Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $\alpha$ and $\beta$ are the roots of the equation $x^2-4x+1=0 (\alpha>\beta)$, then find following If $\alpha$ and $\beta$ are the roots of the equation $x^2-4x+1=0 (\alpha>\beta)$, then find the value of $f(\alpha,\beta)=\dfrac{\beta^3}{2}\mathrm{cosec}^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\beta}{\alpha}\right)+\dfrac{\alpha^3}{2}\sec^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\alpha}{\beta}\right)$
$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\left(1+\cot^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\beta}{\alpha}\right)\right)+\dfrac{\alpha^3}{2}\left(1+\tan^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\alpha}{\beta}\right)\right)$$
As $\dfrac{\alpha}{\beta}$ is positive
$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\mathrm{cosec}^2\left(1+\cot^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\beta}{\alpha}\right)\right)+\dfrac{\alpha^3}{2}\left(1+\tan^2\left(\dfrac{1}{2}\cot^{-1}\dfrac{\beta}{\alpha}\right)\right)$$
$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\left(1+\cot^2\left(\dfrac{1}{2}\tan^{-1}\dfrac{\beta}{\alpha}\right)\right)+\dfrac{\alpha^3}{2}\left(1+\tan^2\left(\dfrac{\pi}{4}-\dfrac{1}{2}\tan^{-1}\dfrac{\beta}{\alpha}\right)\right)$$
Let $\tan^{-1}\dfrac{\beta}{\alpha}=\theta$
$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\left(1+\cot^2\dfrac{\theta}{2}\right)+\dfrac{\alpha^3}{2}\left(1+\tan^2\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)\right)$$
As we have standard identity: $\tan^2\theta=\dfrac{1-\cos2\theta}{1+\cos2\theta}$
$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\left(1+\dfrac{1+\cos\theta}{1-\cos\theta}\right)+\dfrac{\alpha^3}{2}\left(1+\dfrac{1-\cos\left(\dfrac{\pi}{2}-\theta\right)}{1+\cos\left(\dfrac{\pi}{2}-\theta\right)}\right)$$
$$f(\alpha,\beta)=\dfrac{\beta^3}{2}\left(1+\dfrac{1+\cos\theta}{1-\cos\theta}\right)+\dfrac{\alpha^3}{2}\left(1+\dfrac{1-\sin\theta}{1+\sin\theta}\right)$$
$$f(\alpha,\beta)=\dfrac{\beta^3}{1-\cos\theta}+\dfrac{\alpha^3}{1+\sin\theta}$$
As $\dfrac{\beta}{\alpha}=\tan\theta$
$$\cos\theta=\dfrac{\alpha}{\sqrt{\alpha^2+\beta^2}},\sin\theta=\dfrac{\beta}{\sqrt{\alpha^2+\beta^2}}$$
$$f(\alpha,\beta)=\dfrac{\beta^3}{1-\dfrac{\alpha}{\sqrt{\alpha^2+\beta^2}}}+\dfrac{\alpha^3}{1+\dfrac{\beta}{\sqrt{\alpha^2+\beta^2}}}$$
$$f(\alpha,\beta)=\dfrac{\beta^3\cdot\sqrt{\alpha^2+\beta^2}}{{\sqrt{\alpha^2+\beta^2}}-\alpha}+\dfrac{\alpha^3\sqrt{\alpha^2+\beta^2}}{\sqrt{\alpha^2+\beta^2}+\beta}$$
But actual answer is $$(\alpha^2+\beta^2)(\alpha+\beta)$$
| $$f(\alpha,\beta)=\dfrac{\beta^3\sqrt{\alpha^2+\beta^2}\left(\sqrt{\alpha^2+\beta^2}+\alpha\right)}{\alpha^2}+\dfrac{\alpha^3\sqrt{\alpha^2+\beta^2}\left(\sqrt{\alpha^2+\beta^2}-\beta\right)}{\alpha^2}$$
$$f(\alpha,\beta)=\beta\left(\alpha^2+\beta^2+\alpha\sqrt{\alpha^2+\beta^2}\right)+\alpha\left(\alpha^2+\beta^2-\beta\sqrt{\alpha^2+\beta^2}\right)$$
$$f(\alpha,\beta)=\beta\left(\alpha^2+\beta^2\right)+\alpha\left(\alpha^2+\beta^2\right)$$
$$f(\alpha,\beta)=(\alpha+\beta)(\alpha^2+\beta^2)$$
So previously I forgot to rationalize, hence closing this.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $1 - 3xy + x^3 + y^3 = 0$ I am trying to solve the equation
$$
1 - 3xy + x^3 + y^3 = 0
$$
in the real numbers. I can see several (infinitely many) solutions:
*
*$x=y = 1$
*$x+ y = -1$
I am trying to show that these are all of them.
For example, if $x = y$, then $x^2(-3+2x) = -1$ forcing $x = y = 1$
I can't seem to make much progress with the case where $x\neq y$ and showing that $x + y = 1$ in that case. (I am pretty sure that this must be true.)
| If you substitute $y$ with $-x-1$ you get the identity
$$x^3+(-x-1)^3-3x(-x-1)+1=0$$
This means that the polynomial $x^3+y^3-3xy+1$ is divisible by $x+y+1$. Performing long division you get the following factorization:
$$x^3+y^3-3xy+1= (x+y+1)(x^2-xy+y^2-x-y+1)$$
so the problem becomes proving that $(x;y)=(1;1)$ is the only zero for
$$x^2-xy+y^2-x-y+1=0$$
Call $X=x-1$ and $Y=y-1$. We want to prove that $X=Y=0$. Substituting $x=X+1$ and $y=Y+1$, we get the equation
$$X^2-XY+Y^2=0$$
whose unique solution is $X=Y=0$ (hope you can find the reason of this by yourself). This concludes the proof.
| {
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Prove by induction that $f(a,b) = \frac{a}{b} + \frac{b}{a} + \frac{1}{ab}$ is a multiple of 3 if it is an integer I'm trying to solve the following problem by induction but I'm getting stuck.
For positive integers $a$ and $b$, define $$f(a,b) = \frac{a}{b} + \frac{b}{a} + \frac{1}{ab}.$$ If $f(a,b)$ is an integer, prove that it is a multiple of 3.
Proof by induction on a:
Base case: $a=1$
$$f(1,b) = \frac{1}{b} + \frac{b}{1} + \frac{1}{b} = \frac{2}{b} + b$$
Since $f(1,b)$ is an integer than $\frac{2}{b}$ must be an integer and $b\in\{1,2\}$ . Then $f(1,b) = 3$.
Inductive hypothesis: Assume for some integer $k \ge 1,$ $$f(k,b) = \frac{k}{b} + \frac{b}{k} + \frac{1}{kb}$$ is an integer and is a multiple of 3.
I want to show that $f(k+1,b)$ is also a multiple of 3. Then
$$f(k+1,b) = \frac{k+1}{b} + \frac{b}{k+1} + \frac{1}{(k+1)(b)} = \cdots.$$
And this is where I get stuck. I know there are other ways to solve this problem but I wanted to try it by induction. Hope someone can help with this! Thanks.
| Note that $$f(a,b)=\frac{a^2+b^2+1}{ab}$$ So, if $ab\mid a^2+b^2+1$, then $f(a,b)$ is an integer.
Note that $a^2\equiv0,1\pmod3$.
If $3\mid a$, then $$a^2\equiv0\pmod3\to a^2+b^2+1\equiv1,2\pmod3$$so $3$ doesn't divide it, but $3\mid ab$. This implies that $f(a,b)$ is not an integer. The same analysis holds when $3\mid b$.
So, $f(a,b)$ can only be an integer if $3\not\mid a,b$. But, that implies that $a^2,b^2\equiv1\pmod3$, which implies that $$a^2+b^2+1\equiv3\equiv0\pmod3$$i.e., $a^2+b^2+1$ is divisible by $3$, but $3\not\mid ab$. So, $f(a,b)$, if it is an integer, divides $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the $n^{th}$ derivative of $y=\frac {x}{x^2+a^2}$ Find the $n^{th}$ derivative of $y=\dfrac {x}{x^2+a^2}$.
My Attempt:
$$y=\dfrac {x}{x^2+a^2}$$
$$y=x.(x^2+a^2)^{-1}$$
Differentiating both sides with respect to $x$
$$y_{1}=x.((-1).(x^2+a^2)^{-2}.2x)+(x^2+a^2)^{-1}$$
$$y_{1}=(-1).(x^2+a^2)^{-2}.2x^2+(x^2+a^2)^{-1}$$
| Hint:
Rewrite the definition of $y$ as $(x^2+a^2)y=x$, and apply Leibniz' formula for the $n$-th derivative of a product of functions: for $n>1$, we have
$$(x^2+a^2)y^{(n)}+2nx y^{(n-1)}=0,$$
whence the recurrence relation
$$y^{(n)}=-\frac{2nx}{x^2+a^2}y^{(n-1)}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $x, y, z$ so that $(x+2y)(y+2z)(z+2x)$ is equal to prime power Find all positive integers $x, y, z$ so that $(x+2y)(y+2z)(z+2x)$ is equal to prime power.
My try
Since $x, y, z$ are positive integers it follows that $x+2y, y+2z, z+2x \geq 3$. We have following system of equations:
$$\begin{align}
x+2y &= p^r \tag{1}\label{eq1} \\
y+2z &= p^s \tag{2}\label{eq2} \\
z+2x &= p^t \tag{3}\label{eq3}
\end{align}$$
Where $p$ is prime number and $r, s, t$ are positive integers.
Solving this system of equations for $x, y, z$ gives us $x=\frac{4p^t-2p^s+p^r}{9}$, $y=\frac{4p^r-2p^t+p^s}{9}$, $z=\frac{4p^s-2p^r+p^t}{9}$. Since $x, y, z$ are integers following equations must hold:
$$\begin{align}
4p^t-2p^s+p^r &\equiv 0 \pmod 9 \\
4p^r-2p^t+p^s &\equiv 0 \pmod 9 \\
4p^s-2p^r+p^t &\equiv 0 \pmod 9
\end{align}$$
I don't know how to continue from there. Can somebody help?
| Assume the prime is $p \neq 3.$ Note that we may keep (simultaneously) dividing out $p$ from $x,y,z$ until at least one of them is not divisible by $p\; , \;$ with the result that $$ \gcd(x,y,z) = 1 $$
We do have such primitive solutions when
$$ (x,y,z) = (1,1,1) \; \; , \; \; (x,y,z) = (13,7,1) $$
when both primes are $3.$
Next, add up the three items that must be powers of $p,$ we find
$$ 3 (x+y+z) \equiv 0 \pmod p. $$
With the assumption $p \neq 3,$ we have $x+y+z \equiv 0 \pmod p.$ Then subtract $x+2y \equiv 0 \pmod p,$ so that $y \equiv z \pmod p.$
Repeat with the other pairs, we get
$$ x \equiv y \equiv z \pmod p $$
However, we already had $x+y+z \equiv 0 \pmod p.$ We get $
3x \equiv 0 \pmod p,$ same for $y,z,$ contradicting $\gcd(x,y,z) = 1.$ In turn, this contradicts the assumption that $p \neq 3.$
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the values of the parameters $a,b \in \mathbb{R}$ such that $\lim\limits_{x \to \infty}(\sqrt{x^2+x+1}+\sqrt{x^2+2x+2}-ax-b)=0.$ I have to find the values of $a$ and $b$ (with $a,b \in \mathbb{R}$) such that the following is true:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
This is what I did:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
$$\lim\limits_{x \to \infty} \bigg (x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + x\sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-ax-b \bigg )=0$$
$$\lim\limits_{x \to \infty} x\bigg (\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + \sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-a-\dfrac{b}{x} \bigg )=0$$
So, we'd have something like:
$$\infty \cdot(2-a) = 0$$
$(*)$If $a \in (-\infty, 2)$ $\Rightarrow$ $(2-a) > 0$, which means:
$$\infty \cdot(2-a) = \infty$$
$(*)$If $a\in (2, +\infty)$ $\Rightarrow$ $(2-a) < 0$, which means:
$$\infty \cdot(2-a) = -\infty$$
So the only option left for the limit to have any chance of being true is:
$$a=2$$
which would result in the indeterminate form:
$$\infty \cdot (2-a) = \infty \cdot 0$$
And now that we have $a=2$, we need to find the value of $b$ for the limit to be true. Substituting $a$ with $2$ in the initial limit we get:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-2x-b)=0$$
Since $b$ is a constant we can pull it out of the limit and get:
$$b=\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-2x)$$
And this is where I got stuck. I tried a bunch of methods and tricks for finding this limit and I got nowhere. That leads me to think that either I made some mistake/mistakes along the way, or I simply don't know how to solve this limit. So, how can I find $b$?
| Note
$$x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}=x(1+\frac1{2x}+O(\frac1{x^2}))=x+\frac12+O(\frac1x)$$
$$x\sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}=x+1+O(\frac1x)$$
Therefore,
$$\lim\limits_{x \to \infty} \bigg (x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + x\sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-ax-b \bigg )$$
$$=\lim\limits_{x \to \infty} (2-a)x+(\frac32-b) + O(\frac1x)$$
Thus, to make the limit zero, we need to have $a=2$ and $b=\frac32$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Know that $\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}$ calculate $\sin\alpha$
Know that $\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}$ calculate
$\sin\alpha$
My proof:
$\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}\\
\frac{\sin\left(\alpha-\frac{\pi}{4}\right)}{\cos\left(\alpha-\frac{\pi}{4}\right)}=\frac{1}{3}\\3\sin\left(\alpha-\frac{\pi}{4}\right)=\cos\left(\alpha-\frac{\pi}{4}\right)\\\sin^2\left(\alpha-\frac{\pi}{4}\right)+9\sin^2\left(\alpha-\frac{\pi}{4}\right)=1\\\sin\left(\alpha-\frac{\pi}{4}\right)=\pm\frac{1}{\sqrt{10}}\\
\sin\left(\alpha-\frac{\pi}{4}\right)=\sin\alpha\cos\frac{\pi}{4}-\sin\frac{\pi}{4}\cos\alpha=\frac{\sqrt{2}}{2}\sin\alpha-\frac{\sqrt2}{2}\cos\alpha=\frac{\sqrt2}{2}\left(\sin\alpha-\cos\alpha\right)=\pm\frac{1}{\sqrt{10}}\\\sin\alpha-\cos\alpha=\pm\frac{1}{\sqrt{5}}\\\sin\alpha=\pm\frac{1}{\sqrt{5}}+\cos\alpha$
I have no idea how to determine $\sin\alpha$
| $\tan (\alpha -\frac {\pi} 4)=\frac {\tan (\alpha)-1} {1+\tan \alpha}=\frac 1 3$ and this gives $\tan \alpha =2$. Can you find $\sin \alpha$ from this?
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving HM-GM inequality using AM-GM I want to prove
$$\frac{n}{\frac{1}{a_1} + \cdots + \frac{1}{a_n}} ≤ \sqrt[n]{a_1\cdots a_n}$$
Using AM-GM inequality.
My attempt:
Consider arbitrary $x_1,\cdots,x_n$ We know that
$$\sqrt[n]{x_1\cdots x_n} ≤ \frac{x_1 + \cdots + x_n}{n}$$
Replacing all $x$ with $\frac{1}{x}$
$$\begin{align}\sqrt[n]{\frac{1}{x_1}\cdots \frac{1}{x_n}} ≤ \frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n} & \implies \frac{1}{x_1}\cdots \frac{1}{x_n} ≤ \biggl(\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\biggr)^{n} \\
& \implies \frac{1}{\frac{1}{x_1}\cdots \frac{1}{x_n}} ≥ \frac{1}{\biggl(\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\biggr)^{n}} \\
& \implies x_1\cdots x_n ≥ \frac{1}{\biggl(\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\biggr)^{n}} \\
& \implies \sqrt[n]{x_1\cdots x_n} ≥ \frac{1}{\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}} \\
& \implies \sqrt[n]{x_1\cdots x_n} ≥ \frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}
\end {align}$$
$\Box$
Is it correct? Are there other (better) alternatives?
| That's pretty much the standard proof that AM-GM implies GM-HM. The only variation among specific presentations of them is how much detail to go into. For example, you could argue you can just write$$\sqrt[n]{\frac{1}{x_1}\cdots \frac{1}{x_n}}\le\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\implies\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}.$$Edit: This is basically the point made in @ViktorGlombik's comment. (Incidentally, you can get $\le,\,\ge$ respectively with \le, \ge.)
| {
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Find the sum of $k/2^k, k=1$ to $n$ Let $S=1/2+2/2^2+3/2^3+...+n/2^n$
I try searching on the internet and see only the version of $k=1$ to infinity. I put this equation on Wolfram Alpha and get $(2^{n+1}-n-2)/2^n$ but I dunno how to do that. Please help
| This is an example of what is called an arithmetico-goemetric series. We can write it more compactly as $$S_n = \displaystyle\sum\limits_{k=1}^n \frac{k}{2^k}$$
The common ratio for the denominators is $2$, so we will multiply the entire series by $2$:
\begin{align}
S_n &= \,\qquad\frac{1}{2} +\frac{2}{4} +\frac{3}{8} +\frac{4}{16} + \cdots + \frac{n-1}{2^{n-1}} + \frac{n}{2^n}\tag{1}\\\\
2S_n&=\,\,1 +\frac{2}{2} + \frac{3}{4} + \frac{4}{8} +\frac{5}{16} +\cdots +\frac{n}{2^{n-1}} \qquad\tag{2}\\
\end{align}
Subtract $(1)$ from $(2)$:
$$2S_n - S_n = \left(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots + \frac{1}{2^{n-1}} \right) - \frac{n}{2^n}$$
Everything on the right hand side, except the last term, is a finite geometric series with common ratio $1/2$.
\begin{align}
S_n &= \left(2-\frac{1}{2^{n-1}}\right) - \frac{n}{2^n}\\\\
S_n &= \left(\frac{2^{n+1}}{2^{n}}-\frac{2}{2^{n}}\right) - \frac{n}{2^{n}}\\\\
\displaystyle\sum\limits_{k=1}^n \frac{k}{2^k} &= \boxed{\frac{2^{n+1}-n-2}{2^{n}}}\\\\
\end{align}
The exact same method works even more cleanly for the corresponding infinite series. We can also take the limit of the partial sums:
$$S = \displaystyle\sum\limits_{k=1}^\infty \frac{k}{2^k} = \lim_{n\to\infty} S_n = \lim_{n\to\infty}\left(\frac{2^{n+1}-n-2}{2^{n}}\right) = 2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\cos^2(C)+\cos^2(A)+2\sin(C)\sin(A)\cos(B)$ in $\Delta \text{ABC}$ I am attempting the following trigonometry problem. Given that in an acute angled triangle $\Delta \text{ABC}$, the following equalities hold true
$$\cos^2(A)+\cos^2(B)+2\sin(A)\sin(B)\cos(C)=\dfrac{15}{8}\\
\cos^2(B)+\cos^2(C)+2\sin(B)\sin(C)\cos(A)=\dfrac{14}{9}$$
Find the value of $\cos^2(C)+\cos^2(A)+2\sin(C)\sin(A)\cos(B)$.
My Attempt:
Let the unknown quantity be $x$. Then we have, by adding all the terms.
$$2\sum_{cyc}\cos^2(A)-2\sum_{cyc}\sin(A)\sin(B)\cos(A+B)=\dfrac{15}{8}+\dfrac{14}{9}+x$$
Also simplifying the second summation term as follows, we get $$\sin(A)\sin(B)\cos(A+B)=\dfrac{\sin(2A)\sin(2B)}{4}-(1-\cos^2(A))(1-\cos^2(B))$$
I'm not sure how to proceed further. Any hints are appreciated. Even hints to other possible pathways to the solution are welcome. Thanks
| Hint: Note that $$2\sin A\sin B\cos C-2\cos A\cos B\cos C=2\cos(180^\circ-A-B)\cdot\cos C=2\cos^2 C$$
So, the first equation becomes $$\cos^2 A+\cos^2 B+\cos^2 C+2\cos A\cos B\cos C=\frac{15}8$$This is symmetric, which makes it quite useful.
Now, we subtract the second equation from this to get $$\cos^2 A+2\cos A\cos B\cos C-2\sin B\sin C\cos A=\frac{15}8-\frac{14}9$$$$=\cos^2 A+2\cos(B+C)\cos A$$You can take it from here!
| {
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"timestamp": "2023-03-29T00:00:00",
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$k=-\sqrt{3}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})$
Let $k$ be equal to:
$$k=-\sqrt{3} \left(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}} \right)$$
I am trying to simplify the expression and express $5+2\sqrt{6}$ and $5-2\sqrt{6}$ as squares. I don't think it's smart to multiply the brackets by $-\sqrt{3}$ at first. Is there any algorithm that I should know to express the expressions as squares?
| We have indeed that
$$5+2\sqrt{6} =(\sqrt 3 + \sqrt 2)^2$$
$$5-2\sqrt{6} =(\sqrt 3 - \sqrt 2)^2$$
and therefore
$$k=-\sqrt{3}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})=-\sqrt{3}(\sqrt 3 + \sqrt 2+\sqrt 3 - \sqrt 2)$$
| {
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"url": "https://math.stackexchange.com/questions/3456305",
"timestamp": "2023-03-29T00:00:00",
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Does $\sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$ converge uniformly on $(-\infty,\infty)$?
Does $\displaystyle \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$
converge uniformly on $(-\infty,\infty)$?
My attempt:
No. Consider the case where $x=0$, then
$\displaystyle \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n} = 0$.
For $x \neq 0$, observe $\displaystyle 0 \lt \frac{1}{(1+x^2)^n} \lt 1$, so by geometric series formula
$\displaystyle \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$
$\displaystyle = \frac{x^2}{1 - \frac{1}{1+x^2}} = 1+x^2$
(1) So clearly the series doesn't even converge for all $x$, let alone converge uniformly.
Now, my question is about the case where $x \neq 0$. Does it converge uniformly to $1 + x^2$?
(2) I think, "yes". By Dini's theorem for series the convergence of the series to $1 + x^2$ must be uniform since $1+x^2$ is continuous and $(-\infty,0) \cup (0,\infty)$ is compact.
Is my reasoning for (1) and (2) correct?
| For (1) you got the idea but the argument is not exactly right.
Indeed the series does converges for all $x$, as you show
$$ \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n} = \begin{cases} 1+ x^2 & \text{if } x\neq 0,\\ 0 & \text{if }x=0.\end{cases}.$$
However it does not converges to a continuous function. Thus the convergence is not uniform (I am using that if $f_n \to f$ uniformly and $f_n$ are continuous for all $n$, then so if $f$.
For (2), it does not converges uniformly to $1+x^2$ on $(-\infty, 0) \cup (0,\infty)$. The set $(-\infty, 0) \cup (0,\infty)$ is not compact.
To check that the convergence is not uniform, write $f_k (x) = \sum_{n=0}^k \frac{x^2}{(1+x^2)^n}$. Then
$$\left| f(x) - f_k(x)\right| = \left| 1+x^2 - x^2 \frac{1-(1+x^2)^{-(k+1)}}{1-(1+x^2)^{-1}}\right| = \frac{1}{(1+x^2)^k}. $$
Thus
$$\sup_{x\neq 0} |f(x) - f_k(x)| = 1$$
for all $k$ and so the convergence is not uniform.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Computing $\text{lim}_{n \rightarrow \infty}\frac{n!}{n^{\lfloor y\sqrt{n} \rfloor}(n-\lfloor y \sqrt{n}\rfloor)!}$ We fix $y \ge 0$. What is the following limit?
$$\text{lim}_{n \rightarrow \infty}\frac{n!}{n^{\lfloor y\sqrt{n} \rfloor}(n-\lfloor y \sqrt{n}\rfloor)!}$$
My attempt so far is as follows:
By Stirling's formula, $n! \approx e^{-n}n^{n+1/2} \sqrt{2\pi}$, so that we can compute:
$$\text{lim}_{n \rightarrow \infty}\frac{e^{-n}n^{n+1/2} \sqrt{2\pi}}{n^{\lfloor y\sqrt{n} \rfloor}e^{-n+\lfloor y\sqrt{n} \rfloor}(n-\lfloor y\sqrt{n} \rfloor)^{n-\lfloor y\sqrt{n} \rfloor+1/2} \sqrt{2\pi}} \\
= \text{lim}_{n \rightarrow \infty}e^{-\lfloor y\sqrt{n} \rfloor} \left( \frac{n}{n-\lfloor y\sqrt{n} \rfloor} \right)^{n-\lfloor y\sqrt{n} \rfloor+1/2}$$
upon cancellation and rearrangement of terms. This is endlessly confusing for me to compute at this point. Does anyone know how to deal with the limit from here? I'm asking this question as part of another: Complicated distribution theory question for reference.
| I've chosen to ignore the floor function; you can probably modify this argument to include the floor function if you wish.
The logarithm is
\begin{align}
&-y\sqrt{n} + (n - y\sqrt{n} + 1/2) \log(\frac{n}{n-y\sqrt{n}})
\\
&=-y\sqrt{n} + (n - y\sqrt{n} + 1/2) \log(1 + \frac{y}{\sqrt{n}-y})
\\
&= -y\sqrt{n} +(n - y\sqrt{n} + 1/2) \left(\frac{y}{\sqrt{n}-y} - \frac{1}{2} \frac{y^2}{(\sqrt{n}-y)^2} + O(n^{-3/2})\right)
\\
&= -y\sqrt{n} + y(\sqrt{n} + \frac{y}{2(\sqrt{n}-y)})
- \frac{y^2}{2} (\frac{\sqrt{n}}{\sqrt{n}-y} + \frac{1}{2(\sqrt{n} - y)^2})
+ O(n^{-1/2})
\\
&\to - y^2/2.
\end{align}
So your limit is $e^{-y^2/2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Paradox in indefinite integral Today,I met a problem about calculating such an indefinite integral$$ I=\int \frac{1}{(1+x^2)(1+x^{2019})}$$ My thought is as follows $$
\begin{aligned} I & = \int \frac { 1 } { \left( 1 + x ^ { 2 } \right) \left( 1 + x ^ { 2019} \right) } d x \\ & = \int \frac { 1 } { \left( 1 + x ^ { - 2 } \right) \left( 1 + x ^ { - 2019} \right) } \left( - \frac { 1 } { x ^ { 2 } } \right) d x \\ & = \int \frac { - x ^ { 2019 } } { \left( 1 + x ^ { 2 } \right) \left( 1 + x ^ { 2019} \right) } d x \\ & = I - \int \frac { 1 } { 1 + x ^ { 2 } } d x \end{aligned}
$$ No doubt,it is wrong.However, I don’t know how to illustrate this paradox?
| In going from the first line to the second line, you make a substitution. Thus, if
$$
I =\int \frac{1}{(1+x^2)(1+x^{2019})}dx = \int f(x)\,dx,
$$
Then
$$
\int \frac { 1 } { \left( 1 + x ^ { - 2 } \right) \left( 1 + x ^ { - 2019} \right) } \left( - \frac { 1 } { x ^ { 2 } } \right) d x = \int f(1/x)\,dx.
$$
We have no reason to believe that the two resulting antiderivatives should be identical.
| {
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"url": "https://math.stackexchange.com/questions/3459491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find: $\lim\limits_{x\to +\infty}x\left(\sqrt{x^{2}+1}-\sqrt[3]{x^{3}+1}\right)$ Calculate the following limit:
$$\displaystyle\lim_{x\to +\infty}x\left(\sqrt{x^{2}+1}-\sqrt[3]{x^{3}+1}\right)$$
I need find this limit without l'Hospital or Taylor series.
Wolfram alpha gives $\frac{1}{2}$
My try is:
Let: $a=\sqrt{1+x^{2}}$ and $b=\sqrt[3]{1+x^{3}}$
And we know that:
$a-b=\frac{a^{3}-b^{3}}{a^{2}+b^{2}+ab}$
But after applied this I find again the problems $0.+\infty$ indeterminate
| Note
$$\sqrt{x^{2}+1}-x=x \left(\sqrt{1+\frac1{x^2}}-1\right)=\frac{\frac1x}{\sqrt{1+\frac1{x^2}}+1}$$
$$\sqrt[3]{x^{3}+1}-x=x \left(\sqrt[3]{1+\frac1{x^3}}-1\right)=\frac{\frac1{x^2}}{\sqrt[3]{\left(1+\frac1{x^2}\right)^2}+\sqrt[3]{1+\frac1{x^2}}+1}$$
Thus,
$$\displaystyle\lim_{x\to +\infty}x\left(\sqrt{x^{2}+1}-\sqrt[3]{x^{3}+1}\right)
=\displaystyle\lim_{x\to +\infty}x\left[\left(\sqrt{x^{2}+1}-x\right) -\left(\sqrt[3]{x^{3}+1}-x \right)\right]$$
$$ =\displaystyle\lim_{x\to +\infty} \frac{1}{\sqrt{1+\frac1{x^2}}+1}
-\frac{\frac1x}{\sqrt[3]{\left(1+\frac1{x^2}\right)^2}+\sqrt[3]{1+\frac1{x^2}}+1}$$
$$=\frac12-0=\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How far can I rewrite an equation before the derivative is not equivalent? I've run into a problem that I can't explain to my class.
We are looking at the derivative for the equation $\frac{x}{y}+\frac{y}{x}=3y$. We calculated it to be $\frac{y(x^2-y^2)}{x(3xy^2+x^2-y^2)}$ and we also verified it with Wolfram Alpha.
A student thought about rewriting the original equation as $x^2+y^2=3xy^2$ by multiplying everything by $xy$. I realize this adds to the domain and adds the point $(0,0)$ as a solution, but it's not differentiable at that point regardless as it's not continuous at that point. When we took the derivative of the rewritten equation and got $\frac{3y^2-2x}{2y-6xy}$, which is not equivalent to our previous calculation.
I can't figure out why the derivatives are so vastly different if all that was added to the original was a new point that's not differentiable to begin with.
Any help is much appreciated!
| Along a curve $f(x,y) = 0$ the derivative $y'$ is given by
$$
y_f'(x,y)
= -\frac{\partial_x f(x,y)}{\partial_y f(x,y)}
.
$$
Given some smooth function $\varphi$ which doesn't vanish on the curve, we can create a new function $g(x,y) = \varphi(x,y) \, f(x,y)$ that also vanishes along the curve. For this we get
$$\begin{align}
y_g'(x,y)
&= -\frac{\partial_x g(x,y)}{\partial_y g(x,y)}
= -\frac{\partial_x \varphi(x,y) \, f(x,y) + \varphi(x,y) \, \partial_x f(x,y)}{\partial_y \varphi(x,y) \, f(x,y) + \varphi(x,y) \, \partial_y f(x,y)} \\
&= -\frac{\partial_x \varphi(x,y) \cdot 0 + \varphi(x,y) \, \partial_x f(x,y)}{\partial_y \varphi(x,y) \cdot 0 + \varphi(x,y) \, \partial_y f(x,y)}
= -\frac{\partial_x f(x,y)}{\partial_y f(x,y)}
= y_f'(x,y)
\end{align}$$
as expected.
In the given case we have $f(x,y) = \frac{x}{y}+\frac{y}{x} - 3y$, $\varphi(x,y) = xy$ and $g(x,y) = x^2+y^2-3xy^2$. With the help of the above calculations I manage to show that the two expressions are equivalent along the curve:
$$\begin{align}
\frac{3y^2-2x}{2y-6xy}
&= - \frac{y\left(\frac{x}{y}+\frac{y}{x} - 3y\right) + xy\left(\frac{1}{y}-\frac{y}{x^2}\right)}{x\left(\frac{x}{y}+\frac{y}{x} - 3y\right) + xy\left(-\frac{x}{y^2}+\frac{1}{x}-3\right)} \\
&= - \frac{y \cdot 0 + xy\left(\frac{1}{y}-\frac{y}{x^2}\right)}{x \cdot 0 + xy\left(-\frac{x}{y^2}+\frac{1}{x}-3\right)}
= - \frac{\frac{1}{y}-\frac{y}{x^2}}{-\frac{x}{y^2}+\frac{1}{x}-3} \\
&= - \frac{x^2y^2\left(\frac{1}{y}-\frac{y}{x^2}\right)}{x^2y^2\left(-\frac{x}{y^2}+\frac{1}{x}-3\right)} = - \frac{x^2y-y^3}{-x^3+xy^2-3x^2y^2} \\
&= \frac{y(x^2-y^2)}{x(3xy^2+x^2-y^2)}
.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
General solutions of $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3$ Note: this actually has nothing to do with probability theory. Fell free to skip ahead and solve the equations below.
Since I am unable to solve this continuous version of a problem about moments, I am now trying to do something simple but related. Let's say $X$ is a random variable which can take three values $x,y,z$ with probability $1/3$ each. For what values of $x,y,z$ does $X$ have its first, second and third central moments equal?
That is, how to solve the system of equations
$$
x+y+z=x^2+y^2+z^2=x^3+y^3+z^3?
$$
The first two equations give a sphere which just covers the unit cube $0<|x|,|y|,|z|<1$. But the second equation is harder to tackle. I guess that we can only have trivial solutions at $x,y,z=0,1$, but I am not sure. Writing $p=x+y+z$, we get a cubic with solution $x,y,z$:
$$
u^3-pu^2+\frac12 p(p-1)u-\frac{1}{12}p(p-1)(p-2) =0,
$$
but that doesn't seem to help.
| Let the values be equal to $t$.
Then $xy+yz+zx = \frac{ (x+y+z)^2 - (x^2+y^2+z^2) }{2} = \frac{t^2-t}{2}$
and $xyz = \frac{ (x+y+z)^3 - (x^3+y^3+z^3) - 3(xy+yz+zx)(x+y+z)}{-3} = \frac{t^3-t - 3(t^2-t)/2 \times t}{-3} = \frac{-t^3+3t^2 - 2t } { 6}$.
So, these are roots to the cubic
$$X^3 - 6tX^2 + 3(t^2-t)X+(t^3-3t^2+2t) = 0.$$
We could then obtain the solutions via the cubic formula:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ Tf(x)= \frac{1}{\sqrt{\pi}}\int_0^x\frac{f(x)}{\sqrt{x-t}}dt$ show that , $Tf(x)\in L^1$ If $Tf(x)= \frac{1}{\sqrt{\pi}}\int_0^x\frac{f(t)}{\sqrt{x-t}}dt$. Show that $T:L^1[0,1]\to L^1[0,1]$ and $TTf(x) = \int_0^1 f(t)dt$
$\textbf{My attempt:}$
Suppose that $\frac{1}{\sqrt{\pi}}\chi_{[0,x]}(t)\frac{f(t)}{\sqrt{x-t}}\in L^1$, then $f\in L^1$
\begin{align}
\|Tf(x)\|_{L^1}
& = \int_0^1\Bigg|\frac{1}{\sqrt{\pi}}\int_0^x\frac{f(t)}{\sqrt{x-t}}dt\Bigg|dx \le \frac{1}{\sqrt{\pi}}\int_0^1\int_0^1\chi_{\{t\le x\}}(t,x)\frac{|f(t)|}{\sqrt{x-t}}dtdx\\
& = \frac{1}{\sqrt{\pi}}\int_0^1\int_0^1 \chi_{\{t\le x\}}(t,x)\frac{|f(t)|}{\sqrt{x-t}}dxdt ~~~~~~~~\text{(Fubini)}\\
& = \frac{1}{\sqrt{\pi}}\int_0^1|f(t)|\int_0^1 \frac{\chi_{\{t\le x\}}(t,x)}{\sqrt{x-t}}dxdt\\
& = \frac{1}{\sqrt{\pi}}\int_0^1|f(t)|\int_t^1 \frac{1}{\sqrt{x-t}}dxdt\\
& = \frac{1}{\sqrt{\pi}}\int_0^1|f(t)|~2~\sqrt{1-t} ~dt\\
& \le \frac{2}{\sqrt{\pi}}\int_0^1|f(t)|~dt<\infty\\
\end{align}
b)
\begin{align}
TTf(x)
&= \frac{1}{\sqrt{\pi}}\int_0^x\frac{Tf(t)}{\sqrt{x-t}}dt = \frac{1}{\sqrt{\pi}}\int_0^x\frac{1}{\sqrt{\pi}}\int_0^x\frac{f(s)}{\sqrt{y-s}}ds\frac{1}{\sqrt{x-t}}dt\\
& =\frac{1}{\pi}\int_0^x\int_0^x\frac{f(s)}{\sqrt{y-s}}\frac{1}{\sqrt{x-t}}ds dt\\
& =\frac{1}{\pi}\int_0^x\int_0^x\frac{f(s)}{\sqrt{y-s}}\frac{1}{\sqrt{x-t}}dt ds\\
& =\frac{1}{\pi}\int_0^1f(s)\int_0^1\frac{\chi_{\{s\le x\}}\chi_{\{t\le x\}}}{\sqrt{(y-s)(x-t})}dt ds\\
& =\frac{1}{\pi}\int_0^1f(s)\int_t^x\frac{1}{\sqrt{(y-s)(x-t})}dt ds\\
& =\frac{1}{\pi}\int_0^1f(s)\int_t^x\frac{1}{\sqrt{(y-s)(x-t})}dt ds\\
\end{align}
(I stuck for the rest)
| Part (a) looks fine to me (I would argue you're using Tonelli's theorem instead of Fubini's, but people often combine them).
Part (b) has many errors. For one, you end up with an expression that has $x,y,t$ as free variables instead of just $x$. Here is one solution:
$$\begin{align*}TTf(x) &= \frac{1}{\sqrt{\pi}} \int_0^x \frac{Tf(t)}{\sqrt{x-t}}\,dt \\ &= \frac{1}{\pi}\int_0^x\int_0^t \frac{f(s)}{\sqrt{(t-s)(x-t)}}\,ds\,dt \\ &= \frac{1}{\pi}\int_0^x f(s)\int_s^x \frac{dt}{\sqrt{(t-s)(x-t)}}\,ds & (\text{Fubini}) \\ &= \frac{1}{\pi}\int_0^x f(s)\int_0^1\frac{du}{\sqrt{u(1-u)}}\,ds & \left(u = \frac{t-s}{x-s}\right) \\ &= \int_0^x f(s)\,ds\end{align*}$$
The step by Fubini needs a little justification, but said justification is basically just running through the entire argument with $|f|$ instead of $f$ and noting the final value is finite since $f \in L^1[0,1]$ and $0 < x < 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Present $\alpha^{-1}\beta$ as a product of disjoint cycles, where $\alpha=(1\ 6)(3\ 2\ 8)$ and $\beta=(1\ 2\ 4)(3\ 5\ 7\ 6)(8\ 9\ 10)$. Let $\alpha=(1\ 6)(3\ 2\ 8)$ and $\beta=(1\ 2\ 4)(3\ 5\ 7\ 6)(8\ 9\ 10)$
Present $\alpha^{-1}\beta$ as a product of disjoint cycles
$$\alpha^{-1} = (8\ 2\ 3)(1\ 6)$$
$$\alpha^{-1}\beta = (8\ 2\ 3)(1\ 6)(1\ 2\ 4)(3\ 5\ 7\ 6)(8\ 9\ 10)$$
In Cauchy notation
$$\alpha^{-1}\beta = \begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\
3 & 4 & 5 & 6 & 7 & 8 & 1 & 9 & 10& 2
\end{pmatrix}$$
I'm struggling taking this back to cycle notation, is my inverse wrong?
Edit: So I have worked on it myself, and corrected my mistake with Cauchy notation. I work right to left for cycles and have reached the answer
$$\alpha^{-1}\beta = (1\ 3\ 5\ 7)(2\ 4\ 6\ 8\ 9\ 10)$$
Now I have to find the order of this, using lcm(4,6) I got $$|\alpha^{-1}\beta|=12$$
| We wish to calculate the following in disjoint cycle notation
\begin{eqnarray*}
\alpha^{-1} \beta = (8\ 2\ 3)(1\ 6)(1\ 2\ 4)(3\ 5\ 7\ 6)(8\ 9\ 10).
\end{eqnarray*}
Start with the element $1$ and reading left to right, $1$ is mapped to $6$ in the second cycle & the $6$ is mapped to $3$ in the fourth cycle, so $1$ maps to $3$. (So far $(1\ 3\ \cdots$).
Next $3$, $3$ is mapped to $8$ in the first cyle, which is mapped to $9$ in the last cycle (So far $(1\ 3\ 9\ \cdots)$).
Next $9$ is mapped to $10$ (So far $(1\ 3\ 9\ 10\ \cdots)$).
Continuing gives
\begin{eqnarray*}
\alpha^{-1} \beta = (1\ 3\ 9\ 10\ 8\ 4)(2\ 5\ 7\ 6).
\end{eqnarray*}
| {
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"url": "https://math.stackexchange.com/questions/3466056",
"timestamp": "2023-03-29T00:00:00",
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If $m \equiv 5\mod 10 $ prove that $1991 \mid 12^m + 9^m + 8^m + 6^m$ I tried to find the remainder of each one of $12$,$9$,$8$ and $6 \mod 5$ and then combine them but I didn’t get the answer
| If $m/5$ is odd, $11|2^5+3^5|2^m+3^m$ while $181|3^5+4^5|3^m+4^m$, so$$\operatorname{lcm}(11,\,181)=1991|(2^m+3^m)(3^m+4^m)=6^m+8^m+9^m+12^m.$$
| {
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"url": "https://math.stackexchange.com/questions/3466565",
"timestamp": "2023-03-29T00:00:00",
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How to prove $\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$? I am trying to prove this inequality:
$$\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$$
if there is condition $a+b+c=1$. $a, b, c$ are positive rational numbers.
I have tried it in several ways, but I can't find a solution. (I think that it may be solved using AM-GM inequality.)
What have I done: (I have done much more, but it isn't useful in this example...)
$\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}=\sqrt{ab+c}+\sqrt{ac+b}+\sqrt{bc+a}$
I have also done some googling, but no success. I think that it may be solved using AM-GM inequality.
I also thought, that $\sqrt{a}+\sqrt{b}+\sqrt{c}\geq a+b+c$ if the condition is true. Then we write:
$$\sqrt{ab+c}+\sqrt{ac+b}+\sqrt{bc+a} \geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$$
And somehow reduce to this:
$$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq a+b+c$$
which is true.
Thank you for the answer and effort!
| Another way for positive variables.
We need to prove that
$$\sum_{cyc}2\sqrt{(1-a)(1-b)}\geq\sum_{cyc}(2a+2\sqrt{ab})$$ or
$$\sum_{cyc}(1-a+1-b-2a-2\sqrt{ab})\geq\sum_{cyc}(1-a-2\sqrt{(1-a)(1-b)}+1-b)$$ or
$$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\geq\sum_{cyc}(\sqrt{1-a}-\sqrt{1-b})^2$$ or
$$\sum_{cyc}\frac{(a-b)^2}{(\sqrt{a}+\sqrt{b})^2}\geq\sum_{cyc}\frac{(a-b)^2}{(\sqrt{1-a}+\sqrt{1-b})^2},$$ which is true because
$$\sqrt{1-a}+\sqrt{1-b}=\sqrt{b+c}+\sqrt{a+c}\geq\sqrt{b}+\sqrt{a}$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$ What is the maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$?
Wolfram Alpha plots this expression on a 3d graph, but I want to solve it algebraicly, by modifying the expression
My Attempts
1) add and substract 2 at the equation and we get $\frac{x+y+1}{y+1}+\frac{x+y+1}{x+1}$ and the numerator is same
=>failed
2) use AM-GM or Cauchy-Schwarz inequality
=>also failed
| It is $$\frac{x}{y+1}+\frac{y}{x+1}\le 1$$ since
$$x(x+1)+y(y+1)\le (x+1)(y+1)$$ This is equivalent to
$$0\le x(1-x)+(1-y)(1-x+y)$$
| {
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"url": "https://math.stackexchange.com/questions/3475979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the center of mass of the liquid with a density of $10 -x^{2}$ in a cone? (drawing included) Problem:
A straight circular cone with the height $2$ m and the radius $1$ m has the pointy end downward and is filled with a liquid. In a flat plane parallel with the bottom surface of the cone and at a distance of $x$ m from the pointy end, the density of the liquid is $\rho$ $kg/m^3$ with:
$\rho = 10- x^2$
Determine the center of mass of the liquid.
Attempted solution:
Not the prettiest picture I have ever drawn:
The basic idea is to get the area of a slice:
$$A(r) = r^2 \pi$$
Congruence gives the radius:
$$\frac{r}{x} = \frac{1}{2} \Rightarrow r = \frac{x}{2}$$
The area becomes:
$$A(x) = \pi \Big(\frac{x}{2}\Big)^2 = \frac{\pi x^2}{4}$$
The volume slice becomes:
$$dV = A(x)dx$$
The dm slice becomes:
$$dm = \rho \pi r^2 dx = \frac{\rho \pi x^2}{4}$$
So left to do is to calculate two integrals from $0$ to $2$ (for $x dm$ and $dm$) and divide the first with the second.
$$\int_0^2 x dm = \frac{\pi}{4} \int_0^2 \rho x^3 dx = \frac{\pi}{4} \int_0^2 (10-x^2) x^3 dx = \frac{\pi}{4} \Big( \frac{10x^4}{4} - \frac{x^6}{6} \Big) \Big|_0^2 = \frac{22\pi}{3}$$
$$\int_0^2 dm = \frac{\pi}{4} \int_0^2 (10-x^2) dx = \frac{\pi}{4} \Big( 10x- \frac{x^3}{3} \Big) \Big|_0^2 = \frac{13\pi}{3}$$
Division produces the final answer of:
$$\frac{22}{13}$$
The expected answer is:
$$\frac{55}{38} \approx 1.45$$
length units from pointy end of the cone.
Where did I go wrong and what are some productive ways to finish off this question?
| Your work looks good but there is a mistake in the integrand:$$\int_0^2 dm = \frac{\pi}{4} \int_0^2 (10-x^2) \color{red}{x^2}dx = \cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int_{0}^{\infty}\frac{x^{s}}{ax^{3}+bx^{2}+cx+d}dx$ I know that the following formulas are true:
$$\int_{0}^{\infty}\frac{x^{s}}{ax+b}dx=\frac{-\pi}{a\sin\left(\pi s\right)}\left(\frac{b}{a}\right)^{s}$$
and
$$\int_{0}^{\infty}\frac{x^{s}}{ax^{2}+bx+c}dx=\frac{\pi\sin\left(s\cos^{-1}\left(\frac{b}{2\sqrt{ac}}\right)\right)}{c\sin\left(\pi s\right)\sqrt{1-\frac{b^{2}}{4ac}}}\left(\frac{c}{a}\right)^{\frac{s+1}{2}}$$
And so I was wondering if there is a generalisation of this integral to the ratio of $x^s$ and a polynomial of general degree. More specificaly, is there is closed form for the below integral?
$$\int_{0}^{\infty}\frac{x^{s}}{ax^{3}+bx^{2}+cx+d}dx$$
| Write $1/(ax^3 + b x^2 + c x + d)$ in partial fractions:
$$ \frac{1}{a x^3 + b x^2 + c x + d} = \sum_\alpha \frac{1}{(3 \alpha^2 a + 2\alpha b + c)(x - \alpha)} $$
where the sum is over the roots of $a x^3 + b x^2 + c x + d$ (assuming these are distinct). Of course you'll also want to assume none of these roots are positive reals. Then, if $s < -1$, you can use your first formula on each of the summands.
| {
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Why should it be $\sqrt[3]{6+x}=x$? Find all the real solutions to:
$$x^3-\sqrt[3]{6+\sqrt[3]{x+6}}=6$$
Can you confirm the following solution? I don't understand line 3. Why should it be $\sqrt[3]{6+x}=x$?
Thank you.
$$
\begin{align}
x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \\
x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \\
x &= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}} \\
\sqrt[3]{6+x} &= x \\
x^3 &= 6+x \\
x^3-2x^2+2x^2-4x+3x-6 &= 0 \\
(x-2)(x^2+2x+3) &= 0 \\
x &= 2
\end{align}
$$
|
Goal: Establish the fact that in order to solve $$x=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}$$
we only need to solve
$$
x=\sqrt[3]{6+x}
$$
Define $f(x)$:
$$
f(x) = \sqrt[3]{x+6}$$
From line 3, we are looking to solve:
$$
x = f(f(f(x)))$$
Notice that $f(x)$ is strictly increasing. In other words, for all $a$ and $b$ in the domain of $f(x)$, if $a > b$, then $f(a) > f(b)$.
This means that the only way in which we can get solutions for $x = f(f(f(x)))$ is if $x = f(x)$. To see why, assume that $x > f(x)$:
$$
x > f(x) \Rightarrow \\
f(x) > f(f(x)) \Rightarrow \\
f(f(x)) > f(f(f(x))) \Rightarrow \\
x > f(x) > f(f(x)) > f(f(f(x)))
$$
The statement $x = f(f(f(x)))$ is now untrue. The exact same reasoning can be used for the assumption that $x < f(x)$.
Since $x < f(x)$ and $x > f(x)$ do not give us any solutions, then $x = f(x)$ is the only case we have left.
FYI, the values of a function's domain that don't change when passed into the function are called the "fixed points" of $f(x)$. In other words, all $x$ such that $x = f(x)$ are called the fixed points of $f(x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1-e}{1+e}$ If P is a point of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Whose foci are S and S’.Let angle PSS’=$\alpha$ and $PS’S=\beta$ then prove that $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1-e}{1+e}$
I know that S(ae,0) and S’(-ae,0) and $b^2=a^2(1-e^2)$ but don’t know how to solve further
| Write the coefficients of $P$ as $x=a\cos t,\,y=b\sin t$ so$$\begin{align}PS^2&=a^2(\cos t-e)^2+b^2\sin^2 t\\&=a^2(1-e\cos t)^2,\\PS^{\prime2}&=a^2(1+e\cos t)^2,\\SS^{\prime2}&=4a^2e^2,\end{align}$$and$$\begin{align}\cos\alpha&=\frac{PS^2+SS^{\prime2}-PS^{\prime2}}{2PS\cdot SS^\prime}\\&=\frac{e-\cos t}{1-e\cos t},\\\cos\beta&=\frac{e+\cos t}{1+e\cos t}.\end{align}$$In terms of $A:=\tan\frac{\alpha}{2},\,B:=\tan\frac{\beta}{2}$,$$\frac{1-A^2}{1+A^2}=\frac{e-\cos t}{1-e\cos t}\implies A^2=\frac{(1-e)(1+\cos t)}{(1+e)(1-\cos t)},$$and similarly$$B^2=\frac{(1-e)(1-\cos t)}{(1+e)(1+\cos t)}$$(note we just have to replace $\cos t$ with $-\cos t$). Now take the geometric mean.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given the matrices $A\in M_{m\times n}, B\in M_{n\times p}: \ B\neq 0$, suppose $AB=0$. Prove: there exists a non-trivial solution for $Ax=0$ Given two matrices: $A_{m\times n}$ and $B_{n\times p}$, such that $B\neq 0$,
suppose $AB=0$.
Prove: there exists a non-trivial solution for $Ax=0$.
My Attempt:
suppose by contradiction that the only solution for the linear equation system $Ax=0$ is the trivial solution, meaning $Ax=0 \iff x=0$.
$$Ax=0\Longrightarrow\left[\begin{array}{cccc}
a_{11} & \cdots & \cdots & a_{1n}\\
\cdots & \cdots & \cdots & \cdots\\
\cdots & \cdots & \cdots & \cdots\\
a_{m1} & \cdots & \cdots & a_{mn}
\end{array}\right]\left[\begin{array}{c}
x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{array}\right]=\left[\begin{array}{c}
0\\
\vdots\\
\vdots\\
0
\end{array}\right]$$
Hence, $$\left[\begin{array}{c}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=0\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=0\\
\cdots\\
a_{m1}x_{1}+a_{m2}x_{2}+\cdots+a_{mn}x_{n}=0
\end{array}\right]$$
According to the assumption: $x_{1}=x_{2}=\cdots=x_{n}=0$
Now, Let $B=\left[\begin{array}{cccc}
b_{11} & \cdots & \cdots & b_{1p}\\
\cdots & \cdots & \cdots & \cdots\\
\cdots & \cdots & \cdots & \cdots\\
b_{n1} & \cdots & \cdots & b_{np}
\end{array}\right]$
Let $AB=(t)_{ij}$, as $\begin{array}{c}
1\le i\le m\\
1\le j\le p
\end{array}$
$AB=0\Longrightarrow\left[\begin{array}{cccc}
a_{11} & \cdots & \cdots & a_{1n}\\
\cdots & \cdots & \cdots & \cdots\\
\cdots & \cdots & \cdots & \cdots\\
a_{m1} & \cdots & \cdots & a_{mn}
\end{array}\right]\left[\begin{array}{cccc}
b_{11} & \cdots & \cdots & b_{1p}\\
\cdots & \cdots & \cdots & \cdots\\
\cdots & \cdots & \cdots & \cdots\\
b_{n1} & \cdots & \cdots & b_{np}
\end{array}\right]=\left[\begin{array}{c}
0\\
\cdots\\
\cdots\\
0
\end{array}\right]$
$\Longrightarrow\left[\begin{array}{cccc}
a_{11}b_{11}+\cdots+a_{1n}b_{n1}\ & \cdots\cdots & \cdots\cdots & a_{11}b_{1p}+\cdots+a_{1n}b_{np}\\
\cdots\cdots & \cdots\cdots & \cdots\cdots & \cdots\cdots\\
\cdots\cdots & \cdots\cdots & \cdots\cdots & \cdots\cdots\\
a_{m1}b_{11}+\cdots+a_{mn}b_{n1}\ & \cdots\cdots\ & \cdots\cdots\ & a_{m1}b_{1p}+\cdots+a_{mn}b_{np}
\end{array}\right]$
This means that for every $i,j:\ t{}_{ij}=0$, therefore $$\begin{array}{c}
a_{11}b_{11}+\cdots+a_{1n}b_{n1}=a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=0\\
\cdots\cdots\\
a_{m1}b_{11}+\cdots+a_{mn}b_{n1}=a_{m1}x_{1}+a_{m2}x_{2}+\cdots+a_{mn}x_{n}=0
\end{array}$$
but $x_{1}=\cdots=x_{n}=0$, then $b_{11}=\cdots=b_{n1}=0$.
this equity applies for every column of $B$, consequently, $B=0$.
But that's a contradiction to the fact that
$B\neq 0$, then we'll get a contradiction to the assumption, therefore there exists a non-trivial solution for
$Ax=0$, as wished.
It this correct?
| If $B = [b_1,...,b_p] \neq 0$, there is some $i \in \{1,...,p\}$ such that $b_i \neq 0$. Then, by definition of the matrix product, $Ab_i = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find value of $a_1$ such that $a_{101}=5075$ Let $\{a_n\}$ be a sequence of real numbers where $$a_{n+1}=n^2-a_n,\, n=\{1,2,3,...\}$$
Find value of $a_1$ such that $a_{101}=5075$.
I have
$$a_2=1^2-a_1$$
$$a_3=2^2-a_2=2^2-1^2+a_1$$
$$a_4=3^2-2^2+1^2-a_1$$
$$a_5=4^2-3^2+2^2-1^2+a_1$$
$$\vdots$$
$$a_{101}=100^2-99^2+98^2-97^2+\ldots+2^2-1^2+a_1.$$
Therefore, $$a_{101}=\sum_{i=1}^{50}(2i)^2-\sum_{i=1}^{50}(2i-1)^2.$$
Thus, $$5075=\sum_{i=1}^{50}(4i^2-4i^2+4i-1)+a_1,$$
and, $$a_1=5075-4\sum_{i=1}^{50}(i)+\sum_{i=1}^{50}(1).$$
Hence, $$a_1=5075-4(\frac{50}{2})(51)+50=25,$$
and $a_1=25$.
Is it correct? Do you have another way? Please check my solution, thank you.
| Another way :
$$a_{n+1}=n^2-a_n=n^2-((n-1)^2-a_{n-1})=a_{n-1}+2n-1$$
$$a_{n-1}=a_{n-3}+2(n-1)-1=a_{n-3}+2(n-2)+1$$
$$a_{n+1}=a_{n+1-2r}+\underbrace{2n-1+2n-3+\cdots}_{r\text{ terms}}$$
Here $n+1=101,n+1-2r=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Compute the $PV\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$ Problem :
Evaluate the closed form of : $PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx$
Wolfram alpha give me :
$I=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos (4x)}{\cos (3x)}dx=\sqrt{3}-\coth^{-1}{\sqrt{3}}$
But i can't get by my try as following :
$\cos (4x)=8\cos^{4} x-8\cos^{2} x+1$
And
$\cos (3x)=4\cos^{3} x-3\cos x$
And I know that
$PV\displaystyle\int_0^{\frac{π}{3}}\frac{1}{\cos (3x)}dx=0$
So I need to find
$J=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\cos^{2} x}{\cos (3x)}dx=PV\displaystyle\int_0^{\frac{π}{3}}\frac{\sin^{2} x}{\cos (3x)}dx$
Now take $y=\cos x$
$J=\displaystyle\int_{\frac{1}{2}}^{1}\frac{\sqrt{1-x^2}}{4x^{3}-3x}dx$
From here I don't know how I complete
| Hint:
$$\dfrac{\cos4x}{\cos3x}=\dfrac{2\cos^22x-1}{\cos x(4\cos^2x-3)}=\dfrac{2(1-2\sin^2x)^2-1}{\cos^2x(4\cos^2x-3)}\cdot\cos x$$
Set $\sin x=u$
$$I=\int_0^{\sqrt3/2}\dfrac{8u^4-8u^2+1}{(1-u^2)(1-4u^2)}du$$
Writing $u^2=v$ and use Partial Fraction Decomposition
$$\dfrac{8v^2-8v+1}{(1-v)(1-4v)}=2+\dfrac a{1-v}+\dfrac b{1-4v}$$
Find $a,b$ and replace $v$ with $u^2$
Can you take it home from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\int \frac{x^2}{1+x^5}dx$ I tried factoring the denominator and performing partial fraction decomposition, but the algebra becomes too cumbersome for me...
Is there a cleaner approach for this integral?
| There is a clean way to integrate. Note that $x^5+1$ factorizes as
$$x^5+1= (1+x)(x^2-2\phi_+x+1)(x^2-2\phi_-x+1)$$
with $\phi_{\pm} = \frac{1\pm\sqrt5}{4}$. So, decompose the integrand,
$$\frac{5x^2}{1+x^5}=\frac1{x+1}-\phi_- \frac{2x+2}{x^2-2\phi_+x+1}
-\phi_+ \frac{2x+2}{x^2-2\phi_-x+1}$$
The integral for the first term is just $\ln(x+1)$. The second and third terms have the same form, whose integral can be obtained as,
$$I(x,\phi)= \int \frac{2x+2}{x^2-2\phi x+1}dx =\int \frac{d[(x-\phi)^2] + 2(1+\phi)dx}{(x-\phi)^2 +(1-\phi^2)} $$
$$=\ln\left[(x-\phi)^2 +(1-\phi^2)\right]
+\frac{2(1+\phi)}{\sqrt{1-\phi^2}}
\tan^{-1}\frac{x-\phi}{\sqrt{1-\phi^2}}$$
$$=\ln\left(x^2-2\phi x+1\right)
+2\sqrt{\frac{1+\phi}{1-\phi}}
\tan^{-1}\frac{x-\phi}{\sqrt{1-\phi^2}}
$$
Thus, the original integral is,
$$\int \frac{x^2}{1+x^5}dx=\frac15\left[\ln(x+1)-\phi_-I(x,\phi_+)-\phi_+I(x,\phi_-)\right] + C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$. Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$.
So far I've got $n^2 = 1000k + n$ which means $n^2 ≡ n \mod 1000$. I don't know how to proceed since 1000 seems to be a bit high.
| Another way:
$(\pm1)^2\equiv1,(\pm2)^2\equiv4,(\pm3)^2\equiv9,(\pm4)^2\equiv6\pmod{10}$
The last digit of $n^2,n$ will be same if $n=10m,10m+1,10m+5,10m+6$
$(10m+6)^2\equiv36+20m\pmod{100}$
We need $10m+6\equiv36+20m\pmod{100}\iff10m\equiv-30\pmod{100}\iff m\equiv-3\pmod{10}$
WLOG $m=10r-3$
$10m+6=10(10r-3)+6=100r-24$
$(100r-24)^2\equiv576-4800r\pmod{1000}\equiv576+200r$
We need $100r-24\equiv576+200r\pmod{1000}\iff100r\equiv-600\equiv400\pmod{1000}\iff r\equiv4\pmod{10}$
$r=10s+4\implies100r-24=100(10s+4)-24=1000s+376$
Similarly for $10m+1,10m+5$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find minimum value of $P = \frac{y + z}{x}$
Let $x,y,z > 0$ such that $3x(x+y+z)=yz$ . Find the minimum value of $$P=\frac{y+z}{x}$$
$$3x(x+y+z)=yz\Leftrightarrow 3(x+y)(x+z)=4yz$$
Or $$3(\frac{x}{y}+1)(\frac{x}{z}+1)=4$$
Let $a=x/y, b=x/z ( a,b>0)$
We have: $$4=3(a+1)(b+1)\ge3 \cdot 2 \sqrt a \cdot 2 \sqrt b=12 \sqrt {ab} \rightarrow \sqrt {ab}\le \frac 1 3$$
And $$P=\frac{y+z}{x}=\frac{1}{a}+\frac{1}{b}\ge \frac{2}{\sqrt {ab}}\ge 6$$
But by WA $Min_P=6+4\sqrt 3$. Is my solution is wrong?
| Let $y+z=2u$ and $yz=v^2$, where $v>0$.
Thus, since by AM-GM $$\frac{y+z}{2}\geq\sqrt{yz},$$ we obtain $u\geq v$,
$$3x^2+6ux-v^2=0$$ or $$x=-u+\sqrt{u^2+\frac{v^2}{3}}$$ and
$$\frac{y+z}{x}=\frac{2u\left(\sqrt{u^2+\frac{v^2}{3}}+u\right)}{\frac{v^2}{3}}\geq\frac{2v\left(\sqrt{v^2+\frac{v^2}{3}}+v\right)}{\frac{v^2}{3}}=6+4\sqrt3.$$
The equality occurs for $u=v$ or $y=z$, which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $r^{2} = 3$ from a geometric series
The sum of the first two terms of geometric sequence $a_n=a_0r^n$ is $7$, and the sum of the first six terms is $91$.
Show that $r^{2} = 3$.
How do we go about this?
Here's what I have tried -
$\displaystyle \Rightarrow \frac{\dfrac{a(r^{6}-1)}{r-1}}{\dfrac{a(r^{2}-1)}{r-1}} = \frac{91}{7}$
$\displaystyle \Rightarrow \frac{(r^{6}-1)}{(r^{2}-1)} = \frac{91}{7}$
$\Rightarrow\;???$
Any help would be highly appreciated.
| Use this identity $a^3- b^3 = (a-b)(a^2 +ab+b^2)$, and then solve a quadratic equation.
$$\Rightarrow \frac{(r^{6}-1)}{(r^{2}-1)} = 13$$
$$\Rightarrow \frac{(r^{2}-1)(r^4+r^2+1)}{(r^{2}-1)} = 13$$
$$\Rightarrow r^4+r^2+ 1= 13$$
$$\Rightarrow r^4+r^2-12=0$$
You'll find something like $(r^2 -3)(r^2 +4)=0$
BINGO!
| {
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"timestamp": "2023-03-29T00:00:00",
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Find min and max of $A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$
Find minimum and maximum of $$A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$$
How to solve this problem without using derivatives?
Michael Rozenberg's edit:
Because with a derivative it's not so easy:
$$A'(x)=\frac{1}{2\sqrt{x-2}}+\frac{1}{\sqrt{x+1}}-1=\frac{1}{2\sqrt{x-2}}-\frac{1}{2}+\frac{1}{\sqrt{x+1}}-\frac{1}{2}=$$
$$=\frac{1-\sqrt{x-2}}{2\sqrt{x-2}}+\frac{2-\sqrt{x+1}}{2\sqrt{x+2}}=$$
$$=(3-x)\left(\frac{1}{2(1+\sqrt{x-2})\sqrt{x-2}}+\frac{1}{2(2+\sqrt{x+1})\sqrt{x+2}}\right),$$
which gives $x_{max}=3.$
| $$ \forall_{y\ge -1}\quad\sqrt{1+y}\ \le\ 1+\frac y2 $$
hence
$$ \forall_{x\ge 2}\quad\sqrt{x-2}\ =\ \sqrt{1+(x-3)}\ \le\ 1 +\frac{x-3}2 $$
and
$$ \forall_{x\ge -1}\quad\sqrt{x+1}\ =\ 2\cdot\sqrt{1+\frac{x-3}4}\ \le
\ 2+\frac{x-3}4 $$
hence
$$ \forall_{x\ge 2}\quad\sqrt{x-2}\ +\ 2\cdot\sqrt{x+1}\ \le\ 2+x $$
i.e.
$$ \forall_{x\ge 2}\quad\sqrt{x-2}\ +\ 2\cdot\sqrt{x+1}\ -\ x\ \le\ 2 $$
where the equality holds only for $\ x=3.\ $ Thus A (see the OP Question) attains its maximum $A=2021$ at $x=3$. Great!
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int\limits ^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)} dx$ I want to show that
$$\int\limits ^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)} dx=\frac{\psi^{(1)}(\frac{t}{2\pi})-4\psi^{(1)}(\frac{t}{\pi})}{8\pi}t+\frac14
$$
where $\psi^{(1)}(z)$ is the first derivative of the digamma function, but I've never come across such an integral and don't know what to do. Any help would be greatly appreciated.
| We first modifiy the integral to be calculated, by remarking that, changing $x=-u$,
\begin{align}
I&=\int^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)}\,dx\\
&=-\int_{-\infty }^{0}\frac{u}{\left( u^{2} +1\right)^2\left( e^{-tu} +1\right)}\,du\\
&=-\int_{-\infty }^{0}\frac{ue^{tu} }{\left( u^{2} +1\right)^2\left( e^{tu} +1\right)}\,du\\
&=-\int_{-\infty }^{0}\frac{u}{\left( u^{2} +1\right)^2}\,du+\int_{-\infty }^{0}\frac{u}{\left( u^{2} +1\right)^2\left( e^{tu} +1\right)}\,du\\
&=\frac{1}{2}+I^-
\end{align}
where
\begin{equation}
I^-=\int_{-\infty }^{0}\frac{u}{\left( u^{2} +1\right)^2\left( e^{tu} +1\right)}\,du
\end{equation}
Then, we have
\begin{align}
I&=I^-+\frac{1}{2}\\
I+I^-&=\int_{-\infty }^{\infty}\frac{u}{\left( u^{2} +1\right)^2\left( e^{tu} +1\right)}\,du
\end{align}
Thus,
\begin{equation}
I=\frac{1}{4}+\frac{1}{2}\int_{-\infty }^{\infty}\frac{u}{\left( u^{2} +1\right)^2\left( e^{tu} +1\right)}\,du
\end{equation}
This integral is evaluated by the residue method, by closing the contour with the upper half circle. It can easily be verified that the half-circle contribution vanishes. Interior poles lie at
*
*$u=i$ with residue
\begin{align}
R_i&=\frac{it\exp(it)}{4(1+\exp(it))^2}\\
&=\frac{it}{16\cos^2\left( t/2 \right)}\\
&=\frac{it}{4}\sum_{n=-\infty}^\infty\frac{1}{\left( t+\pi(2n+1) \right)^2}\\
&=\frac{it}{2}\sum_{n\ge0}\frac{t^2+\pi^2(2n+1)^2}{\left( t^2-\pi^2(2n+1)^2 \right)^2}
\end{align}
(using the series representation for $\sec^2(z)$).
*$u=i(2n+1)\pi/t$, where $n=0,1,2,\ldots$ with residues
\begin{align}
R_n&=\frac{-i\pi\left( 2n+1\right)t^2 }{\left(\left( (2n+1)i\pi \right)^2+t^2\right)^2}\\
&=\frac{-i\pi\left( 2n+1\right)t^2 }{\left(t^2-\pi^2(2n+1)^2\right)^2}
\end{align}
Then
\begin{align}
I&=\frac{1}{4}+\frac{1}{2}2i\pi\frac{it}{2}
\sum_{n\ge0}\left[\frac{t^2+\pi^2(2n+1)^2}{\left( t^2-\pi^2(2n+1)^2 \right)^2}-\frac{2\pi\left( 2n+1\right)t }{\left(t^2-\pi^2(2n+1)^2\right)^2}
\right]\\
&=\frac{1}{4}-\frac{\pi t}{2}\sum_{n\ge0}\frac{\left( t-\pi(2n+1) \right)^2}{\left( t^2-\pi^2(2n+1)^2 \right)^2}\\
&=\frac{1}{4}-\frac{\pi t}{2}\sum_{n\ge0}\frac{1}{\left( t+\pi(2n+1) \right)^2}
\end{align}
The summation can be written as
\begin{align}
I&=\frac{1}{4}-\frac{ t}{8\pi}\sum_{n\ge0}\frac{1}{\left( \frac{t}{2\pi}+n+\frac{1}{2} \right)^2}\\
&=\frac{1}{4}-\frac{ t}{8\pi}\psi^{(1)}\left( \frac{t}{2\pi}+\frac{1}{2} \right)
\end{align}
From the classical identity
\begin{equation}
\psi\left(2z\right)=\tfrac{1}{2}\left(\psi\left(z\right)+\psi\left(z+\tfrac{1}{2}\right)\right)+\ln 2
\end{equation}
we obtain, after differentiation
\begin{equation}
\psi^{(1)}\left(z+\tfrac{1}{2}\right)=4\psi^{(1)}\left(2z\right)-\psi^{(1)}\left(z\right)
\end{equation}
and thus
\begin{equation}
I=\frac{1}{4}-\frac{t}{8\pi}\left[4\psi^{(1)}\left( \frac{t}{\pi}\right)-\psi^{(1)}\left( \frac{t}{2\pi}\right)\right]
\end{equation}
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate $\iint_A \frac{1}{x^2+y^2}\,\mathrm dx\,\mathrm dy,$ where $A=[\frac{1}{a},a]\times[0,1]$ I want to compute
$$\iint_A \frac{1}{x^2+y^2}\,\mathrm dx\,\mathrm dy$$ where $A:=\left[\frac{1}{a},a\right]\times[0,1]$.
I got that this is equal to $\int_{1/a} ^a \frac{1}{x}\arctan \Big(\frac{1}{x}\Big) \mathrm dx\ $ and I don't know what to do from here. Can somebody help me, please?
| Let's catch up with you:
$$\begin{align}
\iint\limits_A \frac{1}{x^2+y^2}dx\,dy &= \int\limits_{\frac{1}{a}}^a\int\limits_0^1\frac{1}{x^2+y^2}dy\,dx\\
&=\int\limits_{\frac{1}{a}}^a\int\limits_0^1\frac{1}{x^2(1 + (y/x)^2)}dy\,dx.
\end{align}$$
Taking $u = y/x, \frac{\partial u}{\partial y} = 1/x$ so the last integral becomes
$$\begin{align}
\int\limits_{\frac{1}{a}}^a\int\limits_0^{\frac{1}{x}}\frac{1}{x(1 + u^2)}du\,dx
&= \int\limits_\frac{1}{a}^a\frac{1}{x}\arctan \left(\frac{1}{x}\right)\,dx \\
&=\int\limits_a^\frac{1}{a} -\frac{1}{u}\arctan u\,du \\
&= \int\limits_\frac{1}{a}^a \frac{1}{u}\arctan u\,du,
\end{align}$$
because $u = 1/x \implies du = -1/x^2\,dx = -u^2 \,dx \implies dx = du/u^2$ and $1/a \mapsto a, a \mapsto 1/a$. Accordingly to WolframAlpha, the indefinite integral for your expression has no closed form (in terms of elementary functions), but Wolfram Mathematica tells us it has a closed form with those integration limits (kudos to Semiclassical)!
So we have to exploit our integration limits. Let $I =\int\limits_\frac{1}{a}^a\frac{1}{x}\arctan \left(\frac{1}{x}\right)\,dx$. Since $\arctan x = \pi/2 - \arctan 1/x$,
$$\begin{align}
I &= \int\limits_\frac{1}{a}^a \frac{1}{u}\arctan u\,du,\\
&= \int\limits_\frac{1}{a}^a \frac{1}{u}\left(\frac{\pi}{2}-\arctan \frac{1}{u}\right)\,du \\
&= \frac{\pi}{2}\left(\int\limits_\frac{1}{a}^a 1/u\,du\right) - I \\
&= \frac{\pi}{2}(\ln a - \ln(1/a)) - I \\
&= \pi \ln a - I.
\end{align}
$$
Therefore $I = \frac{\pi}{2}\ln a$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{\sqrt{\sin^4x+\cos^4x}}{\sin^3x\cos x}dx$ $$\int \dfrac{\sqrt{\sin^4x+\cos^4x}}{\sin^3x\cos x}$$
My multiple attempts are as follows:-
Attempt $1$:
$$\int \dfrac{\sqrt{1-2\sin^2x\cos^2x}\cdot\sin x}{\sin^4x\cos x}dx$$
$$\cos x=t$$
$$-\sin x=\dfrac{dt}{dx}$$
$$\int -\dfrac{\sqrt{1-2t^2(1-t^2)}}{t(1-t^2)^2}dt$$
$$\int -\dfrac{\sqrt{2t^4+1-2t^2}}{t(1+t^4-2t^2)}dt$$
$$\int -\dfrac{\sqrt{2t^2+\dfrac{1}{t^2}-2}}{1+t^4-2t^2}dt$$
$$\int -\dfrac{\sqrt{2t^2+\dfrac{1}{t^2}-2}}{\left(\dfrac{1}{t^2}+t^2-2\right)t^2}dt$$
$$\int -\dfrac{\sqrt{\dfrac{2}{t^2}+\dfrac{1}{t^6}-\dfrac{2}{t^4}}}{\left(\dfrac{1}{t^2}+t^2-2\right)}dt$$
Not finding the way to proceed from here.
Attempt $2$:
$$\int \dfrac{\sqrt{\tan^4x+1}\cos x}{\sin^3 x}dx$$
$$\int \dfrac{\sqrt{\tan^4x+1}}{\tan^3 x\cos^2x}dx$$
$$\tan x=t$$
$$\int \dfrac{\sqrt{t^4+1}}{t^3}dt$$
$$t=\sqrt{\tan\theta}$$
$$\dfrac{dt}{d\theta}=\dfrac{1}{2\sqrt{\tan\theta}}\sec^2\theta$$
$$\int \dfrac{\sec\theta\sec^2\theta}{2\tan^2\theta}d\theta$$
$$\int \dfrac{\sec\theta}{2\sin^2\theta}d\theta$$
$$\int \dfrac{\cos\theta}{2\sin^2\theta\cos^2\theta}d\theta$$
$$y=\sin\theta$$
$$\int \dfrac{dy}{2y^2(1-y^2)}$$
$$-\dfrac{1}{2}\cdot\int\dfrac{y^2-(y^2-1)}{y^2(y^2-1)}dy$$
$$-\dfrac{1}{2}\cdot\left(\dfrac{1}{2}\cdot\ln\left|\dfrac{y-1}{y+1}\right|+\dfrac{1}{y}\right)+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{\sin\theta-1}{\sin\theta+1}\right|-\dfrac{1}{2\sin\theta}+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{t^2-\sqrt{1+t^4}}{t^2+\sqrt{1+t^4}}\right|-\dfrac{\sqrt{1+t^4}}{2t^2}+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{\tan^2x-\sqrt{1+\tan^4x}}{\tan^2x+\sqrt{1+\tan^4x}}\right|-\dfrac{\sqrt{1+\tan^4x}}{2\tan^2x}+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{1-\sqrt{\cot^4x+1}}{1+\sqrt{\cot^4x+1}}\right|-\dfrac{\sqrt{1+\cot^4x}}{2}+C$$
But real answer is $-\dfrac{\mathrm{cosec}x}{2}-\dfrac{1}{4}\cdot\ln\left|\dfrac{\mathrm{cosec}x-1}{\mathrm{cosec}x+1}\right|+C$
What am I missing here?
| Continue the second approach below with integration by parts
$$I=\int \dfrac{\sqrt{t^4+1}}{t^3}d
=-\frac12 \frac{1+t^4}{t^2} +\int \frac{tdt}{\sqrt{1+t^4}}dt$$
where
$$\int \frac{tdt}{\sqrt{1+t^4}}=\frac12 \int \frac{d(t^2)}{\sqrt{1+(t^2)^2}}=\frac12 \sinh^{-1}t^2
$$
Thus,
$$I =- \frac{1+t^4}{2t^2} +\frac12 \sinh^{-1}t^2+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the positive integers $a,b,$ and $c$ s.t. $\sqrt{\sqrt[3]{5} - \sqrt[3]{4}}\times 3 = \sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c}.$
Let $a,b,c$ be positive integers such that $\sqrt{\sqrt[3]{5} - \sqrt[3]{4}}\times 3 = \sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c}$. Determine the values of $a, b,$ and $c$.
To solve this simplification problem, it seems that it will be useful to use the difference of cubes formula and/or difference of squares formula as well as some substitutions. I wasn't able to determine a useful method of solving this, however. Using a computer program, I was able to deduce that the desired values $(a,b,c)$ are $(2,20,25)$.
| Let $x,y,z\in\mathbb{R}$ such that $x^3 = 5, y^3 = 4$, and $z^3 = 2$. Then $y=z^2, y^2 = 2z$, and $yz = 2.$ We want to simplify $3\sqrt{x-y}$. Note that
$$\begin{align}
(x-y)(x+y)^2&=(x-y)(x^2+2xy+y^2)\\
&=x^3-y^3+x^2y-xy^2\\
&=1+(xz)^2-2xz\\
&=(xz-1)^2.
\end{align}$$
Hence
$$\begin{align}
3\sqrt{x-y} &= 3\cdot\dfrac{(xz-1)(x^2-xy+y^2)}{(x+y)(x^2-xy+y^2)}\\
&=3\cdot\dfrac{x^3z-x^2-x^2(yz)+xy+x(y^2)z-y^2}{x^3+y^3}\\
&=\dfrac{1}{3}(3z-3x^2+3xz^2)=z+xz^2-x^2\\
&=\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25}\\
&\Rightarrow (a,b,c) = (2,20,25).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why does this pattern occur: $123456789 \times 8 + 9 = 987654321$ I came across the following:
$\begin{align} 1 \times 8 + 1 &= 9 \\ 12 \times 8 + 2 & = 98 \\ 123 \times 8 + 3 & = 987 \\ 1234 \times 8 + 4 & = 9876 \\ 12345 \times 8 + 5 & = 98765 \\ 123456 \times 8 + 6 & = 987654 \\ 1234567 \times 8 + 7 & = 9876543 \\ 12345678 \times 8 + 8 & = 98765432 \\ 123456789 \times 8 + 9 & = 987654321. \\ \end{align}$
I'm looking for an explanation for this pattern. I suspect that there is some connection to the series $\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + \cdots$.
This post asks the same question but has no answers posted.
| If I consider the equations you provide with your "ideas so far":
\begin{align}
1 \times 9 + 1 &= 10 \\
12 \times 9 + 2 & = 110 \\
123 \times 9 + 3 & = 1110 \\
\vdots\\
123456789 \times 9 + 9 & = 1111111110, \\
\end{align}
The first equation being true, this system is equivalent to the system composed of their successive differences all of them having the common pattern :
$$\underbrace{11...1}_{k \ \text{digits}} \times 9 + 1 = 10^k$$
which is an (almost) evident fact.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 4
} |
How to factorize $X^5-1$ into irreductible factors? I have to factorize the polynomial $P(X)=X^5-1$ into irreducible factors in $\mathbb{C}$ and in $\mathbb{R}$, this factorisation happens with the $5$th roots of the unity.
In $\mathbb{C}[X]$ we have $P(X)=\prod_{k=0}^4 (X-e^\tfrac{2ki\pi}{5})$.
In $\mathbb{R}[X]$ the solution states that by gathering all complex conjugate roots we find that $P(X)=(X-1)(X^2-2\cos(\frac{2\pi}{5})+1)(X^2-2\cos(\frac{4\pi}{5})+1)$, but I can't figure out how. Another problem I ran into was trying to figure out where the $2\cos(\frac{2\pi}{5})$ and $2\cos(\frac{4\pi}{5})$ come from so I tried these two methods:
The sum of the roots of unity is zero so we have:
$1+e^\tfrac{2i\pi}{5}+e^\tfrac{4i\pi}{5}+e^\tfrac{6i\pi}{5}+e^\tfrac{8i\pi}{5}=0$
In the $5$th roots of unity circle P3 and P4 are images according to the x-axis the same goes to P2 and P5, therefore $e^\tfrac{6i\pi}{5}=e^\tfrac{-4i\pi}{5}$ and $e^\tfrac{8i\pi}{5}=e^\tfrac{-2i\pi}{5}$ afterwards by using Euler's formula we find $1+2\cos(\frac{2\pi}{5})+2\cos(\frac{4\pi}{5})=0$.
Another method is that $\cos(6\pi/5) = \cos(-6\pi/5) = \cos(-6\pi/5 + 2\pi) = \cos(4\pi/5)$ $\cos(8\pi/5) = \cos(-8\pi/5) = \cos(-8\pi/5 + 2\pi) = \cos(2\pi/5)$
therefore $1 + \cos(2\pi/5) + \cos(4\pi/5) + \cos(4\pi/5) + \cos(2\pi/5) = 0$ and we find $1+2\cos(\frac{2\pi}{5})+2\cos(\frac{4\pi}{5})=0$
I don't know if both of these methods are correct on their own and I don't know if they will help in the factorisation since I don't know how to go from there and find $P(X)=(X-1)(X^2-2\cos(\frac{2\pi}{5})+1)(X^2-2\cos(\frac{4\pi}{5})+1)$
| A way to obtain an explicit factorisation using Chebyshev polynomials:
Using the recurrence relation:
$$P_{n+1}(t)=2tP_n(t)-P_{n-1}(t),\qquad P_0(t)=1,\;P_1(t)=t,$$
we readily obtain
$$P_5(t)=2t\, P_4(t)-P_3(t)=16t^5-20t^3+5t,$$
so that, for $t=\cos\frac\pi{10}$, we obtain the equation
$$\cos5\Bigl(\frac{\pi}{10}\Bigr)=0=16t^5-20t^3+5t,$$
and as $\cos\frac{\pi}{10}>0$, it is a root of the biquadratic equation $\;16t^4-20t^2+5=0$. So $\cos^2\frac\pi{10}$ is a root of the quadratic equation
$$ f(u)=16u^2 -20 u+5=0. $$
To determine which root it is, we need to find a number which separates the roots $u_1$ and $u_2$.
Now $\;\cos^2\dfrac\pi 6=\dfrac34 <\cos^2\dfrac\pi{10}$, and it happens that $
f(3/4)=-1<0$, so $u_1<\dfrac 34<u_2$, and finally the standard formulæ for the solutions of a quadratic equation yield
$$\cos^2\frac\pi{10}=u_2=\frac{5+\sqrt 5}8\quad\text{whence }\;\cos\frac\pi5=2\cos^2\frac\pi{10}-1=\frac{1+\sqrt 5}4$$
Last step, with the same duplication formula
\begin{align}
\cos\frac{2\pi}5&=2\biggl(\frac{1+\sqrt 5}4\biggr)^2-1=\frac{-1+\sqrt 5}4 \\
\cos\frac{4\pi}5&=2\biggl(\frac{-1+\sqrt 5}4\biggr)^2-1=\frac{-1-\sqrt 5}4,
\end{align}
we obtain the factorisation:
$$X^5-1=(X-1)\Bigl(X^2+\frac{1-\sqrt 5}2 X+1\Bigr)\Bigl(X^2+\frac{1+\sqrt 5}2 X+1\Bigr).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to derive this series
The function $\dfrac1{1-x}$, equal to $$1 + x + x^2 + x^3 + \cdots,$$ can also be developed according to the series $$1 + \frac{x}{1 + x} + \frac{1\cdot2\cdot x^2}{(1 + x)(1 + 2x)} + \frac{1\cdot2\cdot3\cdot x^3}{(1 + x)(1 + 2x)(1 + 3x)} + \cdots $$ when $x$ is positive and smaller than $1$.
I know the first series and it is easy to obtain it. But the second series is strange. It is not a power series, not a Taylor series. How does one obtain this series?
| Using @Andrew Chin's comments, we face for the second series
$$S_\infty=\sum_{n=0}^\infty\frac{\Gamma(n+1)\, x^n}{\prod\limits_{m=0}^n(1+mx)}$$
Let $$a_n=\frac{\Gamma(n+1)\, x^n}{\prod\limits_{m=0}^n(1+mx)}=\frac{\Gamma (n+1)}{\left(1+\frac{1}{x}\right)_n}$$ where appear Pochhammer symbols. So
$$S_p=\sum_{n=0}^p a_n=\frac{\frac{\Gamma (p+2)\, \Gamma \left(\frac{1}{x}\right)}{\Gamma
\left(p+1+\frac{1}{x}\right)}-1}{x-1}$$ and, assuming $0< x <1$
$$\lim_{p\to \infty } \, \frac{\Gamma (p+2) \Gamma \left(\frac{1}{x}\right)}{\Gamma
\left(p+1+\frac{1}{x}\right)}=0$$ since, using expansion for large $p$
$$\frac{\Gamma (p+2) \Gamma \left(\frac{1}{x}\right)}{\Gamma
\left(p+1+\frac{1}{x}\right)}=\Gamma \left(\frac{1}{x}\right) p^{1-\frac{1}{x}}\left(1+ \frac{(x-1) (2 x+1)}{2 x^2}\frac 1p+O\left(\frac{1}{p^2}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 3
} |
Solve in $\mathbb{R^{3}}$ : $\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$ Solve in $\mathbb{R^{3}}$ the following system :
$$\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$$
My try but I think not complete :
The system equivalent :
$$\begin{cases}(x+1)(y+1)=20\\(y+1)(z+1)=12\\(x+1)(z+1)=15\end{cases}$$
Then from first equation we have :
$$x+1=\frac{20}{y+1}$$
So third equation give :
$$z+1=3\frac{y+1}{4}$$
Second equation $\implies $
$(y+1)^{2}=16$ then $y=3,-5$
This mean $x=\frac{17}{3}$ or $x=-6$ also $z=3$ or $z=-3$
I'm correct or no ??
| The procedure you followed was correct, including getting the right values of $y$. However, you made mistakes for one value of $x$ (i.e., $\frac{17}{3}$ is not correct) and for both your values of $z$.
However, here's a slightly easier way to deal with the equations instead. You can use the symmetry to multiply your $3$ equations together to get
$$\begin{equation}\begin{aligned}
\left((x+1)(y+1)(z+1)\right)^2 & = 3600 \\
(x+1)(y+1)(z+1) & = \pm 60
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
You can then divide this by each of your $3$ equations to get
$$z + 1 = \pm 3 \implies z = -4, 2 \tag{2}\label{eq2A}$$
$$x + 1 = \pm 5 \implies x = -6, 4 \tag{3}\label{eq3A}$$
$$y + 1 = \pm 4 \implies y = -5, 3 \tag{4}\label{eq4A}$$
This leads to the solutions $(-6,-5,-4)$ and $(4,3,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Nature of the critical point $(0,0)$ of the function $f(x,y)=x^6-2x^2y-x^4y+2y^2$ Consider the function $$f(x,y)=x^6-2x^2y-x^4y+2y^2.$$ The point $(0,0)$ is a critical point. Observe,
\begin{align*}
f_x & = 6x^5-4xy-4x^3y, f_x(0,0)=0\\
f_y & = 2x^2-x^4+4y. f_y(0,0)=0\\
f_{xx} & = 30x^4-4y-12x^2y, f_{xx}(0,0)=0\\
f_{xy} & = 4x-4x^3, f_{xy}(0,0)=0\\
f_{yy} & = 4, f_{yy}=4
\end{align*}
So, in order to determine the nature of the above critical point, we need to check the Hessian at $(0,0)$ which is $0$ and hence the test is inconclusive. $$ H(x,y)= \det \begin{pmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{pmatrix}=\det \begin{pmatrix} 0 & 0 \\ 0 & 4 \end{pmatrix}=0$$So, I tried to see the function on slices like $y=0$ and $y=x$ but nothing worked. So please suggest me how do I find the nature of the critical point in this case?
| You have that
$$
g_a(x)=f(x,ax^2 ) = 2\left( {a^2 - a} \right)x^4 + \left( {1 - a} \right)x^6
$$
With $0<a<1$ the function $g_a(x)$ has a local maximum. With $a>1$ the function has a local minimum. This means that $(0,0)$ is a saddle point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$(a_1+2a_2+\cdots+na_n)(a_1^2+\cdots+a_n^2)\geq \frac49(a_1+\cdots+a_n)^3$ for non-negative real $a_i$ For any integer $n$ and any nonnegative real numbers $a_1,\ldots,a_n$ we have
$$(a_1+2a_2+\cdots+na_n)(a_1^2+\cdots+a_n^2)\geq \frac49(a_1+\cdots+a_n)^3$$
It seems to use Holder inequality, but I can't get the $\frac{4}{9}$.
Thanks.
| Remark1: @fedja gave an excellent solution by proving a continuous version of the OP which implies the OP (see the link @Martin R pointed out in comment for the OP, which was deleted now). I rewrote fedja's solution for the OP (discrete version).
Remark2: Actually, although @fedja proved a continuous version, his proof can be easily modified to corresponding discrete version (essentially the same).
Proof: For any $b, k > 0$, clearly the following inequality holds:
$$b^2 k + \frac{1}{b}(b - b^3k)^{+} \ge 1$$
where $(x)^{+} = \max(0, x)$. Thus, for any $b>0$, we have
\begin{align}
\sum a_k &\le \sum \Big(b^2 k + \frac{1}{b}(b - b^3k)^{+}\Big)a_k\\
&= b^2\sum k a_k + \frac{1}{b} \sum (b - b^3k)^{+}a_k\\
&\le b^2\sum k a_k + \frac{1}{b}\sqrt{\sum ((b - b^3k)^{+})^2}\sqrt{\sum a_k^2}\\
&\le b^2\sum k a_k + \frac{1}{b}\sqrt{\frac{1}{3}}\sqrt{\sum a_k^2}\tag{1}
\end{align}
where we have used the following result (the proof is given later)
$$\sum ((b - b^3k)^{+})^2 \le \frac{1}{3}.\tag{2}$$
Since (1) holds for any $b > 0$, by letting
$$b = \sqrt[3]{\frac{1}{2}\sqrt{\frac{1}{3}}\sqrt{\sum a_k^2} \frac{1}{\sum ka_k}},$$
we have
$$\sum a_k \le \frac{3}{2}\sqrt[3]{\frac{2}{3}}\sqrt[3]{\sum k a_k }\sqrt[3]{\sum a_k^2}.$$
The desired result follows. We are done.
$\phantom{2}$
Proof of (2): (by @Martin R) We have
$$\sum_{k=1}^n ((b-b^3k)^{+})^2 \le \sum_{k=1}^{\lfloor 1/b^2\rfloor} (b-b^3k)^2
\le \int_0^{1/b^2} (b-b^3x)^2 dx = \frac{1}{3}$$
where we have used the fact that $x\mapsto (b-b^3x)^2$ is decreasing on $[0,\frac{1}{b^2}]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Distance from point $D$ to $BI$
I found that : $BI = 2\sqrt 2, DI = \sqrt 5, DB = 2\sqrt 2$. It means $\triangle BDI$ is isosceles triangle. Lets say $B$ to $DI$ is $a = \sqrt {BI^2 - ( \frac 12 DI)^2} = \sqrt{\frac{27}{4}}$. Lets say $t$ is distance from D to BI.
$t. BI = a. DI$
$t = \frac{\sqrt{27}}{4}$.
But why my answer not in the options?
| Note that $BI = \sqrt5$. So, △BDI is isosceles with DI = BI instead. Then, I to DB is
$a = \sqrt {BI^2 - ( \frac 12 DB)^2} = \sqrt{3}$ and
$$t = \frac{DB}{BI}a = \frac{2\sqrt2}{\sqrt5}\cdot\sqrt3=\frac{\sqrt{24}}5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
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Closed form of recurrence relation $a_{n}=a_{n-1}(a_{n-2}^{2}-a_{0})-a_{1}$ Given the following sequence:
$$\begin{cases}a_{0}=2\\a_{1}=\frac{5}{2}\\a_{n}=a_{n-1}(a_{n-2}^{2}-a_{0})-a_{1}\end{cases}$$
Prove that
$$a_{n}=2^{\frac{2^{n}-(-1)^{n}}{3}}+2^{\frac{(-1)^{n}-2^{n}}{3}}$$
For all natural numbers $n$.
My try: (using induction)
Let $b_{n}=\frac{2^{n}-(-1)^{n}}{3}$
For $n=0$: correct $\checkmark$
Now for $n+1$:
$$
\begin{aligned}
a_{n+1} &= a_{n}(a_{n-1}^{2}-a_{0})-a_{1})=(2^{b_{n}}+2^{-b_{n}})(2^{2b_{n}}+2^{-2b_{n}}+2) \\
&=2^{3b_{n}}+2^{-3b_{n}}+2^{b_{n}}+2^{-b_{n}}+2^{b_{n}+1}+2^{-b_{n}+1}
\end{aligned}
$$
How do I proceed from here?
| You want to proof that $a_n = f(n)$ for any $n\inℕ$ with
$$ f(n) := 2^{\frac{2^{n}-(-1)^{n}}{3}}+2^{\frac{(-1)^{n}-2^{n}}{3}}.$$
Check of two (!) initial conditions:
*
*$f(0) = 2 = a_0 \checkmark$
*$f(1) = \frac52 = a_1 \checkmark$
Inductive step:
$$\begin{aligned}
a_{n+1} &= a_n (a_{n-1}^2 - a_0) - a_1\\
&= f(n) (f(n-1)^2 - f(0)) - f(1)\\
&= (2^{b_n} + 2^{-b_n}) \cdot ((2^{b_{n-1}} + 2^{-b_{n-1}})^2 - 2) - \frac52\\
&= (2^{b_n} + 2^{-b_n}) \cdot (2^{2b_{n-1}} + 2^{-2b_{n-1}}) - \frac52 \\
&= 2^{b_n+2b_{n-1}} + 2^{b_n-2b_{n-1}} + 2^{-b_n+2b_{n-1}} + 2^{-b_n-2b_{n-1}} - \frac52 \end{aligned}$$
Simple computations show that:
$$\begin{aligned}
b_n+2b_{n-1} &= \frac13(2^{n+1} + (-1)^n) = b_{n+1}\\
b_n-2b_{n-1} &= (-1)^{n+1}.
\end{aligned}$$
Thus
$$a_{n+1} = f(n+1) + 2^{(-1)^{n+1}} + 2^{(-1)^n} - \frac52$$
With the equality (for $n\in\mathbb N$)
$$2^{(-1)^{n+1}} + 2^{(-1)^n} = \frac52$$
we can finish the proof:
$$a_{n+1} = f(n+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Without using characteristic equation, how can I show genenal formula of this sequence? I know that, solution of the sequence $x_n$ so that
$$x_{n+2}-2 x_{n+1}= 2 x_{n+1} - 4 x_{n}$$
is $x_n = (A+ Bn)2^n$. I want to show this without using characteristic equation. I tried
\begin{align*}
x_{n+2}-2 x_{n+1} &= 2 x_{n+1} - 4 x_{n} \\
&= 2(x_{n+1}-2x_{n}) \\
&= 2^2(x_{n}-2x_{n-1}) \\
&= \ldots \ldots \\
&= 2^n (x_2 - 2x_1)
\end{align*}
From here, I cannot get the solution. Without using characteristic equation, how can I show general formula of this sequence?
| $\begin{align}x_{n+2}=2x_{n+1}+2^n(x_2-2x_1)&&2x_{n+1}=2[2x_n+2^{n-1}(x_2-2x_1)]=4x_{n}+2^n(x_2-2x_1)\\
x_{n+2}=4x_n+2^n(x_2-2x_1)2&&4x_n=4[2x_{n-1}+2^{n-2}(x_2-2x_1)]=8x_{n-1}+2^n(x_2-2x_1)\\
x_{n+2}=8x_{n-1}+2^n(x_2-2x_1)3&&8x_{n-1}=8[2x_{n-2}+2^{n-3}(x_2-2x_1)]=16x_{n-2}+2^n(x_2-x_1)\end{align}$
$\dots$
$x_{n+2}=2^nx_2+2^n(x_2-2x_1)n$
i.e., $x_n=2^{n-2}x_2+2^{n-2}(x_2-2x_1)(n-2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501053",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
Help proving this inequality Give $x$ , $y$ , with $x,y \in \Re$ , and $x^2 + y^2 = 1$ , show that .
$$ \left(\frac{1}{1+x^2}\right) + \left(\frac{1}{1+y^2}\right) + \left(\frac{1}{1+xy}\right) \ge \left(\frac{3}{1+((x+y)/2))^2}\right)$$
This is my approach:
from $x^2 + y^2 = 1$ , we can see that $xy \le \frac{1}{2}$ , which says $1+xy$ is always positive
Since we know $x^2 \ge 0$ and $y^2 \ge 0$ , we can say that $x^2 + 1 $ and $y^2 +1 $ , the denominators of the expression are always positive .
Now evaluating $\left(\frac{3}{1 + \left(\frac{x+y}{2}^2\right) }\right)$ , we can see it has a lower bound of 2 , following from $xy \le \frac{1}{2}$ .
Thus we are left to show that $$ \left(\frac{1}{1+x^2}\right) + \left(\frac{1}{1+y^2}\right) + \left(\frac{1}{1+xy}\right) \ge 2$$
Which follows directly from Titu's lemma .
Now this is wrong , since this concludes nothing .Can anyone give me some hints on which way to proceed ?
| Let $xy=t$.
Thus, by AM-GM $\frac{1}{2}\geq t\geq-\frac{1}{2}$ and we need to prove that $$\frac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}+\frac{1}{1+xy}\geq\frac{3}{1+\frac{x^2+y^2+2xy}{4}}$$ or
$$\frac{3}{2+t^2}+\frac{1}{1+t}\geq\frac{3}{1+\frac{1+2t}{4}}$$ or
$$(1-2t)(5t^2+3t+1)\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
show that $x_1^2+x_2^2+\cdots+x_{2020}^2\geqslant x_1x_2+x_2x_3+\cdots+x_{2020}x_1$ how can i use AM-GM inequality to show that $$x_1^2+x_2^2+\cdots+x_{2020}^2\geqslant x_1x_2+x_2x_3+\cdots+x_{2020}x_1 ?$$
Where $x_1,x_2,\ldots,x_n$
are any real numbers.
to get an idea I tried to do for particular cases with $n=3,4.$
for $n=3$, I could use AM-GM doing $$x_1^2+x_2^2+x_3^2=\frac{x_1^2+x_2^2}{2}+\frac{x_2^2+x_3^2}{2}+\frac{x_3^2+x_1^2}{2}\geqslant x_1x_2+x_2x_3+x_3x_1.$$
for $n=4$ couldn't with that same idea.
but it is possible to show
developing the inequality below $$(x_1-x_2)^2+(x_2-x_3)^2+(x_3-x_4)^2+(x_4-x_1)^2 \geqslant 0.$$
| The idea is the same:
$$
x_1x_2 + x_2x_3 + \cdots + x_{2020}x_1 \leq \frac{x_1^2 + x_2^2}{2} + \frac{x_2^2 + x_3^2}{2} + \cdots + \frac{x_{2020}^2 + x_1^2}{2} = x_1^2 + x_2^2 + \cdots + x_{2020}^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Different result of counting number between 1 and 2000 which the sum of all the digits is 25. How many number between 1 and 2000 which the sum of all the digits is 25?
I have tried to answer in two ways.
First way
Let the digits is $A$, $B$, $C$, and $D$.
$A+B+C+D=25$ with $A\leq 1$, $B\leq 9$, $C\leq 9$, and $D\leq 9$.
I am using transformation $E=1-A$, $F=9-B$, $G=9-C$, $H=9-D$. We have new equation
$$E+F+G+H=28-25=3,$$
with $E\geq 0$, $F\geq 0$, $G\geq 0$, and $H\geq 0$.
So, we have the number between 1 and 2000 which the sum of all the digits is 25 is
$$
\begin{pmatrix}
4+3-1\\
3
\end{pmatrix}
=
\begin{pmatrix}
6\\
3
\end{pmatrix}
=
20 \text{ numbers}.
$$
Second way
Compute 3 digits number which the sum is 25.
Let $A=0$.
$B+C+D=25$ with $B\leq 9$, $C\leq 9$, and $D\leq 9$
I am using transformation $F=9-B$, $G=9-C$, $H=9-D$. We have new equation
$$F+G+H=27-25=2,$$
with $F\geq 0$, $G\geq 0$, and $H\geq 0$.
So, we have the number 3 digits which the sum of all the digits is 25 is
$$
\begin{pmatrix}
3+2-1\\
2
\end{pmatrix}
=
\begin{pmatrix}
4\\
2
\end{pmatrix}
=
6\text{ numbers}.
$$
Let $A=1$.
$1+B+C+D=25\iff B+C+D=24$ with $B\leq 9$, $C\leq 9$, and $D\leq 9$
I am using transformation $F=9-B$, $G=9-C$, $H=9-D$. We have new equation
$$F+G+H=27-24=3$$
with $F\geq 0$, $G\geq 0$, and $H\geq 0$.
So, we have the number 4 digits (1000-1999) which the sum of all the digits is 25 is
$$
\begin{pmatrix}
3+3-1\\
3
\end{pmatrix}
=
\begin{pmatrix}
5\\
3
\end{pmatrix}
=
10\text{ numbers}.
$$
So, we have the number between 1 and 2000 which the sum of all digits is 25 is
$$6+10=16\text{ numbers}.$$
My question
Why the result first way and second way is different?
I try to make list:
1996, 1987, 1978, 1969, 1897, 1888, 1879, 1798, 1789, 1699, 997, 988, 979, 898, 889, 799. There are 16 numbers.
Is the first way wrong?
| The first way is incorrect.
Every combination of $A,B,C,D$ maps to a valid combination of $E,F,G,H$, such as $A=0, E=1$ and $A=1, E=0$. However, every combination of $E,F,G,H$ does not map to a combination of $A,B,C,D$: for example, $(E,F,G,H)=(3,0,0,0)$ leads to $A = -2$.
This means that you are counting solutions that are not valid in the original problem. In other words, you are overcounting in the first method.
In fact, there are four overcountings because when $E=3$, there is only one possibility for $F, G,H$, and when $E=2$ one of $F,G,H$ can be equal to $1$. When $E=1$ there is no issue. Thus there are four cases where you overcount.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Does the series converge, problem with $(-1)^n$, but Leibnitz impossible. Hi my problem is connected with this series
$$
\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt[3]n + (-1)^{n(n+1)/2}}
$$
I was trying to sum it by $4$ elements, but it didn't end up well.
I don't know how to deal with this problem, Leibnitz, Dirichlet, Abel is pointless here. Can you please help me?
| We consider the series
$$
\sum\limits_{n = 4}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}
{{\sqrt[3]{n} + \left( { - 1} \right)^{\frac{{n\left( {n + 1} \right)}}
{2}} }}}
$$
instead your series. If we prove that this series is convergent then your series is convergent also. We can rewrite the series as
$$
\sum\limits_{k = 1}^\infty {\left( {a_{4k} + a_{4k + 1} + a_{4k + 2} + a_{4k + 3} } \right)}
$$
where
$$
a_{4k} = \frac{1}
{{\sqrt[3]{{4k}} + 1}},a_{4k + 1} = - \frac{1}
{{\sqrt[3]{{4k + 1}} - 1}},a_{4k + 2} = \frac{1}
{{\sqrt[3]{{4k + 2}} - 1}},a_{4k + 3} = - \frac{1}
{{\sqrt[3]{{4k + 3}} + 1}}
$$
Now, write down
$$
s_1 \left( k \right) = \frac{1}
{{\sqrt[3]{{4k}} + 1}} - \frac{1}
{{\sqrt[3]{{4k + 1}} - 1}}
$$
and rewrite is as function of $\frac{1}{k}$. After that using the Taylor series also, you can develop this function in powers of $\frac{1}{k}$. You will get
$$
s_1 \left( k \right) = - \frac{{\left( {\frac{1}
{k}} \right)^{2/3} }}
{{\sqrt[3]{2}}} - \frac{{5\left( {\frac{1}
{k}} \right)^{4/3} }}
{{12\;2^{2/3} }} + \frac{{\left( {\frac{1}
{k}} \right)^{5/3} }}
{{12\sqrt[3]{2}}} + O\left( {\frac{1}
{{k^2 }}} \right)
$$
To be more clear, you have that
$$
\begin{gathered}
s_1 \left( k \right) = \frac{1}
{{\sqrt[3]{{4k}} + 1}} - \frac{1}
{{\sqrt[3]{{4k + 1}} - 1}} \hfill \\
\hfill \\
= \frac{{\sqrt[3]{{4k + 1}} - \sqrt[3]{{4k}} - 2}}
{{\left[ {\sqrt[3]{{4k}} + 1} \right]\left[ {\sqrt[3]{{4k + 1}} - 1} \right]}} \hfill \\
\hfill \\
= \frac{{\sqrt[3]{{4k}}\left[ {\sqrt[3]{{1 + \frac{1}
{{4k}}}} - 1} \right] - 2}}
{{\left[ {\sqrt[3]{{4k}} + 1} \right]\left[ {\sqrt[3]{{4k}}\left( {\sqrt[3]{{1 + \frac{1}
{{4k}}}}} \right) - 1} \right]}} \hfill \\
\end{gathered}
$$
and then, from here, you can use the Taylor expansion with
$$
{\sqrt[3]{{1 + \frac{1}
{{4k}}}}}
$$
Consider now
$$
s_2 \left( k \right) = \frac{1}
{{\sqrt[3]{{4k + 2}} - 1}} - \frac{1}
{{\sqrt[3]{{4k + 3}} + 1}}
$$
and repeat the above procedure. You will get
$$
s_2 \left( k \right) = \frac{{\left( {\frac{1}
{k}} \right)^{2/3} }}
{{\sqrt[3]{2}}} + \frac{{7\left( {\frac{1}
{k}} \right)^{4/3} }}
{{12\;2^{2/3} }} - \frac{{5\left( {\frac{1}
{k}} \right)^{5/3} }}
{{12\sqrt[3]{2}}} + O\left( {\frac{1}
{{k^2 }}} \right)
$$
Therefore
$$
s\left( k \right) = s_1 \left( k \right) + s_2 \left( k \right) = \frac{{\left( {\frac{1}
{k}} \right)^{4/3} }}
{{6 \cdot \;2^{2/3} }} - \frac{{\left( {\frac{1}
{k}} \right)^{5/3} }}
{{3\sqrt[3]{2}}} + O\left( {\frac{1}
{{k^2 }}} \right)
$$
so that
$$
s\left( k \right) = s_1 \left( k \right) + s_2 \left( k \right) = \frac{A}
{{k^{4/3} }} + O\left( {\frac{1}
{{k^{5/3} }}} \right)
$$
and the considered series is convergent as well as yours.
By the way: the above expansions which contain non integer powers of $\frac{1}{k}$ are called Puiseux expansions.
| {
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"url": "https://math.stackexchange.com/questions/3505382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of solutions via generating function How to find generating function for number of solutions
$x_1+x_2+x_3=n$ in set of positive integers such that $x_1 \ge x_2 \ge x_3$ and $x_1<x_2+x_3$.
|
We are looking for a generating function $G(z)$ where the coefficient of $z^n$ gives the wanted number of solutions. We want to find a proper representation for
\begin{align*}
G(z)=\sum_{{{x_1,x_2,x_3\geq 1}\atop{x_1\geq x_2\geq x_3}}\atop{x_1<x_2+x_3}}z^{x_1+x_2+x_3}\tag{1}
\end{align*}
In order to simplify $x_1\geq x_2\geq x_3\geq 1$ (and also $x_1,x_2,x_3\geq 1$) we write
\begin{align*}
x_3&=y_3+1\qquad\qquad\ \, y_3\geq 0\\
x_2&=y_2+x_3\qquad\qquad y_2\geq 0\tag{2}\\
x_1&=y_1+x_2\qquad\qquad y_1\geq 0
\end{align*}
We obtain from (2):
\begin{align*}
x_1&<x_2+x_3\\
y_1+y_2+y_3+1&<(y_2+y_3+1)+(y_3+1)\\
y_1&<y_3+1\\
\color{blue}{y_1}&\color{blue}{\leq y_3}\tag{3}\\
\\
x_1+x_2+x_3&=(y_1+y_2+y_3+1)+(y_2+y_3+1)+(y_3+1)\\
&\,\,\color{blue}{=y_1+2y_2+3y_3+3}\tag{4}
\end{align*}
With the help of (2) and (3) we can write $G(z)$ as
\begin{align*}
G(z)=\sum_{{y_1,y_2,y_3\geq 0}\atop{y_1\leq y_3}}z^{y_1+2y_2+3y_3+3}\tag{5}
\end{align*}
In order to simplify $y_1\leq y_3$ in (5) we write
\begin{align*}
y_3&=t_3+y_1\qquad\quad t_3\geq 0\\
y_2&=t_2\qquad\qquad\quad\, t_2\geq 0\tag{6}\\
y_1&=t_1\qquad\qquad\quad\, t_1\geq 0
\end{align*}
and we obtain from (6)
\begin{align*}
y_1+2y_2+3y_3+3&=t_1+2t_2+3(t_3+t_1)+3\\
&\,\,\color{blue}{=4t_1+2t_2+3t_3+3}\tag{7}
\end{align*}
We obtain from (5) - (7)
\begin{align*}
\color{blue}{G(z)}&=\sum_{t_1,t_2,t_3\geq 0}z^{4t_1+2t_2+3t_3+3}\\
&=z^3\sum_{t_2\geq 0}z^{2t_2}\sum_{t_3\geq 0}z^{3t_3}\sum_{t_1\geq 0}z^{4t_1}\\
&\,\,\color{blue}{=\frac{z^3}{(1-z^2)(1-z^3)(1-z^4)}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
If real $a$ and $b$ satisfy $17(a^2+b^2)-30ab-16=0$, then find the maximum value of $\sqrt{16a^2+4b^2-16ab-12a+6b+9}$ If $ a $ and $ b $ are real numbers that satisfy the equation $ 17 (a ^ 2 + b ^ 2) -30ab-16 = 0 $, the maximum value of the expression$$\sqrt{16a^2+4b^2-16ab-12a+6b+9}$$goes:
a) $1$
b) $3$
c) $5$
d) $7$
e) $9$
Can anyone give a hint?
Attempt: I tried Variation, Jacobian, Hessian matrices.
| The hint.
Let $2a-b=k.$
Thus, we need to find a maximal value of $\sqrt{4k^2-6k+9}.$
Now, find all values of $k$, for which the equation
$$k^2(17(a^2+b^2)-30ab)=16(2a-b)^2$$ has solutions and choose from these values such that $\sqrt{4k^2-6k+9}$ will get a maximal value.
I got that it happens for $k=-\frac{5}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $\frac1a+\frac3b+2(a+\sqrt{b^2+8})>11$ Hello everyone: I want to prove that the following holds for all $a,b>0$: $$\frac1a+\frac3b+2(a+\sqrt{b^2+8})>11.$$
My attempt: We notice by AM-GM that $\frac1a+2a\geq 2\sqrt 2$ for all $a>0$. So we only need to prove that $$\frac3b+2\sqrt{b^2+8}>11-2\sqrt{2}.$$
This seems to be true numerically, but I don't know how to prove it.
| Firstly, it should be $$\frac{3}{b}+4\sqrt{b^2+8}\geq11-2\sqrt2,$$ which is true by C-S and AM-GM:
$$\frac{3}{b}+4\sqrt{b^2+8}=\frac{3}{b}+\frac{4}{3}\sqrt{(1+8)(b^2+8)}\geq$$
$$\geq\frac{3}{b}+\frac{4(b+8)}{3}\geq4+\frac{32}{3}>11-2\sqrt2.$$
Secondly, we can prove your second inequality by the similar way:
$$\frac{3}{b}+2\sqrt{b^2+8}=\frac{3}{b}+\frac{2}{\sqrt{11}}\sqrt{(3+8)(b^2+8)}\geq$$
$$\geq\frac{3}{b}+\frac{2(\sqrt3b+8)}{\sqrt{11}}\geq2\sqrt{\frac{6\sqrt3}{\sqrt{11}}}+\frac{16}{\sqrt{11}}>11-2\sqrt2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving differential equation by separating variables Solve the following differential equation:
$$\frac yx \frac {dy}{dx}=\sqrt {1+x^2+y^2+x^2y^2}$$
I have tried like this:
$$\frac yx \frac{dy}{dx}=\sqrt {1+x^2+y^2(1+x^2)}\\
\implies \frac yx \frac {dy}{dx}=\sqrt {(1+x^2)(1+y^2)}\\
\implies \int \frac {ydy}{\sqrt {1+y^2}}=\int \frac {xdx}{\sqrt{1+x^2}}\\
\implies \log \sqrt{1+y^2}=\log \sqrt {1+x^2}+c_1\\
\implies \frac {1+y^2}{1+x^2}=c\\$$
But in my book the answer is:
$$(1+x^2)^{3/2}-3\sqrt {1+y^2}=c$$
I can't understand how they did it,please check this..
| $$\frac yx \frac {dy}{dx}=\sqrt {(1+x^2)(1+y^2)}$$
$$\implies \int \frac {ydy}{\sqrt {1+y^2}}=\int {x}{\sqrt{1+x^2}}~dx$$
$$\implies \sqrt {1+y^2}=\dfrac{\left(x^2+1\right)^\frac{3}{2}}{3}+c$$where $c$ is integrating constant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Check Proof by Contradiction Please check the following proof.
Result to Prove: If $a$ and $b$ are odd integers, then $4\not| (a^2+b^2)$
Assume, to the contrary, that there exists two odd integers $a$ and $b$ such that $4|(a^2+b^2)$.
Then, $(a^2+b^2)=4n$ for some $n\in\Bbb Z$.
Let $a=2x+1$ and $b=2y+1$ for some $x,y\in \Bbb Z$.
$a^2+b^2=(2x+1)^2+(2y+1)^2=4x^2+4y^2+4x+4y+2=2(2x^2+2y^2+2x+2y+1)=4n$.
$\implies 2x^2+2y^2+2x+2y+1=4n$
$\implies 2(x^2+y^2+x+y)+1=2n$.
Since $(x^2+y^2+x+y), n\in\Bbb Z$, $2(x^2+y^2+x+y)+1$ is odd and $2n$ is even, which is a contradiction.
| $2(2x^2+2y^2+2x+2y+1)=4n$ does not imply $2x^2+2y^2+2x+2y+1=4n$, but $2x^2+2y^2+2x+2y+1=2n$. Other than that, it's fine.
| {
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Find a, b, c and d of a function $f(x) = ax^4 + bx^3 + cx^2 + dx + 1$, given values f(1), f(2), f(3), ... I have a (very-likely) 4-grade function, and I know all the values, but I am having problems to find the formula of the function itself.
The number without the x is a one for sure.
The functions points are:
(0 | 1)
(1 | 9)
(2 | 41)
(3 | 129)
(4 | 321)
(5 | 681)
(6 | 1289)
I have been calculated with the following starting-grid:
$$
\begin{cases}
a + b + c + d + 1 = 9\\
16a + 8b + 4c + 2d + 1 = 41\\
81a + 27b + 9c + 3d + 1 = 129\\
256a + 64b + 16c + 4d + 1 = 321\\
\end{cases}
$$
I came to a = $2\over3$, b = $4\over3$, c = $16\over3$, d = $2\over3$, yet it's another function.
I have over 2 pages of calculations and handwriting only I can read, so my question is: Can I calculate it much faster/ simpler/ cleaner? - Checking through every addition to find one mistake just seems lame.
Is there an online (free) calculator for stuff like these?
PS. Be nice to add the tag find-formula :)
| Given $7$ distinct data points , we can use polynomial interpolation in the Lagrange from such that:
$\color{red}{\left(0,1\right)},\color{green}{\left(1,9\right)},\color{pink}{\left(2,41\right)},\color{blue}{\left(3,129\right)},\left(4,321\right),\color{olive}{\left(5,681\right)},\color{orange}{\left(6,1289\right)}$
$$\sum_{j=0}^{6}y_{j}l_{j}\left(x\right)=\sum_{j=0}^{6}y_{j}\prod_{0\le m\le6}^{ }\frac{x-x_{m}}{x_{i}-x_{m}}$$$$=\frac{x-1}{0-1}\frac{x-2}{0-2}\frac{x-3}{0-3}\frac{x-4}{0-4}\frac{x-5}{0-5}\frac{x-6}{0-6}\color{red}{\cdot1}+$$$$\frac{x-0}{1-0}\frac{x-2}{1-2}\frac{x-3}{1-3}\frac{x-4}{1-4}\frac{x-5}{1-5}\frac{x-6}{1-6}\color{green}{\cdot9}+$$$$\frac{x-0}{2-0}\frac{x-1}{2-1}\frac{x-3}{2-3}\frac{x-4}{2-4}\frac{x-5}{2-5}\frac{x-6}{2-6}\color{pink}{\cdot41}+$$$$\frac{x-0}{3-0}\frac{x-1}{3-1}\frac{x-2}{3-2}\frac{x-4}{3-4}\frac{x-5}{3-5}\frac{x-6}{3-6}\color{blue}{\cdot129}+$$$$\frac{x-0}{4-0}\frac{x-1}{4-1}\frac{x-2}{4-2}\frac{x-3}{4-3}\frac{x-5}{4-5}\frac{x-6}{4-6}\cdot321+$$$$\frac{x-0}{5-0}\frac{x-1}{5-1}\frac{x-2}{5-2}\frac{x-3}{5-3}\frac{x-4}{5-4}\frac{x-6}{5-6}\color{olive}{\cdot681}+$$$$\frac{x-0}{6-0}\frac{x-1}{6-1}\frac{x-2}{6-2}\frac{x-3}{6-3}\frac{x-4}{6-4}\frac{x-5}{6-5}\color{orange}{\cdot1289}$$
Another way would be using Horner's method.
| {
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Trouble finding limits at $-\infty$. I'm being asked to evaluate $\lim\limits_{x\rightarrow\infty}f(x)$ and $\lim\limits_{x\rightarrow -\infty}f(x)$ for the following funtion: $f(x)=\frac{4x^3+2}{x^3+\sqrt{4x^6+3}}$.
For the first limit I proceed as follows:
$\begin{align*}
\lim\limits_{x\rightarrow\infty}f(x)&=\lim\limits_{x\rightarrow\infty}\frac{4x^3+2}{x^3+\sqrt{4x^6+3}}\\&=\lim\limits_{x\rightarrow\infty}\frac{(4+\frac{2}{x^3})}{(1+\sqrt{4+\frac{3}{x^6})}}\\&=\frac{4}{3}
\end{align*}$
This is the horizontal asymptote as $x\rightarrow\infty$, however, when I work through it again for $-\infty$, I get the same answer, but I know that as $x\rightarrow -\infty$, $f(x)$ approaches $-4$. Obviously there is some hole in my understanding. Any clarification would be greatly appreiciated! Thank you.
| When $x<0$ we have $x^{-3} \sqrt{4x^6+3} = -\sqrt{x^{-6}(4x^6+3)} = - \sqrt{4 + 3x^{-6}}$.
| {
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Divide $ yy' = (y + 1)^2 $ by $ (y + 1)^2 $ I was reading from Ordinary Differential Equations (Lesson 4C page 36) and came across this question:
Find a 1-parameter family of solutions of the differential equation
$$ (a) \quad yy' = (y + 1)^2 $$
and the particular solution for which $\quad y(2) = 0$
The first step was to divide $ (a) $ by $ (y + 1)^2 $
and they had
$$ \int{\frac{y}{(y + 1)^2} dy } = \int dx \quad,\quad y \neq -1$$
but when I was trying to divide $ (a) $ by $ (y + 1)^2 $
I had
$$ \frac{yy'}{(y+1)^2} = \frac{(y+1)^2}{(y+1)^2} $$
$$ => \quad \frac{yy'}{(y+1)^2} = 1$$
| In differential equations and calculus it is often assumed that
$$y'=\frac{dy}{dx}$$
where $y$ is the dependent variable and $x$ is the independent variable. In your case, given $y \neq 1$
$$\frac{yy'}{(y+1)^2} = \frac{(y+1)^2}{(y+1)^2}=1 \implies \frac{y}{(y+1)^2}\frac{dy}{dx}=1 \implies \frac{y}{(y+1)^2}\,dy=dx$$
by which you can integrate
$$\int \frac{y}{(y+1)^2}\,dy=\int dx$$
by observing that the left-hand side has the following partial fraction decomposition
$$\frac{y}{(y+1)^2}=\frac{1}{y+1}-\frac{1}{(y+1)^2}$$
so that
$$\int \frac{y}{(y+1)^2}\,dy=\int \left(\frac{1}{y+1}-\frac{1}{(y+1)^2}\right)dy=\frac{1}{y+1}+\ln|y+1|+C$$
therefore the implicit solution is
$$\frac{1}{y+1}+\ln|y+1|=x+C$$
| {
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If $A$ and $B$ are linear transformations on a finite-dimensional inner product space, and if $\textbf{0} \leq A \leq B$, then det $A \leq$ det $B$. Exercise 12 from SEC. 82 of Finite-Dimensional Vector Spaces - 2nd Edition by Paul R. Halmos.
If $A$ and $B$ are linear transformations on a finite-dimensional inner product space, and if $\textbf{0} \le A \leq B$, then $\det A \leq \det B$.
Notation: $0 \le A$ means that $A$ is positive.
If $\det B = 0$, choose $x \in \ker B$, then $0 \le ((B-A)x,x) = (Bx, x) - (Ax, x) = -(Ax, x) \le 0$. Therefore $\ker B \subseteq \ker A$ and $0 = det A \le det B = 0$.
How to prove for the case when $B$ is invertible? Intuitively, I feel that since both $A$ and $B$ are positive their proper values are also positive and $B$'s proper values should be greater the $A$'s and same for determinants.
| A verbose ("for dummies") version of Omnomnomnom's and Andreo's proof is as follows.
We have two cases: (a) $B$ is non-invertible, and (b) $B$ is invertible.
Case (a): since $B$ is not invertible, det $B = 0$. Also, the null-space of $B$ has some non-zero vector $x$. It follows from the positivity of $(B-A)$ that $\big\langle(B-A)x, x \big\rangle \geq 0$
$\implies \langle Bx-Ax, x\rangle \geq 0 \implies \langle Bx, x\rangle - \langle Ax, x\rangle \geq 0\\ \implies 0 - \langle Ax, x\rangle \geq 0 \implies \langle Ax, x\rangle = 0$ [$\because \langle Ax, x\rangle \geq 0$ due to the positivity of $A$]
$\implies \langle \sqrt A \sqrt Ax, x\rangle = 0$ [$\because$ positivity of $A$ implies that it has a positive square-root]
$\implies \langle \sqrt Ax, {\sqrt A}^*x \rangle = 0$ $\implies \langle \sqrt Ax, \sqrt Ax \rangle = 0$ [$\because \sqrt A$ is positive, it is self-adjoint too] $\implies \left \Vert \sqrt Ax \right \Vert^2 = 0 \implies \sqrt Ax = 0$ $\implies \sqrt A \sqrt Ax = 0 \implies Ax = 0$
$\implies$ (non-zero) $x$ is in the null-space of $A \implies A$ is not invertible $\implies$ det $A = 0$
$\implies$ det $A =$ det $B$ (since both det are $0$).
Case (b): in finite dimensions, the invertibility of $B \geq \textbf{0}$ implies that $B > \textbf{0}$ (positive-definite). Thus, $B$ has a positive-definite square root, say $B^\frac{1}{2}$ which is also invertible. If $B^{-\frac{1}{2}}$ is the inverse of $B^\frac{1}{2}$, then $B^{-\frac{1}{2}}$ is also positive-definite.
Now, because $B-A$ is given as positive, $C^*(B-A)C$ is positive for an arbitrary $C$, i.e, $C^*(B-A)C \geq \textbf{0}$. If we let $C = B^{-\frac{1}{2}}$, then we have $C^* = {B^{-\frac{1}{2}}}^* = B^{-\frac{1}{2}}$. Substituting $C$ and $C^*$ in the inequality on hand yields $\textbf{1}- B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \geq \textbf{0}$, i.e., $B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \leq \textbf{1}$. We observe that $B^{-\frac{1}{2}}AB^{-\frac{1}{2}}$ is positive transformation, and thus unitarily diagonalizable. This fact, together with the inequality $B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \leq \textbf{1}$, reveals that each Eigenvalue (diagonal entry in the diagonal form) of $B^{-\frac{1}{2}}AB^{-\frac{1}{2}}$ is $\leq 1$. This implies that det $B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \leq 1$. Further, because det $CD$ = det $C \cdot$ det $D$ holds for invertible transformations $C$ and $D$, it follows that det $B^{-\frac{1}{2}}\cdot$ det $A \cdot$ det $B^{-\frac{1}{2}} \leq 1$ where det $B^{-\frac{1}{2}} \neq 0$. This fact, together with the finding that det $B^{-\frac{1}{2}}\cdot$ det $B \cdot$ det $B^{-\frac{1}{2}} =$ det $B^{-\frac{1}{2}}BB^{-\frac{1}{2}} =$ det $B^{-\frac{1}{2}} B^\frac{1}{2} B^\frac{1}{2} B^{-\frac{1}{2}} = $ det $\textbf{1} = 1$, informs that det $A \leq$ det $B$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $n $ such that $n^2+n+7$ devided by 13. An idea please Find all positive integers $n $ such that $n^2+n+7$ devided by 13.
An idea please
| I'm assuming you want the value to be divisible by 13.
Since the values mod 13 (remainder when divided by 13) repeat every 13, we just need to test 1, 2, 3, 4, 5, 6, 7, 8, 9, ..., 13.
1: 1 + 1 + 7 = 9
2: 4 + 2 + 7 = 13 works
3: 9 + 3 + 7 = 19
4: 16 + 4 + 7 = 27
5: 25 + 5 + 7 = 37
6: 36 + 6 + 7 = 49
7: 49 + 7 + 7 = 63
8: 64 + 8 + 7 = 79
9: 81 + 9 + 7 = 97
10: 100 + 10 + 7 = 117 works
11: 121 + 11 + 7 = 139
12: 144 + 12 + 7 = 163
13: 169 + 13 + 7 = 189
So, the values that work should be:
2, 15, 28, 41, 54, etc.
and 10, 23, 36, 49, 62, etc.
Alternate solution:
$n^2+n+7$ in mod 13 is just $n^2+n-6$. This factors as $(n+3)(n-2)$, and then $n+3$ or $n-2$ is a multiple of 13, giving 10 and 2. We can continue as above.
| {
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Condition for $\frac{1}{x^2}+\frac{1}{2x^4}+\frac{1}{3x^6}+ . . . =\frac{2}{y}+\frac{2}{3y^3}+\frac{2}{5y^5}+ . . .$ I saw this problem in a 'classical algebra' textbook:
If $y=2x^2-1$ prove that under certain condition we have:
$\frac{1}{x^2}+\frac{1}{2x^4}+\frac{1}{3x^6}+ . . . =\frac{2}{y}+\frac{2}{3y^3}+\frac{2}{5y^5}+ . . .$
My attempt:
$\ln (1-\frac{1}{x^2})=-(\frac{1}{x^2}+\frac{1}{2x^4}+\frac{1}{3x^6} + . . .)$
$2\ln(1+\frac{1}{y})=-(\frac{1}{y^2}+\frac{1}{2y^4}+\frac{1}{3y^6} + . . .) + (\frac{2}{y}+\frac{2}{3y^3}+\frac{2}{5y^5}+ . . .)$
Therefore:
$ \ln (1+\frac {1}{y})^2+\ln(1-\frac{1}{x^2})+\ln(1-\frac{1}{y^2})=0$
$\ln (1+\frac {1}{y})^2(1-\frac{1}{y^2})=\ln(1-\frac{1}{x^2})^{-1}$
$(1+\frac {1}{y})^2(1-\frac{1}{y^2})=(\frac{x^2}{x^2-1})$
Now to find condition should I substitute $y=2x^2-1$ and find a relation for x?
| By What is the correct radius of convergence for $\ln(1+x)$?, we need $$\dfrac1{x^2},\dfrac1{y^2}<1$$
Now,
$$2\sum_{r=0}^\infty\dfrac{(1/y)^{2r+1}}{2r+1}=\ln\dfrac{1+1/y}{1-1/y}=\ln\dfrac{y+1}{y-1}$$
$$\implies-\ln\left(1-\dfrac1{x^2}\right)=\ln\left(\dfrac{y+1}{y-1}\right)=-\ln\left(\dfrac{y-1}{y+1}\right)$$
$$\implies1-\dfrac1{x^2}= \dfrac{y-1}{y+1} $$
Now simplify
| {
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Cyclic Inequality $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$ I have been challenged to prove the following:
If $xy+yz+zx=3$ for positive $x,y,z$, then $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$.
Initial attempts at AM-GM are not very helpful for me, I don't think.
$x^8+1≥2x^4,$ and $x^4+1≥2x^2$, so $(x^8+1)(x^4+1)≥4x^2(2x^2-1)$. It is easy to show $(x^8+1)(x^4+1)≥4x^2(2x^2-1)≥(x^2+1)^2$ if $x≥\sqrt{\frac{3}{7}},$ but that's insufficient.
I doubt that this problem turns into a Vieta bash due to the nature of having $x^8$ on the LHS. Initial attempts at Newton's sums were giving what looked like annoying numbers and something more bashy than what I would guess is a more elegant solution.
| COMMENT.-I find this inequality quite difficult to prove by ordinary means. However it seems to me we could do something valuable as follows: The maximum of $f(x)=\dfrac{(x^2+1)^2}{(x^8+1)(x^4+1)}$ is equal to $1.695$ and it is taken at $x=0.717$ Putting $f(x)$ and $f(y)$ maximum we must have $z=\dfrac{3-xy}{x+y}=1.7335502$ so $f(z)=0.01936916$. This way we verify that in fact
$$1\ge f(0.717)f(0.717)f(1.73355)=0.0556392$$ what makes quite plausible that $$1\ge f(x)f(y)f(z)$$ which is the proposed inequality.
| {
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Minimum value of $p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$ where $x,y,z\in \mathbb{R}^+$.
Find the minimum value of $$p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$$ where $x,y,z\in \mathbb{R}^+.$
Applying the AM-GM inequality,
$$
\begin{aligned}\frac{p}{6} & \geqslant\left(3x\cdot\frac{1}{15x}\cdot 5y\cdot\frac{25}{y}\cdot z\cdot \frac{1}{36z}\right)^{1/6} \\
\frac{p}{6} & \geqslant \left(\frac{5}{6}\right)^{1/3}\\
p & \geqslant 6\left(\frac{5}{6}\right)^{1/3}
\end{aligned}$$
$$\implies \text{The minimum value of the expression is } 6\left(\frac{5}{6}\right)^{1/3}
$$
Now, consider $f(x)=3x+\dfrac{1}{15x},\ g(y) = 5y+\dfrac{25}{y}$ and $h(z)=z+\dfrac{1}{36z}.$
$$\begin{aligned}f'(x) &= \frac{\mathrm d}{\mathrm dx}\left(3x+\frac{1}{15x}\right) = 3 - \frac{1}{15x^2}\\f''(x)&=\frac{2}{15x^3}\end{aligned}$$
At the critical points, $f'(x) = 0 \implies x = \dfrac{\pm1}{3\sqrt{5}}.$
$f''\left(\dfrac{1}{3\sqrt5}\right) > 0\implies f(x)$ has a local minima at $x = \dfrac{1}{3\sqrt5}.$
Similarly, the local minima of $g(y)$ is at $y = \sqrt5$ and the local minima of $h(z)$ is at $z=1/6.$
Substituting these values into the original expression, the minimum value of the expression comes out to be
$$
\begin{aligned}
p & = f\left(\dfrac{1}{3\sqrt5}\right)+g\left(\sqrt5\right)+h\left(\frac{1}{6}\right) \\
& = \frac{1}{3} + \frac{52}{\sqrt5}.
\end{aligned}
$$
The answer according to the AM-GM inequality is $\approx5.646$ and according to calculus is $\approx23.588$, which are way off. Also, the functions have only two points of inflection, one is the maxima (for values less than $0$) and the other is the minima (for values greater than $0$). As the question clearly states "for $x,y,z \in \mathbb{R}^+$", the "correct" answer should be $\approx23.588$, shouldn't it?
Why are the answers different? Also, is it possible to figure out the individual $x, y$ and $z$ values for which the expression has the minimum value (in case of AM-GM inequality)?
Note: This question is from a Test and the correct answer according to the "test creators" is $$6\left(\dfrac{5}{6}\right)^{1/3}.$$
$\text{Graph made using Desmos.}$
| Use Binary AM-GM as
$$3x+\frac{1}{15x}\ge \frac{1}{\sqrt{5}}, ~\text{equality when}~ x=\frac{1}{\sqrt{45}}\tag1$$
$$5y+\frac{25}{y} \ge 2\sqrt{125}, ~\text{equality when}~ y=\sqrt{5} \tag2$$
$$z+\frac{1}{36z} \ge \frac{1}{3} ~ \text{equality when} z=\frac{1}{6} \tag3$$
Adding them, we get
$$3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z} \ge \frac{1}{\sqrt{5}}+ 10 \sqrt{5}+\frac{1}{3}.$$
The equality exists for afore mentioned values of $x,y,z$
| {
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Trisected sides of a scalene triangle Scalene triangle $\bigtriangleup ABC$ has area 45. Points $P_1$ and $P_2$ are located on side $AB$ such that $AP_1 = P_1P_2 = BP_2$. Additionally, the points $Q_1$ and $Q_2$ are located on side $AC$ such that $AQ_1 = Q_1Q_2 = CQ_2$. The area of the intersection of triangles $BQ_1Q_2$ and $CP_1P_2$ can be expressed as a common fraction $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
$\textbf{(A) } 15 \qquad \textbf{(B) } 47 \qquad \textbf{(C) } 79 \qquad \textbf{(D) } 95 \qquad \textbf{(E) } 257 $
So to solve this, I assumed that an equilateral triangle wouldn't change the answer (I'm very lazy), so I fakesolved with an equilateral triangle.
I scaled it down so that the area was $\sqrt{3}$.
Then, the coordinates of the kite in the middle (in counter-clockwise order) are:
$(1,\frac{\sqrt{3}}{5})$
$(\frac{4}{7},\frac{2\sqrt{3}}{7})$
$(1,\frac{\sqrt{3}}{2})$
$(\frac{10}{7},\frac{2\sqrt{3}}{7})$
The product of the two diagonals divided by 2 is: $\frac{3}{7} \cdot \frac{3\sqrt{3}}{10}=\frac{9\sqrt{3}}{70}$. Multiplying by $\frac{45}{\sqrt{3}}=15\sqrt{3}$ gives $\frac{9 \cdot 15 \cdot 3}{70}=\frac{405}{70}=\frac{81}{14} \implies 95$.
No answer, can you guys check this? Thanks.
| I'm not sure I understand what you mean by "No answer". If every scalene triangle gives the same fraction (strongly implied by the problem), then by taking a limit (scalene triangles that approximate an equilater one better and better), we would expect an equilateral triangle to give the same fraction as well, so that you've found the answer is 95.
If you're just looking for a different approach, I like barycentric coordinates.
*
*$P_1$ is $\left[\dfrac23,\dfrac13,0\right]$
*$Q_1$ is $\left[\dfrac23,0,\dfrac13\right]$
*So $\overleftrightarrow{BQ_1}$ is $\left[\dfrac{2t}3,1-t,\dfrac{t}3\right]$ and $\overleftrightarrow{CP_1}$ is $\left[\dfrac{2s}3,\dfrac{s}3,1-s\right]$, with intersection $\left[\dfrac12,\dfrac14,\dfrac14\right]$.
*$Q_2$ is $\left[\dfrac13,0,\dfrac23\right]$
*So $\overleftrightarrow{BQ_2}$ is $\left[\dfrac{t}3,1-t,\dfrac{2t}3\right]$ and the intersection with $\overleftrightarrow{CP_1}$ is $\left[\dfrac27,\dfrac17,\dfrac47\right]$.
*$P_2$ is $\left[\dfrac13,\dfrac23,0\right]$
*So the intersection of $\overleftrightarrow{BQ_2}$ with $\overleftrightarrow{CP_2}$ ($\left[\dfrac{s}3,\dfrac{2s}3,1-s\right]$) is $\left[\dfrac15,\dfrac25,\dfrac25\right]$.
Now we can use the barycentric area formula to find the area of the right half of the quadrilateral, which is half of the area of the whole quadrilateral by symmetry. So the whole quadrilateral has area:
$$2*45*\det\begin{bmatrix}\dfrac15&\dfrac25&\dfrac25\\\dfrac27&\dfrac17&\dfrac47\\\dfrac12&\dfrac14&\dfrac14\end{bmatrix}$$
$$=9*\det\begin{bmatrix}1&2&2\\\dfrac27&\dfrac17&\dfrac47\\1&\dfrac12&\dfrac12\end{bmatrix}$$
$$=\dfrac{9}{14}*\det\begin{bmatrix}1&2&2\\2&1&4\\2&1&1\end{bmatrix}$$
$$=\dfrac{9}{14}*\det\begin{bmatrix}1&2&2\\0&0&3\\2&1&1\end{bmatrix}$$
$$=-\dfrac{27}{14}*\det\begin{bmatrix}1&2\\2&1\end{bmatrix}$$
$$=\dfrac{81}{14}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sin^220^\circ\sin40^\circ = \sin10^\circ \sin30^\circ \sin60^\circ$. This is the final step of one old competition problem, and in textbook it simply says "reader can prove this", but I found this a bit nontrivial.
Show that $\sin^220^\circ\sin40^\circ = \sin10^\circ \sin30^\circ \sin60^\circ$.
I tried a lot of things and formula on this, but never succeeded.
Can someone help?
| Trigonometric Proof
This follows almost immediately from $$\sin 20^\circ \sin 40^\circ \sin 80^\circ =\frac{\sqrt 3}{8},\tag{1}$$
which is a corollary of
$$\sin x \sin (60^\circ-x) \sin (60^\circ+x)=\frac{\sin(3x)}{4}.$$
My hint is to write one of $\sin 20^\circ$ from $\sin^2 20^\circ$ as $2\sin 10^\circ \cos 10^\circ$, and to note that $\cos 10^\circ=\sin 80^\circ$.
Postscript: We also have $$\cos x\cos(60^\circ-x)\cos(60^\circ+x)=\frac{\cos(3x)}{4},$$
making
$$\cos 20^\circ\cos 40^\circ \cos 80^\circ= \frac18.$$ Consequently
$$\tan x \tan(60^\circ -x )\tan(60^\circ+x)=\tan (3x)$$
and
$$\tan 20^\circ \tan 40^\circ \tan 80^\circ =\sqrt3.$$
We also have the following identity.
$$\prod_{j=1}^{n}\sin\left(x+\frac{(j-1)\pi}{n}\right)=\frac{\sin (nx)}{2^{n-1}}$$
for all positive integers $n$. From this identity,
$$\prod_{j=1}^{n}\cos\left(x+\frac{(j-1)\pi}{n}\right)=(-1)^{\frac{n-1}{2}}\frac{\cos (nx)}{2^{n-1}}$$
and
$$\prod_{j=1}^n\tan\left(x+\frac{(j-1)\pi}{n}\right)=(-1)^{\frac{n-1}{2}}\tan(nx)$$
for all odd positive integers $n$. We also have
$$\prod_{j=1}^n\tan\left(x+\frac{(j-1)\pi}{n}\right)=(-1)^{\frac{n}{2}}$$
if $n$ is an even positive integer, which implies
$$\prod_{j=1}^{n}\cos\left(x+\frac{(j-1)\pi}{n}\right)=(-1)^{\frac{n}{2}}\frac{\sin (nx)}{2^{n-1}}$$
for each even positive integer $n$.
Geometric Proof
There is also a geometric proof without using $(1)$. Let $ABC$ be a triangle with $\angle BAC=90^\circ$ and $\angle ABC=20^\circ$. Let $E$ and $F$ be internal points of $AC$ and $AB$ such that $BE$ bisects $\angle ABC$ and $\angle ACF=30^\circ$. We want to show that $\angle CFE=20^\circ$.
Note that $\angle BCF=40^\circ$. If $D$ is an internal point of $BF$ such that $CD$ bisects $\angle BDC$, then $\triangle DBC$ is isosceles. Therefore, $DM\perp BC$ if $M$ is the midpoint of $BC$.
By the angular bisector theorem on $\triangle ABC$, $\frac{AE}{CE}=\frac{AB}{BC}$. Since $\triangle MBD\sim \triangle ABC$, we have
$$\frac{DM}{BM}=\frac{BC}{AB}=\frac{CE}{AE}.$$
Since $BM=\frac12 BC$, we get
$$DB=\frac{CE}{AE} \cdot BM=\frac{BC\cdot CE}{2\cdot AE}.$$
Using the angular bisector theorem with $\triangle FCB$, we get $\frac{DF}{DB}=\frac{CF}{BC}$ so that
$$DF=\frac{CF}{BC}\cdot DB=\frac{CE}{AE}\left(\frac{CF}{2}\right).$$
Since $\triangle ACF$ is a right triangle with $\angle ACF=30^\circ$, we get $AF=\frac{CF}{2}$ so that $DF=\frac{CE}{AE}\cdot AF$ or
$$\frac{DF}{AF}=\frac{CE}{AE}.$$
This means $EF\parallel CD$. Hence $$\angle CFE=\angle FCD=\frac{1}{2}\angle BCF=20^\circ.$$
From this result, we see using the law of sines on $\triangle CFE$ that
$$\frac{CE}{\sin 20^\circ}=\frac{CF}{\sin 130^\circ}=\frac{CF}{\sin 50^\circ}=\frac{CF}{\cos 40^\circ}.$$
Also the law of sines on $\triangle BCF$ yields
$$\frac{CF}{\sin 20^\circ}=\frac{BC}{\sin 120^\circ}=\frac{BC}{\sin 60^\circ}.$$
Therefore
$$\frac{CE}{\sin 20^\circ}=\frac{BC\sin 20^\circ}{\sin 60^\circ \cos 40^\circ}.\tag{2}$$
But using the law of sines with $\triangle BCE$ gives us
$$\frac{CE}{\sin 10^\circ}=\frac{BC}{\sin 100^\circ}=\frac{BC}{\sin 80^\circ}=\frac{BC}{2\sin 40^\circ \cos40^\circ}.\tag{3}$$
From $(2)$ and $(3)$ we obtain
$$\frac{BC \sin 10^\circ}{2\sin 40^\circ \cos 40^\circ}= CE=\frac{BC\sin^2 20^\circ}{\sin 60^\circ \cos 40^\circ}.$$
This proves that
$$\sin^2 20^\circ \sin 40^\circ =\frac{1}{2}\sin 10^\circ \sin 60^\circ =\sin 10^\circ \sin 30^\circ \sin 60^\circ.$$
And from this result, we can prove $(1)$ as well.
| {
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"url": "https://math.stackexchange.com/questions/3523754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$ If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$
My attempt is as follows:-
$$a_0+a_1(x-2)+a_2(x-2)^2+\cdots a_{2n}(x-2)^{2n}=b_0+b_1(x-3)+b_2(x-3)^2+\cdots b_{2n}(x-3)^{2n}$$
Let's see the term on R.H.S with coefficient as $b_n$
$$T_{n+1}=b_n(x-3)^n$$
Comparing the coefficients of $x^0$
$$a_0-2a_1+(-2)^2a_2+(-2)^3a_3+\cdots a_{2n}(-2)^{2n}=b_0-3b_1+(-3)^2b_2+(-3)^3b_3+\cdots+b_n(-3)^n+\cdots b_{2n}(-3)^{2n}$$
Comparing the coefficients of $x^1$
$$a_1+2a_2(-2)^1+3a_3(-2)^2\cdots 2n\cdot a_{2n}(-2)^{2n-1}=b_1+2a_2(-3)^1+3a_3(-3)^2\cdots nb_n(-3)^{n-1} \cdots (2n)a_{2n}(-3)^{2n-1}$$
Now I am not getting any way to find $b_n$ , its so mixed up in these expressions, not able to get any breakthrough. Any inputs?
| Let $(x-3)=y$ Then we get
$$\sum_{r=0}^{2m} A_r(1+y)^r=\sum_{r-0}^{2n} B_r~ y^r$$
Compare the coefficient of $y^n$ both sides, then
$${n \choose n}+{n+1 \choose n}+{n+2 \choose n}+....+{2n \choose n}=B_n$$ or
$${n+1 \choose n+1}+{n+1 \choose n}+{n+2 \choose n}+....+{2n \choose n}$$
Next use $${n \choose r}+{n \choose r-1}={n+1 \choose r}$$ successively to get
The LHS in above as $${2n+1 \choose n+1}=B_n$$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality $\frac{1}{64}(a^{15} + b^{15} + c^{15}+ d^{15})^4 \ge \sum_{\rm {cyc}} (a b^2 c^4 d^8 )^4 $ Let real $a,b,c,d > 0$. Show that
\begin{align}
\frac{1}{64}(a^{15} + b^{15} + c^{15}+ d^{15})^4 &\ge \sum_{\rm {cyc}} (a b^2 c^4 d^8 )^4 \\
&= (a b^2 c^4 d^8 )^4 +(b c^2 d^4 a^8 )^4 +(c d^2 a^4 b^8 )^4 +(d a^2 b^4 c^8 )^4
\end{align}
This is obviously homogeneous. Equality appears at $a=b=c=d$, I haven't found other equality points. Applying AM-GM inequality to the LHS is too crude because it doesn't take into account the rising exponentials $(1,2,4,8)$ on the RHS. Indeed, AM-GM leaves to prove
\begin{align}
4 &\ge \sum_{\rm {cyc}} \frac{c\cdot d^{17}}{b^7 \cdot a^{11}}
\end{align}
which can be made to fail easily, e.g. by setting $a=b=c =0.01 \cdot d$.
| Let $a b^2 c^4 d^8=t,$ $b c^2 d^4 a^8=z,$ $c d^2 a^4 b^8=y$ and $d a^2 b^4 c^8=x$.
Thus, $$\sum_{cyc}\frac{x^2}{y}=\sum_{cyc}\frac{d^2a^4b^8c^{16}}{cd^2a^4b^8}=\sum_{cyc}c^{15}$$ and we need to prove that
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{t}+\frac{t^2}{x}\geq\sqrt[4]{64(x^4+y^4+z^4+t^4)},$$ which is a known problem.
For example, see here: How to prove $\frac{1}{4}(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a})\ge \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Which one of the following is true?
The image of the half plane $\text {Re} (z) + \text {Im} (z) >0$ under the map $w = \frac {z-1} {z+i}$ is given by
$(\text A)$ $|w| > 1$
$(\text B)$ $|w| < 1$
Let $z = x + iy$ with $x + y >0.$ Then I found that $$|w| = \frac {\sqrt {\left (x^2 + y^2 - x + y \right )^2 + \left (y - x + 1 \right)^2}} {x^2 + \left (y + 1 \right )^2}.$$
By taking $z = 1$ we have $|w| = 0 <1.$ So I believe that $(\text B)$ is the correct option. So we need only to prove that for any $x,y \in \Bbb R$ with $x + y > 0$ $$\sqrt {\left (x^2 + y^2 - x + y \right )^2 + \left (y - x + 1 \right)^2} < x^2 + \left (y + 1 \right )^2.$$
But how do I show that? Any help in this regard will be highly appreciated.
Thank you very much for your valuable time.
| To continue from where you got up to, you could have tried
$$\begin{align}&\sqrt{\left(x^{2}+y^{2}-x+y\right)^{2}+\left(y-x+1\right)^{2}}<x^{2}+\left(y+1\right)^{2}
\\
&\iff \left(x^{2}+y^{2}-x+y\right)^{2}+\left(y-x+1\right)^{2}-\left(x^{2}+\left(y+1\right)^{2}\right)^{2}<0
\\
&\iff \left(x^{2}+\left(y+1\right)^{2}\right)\left(\frac{\left(x^{2}+y^{2}-x+y\right)^{2}+\left(y-x+1\right)^{2}-\left(x^{2}+\left(y+1\right)^{2}\right)^{2}}{x^{2}+\left(y+1\right)^{2}}\right)<0\tag{*}
\\
&\iff \tiny\left(x^{2}+\left(y+1\right)^{2}\right)\left(\frac{\left(x^{4}+2x^{2}y^{2}+y^{4}-2x^{3}+2x^{2}y-2xy^{2}+2y^{3}+2x^{2}-4xy+2y^{2}-2x+2y+1\right)-\left(x^{4}+2x^{2}y^{2}+y^{4}+4x^{2}y+4y^{3}+2x^{2}+6y^{2}+4y+1\right)}{x^{2}+\left(y+1\right)^{2}}\right)<0
\\
&\iff -2\left(x^{2}+\left(y+1\right)^{2}\right)\left(\frac{\left(x^{3}+xy^{2}+2xy+x\right)+\left(x^{2}y+y^{3}+2y^{2}+y\right)}{x^{2}+\left(y+1\right)^{2}}\right)<0
\\
&\iff -2\left(x^{2}+\left(y+1\right)^{2}\right)\left(x\cdot\frac{x^{2}+\left(y+1\right)^{2}}{x^{2}+\left(y+1\right)^{2}}+y\cdot\frac{x^{2}+\left(y+1\right)^{2}}{x^{2}+\left(y+1\right)^{2}}\right)<0
\\
&\iff {-2\left(x^{2}+(y+1)^{2}\right)\left(x+y\right)<0}
\\
&\iff x+y>0\quad \square
\end{align}$$
where, in $(*)$, we use the ansatz that the LHS is divisible by $\left(x^{2}+(y+1)^{2}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3528445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Expectation and variance of the squared distance between $X$ and $Y$ Given that $X$ and $Y$ are two independent univariate random variables sampled uniformly from the unit interval [0,1]. I am trying to find the expected value and the variance of the random variable $Z = \mid X - Y \mid ^2 $ (the squared distance between $X$ and $Y$.
So far, I'm assuming that $X$ and $Y$ are uniformly distributed and this is what I have:
\begin{align}
Z &= \mid X - Y\mid ^2 \\
Z &= \mid X^2 - 2XY + Y^2 \mid \\
E[Z] &= \mid E[X^2] - 2E[XY] + E[Y^2]\mid \\
&= \mid E[X^2] - 2E[X]E[Y] + E[Y^2]\mid \\
&= \frac{1}{3}\frac{1}{b-a}(b^3-a^3)-2(\frac{a+b}{2})(\frac{a+b}{2})+\frac{1}{3}\frac{1}{b-a}(b^3-a^3)
\end{align}
But I am not sure if this is the right approach or the right direction and what is E[Z] = $\mid X - Y\mid ^2$. If so, can I do something similar to the variance?
| If I substitute 0 and 1 for a and b respectively to:
\begin{align}
&= \frac{1}{3}\frac{1}{b-a}(b^3-a^3)-2(\frac{a+b}{2})(\frac{a+b}{2})+\frac{1}{3}\frac{1}{b-a}(b^3-a^3) \\
&=\frac{1}{6}
\end{align}
Would the answer (the expected value of Z) be $\frac{1}{6}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the series representation of $\ln\left(\frac{1+x}{1-x}\right)$
Given that $\frac{1}{1-x}=\sum^{\infty}_{n=0}x^n$, what is the series representation of $\ln\left(\frac{1+x}{1-x}\right)$?
Differentiating $\ln\left(\frac{1+x}{1-x}\right)$ results in: $\frac2{(1-x)^2}$. This means that $\ln\left(\frac{1+x}{1-x}\right) = 2\int\frac1{(1-x)^2}\ dx$
Can I now say that $\ln\left(\frac{1+x}{1-x}\right)=2\left(\sum^{\infty}_{n=0}\int x^n\right)^2$ ? This would get: $2\sum^{\infty}_{n=0}\frac{x^{2(n+1)}}{(n+1)^2}$.
I'm assuming the answer is wrong because the answer key did not agree with me. Did I mess up somewhere in this problem?
The answer key only doubles the $x$ in this step: $\ln\left(\frac{1+x}{1-x}\right) = 2\int\frac1{(1-x)^2}\ dx$ so instead of $2\int\frac1{(1-x)^2}\ dx$ they get just $2\int\frac1{1-x^2}\ dx$. Why is this correct? (or is it incorrect?)
| Your derivative of the logarithm is wrong. It is not $\frac 2{(1-x)^2}$. It is $\frac{2}{1-x^2}$.
$$\left[\ln\frac{1+x}{1-x}\right]'=\frac{\left[\frac{1+x}{1-x}\right]'}{\frac{1+x}{1-x}}=\frac{\frac 2{(1-x)^2}}{\frac{1+x}{1-x}}=\frac{2}{(1+x)(1-x)}=\frac{2}{1-x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Evaluate $\lim_{x\to 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$ $$\lim_{x\to 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$$
$$\lim_{x\to 0}\frac{xe^x+x-2e^x+2}{x^3}$$
Limit is in the form "$\frac{0}{0}$", using L'hopital once:
$$\lim_{x\to 0}\frac{e^x+xe^x+1-2e^x}{3x^2}=\lim_{x\to 0}\frac{-e^x+xe^x+1}{3x^2}$$
Again the limit is of the form "$\frac{0}{0}$", using L'hopital again:
$$\lim_{x\to 0}\frac{-e^x+e^x+xe^x}{6x}=\lim_{x\to 0}\frac{xe^x}{6x}$$
Once again the limit is of the form "$\frac{0}{0}$", using L'hopital one more time:
$$\lim_{x\to 0}\frac{e^x+xe^x}{6}=\frac{1}{6}$$
Is there a way to solve it without L'hopital?
| series expansion of $\displaystyle e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots $
around $x=0$
$$\lim_{x\rightarrow 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$$
$\displaystyle \lim_{x\rightarrow 0}\frac{x\bigg(\bigg(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\bigg)+1\bigg)-2\bigg(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots -1\bigg)}{x^3}$
$$\lim_{x\rightarrow 0}\frac{x^3(\frac{1}{2!}-\frac{2}{3!})+x^4()+\cdots}{x^3}=\frac{1}{6}$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_id": 0
} |
Prove $a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$ for the recursive sequence $a_{n+1}=\frac{3a_n-1}{3-a_n}$
Prove the statement:
$$a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$$
for the given recursive sequence:
$$a_{n+1}=\frac{3a_n-1}{3-a_n}. $$
My attempt:
Proof by induction:
(1) Base: $\tau(1)$.
$$
\begin{align}
a_2 &=\frac{(2+1)a_1-2+1}{2+1-(2-1)a_1} \\
&=\frac{3a_1-1}{3-a_1}.
\end{align}
$$
(2) Assumption: Let
$$ a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1} $$
hold for some $n\in\mathbb N$.
(3) Step $\tau(n+1)$:
$$
\begin{align}
a_n &= \frac{3\cdot\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}-1}{3-\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}} \\
&= \frac{3\cdot2^{n-1}a_1+3a_1-3\cdot2^{n-1}+3-2^{n-1}-1+2^{n-1}a_1-a_1}{3\cdot 2^{n-1}+3-3\cdot 2^{n-1}a_1+3a_1-2^{n-1}a_1-a_1+2^{n-1}-1} \\
&= \frac{(4\cdot 2^{n-1}+2)a_1-4\cdot 2^{n-1}+2}{4\cdot 2^{n-1}+2-(4\cdot 2^{n-1}+2)a_1} \\
&= \frac{(2^n+1)a_1-2^n+1}{2^n+1-(2^n+1)a_1}
\end{align}
$$
Is this correct and is there a more efficient method than induction?
| For each $n = 1, 2, 3, \ldots$, we have
$$
\begin{align}
a_{n+1} &= \frac{3a_n -1 }{3 - a_n } \\
&= \frac{ 3a_n - 9 + 8 }{ 3 - a_n } \\
&= -3 + \frac{8}{ 3 - a_n }.
\end{align}
$$
Thus we obtain
$$ a_2 = - 3 + \frac{8}{3-a_1} = \frac{3a_1 - 1}{3 - a_1}. $$
$$ a_3 = -3 + \frac{ 8 }{ 3 - a_2 } = -3 + \frac{8}{3 - \frac{3a_1 - 1}{3 - a_1} } = ... $$
This approach might make it easier for you to proceed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Asymptotic expression to $e^{-k^2/2n}$ I want to show that $$\Bigl(1-{k\over n}\Bigr)^{-(1+n-k)/2}\Bigl(1+{k\over n}\Bigr)^{-(1+n+k)/2}$$ is asymptotic to $e^{-k^2/2n}$.
I, in my earlier post, was satisfied with the given answer. But only now I noticed that the answer provided has a flaw since though it claims to produce $e^{-k^2/2n}$, it actually produces $e^{-k^2/n}$.
Any thoughts?
| You have
\begin{align}
\Bigl(1-{\tfrac kn}\Bigr)^{-\frac{1+n-k}2}\Bigl(1+{\tfrac kn}\Bigr)^{-\frac{1+n+k}2}
&=
\exp\left[-\tfrac{1+n-k}2\,\log\left(1-\tfrac kn\right)-\tfrac{1+n+k}2\,\log\left(1+\tfrac kn\right)\right]\\ \ \\
&=\exp\left[-\tfrac{1+n-k}2\,\left(-\tfrac kn-\tfrac{k^2}{2n^2}+o(\tfrac{k^3}{n^3})\right)-\tfrac{1+n+k}2\, \left(\tfrac kn-\tfrac{k^2}{2n^2}+o(\tfrac{k^3}{n^3})\right)\right]\\ \ \\
&=\exp\left[-\tfrac{1+n-k}2\,\left(-\tfrac kn-\tfrac{k^2}{2n^2}+o(\tfrac{k^3}{n^3})\right)-\tfrac{1+n+k}2\, \left(\tfrac kn-\tfrac{k^2}{2n^2}+o(\tfrac{k^3}{n^3})\right)\right]\\ \ \\
&=\exp\left[-\tfrac{k^2}n+(1+n)\tfrac{k^2}{2n^2}+o(\tfrac1{n^2}) \right]\\ \ \\
&=\exp\left[-\tfrac{k^2}{2n}+\tfrac{k^2}{2n^2}+o(\tfrac1{n^2}) \right]\\ \ \\
&=\exp\left[-\tfrac{k^2}{2n}+o(\tfrac1{n^2}) \right]\\ \ \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$ I need to solve the following problem:
$$\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$$
I tried with trigonometric substitution, but I couldn't solve it. The square root is messing me up. I searched it up and nothing appears to work well so I hope someone can help. Thanks.
| Disclaimer. In all general cases below, you need to choose your (real) constants such that all terms are well-defined.
First note $2x^2-3=\frac12(2x-\sqrt6)(2x+\sqrt6)$ so we can decompose the integrand as $$\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}=\frac52\cdot\left(\frac{1}{\left(2 x-\sqrt{6}\right) \sqrt{3 x^2-2 x+1}}+\frac{1}{\left(2 x+\sqrt{6}\right) \sqrt{3
x^2-2 x+1}}\right).$$ For an integral of the type $$\int \frac{1}{(d+ex)\sqrt{a+bx+cx^2}}\,\mathrm dx$$ we can use the substitution $u=\frac{2 ae-bd-(2cd-be)x}{\sqrt{a+bx+cx^2}}$, leading to $$\int \frac{1}{\underbrace{4cd^2-4bde+4ae^2}_{\text{constant}}-u^2}\,\mathrm du.$$ But this integral is well-known as $$\frac{\tanh ^{-1}\left(\frac{u}{2 \sqrt{a e^2-b d e+c d^2}}\right)}{2 \sqrt{a e^2-b d e+c d^2}}$$ and using the definition of $u$ we get $$\int \frac{1}{(d+ex)\sqrt{a+bx+cx^2}}\,\mathrm dx=-\frac{\tanh ^{-1}\left(\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c
d^2}}\right)}{\sqrt{e (a e-b d)+c d^2}}.$$
In our special case, this leads to $$\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}\,\mathrm dx=-\frac{5 \tanh ^{-1}\left(\frac{-\left(2-3 \sqrt{6}\right) x-\sqrt{6}+2}{\sqrt{2 \left(11-2 \sqrt{6}\right)} \sqrt{3 x^2-2 x+1}}\right)}{2 \sqrt{2 \left(11-2 \sqrt{6}\right)}}-\frac{5 \tanh ^{-1}\left(\frac{-\left(2+3 \sqrt{6}\right) x+\sqrt{6}+2}{\sqrt{2 \left(11+2 \sqrt{6}\right)} \sqrt{3 x^2-2 x+1}}\right)}{2 \sqrt{2 \left(11+2 \sqrt{6}\right)}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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2018 MathCounts: Let $D(k)$ be the number of diagonals for a polygon with $k$ sides. If $D(m) + D(n) = 125$, what is the value of $m+n$? The question was asked in the 2018 Raytheon MATHCOUNTS National Competition. It appears in the video around the 35:34 mark:
https://www.youtube.com/watch?v=dSnOLW_W6og&t=2134
Let $D(k)$ be the number of diagonals for a polygon with $k$ sides. If
$D(m) + D(n) = 125$, what is the value of $m+n$?
The formula for the number of diagonals is well-known and easily derived as $k(k-3)/2$.
I then computed several values, and found a suitable pair of numbers to solve the problem.
But is there any easier way to solve this equation:
$$\frac{n(n-3)}{2} + \frac{m(m-3)}{2} = 125$$
$$n,m \in {3, 4, ...}$$
The students were supposed to solve this in minutes without a calculator.
My guess is there is not a simpler way--in MathCounts the students may have just memorized many values.
| We want to solve $n^2-3n+m^2-3m = 250$ for $n,m\ge 3$.
Multiplying by $8$ and symmetrizing the polynomials, we get that the equation is equivalent to solving
$$j^2-9 + k^2-9 = 1000,$$
with $j=2n-3$, $k=2m-3$, or
$$j^2+k^2=1018.$$
Mod $3$ we have $$j^2+k^2\equiv 1\pmod{3},$$ so one of $j$ or $k$ is divisible by $3$. Without loss of generality then, $j=3x$.
Now we want to solve
$$9x^2 + k^2 =1018.$$
Mod $9$, we get
$$k^2 \equiv 1\pmod{9}.$$ Then $k=9y+10$ or $9y+8$ for some $y>0$, since the units mod $9$ are cyclic.
Now we have either
$$9x^2 + 81y^2+180y+100 = 1018$$
which reduces to
$$ x^2 + 9y^2 + 20y = 102$$
or
$$9x^2+81y^2+144y + 64 = 1018,$$
which reduces to
$$ x^2 + 9y^2 + 16y = 106$$
In either case, $1\le y\le 2$.
Thus we are reduced to computing four values less than $100$ and checking whether they are squares:
For $y=1$, computing
$102- (9y^2+20y)$ and $106-(9y^2+16y)$ gives
$102-29 = 73$, and $106-25 = 81$.
$81=9^2$, so we are already done without checking $y=2$. This gives $x=9$, $y=1$, corresponding to $j=27$, $k=17$, and thus $n=15$, $m=10$.
Checking our work yields
$$15\cdot 6 + 5\cdot 7 = 90+35=125.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given positives $x, y , z$ such that $x + y + z = xyz$. Calculate the minimum value of $\frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$.
Given positives $x, y , z > 1$ such that $x + y + z = xyz$. Calculate the minimum value of $$\large \frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$$
We have that $x + y + z = xyz \implies \dfrac{1}{yz} + \dfrac{1}{zx} + \dfrac{1}{xy} = 1$ and $$\left(\frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}\right) \cdot \left(\frac{1}{x - 1} + \frac{1}{y - 1} + \frac{1}{z - 1}\right)$$
$$ \ge \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)^2 \ge 3 \cdot \left(\frac{1}{yz} + \frac{1}{zx} + \frac{1}{xy}\right) = 3$$
Now we need to find the maximum value of $\dfrac{1}{x - 1} + \dfrac{1}{y - 1} + \dfrac{1}{z - 1}$.
Let $\dfrac{1}{x} = a, \dfrac{1}{y} = b, \dfrac{1}{z} = c$ which implies that $a, b, c \in (0, 1)$.
It could be observed that $ab + bc + ca = 1$ and $$\dfrac{1}{x - 1} + \dfrac{1}{y - 1} + \dfrac{1}{z - 1} = \frac{a}{1 - a} + \frac{b}{1 - b} + \frac{c}{1 - c}$$
$$ = 3 - \left(\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}\right) \le 3 - \frac{9}{3 - (a + b + c)} = \frac{-3(a + b + c)}{3 - (a + b + c)}$$
Then I'm stuck.
| For $x=y=z=\sqrt3$ we'll get a value $\sqrt3-1$.
But, $$\sum_{cyc}\frac{x - 1}{y^2}-\sqrt3+1=\sum_{cyc}\left(\frac{x}{y^2}-\frac{2}{y}+\frac{1}{x}\right)-\sum_{cyc}\left(\frac{1}{x^2}-\frac{1}{xy}\right)+\frac{xy+xz+yz}{xyz}-\sqrt3=$$
$$=\sum_{cyc}\frac{(x-y)^2}{y^2x}-\sum_{cyc}\frac{(x-y)^2}{2x^2y^2}+\frac{xy+xz+yz}{xyz}-\sqrt{\frac{3(x+y+z)}{xyz}}=$$
$$=\sum_{cyc}\frac{(x-y)^2(2x-1)}{2x^2y^2}+\frac{\sum\limits_{cyc}z^2(x-y)^2}{2xyz\left(xy+xz+yz+\sqrt{3xyz(x+y+z)}\right)}\geq0.$$
Now, we see that $\sqrt3-1$ is a minimal value even for $\{x,y,z\}\subset\left[\dfrac{1}{2},+\infty\right)$ such that $x+y+z=xyz.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int_{0}^{\infty} \frac{1}{x^2}\int_{0}^{x}\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt \, dx $ I'm trying to solve $$\int_{0}^{\infty} \frac{1}{x^2}\int_{0}^{x}\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt \, dx $$
The value should be $\displaystyle \frac{\pi}{2}\ln(2)$. I tried different substitutions, but it doesn't work out. Changing the integration order seems not really nice as below:
$$\int_1^{\infty} \frac{2\sqrt{z+1}\arctan\left(\sqrt{\frac{z-1}{z+1}}\right)+\sqrt{z-1}(\ln(2z)-\ln(z-1))}{4z(z-1)(z+1)} dz$$
Has anyone a good start for me? Maybe Feynman? Thanks!
| Suppose $\displaystyle\int\limits_0^x\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt$. It's well known that $\arctan z=\dfrac{\pi}{2}-\text{arccotan} z$, so our integral is $$\dfrac{\pi}{2}\int\limits_0^x\frac{dt}{(t^2+1)\sqrt{t^2+2}}\, dt-\int\limits_0^x\frac{\text{arccotan}\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt.$$
It's obvious that $\dfrac{1}{z}\text{arccotan}\dfrac{x}{z}=\dfrac{\pi}{2z}-\displaystyle\int_0^x\dfrac{dy}{y^2+z^2}$. Given that $z=\sqrt{2+t^2}$ and substituting the last equality in second integral, we get that our two integrals turns into $$\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+t^2+2)}=I.$$
Further, it's easy to see that $I=\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+1)}-\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(y^2+1)(y^2+t^2+2)}$.
Since in the second integral everything is symmetric, $\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(y^2+1)(y^2+t^2+2)}=I$ (we can also make a replacement $t\to y$ and $y\to t$). Thus $$I=\dfrac{1}{2}\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+1)}=\dfrac{1}{2}\left(\displaystyle\int\limits_0^x\dfrac{dt}{t^2+1}\right)^2=\dfrac{\arctan^2x}{2}.$$
It remains to calculate the integral $\dfrac{1}{2}\displaystyle\int\limits_0^\infty\dfrac{\arctan^2x}{x^2}dx=\displaystyle\int\limits_0^\infty\dfrac{\arctan x}{x(1+x^2)}dx$ (here we applying integrate by parts). Can you finish it yourself? Hint: you can consider the integral $\displaystyle\int\limits_0^\infty\dfrac{\arctan ax}{x(1+x^2)}dx$ and differentiate it by the parameter.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac b{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}$ when $a,b,c>0$ I want to prove that for $a,b,c>0$ we have
$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}=
\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac{b}{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}.$$
My first attempt: By Cauchy-Schwarz we have $$\left(\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}\right)^2\le3\sum_{cyc}\frac{a^2}{a^2+3b^2+3c^2}$$ so we only need to prove that the right-hand side is always less than $\frac{9}{7}$, but this is false. Failed
Second attempt: By Cauchy-Schwarz
$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}=\sum_{cyc} \frac 1{\sqrt{1+3\frac{b^2}{a^2}+3\frac{c^2}{a^2}}}\le\sum_{cyc} \frac{\sqrt 7}{1+3\frac{b}{a}+3\frac{c}a}$$
so it remains to prove that $$\sum_{cyc} \frac{a}{1+3b+3c}\le\frac37$$ but this is wrong for example for $a=1,b=1,c=2$. Failed
Third attempt: Let $S=3(a^2+b^2+c^2)$. We need to prove $$\sum_{cyc} \frac{a}{\sqrt{S-2a^2}}\le \frac37.$$ But $x\mapsto \frac{x}{\sqrt{S
-2x^2}}$ is convex so Jensen has the wrong direction...
| We need to prove that:
$$\sum_{cyc}\sqrt{\frac{a}{a+3b+3c}}\leq\frac{3}{\sqrt7},$$ where $a$, $b$ and $c$ are positive numbers.
Indeed, by C-S $$\left(\sum_{cyc}\sqrt{\frac{a}{a+3b+3c}}\right)^2\leq\sum_{cyc}\frac{a}{(a+3b+3c)(17a+2b+2c)}\sum_{cyc}(17a+2b+2c).$$
Thus, it's enough to prove that:
$$\sum_{cyc}\frac{a}{(a+3b+3c)(17a+2b+2c)}\leq\frac{3}{49(a+b+c)}.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$\sum_{cyc}\frac{a}{(9u-2a)(6u+15a)}\leq\frac{1}{49u}$$ or
$$49u\sum_{cyc}a(9u-2b)(9u-2c)(2u+5b)(2u+5c)\leq3\prod_{cyc}((9u-2a)(2u+5a)).$$
We'll prove that the last inequality is true even for any reals $a$, $b$ and $c$.
Indeed, since $\sum\limits_{cyc}a(9u-2b)(9u-2c)(2u+5b)(2u+5c)$ is a fifth degree polynomial,
the last inequality is equivalent to $f(w^3)\geq0,$ where
$$f(w^3)=-3000w^6+A(u,v^2)w^3+B(u,v^2).$$
But $f$ is a concave function.
Thus, $f$ gets a minimal value for an extreme value of $w^3$, which happens for equality case of two variables.
Since the last inequality is an even degree, homogeneous and symmetrical, it's enough to assume $b=c=1,$ which gives $$\frac{a}{(a+6)(17a+4)}+\frac{2}{(3a+4)(2a+19)}\leq\frac{3}{49(a+2)}$$ or $$(a-1)^2(a-8)^2\geq0$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum of $a + b + c$ Let $a, b, c$ be non-negative real numbers such that
$$abc[(a − b)(b − c)(c − a)]^2 = 1$$
Find the minimum of $a + b + c$.
Source: https://cms.math.ca/crux/v43/n5/public_Chow_et_al_43_5.pdf
My Attempt:
$a \mapsto \frac{x+y}{2}$
$b \mapsto \frac{y+z}{2}$
$c \mapsto \frac{x+z}{2}$ $\;$ without any success
| Remarks: @youthdoo said there is a AM-GM solution. I found it.
Problem: Let $a, b, c\ge 0$ with $abc[(a-b)(b-c)(c-a)]^2 = 1$. Prove that
$$a + b + c \ge \sqrt[9]{\frac{531441}{16}}.$$
Sketch of a proof:
WLOG, assume that $a > b > c$.
Let $0 < x_1 < x_2 < x_3$ be the three real roots of $2x^3 - 54x^2 + 243x - 243 = 0$.
Let
$$x_4 = \frac{x_1x_2}{x_2 - x_1}, \quad x_5 = \frac{9 - x_2}{2} + \frac{x_1x_2}{x_2 - x_1}, \quad
x_6 = \frac{9 - x_1}{2} - \frac{x_1x_2}{x_2 - x_1}.$$
It is easy to prove that $x_4, x_5, x_6 > 0$.
By AM-GM, we have
\begin{align*}
&x_1x_2x_3x_4^2x_5^2x_6^2\cdot abc[(a-b)(b-c)(c-a)]^2\\
={}&x_1a \cdot x_2b \cdot x_3c \cdot x_4(a - b) \cdot x_4(a - b) \cdot x_5(b - c) \cdot x_5(b - c) \cdot x_6(a - c) \cdot x_6 (a - c) \\
\le{}& \left(\frac{x_1a + x_2b + x_3c + x_4(a-b) \times 2 + x_5(b - c) \times 2 + x_6(a-c)\times 2}{9}\right)^9\\
={}& \left(\frac{(x_1+2x_4+2x_6)a + (x_2-2x_4+2x_5)b + (x_3-2x_5-2x_6)c}{9}\right)^9\\
={}& (a + b + c)^9
\end{align*}
where we have used $x_1 + 2x_4 + 2x_6 = 9$, $x_2 - 2x_4 + 2x_5 = 9$, and
$x_3 - 2x_5 - 2x_6 = x_1 + x_2 + x_3 - 18 = 9$ (by Vieta's theorem).
It suffices to prove that
$$x_1x_2x_3x_4^2x_5^2x_6^2 = \frac{531441}{16}.$$
Using $x_1x_2x_3 = \frac{243}{2}$ (by Vieta's theorem), it suffices to prove that
$$(x_4x_5x_6)^2 = \frac{2187}{8}$$
which is true. The proof is omitted here.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $m_a\geq \dfrac{b^2+c^2}{4R}$ Let triangle ABC, $m_a$ the lenght of the median from A, $b,c$ the lenghts of the segments AC and AB respectively and R the circumradius. Prove that:
$m_a\geq \dfrac{b^2+c^2}{4R}$.
I found this in a book and the hint was to denote M the midpoint of BC and $A_2$ the second intersection of the median and the circumcircle of the triangle. Then by power of a point we have that $AM \cdot MA_2=a^2/4$. Also $AM+MA_2 \leq 2R$. And then they said this to imply the conclusion.
Please help me understand. Thank you in advance.
| As your hint says, from the power of $M$ to $(O)$ we get
$$AM \cdot MA_2 = BM \cdot MC = \frac{1}{4}a^2$$
$$\Longrightarrow AA_2 = AM + MA_2 = \frac {AM^2 + \frac {1}{4}a^2}{AM} = \frac {\frac {1}{2}(b^2 + c^2) - \frac {1}{4}a^2 + \frac {1}{4}a^2}{m_a} = \frac {b^2 + c^2}{2m_a}$$
Since a diameter is greater or equal than the chord $ AA_2,$ we have
$$2R \ge \frac {b^2 + c^2}{2m_a}$$
and the conclusion follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Finding constants $a$ and $b$ using derivatives 'Suppose $f(x)= 3(ax-b/x)^3$. Given that $f(3/2)= 3$ and $f'(3/2)= 30$, find $a$ and $b$.'
I've tried chain rule and getting $a$ or $b$ on its own and substituting back into the function, but I feel like I'm overcomplicating it.
Thanks for any help.
| We have $f'(x)=9(a+\frac{b}{x^2})(ax-\frac{b}{x})^3$
$f(\frac{3}{2})=3$ so $3(a.\frac{3}{2}-b\frac{2}{3})^3=3$; $f'(\frac{3}{2})=30$ so $9(a+b\frac{4}{9})(a\frac{3}{2}-b\frac{2}{3})^3$
Implies $9a-4b=6$ and $9a+4b=30$, so $a=2$ and $b=3$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find expression for following summation
Find expression of:
$$
\sum_{r=1}^n \dbinom{n}{r} \sin {r\theta} = ???
$$
|
$$X=\sum_{r=1}^n \dbinom{n}{r} \sin {r\theta} = ?$$
Consider
$$\begin{align}
S:=\sum_{r=1}^n \dbinom{n}{r} e^{ir\theta}
&= \sum_{r=1}^n \dbinom{n}{r} (\underbrace{e^{i\theta}}_{:=a})^r \\
&= \sum_{r=1}^n \dbinom{n}{r} a^r \\
&= -1+\sum_{r=0}^n \dbinom{n}{r} a^r1^{n-r} \\
&= -1+(1+a)^n \\
&= (1+e^{i\theta})^n-1\\
\end{align}$$
Because $e^{ix}=\cos x+i\sin x$, we also have:
$$\begin{align}
S&= \sum_{r=1}^n \dbinom{n}{r} \big(\cos r\theta+i\sin r\theta\big) \\
&= \sum_{r=1}^n \dbinom{n}{r} \cos r\theta
+ i\underbrace{\sum_{r=1}^n \dbinom{n}{r} \sin r\theta}_{=X} \\
\end{align}$$
Hence: $\def\Im{\operatorname{Imag}}$
$$\begin{align}
X&= \Im(S) \\
&= \Im\big((1+e^{i\theta})^n-1\big) \\
&= \Im\big((1+e^{i\theta})^n\big)
\end{align}$$
where $\Im$ denotes the imaginary part of a complex number.
| {
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If x, y and z are rational and strictly positive and if $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$, show that $\sqrt{x^2+y^2+z^2}$ is rational This is a question from "Pathfinder to Mathematical Olympiads", I know that we need to prove that $x^2+y^2+z^2$ is a perfect square of a rational number. I did prove it but I'm wondering if there's a better solution than mine. And there is no solution given in the textbook.
I'm not able to add a pic of my solution so I'll briefly explain it.
$\ z = \frac{xy}{x+y}$, so I found the value of $\ x^2+y^2+z^2$ as a fraction in terms of x and y.
The denominator turned out to be $\ (x+y)^2$ which is a perfect square. So I factored the numerator and showed that it was a perfect square. Hence the whole fraction is a perfect square.
I hope you understood my solution.
P.S. I'm only 13 years old, so please don't put too advanced solutions.
| Replace $z$ by $-z$, we get $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ and do not change $x^2 + y^2 + z^2$, then we can calculate that $xy + yz + zx = 0$, so $x^2 + y^2 + z^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) = (x + y + z)^2$, and $\sqrt{x^2 + y^2 + z^2} = x + y + z$ is rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding all angles that satisfy $8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
$\text { Hence, solve the equation } 8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
The previous part was to prove that $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \quad \text { by replacing } 3 \theta \text { by }(2 \theta+\theta)$.
So, I used this to simplify the equation to
$2 \cos 3 \theta +1 = 0$
$\implies \cos 3 \theta =\frac{-1}{2}$
Since $\cos^{-1} \frac{-1}{2} = \frac{2\pi}{3} + 2 \pi n$,$\implies \theta =\frac{2 \pi}{9},\frac{8 \pi}{9},\frac{-8 \pi}{9}$ or $\frac{-2 \pi}{9}$. However, the graph seems to be showing another root which is $\frac{4 \pi}{9}$. Why did I miss this root? How should I find more angles that satisfy the equation in a given range. In general, how does one find all angles that satisfy an equation even after adding $2 \pi n$
| By your work $$\theta=\pm40^{\circ}+120^{\circ}k,$$ where $k\in\mathbb Z,$ which gives $$\{\pm40^{\circ},\pm80^{\circ},\pm160^{\circ}\}.$$
We see these roots in your picture.
| {
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If I take a line, take the cube, add a constant, over what range is the cube root roughly linear? If a line is linear
$$y= mx+b$$
If I cube the expression
$$y^{3} = (mx +b)^{3}$$
And now add a constant
$$y^{3} = (mx +b)^{3}+C$$
Roughly where is the line still linear if I take the cube root again?
$$y = ((mx +b)^{3}+C)^{1/3}$$
$$ y = (b^3+3 b^2 m x+3 b m^2 x^2+m^3 x^3 + C)^{1/3}$$
I plot it, I have something that looks very linear after a certain point
| We have that
$$\begin{align*}
y &= \sqrt[3]{(mx+b)^3+C} \\
&= \sqrt[3]{(mx+b)^3\cdot\left(1+\dfrac{C}{(mx+b)^3}\right)} \\
&= (mx+b)\cdot\sqrt[3]{1+\dfrac{C}{(mx+b)^3}} \\
\end{align*}$$
As $x\to\infty$, or as $x\to-\infty$, we have that $\dfrac{C}{(mx+b)^3}\to0$ and $\sqrt[3]{1+\dfrac{C}{(mx+b)^3}}\to1$.
So $\sqrt[3]{(mx+b)^3+C}$ will get closer to $mx+b$ if we make $x$ very big or very negative.
Edit: When I wrote the above, I was assuming that $m\ne0$. Of course, if $m=0$ and $y=\sqrt[3]{(mx+b)^3+C}$, then $y$ is constant.
| {
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why is $e^{\frac{\pi i}{3}}$ a root of $f(x) = x^{2}-x+1$? I am not sure how this comes about since, $f(e^{\frac{\pi i}{3}})=(e^{2 \pi i})^{\frac{1}{3}}-(e^{\pi i})^{\frac{1}{3}}+1 = (1)^{\frac{1}{3}}-(-1)^{\frac{1}{3}}+1 = 1+1+1 = 3$
| $x=e^{i\pi/3}=1/2+\sqrt{3}i/2.$
Then $$f(e^{i\pi/3})=[1/2+i\sqrt{3}/2]^2-[1/2+i\sqrt{3}/2]+1=1/4-3/4+i\sqrt{3}/2-1/2-i\sqrt{3}/2+1=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int\limits_{-\infty}^{\infty}\frac{\sin x}{x}\cdot\frac{\sin\frac{x}{3}}{\frac{x}{3}}\cdot\frac{\sin\frac{x}{5}}{\frac{x}{5}}dx$ I would greatly appreciate it if someone could find a mistake in my solution to the following problem.
Evaluate the integral
$$
I=\int\limits_{-\infty}^{+\infty}\frac{\sin x}{x}\cdot\frac{\sin\frac{x}{3}}{\frac{x}{3}}\cdot\frac{\sin\frac{x}{5}}{\frac{x}{5}}dx
$$
I know that it must equal $\pi$. However, I can't understand what is wrong in my calculations. I used the following Fourier transform:
$$
\hat{f}(y)=F[f(x)]=\int\limits_{-\infty}^{+\infty} f(x)e^{ixy}dx
$$
Thus, we have ($I$ is an Indicator function):
$$
F\left[\frac{\sin \frac{x}{a}}{\frac{x}{a}}\right]=a\pi\cdot I_{\left[-\frac{1}{a}, \frac{1}{a}\right]}(y)
$$
Two other formulas that I used (The sign '$*$' is convolution):
$$
\begin{aligned}
&\int\limits_{-\infty}^{+\infty} f(x)g(x)dx=\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty} \hat{f}(y)\hat{g}(y)dy\\
&F[fg]=\frac{1}{2\pi}F[f]*F[g]
\end{aligned}
$$
Also, I used one of the properties of convolution:
$$
\text{if}\ \ \ \int\limits_{-\infty}^{+\infty} g(x)dx=1\ \ \ \text{then}\ \ \ \int\limits_{-\infty}^{+\infty} f(x)*g(x)dx=\int\limits_{-\infty}^{+\infty} f(x)dx
$$
So, here is my solution:
$$
\begin{aligned}
&\text{let}\ \ \ \frac{\sin x}{x}=f(x)\ \ \ \text{and}\ \ \ \frac{\sin\frac{x}{3}}{\frac{x}{3}}\cdot\frac{\sin\frac{x}{5}}{\frac{x}{5}}=g(x)\\
&I=\int\limits_{-\infty}^{+\infty}f(x)g(x)dx=\int\limits_{-\infty}^{+\infty} \hat{f}(y)\cdot\frac{1}{2\pi}\cdot\hat{g}(y)dy=\frac{1}{2}\int\limits_{-\infty}^{+\infty} I_{[-1, 1]}(y)\cdot\hat{g}(y)dy\\
&\hat{g}(y)dy=F\left[\frac{\sin\frac{x}{3}}{\frac{x}{3}}\cdot\frac{\sin\frac{x}{5}}{\frac{x}{5}}\right]=\frac{1}{2\pi}\cdot 3\pi I_{\left[-\frac{1}{3}, \frac{1}{3}\right]}(y)*5\pi I_{\left[-\frac{1}{5}, \frac{1}{5}\right]}(y)=\\
&=3\pi I_{\left[-\frac{1}{3}, \frac{1}{3}\right]}(y)*\frac{5}{2}\cdot I_{\left[-\frac{1}{5}, \frac{1}{5}\right]}(y)\Rightarrow\text{Here I applied that convolution property}\Rightarrow\\
&\Rightarrow \int\limits_{-\infty}^{+\infty}\hat{g}(y)dy=\int\limits_{-\infty}^{+\infty} 3\pi\cdot I_{\left[-\frac{1}{3},\frac{1}{3}\right]}(y)dy=3\pi\cdot\frac{2}{3}=2\pi
\end{aligned}
$$
Therefore, we finally get
$$
I=\frac{1}{2}\int\limits_{-\infty}^{+\infty} I_{[-1,1]}(y)\cdot 2\pi dy=2\pi
$$
So, what's wrong?
| What’s wrong is that you replaced $\hat g(y)$ in the integrand by $\int_{-\infty}^\infty\hat g(y)\mathrm dy$, and there’s no reason why you should be able to do that. What you can do, though, is to argue that the convolution of two rectangular pulses of widths $\frac13$ and $\frac15$ has width less than $1$, so you can omit the indicator function for $[-1,1]$ and write
$$
I=\frac12\int\limits_{-\infty}^{+\infty} I_{[-1, 1]}(y)\cdot\hat g(y)\mathrm dy=\frac12\int\limits_{-\infty}^{+\infty} \hat g(y)\mathrm dy=\frac12\cdot2\pi=\pi\;.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Can a Sum of distinct squares ever equal power of two? Does there exist $2^t,\ t\in\mathbb{Z}_+$ which can be express as Sum of two or more distinct square number.
Or
Can it be shown that
$$\begin{split}2^t &\ne \sum a_i^2 = a_1^2+ a_2^2+\cdots+a_n^2\end{split}$$
Where $n\ge 2$ and $\{a_i,t\}\in\mathbb{Z}_+$ and $a_i \ne a_j$ for $1\le i,j \le n$
Example: $2^6=64=7^2+3^2+2^2+1^2+1^2$ here $1^2$ repeat two times so this is not allowed.
My incomplete attempt for arithmetic squares
Edit: check related new post, Can a sum arithmetic square ever equal to power of two?
Let $n,u,d\in\mathbb{Z}_+$
$$\begin{split}\sum_{q=0}^u (n+qd)^2 &=n^2+(n+d)^2+(n+2d)^2+\cdots+(n+ud)^2\\ &=n^2(u+1)+\frac{(u+1)u}{2}(2nd+d^2)+\frac{(u+1)u(u-1)}{3}d^2 \end{split}$$
Let
$$\begin{split}2^t &=\sum_{q=0}^u (n+qd)^2 \\ \implies 3\cdot 2^{t+1}&=6n^2(u+1)+3(u+1)u(2nd-d^2)+(u+1)u(u-1)2d^2 \\ &= (u+1)(6n^2+3u(2nd+d^2)+u(u-1)2d^2)\\ &(in\ case,\ u+1= 3) \\
\implies 2^t&= 3n^2+3(2nd+d^2)+2d^2\\ &= n^2+(n+1)^2+(n+2d)^2 \end{split}$$
Now we need to simplify for case, $6n^2+3u(2nd+d^2)+u(u-1)2d^2=3\cdot2^x$ and $u+1=2^y$ where $x+y=t+1$ but I'm stuck here. Thank you.
Related post:
Can a sum of consecutive $n$th powers ever equal a power of two?
| We have, for instance,
$$
169 + 49 + 25 + 9 + 4=256
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq\frac{3n-2}{2n-1}$. Let $$x_n=\frac{1}{1^2}+\frac{1}{2^2}+\ldots+\frac{1}{n^2},\;\forall n\geq 1.$$ Show that $$\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq \frac{3n-2}{2n-1}.$$
I tried to prove it by induction but I got a contradiction.
| First note that
\begin{align}x_k&\geq 1+\frac1{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots+\frac{1}{k\cdot(k+1)}\\&=1+\frac{1}{2}-\frac{1}{k+1}=\frac{3}{2}-\frac{1}{k+1}=\frac{3k+1}{2(k+1)}.\end{align}
Therefore,
$$k^2x_k^2\geq \frac{k^2(3k+1)^2}{4(k+1)^2}.$$
For $k\ge 2$,
$$\frac{1}{k^2x_k^2}\leq \frac{4(k+1)^2}{k^2(3k+1)^2}<\frac{4}{9}\left(\frac{1}{k-7/6}-\frac{1}{k-1/6}\right).$$
This is because
\begin{align}4k^2(3k+1)^2-36(k+1)^2(k-7/6)(k-1/6)&=\frac{(57k+17)^2-688}{57}\\&\geq\frac{131^2-688}{57}>0. \end{align}
Therefore for $n\ge 2$,
\begin{align}\sum_{k=1}^n\frac{1}{k^2x_k^2}&=1+\frac{1}{5}+\sum_{k=3}^n\frac{1}{k^2x_k^2}\le 1+\frac15+\frac{4}{9}\sum_{k=3}^n\left(\frac{1}{k-7/6}-\frac{1}{k-1/6}\right)
\\&\le 1+\frac{1}{5}+\frac{4}{9}\left(\frac{1}{3-\frac76}-\frac{1}{n-\frac16}\right)=\frac{238}{165}-\frac{8}{3(6n-1)}\\&=\frac{2(238n-113)}{55(6n-1)}<\frac{3n-2}{2n-1}\end{align}
because
\begin{align}55(6n-1)(3n-2)-2(238n-113)(2n-1)&=\frac{(76n+103)^2-28241}{152}
\\&\geq \frac{255^2-28241}{152}>0.\end{align}
For $n=1$, the inequality to be proven is an equality. Therefore, the desired ineq holds for all positive integers $n$.
| {
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"url": "https://math.stackexchange.com/questions/3565070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit with or without lhopital I am having trouble seeing what is the limits for the following expression. I tried L'hopital without any major success.
$$
\lim_{x \to 0} (1+\frac{1}{x^\frac{1-a}{a}})^{\frac{a}{1-a}} \left(x^{\frac{1}{a}}+x \right)
$$
where $0 < a <1$.
Any views? Thanks
| As Brian Moehring's question comment indicates, it's reasonable to assume that $x \to 0+$ as you will get undefined behavior if you allow $x$ to be negative for fractional powers. Thus, this is what this answer will use.
You have
$$\begin{equation}\begin{aligned}
\lim_{x \to 0^{+}} \left(1+\frac{1}{x^\frac{1-a}{a}}\right)^{\frac{a}{1-a}} \left(x^{\frac{1}{a}}+x \right) & = \lim_{x \to 0^{+}} \left(\frac{x^\frac{1-a}{a} + 1}{x^\frac{1-a}{a}}\right)^{\frac{a}{1-a}} \left(x^{\frac{1}{a}}+x \right) \\
& = \lim_{x \to 0^{+}} \left(\frac{\left(x^\frac{1-a}{a} + 1\right)^{\frac{a}{1-a}}}{\left(x^\frac{1-a}{a}\right)^{\frac{a}{1-a}}}\right) \left(x^{\frac{1}{a}}+x \right) \\
& = \lim_{x \to 0^{+}} \frac{\left(x^\frac{1-a}{a} + 1\right)^{\frac{a}{1-a}}}{x} \left(x^{\frac{1}{a}}+x \right) \\
& = \lim_{x \to 0^{+}} \left(x^\frac{1-a}{a} + 1\right)^{\frac{a}{1-a}} \left(x^{\frac{1-a}{a}} +1\right) \\
& = \lim_{x \to 0^{+}} \left(x^\frac{1-a}{a} + 1\right)^{\frac{1}{1-a}} \\
& = 1
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
The last line comes from $0 \lt a \lt 1$ meaning $1 - a \gt 0$, so the powers involved are positive. Thus, as $x \to 0^{+}$, you have $x^\frac{1-a}{a} + 1 \to 1$, so the power of it goes to $1$.
| {
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Find $a,b$ and $c$ if $(1+\sqrt[3]{2})^{-1}$ in the form of $a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$ Given that
$$ (1+\sqrt[3]{2})^{-1} =a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$$
find the value of rationals $a,b,c.$
Solution I tried: I tried to rationalize it, but I'm not getting the answer:
$$\frac{1}{(1+\sqrt[3]{2})}\times \frac{(1-\sqrt[3]{2})}{(1-\sqrt[3]{2})}$$
$$\frac{(1-\sqrt[3]{2})}{1-2^{\frac{2}{3}}}$$
so doing so not getting answer; also, I tried to expand it, but we have condition that $(1+x)^n$ where $n$ is in fraction can be expandable only when $x < 1$, but here the cube root of $2$ is not less than $1.$
Thank you
| $\alpha=1+\sqrt[3]{2}$ is a root of $(x-1)^3=2$, and so
$$
3 = \alpha^3 - 3 \alpha^2 + 3 \alpha = \alpha(\alpha^2 - 3 \alpha + 3)
$$
Therefore,
$$
\alpha^{-1}=\frac13(\alpha^2 - 3 \alpha + 3)
$$
Now use that
$$
\alpha^2 = 1+2\sqrt[3]2+\sqrt[3]4
$$
This solution works in general because every nonzero algebraic number is a root of a polynomial with nonzero independent term.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\cos(z) =3/4+i/4$ I need to solve the complex trinometric equation
$$\cos(z) =\frac{3}{4}+\frac{i}{4} $$
What I've done so far is:
Using the cos formula I got $e^{iz} +e^{-iz} =\frac{3}{2}+\frac{i}{2}$
Making $t=e^{iz} $ we have $t+\frac{1}{t}=\frac{3}{2}+\frac{i}{2}$
Multiplying by $t^2$ we get
$$t^2-\frac{3+i}{2}t+1=0$$
Solving that we get
$$t=\frac{(\frac{3}{2}+\frac{i}{2}) \pm \sqrt{(\frac{3}{2}+\frac{i}{2}) ^2-4}} {2} = \frac{3+i \pm \sqrt{-8+6i} } {4} $$
Converting 3+i to polar we get $\sqrt{10} e^{0.3218i}$
Converting $\sqrt{-8+6i}$ to polar we get $\sqrt{10} e^{-0.3218i}$
So $t=\frac{\sqrt{10} e^{0.3218i} \pm \sqrt{10} e^{-0.3218i}} {4} $
Which means $e^{iz} =\frac{\sqrt{10} e^{0.3218i} \pm \sqrt{10} e^{-0.3218i}} {4} = \frac{\sqrt{10}} {2} (\frac{e^{0.3218i} \pm e^{-0.3218i} }{2})$
And I dont know where to go from there
| Taking a hint from @ChiefVS, one solution is $\alpha\approx 0.785398-i\, 0.346574$
All solutions are then:
$\{\alpha+2k\pi, -\alpha+2k\pi\}, \quad k\in \mathbb{Z}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $2^n = 10^6$ for $n$ I am trying to solve $2^n = 10^6$ for $n$. Is there some way that I can solve this analytically (say, using base $2$ logarithms), or must I solve it numerically? Thank you.
| Well, sure. $\log_2 2^n = \log_2 10^6$
$n = 6\log_2 10$ or
$\log_{10} 2^n = \log_{10} 10^6$
$n \log_{10} = 6$
$n = \frac {6}{\log_{10}2}$ or
$\log_b 2^n = \log_b 10^6$
$n\log_b 2^n = 6\log_b 10$
$n = 6\frac {\log_b 10}{\log_b 2}$ for any $b > 0; b\ne 1$ are all acceptable answers.
But it's worth noting that
1) $\frac {\log_b 10}{\log_b 2} = \log_2 {10}$ always. that's called the change of base formula and a special case is $\frac 1{\log_{10} 2}=\frac {\log_{10} 10}{\log_{10} 2} = \log_2 {10}$.
2) $\log_2 10= \log_2 (2*5) = \log_2 2 + \log_2 5 = 1 + \log_2 5$.
So $2^n = 10^6 = 2^6*5^6$
$\log_2 2^n = \log_2 (2^6*5^6)$
$n = \log_2 2^6 + \log_2 5^6 = 6 + 6\log_2 5=6(1 + \log_2 5)$.
All acceptable answers.
| {
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If the roots of $x^2-7x+1=0$ are $p$ and $q$, find a quadratic equation with roots $p^2+q^2$ and $\sqrt{p}+\sqrt{q}$
Let $p$ and $q$ be the roots of $x^2-7x+1=0$. Find a quadratic equation with roots of $p^2+q^2$ and $\sqrt{p} + \sqrt{q}$.
Attempt to solve:
What I got from the previous quadratic equation
$$p+q=-\frac{b}{a}=7 \text{ and } p\cdot q=\frac{c}{a}=1$$
I've got no problem for solving the first root, all I need is doing a little expansion
$$p^2+q^2=(p+q)^2-2pq=7^2-2(1)=47$$
The problem starts when I'm trying to find the second root. I tried to do this
$$p-q=\left(\sqrt{p} + \sqrt{q}\right)\left(\sqrt{p} - \sqrt{q}\right)$$
Well, how do I get the value of both $p-q$ and $\left(\sqrt{p} - \sqrt{q}\right)$ or maybe there are other method(s) to get the second root.
| Let $x^2=y$ and $x$ be a root of $$ax^2+bx+c=0$$
$$(ax^2+c)^2=(-bx)^2$$
$$\implies(ay+c)^2=b^2y\iff a^2y^2+y(2ca-b^2)y+c^2=0$$
So, in the first case, $$a=c=1,b=-7$$
For the second, $$\dfrac{a^2}1=\dfrac{b^2-2ca}7=\dfrac{c^2}1$$
$\implies a^2=c^2\iff c=\pm a$
But $\dfrac ca=\sqrt p\sqrt q\ge0$ for real $p,q$
So, $c=a,$ $$\dfrac{b^2-2ca}7=c^2\implies\dfrac{b^2-2c(a)}7=(-a)^2\iff b^2=9a^2\iff b=? $$
But $-\dfrac ba=\sqrt p+\sqrt q\ge0$ for real $p,q$
Hope you can take it from here?
| {
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Two projectiles at an initial height with a missing angle of projection
David and Goliath are testing what they learned in school, that the
distance travelled by launching a projectile is dependent on a range
of variables; the launch angle, height and velocity of projection.
As Goliath is significantly taller, David decides to throw a small
projectile while standing on a ladder so that he is throwing from the
same vertical height and same spot as Goliath.
Goliath launches first with an angle of 60$^{\circ}$ to the
horizontal. David will then launch the projectile next with the same
velocity of projection as Goliath.
If David wants to match Goliath’s attempt and achieve the same
horizontal distance, at what angle should he launch the projectile if
he can’t do it at an angle of 60$^{\circ}$?
(Use $ = −10ms^{-1}$ in your calculations and use this as the only
force acting upon the projectile.)
If David's projectile's displacement is given by $r_d = v t \cos \theta \hat{i} + (h + v t \sin \theta - 5t^2) \hat{j}$ then David's projectile hits the ground when $\frac{v \sin \theta \pm \sqrt{v^2 \sin^2\theta+20h}}{10} = 0$. Solving this, the time it takes David's projectile to reach the ground is $T_d = \frac{v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}}{10}$.
If Goliath's projectile's displacement is given by $r = \tfrac{v}{2} t \hat{i} + (h + \tfrac{\sqrt{3}v}{2}t - 5t^2) \hat{j}$ then Goliath's projectile hits the ground when $\frac{\sqrt{3}v \pm \sqrt{3v^2 + 80h}}{20} = 0$. Solving this, the time it takes Goliath's projectile to reach the ground is $T_g = \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{20}$.
For the horizontal distance travelled to match, their terminal horizontal components must match. Hence,
\begin{align*}
v T_d \cos \theta &= \tfrac{v}{2} T_g\\
T_d \cos \theta &= \tfrac{1}{2} T_g\\
\cos \theta &= \frac{T_g}{2} \frac{1}{T_d}\\
&= \frac{\frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{20}}{2} \frac{10}{v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}}\\
\cos \theta &= \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{4 (v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}) }\\
\end{align*}
I'm having a major mental block on how to obtain an expression for $\theta$ in terms of $h$ and $v$. For what its worth, wolframalpha gives $\theta = -\cos^{-1}\left(-\frac{\sqrt{200 h^2 + 70 h v^2 + 10 \sqrt{3} h v \sqrt{80 h + 3 v^2} + \sqrt{3} v^3 \sqrt{80 h + 3 v^2} + 3 v^4}}{\sqrt{800 h^2 + 80 h v^2 + 8 v^4}}\right)$. I seem to missing out some details in my calculations.
What's the simplest way to obtain the formula for $\theta$ in terms of $v$ and $h$?
| From
\begin{equation*}
\cos \theta = \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{4 (v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}) },
\end{equation*}
multiplying both numerator and denominator by $v\sin{\theta} - \sqrt{v^{2}\sin^{2}{\theta} + 20h}$ gives
\begin{align*}
\cos \theta &= \frac{(\sqrt{3}v + \sqrt{3v^2 + 80h})(v\sin{\theta} - \sqrt{v^{2}\sin^{2}{\theta} + 20h})}{4 (-20h) }\\
&= A(v\sin{\theta} - \sqrt{v^{2}\sin^{2}{\theta}+20h}),
\end{align*}
where we have set $A= -\dfrac{\sqrt{3}v + \sqrt{3v^{2} + 80h}}{80h}.$ We then get
\begin{align*}
\Big(\frac{\cos{\theta}}{A} - v\sin{\theta}\Big)^{2} &= v^{2}\sin^{2}{\theta} + 20h,
\end{align*}
or equivalently
\begin{align*}
\frac{\cos^{2}{\theta}}{A^{2}} - 2\frac{v\cos{\theta}\sin{\theta}}{A} = 20h.
\end{align*}
Thus
\begin{equation*}
(\cos^{2}{\theta} - 20A^{2}h)^{2} = 4A^{2}v^{2}\cos^{2}{\theta}(1-\cos^{2}{\theta}).
\end{equation*}
Expanding this equation gives the quadratic equation in $\cos^{2}{\theta}$
\begin{equation*}
(1+4A^{2}v^{2})\cos^{4}{\theta}- 4A^{2}(10h + v^{2})\cos^{2}{\theta} + 400 A^{4}h^{2} = 0,
\end{equation*}
and solving for $\cos^{2}{\theta}$ yields
\begin{align*}
\cos^{2}{\theta} &= \frac{4A^{2}(10h+v^{2})\pm\sqrt{16A^{4}(10h+v^{2})^{2}-4(400A^{4}h^{2})(1+4A^{2}v^{2})}}{2(1+4A^{2}v^{2})} \\[1ex]
&= \frac{2A^{2}(10h+v^{2})\pm 2A^{2}\sqrt{(10h+v^{2})^{2}-100h^{2}(1+4A^{2}v^{2})}}{1+4A^{2}v^{2}}.
\end{align*}
There will be two solutions for $\theta$, one of them being at $\theta = 60^{\circ}$ which we aren't interested in, and the other being at some angle less than $60^{\circ}$. To get the smaller angle, we take the positive sign in the above expression since $\cos^{2}{\theta}$ is decreasing on $\Big(0, \dfrac{\pi}{2}\Big)$. Finally, as $\cos{\theta} > 0$, we obtain
\begin{align*}
\cos{\theta} = \sqrt{\frac{2A^{2}(10h+v^{2})+2A^{2}\sqrt{(10h+v^{2})^{2}-100h^{2}(1+4A^{2}v^{2})}}{1+4A^{2}v^{2}}}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3571771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$ Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$
I don't really know how to approach this. I was thinking doing something like squaring or cubing $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$ would help, but it doesn't really work out...Any help???
| Use the relationship,
$$ x^3+y^3+z^3 = (x+y+z)^3 - 3(x+y)(y+z)(z+x)$$
to evaluate,
$$a+b+c= (\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c})^3
-3(\sqrt[3]{a} + \sqrt[3]{b})(\sqrt[3]{b} + \sqrt[3]{c})(\sqrt[3]{c} + \sqrt[3]{a} )$$
$$= 0^3 -3(0- \sqrt[3]{c})(0- \sqrt[3]{a})(0- \sqrt[3]{b} )=3\sqrt[3]{abc}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Differentiating secant inverse I want to show that $\cfrac{d}{dx}[\operatorname{arcsec} x]=\cfrac{1}{\vert x \vert\sqrt{x^2-1}}$. Here is my attempt:
\begin{align*} y=\sec^{-1}x &\iff x=\sec y \\ \frac{d}{dx}[x]&=\frac{d}{dx}[\sec y] \\ 1 &=\frac{\cos^2y}{\sin y} \frac{dy}{dx} \\ \frac{dy}{dx} &= \frac{\sin y}{\cos^2y} \end{align*}
I tried using the Pythagorean identity for 1 on the lefthand side, but couldn't see what else that would bring. I'm starting to think implicit differentiation isn't going to let me derive the expression I'm looking for.
Any ideas?
| You have
$$x = \sec y = \frac{1}{\cos y} \implies \cos y = \frac{1}{x} \tag{1}\label{eq1A}$$
Also, you have, assuming that $\sin y \ge 0$, that
$$\sin^2 y + \cos^2 y = 1 \implies \sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{|x|} \tag{2}\label{eq2A}$$
Note you have a mistake in your differentiation where you got the reciprocal of what you should have had. In particular,
$$\begin{equation}\begin{aligned}
\frac{d}{dx}[\sec y] & = \frac{d}{dx}\left(\frac{1}{\cos y}\right) \\
& = \left(-\frac{(-\sin y)}{\cos^2 y}\right)\frac{dy}{dx} \\
& = \left(\frac{\sin y}{\cos^2 y}\right)\frac{dy}{dx}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This then gives
$$\frac{dy}{dx} = \frac{\cos^2 y}{\sin y} \tag{4}\label{eq4A}$$
Using \eqref{eq1A} and \eqref{eq2A} in \eqref{eq4A} then gives the expression you're trying to prove, i.e, you get
$$\begin{equation}\begin{aligned}
\frac{dy}{dx} & = \frac{\frac{1}{x^2}}{\frac{\sqrt{x^2 - 1}}{|x|}} \\
& = \frac{|x|}{x^2\sqrt{x^2 - 1}} \\
& = \frac{1}{|x|\sqrt{x^2 - 1}}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $a_n = \frac{\sqrt{n+1} + \sqrt{n+2} +...+ \sqrt{2n}}{n^{3/2}}$ calculate $\displaystyle \lim_{n \to \infty}a_n$ using integrals
Let
$$
a_n = \frac{\sqrt{n+1} + \sqrt{n+2} +...+ \sqrt{2n}}{n^{3/2}}
$$
My goal is to calculate $\displaystyle \lim_{n \to \infty}a_n$ using integrals.
What I have tried so far and got stucked:
Let:
$$f(x) = \sqrt{n + x}$$
Therefore:
$$a_n = \frac{\sqrt{n+1} + \sqrt{n+2} +...+ \sqrt{2n}}{n^{3/2}} = \frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{\sqrt{n}}\right)$$
Now, I know that there is a relation between the Riemann sum and integrals generally speaking:
$$
\lim_{n \to \infty} \frac{b-a}{n} \sum_{k = 1}^n f\left(a + \frac{k(b-a)}{n}\right) = \int^{b}_{a}f(x) dx
$$
Therefore, what I think I can conlude so far from $\displaystyle \frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{\sqrt{n}}\right)$ is:
$$
b = 1, a = 0, f(x) = \sqrt{n + x}
$$
But it seems wrong, $f(x)$ has $n$ in it, which is weird as a constant here, so i am not sure i did the whole correlation quite right.
| if we have $f(x) > 0$ and $f'(x) > 0,$ then
$$ \int_{a-1}^{b} \; f(x) \; dx \; < \; \sum_{j=a}^b \; f(j) \; < \; \int_{a}^{b+1} \; f(x) \; dx $$
Taking $a=n+1,$ $ b = 2n,$
$$ \int_{n}^{2n} \; \sqrt x \; dx \; < \; \sum_{j=n+1}^{2n} \; \sqrt j \; < \; \int_{n+1}^{2n+1} \; \sqrt x \; dx $$
and a suitable antiderivative is $$ \frac{2}{3} n^{3/2} \; . $$
$$ \frac{2}{3} (2n)^{3/2} - \frac{2}{3} (n)^{3/2} < \; \sum_{j=n+1}^{2n} \; \sqrt j \; < \; \frac{2}{3} (2n+1)^{3/2} - \frac{2}{3} (n+1)^{3/2} $$
The left hand side is $$ \frac{2n^{3/2}}{3} (2^{3/2} - 1) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3575261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$
Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$
My Attempt:
$c$ cannot be $5$ since $70 \times 5 > 343$.
It can't be $4$ either since if we put $c = 4$, we get $30a + 50b < 63$, which would mean there are no positive integer solutions for at least one of $a$ or $b$.
So, there are 3 options for $c$.
Proceeding similarly, we will get that $b = 1,2,3,4$ or there are $4$ options for b.
Similarly, there are $7$ options for $c$.
Since any value of $a$ can be paired with any value of $b$ and $c$, we get $3 \times 4 \times 7=84$ total triplets which is not the answer
| You are correct in saying that $c=1,2,3$ but this is where you then made the mistake. We need to consider each of these cases separately
When $c=3$, we have $30a+50b<133$. Here, $b$ cannot be greater than $3$ as $50\times3=150>133$. So we have $b=1,2$.
When $b=1$, then we have $30a<83$ meaning $a<2.766\ldots$ so $a=1,2$ are the integer solutions
When $b=2$, then we have $30a<33$ meaning $a<1.1$ so $a=1$ is the only integer solution
So, for $c=3$, we have $3$ possible solutions:
\begin{align}(a,b,c)&=(1,2,3)\\
&=(1,1,3)\\
&=(2,1,3)\end{align}
Now we can do similar things for $c=2$ and $c=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
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