Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to express $2x^3-x^2-3x+2$ as a linear combination of Legendre polynomials I have used the formula
\begin{align}p_0(x)&=1\\
p_1(x)&=x\\
p_2(x)&=\frac12(3x^2−1)\\
p_3(x)&=\frac12(5x^3−3x)
\end{align}
$$2x^3-x^2-3x+2=Ap_3(x)+Bp_2(x)+Cp_1(x)+Dp_0(x)$$
EDIT-
\begin{align}&=\frac A2(5x^3−3x) + \frac B2(3x^2−1) + Cx +D \end{align}
but don't know how to continue after this.
| In matrix form,
$$\begin{bmatrix} p_0\\ p_1\\ p_2\\ p_3\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ -\frac{1}{2} & 0 & \frac{3}{2} & 0\\ 0 & -\frac{3}{2} & 0 & \frac{5}{2}\end{bmatrix} \begin{bmatrix} 1\\ x\\ x^2\\ x^3\end{bmatrix}$$
We want to find a weight vector $\mathrm{w}$ such that
$$2 x^3 - x^2 - 3 x + 2 = \begin{bmatrix} w_0\\ w_1\\ w_2\\ w_3\end{bmatrix}^T \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ -\frac{1}{2} & 0 & \frac{3}{2} & 0\\ 0 & -\frac{3}{2} & 0 & \frac{5}{2}\end{bmatrix} \begin{bmatrix} 1\\ x\\ x^2\\ x^3\end{bmatrix}$$
and, thus, we have to solve the linear system
$$\begin{bmatrix} 2\\ -3\\ -1\\ 2\end{bmatrix}^T = \begin{bmatrix} w_0\\ w_1\\ w_2\\ w_3\end{bmatrix}^T \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ -\frac{1}{2} & 0 & \frac{3}{2} & 0\\ 0 & -\frac{3}{2} & 0 & \frac{5}{2}\end{bmatrix}$$
or, in a more standard form,
$$\begin{bmatrix} 1 & 0 & -\frac{1}{2} & 0\\ 0 & 1 & 0 & -\frac{3}{2}\\ 0 & 0 & \frac{3}{2} & 0\\ 0 & 0 & 0 & \frac{5}{2}\end{bmatrix} \begin{bmatrix} w_0\\ w_1\\ w_2\\ w_3\end{bmatrix} = \begin{bmatrix} 2\\ -3\\ -1\\ 2\end{bmatrix}$$
The solution is
$$\frac{5}{3} \, p_0 (x) - \frac{9}{5} \, p_1 (x) - \frac{2}{3} \, p_2 (x) + \frac{4}{5} \, p_3 (x) = 2 x^3 - x^2 - 3 x + 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Continuity of $\frac{x^3y^2}{x^4+y^4}$ at $(0,0)$? Suppose a function $f$ is defined as follows:
$$f(x,y)=\begin{cases} \frac{x^3y^2}{x^4+y^4}&\text{ when }(x,y)\neq(0,0),\\0 & \text{ when }(x,y)=(0,0).\end{cases}$$
Is this function continuous at $(0,0)$? How is this shown? I've tried considering limits for different $y=g(x)$ functions and I am unable to find a counterexample. But I do not see how to prove continuity in general.
| Since $x^4-2x^2y^2+y^4= (x^2-y^2)^2 \ge 0$, we have $2x^2y^2 \le x^4+y^4$.
Therefore, $\dfrac{x^2y^2}{x^4+y^4} \le \dfrac{1}{2}$ for all $(x,y) \neq (0,0)$.
Also, $\dfrac{x^2y^2}{x^4+y^4} \ge 0$ for all $(x,y) \neq (0,0)$.
From the above inequalities, we have that $\left|\dfrac{x^2y^2}{x^4+y^4}\right| \le \dfrac{1}{2}$ for all $(x,y) \neq (0,0)$
Now, multiply both sides by $|x|$ to get $\left|\dfrac{x^3y^2}{x^4+y^4}\right| \le \dfrac{1}{2}|x|$.
Can you finish the problem from here?
The Squeeze Theorem will be useful.
| {
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"url": "https://math.stackexchange.com/questions/1822811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this 1st order ODE $$ \frac{dy}{dx} = \frac{2x^2 + 3y^2 - 7}{3x^2 + 4y^2 + 8}$$
This does not satisfy the exactness and I can't find any integrating factor to transform it. I can't make it homogeneous too.
Thanks in advance for the replies.
edit: The question was written incorrectly by the instructor, so in this version there is no analytic solution for it.
| Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=164:
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\dfrac{2x^2+3(xu)^2-7}{3x^2+4(xu)^2+8}$
$x\dfrac{du}{dx}=\dfrac{(3u^2+2)x^2-7}{(4u^2+3)x^2+8}-u$
$x\dfrac{du}{dx}=-\dfrac{(4u^3-3u^2+3u-2)x^2+8u+7}{(4u^2+3)x^2+8}$
$((4u^3-3u^2+3u-2)x^2+8u+7)\dfrac{dx}{du}=-(4u^2+3)x^3-8x$
Let $v=\dfrac{1}{x^2}$ ,
Then $\dfrac{dv}{du}=-\dfrac{2}{x^3}\dfrac{dx}{du}$
$\therefore-\dfrac{((4u^3-3u^2+3u-2)x^2+8u+7)x^3}{2}\dfrac{dv}{du}=-(4u^2+3)x^3-8x$
$\left(\dfrac{8u+7}{x^2}+4u^3-3u^2+3u-2\right)\dfrac{dv}{du}=\dfrac{8}{x^4}+\dfrac{4u^2+3}{x^2}$
$((8u+7)v+4u^3-3u^2+3u-2)\dfrac{dv}{du}=8v^2+(4u^2+3)v$
This belongs to an Abel equation of the second kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Partial fraction decomposition of $\pi\cdot \tan(\pi z)$
Evaluate the partial fraction decomposition of $\pi \tan(\pi z)$
$$2\pi \tan(\pi z)=\cot\left(\frac{\pi}{2}-\pi z\right)-\cot\left(\frac{\pi}{2}+\pi z\right)$$ $$=\frac{2}{1-2z}+\sum_{k=1}^\infty \frac{1-2z}{(\frac{1}{2}-z)^2-k^2}-\frac{2}{1+2z}-\sum_{k=1}^\infty \frac{1+2z}{(\frac{1}{2}-z)^2-k^2}$$
$$=\frac{8z}{1-4z^2}+\sum_{k=1}^\infty \frac{4-8z}{1-2z+4z^2-4k^2}-\sum_{k=1}^\infty \frac{4+8z}{1+2z+4z^2-4k^2}$$
I know that the result should be $$\pi \tan(\pi z)=\sum_{k=1}^\infty \frac{8z}{(2k+1)^2-4z^2}$$ but I dont know how to get there.
Any suggestions?
| You may exploit the fact that $\tan(z)=\frac{d}{dz}\log(\cos z)$ and:
$$ \cos(z)=\prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2\pi^2}\right)\tag{1}$$
(Weierstrass product) so:
$$ \tan(z) = \sum_{n\geq 0}\frac{-8z}{4z^2-(2n+1)^2\pi^2}\tag{2}$$
and:
$$\begin{eqnarray*}\pi\tan(\pi z) = \sum_{n\geq 0}\frac{-2z}{z^2-\left(\frac{2n+1}{2}\right)^2}&=&-\sum_{n\geq 0}\left(\frac{1}{z-\frac{2n+1}{2}}+\frac{1}{z+\frac{2n+1}{2}}\right)\\&=&\color{red}{\sum_{m\in\frac{1}{2}+\mathbb{Z}}\frac{-1}{z-m}}.\tag{3}\end{eqnarray*}$$
On the other hand, every singularity of $\pi\tan(\pi z)$ is a simple pole and lie in $\frac{1}{2}+\mathbb{Z}$, and for every $m$ that belongs to such a set:
$$ \text{Res}\left(\pi\tan(\pi z),z=m\right)=\lim_{z\to m}\frac{\pi\sin(\pi z)(z-m)}{\cos(\pi z)}\stackrel{DH}{=}\lim_{z\to m}\frac{\pi\sin(\pi z)}{-\pi \sin(\pi z)}=\color{red}{-1},\tag{4}$$
giving another proof of $(3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$ How to prove this inequality?
If $a^{2}+b^{2}+c^{2}\leq 3$ and $a,b,c\in \Bbb R^+$, then
$$\left( a+b+c\right) \left( a+b+c-abc\right)\geq 2\left( a^{2}b+b^{2}c+c^{2}a\right) $$
I tried AM>GM but I couldn't get result
| Another way.
We need to prove that
$$(a+b+c)^2\geq2\sum_{cyc}a^2b+abc(a+b+c)$$ and since $$\frac{3}{a^2+b^2+c^2}\geq1$$ and $$1\geq\frac{a^2+b^2+c^2}{3},$$ it's enough to prove that:
$$(a+b+c)^2\sqrt{\frac{a^2+b^2+c^2}{3}}\geq2\sum_{cyc}a^2b+\sqrt{\frac{3}{a^2+b^2+c^2}}abc(a+b+c)$$ or
$$(a+b+c)^2(a^2+b^2+c^2)\geq2\sum_{cyc}a^2b\sqrt{3(a^2+b^2+c^2)}+3abc(a+b+c).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$, $abc=w^3$ and $u=tv$.
Thus, $t\geq1$ and since by Rearrangement $\sum_{cyc}a^2b\leq4u^3-w^3$, it's enough to prove that:
$$9u^2(9u^2-6v^2)\geq2(4u^3-w^3)\sqrt{3(9u^2-6v^2)}+9uw^3$$ or
$$9u^2(3u^2-2v^2)-8u^3\sqrt{3u^2-2v^2}\geq(3u-2\sqrt{3u^2-2v^2})w^3.$$
Now, if $$(3u-2\sqrt{2u^2-2v^2}<0$$ so our inequality is obviously true because
$$9u^2(3u^2-2v^2)-8u^3\sqrt{3u^2-2v^2}=$$
$$=u^2\sqrt{3u^2-2v^2}(9\sqrt{3u^2-2v^2}-8u)\geq u^2\sqrt{3u^2-2v^2}(9\sqrt{u^2}-8u)>0.$$
But for $3u-2\sqrt{2u^2-2v^2}\geq0$ since $w^3\leq v^3,$ it's enough to prove that
$$9t^2(3t^2-2)-8t^3\sqrt{3t^2-2}\geq3t-2\sqrt{3t^2-2}$$ or
$$27t^4-18t^2-3t\geq(8t^3-2)\sqrt{3t^2-2}$$ or
$$(t-1)(537t^7+537t^6-307t^5-373t^4-49t^3-5t^2-8t-8)\geq0,$$ which is obvious for $t\geq1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
How can I solve this hard system of equations? Solve the system below
\begin{align}
&\sqrt {3x} \left( 1+\frac {1}{x+y} \right) =2\\
&\sqrt {7y} \left( 1-\frac{1}{x+y} \right) =4\sqrt{2}
\end{align}
Frankly I am disappointed, because I spent around 2 hours in solving this equation, but, finally I didn't do it so, I hope you can help me in approaching this problem.
| Square both equations. Then let $u=\frac{1}{x+y}$. Note then $x+y=\frac{1}{u}$
$$3x(1+\frac{1}{x+y})^2=4$$
$$7y(1-\frac{1}{x+y})^2=32$$
So.
$$x(1+u)^2=\frac{4}{3}$$
$$y(1-u)^2=\frac{32}{7}$$
And for $u \neq 1$
$$x=\frac{4}{3}(1+u)^{-2}$$
$$y=\frac{32}{7}(1-u)^{-2}$$
Adding both equations we have:
$$\frac{1}{u}=\frac{4}{3(1+u)^2}+\frac{32}{7(1-u)^2}$$
$$\frac{1}{u}=\frac{28(1-u)^2+96(1+u)^2}{21(1-u)^2(1+u)^2}$$
Cross multiplication, and moving terms to one side yields:
$$28u(1-u)^2+96u(1+u)^2-21(1-u)^2(1+u)^2=0$$
$$-21u^4+124u^3+178u^2+124u-21=0$$
This factors:
$$-(3-22u+3u^2)(7+10u+7u^2)=0$$
So:
$$u=\frac{11 \pm 4\sqrt{7}}{3}$$
$$x+y=\frac{3}{11 \pm 4\sqrt{7}}$$
Back substitution yields:
$$\sqrt{3x}(\frac{14 \pm 4\sqrt{7}}{3})=2$$
$$x=\frac{12}{(14 \pm 4\sqrt{7})^2}$$
$$y=\frac{3}{11 \pm 4\sqrt{7}}-\frac{12}{(14 \pm 4\sqrt{7})^2}$$
We introduced an extraneous solution when squaring so checking the possible solutions, the final solution is:
$$x=\frac{12}{(14-4\sqrt{7})^2}$$
$$y=\frac{3}{11-4\sqrt{7}}-\frac{12}{(14- 4\sqrt{7})^2}$$
If you wish, from the second equation, you can express $y$ as:
$$y=\frac{288}{7(4\sqrt{7}-8)^2}$$
In fact through the method of multiplying by conjugates we may get:
$$x=\frac{1}{21}(11+4\sqrt{7})$$
$$y=\frac{2}{7}(11+4\sqrt{7})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$? I came around this expression when solving a problem.
$$\sqrt{7+4\sqrt{3}}$$
WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this
$$\left(2+\sqrt{3}\right)^2 \;=\; 4+4\sqrt{3} + 3 \;=\; 7 + 4\sqrt{3}.$$
However, the only way I can think of how to simplify that expression in hand is guessing. Is there a better way of calculating square root of a sum like that one?
| $\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$
$a+\sqrt{b}=x+y+2\sqrt{xy}$
$a+\sqrt{b}=(x+y)+\sqrt{4xy}$
Then:
$a=x+y$ and $b=4xy$
$a^2=x^2+y^2+2xy$
$a^2-b=(x-y)^2$
If $x>y$ then:
$\sqrt{a^2-b}=x-y$
Let's take $c=\sqrt{a^2-b}$ then $c=x-y$
$c=x-y$ and $a=x+y$ then:
$x=\frac{a+c}{2}$ and $y=\frac{a-c}{2}$
Then:
$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a-c}{2}}$
Now applly it for $\sqrt{7+4\sqrt{3}}$.
$\sqrt{7+4\sqrt{3}}=\sqrt{7+\sqrt{48}}$
$c=\sqrt{49-48}$
$c=1$
$\sqrt{7+\sqrt{48}}=\sqrt{\frac{8}{2}}+\sqrt{\frac{6}{2}}$
Then:
$\sqrt{7+4\sqrt{3}}=2+\sqrt{3}$
solved!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Upper bound for $\gcd(a,b)$ if $\frac{a+1}{b}+\frac{b+1}{a}\in\Bbb{N}$
Suppose that $a,b$ are two positive integers so that $\frac{a+1}{b}+\frac{b+1}{a}$ is also a positive integer.Find the best upper bound for $\gcd(a,b)$.
My work:
$\frac{a+1}{b}+\frac{b+1}{a}=\frac{a(a+1)+b(b+1)}{ab} \in \Bbb{N} \implies ab|a^2+b^2+a+b , 2ab \implies ab|(a+b)(a+b+1)\implies ab|a+b\ or\ ab|a+b+1$
Now may I infer that $a=b=1\ or\ a=b=2$? If yes how may we set an upper bound for $\gcd(a,b)$??!!
| Fortunately with the help of one of my friends I got to the right answer as follows:
Assuming $d=\gcd(a,b),a=a'd,b=b'd$ we have: $$\frac{a+1}{b}+\frac{b+1}{a}=\frac{a'^2d+b'^2d+a'+b'}{a'b'd}$$
Now it's required that $d|a'+b'$ so that the expression is an integer , then:
$$d^2|a'd+b'd\implies d^2|a+b \implies d^2\leqslant a+b \implies \color{red}{d\leqslant \sqrt{a+b}}$$
Tha's all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the Fourier Sine Transform of $\frac{1}{x(a^2+x^2)}$? (Please check my solution) My textbook says that the answer is $\frac{\pi}{2a^2}(1-e^{-as})$ while my answer is $\frac{1}{a^2}\sqrt{\frac{\pi}{2}}(1-\frac{e^{-as}}{2}+\frac{e^{as}}{2})$. Here is my solution:
\begin{align}
\hat{f_s} & = \sqrt{\frac{2}{\pi}} \int_0^\infty \frac{1}{t(a^2+t^2)} \sin{st}dt \\
& = \sqrt{\frac{2}{\pi}} \int_0^\infty \frac{t}{a^2}[\frac{1}{t^2} -\frac{1}{(a^2+t^2)}] \sin{st}dt \\
& = \frac{1}{a^2}\sqrt{\frac{2}{\pi}} \int_0^\infty [\frac{\sin{st}}{t} -\frac{t}{(a^2+t^2)} \sin{st}] dt \\
& = \frac{1}{a^2}\sqrt{\frac{2}{\pi}} \left( \frac{\pi}{2} - \int_0^\infty \frac{t}{(a^2+t^2)} \sin{st} dt \right)
\end{align}
Let this integral be $y$. Therefore,
\begin{align}
y & = \int_0^\infty \frac{t}{(a^2+t^2)} \sin{st} dt \\
& = \int_0^\infty \frac{1}{t}[1 -\frac{a^2}{(a^2+t^2)}] \sin{st}dt \\
& = \int_0^\infty [\frac{\sin{st}}{t} -\frac{a^2}{t(a^2+t^2)} \sin{st}] dt \\
& = \left( \frac{\pi}{2} - \int_0^\infty \frac{a^2}{t(a^2+t^2)} \sin{st} dt \right)
\end{align}
Differentiating $y$ w.r.t. $s$ twice, we get
\begin{align}
y'' & = a^2\int_0^\infty \frac{t}{(a^2+t^2)} \sin{st} dt \\
& = a^2 y \\
\end{align}
This is a second order differential equation, whose solution is
$$y= C_1e^{-as}+C_2e^{as}$$
Put $s=0$ in $y$ and $y'$, we get $C_1 = \frac{\pi}{4} = - C_2$. Thus, we get $$y = \frac{\pi}{4}(e^{-as}-e^{as})$$ or
$$\hat{f_s}=\frac{1}{a^2}\sqrt{\frac{2}{\pi}}\left(\frac{\pi}{2} - \frac{\pi}{4}(e^{-as}-e^{as})\right)$$ which is nothing but $$\hat{f_s}=\frac{1}{a^2}\sqrt{\frac{\pi}{2}}(1-\frac{e^{-as}}{2}+\frac{e^{as}}{2})$$
| When doing the change of variable $u = sx$, it is implicitly assumed that $s \neq 0$ (otherwise, the integral evaluates to $0$). This means that the middle expression of $y$ and the subsequent chain of differentiations (hence, also, the obtained expression for $y''$) are only valid for $s \neq 0$. To obtain a formula for all $s$, we must differentiate:
$$y = \int_0^{\infty} \frac{t}{t^2 + a^2} \sin(st)dt$$
Now there is a problem if we want to differentiate under the integral sign, because the integrand is not integrable in the sense of Lebesgue
Edit
The added content is elaboration on Olivier Oloa's comment on the original post.
As suggested, we can differentiate:
$$I(s) = \int_0^{\infty} \frac{1}{t(t^2 + a^2)} \sin(st)dt$$
to get:
$$I'(s) = \int_0^{\infty} \frac{\cos(st)}{a^2 + t^2}dt$$
(because the integrand is Lebesgue integrable and its partial derivative wrt $s$ is uniformly bounded, wrt $t$, by an integrable function)
One way to obtain $I'(s)$ is by using the Fourier transform of $g(t) = e^{-a|t|}$. It is given by:
$$\mathcal{F} (g)(t) = \sqrt{\frac{2}{\pi}}\frac{a}{a^2 + t^2}$$
Using the inversion formula, we get:
$$\frac1{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \sqrt{\frac{2}{\pi}}\frac{a}{a^2 + t^2} e^{-it s}dt = e^{-a|s|}$$
Hence:
$$\int_{-\infty}^{\infty} \frac{e^{-it s}}{a^2 + t^2} dt = \frac{\pi}a e^{-a|s|}$$
Assuming $s \ge 0$, we finally get:
$$I'(s) = \frac{\pi}{2a} e^{-as}$$
So
$$I(s) = \frac{\pi}{2a^2} (C - e^{-as})$$
As $I(0) = 0$, we get $C = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution of given equation for $x$ Solve the given equation for $x$
$$\sqrt{x^2-2x+8}+\sqrt{x^2-2x+3}=125$$
I solved the question by taking ${x^2-2x+3}=t$, and squaring twice and finally solving ${x^2-2x+3}=t$ but it required very hectic calculations. I wonder if someone can suggest better approach.
| As $8-3=x^2-2x+8-(x^2-2x+3)$
$=(\sqrt{x^2-2x+8}+\sqrt{x^2-2x+3})(\sqrt{x^2-2x+8}-\sqrt{x^2-2x+3})$
$$\sqrt{x^2-2x+8}+\sqrt{x^2-2x+3}=125\iff(\sqrt{x^2-2x+8}-\sqrt{x^2-2x+3})=\dfrac{8-3}{125}$$
Adding we get $2\sqrt{x^2-2x+8}=125+\dfrac1{25}=?$
Now square both sides.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $|z|$ , $z = \frac{(2+i)^7(1-2i)^3}{(1+2i)^8}$ Compute $|z|$ , $z = \frac{(2+i)^7(1-2i)^3}{(1+2i)^8}$,
if $z = a+ib$ then,
I tried to do that with $|z| = (a+ib)(a-ib)$
then i multipled it $z$ with $z^-$ and then I got stuck. answer is $|z| = 5$
| Hint. One may write
$$
|z|^2=z\cdot \bar z=\frac{|2+i|^{14}|1-2i|^6}{|1+2i|^{16}}=\frac{(2^2+1^2)^7(1^2+2^2)^3}{(1^2+2^2)^8}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways
Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways.
I took extremely long to solve this
I got
$50= 7^2 + 1^2 $
$50= 5^2 + 5 ^2 $
I did it with the trial and error method.
I'm just curious if this came out during my exams, is there a quicker way to find the answer?
| If $p$ is a prime and $p \equiv 1 \bmod 4$, then $p=a^2+b^2$, for some $a,b \in \mathbb N$.
In this case, $2p^2$ has two representations: $p^2+p^2$ and $(a-b)^2+(a+b)^2$.
The second representation ones from $2p^2=|(1+i)(a+bi)^2|^2$.
So, the smallest solution is with $p=5$, for which $a=2$ and $b=1$ and $(a+bi)^2=3-4i$, which give $50=5^2+5^2=1^2+7^2$.
| {
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"url": "https://math.stackexchange.com/questions/1833292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find all pairs such that $x^2 y + x + y$ is divisible by $xy^2 + y + 7$
Find all pairs $(x, y)$ of positive integers such that $x^2 y + x + y$ is divisible by $xy^2 + y + 7$. If there are too many to write, write a generic form.
I was thinking of rewriting the divisibility in a simpler way such as $xy^2 + y + 7\mid x^2 y + x + y \implies xy^2 + y + 7\mid x^2y-xy^2+x-y$ but I am not sure if that helps. Is there an easier way?
| We have $x=7n^2,y=7n$ as a class of solutions because then $n(xy^2+7+y)=x^2y+y+x$.
There is one other solution $x=11,y=1$, giving $xy^2+7+y=19,x^2y+y+x=7\cdot19$.
We have $y(x^2y+x+y)-x(xy^2+y+7)=y^2-7x$, so $xy^2+y+7$ must divide $y^2-7x$.
One possibility is $y^2=7x$. That implies 7 divides $y$ and hence also $x$. So we get the solutions in the first line.
Otherwise, since $y^2-7x<y^2<xy^2+y+7$ we must have $y^2<7x$. But then if $y\ge3$ we have $xy^2+y+7>9x>7x-y^2>0$ which is impossible if $xy^2+y+7$ is a factor of $7x-y^2$. So $y=1$ or 2. It is easy to check that $y=1$ gives the solution $x=11,y=1$, whilst $y=2$ does not give a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $\sum_{n=1}^\infty \frac{2^n+n^2+n}{2^{n+1}n(n+1)}$ is convergent and find the limit when $n \to \infty$ Does the following sumatory converges? If yes find the limit when $n \to \infty$ $$\sum_{n=1}^\infty \frac{2^n+n^2+n}{2^{n+1}n(n+1)}$$
ideas?
I have tried by the comparison test.
| Observe that we can break up the sum to:
$$ \sum_{n=1}^{\infty} \frac{2^n}{2^{n+1} n(n+1)} + \frac{n^2+n}{2^{n+1}n(n+1)} $$
Which then becomes two sums:
$$ \sum_{n=1}^{\infty} \frac{1}{2n(n+1)} + \sum_{n=1}^{\infty}\frac{1}{2^{n+1}} $$
Now observe the left sum decays quadratically (any function f(n) which has the property that $\lim_{n\rightarrow \infty} f(n)/n = \infty$ similarly has the property that $\sum \frac{1}{f(n)}$ converges).
Likewise the right hand sum is an easily recognized geometric sum. So now we can proceed to conclude the sum of the sums, itself must converge, and now the burden is of finding a closed form.
RIGHT SUM:
This is much easier so we warm up with it:
$$ \sum_{n=1}^{\infty} \frac{1}{2^{n+1}} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2} \left( \frac{1}{1 - \frac{1}{2}} - 1 \right) = \frac{1}{2}$$
LEFT SUM:
We can factor out the 2 to yield:
$$ \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n(n+1)}$$
Which is
$$ \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right)$$
So now evluating this sum: is evaluation
$$ 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} ... $$
Which at $n^{th}$ sum is $1 - \frac{1}{n+1}$ so as $n \rightarrow \infty$ this is just 1. Meaning:
$$ \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{2}$$
thus:
$$\sum_{n=1}^{\infty} \frac{1}{2n(n+1)} + \sum_{n=1}^{\infty}\frac{1}{2^{n+1}} = \frac{1}{2} + \frac{1}{2} = 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$
$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\frac{1}{10}$$ gives:
$$\frac{39}{10}<x+3<\frac{41}{10}$$
$$|x+3|<\frac{39}{10}$$
Plugging those results in $$\frac{|x-1|*|x+1|}{|x+3|}<\frac{1}{13}$$
We get: $$\frac{\frac{1}{10}*\frac{19}{10}}{\frac{39}{10}}<\frac{1}{13}$$
$$\frac{19}{390}<\frac{1}{13}$$ Which is true, is this proof is valid as I took the smallest intervals, like $|x+3|<\frac{39}{10}$ and not $|x+3|<\frac{41}{10}$?
| You looked at each of the three terms separately. You have $|x-1|<\frac{1}{10}$. You should have $|x+1|<\frac{21}{10}$ and you should have $\frac{1}{|x+3|}<\frac{10}{39}$. The critical point for your method is that you have to take the worst case each time.
The three inequalities then multiply together to give $\left|\frac{x^2-1}{x+3}\right|<\frac{21}{390}<\frac{1}{13}$.
---------- ASIDE ---------
But note that this method only works because you were given a limit which is substantially larger than the largest possible value of the expression. If you had been asked to show that $$\left|\frac{x^2-1}{x+3}\right|<\frac{1}{19}$$ which is still true for $|x-1|<\frac{1}{10}$ the method would fail.
For a tougher limit the tool of choice is calculus, but one can often solve problems without it. For example in this case we could prove the $\frac{1}{19}$ limit by using the properties of quadratics.
Note that $x+3$ is positive throughout the range $0.9<x<1$, whereas $x^2-1<0$ for $0.9<x<1$ and $x^2-1>0$ for $1<x<1.1$ so we need to show that $19(1-x^2)<x+3$ for $0.9<x<1$ and $19(x^2-1)<x+3$ for $1<x<1.1$. We have
$$19(1-x^2)-(x+3)=-19x^2-x+16=-19\left(x+\frac{1}{38}\right)^2+16\frac{1}{76}$$ which is strictly decreasing for $x>\frac{1}{38}$. So its largest value in the range $0.9\le x\le 1.1$ is $19(1-0.9^2)-3.9=-0.29<0$ at $x=0.9$.
Similarly $$19(x^2-1)-(x+3)=19\left(x-\frac{1}{38}\right)^2-22\frac{1}{76}$$ which is strictly increasing in the range $1\le x\le 1.1$, so its largest value in the range is $19(1.21^2-1)-4.1=-0.11<0$ at $x=1.1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Value of $\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
If $a+b+c=0,$ Then value of $\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\bf{My\; Try::}$ Given $$\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right) = 1+1+1+\frac{b-c}{a}\left(\frac{b}{c-a}+\frac{c}{a-b}\right)+\frac{c-a}{b}\left(\frac{a}{b-c}+\frac{c}{a-b}\right)+\frac{a-b}{c}\left(\frac{a}{b-c}+\frac{b}{c-a}\right)$$
Now How can I Solve after that, Is there is any less complex method plz explain here , Thanks
| We have $$A = \left(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b}\right)\frac{bc(b-c) + ca(c-a) + ab(a-b)}{abc} = \frac{-3abc - a^3 -b^3 -c^3+ a^2(c-b) + b^2(a-c) +c^2(b-a)}{abc}.$$
Note that $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)$ or $a^3+b^3+c^3 = 3abc$.
Then $$A = \frac{(a-b)(c-b)(a-c)}{abc}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find the value of $h$ if $x^2 + y^2 = h$
Consider equation $x^2 + y^2 = h$ that touches the line $y=3x+2$ at some point $P$. Find the value of $h$
I know that $x^2 + y^2 = h$ is a circle with radius $\sqrt{h}$.
Also, since $y = 3x + 2 $ is a tangent, we know that the slope of the radius perpendicular to the tangent is $M_{OP}= -\frac{1}{3}$. I'm not sure how I can determine the value of $h$ though?
| There are couple of approaches already mentioned in other answers, but I will try to summarize them and add (probably unwarranted) generality. Also, I will describe approach using calculus.
If you are not interested in this generalities, please skip to the last paragraph.
We are given a line $y = ax + b$ and a circle centered $(p,q)$ given
with $(x-p)^2+(y-q)^2 = h$ and we are looking for $h$ such that given
line is tangent to the circle.
1. Calculus
We have function $y(x)$ for which relation $(x-p)^2+(y-q)^2 = h$ holds. If we differentiate the relation with respect to $x$ we get $$2(x-p)+2(y-q)y'=0\implies y'=-\frac{x-p}{y-q}$$ Now, since we are looking for a tangent given with $y = ax +b$, we are looking for a point $x$ such that $y'(x) = a$ and $y(x) = ax +b$. Thus we get $$a = -\frac{x-p}{ax + b - q}\implies x =\frac{p-ab+aq}{1+a^2}$$ and also $$y = ax + b = \frac{b + ap + a^2 q}{1+a^2}$$ Now, we get $$\boxed{h = (x-p)^2 + (y-q)^2 = \frac{(b+ap-q)^2}{1+a^2}}$$
2. Tangent of a circle is perpendicular to its radius
This is your original approach and is completely covered in the answer by Adriano.
Now, the line passing through center $(p,q)$ and is perpendicular to $y = ax + b$ is given by formula $$y - q = -\frac 1 a(x-p)$$ and we can find the intersection with the original line by solving linear system
\begin{align}
y &= ax + b \\
y - q &= -\frac 1 a(x-p)\end{align}
to get $x = \frac{p - ab + aq}{a^2 + 1},\ y = \frac{b + ap + a^2 q}{a^2 + 1}$ and if we plug this into $(x-p)^2+(y-q)^2 = h$ we get $$\boxed{h = \frac{(b+ap-q)^2}{1+a^2}}$$
3. $h$ is the square of distance of the center to the line
We find this approach in answers by Senex Ægypti Parvi and pajonk.
We start with line $y = ax + b$ and point $(p,q)$. Square of (Euclidean) distance of any point on the line $y = ax + b$ to the point $(p,q)$ is given by
\begin{align}
(x-p)^2 + (y-q)^2 &= (x-p)^2 +(ax+b-q)^2 \\ &= (1 + a^2) x^2 + 2( a b - p - a q) x+ (b^2 + p^2 - 2 b q + q^2)\end{align} and $h$ is given by minimal distance.
Now, remember that any quadratic $\alpha x^2 + \beta x + \gamma$ can be written in the form $$\alpha x^2 + \beta x + \gamma = \alpha(x + \frac\beta{
2 \alpha})^2 + \gamma - \frac{\beta^2} {4 \alpha}\geq \gamma - \frac{\beta^2}{4\alpha}$$ and thus the minimum of quadratic polynomial ($\alpha > 0$) is given with $\gamma - \frac{\beta^2}{4\alpha}$ and if we set $\alpha = 1+ a^2,\ \beta = 2( a b - p - a q),\ \gamma = b^2 + p^2 - 2 b q + q^2$ we get $$\boxed{h = \gamma - \frac{\beta^2}{4\alpha} = \frac{(b+ap-q)^2}{1+a^2}}$$
4. Tangent is a line that intersects circle at exactly one point
I personally like this approach the best because it beautifully intertwines geometry and algebra. This is suggested in answers by cosmin, Emilio Novati and DeepSea.
We want to find intersections of the line $y = ax + b$ and circle $(x-p)^2+(y-q)^2 = h$ so we solve the appropriate system
\begin{align}
(x-p)^2+(y-q)^2 &= h\\
y &= ax + b
\end{align}
and by substituting $y = ax + b$ in the first equation we get $$(x-p)^2+(ax + b - q)^2 - h = 0$$ and analogously to the previous case we get quadratic equation $$ (1 + a^2) x^2 + 2( a b - p - a q) x+ (b^2 + p^2 - 2 b q + q^2 - h) = 0 $$
It is well known that there are 3 general cases for the solutions of quadratic equation (with real coefficients) depending on the discriminant $D$:
\begin{array}{ c | c | c }
\textbf{discriminant} & \textbf{algebraic interpretation} & \textbf{geometric interpretation}\\
\hline
D>0 & \text{there are 2 real solution} & \text{line intersects the circle at 2 points} \\
D=0 & \text{there is a double real solution} & \text{line intersects the circle at 1 point, i.e. it is a tangent} \\
D<0 & \text{there are 2 complex solutions} & \text{line does not intersect the circle }
\end{array}
We are, thus, interested in case 2), i.e. we want the discriminant of our quadratic equation to be $0$, i.e.:
$$(2 a b - 2 p - 2 a q)^2 = 4 (1 + a^2) (-h + b^2 + p^2 - 2 b q + q^2) \iff \boxed{h = \frac{(b+ap-q)^2}{1+a^2}}$$
TL;DR
What we actually calculated is a well known formula from high school:
For a line $Ax + By + C = 0$, distance of point $(p,q)$ from the line
is given by formula $$d = \frac{| Ap + Bq + C|}{\sqrt{A^2 + B^2}}$$
How to get the formula from our calculation of $h$? Well:
$$Ax + By + C = 0\implies y = -\frac AB x -\frac CB\implies h = \frac{(-C-Ap-Bq)^2}{B^2(1+\frac{A^2}{B^2})} = \frac{(Ap+Bq+C)^2}{A^2 + B^2}$$ Thus, probably the quickest way to solve your problem is
5. High school mathematics
Transform $y = 3x + 2$ into $3x - y + 2 = 0$, and put it in the above formula with $p = q = 0$ to get $$h = \frac{(3\cdot 0 - 1\cdot 0 + 2)^2}{3^2 + (-1)^2} = \frac 2 5$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Difference between $\cos(x)$= positive and $\cos(x)$= negative I'm confusing myself here so need some clarification. If I was to work out the solutions to $\cos(x) = 1/\sqrt2$ , I know the solutions would be $\pi/4, 2\pi - \pi/4, 2\pi + \pi/4...$ etc.
What is the difference when working out the solution to $\cos(x) = -1/\sqrt2$. What values need to be added to what, and what are the solutions significance to the solutions when the value is positive (above) ?
| If $\cos x > 0$, then the terminal side of angle $x$ lies in the first quadrant, the fourth quadrant, or the positive $x$-axis. If $\cos x < 0$, then the terminal side of angle $x$ lies in the second quadrant, the third quadrant, or the negative $x$-axis.
One solution of the equation $\cos x = -\dfrac{1}{\sqrt{2}}$ is
$$x = \arccos\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4}$$
By symmetry, $\cos(-x) = \cos x$, so another solution is
$$x = -\frac{3\pi}{4}$$
Any angle coterminal with one of those angles will have the same cosine. Hence, the general solution is
$$x = \pm \frac{3\pi}{4} + 2k\pi, k \in \mathbb{Z}$$
As the diagram indicates,
\begin{align*}
\cos(\pi + \theta) & = -\cos\theta\\
\cos(\pi - \theta) & = -\cos\theta
\end{align*}
which you can verify using the sum and difference of angle formulas for cosine. You know that one solution of $\cos\theta = \dfrac{1}{\sqrt{2}}$ is
$$\theta = \arccos\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$$
so the solutions to the equation $\cos x = -\dfrac{1}{\sqrt{2}}$ include
\begin{align*}
x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\\
x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}
\end{align*}
Observe that
$$x = \frac{3\pi}{4} = \arccos\left(-\frac{1}{\sqrt{2}}\right)$$
and that
$$x = \frac{5\pi}{4} = -\frac{3\pi}{4} + 2\pi$$
| {
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Proving $\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$ I've been trying to prove
$$\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$$
I've tried perturbation and inversion but still nothing. I've even tried expanding the sum to try and find some pattern that could relate this to the harmonic series but this just seems surreal. I can't find any logical reasoning behind this identity. I don't understand Wikipedia's proof. Is there any proof that doesn't require the integral transform?
| Suppose we seek to verify that
$$\sum_{k=1}^n {n\choose k} (-1)^{k+1} \frac{1}{k} = H_n.$$
Now observe that
$$\frac{1}{k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} \log\frac{1}{1-z} \; dz$$
where we take $\epsilon < 1$ so that the logarithmic term is analytic.
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z} \log\frac{1}{1-z}
\sum_{k=1}^n {n\choose k} (-1)^{k+1} \frac{1}{z^k}
\; dz$$
The term for $k=0$ yields
$$-\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z} \log\frac{1}{1-z} = 0$$
and we may lower $k$ to zero, getting for the sum
$$-\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z} \log\frac{1}{1-z}
\sum_{k=0}^n {n\choose k} (-1)^{k} \frac{1}{z^k}
\; dz
\\ = -\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z}
\left(1-\frac{1}{z}\right)^n
\log\frac{1}{1-z}
\; dz
\\ = -\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
(z-1)^n \log\frac{1}{1-z} \; dz.$$
Now put $$\frac{z}{z-1} = w
\quad\text{so that}\quad
z = -\frac{w}{1-w}
\\ \text{and}\quad
dz = - \frac{1}{(1-w)^2} dw
\quad\text{and}\quad
\frac{1}{1-z} = 1-w.$$
to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
(z-1)^{n+1} \frac{1}{1-z} \log\frac{1}{1-z} \; dz
\\ = - \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}}
(1-w) \log(1-w) \frac{1}{(1-w)^2} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}}
\frac{1}{1-w} \log\frac{1}{1-w} \; dw.$$
This evaluates to $$H_n$$ by inspection since
$$\log\frac{1}{1-z} = \sum_{q\ge 1} \frac{z^q}{q}$$
as seen earlier.
| {
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"timestamp": "2023-03-29T00:00:00",
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Induction Sum of all odd Numbers Show that
$\sum_{k=1}^{n}(2k-1)=n^2$
Beginning: n=1
$\sum_{k=1}^{1}(2k-1)=(2*1-1)=1=1^2$
Let $\sum_{k=1}^{n}(2k-1)=n^2$ be true, then for n=p+1
$\sum_{k=1}^{p+1}(2k-1)=(p+1)^2$ has to be true too.
$\sum_{k=1}^{p+1}(2k-1)=1+3+...+(2(p+1)-3)+(2(p+1)-1)$
$=1+3+...+(2p-1)+(2p+1)$
$=1+3+...+(2p-3)+(2p-1)+(2p+1)$
(Using our assumption that $1+3+...+(2p-3)+(2p-1)=p^2$)
$=p^2+2p+1$
$=(p+1)^2$ which is exactly what we wanted to show.
Is this ok?
| Induction step:
Assume $\sum_{k=1}^n$ $(2k - 1)$ = $n^2$. Then $\sum_{k=1}^{n+1}$ $(2k - 1)$ = $\sum_{k=1}^n$ $(2k - 1)$ + [$(2(n + 1) - 1)$] = $n^2 + [2(n + 1) - 1]$ = $n^2 + 2n + 1$ = $(n + 1)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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On the proof that $\sum\limits_{k=0}^{n-1}\frac {a^k}{(1+a^k x) (1+ a^{k+1}x)}=\frac 1 {1-a} \left( \frac 1 {1+x} -\frac {a^n}{ 1+a^n x }\right)$ Question:-
Find the sum to $n$ terms of the following series
$$\frac{1}{(1+x) (1+ax)} + \frac{a}{(1+ax) (1+a^2 x)} + \frac{a^2}{(1+a^2 x) (1+a^3 x)} + \cdots$$
My solution:-
First of all I found out the general term of the series
$$t_n=\frac {a^n}{(1+a^n x) (1+ a^{n+1}x)} $$
Now to find out the partial fraction, I multiplied and divided the $t_n$ by $(a-1) x$. The partial fraction of the $t_n$ looks like this
$$t_n =\frac {a^n (a-1) x }{(( a-1) x) [ (1+a^n x) (1+ a^{n+1} x)] } =\frac{( 1+ a^{n+1} x ) -( 1+ a^n x )}{((a-1)x) [(1+ a^n x) (1+ a^{n+1}x)]} =\frac 1 {(a-1) x} \left(\frac{1}{1+ a^n x} -\frac {1}{1+a^{n+1} x}\right) $$
As, $n \ge 0$,son on summing till all the $n$ terms the diagonal terms start to cancel out and we are left with
$$ S_n=\sum_{n=0}^{n-1} t_n = \sum_{n=0}^{n-1}{ \frac 1 {(a-1) x } \left( \frac 1 { 1+ a^n x } -\frac 1 { 1+ a^{n+1} x } \right) } = \left( \frac 1 {(a-1) x } \right) \left( \frac 1 {1+x} -\frac 1 { 1+a^n x} \right) $$
The answer given in the book:-
$$\left( \frac 1 {1-a} \right) \left( \frac 1 {1+x} -\frac {a^n}{ 1+a^n x }\right)$$
Now, where did the $x$ in the $\dfrac{1}{(a-1)x}$ go and from where did the $a^{n}$ come in $\displaystyle \dfrac { { a }^{ n } }{ 1+{ a }^{ n }x }$
| Your result is the same as the answer from the book. Proof below :
$$f(a,x)=\left( \frac 1 {1-a} \right) \left( \frac 1 {1+x} -\frac {a^n}{ 1+a^n x }\right)$$
$$f(a,x)=\left( \frac 1 {(1-a)x} \right) \left( \frac x {1+x} -\frac {a^n x}{ 1+a^n x }\right)$$
$\frac x {1+x}=1-\frac 1 {1+x}$
$\frac {a^n x}{ 1+a^n x } = 1-\frac {1}{ 1+a^n x }$
$$f(a,x)=\left( \frac 1 {(1-a)x} \right) \left( \left(1-\frac 1 {1+x}\right) -\left(1-\frac {1}{ 1+a^n x }\right) \right)$$
$$f(a,x)=\left( \frac 1 {(1-a)x} \right) \left(-\frac 1 {1+x} +\frac {1}{ 1+a^n x } \right) = \left( \frac 1 {(a-1)x} \right) \left(\frac 1 {1+x} -\frac {1}{ 1+a^n x } \right)$$
$$\left( \frac 1 {1-a} \right) \left( \frac 1 {1+x} -\frac {a^n}{ 1+a^n x }\right)=\left( \frac 1 {(a-1)x} \right) \left(\frac 1 {1+x} -\frac {1}{ 1+a^n x } \right)$$
Remark :
$\sum_{n=0}^{n-1}$ is a bad notation because $n$ is a constant, not a variable. Better use two different symbols.
COMMENT :
Sorry, I just see the answer of Did in the comment section. It's the same proof as mine but more concise and earlier.
| {
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"timestamp": "2023-03-29T00:00:00",
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Eliminate $\theta$
Eliminate $\theta$ in
$$\sin \theta + \mbox{cosec} \, \theta = m$$
$$\sec \theta - \cos \theta = n$$
My approach-
I multiplied the first equation by $\sin \theta$ and the second equation by $\cos \theta$ but it doesn't give me the desired answer..
| We have $\dfrac{1-\cos^2\theta}{\cos\theta}=n\implies\sin^2\theta=n\cos\theta\ \ \ \ (1)$
$\iff\cos^2\theta+n\cos\theta-1=0\ \ \ \ (2)$
and multiply $\sin\theta$ with the both sides of $\sin \theta +\csc\theta = m$, $$\sin^2\theta=m\sin\theta-1\ \ \ \ (3)$$
$(1),(3)\implies n\cos\theta+1=m\sin\theta$
Squaring & on rearrangement, $(m^2+n^2)\cos^2\theta+2n\cos\theta+1-m^2=0\ \ \ \ (4)$
Solve $(2),(4)$ for $\cos^2\theta,\cos\theta$ and utilize the fact $\cos^2\theta=(\cos\theta)^2$
| {
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Find a polynomial with integer coefficients which has a global minimum equal to (a)$- \sqrt{2}$, (b)$\sqrt{2}$ Find a polynomial with integer coefficients which has a global minimum equal to (a)$- \sqrt{2}$, (b)$\sqrt{2}$.
It it a high-school math contest problem. The answer is given:
$$(a) ~~~~~~~P(x)=N(2x^2-1)^2-2x^3+3x,~~~~N>1$$
$$(b) ~~~~~~~Q(x)=P(x^2)=N(2x^4-1)^2-2x^6+3x^2$$
No solution is given.
My initial approach was (before I saw the answer) to try finding the minimun of a general polynomial starting with degree $3$ and then try to match the coefficients to a given value of the minimum. However, the answer clearly shows that we require degrees $4$ and $8$ respectively, which makes my method of solution practically impossilbe.
It's easy enough to show that the answers are true, for example the case (a):
$$P'(x)=8N(2x^2-1)x-6x^2+3=(2x^2-1)(8Nx-2)=0$$
$$x_1=\frac{1}{\sqrt{2}},~~~~~~~x_2=-\frac{1}{\sqrt{2}},~~~~~~~x_3=\frac{1}{4N}$$
$$P''(x)=8N(2x^2-1)+4x(8Nx-2)$$
$$P''(x_1)=2\sqrt{2}(4N\sqrt{2}-2)>0$$
$$P''(x_2)=-2\sqrt{2}(-4N\sqrt{2}-2)>0$$
$$P''(x_3)=\frac{1}{N}-8N<0$$
So, $x_1$ and $x_2$ are minimum points, $x_3$ is a maximum point.
$$P(x_1)=\frac{3\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=\sqrt{2}$$
$$P(x_2)=-\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=-\sqrt{2}$$
The case (b) follows trivially, if we replace $x \to x^2$.
How is this problem supposed to be solved? How are we supposed to find the degree of $P(x)$ and match the coefficients? Is there some theorem about a minimal value of a polynomial which might help?
| For part (a), it can be easily deduced that $f$ does not have degree 2. Further, $f$ cannot have degree 3, as an odd degree polynomial has no global minimum. So assume
$$f(x) = sx^4 + bx^3 + cx^2 + dx + e$$
with $f(a) = -\sqrt{2}, f'(a) = 0$. As one last piece of guesswork, let $a = k\sqrt{2}$. Then we have
$$4sk^4 + 2bk^3\sqrt{2}+2ck^2+dk\sqrt{2}+e = -\sqrt{2}$$
$$8sk^3\sqrt{2}+6bk^2+2ck\sqrt{2}+d=0$$
from which we derive
$$4sk^4 + 2ck^2 + e = 0$$
$$2bk^3 + dk = -1$$
$$8sk^3 + 2ck = 0$$
$$6bk^2 + d = 0$$
Then $c = -4sk^2, e = 4sk^4, d = -1.5/k, b = 0.25/k^3$. Now we just need to find $k, s$ that makes all of these integers. $k = 1/2, s = 4$ does the trick. Then we have
$$f(x) = 4x^4 + 2x^3 - 4x^2 - 3x + 1$$
Which has all of our desired properties.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive
$$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$
$$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$
Is the proof valid?
| Alternatively, apply AM-GM inequality with $a = \sqrt{x}, b = \sqrt{y}$:
$\dfrac{a^2}{b} + b \geq 2\sqrt{\dfrac{a^2}{b}\cdot b} = 2a, \dfrac{b^2}{a} + a \geq 2\sqrt{\dfrac{b^2}{a}\cdot a} = 2b$. Adding these $2$ inequalities to get the answer.
Generalization: The same technique could be used to prove: If $x, y, z > 0 \implies \dfrac{x^2}{y} + \dfrac{y^2}{z} + \dfrac{z^2}{x} \geq x+y+z$.
| {
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Is difference of two consecutive sums of consecutive integers (of the same length) always square? I am an amateur who has been pondering the following question. If there is a name for this or more information about anyone who has postulated this before, I would be interested about reading up on it. Thanks.
If $x$ is the sum of $y$ integers, and $z$ is the sum of the next $y$ integers, then is it always true that $z$ minus $x$ equals $y$ squared? Perhaps only starting at one.
For example:
$y = 3$
$x = (1+2+3) = 6$
$z = (4+5+6) = 15$
$z - x = 9 = y^2$
Brad
| Consider the $n$th triangular number $T(n)$, which is the sum of $n$ natural numbers $1$ to $n$, and which has the closed form $T(n)=\tfrac{1}{2}n(n+1)=\binom{n+1}{2}$ where $\binom{n+1}{2}$ is a binomial coefficient (see https://en.wikipedia.org/wiki/Triangular_number). Then we can look at the differences of triangular numbers to prove the hypotheses and from which formulate a binomial identity that always gives $n^2$.
Starting at integer $k\ge1$ and adding $n$ consecutive integers gives:
\begin{align}
k+(k+1)+(k+2)+\dotsb+(k+n-1)&=\\
\tfrac{1}{2}(k+n-1)(k+n)-\tfrac{1}{2}(k-1)k&=\tfrac{1}{2}(2kn+n^2-n)\qquad(A)
\end{align}
Next starting at integer $k+n$ and adding $n$ consecutive integers gives:
\begin{align}
(k+n)+(k+n+1)+(k+n+2)+\dotsb+(k+2n-1)&=\\
\tfrac{1}{2}(k+2n-1)(k+2n)-\tfrac{1}{2}(k+n-1)(k+n)&=\tfrac{1}{2}(2kn+3n^2-n)\qquad (B)
\end{align}
Then equation (B) minus equation (A) gives $n^2$.
Written out as a sum of triangular numbers gives us a sum of binomial coefficients which we can explicitly write out and simplify to give us $n^2$ showing the equivalence between the two (and maybe inspiring a binomial proof):
\begin{align}
\big(T(k+2n-1)-T(k+n-1)\big)-\big(T(k+n-1)-T(k-1)\big)&=\\
T(k+2n-1)-2T(k+n-1)+T(k-1)&=\\
\binom{k+2n}{2}-2\binom{k+n}{2}+\binom{k}{2}&=\\
\frac{(k+2n)!}{2!(k+2n-2)!}-2\frac{(k+n)!}{2!(k+n-2)!}+\frac{k!}{2!(k-2)!}&=\\
\tfrac{1}{2}\big((k+2n)(k+2n-1)-2(k+n)(k+n-1)+k(k-1)\big)&=\\
\tfrac{1}{2}\cdot 2n^2&=n^2.
\end{align}
Hence we see that whatever value of $k\ge1$ we take as our starting point is irrelevant to the final value of $n^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all $a, b, c, d$ Find all $a, b, c, d$ satisfying
$$\frac{x^4+ax^3+bx^2-8x+4}{(x^2+cx+d)^2} = 1$$
My answers are:
$$\begin{cases}
d=2\\
d=-2
\end{cases}
\quad \begin{cases}
a=4\\
a=-4
\end{cases} \quad
\begin{cases}
c=2\\
c=-2
\end{cases} \quad
\begin{cases}
b=8\\
b=0
\end{cases}$$
Is it right?
| You can write it as $x^4+ax^3+bx^2-8x+4=(x^2+cx+d)^2$, which you want to be true for all $x$. This requires that the coefficients of each power match. You should expand the square on the right and match the coefficients. This will give you each of $a,b,c,d$
| {
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Finding the 11th term of $\frac{7x^2+2x+6}{(x+2)(x^2+1)}$ Taylor expansion
Given $f(x)=\frac{7x^2+2x+6}{(x+2)(x^2+1)}$, find $f^{(11)}(0).$
I understood that we first need to use partial fractions to simplify the function.
$$\frac{7x^2+2x+6}{(x+2)(x^2+1)}=\frac{A}{(x+2)}+\frac{Cx+B}{(x^2+1)}=\frac{6}{(x+2)}+\frac{x}{(x^2+1)}$$
That can be written as $$\frac{6}{(x+2)}+\frac{x}{(x^2+1)}=\frac{1}{2}*\frac{6}{(1-(-\frac{x}{2}))}+\frac{x}{(1-(-x^2))}=\frac{3}{(1-(-\frac{x}{2}))}+\frac{x}{(1-(-x^2))}$$
The taylor expansion of $\frac{1}{1-x}=1+x+x^2+...+x^{11}+o(x^{12})$
So we have $$\frac{3}{(1-(-\frac{x}{2}))}+\frac{x}{(1-(-x^2))}$$
Taking the 11th derivative and plugin x=0 will give $$\frac{3}{\frac{11!}{2^11}}+\frac{x}{0}$$ and that can not be, where did I get it wrong?
| We have
$$ f(x)=\frac{7x^2+2x+6}{(x+2)(x^2+1)}=\frac{6}{x+2}+\frac{x}{1+x^2}\tag{1}$$
through the residue theorem, then:
$$ f(x) = 3\sum_{n\geq 0}\frac{(-1)^n}{2^n}x^{n}+\sum_{n\geq 0}(-1)^n x^{2n+1} \tag{2}$$
by exploiting geometric series. It follows that:
$$ [x^{11}]\,f(x) = -\frac{3}{2^{11}}-1 \tag{3} $$
so:
$$ f^{(11)}(0) = \color{red}{- 11!\cdot\left(1+\frac{3}{2^{11}}\right)}.\tag{4}$$
| {
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chain rule for multi-variables The monthly demand for the Instant Pie Maker is given by
$$D(x,y)= \frac1{125}xe^{xy/1000} \text{ units}$$
where $x$ dollars are spent on infomercials and $y$ dollars are spent on in-person demonstrations. If $t$ months from now $$x=20+t^{2/3}$$ dollars are spent on infomercials and $$y=t\ln(1+t)$$ dollars are spent on demonstrations, at approximately what rate will the demand be changing with respect to time $8$ months from now?
(Round your answer to $3$ decimal places).
I started this problem by finding $dz/dx=$ $$\frac{e^{xy/1000}(xy+1000)}{125000}$$
and multiplying it by $dx/dt$
$$\frac{2}{3t^{1/3}}$$
and repeated for $y$ getting $$\frac{x^2 e^{xy/1000}}{125000}\ln(t+1)+\frac{t}{t+1}$$
finally I plugged in $$t=8, x=24, y=8\ln 9$$
and I got $.526$ which is wrong. can anyone help me see where I went wrong or what i might be missing here?
| If I understood your problem correctly, then you have to first plug $x = 20 + t^{\frac{2}{3}}$ and $y = t\ln(1 + t)$ into $D(x,y)$ and then calculate it's derivative with respect to $t$ at the point $t = 8$. Here is a possible solution:
First we define
\begin{equation}
\begin{split}
\tilde D(t) := D(20 + t^{\frac{2}{3}}, t\ln(1+t)) &= \frac{1}{125}(20 + t^{\frac{2}{3}})\text{e}^\frac{(20t + t^{\frac{5}{3}})\ln(1 + t)}{1000}\\ &= \frac{1}{125}(20\text{e}^\frac{(20t + t^{\frac{5}{3}})\ln(1 + t)}{1000} + t^{\frac{2}{3}}\text{e}^\frac{(20t + t^{\frac{5}{3}})\ln(1 + t)}{1000})
\end{split}
\end{equation}
Next we want to calculate it's derivative. To keep things simple, we set
\begin{equation}
g(t) := \text{e}^\frac{(20t + t^{\frac{5}{3}})\ln(1 + t)}{1000}
\end{equation}
Because
\begin{equation}
\frac{\text{d}}{\text{d}t} \frac{(20t + t^{\frac{5}{3}})\ln(1 + t)}{1000} = \frac{1}{1000}((20 + \frac{5}{3}t^{\frac{2}{3}})\ln(1 + t) + \frac{20t + t^{\frac{5}{3}}}{1+t})
\end{equation}
we have
\begin{equation}
g'(t) =\frac{1}{1000}((20 + \frac{5}{3}t^{\frac{2}{3}})\ln(1 + t) + \frac{20t + t^{\frac{5}{3}}}{1+t})g(t)
\end{equation}
Finally, we get
\begin{equation}
\tilde D'(t) = \frac{1}{125}(20g'(t) + \frac{2}{3}t^{-\frac{1}{3}}g(t) + t^{\frac{2}{3}}g'(t))= \frac{1}{125}((20 + t^{\frac{2}{3}})g'(t) + \frac{2}{3}t^{-\frac{1}{3}}g(t))
\end{equation}
All we have to do now is to evaluate $\tilde D'(t)$ at $t = 8$, and according to Mathematica this is
\begin{equation}
\tilde D'(8) = 0.0274655 \approx 0.027
\end{equation}
| {
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where is my mistake with this equation ot hyperbolic problem? Let parabola $\Gamma_{1}:$$y^2=4x$,and hyperbolic curve $\Gamma_{2}$: $x^2-y^2=1$.it is well known this two is symmetric.so the two point $A$ and $B$ about $x$ axial symmetry,or mean $x_{A}=x_{B}$.see figure,
but we use
$$\begin{cases}
y^2=4x\\
x^2-y^2=1
\end{cases}$$
we have
$$x^2-4x-1=0\Longrightarrow x=2+\sqrt{5},2-\sqrt{5}$$ so the two point $\color{red}{x_{A}=2+\sqrt{5},x_{B}=2-\sqrt{5}}?$,then
contradiction it is well known$x_{A}=x_{B}$,where is mistake?and the second root $\color{blue}{x=2-\sqrt{5}}$ where is from?or where is point to this figure
Thanks help
| The confusion arises from ignoring an implicit condition of the first equation, $y^2 = 4x$. Since $y^2 \geq 0$, we have the condition $x \geq 0$.
Thus of the two solutions for $x$ you have, the positive $x = 2 + \sqrt{5}$ is the only solution.
We see that there are two values of $y$ that satisfy the equations with the $x$ value above, and this agrees with the graphical representation.
| {
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Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$
Prove that
$$\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$$
My attempt
\begin{align}
\text{LHS}&=\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{8\pi}7\\
&=-2\cos\frac{4\pi}7\cos\frac\pi7+2\cos^2\frac{4\pi}7-1\\
&=-2\cos\frac{4\pi}7\left(\cos\frac\pi7-\cos\frac{4\pi}7\right)-1
\end{align}
Now, please help me to complete the proof.
| $cos(2\pi/7)$+$cos(4\pi/7)$+$cos(8\pi/7)$
= $cos(2\pi/7)$+$cos(4\pi/7)$+$cos(6\pi/7)$ (angles add to give $2\pi$, thus one is $2\pi$ minus the other)
At this point, we'll make an observation
$cos(2\pi/7)$$sin(\pi/7)$ = $\frac{sin(3\pi/7) - sin(\pi/7)}{2}$ ..... (A)
$cos(4\pi/7)$$sin(\pi/7)$ = $\frac{sin(5\pi/7) - sin(3\pi/7)}{2}$ ..... (B)
$cos(6\pi/7)$$sin(\pi/7)$ = $\frac{sin(7\pi/7) - sin(5\pi/7)}{2}$ ..... (C)
Now, add (A), (B) and (C) to get
$sin(\pi/7)*(cos(2\pi/7)+cos(4\pi/7)+cos(6\pi/7))$ = $\frac{sin(7\pi/7) - sin(\pi/7)}{2}$ = -$sin(\pi/7)/2$
The $sin(\pi/7)$ cancels out from both sides to give you your answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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how do I integrate a modulus function. is it possible? I am solving a question in which after calculating, my instantaneous velocity is $f(t)=\frac{ab\sin(bt)}{\sqrt{2-2\cos(bt)}}$.
So I need to find distance in time $T$. For finding distance I need to take its modulus and integrate
$|f(t)|$dt.
So how do I find integral of
$|f(t)|$dt.
Any hints are appreciable.
Thanks in advance.
| \begin{align*}
\left| \frac{ab\sin bt}{\sqrt{2-2\cos bt}} \right| &=
\left|
\frac{2ab\sin \frac{bt}{2} \cos \frac{bt}{2}}{\sqrt{4\sin^{2} \frac{bt}{2}}}
\right| \\
&= \left| ab\cos \frac{bt}{2} \right| \\
s &= \int_{0}^{t} \left| ab\cos \frac{bt}{2} \right| dt \\
&= 2a E \left( \frac{bt}{2}, 1 \right)
\end{align*}
where $\displaystyle E(\theta,k)=\int_{0}^{\theta} \sqrt{1-k^{2} \sin^{2} \phi} \, d\phi$ is the elliptic integral of the second kind.
In this case, $$E \left( \frac{bt}{2},1 \right)=
2\left \lfloor \frac{bt}{2\pi}+\frac{1}{2} \right \rfloor+
(-1)^{\left \lfloor \frac{bt}{2\pi}+\frac{1}{2} \right \rfloor} \sin \frac{bt}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $\tan x $ from $(a+1)\cos x + (a-1)\sin x=2a+1$? How do I find $\tan x$ from this equation?
$$(a+1)\cos x + (a-1)\sin x=2a+1$$
Thanks for any help!!
| Expressing the trigonometric functions $\sin(x)$ and $\cos(x)$ as a function of $\tan(\frac{x}{2})$, we get:
$\tan(\frac{x}{2})=+\frac{a-1+\sqrt{1-4a-2a^2}}{3a+2}$,
$\tan(\frac{x}{2})=-\frac{-a+1+\sqrt{1-4a-2a^2}}{3a+2}$.
Applying the tangent duplication formula:
$\tan(x)=\tan(\frac{x}{2}+\frac{x}{2})=\frac {\tan (\frac{x}{2})+\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})\tan(\frac{x}{2})}$
$\tan(x)=\frac{a^2-1+(2a+1)\sqrt{1-4a-2a^{2}}}{3a(a+2)}$,
$\tan(x)=-\frac{-a^2+1+(2a+1)\sqrt{1-4a-2a^{2}}}{3a(a+2)}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Applying the Cartesian Coordinate Find the surface area of the portion of $2x+y+z=8$ in the first octant.
I know that $f(x,y)=8-y-2x$, $x=0$ to $4$ and $y= 0$ to $8-2x$, but I'm having trouble solving the problem in Cartesian form. The answer is supposedly $16\sqrt6$
| This is an analytic proof:
$$\int_0^4\int_0^{8-2x} \sqrt{(-1)^2+(-2)^2+1} \ dydx=\sqrt{6}\int_0^4 8-2x dx=16\sqrt6$$
This is a geometric proof: the surface is a triangle, whose three vertices lie on the three axes. The three vertices have the following coordinates: $(4,0,0);(0,8,0);(0,0,8)$ which yields the following lengths for the sides: $$a=4\sqrt{5}\quad b=4\sqrt{5}\quad c=8\sqrt{2}$$
Applying Heron's formula yields:$$s=\frac{a+b+c}2=4\sqrt5+4\sqrt2$$
$$A^2=s(s-a)(s-b)(s-c)=(4\sqrt5+4\sqrt2)\times(4\sqrt5+4\sqrt2-4\sqrt5)^2\times(4\sqrt5-4\sqrt2)$$
$$A^2=16\times16\times(5-2)\times2$$
$$A=16\sqrt{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how to partial fraction $\frac{1}{(x+1)^2}$ I need to integrate $\frac{1}{(x^2+2x+1)}$, so I need to use partial fraction as the polynomial can be factored as $\frac{1}{(x+1)^2}$. This is what I've tried:
$$\frac{A}{(x+1)} + \frac{B}{(x+1)^2}$$
$$A\cdot(x+1)^2 + B\cdot(x+1)$$
$$Ax^2+2Ax+A+Bx+B$$
$$Ax^2+(2A+B)x+(A+B)$$
So,
$$A=0$$
$$2A+B=0$$
$$A+B=1$$
but that does not make sense, because if $A=0$, then $2A+B$ can't be zero, could you please tell me what's the problem?
| Just lookoing at the line $\frac{A}{(x+1)} + \frac{B}{(x+1)^2} = \frac{1}{(x+1)^2}$ we clearly see that $A = 0, B = 1$. As for your mistake, you multiplied by $(x+1)$ one time too many when removing the denominators.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$ Integrate using integrating by parts :
$$\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$
My attempt :
$$I=\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$
$$I=\int \frac{e^x}{(1+x^2)^2}\cdot(1-x)^2dx$$
$$I=\frac{e^x}{(1+x^2)^2}\cdot\frac{(1-x)^3}{-3}-\int \frac{(1-x)^3}{-3}\cdot\frac{(1+x^2)^2e^x-e^x\cdot2(1+x^2)\cdot2x}{(1+x^2)^4}dx$$
Now it is little bit confusing. How can I simplify this ? Is there another method to do it differently ?
| Hint
$$\frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}=\frac{e^x}{1+x^2}-\frac{2xe^x}{(1+x^2)^2}$$
Indeed
$$(1-x)^2=(1+x^2)-2x$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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} |
If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, and $k=1$, does it follow that $\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$? Let $\sigma=\sigma_{1}$ be the classical sum-of-divisors function.
If $N = q^k n^2$ is an odd perfect number with Euler prime $q$ (i.e., $q$ satisfies $q \equiv k \equiv 1 \pmod 4$, and $\gcd(q,n)=1$), and $k=1$ holds, does it follow that $$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}?$$
Note that, since the Euler prime $q$ satisfies $q \equiv 1 \pmod 4$, then
$$q \geq 5 \implies \frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q} \geq 2 - \frac{1}{3} = \frac{5}{3},$$
which agrees with what is known about $N = q^k n^2$ when $k=1$.
In fact, when $k=1$,
$$\frac{\sigma(q^k)}{q^k} = \frac{\sigma(q)}{q} = \frac{q+1}{q} = 1 + \frac{1}{q} \leq 1 + \frac{1}{5} = \frac{6}{5}.$$
$\sigma(n^2)/n^2 \geq 5/3$ then follows from
$$\left(\frac{\sigma(q^k)}{q^k}\right)\cdot\left(\frac{\sigma(n^2)}{n^2}\right)=\left(\frac{\sigma(q)}{q}\right)\cdot\left(\frac{\sigma(n^2)}{n^2}\right)=2.$$
| If $N$ is perfect then $2N = 2q^k n^2 = \sigma(q^k n^2) = \sigma(q^k)\sigma(n^2) = \frac{q^{k+1}-1}{q-1} \sigma(n^2)$, which gives
\begin{equation}
\frac{\sigma(n^2)}{n^2} = \frac{2q^k(q-1)}{q^{k+1}-1} = 2-2\frac{q^k-1}{q^{k+1}-1}
\end{equation}
This being $\ge 2-\frac{5}{3q}$ works out to being equivalent to $q^{k+1}-6q+5 \le 0$, which can't work if $k > 1$ for any prime.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $n$ is a positive integer, then $(-2^n)^{-2} + (2^{-n})^2 = 2^{-2n+1}$ I'm not sure why $$(-2^n)^{-2} + (2^{-n})^2=2^{-2n+1}$$
I have been going over this equation for a while now, noticing, and have successfully got quite far in the equation, finding that
$$ (-2^n)^{-2} + (2^{-n})^2 \implies \frac{1}{(-2^n)^2} + \frac{1}{(2^n)^2} $$
which I think then becomes $ \dfrac{2}{2^{2n}}$
But then I get stuck.
| $(-2^n)^{-2} + (2^{-n})^2=$
$\frac 1{(-2^n)^2} + (\frac{1}{2^n})^2 = $
$\frac 1{(2^n)^2} + \frac 1 {(2^n)^2} = $
$\frac 1{2^{2n}} + \frac 1{2^{2n}} = $
$\frac 2{2^{2n}} = $
$\frac 1{2^{2n-1}} = $
$2^{-(2n-1)} = $
$2^{-2n+1}$
Or if you are more comfortable with the rules of exponents:
$(-a)^{b_{\text{even}}} = a^b$ so
$(-2^n)^{-2} + (2^{-n})^2 = (2^n)^{-2} + (2^{-n})^2 = 2^{-2n} + 2^{-2n} = 2*2^{-2n} = 2^{-2n + 1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $5\sqrt [ x ]{ 125 } =\sqrt [ x ]{ { 5 }^{ -1 } } $, then $x$ equals $-4$ For the equation
$$5\sqrt [ x ]{ 125 } =\sqrt [ x ]{ { 5 }^{ -1 } } $$
$x$ is equal to $-4$, but I'm not sure why.
I've taken the right side of the equation ${ \left( \frac { 1 }{ 5 } \right) }^{ \frac { 1 }{ x } }$ and converted it to $5^{-(1/x)}$, but haven't been able to perform the mathematical gymnastics to prove this.
| $$5\cdot125^{\frac{1}{x}}=\left(\frac{1}{5}\right)^{\frac{1}{x}}\Longleftrightarrow5\cdot125^{\frac{1}{x}}=\frac{1^{\frac{1}{x}}}{5^{\frac{1}{x}}}\Longleftrightarrow5\cdot125^{\frac{1}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow$$
Use (real solution) $5^y=125\Longleftrightarrow y=\log_5(125)=3$:
$$5\cdot(5^3)^{\frac{1}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow5\cdot5^{\frac{3}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow5^1\cdot5^{\frac{3}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow$$
$$5^{1+\frac{3}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow5^{1+\frac{3}{x}}\cdot5^{\frac{1}{x}}=1\Longleftrightarrow5^{1+\frac{1}{x}+\frac{3}{x}}=1\Longleftrightarrow$$
$$5^{1+\frac{4}{x}}=1\Longleftrightarrow\log_5\left(5^{1+\frac{4}{x}}\right)=\log_5(1)\Longleftrightarrow1+\frac{4}{x}=0\Longleftrightarrow x=-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding total elements in a series I have a confusion.
How many terms in the following series are needed to make a sun greater than 5/2?
1/2 + 1/3 + 1/4 + 1/5 + 1/6 + .........
Is there any shortcut technique to find this out?
| Observe that, for $k\geq 1$,
\begin{align*}
\sum_{n=2}^{2^k}\frac{1}{n}&=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\cdots+\frac{1}{2^k}\\
&>\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\cdots+\frac{1}{2^k} \\
&=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots+\frac{1}{2} \qquad (\textrm{k times}) \\
&=\frac{k}{2}.
\end{align*}
That is,
$$ \sum_{n=2}^{2^k}\frac{1}{n}>\frac{k}{2} $$
Using this (crude) estimate, we need $2^5=32$ terms.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Monotonicity of the sequence $(a_n)$, where $a_n=\left ( 1+\frac{1}{n} \right )^n$ Define $a_n=\left ( 1+\frac{1}{n} \right )^n$ for $n\geq 1$. I want to show that it is increasing. First, we have
$$\frac{a_{n+1}}{a_n}=\left ( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right )^n\left ( 1+\frac{1}{n+1} \right )=\left ( 1-\frac{1}{(n+1)^2} \right )^n\left ( \frac{n+2}{n+1} \right )$$
Using the Bernoulli inequality, we see that, for all $n\geq 1$,
$$\left ( 1-\frac{1}{(n+1)^2} \right )^n\geq 1-\frac{n}{(n+1)^2}=\frac{n^2+n+1}{n^2+2n+1}.$$
How do you then show that
$$\left ( 1-\frac{n}{(n+1)^2} \right )\left ( \frac{n+2}{n+1} \right )>1?$$
Edit: This is not a duplicate question; the question is not to show about the existence of the definition of Euler number $e$. The question is about showing that it is increasing, the way I have shown that I have been stuck with, not other ways. It seems that the question is too easy that I have been too tired to think at this late.
| $$\frac { a_{ n+1 } }{ a_{ n } } =\frac { \left( 1+\frac { 1 }{ n+1 } \right) }{ { \left( 1+\frac { 1 }{ n } \right) }^{ n } } ^{ n+1 }=\left( 1-\frac { 1 }{ (n+1)^{ 2 } } \right) ^{ n+1 }\frac { n+1 }{ n } >\left( 1-\frac { 1 }{ n+1 } \right) \frac { n+1 }{ n } =1$$
| {
"language": "en",
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"source": "stackexchange",
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} |
Solve the following using AM-GM inequality The least value of $a \in R$ for which $4ax^2 + \frac{1}{x} \ge 1 $for all $x \gt 0 $, is
Using AM-GM inequality
$$\frac{4ax^2 + \frac{1}{2x} + \frac{1}{2x}}{3} \ge \sqrt[3]{a}$$
$$4ax^2 + \frac{1}{x} \ge 3\sqrt[3]{a}$$
Now my question start from here .
Can I do that for least value of $a$, the value of $3\sqrt[3]{a} $ must greater than minimum value of $4ax^2 + \frac{1}{x}$
|
Can I do that for least value of $a$, the value of $3\sqrt[3]{a} $ must greater than minimum value of $4ax^2 + \frac{1}{x}$
I find this strange because "minimum value of $4ax^2 + \frac{1}{x}$" is $3\sqrt[3]{a}$. So I guess that you meant that "the value of $3\sqrt[3]{a} $ must greater than $1$". Then, it is correct. I'll write the details in the following.
As some comment, $a$ has to be positive. (setting $x=1$ gives $a\ge 0$, but for $a=0$, the inequality does not hold for $x=2$.)
You have
$$4ax^2+\frac 1x\ge 3\sqrt[3]{a}$$
The equality holds when $x=\frac{1}{2\sqrt[3]{a}}\ (\gt 0)$.
If $3\sqrt[3]{a}\ge 1$, i.e. $a\ge\frac{1}{27}$,
$$4ax^2+\frac 1x\ge 3\sqrt[3]{a}\ge 1$$
holds for all $x\gt 0$.
If $0\lt 3\sqrt[3]{a}\lt 1$, i.e. $0\lt a\lt \frac{1}{27}$,
$$4ax^2+\frac 1x\ge 1$$
does not hold for $x=\frac{1}{2\sqrt[3]{a}}$.
Therefore, the least value of $a$ is $\frac{1}{27}$.
Another way :
We can have
$$a\ge\frac{x-1}{4x^3}$$
Now consider the graph of $y=\frac{x-1}{4x^3}$ for $x\gt 0$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve $\sec (x) + \tan (x) = 4$ $$\sec{x}+\tan{x}=4$$
Find $x$ for $0<x<2\pi$.
Eventually I get $$\cos x=\frac{8}{17}$$
$$x=61.9^{\circ}$$
The answer I obtained is the only answer, another respective value of $x$ in $4$-th quadrant does not solve the equation, how does this happen? I have been facing the same problem every time I solved this kind of trigonometric equation.
| If $\sec x+ \tan x=4 \\
\implies \frac{1+\sin x}{\cos x}=4\\
\implies 1+\sin x=4\cos x$
That does not help! So let's square it both sides
$\implies 1+\sin^2x+2\sin x=16\cos^2x=16-16\sin^2x
\implies 17\sin^2x+2\sin x-15=0$
Assuming, $\sin x=y$, makes the above equation quadratic in y having 2 solutions. So,
$17y^2+17y-15y-15=0\\
\implies 17y(y+1)-15(y+1)=0\\
\implies (17y-15)(y+1)=0
$
So, $y=\frac{15}{17}$ and $y=-1$
So either $x =270^\circ$ or $x\approx 62^{\circ}$
EDIT: I looked at other answers and was wondering what solid reasoning is behind not accepting $x =270^\circ$.
This below.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 12,
"answer_id": 11
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Evaluating the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer
How to evaluate the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer?
For $k=1$, the series does not converge.
When $k=2$, I can prove that:
$$\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}=\frac{\pi}{3\sqrt{3}}$$
Usually, this can be proven by differentiating $\sum_{n=1}^\infty \frac{x^{2n}}{n^2 \binom{2n}{n}}=2(\arcsin{\frac{x}{2}})^2$, but I have an alternative proof.
Using the result of:
$$\int_0^\infty \frac{x^ndx}{(x+1)^{y+n+1}}=\frac{1}{y \binom{y+n}{n}} \tag1$$
, which can be easily proved.
I can substitute $y=n$ to obtain
$$\int_0^\infty \frac{x^ndx}{(x+1)^{2n+1}}=\frac{1}{n \binom{2n}{n}}$$
$$\sum_{n=1}^\infty\int_0^\infty \frac{x^ndx}{(x+1)^{2n+1}}=\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}$$
Therefore,
\begin{align}
\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}} & = \int_0^\infty \frac{xdx}{(x+1)(x^2+x+1)} \\
& = \lim_{L\to \infty} \frac{1}{2}\ln(x^2+x+1)-\ln(x+1)+\frac{\tan^{-1}(\frac{2x+1}{\sqrt{3}})}{\sqrt{3}}\large{|_0^L} \\
&= \frac{\pi}{3\sqrt{3}}
\end{align}
Now for $k=3$, I tried to substitute $y=2n$ into $(1)$:
$$\int_0^\infty \frac{x^ndx}{(x+1)^{3n+1}}=\frac{1}{2n \binom{3n}{n}}$$
$$\sum_{n=1}^\infty\int_0^\infty \frac{x^ndx}{(x+1)^{3n+1}}=\sum_{n=1}^\infty\frac{1}{2n \binom{3n}{n}}$$
So we can have
$$\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}=\int_0^\infty\frac{2xdx}{(x+1)(x^3+3x^2+2x+1)}$$
However, by partial fraction $$\frac{2xdx}{(x+1)(x^3+3x^2+2x+1)}=-\frac{2}{1+x}+\frac{2x^2+4x+2}{x^3+3x^2+2x+1}$$
The left part does not seem to converge.
Feeling frustrated, Wolfram Alpha plays its part. It spits out these results:
$$\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}=\frac{1}{3}{}_3F_2\left(\left.\begin{array}{c} 1,1,\frac{3}{2}\\ \frac{4}{3}, \frac{5}{3} \end{array}\right| \frac{4}{27}\right)$$
$$\sum_{n=1}^\infty\frac{1}{n \binom{4n}{n}}=\frac{1}{4}{}_4F_3\left(\left.\begin{array}{c} 1,1,\frac{4}{3},\frac{5}{3}\\ \frac{5}{4}, \frac{6}{4}, \frac{7}{4} \end{array}\right| \frac{27}{256}\right)$$
However, I am not very familiar with hypergeometric function.
The pattern suggests that $$\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}=\frac{1}{2}{}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)=\frac{\pi}{3\sqrt{3}}$$
Thus, $${}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)=\frac{2\pi}{3\sqrt{3}}$$
Arriving these results, I have the following questions:
How can ${}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)$ be expressed into this simple elementary form?
How can we arrive to the result for $\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}$ given by Wolfram Alpha?
Ultimately, can we evaluate $\sum_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ for all integers k $\ge$ $2$?
| Again, this is not going to be a full answer. However since my approach is slightly different from what was presented here so far and since my result is similar to the result given by Tito Piezas III I will post my answer.
\begin{eqnarray}
&&S_k:=\sum\limits_{n=1}^\infty \frac{1}{n \binom{ k n}{n}}=\\
&&\sum\limits_{n=1}^\infty \frac{k n+1}{n} \cdot \int\limits_0^1 t^n \cdot (1-t)^{(k-1) n} dt=\\
&&\int\limits_0^1 \left[ k \frac{t\cdot (1-t)^{k-1}}{1-t \cdot (1-t)^{k-1}} - \log(1-t \cdot (1-t)^{k-1})\right] dt=\\
&&-\int\limits_0^1 t \cdot \frac{ (1-t)^{k-3} (k t-1) \left(t (1-t)^k+k (t-1)+t-1\right)}{\left(t (1-t)^{k-1}-1\right)^2} dt=\\
&&\int\limits_0^1 \frac{(u-1) (k (u-1)+1) u^{k-2} \left(-u^{k-1}+u^k+k+1\right)}{\left(-u^{k-1}+u^k+1\right)^2}du=\\
&&\int\limits_1^\infty \left(\frac{\left(k^2-1\right) (v-1)}{v \left(v^k-v+1\right)}-\frac{k (k+1) (v-1)}{v^2 \left(v^k-v+1\right)}-\frac{k^2 (v-1)^2}{v^2 \left(v^k-v+1\right)^2}+\frac{(k-1) k (v-1)^2}{v \left(v^k-v+1\right)^2}\right) dv=\\
&&\int\limits_1^\infty \frac{(1-v) (k-(k-1) v) \left((k+1) v^k-v+1\right)}{v^2 \left(v^k-v+1\right)^2} dv
\end{eqnarray}
Let us discuss the steps from the very top to the bottom.
In the second line we used the integral representation of the beta function and in the third line we did the sum in question. In the fourth line we integrated by parts to get rid of the logarithm.In the fifth line we substituted for $u:=1-t$ and finally in the sixth and seventh lines we substituted for $v:=1/u$ and simplified the integrand.
Now, oddly enough the equation in the last line is similar to that given by Tito Piezas III. This would suggest that maybe the integrands of both results are the same, which they are not, as I have checked. It would be interesting to simplify my result further and bring it to the form given by Tito Piezas III.
Update: Note that:
\begin{equation}
\frac{(1+k) v^k-v+1}{v^2(1-v+v^k)^2}= \frac{d}{d v} \left[\frac{1}{v \cdot(1-v+v^k)} \right] + \frac{1}{v (1-v+v^k)^2}
\end{equation}
Inserting this into the last line above and integrating by parts we get:
\begin{eqnarray}
S_k&=&\int\limits_1^{\infty} \frac{1}{v \cdot (1-v+v^k)} dv + \int\limits_1^\infty \frac{(k-(k-1)v)}{v \cdot (1-v+v^k)^2} \cdot (-2 v^k-1+v) dv\\
&=&\int\limits_1^{\infty} \frac{1}{v \cdot (1-v+v^k)} dv + \left.\left( \frac{1-v}{1-v+v^k} + \log[\frac{1}{v^k}-\frac{1}{v^{k-1}}+1]\right)\right|_{1}^\infty\\
&=&\int\limits_1^{\infty} \frac{1}{v \cdot (1-v+v^k)} dv+0\\
&=&\int\limits_1^{\infty} \sum\limits_{p=1}^k \frac{1}{v\cdot(v-\xi_p)} \cdot \prod\limits_{q=1,q\neq p}^k\frac{1}{\xi_p-\xi_q)}dv=\\
&=&-\sum\limits_{p=1}^k \frac{\log(1-\xi_p)}{\xi_p} \cdot \prod\limits_{q=1,q\neq p}^k\frac{1}{(\xi_p-\xi_q)}
\end{eqnarray}
Here $\left\{ \xi_p \right\}_{p=1}^k$ are roots of the polynomial $v^k-v+1$.
As a final comment we note that in general the following holds:
\begin{equation}
\sum\limits_{n=1}^\infty \frac{x^n}{n} \cdot \frac{1}{\binom{k\cdot n}{n}} =x \cdot \int\limits_1^\infty \frac{1}{v\cdot(v^k+x-x v)} dv
\end{equation}
where again the integral on the right hand side can be expressed through roots of the polynomial in the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 2
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Iterated Integral with variable substitution I need to calculate the double integral of the function $f(x,y) = (x+y)^9(x-y)^9$:
$\int_0^{1/2} \int_x^{1-x} (x+y)^9(x-y)^9 dydx$
I have a solution but I definitely arrived at it after a sloppy attempt. I got -0.0025. To begin with, I know you need to separate the variables. I tried integration by parts, but got lost in it. How can I separate the variables to continue the integration?
| It's convenient to solve this integral with substitution. Since we want to separate the variables in the Integral
\begin{align*}
\int_0^{1/2} \int_x^{1-x} (x+y)^9(x-y)^9 dy\,dx\tag{1}
\end{align*}
it's reasonable to use the substitution
\begin{align*}
\left.
\begin{matrix}
u=x+y\\
v=x-y\\
\end{matrix}
\right\}
\quad\Longleftrightarrow\quad
\left\{
\begin{matrix}
x=\frac{1}{2}(u+v)\\
y=\frac{1}{2}(u-v)
\end{matrix}
\right.\tag{2}
\end{align*}
According to the Change of variable theorem we want to calculate
\begin{align*}
\int_{x_0}^{x_1}\int_{y_0}^{y_{1}}f(x,y)\,dy\,dx
=\int_{u_0}^{u_1}\int_{v_0}^{v_{1}}f(x(u,v),y(u,v))
\left|\operatorname{det}
\begin{pmatrix}
x_u&x_v\\
y_u&y_v
\end{pmatrix}
\right|
\,dv\,du\tag{3}
\end{align*}
Jacobian
At first we calculate the absolute value of the determinant of the Jacobian matrix
\begin{align*}
\left|\operatorname{det}
\begin{pmatrix}
x_u&x_v\\
y_u&y_v
\end{pmatrix}
\right|=
\left|\operatorname{det}
\begin{pmatrix}
\frac{1}{2}&\frac{1}{2}\\
\frac{1}{2}&-\frac{1}{2}
\end{pmatrix}
\right|
=\frac{1}{2}
\end{align*}
Area transformation
We observe from (1) the region of integration is
\begin{align*}
&0\leq x\leq \frac{1}{2}\\
&x\leq y\leq 1-x
\end{align*}
This is the area of a triangle with three lines as boundary lines, the $y$-axis $x=0$, the major diagonal $x=y$ and a parallel to the minor diagonal through $(1,0)$, $y=1-x$.
Since the transformation (2) is linear, lines are transformed to lines. So, the transformed $(u,v)$-area is again enclosed by three lines.
\begin{array}{lclcl}
x=0\qquad&\rightarrow&\qquad \frac{1}{2}(u+v)=0
&\qquad\rightarrow&u=-v\\
x=y\qquad&\rightarrow&\qquad \frac{1}{2}(u+v)=\frac{1}{2}(u-v)
&\qquad\rightarrow&v=0\\
y=1-x\qquad&\rightarrow&\qquad \frac{1}{2}(u-v)=1-\frac{1}{2}(u+v)
&\qquad\rightarrow&u=1\\
\end{array}
We see the $(u,v)$-triangle area is given by the three lines $u=-v, v=0$ and $u=1$ which can be written as
\begin{align*}
0\leq u \leq 1\\
-u\leq v\leq 0
\end{align*}
$$ $$
Integration
Now we have all ingredients to perform the integration according to (3). We obtain
\begin{align*}
\int_0^\frac{1}{2}\int_x^{1-x} (x+y)^9(x-y)^9 dy\,dx
&=\int_0^1\int_{-u}^0(uv)^9\cdot\frac{1}{2}\,dv\,du\\
&=\frac{1}{2}\int_0^1u^9\left(\int_{-u}^0v^9\,dv\right)\,du\\
&=\frac{1}{2}\int_0^1u^9\left(\left.\frac{1}{10}v^{10}\right|_{-u}^0\right)\,du\\
&=-\frac{1}{20}\int_0^1u^{19}\,du\\
&=-\frac{1}{20}\left.\left(\frac{1}{20}u^{20}\right)\right|_{0}^1\\
&=-\frac{1}{400}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$ $$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.
| Let $\alpha = \sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}$. Then
$$ \alpha^2 = 12+2\sqrt{36-20} = 20 $$
hence $\color{red}{\alpha=2\sqrt{5}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Chemical reaction modeled by a differential equation I am badly stuck on the following question. Thus, I am asking for some help :)
Consider a chemical reaction in which compounds $A$ and $B$ combine to form a third compound $X$. The reaction can be written as
$$A + B \xrightarrow{k} X$$
If $2 \, \rm{g}$ of $A$ and $1 \, \rm{g}$ of $B$ are required to produce $3g$ of compound $X$, then the amount of compound $x$ at time $t$ satisfies the differential equation
$$ \frac{dx}{dt} = k \left(a - \frac{2}{3}x\right)\left(b - \frac{1}{3}x\right) $$
where $a$ and $b$ are the amounts of $A$ and $B$ at time $0$ (respectively), and initially none of compound $X$ is present (so $x(0) = 0$). Time is in units of minutes, and $k$ is the reaction rate, per minute per gram.
Use separation of variables (and integration by partial fractions) to show that solution can be expressed in the form
$$ \ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) = ct $$
where the constant c depends on k,a,b. Now suppose that a = 15g, b = 20g and after 10min, 15g of compound X has been formed, find the amount of X after 20 mins.
So, first of all, showing this solution can be expressed as second equation,
by using separation of variables ( and integration by partial fractions), i got
$$ \ln \left(\frac{a - \frac{2}{3}x}{b - \frac{1}{3}x}\right) * \frac{3}{a-2b} = kt + c $$
how can this can be expressed in the form of the second equation ? am I doing something wrong ?, I dont get how I should get $$ \frac{b}{a} $$ inside the ln... and on the RHS, how kt + c becomes just ct ?
and by the second euqation,
"""
where the constant $c$ depends on $k$, $a$ and $b$. Now suppose that a = 15g, b = 20g and after 10min, 15g of compound X has been formed, find the amount of X after 20 mins.
"""
how do we find the amount of $X$ after $20$ minutes?
Thank you !
| The given ODE can be rewritten in the form
$$\dot x = \frac{2 \kappa}{9} \left( x - \frac{3 a}{2} \right) \left( x - 3 b \right)$$
or, alternatively, in the form
$$\left( \frac{1}{x - \frac{3 a}{2}} - \frac{1}{x - 3 b} \right) \, \mathrm{d} x = \left(\frac{a - 2 b}{3}\right) \kappa \,\mathrm{d} t$$
Integrating,
$$\ln \left[ \left( \frac{x - \frac{3 a}{2}}{x - 3 b} \right) \left( \frac{x_0 - 3 b}{x_0 - \frac{3 a}{2}} \right) \right] = \left(\frac{a - 2 b}{3}\right) \kappa t$$
If $x_0 = 0$, then
$$\ln \left[ \left( \frac{2 x - 3 a}{x - 3 b} \right) \frac{b}{a} \right] = \left(\frac{a - 2 b}{3}\right) \kappa t$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integrate the following : $\displaystyle\int \frac{x^n+1}{(x^n-1) \sqrt {x^4+1}} dx$ ; n is odd. My original Question :
$\displaystyle\int \frac{x^3+1}{(x^3-1) \sqrt {x^4+1}} dx $
What to substitute? How to do it?
I want to add that I have tried doing this using substitutions as :
$x = \frac{1}{t}$ and considered the equation $\displaystyle \frac{1-tan^2 \theta}{1+tan^2 \theta} = \cos 2\theta$ but none of these came to any use.
Also I'm aware of a brother of this problem : $\displaystyle\int \frac{x^2+1}{(x^2-1) \sqrt {x^4+1}} dx$
which also seems like one of Elliptic family as found here : How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$
So now I want to re-phrase my question into some general one :
What is the general solution of $\displaystyle\int \frac{x^n+1}{(x^n-1) \sqrt {x^4+1}} dx$
(Specially when n is odd.)
| First, introduce variables $y, z$ such that $x + x^{-1} = y = \frac{1}{\sqrt{2}}(z+z^{-1})$, we have
$$\begin{align}
& \frac{x+1}{x-1}\frac{dx}{\sqrt{x^4+1}}
= \frac{x+1}{x-1}\frac{1}{\sqrt{x^2+x^{-2}}}\frac{dx}{x}
= \frac{x+1}{x-1}\frac{1}{\sqrt{(x+x^{-1})^2-2}}\frac{d(x+x^{-1})}{x-x^{-1}}\\
= & \frac{dy}{(y-2)\sqrt{y^2-2}}\\
= & \frac{\sqrt{2}}{z+z^{-1}-2\sqrt{2}}\frac{d(z+z^{-1})}{\sqrt{(z+z^{-1})^2-4}}
= \frac{\sqrt{2}}{z+z^{-1}-2\sqrt{2}}\frac{d(z+z^{-1})}{z-z^{-1}}
= \frac{\sqrt{2}dz}{z^2-2\sqrt{2}z+1}
\end{align}
$$
Next, for any $n > 1$, we have
$$\frac{x^n - 1}{x-1} = x^{\frac{n-1}{2}}\frac{x^{\frac{n}{2}} - x^{-\frac{n}{2}}}{x^{\frac12}-x^{-\frac12}} = x^{\frac{n-1}{2}} U_{n-1}\left(\frac{x^{\frac12} + x^{-\frac12}}{2}\right)$$
where $U_m(x)$ is the $m^{th}$ Chebyshev polynomial of the $2^{nd}$ kind. When $n = 2k+1$ is an odd number $U_{n-1}(t)$ is an even polynomial of degree $2k$. This means there is a polynomial $f_k(t)$ of degree $k$ such that
$$\frac{x^{2k+1} - 1}{x-1} = x^{k}f_k\left((x^{\frac12} + x^{-\frac12})^2\right)
= x^{k} f_k(y+2)$$
Replace $x$ by $-x$, we get
$$\frac{x^{2k+1} + 1}{x+1} = (-x)^k f_k(-y+2)$$
Combine these, we find
$$\frac{x^{2k+1}+1}{x^{2k+1}-1} = \frac{x+1}{x-1} g_k(y)$$ where
$\displaystyle\;g_k(y) = (-1)^k\frac{f_k(2-y)}{f_k(2+y)}$ is a rational function in $y$.
As a result, when $n = 2k+1$ is odd, the integral can be transformed to
a integral over a rational function in $z$.
$$\mathcal{I}_n \stackrel{def}{=}\int \frac{x^n+1}{x^n-1}\frac{dx}{x^4+1} = \int g_k\left(\frac{z + z^{-1}}{\sqrt{2}}\right)\frac{\sqrt{2}{dz}}{z^2-2\sqrt{2}z+1}$$
For example, when $n = 3$, $U_{2k}(t) = 4t^2 - 1 \implies f_k(t) = t - 1$.
This implies
$$g_1(y) = (-1)^1\frac{(2-y)-1}{(2+y)-1} = \frac{y-1}{y+1}$$
and the integral becomes
$$\begin{align}\mathcal{I}_3
&=\int \frac{y-1}{(y+1)(y-2)}\frac{dy}{\sqrt{y^2-2}}
= \frac13 \int \left(\frac{2}{y+1} + \frac{1}{y-2}\right)\frac{dy}{\sqrt{y^2-2}}\\
&= \frac{\sqrt{2}}{3}\int \left(\frac{2}{z^2 + \sqrt{2}z + 1} + \frac{1}{z^2 - 2\sqrt{2}z + 1}\right) dz\\
&= \frac{4}{3}\tan^{-1}(1+\sqrt{2}z) + \frac{1}{3\sqrt{2}}\log\left(\frac{z-\sqrt{2}-1}{z-\sqrt{2}+1}\right) + \text{const...}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A Problem with the Generating Function of Fibonacci. So basically I want to find the closed form of $G_n = \sum_{k = 1}^n \binom{n+k - 1}{2k-1}$.
After checking for $n = 1,2,3,4$ the values are $1, 3, 8, 21$ respectively. I claim that it is $F_{2n}$ where $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$. My idea was to prove this using generating functions.
We have that $\binom{n+k-1}{2k-1}$ is the $n$th coefficient of $X^k \sum_{i \geq 0} \binom{i + 2k - 1}{i} X^i = \frac{X^k}{(1-X)^{2k}}$ so the $$\sum_{n \geq 0} G_n X^n = \sum_{k = 1}^n \frac{X^k}{(1-X)^{2k}}$$
Now we want to find what $F_0 + F_2X + F_4X^2 + ...$ is and we know that $F_0 + F_1X + ... = \frac{X}{1-X-X^2}$.
So
\begin{align*}
F_0 + F_2X^2 + F_4X^4 + ... & = \frac{1}{2} \left ( \frac{X}{1-X-X^2} + \frac{-X}{1 + X - X^2} \right ) \\
& = \frac{X^2}{(1-X^2)^2 - X^2}
\end{align*}
which means $$F_0 + F_2X + F_4X^2 + ... = \frac{X}{(1-X)^2 - X}.$$
So if suffices to show
\begin{align*}
\frac{X}{(1-X)^2 - X} & = \sum_{k = 1}^n \frac{X^k}{(1-X)^{2k}} \\
\frac{1}{(1-X)^2 - X} & = \sum_{k = 1}^n \frac{X^{k-1}}{(1-X)^{2k}} \\
\frac{(1-X)^{2n}}{(1-X)^2 - X} & = ((1-X)^2)^{n-1} + ((1-X)^2)^{n-2}X + ... + X^{n-1} \\
\frac{(1-X)^{2n}}{(1-X)^2 - X} & = \frac{(1-X)^{2n} - X^n}{(1-X)^2 - X}
\end{align*}
which is not true. I don't see where I went wrong with this.
| Suppose we seek the closed form of
$$G_n = \sum_{k=1}^n {n+k-1\choose 2k-1}.$$
Observe that the binomial coefficient
$${n+k-1\choose 2k-1} = {n+k-1\choose n-k}$$
has the property that its integral representation
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (1+z)^{n+k-1} \; dz$$
vanishes when $k\gt n$ and also when $k=0$ ($[z^n] (1+z)^{n-1} =
0$). Therefore we can set the range of the sum from $k=0$ to infinity,
getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n-1}
\sum_{k\ge 0} z^k (1+z)^k
\; dz$$
which converges for $|z(1+z)|<1$ to yield
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n-1}
\frac{1}{1-z(1+z)}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n+1}
\frac{1}{(1+z)^2}
\frac{1}{1-z(1+z)}
\; dz.$$
Putting $z/(1+z) = w$ we get $z = w/(1-w),$ $1+z=1/(1-w),$ $z(1+z) =
w/(1-w)^2,$ and $dz = 1/(1-w)^2 dw$ which yields
$$\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}}
(1-w)^2
\frac{1}{1-w/(1-w)^2}
\frac{1}{(1-w)^2}
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}}
\frac{1}{1-w/(1-w)^2}
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}}
\frac{1-2w+w^2}{1-3w+w^2}
\; dw.$$
Recognizing the generating function of $F_{2n}$ which is
$$\frac{w}{1-3w+w^2}$$
we get
$$F_{2n+2} - 2 F_{2n} + F_{2n-2}
\\ = F_{2n+1} + F_{2n}
- 2 F_{2n} + F_{2n} - F_{2n-1}
\\ = F_{2n} + F_{2n-1} + F_{2n}
- 2 F_{2n} + F_{2n} - F_{2n-1}
= F_{2n}$$
and we are done. The difference in these two generating function is
the constant term which produces a value of one in the second
generating function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Landau symbol and taylor series $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=\frac{1}{2}$$
I saw the calculation $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=$$...$$=\sqrt x (\sqrt{1+\sqrt{1/x+\sqrt{1/x^3}}}-1)= $$ $$\sqrt x(1+\frac{1}{2}\sqrt{1/x+\sqrt{1/x^3}}+ \color{red}{\mathcal o \left(1/x+\sqrt{1/x^3}\right)}-1)=...$$
Main question: Now wouldn't it have to be $\mathcal o(\sqrt{1/x+\sqrt{1/x^3}})$ instead? So I could write $\mathcal o(\sqrt{1/x+\sqrt{1/x^3}})$ or $\mathcal O(1/x+\sqrt{1/x^3})$?
Are there other ways to calculate the limit?
| If you want to do it with an asymptotic expansion, you have to be cautious with the order of expansion. The scale of comparison will be the powers of $\sqrt x$, and we'll expand $\sqrt{1+\sqrt{\dfrac1x+\sqrt{\dfrac1{x^3}}}}$ at order $2\;$ (i.e. up to a term in $1/x$).
*
*$\sqrt{\dfrac1x+\sqrt{\dfrac1{x^3}}}=\dfrac1{\sqrt x}\biggl(\sqrt{1+\dfrac1{\sqrt{x}}} \biggr)=\dfrac1{\sqrt x}\biggl(1+\dfrac1{2\sqrt{x}}+o\Bigl(\dfrac1{\sqrt x}\Bigr) \biggr)=\dfrac1{\sqrt x}+\dfrac1{2x}+o\Bigl(\dfrac1x\Bigr)$, so that:
*$\sqrt{1+\sqrt{\dfrac1x+\sqrt{\dfrac1{x^3}}}}=\sqrt{1+\dfrac1{\sqrt x}+\dfrac1{2x}+o\Bigl(\dfrac1x\Bigr)}=$
\begin{align*}&=1+\dfrac{1}{2\sqrt x}+\dfrac1{4x}-\dfrac18\Bigl(\dfrac1{\sqrt x}+\dfrac1{2x}\Bigr)^2+o\Bigl(\dfrac1x\Bigr)=1+\dfrac{1}{2\sqrt x}+\dfrac1{4x}-\dfrac1{8x}+o\Bigl(\dfrac1x\Bigr)\\
&=1+\dfrac{1}{2\sqrt x}-\dfrac1{8x}+o\Bigl(\dfrac1x\Bigr)
\end{align*}
Thus,
\begin{align*}\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{\mathstrut x}&=\sqrt{\mathstrut x} \biggl(\sqrt{1+\sqrt{1/x+\sqrt{1/x^3}}}-1\biggr)=\sqrt{\mathstrut x} \biggl(1+\dfrac{1}{2\sqrt x}-\dfrac1{8x}+o\Bigl(\dfrac1x\Bigr)-1\Biggr)\\&=\frac12-\frac1{8\sqrt x}+o\Bigl(\dfrac1{\sqrt x}\Bigr).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
What is the value of the following determinant: given that $q^2- pr<0, p>0$ then evaluate \begin{vmatrix}
p &q &px+qy\\
q &r &qx+ry\\
px+qy &qx+ry &0
\end{vmatrix}
I've tried this:
\begin{vmatrix}
p &q &px+qy\\
q &r &qx+ry\\
px+qy &qx+ry &0
\end{vmatrix}
then, i modified the rows as $R_3=R_3-(xR_1+yR_2)$ to get \begin{vmatrix}
p &q &px+qy\\
q &r &qx+ry\\
0 &0 &px^2+2qxy+ry^2
\end{vmatrix}
then I've simplified it as $(px^2+2qxy+ry^2)(pr-q^2)$, but i cannot do further. can you please give me hints?
| First, $pr-q^2$ is positive. What about $px^2+2qxy+ry^2$? This is the quadratic form $\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}p&q\\q&r\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$. Whether this is a positive quantity comes down to whether the eigenvalues of the matrix are positive (and whether $x,y$ are nonzero). The eigenvalues are the roots of $\lambda^2-(p+r)\lambda-(q^2-pr)$, which are $\frac{1}{2}(p+r\pm\sqrt{p^2+4q^2-2pr+r^2})$. The expression $p^2+4q^2-2pr+r^2$ can be written as $(p+r)^2+4(q^2-pr)<(p+r)^2$, hence the roots are both between $0$ and $p+r$ (exclusive). The condition $p>0$ and $q^2<pr$ together imply that $r>0$ as well, so the roots are positive.
Thus, the determinant of the matrix is nonnegative, and positive if and only if $x,y$ are not both $0$.
Alternatively, instead of examining eigenvalues of the quadratic form, we can use the quadratic formula along with the intermediate value theorem. Let's see for which $x,y$ the factor $px^2+2qxy+ry^2$ is $0$. The quadratic formula gives $x=\frac{-2qy\pm\sqrt{4q^2y^2-4pry^2}}{2p}=y\frac{-q\pm \sqrt{q^2-pr}}{p}$. The discriminant is negative, so $y$ must be $0$, and so $x$ must be $0$ as well. Since the substitution $(x,y)=(1,0)$ gives $px^2+2qxy+ry^2=p>0$, the factor is always non-negative, and positive when $(x,y)\neq (0,0)$. (This is because if it were ever negative, a path between $(1,0)$ and the point where it is negative would contain a point where the factor is $0$, by the intermediate value theorem.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$ Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$
My attempt:First solve when $n$ is not infinity then put infinity in.
$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}{2^n}}$$
$$=2^{\frac{1}{2}}\cdot 2^{\frac{2}{4}}\cdot \dots\cdot 2^{\frac{n}{2^n}}$$
Now calculate the sum of the powers:
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}$$
$$=\frac{2^{n-1}+2\cdot2^{n-2}+3\cdot2^{n-3}+\dots+n\cdot2^0}{2^n}$$
Now calculate the numerator:
$$2^0+2^1+2^2+\dots+2^{n-1}=2^n-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-2}=2^{n-1}-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-3}=2^{n-2}-1$$
$$+$$
$$\vdots$$
$$+$$
$$2^0=2^1-1$$
$$=2^1+2^2+2^3+\dots+2^n-n=2^{n+1}-n-1$$
Now put the numerator on the fraction:
$$\frac{2^{n+1}-n-1}{2^n}=2-\frac{n}{2^n}-\frac{1}{2^n}$$
Now we can easily find $\lim_{n \to \infty}\frac{1}{2^n}=0$
Then we just have to find $\lim_{n \to \infty }\frac{n}{2^n}$, that by graphing will easily give us the answer zero.
That gives the total answer is $4$.
But now they are two problems:
1.I cannot find $\lim_{n \to \infty }\frac{n}{2^n}$ without graghing.
2.My answer is too long.
Now I want you to help me with these problems.Thanks.
| $$I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\cdots$$
$$2I=1+1+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\frac{6}{32}+\cdots$$
$$2I-I=1+\left(1-\frac 12 \right)+\left(\frac 34 -\frac 24 \right)+\left(\frac 48 -\frac 38 \right)+\left(\frac {5}{16} -\frac {4}{16} \right)+\cdots$$
$$I=1+\frac 12+\frac 14+\frac 18+\cdots=2$$
therefore
$$\lim_{n\to \infty }\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots\times\sqrt[2^n]{2^n}=2^2=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 1
} |
Prove that $\frac{1}{\sqrt{nx+1}}+\frac{1}{\sqrt{nx+2}}+\ldots+\frac{1}{\sqrt{nx+n}}=\sqrt n$ has a unique positive solution Prove that for each $n\in \mathbb Z$, $n\ge 1$ there is exactly one $x>0$ satisfying
$$\frac{1}{\sqrt{nx+1}}+\frac{1}{\sqrt{nx+2}}+\ldots+\frac{1}{\sqrt{nx+n}}=\sqrt n$$
Please give me a hint. In fact, after changing the expression, I get the equation:
$$\frac1n\sum\limits_{i=1}^n\dfrac{1}{\sqrt{x+\frac{i}{n}}}=1$$
I don't know what next step I should do. Thank you in advance for your help.
| Note that $n=1$ gives $x=0$.
Considering
$$\int_{0}^{1} \frac{dt}{\sqrt{\frac{9}{16}+t}} = 1$$
Since $\frac{1}{\sqrt{\frac{9}{16}+t}}$ is monotonic decreasing,
the integral is sandwiched by the upper and lower Riemann sums: $U_{n}<U_{n+1}<1<L_{n+1}<L_{n}$
$$\sum_{k=1}^{n} \frac{1}{n\sqrt{\frac{9}{16}+\frac{k}{n}}} < 1 <
\sum_{k=0}^{n-1} \frac{1}{n\sqrt{\frac{9}{16}+\frac{k}{n}}}$$
$$\sum_{k=1}^{n} \frac{1}{n\sqrt{\frac{9}{16}+\frac{k}{n}}} < 1 <
\sum_{k=1}^{n} \frac{1}{n\sqrt{\frac{9}{16}-\frac{1}{n}+\frac{k}{n}}}$$
Hence for $n\ge 2$, $\displaystyle \exists x \in \left( \frac{9}{16}-\frac{1}{n}, \frac{9}{16} \right)$ satisfies the equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
$a_1 + a_2 + \dots a_n = 1$ find min of $a_1^2 +\frac{a_2^2}{2} + \dots + \frac{a_n^2}n.$ Given $n$ numbers $a_1$, $\cdots$ and such that $a_1$, $a_2$, $\cdots$, $a_n > 0$ and their sum is $1$, I want to find the minimum value of
$$a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}.$$
I have tried using weighted AM-GM inequality, like this:
$$\frac{a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}}{a_1 + a_2 + \cdots + a_n } \geqslant \frac{a_1^{a_1} \cdots a_n^{a_n}}{2^{a_2} \cdots n^{a_n}}$$
but was unable to make progress on the right hand side.
Is there a better way to apply AM-GM inequality? Or is there some different way altogether to solve this?
| By the Cauchy-Schwarz inequality we have
$$ \left(\sum_{i=1}^{n}i\right)\left(\sum_{i=1}^{n}\frac{a_i^2}{i}\right) \geq \left(\sum_{i=1}^{n}\sqrt{i}\cdot\frac{a_i}{\sqrt{i}}\right)^2 = 1 \tag{1}$$
hence with our hypothesis we have:
$$ \sum_{i=1}^{n}\frac{a_i^2}{i}\geq \frac{2}{n(n+1)}\tag{2} $$
and equality is achieved iff $\frac{a_i}{\sqrt{i}}=\lambda\sqrt{i}$, i.e. iff $a_i = \frac{2i}{n(n+1)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Prove the inequalitiy
If $a, b, c$ are positive reals such that:
$\frac a {b + c + 1} + \frac b {a + c + 1} + \frac c {b + a + 1} \le
1$ then:
$\frac 1 {b + c + 1} + \frac 1 {a + c + 1} + \frac 1 {b + a + 1} \ge
1$
I tried using inequality of arithmetic and harmonic means, to no avail.
Any help is appreciated.
| We'll write the condition in the following form:
$$a+b+c+1\geq a^3+b^3+c^3+abc+\sum\limits_{cyc}(a^2-ab)$$
and since $\sum\limits_{cyc}(a^2-ab)\geq0$ we obtain
$$a+b+c+1\geq a^3+b^3+c^3+abc$$
In another hand, we need to prove that
$$2(a+b+c+1)\geq(a+b)(a+c)(b+c)$$
Thus, it remains to prove that
$$2(a^3+b^3+c^3+abc)\geq(a+b)(a+c)(b+c)$$
which is $\sum\limits_{cyc}(a+b)(a-b)^2\geq0$. Done!
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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} |
Evaluate $\int_\frac12 ^2 \frac1x\tan(x-\frac1x)dx $ $$\int\limits_\frac{1}{2} ^2 \frac{1}{x}\tan\left(x-\frac{1}{x}\right)\mathrm{d}x$$
I have tried substitution and by parts and it seems failed at all. Can anyone give me some hints?
| Let $$ I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\cdot \tan \left(x-\frac{1}{x}\right)\,dx$$
Now Let $\displaystyle\left(x-\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)\,dx = dt\Rightarrow \left(x+\frac{1}{x}\right)\,dx = x\,dt$
and Changing Limits
Now Using $$ \left(x+\frac{1}{x}\right)^2 -\left(x-\frac{1}{x} \right)^2 = 4\Rightarrow \left(x+\frac{1}{x}\right)=\sqrt{\left(x-\frac{1}{x}\right)^2+4}=\sqrt{t^2+4}$$
So Integral $$ I = \int_{-\frac{3}{2}}^{\frac{3}{2}}\tan t\cdot \frac{1}{\sqrt{t^2+4}}\cdot \frac{x}{x} \, dt = \int_{-\frac{3}{2}}^{\frac{3}{2}}\tan t\cdot \frac{1}{\sqrt{t^2+4}} \, dt$$
So we get $$ I = \int_{-\frac{3}{2}}^{\frac{3}{2}} \underbrace{\frac{\tan t}{\sqrt{t^2+4}}}_{\textbf{odd function}} \, dt = 0$$
Above we have used the formula $$\displaystyle \int_{-a}^a f(x) \, dx = 0\;,$$ If $f(x)$ is odd function.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Elementary matrix operations I am taking an introductory/first year course in Linear Algebra and I am at my wits' end with the following problem.
I am asked to find both the nullspace and the general solution of the following systems $A\tilde{x}=\tilde{b}$.
The first is
$$
A=
\begin{pmatrix}
1 & 1 & 0 & 2\\
2 & 1 & 1 &-2\\
2 & 2 & 2 & 1
\end{pmatrix},\quad \tilde{b}=
\begin{pmatrix}
12\\0\\14
\end{pmatrix}
$$
Proceeding with an augmented matrix,
I have
\begin{align*}
\left(
\begin{array}{rrrr|c}
1 & 1 & 0 & 2 & 12\\
2 & 1 & 1 &-2 & 0\\
2 & 2 & 2 & 1 & 14
\end{array}
\right)&\sim
\left(
\begin{array}{rrrr|r}
1 & 1 & 0 & 2 & 12\\
0 &-1 & 1 &-6 & -24\\
0 & 0 & 2 & -3 &-10
\end{array}
\right)\\
&\sim
\left(
\begin{array}{rrrr|r}
1 & 1 & 0 & 2 & 12\\
0 & 1 &-1 & 6 & 24\\
0 & 0 & 1 & -3/2 &-5
\end{array}
\right)
\end{align*}
I've checked this multiple times and can't find a mistake, plus it agrees with the solutions my professor has provided.
Solving this now, I have
\begin{align*}
x_4&=\alpha\in\mathbb{R}\\
x_3&=-5+\frac{3}{2}\alpha\\
x_2&=19-\frac{9}{2}\alpha\\
x_1&=-7+\frac{5}{2}\alpha
\end{align*}
So from this, my solution will be
$$
\tilde{x}=
\begin{pmatrix}
-7+\frac{5}{2}\alpha\\
19-\frac{9}{2}\alpha\\
-5+\frac{3}{2}\alpha\\
\alpha
\end{pmatrix}
=
\begin{pmatrix}
-7\\19\\-5\\0
\end{pmatrix}+\beta
\begin{pmatrix}
5\\-9\\3\\2
\end{pmatrix},\quad\beta\in\mathbb{R}.
$$
This does not agree with the solutions given, and the solutions given are just solutions without any steps in between so I'm a little confused as to where I went wrong. The weird part is, my nullspace agrees with the solution provided but the particular solution certainly does not.
Edit: My professor's solution is
$$
\tilde{x}=
\begin{pmatrix}
3\\1\\1\\4
\end{pmatrix}+\alpha
\begin{pmatrix}
5\\-9\\3\\2
\end{pmatrix},\quad\beta\in\mathbb{R}.$$
Can somebody please let me know what I've done incorrectly?
Thank you.
| This is correct! You can check if you found solutions of the system simply by replacing $x$ in $Ax=b$ by what you found.
| {
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} |
How to break $\frac{1}{z^2}$ into real and imaginary parts? $$ \frac{1}{(x+iy)^2}=\frac{1}{x^2+i2xy-y^2}=\frac{x^2}{(x^2+y^2)}-\frac{2ixy}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
So I thought I could just say:
$$ Re(\frac{1}{z^2})=\frac{x^2}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
and
$$ Im(\frac{1}{z^2})=-\frac{2ixy}{(x^2+y^2)}$$
But I know that is wrong because it looks nothing like the graph of the real part of 1/z^2 on wolfram alpha found here: http://www.wolframalpha.com/input/?i=1%2F(x%2Bi*y)%5E2
Then I thought I could must multiply $1/z^2$ by $z/z$ to get $\frac{x}{z^3}$ and $\frac{iy}{z^3}$ however graphing these again shows that they are not the real and complex parts of $\frac{1}{z^2}$.
|
Notice, when $z\in\mathbb{C}$:
$$z=\Re[z]+\Im[z]i$$
So, we get (in steps):
*
*$$z^2=\left(\Re[z]+\Im[z]i\right)^2=\Re^2[z]-\Im^2[z]+2\Re[z]\Im[z]i$$
*$$\overline{z^2}=\overline{\Re^2[z]-\Im^2[z]+2\Re[z]\Im[z]i}=\Re^2[z]-\Im^2[z]-2\Re[z]\Im[z]i$$
*$$z^2\cdot\overline{z^2}=|z|^4=\left(\sqrt{\Re^2[z]+\Im^2[z]}\right)^4=\left(\Re^2[z]+\Im^2[z]\right)^2$$
Now, we get:
$$\frac{1}{z^2}=\frac{\overline{z^2}}{z^2\cdot\overline{z^2}}=\frac{\Re^2[z]-\Im^2[z]-2\Re[z]\Im[z]i}{\left(\Re^2[z]+\Im^2[z]\right)^2}$$
So:
*
*$$\color{red}{\Re\left[\frac{1}{z^2}\right]=\frac{\Re^2[z]-\Im^2[z]}{\left(\Re^2[z]+\Im^2[z]\right)^2}}$$
*$$\color{red}{\Im\left[\frac{1}{z^2}\right]=-\frac{2\Re[z]\Im[z]}{\left(\Re^2[z]+\Im^2[z]\right)^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
How does $\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$? How does $$\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$$
| Use the fact that for $t\not=0$,
$\displaystyle \arctan(t)+\arctan(1/t)=\mbox{sgn}(t)\cdot\frac{\pi}{2}$.
See here for the details: $\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$
Then for $x\not=0$ and $x\not=20$,
$$\arctan(\frac{8}{x})+\arctan(\frac{13}{20-x})+\arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})\\
=\mbox{sgn}(x)\cdot\frac{\pi}{2}+ \mbox{sgn}(20-x)\cdot\frac{\pi}{2}=\begin{cases}
\pi\quad \mbox{if $0<x<20$,}\\
0 \quad \mbox{if $x<0$ or $x>20$.}\\
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Complex number fraction Find the real and imaginary number:
$$\frac{1}{z}=\frac{2}{2+j3}+\frac{1}{3-j2}$$
How do I invert $\frac{1}{z}$ to $z$ so that I can start solving it?
| The denominators of the right hand side are closely related.
We obtain
\begin{align*}
\frac{1}{z}&=\frac{2}{2+j3}+\frac{1}{3-j2}=\frac{2}{2+j3}+\frac{j}{2+j3}
\end{align*}
It follows
\begin{align*}
z=\frac{2+j3}{2+j}=\frac{1}{5}(2+j3)(2-j)=\frac{1}{5}(7+j4)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Continuity and differentiability of $f(x, y) = \frac{2x^3 + 3y^3}{x^2 + y^2}$ I'm learning multivariable calculus, specifically multivariable continuity and differentiability, and need help with the following exercise:
Let
$$f(x,y) = \begin{cases} \frac{2x^3 + 3y^3}{x^2 + y^2}, & (x,y) \neq (0, 0) \\ 0, & (x,y) = (0, 0). \end{cases}$$
$(a)$ Examine if $f$ is continuous at $(0, 0)$.
$(b)$ Examine if $f$ is differentiable at $(0, 0)$.
Since I'm having difficulties for $(b)$, I'm going to show my work for $(a)$.
$(a)$ We note that
$$0 \leq \left| \frac{2x^3 + 3y^3}{x^2 + y^2} \right| \leq \frac{2|x|^3}{x^2 + y^2} + \frac{3|y|^3}{x^2 + y^2} \leq \frac{2|x|^3}{x^2} + \frac{3|y|^3}{y^2} = 2|x| + 3|y|$$
and $$\lim_{(x,y) \to (0, 0)} 2|x| + 3|y| = 0.$$
Therefore, by the Squeeze theorem, we conclude that
$$\lim_{(x, y) \to (0, 0)} f(x, y) = f(0, 0)$$
that is
$$\lim_{(x, y) \to (0, 0)} \frac{2x^3 + 3y^3}{x^2 + y^2} = 0$$
so $f$ is continuous at $(0, 0)$.
Is my work correct for $(a)$? How do I examine differentiability of $f$ at $(0, 0)$? More importantly: what is the method to examine differentiability in this case? What do I need to show?
| Your proof about continuity is correct.
In order to check differentiability fistly find partial derivatives so that
$$ { f }_{ x }^{ \prime }\left( 0,0 \right) =\lim _{ x\rightarrow 0 }{ \frac { f\left( x,0 \right) -f\left( 0,0 \right) }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { \frac { 2x^{ 3 } }{ x^{ 2 } } }{ x } } =2\\ { f }_{ y }^{ \prime }\left( 0,0 \right) =\lim _{ y\rightarrow 0 }{ \frac { f\left( y,0 \right) -f\left( 0,0 \right) }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { \frac { 3y^{ 3 } }{ y^{ 2 } } }{ y } } =3$$
differentiability should be shown as
$$f\left( x,y \right) -f\left( 0,0 \right) =\frac { 2x^{ 3 }+3y^{ 3 } }{ x^{ 2 }+y^{ 2 } } =2x+3y+\left( \frac { 2x^{ 3 }+3y^{ 3 } }{ x^{ 2 }+y^{ 2 } } -2x-3y \right) =\\={ f }_{ x }^{ \prime }\left( 0,0 \right) x+{ f }_{ y }^{ \prime }\left( 0,0 \right) y+\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } $$ where $$\alpha \left( x,y \right) =\frac { -3{ x }^{ 2 }y-2{ y }^{ 2 }x }{ \left( x^{ 2 }+y^{ 2 } \right) \sqrt { x^{ 2 }+y^{ 2 } } } $$
We should check whether is this infinity small?
Let $x=\frac { 1 }{ n } ,y=\frac { 1 }{ n } $ so that
$$\\ \alpha \left( \frac { 1 }{ n } ,\frac { 1 }{ n } \right) =\frac { -\frac { 5 }{ { n }^{ 3 } } }{ \frac { 2\sqrt { 2 } }{ { n }^{ 3 } } } =-\frac { 5 }{ 2\sqrt { 2 } } \neq 0,n\rightarrow \infty ,n\in \ \mathbb{N} $$
it shows $\alpha \left( x,y \right) $ is not infinitly small (when $x\rightarrow 0,y\rightarrow 0$) in other words
$$\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \neq o\left( \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right) $$
so it not differentiable
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find the fourier series of a piecewise continuous function
Consider the function
$f(x) =
\begin{cases}
0 & & x \in [-\pi, 0) \\
1 & & x \in [0, \frac{\pi}{2})\\
0 & & x \in [\frac{\pi}{2}, \pi)\\
\end{cases}$
Calculate its fourier series
I am unsure if this is correct but i have:
$a_0=\frac{1}{2\pi} \int_{-\pi}^{\pi}f(x)dx=\frac{1}{2\pi} \int_{0}^{\frac{\pi}{2}}1dx=\frac{1}{2\pi}[x]_{0}^{\frac{\pi}{2}}=\frac{1}{4}$
$a_n=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)\cos(nx)dx=\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}}\cos(nx)dx=\frac{1}{\pi}[\frac{1}{n}\sin(nx)]_{0}^{\frac{\pi}{2}}=\frac{1}{n\pi}$
$b_n=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)\sin(nx)dx=\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}}\sin(nx)dx=\frac{1}{\pi}[\frac{1}{n}(-\cos(nx))]_{0}^{\frac{\pi}{2}}=\frac{-1}{n\pi}$
So we have $f(x)=\frac{1}{4}+\sum_{n=0}^{\infty}[\frac{1}{n\pi} \cos(nx) - \frac{1}{\pi} \sin(nx)]$
Is this correct?
| The setup is alright but the evaluations are not correct. For example
$$
\dfrac{1}{\pi} \int_0^{\pi/2} \sin(nx) \; dx = \dfrac{1 - \cos(n \pi/2)}{n \pi} \neq - \dfrac{1}{n \pi}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proving that a given line is tangent to a hyperbola The question is: a line $x \cos\theta + y\sin\theta = p$ is given such that $a^2\cos^2\theta - b^2\sin^2\theta =p^2$.
I have to prove that it touches a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. I do not see how to proceed. I thought I could proceed by eliminating $p$ by squaring the line equation and equate both but I don't get an idea what to do next. What do I need to show to prove that it touches that hyperbola?
| The tangent at $(x_1,y_1)$ to the hyperbola is
\begin{align*}
\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1
\end{align*}
If this is the given line, we have
\begin{align*}
\frac{x_1/a^2}{\cos \theta} = \frac{-y_1/b^2}{\sin \theta} = \frac{1}{p}
\end{align*}
Hence $x_1 = \frac{a^2\cos\theta}{p}, y_1 = -\frac{b^2\sin\theta}{p}$. Since this lies on the curve, we have
\begin{align*}
a^2\cos^2\theta - b^2\sin^2\theta = p^2
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Loxodromics (curves with fixed angle with meridians) of a revolution surface I am trying to find the curves with a fixed angle $\phi$ with meridians for the revolution surface given by the revolution of the graph of $z=\frac{1}{x}$ around the $z$ axis. The surface is easily parametrized as
$$(x,\theta)\mapsto(x \cos \theta, x\sin \theta,\frac{1}{x}) \qquad x\in \mathbb R_{>0},\theta \in (0,2\pi)$$
Now, some tedious computations (I hope they are correct) show that the first and the second fundamental form for this parametrization are
$$G_{I}=\begin{pmatrix}1+\frac{1}{x^4} &0 \\ 0 & x^2 \end{pmatrix} \qquad G_{II}=\frac{1}{\sqrt{x^4+1}}\begin{pmatrix}\frac{1}{x} & 0 \\ 0 & -x \end{pmatrix}$$
Recall that the angle $\phi$ between two $C^1$ curves $\gamma_1,\gamma_2$ is given by
$$\cos \phi=\frac{\gamma_1'^TG_I \gamma_2'}{\sqrt{\gamma_1'^T G_I \gamma_1'}\sqrt{\gamma_2'^TG_I\gamma_2'}}$$
Since we are looking for curves $(x(t),\theta(t))^T$ with constant angle with meridians, we want that
$$\frac{\begin{pmatrix}1 & 0 \end{pmatrix}\begin{pmatrix} 1+\frac{1}{x^4(t)} &0 \\ 0 & x^2(t)\end{pmatrix}\begin{pmatrix} x'(t) \\ \theta'(t) \end{pmatrix}}{\sqrt{\begin{pmatrix} x(t) & \theta(t)\end{pmatrix}\begin{pmatrix}1+\frac{1}{x^4(t)} &0 \\ 0 & x^2(t) \end{pmatrix}\begin{pmatrix} x'(t)\\ \theta'(t) \end{pmatrix}}\sqrt{\begin{pmatrix} 1 & 0\end{pmatrix}\begin{pmatrix}1+\frac{1}{x^4(t)} &0 \\ 0 & x^2(t) \end{pmatrix}\begin{pmatrix} 1 \\ 0\end{pmatrix}}}=K$$
which, dropping the dependency on $t$, yields the differential equation
$$x'\left (1+\frac{1}{x^4}\right )=K\sqrt{1+\frac{1}{x^4}}\sqrt{x'^2\left ( 1+\frac{1}{x^4}\right)+x^2\theta'^2}$$
Not looking very good. Through a few manipulations, I brought this to the form
$$x'(t)=K\sqrt{x'^2+\frac{x^6\theta'^2}{x^4+1}}$$
Since this is an exercise from an exam, I would expect the solution to have a nice closed form, but maybe I am mistaken.
Is there any elementary way of solving such differential equation or of finding an explicit equation for the loxodromics?
| Well, actually you can also use the vector $(0,1)$, since the tangent vectors to the coordinate lines $x$ and $\theta$ are orthogonal at each point. Then your curve measures a constant angle $\alpha_0$ with the tangent $x-$vector if and only if it also measures a constant angle $\pi/2 - \alpha_0$ with the tangent $\theta-$vector, so you can go this way too. But anyway, the result is going to be virtually the same. You end up with the dot product expression (i.e. the dot product formula $(a \cdot b) = |a||b|\cos{\alpha_0} \,\,$):
\begin{align} (1,0) \,\, G_I(x) \, \, (x',\theta')^T &= \sqrt{\Big((1,0) \,\, G_I(x) \,\, (1,0)^T\Big)} \sqrt{\Big((x',\theta') \,\, G_I(x) \,\, (x',\theta')^T\Big)} \, \cos{(\alpha_0)} \\
\left(1+ \frac{1}{x^4}\right) x' &= \cos{(\alpha_0)}\sqrt{\left(1+ \frac{1}{x^4}\right)} \sqrt{\left(1+ \frac{1}{x^4}\right)(x')^2 + x^2 (\theta')^2}
\end{align}
Notice that all expressions, apart from the derivative $\theta'$, depend only on the variable $x$! This is not a coincidence as the surface is a surface of revolution and is therefore rotatiationally invariant with respect to rotations around the $z$ axis, which in the $x,\theta$ coordinates manifest themselves as translations of the $\theta$ coordinate, i.e. $\theta \mapsto \theta + c$. Consequently, when you have one curve on the surface of constant angle $\alpha_0$ with respect to the $x-$coordinate curves, if you rotate it around $z$, you get again a curve on the surface of the same constant angle $\alpha_0$ with respect to the $x-$coordinate curves. Therefore, your equations for the curves of angle $\alpha_0$ should be invariant with respect to the symmetries $\theta \mapsto \theta + c$, which means that the equations do not depend explicitly on $\theta$. Thus, we can simply write the curve as $\theta = \theta(x)$, i.e. it is parametrized as $x \mapsto (x, \theta(x))$, so $x' = 1$ and $\theta' = \frac{d\theta}{dx}$. So the equation becomes
\begin{align}
\left(1+ \frac{1}{x^4}\right) &= \cos{(\alpha_0)}\sqrt{\left(1+ \frac{1}{x^4}\right)} \sqrt{\left(1+ \frac{1}{x^4}\right) + x^2 \Big(\frac{d\theta}{dx}\Big)^2};\\
\sqrt{1+ \frac{1}{x^4}} &= \cos{(\alpha_0)} \sqrt{\left(1+ \frac{1}{x^4}\right) + x^2 \Big(\frac{d\theta}{dx}\Big)^2};\\
{1+ \frac{1}{x^4}} &= \cos^2{(\alpha_0)}\left( 1+ \frac{1}{x^4}+ x^2 \Big(\frac{d\theta}{dx}\Big)^2\right);\\
\frac{1}{\cos^2{(\alpha_0)}}\left({1+ \frac{1}{x^4}}\right) &= 1+ \frac{1}{x^4}+ x^2 \Big(\frac{d\theta}{dx}\Big)^2;\\
x^2 \Big(\frac{d\theta}{dx}\Big)^2 &= \left(\frac{1}{\cos^2{(\alpha_0)}} - 1\right)\left(1+ \frac{1}{x^4}\right) \\
\Big(\frac{d\theta}{dx}\Big)^2 &= \tan^2{(\alpha_0)}\,\frac{1}{x^2}\left(1+ \frac{1}{x^4}\right) = \tan^2{(\alpha_0)}\,\left(\frac{x^4+1}{x^6}\right);\\
\frac{d\theta}{dx} &= \pm \tan{(\alpha_0)}\,\left(\frac{\sqrt{x^4+1}}{x^3}\right);
\end{align}
So as you can see, there is almost no actual differential equation, the problem is solved by simply explicitly integrating a derivative
\begin{align}
\theta(x) &= \theta_0 \pm k_0 \,\int \frac{\sqrt{x^4+1}}{x^3}dx,
\end{align} where $k_0 = \tan{\alpha_0}$.
| {
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Constant when integrating I was integrating $(t+2)^2$ by using the substitution method and by expanding the function. By expanding it then integrating I got the same answer as the book ($\frac{1}{3}(t^3+6t^2+12t)+C$), but when I substituted I got $\frac{1}{3}\cdot(t+2)^3 + C = \frac{1}{3} \cdot (t^3+6t^2+12t+8) + C$.
Am I missing something here? Got an 8 at the end when substituting, but no constant when expanding.
| No worries,
$$\left(\frac{1}{3}(t^3+6t^2+12t)+C\right)'=t^2+4t+4$$
and
$$\left(\frac{1}{3}(t+2)^3 + C \right)'=(t+2)^2.$$
| {
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If $A$ and $B$ are two matrices such that $AB=B$ and $BA=A$ then how to show that $A^2+B^2$ equals $A+B$?
If $A$ and $B$ are two matrices such that $AB=B$ and $BA=A$ then $A^2+B^2$ equals ?
(a) $2AB$
(b) $2BA$
(c) $A+B$
(d) $AB$
I tried
$(A+B)^2=A^2+B^2+AB+BA$
or,$A^2+B^2=(A+B)^2-AB-BA$
$=(A+B)^2-A-B$
$
=(A+B)^2-(A+B)=(A+B)(A+B-1)$
How to reach the answer from here?
| First, here is how you can figure out the answer by process of elimination. We can see that if $A$ and $B$ are both identity, then the condition is satisfied, and $A^2 + B^2 = 2I$ in that case. This rules out option (d). Next, note that $A^2 + B^2$ is symmetric in $A$ and $B$, so if (a) were correct, then by symmetry (b) must be correct also. So that leaves (c).
Now, this doesn't establish that (c) is actually correct. To solve the problem properly, note that
$$A = BA = (AB)A = A(BA) = A^2$$,
and similarly $B = B^2$. Thus, $A^2 + B^2 = A + B$.
| {
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$(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ I need to show that $(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ and $a,b,c > 0$ using means of univariate Analysis.
It is intuitively clear that $(a+2)^3+(b+2)^3+(c+2)^3$ is at its minimum (when $a+b+c=3$) if $a,b,c$ have "equal weights", i.e. $a=b=c=1$.
To show that formally one can firstly fix $0<c \le 3$, introduce a variable $0\le\alpha\le1$ and express $a$ and $b$ through $\alpha$
$$a= \alpha(3-c)$$ $$b=(1-\alpha)(3-c)$$
Then we minimize $(a+2)^3+(b+2)^3+(c+2)^3$ w.r.t $\alpha$ treating $c$ as a parameter. We get $\alpha=\frac{1}{2}(3-c)$. Further we maximize $(a+2)^3+(b+2)^3+(c+2)^3$ one more time w.r.t. $c$.
That is quite tedious. I am wondering whether their is a more elegant way. Probably using idea of norms. Basically we need to show that $\lVert(a,b,c)+(2,2,2)\rVert_3 \ge (81)^{1/3}$ while $\lVert(a,b,c)\rVert_1=3$ and $a,b,c>0$.
| Alternately, here's a hint
$$(x+2)^3-27 -27(x-1) =(x-1)^2(x+8)\geqslant 0$$
| {
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Find $ \int\frac{\mathrm{d}x}{(1-x^2)^{3/2}}$ $$
\int\frac{1}{(1-x^2)^{3/2}}\mathrm{d}x=?
$$
I am aware that you can do it with $x=\sin u$ but how do you do it with $x=\tanh u$? I kept ending with with $x^2+1$ in my denominator instead of the correct $x^2-1$.
| If you set $x=\tanh u$, $\;\mathrm d\mkern1mu x=\dfrac1{\cosh^2u}\,\mathrm d\mkern1mu u$, you get
\begin{align*}\int\frac{1}{(1-x^2)^{3/2}}\mathrm{d}x&=\int\cosh^3u\dfrac1{\cosh^2u}\,\mathrm d\mkern1mu u=\int\cosh u\,\mathrm d\mkern1mu u\\
&=\sinh u =\tanh u\cosh u=\frac{\tanh u}{\sqrt{1-\tanh^2 u}}\\
&=\frac x{\sqrt{1-x^2}}.
\end{align*}
Other method: integration by parts:
We'll integrate $I=\displaystyle\int\dfrac{\mathrm d\mkern1mu x}{\sqrt{1-x^2}}$ by parts. Set
\begin{align*}
u&=\dfrac{1}{\sqrt{1-x^2}},&\mathrm{d}x&=x\\
\text{whence}\qquad\mathrm d\mkern1mu u&=\frac x{(1-x^2)^{3/2}},& v&=x.
\end{align*}
We obtain
\begin{align*}
I&=uv-\int v\,\mathrm d\mkern1mu u=\frac{x}{\sqrt{1-x^2}}-\int\frac{x^2}{(1-x^2)^{3/2}}\,\mathrm d\mkern1mu x\\
&=\frac{x}{\sqrt{1-x^2}}+\int\frac{1-x^2}{(1-x^2)^{3/2}}\,\mathrm d\mkern1mu x
-\int\frac{\mathrm d\mkern1mu x}{(1-x^2)^{3/2}}\\
&=\frac{x}{\sqrt{1-x^2}}+I-\int\frac{\mathrm d\mkern1mu x}{(1-x^2)^{3/2}}
\end{align*}
whence the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to simplify this expression of square roots? If one simplifies the infinite nested radical
$$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}$$
Does one get $$\sqrt x\frac{1+\sqrt5}2$$
Any help would be appreciated!
| $$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}=\sqrt{x+x\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
$$\sqrt{x+x\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}=\sqrt{x}\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
now let
$$y=\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
so
$$y^2-1=y$$
$$y=\frac{1\pm\sqrt{5}}{2}$$
select the $$y=\frac{1+\sqrt{5}}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Study extreme values of functions of several variables. Well, I have to solve the following problem:
$\textit{Study the extreme values of}$ $f(x,y)=x^2+y^2+\alpha x^3y^3$ $\textit{depending on}$ $\alpha$, $where $ $\alpha\in \mathbb{R}$.
It's easy to find that the extreme values of the function must verify that
$$\left\{
\begin{array}{ll}
2x+3\alpha x^2y^3 =0\\
2y+3\alpha x^3y^2=0
\end{array}
\right.$$
so this extreme values could be
*
*If $\alpha=0$, only $(0,0)$.
*If $\alpha>0$, $(0,0),(\sqrt[4]{\frac{2}{3\alpha}},-\sqrt[4]{\frac{2}{3\alpha}}),(-\sqrt[4]{\frac{2}{3\alpha}},\sqrt[4]{\frac{2}{3\alpha}})$.
*If $\alpha<0$, $(0,0),(\sqrt[4]{\frac{2}{3\alpha}},\sqrt[4]{\frac{2}{3\alpha}}),(-\sqrt[4]{\frac{2}{3\alpha}},-\sqrt[4]{\frac{2}{3\alpha}})$.
Now, we can prove that $(0,0)$ is a relative minimum. Applying the Sylvester theorem, I also proved that if $\alpha>0$, $(\sqrt[4]{\frac{2}{3\alpha}},-\sqrt[4]{\frac{2}{3\alpha}}),(-\sqrt[4]{\frac{2}{3\alpha}},\sqrt[4]{\frac{2}{3\alpha}})$ are saddle points. My problem is to determine what are $(\sqrt[4]{\frac{2}{3\alpha}},\sqrt[4]{\frac{2}{3\alpha}}),(-\sqrt[4]{\frac{2}{3\alpha}},-\sqrt[4]{\frac{2}{3\alpha}})$ for $\alpha<0$, whose hessian matrix is semidefinite positive.
| For the case $\alpha<0$, let $t=\sqrt[4]\frac{-2}{3\alpha}$, so the critical points are given by $(0,0), (t,t), \text{ and } (-t,-t)$.
As remarked above, there is a relative minimum at $(0,0)$,
and there are saddle points at $(t,t)$ and $(-t,-t)$ since
$\;\;\;f_{xx}=2+6\alpha xy^3=2+6\alpha\big(\frac{-2}{3\alpha}\big)=-2$,
$\;\;\;f_{yy}=2+6\alpha x^3y=2+6\alpha\big(\frac{-2}{3\alpha}\big)=-2$,
$\;\;\;f_{xy}=9\alpha x^2y^2=9\alpha\big(\frac{-2}{3\alpha}\big)=-6$, and
$\;\;\;D=f_{xx}f_{yy}-f_{xy}^2=4-36=-32<0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
By really long division i got :-
$$Q=ax^{15} + (a+b)x^{14} + \dots +(610a + 377b) $$
$$R = x(987a+610b)+1+610a+377b$$
since remainder is $0 $,
$$987a+610b = 0$$
$$1+610a+377b = 0$$
from which i got $a = -610, b = 987$
but from wolfram alpha the remainder of $-610x^{17}+987x^{16}+1 \over x^2-x-1$ is $x-1$
Somebody please show me where i went wrong ?
Thanks.
| Alternatively, observe that $x^2-x-1$, $x^3-2x^2+1$, and $2x^4-3x^3-1$ are the only polynomials degree less than $5$ divisible by $x^2-x-1$ which are of the form $A\,x^{n+1}+B\,x^{n}+C$ with $A,B,C\in\mathbb{Z}$, $A>0$, and $\gcd(A,B,C)=1$. We suppose that $a_n\,x^{n+1}-b_n\,x^n+(-1)^n$ is divisible by $x^2-x-1$ for each $n\in\mathbb{N}$. Then, for all integera $n>1$,
$$
\begin{align}
\left(a_n\,x^{n+1}-b_n\,x^n+(-1)^n\right)+\left(a_{n-1}\,x^n-b_{n-1}\,x^{n-1}+(-1)^{n-1}\right)\phantom{aaaaaaaaaaaa}
\\\phantom{aaaaaaaaaaaa}=\left(a_n\,x^2-\left(b_{n}-a_{n-1}\right)\,x-b_{n-1}\right)x^{n-1}
\end{align}$$
must be divisible by $x^2-x-1$. This means $a_n=b_{n}-a_{n-1}=b_{n-1}$ for each integer $n>1$. Therefore, $b_{n}=a_n+a_{n-1}$ and $b_{n}=a_{n+1}$, whence
$$a_{n+1}=a_n+a_{n-1}\text{ and }b_n=a_{n+1}\,.$$
Since $a_1=F_1$ and $a_2=F_2$, we conclude that $a_n=F_n$ and $b_n=F_{n+1}$ for all $n\in\mathbb{N}$. Here, $\left(F_n\right)_{n=1}^\infty$ is the standard Fibonacci sequence: $F_0=F_1=1$ and $F_{n+1}=F_n+F_{n-1}$ for $n-1,2,3,\ldots$. In particular, $F_{16}\,x^{17}-F_{17}\,x^{16}+1$ is the only polynomial of the form $A\,x^{n+1}+B\,x^{n}+C$ with $A,B,C\in\mathbb{Z}$, $A>0$, and $\gcd(A,B,C)=1$ divisible by $x^2-x-1$.
In general, if complex numbers $\alpha$ and $\beta$ are such that $\alpha\neq 0$ and $\beta^2\neq 4\alpha$, then there exist unique $a_n,b_n\in\mathbb{C}$ such that $a_n\,x^{n+1}+b_n\,x^n+1$ is divisible by $\alpha\,x^2+\beta\,x+1$. If the complex numbers $u$ and $v$ are the two roots of $\alpha\,x^2+\beta\,x+1$, say, $u=\frac{-\beta+\sqrt{\beta^2-4\alpha}}{2\alpha}$ and $v=\frac{-\beta-\sqrt{\beta^2-4\alpha}}{2\alpha}$, then
$$a_n=\alpha^n\,\left(\frac{u^n-v^n}{u-v}\right)$$
and
$$b_n=-\frac{a_{n+1}}{\alpha}=-\alpha^n\,\left(\frac{u^{n+1}-v^{n+1}}{u-v}\right)$$
for all $n=0,1,2,\ldots$. If $\beta^2=4\alpha$, then there is a unique root $z:=-\frac{\beta}{2\alpha}$ of $\alpha\,x^2+\beta\,x+1$. In this case,
$$a_n=n\,\alpha^{n}\,z^{n-1}\text{ and }b_n=-(n+1)\,\alpha^{n}\,z^n$$
for all $n=0,1,2,\ldots$. (In fact, $\mathbb{C}$ can be replaced by any field, and we just have to work in the splitting field of $\alpha\,x^2+\beta\,x+1$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Complex Numbers - Given $(a+b) +i(a-b) = (1+i)^2 + i(2+i)$ obtain the values of $a$ and $b$. This would be a very easy complex number question to someone I understand of most of it its just one of those questions I should know but I've stared at it so much I'm stuck! could someone please explain it to me! here it is:
Question - Given $(a+b) +i(a-b) = (1+i)^2 + i(2+i)$ obtain the values of $a$ and $b$. Note $i=\sqrt{(-1)}$
Thank you all! hope to hear from you soon!
| Just do it.
$(1+i)^2 + i(2+i) =$
$(1 + 2i + i^2)+(2i + i^2)=$
$(1 + 2i -1) + (2i - 1)=$
$-1 + 4i$
So $-1 + 4i = (a+b) + (a+b)i$
So $a + b = -1$ and $a+b = 4$.
Well, no wonder you are stuck! You were given an impossible equation.
====
From your commments it sounds like what you we actually given was
$(a+b) + (a-b)i = (1+i)^2 + i(2+i)$
or maybe something else?
You still do it the same way:
$(1+i)^2 + i(2+i) = -1 + 4i$
so $(a+b) +(a-b)i = -1 + 4i$
so $a+b = -1$ and $a-b = 4$.
This is solvable.
$a = - 1 - b$
$-1-b -b = 4$
$-2b = 5$
$b = -5/2$
$a = -1 -(-5/2) = 3/2$.
or....
$a + b = -1$
$a - b = 4$
$(a + b) + (a-b) = -1 + 4$
$2a = 3$
$a = 3/2$
$(a + b) - (a-b) = -1 -4$
$2b = -5$
$b = -5/2$.
| {
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Prove this inequality $\sum\limits_\text{cyc}\sqrt{1-xy}\ge 2$ Let $x,y,z\ge 0$, and $x+y+z=2$, show that
$$\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\ge 2.$$
Mt try: $$\Longleftrightarrow 3-(xy+yz+zx)+2\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge 4$$
or
$$\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge\dfrac{1}{2}(1+xy+yz+zx)$$
| Lets assume $z=\min\{x,y,z\}$ then note $\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\geq\sqrt{1-\frac{(2-z)^2}{4}}+2\sqrt[4]{(1-yz)(1-xz)}=\sqrt{z-\frac{z^2}{4}}+2\sqrt[4]{1-z(2-z)+z^2xy}\geq\sqrt{z-\frac{z^2}{4}}+2\sqrt[4]{(1-z)^2+z^4}\geq 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to evaluate the integral $\int \sqrt{1+\sin(x)} dx$ To find:
$$\int \sqrt{1+\sin(x)} dx$$
What I tried:
I put $\tan(\frac{x}{2}) = t$, using which I got it to:
$$I = 2\int \dfrac{1+t}{(1+t^2)^{\frac{3}{2}}}dt$$
Now I am badly stuck. There seems no way to approach this one. Please give a hint. Also, can we initially to some manipulations on the original integral to make it easy? Thank you.
| \since, $$1=\sin^2\left(\frac{x}{2}\right)+\cos^2\left(\frac{x}{2}\right)$$
$$\sin (x)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$
Therefore, we can put the above values of $1$ and $\sin (x)$ in the question.
Now, we have. $$\int \sqrt{\sin^2\left(\frac{x}{2}\right)+ \cos^2\left(\frac{x}{2}\right)+2\sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)} \,dx$$
And,$$\sin^2\left(\frac{x}{2}\right)+\cos^2\left(\frac{x}{2}\right)+2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)=\left(\sin\left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)\right)^2$$
By putting the above value, we get
$$\int\sqrt{\left(\sin\left(\frac{x}{2}\right)+ \cos\left(\frac{x}{2}\right)\right)^2}\, dx$$
$$=\int \sin\left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)\,\,dx$$
$$=-2\cos\left(\frac{x}{2}\right)+2\sin\left(\frac{x}{2}\right)+C$$
| {
"language": "en",
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Sum of the series $1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$ Find the sum of $n$ terms of following series:
$$1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$$
I was trying to use $S_n=\sum T_n$, but while writing $T_n$ I get a term having $n^4$ and I don't know $\sum n^4$. Is there any other way find the sum?
| An alternative to finding the formula for $\sum k^4$ is to instead write the $k$th term as $$(k+2)(k+1)^2k=(k+3)(k+2)(k+1) k-2(k+2)(k+1)k.$$
That is, one trades sums of integer powers for sums of consecutive products. To see the advantage, note for example that
$$1\cdot 2\cdot 3 +2\cdot 3\cdot 4+3\cdot 4\cdot 5+4\cdot 5\cdot 6=210= \frac{1}{4}(4\cdot 5\cdot 6\cdot 7),$$
i.e. the sum of products of three consecutive integers can be expressed in terms of a product of four consecutive integers. A similar pattern holds for other such sums, allowing the required formulas to be readily found. (By comparison, the pattern for sums of integer powers is much more involved.) From such computations one can then obtain the desired formula for the original sum.
| {
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"timestamp": "2023-03-29T00:00:00",
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Faster way to get a rational integral $\int_2^3 \frac{x^{100}+1}{x^3+1} \, dx$ So my problem is calculating this integral:
$$
\int_{2}^{3} \frac{x^{100}+1}{x^3+1} \, dx
$$
I know it can be done by polynomial division but that is really tedious I would have to divide $\approx 30$ times.
And i know that Mathematica would give me the answer in the blink of an eye.
Is there a clever way to do it pen&paper?
| Well...maybe the division isn't that bad. Do you know about geometric series? What's
$$
1 + r + r^2 + \ldots + r^{32}?
$$
It's
$$
\frac{r^{33} - 1}{r-1}
$$
If you apply this to $r = -x^3$, you get
$$
\frac{-x^{99} - 1}{-(x^3)-1}
= \frac{x^{99} + 1}{x^3 + 1}
$$
I know that's not what you wanted, but if you do a little algebra, you can do this:
\begin{align}
\frac{x^{100} + 1}{x^3 + 1}
&= \frac{x^{100} + x - x + 1}{x^3 + 1}\\
&= \frac{x^{100} + x}{x^3 + 1} + \frac{-x + 1}{x^3 + 1}\\
&= x\frac{x^{99} + 1}{x^3 + 1} + \frac{-x + 1}{x^3 + 1}.
\end{align}
Now integrating the first fraction is easy because of the geometric series, and all you have to deal with is the last one.
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if the sum of the digits of $a+b$, $b+c$ and $a+c$ is $<5$, then is the sum of the digits of $a+b+c$ less or equal than 50? This is a problem from a math olympiad:
Let $S(x)$ denote the sum of the digits of a natural number $x$ in its decimal representation. Do there exist three natural numbers $a$, $b$ and $c$ with $S(a+b)<5$, $S(a+c)<5$, $S(b+c)<5$ and $S(a+b+c)>50$?
My attempt:
NOTATION: Given a real number $x$, $[x]$ is the integer part of $x$.
REMARK 1: Let $x$ be a natural number with $2$ digits. Then the sum of its two digits is
$$ \left[\frac{x}{10}\right]+x-\left[\frac{x}{10}\right]\cdot 10=x-9\cdot \left[\frac{x}{10}\right]. $$
REMARK 2: If $x$ is a real number with $\left[ x\right]\geq1$, then $\left[ 2x\right]\leq 3\left[ x\right]$.
Let $a$, $b$ and $c$ be any natural numbers. Write
$$ a=\sum_{i=0}^{n-1}a_i 10^i,\quad b=\sum_{i=0}^{n-1}b_i 10^i, \quad c=\sum_{i=0}^{n-1}c_i 10^i, $$
where $a_i,b_i,c_i\in\{0,\ldots,9\}$ and $a_n\neq0$ or $b_n\neq0$ or $c_n\neq0$. By Remark 1,
$$ S(a+b+c)=\sum_{i=0}^{n-1}\left(a_i+b_i+c_i-9\cdot \left[ \frac{a_i+b_i+c_i}{10}\right]\right), $$
$$ S(a+b)=\sum_{i=0}^{n-1}\left(a_i+b_i-9\cdot \left[ \frac{a_i+b_i}{10}\right]\right), $$
$$ S(b+c)=\sum_{i=0}^{n-1}\left(b_i+c_i-9\cdot \left[ \frac{b_i+c_i}{10}\right]\right), $$
$$ S(a+c)=\sum_{i=0}^{n-1}\left(a_i+c_i-9\cdot \left[ \frac{a_i+c_i}{10}\right]\right). $$
By Remark 2, if $\left[ (a_i+b_i+c_i)/10\right]\geq1$, then $$3\left[(a_i+b_i+c_i)/10\right]\geq \left[ 2(a_i+b_i+c_i)/10\right]\geq \left[ (a_i+b_i)/10\right]+\left[ (b_i+c_i)/10\right]+\left[ (a_i+c_i)/10\right].$$ Thus,
$$ \begin{align*}
S(a+b+c)= {} & \sum_{i=0}^{n-1}\left(a_i+b_i+c_i-9\cdot \left[ \frac{a_i+b_i+c_i}{10}\right]\right) \\\\
= {} & \frac12\sum_{i=0}^{n-1}(a_i+b_i)+\frac12\sum_{i=0}^{n-1}(b_i+c_i)+\frac12\sum_{i=0}^{n-1}(a_i+c_i)\\
&\qquad-9\sum_{\substack{{0\leq i\leq n-1}\\{(a_i+b_i+c_i)/10\geq1}}}\left[ \frac{a_i+b_i+c_i}{10}\right] \\\\
\leq {} & \frac12\sum_{i=0}^{n-1}(a_i+b_i)+\frac12\sum_{i=0}^{n-1}(b_i+c_i)+\frac12\sum_{i=0}^{n-1}(a_i+c_i)-3\sum_{i=0}^{n-1}\left[ \frac{a_i+b_i}{10}\right]\\
&\qquad-3\sum_{i=0}^{n-1}\left[ \frac{b_i+c_i}{10}\right]-3\sum_{i=0}^{n-1}\left[ \frac{a_i+c_i}{10}\right] \\\\
= & \frac12\sum_{i=0}^{n-1}\left(a_i+b_i-6\left[ \frac{a_i+b_i}{10}\right]\right)+\frac12\sum_{i=0}^{n-1}\left(b_i+c_i-6\left[ \frac{b_i+c_i}{10}\right]\right)\\
&\qquad+\frac12\sum_{i=0}^{n-1}\left(a_i+c_i-6\left[ \frac{a_i+c_i}{10}\right]\right) \\\\
= {} & \frac{S(a+b)+S(b+c)+S(a+c)}{2}+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{a_i+b_i}{10}\right]\\
&\qquad+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{b_i+c_i}{10}\right]+\frac{3}{2}\sum_{i=0}^{n-1}\left[ \frac{a_i+c_i}{10}\right] \\\\
\leq {} & \frac{S(a+b)+S(b+c)+S(a+c)}{2}+\frac{3}{2}S(a+b)+\frac{3}{2}S(b+c)+\frac{3}{2}S(a+c) \\\\
= {} & 2(S(a+b)+S(b+c)+S(a+c))\;.
\end{align*}
$$ Then, if $S(a+b)<5$, $S(a+c)<5$, $S(b+c)<5$, we have $S(a+b+c)<2(5+5+5)=30$.
| The answer to the question in the title is no -- it is possible that $S(a+b+c) = 51$. In particular, your proof is incorrect.
Take
\begin{align*}
a &= 545\;545\;545\;5 \\
b &= 554\;554\;554\;5 \\
c &= 455\;455\;455\;5.
\end{align*}
(Spaces to aid readability, so you can tell what's going on here.) Then we have
\begin{align*}
a+b &= 1\;100\;100\;100\;0 \\
a+c &= 1\;001\;001\;001\;0 \\
b+c &= 1\;010\;010\;010\;0 \\
\end{align*}
So $S(a+b) = S(a+c) = S(b+c) = 4 < 5$.
But
$$
a+b+c = 1\;555\;555\;555\;5
$$
so $S(a+b+c) = 51$.
Why your proof fails
The error is in the beginning, in asserting that the sum of digits of $S(a+b+c)$ is equal to $\sum_i S(a_i + b_i + c_i)$, and similar for $a+b$,$a+c$, and $b+c$. While $S(a+b+c) \le \sum_i S(a_i + b_i + c_i)$, they certainly need not be equal. Replacing it with $\le$ instead of $=$, you can see that your proof no longer goes through.
Also note that if you use $S(a+b), S(a+c), S(b+c) \le 4$ at the end instead of $S(a+b), S(a+c), S(b+c) < 5$, then your proof would purport to show $S(a+b+c) \le 2 (12) = 24$, which has even simpler counterexamples.
Additional notes
While it is not true that $S(a+b+c) \le 50$, it is still true that $S(a+b+c) \le 60$. Here is a proof.
First, we can note that
$$
S(x + y) \le S(x) + S(y). \tag{1}
$$
To prove it, first imagine adding $x$ and $y$ in base 10 without carrying. Then the sum of digits stays the same. But each time we carry, we reduce the sum. So $x+y$ after all carrying is completed must have a smaller sum of digits.
A second observation is that
$$
S(2x) \ge \frac15 S(x) \tag{2}
$$
To prove it, first consider what multliplication by $2$ does to single digits: $1 \mapsto 2$, $2 \mapsto 4$, $3 \mapsto 6$, $4 \mapsto 8$, $5 \mapsto 10$, $6 \mapsto 12$, $7 \mapsto 14$, $8 \mapsto 16$, $9 \mapsto 18$.
In every case, the sum of digits is at least $\frac{1}{5}$ of the original digit (the worst case being the digit $5$ becoming $10$).
Then notice (and this is special to multiplication by $2$ in particular) that to find $2x$, we may multiply each digit by $2$ individually, resulting in a one or two digit number for each digit, and then add all these up with no carrying.
In other words, $S(2x)$ is the sum over digits $d$ of $x$ of $S(2d)$. Since $S(2d) \ge \frac{1}{5}$, it follows that $S(2x) \ge \frac{1}{5} S(x)$.
Putting (2) and (1) together, in that order:
$$
S(a+b+c) \le 5S(2(a+b+c)) \le 5\left[S(a+b) + S(b+c) + S(a+c)\right] \le 5\left[4 + 4 + 4\right] = 60.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $(A-B)^2=O_2$ then $\det(A^2 - B^2)=(\det(A) - \det(B))^2$
Let $A,B \in M_2(\mathbb{R})$ be two matrices such that $(A-B)^2=O_2$.
Prove $\det(A^2 - B^2)=(\det(A) - \det(B))^2$.
OBS. $O_2$ is the zero matrix
Let $D=A-B$ then $D^2=O_2$ therefore, using Cayley Hamilton theorem, we get $tr(D)=\det(D)=0$. It follows $tr(A)=tr(B)=t$ and, from here, $A^2 -B^2=t(A-B) + (\det(A) - \det(B))I_2$
This is all I could get.
UPDATE
The matrices are from $M_2(\mathbb{R})$. Sorry for misleading you.
| Since $D=A-B$ is nilpotent, it is similar to the matrix $\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$. So without loss of generality assume $D = \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$. If $B = \left(\begin{matrix} a & b \\ c & d \end{matrix}\right)$, then $A = B+D = \left(\begin{matrix} a & b+1 \\ c & d \end{matrix}\right)$. Now
$$ A^2-B^2 = (B+D)^2 - B^2 = (B^2 + BD + DB + D^2) - B^2 = BD+DB$$
since $D^2 = 0$. Notice that
$$ BD+DB = \left(\begin{matrix} a & b \\ c & d \end{matrix}\right)\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right) + \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)\left(\begin{matrix} a & b \\ c & d \end{matrix}\right) = \left(\begin{matrix} c & a + d \\ 0 & c\end{matrix}\right) $$
and hence
$$ \det(A^2-B^2) = \det(BD+DB) = c^2. $$
On the other hand,
$$ \det(A) = \det\left(\begin{matrix} a & b+1 \\ c & d \end{matrix}\right) = ad - (b+1)c = \det(B) - c$$
and hence $(\det(A) - \det(B))^2 = c^2$ as well.
Thus $\det(A^2-B^2) = c^2 = (\det(A)-\det(B))^2$, as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve an inequality with absolute values on both sides? I have the following inequality:
$$|x+3| \geq |x-1| $$
Following this answer I get:
$$
|x+3|=\left\{ \begin{align}
x+3 & \text{ , if }x\geq -3 \\
-x-3 & \text{ , if }x <-3
\end{align}
\right\}
$$
$$
|x-1|=\left\{ \begin{align}
x-1 & \text{ , if }x\geq 1 \\
-x+1 & \text{ , if }x < 1
\end{align}
\right\}
$$
Putting those together I get 3 sets of equations:
For $x<-3$:
$$-x-3\ge-x+1$$
For $-3 \le x< 1$:
$$x+3\ge-x+1$$
For $x\ge1$:
$$x+3\ge x-1$$
The first inequality however gives me:
$$-3 \ge 1$$
What am I doing wrong here? The answer to the problem is $x \ge -1$ by the way.
| You didn't do anything wrong. You just misinterpreted what you were trying to do and what the information tells you:
You have 3 ranges to consider and in each range you have a set of inequalities to interpret.
If $x \le -3$ (which it might or might not be).
We have $-x - 3 \ge -x + 1$ so $-3 \ge 1$. This is impossible so this is not an option.
So we know $x > -3$.
If $-3 < x \le 1$ (which it might or might not be).
We have $x + 3 \ge -x + 1$ so $x \ge -1$.
So if $-3< x \le 1 \implies x \ge -1 \implies -1 \le x \le 1$. IF $-3 < x \le -1$ which it might or might not be.
And if $x > 1$ (which it might or might not be)
We have $x + 3 \ge x -1$ so $3 \ge -1$ which is trivially true.
So if $x > 1$ then SOMETHING TRUE. So $x > 1$ is possible. But it might or might not be true. But it's possible and if it is true ... we conclude nothing further.
Putting those together we conclude:
$-1 \le x \le 1$ OR $x > 1$ or
$x \in [-1,1] \cup [1,\infty) = [-1,\infty)$ or
$x \ge -1$.
.....
But another way, maybe better, to do it is:
$|x+3| \geq |x-1|$
$-|x+3| \le x-1 \le |x+3|$
$\min(x+3,-x-3) \le x-1 \le \max (x+3, -x - 3)$
$\min(x+4, -x -2) \le x \le \max (x+4, -x-2)$
So either $x+4 \le x \le -x-2$ or $-x-2 \le x \le x+4$.
The first one is clearly impossible so
$-x -2 \le x \le x+4$ ($x \le x + 4$ is redundant.)
so $-x-2 \le x$ so $x \ge -1$.
.....
I don't know. Maybe that is harder.
====
Oh.
David Quinn offers a third option that is probably the easiest to solve, although intuitively I would have assumed it'd give the most extraneous impossible ranges. As it turns out, it gives the least.
$|x + 3| \ge |x-1|$
$(x+3)^2 \ge (x-1)^2$
$x^2 + 6x + 9 \ge x^2 - 2x + 1$
$6x + 9 \ge -2x + 1$
which will give one and only one unambiguous solution range.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does this constitute a valid proof that $\frac{x^2}{1+x^4} \leq \frac{1}{2}$?
Prove that $$\frac{x^2}{1+x^4} \leq \frac{1}{2}.$$
First off, we observe that the expression on the LHS is positive for all $x \in \Bbb R,$ and equality is achieved iff $x \in \{-1, 1 \}$. That being said, we start by manipulating the expression as below;
\begin{align}
\frac{x^2}{1+x^4} - \frac{1}{2} &\leq 0\\\\
\impliedby\frac{2x^2 - (1+x^4)}{2(1+x^4)} &\leq 0\\\\
\impliedby -\frac{(x^4 - 2x^2 + 1)}{2(1+x^4)}&\leq 0\\\\
\impliedby \frac{x^4 - 2x^2 + 1}{2(1+x^4)} &\geq 0\\\\
\end{align}
We then make the observation that $x^4 - 2x^2 + 1 = (x^2 - 1)^2 \geq 0$ for all $x \in \Bbb R$ and that $1 + x^4 \geq 1$ for all $x \in \Bbb R$, and so the final inequality holds for all $x \in \Bbb R. \\\hspace{92pt}\square$
Is it okay to prove it in this way? I have no solutions so it would be nice if I could have some validation!
| Yes it's Ok, you could also just multiply both sides by $2(1+x^4)$ having to prove that
$$
2x^2\le x^4+1
$$ that is
$$
0\le(x^2-1)^2.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Largest divisor of $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$
Let $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$. What is the largest integer that is a divisor of $P(n)$ for all positive even integers $n$?
I can see that obviously it is divisible by $3$ and $5$ and hence answer is $15$ but I can't prove it.
| We have that $\gcd(P(1),P(2))=15$, hence a number $N$ with the property that $N\mid P(n)$ for every $n$ has to be a divisor of $15$. We may notice that $3\mid P(n)$ for every $n$ since exactly one number among $n+3,n+1,n+5$ is a multiple of three. In a similar way, exactly one number among $n+5,n+1,n+7,n+3,n+9$ is a multiple of $5$, hence $\color{red}{15}$ is the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a$ and $b$ are consecutive integers, prove that $a^2 + b^2 + a^2b^2$ is a perfect square. Problem is as stated in the title. Source is Larson's 'Problem Solving through Problems'. I've tried all kinds of factorizations with this trying to get it to the form $$k^2l^2$$ but nothing's clicking. I tried Bézout but the same expression can be written as $$a^2 + (a^2 + 1)(a+1)^2$$ which would imply that there is no real root. Would really appreciate some help, thanks.
| Let $p=\frac{a+b}{2}$, that is if $a<b$, $p=(2a+1)/2$.
\begin{align}(p-\frac{1}{2})^2+(p+\frac12)^2+(p-\frac12)^2(p+\frac12)^2&=(2p^2+\frac12)+(p^2-\frac14)^2\\
&=p^4+\frac32 p^2+\frac{9}{16}\\
&=(p^2+\frac{3}{4})^2
\end{align}
since $p^2=\frac{4a^2+4a+1}{4}$,
$$p^2+\frac{3}{4}=a^2+a+1.$$
Hence, $$a^2+b^2+a^2b^2=(a^2+a+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909974",
"timestamp": "2023-03-29T00:00:00",
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if $AB=AC=EF,AE=BE,\angle BAC=120^{\circ}$ then find $\angle ABE$ If $$AB=AC=EF, AE=BE, \angle BAC=120^{\circ}, \angle ADB=\angle BEF=90^{\circ}$$
find $\angle ABE$
Following is one method:
Let $\angle ABE=a, \angle AFE=\angle DBE=30^{\circ}-a, \angle EAF=60^{\circ}-a$. Use sine thereom, we have
$$\dfrac{\sin{a}}{\sin{(180^{\circ}-2a)}}=\dfrac{AE}{AB}=\dfrac{AE}{EF}=\dfrac{\sin{(30^{\circ}-a)}}{\sin{(60^{\circ}-a)}}$$
$$\Longrightarrow 2\cos{a}\sin{(30^{\circ}-a)}=\sin{(60^{\circ}-a)}$$
$$\Longrightarrow \dfrac{1}{2}+\sin{(30^{\circ}-2a)}=\sin{(60^{\circ}-a)}$$
$$\Longrightarrow 2\cos^2{(a+30^{\circ})}-\cos{(a+30^{\circ})}-\dfrac{1}{2}=0$$
$$\Longrightarrow \cos{(a+30^{\circ})}=\dfrac{\sqrt{5}+1}{4}=\cos{36^{\circ}}$$
so we have
$$a=6^{\circ}$$
Question: Can find geometry methods or other simple methods?
|
Let $\angle EBA = \alpha$.
Let $A$ and $X$ be symmetric with respect to $BE$. Then $BX=AB$. Also note that $BE=AE=EX$, so $A,B,X$ lie on the circle with center $E$ and radius $AE$.
Let $Y$ be on $AF$ and $AY=AB$. Since $\angle BAY = 60^\circ$, $\triangle BAY$ is equilateral. Thus $BY=AB$. Therefore $A,X,Y$ lie on the circle with center $B$ and radius $AB$.
Since $AE=BE$ and $AY=BY$, triangles $AYE$, $BYE$ are congruent (sss). In particular $\angle AYE = \angle EYB = \frac 12 \angle AYB = 30^\circ$.
We have
$$\angle AXY = \frac 12 (360^\circ - \angle YBA) = 150^\circ = 180^\circ - \angle AYE = \angle EYF.$$
Moreover
$$\angle XYA = \frac 12 \angle XBA = \alpha$$
and
$$\begin{align*}\angle FEY & = \angle BEY - \angle BEF = \\
& = (180^\circ - \angle YBE - \angle EYB) - 90^\circ = \\
& = 180^\circ - (60^\circ - \alpha) - 30^\circ - 90^\circ = \\
& = \alpha,
\end{align*}$$
therefore
$$\angle XYA = \angle FEY.$$
It follows that triangles $AXY$ and $FYE$ are similar. Their similarity ratio is $\dfrac{AY}{EF} = 1$ so in fact they are congruent. Thus $EY=XY$.
Using the fact that $EXY$ is isosceles we get
\begin{align*}180^\circ & = \angle XYE + 2 \angle YEX = \\
& = 30^\circ + \alpha + 2(\angle YEA - \angle XEA) = \\
& = 30^\circ + \alpha + 2\angle BEY - 2\angle XEA = \\
& = 30^\circ + \alpha + 180^\circ + 2 \alpha - 8 \alpha = \\
& = 210^\circ - 5 \alpha,
\end{align*}
therefore $$\alpha=6^\circ.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Partial Fraction Expansion of $12\frac{x^3+4}{(x^2-1)(x^2+3x+2)}$ Find the vector $(A,B,C,D)$ if $A$, $B$, $C$, and $D$ are the coefficients of the partial fractions expansion of
$$12\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2}.$$
After plugging in a few values for $x,$ I got these equations:
$$6A+D=12$$
$$4A+B+2C+D=0$$
$$5A+D=B+C$$
$$2A+B=2C+2D+48$$
But now I'm stuck. Thanks in advance for answering!
| You should not begin with solving systems of linear equations. Rewrite the partial fractions decomposition so as to remove all denominators. Multiplying both sides with $(x-1)(x+2)(x+1)^2$, you get
$$12(x^3+4)=A(x+2)(x+1)^2+B(x-1)(x+1)^2 +C(x-1)(x+2)(x+1)+D(x-1)(x+2).$$
Now set
*
*$x=1$: it yields at once $12\cdot 5= 12A$, whence $A=5$;
*$x=-2\;$ yields $\;12^2=-3B$, whence $B=-48$;
*$x=-1\;$ yields $\;12\cdot 5=-2D$, whence $D=-30$.
There remains to calculte $C$. The simplest is to set $x=0$, and you get $48=2A-B-2C-2D$, i.e. $C=35$.
| {
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Prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$ I'm trying to prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$, which is part of a larger proof. We can write:
$n^3+2n^2+n=2(1+2+...+n)(n+1)$
$n^3+2n^2+n=(1+2+...+n)(2n+2)$
$n^3+2n^2+n=(2n+4n+6n+...+2n^2)+(2+4+6+...+n)$
I have no idea how to proceed, though.
| The sum $P(n):=1+2+\cdots n$ must be a quadratic polynomial in $n$, because $P(n)-P(n-1)=n$ is a linear polynomial.
Then as both members are cubic polynomials, it suffices to check equality for four distinct values of $n$.
$$\begin{align}n=0&\to 1^3=2()\cdot1+1^2&=1,\\
n=1&\to 2^3=2(1)\cdot2+2^2&=8,\\
n=2&\to 3^3=2(1+2)\cdot3+3^2&=27,\\
n=3&\to 4^3=2(1+2+3)\cdot4+4^2&=64.\end{align}
$$
This proves the identity for all $n$.
With the simplified identity (in my other answer), checking for three values suffices
$$\begin{align}n=0&\to 1^2=2()+1&=1,\\
n=1&\to 2^2=2(1)+2&=4,\\
n=2&\to 3^2=2(1+2)+3&=9.\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{9}{2}$$
I only use Nesbitt's inequality. Then we only need that $3(ab+bc+ca)\geq 1$, but it is not correct, because $3(ab+bc+ca)\leq 1$.
Maybe some ideas will be more that grateful.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
We need to prove that
$$\frac{(a+b+c)^3\sum\limits_{cyc}(a^2+3ab)}{\prod\limits_{cyc}(a+b)}+3(ab+ac+bc)\geq\frac{11}{2}(a+b+c)^2$$
which is fifth degree, which says that we need to prove a linear inequality of $w^3$,
which says that it's enough to prove the last inequality for an extremal value of $w^3$.
Since $a$, $b$ and $c$ are positive roots of the equation
$(x-a)(x-b)(x-c)=0$ or $w^3=x^3-3ux^2+3v^2x$,
we see that $w^3$ gets an extremal value for equality case of two variables
or maybe for $w^3\rightarrow0^+$.
*
*$w^3\rightarrow0^+$.
Let $c\rightarrow0^+$ and since the last inequality is homogeneous, we can assume $b=1$,
which gives $2a^4-a^3-a+2\geq0$, which is obvious;
*$b=c=1$, which gives $(a-1)^2(a^2+2a+2)\geq0$.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
How to find the locus of this equation? I am a beginner in math, and am stuck by this problem. The problem is,
Find the locus of the point of intersection of the lines, $x\cos\alpha$ + $y\sin\alpha$ = a and $x\sin\alpha$ - $y\cos\alpha$ = b , where "alpha" is a variable.
What they did, is that they eliminated alpha from the equations and got a relation that x^2+y^2=a^2+b^2.
But what i did is , i found out the value of x from equation 1, then i put that in equation 2 and i get y from there, then i again put y back in equation 1 and i get a relation between x and y, but i also get $\sin\alpha$ and $\cos\alpha$ in my answer with that! What does this mean? Howard where is my answer different from the answer that’s given in the book?
Thanks!
| You goal is to find the set of all $\{x,y\}$ that meet the following criteria:
$x \cos \alpha + y \sin \alpha = a\\
x \sin \alpha - y \cos \alpha = b$
These are perpendicular lines that intersect in a point. But as we allow $\alpha$ to move, that point of intersection moves.
If we can eliminate $\alpha$ then we would have a set of points for all $\alpha$
$x \cos^2 \alpha + y \cos \alpha \sin \alpha = a \cos \alpha\\
x \sin^2 \alpha - y \cos \alpha \sin \alpha= b\sin \alpha$
$x = a \cos\alpha + b \sin \alpha$
Now here is where I would get a little tricky
$x = \sqrt {a^2 + b^2} \cos\alpha \cos \phi + b \sin \alpha \sin \phi $
where $\phi$ is a function of $a,b$ i.e. $\phi = tan^{-1} \frac ba$
but since $a,b$ are constants, $\phi$ is also constant.
$x = \sqrt {a^2 + b^2} \cos(\alpha-\phi)$
and if we solve for $y$ then:
$y = \sqrt {a^2 + b^2} \sin(\alpha-\phi)$
and that is the parametric equation of a circle.
But supposing you don't subscribe such witchcraft.
$x = a \cos\alpha + b \sin \alpha\\
y = a \sin\alpha - b \cos \alpha\\
x^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha + 2ab \cos\alpha \sin\alpha\\
y^2 = a^2 \sin^2 \alpha + b^2 \cos^2 \alpha - 2ab \cos\alpha \sin\alpha\\
x^2 + y^2 = (a^2 + b^2)$
| {
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How to show that the roots of $-x^3+3x+\left(2-\frac{4}{n}\right)=0$ are real (and how to find them) I'm trying to find the three distinct and real roots of
$$-x^3+3x+\left(2-\frac{4}{n}\right)=0,$$
where $n>0$ (we could say $n\geq 2$ if that helps), but I'm not able to get very far:
Using the notation of the Wikipedia-page, I find that the discriminant is
$$\Delta=27\cdot 4\left(\frac{1}{n}-\frac{1}{n^2}\right),$$
which gives
$$C=3\left( 1-\frac{2}{n} \pm 2 i \sqrt{\frac{1}{n}-\frac{1}{n^2}} \;\right)^{1/3},$$
which is then used to find $x$, as
$$x=\frac{C}{3}+\frac{3}{C}.$$
Two things confuse me:
*
*$C$ looks like a non-real number, but $x$ should be real (since $\Delta >0$). How can one further reduce the expression for $x(C)$ to show that the imaginary part is zero? I'm having trouble evaluating that cube root.
*When I use my expression for $x(C)$ in Mathematica and evaluate numerically for some $n$, I only find one of the three solutions that Mathematica finds if I just ask it to give the roots of the original equation (the change of sign in $C$, i.e. the $\pm$, doesn't even give two different solutions). What have I done to exclude the two other solutions (or is it just Mathematica excluding them somehow)?
Context:
Actually, I'm only trying to find the root where $-1\leq x\leq 1$. I'm trying solve (the first) part of system of equations that I solved numerically in order to make this answer analytically, so that I can play with the limit of the expression (as $n\rightarrow \infty$).
Thank you.
| For the calculation of the roots of the depressed cubic
$$
y^{\,3} + p\,y + q = 0
$$
where $p$ and $q$ are real or complex,
I personally adopt a method indicated in this work by A. Cauli, by which putting
$$
u = \sqrt[{3\,}]{{ - \frac{q}
{2} + \sqrt {\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}}} }}\quad v = - \frac{p}
{{3\,u}}\quad \omega = e^{\,i\,\frac{{2\pi }}
{3}}
$$
where for the radicals you take one value, the real or
the first complex one (but does not matter which)
then you compute the three solutions as:
$$
y_{\,1} = u + v\quad y_{\,2} = \omega \,u + \frac{1}
{\omega }\,v\quad y_{\,3} = \frac{1}
{\omega }\,u + \omega \,v
$$
In your case:
$$
y^{\,3} - 3\,y - 2\left( {\frac{{n - 2}}
{n}} \right) = 0
$$
we obtain
$$
\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}} = \left( {\frac{{n - 2}}
{n}} \right)^{\,2} - 1 = - 4\frac{{\left( {n - 1} \right)}}
{{n^{\,2} }} < 0
$$
which confirms that there are three real solutions, and
$$
\begin{gathered}
u = \sqrt[{3\,}]{{\frac{{n - 2}}
{n} + i\,\frac{2}
{n}\sqrt {\left( {n - 1} \right)} }} = \frac{1}
{{\sqrt[{3\,}]{n}}}\;\sqrt[{3\,}]{{n - 2 + i\,2\sqrt {\left( {n - 1} \right)} }} = \hfill \\
= \frac{1}
{{\sqrt[{3\,}]{n}}}\;\sqrt[{3\,}]{{n\,e^{\,i\,\alpha } }} = e^{\,i\,\alpha /3} \quad \left| {\,\alpha = \arctan \left( {\frac{{2\sqrt {\left( {n - 1} \right)} }}
{{n - 2}}} \right)} \right. \hfill \\
v = - \frac{p}
{{3\,u}} = \frac{1}
{u} = e^{\, - \,i\,\alpha /3} \hfill \\
\end{gathered}
$$
with the understanding that for $n=1,\; 2$, $\alpha= \pi , \; \pi /2$, i.e. that we use the 4-quadrant $arctan$.
So that in conclusion, for $0<n$, we have:
$$
\left\{ \begin{gathered}
y_{\,1} = e^{\,i\,\alpha /3} + e^{\, - \,i\,\alpha /3} = 2\cos \left( {\frac{\alpha }
{3}} \right) \hfill \\
y_{\,2} = e^{\,i\,\alpha /3 + 2\pi /3} + e^{\, - \,i\,\alpha /3 - 2\pi /3} = 2\cos \left( {\frac{{\alpha + 2\pi }}
{3}} \right) \hfill \\
y_{\,3} = e^{\,i\,\alpha /3 - 2\pi /3} + e^{\, - \,i\,\alpha /3 + 2\pi /3} = 2\cos \left( {\frac{{\alpha - 2\pi }}
{3}} \right) \hfill \\
\end{gathered} \right.
$$
Concerning the range spanned by the solutions, apart for $n=1$ where we get the solutions (1,-2,1), then
for $2 \le\; n$ we have
$$
\frac{{\alpha (n)}}
{3}\quad \left| {\;2 \leqslant n} \right.\quad = \frac{1}
{3}\arctan _{\,4\,Q} \left( {n - 2,\;2\sqrt {\left( {n - 1} \right)} } \right) = \left\{ {\frac{\pi }
{6},\frac{\pi }
{{7.66}},\; \cdots } \right\}
$$
which means:
$$
\left\{ \begin{gathered}
\quad \quad 2 \leqslant n \hfill \\
0 < \frac{{\alpha (n)}}
{3} \leqslant \frac{\pi }
{6}\quad \Rightarrow \quad \sqrt 3 \leqslant y_{\,1} < 2 \hfill \\
2\frac{\pi }
{3} < \frac{{\alpha (n)}}
{3} + 2\frac{\pi }
{3} \leqslant \frac{5}
{6}\pi \quad \quad \Rightarrow \quad - 2 < y_{\,2} \leqslant - \sqrt 3 \hfill \\
- 2\frac{\pi }
{3} < \frac{{\alpha (n)}}
{3} - 2\frac{\pi }
{3} \leqslant - \frac{\pi }
{2}\quad \quad \Rightarrow \quad - 1 < y_{\,3} \leqslant 0 \hfill \\
\end{gathered} \right.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Number of roots of $x^4 + x^3 \sin(x) + x^2 \cos(x) = 0$ How many roots do the following polynomial have?
$$x^4 + x^3 \sin(x) + x^2 \cos(x)$$
Obviously $0$ is a root. Dividing by $x^2$, I am left with
$$x^2 + x\sin(x) + \cos(x)$$
How do I know this polynomial has no root?
| Look for extremum of $f(x)=x^2 + x\sin(x) + \cos(x)$
$$
\frac{d}{dx}(x^2 + x\sin(x) + \cos(x))=2x+x\cos(x)=x(2+\cos(x))=0
$$
Then the only extremum we have is at $x=0$
We know that $x(2+\cos(x))<0$ for $x<0$ and $x(2+\cos(x))>0$ for $x>0$, also $f(0)=1$, which means that $f(x)\ge1$ for all $x$, hence it have no roots.
| {
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"answer_count": 2,
"answer_id": 0
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Why $4a^3+27b^2<0 \iff x^3+ax+b=0$ has exactly three solutions? According to Exc. 8 Sec. 4.3 of the book Advanced Calculus by Fitzpatrick,
For numbers $a$ and $b$, prove that the following equation has exactly three solutions if and only $4a^3+27b^2<0$: $$ x^3+ax+b=0, \ \ \ \ \ x \ \text{in} \ \mathbb R.$$
Long time struggling, I can't prove neither directions. The book is pretty rudimentary and discussions like here are beyond the scope of the book.
Let $f(x) = x^3 + ax + b$. So $f'(x) = 3x^2 + a$. Set $3x^2 + a = 0$, so that at the points $\sqrt{-a/3}$ and $-\sqrt{-a/3}$, $f'(x) = 0$. There exists exactly two maximum/minimum due to the existence of $3$ solutions. These $2$ max/min must be at $x = \sqrt{-a/3}$ and $x = -\sqrt{-a/3}$. Note that must $a<0$. Also must $f(-\sqrt{-a/3}) > 0$ and $f(\sqrt{-a/3})<0$. Putting $\pm \sqrt{-a/3}$ into $f(x)$, I fail to conclude $4a^3 + 27b^2 < 0$. As, $$-(-a/3)^{3/2}-a(-a/3)^{1/2}+b>0 \\ (-a/3)^{3/2}+a(-a/3)^{1/2}+b<0$$ results in $a<0$ and no more thing!
Please help!
| A third degree-polynomial $p(x)=x^3+ax+b$ has three real roots iff it has two stationary points $x_1,x_2$ and $p(x_1)\,p(x_2)<0$. Such stationary points are located at the roots of $p'(x)=3x^2+a$, hence at $\pm\sqrt{-\frac{a}{3}}$ ($a<0$ is a necessary condition, otherwise $p(x)$ is an increasing function and has only one real root). On the other hand,
$$\begin{eqnarray*} p(x_1)\,p(x_2)&=&p\left(\sqrt{-\frac{a}{3}}\right)p\left(-\sqrt{-\frac{a}{3}}\right)\\&=&\left(b+\sqrt{-\frac{a}{3}}\left(a-\frac{a}{3}\right)\right)\left(b-\sqrt{-\frac{a}{3}}\left(a-\frac{a}{3}\right)\right)\\&=&b^2+\frac{a}{3}\left(\frac{2a}{3}\right)^2=\color{red}{b^2+\frac{4a^3}{27}}.\end{eqnarray*} $$
| {
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"source": "stackexchange",
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Proof for formula for sum of sequence $3 + 9 + 27 + 81...+ 3^n$ Trying to formulate a proof for that sequence as practice. After reading this question's answer and lecture on this, I decided to try and practice with this sequence.
My try:
Base case $n = 1$
$S(n) = 3^n $
Induction step
$S(n+1) = 3^{n+1}$
Then
$Sn + n+1 = S(n+1)$
| $$\begin{align}
\require{cancel}S & =3+9+27+\dots3^n \\
3S & =3(3+9+27+\dots3^n) \\
& = 9+27+81+\dots3^{n+1} \\
\hline S-3 & =(3+9+27+\dots3^n)-3 \\
& = 9+27+\dots3^n \\
\hline 3S-(S-3) & = \left(\color{blue}{9+27+\dots3^n}+3^{n+1}\right)-\left(\color{blue}{9+27+\dots3^n}\right) \\
2S+3 & = \left(\xcancel{9+27+\dots3^n}+3^{n+1}\right)-\left(\xcancel{9+27+\dots3^n}\right) \\
2S+3 & = 3^{n+1} \\
2S & = 3^{n+1}-3 \\
S & = \frac12\left(3^{n+1}-3\right)
\end{align}$$
$$3+9+27+\dots3^n=\frac12\left(3^{n+1}-3\right)$$
| {
"language": "en",
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"source": "stackexchange",
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How to evaluate $\sum\limits_{k=0}^{n-1} \sin^t(\pi k/2n)$? How to evaluate $\displaystyle\sum_{k=0}^{n-1} \sin^t\left(\frac{\pi k}{2n}\right)$?
$t,n$ are constants $\in \Bbb{Z}$.
My try:
$$\begin{align}
\zeta:=e^{i\pi/2n} \implies & \sum_{k=0}^{n-1}\sin^t\left(\frac{\pi k}{2n}\right) \\
= &\frac{1}{2^t}\sum_{k=0}^{n-1}\left(\zeta^k- \zeta^{-k}\right)^{t} \\
= &\frac{1}{2^t}\sum_{k=0}^{n-1}\left(\zeta^{-kt}\right)\left(\zeta^{2k}-1\right)^{t}\\
\end{align}$$
How to proceed from here?
If a closed-form solution is not possible, can we still work out cases such as $t = n, t = 2n, $ etc?
EDIT: Sangchul Lee provided an answer for even t. Still looking for a solution for odd t as well.
| This integral remind me of Matsubara sum. Let me use contour integral to provide another perspective of the case $t = 2s$.
Define $a_n = \sum_{k = 1}^{n-1} \sin ^{t} \left( \frac{\pi k }{n} \right) $. What OP is looking for is
\begin{equation}
\sum_{k=1}^{n-1} \sin^{t} \left( \frac{\pi k }{2n} \right) = \frac{1}{2} \left( \sum_{k=1}^{2n-1} \sin^{t} \left( \frac{\pi k }{2n} \right) -1 \right) = \frac{1}{2} ( a_{2n } - 1 )
\end{equation}
One can extend the summation to all the roots of $\zeta^{2n} = 1$ if $t = 2s$
\begin{equation}
a_n = \frac{1}{2} \sum_{k=1}^{2n} \sin^{2s} \left( \frac{\pi k }{n} \right ) = \frac{1}{2} \sum_{ k = 1}^{2n} f( \zeta_k ) \qquad \zeta_k = \exp( \frac{ik\pi }{n})
\end{equation}
where
\begin{equation}
f(z ) = \left( \frac{z - \frac{1}{z}}{2i} \right)^{2s}
\end{equation}
The summation can be regarded as the sum of the residue of the follow contour integral
\begin{equation}
\sum_{ k = 1}^{2n} f( \zeta_k ) = \frac{2n}{2\pi i }\oint_{C} \frac{z^{2n-1}}{ z^{2n} - 1} f( z) dz
\end{equation}
where the integrand is specifically designed such that
\begin{equation}
\text{Res} ( \frac{ 2n z^{2n-1}}{z^{2n} - 1} f(z) , \zeta_k ) = 2n \frac{z^{2n-1}(z - \zeta_k)}{z^{2n} - \zeta^{2n}_k} f(\zeta_k ) \Big|_{z = \zeta_k } = f(\zeta_k )
\end{equation}
Here the closed contour should enclose all $\zeta_k$ but avoid the singularities of $f$, which can be taken as two concentric circles as shown in the following figure(red spots are those roots $\xi_k$).
The inner contour gives (minus) the residue at $0$, and the outer contour gives (minus) the residue at $\infty$, hence
\begin{equation}
\begin{aligned}
a_n &= n \left[ -\text{Res}( \frac{z^{2n-1}}{ z^{2n} - 1} f( z) , 0 ) - \text{Res}( \frac{z^{2n-1}}{ z^{2n} - 1} f( z) , \infty ) \right] \\
\end{aligned}
\end{equation}
Convert the residue of $\infty$ to $0$
\begin{equation}
\begin{aligned}
a_n =& n \left[ \text{Res}( \frac{z^{2n-1}}{ 1 - z^{2n} } f( z) , 0 ) + \text{Res}( \frac{\frac{1}{z}}{1 - z^{2n} } f( \frac{1}{z}) , 0 ) \right] \\
=& n \left[ \text{Res}( \frac{1}{z}\frac{z^{2n} + 1 }{ 1 - z^{2n} } f(z) , 0 ) \right] \\
=& n \left[ \text{Res}( \frac{1}{z} ( 1 + 2\sum_{k=1}^{\infty} z^{2nj} ) f(z) , 0 ) \right] \\
\end{aligned}
\end{equation}
Finally, expand $f(z)$ and extract the terms with power $0, -2nj, -4nj$ etc,
\begin{equation}
\begin{aligned}
a_n &= n \frac{(-1)^s}{2^{2s}} \left( (-1)^s {2s \choose s } + 2 \sum_{0 < nj < s } (-1)^{s+ nj} {2s \choose s+nj } \right)\\
&= \frac{n}{2^{2s}} \sum_{ -s < nj < s } (-1)^{nj} {2s \choose s + nj } \\
\end{aligned}
\end{equation}
This is the same as what @Sangchul Lee obtained.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "16",
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How to solve $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
Solve the equation $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
I was thinking on it for a few minutes and came up with a few ideas (none of them worked).
My first idea: Use the quadratic formula. Is that allowed? If so, I got to this:
$$
z = \frac{-(2i-3) \pm \sqrt{(2i-3)^2-4(5-i)}}{2} = \cdots = \frac{3-2i \pm \sqrt{-8i-15}}{2}
$$
I am guessing I cannot go further than this.
Other idea: Being $z = ai+b$,
$$
(ai+b)^2 + (2i-3)(ai+b) + 5 - i = 0
$$
Which will give me a system of two equations and two variables
$$
\begin{cases}
-a^2-2a+b^2-3b = -5\\
2ab+2b-3a = 1
\end{cases}
$$
which seems almost impossible to solve.
What am I missing? Is there a simpler solution to this that I am not thinking about?
| For a general approach, note that
$$
\begin{array}{l}
\sqrt z = \sqrt {x + i\,y} = \sqrt {\left| z \right|} \;e^{\,i\,\arg (z)/2} = \\
= \pm \left( {\sqrt {\frac{{\left| z \right| + {\mathop{\rm Re}\nolimits} (z)}}{2}} + \;i\,{\rm sign}^{\rm *} \left( {{\mathop{\rm Im}\nolimits} (z)} \right)\sqrt {\frac{{\left| z \right| - {\mathop{\rm Re}\nolimits} (z)}}{2}} } \right) = \\
= \pm \left( {\sqrt {\frac{{\sqrt {x^{\,2} + y^{\,2} } + x}}{2}} + i\,{\rm sign}^{\rm *} \left( y \right)\sqrt {\frac{{\sqrt {x^{\,2} + y^{\,2} } - x}}{2}} } \right) \\
\end{array}
$$
where the star on the $sign$ function means a definition modified with respect to the "common one", for which
$sign^ * \left( 0 \right) = 1$
i.e.:
$$
sign^ * \left( x \right) = \left\{ {\begin{array}{*{20}c}
{ - 1} & {x < 0} \\
1 & {0 \leqslant x} \\
\end{array} } \right.
$$
(refer to [Wikipedia article][1])
The quadratic equation $z^{\,2} + b\,z + c = 0$ can be solved by the method of completing the square
$$
z^{\,2} - 2\left( { - \frac{b}
{2}} \right)\,z + \left( {\frac{b}
{2}} \right)^2 = \left( {z - \left( { - \frac{b}
{2}} \right)} \right)^2 = \left( {\frac{b}
{2}} \right)^2 - c
$$
Thus providing
$$
z = - \frac{b}
{2} \pm {}_{\text{(complex)}}\sqrt {\left( {\frac{b}
{2}} \right)^2 - c}
$$
| {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Cube root of a binomial The cube of a certain binomial is $8y^3-36y^2+54y-27$. Find the binomial.
I know that $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and that$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ but don't know how to go further...
| Simply look at the last two terms. $$(a-b)^3=a^3-3a^2b+3ab^2-b^3\tag{1}$$ and from the binomial $$8y^3-36y^2+54y-27\tag{2}$$
We see that $$\begin{cases}8y^3=a^3\\27=b^3\end{cases}\tag{3}$$
Solving, we see that $a=2y$ and $b=3$. So the binomial factors into $$(2y-3)^3\tag{4}$$
Expanding out $(4)$ to check, we get: $$8y^3-27-3(2y)^2(3)+3(2y)(9)\tag{5}\\=8y^3-9\cdot4y^2+54y-27\\=8y^3-36y^2+54y-27$$
Which is equal to $(2)$.
| {
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Minimum value of $ \frac{((x-a)^2 + p )^{\frac 12}}{v_1} + \frac{( (x-b)^2 + q )^{\frac12}} {v_2}$ I need to calculate the minimum value of the expression
$ y = \frac{((x-a)^2 + p )^{\frac 12}}{v_1} + \frac{( (x-b)^2 + q )^{\frac12}} {v_2}$
I'm calculating the first derivative , then calculting the corresponding value for $x$ . But the calculation is getting complex.
Here $v_1 , v_2 , p$ and $q$ are positive integers greatest than $1$. I have also tried using $AM >= GM$ . but that too reduces finding maximum of a degree $4$ polynomial
| It's derivation of Snell's Law from Fermat's Principle
Assume $a<b$,
\begin{align}
T(x) &= \frac{\sqrt{(x-a)^2+p}}{v_1}+\frac{\sqrt{(x-b)^2+q}}{v_2} \\
T'(x) &=\frac{x-a}{v_1 \sqrt{(x-a)^2+p}}-
\frac{b-x}{v_2 \sqrt{(x-b)^2+q}} \\
0 &= \frac{\sin \theta_{1}}{v_{1}}-\frac{\sin \theta_{2}}{v_{2}} \\
s &= \sin \theta_{1} \\
\sin \theta_{2} &= \frac{v_{2} s}{v_{1}} \\
b-a &= \sqrt{p} \tan \theta_{1}+\sqrt{q} \tan \theta_{2} \\
b-a &=
\frac{s\sqrt{p}}{\sqrt{1-s^{2}}}+
\frac{v_{2} s\sqrt{q}}{\sqrt{v_{1}^{2}-v_{2}^2 s^2}} \tag{$\star$} \\
T_{\min} &=
\frac{\sqrt{p}}{v_{1}\cos \theta_{1}}+\frac{\sqrt{q}}
{v_{2}\cos \theta_{2}} \\
&= \frac{\sqrt{p}}{v_{1}\sqrt{1-s^{2}}}+
\frac{v_{1} \sqrt{q}}{v_{2}\sqrt{v_{1}^{2}-v_{2}^2 s^2}}
\end{align}
We seldom calculate the exact value of minimal $T$ but condition for minimum instead.
See also the link here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Differentiation and proving Prove that if the curve $y=\frac{x^3}{3} + px + q$ is tangent to the straight line $y = x$, then $4(p-1)^3 + 9q^2 = 0$
I have differentiated the equation of the curve to get $x^2 + p$ and equated it to $1$.
But i have no idea how to proceed to get the subsequent equation.
| $$y=\frac{x^3}{3} + px + q$$
Therefore $$y'=x^2 + p $$
Thus the gradient of $y=x$ must be equal to $x^2+p$.
That is $$x^2+p=1$$.
Since $y=x$ at each tangent points $x=\sqrt{1-p}$ and $x=-\sqrt{1-p}$ ,
$$x=\frac{x^3}{3} + px + q$$
$$x^2=\frac{x^4}{3} + px^2 + qx$$
$$(1-p)=\frac{{(1-p)}^2}{3} + p(1-p) + qx$$
$$(1-p)-p(1-p)=\frac{{(1-p)}^2}{3} + qx$$
$$(1-p)(1-p)=\frac{{(1-p)}^2}{3} + qx$$
$$(1-p)^2=\frac{{(1-p)}^2}{3} + qx$$
$$\frac{{2(1-p)}^2}{3} = qx$$
Thus $$\frac{{4(1-p)}^4}{9} = q^2x^2$$
$$\frac{{4(1-p)}^4}{9} = q^2(1-p)$$
For $p \neq 1$ ,
$$4(1-p)^3 = 9q^2$$
$$-4(p-1)^3 = 9q^2$$
$$4(p-1)^3 + 9q^2=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I go from $33/64 - 2x + x^2 \longrightarrow \bigg(x - \frac{2 + \sqrt{4 +33/16}}{2}\bigg)\bigg(x - \frac{2 - \sqrt{4 +33/16}}{2}\bigg)$?
How do I go from $\frac{33}{64} - 2x + x^2 \longrightarrow\bigg(x - \frac{2 + \sqrt{4 +33/16}}{2}\bigg)\bigg(x - \frac{2 - \sqrt{4 +33/16}}{2}\bigg)$ and then to $\big(x - 1-\frac{\sqrt{33}}{8}\big)\big(x - 1 + \frac{\sqrt{31}}{8}\big)?$
For the first, I know that it has something with the quadratic equation to do, but what is that? And how do I go from the second to the third?
| The quadratic eqation:
If $ax^2 +bx +c =0$ then
$x^2 +\frac {b }{a}x =-c/a$
$x^2 + 2\frac {b}{2a}x +\frac {b}{2a}^2 = -c/a +\frac {b}{2a}^2=\frac {b^2 - 4ac}{4a^2} $
$(x+b/2a)^2 = \frac {b^2 - 4ac}{4a^2} $
$x+ b/2a = \pm \frac {\sqrt { b^2 - 4ac }}{2a} $
$x=\frac {-b\pm \sqrt { b^2 - 4ac } }{2a}$
And therefore:
$ax^2 + bx+ c= a (x + \frac {b +\sqrt { b^2 - 4ac } }{2a} )(x + \frac {b- \sqrt { b^2 - 4ac } }{2a})$
So if $ax^2 + bx + c >0; a >0$ then $ (x + \frac {b +\sqrt { b^2 - 4ac } }{2a} )(x + \frac {b- \sqrt { b^2 - 4ac } }{2a}) > 0$ and either both terms are positive or both terms are negative.
The final bit is just arithmetic on fractions.
There is an error. It should be $1 \pm \sqrt {97}/8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Principal value of Carlson elliptic integral $R_C$ $$R_C(x,y)={1\over2}\int_0^\infty\frac{dt}{(t+y)\sqrt{t+x}},\quad(x\ge0)$$
$R_C$ has a singularity in the denominator when $y$ is negative. Wikipedia says in that case $R_C$ can be evaluated as follows
$$
PV\;R_C(x,-y)=\sqrt{\frac{x}{x+y}}R_C(x+y,y),\quad(y>0)
$$
Q: How can we prove the above equality?
The integrand has an easily computable anti-derivative but I don't know if it's useful in this context.
| One proves this identity by doing two things: 1) evaluating the integral for $y \gt 0$ and then 2) setting up the Cauchy PV integral in the complex plane.
1) The evaluation of the Carson integral is straightforward. Let's consider the case $x \gt y$:
$$\begin{align}R_C(x,y) &= \frac12 \int_0^{\infty} \frac{dt}{(t+y) \sqrt{t+x}} \\ &=\int_{\sqrt{x}}^{\infty} \frac{dt}{t^2-(x-y)}\\ &= \frac1{\sqrt{x-y}} \operatorname{arctanh}{\left (\frac{\sqrt{x-y}}{\sqrt{x}} \right )}\end{align} $$
Then we want to show that, for $y \lt 0$:
$$PV \left [R_C(x,-y)\right ] = \sqrt{\frac{x}{x+y}} R_C(x+y,y) = \sqrt{\frac{x}{x+y}} \frac1{\sqrt{x}} \operatorname{arctanh}{\left (\frac{\sqrt{x}}{\sqrt{x+y}} \right )} = \frac1{\sqrt{x+y}} \operatorname{arctanh}{\left (\frac{\sqrt{x}}{\sqrt{x+y}} \right )}$$
2) We will now proceed along those lines. Consider the following contour integral in the complex plane for $x$ and $y \gt 0$:
$$\oint_C dz \frac{\log{z}}{(z-y)\sqrt{z+x}} $$
where $C$ is the following contour:
Note that $C$ deals with the asymmetric integral over $[0,\infty)$ by introducing a log with a branch cut along the positive real axis. The contour $C$ avoids the pole at $z=y$ by introducing semicircular detours about that pole of radii $\epsilon$. $C$ also goes around the branch cut of the square root along the negative real axis.
I will take the limit as $\epsilon \to 0$ and the radius of the large circle $R \to \infty$. The result is, for the contour integral in this limit:
$$PV \int_0^{\infty} dt \frac{\log{t}}{(t-y) \sqrt{t+x}} -PV \int_0^{\infty} dt \frac{\log{t}+i 2 \pi}{(t-y) \sqrt{t+x}} - i \pi \frac{\log{y}}{\sqrt{x+y}} - i \pi \frac{\log{y}+i 2 \pi}{\sqrt{x+y}} \\ +i 2 \int_x^{\infty} dt \frac{\log{t}+i \pi}{(t+y) \sqrt{t-x}} $$
By Cauchy's theorem, the contour integral is zero. Simplifying, we get that
$$-i 2 \pi PV \int_0^{\infty} \frac{dt}{(t-y) \sqrt{t+x}} = i 2 \pi \frac{\log{y}}{\sqrt{x+y}} +i 2 \pi \frac{i \pi}{\sqrt{x+y}} - i 2 \int_x^{\infty} dt \frac{\log{t}+i \pi}{(t+y) \sqrt{t-x}}$$
or
$$\begin{align}PV \left [R_C(x,-y) \right ] &= -\frac{\log{y}}{2 \sqrt{x+y}} - \frac{i \pi}{2 \sqrt{x+y}} + \frac{1}{\pi} \int_0^{\infty} dt \frac{\log{(t^2+x)}}{t^2+x+y} + i \int_0^{\infty} \frac{dt}{t^2+x+y} \end{align} $$
Note that the second and fourth terms cancel. Thus,
$$\begin{align}PV \left [R_C(x,-y) \right ] &= -\frac{\log{y}}{2 \sqrt{x+y}} + \frac{1}{\pi} \int_0^{\infty} dt \frac{\log{(t^2+x)}}{t^2+x+y} \end{align} $$
The latter integral is
$$\int_0^{\infty} dt \frac{\log{(t^2+x)}}{t^2+x+y} = \frac{\pi}{\sqrt{x+y}} \operatorname{arctanh}{\left (\frac{\sqrt{x}}{\sqrt{x+y}} \right )} + \frac{\pi \log{y}}{2 \sqrt{x+y}} $$
Thus, we have shown that
$$PV \left [R_C(x,-y) \right ] = \frac{1}{\sqrt{x+y}} \operatorname{arctanh}{\left (\frac{\sqrt{x}}{\sqrt{x+y}} \right )} = \sqrt{\frac{x}{x+y}} R_C(x+y,y) $$
| {
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"url": "https://math.stackexchange.com/questions/1940298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Show that $p$ is a divisor of numbers $m^2 + 1$, $n^2 + 2$, if and only if it is a divisor of some number of the form $k^4 + 1$. Show that $p$ is a divisor of numbers $m^2 + 1$, $n^2 + 2$, if and only if it is a divisor of some number of the form $k^4 + 1$.
Since $p \mid m^2 + 1$, we have $m^2 \cong -1( \mod p)$ and hence $\left ( \frac{-1}{p} \right)$ (Legendre Symbol) is $1$.
Again, $p \mid n^2 + 2$, we have $n^2 \cong -2( \mod p)$ and hence $\left ( \frac{-2}{p} \right)$ (Legendre Symbol) is $1$.
I am stuck here...How to proceed with the problem...Help Needed.
Thank You.
| We have in $\Bbb F_p$
$$x^2=-1\iff (-1)^{\frac{p-1}{2}}=1\Rightarrow p=4s+1\\x^2=-2\iff (-2)^{\frac{p-1}{2}}=1\Rightarrow p=8t\pm 1$$
Hence we must have $p=8k+1$ (the cases $p=8t-1$ must be discarded)
On the other hand $$(-1)^{\frac{p-1}{2}}=((-1)^{\frac{p-1}{4}})^2$$
This means, since $\frac{8t+1-1}{4}$ is a natural integer, that there exists $k\in\Bbb F_p$ such that $$k^4+1=0\iff k^4+1=Mp$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $A+B+C=\pi$, prove: $\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$
If $A+B+C=\pi$, prove that:
$$\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$$
My Attempt:
Here, $A+B+C=\pi$
Now,
$$\begin{align}
LHS &=\cos(B+2C)+\cos(C+2A)+\cos(A+2B) \\
&=\cos(B+C+C)+\cos(C+A+A)+\cos(A+B+B) \\
&=\cos(\pi-(A-C))+\cos(\pi-(B-A))+\cos(\pi-(C-B)) \\
&=-\cos(A-C)-\cos(B-A)-\cos(C-B)
\end{align}$$
Please help to continue from here.
| Let $C-A=2x,B-C=2y,A-B=2z\implies2(x+y+z)=0$
$$F=\cos2x+\cos2y+\cos2z=2\cos(x+y)\cos(x-y)+2\cos^2z-1$$
Now as $\cos(x+y)=\cos(-z)=\cos z,$
$$F=2\cos z\cos(x-y)+2\cos z\cdot\cos(x+y)-1$$
$$=2\cos z\{\cos(x+y)+\cos(x-y)\}-1=2\cos z\{2\cos x\cos y\}-1=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the sum $\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \cdots$ Suppose $|x| < 1$. Can you give any ideas on how to find the following sum?
$$
\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \frac{8x^8}{x^8+1} + \cdots
$$
| Evaluating $$\sum_{q\ge 0} \frac{2^q x^{2^q}}{1+x^{2^q}}$$
we obtain
$$\sum_{q\ge 0} 2^q \sum_{k\ge 0} (-1)^k x^{(k+1)2^q}
= \sum_{n\ge 1} x^n \sum_{2^q|n} 2^q (-1)^{n/2^q-1}.$$
Now observe that
$$\sum_{2^q|n} 2^q (-1)^{n/2^q-1}
= \sum_{p=0}^{v_2(n)} 2^p (-1)^{n/2^p-1}$$
where $v_2(n)$ is the exponent of the highest power of $2$ that
divides $n.$ This is
$$-\sum_{p=0}^{v_2(n)-1} 2^p + 2^{v_2(n)}
= - (2^{v_2(n)}-1) + 2^{v_2(n)} = 1.$$
because $n/2^p$ is even unless $p=v_2(n).$
(This also goes through when $n$ is odd and we have one value for $p$, namely zero.)
Hence the end result is
$$\sum_{n\ge 1} x^n = \frac{x}{1-x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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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.