complex number: Roots
Solve all the roots of the following equation: $$(z-i)^2(z+i)^2=\frac{1}{4}.$$ Find the set of complex numbers $z$ such that $$\left|\frac{z-3}{z+3}\right|=2.$$
Would anyone mind telling me how to solve the above problems? I really have no idea.
| The first one:
$$(z-i)^2(z+i)^2 = \left( (z-i)(z+i) \right)^2 = (z^2 - i^2)^2 = (z^2+1)^2,$$
so
$$z^2+1 = \pm\frac{1}{2}.$$
For the second one, write $z = x + iy$, so $|z-3| = 2|z+3|$, which is equivalent to $|z-3|^2 = 4|z+3|^2$, gives you
$$(x-3)^2 + y^2 = 4(x+3)^2 + 4y^2.$$
| {
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Probability task (Find probability that the chosen ball is white.) I have this task in my book:
First box contains $10$ balls, from which $8$ are white. Second box contains $20$ from which $4$ are white. From each box one ball is chosen. Then from previously chosen two balls, one is chosen. Find probability that the chosen ball is white.
The answer is $0.5$. Again I get the different answer:
There are four possible outcomes when two balls are chosen:
$w$ -white, $a$ - for another color
$(a,a),(w,a),(a,w),(w,w)$.
Each outcome has probability:
$\frac{2}{10} \cdot \frac{16}{20}; \frac{8}{10} \cdot \frac{16}{20}; \frac{2}{10} \cdot \frac{4}{20}; \frac{8}{10} \cdot \frac{4}{20};$
In my opinion the probability that the one ball chosen at the end is white is equal to the sum of last three probabilities $\frac{8}{10} \cdot \frac{16}{20} + \frac{2}{10} \cdot \frac{4}{20} + \frac{8}{10} \cdot \frac{4}{20}=\frac{21}{25}$. Am I wrong or there is a mistake in the answer in the book?
| The desired probability is
$$\frac{8}{10}\frac{16}{20}\frac{1}{2}+\frac{2}{10}\frac{4}{20}\frac{1}{2}+\frac{8}{10}\frac{4}{20}$$
You will notice that in $(a, w)$ or $(w, a)$ cases, there is only $\frac12$ chance of selecting the white ball.
| {
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} |
How find this $\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)$ Find this limit
$$\lim_{n\to\infty}n^2\left(\dfrac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\dfrac{1}{k+1}-\dfrac{1}{2n}\right)\tag{1}$$
I can only solve this limit
$$I=\lim_{n\to\infty}\left(\dfrac{1^k+2^k+\cdots+n^k}{n^k}-\dfrac{n}{k+1}\right)=\dfrac{1}{2}$$
proof:
use Stolz therom:let
$$a_{n}=(k+1)(1^k+2^k+\cdots+n^k)-n^{k+1},b_{n}=(k+1)n^k$$
then
$$I=\lim_{n\to\infty}\dfrac{a_{n}}{b_{n}}=\lim_{n\to\infty}\dfrac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\dfrac{1}{k+1}\dfrac{(k+1)(n+1)^k-(n+1)^{k+1}+n^{k+1}}{(n+1)^k-n^k}$$
so
$$I=\lim_{n\to\infty}\dfrac{1}{k+1}\dfrac{[k(k+1)/2]n^{k-1}+\cdots+k}{kn^{k-1}+\cdots+1}=\dfrac{1}{2}$$
But for (1),I can't,Thanks.
| We know that $\sum_{i=1}^n i^k=s_k(n)$ for some polynomial $s_k(x)=a_0+a_1x+\ldots +a_{k+1}x^{k+1}$. (We know this because the map $\Delta\colon f\mapsto f(X)-f(X-1)$ is a linear map from the $\mathbb Q$-vector space of polynomials of degree $\le k+1$ to the vector space of polynomials of degree $\le k$ and the kernel consists precisely of the onedimensional space of constant polynomials, hence $\Delta$ is onto).
We want to take the limit of
$$ \begin{align}&n^2\left(\frac{p_k(n)}{n^{k+1}}-\frac{1}{k+1}-\frac1{2n}\right)\\=&n^2\left[\left(a_{k+1}-\frac1{k+1}\right)+\left(a_k-\frac12\right)\frac1n+a_{k-1}\frac1{n^2}+O\left(\frac1{n^3}\right)\right]\\
\end{align}$$
and observe that the result is $a_{k-1}$ (provided that we can confirm that $a_{k+1}=\frac1{k+1}$ and $a_k=\frac12$).
One can determine the highest $a_j$ step by step by comparing coefficients:
$$\begin{align}n^k &=s_k(n)-s_k(n-1)\\
&=a_{k+1}(n^{k+1}-(n-1)^{k+1})+a_k(n^k-(n-1)^k)+a_{k-1}(n^{k-1}-(n-1)^{k-1})+\ldots \\
&=\hphantom{+\;}a_{k+1}\left((k+1)n^k-\tfrac{(k+1)k}{2}n^{k-1}+\tfrac{(k+1)k(k-1)}{6}n^{k-2}+O(n^{k-3})\right)\\
&\quad+\;a_{k}\left(kn^{k-1}-\tfrac{k(k-1)}{2}n^{k-2}+O(n^{k-3})\right)\\
&\quad+\;a_{k-1}\left((k-1)n^{k-2}+O(n^{k-3})\right)\\
&\quad +\;O(n^{k-3})\\
&=\hphantom{+\;}a_{k+1}(k+1)\cdot n^k\\
&\quad+\;\left(a_kk-a_{k+1}\tfrac{(k+1)k}{2}\right)\cdot n^{k-1}\\&\quad+\;\left(a_{k-1}(k-1)-a_k\tfrac{k(k-1)}{2}+a_{k+1}\tfrac{(k+1)k(k-1)}{6}\right)\cdot n^{k-2}\\&\quad +\;O(n^{k-3}),\end{align}$$
that is
$$\begin{align}a_{k+1}&=\frac1{k+1}\\
a_k&=\frac1k\cdot a_{k+1}\frac{(k+1)k}{2}=\frac12\\
a_{k-1}&=\frac1{k-1}\left(a_k\frac{k(k-1)}{2}-a_{k+1}\frac{(k+1)k(k-1)}{6}\right)=\frac k{12}.\end{align} $$
Thus the final result is
$$ \lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\ldots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac1{2n}\right)=\frac k{12}.$$
| {
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Prove that $ \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$. Prove that $\displaystyle \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$.
$\bf{My\; Try}::$ Let $x = \sin \theta$, Then $dx = \cos \theta d\theta$
$\displaystyle = \int _{0}^{1}\frac{\cos \theta }{1-\sin^2 \theta \cdot \sin^2 \alpha}\cdot \cos \theta d\theta = \int_{0}^{1}\frac{\cos ^2 \theta }{1-\sin^2 \theta \cdot \sin ^2 \alpha}d\theta$
$\displaystyle = \int_{0}^{1}\frac{\sec^2 \theta}{\sec^4 \theta -\tan^2 \theta \cdot \sec^2 \theta \cdot \sin^2 \alpha}d\theta = \int_{0}^{1}\frac{\sec^2 \theta }{\left(1+\tan ^2 \theta\right)^2-\tan^2 \theta \cdot \sec^2 \theta\cdot \sin^2 \alpha}d\theta$
Let $\tan \theta = t$ and $\sec^2 \theta d\theta = dt$
$\displaystyle \int_{0}^{1}\frac{1}{(1+t^2)^2-t^2 (1+t^2)\sin^2 \alpha}dt$
Now How can I solve after that
Help Required
Thanks
| Letting $x=\sin \theta$ yields
$$
\begin{aligned}
I &=\int_0^{\frac{\pi}{2}} \frac{d t}{\left(1+t^2\right)^2-\left(1+t^2\right) t^2 \sin ^2 \alpha} \\
&=\int_0^{\infty} \frac{d x}{t^4 \cos ^2 \alpha+\left(1+\cos ^2 \alpha\right) t^2+1}
\end{aligned}
$$
Dividing both denominator and numerator of the integrand gives
$$
\begin{aligned}
I&=\int_0^{\infty} \frac{\frac{1}{t^2}}{t^2 \cos ^2 \alpha+\frac{1}{t^2}+\left(1+\cos ^2 \alpha\right)} d t \\
&=\frac{1}{2} \int_0^{\infty} \frac{\left(\cos \alpha+\frac{1}{t^2}\right)-\left(\cos \alpha-\frac{1}{t^2} \right)}{t^2 \cos ^2 \alpha+\frac{1}{t^2}+\left(1+\cos ^2 \alpha\right)} d t \\
&=\frac{1}{2}\left[\int_0^{\infty} \frac{d\left(t \cos \alpha-\frac{1}{t}\right)}{\left(t \cos \alpha-\frac{1}{t}\right)^2+(1+\cos \alpha)^2}+\int_0^{\infty} \frac{d\left(t \cos \alpha+\frac{1}{t}\right)}{\left.\left(t \cos \alpha+\frac{1}{t}\right)^2+(1-\cos \alpha t)\right]}\right] \\
&=\frac{1}{2(1+\cos \alpha)}\left[\tan ^{-1}\left(\frac{t \cos \alpha-\frac{1}{t}}{1+\cos \alpha}\right)\right]_0^{\infty} \\
&=\frac{1}{4 \cos ^2 \frac{\alpha}{2}}\left(\frac{\pi}{2}+\frac{\pi}{2}\right) \\
&=\frac{\pi}{4 \cos ^2 \frac{\alpha}{2}}
\end{aligned}
$$
| {
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} |
Limit of $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$ What is the limit of this sequence $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$?
Where $a$ is a constant and $n \to \infty$.
If answered with proofs, it will be best.
| The limit does not exist if $|a|\le 1$, and does if $|a|\gt 1$. The fact that it exists if $|a|\gt 1$ can be shown using the Ratio Test.
So we concentrate on the value of the limit, when it exists.
Let $S_n$ be our sum, and let $S$ be its limit. Then
$$aS_{n+1}-S_n=1+\frac{1}{a}+\frac{1}{a^2}+\cdots+\frac{1}{a^n}.$$
Let $n\to\infty$. The left-hand side approaches $(a-1)S$, while the right-hand side approaches the sum of an infinite geometric series. That sum is $\frac{a}{a-1}$. Thus $S=\frac{a}{(a-1)^2}$.
| {
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Calculation of integers $b,c,d,e,f,g$ such that $\frac{5}{7} = \frac{b}{2!}+\frac{c}{3!}+\frac{d}{4!}+\frac{e}{5!}+\frac{f}{6!}+\frac{g}{7!}$ There are unique integers $b,c,d,e,f,g$ such that $\displaystyle \frac{5}{7} = \frac{b}{2!}+\frac{c}{3!}+\frac{d}{4!}+\frac{e}{5!}+\frac{f}{6!}+\frac{g}{7!}$
Where $0\leq b,c,d,e,f,g <i$ for $i=2,3,4,5,6,7$. Then the value of $b+c+d+e+f+g = $
$\bf{My\; Try}::$ $\displaystyle \frac{5}{7} = \frac{2520\cdot b+840\cdot c+210\cdot d+42\cdot e+7\cdot f+g}{7\times 720}$
$\displaystyle 2520\cdot b+840\cdot c+210\cdot d+42\cdot e+7\cdot f+g = 720\times 5 = 3600$
Now I did not understand how can I solve after that.
Help Required
Thanks
| Look at this equality that you've got: $$ 2520 \cdot b + 840 \cdot c + 210 \cdot d + 42 \cdot e + 7 \cdot f + g = 3600.$$
Note that if you consider everything modulo $7$, then most of the summands disappear, because $2520,840,210,42$ and $7$ are all multiples of $7$. So, taking remainders modulo $7$, we get $g \equiv 2 \pmod 7$. Since $0 \leq g < 7$, it follows that $g = 2$. Now substitute $2$ for $g$ in your equality, subtract $2$ from both sides and divide everything by $7$. You get
$$
360 \cdot b + 120 \cdot c + 30 \cdot d + 6 \cdot e + f = 514.
$$
Now consider both sides modulo $6$, and go on in a similar fashion. You will eventually find the values for all variables.
| {
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} |
Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$ How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).
| Starting with $\displaystyle \dfrac{H_n}{n} = \sum_{k=1}^{\infty} \dfrac{1}{k(k+n)}$ we have,
\begin{align*}\sum_{n=1}^{\infty} \dfrac{H_n^2}{n^2} &= \sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty}\dfrac{1}{k(k+n)}\right)^2\\&= \sum_{n=1}^{\infty}\sum_{k,j=1}^{\infty} \dfrac{1}{jk(j+n)(k+n)} \\&= \sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty} \dfrac{1}{k^2(n+k)^2}+ 2\sum_{1 \le k<j} \dfrac{1}{jk(j+n)(k+n)}\right) \\&= \sum_{1 \le k < j} \dfrac{1}{k^2j^2} + 2\sum_{n=1}^{\infty}\sum_{k,m=1}^{\infty} \dfrac{1}{k(k+m)(k+m+n)(k+n)}\\&= \dfrac{1}{2}\left(\left(\sum_{k=1}^{\infty}\dfrac{1}{k^2}\right)^2 – \sum_{k=1}^{\infty}\dfrac{1}{k^4}\right) + 2\sum_{k,m,n=1}^{\infty}\dfrac{(k+m)(k+n) – k(k+m+n)}{kmn(k+m)(k+m+n)(k+n)} \\&= \dfrac{1}{2}\left(\zeta^2(2) – \zeta(4)\right) + 2\sum_{k,m,n=1}^{\infty} \dfrac{1}{kmn(k+m+n)} – 2\sum_{k,m,n=1}^{\infty} \dfrac{1}{mn(m+k)(n+k)}\end{align*}
Therefore, $\displaystyle 3\sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty}\dfrac{1}{k(k+n)}\right)^2 = \dfrac{1}{2}\left(\zeta^2(2) - \zeta(4)\right) + 2\sum_{k,m,n=1}^{\infty} \dfrac{1}{kmn(k+m+n)}$
Using, $\displaystyle \sum_{k,m,n=1}^{\infty} \dfrac{1}{kmn(k+m+n)} = 6\zeta(4)$ we conclude,
$$\sum_{n=1}^{\infty} \dfrac{H_n^2}{n^2} = \dfrac{1}{6}\zeta^2(2) + \dfrac{23}{6}\zeta(4)$$
To see the last result, \begin{align*} \sum_{k,m,n=1}^{\infty} \dfrac{1}{kmn(k+m+n)} &= \sum_{k,m,n=1}^{\infty} \int_0^1 \dfrac{x^{k+m+n}}{kmn}\,\dfrac{dx}{x} \\&= \int_0^1 \log^3(1-x)\,\dfrac{dx}{x} \\&= -\int_0^1 \dfrac{\log^3 x}{1-x}\,dx \\&= -\sum_{n=0}^{\infty} \int_0^1 x^n\log^3 x\,dx \\&= 6\sum_{n=0}^{\infty}\dfrac{1}{(n+1)^4} = 6\zeta(4)\end{align*}
| {
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"question_score": "49",
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Number of ways a sum of r can occur if a die is rolled a number of times. Show that $(1 - x - x^2 - x^3 - x^4 - x^5 -x^6)^{-1}$ is the generating function for the umber of ways a sum of r can occur if a die is rolled a number of times.
| Consider the product
$$
\left(x+x^2+x^3+x^4+x^5+x^6\right)^n\tag{1}
$$
The choice of a term in each factor of $(1)$ corresponds to the choice of a face on each of $n$ dice. The coefficient of $x^k$ in $(1)$ is the number of ways for the faces of $n$ dice to sum to $k$.
The coefficient of $x^k$ in the sum of $(1)$ over all $n$ gives the number of ways to roll a $k$ on the faces of any number of dice.
Using the formula for the sum of a geometric series, we get
$$
\sum_{n=0}^\infty\left(x+x^2+x^3+x^4+x^5+x^6\right)^n=\frac1{1-\left(x+x^2+x^3+x^4+x^5+x^6\right)}\tag{2}
$$
| {
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Proof: $2^{n-1}(a^n+b^n)>(a+b)^n$ If $n \in \mathbb{N}$ with $n \geq 2$ and $a,b \in \mathbb{R}$ with $a+b >0$ and $a \neq b$, then $$2^{n-1}(a^n+b^n)>(a+b)^n.$$
I tried to do it with induction. The induction basis was no problem but I got stuck in the induction step: $n \to n+1$
$2^n(a^{n+1}+b^{n+1})>(a+b)^{n+1} $
$ \Leftrightarrow 2^n(a\cdot a^n + b\cdot b^n)>(a+b)(a+b)^n$
$\Leftrightarrow a(2a)^n+ b(2b)^n>(a+b)(a+b)^n$
dont know what to do now :/
| Start from the other side:
$$(a+b)^{n+1} < (a+b)(2^{n-1}(a^n+b^n))= 2^{n-1}a^{n+1}+2^{n-1}a^{n}b+2^{n-1}ab^{n}+2^{n-1}b^{n+1}$$
Now, prove that $(a^{n}-b^{n})(a-b)>0$
which implies
$$2^{n-1}a^{n}b+2^{n-1}ab^{n}< 2^{n-1}a^{n+1}+2^{n-1}b^{n+1} \,.$$
P.S. If you are familiar, the last inequality:
$$a^{n}b+ab^{n}< a^{n+1}+b^{n+1} $$
also follows immediately from the AM-GM inequality.
| {
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The number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ Find the number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ and the equation of type $x+y = x^2 + y^2 - xy$
| This equation can be rewritten just in another form: $\frac{m+1}{n}+\frac{n+1}{m}=a$
Can be solved using the equation Pell:
$p^2-(a^2-4)s^2=1$
Solutions have the form:
$n=2(p-(a+2)s)s$
$m=-2(p+(a+2)s)s$
And more:
$n=\frac{2p(p+(a-2)s)}{a-2}$
$m=\frac{2p(p-(a-2)s)}{a-2}$
If this equation has a solution: $p^2-(a^2-4)s^2=4$ Formulas are: :
$n=\frac{p-(a-2)s+2}{2(a-2)}$
$m=\frac{p+(a-2)s+2}{2(a-2)}$
Anyway I'll write more general equation: $\frac{X^2+aX+Y^2+bY+c}{XY}=j$
If this square a:
$t=\sqrt{(b+a)^2+4c(j-2)}$
Then using the equation Pell:
$p^2-(j^2-4)s^2=1$
Solutions can be written:.
$X=\frac{(b+a\pm{t})}{2(j-2)}p^2+(t\mp{(b-a)})ps-\frac{(b(3j-2)+a(6-j)\mp{(j+2)t})}{2}s^2$
$Y=\frac{(b+a\pm{t})}{2(j-2)}p^2+(-t\mp{(b-a)})ps-\frac{(b(6-j)+a(3j-2)\mp{(j+2)t})}{2}s^2$
We must also take into account that the number $p,s$ may still be different signs.
| {
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Test for convergence/divergence of $\sum_{k=1}^\infty \frac{k^2-1}{k^3+4}$ Given the series
$$\sum_{k=1}^\infty \frac{k^2-1}{k^3+4}.$$
I need to test for convergence/divergence. I can compare this to the series $\sum_{k=1}^\infty \frac{1}{k}$, which diverges.
To use the comparison test, won't I need to show that $\frac{k^2-1}{k^3+4}>\frac{k^3}{k^4}=\frac{1}{k}$, in order to state the original series diverges?
This doesn't seem to hold, I feel like I'm missing the obvious.
Any help is appreciated.
Thanks.
| To show that the given series is (ultimately) greater than the harmonic series for some k and beyond, We could consider this limit:
$$\lim_{k\rightarrow\infty}\frac{\frac{1}{k}}{\frac{k^2-1}{k^3+4}} = \lim_{k\rightarrow\infty}\frac{k^3+4}{k^3-k}= 1$$
Hence, by the definition of limit,
$$\forall\varepsilon\gt0,\exists N\gt0,\text{such that} ~k \gt N,~~~~~\left|\frac{\frac{1}{k}}{\frac{k^2-1}{k^3+4}} - 1 \right| \lt \varepsilon$$
Let $\varepsilon = 1$, for $k \gt N$, we have
\begin{align*}
\ \left|\frac{\frac{1}{k}}{\frac{k^2-1}{k^3+4}} - 1\right| \lt 1\rightarrow 0 \lt \frac{\frac{1}{k}}{\frac{k^2-1}{k^3+4}} \lt 2 \rightarrow \frac{k^2-1}{k^3+4} \gt \frac{1}{2k}\rightarrow \sum_{k=N+1}^{\infty}\frac{k^2-1}{k^3+4} \gt \sum_{k=N+1}^{\infty}\frac{1}{2k} \space\space\space\text{(for k > N)}
\end{align*}
Since the series $\sum_{k=1}^{\infty}\frac{1}{2k}$ is divergent,by the comparison test, $ \sum_{k=1}^{\infty}\frac{k^2-1}{k^3+4}$ is divergent
| {
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How prove this inequality $\sum_{i=1}^{n}\frac{\ln{i}}{i^4}<\frac{1}{14}$ I need to show that
$$\sum_{i=2}^{n}\dfrac{\ln{i}}{i^4}<\dfrac{1}{14}$$
This problem from high school competition, so usage of integrals and infinite serries is forbidden.
My try: let $x\in (n-1,n)$,then
$$\dfrac{ln{x}}{x^4}<\dfrac{\ln{(n-1)}}{(n-1)^4}?$$
and then I can't proceed.
Thank you
| Since
$$\frac{d}{dx}\left[\frac{\log x}{x}\right] = \frac{1 - \log x}{x^2} < 0
\quad\text{ when } x > e$$
We have
$$\frac{\log k}{k} \le \frac{\log 4}{4} = \frac{\log 2}{2}\quad\text{ for any } k \ge 4.$$ As a result,
$$\sum_{k=2}^\infty\frac{\log k}{k^4}
= \sum_{k=2}^\infty\left(\frac{\log k}{k}\right)\frac{1}{k^3}
\le \frac{\log 2}{2^4} + \frac{\log 3}{3^4} + \frac{\log 2}{2}\sum_{k=4}^\infty\frac{1}{k^3}
$$
Notice the last term is bounded by a telescoping series:
$$\sum_{k=4}^\infty\frac{1}{k^3} < \sum_{k=4}^\infty\frac{1}{k^3-k}
= \sum_{k=4}^\infty\frac{1}{(k-1)k(k+1)}
= \frac12 \sum_{k=4}^\infty\left( \frac{1}{(k-1)k} - \frac{1}{k(k+1)}\right)
= \frac12 \frac{1}{(4-1)4} = \frac{1}{24}
$$
We get
$$\sum_{k=2}^\infty\frac{\log k}{k^4} < \frac{\log 2}{16} + \frac{\log 3}{81} + \frac{\log 2}{48} = \frac{\log 2}{12} + \frac{\log 3}{81} \sim 0.071325 < \frac{1}{14}$$
| {
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Divisibility by $10^6$? Let $p_k$ be the $k^{th}$ prime number.
Find the least $n$ for which $(p_1^2+1)(p_2^2+1) \cdots (p_n^2+1)$ is divisible by $10^6$.
I have no idea where to start on this problem. Any help would be appreciated.
| Let's evaluate each $(p^2 + 1)$ expression, counting up the factors of $2$ and $5$, stopping when they both are at least $6$:
$n = 1$
$p = 2$, and $(p^2 + 1) = 5$. 2s so far = $0$, 5s so far = $1$.
$n = 2$
$p = 3$, and $(p^2 + 1) = 10$. 2s so far = $1$, 5s so far = $2$.
$n = 3$
$p = 5$, and $(p^2 + 1) = 26$. 2s so far = $2$, 5s so far = $2$.
$n = 4$
$p = 7$, and $(p^2 + 1) = 50$. 2s so far = $3$, 5s so far = $4$.
$n = 5$
$p = 11$, and $(p^2 + 1) = 122$. 2s so far = $4$, 5s so far = $4$.
$n = 6$
$p = 13$, and $(p^2 + 1) = 170$. 2s so far = $5$, 5s so far = $5$.
$n = 7$
$p = 17$, and $(p^2 + 1) = 290$. 2s so far = $6$, 5s so far = $6$.
We have 6 $2$s and 6 $5$s, enough for the product to be divisible by $1000000$. $n = 7$. (The actual product here is $390,949,000,000$.)
| {
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Non isolated minimum Consider the $C^2$ function $F:\mathbb{R}^k \rightarrow \mathbb{R}^k$
is it possible for it to be such that $x_n$ is a strict local minimizer for all $n$ and $x_n \rightarrow x$ where $x$ is a strict local minimizer itself.
I know that $\sin(\frac{1}{x})$ has a somewhat similar property, but it is not continuous at zero.
Thanks!
| Here is an example: $F(x) = x^6 (\sin \frac{1}{x})^2 + x^8$.
Establishing that this satisfies your criteria takes a little work:
First, let us establish the relevant smoothness properties:
Note that $F$ is smooth for $x \ne 0$, and near $x=0$ we have some $K_0$ such that $F(x) \le K_0x^6$, we have $|F(x)-F(0) - 0.x| \le K x^6$, hence $F'(0) = 0$. Some differentiation gives $F'(x) = 8{x}^{7}+6 {\mathrm{sin}\frac{1}{x}}^{2}{x}^{5}-2\mathrm{cos} \frac{1}{x} \mathrm{sin} \frac{1}{x} {x}^{4}$ for $x \neq 0$, and since near $x=0$ we have some $K_1$ such that $|F'(x)| \le K_1 x^4$, hence $F'$ is continuous.
Since $|F'(x)-F'(0)| \le K_1 x^4$, we see that $F''(0) = 0$.
Some monotonous work shows that for $x \neq 0$ we have $|F''(x)| \le K_2 x^2$, from which it follows that $F$ is $C^2$. Finally, we note that $F(z) = z^6 (\sin \frac{1}{z})^2 + z^8$ is an analytic function on $\operatorname{re}z > 0$.
Since $F(x) \ge x^8 \ge 0 = F(0)$, we see that $0$ is a strict local minimizer.
Let $\mu_n = \frac{1}{\frac{\pi}{2}+n\pi}$, $\zeta_n = \frac{1}{n\pi}$. We see that $\mu_{n+1} < \zeta_n < \mu_n$ and
$F(\mu_n) = (\frac{1}{\frac{\pi}{2}+n\pi})^6 (1+ (\frac{1}{\frac{\pi}{2}+n\pi})^2) \ge (\frac{1}{\frac{\pi}{2}+n\pi})^6 = (\frac{1}{n\pi})^6 \left( \frac{n \pi}{\frac{\pi}{2}+n\pi} \right)^6$,
$F(\zeta_n) = (\frac{1}{n\pi})^6 (\frac{1}{n\pi})^2$, and
$F(\mu_{n+1}) = (\frac{1}{\frac{3\pi}{2}+n\pi})^6 (1+ (\frac{1}{\frac{3\pi}{2}+n\pi})^2) \ge (\frac{1}{\frac{3\pi}{2}+n\pi})^6 = (\frac{1}{n\pi})^6 \left( \frac{n \pi}{\frac{3\pi}{2}+n\pi} \right)^6$.
Since $\frac{n \pi}{\frac{\pi}{2}+n\pi} > \frac{n \pi}{\frac{3\pi}{2}+n\pi} > \frac{1}{n\pi}$ for $n \ge 1$, we see that $F(\mu_n) > F(\zeta_n), F(\mu_{n+1}) > F(\zeta_n)$, and so $F$ has a minimizer $x_n$ in $(\mu_{n+1}, \mu_n)$. To see that this is a strict minimizer, suppose $y_k \to x_n$ is a sequence such that $F(y_k) = F(x_n)$, then since $F$ is analytic, this would imply that $F$ is constant on $\operatorname{re}z > 0$, which is a contradiction. Hence $x_n$ is a strict local minimizer.
Since $x_n \to 0$, we have the desired result.
| {
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homework - Show a matrix as a combination of other matrices and long division The topic we are dealing with here is polynomial division.
The question is:
We are given a polynomial: $f(x) = (x+1)(x-1)^2$, and a matrix $D \in R^{nxn}$ such that $f(D)=0$
Using only $I, D, D^2$ find:
1) $D^5+3D^2-D+I$
2) $D^{2013}$
3) $D^{-1}$
What I did: For question (1) I divided the polynomial $x^5 +3x^2-x+1$ by $x^3-x^2-x+1 = (x+1)(x-1)^2$ and I got:
$x^5 +3x^2-x+1 = (x^2+x+2)*(x^3-x^2-x+1)+5x^2-1$ but $(x^3-x^2-x+1)=$ when $x=D$ and so:
$D^5+3D^2-D+I = 5D^2-I$
So that solves question (1). But question (2)...I have no clue.
| Hint: Suppose $x^{2013} = (x+1)(x-1)^2q(x) + r(x)$ with $\operatorname{deg}(r)\le2$. Then $r(1)=1,\ r(-1)=-1$ and $r'(1)=2013$.
| {
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Express $\cos2\theta$ in terms of $\cos$ and $\sin$ (De Moivre's Theorem) Use De Moivre's to express $\cos2\theta$ in terms of powers of $\sin$ and $\cos$.
What I have is:
$\cos2\theta + i\sin2\theta\\
= (\cos\theta + i \sin\theta)^2\\
= \cos^2\theta + 2 \cos\theta ~i \sin\theta + (i \sin)^2\theta\\
= \cos^2\theta + i(2\cos\theta \sin\theta) - \sin^2\theta\\
= \cos^2\theta - \sin^2\theta + i(2\cos\theta \sin\theta)
$
so $\cos2\theta = \cos^2\theta - \sin^2\theta$
Is this correct?
| Yes,
$$\cos^2\theta - \sin^2\theta = (\cos\theta)(\cos\theta) - (\sin\theta)(\sin\theta)$$
$$=\cos(\theta + \theta) = \cos(2\theta)$$
| {
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In $\mathbb{Z_2[x]}/(x^2 + x +1)$, what is $\overline{x} \cdot \overline{x}$? In $\mathbb{Z_2[x]}/(x^2 + x +1)$, what is $\overline{x} \cdot \overline{x}$?
$\overline{x} \cdot \overline{x} = \overline{x^2}$
If I divide $\overline{x^2}$ by $(x^2 + x +1)$ I get an answer of $1$ with remainder $(-x-1)$.
$\overline{-1}$ in $\mathbb{Z_2}$ is $\overline{1}$
and
$\overline{-x}$ in $\mathbb{Z_2}$ is $\overline{x}$
So $\overline{x} \cdot \overline{x} = \overline{x + 1}$
Have I got that right?
| Observe that $$\overline{x^2 + x + 1} = \overline 0$$ since $x^2 + x + 1 \in (x^2 + x + 1)$, which means that $$\overline{x^2} =\overline{ - (x + 1)},$$ but since $-1 \equiv 1 \pmod 2,$ we get $$\overline{x^2} = \overline{x + 1}.$$
| {
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how to find the barycentric coordinates of the orthocenter $A = (0,0),B = (4,0),C = (1,2)$
How can I find the barycentric coordinates of the orthocenter of $\triangle ABC$?
| This Wikipedia page describes the orthocentre in barycentric coordinates of the triangle $ABC$ as
$$
\left((a^2+b^2-c^2)(a^2-b^2+c^2):(a^2+b^2-c^2)(-a^2+b^2+c^2):(a^2-b^2+c^2)(-a^2+b^2+c^2)\right),
$$ where $a=|BC|,b=|CA|$ and $c=|AB|$ are the side lengths of the triangle.
| {
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Find the sum of $\sum_{n=1}^\infty (-1)^{n+1} (2n-1)x^{2n-1}$ Find the sum of $S(x) = \sum_{n=1}^\infty (-1)^{n+1} (2n-1)x^{2n-1}$.
I know convergence radius is $1$ because $\frac{1}{\sqrt[n] {(2n-1)}} = \frac{1}{1} = 1.$
Then:
$$x^{-1} S(x) = \sum_{n=1}^\infty (-1)^{n+1}(2n-1)x^{2n+2}. $$
$$\int x^{-1}S(x) = \sum_{n=1}^{\infty} (-1)^{n+1}(2n-1) \frac{x^{2n-1}}{2n-1} = $$
$$ \sum_{n=1}^\infty (-1)^{n+1}x^{2n+1} = $$
$$ -\frac{x}{1-x} = \frac{x}{-1+x}. $$
Now getting back our original series:
$$x^{-1} S(x) = (\frac{x}{-1+x})' = \frac{(-1+x)-x}{(-1+x)^2} = \frac{-1}{1-2x+x^2}$$
$$xx^{-1}S(x) = S(x) = \sum_{n=1}^\infty (-1)^{n+1} (2n-1)x^{2n-1} = \frac{-x}{1-2x+x^2} = \frac{1}{-\frac{1}{x} + 2 - x}.$$
What is my problem?
wolframalpha.com says the answer is wrong (it doesn't specify sum of series with $x$).
Thanks in advance!
| $$S=\sum_{n=0}^\infty (-1)^n (2n+1)x^{2n+1}$$
Thus, $$\begin{align}
S&=\sum_{n=0}^\infty (-1)^n(2n)x^{2n+1}+\sum_{n=0}^\infty (-1)^nx^{2n+1}\\
&=2x\sum_{n=0}^{\infty}n(-x^2)^{n}+\frac{x}{1+x^2}\\
&=2x\left(\sum_{n=0}^\infty(n+1)(-x^2)^n-\sum_{n=0}^\infty (-x^2)^n\right)+\frac{x}{1+x^2}\\
&=2x\left(\frac{1}{(1+x^2)^2}-\frac{1}{1+x^2}\right)+\frac{x}{1+x^2}\\
&=\frac{2x}{(1+x^2)^2}-\frac{x}{1+x^2}=\frac{x-x^3}{(1+x^2)^2}
\end{align}$$
Well, that's my way. In reference to your calculus,
$$\begin{align}
S&=x\frac{\mathrm d}{\mathrm dx}\left(\sum_{n=0}^\infty (-1)^nx^{2n+1}\right) \\
&=x\frac{\mathrm d}{\mathrm dx}\left(x\sum_{n=0}^\infty(-x^2)^n\right) \\
&=x\frac{\mathrm d}{\mathrm dx}\left(\frac{x}{1+x^2}\right) \\
&=x\left(\frac{1(1+x^2)-x(2x)}{(1+x^2)^2}\right)=\frac{x(1-x^2)}{(1+x^2)^2}
\end{align}$$
| {
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If $x_1, \ldots, x_6$ are positive real numbers that add up to $2$. Show that:
If $x_1,x_2,x_3,x_4,x_5$ and $x_6$ are positive real numbers that add up to $2$, then:
$$2^{12} \leq \left(1+\dfrac{1}{x_1}\right) \left(1+\dfrac{1}{x_2}\right)\left(1+\dfrac{1}{x_3}\right)\left(1+\dfrac{1}{x_4}\right)\left(1+\dfrac{1}{x_5}\right)\left(1+\dfrac{1}{x_6}\right) .$$
Factoring out the $\dfrac{1}{x_1}, \ldots, \dfrac{1}{x_6}$ from each term, I get $$\dfrac{1}{x_1}(x_1 + 1)\dfrac{1}{x_2}(x_2 + 1)\dfrac{1}{x_3}(x_3 + 1)\dfrac{1}{x_4}(x_4 + 1)\dfrac{1}{x_5}(x_5 + 1)\dfrac{1}{x_6}(x_6 + 1)$$
We know that: $x_1 + \cdots + x_6 = 2$
Also, I see that $x_1, \ldots, x_6$ are small numbers and $1$ over something small will give something big.
How can I keep going?
| The inequality is equivalent to
$$ \frac{1}{ \left(1+\dfrac{1}{x_1}\right) \left(1+\dfrac{1}{x_2}\right)\left(1+\dfrac{1}{x_3}\right)\left(1+\dfrac{1}{x_4}\right)\left(1+\dfrac{1}{x_5}\right)\left(1+\dfrac{1}{x_6}\right)} \leq \frac{1}{2^{12}}$$
or
$$\sqrt[6]{\prod \frac{x_i}{x_i+1}} \leq \frac{1}{4}$$
Now, by AM-GM we have
$$\sqrt[6]{\prod \frac{x_i}{x_i+1}} \leq \frac{\sum \frac{x_i}{x_i+1}} {6}$$, so it suffices to prove
$$\sum \frac{x_i}{x_i+1} \leq \frac{3}{2}$$
This is equivalent to
$$6- \sum \frac{1}{x_i+1} \leq \frac{3}{2} \Leftrightarrow \sum \frac{1}{x_i+1} \geq \frac{9}{2}$$
But this is Just Cuachy Schwartz
$$6^2 \leq \left( \sum \frac{1}{x_i+1} \right) \left(\sum (x_i+1)\right)=\left( \sum \frac{1}{x_i+1} \right) \left(2+6\right)$$
| {
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If 4n+1 and 3n+1 are both perfect sqares, then 56|n. How can I prove this?
Prove that if $n$ is a natural number and $(3n+1)$ & $(4n+1)$ are both perfect squares, then $56$ will divide $n$.
Clearly we have to show that $7$ and $8$ both will divide $n$.
I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd perfect square.
- so we have $4n+1\equiv 1\pmod{8}$; from this $2|n$ so $3n+1$ is a odd perfect square.
- so $3n+1\equiv 1\pmod{8}$ so $8|n$ but I can't show $7|n$. How do I show this?
Thanks for the help.
| $3n+1=x^2$
$4n+1=y^2$
$4y^2-3x^2=1$
Put $z=2y $
$z^2-3x^2=1$
One of solution is: $(z_0,x_0)=(2,1) $
Other solutions are given by:
$(z_n+ \sqrt3 y_n)=(2+ \sqrt 3)^{n} $
From above we get ,
$z_{n+1}=7z_n+12y_n $
And
$y_{n+1}=4z_n+7y_n $
Or (in terms of $x_n $);
$x_{n+1}=7x_n+6y_n $
$y_{n+1}=8x_n+7y_n $
Now,
$x_{n+1} \equiv -y_n $ mod 7
$y_{n+1} \equiv x_n $ mod 7
${y_{n+1}}^2-{x_{n+1}}^2 \equiv n $
$\equiv {x_n}^2 -{y_n}^2 \equiv -n $
$\equiv 0 $
| {
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Indefinite Integral of $(x^2+1)\over (x^3 + 8)$ I need help in solving this indefinite integral: $$\int {(x^2+1)\over (x^3 + 8)} $$
I know it needs to be reduced to the form $A\over (x+2)$ + $B\over (x+2)^2$ where A and B are constants, but I cannot seem to solve this integral.
Thanks in advance for any help.
| $$(x^3 + 8) = (x + 2)(x^2 - 2x + 4) \neq (x+2)^3$$
To break the integrand into partial fractions, use $$\int {x^2+1\over x^3 + 8}\,dx = \int \left(\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+4}\right)\,dx$$
The first term will integrate as $\;A\log|x + 2| + \text{constant}$.
Once you solve for $A, B, C$, you'll get a better idea how to integrate the last term: whether the numerator becomes the derivative of the denominator (if $B = 2$ and $C = -2$), or if you need to split it into the sum of two terms and use some other approach, for example, completing the square to get $\;x^2 - 2x + 4 = (x - 1)^2 + 3\;$ and using trig substitution if/when needed.
Let's pin down our constants $A, B, C$:
We need for $$A(x^2 - 2x + 4) + (Bx + C)(x + 2) = x^2 + 1$$
Setting $x = -2$, we can solve for $A$: $$A(4+4+4) + (Bx + C)\cdot (0) = 4 + 1 \implies 12A= 5 \iff A = \dfrac 5{12}.$$
Now, we can expand the left-hand side, equate coefficients, and solve for $B, C$ knowing $A$:
$$\begin{align} Ax^2 - 2Ax + 4A + Bx^2 + (2B + C) x + 2C & = (A + B)x^2 + (-2A + 2B + C)x + (4A + 2C) \\ \\ & = 1\cdot x^2 + 0\cdot x + 1 \iff \end{align}$$
$$ (A+B) = 1 \iff B = 1 - A = 1 - \dfrac 5{12} = \dfrac 7{12}$$
$$4A + 2C = 1 \iff C = \frac 12(1 - 4A) = \frac 12\left(1 - 4\cdot \frac 5{12}\right) = -\frac 13$$
| {
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Proof of Heron's Formula for the area of a triangle
Let $a,b,c$ be the lengths of the sides of a triangle. The area is given by Heron's formula:
$$A = \sqrt{p(p-a)(p-b)(p-c)},$$
where $p$ is half the perimeter, or $p=\frac{a+b+c}{2}$.
Could you please provide the proof of this formula?
Thank you in advance.
| Consider a triangle in the 3D cartesian system whose side lengths are a, b and c. Assume the three vertices of the triangle are on three coordinate axes. The area of this triangle can be represented as the sum of three area vectors where each vector corresponds to the projections of the given triangle in each of the three planes XY, YZ and ZX respectively. Let x,y be the base and altitude respectively for the XY-plane triangle, y,z for the YZ-plane triangle and x, z for the XZ-plane triangle. By Pythagoras theorem,
$$ a^2=x^2+y^2 $$
$$ b^2=y^2+z^2 $$
$$ c^2=z^2+x^2 $$
The area vector of the first triangle would be
$ \frac{1}{2} xy \hat{k} $
Similarly, we can form the area vectors of the other two projected triangles. The resultant vector, i.e., the area vector of the triangle in consideration would be
$ \frac{1}{2} xy \hat{k} + \frac{1}{2} yz \hat{i} + \frac{1}{2} zx \hat{j} $.
The magnitude of this vector(in effect the area of the triangle in consideration) is given by
$$ A=\frac{1}{2} \sqrt{x^2y^2+y^2z^2+z^2x^2} $$
By using the three relations we derived earlier connecting x, y and z with a, b and c as well as some expansions and cancellations, we can rewrite the above equation as
$$ A=\frac{1}{4} \sqrt{2c^2a^2+2c^2b^2+2a^2b^2-b^4-a^4-c^4} $$
This is what we obtain on expanding the original heron's formula in terms of s, a, b and c after substituting
$$ s=\frac{(a+b+c)}{2} $$
| {
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Limit Proof Check: $\lim _{x \to a} x^4 = a^4$ Reviewing limits and I'm afraid I may be making mistakes, just looking for a quick proof check.
$f(x)=x^4$, prove that $\lim _{x \rightarrow a}f(x)=a^4$ by showing how to find $\delta$ .
This is my work. $|x^4 - a^4 | < \epsilon$ and $0<|x-a|<\delta$
Factoring: $|x^2 +a^2||x-a||x+a|< \epsilon$ next we set $|x-a|<1$ which means $-1+a<|x|<1+a$ so that
\begin{align}
|x^2 +a^2||x-a||x+a|& =((-1+a)^2 +a^2 )((1+a)+a)(x-a))\\
& = |2a(1+2a^2 -2a)||x-a|\\
& <\epsilon \\
\end{align}
We then have $$\delta =\min\lbrace 1, \frac{\epsilon}{ |2a(1+2a^2 -2a)|} \rbrace$$
How's it look?
| Alternate Solution.
Need to find a bound for $|x^2 + a^2||x + a|$. Take for instance, $|x - a| < 1$ then
$$|x^2 + a^2||x + a| = |x^2 - a^2 + 2a^2 ||x - a + 2a| = (|x^2 - a^2 + 2a^2 |)(|x - a + 2a|)$$
This does not change the expression since $a^2 - a^2 = 0$ and $a - a = 0$
Using the triangle inequality $|x + y| \le |x| + |y|$
$$(|x^2 - a^2 + 2a^2 |)(|x - a + 2a|)$$
$$\le (|x^2 - a^2| + |2a^2 |)(|x - a| + |2a|)$$
$$\le (|x - a||x + a| + |2a^2 |)(|x - a| + |2a|)$$
$$\le (|x - a||x -a + 2a| + |2a^2 |)(|x - a| + |2a|)$$
$$\le (|x - a| \cdot (|x -a| + |2a|) + |2a^2 |)(|x - a| + |2a|)$$
$$|x - a| < 1$$
$$< (1 \cdot (1 + 2|a|) + 2|a^2|)(1 + 2|a|)$$
$$< (1 + 2|a| + 2|a^2|)(1 + 2|a|)$$
Therefore
$$|x^2 + a^2||x + a| < (1 + 2|a| + 2|a^2|)(1 + 2|a|)$$
Therefore
$$|x^2 + a^2||x + a||x - a| < (1 + 2|a| + 2|a^2|)(1 + 2|a|)|x - a| < \epsilon$$
$$|x - a| < \frac{\epsilon}{(1 + 2|a| + 2|a^2|)(1 + 2|a|)}$$
Take $$\delta = min \lbrace 1, \frac{\epsilon}{(1 + 2|a| + 2|a^2|)(1 + 2|a|)} \rbrace$$
| {
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If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then to prove $x=\sqrt{3}+\sqrt{2}$
If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then we have to prove $x=\sqrt{3}+\sqrt{2}$
The question would have been simple if it asked us to prove the other way round.
We can multiply by $x^3$ and solve the quadratic to get $x^3$ but that would be unnecessarily complicated.Also, as $x^3$ has 2 solutions,I can't see how x can have only 1 value. But the problem seems to claim that x can take 1 value only.Nevertheless,is there any way to get the values of x without resorting to unnecessarily complicated means?
NOTE: This problem is from a textbook of mine.
| $x^3+x^{-3}=18\sqrt{3}\Rightarrow x^6+1=18\sqrt{3}x^3\Rightarrow x^6-18\sqrt{3}x^3+1=0$
From there if you look at it like a quadratic equation, you can find 2 solutions for $x^3$ So x would simply be the cube roots of that. and $x^3=a$ has only 1 real root for any real-valued a.
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Find the residue of $\frac{e^{iz}}{(z^2+1)^5}$ at $z = i$ and evaluate $\int_0^{\infty} \cos x/(x^2+1)^5 dx$ I know the evaluation of $\int_0^{\infty} \cos x/(x^2+1)^5 dx$ requires that I solve the first part, but for some reason I'm stumped. I get that I should use $\lim_{z \to i}\frac{1}{24}\frac{d^4}{dz^4}((z-i)^5\frac{e^{iz}}{(z^2+1)^5})$, but I can't seem to evaluate that for the life of me.
| When you have such a high order pole, it is likely better to evaluate the residue directly from the Laurent series about the pole. That is, rewrite
$$\frac{e^{i z}}{(z^2+1)^5} = \frac{1}{e} \frac{e^{i (z-i)}}{(z-i)^5 [(z-i)+2 i]^5} = \frac{1}{i 2^5 e}\frac{e^{i (z-i)}}{(z-i)^5 [1-i(z-i)/2]^5}$$
So we want to find the coefficient of $(z-i)^4$ of the expression
$$\frac{1}{i 2^5 e}\frac{e^{i (z-i)}}{ [1-i(z-i)/2]^5}$$
Note that
$$(1-y/2)^{-5} = 1+(-5)\left (\frac{-y}{2} \right ) + \frac{(-5)(-6)}{2!}\left (\frac{-y}{2} \right )^2 + \frac{(-5)(-6)(-7)}{3!}\left (\frac{-y}{2} \right )^3 + \frac{(-5)(-6)(-7)(-8)}{4!}\left (\frac{-y}{2} \right )^4+\cdots\\ = 1+\frac{5}{2} y+\frac{15}{4} y^2+\frac{35}{8} y^3 + \frac{35}{8} y^4 + \cdots$$
where $y=i(z-i)$. So we want the coefficent of $y^4$ of
$$\frac{1}{i 2^5 e}\left (1+y+\frac12 y^2+\frac16 y^3+\frac{1}{24} y^4+\cdots \right )\left (1+\frac{5}{2} y+\frac{15}{4} y^2+\frac{35}{8} y^3 + \frac{35}{8} y^4 + \cdots \right )$$
which is
$$\frac{1}{i 32 e} \left (\frac{35}{8}+\frac{35}{8}+\frac{15}{8}+\frac{5}{12}+\frac{1}{24} \right ) = \frac{133}{i 384 e}$$
That is the sought-after residue. The integral is simply $i \pi$ times this residue, or
$$\int_0^{\infty} dx \frac{\cos{x}}{(1+x^2)^5} = \frac{133 \pi}{384 e}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/583318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Math induction sum of even numbers I need to prove by induction this thing:
$2+4+6+........+2n = n(n+1)$
so, this thing is composed by sum of pair numbers, so its what I do, but I'm stucked.
$2+4+6+\cdots+2n = n(n+1)$
$(2+4+6+\cdots+2n)+(2n+2) = n(n+1) + (2n+2) $
$n(n+1)+(2n+2) = n(n+1)+(2n+2) $
$n^2 + 3n + 2$
$n(n+2+1)+2$
I don't know how to move forward from this.
Thanks.
| $2+4+6+........+2n = n(n+1)$
$2+4+6+........+2n+2(n+1) = n(n+1)+2(n+1)=(n+1)(n+2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/583884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Find $a,b \in \mathbb{Z}^+$ such that : $\frac{a^{2}-2}{ab+2}\in \mathbb{Z}$
$1$. Find $a;b\in \mathbb{Z}^+$ such that : $\frac{a^{2}-2}{ab+2}\in \mathbb{Z}$
$2$. Find $m;n>1$ such that : $2^m+3^n=k^2$ $(k\in \mathbb{Z})$
Problem 1. I thought :
$\frac{a^{2}-2}{ab+2}\in \mathbb{Z}\Rightarrow a^{2}-2\vdots ab+2\Rightarrow a^{2}+ab\vdots ab+2\Rightarrow ab(a+b)\vdots ab+2\Rightarrow 2(a+b)\vdots ab+2$
But i don't know how to do next!!
| 1
Case 1 : $\frac{a^2-2}{ab+2} =0$ no solution
Case 2 : $\frac{a^2-2}{ab+2} >0 $ and $a < b$ : So $$
\frac{a^2-2}{ab+2} < 1$$
no solution.
Case 3 : $\frac{a^2-2}{ab+2} >0 $ and $a \geq b\ (a>1,\ a>b)$ : So $$
\frac{a^2-2}{ab+2} < \frac{a^2}{ab} = \frac{a}{b},\ ba^2-2b < a^2b +2a$$
So $$ ab\leq ab(a-b)< 2(a+b),\ (a-2)(b-2)< 4$$
Hence if $b \geq 3$, then $a\leq 5$. In this case by computation we have $(a,b)=(4,3)$ (There was a solution-finding of questioner)
Case 4 : $\frac{a^2-2}{ab+2} <0 $ : no solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $a = b = c$, given $P_1(x) = ax^2-bx-c$ , $P_2(x) = bx^2-cx-a$, $P_3(x)=cx^2-ax-b$ and $P_1(v)=P_2(v)=P_3(v)$ Prove $a = b = c$, given $P_1(x) = ax^2-bx-c$, $P_2(x) = bx^2-cx-a$, $P_3(x)=cx^2-ax-b$ and $P_1(v)=P_2(v)=P_3(v)$
where $v$ is a real number.
$a,b,c$ are non zero real numbers.
| Let us put $w=P_1(v)=P_2(v)=P_3(v)$. We have
$$
\begin{array}{lcl}
P_1(v)=av^2-bv-c, & -(v+v^2)P_1(v) &=& a(-v^4-v^3)+b(v^3+v^2)+c(v^2+v) \\
P_2(v)=bv^2-cv-a, & (v^2+1)P_2(v) &=& a(-v^2-1)+b(v^4+v^2)+c(-v^3-v) \\
P_3(v)=cv^2-av-b, & (v-1)P_3(v) &=& a(-v^2+v)+b(-v+1)+c(v^3-v^2) \\
\end{array}
$$
So if we put $z_1=-(v+v^2)P_1(v)+(v^2+1)P_2(v)+(v-1)P_3(v)$, we have
$z_1=-(v+v^2)w+(v^2+1)w+(v-1)w=0$ but on the other hand
$z_1=(b-a)(v^4+v^3+2v^2-v+1)$. So
$$
(v^4+v^3+2v^2-v+1)(b-a)=0 \tag{1}
$$
Similarly,
$$
\begin{array}{lcl}
P_1(v)=av^2-bv-c, & -(v^2+1)P_1(v) &=& a(-v^4-v^2)+b(v^3+v)+c(v^2+1) \\
P_2(v)=bv^2-cv-a, & (-v+1)P_2(v) &=& a(v-1)+b(-v^3+v^2)+c(v^2-v) \\
P_3(v)=cv^2-av-b, & (v^2+v)P_3(v) &=& a(-v^3-v^2)+b(-v^2-v)+c(v^4+v^3) \\
\end{array}
$$
So if we put $z_2=-(v^2+1)P_1(v)+(-v+1)P_2(v)+(v^2+v)P_3(v)$, we have
$z_2=-(v^2+1)w+(-v+1)w+(v^2+v)w=0$ but on the other hand
$z_2=(c-a)(v^4+v^3+2v^2-v+1)$. So
$$
(v^4+v^3+2v^2-v+1)(c-a)=0 \tag{2}
$$
In view of (1) and (2), it will suffice to show that $u=v^4+v^3+2v^2-v+1$ is nonzero.
But
$$
u=\frac{15}{4}v^2+(2v^2+v-2)^2 \tag{3}
$$
so we are done.
| {
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"url": "https://math.stackexchange.com/questions/586558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Doubts about a nested exponents modulo n (homework) As part of my homework I am supposed to find the remainder of the division of $2^{{14}^{45231}}$ by $31$.
Using the ideas explained in calculating nested exponents modulo n I tried the following:
$\phi(31)=30$ since $31$ is prime. Then:
$2^{{14}^{45231}}$ (mod $31$) = $2^{{14}^{45231} (\textrm{mod}30)}$ (mod $31$)
Since $30 = 2 \cdot 3 \cdot 5 \quad $ I need to solve:
$$\begin{matrix} 14^{45231} \equiv a_1 \; (\textrm{mod} \; 2) \\ 14^{45231} \equiv a_2 \; (\textrm{mod} \; 3) \\ 14^{45231} \equiv a_3 \; (\textrm{mod} \; 5) \end{matrix}$$
That means to solve:
$$ \begin{matrix} x \equiv 0 \; (\textrm{mod} \; 2) \\ x \equiv -1 \equiv 2 \; (\textrm{mod} \; 3) \\ x \equiv -1 \equiv 4 \; (\textrm{mod} \; 5) \end{matrix}$$
Let $N = 30 = 2 \cdot 3 \cdot 5$; $n_1=2$; $n_2=3$; $n_3=5$, since $\textrm{gdc}(n_i,n_j) = 1 \; \forall i \neq j \; ; \; i,j \in (1,2,3)$, I can use the chinese remaider theorem which gives me that:
$$x = \sum_{i=1}^{3}a_i \cdot N_i \cdot y_i \; (\textrm{mod} \; n_1 \cdot n_2 \cdot n_3)$$
Where $N_i = \frac{N}{n_i}$; $y_i=N_i^{-1} \; (\textrm{mod} \; n_i)$; $a_1=0$; $a_2=2$ and $a_3=4$.
Applying the formula I found that: $x=44 \equiv 14 \; (\textrm{mod} \; 30)$
Going back to the beginning I made:
$2^{{14}^{45231}}$ (mod $31$) = $2^{{14}^{45231} (\textrm{mod}30)}$ (mod $31$) = $2^{{14}} \equiv 2^4 \equiv 16$ (mod $31$)
And because of that I can say that the remainder of the division of $2^{{14}^{45231}}$ by $31$ is $16$.
I have two doubts about this resolution:
1) Is it right? If not, could you please point out where I made a mistake?
2) Is there another way to solve nested exponents other than using Euler's Totient Function? Even better, is there any other way to solve this using less step's or arguments? (Despite of the obvious fact that they are valid answer's to the second question, I think its better to mention that I am not looking for answer's like: Yes, using Mathematica. (or similar ones))
| Note that $14 = 2 \cdot 7$. We have $2^5 \equiv 2 \pmod{30}$. Similarly, $7^4 \equiv 1 \pmod{30}$. Hence, $14^{45231} = 2^{45231} \cdot 7^{45231}$.
$$2^{45231} \equiv \pmod{30} \equiv 2^{5(9046)+1} \equiv \pmod{30} \equiv 2^{9047} \cdot 2^{5(1809)+2} \pmod{30} \equiv 2^{1811} \pmod{30} \equiv 2^{5(362)+1} \pmod{30} \equiv 2^{362+1} \pmod{30} \equiv 2^{5(72)+3} \pmod{30} \equiv 2^{72+3} \pmod{30} \equiv 2^{3} \pmod{30} \equiv 8 \pmod{30}$$
$$7^{45231} \pmod{30} \equiv 7^{4M+3} \pmod{30} \equiv 7^3 \pmod{30} \equiv 13 \pmod{30}$$
Hence,
$$14^{45231} \equiv 14 \pmod{30}$$
Now conclude what you want.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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struggle simplifying $\sqrt{9+\sqrt{5}}$ I need to simplify $\sqrt{9+\sqrt{5}}$
I already do this (proven it) $\sqrt{9-4\sqrt{5}}=2-
\sqrt{5}$
But I couldn't when apply to $\sqrt{9+\sqrt{5}}=\sqrt{9-4\sqrt{5}+5\sqrt{5}}=\sqrt{(2+\sqrt{5})^2+5\sqrt{5}}$
PLEASE help me out
| Suppose
$\sqrt{9+\sqrt{5}}
= a+\sqrt{b}
$.
Squaring both sides,
$9+\sqrt{5}
= a^2+b+2a\sqrt{b}
= a^2+b+\sqrt{4a^2b}
$.
Equating the parts,
$9 = a^2+b$
and
$5 = 4a^2b$.
From the second,
$a^2 = 5/(4b)$,
so, from the first,
$9 = 5/(4b)+b$,
or $4b^2-36b+5 = 0$.
The discriminant is
$d = 36^2-4\cdot 4\cdot 5
=16(9^2-5)
=16\cdot 76
=64\cdot 39
$.
This is not a square of an integer,
so there is no integer (or rational)
expression in a simplified form.
You could, of course,
write
$\sqrt{9+\sqrt{5}}
= 3\sqrt{1+\sqrt{5}/9}
$,
but this doesn't seem
to be worth much.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/591547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to solve this reccurence relation? Let a,b,c be real numbers. Find the explicit formula for $f_n=af_{n-1}+b$ for $n \ge 1$ and $f_0 = c$
So I rewrote it as $f_n-af_{n-1}-b=0$ which gives the characteristic equation as $x^2-ax-b=0$. The quadratic formula gives roots $x= \frac{a+\sqrt{a^2+4b}}{-2}, \frac{a-\sqrt{a^2+4b}}{-2}$
Then $f_n=P_1(\frac{a+\sqrt{a^2+4b}}{-2})^n+P_2(\frac{a-\sqrt{a^2+4b}}{-2})^n$ and using the initial condition $t_0=c$ gives $C=P_1+P_2 \Rightarrow P_1=C-P_2$
So $(C-P_2)(\frac{a+\sqrt{a^2+4b}}{-2})^n+P_2(\frac{a-\sqrt{a^2+4b}}{-2})^n$ what next? I tried expanding but that didn't help. I know the answer is something like $cd^n-\frac{b}{a-1}+\frac{bd^n}{a-1}$
| Define $F(z) = \sum_{n \ge 0} f_z z^n$, and write your recurrence as:
$$
f_{n + 1} = a f_n + b \quad f_0 = c
$$
Multiply by $z^n$, sum over $n \ge 0$, and recognize:
\begin{align}
\sum_{n \ge 0} f_{n + 1} z^n &= \frac{F(z) - f_0}{z} \\
\sum_{n \ge 0} b z^n &= \frac{b}{1 - z}
\end{align}
to get:
$$
\frac{F(z) - c}{z} = a F(z) + \frac{b}{1 - z}
$$
This gives:
$$
F(z) = \frac{c + (c - b) z}{1 - (a + 1) z + a z^2}
= \frac{b + (a - 1) c}{(a - 1) (1 - a z)} - \frac{b}{(a - 1) (1 - z)}
$$
This is just two geometric series:
$$
f_n = \frac{b + (a - 1) c}{a - 1} \cdot a^n - \frac{b}{a - 1}
$$
As a check, $f_n = c$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $y = \frac{1}{x-1} +\frac {1}{x-5}$ for $x$ I'm stuck on this one (too long ago for me I guess).
I expanded the fractions coming to $y = \frac{2x-6}{x^2-6x+5}$ and even tried to apply a polynom division (translation?) but this came to nothing.
What's the proper approach on this one?
| You have $y (x^2 - 6 x + 5) = 2x - 6$, or $y x^2 - (6y + 2)x + (5y + 6) = 0$, so applying the quadratic formula yields
$$
x = \frac{(6y + 2) \pm \sqrt{(6y +2)^2 - 4 y (5y + 6)}}{2y}.
$$
This simplifies to
$$
x = \frac{(3y + 1) \pm \sqrt{4y^2 + 1}}{y}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/592467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find $x$ for $\left(\frac1{1\times101} + \frac1{2\times102} + \dots +\frac1{10\times110}\right)x = \frac1{1\times11} + \frac1{2\times12}...$ $$\left(\frac1{1\times101} + \frac1{2\times102} + \dots +\frac1{10\times110}\right)x = \frac1{1\times11} + \frac1{2\times12} + \dots +\frac1{100\times110}$$
Find x
My younger sister in grade 5 had this question in a test. But I, a college student, still can't solve this. What a shame :<
| One can group the $100$ summands on the right side by the ending digit of the denominators. For example, taking the numbers ending in 2:
$$\frac{1}{2\times 12}+\frac{1}{12 \times 22}+ ... +\frac{1}{92 \times 102}=\\
= \frac{1}{10}\left(\frac{1}{2} -\frac{1}{12}\right) +\frac{1}{10}\left(\frac{1}{12} -\frac{1}{22}\right) +\cdots +\frac{1}{10}\left(\frac{1}{92}- \frac{1}{ 102}\right) =\\ =\frac{1}{10}\left(\frac{1}{2}-\frac{1}{102}\right)=\\ =10\left(\frac{1}{2 \times 102}\right)\\
$$
Therefore, $x=10$
This can be formalized, of course. But I don't think that this (even without the formalization) is appropiate for a 10 years old...
| {
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"question_score": "4",
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How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know what to do with the $n^2$.
| We will use the same technique found here to show that $\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$. But we'll need the following theory extension (see this link) to that result. For $k \ge 1$,
$\tag 1 \sum_{n=k}^{\infty}\frac{n}{2^n} = 2^{1 - k} \, (k + 1)$
We decompose the summands of $\sum_{n=1}^{\infty}\frac{n^2}{2^n}$ in a natural/straightforward manner (so that $\text{(1)}$ applies) , arranging these numbers into a table:
$$\begin{pmatrix}
\frac{1}{2} & \frac{2}{4} & \frac{3}{8} & \frac{4}{16} & \frac{5}{32} & \dots \\
0 & \frac{2}{4} & \frac{3}{8} & \frac{4}{16} & \frac{5}{32} & \dots \\
0 & 0 & \frac{3}{8} & \frac{4}{16} & \frac{5}{32} & \dots \\
0 & 0 & 0 & \frac{4}{16} & \frac{5}{32} & \dots \\
0 & 0 & 0 & 0 & \frac{5}{32} & \dots \\
0 & 0 & 0 & 0 & 0 & \dots \\
. \\
. \\
. \\
\end{pmatrix}$$
Using $\text{(1)}$ we add up the entries in each row, giving the sequence of numbers
$\tag 2 \bigr ( \, 2^{1 - k} \, (k + 1) \, {\bigr )}_{\,k \ge 1}$
Performing the summation,
$\quad \sum_{k=1}^{\infty} 2^{1 - k} \, (k + 1) = 2 \sum_{k=1}^{\infty} \frac{k + 1}{2^k} = 2 \big ( \sum_{k=1}^{\infty} \frac{k}{2^k} + \sum_{k=1}^{\infty} \frac{1}{2^k} \big)$
$\quad \quad 2 \sum_{k=1}^{\infty} \frac{k + 1}{2^k} = 2(2+1) = 6$
So adding up all those numbers in the table gives
$\quad \text{ANS: } 6$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
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How do you simplify this expression? $$\lim_{h\to0}(\frac{x}{h(x+h+1)} + \frac{1}{x+h+1} - \frac{x}{h(x+1)})$$
I know the answer is $$\frac{1}{(1+x)^2}$$
But I can't get there
| Using fraction addition rules, we have
$$\lim_{h\to0}\left(\frac{x}{h(x+h+1)} + \frac{1}{x+h+1} - \frac{x}{h(x+1)}\right)\\
=\lim_{h\to0}\left(\frac{x(x+1)}{h(x+h+1)(x+1)} + \frac{h(x+1)}{h(x+h+1)(x+1)} - \frac{x(x+h+1)}{h(x+1)(x+h+1)}\right)\\
=\lim_{h\to0}\frac{(x+h)(x+1)-x(x+h+1)}{h(x+1)(x+h+1)}\\
=\lim_{h\to0}\frac{x^2+x+hx+h-x^2-xh-x}{h(x+1)(x+h+1)}\\
=\lim_{h\to0}\frac h{h(x+1)(x+h+1)}=\frac 1{(x+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/596253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\triangle ABC;AB=c;AC=b;BC=a$ such that $a\geq b\geq c$. Prove : $\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}$ $\triangle ABC;AB=c;AC=b;BC=a$ such that $a\geq b\geq c$. Prove :
$$\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}$$
I have tried that :
$a\geq b\geq c\Rightarrow \frac{a^{2}-b^{2}}{c}\geq 0;\frac{b^{2}-c^{2}}{a}\geq 0;\frac{3a^{2}}{b}\geq \frac{3ac}{b}$
$\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}\Leftrightarrow \frac{c^{2}-a^{2}}{b}+\frac{3a^{2}}{b}\geq 2(a-c)+\frac{3ac}{b}\Leftrightarrow \frac{(c-a)(c+a)}{b}\geq 2(a-c)\Leftrightarrow c+a\geq -2b$
!!??
| Another way to look at it would be:
\begin{align}
LHS &= \frac{(a - b)(a + b)}{c} + \frac{(b - c)(b + c)}{a} + \frac{c^2 + 2a^2}{b} \\
&> (a - b) + (b - c) + \frac{c^2 + 2a^2}{b} \\
&= a - c + \frac{c^2 + 2a^2}{b}.
\end{align}
And
RHS $= 2a - 2c + \dfrac{3ca}{b}$.
\begin{align}
\text{So, }LHS > RHS &\iff \frac{c^2 + 2a^2 - 3ac}{b} > a - c \\
&\iff c^2 + 2a^2 - 3ac > ab - bc \\
&\iff (a - c)^2 + a(a - c) > b(a - c) \\
&\iff a - c + a > b \\
&\iff 2a > b + c
\end{align}
and this last inequality is true since $a > b > c$.
| {
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Limit of $\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$ I have to determine the following:
$\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$
$\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)=\lim\limits_{x \rightarrow \infty}(\sqrt{x^8(1+\frac{4}{x^8})}-x^4 = \lim\limits_{x \rightarrow \infty}(x^4\sqrt{1+\frac{4}{x^8}}-x^4 = \lim\limits_{x \rightarrow \infty}(x^4(\sqrt{1+\frac{4}{x^8}}-1)= \infty$
Could somebody please check, if my solution is correct?
| We have
$$
x^4 (\sqrt{1 + \frac{4}{x^8}}-1)=
x^4 (\frac{2}{x^8} + O(x^{-8}))
=\frac{2}{x^4} + O (x^{-4}) \rightarrow 0
$$
when $x \rightarrow \infty$.
| {
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"question_score": "3",
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Is it a Fermat polynomial? A Fermat polynomial is a polynomial which can be written as the sum of squares of two polynomials with integer coefficients. Let $f(x)$ be a Fermat polynomial such that $f(0)=1000$. Prove that $f(x)+2x$ is not a Fermat polynomial.
| $1000$ can be expressed as a sum of two squares in two different ways: either $1000 = 10^2 + 30^2$ or $1000 = 18^2 + 26^2$. Thus if $f(x)$ is a Fermat polynomial with $f(0) = 1000$ and $f(x) = [a(x)]^2 + [b(x)]^2$, then $a(0)^2 + b(0)^2 = 1000$ so that (wlog) either $a(0) = 10$ and $b(0) = 30$, or $a(0) = 18$ and $b(0) = 26$.
Writing $a(x) = xp(x) + 10$ and $b(x) = xq(x) + 30$ for $p,q \in \mathbb Z[x]$ yields $$f(x) = a(x)^2 + b(x)^2 = x^2 (p(x)^2 + q(x)^2) + 80x(p(x) + q(x)) + 1000,$$ and writing $a(x) = xp(x) + 18$ and $b(x) = xq(x) + 26$ for $p,q \in \mathbf Z[x]$ yields $$f(x) = a(x)^2 + b(x)^2 = x^2 (p(x)^2 + q(x)^2) + 88x (p(x) + q(x)) + 1000.$$
In both cases we have that $\displaystyle \left.\frac{f(x) - 1000}{x}\right|_{x = 0}$ is a multiple of $8$.
$f(x)$ and $f(x) + 2x$ can't both have this property.
| {
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"source": "stackexchange",
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For a prime $p$ find integers $n,m$ s.t $ p > n>m>0$ and $n^3 \equiv m^3 \pmod p$ Sitting at home on daddy leave i have decided to try to learn more number theory. I was playing around with some equations and got to this, which i don't know how to proceed with:
"
For a prime $p$, how to find (if there is) the smallest integers $p > n > m > 0$ such that
$n^3 \equiv m^3 \pmod p$"
| $$n^3\equiv m^3\pmod p\implies p|(n^3-m^3)=(n-m)(n^2+mn+m^2)$$
Since $p$ is prime, $p|(n-m)$ or $p|(n^2+mn+m^2)$ but $n\ne m$ and $0<m<n<p$ so we can see $p\not\mid(n-m)$.
Now I'm just going to kind of pull something out of a hat and say that $m^2+mn+n^2$ is the norm on the Euclidean domain $\Bbb Z[\omega]$ where $\omega=e^{2\pi i/3}$, also known as the Eisenstein integers, which are numbers of the form $m+n\omega$ where $m,n\in \Bbb Z$.
The norm of $a+b\omega$ is actually defined as $(a+b\omega)(a+b\overline \omega)=a^2-ab+b^2$. The norm is multiplicative (you can check this yourself!), which means $N(\alpha)N(\beta)=N(\alpha\beta)$ for all $\alpha,\beta\in\Bbb Z[\omega]$.
Okay, so now we have $$m^2+mn+n^2=(m-n\omega)(m-n\overline\omega)=(m-n\omega)(m+n-n\omega)$$
And we want $p|(m-n\omega)(m+n-n\omega)$. Question: is $p$ prime in $\Bbb Z[\omega]$? A: it depends.
Note that if $p$ is not prime on the Eisensteins, there exist $\alpha,\beta\in \Bbb Z[\omega]$ such that $N(\alpha),N(\beta)\ne 1$ (i.e. $\alpha$ and $\beta$ are not units) and $\alpha\beta=p$. Thus $N(\alpha)N(\beta)=p^2$ and since $p$ is prime we can conclude that $N(\alpha)=N(\beta)=p$ (note that, upon inspection, this also implies they are conjugate). Thus we need to find some $a,b$ such that $a^2-ab+b^2=p$. Consider $\mod 3$:
$$ \begin{align} (-1)^2-(-1)(-1)+(-1)^2&\equiv 1\pmod3\\
(-1)^2-(-1)0+0^2&\equiv 1\pmod 3\\
(-1)^2-(-1)1+1^2&\equiv 0\pmod 3\\
0^2-0\cdot 0+0^2&\equiv 0\pmod 3\\
0^2-0\cdot1+1^2&\equiv 1\pmod 3\\
1^2-1\cdot 1+1^2&\equiv 1\pmod 3 \end{align} $$
So we see if $p\equiv 2\pmod 3$ then $p$ is prime in the Eisensteins.
By quadratic reciprocity, $\left(\frac{3}{p}\right)\left(\frac{p}{3}\right)=(-1)^{\frac{p-1}{2}\frac{3-1}{2}}$. If $p\equiv 1\pmod 3$,
$$\begin{align}\left(\frac{3}{p}\right)&=(-1)^{(p-1)/2}\\
\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)&=(-1)^{(p-1)/2}(-1)^{(p-1)/2}\\
\left(\frac{-3}{p}\right)&=(-1)^{p-1}=1\end{align}$$
So there exists $k$ such that $k^2+3\equiv 0\pmod p$. Now let $$\mathcal L=\{(a,b)\in \Bbb Z^2\mid a\equiv \ell b\pmod p\}=\{\lambda (\ell,1)+\mu (0,p)\mid \lambda,\mu\in\Bbb Z\}$$
Where $\ell\equiv\frac{d-1}{2}\pmod p$ (note $\ell^3\equiv 1$). By Minkowski's Theorem, there is at least one lattice point in the region $\{(a,b)\in\Bbb Z^2\mid a^2+ab+b^2<(2p)^2\}$. Since $a^2+ab+b^2\equiv b^2(z^2+z+1)\equiv b^2 \frac{z^3-1}{z-1}\equiv 0\pmod p$, we can deduce that $a^2+ab+b^2=p$. Thus given a prime $p\in\Bbb N$, $p$ is an Eisenstein prime if and only if $p\equiv 2\pmod 3$.
If $p$ is an Eisenstein prime, then $p|(m-n\omega)(m+n-n\omega)\implies p|(m-n\omega)\lor p|(m+n-n\omega)$. Since $p$ is real, $p$ must divide the real and imaginary parts separately, so we find $p|n$, which contradicts the supposition that $0<n<p$.
If $p\equiv 1\pmod 3$, the above proof illustrates that we can actually find a lattice of points $m,n$ such that $p|(m^2+mn+n^2)\implies m^3\equiv n^3\pmod p$, generated by the vectors $\begin{pmatrix}\frac{\sqrt{-3}-1}{2}\\1\end{pmatrix}$ and $\begin{pmatrix}0\\p\end{pmatrix}$.
| {
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"source": "stackexchange",
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If $a+b+c \ge 1$ and $a,b,c>0$, prove that $\frac{1}{2a+s}+\frac{1}{2b+s}+\frac{1}{2c+s} \ge \frac{1}{s}$, where $s=ab+bc+ca$. What I know is that $$s \le \frac{(a+b+c)^2}{3} \ge \frac{1}{3}$$But as you can see the sign is pointing to different sides. So I can't see how this could be helpful. Just a small observation. I don't have any other ideas and I'd like to get some help. Thanks.
| first we prove $a+b+c=1$ case:
$\dfrac{1}{2a+s}+\dfrac{1}{2b+s}+\dfrac{1}{2c+s} \ge \dfrac{1}{s} \iff s^3+s^2-4abc\ge 0$
let $x=3a,y=3b,z=3c,3u=x+y+z=3,3v^2=xy+yz+xz=9s,w^3=xyz=27abc$
$s^3+s^2-4abc\ge 0 \iff v^6+3v^4-4w^3 \ge 0 \iff v^6+3v^4\ge 4w^3$
accoding to $uvw$ method, $w^3\le 3v^2-2+2\sqrt{(1-v^2)^3}$
$\iff v^6+3v^4 \ge 12v^2-8+8\sqrt{(1-v^2)^3} \iff v^6+3v^4 -12v^2+8\ge 8\sqrt{(1-v^2)^3} \iff (v^6+3v^4 -12v^2+8)^2 \ge 64(1-v^2)^3 \iff v^{12}+6v^{10}-15v^8+8v^6 \ge 0 \iff v^6+6v^4-15v^2+8 \ge 0 \iff (v-1)^2(v^2+8)(v+1)^2\ge 0$
when $v=1$ the "="hold $\implies x=y=z=1 \implies a=b=c=\dfrac{1}{3}$
for $a'+b'+c'>1$, let$ a'+b'+c'=k(a+b+c) \to k > 1, ,a'=ka,b'=kb,c'=kc,s'=k^2s$
$\dfrac{1}{2a'+s'}+\dfrac{1}{2b'+s'}+\dfrac{1}{2c'+s'} \ge \dfrac{1}{s'} \iff \dfrac{1}{2\frac{a}{k}+s}+\dfrac{1}{2\frac{b}{k}+s}+\dfrac{1}{2\frac{c}{k}+s} \ge \dfrac{1}{s}$
$\frac{a}{k} < a \implies \dfrac{1}{2\frac{a}{k}+s}>\dfrac{1}{2a+s} $ so the inequlity hold.
QED.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the lim $\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}$ $$
\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}
$$
I multiply it with:
$$
\frac{ \sqrt{x} + 1}{\sqrt{x} + 1}
$$
And I get :
$$
\lim_{x\to 1} \frac{ \sqrt{x}^2 - 1^2}{(\sqrt[3]{x} - 1) * (\sqrt{x} + 1)}
$$
But the solution is still division by 0 and not possible, so I think I've made a mistake somewhere..
| You basically want to insert $1$ written as a “conjugate” that removes the radical divided by itself. For $\sqrt{x}-1$ it is $\sqrt{x}+1$, for $\sqrt[3]{x}-1$ it is $\sqrt[3]{x^2}+\sqrt[3]{x}+1$. So you get
\begin{align}
\lim_{x\to1}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}&=
\lim_{x\to1}\left((\sqrt{x}-1)\frac{\sqrt{x}+1}{\sqrt{x}+1}\right)
\left(\frac{1}{\sqrt[3]{x}-1}
\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\right)
\\[2ex]
&=\frac{x-1}{\sqrt{x}+1}\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{x-1}
\end{align}
and the indetermination goes away.
It's the same as Bill Dubuque's answer, actually, but without substitution.
| {
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} |
Diagonalization of Matrix with Trigonometric Functions Problem statement:
Diagonalize the following matrix:
$$
\begin{pmatrix}
\cos \theta & \sin \theta \\
\sin \theta & -\cos \theta\\
\end{pmatrix}
$$
My attempt:
Ie found that the eigenvalues of this matrix are $\lambda = -1$ or $\lambda = 1$, so I plugged in $\lambda = -1$:
$$
\begin{pmatrix}
\cos\theta+1 & \sin\theta \\
\sin\theta & -\cos\theta+1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
= 0
$$
I tried to solve this, but I got $2x\sin\theta=0$. How do I find the eigenvectors?
| For $\;\lambda=-1\;$ :
$$\begin{align*}I&\;\;(\cos\theta+1)x+\sin\theta y=0\iff y=-\frac{\cos\theta+1}{\sin\theta}x\\II&\;\;\sin\theta x-(\cos\theta-1)y=0\end{align*}$$
We need only one equation as these two are linearly dependent (proof?), so
$$II\longrightarrow x=\frac{\cos\theta-1}{\sin\theta}y\;\implies\binom {\frac{\cos\theta-1}{\sin\theta}}{y}\;\;\text{is an eigenvector,}\;\;\theta\neq k\pi\;,\;k\in\Bbb Z$$
Something similar happens if $\;\lambda=1\;$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
solving an expression based on sin $\theta$ If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$, then $x$ must be equal to what?
What does the following solution mean?
$0 \le \sin^2 \le 1$
This implies $0 \le \frac{x^2 + y^2 + 1 }{ 2x } \le 1$
This implies $\frac{(x - 1)^2 + y^2 }{2x} \le 0 $
This implies $x = 1$.
Can you please explain me the solution?
| On rearrangement we have $$x^2-2x\sin^2\theta+y^2+1=0\ \ \ \ (1)$$
which is clearly a Quadratic Equation in $x$
As $x$ is real, the discriminant must be $\ge0$
But actually the discriminant,
$\displaystyle (2\sin^2\theta)^2-4(y^2+1)=-4(y^2+1-\sin^4\theta)$
$\displaystyle=-4\{y^2+(1-\sin^2\theta)(1+\sin^2\theta)\}=-4\{y^2+\cos^2\theta(1+\sin^2\theta)\}$ which is $\not>0$
So for real $\displaystyle x, y$ and $\cos\theta$ must be individually equal to zero
$\displaystyle \implies \sin^2\theta=1-0^2=1$
So, $(1)$ reduces to $\displaystyle x^2-2x+1=0$
| {
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"source": "stackexchange",
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How to integrate $\int{ dx \over \sqrt{1 + x^2}}$ How to integrate $dx \over \sqrt{1 + x^2}$?
Answer should be $\log ( x + \sqrt{1 + x^2})$
Please help as possible...
Thank you
| We have
$$\int \frac{1}{\sqrt{1+x^2}}\, dx.$$
We make the following substitution. Let
$$
\begin{align*}
x&=\tan \theta \\
dx &= \sec^2 \theta \, d \theta\\
1+x^2&=1+\tan^2 \theta \\
&=\sec^2 \theta.
\end{align*}
$$
Hence our integral becomes
$$
\begin{align*}
\int \frac{1}{\sqrt{1+x^2}}\, dx &= \int\frac{\sec^2 \theta \, d \theta}{\sqrt{\sec^2 \theta}}\\
&=\int \sec \theta \, d \theta \\
&=\ln |\sec \theta + \tan \theta|+c.
\end{align*}
$$
For the back substitution, since
$$\tan \theta = \frac{x}{1},$$
we can form a right triangle with side opposite $\theta$ equal to $x$, and side adjacent to $\theta$ equal to $1$. Hence the hypoteneuse will have length $\sqrt{1+x^2}$. We can now read straight from the right triangle and back substitute,
\begin{align*}
\ln |\sec \theta + \tan \theta|+c &= \ln \left | \frac{\sqrt{1+x^2}}{1}+\frac{x}{1} \right |+c\\
&=\ln \left | \sqrt{1+x^2}+x \right |+c.
\end{align*}
| {
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Motivation for $y(x - y) = \frac{1}{2}(x^2 - y^2) - \frac{1}{2}(x-y)^2$ It's a really simple identity and according to the text, I am reading, this is an obvious identity
$$y(x - y) = \frac{1}{2}(x^2 - y^2) - \frac{1}{2}(x-y)^2$$
Without expanding the RHS, I don't know how they came up with this.
| Factor the first term on the right hand side: $\frac{1}{2} (x^2-y^2) = \frac{1}{2} (x+y)(x-y).$ Factor the second term: $(x-y)^2=(x-y)(x-y)$ since squaring a number is defined to be the product of that number with itself.
Then factor out $(x-y)$ from $\frac{1}{2} (x+y)(x-y)-\frac{1}{2}(x-y)(x-y)$ since it is a common term (remember to think of parenthetical expressions like $(x-y)$ as SINGLE terms). You should get the left hand side from there.
| {
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"source": "stackexchange",
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Find all postive integer numbers $x,y$,such $x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$ Find all postive integer $x$ and $y$ such that
$x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$
My try: since
$$(x+y)^2-2xy=x^2+y^2$$
I know this well know reslut:
$$xy|(x^2+y^2+1),\Longrightarrow \dfrac{x^2+y^2+1}{xy}=3,x,y\in N$$
and such condition $$(x,y)=(u_{n},u_{n+1})$$
where
$$u_{0}=u_{1}=1,u_{n+2}=3u_{n+1}-u_{n}$$
so I can't,Thank you
| Noting that
$\left(x+y+1\right)\left(x+y-1\right)=x^2+y^2-1+2xy\quad * $
Use the above expression to prove that $x+y-1$ divides $2xy$ and $x+y+1$ divides $x^2+y^2-1$
Note that $x+y-1$ and $x+y+1$ have the same parity.
Let both of them be odd, then they are coprime
$\implies \left(x+y+1\right)\left(x+y-1\right)$ divides each of $2xy$ and $x^2+y^2-1$
Let the quotient of the division be $a$ and $b$ respectively. Use $*$ to arrive at $a+b=1$
$$a\left(x+y+1\right)\left(x+y-1\right)=2xy$$ and $$b\left(x+y+1\right)\left(x+y-1\right)=x^2+y^2-1$$ adding both of them gives $\left(a+b\right)\left(x+y+1\right)\left(x+y-1\right)=x^2+y^2-1+2xy=\left(x+y+1\right)\left(x+y-1\right)$ as we are speaking of $2xy$... being divided by $x+y+1$ or $x+y-1$ and as we don't say is divisible by zero, we can neglect the case in which $\left(x+y+1\right)\left(x+y-1\right)=0$. Which yields the condition $a+b=1$
let both of them be even.
$\implies \frac{\left(x+y+1\right)\left(x+y-1\right)}{2}$ divides each of $2xy$ and $x^2+y^2-1$ . Let the quotient of the division be $c$ and $d$ respectively. Use $*$ to arrive at $c+d=2$.
$$c\frac{\left(x+y+1\right)\left(x+y-1\right)}{2}=2xy$$ and $$d\frac{\left(x+y+1\right)\left(x+y-1\right)}{2}=x^2+y^2-1$$ adding both of them gives $\left(a+b\right)\frac{\left(x+y+1\right)\left(x+y-1\right)}{2}=x^2+y^2-1+2xy=\left(x+y+1\right)\left(x+y-1\right)$ as we are speaking of $2xy$... being divided by $x+y+1$ or $x+y-1$ and as we don't say is divisible by zero, we can neglect the case in which $\left(x+y+1\right)\left(x+y-1\right)=0$. Which yields the condition $a+b=2$
Since $a,b,c$ and $d$ are positive integers, there cannot exist a case in which $a+b=1$. And $c=d=1$. This causes either of $x+y-1$ or $x+y+1$ zero which is absurd as we are talking about division.
For the second case we have $x^2+y^2-1=2xy=\frac{\left(x+y+1\right)\left(x+y-1\right)}{2}$
Which gives $\left(x-y\right)^2=1$ $\implies x=y+1$ or $x=y-1$
| {
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Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(2,6,3),(3,6,2),(3,2,6),(6,2,3),(6,3,2)}$. Are these the only 10 solutions?
First, none of $x$, $y$ or $z$ can be $1$ ($x$, $y$ and $z$ are positive integers)
If I let $x=2$, then finding all solutions to $1/y+1/z = 1/2$ leads to $(4,4), (3,6)$ and $(6,3)$ which gives me $(x,y,z)$ as $(2,4,4), (2,3,6), (2,6,3)$ but this also means $(4,4,2), (4,2,4), (3,2,6), (3,6,2), (6,2,3), (6,3,2)$ are all valid triples for this equation.
If I let $x=3$, the only different values of $y$ and $z$ are $(3,3)$
How do I prove these are the only ten solutions? (without using any programming)
Known result: If we denote $d(n^2)$ as the number of divisors of $n^2$, then the number of solutions of ${\frac {1} {x} }+{\frac {1} {y} } = {\frac {1} {n} }$ = $d(n^2)$ (For positive $x$, $y$)
For ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$
$z = \frac{xy}{y(x-1)-x}$ where $xy \neq 0$
What happens after that?
Question is: how do we sho there are the only ten solutions? I'm not asking for a solution.
Assuming $x \le y \le z$
$1 \le y \le \frac{xy}{y(x-1)-x}$
$\Longrightarrow 1 \le x \le y \le \frac{2x}{x-1} $
Got the answer. I'll probably call @mathlove's answer. (Any additional answers I'll view later)
Liked @user44197 answer.
| Hint: Deduce that none of $x, y, z \in \mathbb{N}$ exceeds $7$. This can be done by mathlove's hint above.
Additionally, you can show that there is only one $(a, 2, 2)$-tuple and thus use $(a, 2, 3)$ to bound the solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$ Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$
how to do it? I think it should be $3^6$ since $(3x^2)^6=3^6x^8$. (this is false)
Is this true?
| Let $S:=(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)$. Then $$\begin{align}S(1+x)&=1-x+x^2-x^3+x^4-x^5+x^6+7x^7\\
&=\frac{1+x^7}{1+x}+7x^7\end{align}$$
So $$\begin{align} S&=\frac{1+x^7}{(1+x)^2}+\frac{7x^7}{1+x} \end{align}$$
Then we have $$ S^6=\sum_{r=0}^6{n\choose r}\frac{(1+x^7)^r(7x^7)^{n-r}}{(1+x)^{2r+n-r}} $$
The observant eye will note that $n-r\le 1$, otherwise $\deg(7x^7)^{n-r}>8$, so we look at:
$$ 6\frac{(1+x^7)^5\cdot 7x^7}{(1+x)^{11}}+\frac{(1+x^7)^6}{(1+x)^{12}} $$
One may note that $\frac{1}{(1-x)^n}=\sum_{k=0}^\infty {k+n-1\choose n-1}x^k$, so in the first term we have $$42x^7(1+x^7)^5(1-11x+\ldots)\to-462x^8$$
And in the second term: $$ (1+6x^7+\ldots)(1-12x+\ldots+{19\choose 11}x^8-\ldots )\to -72x^8+75582x^8 $$
So $75582-462-72=75048$ is the coefficient of $x^8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
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Dealing with Generating Functions accurately I'm currently working through Iven Niven's "Mathematics of Choice." In the chapter on Generating Functions, the exercises include problems like:
How many solutions in non-negative integers does the equation $2x+3y+7z+9r=20$ have?
This of course ends up being the same as finding the coefficient of $x^{20}$ in $(1+x^2+x^4+x^6+\cdots+x^{20})(1+x^3+x^6+\cdots+x^{18})(1+x^7+x^{14})(1+x^9+x^{18})$. Maybe I'm just being a big weenie, but dealing with polynomials this large ends up being really tedious and error-prone.
I've realized that you can save the most involved multiplication for last, since that requires the least detail, and I'm of course not bothering to keep track of any powers greater than the one I'm interested in.
Any tips for keeping these giant polynomials manageable?
| Dealing with coefficients of generating functions is either synthetically (by learning patterns from experience what the formula for coefficients might be) or directly by manipulating the coefficients of constituent gfs (which helps with the first one). Usually the latter ends up by resulting in the multiplication of polymonials.
For multiplying large complicated polynomials in order to get a single coefficient:
*
*throw away anything above the degree you care about.
*use grid paper to manage coefficients in a matrix
*shift successive lines so you can add by columns instead of diagonals.
For example, for $(1 + 3x^2 + x^4 + 2x^6)(1 + 5x^3 + x^6)$:
$$
\begin{array}{c|cccccccc}
\hline
& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\
\hline
& 1 & 0 & 3 & 0 & 1 & 0 & 2 \\
\hline
1 & 1 & & 3 & & 1 & & 2 \\
0 & & & & & & & \\
0 & & & & & & & \\
5 & & & & 5 & & 15& & 5 & & 10\\
0 & & & & & & & \\
0 & & & & & & & \\
1 & & & & & & & 1 & & 3 & & 1 & & 2 \\
\hline
& 1 & 0 & 3 & 5 & 1 & 15& 3 & 5 & 3 & 10& 1 & 0 & 2
\end{array}
$$
for $1 + x^2 + 5x^3 + x^4 + 15 x^5 + 3 x^6 + ...$
If all we wanted was the coefficient of $x^6$, we could have avoided any column beyond 6.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/617907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Help with finding tangent to curve at a point
Find an equation for the tangent to the curve at $P\left( \dfrac{\pi}{2},3 \right )$ and the horizontal tangent to the curve at $Q.$
$$y=5+\cot x-2\csc x$$
$y\prime=-\csc ^2 x -2(-\csc x \cot x)$
$y\prime= 2\csc x \cot x - \csc ^2 x\implies$ This is the equation of the slope.
Now I find the slope at $x=\dfrac{\pi}{2}:$
$$y^{\prime}=2\csc \left( \frac{\pi}{2} \right) \cot \left( \frac{\pi}{2} \right)- \csc ^2 \left( \frac{\pi}{2} \right)$$ $$y^{\prime}= -1$$
Equation for the tangent to the curve at $P$:
$$y-3=-1\left( x-\frac{\pi}{2} \right)$$ $$y=-x+\frac{6+\pi}{2}$$
Then I found the horizontal tangent at $Q$, which is where the slope is $0.$ So I set the slope equation equal to $0:$
$$y^{\prime}= 2\csc x \cot x - \csc ^2 x=0$$ $$-\csc ^2 x = -2\csc x \cot x$$
$$-\frac{1}{\sin^2 x}=-2\left( \frac{1}{\sin x} \frac{\cos x}{\sin x} \right )$$
$$\cos x = \frac{1}{2}$$
I don't know how to proceed from here. Since $\cos x = \dfrac{1}{2}$, $\cos^{-1}\left (\dfrac{1}{2}\right)=\dfrac{\pi}{3}$.
Plugging this into the equation would give the $y$ value, and I'd be able to find the equation of the tangent line, but I'm confused because there is not just one $x$ to consider, since $x=2n \pm \dfrac{\pi}{3}$.
The answer for equation of tangent at $Q$ is $y=5-\sqrt{3}$, and I don't know how to get this.
Can you please show how to work this last part in details? Thank you.
| This means that there can be infinitely many horizontal tangents. As simple as that .
Just pick up the one you want.
$x = 2n \pm \dfrac{\pi}3$ is one solution.
Corresponding $y$ value
$y = 5 + \cot\left(2n \pm \dfrac{\pi}3\right) -2\csc\left(2n \pm \dfrac{\pi}3\right)$
$y = 5 \pm \dfrac1{\sqrt3} \pm \dfrac4{\sqrt3} $
$y = 5 \pm \sqrt3$
So one solution is $ y = 5 - \sqrt3$
It's better you solve $x = 2n + \dfrac{\pi}3$ and $x = 2n - \dfrac{\pi}3$ separately so as to not get yourself confused.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $a,b,c\in(0;+\infty)$, prove that $\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$. If $a,b,c\in(0;+\infty)$ and $$\frac{c}{1+a+b}+\frac{a}{1+b+c}+\frac{b}{1+c+a}\ge\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ca}{1+c+a}$$Prove that $$\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$
I know that $$a^2+b^2+c^2\ge\frac{(a+b+c)^2}{3}\ge ab+bc+ac$$
So we could prove that $$a+b+c+3\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$
I.e. $$2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\le a+b+c+3$$
By using AM-GM we see that we could prove that: $$2(a+b+c)\le a+b+c+3\Rightarrow a+b+c\le3$$
So this didn't work.
| The given condition is equivalent to:
$$\sum_{cyc}\frac{c-ab}{1+a+b} \ge 0$$
$$\implies \sum_{cyc}\frac{c(1+a+b)-ab}{1+a+b} \ge \sum_{cyc}\frac{c(a+b)}{1+a+b} \qquad \text{adding to both sides}$$
$$\implies \sum_{cyc}c \ge \sum_{cyc}\frac{c(a+b)+ab}{1+a+b} = (ab+bc+ca)\sum_{cyc}\frac{1}{1+a+b} \tag{1}$$
By Cauchy-Schwarz, we also have
$$\left(\sum_{cyc}\frac{1}{1+a+b}\right)\left(\sum_{cyc}(c+ca+bc)\right)\ge \left(\sum_{cyc} \sqrt{c} \right)^2 \tag{2} $$
Using this in $(1)$, we have:
$$a+b+c \ge (ab+bc+ca)\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{a+b+c+2(ab+bc+ca)}$$
Cross multiplying and expanding we get,
$$(a+b+c)^2 + 2(ab+bc+ca)(a+b+c)\ge (ab+bc+ca)(a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca})$$
Expanding the first term, cancelling part of the second term with RHS and dividing throughout by $ab+bc+ca$, we get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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proving the inequality $\triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$ If $\triangle$ be the area of $\triangle ABC$ with side lengths $a,b,c$. Then show that $\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$
and also show that equality hold if $a=b=c$.
$\bf{My\; Try}::$ Here we have to prove $4\triangle\leq \sqrt{(a+b+c)\cdot abc}$
Using the formula $$\triangle = \sqrt{s(s-a)(s-b)(s-c)},$$ where $$2s=(a+b+c)$$
So $$4\triangle = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{(a+b+c)\cdot(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$
Now using $\bf{A.M\geq G.M}$ for $(b+c-a)\;,(c+a-b)\;,(a+b-c)>0$
$$\displaystyle \frac{(b+c-a)+(c+a-b)+(a+b-c)}{3}\geq \sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$
So we get $\displaystyle (a+b+c)\geq 3\sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$
But I did not understand how can I prove above inequality
help Required
Thanks
| Thanks mathlove,phira,and user for solution.
My prove for $(b+c-a)\cdot(c+a-b)\cdot(a+b-c)\leq abc$, where $a,b,c$ are the sides of a $\triangle$.
Using $\;\;\;\; \{b+(c-a)\}\cdot \{b-(c-a)\} = b^2-(c-a)^2\leq b^2$
similarly $ \{c+(a-b)\}\cdot \{c-(a-b)\} = c^2-(a-b)^2\leq c^2$
similarly $\{a+(b-c)\}\cdot \{a-(b-c)\} = a^2-(b-c)^2\leq a^2$
Multimply these three equations, we get $(b+c-a)\cdot(c+a-b)\cdot(a+b-c)\leq abc$
and equality hold when $a=b=c$
| {
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"url": "https://math.stackexchange.com/questions/621182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Proving that if $a^2+b^2=c^2$, then $a+b\ge c$. Hello, I'm trying to prove this statement.
Let a,b & c be three positive real numbers and if $a^2+b^2=c^2$ then $a+b\ge c$
Any help, please?
| $(a+b)^2=a^2+b^2+2ab\geq a^2+b^2= c^2\rightarrow (a+b)^2\geq c^2\rightarrow a+b\geq c$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Closed form of $\arccos\left(\frac{2 \pi}{2^N}\right)$ Is there a closed form for $\arccos\left(\dfrac{2 \pi}{2^N}\right)$ in terms of $N \in \mathbb{Z}, N \ge 3$?
I'm not super optimistic, but I'm not sure how to really start exploring the problem, either.
| $$\arccos z= \frac {\pi} {2} - \left( z + \left( \frac {1} {2} \right) \frac {z^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^5} {5} + \cdots\ \right)
= $$
$$=\frac {\pi} {2} - \sum_{n=0}^\infty \frac {\binom{2n} n z^{2n+1}} {4^n (2n+1)}; \qquad | z | \le 1 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
x-4=x-5-\sqrt{x-5}+1\\
x-4=x-4-\sqrt{x-5}\\
\text{substracting x, and adding 4 to both sides}\\
0=-\sqrt(x-5)\\
\text{switching both sides}\\
\sqrt{x-5}=0\\
\text{sqaring both sides}\\
x-5=0\\
x=5\\
\text{When I place 5 in the equation, I get:}\\
\sqrt{5-4}-\sqrt{5-5}+1=0\\
\sqrt{1}-\sqrt{0}+1=0\\
1-0+1=0\\
2=0\\
\text{this means that the equation dosent have any solution, right??}\\
$$
Any advice and suggestion is helpful.
Thanks!!!
| It is easily seen that the equation $\sqrt{x-4}-\sqrt{x-5}=-1$ has no solutions since $\sqrt{x-4}>\sqrt{x-5}$ for all $x>=5$.
Your method is correct but you made a mistake in going from step $3$ to step $4$. After correcting that, you will still reach the same conclusion that there are no solutions.
| {
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"url": "https://math.stackexchange.com/questions/624974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
calculus limit question: another difficult limit problem I have posted previously on a problem in a similar vein here:
Limit evaluation: very tough question, cannot use L'hopitals rule
I believe this problem is very similar, but it has stumped me.
$$\lim_{x \to 0}\frac{1-\frac12 x^2 - \cos\left(\frac{x}{1-x^2}\right)}{x^4}$$
Really appreciate it if someone has some insight on this.This comes out to be indeterminate if one plugs in zero.
Following the idea from the link above, I tried to recognize this as derivative evaluated at zero of a function, BUT I could not find the function, because I tried to make this all over x, so that means the function I would create would generate a rational type with x^3 on the bottom.
I guess I should also try to look at some trig limit identities as well.
Hope someone out there can see how to navigate this problem.
P
| Like what Emanuele said, asymptotic expansions are useful for this kind of limits, and in fact better than L'Hopital (which fails miserably for some limits):
$\frac{x}{1-x^2} \in x + x^3 + O(x^5) \to 0$ as $x \to 0$
[We keep the error term so that at the end we know the error of the final approximation.]
$\cos( \frac{x}{1-x^2} ) \in 1 - \frac{1}{2} ( \frac{x}{1-x^2} )^2 + \frac{1}{24} ( \frac{x}{1-x^2} )^4 + O( ( \frac{x}{1-x^2} )^6 ) \\
\subset 1 - \frac{1}{2} (x+x^3+O(x^5))^2 + \frac{1}{24} (x+x^3+O(x^5))^4 + O(x^6) \text{ as } x \to 0 \\
\subset 1 - \frac{1}{2} x^2 - \frac{23}{24} x^4 + O(x^6) \text{ as } x \to 0$
[We can make the substitution into the Taylor expansion only because the input to $\cos$ tends to 0.]
$\frac{ 1 - \frac{1}{2} x^2 - \cos( \frac{x}{1-x^2} ) }{ x^4 } \in \frac{ \frac{23}{24} x^4 + O(x^6) }{ x^4 } = \frac{23}{24} + O(x^2) \to \frac{23}{24}$ as $x \to 0$
But you must make sure you understand the meaning of the Big-O notation and when and why they can be used. To make it more concrete, you can in many cases find explicit constants for bounds instead of using Big-O notation. For example:
$1 - \frac{1}{2} x^2 + \frac{1}{24} x^4 - \frac{1}{720} x^6 \le cos(x) \le 1 - \frac{1}{2} x^2 + \frac{1}{24} x^4$ [obtained by repeated differentiation and Mean-value theorem]
$x + x^3 \le \frac{x}{1-x^2} \le x + x^3 + 2 x^5$ for sufficiently small $x \ge 0$
$x + x^3 \ge \frac{x}{1-x^2} \ge x + x^3 + 2 x^5$ for sufficiently small $x \le 0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find solutions of this equation If $a+b+c+d = 30$ and $a,b,c,d$ lie between $0$ and $9$. How to find number of solutions of this equation.
| Hint: Consider the coefficient of $x^{30}$ in
$$
\left(\frac{x^{10}-1}{x-1}\right)^4
$$
Since
$$
(1-x)^{-4}=\sum_{n=0}^\infty\binom{n+3}{3}x^n
$$
and
$$
\left(1-x^{10}\right)^4=\sum_{k=0}^4(-1)^k\binom{4}{k}x^{10k}
$$
we get the coefficient of $x^n$ to be
$$
\sum_{k=0}^4(-1)^k\binom{4}{k}\binom{n-10k+3}{n-10k}
$$
Note that $\binom{n-10k+3}{n-10k}=\binom{n-10k+3}{3}$ when $n\ge10k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/626837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
| $$2^{10} + 2^{9} + 2^{8} +2^7+2^6+2^5+2^4+ 2^{3} + 2^2 + 2^1 = 2^{11} - 2$$
Add $(2-2)$ to the left-hand side, obtaining:
$$\begin{align}
2^{10} + 2^{9} + 2^{8} + 2^7+2^6+2^5+2^4 + 2^{3} + 2^2 + 2^1 \color{green}{+ 2 - 2} & = 2^{11} -2 \\
2^{10} + 2^{9} + 2^{8} + 2^7+2^6+2^5+2^4 + 2^{3} + 2^2 + \color{maroon}{2^1 + 2^1} - 2 & = 2^{11} -2 \end{align}$$
Since $2^1 + 2^1 = 2\cdot(2^1) = 2^2$, the left side simplifies to:
$$2^{10} + 2^{9} + 2^{8} +2^7+2^6+2^5+2^4+ 2^{3} + \color{maroon}{2^2 + 2^2} - 2 = 2^{11} - 2$$
Since $2^2 + 2^2 = 2\cdot(2^2) = 2^3$, the left side simplifies to:
$$2^{10} + 2^{9} + 2^{8} + 2^7+2^6+2^5 + 2^{4} + \color{maroon}{2^3 + 2^3} - 2 = 2^{11} - 2$$
Repeat this simplification several more times:
$$2^{10} + 2^{9} + 2^{8} + 2^7+2^6+2^5 + \color{maroon}{2^4 + 2^4} - 2 = 2^{11} - 2 \\
\\ \vdots \\\color{maroon}{2^{10} + 2^{10}} - 2 = 2^{11} - 2\\
\color{maroon}{2^{11}} - 2= 2^{11} - 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to memorize the families that are $\sin$, $\cos$, and $\tan$ of $\pi$ over something? Is there any way to easily memorize these? Such as $\sin \frac{\pi}{6} = 1/2$. Any help needed!!!
| Here's a nice trick:
$$
\sin\left(0\right) = \sqrt{\frac{0}{4}}\\
\sin\left(\frac{\pi}{6}\right) = \sqrt{\frac{1}{4}}\\
\sin\left(\frac{\pi}{4}\right) = \sqrt{\frac{2}{4}}\\
\sin\left(\frac{\pi}{3}\right) = \sqrt{\frac{3}{4}}\\
\sin\left(\frac{\pi}{2}\right) = \sqrt{\frac{4}{4}}
$$
where the angles are in increasing order.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Basis of Kernel of a matrix Given $\theta>0$. Let $H$ be $5 \times 6$ matrix
$$\left[\begin{matrix}
1 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & -1 & 0 & 0 & 0 \\
1 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
\theta & 0 & 0 & 0 & 0 & -1
\end{matrix}\right]$$
Consider the subspace $S=\{x\in\mathbb{R}^6:Hx=0$}. I know the subspace $S$ has dimension $1$. However, I couldn't find the basis of $S$. Could anybody help me please. Thanks in advance.
| We perform the row operations:
$$\begin{bmatrix}
1 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & -1 & 0 & 0 & 0 \\
1 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
\theta & 0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}
\xrightarrow{R_5 \gets \tfrac{1}{\theta} R_5}
\begin{bmatrix}
1 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & -1 & 0 & 0 & 0 \\
1 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
1 & 0 & 0 & 0 & 0 & -\tfrac{1}{\theta} \\
\end{bmatrix}
\xrightarrow{R_i \gets R_{\alpha(i)} \text{ where } \alpha=(15432)}
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & -\tfrac{1}{\theta} \\
1 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & -1 & 0 & 0 & 0 \\
1 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
\end{bmatrix}
\xrightarrow{R_i \gets R_i-R_1 \text{ for } i \in \{2,3,4\}}
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & -\tfrac{1}{\theta} \\
0 & -1 & 0 & 0 & 0 & \tfrac{1}{\theta} \\
0 & 0 & -1 & 0 & 0 & \tfrac{1}{\theta} \\
0 & 0 & 0 & -1 & 0 & \tfrac{1}{\theta} \\
0 & 0 & 0 & 0 & 1 & -1 \\
\end{bmatrix}
\xrightarrow{R_i \gets -R_i \text{ for } i \in \{2,3,4\}}
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & -\tfrac{1}{\theta} \\
0 & 1 & 0 & 0 & 0 & -\tfrac{1}{\theta} \\
0 & 0 & 1 & 0 & 0 & -\tfrac{1}{\theta} \\
0 & 0 & 0 & 1 & 0 & -\tfrac{1}{\theta} \\
0 & 0 & 0 & 0 & 1 & -1 \\
\end{bmatrix}
$$
which is in reduced row echelon form. So the matrix has rank $5$ and hence indeed the nullity is $6-5=1$, by the Rank-Nullity Theorem.
By inspection the null space is: $$\mathrm{span}\{(1,1,1,1,\theta,\theta)\}.$$
| {
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Calculation of $\bf{Max.}$ value of $\sqrt{10x-x^2}-\sqrt{18x-x^2-77}\;\;\forall x\in \mathbb{R}$ (1) Calculation of Max. and Min. value of $\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\;\; \forall x\in \mathbb{R}$
(2) Calculation of Max. value of $\sqrt{10x-x^2}-\sqrt{18x-x^2-77}\;\;\forall x\in \mathbb{R}$
My Try:: for $(1)$ one:: Using the Cauchy-Schwarz inequality
$$\Rightarrow \displaystyle \left\{\left(\sqrt{x^2-3x+2}\right)^2+\left(\sqrt{2+3x-x^2}\right)^2\right\}\cdot \left(1^2+1^2\right)\geq \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}^2$$
$$\Rightarrow \displaystyle \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}^2\leq 8$$
$$\Rightarrow \left\{\sqrt{x^2-3x+2}+\sqrt{2+3x-x^2}\right\}\leq 2\sqrt{2}$$
and equality hold when $\displaystyle \sqrt{x^2-3x+2}=\sqrt{2+3x-x^2}\Rightarrow x=0\;,3$
Is my solution right for Max.? If not then please help me, and how can I calculate for Min. and also help required in $(2)$ one.
| (1):the approach is not right ,you can't make sure $a\ge b,b\le c \to a \ge c $
the easy way is let$ u=x^2-3x, f(x)=f(u)=\sqrt{2+u}+\sqrt{2-u},f^2=4+2\sqrt{4-u^2}$
$0\le\sqrt{4-u^2} \le 2 \implies 4 \le f^2 \le 8 \implies 2 \le f \le 2\sqrt{2}$ when $u=0$ and $u=2$ get Max and Min.
(2): $\sqrt{10x-x^2}$ domain is [$0,10$], $\sqrt{18x-x^2-77}$ domain is [$7,11$] so $f(x)=\sqrt{10x-x^2}-\sqrt{18x-x^2-77}$ domain is [$7,10$]
$\sqrt{10x-x^2}$ is mono decreasing on [$7,10$], $\sqrt{18x-x^2-77}$ is increasing on [$7,9$] so we can sure $f(x)$ will get Max when $x=7$ when $x$ on [$7,9$], the remain is [$9,10$],but it is impossible for any $f(x) \ge f(7)$ because it is always $ <f(7)- a, a >0$
| {
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How to prove $(b-2)^2 > 12a(5c + 2)$ provided $(3a + b + 5c)(5c + 2) < 0$? $a$, $b$, $c$ are rational numbers. It is known that $(3a + b + 5c)(5c + 2) < 0$.
How do I prove that $(b-2)^2 > 12a(5c + 2)$?
According to Bernoulli's inequality, I got this:
$(1 + b - 3)^2 \ge 1 + 2(b - 3) > 12a(5c + 2)$
$2b - 5 > 12a(5c + 2)$
What should I do next?
| It seems the following.
Put $x=3a$, $y=b-2$, and $z=5c+2$. Given $(x+y+z)z<0$ we need to show that $y^2>4xz.$ Assume the converse. Then $(x+z)^2\ge 4xz>y^2$. So $|x+z|>|y|$ and the sign of $x+y+z$ is equal to the sign of $x+z$. Then $(x+z)z<0$, but $xz>y^2/4\ge 0$ and $z^2\ge 0$, a contradiction.
| {
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A claim that a is a square We have integers $a,b,c,d$ such that $a<b\le c<d$ and $ad=bc$ and $\sqrt{d}-\sqrt{a}\le 1$.Show that $a$ is a perfect square.This question is pretty unbelievable for me.anyway I don't know if I am reposting this here.
| Let us show that $a = n^2 ,\; b = n^2+n = c,\; d = (n+1)^2$ is the only solution.
The condition $\sqrt{d}-\sqrt{a} \leq 1$ shows that $$ d\leq a+2\sqrt{a}+1.$$
Let $b = a+x,\; c = a+y,\; d = a+z$ with $x,y,z \in \Bbb N$. Then $0<x\leq y < z \leq 2\sqrt{a}+1$, showing $0<x\leq y \leq 2 \sqrt{a}$. Furthermore:
$$ad = bc = (a+x)(a+y) = a^2+(x+y)a+xy$$ and reducing $\bmod a$ shows $a \mid xy$. Thus $$a\leq xy$$ and there is $k \in \Bbb N$ such that $xy = ak$. This leads to $ad = bc= (a+x)(a+y) = a(a+x+y+k)$ and dividing by $a$ yields $ a+x+y+k = d \leq a+2\sqrt{a}+1$. As $k\geq 1$, we conclude $$x+y\leq 2\sqrt{a}.$$
We find $4a \leq 4xy \leq 4xy +(x-y)^2 = (x+y)^2 \leq 4a$. Hence $x=y$ by $(x-y)^2 = 0$. Finally $x+y = 2\sqrt{a}$ implies that $a$ is necessarily a perfect square.
| {
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Prove that $\lim\limits_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)=0$
Prove that:
$$\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}
\,\right)=0$$
Epsilon>0. According to the Archimedean Property of reals, we have n1 element N with epsilon*n1>11. (Why 11? Seems so random...)
n0:=max{3,n1} (What's the point of that? It doesn't appear anywhere in the proof...).
For each n element N with n>=n0 we have:
\begin{align}
0&\lt\sqrt{n^4+n^2+20n+7}-\sqrt{n^4+n^2+1} \\\,\\
&=\dfrac{\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}\right)\cdot\left(\sqrt{n^4+n^2+20n+7}+\sqrt{n^4+n^2+1}\right)}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}} \\\,\\
&=\dfrac{20n+6}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}}\leqslant\dfrac{20n+2n}{\sqrt{n^4}+\sqrt{n^4}}=\dfrac{11}n\leqslant\dfrac{11}{n_1}\lt\epsilon
\end{align}
(I don't get the circled part. Why 20n+2n, why "cut off" the roots? And why should it equal 11/n?)
From this we get
$$\left|\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right|\lt\epsilon$$
, thus the proof is complete.
Thanks for the clarifications!
| The numerator was made bigger (or equal, if $n=3$) and the denominator was made smaller, so the fraction becomes bigger. This only works, when $n\geqslant3$ is assumed somewhere.
| {
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"source": "stackexchange",
"question_score": "2",
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Inverse Function (and WolframAlpha gives different Result) I wanted to calculate the inverse function of
$$
f(x) = \frac{1}{x} + \frac{1}{x-1}
$$
Quite simple I thought, put
$$
y = \frac{1}{x} + \frac{1}{x-1} = \frac{2x-1}{x(x-1)}
$$
rearrange and solve
$$
y(x(x-1)) - 2x + 1 = 0
$$
which give the quadratic equation
$$
yx^2 - (y + 2)x + 1 = 0
$$
Using the Solution Formula we got
$$
x = \frac{(y+2) \pm \sqrt{y^2+4}}{2y}
$$
So the inverse function is
$$
f^{-1}(x) = \frac{(x+2) \pm \sqrt{x^2+4}}{2x}
$$
Just to confirm I put in WolframAlpha and it gives me
$$
\frac{-x-2}{2x} \pm \frac{\sqrt{x^2+4}}{2x}
$$
(just click on the link to start WolframAlpha with this parameter), which
is different up to a sign in the first summand, can not see an error, do you (or is WolframAlpha wrong...)?
EDIT: If the link is not working for you:
| Your error is in the solution formula. You have $(y+2)^2 - 4\cdot y\cdot 1 = y^2+4 \neq (y+2)(y-2)$. It would be $y^2-4 = (y+2)(y-2)$.
| {
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prove that $a^{nb}-1=(a^n-1)((a^n)^{b-1}+...+1)$ Prove that $a^{nb}-1=(a^n-1)\cdot ((a^n)^{b-1}+(a^n)^{b-2}+...+1)$
We can simplify it, like this:
$$(a^{n})^b-1=(a^n-1)\cdot \sum_{i=1}^{b}(a^{n})^{b-i}$$
How can we prove this?
| For $\ x = a^b\ $ it is $\ f_n \,:=\, \dfrac{x^n-1}{x-1}\, =\, x^{n-1}+ x^{n-2} +\cdots + x + 1.\,$ Here is a proof by telescopy.
Notice that $\ \color{#0a0}{f_{n} - f_{n-1} = x^{n-1}}\ $ since $\ \dfrac{x^n-1}{x-1} - \dfrac{x^{n-1}-1}{x-1}\, =\, \dfrac{x^n-x^{n-1}}{x-1\quad }\, =\, x^{n-1}$
$\begin{eqnarray}{\rm\! Hence}\ \ &&\qquad\quad\ \ \, \underbrace{\color{#0a0}{f_n}} \\
\,&=&\ \ \overbrace{\color{#0a0}{x^{n-1}}\ \ \ \ \ \ +\ \ \ \ \ \ \underbrace{\color{#0a0}{f_{n-1}}}}\\
\,&=&\ \ x^{n-1} +\ \overbrace{{x^{n-2}\ \ \ \ \ \,+\,\ \ \ \ \ \underbrace{f_{n-2}}}}\\
\,&=&\ \ x^{n-1} +\ x^{n-2}+\ \overbrace{x^{n-3}\ \ \ \ +\ \ \ \ {f_{n-3}}}\\
\,&&\qquad\qquad\ \ \vdots\qquad\qquad\qquad\qquad\qquad\ddots
\end{eqnarray}$
Alternatively we can write it in telescopic form
$\ \ \begin{eqnarray}
\color{#c00}{f_n-f_k}\, =\,
\underbrace{\phantom{f_n-f_{n-1}}}_{\Large x^{n-1}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\color{#c00}{f_n}}&&\!\!\overbrace{{-f_{n-1}}
+\underbrace{\phantom{f_{n-1} - f_{n-2}}}_{\Large x^{n-2}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!f_{n-1}}^{=\, 0}&&\!\!\overbrace{-f_{n-2} + f_{n-2} }^{=\, 0}
- \,\cdots\, \overbrace{-f_{k+2}
+ \underbrace{\phantom{f_{k+2} - f_{k+1}}}_{\Large x^{k+1}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!f_{k+2}}^{=\, 0}&&\!\!\overbrace{-f_{k+1}
+\underbrace{\phantom{f_{k+1} - f_{k}}}_{\Large x^{k}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!f_{k+1}}^{=\,0}&&\!\!\!\color{#c00}{-\,f_k}\\
\end{eqnarray}$
| {
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Linear algebra question on rank and null space This is an exercise from linear algebra and optimization by Gill, I do exercises to be prepared for my final exam and these are not homework questions!
Exercise $\mathbf{6.1.}\,$ Consider the following matrix $A$ and vector $b$:
$$
A=\begin{pmatrix}
2 & 4 \\
1 & 2 \\
1 & 2 \\
\end{pmatrix},\quad
b=\begin{pmatrix}
3 \\
2 \\
1 \\
\end{pmatrix}.
$$
$\textbf{(a)}$ What is the rank of $A$? Give a general form for any vector in the range of $A$.
$\textbf{(b)}$ Show that the dimension of the null space of $A^T$ is two, and display two linearly independent vectors $z_1$ and $z_2$ in $\operatorname{null}(A^T)$. Give a general form for every vector in $\operatorname{null}(A^T)$.
$\textbf{(c)}$ Find the vectors $b_R\in\operatorname{range}(A)$ and $b_N\in\operatorname{range}(A^T)$ such that $b=b_R+b_N$.
$\textbf{(d)}$ Give the general form of $b_A$ such that $b_R=Ab_A$. (Hint: consider all vectors $q$ such that $Aq=0$.)
For part (a), I think $rank(A)=1$ since all the columns are linear combination of each other. As for a general form for any vector in the range of $A$, when I write $Ax=b$ I get an over-determined system:
$$2x_1+4x_2=3$$
$$x_1+2x_2=2$$
$$x_1+2x_2=1$$
So I'm not sure about the general form.
As for part (b), since the rank is two, dimension of $N(A^T)$ must be $3-1=1$, right?
And no idea about the rest. Any help would be greatly appreciated.
| If you do row reduction, you find
$$
\begin{pmatrix}
2 & 4 \\
1 & 2 \\
1 & 2
\end{pmatrix}
\to
\begin{pmatrix}
1 & 2 \\
0 & 0 \\
0 & 0
\end{pmatrix}
$$
which means that the first column is a basis for the column space of $A$ (which is better terminology than “range of $A$”, in my opinion). So the general form of the vectors in the column space of $A$ is
$$
\begin{pmatrix}
2a \\
a \\
a
\end{pmatrix},\quad \text{$a$ any scalar}
$$
By the rank nullity theorem, the null space of $A$ has dimension $1$; the equation defining it is
$$
x_1+2x_2=0
$$
so a basis for it is the single vector
$$
\begin{pmatrix}
-2 \\
1
\end{pmatrix}
$$
The null space of $A^T$ has indeed dimension $2$; the row reduction on $A^T$ is
$$
\begin{pmatrix}
2 & 1 & 1 \\
4 & 2 & 2
\end{pmatrix}
\to
\begin{pmatrix}
2 & 1 & 1 \\
0 & 0 & 0
\end{pmatrix}
$$
so the equation defining the null space is $2x_1+x_2+x_3=0$ and a basis for it is
$$
\left\{
\begin{pmatrix}
-1 \\ 2 \\ 0
\end{pmatrix}
\,,
\begin{pmatrix}
-1 \\ 0 \\ 2
\end{pmatrix}
\right\}
$$
Writing $b=b_R+b_N$ should now be easy: the system to solve is
$$
\left(\begin{array}{ccc|c}
2 & -1 & -1 & 3 \\
1 & 2 & 0 & 2 \\
1 & 0 & 2 & 1
\end{array}\right)
$$
but you can as well find the orthogonal projection of $b$ on the column space of $A$:
$$
b_R=
\frac{(2\ 1\ 1)\begin{pmatrix}3\\2\\1\end{pmatrix}}
{(2\ 1\ 1)\begin{pmatrix}2\\1\\1\end{pmatrix}}
\begin{pmatrix}2\\1\\1\end{pmatrix}
=\frac{9}{6}\begin{pmatrix}2\\1\\1\end{pmatrix}
=\begin{pmatrix}3\\3/2\\3/2\end{pmatrix}
$$
and $b_N=b-b_R$.
With this last idea it's easy to solve the last point.
| {
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Help with local extrema of $f(x)=x^4-5x^2$
Find the coordinates of any local extreme points and inflection points of the function $f(x)=x^4-5x^2$
My try:
Find critical points: $f^{\prime}(x)=4x^3-10x=0$
$f^{\prime}(x)=2x(2x^2-5)=0 \implies x=0, x=\pm\sqrt{\dfrac{5}{2}}$
I would then use the critical points to determine where the function is increasing/decreasing and by inputing critical point $c$ into $f^{\prime\prime}(c)$, I would determine local min/max.
This is wrong though, because the answers are:
local min: $\left( \dfrac{-\sqrt{10}}{2},\dfrac{-25}{4} \right)$, $\left( \dfrac{\sqrt{10}}{2},\dfrac{-25}{4} \right)$
inflection points: $\left( \dfrac{-\sqrt{30}}{6},\dfrac{-125}{36} \right)$, $\left( \dfrac{\sqrt{30}}{6},\dfrac{-125}{36} \right)$
What am I doing wrong and how do I do it correctly? Thanks.
| From where you left off with the critical points you correctly found, we determine their $y$-coordinates as followed
$$\begin{aligned}
f(0) &= 0\\
f\left(-\dfrac{\sqrt{10}}{2} \right) &= -\dfrac{25}{4}\\
f\left(\dfrac{\sqrt{10}}{2} \right) &= -\dfrac{25}{4}
\end{aligned}$$
To determine the nature of those critical points, we use the Second Derivative Test. First, we find the second derivative of $f(x)$. Then, we check each critical point. Here is how one uses the test:
$$\begin{aligned}
f'(x) &= 4x^3 - 10x\\
f''(x) &= 12x^2 - 10\\
f''(0) &= -10 < 0\\
f''\left(-\dfrac{\sqrt{10}}{2} \right) = f''\left(\dfrac{\sqrt{10}}{2} \right) &= 20 > 0
\end{aligned}$$
This shows that the points $\left(\dfrac{\sqrt{10}}{2}, -\dfrac{25}{4}\right)$ and $\left(-\dfrac{\sqrt{10}}{2}, -\dfrac{25}{4} \right)$ are local minima while $(0,0)$ is local maximum.
Finding the inflection points should be extremely easy. All you need to do is to use $f''(x)$ and set it equal to $0$. Then, find the values of $x$.
| {
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How to prove that $\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0?$
How to prove that $\lim_{(x,y) \to (0,0)} \dfrac{x^3y}{x^4+y^2} = 0?$
First I tried to contradict by using $y = mx$ , but I found that the limit exists.
Secondly I tried to use polar coordinates, $x = \cos\theta $ and $y = \sin\theta$,
And failed .. How would you prove this limit equals $0$?
| With polar coordinates we're cool, too:
$$x=r\cos\theta\;\;,\;\;y=r\sin\theta$$
$$\frac{x^3y}{x^4+y^2}=\frac{r^4\cos^3\theta\sin\theta}{r^4\cos^4\theta+r^2\sin^2\theta}=\frac{r^2\cos^3\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}\xrightarrow[r\to 0]{}\frac0{0+\sin^2\theta}=0$$
But what if $\;\sin^2\theta=0\;$ ? Well, then also $\;\sin\theta=0\;$ and the first expression above's already zero from the beginning...
| {
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mixed numbers subtraction vertically In the following subtraction we are subtracting $2$ mixed numbers vertically. I know how it works except the last step.
$$ 7 \frac{1}{3} - 4 \frac{1}{2} = 3 + \frac{-1}{6} = 2 + \frac{5}{6} = 2 \frac{5}{6}$$
I am confused about this step: $ 3 + \frac{-1}{6} = 2 + \frac{5}{6}$
How does this work?
| Write it as
$$ 3 - \frac{1}{6}= 2 + 1 - \frac{1}{6}= 2 + \frac{6}{6} -\frac{1}{6} = 2 + \frac{6-1}{6} = 2+\frac{5}{6} $$
| {
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The sum of areas of 2 squares is 400and the difference between their perimeters is 16cm. Find the sides of both squares. The sum of areas of 2 squares is 400and the difference between their perimeters is 16cm. Find the sides of both squares.
I HAVE TRIED IT AS BELOW BUT ANSWER IS NOT CORRECT.......CHECK - HELP!
Let side of 1st square=x cm.
∴ Area of 1st square=x²cm²
GIVEN,
Sum of areas =400cm²
∴ Area of 2nd square=(400-x²)cm²
AND side of 2nd square=√[(20-x)² i.e.20-x .......(1)
Difference of perimeters=16cm.
THEN-
4x-4(20-x)=16 (ASSUMING THAT 1ST SQUARE HAS LARGER SIDE)
X=12
HENCE - SIDE OF 1ST SQUARE = 12CM ;
SIDE OF 2ND SQUARE=20-12=8CM. [FROM (1)]
WHICH IS NOT THE REQUIRED ANSWER AS SUM OF AREAS OF SQUARES OF FOUNDED SIDES IS NOT 400CM²
| $x^2+y^2=400$
$4*x-4*y=16$
could you continue?
ok we have $x^2+y^2=400$ and $x-y=4$
from there $x=4+y$
$(4+y)^2+y^2=400$
$16+8*y+y^2+y^2=400$
$2*y^2+8*y-384=0$
$y^2+4*y-192=0$
$D=4+192=196$
$y_1=(-2+\sqrt{196})$
$y_2=(-2-\sqrt{196})$
now calculate $x_1$ and $x_2$
$y=12$
$x=16$
now check $16*16+12*12=256+144=400$
$64-48=16$
| {
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Computing $\lim_{n\to \infty}{\frac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}}$ Let $\{a_n\}_{n\ge1}^{\infty}=\bigg\{\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}\bigg\}$. Find $\lim_{n\to \infty}{a_n}$.
I.) In the first step I studied monotony:
$a_{n+1}-a_{n}=\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}\cdot\cfrac{-2}{4n+7}<0$, $\{a_n\}$ is decreasing.
II.) In the second step I studied boundary.
$$1>a_{1}=\cfrac{5}{7}>a_{2}=\cfrac{45}{77}>\dots>a_{n}>0$$
III.) In the last step I know that $\{a_n\}$ converges to $a\in\mathbb R$.
$$a_{n+1}=a_{n}\cdot\cfrac{4n+1}{4n+3}$$
Taking the limit as $n\to\infty$:
$$a=a$$
No conclusion.
But if I apply Cesaro-Stolz?
IV.) Let $\{x_n\}_{n\ge1}^{\infty}=\{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)\}$ and $\{y_n\}_{n\ge1}^{\infty}=\{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)\}$. Then
$$\lim_{n\to \infty}{\cfrac{x_{n+1}-x_{n}}{y_{n+1}-y_{n}}=\lim_{n\to \infty}{\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)^2}{7\cdot11\cdot15\cdot\dots.\cdot(4n+5)}=?}}$$
If you have a simple solution, I would appreciate it. Thank you!
| One can check that
$$
\bigg( \frac{4n+1}{4n+3} \bigg)^2 < \frac{n+1}{n+2}
$$
for all $n\ge0$. Therefore
$$
0 < a_n = \frac57 \frac9{11} \cdots \frac{4n+1}{4n+3} < \bigg( \frac23 \frac34 \cdots \frac{n+1}{n+2} \bigg)^{1/2} = \sqrt{\frac2{n+2}},
$$
and so $a_n\to0$ by the squeeze theorem.
| {
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Pre-cal trigonometric equation problem am i correct to factor out a 2 first?
$$2\sin^2 2x =1$$
$$2(\sin^2 x-1) =1$$
$$2\cos^2 x =1$$
$$\cos^2 x ={1\over2}$$
$$\cos x =\pm{\sqrt{2}\over 2 }$$
i'm only looking for solutions from $$0≤ x ≤ 2\pi $$
$$x = {\frac{\pi}{4}},{\frac{7\pi}{4}},{\frac{3\pi}{4}},{\frac{5\pi}{4}}$$
thanks for any corrections
| $$
\begin{align*}
& 2\sin^2(2x)=1 \\
\Rightarrow & 2\left( \frac{1-\cos(4x)}{2} \right)=1\\
\Rightarrow&1-\cos(4x)=1\\
\Rightarrow & \cos(4x)=0 \\
\Rightarrow & 4x=\frac{n\pi}{2}, \, n \in \{ 1,3,5,7,\ldots \}\\
\Rightarrow & x=\frac{n\pi}{8}, n \in\{ 1,3,5,7,\ldots \}
\end{align*}
$$
Hence the set is
$$x\in\left\{\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\ldots\right\}$$
| {
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Why $\lim_{n\to \infty}{\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\dots+\frac{1}{n\cdot(n+2)}}=\frac{3}{4}$? Let $\{a_n\}_{n\ge1}^{\infty}=\bigg\{\cfrac{1}{1\cdot3}+\cfrac{1}{2\cdot4}+\dots+\cfrac{1}{n\cdot(n+2)}\bigg\}$. Find $\lim_{n\to \infty}{a_n}$.
I write:
$$\lim_{n\to \infty}{a_n}=\sum_{n=1}^{\infty}{\frac{1}{n\cdot(n+2)}}=\sum_{n=1}^{\infty}{\frac{1}{n^2+2n}}\approx\sum_{n=1}^{\infty}{\cfrac{1}{n^2}}$$
I put some values of $n$ for finding a pattern:
$$
\begin{array}{c|lcr}
n & \text{1}&\text{2}&\text{3}&\text{4}\\
\hline
\sum &\cfrac{1}{3}&\cfrac{11}{24}&\cfrac{21}{40}&\cfrac{17}{30}
\end{array}
$$
... but no hope.
I know that the limit/series converges, because $\forall n\in\mathbb N^*$:
*
*$\{a_n\}$ is increasing by the test of monothony : $a_{n+1}-a_{n}=\cfrac{1}{(n+1)(n+3)}>0$
*$\{a_n\}$ is bounded : $0\le a_1=\frac{1}{3}\le a_2=\frac{11}{24}\le \dots \le a_n \le1$
Wolfram says that the summation can be written as follows:
$$\cfrac{3}{4}-\cfrac{2n+3}{2(n+1)(n+2)}$$
How did it end up at this formula? Thank you.
| Telescope! Note that $$ \frac1{n(n+2)}=\frac12\cdot\left(\frac1n-\frac1{n+2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/642443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Complex numbers - Exponential numbers - Proof Let $z$ be a complex number, and let $n$ be a positive integer such that $z^n = (z + 1)^n = 1$. Prove that $n$ is divisible by 6.
For this problem I am stumped...how should I begin?
Also there's a hint for it:
From $z^n = 1$, prove that $|z| = 1$. What does the equation $(z + 1)^n = 1$ tell you? What do the resulting equations tell you about $z$?
Could someone give me a hint on where to begin? thanks in advance
| A problem way too cool and cute to pass up, so check this out:
$(1.) \; z^n = 1 \Rightarrow \vert z \vert^n = 1, \tag{1}$
$(2.) \; \vert z \vert^n = 1 \Rightarrow \vert z \vert = 1 \Rightarrow \exists \theta \in \Bbb R \;\text{such that} \; z = e^{i\theta}, \tag{2}$
$(3.) \; \vert z \vert = 1 \Rightarrow z \bar z = 1, \tag{3}$
$(4.) \; (z + 1)^n = 1 \Rightarrow \vert z + 1 \vert^n = 1 \Rightarrow \vert z + 1\vert = 1, \tag{4}$
$(5.) \; \vert z + 1 \vert = 1 \Rightarrow (1 + z)(1 + \bar z) = 1, \tag{5}$
$(6.) \; (1 + z)(1 + \bar z) = 1 \Rightarrow z \bar z + z + \bar z + 1 = 1 \Rightarrow z \bar z + z + \bar z = 0, \tag{6}$
$(7.) \; \text{by (3) and (6),} \; z + \bar z = -1, \tag{7}$
$(8.) \; \text{by (2) and (7),} \; 2 \Re{z} = 2 \cos \theta = -1 \Rightarrow \cos \theta = -\dfrac{1}{2}, \tag{8}$
$(9.) \; \cos \theta = -\dfrac{1}{2} \Rightarrow \Re{(1 + z)} = 1 + \cos \theta = \dfrac{1}{2}, \tag{9}$
$(10.) \; \Re{(1 + z)} = \dfrac{1}{2} \; \text{and} \; \vert 1 + z \vert = 1 \Rightarrow 1 + z = e^{(\pm 2\pi i/ 6) + 2k \pi}, k \in \Bbb Z, \tag{10}$
$(11.) \; 1 + z = e^{(\pm 2\pi i/ 6) + 2k \pi} \Rightarrow 1 = (1 + z)^n = e^{\pm 2n \pi i / 6}, \tag{11}$
$(12.) \; e^{\pm 2n \pi i / 6} = 1 \Rightarrow 6 \mid n. \tag{12}$
QED
| {
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"url": "https://math.stackexchange.com/questions/643024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Numerical Analysis - Richardson Extrapolation Question: Suppose that N(h) is an approximation to $M$ for every $h > 0$ and that
$M = N(h) + K_1 h + K_2 h^2 + K_3 h^3 +\cdots$, for some constants $K_1, K_2, K_3,\cdots$. Use the values $N(h), N( h/3),$ and $N (h/9)$ to produce an $O(h^3)$
approximation to $M$.
My work:
$$M=N(h)+K_1 h+K_2 h^2 + K_3 h^3+\cdots$$
$$M=N(\frac{h}{3})+\frac{K_1}{3} h+\frac{K_2}{9}h^2 + \frac{K_3}{27}h^3+\cdots$$
$$M=N(\frac{h}{9})+\frac{K_1}{9}h+\frac{K_2}{81}h^2 + \frac{K_3}{729}h^3+\cdots$$
I'm not sure what to do next.
I was told that $N_1(h)$ gives $O(h^2)$, $N_2(h)$ gives $O(h^4)$, $N_3(h)$ gives $O(h^6)$, $N_4(h)$ gives $O(h^8)$, so I'm not sure how to get $O(h^3)$.
| It is not true that $N_1(h)=O(h^2), N_2(h)=O(h^4), \cdots$ for the formula
$$
M=N(h)+K_1 h+K_2 h^2+K_3 h^3+\cdots
$$
but it is true for
$$
M=N(h)+K_1 h^2+K_2 h^4+K_3 h^6+\cdots
$$
It's easy to get an $O(h^3)$ formula. Subtracting the first formula from $3$ times the second formula eliminates the $h$ term
$$
2M=3N(\frac{h}{3})-N(h)+(\frac{K_2}{3}h^2-K_2 h^2)+(\frac{K_3}{9}h^3-K_3 h^3)+\cdots\\
$$
Dividing this equation by $2$ produces an $O(h^2)$ formula
$$
M=\frac{3}{2}N(\frac{h}{3})-\frac{1}{2}N(h)-\frac{K_2}{3}h^2-\frac{4K_3}{9}h^3+\cdots\\
$$
Subtracting the second formula from $3$ times the third formula eliminates the $h$ term
$$
2M=3N(\frac{h}{9})-N(\frac{h}{3})+(\frac{K_2}{27}h^2-\frac{K_2}{9}h^2)+(\frac{K_3}{243}h^3-\frac{K_3}{27}h^3)+\cdots
$$
Dividing this equation by $2$ produces an $O(h^2)$ formula
$$
M=\frac{3}{2}N(\frac{h}{9})-\frac{3}{2}N(\frac{h}{3})-\frac{2K_2}{27}h^2-\frac{8K_3}{243}h^3+\cdots
$$
Multiplying this by $9/2$ gives
$$
\frac{9}{2}M=\frac{27}{4}N(\frac{h}{9})-\frac{27}{4}N(\frac{h}{3})-\frac{K_2}{3}h^2+\frac{4K_3}{27}h^3+\cdots
$$
Subtracting the first $O(h^2)$ formula from this $O(h^2)$ formula eliminates the $h^2$ term
$$
\frac{7}{2}M=\frac{27}{4}N(\frac{h}{9})-\frac{3}{2}N(\frac{h}{3})-\frac{27}{4}N(\frac{h}{3})+\frac{1}{2}N(h)+(\frac{4K_3}{27}h^3+\frac{4K_3}{9}h^3)+\cdots
$$
Multiplying this equation by $2/7$ produces an $O(h^3)$ formula
$$
M=\frac{27}{14}N(\frac{h}{9})-\frac{201}{28}N(\frac{h}{3})+\frac{1}{7}N(h)+\frac{32 K_3}{189}h^3+\cdots
$$
then $N_3(h)=\frac{27}{14}N(\frac{h}{9})-\frac{201}{28}N(\frac{h}{3})+\frac{1}{7}N(h)$ with the truncation error $O(h^3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/644446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this inequality: $\frac{x^3+1}{\sqrt{x^4+y+z}}+\frac{y^3+1}{\sqrt{y^4+z+x}}+\frac{z^3+1}{\sqrt{z^4+x+y}}\geq2\sqrt{xy+yz+zx}$ Let $x,y,z>0$ such that $xyz=1$. Show that: $$\dfrac{x^3+1}{\sqrt{x^4+y+z}}+\dfrac{y^3+1}{\sqrt{y^4+z+x}}+\dfrac{z^3+1}{\sqrt{z^4+x+y}}\geq2\sqrt{xy+yz+zx}$$
I've tried many things but all failed. Please help. Thank you.
| proof: since
\begin{align*}2\sqrt{(x^4+y+z)(xy+zx+yz)}&=2\sqrt{[x^4+xyz(y+z)][xy+yz+xz]}\\
&=2\sqrt{(x^3+y^2z+yz^2)(x^2y+
x^2z+xyz)}
\end{align*}
Use AM-GM inequality we have
$$2\sqrt{(x^3+y^2z+yz^2)(x^2y+x^2z+xyz)}\le x^3+y^2z+yz^2+x^2y+x^2z+xyz=(x+y+z)(x^2+yz)=\dfrac{(x+y+z)(x^3+1)}{x}$$
so
$$\dfrac{x^3+1}{\sqrt{x^4+y+z}}\ge\dfrac{2x\sqrt{xy+yz+xz}}{x+y+z}$$
so
$$\sum_{cyc}\dfrac{x^3+1}{\sqrt{x^4+y+z}}\ge\sum_{cyc}\dfrac{2x\sqrt{xy+yz+xz}}{x+y+z}=2\sqrt{xy+yz+xz}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/648530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the error in the simultaneous equation. The question is:
$$\begin{align}
\tag{1} 2x - 3y &= 3\\
\tag{2} 4x^2 - 9y^2 &= 3
\end{align}$$
From equation (1):
$$\tag{3} 2x = 3 - 3y.$$
Substitute equation (3) in (2):
$$\begin{align}
4x^2 - 9y^2 &= 3\\
(2x)^2 - 9y^2 &= 3\\
(3 - 3y)^2 - 9y^2 &= 3\\
[(3)^2 - 2\cdot 3\cdot 3y + (-3y)^2] - 9y^2 - 3 &= 0\\
9 +18y + 9y^2 - 9y^2 - 3 &= 0\\
-18y + 6 &= 0\\
-18y &= -6\\
y &= \frac 1 3
\end{align}$$
But the answer for y in my book is $ - \frac 1 3\ $
Which one is right?
| There is other way:
$\ 2x - 3y = 3 \tag{1}$
$(2x - 3y)(2x + 3y) = 3 \tag{2} $
Using $(1)$ we have
$$ 2x - 3y = 3 $$
$$ 2x + 3y = 1 $$
$$(1)+(2)$$
$$4x=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/651223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove all $c_{n},d_{n}$ to be integers if $(n+1)c_{n}=nc_{n-1}+2nd_{n-1}$ and $d_{n}=2c_{n-1}+d_{n-1}$? Let sequences $(c_n)$ and $(d_n)$ be given by
$$c_0=0,\:d_0=1$$
and recursively for $n\ge 1$ by
$$\begin{align}
c_n & =\frac{n}{n+1}c_{n-1}+\frac{2n}{n+1}d_{n-1} \\[2ex]
d_n & =2c_{n-1}+d_{n-1}
\end{align}$$
I'd like to show that all $c_{n},d_{n}$ are integers. (Creat by wang yong xi）
My try： Since
$$\begin{align}(n+1)c_n & =nc_{n-1}+2nd_{n-1}\\[1ex]
d_n & =2c_{n-1}+d_{n-1}
\end{align}$$
we easily find $$c_{1}=1,\:d_{1}=1,\\
c_{2}=2,\:d_{2}=3,\\
c_{3}=6,\:d_{3}=7,$$
a.s.o. How to prove that all the $c_{n},d_{n}$ are integers?
| Consider the polynomials $(1+x+x^2)^n$; the only coefficients that will be non zero are terms whose $x$ exponent is between $0$ and $2n$. Note two other things; firstly they will all be whole numbers and secondly they satisfy $a_{i}= a_{2n-i}$ (i.e they will be symmetrical around the central term).
Now we have define sequences $d_n$ to be the central coefficient and $c_n$ to $1$ off central. So we have
\begin{eqnarray*}
(1+x+x^2)^{n-1} =1+ \cdots +c_{n-1}x^{n-2}+ d_{n-1} x^{n-1} +c_{n-1} x^{n} + \cdots +x^{2n-2} \\
(1+x+x^2)^n =1+ \cdots +c_{n}x^{n-1}+ d_{n} x^{n} +c_{n} x^{n+1} + \cdots +x^{2n}. \\
\end{eqnarray*}
Multiply the first equation by $(1+x+x^2)$ and consider the coefficient of $x^{n}$ and we have
\begin{eqnarray*}
d_n=2c_{n-1}+d_{n-1}.
\end{eqnarray*}
Next differentiate both sides of the second equation
\begin{eqnarray*}
n(1+2x)(1+x+x^2)^{n-1} = \cdots +(n-1)c_{n}x^{n-2}+ nd_{n} x^{n-1} +(n+1)c_{n} x^{n} +\cdots +2nx^{2n-1} \\
\end{eqnarray*}
use the first equation and consider the coefficient $x^{n}$ and we have
\begin{eqnarray*}
(n+1)c_n=nc_{n-1}+2_nd_{n-1}.
\end{eqnarray*}
It suffices to show that the sequences $c$ and $d$ satisfy the initial conditions and remind ourselves of the observation made earlier that all of these values are whole numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Proving a property of about the Fermat numbers
Show that the last digit in the decimal expansion of $F_n=2^{2^n}+1$ is $7$ for $n \geq 2$.
For our base step we let $n=2$. Now we have $2^{2^2}=16$. So the assertion holds for our base case. Then we assume it holds for $n$. For the $n+1$ case, is there a way to demonstrate this without resorting to this: $$(2^{2^n})^{{2^{n+1}}-2^n}.$$
That is the solution available in the back of the book. My attempt was to look at $2^{2^{n+1}}=2^{2^n}2^2.$
| Continuing from where you left, we just need to prove that $F_{n+1}$ holds true.
$$F_{n+1} = 2^{2^{n+1}}+1$$
$$F_{n+1} = 2^{2^n.2}+1$$
$$F_{n+1} = (2^{2^n})^2 + 1$$
$$F_{n+1} = (2^{2^n} + 1)^2 - 2.2^{2^n}$$
First term on the right hand side is $F_n^2$ which has a last digit of $7^2=9$.
Second term on the right hand side is $2(F_n-1)$, which has a last digit of $2(7-1)=2$.
Hence on a whole the last digit of right hand side, i.e., $F_{n+1}$ is :
$$9-2 = 7$$
Hope the answer is clear !
NOTE: You can cut short the above solution at Step:$3$ by saying that $(F_n-1)^2$ has a last digit of $6$ and hence $(F_n-1)^2+1 = (2^{2^n})^2 + 1$ has a last digit of $7$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Legendre polynomial problem (please help!)
Problem: Show that $P^{'}_{2n+1}(0)= \frac{(-1)^n (2n+1){^{2n}}C_n}{2^n}$.
My attempt:
Given: $P_{n}(x)=\frac{1}{2^n}\sum_{k=0}^{\frac{1}{2}n}(-1)^{k} {{^n}}C_k{^{2n-2k}C_n}x^{n-2k}$
Let $n = 2n+1$.
We have $P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{k=0}^{{n}+\frac{1}{2}}(-1)^{k} {{^{2n+1}}}C_k{^{2(2n+1)-2k}C_{2n+1}}x^{2n+1-2k}$
Now, let $k=n$.
$P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+\frac{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2(2n+1)-2n}C_{2n+1}}x^{2n+1-2n}$
$P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+\frac{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2n+2}C_{2n+1}}x^{1}$
$P_{2n+1}^{'}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+\frac{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2n+2}C_{2n+1}}$
Upon simplifying, we have
$P^{'}_{2n+1}(0)= \frac{(-1)^n (2n+1){^{2n}}C_n}{2^{2n}}$
This result is different from the one given. I have missed out something, please help!
| Your result is correct although it is difficult to read the derivation: I strongly suggest you use the binomial coefficient notation $\binom{a}{b}$ instead of $^aC_b$.
My own derivation is as follows: The recurrence relation
$$(x^2-1)P_n'=n(xP_n-P_{n-1})$$
implies that
$$P_n'(0)=nP_{n-1}(0)$$
so it remains to find the value of $P_n(0)$.
As
$$P_n(x)=\frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j}^2(x+1)^{n-j}(x-1)^j$$
we deduce that
$$P_n(0)=\frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j}^2(-1)^j$$
Now from the fact that $(1-x)^m(1+x)^m=(1-x^2)^m$, equating the coefficients for $x^m$, we derive that
$$
\begin{equation}
\sum_{j=0}^{m}\binom{m}{j}^2(-1)^j=\sum_{j=0}^{m}\binom{m}{j}(-1)^j\binom{m}{m-j}\cdot (+1)^j=
\begin{cases}0 &\text{if m is odd}\\
\binom{m}{m/2}(-1)^{m/2} & \text{if m is even}
\end{cases}
\end{equation}$$
And thus
$$P_{2n+1}'(0)=(2n+1)P_{2n}(0)=\frac{2n+1}{2^{2n}}\binom{2n}{n}(-1)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/655808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Intuitively, why is the Euler-Mascheroni constant near $\sqrt{1/3}$? Questions that ask for "intuitive" reasons are admittedly subjective, but I suspect some people will find this interesting.
Some time ago, I was struck by the coincidence that the Euler-Mascheroni constant $\gamma$ is close to the square root of $1/3$. (Their numerical values are about $0.57722$ and $0.57735$ respectively.)
Is there any informal or intuitive reason for this? For example, can we find a series converging to $\gamma$ and a series converging to $\sqrt{1/3}$ whose terms are close to each other?
An example of the kind of argument I have in mind can be found in Noam Elkies' list of one-page papers, where he gives a "reason" that $\pi$ is slightly less than $\sqrt{10}$. (Essentially, take $\sum\frac1{n^2}=\pi^2/6$ as known, and then bound that series above by a telescoping series whose sum is $10/6$.)
There are lots of ways to get series that converge quickly to $\sqrt{1/3}$. For example, taking advantage of the fact that $(4/7)^2\approx1/3$, we can write
$$
\sqrt{\frac{1}{3}}=(\frac{16}{48})^{1/2}
=(\frac{16}{49}\cdot\frac{49}{48})^{1/2}=\frac{4}{7}(1+\frac{1}{48})^{1/2}
$$
which we can expand as a binomial series, so $\frac{4}{7}\cdot\frac{97}{96}$ is an example of a good approximation to $\sqrt{1/3}$. Can we also get good approximations to $\gamma$ by using series that converge quickly, and can we find the "right" pair of series that shows "why" $\gamma$ is slightly less than $\sqrt{1/3}$?
Another type of argument that's out there, showing "why" $\pi$ is slightly less than $22/7$, involves a particular definite integral of a "small" function that evaluates to $\frac{22}{7}-\pi$. So, are there any definite integrals of "small" functions that evaluate to $\sqrt{\frac13}-\gamma$ or $\frac13-\gamma^2$?
| From the continuous fraction expansion, the seventh convergent is
$$\gamma \approx \frac{15}{26}$$
From the limit definition
$$\begin{align}
\gamma
&=
\lim_{n \to \infty} {\left(2H_n-\frac{1}{6}H_{n^2+n-1}-\frac{5}{6}H_{n^2+n}\right)} \\
&=
\frac{7}{12}+\sum_{n=1}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\
&=\frac{7}{12}-\frac{1}{180}+\sum_{n=2}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\
&=\frac{26}{45}+\sum_{n=2}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\
\end{align}
$$
so
$$\gamma \approx \frac{26}{45}$$
Multiplying both approximations,
$$\gamma^2 \approx \frac{1}{3}$$
The origin of this limit definition is improving the convergence of Macys formula from $o(n^{-2})$ to $o(n^{-4})$ downweighting the last term, without additional fractions (https://math.stackexchange.com/a/129808/134791)
In fact, the convergent approximation is not necessary.
Given $$\gamma \approx \frac{26}{45}$$
we have
$$3\gamma^2\approx3\left(\frac{26}{45}\right)^2=3\frac{676}{2025}=3\frac{676}{3\cdot675}=\frac{676}{675}\approx 1$$
which also yields
$$\gamma^2 \approx \frac{1}{3}$$
Towards proving that $\gamma < \frac{1}{ \sqrt {3} }$, we may take one more term out of the summation:
$$\gamma \approx \frac{7}{12}-\frac{1}{180}-\frac{1}{2310}=\frac{4001}{6930}<\frac{1}{\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Evaluating $\int_{0}^{\infty} \frac{x^{3}- \sin^{3}(x)}{x^{5}} \ dx $ using contour integration EDIT: Instead of expressing the integral as the imaginary part of another integral, I instead expanded $\sin^{3}(x)$ in terms of complex exponentials and I don't run into problems anymore.
\begin{align} \int_{0}^{\infty} \frac{x^{3}-\sin^{3}(x)}{x^{5}} \ dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^{3}-\sin^{3}(x)}{x^{5}} \ dx \\ &= \frac{1}{2} \ \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i}(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix})}{x^{5}} \ dx \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \ \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i}(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix})}{(x-i \epsilon)^{5}} \ dx \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i} (e^{3ix}-3e^{ix})}{(x-i \epsilon)^{5}} + \frac{1}{16i} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{3e^{-ix}-e^{-3ix}}{(x-i \epsilon)^{5}} \ dx \end{align}
Then I integrated $ f(z) = \frac{z^{3}+ \frac{1}{8i}(e^{3iz}-3e^{iz})}{(z-i \epsilon)^{5}}$ around the upper half of $|z|=R$ and $ g(z) = \frac{3e^{-iz}-e^{-3iz}}{(z-i \epsilon)^{5}}$ around the lower half of $|z|=R$ and applied Jordan's lemma.
\begin{align} \int_{0}^{\infty} \frac{x^{3}-\sin^{3}x}{x^{5}} \ dx &= \frac{1}{2} \lim_{\epsilon \to 0^{+}}2 \pi i \ \text{Res}[f(z),i \epsilon] + \frac{1}{16i} \lim_{\epsilon \to 0^{+}} 2 \pi i (0) \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \frac{2 \pi i}{4!} \lim_{z \to i \epsilon} \frac{d^{4}}{dz^{4}} \Big(z^{3}+\frac{1}{8i}e^{3iz}-\frac{3}{8i}e^{iz} \Big) \\ &= \frac{\pi i}{24} \lim_{\epsilon \to 0^{+}} \ \lim_{z \to i \epsilon}\Big( \frac{1}{8i}(3i)^{4}e^{3iz}- \frac{3}{8i} (i)^{4} e^{iz} \Big) \\ &= \frac{\pi i}{24} \lim_{\epsilon \to 0^{+}} \Big( \frac{81}{8i}e^{- 3\epsilon} - \frac{3}{8i}e^{- \epsilon} \Big) \\ &= \frac{\pi i}{24} \Big(\frac{81}{8i}-\frac{3}{8i} \Big) \\ &= \frac{13 \pi}{32} \end{align}
| Another approach :
Sorry Random Variable, this is not using contour integration technique since I don't know how to approach the integral using that way. $\ddot\smile$
Consider
$$
\mathcal{I}(\alpha)=\int_0^\infty\frac{(\alpha x)^3-\sin^3\alpha x}{x^5}dx.\tag1
$$
Differentiating $(1)$ four times yields
\begin{align}
\frac{d^4\mathcal{I}}{d\alpha^4}&=\int_0^\infty\frac{\partial^4}{\partial\alpha^4}\left(\frac{(\alpha x)^3-\sin^3\alpha x}{x^5}\right)dx\\
&=\color{green}{\int_0^\infty\left(\frac{81\sin3\alpha x-3\sin\alpha x}{4x}\right)dx}\\
&=\frac{81}{4}\cdot\frac\pi2-\frac{3}{4}\cdot\frac\pi2\\
&=\frac{39\pi}{4},\tag2
\end{align}
where
$$
\int_0^\infty\frac{\sin\alpha x}{x}dx=\frac\pi2\qquad\text{for }\alpha\neq0.
$$
Then from $(2)$ we obtain
$$
\large\color{blue}{\mathcal{I}(\alpha)=\frac{13\pi}{32}a^4},\tag3
$$
where $\mathcal{I}(0)=\mathcal{I'}(0)=\mathcal{I''}(0)=\mathcal{I'''}(0)=0$. Thus
$$
\mathcal{I}(1)=\int_0^\infty\frac{x^3-\sin^3 x}{x^5}dx=\large\color{blue}{\frac{13\pi}{32}}.
$$
P.S.
I think you can easily apply the contour integration technique in line $2$ (green-colored) equation $(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 4,
"answer_id": 1
} |
Prove that $2xy\mid x^2+y^2-x$ implies $x$ is a perfect square. Prove that $2xy\mid x^2+y^2-x$ implies $x$ is a perfect square.
My work:
$2xy\mid x^2+y^2-x \implies x^2+y^2-x=2xy\cdot k$
So,$x^2+y^2+2xy-x=(x+y)^2-x=2xy \cdot (k+1)$
And,$x^2+y^2-2xy-x=(x-y)^2-x=2xy \cdot (k-1)$
I found that for $x,y$ both odd, no solution exists. For $x$ even, and $y$ odd,no solution exists. Solution exists only for $x$ odd, $y$ even and $x$ even and $y$ even solution exists. Cannot do anything more. Please help!
| We use the following
Fact: A non-zero integer is a perfect square (by that I mean a number of the form $k^2$ or $-k^2$) if and only if in its prime factorization, the exponent of every prime factor is even.
Now let $p$ be any prime factor of $x$ and $k$ the exponent of $p$ in the prime factorization of $x$. If $k$ is even, there is nothing to show. So assume that $k$ is odd: $k=2j+1$.
Then $p^k|x|2xy|x^2+y^2-x$. Since $p^k|x^2-x$, it must also hold that $p^{2j+1}=p^k|y^2$. Since $y^2$ is a square, also $p^{k+1}=p^{2j+2}|y^2$.
But then $p^{k+1}|2xy|x^2+y^2-x$. But since $p^{k+1}|x^2+y^2$, it also follows that $p^{k+1}|x$. This is a contradiction (we assumed that $k$ is the exponent of $p$ in the factorization of $x$).
Since $p$ was an arbitrary prime factor, the exponents of all prime factors in the prime factorization of $x$ are even, i.e. $x$ is a square in the above sense.
Edit: Made precise what is meant by a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/663283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Method of moments on uniform distributions I need help on how to find the estimates $a$ and $b$ in the uniform distribution $\mathcal U[a,b]$ using the method of moments.
This is where I am at:
I have found $U_1=\overline X$ and $m_1=\frac{a+b}2$ Also, $m_2=\frac1{n(E(X_i^2)}$ and $u_2=E(X_i^2)$ when I equate $m_1$ to $u_1$ and $m_2$ to $u_2$ I should find the estimates,right? I am unable to solve that.Kindly help.
| The population mean is $\dfrac{a+b}{2}$. The population variance is $\dfrac{(b-a)^2}{12}$. If you have already found that the popuation variance of the $\mathcal U[0,1]$ distribution is $1/12$, just notice that the length of the interval has been multiplied by $b-a$, and that is a scale factor, so you multiply the variance by $(b-a)^2$. Hence the second moment is
$$
\frac{(b-a)^2}{12}+\left(\frac{a+b}{2}\right)^2 = \frac{a^2+b^2+ab}{3}.
$$
So the equations to be solved for $a$ and $b$ are:
\begin{align}
\frac{x_1+\cdots+x_n}{n} & = \frac{a+b}{2} \\[10pt]
\frac{x_1^2+\cdots+x_n^2}{n} & = \frac{a^2+b^2+ab}{3}
\end{align}
One way would be to solve the first equation for $b$ and then substitute that for $b$ in the second equation, getting a quadratic equation in $a$:
$$
b = 2\bar x - a,
$$
$$
\frac{x_1^2+\cdots+x_n^2}{n} = \frac{a^2+(2\bar x - a)^2+a(2\bar x-a)}{3}
$$
$$
\frac{x_1^2+\cdots+x_n^2}{n} = \frac{a^2 + 2\bar x a + 4\bar x^2}{3}
$$
$$
a^2 + 2\bar x a + \left(4\bar x^2 - 3\frac{x_1^2+\cdots+x_n^2}{n}\right)=0.
$$
Then proceed the way you usually do with quadratic equations.
PS: Completing the square gives
$$
\begin{align}
a^2 + 2\bar x a + \bar x^2 + 3\left(\bar x^2 - \frac{x_1^2+\cdots+x_n^2}{n}\right)& = 0\\[10pt]
(a+\bar x)^2 & = 3\left(\frac{x_1^2+\cdots+x_n^2}{n} - \bar x^2\right) \\[10pt]
(a+\bar x)^2 & = 3s^2
\end{align}
$$
where this is taken to define the notation $s^2$. That means $s^2$ is the sample variance if that is taken to mean the version with $n$ rather than $n-1$ in the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/664811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\left(\frac12(x+y)\right)^2 \le \frac12(x^2 + y^2)$ Prove that $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
I've gotten that
$$\left(\frac12(x+y)\right)^2 \ge 0 $$
but stumped on where to go from here...
| $\frac12 (x^2+y^2)-(\frac12 (x+y))^2$
$=\frac12 (x^2+y^2)-\frac14 (x^2+y^2+2xy)$
$=\frac14 (2x^2+2y^2-x^2-y^2-2xy)$
$=\frac14 (x^2+y^2-2xy)$
$=\frac14 (x-y)^2\ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/665206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
| $(a+b)^2+(a-b)^2=2(a^2+b^2)\leq 4$, so $|a+b|\leq 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/666217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
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