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Find the domain of $f(x)=\log_{10}(1-\log_7(x^2-5x+13))+\arccos\left(\frac{3}{2+\sin\frac{9\pi x}{2}}\right)$ Find the domain of definition of $$f(x)=\log_{10}(1-\log_7(x^2-5x+13))+\arccos\left(\frac{3}{2+\sin\frac{9\pi x}{2}}\right)$$
I found the domain of $\log_{10}(1-\log_7(x^2-5x+13))$ to be $x\in(2,3)$ and the domain of $\arccos\left(\frac{3}{2+\sin\frac{9\pi x}{2}}\right)$ to be $\sin\frac{9\pi x}{2}=1$ i.e. $x=1,5,9$ etc but the answer given is $\frac{21}{9},\frac{25}{9}$.I dont know where i am wrong.
| Since $\sin(\frac{9\pi x}{2})=1$,
$$\frac{9\pi x}{2}\in(\frac{\pi}{2}, \frac{5\pi}{2}...\frac{21\pi}{2}...) \quad or \quad x\in(\frac{1}{9}, \frac{5}{9},..., \frac{21}{9}, \frac{25}{9}...)$$
We have seen that the domain of the first part of equation should be in $(2,3)$, So the only feasible solution is $\frac{21}{9}$ and $\frac{25}{9}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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let $a,b,c \in \mathbb{R^+} \ \ a+b+c =1$ Then prove that : $a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$
Let $\{a,b,c\}\subset\mathbb{R^+}$ such that $a+b+c =1$. Prove that :
$$a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$$
$$a^3+b^3+c^3-3abc=(x+b+c)(a^2+b^2+c^2-(ab +ac+bc))$$
$$a^2+b^2+c^2=(a+b+c)^2-2(ac+bc+ab)$$
Now what ?
| Let us consider the vectors $u=(\sqrt{a},\sqrt{b},\sqrt{c})$, $v=(a\sqrt{a},b\sqrt{b},c\sqrt{c})$.
By the Cauchy-Schwarz inequality
$$ a^2+b^2+c^2 = \left|u\cdot v\right|\leq \|u\|\cdot\|v\|=\sqrt{a+b+c}\sqrt{a^3+b^3+c^3} $$
but we also have
$$ 3(a^2+b^2+c^2)=(1+1+1)(a^2+b^2+c^2)\geq (a+b+c)^2 = 1 $$
hence
$$ a^3+b^3+c^3 \geq (a^2+b^2+c^2)^2 \geq \frac{1}{3}(a^2+b^2+c^2).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show $\exp(ix)=\cos x+i\sin x$ using $\exp(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$ I have the following problem:
How to show $\exp(ix)=\cos x+i\sin x$ using $\exp(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$
My attempt:
\begin{align*}
\exp(ix)=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}
\end{align*}
Now split the sum in 4 sums because $i^n$ has period 4, more precisely $i^{4n}=1$, $i^{4n+1}=i$, $i^{4n+2}=-1$ and $i^{4n+3}=-i$ for $n\geq 0$.
\begin{align*}
\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}&=\sum_{n=0}^{\infty}\frac{(ix)^{4n}}{(4n)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+1}}{(4n+1)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+2}}{(4n+2)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+3}}{(4n+3)!}\\
&=\sum_{n=0}^{\infty}\left(\frac{x^{4n}}{(4n)!}-\frac{x^{4n+2}}{(4n+2)!}\right)+i\sum_{n=0}^{\infty}\left(\frac{x^{4n+1}}{(4n+1)!}-\frac{x^{4n+3}}{(4n+3)!}\right)
\end{align*}
Then $4n$ and $4n+2$ are nonnegative and even numbers and $4n+1$ and $4n+3$ are odd. How can I use this to convert the two last expressions into
\begin{align*}
\cos x&=\sum_{m=0}^{\infty}(-1)^m \frac{x^{2m}}{(2m)!}\\
\sin x&=\sum_{m=0}^{\infty}(-1)^m \frac{x^{2m+1}}{(2m+1)!}
\end{align*}
I don't want to see proofs like
\begin{align*}
\exp(ix)&=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\frac{(ix)^5}{5!}\ldots\\
&=\Big(1-\frac{x^2}{2!}+\frac{x^4}{4!}\mp\ldots\Big)+i\Big(x-\frac{x^3}{3!}+\frac{x^5}{5!}\mp\ldots\Big)\\
&=\cos x+i\sin x
\end{align*}
| You split the sum like that:
\begin{align*}
\exp(ix) = \sum_{n=0}^{\infty}\frac{(ix)^n}{n!}&=\sum_{n=0}^{\infty}\frac{(ix)^{4n}}{(4n)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+1}}{(4n+1)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+2}}{(4n+2)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+3}}{(4n+3)!}\\
&=\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}- \sum_{n=0}^{\infty}\frac{x^{4n+2}}{(4n+2)!} + i\left( \sum_{n=0}^{\infty}\frac{x^{4n+1}}{(4n+1)!} - \sum_{n=0}^{\infty}\frac{x^{4n+3}}{(4n+3)!}\right)
\end{align*}
Now we join the sums as follows:
\begin{align*}
\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}- \sum_{n=0}^{\infty}\frac{x^{4n+2}}{(4n+2)!} &= \sum_{n=0}^{\infty} (-1)^{2n}\frac{x^{4n}}{(4n)!} + \sum_{n=0}^{\infty}(-1)^{2 n+ 1}\frac{x^{4n+2}}{(4n+2)!} \\
&= \sum_{n=0}^{\infty} (-1)^{2n}\frac{x^{2(2n)}}{(2(2n))!} + \sum_{n=0}^{\infty}(-1)^{2n+ 1}\frac{x^{2(2n+1)}}{(2(2n+1))!} \\
&= \sum_{m \in \mathbb N_0 \text{ even}} (-1)^{m}\frac{x^{2m}}{(2m)!} + \sum_{m \in \mathbb N_0 \text{ odd}} (-1)^{m}\frac{x^{2m}}{(2m)!} \\
&= \sum_{m=0}^{\infty} (-1)^{m}\frac{x^{2m}}{(2m)!} = \cos(x).
\end{align*}
The same trick works for $\sin(x)$, so you get
$$ \exp(ix) = \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}- \sum_{n=0}^{\infty}\frac{x^{4n+2}}{(4n+2)!} + i\left( \sum_{n=0}^{\infty}\frac{x^{4n+1}}{(4n+1)!} - \sum_{n=0}^{\infty}\frac{x^{4n+3}}{(4n+3)!}\right) = \cos(x) + i \sin(x).$$
I hope that helps you :)
| {
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"timestamp": "2023-03-29T00:00:00",
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let $a_1,a_2,. . . ,a_n \in \mathbb{R^+} $ then prove that :
Let $a_1,a_2,. . . ,a_n \in \mathbb{R^+} $. Prove that :
$$\frac{ a_{1} }{ a_{2} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}}
+\frac{ a_{2} }{ a_{1} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}}
+\cdots+\frac{ a_{n} }{ a_{1} ^{2}+ a_{2} ^{2}+\cdots+ a_{n-1} ^{2}}
\geq \frac{ 4}{ a_{1}+ a_{2} + a_{3}+\cdots+ a_{n}}$$
I do not know where to start. please help me .
| I think we can start by C-S:
$$\sum_{k=1}^n\frac{a_k}{\sum\limits_{i\neq k}a_i^2}=\sum_{k=1}^n\frac{a_k^2}{a_k\sum\limits_{i\neq k}a_i^2}\geq\frac{\left(\sum\limits_{k=1}a_k\right)^2}{\sum\limits_{k=1}^na_k\sum\limits_{i\neq k}a_i^2}.$$
Thus, it remains to prove that
$$\left(\sum\limits_{k=1}a_k\right)^3\geq4\sum\limits_{k=1}^n\left(a_k\sum\limits_{i\neq k}a_i^2\right)$$ or
$$\left(\sum\limits_{k=1}a_k\right)^3\geq4\sum\limits_{k=1}^n\left(a_k^2\sum\limits_{i\neq k}a_i\right).$$
Now, let $\sum\limits_{k=1}^na_k=1$.
Hence, we need to prove that $$\sum\limits_{k=1}^nf(a_k)\leq\frac{1}{4},$$
where $f(x)=x^2-x^3$.
But $f''(x)=2-6x$, which says that $f$ has an unique inflection point on $(0,1)$.
Thus, by Vasc's HCF Theorem it's enough to prove our inequality for
$a_1=a_2=...=a_{n-1}=x$ and $a_n=1-(n-1)x$, where $0<x<\frac{1}{n-1}$.
Thus, it's enough to prove that $g(x)\geq0,$ where
$$g(x)=\frac{1}{4}-(n-1)(x^2-x^3)-(1-(n-1)x)^2+(1-(n-1)x)^3.$$
We have
$$g'(x)=(n-1)(1-nx)(3(n-2)x-1).$$
We see that $0<\frac{1}{3(n-2)}<\frac{1}{n}<\frac{1}{n-1}$, $x_{min}=\frac{1}{3(n-2)}$ and $x_{max}=\frac{1}{n}$, which says
$$g(x)\geq\min\left\{g\left(\frac{1}{3(n-2)}\right),g\left(\frac{1}{n-1}\right)\right\}=\min\left\{\frac{11n^2-56n+72}{108(n-2)^2},\frac{(n-3)^2}{4(n-1)^2}\right\}\geq0.$$
Done!
For the proof we can use also the Vasc's EV Method.
| {
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"timestamp": "2023-03-29T00:00:00",
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System of ellipses with same foci Consider a system of ellipses $E_u$ with same foci $(\pm u,0)$, here $u>0$ is a constant. In other words, we consider a system of equations
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a^2-b^2=u^2.$$
Q: How to show that any ellipse in $E_u$ satisfies
$$xyy'^2+(x^2-y^2-u^2)y'-xy=0,$$
here $y'=\frac{dy}{dx}$?
| We have $$\frac{2x}{a^2}+\frac{2yy'}{b^2}=0$$ or
$$y'=-\frac{b^2x}{a^2y}.$$
Thus, we need to prove that
$$\frac{xy\cdot b^4x^2}{a^4y^2}-\frac{(x^2-y^2-a^2+b^2)b^2x}{a^2y}-xy=0$$ or
$$x(b^2-a^2)(b^2x^2-a^2y^2-a^2b^2)=0$$ or
$$(b^2-a^2)x\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}-1\right)=0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\sum_{cyc} \frac{a}{c+a^3+b^2} \leq \frac{1}{2}$
Let $a$, $b$ and $c$ be positive real numbers such that $abc \geq 10+6\sqrt{3}$. Prove that: $$ \displaystyle\sum_{cyc} \frac{a}{c+a^3+b^2} \leq \frac{1}{2}.$$
I've tried to bound the LHS of Michael Rozenberg's last inequality,
$\frac{3\left(7.5^{7.5}\cdot27\cdot6^6\right)^{\frac{1}{16.5}}}{16.5(10+6\sqrt3)^{\frac{6}{16.5}}} = \frac{3}{16.5}\cdot\left(\frac{7.5^{7.5}\cdot27\cdot6^6}{(10+6\sqrt3)^6}\right)^{\frac{1}{16.5}} < \frac{3}{16.5}\cdot\left(\frac{7.5^{7.5}\cdot27\cdot6^6}{20^6}\right)^{\frac{1}{16.5}}$
$7.5^{7.5}=7.5^6\cdot7.5^{1.5}<7.5^6\cdot7.5^2 <7.5^6\cdot2^6$
$<\frac{3}{16.5}\cdot\left(\frac{15^6\cdot(\sqrt{3})^6\cdot6^6}{20^6}\right)^{\frac{1}{16.5}}=\frac{3}{16.5}\cdot\left(\frac{90\sqrt{3})^6}{20^6}\right)^{\frac{1}{16.5}}$
$=\frac{3}{16.5}\cdot\left(\frac{90\sqrt{3}}{20}\right)^{\frac{6}{16.5}}=\frac{3}{16.5}\cdot(4.5\sqrt{3})^{\frac{6}{16.5}}$
$<\frac{3}{16.5}\cdot(4.5\sqrt{3})^{\frac{1}{2}}<\frac{3}{16.5}\cdot(8)^{\frac{1}{2}}<\frac{3}{16.5}\cdot3=0.54$
| Let $a=(1+\sqrt3)kx^2$, $b=(1+\sqrt3)ky^2$ and $c=(1+\sqrt3)kz^2$, where $k>0$ and $xyz=1$.
Hence, $10+6\sqrt3\leq abc=(10+6\sqrt3)k^3xyz$, which gives $k\geq1$ and
$$\sum_{cyc}\frac{a}{a^3+b^2+c}=\sum_{cyc}\frac{x}{k^2(1+\sqrt3)^2x^3+(1+\sqrt3)ky^2+z}\leq$$
$$\leq\sum_{cyc}\frac{x}{(1+\sqrt3)k(2\sqrt{x^3z}+y^2)}\leq\sum_{cyc}\frac{x}{3x^3+2y^2+z}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{x}{3x^3+2y^2+z}\leq\frac{1}{2},$$
which is true, but it's very ugly.
In another hand, by AM-GM we obtain:
$$\sum_{cyc}\frac{a}{a^3+b^2+c}=\sum_{cyc}\frac{a}{7.5\cdot\frac{a^3}{7.5}+3\cdot\frac{b^2}{3}+6\cdot\frac{c}{6}}\leq$$
$$\leq\sum_{cyc}\frac{a}{16.5\left(\frac{a^3}{7.5}\right)^{\frac{7.5}{16.5}}\cdot\left(\frac{b^6}{27}\right)^{\frac{1}{16.5}}\cdot\left(\frac{c^6}{6^6}\right)^{\frac{1}{16.5}}}=$$
$$=\frac{\left(7.5^{7.5}\cdot27\cdot6^6\right)^{\frac{1}{16.5}}}{16.5}\sum_{cyc}\frac{a}{a^{\frac{22.5}{16.5}}b^{\frac{6}{16.5}}c^{\frac{6}{16.5}}}=\frac{\left(7.5^{7.5}\cdot27\cdot6^6\right)^{\frac{1}{16.5}}}{16.5}\sum_{cyc}\frac{1}{a^{\frac{6}{16.5}}b^{\frac{6}{16.5}}c^{\frac{6}{16.5}}}=$$
$$=\frac{3\left(7.5^{7.5}\cdot27\cdot6^6\right)^{\frac{1}{16.5}}}{16.5(abc)^{\frac{6}{16.5}}}\leq\frac{3\left(7.5^{7.5}\cdot27\cdot6^6\right)^{\frac{1}{16.5}}}{16.5(10+6\sqrt3)^{\frac{6}{16.5}}}=0.3555...\leq\frac{1}{2}.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Possibilities for the side and angle of the triangle We have the triangle ABC with $c=18$, $\alpha=\frac{\pi}{6}$ and the median through $C$ has the length $5$. I want to determine all the possibilities for $a$ and $\gamma$.
I have done the following:
From the median we have the following:
$$m_c=\sqrt{\frac{2a^2+2b^2-c^2}{4}}\Rightarrow m_c^2=\frac{2a^2+2b^2-c^2}{4} \Rightarrow 4m_c^2=2a^2+2b^2-c^2 \\ \Rightarrow 4\cdot 25=2a^2+2b^2-18^2 \Rightarrow 2a^2+2b^2=100+324 \Rightarrow 2a^2+2b^2=424 \\ \Rightarrow a^2+b^2=212 \Rightarrow b^2=212-a^2$$
From the cosine law we have that $$a^2=b^2+c^2-2\cdot b\cdot c\cdot \cos \alpha \Rightarrow a^2=212-a^2+c^2-2\sqrt{212-a^2}\cdot c\cdot \frac{\sqrt{3}}{2} \\ \Rightarrow 2a^2=212+324-18\sqrt{3}\sqrt{212-a^2}\Rightarrow 2a^2=536-18\sqrt{3}\sqrt{212-a^2} \\ \Rightarrow a^2=268-9\sqrt{3}\sqrt{212-a^2} \ \ \
\ (1)$$
From the sine law we have that $$\frac{\sin\alpha}{a}=\frac{\sin \gamma}{c} \Rightarrow c\sin \alpha=a\sin \gamma \Rightarrow 18\cdot \frac{1}{2}=a\sin \gamma \Rightarrow a\sin \gamma=9 \ \ \ \ (2)$$
So from the equation we solve for $a$ and from the equation we solve for $\gamma$, right?
Are they then all the posiibilities for $a$ and $\gamma$ ?
| Let $CM$ is a median of our triangle.
Thus, $$\frac{5}{\sin30^{\circ}}=\frac{9}{\sin\measuredangle ACM}$$
Hence, we have to cases.
*
*$\measuredangle ACM=\arcsin\frac{9}{10}$.
Thus, $$AC=\sqrt{9^2+5^2-2\cdot9\cdot5\cdot\cos\left(150^{\circ}-\arcsin\frac{9}{10}\right)}=$$
$$=\sqrt{106-90\left(-\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{19}}{10}+\frac{1}{2}\cdot\frac{9}{10}\right)}=$$
$$=\sqrt{65.5-4.5\sqrt{57}}=\frac{1}{2}\sqrt{262-18\sqrt{57}}=$$
$$=\frac{1}{2}\sqrt{243-18\sqrt{57}+19}=\frac{1}{2}(9\sqrt{3}-\sqrt{19})$$ and
$$BC=\sqrt{18^2+65.5-4.5\sqrt{57}-2\cdot18\cdot\frac{1}{2}(9\sqrt{3}-\sqrt{19})\cdot\frac{\sqrt3}{2}}=$$
$$=\frac{1}{2}\sqrt{586+18\sqrt{57}}.$$
*$\measuredangle ACM=180^{\circ}-\arcsin\frac{9}{10}$.
This case is the same.
| {
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Show that $\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
I want to prove that $\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
My ideas, I don't know if they lead anywhere:
Let's substitute $\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$ and $z=e^{i\theta}$ right after:
$\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=-i\cdot\int_1^{-1}\frac{z^2+z^{-2}+\frac{1}{2}z^3+\frac{1}{2}z^{-3}}{5z+2z^2+2}dz$
This now gives me 4 new integrals, for example
$\displaystyle-i\int_1^{-1}\frac{z^2}{2z^2+5z+2}dz$, $\displaystyle-i\int_1^{-1}\frac{1}{2z^4+5z^3+2z^2}dz$ and so on.
But since I haven't been able to solve any of the new integrals, I'm a little lost.
Edit: Can't I do a partial fractions decomposition of all the 4 integrals and solve them seperately?
| If you wish to exploit the residue theorem, then first exploit the fact that the integral is even. In addition, use Euler's formula to write $2\cos(2\theta)+\cos(3\theta)=\text{Re}(2e^{i2\theta}+e^{i3\theta})$.
Then, we have
$$\begin{align}
\int_0^\pi \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,d\theta&=\frac12\text{Re}\left(\oint_{|z|=1}\frac{2z^2+z^3}{5+2(z+z^{-1})}\,\frac{1}{iz}\,dz\right)\\\\
&=\frac12\text{Re}\left(\frac1i\oint_{|z|=1}\frac{z^2(2+z)}{(2z+1)(z+2)}\,dz\right)\\\\
&=\frac12\text{Re}\left(\frac1i\oint_{|z|=1}\frac{z^2}{2z+1}\,dz\right)\\\\
&=\frac{\pi}{8}
\end{align}$$
And we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Categorizing the conic section given by $x^2-4xy+y^2+8x+2y-5=0$
Find the conic section of $x^2-4xy+y^2+8x+2y-5=0$
I came to the following using diagonalization and bi-linear form
$$-\left(x-\frac{5}{\sqrt{2}}\right)^2+\left(\sqrt{3}y-\frac{\sqrt{3}}{\sqrt{2}}\right)^2-19=0$$
So I can conclude (using eigenvalues)that $\lambda_{1}$ have the same sign as $k=-19$ so it is an empty set?
| First thing is to remove the offset and center the curve on the origin. The easiest way I do this is to take $f(x,y) = x^2-4xy+y^2+8x+2y-5$ and solve the system $$\left. \begin{align} \frac{\partial f}{\partial x} & = 0 \\ \frac{\partial f}{\partial y} & = 0 \end{align} \right\} \begin{aligned} x &= 2 \\
y &= 3 \end{aligned}$$
So the center is $(2,3)$ and the centered curve is found with the substitution
$$ \pmatrix{x \\y} \rightarrow \pmatrix{x+2 \\y+3} $$
$$f(x+2,y+3) = x^2 - 4 x y + y^2 -6 = 0$$
Now we need to rotate the coordinates such as to make the coefficient of $x y$ zero.
$$ \pmatrix{x \\y} \rightarrow \pmatrix{x \cos \theta - y \sin\theta \\ x \sin \theta + y \cos \theta} $$ which yields the curve
$$ x^2 +y^2 -2 (x^2-y^2) (\sin 2\theta) -4 x y (\cos 2 \theta) + 6 =0 $$
and the corresponding condition $\cos 2\theta =0 \Rightarrow \theta = \pm \frac{\pi}{4} $. Using the angle $\theta=\frac{\pi}{4}$ the above is
$$ -x^2 +3 y^2 + 6 = 0 $$ or in $a x^2+b y^2 =1$ form:
$$ \boxed{ \frac{1}{6}x^2 - \frac{1}{2}y^2 = 1 }$$
Immediately I see this as a hyperbola laying along the x-axis since $a>0$ and $b<0$.
| {
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Computing $\sum\limits_{n=2}^{\infty }\frac{1}{n^3-n}$ I don't understand why I can't get the telescopic sum after the partial fraction decomposition:
$$\frac{1}{n^3-n}= \frac{1}{n(n-1)(n+1)}=\frac{-1}{n}+\frac{1}{2(n-1)}+\frac{1}{2(n+1)}=\frac{-1}{n}+\frac{n}{(n-1)(n+1)}.$$
I have $\sum\limits_{n=2}^{\infty }\frac{1}{n^3-n}=\phi(n+1)-\phi(0).$
The answer should be $1/4$.
| $$\sum _{ n=2 }^{ \infty } \frac { 1 }{ n^{ 3 }-n } =\sum _{ n=2 }^{ \infty } \frac { 1 }{ \left( n-1 \right) n\left( n+1 \right) } =\frac { 1 }{ 2 } \sum _{ n=2 }^{ \infty } \left[ \frac { 1 }{ \left( n-1 \right) } +\frac { 1 }{ \left( n+1 \right) } -\frac { 2 }{ n } \right] =\\ =\frac { 1 }{ 2 } \sum _{ n=2 }^{ \infty } \left[ \frac { 1 }{ \left( n-1 \right) } -\frac { 1 }{ n } +\frac { 1 }{ \left( n+1 \right) } -\frac { 1 }{ n } \right] =\color{red}{\frac { 1 }{ 4 }} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability that at least $2$ people will not receive any ace.
I've a deck with 52 french cards ($13$ values for each of $4$ suits) and $4$ players. Randomly dealing out all cards, what's the probability that at least
$2$ people will not receive any ace?
My try:
$$p=\frac{\frac{4!}{2!}\binom{48}{13,13,12,10}+\binom{4}{2}\binom{48}{13,13,11,11}+\frac{4!}{3!}\binom{48}{13,13,13,9}}{\binom{52}{13,13,13,13}}$$
Where:
$\binom{48}{13,13,13,9}$ is the case $A$ has all $4$ aces,
$\binom{48}{13,13,12,10}$ the case $A$ has $1$ ace $B$ has $3$ aces,
$\binom{48}{13,13,11,11}$ the $A$ has $2$ aces and same for $B$,
$\frac{4!}{2!}$ arrangements of $4$ people to be $A$ and $B$,
$\binom{4}{2}$ combination of $4$ people to be $A$ and $B$,
$\frac{4!}{3!}$ arrangements of $4$ people to be $A$
Am I right? If yes, is there a more elegant solution than mine?
| With reference with my last comment against your question.
Imagine $4$ labelled rooms, each with $13$ labelled beds, to be occupied by $4$ travellers (aces)
As you have found, either $4$ are in one room, or $3-1 \;or\; 2-2$ in two rooms to satisfy the question's constraints
$Pr = \dfrac{4\binom{13}4 + (4\cdot3)\binom{13}3\binom{13}1 + \binom42\binom{13}2\binom{13}{2}}{\binom{52}4} = \dfrac{76}{245},\; \approx0.310204 $
Note that inclusion-exclusion has not been resorted to.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
"answer_id": 1
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How to solve the following Cauchy problem. Solve the following Cauchy problem for a first order PDE:
$$(2x_1 + x_2)u_{x_1} + (x_2 + 1)u_{x_2} = u^2, \ \ u(x_1, 1) = x_1^2 + 1, \ \ x_1 \ge 0, x_2 \ge 1$$
and find an implicit conldition over $x_1$ and $x_2$ under which this Cauchy problem has a bounded solution.
Attempt at the problem:
Our characteristic ODEs are:
\begin{align}
\frac{dx_1}{dt} &= 2x_1 +x_2 &x_1(s, 0) =& s\\
\frac{dx_2}{dt} &= x_2 + 1 &x_2(s, 0) = 1 \\
\frac{dx_3}{dt} &= x_3^2 &x_3(s, 0) = s^2 + 1 \\
\end{align}
which have solutions:
\begin{align}
x_1(s, t) &= (s+\frac{3}{2})e^{2t} - 2 e^t+\frac{1}{2} \\
x_2(s, t) &= 2e^t - 1 \\
x_3(s, t) &= \frac{1}{\frac{1}{s^2 + 1} - t}\\
\end{align}
Which implies:
\begin{align}
s &= \frac{4x_1 + 4x_2 + 2}{(x_2 + 1)^2} - \frac{3}{2} \\
t &= \ln\frac{x_2+1}{2}
\end{align}
Substituting $s,t$ into $x_3$ gives a complex expression that seems wrong? Is this the correct way to go about solving this problem?
| Consider writing your system as follows
$$\left( 2x+t \right){{u}_{x}}+\left( t+1 \right){{u}_{t}}={{u}^{2}}$$ where $$u\left( x,1 \right)={{x}^{2}}+1,\,\,t\ge 1$$
Let $u\left( x,t \right)=u\left( x\left( s \right),t\left( s \right) \right)\Rightarrow \frac{du}{ds}=\frac{\partial u}{\partial x}\frac{dx}{ds}+\frac{\partial u}{\partial t}\frac{dt}{ds}$ .
Equating ‘coefficients’ against ${{u}_{t}}$implies
$$\frac{dt}{ds}=t+1\Rightarrow \ln \left( \frac{t\left( s \right)+1}{t\left( 0 \right)+1} \right)=s$$ where $t\left( 0 \right)=1$, so $t\left( s \right)=2{{e}^{s}}-1$.
We also have by the same process
$$\frac{dx}{ds}=2x+t=2x+2{{e}^{s}}-1$$
Or
$$\frac{dx}{ds}-2x=2{{e}^{s}}-1$$
which we can write as
$$\frac{d}{ds}\left( x{{e}^{-2s}} \right)=\left( 2{{e}^{s}}-1 \right){{e}^{-2s}}$$
Integrating
$$x\left( s \right){{e}^{-2s}}-x\left( 0 \right)=\frac{1}{2}{{e}^{-2s}}-2{{e}^{-s}}+\frac{3}{2}$$
Or
$$x\left( 0 \right)=x\left( s \right){{e}^{-2s}}+2{{e}^{-s}}-\frac{1}{2}{{e}^{-2s}}-\frac{3}{2}$$
which is constant along the characteristics. Now we also have
$$\frac{du}{ds}={{u}^{2}}\Rightarrow \frac{1}{u\left( x\left( 0 \right),t\left( 0 \right) \right)}-\frac{1}{u\left( x\left( s \right),t\left( s \right) \right)}=s$$
Or
$$u\left( x\left( s \right),t\left( s \right) \right)=\frac{u\left( x\left( 0 \right),t\left( 0 \right) \right)}{1-su\left( x\left( 0 \right),t\left( 0 \right) \right)}$$
Note the initial condition being $u\left( x\left( 0 \right),t\left( 0 \right) \right)=u\left( x\left( 0 \right),1 \right)=x{{\left( 0 \right)}^{2}}+1$, therefore
$$u\left( x\left( s \right),t\left( s \right) \right)=\frac{x{{\left( 0 \right)}^{2}}+1}{1-s\left( x{{\left( 0 \right)}^{2}}+1 \right)}$$
Now from $t\left( s \right)=2{{e}^{s}}-1\Rightarrow s=\ln \left( \frac{1}{2}\left( t+1 \right) \right)$. we have
$$x\left( 0 \right)=\frac{2\left( 2x-1 \right)+4\left( t+1 \right)}{{{\left( t+1 \right)}^{2}}}-\frac{3}{2}$$
Substituting we obtain finally
$$u\left( x,t \right)=\frac{1+{{\left( \frac{2\left( 2x-1 \right)+4\left( t+1 \right)}{{{\left( t+1 \right)}^{2}}}-\frac{3}{2} \right)}^{2}}}{1-\left( 1+{{\left( \frac{2\left( 2x-1 \right)+4\left( t+1 \right)}{{{\left( t+1 \right)}^{2}}}-\frac{3}{2} \right)}^{2}} \right)\ln \left( \frac{1}{2}\left( t+1 \right) \right)}$$
Which is a great big mess (pretty much anyway you write it) but can be verified to solve the initial condition and the PDE (I suggest you use a CAS to verify it).
| {
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Showing that $S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000} \gt 1$
If $$S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000}$$ then
$$S\gt 1,$$
but how?
I understood that there are $451$ pair of terms. So clubbed two terms together.
$\frac{1}{100}+\frac{1}{1000}+\frac{1}{101}+\frac{1}{999}.....$
But I am not able to solve it further. Such a tricky question for me.
| numbers from $\frac{1}{100}$ to $\frac{1}{200}$ are larger than $\frac{1}{200}$
and there are $100$ of them (101 actually, but doesn't matter) so their sum is larger than $100\times \frac{1}{200}=\frac12$
You can repeat a similar argument to show that from $\frac{1}{200}$ to $\frac{1}{300}$ the sum is larger than $\frac13$
until you arrive to numbers from $\frac{1}{900}$ to $\frac{1}{1000}$ which have sum larger than $\frac{1}{10}$
so the numbers from $\frac{1}{100}$ to $\frac{1}{1000}$ are less than $\frac12+\frac13+\ldots+\frac{1}{10}\approx 1.93 $
Our sum is less than the actual sum ($\approx 2.308$) and is larger than $1$ so we have proved what was required to prove
hope this helps
| {
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"url": "https://math.stackexchange.com/questions/2381645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $(a^2+b^2),$ where $(a+b)=\dfrac{a}{b}+\dfrac{b}{a}$ The question is the same .Find $a^2+b^2$. I think we have to find $a$ and $b$ firstly. Given that $a$ and $b$ are integers.
| Lemma(1): Let $a$ & $b$ to be integers such that $ab \mid a^2+b^2$.
If $\gcd(a,b)=1$, then prove that $a=\pm b$.
Proof: We claim that $ab=\pm 1$.
*
*Proof of the claim: Suppose on contrary; that $1 < |ab|$.
So there exist a prime number $p$, which divides $ab$; i.e. $p \mid ab$.
Without loss of generality we can assume that $p \mid a$.
So $p$ must divides $b^2=(a^2+b^2)-a^2$.
[Because $p$ divides both of the $(a^2+b^2)$ & $a^2$.]
So we can conclude that $p$ must divides $b$;
which is an obvious contradiction with the assumption that $\gcd(a,b)=1$.
So we can conclude that $a=\pm 1$ & $b=\pm 1$; which implies that $a=\pm b$.
Lemma(2): Let $a$ & $b$ to be integers such that $ab \mid a^2+b^2$.
Prove that $a=\pm b$.
Proof: Let $d:=\gcd(a,b)$,
so there exist integers $a^{\prime}$ & $b^{\prime}$ such that:
$$
a=da^{\prime} \ , \ \ \ \ \ \ \
b=db^{\prime} \ , \ \ \ \ \ \ \
\gcd(a^{\prime},b^{\prime})=1 . $$
The relation $ab \mid a^2+b^2$, implies that there is an integer $k$,
such that:
$$
k(ab) = a^2+b^2
\Longrightarrow
k\big( (da^{\prime})(db^{\prime}) \big) = (da^{\prime})^2+(db^{\prime})^2
\Longrightarrow
k\big( a^{\prime}b^{\prime} \big) = (a^{\prime})^2+(b^{\prime})^2 ,
$$
so we obtain a pair $(a^{\prime},b^{\prime})$ such that:
$$a^{\prime}b^{\prime} \mid (a^{\prime})^2+(b^{\prime})^2 \ ,
\ \ \ \ \ \ \ \ \ \ \ \
\gcd(a^{\prime},b^{\prime})=1 .$$
So by Lemma(1) we have:
$$a=d(a^{\prime})=d(\pm b^{\prime})=\pm d(b^{\prime})=\pm b .$$
The relation
$$a+b=\frac{a}{b} + \frac{b}{a} \ \ , \ \ \ \ \ \ \ (*) $$
implies $ab(a+b)=a^2+b^2$,
so we have: $ab \mid a^2+b^2$,
so by the Lemma(2) we have one of the two folowing cases:
*
*$a= + b$ , in this case both of the fractions
on the Right-hand-side of the relation $(*)$,
both are equal to $+1$,
so the R-H-S is equal to $+ 2$.
On the other hand in this case the Left-hand-side of the relation $(*)$ is equal to $2a$, so must have: $2a=+2$, which yields the solution $a=b=+1$.
*$a= - b$ , in this case both of the fractions
on the Right-hand-side of the relation $(*)$,
both are equal to $-1$,
so the R-H-S is equal to $- 2$.
On the other hand in this case the Left-hand-side of the relation $(*)$ is equal to $0$, so must have: $0=-2$, which is impossible.
So the quantity $a^2+b^2$ will be equal to $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$, what is the value of $a+b$?
Given that
$$\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$$
What is the value of $a+b$
My try
$\lim\limits_{x\to 0} (\frac {x(1+a\cos x)}{x^3}-\frac {b\sin x}{x^3})=1$
$\lim\limits_{x\to 0} (\frac {(1+a×cos(x)}{x^2}-\frac {b}{x^2})=1$
$\lim\limits_{x\to 0} \frac {1+a\cos x-bx}{x^2}=1$
Apply L'Hôpital's rule:
$\lim\limits_{x\to 0} \frac {-a\cos x-b}{2x}=1$
Apply L'Hôpital's rule again:
$\lim\limits_{x\to 0} \frac {-a\sin x}{2}=1$ $\to$ $a=-2$
Is my approach right?
| As $x(1+a\cos x)-b\sin x\sim (1+a-b)x$ as $x\to0$ then the given limit
can only exist when $b=1+a$. A series expansion then gives
$$x(1+a\cos x)-(1+a)\sin x\sim-\frac{a}2x^3+(1+a)\frac{x^3}6.$$
One needs then $(1+a)/6-a/2=1$ for the given limit to equal $1$.
Now one can obtain $a$ and $b$.
| {
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"source": "stackexchange",
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Given $a+b+c=0$ find the value of $\big(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\big)\big(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\big)$ I already have a solution, which is correct, it is $9$. I'm just wondering if there is a simpler method.
First we just expand and find
$3+\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big)+\frac{c-a}{b}\big(\frac{a}{b-c}+\frac{c}{a-b}\big)+\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big)$
Each of these "not yet a number" terms can be expanded to give
$\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-b}\big) = \frac{2(c-b)^2}{(c-a)(a-b)}$
$\frac{c-a}{b}\big(\frac{a}{b-c}\big)= \frac{2(a-c)^2}{(c-a)(a-b)}$
$\frac{a-b}{c}\big(\frac{a}{b-c}+\frac{b}{c-a}\big) = \frac{2(b-a)^2}{(b-c)(c-a)}$
Adding each of these terms together and factoring out the two we find that they equal
$2\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=6$
So our total is $9$.
This is problem 160 of The USSR olympiad problem book by Shklarsky, Chentzov and Yaglom, they give the answer and $9$ is correct. Obviously they don't give the method.
| Observe that $(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)=a^3+b^3+c^3-3abc$ so $a^3+b^3+c^3=3abc$. Now
\begin{eqnarray*}
&&\left( \frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c} \right) \left( \frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b} \right)\\
&=&3+\frac{a}{b-c}\left(\frac{c-a}{b}+\frac{a-b}{c}\right)+
\frac{b}{c-a}\left(\frac{a-b}{c}+\frac{b-c}{a}\right)+
\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\\
&=&3+\frac{a}{b-c}\frac{c^2-ac+bc-b^2}{cb}+.+.\\
&=&3+\frac{a}{b-c}\frac{(a-b-c)(b-c)}{cb}+.+.\\
&=&3+\frac{2(a^3+b^3+c^3)}{abc}=\color{red}{9}.
\end{eqnarray*}
| {
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"url": "https://math.stackexchange.com/questions/2382672",
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"source": "stackexchange",
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What significance do squares of trigonometric ratios have algebraically and geometrically? I recently came upon a problem which asked me to:
"Show that $(\cos x + 1)^2 = \frac{\cos2x}{2} + 2\cos x + \frac32$ and hence determine $\int(\cos x + 1)^2 dx$.";
from Mathematics for the International Student-IB Diploma: SL. Being given a value with which to substitute the argument of the integral meant to be found, the second task proved manageable:
Since $(\cos x + 1)^2 = \frac{\cos2x}{2} + 2\cos x + \frac32$, it is evident that $\int(\cos x + 1)^2 dx = \int(\frac{\cos2x}{2} + 2\cos x + \frac32)dx$, which is solved quite easily.
However, I struggled at finding a way to show that the initial argument made sense. Here is the work I did:
$$\frac{\cos 2x}{2} + 2 \cos x + \frac32 = (\cos x + 1)^2$$
$$\frac{\cos 2x}{2} + 2 \cos x + \frac32= \cos^2 x + 2 \cos x + 1$$
$$\frac{\cos 2x + 3}{2} = \cos^2 x + 1$$
$$\cos 2x + 3 = 2 \cos^2 x + 2$$
$$\cos 2x = 2\cos^2 x - 1$$
It is at this point that I get stuck. I realize that trigonometric identities will facilitate the above statement, but I do not know which ones "fit". Therefore, I come to my question: what significance do squares of trigonometric ratios have algebraically and geometrically?
| Expanding the expression we have $$(\cos x + 1)^2 = \cos ^2 x + 2 \cos x + 1$$
However, $\cos^2 x = \frac {\cos 2x + 1}{2}$ so we have $$\frac {\cos 2x + 1}{2} + 2 \cos x + 1$$ Factoring $1/2$ from the first term $\frac {\cos 2x + 1}{2}$, we then have $$\frac {\cos 2x}{2} + \frac {1}{2} + 2 \cos x + 1 $$ or $$\frac {\cos 2x}{2} + 2 \cos x + \frac {3}{2}$$
The integral for $\int (\frac {\cos 2x}{2} + 2 \cos x + \frac {3}{2}) dx$ is $$\frac {\sin 2x}{4} + 2 \sin x + \frac {3x}{2} + C$$
| {
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Evaluating a Discontinuous Integral How do you evaluate this integral? $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$
My work:
The expression $x^2 - 2x - 3$ is discontinuous at $x = 3$ in the interval $x = 0$ to $x = 4$, so I got to integrate the expression like this:
$$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ is equal to
$$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3} $$.
Then recall that in evaluating a discontinuous integrand, the integral is defined by the relations $$\int_a ^b f(x) = lim_{x->b^-} \int_a ^x f(x) dx $$
if $x = b$ is the discontinuous point or $$\int_a ^b f(x) = lim_{x->a^+} \int_x ^b f(x) dx $$
if $x = a$ is the discontinuous point.
With that in mind:
$$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3} $$
$$\int _0 ^3\frac{dx}{(x^2 - 2x + 1) + (-3)(-1)} + \int _3 ^4\frac{dx}{(x^2 - 2x + 1) + (-3)(-1)} $$
$$\int _0 ^3\frac{dx}{(x-1)^2 -(2)^2} + \int _3 ^4\frac{dx}{(x-1)^2 -(2)^2} $$
Remembering that $\int \frac{du}{u^2 +a^2} = \frac{1}{a} \arctan \left( \frac{u}{a}\right) + C, $
we get:
$$\int _0 ^3\frac{dx}{(x-1)^2 -(2)^2} +\int _3 ^4\frac{dx}{(x-1)^2 -(2)^2}$$
equals
$$lim_{x->3^+} \int_0 ^x \frac{dx}{(x-1)^2 -(2)^2} + lim_{x->3^-} \int_x ^4 \frac{dx}{(x-1)^2 -(2)^2} $$
equals
$$lim_{x->3^+} \left( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right)|_3 ^x + lim_{x->3^-} \left(\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right)|_x ^4$$
equals
$$\left ( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right)-\frac{1}{2} \arctan \left( \frac{3-2}{2}\right) \right ) + \left ( \frac{1}{2} \arctan \left( \frac{4-2}{2}\right)-\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right )$$
equals
$$\left ( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right)- 0.2318 \right ) + \left ( \frac{\pi}{8}-\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right )$$
which makes $$\int _0 ^4\frac{dx}{x^2 - 2x - 3} = 0.1609$$
But in my book, it said there is no value of $\int _0 ^4\frac{dx}{x^2 - 2x - 3}.$
How do you prove that the integral $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ has no value?
| The indefinite integral is
$$\int \frac{1}{x^2-2 x-3} \, dx=\frac{1}{4} \left[\log |3-x|-\log (x+1)\right]+C$$
Indeed the fraction can be split in the sum of two fractions
$\dfrac{1}{(x-3) (x+1)}=\dfrac{a}{x-3}+\dfrac{b}{x+1}=\dfrac{(a+b) x+a-3 b}{(x-3) (x+1)}$
which is equal to the given fraction if
$\left\{ \begin{gathered}
a + b = 0 \hfill \\
a - 3b = 1 \hfill \\
\end{gathered} \right.\rightarrow$
$\left\{ \begin{gathered}
a = \frac{1}{4} \hfill \\
b = - \frac{1}{4} \hfill \\
\end{gathered} \right.\rightarrow \dfrac{1}{(x-3) (x+1)}=\dfrac{1}{4}\left(\dfrac{1}{x-3}-\dfrac{1}{x+1}\right)$
If we write the given definite integral as you did
$$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3}=\\=\frac{1}{4} \left[\log |3-x|-\log (x+1)\right]_0^3+\frac{1}{4} \left[\log |3-x|-\log (x+1)\right]_3^4$$
we see that the first tends to $+\infty$ as $x\to 3$ because of the logarithm
So the integral diverges
Hope this helps
| {
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Find the value of expression $^2+^2+^2$ given $++=1$ and $\frac 1x +\frac 1y +\frac 1z=0$ Find the value of expression $$^2+^2+^2$$ given $$++=1$$ and $$\frac 1x +\frac 1y +\frac 1z=0$$
I am sorry I don't really know how to tag this question.
| Assume that $x,y,z$ are the roots of a monic, cubic polynomial $p(t)$
$$ p(t) = (t-x)(t-y)(t-z) = t^3-e_1 t^2+e_2 t-e_3. $$
Since $x+y+z=1$ we have $e_1=1$. Similarly, from $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$ we get $e_2=0$.
$$x^2+y^2+z^2 = (x+y+z)^2-2(xy+xz+yz) = e_1^2-e_2$$
then leads to $x^2+y^2+z^2=\color{blue}{1}$.
| {
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If $a_n = \sum\limits_{r=0}^n \binom{n}{r} b_r$, prove $(-1)^n b_n = \sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s$ Suppose that sequences of real numbers satisfy:
\begin{align*}
a_n &= \sum\limits_{r=0}^n \binom{n}{r} b_r \\
\end{align*}
Prove that:
\begin{align*}
(-1)^n b_n &= \sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s \\
\end{align*}
My work:
The $n=0$ case:
\begin{align*}
a_0 &= b_0 \\
b_0 &= a_0 \\
\end{align*}
Then $n=1$ case:
\begin{align*}
a_1 &= \binom{1}{0} b_0 + \binom{1}{1} b_1 = b_0 + b_1 \\
-b_1 &= \binom{1}{0} a_0 - \binom{1}{1} a_1 = a_0 - a_1 \\
b_1 &= a_1 - a_0 = b_0 + b_1 - a_0 = b_1 \\
\end{align*}
The inductive step:
\begin{align*}
(-1)^{n+1} b_{n+1} &= \sum\limits_{s=0}^{n+1} \binom{n+1}{s} (-1)^s a_s \\
\end{align*}
I'm not sure where to go from here. I tried substituting $a_s$ into that last equation and it didn't help. Any ideas?
| Here is a direct proof
that uses only
$\sum\limits_{s=0}^{m} (-1)^{s}\binom{m}{s}
=1$
when $m=0$
and
$=0$ when
$m \ge 1$,
This last is
from the expansion
$0
=(1-1)^m
=\sum\limits_{s=0}^{m} (-1)^{s}\binom{m}{s}
$.
$\begin{array}\\
\sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s
&=\sum\limits_{s=0}^n \binom{n}{s} (-1)^s
\sum\limits_{r=0}^s \binom{s}{r} b_r \\
&=\sum\limits_{r=0}^n\sum\limits_{s=r}^n \binom{n}{s} (-1)^s
\binom{s}{r} b_r \\
&=\sum\limits_{r=0}^nb_r\sum\limits_{s=r}^n \binom{n}{s} (-1)^s
\binom{s}{r} \\
&=\sum\limits_{r=0}^nb_r\sum\limits_{s=r}^n (-1)^s \binom{n}{s}
\binom{s}{r} \\
&=\sum\limits_{r=0}^nb_r\sum\limits_{s=r}^n (-1)^s \dfrac{n!s!}{s!(n-s)!r!(s-r)!}\\
&=\sum\limits_{r=0}^nb_r\sum\limits_{s=r}^n (-1)^s \dfrac{n!}{(n-s)!r!(s-r)!}\\
&=\sum\limits_{r=0}^nb_r\dfrac{n!}{r!}\sum\limits_{s=r}^n (-1)^s \dfrac{1}{(n-s)!(s-r)!}\\
&=\sum\limits_{r=0}^nb_r\dfrac{n!}{r!(n-r)!}\sum\limits_{s=r}^n (-1)^s \dfrac{(n-r)!}{(n-s)!(s-r)!}\\
&=\sum\limits_{r=0}^nb_r\dfrac{n!}{r!(n-r)!}\sum\limits_{s=0}^{n-r} (-1)^{s+r} \dfrac{(n-r)!}{(n-(s+r))!(s+r-r)!}\\
&=\sum\limits_{r=0}^nb_r\binom{n}{r}\sum\limits_{s=0}^{n-r} (-1)^{s+r} \dfrac{(n-r)!}{(n-s-r)!s!}\\
&=\sum\limits_{r=0}^nb_r(-1)^r\binom{n}{r}\sum\limits_{s=0}^{n-r} (-1)^{s}\binom{n-r}{s}\\
&=(-1)^nb_n\\
\end{array}
$
To rephrase the last statement
in terms of the sum,
$\sum\limits_{s=0}^{n-r} (-1)^{s}\binom{n-r}{s}
=1$
when $n=r$
and
$=0$ when
$n > r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Maximum of sum of fractions with tans Let $ABC$ be an acute triangle. What is the maximum value of
$$\frac{\tan^2A+\tan^2B}{\tan^4A+\tan^4B}+\frac{\tan^2B+\tan^2C}{\tan^4B+\tan^4C}+\frac{\tan^2C+\tan^2A}{\tan^4C+\tan^4A}?$$
For the equilateral triangle, $\tan A=\tan B=\tan C=\sqrt{3}$, and the sum is $1$. For the isosceles triangle with $\angle A=\angle B\rightarrow\pi/2$ and $\angle C\rightarrow 0$, the sum approaches zero.
| Say $x= \tan A$, $y= \tan B$ and $z = \tan C$. It is
well-known that $x+y+z=xyz$. Then we have $$E = {x^2+y^2\over
x^4+y^4} +{y^2+z^2\over y^4+z^4} + {z^2+x^2\over
z^4+x^4}$$
By Cauchy inequality and then by $AM-GM$ we have
$${x^2+y^2\over x^4+y^4} \leq {2\over x^2+y^2}\leq {1\over xy}$$
So $$ E \leq {1\over xy} +{1\over yz}+{1\over zx} = 1$$
We get equality exactly when $x=y=z$, thus $3x=x^3 \Longrightarrow x=\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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} |
Formulation of the Wallis product for $\pi/2$ using the $\Gamma$ function I am trying to write the Wallis Product (WP) using the $\Gamma$ function, thereby hoping for either a new identity or a simple derivation of the formula, but I get a result I am not sure its correct.
This is the WP
$$ \frac{\pi}{2} = \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}\cdots\frac{2n}{2n-1}\frac{2n}{2n+1}\cdots $$
Now
$$ \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}\cdots\frac{2n}{2n-1}\frac{2n}{2n+1}\cdots = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdot \dots\ (2n)(2n)}{1\cdot1\cdot3\cdot3\cdot5\cdot5\cdot\dots\ (2n-1)(2n-1)(2n+1)}\cdots $$
$$=\frac{1\cdot2\cdot1\cdot2\cdot2\cdot2\cdot2\cdot2\cdot3\cdot2\cdot3\cdot2\cdot \hspace{12pt} \dots\hspace{12pt}n\cdot2\hspace{12pt}\cdot\hspace{12pt} n\cdot2\hspace{24pt}}{\frac{1}{2}\cdot2\cdot\frac{1}{2}\cdot2\cdot\frac{3}{2}\cdot2\cdot\frac{3}{2}\cdot2\cdot\frac{5}{2}\cdot2\cdot\frac{5}{2}\cdot2\cdot\dots\ (n-\frac{1}{2})\cdot2\cdot(n-\frac{1}{2})\cdot2\cdot(2n+1)}\cdots $$
$$=\frac{1\cdot1\cdot2\cdot2\cdot3\cdot3\cdot \hspace{12pt} \dots\hspace{12pt}n\hspace{12pt}\cdot\hspace{12pt} n\hspace{24pt}}{\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdot\frac{5}{2}\cdot\dots\ (n-\frac{1}{2})\cdot(n-\frac{1}{2})\cdot(2n+1)}\cdots $$
$$=\big(\frac{1\cdot2\cdot3\cdot\ \dots\ n\hspace{24pt}}{\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdot\dots\ (n-\frac{1}{2})\cdot\sqrt{2n+1}}\cdots\big)^2 $$
$$ = \lim_{n\rightarrow\infty}\;\;\Bigg(\frac{\Gamma{(n+1)}}{\frac{\Gamma{(n+\frac{1}{2})}}{\sqrt{\pi}}}\Bigg)^2\frac{1}{(2n+1)}$$
$$=\lim_{n\rightarrow\infty}\;\; \frac{\pi}{2n+1}\Big(\frac{\Gamma{(n+1)}}{\Gamma{(n+\frac{1}{2})}}\Big)^2$$
$$= \frac{\pi}{2}\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2.$$
Question 1) Is this a correct fully equivalent reformulation of rhs of WP?
When using WP we would get:
$$\frac{\pi}{2} = \frac{\pi}{2}\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2,$$
$$\lim_{n\rightarrow\infty}\; \frac{1}{n}\Big(\frac{\Gamma{(n)}}{\Gamma{(n-\frac{1}{2})}}\Big)^2 = 1,$$ which implies
$$\frac{\Gamma(n)}{\Gamma(n-\frac{1}{2})} \overset{n\rightarrow \infty}{\longrightarrow} \sqrt{n}. \tag{1}$$
Alternatively if (1) is known otherwise (e.g. from the Sterling approximation (Question 2) Is it possible to derive (1) from the Sterling approximation or from any other elementary considerations?)) one could derive the WP easily from that.
Question 3) Is this conclusion correct?
| Let's say $\rho(n) = \Gamma(n)/\Gamma(n-1/2).$ Then, yes, $\rho(n) \approx \sqrt n$ for large $n$. More precisely, $\rho(n)/\sqrt n \to 1$ as $n \to \infty$.
You should expect this. After all, $$\rho(n)\cdot \rho(n-1/2) = \frac{\Gamma(n)}{\Gamma(n-1/2)} \cdot \frac{\Gamma(n-1/2)}{\Gamma(n-1)} = \frac{\Gamma(n)}{\Gamma(n-1)} \approx n$$
for large $n$, so if $\rho(n) \approx \rho(n-1/2)$ for large $n$, then we must have $\rho(n) \approx \sqrt n$ for large n.
Indeed, this observation is one way to pick out the Gamma function as the "best" generalization of factorial to non-integer arguments. For any fixed $t$ between $0$ and $1$ this sort of heuristic argument suggests that you want $$\Gamma(n+t) \approx n^t \cdot \Gamma(n) $$ for large integer $n$. Combined with the functional equation $\Gamma(x+1) = x \cdot \Gamma(x)$, that means for large $n$ you want
$$(n-1+t)\cdot\dots\cdot(1+t) \cdot \Gamma(1+t) \approx n^t \cdot (n-1)\cdot \dots \cdot 2\cdot 1.$$
And it turns out that taking the definition
$$\Gamma(1+t) = \lim_{n\to \infty} n^t \frac{(n-1)\cdot \dots \cdot 2\cdot 1}{(n-1+t)\cdot\dots\cdot(2+t)\cdot(1+t)}$$
gives you the same Gamma function that we know and love.
This is closely related to the observation that the Gamma function is the unique log-convex extension of the factorial function which satisfies the right functional equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve this problem on complex numbers? Let $ Z $ be a complex number with nonzero imaginary part such that
$$ (2Z + 1)(3Z + 1)(5Z + 1)(30 Z + 1) = 10 $$
Then compute $$ \frac{\text {sum of all values of Z}}{\text {product of all values of Z} } $$
| Hint: $1/2+1/30=1/3+1/5 = 2 \cdot 4/15\,$, which suggests the substitution $z = x - 4/15\,$, which gives a biquadratic in $x\,$:
$$
2\cdot3\cdot5\cdot30\left(z+\frac{1}{2}\right)\left(z+\frac{1}{3}\right)\left(z+\frac{1}{5}\right)\left(z+\frac{1}{30}\right) \\ = 900\left(x+\frac{7}{30}\right)\left(x+\frac{1}{15}\right)\left(x-\frac{1}{15}\right)\left(x-\frac{7}{30}\right) \\ = 900\left(x^2-\frac{7^2}{30^2}\right)\left(x^2-\frac{1}{15^2}\right) \\ = \frac{1}{225}(900x^2-49)(225x^2-1)
$$
Then $225(\text{LHS} - 10)$ factors into two qadratics, one of which has the complex roots being sought:
$$
(900x^2-49)(225x^2-1)-2250=(900 x^2 + 71) (225 x^2 - 31)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Reference (database) for polynomials
Suppose I want to know what is known about the following polynomial
$$x^4 - 2x^2y^2 - 2x^2z^2 + y^4 - 2y^2z^2 + z^4.$$
Where can one find such information?
| It's just
$$(x+y+z)(x-y-z)(x+y-z)(x-y+z).$$
We can get it by the following way.
$$x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2=$$
$$=x^4+y^4+z^4-2x^2y^2-2x^2z^2+2y^2z^2-4y^2z^2=$$
$$=(x^2-y^2-z^2)^2-(2yz)^2=$$
$$=(x^2-y^2-z^2-2yz)(x^2-y^2-z^2+2yz)=$$
$$=(x^2-(y+z)^2)(x^2-(y-z)^2),$$
which gives the final factorization.
Also, the area of a triangle with sides-lengths $x$, $y$ and $z$ it's
$$\frac{1}{4}\sqrt{2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$ for $A$, $B$, $C$ the angles of a triangle Given that $A$,$B$ and $C$ are angles of a triangle, show that
$$1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$$
| We have: $1+\cos(2C) = 2\cos^2C, \cos(2A)+\cos(2B) = 2\cos(A+B)\cos(A-B)=-2\cos C\cos(A-B)\implies LHS = 2\cos C(\cos C+ \cos(A-B))= 2\cos C(\cos(A-B) - \cos(A+B)) = 2\cos C(-2\sin A\sin (-B)) = RHS$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing determinant without expansion $$\begin{align}\mathrm D &= \left|\begin{matrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (a+c)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{matrix}\right|\\ &= (a+b+c)\left|\begin{matrix} b+c - a & a^2 & a^2 \\ b - a -c & (a+c)^2 & b^2 \\ 0 & c^2 & (a+b)^2 \end{matrix}\right| \\ &= (a+b+c)^2\left|\begin{matrix} b+c - a & 0 & a^2 \\ b - a -c & a+c - b & b^2 \\ 0 & c - a-b & (a+b)^2 \end{matrix}\right|\\ &= (a+b+c)^2\left|\begin{matrix} b+c - a & 0 & a^2 \\ 0 & a+c - b & b^2 \\ c - a-b & c - a-b & (a+b)^2 \end{matrix}\right|\end{align}$$
Can $\rm D$ be further simplified without expanding ? I feel it should be because this was competition question.
| We already reduced the problem to calculate
$$D’=\left|\begin{matrix}
b+c - a & 0 & a^2 \\
0 & a+c - b & b^2 \\
b & a & -ab
\end{matrix}\right|$$
If $a=0$ then
$$D’=\left|\begin{matrix}
b+c & 0 & 0\\
0 & c - b & b^2 \\
b & 0 & 0
\end{matrix}\right|=0.$$
If $b=0$ then
$$D’=\left|\begin{matrix}
c - a & 0 & a^2 \\
0 & a+c & 0 \\
0 & a & 0
\end{matrix}\right|=0.$$
Otherwise put $R’_1=R_1+\frac abR_3$ and $R’_2=R_2+\frac baR_3$. Then
$$D’=\left|\begin{matrix}
b+c & \frac {a^2}b & 0\\
\frac {b^2}a & a+c & 0 \\
b & a & -ab
\end{matrix}\right|=-ab\left|\begin{matrix}
b+c & \frac {a^2}b \\
\frac {b^2}a & a+c \\
\end{matrix}\right|=-ab[(a+c)(b+c)-ab]=-ab[ac+bc+c^2]=-abc(a+b+c).$$
The latter formula holds also when $a=0$ or $b=0$. Finally,
$$D=(a+b+c)(-2)D’=2(a+b+c)^3abc.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
prove that $(x^2 + y^2 - z^2)^2 = 4x^2y^2$. We have
$$x\cos \theta+y\cos \phi = -z\cos \psi \tag 1$$
$$x\sin \theta+y\sin \phi = -z\sin \psi \tag 2$$
$$x\sec \theta+y\sec \phi = -z\sec \psi \tag 3$$
and we have to prove that
$$(x^2 + y^2 - z^2)^2 = 4x^2y^2$$
squaring (1) & (2), and adding them we have
$$x^2 + y^2 - z^2 = -2xy \cos(\theta - \phi)$$
multiplying (1) & (3), we have
$$x^2 + y^2 - z^2 = -xy (\cos\theta \sec\phi + \cos\phi \sec\theta)$$
from here I can't move forward
help me
| multiplying (2) & (3), we have
$$x^2\tan\theta+y^2\tan\phi+xy\left(\sin\theta \sec\phi+\sin\phi\sec\theta \right)=z^2\tan\psi$$
Therefore in simplest terms $\tan\theta=1$, $\tan\phi=1$ and $\tan\psi=1$ whilst at the same time
$\cos(\theta - \phi) =-1$ from squaring (1) and (2).
One solution that meets these constraints is $\theta=\frac{\pi}{4}$, $\phi=-\frac{3\pi}{4}$ and $\psi=-\frac{3\pi}{4}$ which just indicates a fixed phase relationship between $\theta$, $\phi$ and $\psi$ I think.
If there is a shared angle of rotation, $\alpha$, the actual solution will be in the form $\theta=\frac{\pi}{4}+\alpha$, $\phi=-\frac{3\pi}{4}+\alpha$ and $\psi=-\frac{3\pi}{4}+\alpha$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $a+b+c=3$ show $a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ I have become interested in constrained relations among simple cyclic sums involving three positive variables. By simple, I mean so simple that they are also fully symmetric. The "building blocks" of the constraints and relations I have been looking at are:
$$
\sum_{\mbox{cyc}} 1 \equiv 3 \\
\sum_{\mbox{cyc}} a \\
\sum_{\mbox{cyc}} ab \\
\sum_{\mbox{cyc}} a^2 \\
\sum_{\mbox{cyc}} 1/a \\
\sum_{\mbox{cyc}} abc \equiv 3abc \\
$$
So an easy sample would be that $$\frac{\sum_{\mbox{cyc}} abc}{\sum_{\mbox{cyc}} a^2} \leq 1$$
The first really tough one I have encountered is:
If $a$, $b$ and $c$ are positives and $a+b+c=3$, show that:
$$a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
I got to this while trying to prove that if $a+b+c=3$ then $a^2+b^2+c^2 \leq 1/a+1/b+1/c$; that turns out to be untrue, but only by a little bit ($27-15\sqrt{3}\approx 1.019)$.
You might show this using BW, but I would hope to find something easier to follow.
EDIT
The maximum ratio is $(27-15\sqrt{3})$ and it occurs at
$$
\left(a = \sqrt{3}, b=c= \frac{3-\sqrt{3}}{2}\right)
$$
and at the two other cyclic permutations of that point.
| (pqr method) Let $p = a + b + c = 3, q = ab + bc + ca, r = abc$.
We need to prove that $p^2 - 2q \le (27 - 15\sqrt{3})\frac{q}{r}$
or $9 - 2q \le (27 - 15\sqrt{3})\frac{q}{r}$.
Since $(a-b)^2(b-c)^2(c-a)^2 = -4p^3r+p^2q^2+18pqr-4q^3-27r^2 \ge 0$,
we have
$r\le q - 2 + \frac{2}{9}\sqrt{3(3-q)^3}$.
It suffices to prove that
$$9 - 2q \le (27 - 15\sqrt{3})\frac{q}{q - 2 + \frac{2}{9}\sqrt{3(3-q)^3}}. \tag{1}$$
Let $q = 3 - \frac{x^2}{3}$ for $x\in [0,3)$ (note: $0 < q \le 3$). Then $\sqrt{3(3-q)} = x$. (1) becomes
$$\frac{(2x + 3 - 3\sqrt{3})^2[2x^2 + (6\sqrt{3} - 9) + 63 - 36\sqrt{3}]}{6(2x+3)(3-x)}\ge 0$$
which is true.
We are done.
| {
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"url": "https://math.stackexchange.com/questions/2393792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that $f(x)$ is an Odd Function
Show that $$f(x) = \ln \left(x+\sqrt{x^2+1}\right)$$ is an odd function.
My attempt:
$$f(-x)=\ln\left(-x+\sqrt{(-x)^2+1}\right)=\ln\left(-x+\sqrt{x^2+1}\right).$$
How should I proceed? I know that if $f(-x)=-f(x)$, the function is odd.
| One more:
$f(x) = \ln (x + \sqrt{x^2+1})$, $x \in \mathbb{R}.$
$\star)$ $f(-x) = \ln(-x + \sqrt{x^2 +1}) =$
$\ln (\frac{1}{x+\sqrt{x^2+1}} )=$
$- \ln (x + \sqrt{x^2+1})$.
Combining:
$f(-x) = - f(x)$.
Used :
$ \ln [( -x +\sqrt{x^2+1}) \frac{x + \sqrt{x^2 +1}}{x+\sqrt{x^2+1}}] =$
$\ln ( \frac{1}{x + \sqrt{x^2+1}} )=$
$- \ln (x + \sqrt{x^2+1})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Proving the sequence is increasing Suppose that a sequence $a_n$ of real numbers satisfies $7a_{n+1 }= a_n^3+6$
for $n ≥ 1$. If $a_1 =\frac{1}{2}$
, prove
that the sequence increases and find its limit.
Now I calculated the difference of two successive terms and factorized it. I got $$a_{n+1}-a_{n}=\frac{(a_n-1)(a_n-2)(a_n+3)}{7}$$ But I am unable to proceed further.
| Use induction you can show $a_n$ is always less than $1$ so the product is positive.
$a_{n+1} = \frac{a_n^3+6}{7}$, so when $a_n<1$, $a_n^3<1$, which implies $a_{n+1} = \frac{a_n^3+6}{7}<1$. Base case is $a_1=\frac{1}{2}$.
So $(a_n-1)(a_n-2)(a_n+3)>0$. Thus $a_{n+1}-a_n>0$.
Now the limit can be found by investigating the solutions to $7x=x^3+6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If $y^3 + 3a^2x + x^3 = 0,$ then prove that $y'' + \frac{2a^2x^2}{y^5} = 0$ I am comfortable with second derivatives but I am just unable to set all the variables up in such a format that I get the latter (the part which needs to be proven). A hint would be a lot of help.
| You can also differentiate explicitely:
$$y=(-3a^2x-x^3)^{1/3}.$$
$$y'=\frac13 (-3a^2x-x^3)^{-2/3}\cdot (-3a^2-3x^2).$$
$$y''=-\frac{2}{3^2}(-3a^2x-x^3)^{-5/3}\cdot (-3a^2-3x^2)^2+\frac13 (-3a^2x-x^3)^{-2/3}\cdot (-6x)=$$
$$-\frac{2(a^2+x^2)^2}{y^5}-\frac{2x}{y^2}=\frac{-2(a^2+x^2)^2-2x(-3a^2x-x^3)}{y^5}=\frac{2a^2x^2-2a^4}{y^5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\dfrac {\pi x}{\sin \pi x} \times \dfrac {1}{\pi^2 x^2})$$
| \begin{align*}\lim_{x\to1}\frac{1+\cos\pi x}{\sin^2\pi x}&=\lim_{x\to1}\frac{-\pi\sin\pi x}{2\pi\sin\pi x\cos\pi x}\\&=-\lim_{x\to1}\frac1{2\cos\pi x}\\&=\frac12\end{align*}and therefore$$\lim_{x\to1}\frac{1+\cos\pi x}{\tan^2\pi x}=\lim_{x\to1}\frac{1+\cos\pi x}{\sin^2\pi x}\times\lim_{x\to1}\cos^2\pi x=\frac12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Correct logic of permuting 5 men and 5 women to find probability of different highest women rank The problem reads like this:
Problem
Five men and $5$ women are ranked according to their scores on an examination. Assume that no two scores are alike and all $10!$ possible rankings are equally likely. Let $X$ denote the highest ranking achieved by a woman. (For instance, $X = 1$ if the top-ranked person is female.) Find $P(X = i),i = 1, 2, 3, . . ., 8, 9, 10$.
Solution given was:
*
*$P(X=1)= \frac{5}{10}= \frac{1}{2}$ because there are 5 women and total of 10 to choose from
*$P(X=2)=\frac{5}{10}\times \frac{5}{9}=\frac{5}{18}$ because for rank1 thereare 5 men and total of 10 to choose from, for rank 2 (we want awoman) we still have 5 women but only a total of 9 to choosefrom.
*$P(X=3)=\frac{5}{10}\times \frac{4}{9}\times \frac{5}{8}=\frac{5}{36}$
*$P(X=4)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{5}{7}=\frac{10}{168}$
*$P(X=5)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{2}{7}\times \frac{5}{6}=\frac{5}{252}$
*$P(X=6)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{2}{7}\times \frac{1}{6}\times \frac{5}{5}=\frac{1}{252}$
My solution was
*
*$P(X=1)=\frac{5\times 9!}{10!}=\frac{1}{2}$ because there are five women to occupy 1st rank and then there remained 9 which can permute in $9!$ ways. There are total $10!$ ways to permute $10$ people
*$P(X=2)=\frac{5\times \binom{5}{4}\times 8!}{10!}=\frac{5}{18}$ because there are five women to occupy 2st rank. The 1st rank will be of a man. So we have to select $4$ out of $5$ men which will be ranked after $2$nd rank. These four men and remaining 4 women can be permuted in $8!$ ways.
*$P(X=3)=\frac{5\times \binom{5}{3}\times 7!}{10!}=\frac{5}{72}$
*$P(X=4)=\frac{5\times \binom{5}{2}\times 6!}{10!}=\frac{5}{504}$
*$P(X=5)=\frac{5\times \binom{5}{1}\times 5!}{10!}=\frac{5}{6048}$
*$P(X=6)=\frac{5\times \binom{5}{0}\times 4!}{10!}=\frac{1}{30240}$
Doubts
*
*Where my logic went wrong?
*When I compared the two approach, I realized that the books solution is permuting ranks higher than the highest ranked girl, while my solution is permuting ranks lower than the highest ranked girl. So I was guessing what makes book solution not permute lower ranks and my solution not permuting higher ranks. Shouldn't we permute on both sides of highest ranked girl?
| In your calculations,
*
*$5$ is the number of possible top-ranked women
*$\binom{5}{k}$ is the number of ways $k$ of the five men can have a lower rank than the top-ranked woman
*$(9 - k)!$ is the number of ways of arranging the $9 - k$ people whose rankings are lower than that of the top-ranked woman
*$10!$ is the number of possible sequences of rankings
In your numerators, you failed to multiply by the number of ways the men who are selected before the first woman can be ranked. Observe that
\begin{align*}
P(X = 1) & = \frac{0! \cdot 5 \cdot \binom{5}{5} \cdot 9!}{10!} = \frac{1}{2}\\
P(X = 2) & = \frac{1! \cdot 5 \cdot \binom{5}{4} \cdot 8!}{10!} = \frac{5}{18}\\
P(X = 3) & = \frac{2! \cdot 5 \cdot \binom{5}{3} \cdot 7!}{10!} = \frac{5}{36}\\
P(X = 4) & = \frac{3! \cdot 5 \cdot \binom{5}{2} \cdot 6!}{10!} = \frac{5}{84}\\
P(X = 5) & = \frac{4! \cdot 5 \cdot \binom{5}{1} \cdot 5!}{10!} = \frac{5}{252}\\
P(X = 1) & = \frac{5! \cdot 5 \cdot \binom{5}{0} \cdot 4!}{10!} = \frac{1}{252}
\end{align*}
The reason you obtained the correct answer for $X = 1$ and $X = 2$ is that $0! = 1! = 1$.
Note: The author(s) of your text are calculating the probability that the top-ranked woman is in the $k$th position. For that to occur, we must select $k - 1$ of the five men to be ranked ahead of her and one of the five women to fill the $k$th position while choosing $k$ of the $10$ people. Hence, the answer in the text is equivalent to
\begin{align*}
P(X = 1) & = \frac{\binom{5}{0}\binom{5}{1}}{\binom{10}{1}} = \frac{1}{2}\\
P(X = 2) & = \frac{\binom{5}{1}\binom{5}{1}}{\binom{10}{2}} = \frac{5}{18}\\
P(X = 3) & = \frac{\binom{5}{2}\binom{5}{1}}{\binom{10}{3}} = \frac{5}{36}\\
P(X = 4) & = \frac{\binom{5}{3}\binom{5}{1}}{\binom{10}{4}} = \frac{5}{84}\\
P(X = 5) & = \frac{\binom{5}{4}\binom{5}{1}}{\binom{10}{5}} = \frac{5}{252}\\
P(X = 6) & = \frac{\binom{5}{5}\binom{5}{1}}{\binom{10}{6}} = \frac{1}{252}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving an infinite product of consecutive square roots Given $a$ and $b$ calculate $ab$
$$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$
$$b=\sqrt{2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}$$
I simplified the terms and further obtained that $ab$ is equal to:
$$ab=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...}\cdot7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...}$$
How can I get a finite value?
| First answer:
Notice that:
$$\color{Blue}{a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}} \ \ \ ;$$
$$\color{Red}{b=}\sqrt{\color{Red}{2}\color{Blue}{\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}} \ \ \ ;$$
which implies that:
$$
\color{Red}{b}=\sqrt{\color{Red}{2}\color{Blue}{a}}
\ \ \ ;
$$
similarly we have:
$$a=\sqrt{7b}
\ \ \ .
$$
So we must have:
$$a=
\sqrt{7b}=
\sqrt{7\sqrt{2a}} =
\sqrt[4]{98a}
\Longrightarrow
a^4=98a
\Longrightarrow
a^4-98a=0
; $$
but notice that
the equation $x(x^3-98)$
has only two real solutions;
$0$ and $\sqrt[3]{98}$.
So we can conclude that $a=\sqrt[3]{98}$.
Also we must have:
$$b=
\sqrt{2a}=
\sqrt{2\sqrt{7b}} =
\sqrt[4]{28b}
\Longrightarrow
b^4=28b
\Longrightarrow
b^4-28b=0
; $$
but notice that
the equation $x(x^3-28)$
has only two real solutions;
$0$ and $\sqrt[3]{28}$.
So we can conclude that $b=\sqrt[3]{28}$.
So we have: $ab=\sqrt[3]{98}\sqrt[3]{28}=\sqrt[3]{2^3.7^3}=\color{Green}{14}.$
Second answer:
Notice that
$$
\color{Green}{\dfrac{1}{2} +
\dfrac{1}{4} +
\dfrac{1}{8} +
... = 1 }
;
$$
so we can conclude that $ab=2^1.7^1=\color{Green}{14}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
How many solutions does the equation $n_1 + n_2 + n_3 + n_4 + n_5 = 20$ have in the positive integers if $n_1 < n_2 < n_3 < n_4 < n_5$? Let $n_1 < n_2 < n_3 < n_4 < n_5$ be positive integers such that $n_1 + n_2 + n_3 + n_4 + n_5 = 20$. Then the number of such distinct arrangements $(n_1, n_2, n_3, n_4, n_5)$ is......
I have no idea how to proceed. Manually, I have done it
$$1+2+3+4+10$$
$$1+2+3+5+9$$
$$1+2+3+6+8$$
$$1+2+4+5+8$$
$$1+2+4+6+7$$
$$1+3+4+5+7$$
$$2+3+4+5+6$$
But is there any way I can do it by Permutation and Combination method?
| Using combinatorics we find that the answer is coefficient of $x^{20} $ in $(x+x^2+x^3+x^4+x^5)(x^2+x^3+..x^6)(x^3..+x^7)(x^4+..+x^8)(x^5+..+x^9)=x^{15}(1+x+x^2+x^3+x^4)^5$ which is $7$ ie ways are $(1,1,1,x,x^4), (1,1,x,x,x^3), (1,1,1,x^2,x^3), (x,x,x,x,x), (1,1,x,x^2,x^2),(1,x,x,x,x^2), (1,1,x,x,x^3) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Find $\lim_{n\to\infty}x_n\sqrt{\log n}$.
Let $0<x_0<\pi$ and consider the sequence defined by $x_{n+1}=x_n-\frac{x_n-\sin(x_n)}{n+1}$. Calculate $\lim_{n\to\infty}x_n\sqrt{\log n}$.
I did a little code in C to guess an answer and it seems like this limit is $\sqrt{3}$.
I proved that $0<x_n<\pi$ for all $n$. However I couldn't do much more.
Can anyone help?
EDIT: I had, unfortunatelly wrote the limit wrong. It was written $\lim_{n\to\infty}x_n\sqrt{\log x_n}$.
| By writing $x_{n+1} = \frac{n}{n+1} x_n + \frac{1}{n}\sin x_n$, we can prove that $(x_n)$ is monotone decreasing. So $(x_n)$ converges to some value $\ell \in [0, x_0)$.
Now introduce the function $f(x) = \frac{x - \sin x}{x^3}$ and notice that (1) this can be continuated at $x = 0$ with value $f(0) = \frac{1}{6}$, and that (2) $f(x) \geq f(\pi) > 0$ on $[0, \pi$]. With this function, we may write
\begin{align*}
\frac{1}{x_{n+1}^2}
&= \frac{1}{x_n^2} \left( 1 - \frac{x_n^2 f(x_n)}{n+1} \right)^{-2} \\
&= \frac{1}{x_n^2} \left( 1 + \frac{2x_n^2 f(x_n)}{n+1} + \mathcal{O}\left( \frac{x_n^4}{n^2} \right) \right) \\
&= \frac{1}{x_n^2} + \frac{2f(x_n)}{n+1} + \mathcal{O}\left( \frac{1}{n^2} \right).
\end{align*}
So it follows from the Stolz-Cesàro theorem that
$$ \lim_{n\to\infty} \frac{x_n^{-2}}{\log n}
= \lim_{n\to\infty} \frac{x_{n+1}^{-2} - x_{n}^{-2}}{\log(n+1) - \log n}
= \lim_{n\to\infty} \frac{\frac{2f(x_n)}{n+1} + \mathcal{O}\left( \frac{1}{n^2} \right)}{\log\left(1+\frac{1}{n}\right)}
= 2f(\ell). $$
But this in particular tells that $x_n$ converges to $0$, i.e. $\ell = 0$. So we have
$$ \lim_{n\to\infty} x_n \sqrt{\log n}
= \sqrt{\frac{1}{2f(0)}}
= \sqrt{3}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is there a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$? I'm asked in an exercise from an algebra textbook if there exists a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$
I'm trying to prove that such pair of numbers does not exist, but I'm not sure my proof and my reasoning are correct. Could anyone please check my proof attempt?
First of all, $\exists\frac{1}{a+b}\in\Bbb{R}\iff a\neq -b$
and $\exists(\frac{1}{a}+\frac{1}{b})\in\Bbb{R}\iff a\neq 0 \text{ and } b \neq 0$
$\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b} \iff \frac{1}{a+b}=\frac{a+b}{ab}$
$\iff \frac{ab}{a+b}=a+b$
$\iff ab=(a+b)^2$
$\iff ab= a^2 + 2ab + b^2$
$\iff 0 = a^2+2ab+b^2-ab$
$\iff 0= a^2+b^2+ab$
Since $a \neq 0$ and $b \neq 0$; $\exists (ab)^{-1}\in\Bbb{R}$. This allows me to continue like this:
$ 0= a^2+b^2+ab \iff 0(ab)^{-1}=(a^2+b^2+ab)(ab)^{-1}$
$\iff 0 = \frac{a^2+b^2}{ab} + 1$
$\iff -1 = \frac{a^2+b^2}{ab}$
$a^2+b^2 > 0$ because $a \neq 0$ and $b \neq 0$
If a and b have the same sign, then $ab>0$.
So $\frac{a^2+b^2}{ab}$ could be negative only if:
*
*$a>0 \text{ and } b<0$ or
*$a<0 \text{ and } b>0$
Taking into account that $a \neq -b$, any of these two options would be possible only if either:
*
*$|a| > |b|$
*$|b| > |a|$
Assuming that $|a| > |b|$ we have the following:
$|a| > |b| \implies |a|*|a| > |b|*|a|$
$\implies |a|^2 > |ab|$
$\implies \frac{|a|^2}{|ab|}>1$
$\implies |\frac{a^2}{ab}|>1$
And $b \neq 0$ so we also have $|\frac{b^2}{ab}|>0$. Therefore $|\frac{a^2}{ab}|+|\frac{b^2}{ab}|> 1 + 0$.
$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions.
This implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$.
Assuming $|b|>|a|$ we arrive at the same conclusion:
$|b| > |a| \implies |b|*|b| > |a|*|b|$
$\implies |b|^2 > |ab|$
$\implies \frac{|b|^2}{|ab|}>1$
$\implies |\frac{b^2}{ab}|>1$
And $a \neq 0$ so we also have $|\frac{a^2}{ab}|>0$. Therefore $|\frac{b^2}{ab}|+|\frac{a^2}{ab}|> 1 + 0$.
$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions.
So this also implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$.
$\blacksquare$
Is this correct??
|
$\iff 0= a^2+b^2+ab$
Multiplying by $a-b$ gives $a^3-b^3=0 \iff a^3 = b^3\,$, then taking the (real) cube roots $a=b\,$. But it is readily verified that $\frac{1}{a+a} \ne \frac{1}{a} + \frac{1}{a}\,$ for any $\forall a \in \mathbb{R}\,$, so there are no real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$, $\frac{P(x)}{x^3+x}$ remainder Given the polynomial:
$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$
What is the remainder of $\frac{P(X)}{x^3+x}$?
I don't think the long division is efficient the way to go, and the remainder theorem doesn't seem to be applicable here as $x^3+x$ is not linear. Could I have some hints on how to approach this? Thank you.
| $x^3+x=x(x+i)(x-i)$
Remainder of $P(x)/(x^3+x)$ is a (at most) second degree polynomial
$R(x)=a x^2+bx+c$
We have $P(i)=-1+2i$ and $R(i)=-a + i b + c$ so $-a + i b + c=-1+2i$
then $P(-i)=-1-2i$ and $R(-i)=-a - i b + c$ so $-a - i b + c=-1-2i$
and $P(0)=1$ and $R(0)=c$
So $c=1$. Substitute in the first two equations
$-a+ib+1=-1+2i;\;-a-ib+1=-1-2i$
add the two equations $-2a+2=-2$ then $a=2$
substitute and get $b=2$
Therefore the remainder is $R(x)=2x^2+2x+1$
Hope this helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
How does this proof that $xyz \le\left (\frac{x+y+z}{3}\right)^3$ work? I have shown that the function $f(x)=x\left(\frac{na-x}{n-1}\right)^{n-1}$, $n \ge 2, a \ > 0$ with domain $0 \le x \le na$ achieves a global maximum at $x=a$. I am then supposed to show, using this function when $n=2$ and $n=3$ that if $x + y + z = 3a$, $x, y, z \ge 0$, then $a^3 \ge xyz$.
I've tried finding $f(xyz), f(x+y+z)$ etc. all to no avail. I have shown that $f(a)=a^n$, but got no further. The solution goes as follows:
By (i) [the results that there is a global maxima at a] with $n = 2$, if $z$ is fixed and $x + y + z = 3a$ with $x, y, z ≥ 0$,
then $xy \le \left(\frac{3a − z}{2}\right)^2$ with equality only if $x = y = (3a − z)/2$.
Thus if $x + y + z = 3a$ with $x, y, z ≥ 0$, (i) with $n = 3$, yields
$xyz \le z\left(\frac{3a-z}{2}\right)^2 \le a^3$ with equality only if $x = y = z = a$.
I'm struggling to even gain a foothold in this solution as I really don't understand it at all. Can anyone explain what values the author inputs into the function and how his solution works?
| First, show that $xy\le\left(\frac{x+y}{2}\right)^2$.
This can be done by applying your result with $n=2$ and $a = \frac{x+y}{2}$. Indeed, we get
$$ x\left(\frac{2a-x}{2-1}\right)^{2-1} \le a\left(\frac{2a-a}{2-1}\right)^{2-1} = a^2 $$
since the function $x\mapsto x\left(\frac{2a-x}{2-1}\right)^{2-1}$ attains a maximum at $x=a$. Notice that $2a-x = y$, so we get $xy\le a^2 = \left(\frac{x+y}{2}\right)^2$, as desired.
In particular, if we now write $x+y+z=3a$, then $x+y = 3a-z$, and so we get $xy\le\left(\frac{3a-z}{2}\right)^2$.
Now, we apply your result with $n=3$, $a = \frac{x+y+z}{3}$, and "$x$" replaced by "$z$". We then get
$$ z\left(\frac{3a-z}{3-1}\right)^{3-1}\le a\left(\frac{3a-a}{3-1}\right)^{3-1} = a^3, $$
i.e. $z\left(\frac{3a-z}{2}\right)^2\le a^3$. Since we already had $xy\le\left(\frac{3a-z}{2}\right)^2$, this implies
$$ xyz\le z\left(\frac{3a-z}{2}\right)^2 \le a^3 = \left(\frac{x+y+z}{3}\right)^3,$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Remainder when the polynomial $1+x^2+x^4+\cdots +x^{22}$ is divided by $1+x+x^2\cdots+ x^{11}$
Question : Find the remainder when the polynomial $1+x^2+x^4+\ldots +x^{22}$ is divided by $1+x+x^2+\cdots+ x^{11}$.
I tried using Euclid's division lemma, I.e.
$$P_1(x)=1+x^2+x^4+\cdots+x^{22}$$
$$P_2(x)=1+x+x^2+\cdots+x^{11}$$
Then for some polynomial $Q(x)$ and $R(x)$; we have
$$P_1(x)=Q(x)\cdot P_2(x)+R(x)$$
Now, we put the values of $x$ such that $R(x)=0$ and form equations, but this method is way too long and solving the 11 set of equations for 11 variable (Since $R(x)$ a polynomial of at most 10 degree) is impossible to do for a competitive exam where the average time for solving a question is 3 minutes.
Another method is using the original long division method, and following the pattern, we can predict $Q(x)$ and $R(x)$, but it's also very hard and time taking.
I am searching for a simple solution to this problem since last a week and now I doubt even we have a simple solution to this question.
Can you please give me a hint/solution on how to proceed to solve this problem in time?
Thanks!
| Let $\,P_n(x)=1+x+\cdots+x^{n-1}=(x^n-1)/(x-1)\,$, then the problem is equivalent to finding the remainder of the division $\,P_{12}(x^2) / P_{12}(x)\,$.
The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
$$
\begin{align}
P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}+1}{x+1} \\[5px]
&= \, \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}-1+2}{x+1} \\[5px]
&= - \, \frac{x^{2n}-1}{x-1} \, \frac{(-x)^{2n}-1}{(-x)-1} + 2 \, \frac{x^{2n}-1}{x^2-1} \\[5px]
&= - \, P_{2n}(x) P_{2n}(-x) + 2 P_n(x^2)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Solve the system $x^2-3xy+2y^2+x-y=0$, $2x^2-2xy-3y^2-2x+5y=0$
Solve the following system of equations:($x,y \in \Bbb R)$
$$\begin{cases} x^2-3xy+2y^2+x-y=0 \\ 2x^2-2xy-3y^2-2x+5y=0
\end{cases}$$
It's easily seen that $(0,0)$ and $(1,1)$ are two set of answers,but I can't analyze it to find all answers...
Also,trying to delete $x^2$ gives an equation in terms of $y^2$ whose delta is always positive
| $$\left\{\begin{array}{l} x^2-3xy+2y^2+x-y=0 \\
2x^2-2xy-3y^2-2x+5y=0
\end{array}\right.$$
substitute $x=X+1;\;y=Y+1$ in the first equation
$(X+1)^2+-3 (X+1) (Y+1)+2 (Y+1)^2+X+1-Y-1=0$
becomes $X^2 - 3 X Y + 2 Y^2=0$
$X=x-1;\;Y=y-1$
Solve $X=Y;\;X=2Y$ and then $x=y;\;x-1=2y-2\to x=2y-1$
Plug in the second equation and find, for $x=y$
$3 y-3 y^2=0\to y=0;\;y=1$
first two solutions $(0,0);\;(1;\;1)$
for $x=2y-1$
$y^2-5 y+4=0\to y=1;\;y=4$
The other two solutions are $(1,1);\;(7,4)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a+b)^2 = 1^2 = 1 \implies (a+b)-(a+b)^2 = 1-1 =0$ (1)
$(a-b)^2 \ge 0$
So by (1) we have:
$(a-b)^2 \ge (a+b)-(a+b)^2$
$(a^2-2ab+b^2) \ge (a+b) - (a^2+2ab+b^2)$
$(a^2-2ab+b^2) + (a^2+2ab+b^2) \ge (a+b) $
$ a^2+a^2+b^2+b^2+2ab-2ab \ge (a+b)$
$2(a^2+b^2) \ge (a+b)$
$2(a^2+b^2) \ge 1$
$(a^2+b^2) \ge \frac{1}{2} $
$\blacksquare$
| Ok, here is a different approach: Using graphs. In the $a,b$ coordinate plane, $a+b=1$ represents a line and $a^2+b^2=\frac{1}{2}$ represents a circle. Now when you graph (You can do that) you notice that the line falls completely outside the circle, except for one point where the line is tangent. Conclusion? Note: While there is absolutely nothing wrong with an algebraic approach, sometimes it is also good to consider a geometric approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
Convergence of the series $\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )$ Given $a, b \in \mathbb{R}$ with $b > 0$, is the series
\begin{equation}
\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )
\end{equation}
convergent or divergent?
If we drop out the $n=0$ term and fold the remaining sum, the question can be equivalently asked for the series
\begin{equation}
\sum_{n = 1}^{\infty} (\sqrt{(a+n)^2+b^2} + \sqrt{(a-n)^2+b^2} - 2n ).
\end{equation}
| Your two expressions are not equivalent. The sum is not uniformly convergent, so the order you sum the terms in can matter. The second has an explicit order of summing while the first does not and will cancel out the constant terms. Even the second one does not converge, however.
We have $\sqrt {(n+a)^2+b^2}-n \sim a+\frac {b^2}{2n}+o(n^{-1})$ as $n$ gets large, so $\sqrt {(n+a)^2+b^2}+\sqrt {(n-a)^2+b^2}-2n \sim\frac {b^2}{n}+o(n^{-1})$ and the sum diverges logarithmically
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Write $(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$ as a sum of (three) squares of quadratic forms The quartic form
$$(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$$
is non-negative for all real $x$, $y$, $z$, as one can check (with some effort). A theorem of Hilbert implies that there exist quadratic forms $Q_1$, $Q_2$, $Q_3$ so that
$$(x^2 + y^2 + z^2)^2 - 3( x^3 y + y^3 z + z^3 x) = Q_1^2 + Q_2^2 + Q_3^2$$
I would like to find an explicit writing of the quartic forms, with rational quadratic forms $Q_i$. Maybe more than $3$ terms are necessary.
| Using Macaulay2 (version 1.16) with package SumsOfSquares:
i1 : needsPackage( "SumsOfSquares" );
i2 : R = QQ[x,y,z];
i3 : f = (x^2 + y^2 + z^2)^2 - 3*( x^3 * y + y^3 * z + z^3 * x);
i4 : sosPoly solveSOS f
1 2 1 1 1 2 2 9 1 2 2 2 1 2 2
o4 = (3)(- -x + x*y - -x*z - -y*z + -z ) + (-)(- -x + -y + x*z - y*z - -z )
2 2 2 2 4 3 3 3
o4 : SOSPoly
Printing:
i5 : tex o4
o5 = $\texttt{SOSPoly}\left\{\texttt{coefficients}\,\Rightarrow\,\left\{3,\,\frac{9}{4}\right\},\,\texttt{generators}\,\Rightarrow\,\left\{-\frac{1}{2}\,x^{2}+x\,y-\frac{1}{2}\,x\,z
-\frac{1}{2}\,y\,z+\frac{1}{2}\,z^{2},\,-\frac{1}{3}\,x^{2}+\frac{2}{3}\,y^{2}+x\,z-y\,z-\frac{1}{3}\,z^{2}\right\},\,\texttt{ring}\,\Rightarrow\,R\right\}$
In $\TeX$:
$$\texttt{SOSPoly}\left\{\texttt{coefficients}\,\Rightarrow\,\left\{3,\,\frac{9}{4}\right\},\,\texttt{generators}\,\Rightarrow\,\left\{-\frac{1}{2}\,x^{2}+x\,y-\frac{1}{2}\,x\,z
-\frac{1}{2}\,y\,z+\frac{1}{2}\,z^{2},\,-\frac{1}{3}\,x^{2}+\frac{2}{3}\,y^{2}+x\,z-y\,z-\frac{1}{3}\,z^{2}\right\},\,\texttt{ring}\,\Rightarrow\,R\right\}$$
Hence,
$$\boxed{ f = 3 \left( -\frac{1}{2}\,x^{2}+x\,y-\frac{1}{2}\,x\,z
-\frac{1}{2}\,y\,z+\frac{1}{2}\,z^{2} \right)^2 + \frac{9}{4} \left( -\frac{1}{3}\,x^{2}+\frac{2}{3}\,y^{2}+x\,z-y\,z-\frac{1}{3}\,z^{2} \right)^2 }$$
polynomials sum-of-squares-method macaulay2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Partial Fractions Decomposition $\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$ explanation repeated factors I am trying to solve the fraction $$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$$
into partial fractions.
Now, I thought it could be solved into the following
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{(x-3)^2}$$
but this is apparently incorrect.
According to the text, the decomposition is
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$
I discussed this with my friend that the fraction first decomposes into
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{Bx +C}{(x-3)^2}$$
but I can't see how he derived this.
I don't understand how he is correct.
| Both decompositions used by your friend and the text are analogous since
$$\frac{Bx+C}{(x-3)^2}=\frac{Bx}{(x-3)^2}+\frac{C}{(x-3)^2}=\frac{B(x-3)+3B}{(x-3)^2}+\frac{C}{(x-3)^2}$$
$$=\frac{B}{x-3}+\frac{3B+C}{(x-3)^2}=\frac{B'}{x-3}+\frac{C'}{(x-3)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the sum $\sum_{n=1}^{b}n\binom{a}{n}\binom{b}{n}$ Let $a$ and $b$ be natural numbers such that $a \geq b \geq 1$. How can we evaluate the following sum?
$$ \sum_{n=1}^{b}n\binom{a}{n}\binom{b}{n}$$
| For any polynomial $p(z) = a_0 + a_1 z + \cdots a_n z^n$, let $[z^k]p(z)$ be the
coefficient $a_k$ for the monomial $z^k$ in $p(z)$. We have
$$\begin{align}\sum_{n=1}^b n \binom{a}{n}\binom{b}{n}
&= \sum_{n=0}^b \binom{a}{a-n}\times n\binom{b}{n}
= \sum_{n=0}^b [z^{a-n}]\{(1+z)^a\} \times [z^n]\{z\frac{d}{dz}(1+z)^b\}\\
&= [z^a]\{(1+z)^a \times bz(1+z)^{b-1}\}
= b[z^{a-1}]\{(1+z)^{a+b-1}\}\\
&= b\binom{a+b-1}{a-1}
= \frac{(a+b-1)!}{(a-1)!(b-1)!}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $x$, for which the function $h(x) = \frac 1{(x-5)^2+4(x-5)+4}$ is undefined Picture of problem
I'm having an issue with a simple SAT question. If you can't see from the picture, I'm being asked to find the value of $x$, for which the function:
$$h(x) = \frac 1{(x-5)^2+4(x-5)+4}$$
is undefined.
I figured that for it to be undefined, it should be equal to 1/0. This therefore means that $(x-5)^2+4(x-5)+4 = 0$. I further simplified it into
$$(x-5)((x-5)+4) = 0$$
However, this gives me either $x = 1$ or $x = -3$, both of which are false. The answer is $3$. Why do I get $-3$, when the answer is $3$?
| The denominator has to be zero for the chosen $x$ values of course;
$(x-5)^2+4(x-5)+4=0$
We can name $(x-5)=a$ to make things easier.
$a^2+4a+4=0$
That is a perfect square;
$(a+2)^2=0$
$a=-2$ so $x-5=-2 \rightarrow x=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Evaluate $\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$ Evaluate
$$I=\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$$
My Try:
$$I=\int \frac{(\cos x-\sin x)(\cos x+\sin x)dx}{3 \sin x+4 \cos x}$$
we have $$\cos x-\sin x=\frac{1}{25}(3 \sin x+4 \cos x)+\frac{7}{25}(3 \cos x-4 \sin x)$$ and
$$\cos x+\sin x=\frac{7}{25}(3 \sin x+4 \cos x)+\frac{-1}{25}(3 \cos x-4 \sin x)$$
So
$$\cos 2x=\frac{7}{625}(3 \sin x+4 \cos x)^2+\frac{48}{625}\left((3 \sin x+4 \cos x)(3\cos x-4 \sin x)\right)-\frac{7}{625}(3 \cos x-4 \sin x)^2$$
So
$$I=\frac{7}{625}\int (3 \sin x+4 \cos x)dx+\frac{48}{625}\int (3 \cos x-4 \sin x)dx-J$$
where
$$J=\frac{7}{625}\int \frac{(3 \cos x-4 \sin x)^2dx}{3 \sin x+4 \cos x}$$
I got stuck up to integrate $J$.
| \begin{align}
Hint:(3\cos x-4\sin x)^2&=25-(3\sin x+4\cos x)^2\\
\int\frac{(3\cos x-4\sin x)^2}{3\sin x+4\cos x}dx&=\int\frac{25-(3\sin x+4\cos x)^2}{3\sin x+4\cos x}dx\\
&=25\int\frac{dx}{3\sin x+4\cos x}-\int(3\sin x+4\cos x)dx\\
&=25\int\frac{dx}{3\sin x-4+4(1+\cos x)}+3\cos x-4\sin x\\
&=25\int\frac{\frac{1}{1+\cos x}dx}{3\frac{\sin x}{1+\cos x}-4\frac{1}{1+\cos x}+4}+3\cos x-4\sin x\\
&=25\int\frac{d(\frac{\sin x}{1+\cos x})}{3\frac{\sin x}{1+\cos x}-2((\frac{\sin x}{1+\cos x})^2+1)+4}+3\cos x-4\sin x\\
&=25\int\frac{dt}{2+3t-2t^2}+3\cos x-4\sin x\\
&=25(\frac{1}{5}\ln \left|\frac{2t+1}{2-t}\right |)+3\cos x-4\sin x\\
&=5\ln \left|\frac{2\sin x+1+\cos x}{2(1+\cos x)-\sin x}\right |+3\cos x-4\sin x+C\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Ordered pairs of positive integers $2^{2x} - y^2 = 60$ How many ordered pairs of positive integers are there that satisfy $2^{2x} - y^2 = 60$?
This can be rewritten as $(2^x)^2-y^2 = 60$ and then $(2^x+y)(2^x-y) = 60$. Then since $2^x$ is always positive and so $2^x+y$ and $2^x-y$ are both positive and the first is always bigger than the latter. This means I only have to account for the positive factors of $60$, and so I had the following pairs $(60,1), (30,2), (20,3), (15,4), (12,5)$, and I plugged them in and solved the systems. Of the 5 possible cases, only $(30,2)$ worked and yielded $x=4, y=14$.
Am I right in having done so? They don't have an answer, and when I tried to graph this, it didn't exactly work out well (I didn't get an integer value for $x$).
| Basic things to note about $(2^x + y)(2^x - y) = ab = 60; a = 2^x + y; b=2^x-y)$.
1) $2^x + y > 0$ so $2^x - y > 0$.
2) $a= 2^x + y > 2^x -y=b > 0$ so $a > \sqrt{60} > 7$ and $b < \sqrt{60} < 8$.
3) $a$ and $b$ can't both be odd so they are both even.
So $b = 2,\color{red}4,6$ and $a=30,\color{red}{15},10$
4) $a+b = 2^{x+1} = 32, 16$ so $x=4,3$ and $y = \frac {a-b}2 = 14, 2$.
So the two possible solutions are $(x,y)=(4,14): (2^4+14)(2^4 -14)=30*2=60$ and $(x,y)=(3,2): (2^3+2)(2^3-2)=10*6=60$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Prove that $\left|\begin{smallmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{smallmatrix}\right|=(a^2+b^2+c^2+d^2)^2$
Let $a, b, c, d \in \mathbb K$ where $\mathbb K$ is a field. Prove that
$$\det \begin{bmatrix}
a & -b & -c & -d\\
b & a & -d & c\\
c & d & a & -b\\
d & -c & b & a
\end{bmatrix} = (a^2+b^2+c^2+d^2)^2$$
I'm looking for a smart way to solve this problem. If we denote
$$A = \begin{bmatrix}
a & -b \\
b & a \\
\end{bmatrix}$$
and
$$B = \begin{bmatrix}
-c & -d \\
-d & c \\
\end{bmatrix}$$
we have that
$$ \begin{bmatrix}
a & -b & -c & -d\\
b & a & -d & c\\
c & d & a & -b\\
d & -c & b & a
\end{bmatrix} = \begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix} $$
So it's sufficient to proof that
$$ \det \begin{bmatrix}
A & B \\
-B & A \\
\end{bmatrix} = (\det A - \det B)^2. $$
Help?
| Calculate $$ P P^T $$ then think about it.
Or
$$ P^T P $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
What special functions do we need to get a closed form for $\sum\limits_{n=0}^\infty \frac{2^{n+1}}{\binom{2 n}{n}} \cdot \frac{1}{(2n+1)^p}$? This question generalizes the considerations in Is there a closed form for $\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2?$ .
Let $p\ge 3$ be an integer. We consider the following sum:
\begin{equation}
S_{p} := \sum\limits_{n=0}^\infty \frac{2^{n+1}}{\binom{2 n}{n}} \cdot \frac{1}{(2n+1)^p}
\end{equation}
By pushing through the approach developed in the aforementioned question we got the following result:
\begin{eqnarray}
S_3 = -2 \imath \sqrt{2} \left(\right.\\
\left.
(\pi-\imath \log(2))\cdot \frac{2 \pi ^2+2 \pi \left(\psi ^{(0)}\left(\frac{3}{8}\right)-\psi ^{(0)}\left(\frac{1}{8}\right)\right)-\psi ^{(1)}\left(\frac{1}{8}\right)-\psi
^{(1)}\left(\frac{3}{8}\right)}{32 \sqrt{2}}+ \right.\\
\left.\frac{\sqrt{2}}{256}\cdot(448 \sqrt{2} \zeta (3)+28 \pi ^3+4 \pi ^2 (\psi ^{(0)}(\frac{3}{8})-\psi ^{(0)}(\frac{1}{8}))-4 \pi (\psi^{(1)}(\frac{1}{8})+\psi ^{(1)}(\frac{3}{8}))+\psi ^{(2)}(\frac{1}{8})-\psi^{(2)}(\frac{3}{8}))+\right.\\
\left.
-\int\limits_1^{\exp(\imath \frac{\pi}{4})} \log(z) \left(\log(z-1)+\log(z+1)\right) \cdot \left( \frac{1}{z-1} - \frac{1}{z+1}\right)dz \right.\\
\left.
\right)
\end{eqnarray}
Here $\psi^{(m)}(z)$ is the poly-gamma function.
Now, the question is twofold. Firstly can we evaluate the remaining integral in the right hand side and express the whole thing a a "closed form". Secondly, can we come up with closed forms for arbitrary values of $p$. If yes what kind of special functions will be involved?
| We provide a partial answer to question number one. It is not hard to see that one can construct anti-derivatives of the integrand by using poly-logarithms.
As a matter of fact we have:
\begin{eqnarray}
\int \log(z) \log(z+1) \frac{1}{z+1} dz&=& \frac{1}{2} \imath \pi [\log(1+z)]^2- \log(1+z) Li_2(1+z)+Li_3(1+z)\\
\int \log(z) \log(z-1) \frac{1}{z-1} dz&=& - \log(-1+z) Li_2(1-z)+Li_3(1-z)
\end{eqnarray}
Now using the above after tedious but elementary simplifications we got the following:
\begin{eqnarray}
&&\int\limits_1^{\exp(\imath \frac{\pi}{4})} \log(z) \log(z+1) \frac{1}{z+1} dz=-\frac{7}{8} \zeta(3)+\frac{1}{512}(-\pi+2 \imath(-2 \log(-1+\sqrt{2})+\log(2)))\cdot\\
&&\left(16 C+\left(4 \sqrt{2}+13 i\right) \pi ^2-4 \pi \left(\log (2)-2 \log \left(\sqrt{2}-1\right)\right)-\sqrt{2} \left(\psi ^{(1)}\left(\frac{1}{8}\right)+\psi
^{(1)}\left(\frac{3}{8}\right)\right)\right)+\\
&&+Li_3[1+\exp(\imath \frac{\pi}{4})]
\end{eqnarray}
Likewise
\begin{eqnarray}
&&\int\limits_1^{\exp(\imath \frac{\pi}{4})} \log(z) \log(z-1) \frac{1}{z-1} dz=\frac{1}{512}(5\pi+2 \imath(-2\log(-1+\sqrt{2})-\log(2)))\cdot\\
&&\left(-16 C+\left(4 \sqrt{2}-i\right) \pi ^2-4 \pi \left(\log (2)+2 \log \left(\sqrt{2}-1\right)\right)-\sqrt{2} \left(\psi ^{(1)}\left(\frac{1}{8}\right)+\psi
^{(1)}\left(\frac{3}{8}\right)\right)\right)+\\
&&+Li_3[1-\exp(\imath \frac{\pi}{4})]
\end{eqnarray}
where $C$ is the Catalan constant.
The remaining tri-logarithms are expressed in terms of the Hurwitz zeta functions $\Psi$ as follows:
\begin{eqnarray}
Li_3[1+\exp(\imath \frac{\pi}{4})]&=&\frac{1}{8^3} \sum\limits_{\xi=1}^8 \exp(\imath \xi \frac{\pi}{8})(2 \cos(\frac{\pi}{8}))^\xi \cdot \Psi(-(2 \cos(\frac{\pi}{8}))^8,3,\frac{\xi}{8})\\
Li_3[1-\exp(\imath \frac{\pi}{4})]&=&\frac{1}{8^3} \sum\limits_{\xi=1}^8 \exp(-\imath \xi \frac{3\pi}{8})(2 \sin(\frac{\pi}{8}))^\xi \cdot \Psi(-(2 \sin(\frac{\pi}{8}))^8,3,\frac{\xi}{8})\\
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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The ellipse $x^2+2y^2=2.$ Question: The ellipse $x^2+2y^2=2$ is given by $x=a\cos{t}, \quad y=b\sin{t}, \quad t\in[0,2\pi],$ for what values of $a$ and $b$?
a) $(a,b)=(2,1)$
b) $(a,b)=(1,2)$
c) $(a,b)=(1,\sqrt{2})$
d) None of the above.
Attempt: Substitute the trig-values for $x$ & $y$ in the elliptic equation:
$$x^2+2y^2=2\Longleftrightarrow a^2\cos^2{t}+2b^2\sin^2{t}=2.$$
Trig identities gives
$$a^2(1-\sin^2{t})+2b^2\sin^2{t}=a^2-a^2\sin^2{t}+2b^2\sin^2{t}=a^2-\sin^2{t}(a^2+2b^2)=2.$$
Solving for $\sin{t}$ gives
$$\sin{t}=\pm\sqrt{\frac{a^2-2}{a^2+2b^2}}=\pm f(a,b).$$
Since $$-1\leq\sin{t}\leq1\Longleftrightarrow -1\leq\pm f(a,b)\leq1.$$
Answer a) gives $f(2,1)=\frac{\sqrt{3}}{3} \in[-1,1]$. OK!
Answer b) gives $f(1,2)=\frac{i}{3}\in \mathbb{C}$. Disregard.
Answer c) gives $f(1,\sqrt{2}) = \frac{i\sqrt{5}}{5}\in \mathbb{C}$. Disregard.
So according to me, the correct answer should be a). But correct answer is d). Where did I go wrong?
| You made a mistake when you wrote $$a^2-a^2\sin^2{t}+2b^2\sin^2{t}=a^2-\sin^2{t}(a^2+2b^2)=2.$$
In fact,
$$a^2-a^2\sin^2{t}+2b^2\sin^2{t}=a^2-\sin^2{t}(a^2\color{red}{-}2b^2)=2.$$
From here, you can conclude that $a^2=2b^2$ as the right-hand side is not a function of $t$. Consequently, $a^2=2$, and $b^2=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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} |
Prove by induction that $ \ \forall n \ge 2, \ 2^{n+1} > n^{2} + 3$ Question:
Prove by induction that $$ \ \forall n \ge 2, \ 2^{n+1} > n^{2} + 3$$
My attempt:
Base case is trivial.
Suppose that $ n \ge 2$ and $\ 2^{n+1} > n^{2} + 3$
WTS $ \ 2^{n+2} > (n+1)^{2} + 3$
$ 2^{n+2} = 2.2^{n+1}> 2(n^2 + 3) = 2n^2 + 6 = n^2 + n^2 + 2n -2n +6 = (n^2+2n+1) + 3 + (n^2 -2n+2) = (n+1)^2 + 3 + (n^2 -2n+2)$.
I am not sure what to do from here. How can I show $ (n^2 -2n+2) > 0?$
| $n^2-2n+2=(n-1)^2+1>0$.
By the way, this reasoning gives also the proof without an induction:
$$2^{n+1}-n^2-3=(1+1)^{n+1}-n^2-3\geq$$
$$\geq1+(n+1)+\frac{(n+1)n}{2}+\frac{(n+1)n(n-1)}{6}-n^2-3=$$
$$=\frac{(n-1)(n^2-2n+6)}{6}>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Polynomials : $P(xy)+P(yz)+P(zx) =P(xy+yz+zx)$ Find all polynomials $P(x) \in \mathbb{R}[x]$ such that $$P(xy)+P(yz)+P(zx) =P(xy+yz+zx)$$
$\forall x, y, z \in \mathbb{R}$ satisfying the equation $x+y+z=0$
My attempt :
Consider $P(0,0,0)$, $3P(0)=P(0)$, so $P(0)=0$
Since $x+y+z=0$, let $x=a-b, \;y=b-c, \;z=c-a, \;\forall a, b, c \in \mathbb{R}$
$P((a-b)(b-c)) + P((b-c)(c-a)) + P((c-a)(a-b))$
$= P((a-b)(b-c)+(b-c)(c-a)+(c-a)(a-b))$
$= P((a-b)(b-c)+(c-a)(a-c))$
$= P(ab+bc+ca-a^2-b^2-c^2)$
| Working first with $y=x$ and $z=-2x$, we obtain
$$P(x^2)+ 2 P(-2 x^2) = P(-3 x^2)$$
Let $a_n x^n$ be the highest order term in $P(x)$, this relation implies
$$a_n x^{2n} + 2 (-2)^n a_n x^{2n} = (-3)^n a_n x^{2n} \quad\Longrightarrow\quad 1 + 2 (-2)^n = (-3)^n$$
The only solutions are $n = 1$ or $n = 2$. So that $P$ has degree $1$ or
$2$. Observe also that $P(0) = 0$ so that there is no constant term in
$P$ and also that the set of solutions is a vector space. It only remains to check that the polynomials $x$ and $x^2$ satisfy the initial property, finally
$$P(x) = \alpha x^2 + \beta x \qquad \alpha, \beta \in {\mathbb R}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\frac12\cdot\frac34\cdot\dots\cdot\frac{2n-1}{2n}\leq\frac1{\sqrt{3n+1}} ,\;\forall n\in\mathbb{N}$ using induction Base case. Let $n=1$, then $\frac12\leq\frac1{\sqrt{3+1}}$.
Induction step. Let's assume the inequality is true for some $k\in\mathbb{N}$.
We need to show that it's true for $k+1$, i.e. $\frac12\cdot\frac34\cdot\dots\cdot\frac{2k-1}{2k}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$.
From the assumption we get that $\frac12\cdot\frac34\cdot\dots\cdot\frac{2k-1}{2k}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}$.
So now I need to show that $\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$. How should I do this?
| After squaring of the both sides you'll get
$$(2k+1)^2(3k+4)\leq4(k+1)^2(3k+1)$$ or
$$19k\leq20k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to show that the deltoid is a plane algebraic curve of degree $4$?
How can I show that a deltoid is a plane algebraic curve of degree 4?
I have searched that the parametric equation for deltoid is given by
$$\begin{align}
x&= 2 \cos t + \cos 2t \\
y&=2 \sin t - \sin 2t
\end{align}$$
So, by using some trigonometric properties, we can write
$$\begin{align}
x = 2 \cos t + 2 \cos^2 t - 1 &\quad\to\quad x^2 = \phantom{-}4 \cos^4t + 8\cos^3 t - 4 \cos t + 1 \\
y =2 \sin t -2\sin t \cos t\;\, &\quad\to\quad y^2=-4 \cos^4t+ 8 \cos^3t -8\cos t+4
\end{align}$$
| In M2 setting $c=\cos(t), s=\sin(t)$ using $c^2+s^2-1=0$:
R=QQ[c,s,x,y,MonomialOrder=>Lex]
I=ideal(c^2+s^2-1,x-(2*c+c^2-s^2),y-(2*s-2*s*c))
gens gb I
yields $x^4-8x^3+2x^2y^2+18x^2+24xy^2+y^4+18y^2-27$ or as wikipedia puts it (for $b=3, a=1$): $$(x^2+y^2)^2+18(x^2+y^2)-27=8(x^3-3xy^2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Computing derivatives with fractional exponents I'm trying to compute the following derivative:
$$
\text{Using first principles, differentiate}: f'(x) = (x)^\frac{1}{4}\\\\
$$
I'm used to the functions being whole numbers or some simple algebra, i'm a little confused with what exactly to do when we're working with $(x)^\frac{1}{4}$.
Below is my attempt at determining $x + h$:
$$
\text{First principle formula}: f(x) = \lim_{h \to 0} \frac{f(x + h) - f (x)}{h}\\\\
\text{determine}:f(x + h)\\
f(x) = (x)^\frac{1}{4}\\\\
f(x) = (\sqrt[4]{x})\\\\
f(x + h) = (\sqrt[4]{x+h})
$$
This is where I get stuck, not sure how to determine it or substitute it into the formula and then simplify.
Any suggestions are welcomed, thanks!
| Note that
\begin{align*}
f(x + h) - f (x)&=\left(\sqrt[4]{x+h}-\sqrt[4]{x}\right)\cdot
\frac{\sqrt[4]{x+h}+\sqrt[4]{x}}{\sqrt[4]{x+h}+\sqrt[4]{x}}\cdot
\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\
&=\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt[4]{x+h}+\sqrt[4]{x}}\cdot
\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\
&=\frac{(x+h)-x}{\sqrt[4]{x+h}+\sqrt[4]{x}}\cdot
\frac{1}{\sqrt{x+h}+\sqrt{x}}.
\end{align*}
Hence as $h\to 0$,
$$\frac{f(x + h) - f (x)}{h}=\frac{1}{\sqrt[4]{x+h}+\sqrt[4]{x}}\cdot
\frac{1}{\sqrt{x+h}+\sqrt{x}}\to \frac{1}{2\sqrt[4]{x}}\cdot
\frac{1}{2\sqrt{x}}=\frac{1}{4x^{3/4}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Eliminate $t$ from the equations: $x = \frac{1}{t} - t \, , \, y = \frac{1}{t^2} - 1$ $$x = \frac{1}{t} - t \tag{1}$$
$$y = \frac{1}{t^2} - 1 \tag{2}$$
$$\frac{1}{t^2} = y + 1$$
$$t^2 = \frac{1}{y + 1}$$
$$t = \pm\sqrt{\frac{1}{y + 1}}$$
$$x = \frac{1 - t^2}{t}$$
$$y = \frac{1 - t^2}{t^2}$$
$$\frac{x}{y} = \frac{1 - t^2}{t} \times \frac{t}{1 - t^2}$$
$$\frac{x}{y} = t \tag{3}$$
Substitute $(3)$ into $(1)$
$$x = \frac{y}{x} - \frac{x}{y}$$
$$x = \frac{y^2 - x^2}{xy}$$
$$x^2y = y^2 - x^2$$
$$x^2y + x^2 = y^2$$
$$x^2(y + 1) = y^2$$
$$x^2 = \frac{y^2}{y + 1}\tag{4}$$
$$x = \sqrt{\frac{y^2}{y + 1}}$$
From $(4):$
$$x^2y + x^2 = y^2$$
$$y^2 - x^2y = x^2$$
$$y(y-x^2) = x^2$$
I have no idea how to progress from that point. I've tried different methods in paper to make $y$ the subject of the formula, but none of them were fruitful.
How do I make $y$ the subject of the formula in $(4)$?
| Assuming that the manipulations up till (4) are correct,
$$y^2 - x^2y=x^2$$
$$(y-\frac{x^2}{2})^2-\frac{x^4}{4}=x^2$$
$$(y-\frac{x^2}{2})^2=x^2+\frac{x^4}{4}$$
$y=\frac{x^2}{2}+\sqrt{x^2+\frac{x^4}{4}}$ or $y=\frac{x^2}{2}-\sqrt{x^2+\frac{x^4}{4}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve the equation $ z^2 + \left\vert z \right\vert = 0 $, where $z$ is a complex number. I've tried solving this, but I'm stuck at one point.
Here's what I did:
Let $ z = x + yi $, where $x, y \in \mathbf R$
Then , $ (x + yi)^2 + \sqrt{x^2 + y^2} = 0 $
Or, $x^2 + {(yi)}^2 + 2xyi + \sqrt{x^2 + y^2} = 0 $
Or, $ x^2 - y^2 + 2xyi + \sqrt{x^2 + y^2} = 0 + 0i$
Thus, $ x^2 - y^2 + \sqrt{x^2 + y^2} = 0\qquad\qquad\qquad\qquad\qquad\qquad\ (i)$
and $2xy = 0 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(ii)$
If $2xy = 0$, then either $x = 0 $ or $y = 0$
Now, if I take $ x = 0$, and subsitute in $(i)$, I get either $y = 0$ or $y = 1$.
So far, so good, but if I take $y = 0$, and substitute in $(ii)$:
We have $x^2 + \sqrt{x^2} = 0$
so $x^2 = -\sqrt{x^2} $
or $x^2 = -x$
or $\frac{x^2}{x} = -1 $
or $x = -1$
However, this solution doesn't satistfy the equation $x^2 + \sqrt{x^2}$ or the original equation.
What am I doing wrong here ?
| Let $x,y\in\mathbb{R};\;z=x+iy$ the given equation $z^2+|z|=0$ becomes
$(x+iy)^2+\sqrt{x^2+y^2}=0$
$x^2-y^2+\sqrt{x^2+y^2}+ixy=0$
which translates in the system
$\left\{
\begin{array}{l}
x^2-y^2+\sqrt{x^2+y^2}=0 \\
y-2 x y=0 \\
\end{array}
\right.
$
$\left\{
\begin{array}{l}
x^2-y^2+\sqrt{x^2+y^2}=0 \\
2xy=0 \\
\end{array}
\right.
$
Which splits into two systems
$\left\{
\begin{array}{l}
x^2-y^2+\sqrt{x^2+y^2}=0 \\
x=0 \\
\end{array}
\right.
$
$\left\{
\begin{array}{l}
x^2-y^2+\sqrt{x^2+y^2}=0 \\
y=0 \\
\end{array}
\right.
$
$\left\{
\begin{array}{l}
-y^2+\sqrt{y^2}=0 \\
x=0 \\
\end{array}
\right.
$
$\left\{
\begin{array}{l}
x^2+\sqrt{x^2}=0 \\
y=0 \\
\end{array}
\right.
$
$\left\{
\begin{array}{l}
y^2=|\,y\,| \\
x=0 \\
\end{array}
\right.
$
$\left\{
\begin{array}{l}
x^2=-|\,x\,| \\
y=0 \\
\end{array}
\right.
$
$\left\{
\begin{array}{l}
y=0\lor y= \pm 1 \\
x=0 \\
\end{array}
\right.
$
$\left\{
\begin{array}{l}
x=0\\
y=0 \\
\end{array}
\right.
$
So we have the solutions
$z=0\lor z=\pm i$
Hope this helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determine all real $x$ such that $\arccos{\frac{1-x^2}{1+x^2}}=-2\arctan{x}.$ I just need you guys to check this solution and tell me where I can improve. The key questions to be answered are:
*
*Are things clear?
*Is there any unnecessary redundancy?
*Any logical fallacies?
*Quality of stringency and mathematical language?
Problem: Determine all real $x$ such that $$\arccos{\frac{1-x^2}{1+x^2}}=-2\arctan{x}.$$
Attempt: Let $f(x)=\arccos{\frac{1-x^2}{1+x^2}}$ and $g(x)=-2\arctan{x}.$ By the definition of the inverse trigonometric functions it now follows that
$$\left\{
\begin{array}{rcr}
D_{f} & ?=? & [-1,1] \\
V_{f} & = & [0,\pi] \\
\end{array}
\right. \quad \text{and} \quad \left\{
\begin{array}{rcr}
D_{g} & = \mathbb{R}\\
V_{g} & = & 2\left(-\frac{\pi}{2},\frac{\pi}{2}\right)=(-\pi,\pi)\\
\end{array}
\right.$$
Where $V$ denotes ranges and $D$ denotes domains. By $D_{f}$ it follows that $-1\leq \frac{1-x^2}{1+x^2}\leq1.$ Upon examination of the inequalities $-1\leq\frac{1-x^2}{1+x^2}$ and $\frac{1-x^2}{1+x^2} \leq 1$ one finds that they get satisfied $\forall x\in\mathbb{R}.$ This implies that $D_{f}=D_{g}=\mathbb{R}.$ The the function values can only be the same in the intersection of their respective ranges; $[0,\pi]\cap(-\pi,\pi)=[0,\pi).$ So $f(x)\rightarrow\pi$ when $x\rightarrow -\infty$ and $g(x)\rightarrow\pi$ when $x\rightarrow -\infty.$ We also note that both $f(x)$ and $g(x)\rightarrow 0$ when $x\rightarrow0^{-}.$ This means that the solutions, if they exist, should be in the interval $(-\infty,0]$. Lets find them:
Taking cosine of both sides we get $$\frac{1-x^2}{1+x^2}=\cos{(2\arctan{x})}=2\cos^2{(\arctan{x})}-1.$$ Using the fact that $\cos{\arctan{x}}=\frac{1}{\sqrt{1+x^2}}$ from
So $$\frac{1-x^2}{1+x^2}=\frac{2}{1+x^2}-1 = \frac{1-x^2}{1+x^2} \Longleftrightarrow x=x.$$
This means that the solutions are all reals, but according to our earlier conclusion the solutions must exist in $(-\infty,0].$ Thus the solution set to the original equation is $$x\in(-\infty,0]\cap(-\infty,\infty)=(-\infty,0].$$
| $\cos (-2\arctan x) = \frac {1-x^2}{1+x^2}$
However
the range of $\arccos \theta$ is $[0,\pi]$
the range of $-2\arctan x$ is $(-\pi,\pi)$
These two functions can equal one another when $x$ is such that $-2\arctan x$ is in $[0,\pi)$
or $x\in(-\infty, 0]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $ax+by=7$, .
If
$$ax+by=7$$
$$ax^2+by^2=49$$
$$ax^3+by^3=133$$
$$ax^4+by^4=406$$
then find the value of
$$2014(x+y-xy) - 100(a+b)$$
My attempt:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$
| Multiplying the first three equations with $x+y$, we get
\begin{eqnarray*}
(ax+by)(x+y) &=& ax^2+by^2 +(a+b)xy \\
(ax^2+by^2)(x+y) &=& ax^3+by^3 +(ax+by)xy \\
(ax^3+by^3)(x+y) &=& ax^4+by^4 +(ax^2+by^2)xy
\end{eqnarray*}
or
\begin{eqnarray*}
7(x+y) &= & \;\;49 +(a+b)xy \\
49(x+y) &= &133 +\;\;7xy \\
133(x+y) &=& 406 +49xy
\end{eqnarray*}
Now you have three equations with three unknowns ($x+y$, $xy\;$ and $a+b$). You get $x+y = \frac{5}{2}$, $xy = -\frac{3}{2}$ and $a+b=21.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi$ in verifying density function Let $(X,Y)$ have a bivariate normal density centered at the origin with $\mathbb{E}(X^2)=\mathbb{E}(Y^2)=1$ and $\mathbb{E}(XY)=\rho$. In polar coordinates, $(X,Y)$ becomes $(R,\Phi)$ where $R^2=X^2+Y^2$. Then it can be shown that $\Phi$ has a density given by $$f_{\Phi}(\phi)=\frac{\sqrt{1-\rho^2}}{2\pi(1-\rho\sin2\phi)},\quad 0<\phi<2\pi$$
But I am having difficulty proving that this is indeed a density, more specifically the fact that $\displaystyle\int_0^{2\pi}f_{\Phi}(\phi)\,\mathrm{d}\phi=1$. Applying the usual substitution $t=\tan\phi$ in the integral $\displaystyle\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi$ seems to be misleading. Should there be any change in the integration limits for proving that this a density. If so, why?
EDIT.
Another related integral is $I=\displaystyle\int_0^{\pi/2}\frac{\mathrm{d}\theta}{1-\rho\sin{2\theta}}=\frac{\pi-\arccos\rho}{\sqrt{1-\rho^2}}$, which arises while calculating $\mathbb{P}(X>0,Y>0)$ for the above distribution once I transform to polar coordinates. The above probability is then given by $\dfrac{\sqrt{1-\rho^2}}{2\pi}I$. Following the suggestions given in the answers below, I was able to evaluate $I$ correctly and hence calculated the probability by direct integration.
| \begin{align}
t & = \tan\varphi \\[10pt]
dt & = \sec^2\varphi\,d\varphi = (1+\tan^2\varphi)\,d\varphi = (1+t^2)\,d\varphi, \\[10pt]
\text{so } \frac{dt}{1+t^2} & = d\varphi \\[10pt]
\text{and } \sin(2\varphi) & = 2\sin\varphi\cos\varphi = 2\sin(\arctan t)\cos(\arctan t) \\[10pt]
& \phantom{{}= 2\sin\varphi\cos\varphi} = 2\frac{t}{\sqrt{1+t^2}} \cdot\frac{1}{\sqrt{1+t^2}} = \frac{2t}{1+t^2}
\end{align}
Then we have
$$
\int_0^{2\pi} \frac{d\varphi}{1 - \rho\sin(2\varphi)} = \left(\int_{-\infty}^\infty + \int_{-\infty}^\infty
\right) \frac{\left(
\dfrac{dt}{1+t^2} \right)}{1 - \rho\left(\dfrac{2t}{1+t^2}\right)}
$$
What do I mean by that?? Simply that as $\varphi$ goes from $0$ to $2\pi,$ then $(\cos(2\varphi),\sin(2\varphi))$ goes around the circle twice, and each revolution causes $t$ to run through the whole real line once. We have a function of period $\pi$ integrated over the interval from $0$ to $2\pi.$ Thus we have
$$
2 \int_{-\infty}^\infty
\frac{\left( \dfrac{dt}{1+t^2} \right)}{1 - \rho\left(\dfrac{2t}{1+t^2}\right)} = 2\int_{-\infty}^\infty \frac{dt}{1+t^2 -2\rho t}.
$$
Now complete the square:
$$
t^2 - 2\rho t + 1 = (t^2 - 2\rho t + \rho^2) + 1 - \rho^2 = (t-\rho)^2 + 1-\rho^2.
$$
We would like a quadratic polynomial whose constant term is $1,$ so this becomes
\begin{align}
& (t-\rho)^2 + 1-\rho^2 = (1-\rho^2)\left( \frac{(t-\rho)^2}{1-\rho^2} + 1 \right) \\[10pt]
= {} & (1-\rho^2) \left( \left( \frac{t-\rho}{\sqrt{1-\rho^2}} \right)^2 + 1 \right) \\[10pt]
= {} & (1-\rho^2) (u^2 + 1) \\[10pt]
\text{so } \sqrt{1-\rho^2}\,\,du & = dt.
\end{align}
Our integral becomes
\begin{align}
2\sqrt{1-\rho^2}\int_{-\infty}^\infty \frac{du}{(1-\rho^2) (u^2+1)} = \frac{2\pi}{\sqrt{1-\rho^2}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How can I solve this equation for real numbers? How can I solve this equation for real numbers?
$$(x+2)^4+x^4=82.$$
I tried $(x+2)^4+x^4-82=
2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$
It is very difficult to continue.
| A different viewpoint, in case you don't like the (smart) change of variables $y:=x+1$:
First a bit of calculus: Observe that $f(x):=(x+2)^4+x^4-82$ is a twice-differentiable function, with second derivative $f''(x)=12(x+2)^2+12x^2$, which is positive everywhere. Hence $f$ is convex, so it has at most two real roots. Therefore there are at most two different values such that $(x+2)^4+x^4=82$. By Descartes' sign of rules we deduce that there is exactly one positive real solution, while by the same rule applied to $f(-x)$ we find that there are either 3 or 1 negative real solutions; therefore we have two solutions, one positive and one negative.
Now a bit of playing-with-numbers: Observe that $82=81+1=3^4+1^4$. Therefore
$$(x+2)^4+x^4 = 3^4+1^4 = (1+2)^4 + 1^4$$
gives us the positive solution $x=1$. For the negative solution, due to the fact that $x^4=(-x)^4$, we can try with $x=-1$ and $x=-3$: the first one is a no-go, but the second one satisfies $x+2=-1$, so that
$$(x+2)^4+x^4 = (-1)^4 + (-3)^4 = 1^4 + 3^4 = 82.$$
Hence the solutions are $x=1$ and $x=-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Given the problem : find the coefficient of $x^{100}$ in $\frac{1}{\left(1-x^5\right)\left(1-x^{10}\right)}$, why this solution works? We will use partial fractions so :
$\\ \frac{1}{\left(1-x^5\right)\left(1-x^{10}\right)}=\frac{1}{\left(1-x^5\right)(1-x^5)\left(1+x^5\right)}=\:\frac{A}{1-x^5}+\frac{B}{\left(1-x^5\right)^2}+\frac{C}{1+x^5}$
We get
$a\left(1-x^5\right)\left(1+x^5\right)+b\left(1+x^5\right)+c\left(1-x^5\right)^2=1$
Now in the solution my friend did this :
$(*)\\x=1\implies b=\frac{1}{2}\:;\\x=-1\implies c=\frac{1}{4}\:;\\x=0\implies a+b+c=1\implies a=\frac{1}{4}\:;$
Why this works? why can I put different values for x?
The rest of the solution I understand :
We need to set i=j=k=20 for the coefficients in
$\frac{1}{4}\sum \:_{i=0}^{\infty \:}x^{5i}\:+\:\frac{1}{2}\sum \:\:_{j=0}^{\infty \:\:}\left(j+1\right)x^{5j}\:+\frac{1}{4}\sum \:\:_{k=0}^{\infty \:\:}\left(-1\right)^kx^{5k}$
To get
$\frac{1}{4}\cdot 1\:+\:\frac{1}{2}\cdot 21\:+\:\frac{1}{4}\cdot \left(-1\right)^{20}\:=\:11$
I solved it in 2 different ways. But in this way above, why can I do what I did in $(*) ?$
| example -
$(1-x^5)^2$ will be zero when $x=1$
if you multiply by $(1-x^5)^2$ on both sides of the equation (considering your second and third terms), you then have
$\frac{1}{1+x^5} = A(1-x^5) + B + \frac{C(1-x^5)^2}{1 + x^5}$
if you then go back to your case where x=1, the first and third terms on the right vanish (since they become zero), conveniently leaving
$\frac{1}{2} = B$
If the value of B 'works' when $x=1$, then so long as our partial fraction technique is sound, it has to work for all general values too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2444393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the PDF of $Y = (X - \frac{1}{\theta})^2$ Let $X$ be a random variable with probability density function
$f(x) = \theta e^{(-\theta x)}$ for $x \geq 0$, otherwise $0$, when $\theta > 0$.
Let $Y = (X - \frac{1}{\theta})^2$. Find pdf of $Y$.
I'm just not sure how to incorporate $\theta$.
| Let $Y = (X -\frac{1}{\theta})^{2}$ we want the PDF of $Y$,
So Lets find the CDF of Y that is $F_{Y}(y)$ and to find the PDF we will differentiate $F_{Y}(y)$ so that $f(y) = F'_{Y}(y)$.
$F_{Y}(y) = P(Y \leq y) = P((X -\frac{1}{\theta})^{2} \leq y) = P(-\sqrt{y}+\frac{1}{\theta} \leq X \leq \frac{1}{\theta}+\sqrt{y})$
$= F_{X}(\frac{1}{\theta}+\sqrt{y}) - F_{X}(-\sqrt{y}+\frac{1}{\theta})$
So $F_{Y} = F_{X}(\frac{1}{\theta}+\sqrt{y}) - F_{X}(-\sqrt{y}+\frac{1}{\theta})$
to obtain th PDF we differentiate the CDF
so differentiating we obtain
$f(y) = \frac{1}{2\sqrt{y}}f(\frac{1}{\theta}+\sqrt{y}) +\frac{1}{2\sqrt{y}} f(-\sqrt{y}+\frac{1}{\theta})$
$f(y) = \frac{1}{2\sqrt{y}}\theta e^{-\theta(\sqrt{y}+\frac{1}{\theta})} + \frac{1}{2\sqrt{y}}\theta e^{-\theta(-\sqrt{y}+\frac{1}{\theta})} = \frac{1}{2\sqrt{y}}\theta e^{-1}(e^{-\theta\sqrt{y}} + e^{\theta\sqrt{y}})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this a Taylor series? $\ln(x) +1 = \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x}$ Can you provide a proof of this identity using only calculus?
$$\ln x + 1 = \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln x)^n}{x}$$
By the way, here is how I arrived at it:
There is string of length $x$ units. Select a point on the string uniformly at random and cut the string at that point. Repeat the process with the string on the left side of the cut until the string you have is shorter than $1$ unit. The problem is to figure out the expected number of cuts.
Here is how I did it:
Let $E(x)$ denote the expected number of cuts to be made on a string of length $x$. If $x<1$, clearly, $E(x)=0$. If $x>1$, we have:
\begin{align} E(x) &= 1 + \int_0^x E(u) \cdot \frac{du}{x} \\
&= 1 + \frac 1x \int_1^x E(u) \ du
\end{align}
Multiplying by $x$ and differentiating (applying the Fundamental Theorem of Calculus),
\begin{align} xE'(x) + E(x) &= 1 + E(x) \\
\Rightarrow E(x) &= \ln x + C \end{align}
Clearly for $x=1, \ E(x)=1$ thus $E(x) = \ln x + 1$
However, we can also calculate $E(x)$ in a different way:
Let $P(n,\ x)$ denote the probability of exactly $n$ cuts being made on a string of length $x$.
If $x<1$, $P(n,\ x) = 0$; if $n=1$ and $x>1$ $P(n,\ x)=\frac 1x$; if $n>1$ and $x>1$:
\begin{align} P(n,\ x) &= \int_0^x P(n-1,\ u) \cdot \frac{du}{x} \\
&= \frac 1x \int_1^x P(n-1,\ u) \ du \end{align}
I calculated that $P(1,\ x) = \frac 1x,\ P(2,\ x) = \frac{\ln x}{x},\ P(3,\ x) = \frac{(\ln x)^2}{2x},\ P(4,\ x) = \frac{(\ln x)^3}{6x}$
This led me to hypothesize that $P(n,\ x)=\frac{(\ln x)^{n-1}}{x(n-1)!}$, which can be proven by induction:
\begin{align} P(n,\ x) &= \frac 1x \int_1^x P(n-1,\ u) \ du \\
&= \frac 1x \int_1^x \frac{(\ln u)^{n-2}}{u(n-2)!} \ du \\ &= \frac 1{x(n-2)!} \int_1^x (\ln u)^{n-2}\ d(\ln u) \\
&= \frac 1{x(n-2)!} \left[\frac {(\ln u)^{n-1}}{n-1} \right]_1^x \\ &= \frac{(\ln x)^{n-1}}{x(n-1)!} \end{align}
But then $E(x)$ can be written as follows:
\begin{align} E(x) &= \sum_{n=1}^{\infty} n \cdot P(n,\ x) \\
&= \sum_{n=1}^{\infty} n \cdot \frac{(\ln x)^{n-1}}{x(n-1)!} \\ &= \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x}
\\ \\
\therefore \ \ln x + 1 &= \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x} \ \blacksquare \end{align}
So can you prove this result by finding an appropriate Taylor series (this would be especially appreciated) or if not using Taylor series then using just the methods of calculus?
| In more steps than maybe necessary:
$$\sum_{n=0}^\infty{n+1\over n!}{(\ln x)^n\over x}={1\over x}\left(\sum_{n=1}^\infty{(\ln x)^n\over(n-1)!}+\sum_{n=0}^\infty{(\ln x)^n\over n!} \right)={1\over x}\left(\ln x\sum_{n=1}^\infty{(\ln x)^{n-1}\over(n-1)!}+e^{\ln x} \right)={1\over x}\left(\ln x\sum_{k=0}^\infty{(\ln x)^k\over k!}+x \right)={1\over x}\left((\ln x)e^{\ln x}+x \right)={1\over x}((\ln x)x+x)=\ln x+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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} |
How to solve :$\operatorname{arcsin}x-\operatorname{arcsin}\left(x/2\right)=\operatorname{arcsin}\left(\frac{x\sqrt3}2\right)$ $$\operatorname{arcsin}x-\operatorname{arcsin}\left(x/2\right)=\operatorname{arcsin}\left(\frac{x\sqrt3}2\right)$$
Why can't I figure this one out??
Is it possible to cancel out the arcsins?
I know from graphing on my calculator that the answers are $-1, 0$, and $1$, but want help getting there by hand.
Thank you!
| this makes sense ony if $|x|<1$
\begin{split}
&&\arcsin x - \arcsin \dfrac x 2 = \arcsin \dfrac{x\sqrt3}2 \\
&&\sin\left(\arcsin x - \arcsin \dfrac x 2\right) =\sin\arcsin \dfrac{x\sqrt3}2 \\
&&\sin(\arcsin x )\cos(\arcsin \frac x 2)- \cos(\arcsin x )\sin (\arcsin \dfrac x 2) = \frac{x\sqrt3}{2} \\
&&x\sqrt{1-\left(\dfrac x2\right)^2} - \dfrac x2 \sqrt{1-x^2} =\dfrac{x\sqrt3}2 \\
&&x\sqrt{4-x^2} - x \sqrt{1-x^2} = x\sqrt3 \\
&& x^2\left(5-2x^2 - 2 \sqrt{(4-x^2)(1-x^2)}\right)= 3x^2 \\
\end{split}
Hence $$x= 0 \qquad or \qquad 5-2x^2 - 2 \sqrt{(4-x^2)(1-x^2) } = 3$$
However,
\begin{split}
&&5-2x^2 - 2 \sqrt{(4-x^2)(1-x^2) } = 3\\
&\implies & \sqrt{(4-x^2)(1-x^2) } = 1-x^2\\
&\implies & \sqrt{(4-x^2) } =\sqrt{(1-x^2) } \qquad or \qquad 1-x^2 = 0
\end{split}
But $ \sqrt{(4-x^2) } =\sqrt{(1-x^2) }$ is impossible
Conclusion $x\in \{-1,0,1\}$ is the set of solutions
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence and limit of $\left((n+3)^{1/(n+3)}-(n+4)^{1/(n+4)}\right)^{1/(n+5)}$ as $n \to \infty$? I was calculating in WA some values of some sequences and constructed this sequence:
$$a_n=\left((n+3)^{1/(n+3)}-(n+4)^{1/(n+4)}\right)^{1/(n+5)}$$
It is immediate that expression in parentheses tends to $0$ since both terms tend to $1$, but the exponent also tends to $0$ so it may be that in the limit this behaves similarly to $x^x$ when $x$ tends to $0$ from the right side, but I am not sure.
How to prove existence of this limit and what is its value?
| Define $f(x) := (1/x)^x$. Then we have
$$
(n + 3)^\frac{1}{n + 3} - (n + 4)^\frac{1}{n + 4} = f\left(\frac{1}{n + 3}\right) - f\left(\frac{1}{n + 4}\right).
$$
By the mean value theorem
$$
f\left(\frac{1}{n + 3}\right) - f\left(\frac{1}{n + 4}\right) = f'(c_n) \left(\frac{1}{n + 3} - \frac{1}{n + 4}\right)
$$
for some $c_n \in (\frac{1}{n + 4}, \frac{1}{n + 3})$. So we have
$$
a_n = \left(\frac{f'(c_n)}{(n + 3)(n + 4)}\right)^{\frac{1}{n + 5}}.
$$
Now we can compute the derivative of $f$ to be
$$
f'(x) = \left(\frac{1}{x}\right)^x \left(\log\left(\frac{1}{x}\right) - 1\right).
$$
If we define
$$
b_n := \left(\frac{1}{(n + 3)(n + 4)}\right)^{\frac{1}{n + 5}},
$$
then we see that $\lim_{n \rightarrow \infty} b_n = 1$, so it remains to compute
$$
\lim_{n \rightarrow \infty} f'(c_n)^{\frac{1}{n +5}}
$$
and this should be straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Matrix whose characteristic polynomial does not splits over field. I need an example of matrix which is non-zero and not nilpotent whose characteristic polynomial does not splits over field $\mathbb{R}$
| $$
\left(
\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}
\right)
$$
Thursday morning: if we try to diagonalize over the reals, we have real numbers $a,b,c,d$ with $ad-bc \neq 0,$ with
$$
\frac{1}{ad-bc} \; \;
\left(
\begin{array}{rr}
d & -b \\
-c & a
\end{array}
\right)
\left(
\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}
\right)
\left(
\begin{array}{rr}
a & b \\
c & d
\end{array}
\right) =
\frac{1}{ad-bc} \; \;
\left(
\begin{array}{rr}
-(ab+cd) & -(b^2 + d^2) \\
a^2 + c^2 & ab+cd
\end{array}
\right)
$$
The resulting matrix cannot be diagonal unless both $b^2 + d^2 = 0$ and $a^2 + c^2 = 0,$ meaning all four of $a,b,c,d = 0,$ with the contradictory outcome that $ad-bc = 0.$
With the same letters, if we now allow complex values, we can simply assign $a=d=1$ and $b=c=i$ to diagonalize over the complexes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Equation of a plane by given 3 points I've read of solutions, with let's say points $P(1,2,3)$, $Q(3,1,2)$, $R(2,3,1)$, where one subtracts point $P-Q$ and $P-R$ to get two vectors:
$\begin{pmatrix}1\\2\\3\end{pmatrix}-\begin{pmatrix}3\\1\\2\end{pmatrix}=\begin{pmatrix}-2\\1\\1\end{pmatrix}$
$\begin{pmatrix}1\\2\\3\end{pmatrix}-\begin{pmatrix}2\\3\\1\end{pmatrix}=\begin{pmatrix}-1\\-1\\2\end{pmatrix}$
to now get the cross-product:
$\begin{pmatrix}-2\\1\\1\end{pmatrix}\times\begin{pmatrix}-1\\-1\\2\end{pmatrix}=\begin{pmatrix}2\\-1\\2\end{pmatrix}$
hence the equation is $2x-y+2z=d$
by inserting, for example, point P we get
$2\times1-2+2\times3=d=6$
and thus
$2x-y+2z-6=0$
which can be transformed to
$z=-x+\frac12y+3$
so when I put in any point it should work right? When I put in R though, I get:
$1=\frac52$
So, what did I do wrong?
| That's because “cross-product” doesn't mean to multiply each entry be the corresponding entry of the other vector. In this case$$\begin{pmatrix}-2\\1\\1\end{pmatrix}\times\begin{pmatrix}-1\\-1\\2\end{pmatrix}=\begin{pmatrix}3\\3\\3\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For what primenumbers, $p$, is $\sqrt{5p +49}$ an integer? For what primenumbers, $p$, is $\sqrt{5p +49}$ an integer?
I managed to figure out:
$5p +49 = n^2$
$5p = (n+7)(n-7)$
But can't think of anything more than that, anyone able to solve this without bruteforcing?
For primenumbers under $10^7$, the solutions are $3$ and $19$.
| Then $n-7=5$ and $n+7=p$, that is $n=12$ and $p=19$.
Or $n-7=-5$ and $n+7=-p$, that is $n=2$ and $p=-9$ which is negative.
Or $n+7=5$ and $n-7=p$, that is $n=-2$ and $p=-9$ which is negative.
Or $n+7=-5$ and $n-7=-p$, that is $n=-12$ and $p=-19$, which is negative.
Or $n-7=1$ and $n+7=5p$, that is $n=8$ and $p=3$.
Or $n-7=-1$ and $n+7 = -5p$, that is $n=6$ and $p=-13/5$ which is not an integer.
Or $n+7=1$ and $n-7=5p$, that is $n=-6$ and $p=-13/5$ which is not an integer.
Or $n+7=-1$ and $n-7=5p$, that is $n=-8$ and $p=-3$ which is negative.
So in conclusion, the only possible numbers are $p= 3, 19$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that $\left(\frac{1+\sqrt{5}}{2}\right)^5 > 10$ I have to show that $\left(\frac{1+\sqrt{5}}{2}\right)^5 > 10$.
I've already proven that if $f_n$ is the nth Fibonnaci number, then,
$$f_{n+1} > \left(\frac{1+\sqrt{5}}{2}\right)^{n-1} $$
But I really don't get how to go from this statement to what I need to prove.
Any advices on how to take from here? Any help is welcome.
Thanks.
| Binomially expand
\begin{eqnarray*}
\left( \frac{1+ \sqrt{5}}{2} \right)^5 &=& \frac{1+5 \sqrt{5} +10 \times 5 +10 \times 5 \sqrt{5}+5 \times 25 +25 \sqrt{5}}{32} \\
&=&\frac{11+5 \sqrt{5}}{2}.
\end{eqnarray*}
Now $125>121 $ square root this and we have the stronger result that
\begin{eqnarray*}
\left( \frac{1+ \sqrt{5}}{2} \right)^5 > 11. \\
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solving this Bernoulli ODE I have the following nonlinear ODE which I am trying to solve.
$$
\begin{align}
\frac{dr}{dt} &= ar(1-r) + b (1-r)^2\\
&= (b-2a) r^2 + (a+2b)r + b
\end{align}
$$
We then let $r = y + z$.
$$
\begin{align}
y '
&= (b-2a) (y-z)^2 + (a+2b)(y-z) + b\\
&= (b-2a) y^2 + y(2z(b-2a)+(a+2b))+((b-2a)z^2+(a+2b)z + b)
\end{align}
$$
We then determine $z$ such that $(b-2a)z^2+(a+2b)z + b=0$, generating the Bernoulli equation:
$$
y'-sy=(b-2a)y^2\qquad \text{where}\qquad s=\pm\sqrt{a^2+12ab}.
$$
Let $h = y^{-1}$ then $h' = -y^{-2}y'$ or $-h' = y^{-2}y'$ before dividing the Bernoulli equation by $y^2$.
$$
\begin{align}
y^{-2} y' + sy^{-1} &= (b-2a)\\
-h' + s h &= (b-2a)\\
h' - s h &= (2a-b)
\end{align}
$$
The integrating factor $\eta(t) = e^{\int dt (-s)} = e^{-st}$
$$
e^{-s t} h' - e^{-s t} sh = e^{-s t}(2a -b)\\
\frac{d}{dt}(e^{-st}h) = e^{-s t}(2a -b)
$$
We then integrate the R.H.S by parts with $u = 2a-b$, $dv = e^{-st}dt$, $v = -\frac 1 s e^{-st}$, $du = 0$.
$$
\int e^{-s t}(2a -b) dt = -\frac 1 s (2a-b)e^{-st}+c
$$
To give,
$$
e^{-st}h = -\frac 1 s (2a-b)e^{-st}+c
$$
Divide this by $e^{-st}$.
$$
h = - \frac {(2a-b)}{s} + ce^{st}
$$
We then simply insert the substitution back in $h=y^{-1}$.
$$
y = \frac{s}{-(2a-b) - cse^{st}}
$$
We can then reconstruct $r$ using $y$ and $z$.
$$
r = \frac{s}{-(2a-b) - cse^{st}} + \frac{-a-2b \pm s}{2(b-2a)}
$$
Any help is appreciated and the final solution is:
$$
r(t) = \frac{e^{at} -1}{ e^{at} - 1 + (a/b)}
$$
| if $$r=y+z$$ it must be
$$y'=(b-2a)(y+z)^2+(a+2b)(y+z)+b$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Computing $\int \frac{1}{\sqrt{25y^2-30y-7}} \,dy$ $$\int \frac{1}{\sqrt{25y^2-30y-7}} \, dy$$
I start by completing the square: so that $(5y-3)^2-16=25y^2-30y-7$.
Then I let $u=5y-3$, and then let $ u=4 \sec \theta$.
This leads to the equation becoming $\frac{1}{4\tan\theta}$ which is the same as $\frac{1}{4}\cot \theta$.
Taking the integral of this is $\frac{1}{4}\ln\left|\sin\theta\right|$.
Substituting back I get $\frac{1}{4}\ln\left| \sin \left( \operatorname{arcsec} \frac{5y-3} 4 \right)\right|$.
I drew a triangle and came up with this: $\frac{1}{4}\ln\left|\frac{\sqrt{25y^2-30y-7}}{5y-3}\right|$. Please, where did I go wrong?
| I realized my mistake! I forgot to put in the du in my substitutions. I ended up with $\frac{1}{5}\int sec \theta$ which ends up being $\frac{1}{5}ln|sec\theta+tan \theta|$ which is $\frac{1}{5}ln|\frac{5y-3+\sqrt(25y^2-30y-7)}{4}$|+C. Hooray!
| {
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"timestamp": "2023-03-29T00:00:00",
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Area enclosed by quarter circle arcs in a square
The problem I have here asks to find the area $A$ enclosed by the two quarter circle arcs in terms of $a$. The first circle has radius $a/2$, and the second has radius $a$. This is what I did.
Firstly I determined that $$A=\frac{a^2}{8}(2\theta_1-\sin(2\theta_1))+\frac{a^2}{2}(2\theta_2-\sin(2\theta_2))$$ $$=\frac{a^2}{4}(\theta_1+4\theta_2-\sin\theta_1\cos\theta_1-4\sin\theta_2\cos\theta_2)$$
Then $$h_1+h_2=a\sqrt2=\sqrt{\frac{a^2}{4}-x^2}+\sqrt{a^2-x^2}$$
$$(a\sqrt2-\sqrt{a^2-x^2})^2=\frac{a^2}{4}-x^2$$
$$12a^2-8a\sqrt{2a^2-2x^2}=a^2$$
$$11a=8\sqrt{2a^2-2x^2}$$
$$121a^2=128a^2-128x^2$$
$$x=a\sqrt{\frac{7}{128}}=\frac{a\sqrt{14}}{16}$$
So
$$h_1=\sqrt{\frac{a^2}{4}-\frac{14a^2}{256}}=\frac{5a\sqrt2}{16}$$
$$h_2=\sqrt{a^2-\frac{14a^2}{256}}=\frac{11a\sqrt2}{16}$$
Therefore
$$\sin\theta_1\cos\theta_1=\frac{\sqrt{14}}{8}\times\frac{5\sqrt2}{8}=\frac{5\sqrt7}{32}$$
$$4\sin\theta_2\cos\theta_2=4\times\frac{\sqrt{14}}{16}\times\frac{11\sqrt2}{16}=\frac{11\sqrt7}{32}$$
So
$$A=\frac{a^2}{4}(\theta_1+4\theta_2-\frac{5\sqrt7}{32}-\frac{11\sqrt7}{32})$$
$$=\frac{a^2}{4}(\theta_1+4\theta_2-\frac{\sqrt7}{2})$$
Now I can find $\theta_1$ and $\theta_2$ with my calculator, but there must be a way to express them in terms of surds but I don't know how. Also, is what I did actually correct and is there another shorter/nicer way to solve this?
| $\sin\theta_1=2\sin\theta_2$
$\frac a 2\cos\theta_1+a\cos\theta_2=a\sqrt 2 \implies\cos\theta_1+2\cos\theta_2=2\sqrt 2\implies\cos\theta_1=2\sqrt 2-2\cos\theta_2$
Then you can square both equations and add to each other
$$\sin^2\theta_1+\cos^2\theta_1=4\sin^2\theta_2+(2\sqrt2-2\cos\theta_2)^2$$
$$8\sqrt2 \cos\theta_2=11$$
$$\cos\theta_2=\frac{11}{8\sqrt2}$$
| {
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"url": "https://math.stackexchange.com/questions/2461843",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove that there exist infinitely many primitive Pythagorean triples $x, y, z$ whose even member $x$ is a perfect square. Prove that there exist infinitely many primitive Pythagorean triples $x, y, z$ whose even member $x$ is a perfect square. [Hint: consider the triple $4n^2, n^4-4, n^4+4$, where $n$ is an arbitraty odd integer.]
What I got:
Using the hint. $4n^2, n^4-4, n^4+4$ is a Phytagorean triple if $x^2+y^2=z^2$. Replacing and solving the equation it is clear that, $(4n^2)^2+(n^4-4)^2=(n^4+4)^2$ where $n$ is odd is indeed a Pythagorean triple with $x=4n^2=(2n^2)^2$ a perfect square.
Now I have to prove that $gcd(4n^2, n^4-4, n^4+4)=1$.
But I'm stuck here. I tried $gcd(n^4-4, n^4+4)=1$ without success. Any ideas?
| All Pythagorean triples can be generated by the formula:
$$A=(m^2-n^2)k\qquad B=2mnk\qquad C=(m^2+n^2)k$$
For a triple to be primitive, it is necessary but not sufficient that $k=1$. This leaves $B=2mn$. The smallest $m$ is $2$; $m,n$ must be of opposite parity; $m>n$, meaning 2mn must be a multiple of $4$
Every side-B is a multiple of $4$, e.g. $\quad (3,4,5)\quad (15,8,17)\quad (5,12,13)\quad (35,16,37)\quad ...$
Among these are all powers of $2$ that are perfect squares so, there being infinite combinations of $(m,n)$, there are an infinite number of triples where $\sqrt{B}=\lfloor\sqrt{B}\rfloor.$
| {
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Napier analogy and algebra in triangle.
If in a $\triangle{ABC}$, we define $x=\tan\frac{B-C}{2}\tan\frac{A}{2}$, $y=\tan\frac{C-A}{2}\tan\frac{B}{2}$ and $z=\tan\frac{A-B}{2}\tan\frac{C}{2}$, then show that $x+y+z=-xyz$.
My attempts:
By Napier analogy,
$x=\frac{b-c}{b+c},\ y=\frac{c-a}{c+a},\ z=\frac{a-b}{a+b}$
Then one can simply put these values in LHS, but that is cumbersome, I need to use some beautiful algebra, please help.
I just need to solve that algebra stuff, if such problem exists somewhere on this site then please comment with that link, I'll delete this, then.
| Let $u = \tan \frac{B}{2}\tan \frac{C}{2}$, $v = \tan \frac{C}{2}\tan \frac{A}{2}$, $t = \tan \frac{A}{2}\tan \frac{B}{2}$. One has $u+v+t = 1$.
$x = \tan \frac{B-C}{2}\tan \frac{A}{2} = \frac{\tan\frac{B}{2}-\tan\frac{C}{2}}{1+\tan\frac{B}{2}\tan\frac{C}{2}}\tan \frac{A}{2} = \frac{t-v}{1+u}$.
$y = \frac{u-t}{1+v}$, $z= \frac{v-u}{1+t}$.
Thus, $x+y+z = \frac{t-v}{1+u} + \frac{u-t}{1+v} + \frac{v-u}{1+t} = \frac{\sum (t-v)(1+v+t+vt)}{(1+u)(1+v)(1+t)} = \frac{\sum vt(t-v)}{(1+u)(1+v)(1+t)} = -\frac{(v-u)(u-t)(t-v)}{(1+u)(1+v)(1+t)} = -xyz.$
| {
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Evaluating limit using taylor expansions How can I evaluate this limit? Trying with Taylor doesn't seem to give me the right result, why ?
$$\lim_{x\to 0} \frac{\sqrt{1+4x} -1 -\sin(2x)}{\log(1+x^2)}$$
with Taylor I can approximate $\sin(2x)$ to $ 2x $
and $\log(1+x^2)$ to $ x^2 $. If I plug in these in the limits it does not give me the right limit, what am I doing wrong?
| $$\log(x^2+1)\sim x^2$$
$$\sin(2x)\sim 2x -\frac{8}{6}x^3$$
$$ \sqrt{1+4x} -1\sim 2x-2x^2 $$
since $$(\sqrt{1+4x} -1)'|_{x=0} = 2~~~and ~~~(\sqrt{1+4x} -1)"|_{x=0} = -4$$
$$\lim_{x\to0} \frac{\sqrt{1+4x} -1 -\sin{2x}}{\log{1+x^2}}\sim \lim_{x\to0}\frac{ -2x^2+\frac{8}{6}x^3}{ x^2} = -2$$
| {
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Prove by Induction that $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{2^n}\leq n$ The question asks to prove by induction that for every integer, $n\geq3$.
We have the sequence: $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2^n}\leq n$$
So from here, I started with a base case scenario where $n=1$. Plugging $n$ into the sequence, we get that $$\frac{1}{2}\leq 1$$ which is indeed a true assumption. From here, the induction hypothesis is that the theorem is true for $n=k$. Rewriting this and plugging in $k$, we get: $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2^k}\leq k$$
Since the whole sequence is less than or equal to $k$, we can take $k$ to represent the whole sequence. So, now we also assume that $n=k+1$ is also true, by doing so, we arrive at the sequence:$$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2^k}+\frac{1}{2^{k+1}}\leq k+1.$$
Rewriting this with $k$, we get that the sequence would then look like this: $$k+\frac{1}{2^{k+1}}\leq k+1.$$
After this step, I seem to not be able to continue further. Are there any tips on how I can continue to prove this?
| Ok, so you want to show that
$$
\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} + \frac{1}{5} + \dots+\frac{1}{2^n}\leq n
$$
for all $n\geq 3$. I assume that the $3$ is there because the sum actually has $2^n$ terms. If
$$
\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2^k}\leq k
$$
for some $k$, you then want to show that
$$
\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2^{k+1}-1} + \frac{1}{2^{k+1}}\leq k +1
$$
But
$$\begin{align}
\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2^{k+1}-1} + \frac{1}{2^{k+1}}&\leq k + \overbrace{\frac{1}{2^{k} + 1} + \dots +\frac{1}{2^{k+1}}}^{2^k \text{ terms}} \\
&\leq k + \frac{2^{k}}{2^{k} +1} \\
&\leq k + 1
\end{align}
$$
| {
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Arithmetic of complex numbers
If $z = \cos x + i\sin x$ , show that
$$\frac{2}{1+z} =1-i\tan\left(\frac{x}{2}\right),$$
I get up to:
$$\frac{1+\cos x -i\sin x}{1+\cos x}.$$
| You can apply the Weierstrass substitutions (a bit of a misnomer, but that's how they're commonly known).
Note that if $\displaystyle t = \tan \frac x2$, then:
$$\sin x = \frac{2t}{1+t^2}$$
and
$$\cos x = \frac{1-t^2}{1+t^2}$$
So:
$$z = \frac{1}{1+t^2}(1-t^2 + i2t)$$
$$1 + z = \frac{1}{1+t^2}(1-t^2 + i2t) + \frac{1+t^2}{1+t^2} = \frac{2}{1+t^2}(1+it)$$
$$\frac{2}{1+z} = (1+t^2)\frac{1}{1+it} = (1+t^2)\frac{1-it}{1+t^2} = 1-it$$
as required.
| {
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Solve $a^2+b^2+c^2+d^2+2ab+2bc+2cd-2ca-2ad-2db=N^2$ I am looking for a characterization of the solutions of
$$a^2+b^2+c^2+d^2+2ab+2bc+2cd-2ca-2ad-2db=N^2$$ in positive integers.
| I suppose that you mean
$$
a^2 + b^2+c^2+d^2 + 2ab - 2ac - 2ad - 2bc - 2bd + 2cd =N^2.
$$
Since the left hand side is $(a+b-c-d)^2$, we are done.
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What is the limit of $2c^{n-1}-c^{n-2}-...-c-1$ when $n\to\infty$ I have a sequence in which the value of the $n^{th}$, $x_n$ is given by
$$x_n=2c^{n-1}-c^{n-2}-...-c-1$$
Where $c>0,$
then in the limit when $n\to\infty$ what is $x_n$?
| First simplify the expression. When $c \ne 1$, we have
$$\begin{align*}
2c^{n-1} - c^{n-2} - \cdots - c - 1
&= 2c^{n-1} - \sum_{i=0}^{n-2} c^i \\
&= 2c^{n-1} - \dfrac{c^{n-1} - 1}{c-1} \\
&= \frac{2c^n-3c^{n-1}+1}{c-1}
\end{align*}$$
Now consider what happens if $0<c<1$ and if $c>1$ separately.
When $c = 1$, the expression is equal to $2-(n-1) = 3-n$, so tends to $-\infty$.
| {
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prove that $\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^1}=0$ I need prove that
$$\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^2}=0$$
Can I use it?
if $\sqrt{x^2+y^2}< \delta$ then $|\frac{x^2y-xy^2}{x^2+y^2}|<\epsilon$
$$\left |\frac{x^2y-xy^2}{x^2+y^2} \right |=\frac{x^2 \left |y \right | -\left |x \right |y^2}{x^2+y^2}<\frac{x^2 \sqrt{x^2+y^2} -y^2\sqrt{x^2+y^2}}{x^2+y^2}=\frac{\sqrt{x^2+y^2}(x^2-y^2) }{x^2+y^2}<\frac{\sqrt{x^2+y^2}(x^2+y^2-y^2) }{x^2+y^2}=\frac{\sqrt{x^2+y^2}(x^2) }{x^2+y^2}<\frac{\sqrt{x^2+y^2}(x^2+y^2) }{x^2+y^2}=\sqrt{x^2+y^2}<\delta=\epsilon$$
I use that $x^2<x^2+y^2$ twice, can I do it? or some identity to use when you have $x^2-y^2$
| There's a mistake in the first equality: it's false that
$$
|x^2y-xy^2|=x^2|y|-|x|y^2
$$
However, by the triangle inequality,
$$
|x^2y-xy^2|\le x^2|y|+|x|y^2<x^2\sqrt{x^2+y^2}+y^2\sqrt{x^2+y^2}
$$
so
$$
\left|\frac{x^2y-xy^2}{x^2+y^2}\right|
<\frac{(x^2+y^2)\sqrt{x^2+y^2}}{x^2+y^2}=\sqrt{x^2+y^2}
$$
Now the squeeze theorem allows to finish.
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Find the constant and limit Here is the limit
$$\lim_{x\rightarrow-2} \frac{4x^2 + ax + a + 12}{x^2 + x - 2}.$$
a) Find the constant $a$
b) find the limit
I dont think it is solvable because it didn't tell me as $x \rightarrow -2$, $y \rightarrow $?
| I think it's saying that if the limit exists, find $a$ and find the limit.
If $x \to -2$ then the denominator $\to 0$, so in order for the limit to exist, then the numerator must also $\to 0$.
$(4x^2 +ax +a+12) \to (16 -2a +a+12) = (28-a)$ therefore $a = 28$
So that expression becomes:
$$\lim_{x \to -2} \frac{4x^2 + 28x +40}{x^2+x-2} = \lim_{x \to -2} \frac{4(x+2)(x+5)}{(x+2)(x-1)} = -4$$
| {
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minimize $x^4 - 6x^2 y^2 + y^4$ given $x^2 + y^2 \leq 1$ I have a constrained optimization problem. Can we maximize / minimize this function on the unit sphere?
$$ f(x,y,z) = x^4 - 6 x^2 y^2 + y^4 \quad\text{given that}\quad x^2 + y^2 + z^2 = 1$$
One idea could be to use the Cauchy-Schwartz inequality. Since I forget the proof:
$$ (x^2 + y^2)^2 \geq 0 \text{ so that }x^4 + y^4 \geq 2 x^2 y^2 \text{ and }f(x,y,z) \geq - 4 x^2 y^2 \geq - 4 $$
I could try other rearrangements as well. This one gives me an upper bound of $3$.
$$ x^4 - 6x^2 y^2 + y^4 \leq x^4 + 6x^2 y^2 + y^4
= (x^2 + y^2)^2 + 4x^2 y^2 \leq 3\, \big( x^2 + y^2 \big)^2 \leq
3\, \big( x^2 + y^2 + z^2 \big)^2 = 3$$
If I use some real analysis we know that the sphere as a subset of Euclidean space is compact, so that:
$$ -\infty < -4 \leq \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) \leq \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) < 3 < +\infty$$
I'm trying to avoid Lagrange multipliers unless the're really natural here. Observer also that:
$$ \left[ x^2 + y^2 + z^2 = 1 \right] \to \left[ x^2 + y^2 \leq 1\right] $$
as the original problem was defined on the unit sphere but the $z$ is extraneous.
They might not be extraneous we could set spherical coordinates:
$$ (x,y,z ) = \big(\cos \theta \, \cos \varphi, \;\cos \theta \sin \varphi, \;\cos \varphi\big)$$
and we could put into our inequality:
\begin{eqnarray*} x^4 - 6x^2 y^2 + y^4 &=& \cos^4 \theta \cos^4 \varphi - 6 \cos^4 \theta \sin^2 \varphi \cos^2 \varphi+ \cos^4 \theta \sin^4 \varphi \\ \\
&=& \cos^4 \theta \,\big( \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi \big) \\ \\
&\leq & \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
\end{eqnarray*}
This looks promising as I have reduced a three-dimensional problem to a problem with only an angle $\varphi$, but I may have lost something with the final "$\leq$" sign.
Just a tiny bit more:
$$
\cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
= (\cos^2 \varphi - \sin^2 \varphi)^2 - 4 \cos^2 \varphi \sin^2 \varphi
= \cos^2 2\varphi - \sin^2 2\varphi
$$
and if we use the double-angle identity.
$$ 1 \geq \cos^2 2\varphi - \sin^2 2\varphi
= \cos^2 2\varphi - (1 -\cos^2 2\varphi)
= 2\, \cos^2 2\varphi - 1 \geq - 1$$
This is very similar to what I obtained before.
| The first approach is good, except for its last step. The correct argument is
$$f(x,y,z)\ge -4x^2y^2\ge -(x^2+y^2)^2\ge -(x^2+y^2+z^2)^2 = -1.$$
Minimum attains at $x=y=\pm\sqrt2,z=0$.
| {
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If $a+b+c=0$ prove that $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $$
I've tried squaring, cubing, etc. the $a+b+c=0$, but I've just dug myself in.
Is there a more elegant way to prove this, or a way to find the right "trick"?
| Substituting $$c=-a-b$$ in $$\frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{6}$$ we obtain $$-(a^2+ab+b^2)ab(a+b)$$ and so is the right-hand side
| {
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Show that $\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}} $ I wonder why this is true
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}} $$
Where the sum omits the case $n = m = 0$ ofcourse.
| (Too long for a comment.)
We have,
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 10n^2} = - \frac{2\pi \ln( \sqrt2\; U_{5})}{\sqrt {10}} $$
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{2\pi \ln( \sqrt2\; U_{29})}{\sqrt {58}} $$
with fundamental units $U_5 = \frac{1+\sqrt5}2$ and $U_{29} = \frac{5+\sqrt{29}}2$.
P.S. Presumably there might be a family for $d = 5,\,13,\,37$.
Added:
Courtesy of a comment by Paramanand Singh, we have the closed-form in terms of the Dedekind eta function $\eta(\tau)$ as,
$$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + pn^2} = - \frac{2\pi \ln(\sqrt2\,g_p^2)}{\sqrt {p}} =- \frac{\pi \ln(2\,g_p^4)}{\sqrt {p}} $$
where,
$$g_p = 2^{-1/4}\frac{\eta(\tfrac12\sqrt{-p})}{\eta(\sqrt{-p})}$$
In particular,
$$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 6n^2} = - \frac{2\pi \ln\big(\sqrt2\,(1+\sqrt2)^{1/3}\big)}{\sqrt{6}}$$
$$\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 22n^2} = - \frac{2\pi \ln\big(\sqrt2\,(1+\sqrt2)\big)}{\sqrt{22}}$$
and more complicated ones for $d=5,\,13,\,37$.
| {
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Euclidean Algorithm for polynomials I know how to use the extended euclidean algorithm for finding the GCD of integers but not polynomials. I can't really find any good explanations of it online. The question here is to find the GCD of
m(x) = $\ x^3+6x+7 $ and n(x) = $\ x^2+3x+2 $.
I try to use it the same way as for integers but don't really get anywhere and just get huge lines without ever reducing it and getting closer to finding the GCD.
| For $m(x) = ^3 + 6 + 7$ and $n(x) = ^2 + 3 + 2$,
Set
$m(x) = ^3 + 6 + 7 = p(x)n(x) + r_1(x) = p(x)(^2 + 3 + 2) + r_1(x)$
Where $r_1(x)$ is your remainder term. You want to find some p(x) that multiplies with n(x) to give you the first term of m(x). You also want $deg r_1(x) < deg n(x)$.
In this case, p(x) should be linear to get the $x^3$ term. So set $p(x) = ax+b$.
So now you have,
$m(x)= p(x)n(x) + r_1(x) = (ax+b)(^2 + 3 + 2) + r_1(x)$
Mulitply out:
$m(x) = (ax+b)(^2 + 3 + 2) + r_1(x) = ax^3 + (3a+b)x^2 + (2a+3b)x + 2b + r_1(x)$
Since the coefficient of $x^3$ in m(x) is 1, set $a = 1$.
Similarly, set $3a+b=0$ since there's no $x^2$ term in m(x). Solve for b, $b = -3$.
Plug $a = 1$ and $b = -3$ into the rest of the polynomial:
$m(x) = x^3 -7x -6 + r_1(x)$
Then rearrange to find $r_1(x)$:
$r_1(x) = m(x) - p(x)n(x) = 13x + 13$
Now repeat the process!
This time set:
$n(x) = p_2(x)r_1(x) + r_2(x)$
Follow the same process to find $p_2(x)$ and $r_2(x)$, keep repeating this, eg once you know $r_2(x)$ the next equation would look like $r_1(x) = p_3(x)r_2(x) + r_3(x)$ etc, until you get a remainder term that equals zero. Then the previous remainder term is your gcd.
| {
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Prove $\frac {2 - \csc^2 A}{\csc^2 A\space + \space2\cot A} \equiv \frac {\sin A \space -\space \cos A} {\sin A \space+\space \cos A}$ So I have this rather simple trigonometric identity that, for the life of me, I cannot solve. I have worked on it for about 2 hours and still can't get it.
Show that
$$\frac {2 - \csc^2 A}{\csc^2 A\space + \space2\cot A} \equiv \frac {\sin A \space -\space \cos A} {\sin A \space+\space \cos A}$$
Here's what I've done so far:
\begin{align}
{2-\csc^2 A \over \csc^2 A+2\cot A} & = {2 - ({1 \over \sin A })^2 \over ({1 \over \sin A })^2 + 2({1 \over \tan A})} \\
&= {{2\sin^2 A \over \sin^2 A} - {1\over \sin^2 A} \over {1\over \sin^2 A} + 2({\cos A \over \sin A})} \\
& = {{2\sin^2 A -1 \over \sin^2 A} \over {1\over sin^2 A}+2({\sin A \cos A \over \sin^2 A})}\\
& = {2 \sin ^2 A -1 \over \sin^2 A} \times {\sin^2 A \over 2\sin A \cos A +1} \\
& = {2\sin^2 A -1 \over 2\sin A \cos A + 1}
\end{align}
Lots of fractions are involved so I fear I may have made a mistake somewhere.
If anyone has any tips on proving these trigonometric identities, could they please add them in their answer? I've been told just to keep trying; though I believe there must be some 'troubleshooting' method to finish the problem.
| You are nearly there! Let us go on with the identity $\sin^2 + \cos^2 = 1$ and some clever factorizations:
$$\frac{2 \sin^2 A - 1}{2 \sin A \cos A + 1} = \frac{\sin^2 A - \cos^2 A}{2\sin A \cos A + \sin^2 A + \cos^2 A} = \frac{(\sin A + \cos A)(\sin A - \cos A)}{(\sin A + \cos A)^2} = \frac{\sin A - \cos A}{\sin A + \cos A}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2474872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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a prime divisor of $1+q+q^2+q^3+q^4$ Suppose $q$ is a prime power. Can we say anything about a prime divisor of $1+q+q^2+q^3+q^4$? Is any prime divisor congruent to 1 modulo 10?
| Let $p$ be a prime divisor of $1+q+q^2+q^3+q^4$, then
$$1+q+q^2+q^3+q^4 \equiv 0 \pmod{p} \implies q^5-1 \equiv 0 \pmod{p}$$
So $q$ has order $1$ or $5$ in the group $(\mathbb{Z}/p\mathbb{Z})^{\times}$. If it has order $1$, then $q\equiv 1 \pmod{p}$, hence $$5 \equiv 1+q+q^2+q^3+q^4 \equiv 0 \pmod{p} \implies p=5$$
If $q$ has order $5$, from $|(\mathbb{Z}/p\mathbb{Z})^{\times}| = p-1$, we obtain $5|(p-1)$.
Hence every prime divisor of $1+q+q^2+q^3+q^4$ is either $5$ or congruent to $1$ modulo $5$. Since the number $1+q+q^2+q^3+q^4$ is always odd, its prime divisor is either $5$ or congruent to $1$ modulo $10$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Understanding $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ by means of determinants. The identity comes from expanding the determinant
$$\begin{vmatrix}
a & b & c \\
c & a & b \\
b & c & a \\
\end{vmatrix}$$
in two ways.
The LHS comes from expanding the determinant by Sarrus' rule. The RHS comes from adding up all columns to the first, factoring (a+b+c), and then expanding the remaining determinant.
The derivation of the RHS is what I don't understand. I'm guessing that there is some elementary determinant operation at work that I'm unfamiliar with.
| Add $\omega$ times the second row and $\omega^2$ times the third row to the first row to get
$\begin{align}
\begin{vmatrix}a & b & c \\ c & a & b \\ b & c & a\end{vmatrix} &= \begin{vmatrix}a+c\omega+b\omega^2 & b+a\omega+c\omega^2 & c+b\omega + a\omega^2 \\ c & a & b \\ b & c & a\end{vmatrix} \\
&=\begin{vmatrix}a+c\omega+b\omega^2 & \omega(a+b\omega^2+c\omega) & \omega^2(c\omega+b\omega^2 + a) \\ c & a & b \\ b & c & a\end{vmatrix}
\end{align}$
and hence $a+c\omega+b\omega^2$ is a factor. Similar arguments can be used to show the other two factors in the proof given by Lord Shark the Unkniwn.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the equations? $\begin{cases}x_1+x_2+x_3+x_4 = 6,\\
x_1^2+x_2^2+x_3^2+x_4^2 = 10,\\
x_1^3+x_2^3+x_3^3+x_4^3 = 18,\\
x_1^4+x_2^4+x_3^4+x_4^4 = 34.\end{cases}$
I have found a solution of $(1,1,2,2)$. And I tried to solve it by MAPLE.
All solutions are in the form of $(1,1,2,2)$ (and any possible arrangement of $(1,1,2,2)$). But how can I prove that the equation has no more solution except these 6 solutions?
How about a general $(a_1,a_2,a_3,a_4)$?
$\begin{cases}x_1+x_2+x_3+x_4 = a_1,\\
x_1^2+x_2^2+x_3^2+x_4^2 = a_2,\\
x_1^3+x_2^3+x_3^3+x_4^3 = a_3,\\
x_1^4+x_2^4+x_3^4+x_4^4 = a_4.\end{cases}$
Do they have an unique solution? Or not?
Do we have a general method to deal with it?
| As Rene comments, generally speaking, to solve equations of this kind, you should first transform it to
\begin{cases}x_1+x_2+x_3+x_4 = c_1,\\
x_1 x_2 + x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 = c_2,\\
x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 = c_3,\\
x_1x_2x_3x_4=c_4.\end{cases}
Then, by Vieta's formulas, you immediately get that $x_1, x_2, x_3, x_4$ are $4$ roots of the equation
$$
(x-x_1)(x-x_2)(x-x_3)(x-x_4) = x^4-c_1x^3+c_2x^2-c_3x^3+c_4=0
$$
Therefore all your problems are solved.
| {
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Conjugate of Quaternion
The conjugate of $$\alpha=\left[ {\begin{array}{cc}
a+bi & c+di \\
-c+di & a-bi \\
\end{array} } \right]$$ is$$\overline{\alpha}=\left[ {\begin{array}{cc}
a-bi & -c-di \\
c-di & a+bi \\
\end{array} } \right]$$
The norm of $\alpha$ is $a^2+b^2+c^2+d^2$ and is written $\lvert\lvert\alpha\rvert\rvert$. Show directly that $$\overline{\alpha}\alpha=\alpha\overline{\alpha}=\left[ {\begin{array}{cc}
t & 0 \\
0 & t \\
\end{array} } \right]$$ where $t= \lvert\lvert\alpha\rvert\rvert$.
Conclude that the multiplicative inverse of $\alpha$ is $(1/t)\overline{\alpha}$.
So to show directly I should do matrix multiplication, so
$\overline{\alpha}\alpha=\left[ {\begin{array}{cc}
a-bi & -c-di \\
c-di & a-bi \\
\end{array} } \right]\cdot \left[ {\begin{array}{cc}
a+bi & c+di \\
-c+di & a-bi \\
\end{array} } \right] = \left[ {\begin{array}{cc}
(a - bi) (a + bi) + (c - di) (c + di) & (a + bi) (-c - di) + (a + bi) (c + di) \\
(a - bi) (c - di) + (a - bi) (di - c) & (a - bi) (a + bi) + (-c - di) (di - c) \\
\end{array} } \right] = \left[ {\begin{array}{cc}
a^2-(bi)^2+c^2-(di)^2 & 0 \\
0 & a^2-(bi)^2+c^2-(di)^2 \\
\end{array} } \right] = \left[ {\begin{array}{cc}
a^2-(bibi)+c^2-(didi) & 0 \\
0 & a^2-(bibi)+c^2-(didi) \\
\end{array} } \right]$.
EDIT: Since complex number commute I am able to conclude that $\overline{\alpha}\alpha=\alpha\overline{\alpha}= \left[ {\begin{array}{cc}
t & 0 \\
0 & t \\
\end{array} } \right]$.
| You did everything right except you need to go one more step:
$a,b,c,d$ are ordinary numbers and they commute multiplicatively, and they commute with $i$ as well. So $bibi = b^2i^2 = -b^2$ and
$$
a^2 - bibi + c^2 -didi = a^2+b^2+c^2+d^2 = t
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}$ I am just trying to calculate
$$\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}.$$
To do this I apply formula for sum of fourth powers of $n$ number. My result: $$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}=1$$. I'm intrested in finding other method to solve the following problem.
| Very short Answer
From this What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression
we have that,
$$ \sum_{k=1}^{n} k^{4} = \frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.$$
Therefore, By polynomial limit we have,
$$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}= \lim_{n\to\infty}\frac{\frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.}{\frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.+(n+1)^4}=1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the probability that player A and B play against each other?
Players with same skill take part in a competition. The probability of
winning each game is 0.5. At first, we divide a group of $2^n$ people
to random pairs that play against each other. Then we will do the same
for $2^{n-1}$ winners and this continues until there is only
one winner. What is the probability that player A and B play against
each other?
I know we should calculate the sum of probabilities that two players play against each other in the first round, then for second round and etc. So how can i calculate the probability that two players play at round K?
| Hmm...I don't see any real difference between this and an ordinary single-elimination tournament where the seeding is set from the start. Maybe I'm missing something.
Anyway, assuming that there isn't any difference, we observe that Player A has $2^n-1$ different possible opponents in the first round. Of those, $2^{n-1}$ are in the opposite half of the draw, and both they and Player A would have to win $n-1$ games to meet; this happens with probability $\frac{1}{4^{n-1}}$. $2^{n-2}$ are in the same half, but opposite quarters, and both they and Player A would have to win $n-2$ games to meet; this happens with probability $\frac{1}{4^{n-2}}$. And so on, until we get to the $2^{n-n} = 2^0 = 1$ single player who meets Player A in the first round. Altogether, the probability of Player A and a given Player B meeting eventually is
\begin{align}
\frac{1}{2^n-1} \sum_{k=1}^n 2^{n-k}\frac{1}{4^{n-k}}
& = \frac{1}{2^n-1} \sum_{k=1}^n \frac{1}{2^{n-k}} \\
& = \frac{1}{2^n-1} \sum_{k=0}^{n-1} \frac{1}{2^k} \\
& = \frac{1}{2^n-1} \left(2-\frac{1}{2^{n-1}}\right) \\
& = \frac{1}{2^n-1} \times \frac{2^n-1}{2^{n-1}} \\
& = \frac{1}{2^{n-1}}
\end{align}
I'll come back to edit this if I think (or someone points out) that there's a substantive difference between random reseeding and not reseeding.
There's also a pretty simple proof by induction: For $n = 1$ (two players), the probability is clearly $\frac{1}{2^n-1} = 1$. For $n > 1$, the probability that they meet in the first round is $\frac{1}{2^n-1}$. If they don't meet in the first round (with probability $\frac{2^n-2}{2^n-1}$), then they must both win their first games (with probability $\frac14$) to get to the next round. With reseeding, this is clearly the case of $n-1$, so with the premise that the probability of them meeting at that point is $\frac{1}{2^{n-1-1}} = \frac{1}{2^{n-2}}$, the overall probability for case $n$ is
\begin{align}
\frac{1}{2^n-1}+\frac{2^n-2}{2^n-1} \times \frac14 \times \frac{1}{2^{n-2}}
& = \frac{1}{2^n-1} \left(1+\frac{2^n-2}{2^n}\right) \\
& = \frac{1}{2^n-1} \times \frac{2^{n+1}-2}{2^n} \\
& = \frac{1}{2^n-1} \times \frac{2^n-1}{2^{n-1}} \\
& = \frac{1}{2^{n-1}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{1}{xy}\ge4$ given that $x+y=1$ and conclude that $(1+\frac1{x^2})(1+\frac1{y^2})\ge2$
Let $x,y\in\mathbb R^+$ and $x+y=1$
1- Prove that $\frac{1}{xy}\ge4$
2- Conclude that $(1+\frac1{x^2})(1+\frac1{y^2})\ge25$
I have tried to start from $x+y=1$ or $x\ge0\land y\ge0$ and reach $\frac{1}{xy}\ge4$ but with no result.
Update:
I've proved the first part of the question (Thanks to Jack's comment).
Since the second question says "Conclude" that means I have to use the first proof. I tried to square the first proof and got it close to the second question, but
again no result. There is probably a trick that I don't know.
| \begin{eqnarray*}
x+y &=& 1 \\
x^2+2xy+y^2 &=& 1 \\
x^2-2xy+y^2 &=& 1-4xy \\
0 \leq (x-y)^2 &=& 1 -4xy.
\end{eqnarray*}
Now rearrange and we have the desired inequality.
For the second part
\begin{eqnarray*}
\left(\frac{1}{x} -\frac{1}{y}\right)^2 \geq 0 \\
\frac{1}{x^2} +\frac{1}{y^2} \geq \frac{2}{xy}
\end{eqnarray*}
So
\begin{eqnarray*}
1+ \frac{1}{x^2} +\frac{1}{y^2} + \left(\frac{1}{xy}\right)^2 \geq 1+ \frac{2}{xy} + \left(\frac{1}{xy}\right)^2 =\left(1+\frac{1}{xy}\right)^2 \geq 25.\\
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of solutions of $\sin (2x)+\cos (2x)+\sin x+\cos x=1$ Find Number of solutions of $\sin (2x)+\cos (2x)+\sin x+\cos x=1$ in $\left [0 \:\: 2\pi\right]$
The equation can be written as:
$$\sin (2x)+1-2 \sin^2x+\sin x+\cos x=1$$ $\implies$
$$\sin x+\cos x=2\sin^2 x-2 \sin x\cos x$$ $\implies$
$$\sin x+\cos x=2\sin x\left(\sin x-\cos x\right)$$
$\implies$
$$\frac{\sin x+\cos x}{\sin x-\cos x}=2\sin x$$ $\implies$
$$\frac{1+\tan x}{1-\tan x}=-2\sin x$$
$$\tan \left(\frac{\pi}{4}+x\right)=-2\sin x$$
Now i have drawn the graphs of $\tan \left(\frac{\pi}{4}+x\right)$ and $-2\sin x$ and observed there are two solutions.
is there any other way?
| Another way would be to let $t=\tan(\frac x2)$ and arrive to $$t^4+2 t^3+8 t^2-6 t-1=0$$ Now, using the formulae for the quartic equation, the discriminant is $\Delta=-309248$ which shows that the equation has two distinct real roots and two complex conjugate non-real roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate the surface area of a cube given a diagonal? I encountered this problem while practicing for a mathematics competition.
A cube has a diagonal length of 10. What is the surface area of the cube? No Calculators Allowed.
(Emphasis mine)
I'm not even sure where to start with this, so I scribbled down some numbers and solved for a square instead of a cube. Presumably, you can calculate the diagonal of a cube using the Pythagoras Theorem somehow, though I'm not sure how.
| This problem has a simple solution: $s=2d^2$ where $s$ is the surface area and $d$ is the spacial diagonal.
Explanation:
I discovered through Google (while writing the question) that the side length of a cube can be calculated with $d = a\sqrt{3}$ where $d$ is the diagonal of the cube and $a$ is the length of a single side. Nonetheless, I think the question still stands as a valid question and could be useful to future users, so I'll answer it myself.
We know that the surface area is equal to $6a^2$, so isolating a:
$$d=a\sqrt{3} \\ d^2=3a^2 \\ \frac{d^2}{3}=a^2 \\ a=\sqrt{\frac{d^2}{3}} \\ a=\frac{d\sqrt{3}}{3}$$ This means that surface area is equal to $6\left(\frac{d\sqrt{3}}{3}\right)^2$ Expanding to simplify further: $$s=6\left(\frac{d\sqrt{3}}{3}\right)\left(\frac{d\sqrt{3}}{3}\right) \\ s=6\left(\frac{3d^2}{9}\right) \\ s = \frac{6d^2}{3} = 2d^2$$
The surface area can be calculated by $2d^2$
Lets test this: Going the long way:
$$10=a\sqrt{3} \\ 100=3a^2 \\ \frac{100}{3}=a^2 \\ a=\sqrt{\frac{100}{3}} \\ a = \frac{10}{\sqrt{3}} \\ a=\frac{10\sqrt{3}}{3}$$ This means the side length of the cube is $\frac{10\sqrt{3}}{3}$. If we square this, we get the area of a single face is $\left(\frac{10\sqrt{3}}{3}\right)^2$
$$s=\left(\frac{10\sqrt{3}}{3}\right)^2 \\ s=\left(\frac{10\sqrt{3}}{3}\right)\left(\frac{10\sqrt{3}}{3}\right) \\ s=\frac{100\sqrt{9}}{9} \\ s=\frac{300}{9} \\ s=\frac{100}{3} $$
This means the surface area of a single face is $\frac{100}{3}$, So we multiply by six to get the total surface area: $\frac{600}{3} = 200$
And going the short way gives us $2(10)^2 = 200$. Both ways give us the same solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximizing a polynomial expression involving the sidelengths of an inscribed pentagon
Given that $ABCDE$ is a convex pentagon and is inscribed in a circle of radius $1$ unit with $AE$ as diameter. If $AB=a,BC=b ,CD=c$ and $DE=d$, then maximum possible integral value of $a^2+b^2+c^2+d^2+abc+bcd$ is ?
What I tried
$AE =2$ , I joined points to form lines $BE,AC,CE$ and $AD$ and tried to use the fact that angle subtended by diameter would be a right angle.
However, I was unable to solve the problem.
I was looking for some hint or an alternate approach to the problem.
| In quadrilateral $ABCE$ By Ptolemy's theorem
$$2b+av=(\sqrt {4-a^2})u$$
And similarly in quadrilateral $CDEA$
$$2c+du=(\sqrt {4-d^2})v$$
On squaring these equations and adding them we get $$4b^2+a^2v^2+4abv+4c^2+d^2u^2+4cdu$$
$$=4u^2-a^2u^2+4v^2-v^2d^2$$
Which simplifies to
$$4(b^2+c^2)+(a^2+d^2)(u^2+v^2)+4(abv+cdu)=4(u^2+v^2)$$
By Pythagoras theorem we get $$u^2+v^2=4$$
Hence the obtained expression simplifies to $$a^2+b^2+c^2+d^2+(abv+cdu)=4$$
In triangle $ABC$ and triangle $CDE$ , By triangle inequality we get $u\lt (a+b)$ and $v\lt (c+d)$
Using all these expressions and inequalities we get
$$a^2+b^2+c^2+d^2+abc+bcd+ ad(b+c)\lt 4$$
Now by AM-GM we have
$$\frac{a+b+c+d}{3}\ge.\sqrt[3] {ad(b+c)}$$
Hence
$$ad(b+c)\le (\frac{a+b+c+d}{3})^3$$
We have triangle inequalities
$$u\lt (a+b)$$ $$v\lt (c+d)$$ $$u+v\gt 2$$
Using the result of AM GM and these inequalities we get
$$a^2+b^2+c^2+d^2+abc+bcd\lt \frac{100}{27}$$
Hence the integral value is 3
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $ \left(\frac{a}{a + 2b}\right)^2 + \left(\frac{b}{b + 2c}\right)^2 + \left(\frac{c}{c + 2a}\right)^2 \geq 1/3 $ I'm trying to prove the following inequality using Holder's Inequality:
$$ \left(\frac{a}{a + 2b}\right)^2 + \left(\frac{b}{b + 2c}\right)^2 + \left(\frac{c}{c + 2a}\right)^2 \geq \frac{1}{3}, $$
where $ a, b ,c $ are positive.
I've tried a number of combinations but I'm stuck - any ideas?
| By C-S twice we obtain:
$$\sum_{cyc}\left(\frac{a}{a+2b}\right)^2=\frac{1}{3}\sum_{cyc}1^2\sum_{cyc}\left(\frac{a}{a+2b}\right)^2\geq\frac{1}{3}\left(\sum_{cyc}\frac{a}{a+2b}\right)^2=$$
$$=\frac{1}{3}\left(\sum_{cyc}\frac{a^2}{a^2+2ab}\right)^2\geq\frac{1}{3}\left(\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+2ab)}\right)^2=\frac{1}{3}.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.