Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Diophantine equation: $(x-y)^2=x+y$ I have to solve the following equation: $(x-y)^2=x+y$, where $x$ and $y$ are non-negative integers. This equation has an infinite number of solutions, but how to prove that there exists a positive integer k such that $x=\frac{(k+1)(k+2)}{2}$ and $y=\frac{k(k+1)}{2}$? Thanks!
| On rearranegment we have $$x^2-x(2y+1)+y^2-y=0$$
$$\implies x=\frac{2y+1\pm\sqrt{(2y+1)^2-4(y^2-y)}}2=\frac{2y+1\pm\sqrt{8y+1}}2$$
As $x$ has to be integer, we need $8y+1$ to be perfect square
As $y$ is also an integer, $8y+1$ is odd, hence can be written as $(2a+1)^2$ where $a$ is any integer
Now simplify and find $x,y$ is terms of $a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/689924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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If $z_{n+1}=\frac{27}{\overline{z_{n}}}+6$ and $z_1 = 3 + 6i$, then find $z_{n}$
Let the complex sequence $\{z_{n}\}$ satisfy $z_{1}=3+6i$, and
$$z_{n+1}=\dfrac{27}{\overline{z_{n}}}+6.$$
Find the $z_{n}$.
My idea: since
$$z_{n+2}=\dfrac{27}{\overline{z_{n+1}}}+6=\dfrac{27}{\dfrac{27}{z_{n}}+6}+6?$$
So I can't. Thank you
| Hint.
Clearly
$$
z_{n+2}=\dfrac{27}{\overline{z_{n+1}}}+6=\dfrac{27}{\dfrac{27}{z_{n}}+6}+6=
\frac{27z_n}{6z_n+27}+6=\frac{63z_n+162}{6z_n+27}.
$$
Setting $x=z_n=z_{n+2}$ and solving the resulting equation we obtain that $x=9$ is a solution (a fixed point). In particular, subtracting the fixed point we get
$$
z_{n+2}-9=\frac{3(z_n-9)}{2(z_n-9)+27}
$$
or
$$
\frac{1}{z_{n+2}-9}=\frac{2(z_n-9)+27}{3(z_n-9)}=\frac{2+\frac{27}{z_n-9}}{3}.
$$
Setting
$$
w_n=\frac{3}{z_n-9},
$$
you obtain that
$$
w_{n+2}=9w_n+2,
$$
and thus
$$
\frac{w_{n+2}}{3^{n+2}}=\frac{w_n}{3^n}+\frac{2}{3^{n+2}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/693911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does $f(x)\in L^1$ imply that $f'(x) \in L^1$? Let $f(x)$ be defined for all real numbers differentiable function of one variable.We know that:
$$\int_{-\infty }^{+\infty } |f(x)| \, dx\neq +\infty$$
Problem is to resolve if it is possible or not that:
$$\int_{-\infty }^{+\infty } |f'(x)| \, dx= +\infty$$
| Counterexample: $f(x)=\frac{\sin(e^x)}{1+x^2}$ - smooth function
$$\int_{-\infty}^{+\infty}\left|\frac{\sin(e^x)}{1+x^2}\right| \,dx\leqslant \int_{-\infty}^{+\infty}\frac{1}{1+x^2} \,dx=\pi$$
Which means that $f(x)\in L^1$
$$f'(x)=\frac{e^x \left(x^2+1\right) \cos \left(e^x\right)-2 x \sin \left(e^x\right)}{\left(x^2+1\right)^2}$$
Note: $\forall_{\varepsilon > 0} \lim_{X\to \infty } \, \int_X^{X+\varepsilon} f'(x) \, dx=0$
$$|f'(x)|=\frac{e^x \left(x^2+1\right) \left|\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right|}{\left(x^2+1\right)^2}\geqslant \frac{1}{2}\frac{e^x \left|\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right|}{\left(x^2+1\right)}\geqslant \frac{1}{4}\frac{e^x \left(\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right)^2}{\left(x^2+1\right)}$$
$$\int_{-\infty}^{+\infty}|f'(x)| \,dx > \frac{1}{4}\int_{0}^{+\infty}\frac{e^x \left(\cos \left(e^x\right)-\frac{2 x}{x^2+1} \frac{\sin \left(e^x\right)}{e^x}\right)^2}{\left(x^2+1\right)} \,dx$$
$\left|\frac{2 x}{x^2+1} \sin \left(e^x\right)\right|\leqslant 1$ which means that if we prove divergence of $\int_{0}^{+\infty}\frac{e^x \cos \left(e^x\right)^2}{\left(x^2+1\right)} \,dx$ we prove divergence of $\int_{-\infty}^{+\infty}|f'(x)| \,dx$. $$\int_0^{+\infty}\frac{e^x \cos \left(e^x\right)^2}{\left(x^2+1\right)} \,dx> \int_0^{+\infty}\cos \left(e^x\right)^2 \,dx=\int_0^{+\infty}\frac{1}{2} \,dx+\frac{1}{2}\int_0^{+\infty}\cos \left(2e^x\right) \,dx$$
Note that:$\left(\frac{e^x}{x^2+1}\right)^{'}=\frac{e^x (-1+x)^2}{\left(1+x^2\right)^2}\geqslant 0$, so finally if $\int_0^{+\infty}\cos \left(2e^x\right) \,dx$ is convergent than $\int_{-\infty}^{+\infty}|f'(x)| \,dx=+\infty$.
$$\int_0^{+\infty}\cos \left(2e^x\right) \,dx=\int_2^{+\infty}\frac{\cos \left(v\right)}{v} \,dv$$
$\int_2^{+\infty}\frac{\cos \left(v\right)}{v} \,dv$ is convergent by Dirichlet's test or even Leibniz test with $a_n=\int_{n\pi+0.5\pi}^{(n+1)\pi+0.5\pi}\left|\frac{\cos \left(v\right)}{v}\right| \,dv$ , which completes the proof.
| {
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"url": "https://math.stackexchange.com/questions/693998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
How do I solve $yy'+x=\sqrt{x^2+y^2}$? I tried this:
$yy'+x=\sqrt{x^2+y^2}$
$y'=-\frac{x}{y}+\frac{1}{y}\sqrt{x^2+y^2}$
$y'=-\frac{x}{y}+\sqrt{(\frac{x}{y})^2+1}$
Substitution: $v=\frac{y}{x}$
$v'x+v=-\frac{1}{v}+\sqrt{(\frac{1}{v})^2+1}$
$v'x=-\frac{1-v^2}{v}+\sqrt{(\frac{1}{v})^2+1}$
$v'x=-\frac{1-v^2+ v\sqrt{(\frac{1}{v})^2+1}}{v}$
$v'x=-\frac{1-v^2+\sqrt{1+v^2}}{v}$
If I have to separate at this point, it's going to be pretty awkward to work with.
Is there a better way?
| The $x^2+y^2$ suggests using polar coordinates. Substitute $x=r\cos\theta$ and $y=r\sin\theta$. The difficult bit is the expression for $dy/dx$.
$$ \frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$$
Then you can substitute into your ODE.
$$\begin{align}
yy' + x &= \sqrt{x^2+y^2}\\
r\sin\theta\frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta} + r\cos\theta &= r\\
\sin\theta(r'\sin\theta + r\cos\theta) + \cos\theta(r'\cos\theta - r\sin\theta) &= r'\cos\theta - r\sin\theta\\
r'\sin^2\theta + r\sin\theta\cos\theta + r'\cos^2\theta - r\sin\theta\cos\theta &= r'\cos\theta - r\sin\theta\\
r' &= r'\cos\theta - r\sin\theta
\end{align}$$
It looks like $\dfrac{d}{d\theta}r\cos\theta$ might be useful on the right. Following this through led me to a solution that satisfied your ODE on substitution (which I will now add as the other answer provides the solution).
$$\begin{align}
r' &= \frac{d}{d\theta}r\cos\theta\\
r &= r\cos\theta + k\\
x^2 + y^2 &= x^2 + 2kx + y^2\\
y &= \sqrt{k^2 + 2kx}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Two PDE Questions 1.
Solve the equation
$$
u_x^3-u_y=0~,
$$ with
$u(x,0)=2x^\frac{3}{2}$
2.
Solve the equation
$$
u=xu_x+yu_y+\frac{1}{2}(u_x^2+u_y^2)~,
$$ with
$u(x,0)=\dfrac{1}{2}(1-x^2)$
How I solve these problems if I learnt PDE three weeks ago?
| $1.$
$u_x^3-u_y=0$
$u_{xy}-3u_x^2u_{xx}=0$
Let $v=u_x$ ,
Then $v_y-3v^2v_x=0$ with $v(x,0)=3\sqrt x$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$\dfrac{dv}{dt}=0$ , letting $v(0)=v_0$ , we have $v=v_0$
$\dfrac{dx}{dt}=-3v^2=-3v_0^2$ , letting $x(0)=f(v_0)$ , we have $x=-3v_0^2t+f(v_0)=-3v^2y+f(v)$ , i.e. $v=F(x+3v^2y)$
$v(x,0)=3\sqrt x$ :
$F(x)=3\sqrt x$
$\therefore v=3\sqrt{x+3v^2y}$
$v^2=9(x+3v^2y)$
$v^2=9x+27v^2y$
$v^2-27v^2y=9x$
$v^2(1-27y)=9x$
$v^2=\dfrac{9x}{1-27y}$
$v=\pm\dfrac{3\sqrt x}{\sqrt{1-27y}}$
$u_x=\pm\dfrac{3\sqrt x}{\sqrt{1-27y}}$
$u(x,y)=\pm\dfrac{2x^\frac{3}{2}}{\sqrt{1-27y}}+g(y)$
$u_y=\pm\dfrac{27x^\frac{3}{2}}{(1-27y)^\frac{3}{2}}+g_y(y)$
$\therefore\pm\dfrac{27x^\frac{3}{2}}{(1-27y)^\frac{3}{2}}\mp\dfrac{27x^\frac{3}{2}}{(1-27y)^\frac{3}{2}}-g_y(y)=0$
$g_y(y)=0$
$g(y)=C$
$\therefore u(x,y)=\pm\dfrac{2x^\frac{3}{2}}{\sqrt{1-27y}}+C$
$u(x,0)=2x^\frac{3}{2}$ :
$C=0$ and the negative part reject
$\therefore u(x,y)=\dfrac{2x^\frac{3}{2}}{\sqrt{1-27y}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $(a-1),a,(a+1)$ are consecutive positive integers, $ (a+1)^3 \neq a^3 + (a-1)^3$ I had to prove the following statement:
If $(a-1),a,(a+1)$ are consecutive positive integers, $(a+1)^3 \neq a^3 + (a-1)^3$
My attempt at the solution was to first expand each side to get
$$a^3 + 3a^2 + 3a + 1 \neq 2a^3 - 3a^2 +3a - 1\\
0 \neq a^3 - 6a^2 - 2$$
However, $a^3 - 6a^2 - 2$ does hit the $x$-axis at $a = 6.0546$.
Does that mean that the statement is incorrect?
| Suppose $n^3-6n^2 = 2$ for some integer $n$. It is easy to check that $n$ cannot be odd. Hence $n = 2k$ for some integer $k$. We have $8k^3-24k^2 = 2$, or equivalently, $k^3-3k^2 = \frac{1}{4}$, which is a contradiction since $k$ was supposed to be an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/701166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Geometric explanation of $\sqrt 2 + \sqrt 3 \approx \pi$ Just curious, is there a geometry picture explanation to show that $\sqrt 2 + \sqrt 3 $ is close to $ \pi $?
| Divide this approximation by four:
$\displaystyle \begin{align*}\pi &\approx \sqrt{2}+\sqrt{3} \\ &\Leftrightarrow \\ \frac{\pi}{4} &\approx \frac{\sqrt{2}+\sqrt{3}}{4}\\ &\Leftrightarrow \\ \frac{\pi}{4} &\approx \frac{\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3}}{2} = \frac{\sin{(\frac{\pi}{4})}+\sin{(\frac{\pi}{3})}}{2} , \end{align*}$
So this approximation tells that $\frac{\pi}{4}$ is roughly the average of $\sin{(\frac{\pi}{4})}$ and $\sin{(\frac{\pi}{3})}$, which is visualized below:
If the green arc segment with length $\frac{\pi}{4}$ is made straight, its red tip follows a cycloid (https://en.wikipedia.org/wiki/Cycloid), like the valve of a running bicycle wheel, and ends up between $\frac12 \sqrt{2}$ and $\frac12 \sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/701822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
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Can be solved without L'Hopital? Can this limit be evaluated without l'hopital's rule?
$$\lim_{h\to0}\frac{\sqrt[3]{8+h}-2}{h}$$
| \begin{align}
\frac{\sqrt[3]{8+h}-2}{h}&=
\frac{\sqrt[3]{(8+h)^2}+2\sqrt[3]{8+h}+4}{\sqrt[3]{(8+h)^2}+2\sqrt[3]{8+h}+4}
\cdot \frac{\sqrt[3]{8+h}-2}{h} \\&=\frac{(8+h)-8}{h}
\cdot\frac{1}{\sqrt[3]{(8+h)^2}+2\sqrt[3]{8+h}+4}
\to\frac{1}{12}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/703149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
$ \dfrac a{\sqrt[4] {16+b-d}} +\dfrac b{\sqrt[4] {16+c-a}}+ \dfrac c{\sqrt[4] {16+d-b}}+\dfrac d{\sqrt[4] {16+a-c}} \ge 8$ Let $a,b,c,d>0 ; a+b+c+d=16$ , then how to prove that
$ \dfrac a{\sqrt[4] {16+b-d}} +\dfrac b{\sqrt[4] {16+c-a}}+ \dfrac c{\sqrt[4] {16+d-b}}+\dfrac d{\sqrt[4] {16+a-c}} \ge 8$ ?
| By AM-GM and C-S
$$\sum_{cyc}\frac{a}{\sqrt[4]{16+b-d}}=\sum_{cyc}\frac{32a}{4\cdot2^3\cdot\sqrt[4]{16+b-d}}\geq\sum_{cyc}\frac{32a}{3\cdot2^4+16+b-d}=$$
$$=\sum_{cyc}\frac{32a}{64+b-d}=\sum_{cyc}\frac{32a^2}{64a+ab-da}\geq\frac{32(a+b+c+d)^2}{\sum\limits_{cyc}(64a+ab-da)}=\frac{32\cdot16^2}{64\cdot16+0}=8.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine real number exists for relation with square roots We have $$\sqrt{x -2} = 3 -2\sqrt{x}$$.
I am to find whether a real number exists for this relation, and the real number that satisfies.
I start by squaring both sides, which yields:
$$x - 2 = 4x - 12\sqrt{x} + 9$$.
Whence:
$$ -3x = -12\sqrt{x} + 11 \\
\sqrt{x} = \frac{x}{4} + \frac{11}{12}.
$$
But once i get here i am stuck. How can i find whether a solution exists for x from here?
| Let $y=\sqrt{x}$. Then,
$$ \sqrt{x} = \frac{x}{4} + \frac{11}{12}\\
\implies 0 = \frac{y^2}{4} -y + \frac{11}{12}
$$
You should be able to use the quadratic formula to solve for $y$ (check for extraneous solutions), and then don't forget to solve for $x$ from $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/704680",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Simplify this, how to? The original function is $$f(x)=\cfrac{3x^3}{\cfrac{1}{x}} $$
It's the quotient rule I'm using so:
$$ f^\prime\left(x\right) =\dfrac{9x^2\cdot\cfrac{1}{x} - 3x^3 \cdot \left(-x^{-2}\right)}{\dfrac{1}{x^2}}$$
I suck at simplifying, so I just need to get the hang of it. I've done some already, but this one is troubling me. Thanks. Oh, and STEP BY STEP please.
| $$\dfrac{9x^2 \cdot \frac 1x - 3x^3 \cdot -x^{-2}}{1/x^2} = \dfrac{9x^2 \cdot \frac 1x - 3x^3 \cdot \frac{-1}{x^2}}{1/x^2} = \dfrac{9x +3x}{1/x^2} = x^2(12x) = 12x^3$$
Note, however that $$f(x) = \dfrac {3x^3}{\frac 1x} = x(3x^3) = 3x^4 \implies f'(x) = 12x^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/705306",
"timestamp": "2023-03-29T00:00:00",
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Find volume between two spheres using cylindrical & spherical coordinates I've got two spheres, one of which is the other sphere just shifted, and I'm trying to find the volume of the shared region. The spheres are $x^2 + y^2 +z^2 = 1$ and $x^2 + y^2 +(z-1)^2 = 1$
I know how to transform the variables into cylindrical and spherical coordinates but I'm having trouble figuring out the bounds.
How do I do this?
EDIT: Based on Kaladin's answer, which helped me realize the bounds for $r$, would it be correct to express the volume of the region as follows? (as cylindrical coordinates)
$$V = 2\int_0^{2\pi} \int_{1/2}^1 \int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} rdzdrd\theta$$
EDIT 2: Assuming I integrated the above integral properly, that equals $\frac{2\pi\sqrt{2}}{3}$, which is obviously not Kaladin's answer. What's the problem?
| Answer using Cylindrical Coordinates:
Volume of the Shared region =
Equating both the equations for z, you get z = 1/2. Now substitute z = 1/2 in in one of the equations and you get r = $\sqrt{\frac{3}{4}}$.
Now the sphere is shifted by 1 in the z-direction, Hence
Volume of the Shared region = $$\int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{3}{4}}} \int_{1-\sqrt{1-r^2}}^{\sqrt{1-r^2}} rdzdrd\theta$$
$$V=2\pi \int_{0}^{\sqrt{\frac{3}{4}}} [2{\sqrt{1-r^2}}-1] rdr$$
substitute $$u = 1-r^2 ; r = 0 => u = \frac{1}{4} ; r = \sqrt{\frac{3}{4}} => u = 1$$
$$V = 2\pi [-\int_1^{\frac{1}{4}} u^{\frac{1}{2}} du - \int_{0}^{\sqrt{\frac{3}{4}}} rdr]$$
$$V= 2\pi (\frac{2}{3}u^{\frac{3}{2}}) - (\frac{r^2}{2})$$
$$V =2\pi*( \frac{2}{3}(1-\frac{1}{8}) - \frac{3}{8})$$
$$V = 2\pi*(\frac{14}{24} - \frac{3}{8}) = 2\pi*\frac{5}{24} = \frac{5}{12} \pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/705935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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if $a,b$ such $a^2+b^2+4=2a+ab+2b$,find $a^2b$ let $a,b\in R$, and such
$$a^2+b^2+4=2a+ab+2b$$
Find $a^2b=?$
My idea: $$a^2-2a+1+b^2-2b+1+2+ab=0$$
$$(a-1)^2+(b-1)^2+ab+2=0$$
then I can't it
maybe this problem can use inequality to solve it,Than you
| The expression can be written as,
$a^2-ab+b^2-2a-2b+4=\frac{3}{4}(a-b)^2 + \frac{1}{4}(a+b-4)^2=0$
Since, $(a-b)^2\ge0$ and $(a+b-4)^2\ge 0$,
So, the only possible real solutions are $a-b=0$ and $a+b-4=0$
That is $a=b=2$.
$a^2b=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/706511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Derivation of factorization of $a^n-b^n$ How does one prove that:
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\dots+a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$$
Better yet, why is $a^n-b^n$ divisible by $a-b$? I would very much appreciate some help on this. Thanks
| As to "why" $a - b$ divides $a^n - b^n$, here's one attempt to "convince" you. Let $c = a - b$. Then $$a - b \mid a^n - b^n \iff c \mid (b + c)^n - b^n.$$
Expanding $(b + c)^n$ we get
$$(b + c)^n = b^n + \binom{n}{1} b^{n-1} \color{red}{c} + \binom{n}{2} b^{n-2} \color{red}{c^2} + \dots + \binom{n}{n-1} b \color{red}{c^{n-1}} + \color{red}{c^n}.$$
Since all terms are divisible by $c$, except for possibly the first one (which is cancelled by $-b^n$), it follows that $c \mid (b + c)^n - b^n$ for all $b,c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/712758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 1
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Identity involving binomial coefficients: $\sum_{j=0}^i (-1)^{i+j}\binom {n-j}{i-j}\binom mj=\sum_{j=0}^i (-1)^{i+j}\binom {n-j+k}{i-j}\binom {m+k}j$ Can you help me prove the following identity? I know it holds because I simulated it.
For positive integers $n,m,k$ and for $i=0,\ldots,n$ and for $n \leq m$ we have:
$$\sum_{j=0}^i (-1)^{i+j}\binom {n-j} {i-j} \binom {m}{j} = \sum_{j=0}^i (-1)^{i+j}\binom {n-j+k} {i-j} \binom {m+k}{j}$$
| This can be done using a basic complex variables technique.
Suppose we seek to verify that
$$\sum_{j=0}^q {m\choose j} (-1)^j {n-j\choose q-j}
= \sum_{j=0}^q {m+k\choose j} (-1)^j {n-j+k\choose q-j} .$$
We will treat the case $q=m$ and $n\le m.$
Introduce the two integral representations
$${n-j\choose q-j}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q-j+1}} (1+z)^{n-j} \; dz$$
and
$${n-j+k\choose q-j}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q-j+1}} (1+z)^{n-j+k} \; dz$$
This gives the following integral for the sum on the LHS
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{j=0}^m {m\choose j} (-1)^j
\frac{1}{z^{q-j+1}} (1+z)^{n-j} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z^{q+1}}
\sum_{j=0}^m {m\choose j} (-1)^j
\frac{z^j}{(1+z)^j} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z^{q+1}}
\left(1-\frac{z}{1+z}\right)^m \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}}
\frac{1}{(1+z)^{m-n}} \; dz.$$
We get the following integral for the sum on the RHS
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{j=0}^m {m+k\choose j} (-1)^j
\frac{1}{z^{q-j+1}} (1+z)^{n-j+k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+k}}{z^{q+1}}
\sum_{j=0}^m {m+k\choose j} (-1)^j
\frac{z^j}{(1+z)^j} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+k}}{z^{q+1}}
\left(
\left(1-\frac{z}{1+z}\right)^{m+k}
-\sum_{j=m+1}^{m+k}
{m+k\choose j} (-1)^j
\frac{z^j}{(1+z)^j}
\right) dz.$$
There are two pieces here inside the parentheses, call them $A$ and $B$.
For $A$ we get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+k}}{z^{q+1}}
\left(1-\frac{z}{1+z}\right)^{m+k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+1}}
\frac{1}{(1+z)^{m-n}} \; dz$$
This is the same as the LHS. Now we just need to show that piece $B$
is zero. It is given by
$$- \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+k}}{z^{q+1}}
\sum_{j=m+1}^{m+k}
{m+k\choose j} (-1)^j
\frac{z^j}{(1+z)^j} \; dz.$$
But we have $j\ge m+1 = q+1$ so the apparent pole at zero vanishes and
this term is analytic in and on the circle $|z|=\epsilon$ with no poles
inside it and piece $B$ is indeed zero.
We can stop here without further evaluation because the integrals for
LHS and RHS are seen to be the same. Moreover, they are trivial to evaluate,
we get $$(-1)^m \times {2m-n-1\choose m-n-1}.$$
The reader is invited to supply a proof for the case $q<m.$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
| {
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How do I prove divisibility by 3 without induction? How do I prove that:
*
*$3$ divides $4^n-1$, where $n$ is a natural number, and
*$3$ divides $n^3-n$, where $n$ is a natural number?
All without induction?(only number theory)
Thanks !
| For the first one, use modulo arithmetic:
$$4^n - 1 \equiv 1- 1 \pmod 3\\
= 0 \pmod 3\\
\implies 3 | (4^n - 1)$$
For the second one, do a simple factorization:
$$n^3 - n = n (n^2 - 1) = (n-1)(n)(n+1)$$
and then notice that one of $n-1, n, n+1$ must be divisible by $3$. This is a clever and very worthy trick : One and only one of $k$ consecutive positive integers must be divisible by $k$.
In fact, using similar arguments, we can deduce that $2$ divides $n^3 - n$ as well, and therefore since both $2$ and $3$ are divisors, then $6$ must also be a divisor of $n^3 - n$.
Just for completeness' sake:
$$\begin{align}4^n &= (1 + 3)^n \\
&= 1 + 3\binom{n}{1} + 3^2\binom{n}{2} +\cdots+3^n\binom{n}{n}\\
&= 1 + 3\left(\binom{n}{1} + 3\binom{n}{2} +\cdots+3^{n-1}\binom{n}{n}\right)\end{align}$$
So we've proved that for integers $n \ge 1$, $4^n$ gives a remainder of $1$ when divided by $3$, i.e. $4^n \equiv 1 \pmod 4$.
| {
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Evaluate $\sum_{n=1}^\infty \frac{n}{2^n}$ This is a homework question; I'm supposed to use power series to find the following sum: $$\sum_{n=1}^\infty \frac{n}{2^n}$$ I took the geometric series $$\frac{1}{1-x}=\sum_{n=0}^\infty {x^n}$$ and differentiated and multiplied both sides by x to get $$\frac{x}{(1-x)^2}=\sum_{n=1}^\infty {nx^n}$$ I'm stuck because I'm not sure how to make the $$\frac{1}{2^n}$$ term appear.
| You are basically all of the way there. In your last step, you have
$$ \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^n, \tag{$\ast$}$$
where the series on the right converges for any $|x| < 1$ to the rational expression on the left. Since we get to pick $x$ to be anything we like (as long as $|x| < 1$), we might as well choose $x = \frac{1}{2}$. If we do this, the series on the right-hand side of ($\ast$) becomes
$$ \sum_{n=1}^{\infty} n \left( \frac{1}{2} \right)^n
= \sum_{n=1}^{\infty} n \cdot \frac{1}{2^n}
= \sum_{n=1}^{\infty} \frac{n}{2^n}, $$
which is the original series that you were given to evaluate. But with $x = \frac{1}{2}$, the left-hand side of ($\ast$) is
$$ \frac{1}{\left( 1 - \frac{1}{2} \right)^2} = \frac{1}{\left(\frac{1}{2}\right)^2} = \frac{1}{\frac{1}{4}} = 4.$$
Therefore, equating the left- and right-hand sides of ($\ast$) when $x = \frac{1}{2}$, we obtain
$$ \sum_{n=1}^{\infty} \frac{n}{2^n} = 4.$$
| {
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Infinite sums and integrals using residues I have no idea how to solve these two, any help?
$\mathtt{i)}$ $$\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\frac{e^{tz}}{\sqrt{z+1}}dz$$ $$ a,t\gt0$$
$\mathtt{ii)}$
$$ \sum_{n=1}^\infty \frac{\coth (n\pi)}{n^3}=\frac{7\pi^3}{180}$$
| For the second problem, let $ \displaystyle f(z) = \frac{\pi\cot (\pi z) \coth(\pi z)}{z^{3}}$ and integrate around a square ($C_{N}$) with vertices at $\pm (N+\frac{1}{2}) \pm (N+\frac{1}{2})$ where $N$ is a positive integer.
Both $\cot (\pi z)$ and $\coth(\pi z)$ are uniformly bounded on the contour.
So $ \displaystyle \int_{C_{N}} f(z) \ dz$ vanishes as $N$ goes to infinity through the integers..
And we have
$$ 2 \sum_{n=1}^{\infty} \frac{\coth (\pi n)}{n^{3}} + \text{Res}[f(z),0] + \sum_{n=1}^{\infty} \text{Res} [f(z),in] + \sum_{n=1}^{\infty} \text{Res} [f(z),-in] =0 $$
where
$$ \text{Res} [f(z),in] = \lim_{z \to in} \frac{\pi \cot (\pi z)}{\frac{d}{dz} (z^{3} \tanh (\pi z) )} = \lim_{z \to in} \frac{\pi \cot (\pi z)}{3z^{2} \tanh (\pi z) + \pi z^{3} \text{sech}^{2} (\pi z) } = \frac{\coth(\pi n)}{n^{3}}$$
and similarly
$$ \text{Res} [f(z),-in] = \frac{\coth (\pi n)}{n^{3}} \ .$$
And expanding $\cot (\pi z)$ and $\coth(\pi z)$ in Laurent expansions about the origin,
$$ \begin{align} \frac{\pi\cot (\pi z) \coth(\pi z)}{z^{3}} &= \frac{\pi}{z^{3}} \left( \frac{1}{\pi z} - \frac{\pi z}{3} - \frac{\pi^{3}z^{3}}{45} + \ldots \right) \left( \frac{1}{\pi z} + \frac{\pi z}{3} - \frac{\pi^{3}z^{3}}{45} + \ldots \right) \\ &= \frac{\pi}{z^{3}} \left(\frac{1}{\pi^{2}z^{2}} + \frac{1}{3} - \frac{\pi^{2}z^{2}}{45} - \frac{1}{3} - \frac{\pi^{2} z^{2}}{9} - \frac{\pi^{2}z^{2}}{45} + \ldots \right) \\ &= \frac{1}{\pi z^{5}} - \frac{7 \pi^{3}}{45 z} + \ldots \ \ .\end{align}$$
Therefore, $$\text{Res}[f(z),0] = - \frac{7 \pi^{3}}{45}$$
and $$\sum_{n=1}^{\infty} \frac{\coth (\pi n)}{n^{3}} = \frac{1}{4} \left( \frac{7 \pi^{3}}{45} \right) = \frac{7 \pi^{3}}{180} .$$
| {
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Prove $2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2)$ for $x, y, z \ge 0$
Let $x,y,z\ge 0$. Show that
$$2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2).$$
my idea: let $$x+y+z=p,xy+yz+xz=q,xyz=r$$
since
$$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=p^2-2q$$
and
$$(xy+yz+xz)^2=x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z)$$
$$\Longrightarrow x^2y^2+y^2z^2+x^2z^2=q^2-2pr$$and
$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+x^2z^2)=(p^2-2q)^2-2(q^2-2pr)$$
so
$$\Longleftrightarrow 2[(p^2-2q)^2-2(q^2-2pr)]+2r+7\ge 5(p^2-2q)$$
$$\Longleftrightarrow 2p^4-8p^2q+4q^2+8pr-5p^2+10q+2r+7\ge 0$$
then I can't
This link has a similar problem:
see this
Maybe this problem can use AM-GM inequality,But I can't.Thank you
| I've found a solution. Hugly, but still a solution. The procedure suggests that maybe a nice solution is possible by changing variables and using $(x-y), (x-z), (z+y)$.
The idea is that at a min point of the function we are looking for, conditions on the Hessian and the differential force $x=y=z$, so that the problem reduce to an easy case.
Here the (some) details:
1) Check that the minimum of the function $2(x^4+y^4+z^4)+2xyz+7-5(x^2+y^2+z^2)$ exists and it is atteined for $x,y,z$ strictly positive.
2) At a minimum point, the Jacobian (the differential) must degenerate and the Hessian must be positive definite (or degenerate). The diagonal entries of the Hessian are $24x^2-10$, $24y^2-10$ and $24z^2-10$, so we get
$$x\geq\sqrt{\frac{5}{12}}
\qquad
y\geq\sqrt{\frac{5}{12}}
\qquad
z\geq\sqrt{\frac{5}{12}}$$
3) The Jacobian equations are
$$\begin{array}{l}
4x^3+yz=5x\\
4y^3+xz=5y\\
4z^3+xy=5z\end{array}
\qquad\text{that give}
\qquad
\begin{array}{l}
4x^4+xyz=5x^2\\
4y^4+xyz=5y^2\\
4z^4+xyz=5z^2\\
\end{array}$$
4) Subtracting the firts two equations (of the right side) we get
$$4(x^4-y^4)=5(x^2-y^2)$$
that is
$$4(x-y)(x^3+x^2y+xy^2+y^3)=5(x-y)(x+y)$$
if $(x-y)\neq 0$ we can divide and obtain
$$4(x^3+x^2y+xy^2+y^3)=5(x+y)$$
simiarly, by repalcing $x$ with $z$ we get
$$ 4(z^3+z^2y+zy^2+y^3)=5(z+y)$$
by subtracting:
$$4((x^3-z^3)+(x^2-z^2)y+(x-z)y^2)=5(x-z)$$
if $(x-z)\neq 0$ we can divide and obtain
$$4(x^2+xz+z^2+(x+z)y+y^2)=5$$
which is
$$(x+y)^2+(x+z)^2+(z+y)^2=5/2$$
but now, from point 2) we get that the terms $(x+y)^2,(x+z)^2,(z+y)^2$ are all at least $5/3$
whence
$$5/2=(x+y)^2+(x+z)^2+(z+y)^2>5$$
Contradiction.
Therefore the minpoint is at $x=y=z$ and we reduce to an easy polynomial in one variable.
| {
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Simplifying Second Derivatives
I can't seem to figure out how my professor simplified this second derivative. Any help is much appreciated. I'm having trouble simplifying the second derivatives of most problems so step by step instructions would be awesome.
| Notice that each term in the numerator has a common factor of $1-x^2$ so cancelling that you are left with
$$\begin{align}\frac{(6x)(1-x^2)^2-3(1+x^2)(2)(1-x^2)(-2x)}{(1-x^2)^4} &= \frac{6x(1-x^2)-3(1+x^2)(2)(-2x)}{(1-x^2)^3} \\ &= \frac{6x(1-x^2)+12x(1+x^2)}{(1-x^2)^3} \\ &= \frac{6x-6x^3+12x+12x^3}{(1-x^2)^3} \\ &= \frac{18x+6x^3}{(1-x^2)^3} \\ &= \frac{6x(3+x^2)}{(1-x^2)^3}\end{align}$$
| {
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How would I solve this mathematical induction proof? I am stuck after the first part of the induction. $$1 + 5 + 5^2 + \ldots + 5^n = \frac{5^{n+1}-1}{4}$$
Basis case $n= 0$:
$1^0 = 1 \;\;\;\;\;\;\;\;\;\;\;\; \frac{5^{1+1}-1}{4}=1$
Assume true for $n=k$:
$$1 + 5 + 5^2 + \ldots + 5^k = \frac{5^{k+1}-1}{4}$$
Need to show for $n=k+1$:
$$1 + 5 + 5^2 + \ldots + 5^{k+1} = \frac{5^{k+1+1}-1}{4} = \frac{5^{k+2}-1}{4}$$
Induction proof
$$\frac{5^{k+1}-1}{4} + 5^{k+1}=\frac{5^{k+1}-1}{4} + \frac{4(5^{k+1})}{4}$$ This is where I am stuck and do not know if I am even right with this at all.
| This is a geometric series, with common ratio = 5.
$$1 + q + q^2 + ... + q^n = (1 - q^{n+1})/(1-q)$$
Proof of this assumption by induction :
The case n=0 is trivial ($ 1 = 1 $)
Assume you have the previous result for n.
$$1+q+...+q^{n+1} = (1-q^{n+1})/(1-q) + q^{n+1}$$
$$1+...+q^{n+1} =(1-q^{n+1} + q^{n+1}\cdot (1-q) ) / (1-q)$$
If you develop, you get the Right hand side :
$$ (1 - q^{n+1} + q^{n+1} - q^{n+2})/(1-q)$$
Thus,
$$(1-q^{n+2})/(1-q)$$
Which is the expected result for n+1
| {
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Proving that $\sqrt{2}+\sqrt{3}$ is irrational This is from Spivak.
Prove that $\sqrt{2}+\sqrt{3}$ is irrational.
So far, I have this:
If $\sqrt{2}+\sqrt{3}$ is rational, then it can be written as $\frac{p}{q}$ with integral $p, q$ and in lowest terms.
$$\sqrt{2}+\sqrt{3}=\frac{p}{q}$$
$$2\sqrt{6}+5 =\frac{p^2}{q^2}$$
$$(2\sqrt{6}+5)q^2=p^2$$
And that's about where I get stuck. In a similar question (prove that $\sqrt{2}+\sqrt{6}$ is irrational,) I was able to show that both $p,q$ had to be even which is impossible. I obviously can't apply this trick here. Any hints?
| We can solve this problem by contradiction. Suppose that it is rational, so it is a non-zero rational number such as $\frac{a}{b}$.
$(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=1\rightarrow \sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{b}{a}$
So we conclude that $\sqrt{3}-\sqrt{2}$ is a rational number as well.
$\sqrt{3}-\sqrt{2}=\frac{b}{a}\rightarrow 5-2\sqrt{6}=(\frac{b}{a})^2$
$\sqrt{3}+\sqrt{2}=\frac{a}{b}\rightarrow 5+2\sqrt{6}=(\frac{a}{b})^2$
Subtract the former equation from the latter one, we conclude $4\sqrt{6}=(\frac{a}{b})^2-(\frac{b}{a})^2$
From above equation we conclude $\sqrt{6}$ is rational, but we can prove the irrationality of $\sqrt{6}$ quite simple(It has been brought in following) which shows contradiction and hence $\sqrt{2}+\sqrt{3}$ is irrational.
proof for irrationality of $\sqrt{6}$
If it is rational then it is equal to $\frac{p}{q}$ such that $p$ and $q$ are relatively prime natural numbers. then $p^2=6q^2\rightarrow q^2|p^2\rightarrow$ $p$ and $q$ are relatively prime if and only if $q=1$ which implies 6 is a square integer which is a contradiction.
| {
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How to get the series expansion? I found a series expansion of $(1+2x)^{1/2x}$ is:
$$1+ \frac{1}{2x} 2x + \frac{\frac{1}{2x} (\frac{1}{2x}-1)}{2!} (2x)^2 + \frac{\frac{1}{2x} (\frac{1}{2x}-1) (\frac{1}{2x}-2)}{3!} (2x)^3 $$
But which law and how they have got this expansion. Is it related to the the binomial theorem $(1+x)^n$?
| Your expression isn't true with the exponent $\frac1{2x}$ since in this case we should write
$$(1+2x)^{1/2x}=\exp\left(\frac1{2x}\log(1+2x)\right)$$
but if the expression has the form $(1+2x)^{1/2\alpha}$ with constant exponent then we have
$$(1+2x)^{1/2\alpha}=1+ \frac{1}{2\alpha} 2x + \frac{\frac{1}{2\alpha} (\frac{1}{2\alpha}-1)}{2!} (2x)^2 + \frac{\frac{1}{2\alpha} (\frac{1}{2\alpha}-1) (\frac{1}{2\alpha}-2)}{3!} (2x)^3+\cdots $$ and it's just the Tayor expansion.
| {
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Is there any other method to solve $\sin x -\sqrt 3 \cos x=1$? Consider
$$
\sin x -\sqrt 3 \cos x=1
$$
for $0\leq x \leq 2\pi$.
I solved it by converting $\sin x -\sqrt 3 \cos x$ to $2\sin(x-\pi/3)$ as follows
\begin{align*}
\sin x -\sqrt 3 \cos x &= r\sin(x-\theta)\\
&= r\sin x\cos\theta -r \cos x\sin\theta\\
\end{align*}
we have
\begin{align*}
r\cos \theta &= 1\\
r\sin \theta &=\sqrt 3
\end{align*}
where $r=2$ and $\theta=\frac{\pi}{3}$ are the solution.
So the equation becomes
$$
2\sin (x-\pi/3) =1
$$
and the solution are $x=\pi/2$ and $x=7\pi/6$.
Is there any other method to solve it?
| Let me introduce the one and the only, the grandest substitution :
$t=\tan(\frac x 2)$
Now, $\sin(x)=\frac{2t}{1+t^2}$ and $\cos(x)=\frac{1-t^2}{1+t^2}$
Now put them back and solve the quadratic. Find value of $t$. Can you now find x?
Another method : Type the equation here
| {
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Matrix Mathematical Notation I am trying to work out the mathematical notation for combining the columns of two matrices, $$A=\begin{pmatrix}1 & 3 & 5 \\ 2 & 4 & 1 \\ 3 & 7 & 9\end{pmatrix}$$ and $$B=\begin{pmatrix}4 & 4 & 3 \\ 9 & 10 & 11 \\ 12 & 15 & 13\end{pmatrix},$$ to form the new matrix $$C=\begin{pmatrix}1 & 4 & 3\\ 2 & 10 & 11 \\ 3 & 15 & 13\end{pmatrix}.$$ $C$ is a matrix which is made up of the first column of $A$ and the last two columns of $B$. The problem I have is expressing $C$ in terms of $A$ and $B$ using appropriate mathematical notation, I can code it, I just don't know the notation for it! Any suggestions?
| Describing the columns of the matrices as vectors, you can also say
$$
\begin{align*}
A &= \begin{bmatrix}\mathbf{a}_1 & \mathbf{a}_2 & \mathbf{a}_3 \end{bmatrix}, \\
B &= \begin{bmatrix}\mathbf{b}_1 & \mathbf{b}_2 & \mathbf{b}_3 \end{bmatrix}, \\
C &= \begin{bmatrix}\mathbf{a}_1 & \mathbf{b}_2 & \mathbf{b}_3 \end{bmatrix}.
\end{align*}
$$
where the reader will instantly see what is happening. But of course this is not a "mathematical operation" to obtain $C$ from $A$ and $B$ as in the other answers.
| {
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Proving that if $\frac{m}{n}<\sqrt{2}$ then there exists $\frac{m'}{n'}$ such that $\frac{m}{n}< \frac{m'}{n'}<\sqrt{2}$ This is from Spivak. The problem is as follows:
Prove that if $\frac{m}{n}<\sqrt{2}$ then there exists $\frac{m'}{n'}$ such that $\frac{m}{n}< \frac{m'}{n'}<\sqrt{2}$
So far, I have that since $\frac{m^2}{n^2}<2$ it follows that:
$$\frac{(m+2n)^2}{(m+n)^2}-2<2-\frac{m^2}{n^2}<2$$
and: $$\frac{2n^2-m^2}{(m+n)^2}<\frac{2n^2-m^2}{n^2}<2$$
I suppose at this point, I would need to prove that both the terms $\sqrt{\frac{2n^2-m^2}{(m+n)^2}}$ and $\sqrt{\frac{2n^2-m^2}{n^2}}$ are rational. This is about where I'm stuck... Any hints would be much appreciated.
Edit:
I should probably add that at this point, I'm mostly interested in a solution that uses the techniques given in Spivak so far. Basically the only things that have been covered are the properties of $\mathbb{R}$ and induction...
| The easiest way to do this is probably just to construct $m'$ and $n'$ and show that they satisfy the requirement. Assume wlog $m > n > 0$. (If $m, n \leq 0$ then try again with $-m$ and $-n$. If $m \leq n$ then $4/3$ is the answer.) Since $m/n < \sqrt{2}$, we have $m^2 < 2n^2$.
Now consider the number $x = \frac{m^2 + 2n^2}{2mn} = \frac{1}{2}(\frac{m}{n} + \frac{2n}{m}) = \frac{m}{2n} + \frac{n}{m}$. (I got this by using Newton's method to approximate $\sqrt{2}$, starting with $m/n$.) Because $2n^2 > m^2$, we have $$x = \frac{m^2 + 2n^2}{2mn} < \frac{2n^2 + 2n^2}{2mn} = \frac{2n}{m}.$$ However, we also have $$x^2 - 2 = \left(\frac{m}{2n} + \frac{n}{m}\right)^2 - 2 = \frac{m^2}{4n^2} + 1 + \frac{n^2}{m^2} - 2 = \frac{m^2}{4n^2} - 1 + \frac{n^2}{m^2} = \left(\frac{m}{2n} - \frac{n}{m}\right)^2 > 0.$$ Therefore $x^2 > 2$ and $x > \sqrt{2}$.
In conclusion, $\sqrt{2} < x < \frac{2n}{m}$, so $2/x = \frac{4mn}{m^2 + 2n^2}$ satisfies the requirements.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $R(x) := \frac {3x^4+5x+2} {2x^4+3x^2+7}$. Prove $R$ has a limit for $x \rightarrow \infty$. Let $$R(x) := \frac {3x^4+5x+2} {2x^4+3x^2+7}$$
Prove $R$ has a limit for $x \rightarrow \infty$.
Using a calculator I've expaneded $\frac {3x^4+5x+2} {2x^4+3x^2+7} = \frac {-9x^2}{2(x^4+3x^2+7)} +\frac {5x}{2x^4+3x^2+7} - \frac {17}{2(x^4+3x^2+7)} + 3/2$.
And I can easily see that three of these fractions has limit $0$ for $x \rightarrow \infty$.
However am I relying a bit to much on a calculator here ? And should I prove for each of these fractions rigorous that the limit is $0$ ?
| There's a simpler way:
$$
\lim_{x\to\infty}\frac{3x^4+5x+2}{2x^4+3x^2+7}=\lim_{x\to\infty}\frac{3+\frac5{x^3}+\frac2{x^4}}{2+\frac3{x^2}+\frac7{x^4}}=\frac32
$$
since
$$
\lim_{x\to\infty}\frac{f(x)}{g(x)}=\frac{\lim_{x\to\infty}f(x)}{\lim_{x\to\infty} g(x)}
$$
and
$$
\lim_{x\to\infty}(f(x)+g(x))=\lim_{x\to\infty}f(x)+\lim_{x\to\infty}g(x)
$$ provided that both limits exist and $\lim_{x\to\infty}g(x)\ne0$ in the former case.
| {
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"timestamp": "2023-03-29T00:00:00",
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Writting a proof for $\lfloor 4x \rfloor = \lfloor x+1/4 \rfloor + \lfloor x+1/2 \rfloor + \lfloor x+3/4 \rfloor + \lfloor x \rfloor$ I am working on an assignment and have to write a proof for the following:
$$\lfloor 4x \rfloor = \lfloor x \rfloor + \lfloor x+\frac{1}4 \rfloor + \lfloor x+\frac{1}2 \rfloor + \lfloor x+\frac{3}4 \rfloor$$
I have to do by proof by case I know the cases to be as follows
$$0 \le x-\lfloor x \rfloor \lt \frac 1 4$$
$$\frac 1 4 \le x-\lfloor x \rfloor \lt \frac 1 2$$
$$\frac 1 2 \le x-\lfloor x \rfloor \lt \frac 3 4$$
$$\frac 3 4 \le x-\lfloor x \rfloor \lt 1$$
The problem is I don't understand how I can express this.
Case 0
In this case I know that it would be $4n$ here
Case 1
In this case I know that it would be $4n+1$
Case 2
Here it would be $4n+2$
Case 3
Now the last case would end up being $4n+3$
Can someone help me express this?
| One way to tackle these sorts of proof is to give a name to the leftover bit: $\epsilon = x-\lfloor x \rfloor$. [I'm sure there is a slicker way to prove this, but here's a straight forward attempt...]
Then, for example, in case 1:
$$0 \le x-\lfloor x \rfloor = \epsilon \lt \frac 1 4$$
Clear the fraction...
$$0 \le 4x-4\lfloor x \rfloor = 4\epsilon \lt 1$$
Then
$$\lfloor 4x \rfloor = \lfloor 4\lfloor x \rfloor +4\epsilon \rfloor = 4\lfloor x \rfloor$$
since $4\lfloor x \rfloor$ is an integer and $0 \leq 4\epsilon \lt 1$.
$$\lfloor x \rfloor + \lfloor x+\frac{1}4 \rfloor + \lfloor x+\frac{1}2 \rfloor + \lfloor x+\frac{3}4 \rfloor = 4\lfloor x \rfloor$$
since $\lfloor x \rfloor \leq x < x + \frac 1 4 < x + \frac 1 2 < x + \frac 3 4 = \lfloor x \rfloor + \epsilon + \frac 3 4$ where $0 \leq \epsilon + \frac 3 4 < \frac 1 4 + \frac 3 4 =1$ so all get rounded down to $\lfloor x \rfloor$.
I hope this helps get you started on the other cases! :)
| {
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Description of the boundary of a ball Solving an optimization problem in multiple variables, I had to examine a function
$$
f(x,y,z)=x^2+2yz
$$
defined on a ball
$$
\{ (x,y,z)\enspace|\enspace x^2+y^2+z^2\leq1 \}.
$$
The boundary is then described by the sphere $x^2+y^2+z^2=1$. According to my textbook, one can search for maxima on the boundary by rewriting $x^2+y^2+z^2=1$ as $x^2=1-y^2-z^2$, then substituting this into $f$ to obtain
$$
f(x,y,z)=1-y^2-z^2+2yz,\quad y^2+z^2\leq1
$$
Now what I don't understand is, how come we need to specify that $y^2+z^2\leq1$? The unit sphere is described by $x^2+y^2+z^2=1$ with no need to specify any restrictions on $y$ or $z$, but once we rewrite this as $x^2=1-y^2-z^2$ and plug it into the function, suddenly it becomes necessary to specify that $y^2+z^2\leq1$. Why?
| When you write $$x^2+y^2+z^2=1$$
The set of solutions implicitly has the restriction that $y^2+z^2\le1$ because if this condition is violated there are no solutions in $\mathbb R$.
$$y^2+z^2\gt1$$
$$x^2+y^2+z^2\gt1+x^2$$
$$1\gt1+x^2\implies x^2\lt0\implies x\in \phi$$
| {
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difference equation( recurrence relation) Let $y_n$ satisfy the nonlinear difference equation:
$$(n+1)y_n=(2n)y_{n-1}+n.$$
Let $u_n=(n+1) y_n$. Show that
$$u_n= 2u_{n-1}+n.$$
Solve the linear difference equation for $u_n$. Hence find $y_n$ subject to the initial condition $y_0=4$.
I have showed that $u_n=2u_{n-1}+n$, but I don't know how to do the next step, can anyone help me with this please?
| Firstly, divide throughout by $2^n$ to get:
$$\frac{u_n}{2^n} = \frac{u_{n-1}}{2^{n-1}} + \frac{n}{2^n}$$
Rearrange to get
$$\frac{u_n}{2^n} - \frac{u_{n-1}}{2^{n-1}} = \frac{n}{2^n}$$
Sum from $1$ to $n$:
$$\sum_{i=1}^n \frac{u_i}{2^i} - \frac{u_{i-1}}{2^{i-1}} = \sum_{i=1}^n\frac{i}{2^i}$$
$$\frac{u_n}{2^n} - \frac{u_0}{2^0} = \left(\frac{1}{2}\right)^1 + 2\cdot\left(\frac{1}{2}\right)^2 + 3\cdot\left(\frac{1}{2}\right)^3 + \dots + n\cdot\left(\frac{1}{2}\right)^n$$
Now, recall that
$$(1 - x^n)(1-x)^{-1} = (1 + x + x^2 + \cdots + x^{n-1})$$
Differentiate both sides to get:
$$(1-x^{n+1})(1-x)^{-2} -(n+1)x^{n}(1-x)^{-1} = 1 + 2x + 3x^2 + \cdots + nx^{n-1}$$
I'll leave the rest to you :)
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
coefficient of $x^2$, in $(1+x+x^2)^{10}$
How to find coefficient of $x^2$, in $(1+x+x^2)^{10}$, without actually expanding it?
I think the fact $\dfrac{1-x^3}{1-x}=1+x+x^2$ may help.
But can't use it!
| Using the binomial theorem,
$(1 + x + x^2)^{10} = \displaystyle \sum_{j=0}^{10} \binom{10}{j} (1 + x)^j (x^2)^{10 - j}$
Using the binomial theorem again,
$\begin{align*}
(1 + x + x^2)^{10} & = \displaystyle \sum_{j=0}^{10} \binom{10}{j} \left( \sum_{k=0}^j \binom{j}{k} x^k \right) (x^2)^{10 - j} \\
& = \sum_{j=0}^{10} \sum_{k=0}^j \binom{10}{j} \binom{j}{k} x^{20 + k - 2j}
\end{align*}$
Now, by parity, $k$ must be even. There are only two possbilities: $k = 0, j = 9$ and $k = 2, j = 10$. Thus, the coefficient is
$\displaystyle \binom{10}{9} \binom{9}{0} + \binom{10}{10} \binom{10}{2} = 10 \cdot 1 + 1 \cdot 45 = 55$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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how to find rational numbers satisfying the binary quadratic equation $x^2+3xy+5y^2=4$ I am looking for a generalisation of the solution of $x,y$ wich are rational numbers,they could be infinite,how can i find such solutions,integer solutions are obvious
I have found that
$\Delta=-11,u=1,d=-11$
also
$(x+\frac{3+\sqrt{-11}y}{2})(x+\frac{3-\sqrt{-11}y}{2})=4$
| This is a variation of Hecke's answer
$$(2x+3y)^2+11y^2=4^2$$
$$\left(\frac{2x}y+3\right)^2+11=\left(\frac4y\right)^2$$
$$11=\left(\frac4y\right)^2-\left(\frac{2x}y+3\right)^2$$
$$11=\left(\frac{4-2x}y-3\right)\left(\frac{4+2x}y+3\right)$$
$$\frac{4-2x}y-3=t, \frac{4+2x}y+3=\frac{11}t$$
$$\implies x = -\frac{2(t^2+6t-11)}{t^2+11}, y = \frac{8t}{t^2+11}, t\neq0$$
My comment contained an imprecision. This works for every equation that can be transformed into the form:
$$(ax+by)^2+cy^2=\left(\frac pq\right)^2$$
Where $a,b,c,p,q\in\mathbb Z, q\neq 0$
| {
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} |
Integrate $\int\frac{\cos^5(x)}{\sin(x)^{1/2}}dx$ I can't find a way to solve this:
$$\int\frac{\cos^5(x)}{\sin(x)^{1/2}}dx$$
I'm not sure how i have to proceed, can you help me? I've tried the substitution u=(sinx), but it got me nowhere.
| Hint:
(Long) METHOD 1: See that
$$\dfrac{d\sqrt{\sin x}}{dx}=\dfrac{1}{2\sqrt{\sin x}}\cos x\\
\implies \dfrac{1}{\sqrt{\sin x}}\cos xdx=2d\sqrt{\sin x}$$
Thus,
$$\int\dfrac{\cos^5x}{\sqrt{\sin x}}dx=2\int{\cos^4x}d\sqrt{\sin x}=\\
2\left(\sqrt{\sin x}\cos^4x+4\int\sqrt{\sin x}\cos^3 x\sin xdx\right)$$
Now, $\int\cos^3 x\sin^{3/2} xdx$ can be calculated by $\dfrac{d\sin^{5/2}x}{dx}=\dfrac{5}{2}\sin^{3/2}x\cos x$, i.e., $\sin^{3/2}x\cos xdx=\dfrac{2}{5}d\sin^{5/2}x$. Continue integration by parts.
(Shorter) METHOD 2: Use the identity $1-\sin^2 =\cos^2$, to get
$$\int\dfrac{(1-\sin^2x)^2}{\sqrt{\sin x}}dx=2\int2{(1-u^4)^2}du,\text{ $u=\sqrt{\sin x}$}\\
\implies \int\dfrac{(1-\sin^2x)^2}{\sqrt{\sin x}}dx=\int(1+u^8-2u^4)du=\\
\boxed{\sqrt{\sin x}+\dfrac{\sqrt{\sin^9 x}}{9}-2\dfrac{\sqrt{\sin^5x}}{5}+C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Use Lagrange Multipliers to find the absolute extrema Use Lagrange Multipliers to find the absolute extrema (if any) of:
$f(x,y) = 4x^2 + 9y^2$; subject to $2x +3y = 6$.
Using Lagrange I end up with one point: $(\frac{3}{2}, 1)$
I'm just not sure how to show if that point is a max or a min?
| Another way to check the nature of the extremum in this situation is to substitute the constraint into the function. Thus, with $ \ 3y \ = \ 6 \ - \ 2x \ $ , we have
$$ f(x,y) \ = \ 4x^2 \ + \ 9y^2 $$
$$ \rightarrow \ \ \phi(x) \ = \ 4x^2 \ + \ (6 \ - \ 2x)^2 \ = \ \ 4x^2 \ + \ 36 \ - \ 24x \ + \ 4x^2 \ = \ 4 \ ( \ 2x^2 \ - \ 6x \ + \ 9 \ ) $$
$$ = \ 4 \ ( \ 2 \ [ \ x^2 \ - \ 3x \ ] \ + \ 9 \ ) \ = \ 4 \ \left( \ 2 \ [ \ x^2 \ - \ 3x \ + \ \frac{9}{4} \ ] \ + \ 9 \ - \ \frac{9}{2} \ \right) $$
$$ = \ 4 \ \left( \ 2 \ [ \ x \ - \ \frac{3}{2}\ ]^2 \ + \ \frac{9}{2} \ \right) \ = \ 8 \ [ \ x \ - \ \frac{3}{2}\ ]^2 \ + \ 18 \ \ . $$
We have thereby shown, since this function describes an "upward-opening" parabola, that the minimum of $ \ \phi(x) \ $ or $ \ f(x,y) \ $ is 18, which occurs at $ \ x \ = \ \frac{3}{2} \ \Rightarrow \ y \ = \ 1 \ . $
A geometrical interpretation of the problem is that, by using the Lagrange-multiplier method, we are looking for level curves of the function $ \ f(x,y) \ $ which are just tangent to the constraint "curve", which is the line $ \ 2x \ + \ 3y \ = 6 \ $ . The level curves $ \ 4x^2 \ + \ 9y^2 \ = \ C \ $ are concentric ellipses, only one of which is large enough to "touch" the constraint line without passing through it. That ellipse is $ \ 4x^2 \ + \ 9y^2 \ = \ 18 \ $ ; ellipses with larger values of $ \ C \ $ cross the line at two points and can be made arbitrarily large. Thus, the Lagrange-multiplier method locates the absolute minimum of $ \ f(x,y) \ $ ; there is no absolute maximum.
| {
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"timestamp": "2023-03-29T00:00:00",
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Implicit differentiation for $x^6-y^6=14$ I am trying to find the $\dfrac{\mathrm d^2 y}{\mathrm d x^2}$ for $x^6-y^6=14$.
I got $$\frac{5x^4\left(y^6-x^6\right)}{y^{11}}$$ but I'm not sure if it's right or not.
I am completely stuck on getting $\dfrac{\mathrm d^2 y}{\mathrm dx^2}$ for $y-\cos y =2x$.
| It would be really helpful if you would also outline your computation. How can we see what your problem is when you do not tell about it?
In the first equation, differentiation of $x^6-y(x)^6=14$ gives
\begin{align}
6x^5-6y(x)^5y'(x)&=0\\
\implies y'(x)&=\frac{x^5}{y(x)^5}=\frac{x^5y(x)}{x^6-14}\\
5x^4-5y(x)^4y'(x)^2-y(x)^5y''(x)&=0\\
\implies
y''(x)&=\frac{5x^4}{y(x)^5}-\frac{5y'(x)^2}{y(x)}\\
&=\frac{5x^4}{y(x)^5}-\frac{5x^{10}}{y(x)^{11}}&
&=\frac{5x^4}{y(x)^{11}}(y(x)^6-x^6)\\
&=-70\frac{x^4y(x)}{(x^6-14)^2}
\end{align}
so yes, the first result is correct.
The second example works in the same fashion
\begin{align}
y(x)-\cos y(x)&=2x\\
y'(x)\,(1+\sin y(x))&=2\\
y''(x)\,(1+\sin y(x))+y'(x)^2\,\cos y(x)&=0\\
\implies\quad
y''(x)\,(1+\sin y(x))^3+4\,\cos y(x)&=0
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
How to find infinite sum How to find infinite sum $$1+\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\dfrac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $$
I can see that 3 cancels out after 1/3, but what next? I can't go further.
| Here is another approach. The answer equals $f(1)$ where
$$\begin{align}f(x)&=1+\frac13x+\frac{1\cdot 3}{3\cdot 6}x^2+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^3+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12}x^4+\cdots\\
\frac{df}{dx}&=\frac13\cdot 1+\frac{1\cdot 3}{3\cdot 6}x\cdot 2+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^2\cdot 3+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12}x^3\cdot 4+\cdots\\
3\frac{df}{dx}&=1+\frac13x\cdot 3+\frac{1\cdot 3}{3\cdot 6}x^2\cdot 5+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^3\cdot 7+\cdots\\
&=f(x)+2x\left(\frac13\cdot 1+\frac{1\cdot 3}{3\cdot 6}x\cdot 2+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^2\cdot 3+\cdots\right)\\
3\frac{df}{dx}&=f(x)+2x\frac{df}{dx}\end{align}$$
Now solve the initial value problem with $f(0)=1$.
$$\begin{align}(3-2x)\frac{df}{dx}&=f \\
\frac{df}{f}&=\frac{dx}{3-2x} \\
\ln f&=\ln [C(3-2x)^{-1/2}]\\
f(x)&=C(3-2x)^{-1/2}\\
f(0)&=1=C\cdot 3^{-1/2} \\
C&=3^{1/2} \\
f(x)&=3^{1/2}(3-2x)^{-1/2}\\
f(1)&=3^{1/2}.\end{align}$$
| {
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Prove that $f(x)=(e^x-1)(x^2+3x-2)+x$ has exactly one positive root, exactly one negative root and one root at $x=0$. Prove that the function
$f(x)=(e^x-1)(x^2+3x-2)+x$
has exactly one positive root, exactly one negative root and one root at $x=0$.
My work so far:
$f(0)=0$
Thus, $x=0$ is a root.
For the positive root,
$f(\frac{1}{4})=(e^{\frac{1}{4}}-1)((\frac{1}{4})^2+3(\frac{1}{4})-2)+(\frac{1}{4})
=(e^{\frac{1}{4}}-1)(\frac{1}{16}+\frac{3}{4}-2)+\frac{1}{4}
=-\frac{19}{16}e^{\frac{1}{4}}+\frac{23}{16}
<0$
$f(1)=(e^1-1)((1)^2+3(1)-2)+(1)
=(e^1-1)(1+3-2)+1=2e^1-1
>0$
Thus, there is at least one positive root, by the intermediate value theorem. I have also done the same thing for the negative root, as $f(-4)<0$ and $f(-2)>0$.
But how do I show that there is only one positive root and only one negative root?
| You can also use the fact that the function is differentiable (and continuous) at all $R$, to find a contradiction:
Suppose that $0 <a <b$ with $a$, $b$, roots of $f(x)$, with $0$, $a$, $b$, consecutive roots, then there exists $c_1\in[0,a]$ and $c_2 \in[a, b]$ such that $f(x)$ reaches maximum or minimum, as the function is twice differentiable we have:
$$f''(x)=e^x(x^2+5x+1)+e^x(2x+5)-2$$
$$f''(x)=e^x(x^2+5x+1)+2xe^x+5e^x-2$$
how $5e^x>2$, $f''(x)$ is always positive for all $x>0$ then $c_1$ and $c_2$ are both minimuns, but this implies that there is no tangent line at $a$ which contradicts the fact that $f(x)$ is differentiable at all $R$.
God bless
| {
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How find this $x,y$ such this equation $x^x+x=y!$
Find all pairs of positive integers $(x,y)$,such
$$x^x+x=y!$$
I find the $$(x,y)=(1,2)$$ is such it. and
$$(x,y)=(2,3)$$
I think this equation have other roots. and maybe use inequality to solve this?
since Use Bernoulli inequality we have
$$x^x=(x-1+1)^x\ge 1+x(x-1)=x^2-x+1$$
Thank you for you help, this problem is from Mathematical contest in china jiangxi province in 2014
| There are no other such pair of integers for $x>2$.
Updated Proof:
*
*If $x$ is not prime:
*
*$x^x \geq x! \implies x^x+x>x!$
*Therefore, we must choose $y>x$
*$x^x+x$ is divisible by $x$ exactly once, because $\frac{x^x+x}{x}=x^{x-1}+1$ is not divisible by $x$
*But since $x$ is not prime, for every $y>x$ that we choose, $y!$ is divisible by $x$ at least twice
*If $x$ is a prime such that $x \equiv 1 \bmod 6$, then let's analyze $x^{x-1}+1$:
*
*$x^K \equiv 1 \bmod 6$ for every non-negative integer $K$
*$x^{x-1} \equiv 1 \bmod 6$, since $x-1$ is a non-negative integer
*$x^{x-1}+1 \equiv 2 \bmod 6$
*$x^{x-1}+1$ is not divisible by $3$
*$x^x+x$ is not divisible by $3$, hence, no $y>2$ exists such that $x^x+x=y!$
*If $x$ is a prime such that $x \equiv 5 \bmod 6$, then let's analyze $x^{x-1}+1$:
*
*$x^K \equiv 1 \bmod 6$ for every non-negative even integer $K$
*$x^{x-1} \equiv 1 \bmod 6$, since $x-1$ is a non-negative even integer
*$x^{x-1}+1 \equiv 2 \bmod 6$
*$x^{x-1}+1$ is not divisible by $3$
*$x^x+x$ is not divisible by $3$, hence, no $y>2$ exists such that $x^x+x=y!$
*Finally, if $x=3$ then $x^x+x=30$, and no $y$ exists such that $y!=30$
| {
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number of terms The following problem maybe tedious if done by hand and requires patience. After factorizing the following variables find the number of terms and the sum of the number of terms.
$(a^0),(a+b)^0,(a+b+c)^0,(a+b+c+d)^0,(a+b+c+d+e)^0,(a+b+c+d+e+f)^0$
$(a^1),(a+b)^1,(a+b+c)^1,(a+b+c+d)^1,(a+b+c+d+e)^1,(a+b+c+d+e+f)^1$
$(a^2),(a+b)^2,(a+b+c)^2,(a+b+c+d)^2,(a+b+c+d+e)^2,(a+b+c+d+e+f)^2$
$(a^3),(a+b)^3,(a+b+c)^3,(a+b+c+d)^3,(a+b+c+d+e)^3,(a+b+c+d+e+f)^3$
$(a^4),(a+b)^4,(a+b+c)^4,(a+b+c+d)^4,(a+b+c+d+e)^4,(a+b+c+d+e+f)^4$
$(a^5),(a+b)^5,(a+b+c)^5,(a+b+c+d)^5,(a+b+c+d+e)^5,(a+b+c+d+e+f)^5$
$(a^6),(a+b)^6,(a+b+c)^6,(a+b+c+d)^6,(a+b+c+d+e)^6,(a+b+c+d+e+f)^6$
| If you can do one, you can probably do them all. Let's look at $(a + b + d + d)^5$. When you distribute the terms, you choose a variable from each of five factors, each of $(a + b + c + d)$. You will generate all possible permutations of: $xxxxx$ where $x$ can be any of $a$, $b$, $c$, or $d$. You have four possible "digits" and five places to put them, therefore you have $4^5 = 1024$ ways of arranging them.
However, if a term contains the same number of $a$, $b$, $c$, or $d$'s then we say it's a "like term" (since multiplication is commutative). How many distinct groups can we get from the $1024$? We can assign a number to each digit (the exponent). This digit represents the number of times we selected this digit, e.g. $a^2b^2c$ means we chose two $a$'s, two $b$'s, and one $c$. Since we must select five things, these numbers will always add to five.
Determining the number of terms is a partitioning problem
The $a$, $b$, $c$, and $d$ are buckets which each hold a number--added together they must equal $5$. Using the stars and bars method, you can find how many ways four numbers can add to give $5$ (assuming they can be zero):
$$
|||*****
$$
First notice that we only need three bars (one less than the four digits). This is because these three bars create four regions: one to the left of all three, two in between the first and last, and another to the right of the last bar. This particular configuration would mean the first three buckets held $0$ and the final (fourth) bucket held all five (this could be $d^5$ for example). Since we have $5 + 3 = 8$ places to put things, deciding how many unique partitions is just figuring out how many ways to pick three places from the $8$: $\binom{8}{3} = 56$. So there will be $56$ distinct terms in the expansion of $(a + b + c + d)^5$. In general, there will be $\binom{n + m - 1}{n - 1} = \binom{n + m - 1}{m}$ in the expansion of $(x_1 + x_2 + ... x_n)^m$.
Finding sum for each Term
To find the sum for each term, assume you have found:
$$
a^{a_n}b^{b_n}c^{c_n}d^{d_n}, a_n, b_n, c_n, d_n \geq 0, a_n + b_n + c_n + d_n = 5
$$
You need to choose $a_n$ from $5$. In this case each choice is which factor it came from: $(a + b + c + d)_1(a + b + c + d)_2(a + b + c + d)_3(a + b + c + d)_4(a + b + c + d)_5$. And then you need to choose $b_n$ from $5$ and so on and so forth, right?
Not quite. Once you choose the $a_n$ factors for $a$, there are only $5 - a_n$ left. So for $b$, you should choose from $5 - a_n$, not $5$! This gives:
$$
\binom{5}{a_n}\binom{5 - a_n}{b_n}\binom{5 - a _n - b_n}{c_n}\binom{5 - a_n - b_n - c_n}{d_n}
$$
The order you do this in should not matter which means there are $4! = 24$ different ways to write this and they should all be equal. In general, you would get something like:
$$
\prod_{i = 1}^{n} \binom{m - \sum_{k = 1}^{i - 1}(x_i)_n}{(x_i)_n}
$$
Where $(x_i)_n$ is the exponent of the $i^\text{th}$ variable for this particular term.
Simplifying the inner product
$$
\binom{5}{a_n}\binom{5 - a_n}{b_n}\binom{5 - a _n - b_n}{c_n}\binom{5 - a_n - b_n - c_n}{d_n}
$$
Writing out the terms explicitly gives:
$$
\frac{5!}{a_n!(5 - a_n)!}\frac{(5 - a_n)!}{b_n!(5 - a_n - b_n)!}...
$$
You can see that the "bottom" term is canceling the next "top" term until you finally get to $\frac{(5 - a_n - b_n - c_n)!}{d_n!(5 - a_n - b_n - c_n - d_n)!}$. The top cancels with the previous bottom and since $a_n + b_n + c_n + d_n = 5$, the bottom factorial is just $0! = 1$, leaving:
$$
\frac{5!}{a_n!b_n!c_n!d_n!}
$$
Or, in general:
$$
\prod_{i = 1}^{n} \binom{m - \sum_{k = 1}^{i - 1}(x_i)_n}{(x_i)_n} = \frac{m!}{\prod_1^n (x_i)_n!}
$$
Which is the term you see in the multinomial theorem.
We could have arrived at this final result by reframing the problem as a permutation problem. Lets say we have a particular sequence (perhaps $aabbb$ which would represent the term $a^2b^3$) and we treat each digit as unique (i.e. $a_1a_2b_1b_2b_3$). There are $5!$ ways to rearrange distinct things. However for each unique ordering there are $2!$ ways to rearrange the $a$'s and $3!$ ways to rearrange the $b$'s. So using that we come up with the equation:
$$
x\cdot 2!3! = 5! \rightarrow x = \frac{5!}{2!3!}
$$
Hopefully then it's easy to see how to extend this to the general case.
| {
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Prove $\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\ge 4+(x-y)^2$ for $4 \le x + y + z \le 5$
Let $x,y,z>0$, and such $$4\le x+y+z\le 5.$$
Show that
$$\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge 4+(x-y)^2.$$
It seems that the condition $\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge 4+(x-y)^2$ is maybe old, and this condition is strange?
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=586357 .
I want to use
$$\dfrac{x^2}{y}=2x-y+\dfrac{(x-y)^2}{y}.$$
If we let $$x\to x'r,y\to y'r,z\to z'r,$$then $$x'+y'+z'=4, r\in [1,\dfrac{5}{4}]$$
But I can't prove that either, because I felt I can't use the condition.
Thank you.
| I cannot give a full proof but here is a piece that might take you closer to what you're looking for.
Using the equality you have given, we can transform the left-hand side of the equation to the form
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x} = 2x - y + \frac{(x-y)^2}{y} + 2y - z + \frac{(y-z)^2}{z} + 2z - x + \frac{(z-x)^2}{x} = x + y + z + \frac{(x-y)^2}{y} + \frac{(y-z)^2}{z} + \frac{(z-x)^2}{x} $$
Since we know that $x+y+z \geq 4$, we get
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x} \geq 4 + \frac{(x-y)^2}{y} + \frac{(y-z)^2}{z} + \frac{(z-x)^2}{x}$$
So, in order to prove the given inequality, it is sufficient to prove that
$$\frac{(x-y)^2}{y} + \frac{(y-z)^2}{z} + \frac{(z-x)^2}{x} \geq (x-y)^2$$
This inequality can be rewritten as
$$(x-y)^2 \left(\frac{1}{y} - 1 \right) + \frac{(y-z)^2}{z} + \frac{(z-x)^2}{x} \geq 0$$
Since any square $x^2 \geq 0$ and $x,y,z > 0$, this inequality certainly holds when
$$\frac{1}{y} - 1 \geq 0$$ or in other words when $0 < y \leq 1$.
However, this is a rather strict condition that was not actually given. Additionally, the condition $x+y+z \leq 5$ was not used anywhere in here - and as @EwanDelanoy pointed out, it seems to be crucial.
| {
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} |
What is the difference between $\arg(z)$ and $\operatorname{Arg}(z)$, where $z=a+bi$ What is the difference between the $\arg(z)$ and the $\operatorname{Arg}(z)$,
where $z$ is a complex number of the form $a+bi$, for example: $z = -2 - 2i$
The angle from the positive x-axis to the vector would be $5π/4$
Does that mean that the $\arg(z)=\dfrac{5π}4$?
If so, is $\operatorname{Arg}(z) = \dfracπ4, -\dfracπ4$, or $\dfrac{3π}4$?
| It varies among authors, but: $-\pi < Arg(z) \leq \pi$ and $\arg(z) = Arg(z) + 2 \pi K$ for $K \in \mathbb{Z}$
To answer the example: $z = -2-2i \Rightarrow r=|z|=\sqrt{4+4}=2\sqrt{2} \Rightarrow z = 2\sqrt{2} (-\frac{2}{2\sqrt{2}}-\frac{2i}{2\sqrt{2}}) = 2\sqrt{2} (-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2} i) $
$\Rightarrow \theta = \frac{5\pi}{4}-2\pi = -\frac{3\pi}{4}$ (this is done to get it in range).
| {
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Solution for $x$ with exponents? I am trying to solve the following,
$$7^{(2x+1)} + (2(3)^x) - 56 = 0$$
Should I put the 56 on the other side and get the log of both sides and is there a better way to solve this.
| \begin{align*}
7^{2x+1}+2\cdot3^x-56 = 0 &\iff 7^{2x}\cdot7+2\cdot 3^x-7\cdot 8=0\\
&\iff 7(7^{2x}-8)+2\cdot3^x=0 \\
&\iff 7(7^{2x}-8)={-2\cdot3^x}\bf<0
\end{align*}
This is possible if and only if $7^{2x}-8<0$ which is equivalent to $x<\dfrac{\log_78}{2}\approx\dfrac{\log_77}2=0.5$. So we can only hope to find a solution using approximation methods. The answer given by Claude shows that $x$ is pretty close to $0.5$, so we can say: $$\color{grey}{\boxed{\color{white}{\underline{\overline{\color{black}{\displaystyle\, x\approx \dfrac{\log_78}{2}\,}}}}}}$$
| {
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William Lowell Putnam Integral Problem Prove That
$$
\frac{22}{7}-\pi
=
\int_{0}^{1}\frac{x^{4}\left(1 - x\right)^{4}}{1 + x^{2}}\,{\rm d}x
$$
| I Tried in this way:
$$ \frac{x^4\,\left(1-x\right)^4}{1+x^2}=\frac{\left(x^4-1+1\right)\,\left(1-x\right)^4}{1+x^2}=\frac{\left(x^4-1\right)\,\left(1-x\right)^4}{1+x^2}+\frac{\left(1-x\right)^4}{1+x^2}\\=\left(x^2-1\right)\left(1-x\right)^4+\frac{\left(1-x\right)^4}{1+x^2}\\=\left(x+1\right)\left(x-1\right)^5+\frac{\left(1-x\right)^4}{1+x^2}\\=\left(x-1\right)^6+2\left(x-1\right)^5+\frac{\left(1-x\right)^4}{1+x^2}$$ First two terms are straight forward to integrate.
$$\int_0^1 \frac{\left(1-x\right)^4}{1+x^2}=\int_0^1\left(x^2-4x+5\right)-\frac{4}{x^2+1}=\frac{10}{3}-\pi$$
Can any one suggest ant alternate ways to do this problem..
| {
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Where is the error in finding the particular solution to this recurrence relation? The question is to write the general solution for this recurrence relation:
$y_{k+2} - 4y_{k+1} + 3y_{k} = -4k$.
I first solved the homogeneous equation $y_{k+2} - 4y_{k+1} + 3y_{k} = 0$, by writing the auxiliary equation $r^2 - 4r + 3 = (r-3)(r-1) = 0$. Thus $y_k^{h} = c_1(1)^k + c_2 (3)^k$. The general solution is just $y_k^{gen} = y_k^{h} + y_k^{p}$. My trouble is coming up with a particular solution. I keep up coming with $y_k^{p} = 2k^2$ when that doesn't work, but $k^2$ works, so my answer is close. I've gone through the artihmetic several times and cannot spot the mistake, here's the work:
The particular solution is of the form $y_k^{p} = a + bk$. Plugging in recurrence relation: $a + b(k+2) - 4(a + b(k+1) + 3(a + bk) = (a - 4a + 3a) + (bk - 4bk + 3bk) + (2b - 4b) = 0 + 0 - 2b = -4k$.
Thus $b = 2k$, and since our $y_k^p = a + bk$, it doesn't matter what pick $a$ to be so choose $a = 0$, which gives us $y_k^p = 2k^2$.
However, $2k^2$ doesn't satisfy the recurrence relation:
$2(k+2)^2 - 8(k+1)^2 + 6k^2 = (2k^2 - 8k^2 + 6k^2) + (8k - 16k) + (8 - 8) = -8k \ne -4k$.
Where is the error in my reasoning? I know $y_k^p = k^2$ works, but why do I keep coming up with $2k^2$.
| As a side remark, it is best to use generating functions and solve the equation in one go. Define $g(z) = \sum_{k \ge 0} y_k z^k$, multiply the recurrence by $z^k$, sum over $k \ge 0$, and recognize a few sums:
\begin{align}
\sum_{k \ge 0} y_{k + r} z^k
&= \frac{g(z) - y_0 - y_1 z - \ldots - y_{r - 1} z^{r - 1}}{z^r} \\
\sum_{k \ge 0} k z^k
&= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \\
&= \frac{z}{(1 - z)^2}
\end{align}
and so get:
$$
\frac{g(z) - y_0 -y_1 z}{z^2}
- 4 \frac{g(z) - y_0}{z}
+ 3 g(z)
= - 4 \frac{z}{(1 - z)^2}
$$
Written as partial fractions:
$$
g(z) = \frac{1 - y_0 - y_1}{2 (1 - 3 z)}
+ \frac{3 + 3 y_0 - y_1}{2 (1 - z)}
- \frac{3}{(1 - z)^2}
+ \frac{2}{(1 - z)^3}
$$
Your particular solution comes from the terms that don't include the initial values.
The generalized binomial theorem for negative integer powers gives for them:
\begin{align}
- 3 \binom{-2}{k} (-1)^k + 2 \binom{-3}{k} (-1)^k
&= -3 \binom{k + 2 - 1}{2 - 1} + 2 \binom{k + 3 - 1}{3 - 1} \\
&= -3 (n + 1) + 2 \frac{(n + 2) (n + 1)}{2} \\
&= n^2 - 1
\end{align}
| {
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Finding (sin(A+B))^2 given roots of a quadratic equation. If tan A and tan B are the roots of the equation x^2 -ax + b = 0, then the value of sin(A+B)^2 is?
Options are:
((a^2)/((a^2)+(1-b)^2),
(a^2)/(a^2+b^2),
a^2/(b+a)^2,
a^2/(b^2*(1-a)^2)
The value is needed in terms of a and b.
The farthest i have gotten is the sum of roots giving me:
(sinAcosB + CosAsinB)/CosAcosB = a
which simplifies to sin^2(A+B) = (a*cosA*cosB)^2
after which i substituted the value obtained from product of roots which was ((sinAsinB)/b)^2 = (CosACosB)^2 which is substituted again to obtain:
(a*SinA*sinB)^2/(b^2)
Please find the value with steps as i am very confused about this question.
| Sum of roots; i.e., $\tan A + \tan B = $ -coefficient of $x /$ coefficient of $x^2 = a$
Product of roots; i.e., $\tan A \tan B = $ constant $/$ coefficient of $x^2 = b$
Now, $\tan (A+B)= \dfrac{\tan A + \tan B}{1 - \tan A \tan B} = \dfrac{a}{1-b}$
Draw a right triangle with opp side as $a$ and adjacent side as $1-b$.
So our hypotenuse will be $\sqrt{a^2 + (1-b)^2}$
$\sin(A+B)=$ opp/ hypotenuse $=\dfrac{a}{\sqrt{a^2 + (1-b)^2}}$
So $\sin^2(A+B)= \dfrac{a^2}{a^2 + (1-b)^2}$
therefore, option A :-)
| {
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Calculate modulo of large numbers I have $2^{2^n}+1$ and i want to calculate ($(2^{2^{^n}} +1 )\mod 19$). How can i do it if for example i choose $n = 19$. Can i use Fermat's Little Theorem?
| Because $\phi(19)=18$, we have $2^a\equiv 2^b\pmod{19}$ if and only if $a\equiv b\pmod{18}$. This is equivalent to Fermat's Little Theorem.
Hence you need to find a small number $m$ such that $2^n\equiv m\pmod{18}$.
We compute and see the pattern mod 18: $2^1\equiv 2, 2^2\equiv 4, 2^3\equiv 8, 2^4\equiv 16, 2^5\equiv 32\equiv 14, 2^6\equiv 28\equiv 10, 2^7\equiv 20\equiv 2$, and then it repeats. So $2\equiv 2^1\equiv 2^7\equiv 2^{13}\equiv 2^{19}$.
Hence $2^{19}\equiv 2\pmod{18}$ and hence $2^{2^{19}}\equiv 2^2\pmod{19}$.
| {
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Proof by induction of a recursive sequence I am studying CIE A levels Further Maths and I am stuck at a question from June 2002:
Q
The sequence of positive numbers $u_1,u_2,u_3,...$ is such that $u_1<4$ and $$u_{n+1}=\frac{5u_n+4}{u_n+2}$$By considering $4-u_{n+1}$, or otherwise, prove by induction that $u_n<4$ for all $n\geq1$.
Prove also that $u_{n+1}>u_n$ for all $n\geq 1$.
Here is my solution
To Prove the Initial Relation $u_n<4$
By using long division:
$$\frac{5u_n+4}{u_n+2}\equiv 5-\frac{6}{u_n+2}$$
For $n=1$
$$u_{2}= 5-\frac{6}{u_1+2}$$
Since $u_1<4$ then,
$$\frac{6}{u_1+2}>1$$
and,
$$5-\frac{6}{u_1+2}<4$$
therefore,
$$u_{2}<4$$
thus the statement is true for $n=1$
Conjecture: The statement is true for $n=k$
$$u_{k}<4$$
Since $u_{k+1}<4$ ... (proved above for $n=1$),
then,
$$\frac{6}{u_{k+1}+2}>1$$
and,
$$5-\frac{6}{u_{k+1}+2}<4$$
therefore,
$$u_{k+2}<4$$
Since the statement is true for $n=k$, $n=k+1$, and $n=1$, the given relation ($u_n<4$) is correct.
To prove that $u_{n+1}>u_n$:
$$u_{n+1}-u_n=\frac{5u_n+4}{u_n+2}-u_n\equiv -\frac{(u_n)^2+3u_n+4}{u_2+2}$$
$$-\frac{(u_n)^2+3u_n+4}{u_2+2}\equiv \frac{(4-u_n)(u_n+1)}{u_n+2}>0$$
Since,
$$u_{n+1}-u_n>0$$
$$u_{n+1}>u_n$$
I am not sure if my solution to the first part is correct, and even if it is correct, I didn't use the relation $4-u_{n+1}$ so it would be really great if someone suggests abetter solution using the relation mentioned in the question.
As for second part, I am pretty confident that it is correct, but I am open to suggestions.
| For the record, this is a Ricatti recurrence, as $5 \cdot 2 - 1 \cdot 4 \ne 0$. There are at least three solution ideas around, I like Mitchell's "An Analytic Riccati solution for Two-Target Discrete-Time Control", Journal of Economic Dynamics and Control 24(4), pp 615-622 (2000).
Take:
$$
u_{n + 1} = \frac{a u_n + b}{c u_n + d}
$$
Define a new variable by:
$$
x_n = \frac{1}{1 + \eta u_n}
$$
Substituting into the recurrence gives:
$$
x_{n + 1}
= \frac{(d \eta - c) x_n + c}
{(b \eta^2 - (a - d) \eta - c) x_n + a \eta + c}
$$
Selecting $\eta$ such that $b \eta^2 - (a - d) \eta - c = 0$ you get a linear first-order recurrence. Here:
$$
4 \eta^2 -3 \eta - 1 = 0
$$
gets $\eta = 1$ or $\eta = -1/4$. First is simpler, pick that one:
$$
x_{n + 1} = \frac{(2 \cdot 1 - 1) x_n + 2}{5 \cdot 1 + 1} = \frac{x_n + 1}{6}
$$
We just have a limit on $u_1$, but we can just shift by one without changing anything. We get:
$$
x_1 = \frac{1}{1 + u_1}
$$
Use generating functions to solve the linear recurrence by defining $X(z) = \sum_{n \ge 0} x_{n + 1} z^n$, multiplying by $z^n$ and summing over $n \ge 0$:
$$
\frac{X(z) - x_1}{z} = \frac{X(z)}{6} + \frac{1}{6} \cdot \frac{1}{1 - z}
$$
to get:
$$
X(z) = \frac{1}{5 (1 - z)} + \frac{5 x_1 - 1}{5} \cdot \frac{1}{1 - z/6}
$$
The coefficients are immediate:
$$
x_{n + 1} = \frac{1}{5} + \frac{5 x_1 - 1}{5} \cdot 6^{-n}
$$
Going back to the original variables:
\begin{align}
u_n
&= \frac{4 (u_1 + 1) \cdot 6^n + (u_1 - 4)}
{(u_1 + 1) \cdot 6^n - (u_1 - 4)}
&= 4 - \frac{5 (4 - u_1)}{(u_1 + 1) \cdot 6^n + (4 - u_1)}
\end{align}
If $u_1 < 4$, the second term is always positive, and $u_n < 4$. In that case, the second term decreases as $n$ increases, so $u_{n + 1} > u_n$.
| {
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Evaluate $\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x$ My try, using $x = \sec(u)$ substitution:
$$
\begin{eqnarray}
\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &=& \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\
&=& \int \tan^2(u) \mathrm{d}u \\
&=& \tan(u) - u + C \\
&=& \tan(arcsec(x)) - arcsec(x) + C
\end{eqnarray}
$$
However, according to Wolfram Alpha, the answer should be:
$$
\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x = \sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C
$$
When I derive this last answer I don't get back the integrand, but rather:
$$
\frac{\mathrm d}{\mathrm d x}\left(\sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C\right) = \frac{x}{\sqrt{x^2-1}}- \frac{x}{(x^2-1)^{3/2}\left(1+\frac{1}{x^2-1}\right)}
$$
I don't know how to simplify this expression more. Also, I am unable to check whether my answer is correct because I don't know how to find the derivative of $arcsec(x)$.
Can someone check my calculations and tell me where I've done something wrong and how one can simplify the last expression to get back the integrand?
| \begin{aligned}
& \int \frac{\sqrt{x^{2}-1}}{x} d x \\
=& \int \frac{x^{2}-1}{x \sqrt{x^{2}-1}} d x \\
=& \int \frac{x^{2}-1}{x^{2}} d\left(\sqrt{x^{2}-1}\right) \\
=& \sqrt{x^{2}-1}-\int{\frac{1}{x^{2}} d\left(\sqrt{x^{2}-1}\right)} \\
=& \sqrt{x^{2}-1}-\int \frac{d\left(\sqrt{x^{2}}-1\right)}{\left(\sqrt{x^{2}-1}\right)^{2}+1}\\
=& \sqrt{x^{2}-1}-\tan ^{-1}\left(\sqrt{x^{2}-1}\right)+C
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Integral of $\frac{1}{x^2+1}$ using complex partial fractions. Is there any way to evaluate the following integral via a complex partial fraction decomposition? $$ \int \dfrac{1}{x^2 + 1} \text{ d}x $$ So far I have: $$ \begin{aligned} \int \dfrac{1}{x^2 + 1} \text{ d}x & = \dfrac{1}{2i} \int \dfrac{1}{x-i} \text{ d}x - \dfrac{1}{2i} \int \dfrac{1}{x+i} \text{ d}x \\ & = \dfrac{1}{2i} \log \left| \dfrac{x-i}{x+i} \right| + \mathcal{C} \end{aligned} $$ I know that the integral of a real valued function is a real valued function, so how do I take the real part of this final result?
| Using the following identity for the logarithm of a complex number:
$$\log(a+ib)=\log|a+ib|+i\arg(a+ib)=\log\sqrt{a^2+b^2}+i\tan^{-1}\left(\frac{b}{a}\right)$$
results in $$\frac{1}{2i}\log\left|\frac{x-i}{x+i}\right|=\frac{1}{2i}(\log|x-i|-\log|x+i|)=-\frac{1}{2i}(2i\tan^{-1}\frac{1}{x})=-\tan^{-1}\left(\frac{1}{x}\right)$$
Using the reciprocal property of the arctangent function, we have for $x>0$
$$-\tan^{-1}\left(\frac{1}{x}\right)=\tan^{-1}x-\frac{\pi}{2}$$
and for $x<0$
$$-\tan^{-1}\left(\frac{1}{x}\right)=\tan^{-1}x+\frac{\pi}{2}$$
Incorporating the $\pm\frac{\pi}{2}$ into the constant term, the integral is thus $$\tan^{-1}x+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Is $(1+2+3+…)=(1+2+2^2+2^3+…)(1+3+3^2+…)(1+5+5^2+…)…$? Are these equal?
$$(1+2+3+…)=(1+2+2^2+…)(1+3+3^2+…)(1+5+5^2+…)…$$
Where the RHS has a series for each prime. Looks like they are the same series by the fundamental theorem of arithmetic.
Every number on the LHS is a product of prime factors which can be obtained from the right hand side by choosing the appropriate numbers from each contributing prime and $1$ from the rest and conversely.
By a similar argument, it looks like:
$$(1+2+3+\dots)=(1+2+3+5+\dots)(1+2+3+5+\dots)(1+2+3+5+\dots)\dots$$
| Maybe you can try proving
$$
(1^s+2^s+3^s+\dots)=(1^s+2^s+2^{2s}+\dots)(1^s+3^s+3^{2s}+\dots)(1^s+5^s+5^{2s}+\dots)\dots
$$
for $s$ such that the series converge.
The left side is $\zeta(-s)$, the factors on the right side are geometric series. I think your result will be a well-known formula. But there are analytic continuations for all of these to other values of $s$ where the series diverge. Put $s=1$ to get a conclusion, $\zeta(-1) = \dots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/758501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Tricky looking integration (after separation of variables)? I've come across something in my notes that jumps from:
$${d\rho \over dz} = \sqrt{\left({\rho \over C}\right)^2 - 1}$$
to:
$$\rho(z) = C \cosh\frac{z-z_0}{C}$$
I know that separation of variables is used so what I'm really asking is how to integrate this part:
$$\int{1 \over \left(\left(\dfrac{\rho}{C}\right)^2 - 1\right)^{1/2}} \, d\rho ,$$
where $\rho$ is a function of $z$ and $C$ is a constant.
Help with this one would be greatly appreciated, thanks!
| \begin{align}
\frac \rho C & = \sec\theta \\[10pt]
\sqrt{\left(\frac\rho C\right)^2 -1} & = \sqrt{\sec^2\theta-1}= \tan\theta \\[10pt]
\frac{d\rho}{C} & = \sec^2\theta\,d\theta
\end{align}
\begin{align}
\int{1 \over \sqrt{\left(\dfrac{\rho}{C}\right)^2 - 1}} \, d\rho & = \int\frac{1}{\tan\theta} \Big( C\sec^2\theta\,d\theta\Big) = C\int \frac{d\theta}{\sin\theta\cos\theta} \\[15pt]
& = C\int\frac{2\,d\theta}{2\sin\theta\cos\theta} = C\int\frac{2\,d\theta}{\sin(2\theta)} = C\int\frac{d\eta}{\sin\eta} \\[15pt]
& = -C\log(\csc\theta+\cot\theta)+\text{constant}.
\end{align}
\begin{align}
\csc\operatorname{arcsec}\frac\rho C & = \frac{\rho}{\sqrt{\rho^2-C^2}} \\[15pt]
\cot\operatorname{arcsec}\frac\rho C & = \frac{C}{\sqrt{\rho^2-C^2}}
\end{align}
So you get
$$
-C\log\frac{\rho+C}{\sqrt{\rho^2-C^2}} = z + \text{constant} = z - z_0.
$$
Consequently
$$
\frac{\rho+C}{\sqrt{\rho^2-C^2}} = \exp\left(- \frac{z-z_0}{C} \right) = A
$$
$$
\rho+C = A \sqrt{\rho^2-C^2}
$$
Now if you square both sides you get an equation that is quadratic in $\rho$.
$$
\rho^2+2\rho C+C^2 = A^2\rho^2 - A^2 C^2
$$
$$
\rho^2(1-A^2)+2\rho C + C^2(1+A^2) = 0
$$
Now apply the "quadratic formula":
$$
\rho = \frac{-2C \pm \sqrt{4C^2 - 4(1-A^4)}}{2(1-A^2)} = \frac{-C\pm\sqrt{C^2+A^4-1}}{1-A^2}.
$$
Notice that
\begin{align}
1-A^2 & = 1 - \exp\left( -2\frac{z-z_0}{C} \right) \\[10pt]
& = 2\exp\left( -\frac{z-z_0}{C} \right)\left( \frac{\exp\left( +\frac{z-z_0}{C}\right) - \exp\left( -\frac{z-z_0}{C} \right)}{2} \right) \\[10pt]
& = 2\exp\left( -\frac{z-z_0}{C} \right) \cosh\left(\frac{z-z_0}{C}\right).
\end{align}
(Maybe I'll be back later; I may want to check details . . . )
[ . . . ]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/759453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Compute $I=\int_0^{+\infty}\frac{\arctan(t)}{e^{\pi t}-1}dt$
I would like to compute $\displaystyle I=\int_0^{+\infty}\frac{\arctan(t)}{e^{\pi t}-1}dt$
Let $D=(0,+\infty)$, I have $\frac{1}{e^{-\pi t}-1}=\frac{e^{-\pi t}}{1-e^{-\pi t}}$
So $$\frac{\arctan(t)}{e^{\pi t}-1}=\sum_{k=1}^{+\infty}\arctan(t)e^{-k \pi t}$$
Now, I can use integration term by term theorem, And finally I have,
$$
I=\sum_{k=1}^{+\infty}\int_0^{+\infty}\arctan(t)e^{-k \pi t}dt
$$
By integration by parts I get,
$$
\int_0^{X}\arctan(t)e^{k \pi t}dt\rightarrow \frac{1}{k\pi}\int_0^{+\infty}\frac{e^{-k \pi t}}{1+t^2}dt
$$
Therefore,
$$
I=\sum_{k=1}^{+\infty}\frac{1}{k\pi}\int_0^{+\infty}\frac{e^{-k \pi t}}{1+t^2}dt
$$
Now, I am stuck.
I would like to find a closed form,
Thank you in advance for your time.
| Let $ \displaystyle I(z) = \int_{0}^{\infty}\frac{\arctan \frac{x}{z}}{e^{\pi x}-1} \ dx$.
Then
$$\begin{align} I(z) &= \int_{0}^{\infty} \int_{0}^{\infty}\frac{1}{e^{\pi x}-1} \frac{\sin (xt)}{t}e^{-zt} \ dt \ dx \tag{1} \\ &= \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty} \frac{\sin (tx)}{e^{\pi x}-1} \ dx \ dt \\&= \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty} \sin (tx) \sum_{n=1}^{\infty} e^{-\pi nx}\ dx \ dt \\ &=\int_{0}^{\infty} \frac{e^{-zt}}{t} \sum_{n=1}^{\infty} \int_{0}^{\infty} \sin(tx) \ e^{-\pi n x} \ dx \ dt \\ &= \int_{0}^{\infty} \frac{e^{-zt}}{t} \sum_{n=1}^{\infty} \frac{t}{t^{2} + \pi^{2}n^{2}} \\&= \frac{1}{2}\int_{0}^{\infty} \frac{e^{-zt}}{t} \Big(\coth t -\frac{1}{t}\Big) \ dt \tag{2} \end{align}$$
Now differentiate inside of the integral with respect to $z$.
$$ I'(z) = - \frac{1}{2} \int_{0}^{\infty}e^{-zt} \Big( \coth t - \frac{1}{t}\Big) \ dt $$
And then integrate by parts.
$$ \begin{align} I'(z) &= -\frac{1}{2} e^{-zt} \Big(\log(\sinh t) - \log t \Big) \Big|^{\infty}_{0} - \frac{z}{2} \int_{0}^{\infty} e^{-zt} \Big( \log (\sinh t) - \log(t) \Big) \ dt \\ &=- \frac{z}{2} \int_{0}^{\infty} e^{-zt} \Big( \log (\sinh t) - \log(t) \Big) \ dt \\ &= - \frac{z}{2} \int_{0}^{\infty} e^{-zt} \Big(t - \log (2) + \log(1-e^{-2t}) - \log(t) \Big) \ dt \\ &= -\frac{z}{2} \Big( \frac{1}{z^{2}} - \frac{\log(2)}{z} + \frac{\log(z) + \gamma}{z} \Big) + \frac{z}{2} \int_{0}^{\infty} e^{-zt} \sum_{n=1}^{\infty} \frac{e^{-2tn}}{n} \ dt \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} - \frac{\gamma}{2} +\frac{z}{2} \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\infty} e^{-(2n+z)t} \ dt \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} - \frac{\gamma}{2} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{z/2}{n(n+z/2)} \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} - \frac{\gamma}{2} + \frac{1}{2}\psi\left( \frac{z}{2}+1 \right) + \frac{\gamma}{2} \tag{3} \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} + \frac{1}{2} \psi \Big( \frac{z}{2}+1 \Big) \end{align}$$
Then integrating back,
$$I(z) = - \frac{\log z}{2} + \frac{z \log 2}{2} - \frac{z \log z}{2} + \frac{z}{2} + \log \Gamma \left( \frac{z}{2} +1 \right) + C$$
where using Stirling's formula the constant of integration is found to be $ \displaystyle -\frac{\log(\pi)}{2}$.
Therefore,
$$ \begin{align} \int_{0}^{\infty} \frac{\arctan x}{e^{\pi x}-1} \ dx &= I(1) \\ &= \frac{\log 2}{2} + \frac{1}{2} + \log \Gamma \left(\frac{3}{2} \right)- \frac{\log \pi}{2} \\ &= \frac{\log 2}{2} + \frac{1}{2} + \log \left( \frac{\sqrt{\pi}}{2} \right) -\frac{\log \pi}{2} \\ &=\frac{1}{2} - \frac{\log 2}{2} \end{align}$$
$ $
$(1)$ $\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x} dx$ Evaluate Integral
$(2)$ Series expansion of $\coth x$ using the Fourier transform
$(3)$ http://en.wikipedia.org/wiki/Digamma_function#Series_formula
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/759747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Permutations and Combinations Show that $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - ...+(-1)^k * \binom{n}{k} = (-1)^k * \binom{n-1}{k}$.
I know this has to do with permutations and combination problems, but I'm not sure how would I start with this problem.
| We have the identity $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$. So we see the series telescope:
$$\sum_{i=0}^{k} (-1)^{i} \binom{n}{i} = \binom{n}{0} + \sum_{i=1}^{k} \binom{n-1}{i-1} + \sum_{i=0}^{k} \binom{n-1}{i}$$
So we see $\binom{n}{0} = 1$. Then $\binom{n}{1} = \binom{n-1}{0} + \binom{n-1}{1}$. For any $x$, $\binom{x}{0} = 1$. So $\binom{n}{0} - \binom{n-1}{0} = 0$.
Now look at $\binom{n}{2} = \binom{n-1}{1} + \binom{n-2}{2}$. By telescoping, $-\binom{n-1}{1} + \binom{n-1}{1} = 0$.
So we are left with the term $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$, with the $\binom{n-1}{k-1}$ cancelling out out the $\binom{n-1}{k-1}$ term from $\binom{n}{k-1}$. Then we are left with $\binom{n-1}{k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/760926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A question about inequality ${(n+1)\over e^n}^nHow to prove the inequality $${(n+1)\over e^n}^n<n!$$
I have tried mathematical induction, but it doesn't work! Are there other methods to solve it?
| Use the binomial theorem on $(1 + n)^n$ and try.
$$(1 + n)^n = 1 + \frac{n!}{1!(n - 1)!} n + \frac{n!}{2!(n - 2)!} n^2 + \frac{n!}{3!(n - 3)!} n^3 + \dots + \frac{n!}{n!(n - n)!} n^n \\ = n! \{\frac{1}{0! n!} + \frac{n}{1! (n-1)!} + \frac{n^2}{2! (n-2)!} + \frac{n^3}{3! (n-3)!} + \dots + \frac{n^n}{n! (n-n)!}\} \\ \le n! \{1 + \frac{n}{1!} + \frac{n^2}{2!} + \frac{n^3}{3!} + \dots\} \\ = n! e^n$$
Hope it is clear now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/761412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Calculate C*-subalgebra generated by $A$, $A^*$ and $\mathbb 1$ I've given a matrix $A=\left(
\begin{array}{ccc}
1-3 \cos (2 \lambda ) & 3 i \sin (2 \lambda ) & 2 i \sin (\lambda ) \\
-3 i \sin (2 \lambda ) & 3 \cos (2 \lambda )+1 & 2 \cos (\lambda ) \\
0 & 0 & 4 \\
\end{array}
\right)$.
I have to calculate the subalgebra generated by $A,A^*$ and $\mathbb{1}$.
How should I proceed? Is it useful to compute the Jordan normal form?
Is there some trick?
Cheers, Peter
| Consider
$$
A-I=\begin{bmatrix}
-3 \cos (2 \lambda ) & 3 i \sin (2 \lambda ) & 2 i \sin (\lambda ) \\
-3 i \sin (2 \lambda ) & 3 \cos (2 \lambda ) & 2 \cos (\lambda ) \\
0 & 0 & 3 \\
\end{bmatrix}
$$
If you square it you get
$$
\begin{bmatrix}
9&0&12i\,\sin\lambda\\
0&9&12\,\cos\lambda\\
0&0&9
\end{bmatrix}.
$$
So
$$
B=(A-I)^2-9I=\begin{bmatrix}
0&0&12i\,\sin\lambda\\
0&0&12\,\cos\lambda\\
0&0&0
\end{bmatrix}.
$$
Now $$E_{33}=\frac1{144}B^*B=\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}\in C^*(A).$$
If you look at
$$
C=\frac1{144}\,BB^*=\begin{bmatrix}\sin^2\lambda&i\cos\lambda\,\sin\lambda&0\\
-i\cos\lambda\,\sin\lambda&\cos^2\lambda&0\\0&0&0\end{bmatrix},
$$
its eigenvalues are $0$ and $1$, so it is a projection.
Also,
$$
X=\frac13\,\left[(A-I)-\frac16\,B-3E_{33}\right]=\begin{bmatrix}
- \cos (2 \lambda ) & i \sin (2 \lambda ) & 0\\
- i \sin (2 \lambda ) & \cos (2 \lambda ) &0 \\
0 & 0 & 0 \\
\end{bmatrix}
=
\begin{bmatrix}
- 1+2\sin ^2 \lambda & 2i \sin \lambda \cos\lambda & 0\\
-2i \sin \lambda \cos\lambda & 1-2\sin^ 2 \lambda &0 \\
0 & 0 & 0
\end{bmatrix}
=\begin{bmatrix}
-1&0&0\\0&-1&0\\0&0&0
\end{bmatrix}
+2C
$$
We have $$A=3X+\frac16\,B+I+3E_{33},$$
with all the summands in $C^*(A)$. I claim that
$$
C^*(A)=\text{span}\,\{X,B,B^*,I,E_{33}\}.
$$
They are clearly linearly independent. And $X^2=I-E_{33}$, $B^2=(B^*)^2=0$, $XB=B$, $BX=0$, $XE_{33}=E_{33}X=0$, $B^*B=144E_{33}$, $BB^*=(X-X^2)/2$, $BE_{33}=B$, $E_{33}B=0$. These relations are enough to show that the span of $X,B,B^*,I,E_{33}$ is a C$^*$-algebra.
Thus $C^*(A)$ has dimension $5$, which means it is isomorphic to $M_2(\mathbb C)\oplus\mathbb C$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding a particular solution to the non-homogenous system I have the following problem $\vec{x}^{'}(t)=\begin{pmatrix} 2 & -5\\1 & -2 \end{pmatrix}\vec{x} + \begin{pmatrix} \csc t\\ \sec t \end{pmatrix}$
Step 1) Find the Eigenvalues $(2-\lambda)(-2-\lambda)+5=0 \implies \lambda= \pm i$
Step 2) The eigenvectors are $e_1=\begin{pmatrix}5 \\ 2-i \end{pmatrix}$ and $e_2= \begin{pmatrix} 5 \\ 2+i \end{pmatrix} $
This means that the fundamental matrix will be $\begin{pmatrix} 5e^{it} & 5e^{-it}\\ (2-i)e^{it} & (2+i)e^{-it}\end{pmatrix}$ and its inverse will be $$\dfrac{1}{10i}\begin{pmatrix} (2+i)e^{-it} & -5e^{-it}\\ (i-2)e^{it} & 5e^{it} \end{pmatrix}$$
This is going to get really really messy.... Please tell me if I am on the right track and if so for the love of god tell me I can omit the imaginary part when apply euler's formula.
So I decided to go forth with this messy problem and I'll show where I am stuck. I decided to leave the $\dfrac{1}{10i}$ to the end. So far I multiply $X^{-1}(t)g(t)$ which is $$\dfrac{-i}{10}\begin{pmatrix} (2+i)(\cos t - i \sin t) & -5\cos t +5i\sin t\\(i-2)(\cos t+i\sin t)& 5\cos t +5i\sin t) \end{pmatrix}\begin{pmatrix}\csc t\\ \sec t \end{pmatrix}=\dfrac{i}{10}\begin{pmatrix} 2\cot t-2i +i\cot t +5i \tan t -4\\ i \cot t -2\cot t -2i+5i\tan t +4 \end{pmatrix}$$ I than proceeded to integrate this monstrosity to obtain $$\begin{pmatrix} 2\ln(\sin t)-2it+i\ln(\sin t)-5i\ln(\cos t)-4t\\ i\ln(\sin t)-2it-2\ln(\sin t)-5i\ln(\cos t) +4t \end{pmatrix}$$ I than factored in the $\dfrac{-i}{10}$ to get $$\begin{pmatrix}-\frac{i}{5}\ln(\sin t)-\frac{t}{5}+\frac{1}{10}\ln(\sin t)-\frac{1}{2}\ln(\cos t) -\frac{2ti}{5}\\ \frac{1}{10}\ln(\sin t)-\frac{t}{5}+\frac{i}{5}\ln(\sin t)-\frac{1}{2}\ln(\cos t)-\frac{2ti}{5} \end{pmatrix}$$
But now I have to multiply THAT matrix by $X$ which is even messier... Am I doing something wrong?
Proceding with the Amzoti's method as follows:
$\phi(t)= \begin{pmatrix} 5\cos t&5 \sin t\\ 2\cos t + \sin t& 2\sin t-\cos t \end{pmatrix}$ $$\phi^{-1}(t)=\dfrac{1}{5}\begin{pmatrix} \cos t -2 \sin t & 5 \sin t\\ 2\cos t + \sin t& -5\cos t \end{pmatrix}$$ I multiply by $g(t)$ and get $$\begin{pmatrix} \cot t-2+5 \tan t\\ 2\cot t +1 -5 \end{pmatrix}$$
Next I take the integral: $$\begin{pmatrix}\int \cot t -2+5 \tan t=\ln(\sin t) -2t -5\ln(\cos t)\\\\ \int 2\cot t+1-5 t=2\ln(\sin t)+t-5t \end{pmatrix}$$ thus $$\dfrac{1}{5} \begin{pmatrix} \ln(\sin t)-2t-5\ln(\cos t)+c_1\\2\ln(\sin t)-4t+c_2\\ \end{pmatrix}$$ The top part isn't matching but the bottom part is.EDIT The moral of the problem... do your work in latex to find silly mistakes....
The solution in the book is $[\frac{1}{5}\ln(\sin t)-\ln(-\cos t)-\frac{2}{5}t+c_1]\begin{pmatrix} 5\cos t \\ 2\cos t +\sin t\end{pmatrix}+[\frac{2}{5}\ln(\sin t)-\frac{4}{5}t+c_2]\begin{pmatrix} 5\sin t\\-\cos t+2\sin t \end{pmatrix}$
| We are given the nonhomogeneous system:
$$x'(t)=\begin{bmatrix} 2 & -5\\1 & -2 \end{bmatrix}\vec{x} + \begin{bmatrix} \csc t\\ \sec t \end{bmatrix}$$
We can write this as $X'(t) = A x(t) + F[t]$, where:
$$A = \begin{bmatrix} 2 & -5\\1 & -2 \end{bmatrix}, ~~ F[t] = \begin{bmatrix} \csc t\\ \sec t \end{bmatrix}$$
We first find the homogeneous solution. The eigenvalues and eigenvectors of $A$ are:
$$\lambda_1 = i, v_1 = (2+i, 1), \lambda_2 = -i, v_2 = (2-i, 1)$$
This gives us a solution of:
$$X_h(t) = \begin{bmatrix} x_h(t) \\ y_h(t) \end{bmatrix} = \begin{bmatrix}c_1(\cos (t)+2 \sin (t)) & -5c_2 \sin (t) \\ c_1\sin (t) & c_2(\cos (t)-2 \sin (t)) \end{bmatrix} $$
Now we need to find the particular solution using the Fundamental Matrix approach, by solving:
$$X(t) = e^{At}X_0 + \int_{t_0}^t e^{A(t-s)}F(s)~ds$$
So, we we use a linear combination from the components of the homogeneous solution to write:
*
*Write $\phi(t) = [x_1(t)~ | ~x_2(t)] $
*Find $\phi^{-1}(t) $
*Find $\phi^{-1}(t) \cdot F(t)$
*Now we integrate the previous result and this gives us: $w$
*Next, we find $X_p(t) = \begin{bmatrix} x_p(t) \\ y_p(t) \end{bmatrix} = \begin{bmatrix}\phi(t) \cdot w \end{bmatrix}$
Write the final solution as:
$X(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} =\begin{bmatrix} -5 c_2 \sin (t)+c_1 (2 \sin (t)+\cos (t))+2 \sin (t) (\log (\sin (t))-2 t)+\cos (t) (-2 t+\log (\sin (t))-5 \log (\cos (t))) \\ \\ c_1 \sin (t)+c_2 (\cos (t)-2 \sin (t))-2 \cos (t) \log (\cos (t))+\sin (t) (-2 t+\log (\sin (t))-\log (\cos (t)))\end{bmatrix}$
Update
If we use your solution with the eigenvalues and eigenvectors, we have a solution of (we want to get rid of these imaginary components):
$$e^{it}\begin{pmatrix}5 \\ 2-i \end{pmatrix} = (\cos t + i \sin t)\begin{pmatrix}5 \\ 2-i \end{pmatrix} = \begin{pmatrix}5 \cos t + i(5 \sin t) \\ 2 \cos t + \sin t + i(2 \sin t - \cos t) \end{pmatrix}$$
So,
$$\phi(t) = \begin{pmatrix}5 \cos t & 5 \sin t \\ 2 \cos t + \sin t & 2 \sin t - \cos t \end{pmatrix}$$
Update 2
$$\phi^{-1}(t)=\dfrac{1}{5}\begin{pmatrix} \cos t -2 \sin t & 5 \sin t\\ 2\cos t + \sin t& -5\cos t \end{pmatrix}$$
When you multiply out with $F(t)$ and do the integration, you get the same exact result as the author.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Maximize $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$ Given three non-negative (as pointed out by Calvin Lin) real numbers $x+y+z = 3$, find the maximum value of $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$.
(Source : Singapore Math Olympiad 2012, Senior section, Round 1, question 29).
I tried using the fact that $2x +13, 8z + 12\ge 0$ to deduce that $y \le 11$, but I couldn't continue from there on. The answer should be an integer, since only integer answers were allowed.
The competition was designed for 15/16 year olds. A simple yet elegant solution would be nice.
For reference of the original problem,
| math110's AM-GM solution is no doubt (to me at least) what the author's wanted. But a less inspired approach is that with Lagrange Multipliers, you need to solve the system of equations:
$$\left\{\begin{aligned}
x+y+z&=3\\
\frac{1}{\sqrt{2x+13}}&=\lambda\\
\frac{1}{\sqrt[3]{(3y+5)^2}}&=\lambda\\
\frac{2}{\sqrt[4]{(8z+12)^3}}&=\lambda\\
\end{aligned}\right.$$
The second and third imply that $\sqrt{2x+13}=\sqrt[3]{(3y+5)^2}$, so $(2x+13)^3=(3y+5)^4$.
Similarly the second and fourth imply that $2\sqrt{2x+13}=\sqrt[4]{(8z+12)^3}$, so $16(2x+13)^2=(8z+12)^3$.
Now you can forget about $\lambda$ and work with the system
$$\left\{\begin{aligned}
x+y+z&=3\\
(2x+13)^3&=(3y+5)^4\\
16(2x+13)^2&=(8z+12)^3\\
\end{aligned}\right.$$
Eliminating $z$:
$$\left\{\begin{aligned}
(2x+13)^3&=(3y+5)^4\\
(2x+13)^2&=4(9-2x-2y)^3\\
\end{aligned}\right.$$
If there are any solutions where the sides of these equations are integers, we would have to have some integer to the $12$th on both sides of the first equation. Trying something nice and small like $2^{12}$ implies that $x$ would have to be $1.5$ and $y$ would have to be $1$. And this is also a solution to the second equation. So one solution is $(x,y,z)=(1.5,1,0.5)$, which yields a value of $8$. Maybe the relative simplicity of the the curves in this last system can show that there are no other solutions:
It's a little tedious, but we can solve for $y$ in each equation, and the difference of the two functions of $y$ is $$\frac{(2x+13)^{3/4}-5}{3}-\left(\frac{\left(\frac{(2x+13)^2}{4}\right)^{1/3}-9+2x}{2}\right)$$
whose derivative is
$$\frac{1}{2(2x+13)^{1/4}}-\frac{2^{1/3}}{3(2x+13)^{1/3}}-1$$
For positve $x$, this is always negative. So the curves from the last system above can only cross once in the domain where $x$ is positive, and we already found where they do.
And then it remains to show that this yields a local maximum value, not a minimum value or a degenerate solution to the Lagrange multiplier problem.
If there are constraints on $x$, $y$, $z$ being positive, then the boundaries of the planes $x=0$ (subject to $y+z=3$), $y=0$ (subject to $x+z=3$), and $z=0$ (subject to $x+y=3$) need to be separately checked, followed by the boundaries of those boundaries: $(3,0,0)$, $(0,3,0)$, and $(0,0,3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/766869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
How to prove this sequence of inequalities The number $c_{g}(n)$ is defined by the recurrence
\begin{equation}
c_{g}(n) = c_{g}(n-1)+ (n-1)(n-2)c_{g-1}(n-2) ,
\end{equation}
with $c_{0}(n)=1$ for any $n\geq 1$ and $c_{g}(n)=0$ if $n \leq 2g$.
My question is how to prove the following inequality
\begin{equation}
\frac{c_{g}(n+1)^2}{c_{g}(n) c_{g}(n+2)} \geq \frac{(n+1)(2n+1)}{(n+2)(2n-1)}.
\end{equation}
for any $g\geq 1$.
I have tried to prove it by induction, but failed.
| It is straighforward to prove that:
$$ c_1(n) = 2 \binom{n}{3}. \tag{1}$$
By induction, it follows that $c_g(n)$ is a $3g$-degree polynomial in the variable $n$. In particular:
$$\begin{eqnarray*} c_2(n)-c_2(n-1)&=&2(n-1)(n-2)\binom{n-2}{3}\\&=&8(n-2)\binom{n-1}{4}=40\binom{n}{5}-16\binom{n-1}{4},\end{eqnarray*}$$
$$ c_2(n) = 40\binom{n+1}{6}-16\binom{n}{5} = 40\binom{n}{6}+24\binom{n}{5}. $$
By replacing $(n-1)(n-2)$ with $n(n-1)$ in the functional equation we get that the "main term" in $c_g(n)$ is just
$$ c_g(n) \sim \frac{(3g)!}{3^g\cdot g!}\binom{n}{3g},\tag{2} $$
and by assuming $c_g(n)=K_g\cdot n^{3g}$ we have:
$$ \frac{c_g(n+1)^2}{c_g(n)c_g(n+2)} = \left(\frac{(n+1)^2}{(n+1)^2-1}\right)^{3g}\geq\left(1+\frac{1}{(n+1)^2}\right)^{3g}\geq 1+\frac{3g}{(n+1)^2}.$$
We have to estimate something that is related with $\frac{d^2}{dx^2}\log c_g(x)$, hence it would be useful to prove that $c_g(n)$ has $3g$ real roots in the interval $[0,2g]$. This may be achieved in the following way:
*
*Prove, by induction, that $c_g(n) = \sum_{h=2g+1}^{3g} K_h \binom{n}{h}$ with $K_h\geq 0$;
*Prove that $c_g(n)$ is strictly increasing when $n\geq 2g$;
*Use Descartes sign rule.
After that, $c_g(n)=K_g \prod_{\rho}(n-\rho)$ and the convexity of $\frac{1}{x^2}$ give:
$$\frac{c_g(n+1)^2}{c_g(n)c_g(n+2)}=\prod_{\rho}\frac{(n+1-\rho)^2}{(n+1-\rho)^2-1}\geq 1+\frac{3g}{(n+1-\frac{1}{3g}\sum_{\rho}\rho)^2}\geq 1+\frac{3g}{(n+1)^2},\tag{3}$$
since $\frac{1}{3g}\sum_{\rho}\rho$ obviously belong to $[0,2g]$ - and may be explicitely computed with a little effort, since it depends only on the coefficients of $n^{3g}$ and $n^{3g-1}$ in $c_g(n)$.
Now we prove $(1.)$ It holds for $g=1$ and $g=2$. Assume that a $K_h\binom{n}{h}$ contributes to $c_{g-1}(n)$. Then:
$$ K_h\left((h+1)(h+2)\binom{n+1}{h+3}-2(h+1)\binom{n}{h+2}\right)$$
contributes to $c_g(n)$, by the same techinique used to find a closed expression for $c_2(n)$. However,
$$\binom{n+1}{h+3}=\binom{n}{h+3}+\binom{n}{h+2},$$
hence the contribute given to $c_g(n)$ can be written as:
$$ K_h\left((h+1)(h+2)\binom{n}{h+3}+h(h+1)\binom{n}{h+2}\right)\tag{4}$$
and $(1.)$ is proved. By the Descartes' sign rule, now we know that $c_g(n)$ has $3g$ real roots, all positive except one root in zero. The inequality:
$$\frac{c_g(n+1)^2}{c_g(n)c_g(n+2)}\geq 1+\frac{3g}{(n+1)^2}$$
yet follows. Now we notice that $(1.)$ implies $(2.)$, since:
$$ c_g(n) = \sum_{h=2g+1}^{3g}K_h\binom{n}{h} $$
with $K_h>0$, gives that $c_g(n)$ is an increasing function over $n\geq 2g$. Since $K_{3g}=\frac{(3g)!}{3^g\cdot g!}$, we just need to find a closed expression for $K_{3g-1}$ to refine our inequality. By $(4)$, given that:
$$ c_{g-1}(n) = A\binom{n}{3g-3}+B\binom{n}{3g-4}+\ldots, $$
$$ c_g(n) = C\binom{n}{3g}+D\binom{n}{3g-1}+\ldots,$$
we have:
$$ C = (3g-1)(3g-2)A,\qquad D=(3g-3)(3g-2)(A+B), $$
and:
$$ \sum_{\rho}\rho = 3g\left(\frac{3g-1}{2}-\frac{D}{C}\right),\qquad \frac{D}{C}=\frac{3g-3}{3g-1}\left(1+\frac{B}{A}\right),$$
hence we have
$$\frac{1}{3g}\sum_{\rho}\rho > \frac{g}{2}$$
and the final refined inequality:
$$\frac{c_g(n+1)^2}{c_g(n)c_g(n+2)}\geq 1+\frac{3g}{\left(n+1-\frac{1}{2}g\right)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Why is $ \int \frac{\sin x (b-a\cos x)}{(b^2+a^2-2ab \cos x)^{3/2}}\,dx = \frac{a-b\cos x}{b^2 \sqrt{a^2-2ab\cos x + b^2}}$? Why is $$ \int \frac{\sin x (b-a\cos x)}{(b^2+a^2-2ab \cos x)^{3/2}}\,dx = \frac{a-b\cos x}{b^2 \sqrt{a^2-2ab\cos x + b^2}}\text{ ?}$$
Constant of integration omitted.
| $\newcommand{\+}{^{\dagger}}
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$\ds{\int{\sin\pars{x}\bracks{b - a\cos\pars{x}} \over
\bracks{b^{2} + a^{2} - 2ab\cos\pars{x}}^{3/2}}\,\dd x
={a - b\cos\pars{x} \over b^{2} \root{a^{2} - 2ab\cos\pars{x} + b^{2}}};\
{\large ?}}$
With $\ds{t \equiv \cos\pars{x}}$:
\begin{align}
&\color{#00f}{\large%
\int{\sin\pars{x}\bracks{b - a\cos\pars{x}} \over
\bracks{b^{2} + a^{2} - 2ab\cos\pars{x}}^{3/2}}\,\dd x}
=\int{b - at \over \pars{b^{2} + a^{2} - 2abt}^{3/2}}\,\pars{-\dd t}
\\[3mm]&=\partiald{}{b}\int\pars{b^{2} + a^{2} - 2abt}^{-1/2}\,\dd t
=\partiald{}{b}\bracks{\pars{b^{2} + a^{2} - 2abt}^{1/2} \over -ab}
\\[3mm]&=-\,{1 \over a}\,{%
\pars{1/2}\pars{b^{2} + a^{2} - 2abt}^{-1/2}\pars{2b - 2at}b
-\pars{b^{2} + a^{2} - 2abt}^{1/2} \over b^{2}}
\\[3mm]&=-\,{1 \over ab^{2}}\,{%
\pars{b - at}b - \pars{b^{2} + a^{2} - 2abt}\over \root{b^{2} + a^{2} - 2abt}}
=-\,{1 \over ab^{2}}\,{%
abt - a^{2}\over \root{b^{2} + a^{2} - 2abt}}
\\[3mm]&={1 \over b^{2}}\,{a - bt\over \root{a^{2} - 2abt + b^{2}}}
=\color{#00f}{\large%
{a - b\cos\pars{x}\over b^{2}\root{a^{2} - 2ab\cos\pars{x} + b^{2}}}}
+ \mbox{a constant}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Collinearity in geometry Let S be the intersection of diagonals in a cyclic quadrilateral. Let p be a circumcircle of a triangle ABS and it intersects BC in M and q is a circumcircle of a triangle ADS and q intersects CD in N. Prove that M, N and S are collinear.
I tried proving that vectors NC and DC are the same, and also that vectors CM and CB are equal but nothing seemed to work. I'm probably not thinking right.
| Brute force computation
You tagged your question as linear algebra, so here you get a linear algebra proof. With a bit of projective geometry thrown in, due to my background.
W.l.o.g. assume $A,B,C,D$ on the unit circle, with the following homogeneous coordinates (using a rational parametrization to avoid square roots and trigonometric functions):
$$
A=\begin{pmatrix}1\\0\\1\end{pmatrix}\qquad
B=\begin{pmatrix}b^2-1\\2b\\b^2+1\end{pmatrix}\qquad
C=\begin{pmatrix}c^2-1\\2c\\c^2+1\end{pmatrix}\qquad
D=\begin{pmatrix}d^2-1\\2d\\d^2+1\end{pmatrix}
$$
Then you get
$$
S=\begin{pmatrix}bc - bd + cd - 1\\2c\\bc - bd + cd + 1\end{pmatrix} \\
p: (bc-bd+cd+1)(x^2+y^2)+2b(d-c)xz+2(d-c)yz+(bc-bd-cd-1)z^2=0 \\
q: (bc-bd+cd+1)(x^2+y^2)+2d(b-c)xz+2(b-c)yz+(cd-bd-bc-1)z^2=0 \\[2ex]
M=\begin{pmatrix}
b c^{3} - b c^{2} d + c^{3} d + b c + 3 c^{2} - b d - 3 c d - 1 \\
2 c^{2} d + 4 c - 2 d \\
b c^{3} - b c^{2} d + c^{3} d + b c + c^{2} - b d + c d + 1
\end{pmatrix} \\[1ex]
N=\begin{pmatrix}
b c^{3} - b c^{2} d + c^{3} d - 3 b c + 3 c^{2} - b d + c d - 1 \\
2 b c^{2} - 2 b + 4 c \\
b c^{3} - b c^{2} d + c^{3} d + b c + c^{2} - b d + c d + 1
\end{pmatrix} \\
\det(M,N,S)=0
$$
which proves the collinearity. Obviously not a proof you'd want to attempt without help from a computer algebra system.
Angle chasing
If the above is not the kind of thing you want to compute yourself, I suggest you turn your known cocircularities into angle equalities, and follow them though. A bit like what I did in the second half of this post.
You have to show that $\angle NSD=\angle MSB$ to show the collineariry. Let's follow that backwards to something which only uses $A,B,C,D$:
\begin{align*}
\angle NSD&=180°-\angle DNS-\angle SDN \\
&=180°-\angle DNA-\angle ANS-\angle BDC \\
&=180°-\angle DSA-\angle ADS-\angle BDC \\
&=\angle SAD-\angle BDC \\
&=\angle CAD-\angle BDC \\
\angle MSB&=180°-\angle SBM-\angle BMS \\
&=\angle CBD-\angle BMA+\angle SMA \\
&=\angle CBD-180°+\angle ASB+\angle SBA \\
&=\angle CBD-\angle BAS \\
&=\angle CBD-\angle BAC \\
&=\angle CAD-\angle BDC=\angle NSD \tag*{$\Box$}
\end{align*}
Note that the computation seems less symmetric than it actually is, due to the fact that I tried to make all angles positive in the following angle. For the proof to be universal, you'll have to consider signed angles, though.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Log Trig $\int_0^{\pi/2}\log^4 \tan \frac{x}{2}dx=\frac{5\pi^5}{32}$ Hi I am trying to integrate a log trigonometric integral given by
$$
I:=\int_0^{\pi/2}\log^4 \tan \frac{x}{2}dx=\frac{5\pi^5}{32}.
$$
This is very similar to a previous integral posted except the power of the logarithm.
Note this integral is also equal to
$$
\int_0^{\pi/2}\log^4 \tan x \, dx=\frac{5\pi^5}{32}.
$$
I have wrote I as
$$
I=\int_0^{\pi/2} \left(\log \sin \frac{x}{2}-\log \cos \frac{x}{2}\right)^4 dx
$$
but got stuck here since factoring this out seems like a mess. Having seen how David H solved a similar integral I posted, I tried another method starting with I and using $t=\tan x/2$, and obtained
$$
I=2\int_0^{1}\log^4 t \frac{dt}{1+t^2}.
$$
Following this I tried $u=-\log t$ but got stuck after this. Thanks, it would be nice to see a solution that doesn't reduce the integral to a difficult sum to evaluate
| \begin{align}
J&=\int_0^{\frac{\pi}{2}}\log^4 \left(\tan x\right) \, dx\\
&\overset{y=\tan x}=\int_0^\infty \frac{\ln^4 x}{1+x^2}\,dx\\
J_n&=\int_0^\infty \frac{\ln^n x }{1+x^2}\,dx\\
J_0&=\frac{\pi}{2}\\
K_n&=\int_0^\infty\int_0^\infty \frac{\ln^n(xy) }{(1+x^2)(1+y^2)}\,dx\,dy\\
&\overset{u(x)=xy}=\int_0^\infty \int_0^\infty \frac{\ln^n u}{(u^2+y^2)(1+y^2)}\,du\,dy\\
&=\int_0^\infty \frac{\ln^{n+1}u }{u^2-1}\,du\\
&=\int_0^1 \frac{\ln^{n+1}u }{u^2-1}\,du+\int_1^\infty \frac{\ln^{n+1}u }{u^2-1}\,du\\
&=\Big(1+(-1)^n\Big)\int_0^1 \frac{\ln^{n+1}u }{u^2-1}\,du\\
&=\Big(1+(-1)^n\Big)\left(\int_0^1 \frac{\ln^{n+1}u }{u-1}\,du-\int_0^1 \frac{u\ln^{n+1}u }{u^2-1}\,du\right)\\
&=\Big(1+(-1)^n\Big)\left(1-\frac{1}{2^{n+2}}\right)\int_0^1 \frac{\ln^{n+1}u }{u-1}\,du\\
&=\Big(1+(-1)^n\Big)\left(1-\frac{1}{2^{n+2}}\right)(-1)^{n+2}(n+1)!\zeta(n+2)\\
&=\Big(1+(-1)^n\Big)\left(1-\frac{1}{2^{n+2}}\right)(n+1)!\zeta(n+2)\\
K_n&=\sum_{k=0}^n \binom{n}{k} J_kJ_{n-k}\\
K_2&=\pi J_2\\
K_2&=\frac{45}{4}\zeta(4)\\
K_4&=\pi J_4+6J_2^2\\
K_4&=\frac{945}{4}\zeta(6)\\
\end{align}Therefore,
\begin{align}
J_4&=\frac{K_4-6{J_2}^2}{\pi}\\
&=\frac{K_4-6\left(\frac{K_2}{\pi}\right)^2}{\pi}\\
&=\frac{K_4-6{J_2}^2}{\pi}\\
&=\frac{\frac{945}{4}\zeta(6)-6\left(\frac{45}{4}\zeta(4)\times\frac{1}{\pi}\right)^2}{\pi}\\
&=\frac{945\zeta(6)}{4\pi}-\frac{6075\zeta(4)^2}{8\pi^3}\\
\end{align}
Moreover, if you assume,
\begin{align}
\zeta(4)&=\frac{\pi^4}{90}\\
\zeta(6)&=\frac{\pi^6}{945}
\end{align}
Therefore,
\begin{align}
J&=J(4)\\
&=\frac{945}{4\pi}\times \frac{\pi^6}{945}-\frac{6075}{8\pi^3}\times \left(\frac{\pi^4}{90}\right)^2\\
&=\boxed{\frac{5}{32}\pi^5}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/770903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
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} |
How prove this inequality $\sum_{k=1}^{n}\frac{2k-1}{k\binom{n}{k}}\ge \frac{n}{2^{n-1}}$
let $1\le k\le n,k,n\in N^{+}$, show that
$$\sum_{k=1}^{n}\dfrac{2k-1}{k\binom{n}{k}}\ge \dfrac{n}{2^{n-1}}$$
I know this $$\sum_{k=1}^{n}(2k-1)=n^2$$
and $$\sum_{k=1}^{n}k\binom{n}{k}=n\cdot 2^{n-1}$$
I want Use Cauchy-Schwarz inequality .
$$\left(\sum_{k=1}^{n}\dfrac{2k-1}{k\binom{n}{k}}\right)\left(\sum_{k=1}^{n}k\binom{n}{k}\right)\ge (\sum_{k=1}^{n}\sqrt{2k-1})^2$$
then
$$\sum_{k=1}^{n}\dfrac{2k-1}{k\binom{n}{k}}\ge\dfrac{(\sum_{k=1}^{n}\sqrt{2k-1})^2}{n\cdot 2^{n-1}}$$
Now we must prove
$$\sum_{k=1}^{n}\sqrt{2k-1}\ge n?$$
maybe can use integral inequality to prove it.
I can't prove this.Thank you
| Another starting point is the inequality $\boxed{\binom{n}{k} \leq 2^n}$ This trivial starting point allows us to deduce that
$$ \sum_{k=1}^n \binom{n}{k}^{-1} \geq \frac{n}{2^n} $$
If we plugged this into the original inequality we fall short of what we're trying to prove:
$$ \sum_{k=1}^n \left( 1 - \frac{1}{2k}\right)\binom{n}{k}^{-1} \geq
2^{-n} \sum_{k=1}^n \left( 1 - \frac{1}{2k}\right) \geq \mathbf{\color{blue}{\frac{n - \tfrac{1}{2}\log n}{2^n}}}
$$
Or we can try it the other way, but we still fall a little bit short.
$$ \sum_{k=1}^n \left( 1 - \frac{1}{2k}\right)\binom{n}{k}^{-1} =
\sum_{k=1}^n \left( k - \frac{1}{2}\right)\frac{1}{n}\binom{n-1}{k-1}^{-1}
= \frac{1}{n 2^{n-1}}\sum_{k=1}^n \left( k - \frac{1}{2}\right)
= \mathbf{\color{blue}{\frac{n- \frac{1}{n}}{ 2^n} }}
$$
I hope improve $\frac{1}{2^n}\binom{n}{k} \leq 1$, perhaps by a constant that depends on $k$.
In fact, we can use the Arithmetic mean - Harmonic mean inequality - or possibly Jensen's inequality - to get:
$$ \sum_{k=1}^n \left( 1 - \frac{1}{2k}\right)\binom{n}{k}^{-1} \geq
\frac{n^2}{\sum_{k=1}^n \left( 1 - \frac{1}{2k}\right)^{-1}\binom{n}{k}}
\geq
\frac{n^2}{2 \sum_{k=1}^n \binom{n}{k}}
> \mathbf{\frac{n^2}{2^{n+1}} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/773611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Proving a reduction formula for $I_n = \int \:e^{ax}\cos^n x\;dx$ (Further Maths: F3)
Given $$
I_n = \int \:e^{ax}\cos^n x\;dx
$$
I have to prove that
$$
\left(a^2+n^2\right)I_n\:=\:e^{ax}\cos^{n-1}x\left(a\cos x+n\sin x\right)+n(n-1)I_{n-2}
$$
I just cannot get it to reduce. I keep ending up with too many species in the next integral to use parts again.
I have an important Further Pure F3 exam in a month and reduction is proving the hardest of the topics. Help is greatly appreciated.
| Recall the integration by parts formula:-
$$\int uv'dx=uv-\int vu'dx$$
where $v'=\frac{dv}{dx}$ and $u'=\frac{du}{dx}$
Applying integration by parts by setting $v'=e^{ax}$ and $u=\cos^nx$ we obtain
$$\int \color{red}{e^{ax}}\color{blue}{\cos ^n\left(x\right)}dx=\color{red}{\frac{1}{a}e^{ax}}\color{blue}{\cos^nx}+\frac{\color{blue}{n}}{\color{red}{a}}\int \color{red}{e^{ax}}\color{blue}{\cos^{n-1}x\sin x}\text{ }dx$$
Apply integration by parts to the integral on the right hand side of the above equation setting $v'=e^{ax}$ and $u=\cos^{n-1}x\sin x$. This results in the following expression (note the use of the product rule for differentiation in the integral on the RHS):-
$$\int \color{red}{e^{ax}}\color{blue}{\cos^{n-1}x\sin x}\text{ }dx=\color{red}{\frac{1}{a}e^{ax}}\color{blue}{\cos^{n-1}x\sin x}-\color{red}{\frac{1}{a}}\int \color{red}{e^{ax}}\color{blue}{[-(n-1)\cos^{n-2}x\sin^2x+\cos^n x]}\text{ }dx\\=\frac{1}{a}e^{ax}\cos^{n-1}x\sin x+\frac{1}{a}\int e^{ax}(n-1)\cos^{n-2}x\color{magenta}{\sin^2x}\text{ }dx-\frac{1}{a}\int e^{ax}\cos^n x\text{ }dx\\=\frac{1}{a}e^{ax}\cos^{n-1}x\sin x+\frac{1}{a}\int e^{ax}(n-1)\cos^{n-2}x\color{magenta}{(1-\cos^2x)}\text{ }dx-\frac{1}{a}\int e^{ax}\cos^n x\text{ }dx\\=\frac{1}{a}e^{ax}\cos^{n-1}x\sin x+\frac{n-1}{a}\int e^{ax}\cos^{n-2}x\text{ }dx+\frac{1-n}{a}\int e^{ax}\cos^{n}x\text{ }dx-\frac{1}{a}\int e^{ax}\cos^n x\text{ }dx\\=\frac{1}{a}e^{ax}\cos^{n-1}x\sin x+\frac{n-1}{a}I_{n-2}-\frac{n}{a}I_n$$
If we substitute this expression into what we obtained after the we integrated by parts for the first time, we obtain:-
$$I_n=\frac{1}{a}e^{ax}\cos^nx+\frac{n}{a^2}\left[e^{ax}\cos^{n-1}x\sin x+(n-1)I_{n-2}-nI_n\right]\\\Rightarrow a^2I_n=ae^{ax}\cos^nx+ne^{ax}\cos^{n-1}x\sin x+n(n-1)I_{n-2}-n^2I_n\\\Rightarrow (a^2+n^2)I_n=e^{ax}\cos^{n-1}x(a\cos x+n\sin x)+n(n-1)I_{n-2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
Finding the largest angle of a triangle The sides of a triangle are $(x^2+x+1), (2x+1)$ and $(x^2-1)$. Then what is the largest of the 3 angles of triangle?
| Suppose $x>0$. As $(x^{2}-1)$ is a side of the triangle then $x^{2}-1>0$ , therefore $x>1$.
Note that
$(x^{2}+x+1)-(x^{2}-1)=x+2>0$, then $(x^{2}+x+1)>(x^{2}-1)$. Another hand, as $x>1$,
$(x^{2}+x+1)-(2x+1)=x^{2}-x>0$. Therefore $(x^{2}+x+1)$ is the largest side of the triangle. The largest angle is the opposite to the side $(x^{2}+x+1).$
As $(x^{2}+x+1)$ is the largest side of the triangle, by the law of cosines,
$$(x^{2}+x+1)^{2}=(2x+1)^{2}+(x^{2}-1)-2(2x+1)(x^{2}-1)\cos\theta$$
Therefore $\cos\theta=-\frac{1}{2}$, then $\theta=120^{\circ}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Integral$\int_0^\infty \ln x\,\exp(-\frac{1+x^4}{2\alpha x^2}) \frac{x^4+\alpha x^2- 1}{x^4}dx$? I am trying to prove
$$
I:=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+\alpha x^2- 1}{x^4}dx=\frac{\sqrt{2\alpha^3 \pi}}{2\sqrt[\alpha]e},\qquad \alpha>0.
$$
Note: The proof below shows how this is just a Gaussian integral!
I am not sure how to start this one. It seems very difficult to me However the answer is very nice.
I thought maybe trying to write $I(\alpha)$ and $I'(\alpha)$ to try and simplify things but it didn't help much. at $x=0$ there seems to be a problem with the integrand also however I am not sure how to go about using this. Perhaps we could try and use a series expansion for $e^x=\sum_{n=0}^\infty x^n / n!$ however the function $e^{-1/x^2}$ is well known that its taylor series is zero despite the function not being. The factor of $x^4+\alpha x^2-1$ has been giving me trouble with simplifying the integrand. Thanks.
To those who just made an edit: If you are looking for a +2, please edit something worthwhile. I edited it back to what I had considering you didn't fix anything as is shown in the Edit History.
| $$\begin{align*}
I&=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+\alpha x^2- 1}{x^4}dx\\
&=\int_0^\infty \ln x\, d\left(-\alpha x^{-1}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right)\\
&=-\alpha\left(\left.\frac{\ln x}{x}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right|_0^\infty-\int_0^\infty \frac{1}{x}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) d\,\ln x\right)\\
&=\alpha\int_0^\infty \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) dx\\
&=\alpha\left(\int_0^1 \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\underbrace{\int_1^\infty \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx}_{x\to1/x}\right) \\
&=\alpha\left(\int_0^1 \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\int_1^0 -\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\right) \\
&=\alpha\int_0^1 (1+\frac{1}{x^2})\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\\
&=\alpha\int_0^1 \exp\left(-\frac{1}{\alpha}-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\
&=\alpha e^{-1/\alpha}\int_0^1 \exp\left(-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\
&=\alpha e^{-1/\alpha}\int_{-\infty}^0 \exp\left(-\frac{y^2}{2\alpha }\right)dy\\
&=\alpha e^{-1/\alpha}\sqrt{\frac{\alpha\pi}{2}}.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Solving recurrence relation: $ f(n) = 3f(n/2) - 2f(n/4) | f(2) = 5, f(1) = 3$
$f(n) = 3f(n/2) - 2f(n/4) | f(2) = 5, f(1) = 3$
I have attempted to solve it by letting
$n = 2^k$
$f(2^k) = 3f(2^{k-1}) - 2f(2^{k-2})$
Then set
$S(k) = f(2^k)$
$S(k) = 3*S(k-1) - 2*S(k-2)$
let S(k) = $x^k$
$x^k = 3x^{k-1} - 2x^{k-2} $ (divide by $x^{k-2}$ and
rearrange)
$x^2 - 3x + 2 = 0$
solving for $(x-1)(x-2)$
Here I get a bit stuck as I try to proceed with the following:
$S(k) = c_1\times 1^k + c_2 \times 2^k$
But cannot go further on my own. Any advice as the substitutions have confused me!
| The reader may be interested to note that there is a closely related
recurrence that we can solve exactly and not just for powers of
two.
Suppose we put
$$f(n) = 3 f(\lfloor n/2 \rfloor) - 2 f(\lfloor n/4 \rfloor).$$
This requires a value for $f(0)$ so we set
$$f(n) = 2n+1 \quad\text{when} \quad n<3.$$
Now observe that the generating function $$g(z) =
\frac{1}{1-(3z-2z^2)}$$ encodes the tree of values visited during the
computation of $f(n)$ in a natural way.
Let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$ be the
binary representation of $n$. We now compute a closed form expression
for $f(n)$ when $n\ge 2.$
Case A. The leading digits are two one digits, i.e.
$(d_{\lfloor \log_2 n \rfloor} d_{\lfloor \log_2 n \rfloor-1})_2 =
(11)_2.$ Then the two terminal values for the recursion are $n=3$ and
$n=1$.
The case $n=3$ has $f(n)=7$ and thus gives the contribution
$$7\times
[z^{\lfloor \log_2 n \rfloor - 1}] \frac{1}{1-(3z-2z^2)}.$$
The case $n=1$ has $f(n)=3$ but the last step must have been on the
$\lfloor n/4 \rfloor$ branch so as not to be routed through $n=3$,
giving
$$3\times (-2)\times
[z^{\lfloor \log_2 n \rfloor - 2}] \frac{1}{1-(3z-2z^2)}.$$
Case B. The leading digits are a one digit followed by a zero
digit i.e. $(d_{\lfloor \log_2 n \rfloor} d_{\lfloor \log_2 n
\rfloor-1})_2 = (10)_2.$ Then the terminal values for the recursion
are $n=2$ and $n=1$.
The case $n=2$ has $f(n)=5$ and thus gives the contribution
$$5\times
[z^{\lfloor \log_2 n \rfloor - 1}] \frac{1}{1-(3z-2z^2)}.$$
The case $n=1$ has $f(n)=3$ but the last step must have been on the
$\lfloor n/4 \rfloor$ branch so as not to be routed through $n=2$,
giving
$$3\times (-2)\times
[z^{\lfloor \log_2 n \rfloor - 2}] \frac{1}{1-(3z-2z^2)}.$$
Evaluation. Note that by partial fractions we have that
$$[z^q] \frac{1}{1-(3z-2z^2)} = 2^{q+1}-1$$
so that we get for case A
$$ 7 \times 2^{\lfloor \log_2 n \rfloor} - 7
- 6 \times 2^{\lfloor \log_2 n \rfloor-1} + 6
= (14-6) \times 2^{\lfloor \log_2 n \rfloor-1} - 1
= 2^{\lfloor \log_2 n \rfloor+2} - 1$$
and for case B
$$ 5 \times 2^{\lfloor \log_2 n \rfloor} - 5
- 6 \times 2^{\lfloor \log_2 n \rfloor-1} + 6
= (10-6) \times 2^{\lfloor \log_2 n \rfloor-1} + 1
= 2^{\lfloor \log_2 n \rfloor+1} + 1 $$
This produces the sequence for $n\ge 2$
$$5, 7, 9, 9, 15, 15, 17, 17, 17, 17, 31, 31, 31, 31, 33,\ldots$$
which perfectly matches $f(n).$
Observe that we may say that $f(n)\in\Theta(n)$ in certain sense
($2^{\lfloor \log_2 n \rfloor}\in\Theta(n).$)
This MSE link shows a more sophisticated application of the trick with the generating function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/777597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Rotate $z = 0$ plane in 3D I have 100 points on $z=0$ plane.
I want to rotate those points, such that they lie on any plane $P(a,b,c,d)$, preserving distances.
Hence, I need a rotation matrix.
For instance, if my points are $(2,5,0)$, $(4,4,0)$, $(1,3,0)$ and $(3,5,0)$, how do I rotate them and put them onto $x + 3y - z + 1 = 0$ plane?
This is what I have tried:
First, I have calculated the angle between the normal vector of $P$ and all the normal vectors of $x=0$, $y=0$ and $z=0$ plane
$(1,0,0)$, $(0,1,0)$, $(0,0,1)$
Let us refer to those angles as $\alpha$, $\beta$ and $\theta$ respectively.
Now, I rotate my points $\alpha$ around $x-axis$, $\beta$ around $y-axis$ and $\theta$ around $z-axis$.
I think my approach is wrong. Distances are preserved but the new plane is not the plane I want.
| No need to compute angles anywhere.
The columns of the transformation matrix are the images of the unit vectors under said transformation. Since you didn't specify placement of your points in that plane, you can use any pair of orthogonal uint vectors in that plane. A nice trick to obtain orthogonal vectors in 3d is using the cross product.
So start with the normal vector $(1,3,-1)$ of the plane, and an arbitrarily chosen second vector $(0,0,1)$ Compute the cross products
$$v_1=\begin{pmatrix}1\\3\\-1\end{pmatrix}\times\begin{pmatrix}0\\0\\1\end{pmatrix}
=\begin{pmatrix}3\\-1\\0\end{pmatrix} \\
v_2=\begin{pmatrix}1\\3\\-1\end{pmatrix}\times\begin{pmatrix}3\\-1\\0\end{pmatrix}
=\begin{pmatrix}-1\\-3\\-10\end{pmatrix}$$
To preserve distances, you'll have to scale them to unit length. Then you can plug them into the columns of a transformation matrix, and choose an offset such that the image of the origin lies within your plane. For example like this:
$$\begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto\begin{pmatrix}
\frac{3}{\sqrt{10}} & -\frac{1}{\sqrt{110}} & \frac{1}{\sqrt{11}} \\
-\frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{110}} & \frac{3}{\sqrt{11}} \\
0 & -\frac{10}{\sqrt{110}} & -\frac{1}{\sqrt{11}}
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}
+\begin{pmatrix}0\\0\\1\end{pmatrix}$$
If you really want to rotate around the axis where both your planes intersect, so that points on that axis remain fixed, you'll need a bit more work. First find the direction of the common line. It is perpendicular to both the normals of the two planes. Since I used $(0,0,1)$ as the arbitrary vector in the first step above, and that's the normal of the $z=0$ plane, you already have $v_1$ as the direction of that axis. You can get the transformation matrix like this:
$$\begin{pmatrix}
\frac{3}{\sqrt{10}} & -\frac{1}{\sqrt{110}} & \frac{1}{\sqrt{11}} \\
-\frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{110}} & \frac{3}{\sqrt{11}} \\
0 & -\frac{10}{\sqrt{110}} & -\frac{1}{\sqrt{11}}
\end{pmatrix}\begin{pmatrix}
\frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} & 0 \\
-\frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} & 0 \\
0 & 0 & 1
\end{pmatrix}^{-1}$$
The right matrix (prior to being inverted) will map the $x$ direction to the direction of the line of intersection, and it will map the $y$ direction to the direction perpendicular to that line within the $z=0$ plane. The $z$ axis is preserved, so this is a rotation around the $z$ axis. The inverse of that matrix will map the direction of the line of intersection to the $x$ direction, from where the left matrix will map it back to the line of intersection. You have to make sure that the second column of the right matrix has the correct sign, otherwise you'll end up describing a reflection (in the angle bisector plane) instead of a rotation. I did this at first, accidentially.
To get the offset right, simply plug in a point on the line of intersection (i.e. one which satisfies $x+3y+1=0$ and $z=0$), and compute the difference between the transformed and the original point. You want to subtract that difference, so that the point remains fixed.
In the end you obtain
$$\begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto
\frac1{110}\begin{pmatrix}
-\sqrt{11} + 99 & -3 \, \sqrt{11} - 33 & 10 \, \sqrt{11} \\
-3 \, \sqrt{11} - 33 & -9 \, \sqrt{11} + 11 & 30 \, \sqrt{11} \\
-10 \, \sqrt{11} & -30 \, \sqrt{11} & -10 \, \sqrt{11}
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}-
\frac1{110}\begin{pmatrix}
\sqrt{11} + 11 \\
3 \, \sqrt{11} + 33 \\
10 \, \sqrt{11}
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/777679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Working out the value of $a^4+b^4$ If $ab = 2$ and $a+b = 5$ then calculate the value of $a^4+b^4$
My approach:
$$a^4+b^4 = (a+b)^4-4a^3b-6a^2b^2-4ab^3$$
$$=(5)^4-6(ab)^2-4ab.a^2-4ab.b^2$$
$$=(5)^4-6(24)-4ab(a^2-b^2)$$
$$=(5)^4-6(24)-8(a+b)(a-b)$$
$$=(5)^4-6(24)-8(5)(a-b)$$
I am a little stuck now and any help will be appreciated.
| An easier approach: Since $ab = 2$, then
$$ (a+b)^2 = 25 $$.
Hence
$$ a^2 + b^2 + 2ab = 25 \iff a^2 + b^2 = 25 - 2 ab = 21 $$
Hence,
$$ (21)^2 = (a^2 + b^2)^2 = a^4 + b^4 + 2 a^2 b^2 = a^4 + b^4 + 2(2)^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/778184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Prove that integral is independant of its parameter We are given intergral $\int_0^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}}$ and task is to prove that it's independent of $\alpha$.
Task was too complicated for me, so I had to stick with solution, recommended in the textbook. It goes on like that:
$\int_0^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = \int_0^1 {\frac {dx} {(1+x^2)(1+x^\alpha)}} + \int_1^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = J_1 + J_2$
In $J_1$ we do replacement $y = \frac 1 x$
$J_1 = \int_1^\infty \frac {dy} {-(1+y^2)(1+y^{-\alpha})}$
Than textbook says that next step is
$J_1+J_2 = \int_1^\infty {(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha}) \frac {dx}{1+x^2}}$
From where it's obvious that inital statement is true. Problem is, I completely fail to understand how that summation of $J_1+J_2$ was done.
Edit
To be more explicit, I can't get this: $\int_1^\infty{\frac {dy}{-(1+y^2)(1+y^{-\alpha})}} + \int_1^\infty {\frac {dx} {(1+x^2)(1+x^\alpha)}} = \int_1^\infty {(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha}) \frac {dx}{1+x^2}}$
So I would appreciate some explanations and pointers.
| There is no need to break the integral into pieces. Simply use the substitution $t=1/x$:
$$
\begin{align}
A
&=\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)(1+x^\alpha)}\tag{1}\\
&=\int_0^\infty\frac{t^\alpha\,\mathrm{d}t}{(1+t^2)(1+t^\alpha)}\tag{2}
\end{align}
$$
Add $(1)$ and $(2)$ to get
$$
\begin{align}
2A
&=\int_0^\infty\frac{(1+x^\alpha)\,\mathrm{d}x}{(1+x^2)(1+x^\alpha)}\\
&=\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)}\\
&=\frac\pi2\tag{3}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/779342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Inverse of this $3\times 3$ matrix using the Cayley–Hamilton theorem Find the inverse of the matrix $$\begin{pmatrix} -1 & 2& 0
\\ 1& 1 &0 \\ 2 & -1& 2 \end{pmatrix}$$ using the Cayley–Hamilton theorem.
Thanks!
| The matrix $A$ is:
$A = \begin{bmatrix} -1 & 2 & 0 \\ 1 & 1 & 0 \\ 2 & -1 & 2 \end{bmatrix}, \tag{1}$
so the characteristic polynomial $p_A(\lambda)$ is
$p_A(\lambda) = \det(A - \lambda I) = \det \begin{bmatrix} -1 - \lambda & 2 & 0 \\ 1 & 1 - \lambda & 0 \\ 2 & -1 & 2 - \lambda \end{bmatrix}$
$= ( -1 - \lambda)(1 - \lambda) (2 - \lambda) - 2(2 - \lambda) = (\lambda^2 - 1)(2 - \lambda) - 4 + 2\lambda$
$= -\lambda^3 + 2\lambda^2 + 3\lambda - 6, \tag{2}$
and by Cayley-Hamilton we have
$0 = p_A(A) = -A^3 + 2A^2 + 3A - 6I; \tag{3}$
(3) may be written as
$A(-A^2 +2A + 3I) = 6I, \tag{4}$
or
$A(\dfrac{1}{6}(-A^2 +2A + 3I)) = I, \tag{5}$
which shows that
$A^{-1} = \dfrac{1}{6}(-A^2 +2A + 3I). \tag{6}$
I leave the explicit calculation of $A^{-1}$ from (6) to any interested readers; it is not difficult.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
| {
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"url": "https://math.stackexchange.com/questions/780160",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Finding an equation of a circle with a given center and a tangent line. My math homework is finding an equation of the circle. Given that the center is at (-3,-5) and tangent to the line 12x + 5y =4.
I don't know how to solve this since our professor didn't teach this to us. :/
| Let $(x+3)^2 + (y+5)^2 = r^2$ be the equation of the circle with center $C = (-3,-5)$ and radius equal to $r$. Rewrite the tangent line's equation as: $y + 5 = \dfrac{29 - 12x}{5}$, substituting this $y$ into the equation of the circle and get:
$(x+3)^2 + \dfrac{(29 -12x)^2}{25} - r^2 = 0 \iff 169x^2 - 546x + 1066 - 25r^2 = 0$. We require that this equation has one and only one real root. This means: $\triangle' = 0 \iff (-273)^2 - 169(1066 - 25r^2) = 0 \iff 4225r^2 - 105625 = 0 \iff r = 5$. So the equation of the circle is: $(x+3)^2 + (y+5)^2 = 25$
| {
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"url": "https://math.stackexchange.com/questions/780578",
"timestamp": "2023-03-29T00:00:00",
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Need to extremize the function $f(x,y)=x^2+y^2$. Determine the points on the curve $$x^4+y^4=1$$ that are closest and furthest away from the origin. Explain why this corresponds to extremizing the function $f(x,y)=x^2+y^2$ under the condition $x^4+y^4=1$.
I don't understand this question at all and would be grateful if someone could provide some explanation to this!
Also when we are asked to extremize a function, given the constraint like in this example, how exactly do we find the function generally?
| Let's find the shortest and longest distances from the origin $O = (0,0)$ of points on the curve $x^4 + y^4 = 1$. The distance is $d = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$. So :$d^4 = (x^2 + y^2)^2 = (1\cdot x^2 + 1\cdot y^2)^2 \leq (1^2 + 1^2)\cdot ((x^2)^2 + (y^2)^2) = 2\cdot (x^4 + y^4) = 2\cdot 1 = 2$ due to Cauchy-Schwarz inequality. So $d_{max} = 2^{\frac{1}{4}} = 1.189$. This value is achieved when $(x,y) = (2^{-1/4}, 2^{-1/4}), (2^{-1/4},-2^{-1/4}),(-2^{-1/4},2^{-1/4}),(-2^{-1/4},-2^{-1/4})$
To find the shortest distance $d_{min}$, we observe that: if $x^4 + y^4 = 1$, then $|x| \leq 1$, and $|y| \leq 1$, implying that $x^2 \geq x^4$, and $y^2 \geq y^4$ with equality when $x = 0, 1$, and $y = 0,1$. So: $x^2 + y^2 \geq x^4 + y^4 = 1$. Thus: $d_{min} \geq \sqrt{1} = 1$, and this happens when $(x,y) = (0,1), (1,0),(-1,0),(0,-1)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Primitive roots used to work out $x^7 \equiv 5 \pmod {11}$ I have a workbook question that doesn't have any example solution, that is as follows:
Primitive roots used to work out $x^7 \equiv 5 \pmod {11}$
Now I can see $\phi(11)=10$ and $2$ has order $10$ so $2$ is the primitive root $\pmod{11}$
I can see aswell that $2^4 \equiv 5 \pmod{11}$, so I can show $x^7 = 2^4 \pmod{11}$, must I compute the powers of all the numbers until I find which yields $5 \pmod{11}$, how do I use primitive roots to solve the prolem?
| Let $x=2^k$. Then solve for $k$.
ADD We have $2^{7k}=2^4\mod 11$, so that $2^{7k-4}=0\mod 11$. Since the order of $2$ modulo $11$ is $10$, $7k=4\mod 10$. Hence $k=3\cdot 4=12=2\mod 10$. It follows $x=4$.
| {
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"source": "stackexchange",
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If $30$ divides $p_1^4 + p_2^4 + \ldots + p_{31}^4$. Prove that $p_1=2$, $p_2=3$ and $p_3=5$. Let $p_1<p_2<\cdots<p_{31}$ be prime numbers such that $30$ divides $p_1^4 + p_2^4 + \cdots + p_{31}^4$. Prove that $p_1=2$, $p_2=3$ and $p_3=5$.
No clue how to start..Hints are welcomed.
| First, if $p_1\neq 2$ then $p_1>2$ and so $p_1$, $p_2$, $\ldots$, $p_{31}$ are all odd, in such a case $p_1^4+p_2^4+_\ldots+p_{31}^4$ is odd, in particular 30 doesn't divide $p_1^4+p_2^4+_\ldots+p_{31}^4$. This shows $p_1=2$.
If $p_2\neq 3$ then $p_2\equiv \pm 1(\mod 3)$ and so $p_i^{4}\equiv 1(\mod 3)$ for $i=1,2,\ldots,31$, hence $p_1^4+p_2^4+_\ldots+p_{31}^4\equiv 1 (\mod 3)$ that means $p_1^4+p_2^4+_\ldots+p_{31}^4$ is not a multiple of $30$. So $p_3=3$. A similar argue show us that $p_3$ should be $5$: $p_3\neq 5 \Longrightarrow p_i^4\equiv 1 (\mod 5)$ for $i=1,2,\ldots,31$ by Little Fermat Theorem and $p_1^4+p_2^4+_\ldots+p_{31}^4\equiv 1 (\mod 5)$.
| {
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Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
| Maybe the answer that gives the direct clue about what is going on is going through a proof that
$$\lim_{n \to +\infty} (\sqrt{n^2+n+c_2}-n)-(\sqrt{n^2+n+c_1}-n)=0$$
or equally
$$\lim_{n \to +\infty} \sqrt{n^2+n+c_2}-\sqrt{n^2+n+c_1}=0$$
Now write this to make it obvious as
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n^2+n+c_1}}(\frac{\sqrt{1+\frac{1}{n}+\frac{c_2}{n^2}}}{\sqrt{1+\frac{1}{n}+\frac{c_1}{n^2}}}-1)=0$$
So we can pick whatever constant we want. We pick $\frac{1}{4}$ and have
$$\lim_{n \to +\infty} \sqrt{n^2+n}-n=\lim_{n \to +\infty} (\sqrt{n^2+n+\frac{1}{4}}-n)=$$ $$\lim_{n \to +\infty} (\sqrt{(n+\frac{1}{2})^2}-n)=\lim_{n \to +\infty} n+\frac{1}{2}-n=\frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Maclaurin series for $\dfrac{\cos(x)-1}{x^2}$ The solution for this is
$$
-\dfrac{1}{x} + \dfrac{x^2}{4!} - \dfrac{x^2}{6!} + \ldots \;,
$$
but I'm not sure how to derive this Maclaurin series from $\cos(x)$. The solution just divided each term in the Maclaurin series for $\cos(x)$ by $x^2$, and then subtracted out the beginning term in $\cos(x)$ (which is $1$ in $1+x^2/2!-\ldots$). Dividing by $x^2$ makes sense, but how come only the first $1$ is subtracted? Shouldn't a $1$ be subtracted for each term and then the term be divided by $x^2$?
| Since
$$
\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dotsb
$$
you have
$$
\cos x - 1 = -\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dotsb
$$
and so
$$
\frac{\cos x - 1}{x^2} = -\frac{1}{2!}+\frac{x^2}{4!}-\frac{x^4}{6!}+\dotsb
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the Square Root of a Polynomial $4x^4 + 4x^3 - 11x^2 -6x + 9$
How do you find the square root of this polynomial? I really don't understand. Please provide an easy-to-understand explanation. Thanks.
| If you know that the polynomial is a perfect square, then the square root algorithm works. For example
$$\sqrt{4x^4 + 4x^3 - 11x^2 -6x + 9}$$
\begin{array}{lcccccccccccccc}
&&2x^2 && +x && -3\\
&&---&---&---&---&---\\
&|& 4x^4 & 4x^3 & -11x^2 & -6x & +9\\
2x^2 && 4x^4\\
&&---&---&---\\
&&& +4x^3 & -11x^2 \\
4x^2+x &|& &+4x^3 &+x^2\\
&&&---&---&---&---\\
&&&& -12x^2 &-6x &+9\\
4x^2+ 2x -3 &|&&&-12x^2 &-6x &+9\\
&&&&---&---&---\\
&&&&&&0\\
\end{array}
\begin{array}{ll}
\text{STEP}\;1. &\text{Compute the square root of the leading term (4x^4) and put it,}\\
& \text{(2x^2), in the two places shown.} \\
\text{STEP}\;2. &\text{Subtract and then append the next two terms ($4x^3-11x^2$).}\\
\text{STEP}\;3. &\text{Double the currently displayed quotient $(2x^2 \to 4x^2)$,} \\
& \text{then add the new term, $x$, so that $x(4x^2 + x)$ will "cancel out" the} \\
& \text{$4x^3$ in $4x^3 -11x^2$.}\\
\text{STEP}\;4. &\text{Subtract and then append the next two terms ($-6x+9$).}\\
\text{STEP}\;5. &\text{Double the currently displayed quotient} \\
& \text{ $(2x^2 + x \to 4x^2 + 2x)$, then add a new term, $x$,} \\
& \text{then add a new term, $-3$, so that $-3(4x^2 + 2x - 3)$ will "cancel out"} \\
& \text{ the $-12x^2$ in $-12x^2 -6x + 9$.}
\end{array}
| {
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Integral $\int_0^1 \frac{x\log x+1-x}{x \log^2 x}\log(1+x)\, dx=\log\frac{4}{\pi}$ Hi I am trying to prove this
$$
I:=\int_{0}^{1}
{x\log\left(\,x\,\right) + 1 - x \over x\log^{2}\left(\,x\,\right)}\,
\log\left(\,1 + x\,\right)\,{\rm d}x=\log\left(\,4 \over \pi\,\right).
$$
Thanks.
This is just a beautiful integral for many reasons. Logs are everywhere and an inspirational solution!!! I am not sure if breaking it up into three separate pieces is of any use, I tried that by writing
$$
I=\int_0^1\frac{ \log(1+x)}{\log x}dx+\int_0^1\frac{\log(1+x)}{x \log^2 x}dx-\int_0^1\frac{\log(1+x)}{\log^2 x}dx
$$
but wasn't sure how to handle these. Also note that
$$
\int_0^1 \frac{x\log x+1-x}{x \log^2 x}dx=1,
$$
in case that happened to come up anywhere along the calculation.
| Note
$$\frac{1}{x\log^2x}=-\frac{d}{dx}\frac{1}{\log x} $$
and hence
\begin{eqnarray}
I&=&\int_{0}^{1}
{x\log\left(\,x\,\right) + 1 - x \over x\log^{2}\left(\,x\,\right)}\,
\log\left(\,1 + x\,\right)\,{\rm d}x\\
&=&-\int_{0}^{1}
(x\log x + 1 - x)\log(1+x)d\frac{1}{\log x}\\
&=&-(x\log x + 1 - x)\log(1+x)\frac{1}{\log x}\bigg|_0^1+\int_{0}^{1}\frac{1}{\log x}d[
(x\log x + 1 - x)\log(1+x)]\\
&=&\int_0^1\frac{1}{\log x}\left(\frac{x \log x+1-x}{x+1}+\log x \log (x+1)\right)dx\\
&=&\int_0^1\log(x+1)dx+\int_0^1\frac{x}{x+1}+\int_0^1\frac{1}{\log x}\frac{1-x}{x+1}dx\\
&=&\log 2+J,
\end{eqnarray}
where
$$ J=\int_0^1\frac{1}{\log x}\frac{1-x}{x+1}dx.$$
Define
$$ f(a)=\int_0^1x^a\frac{1-x}{x+1}dx. $$
Then
\begin{eqnarray}
f(a)&=&\int_0^1\sum_{n=0}^\infty(-1)^nx^{a+n}(1-x)dx
&=&\sum_{n=0}^\infty(-1)^n\left(\frac{1}{n+a+1}-\frac{1}{n+a+2}\right)
\end{eqnarray}
and hence
$$ \int_0^a f(a)da=\sum_{n=0}^\infty(-1)^n\log\frac{n+a+1}{n+a+2},$$
and
$$ J=\lim_{a\to0}\sum_{n=0}^\infty(-1)^n\log\frac{n+a+1}{n+a+2}=\sum_{n=0}^\infty(-1)^n\log\frac{n+1}{n+2}=\log\frac{2}{\pi}.$$
Thus
$$ I=\log\frac{4}{\pi}. $$
| {
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"source": "stackexchange",
"question_score": "9",
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How prove $abc+1>3a$ if $a^2+b^2+c^2=9,0Let $a,b,c\in\mathbb{R}$ and such $0<a\le b\le c$, and $ a^2+b^2+c^2=9$. Show that
$$abc+1>3a$$
Since $$9=a^2+b^2+c^2\ge 3a^2\Longrightarrow a^2\ge 3$$
then I can't, maybe can use AM-GM inequality to solve it?
| From $b \le c$ and $a^2+b^2+c^2= 9$, we have
$$b^2c^2 \ge a^2c^2 = a^2(9-a^2-b^2) \ge a^2(9-2a^2)$$
Now $2a^2 \le 6 < 9 \implies 9-2a^2 > 0$, so the above gives us $bc \ge a\sqrt{9-2a^2}$. As $a$ is positive, $\implies abc \ge a^2\sqrt{9-2a^2}$.
Thus it is enough to show for $a \in (0, \sqrt3]$, we must have
$$a^2\sqrt{9-2a^2} \ge 3a-1 \iff f(a) = 2a^6-9a^4-(3a-1)^2 \le 0 \tag{1}$$
Case 1: $0 < a \le 1$
We note that $f$ can be written as $f(a) = 2a^4(a^2-1)-7a^4-(3a-1)^2 < 0$
Case 2: $1 < a \le \sqrt{3 \over 2}$
Note $f(a) = \left(a^4+\frac32 \right)(2a^2-3)-6a^2(a^2-1)-6\left(a-\frac{11}{12}\right)< 0 $
Case 3: $\sqrt{3 \over 2} < a \le \sqrt{3}$
Note $f(a)=a^2(a^2-3)(2a^2-3)+(1-6a) < 0 $
Thus in all cases, we find $abc+1 > 3a$.
P.S. The inequality $abc+1 \ge 3a$ would hold for all real $a, b, c$ s.t. $a \le b \le c$ and $a^2+b^2+c^2=9$, positivity is needed only for the strict inequality version.
| {
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Perfect Square relationship with no solutions I would like to show that for positive integers $a>b,c$ all greater than 1 such that $c\nmid a$, there are no solutions to the following equation:
$$a^2+1=b^2(c^2+1)$$
As was pointed out in the comments, it might help to rearrange and factor:
$$(a-b)(a+b)=(bc+1)(bc-1)$$
Another interesting approach to the question, if we rewrite:
$$a^2-b^2c^2=b^2-1$$
Then we are trying to argue that $b^2-1$ cannot be squarefree (in particular, $c^2|b^2-1$)
| No. On the other hand, it appears that $c | a.$
I am watching a longer computer run, one infinite family turns out to be
$$ a = 4 c^3 + 3 c, \; \; \; b = 4 c^2 + 1$$
for any positive integer $c.$ You can check that $a^2 + 1 = b^2 (c^2 + 1).$
There are a few others, though, with especially small $c,$ which makes me think they might be derivable from the infinite sequence.
EEDDIITTTTT: that worked. I have a doubly infinite sequence, one infinite sequence for a given $c.$
$$ a_0 = c, \; \; \; b_0 = 1, $$
$$ a_1 = 4 c^3 + 3 c, \; \; \; b_1 = 4 c^2 + 1,$$
$$ a_2 = 16 c^5 + 20 c^3 + 5 c , \; \; \; b_2 = 16 c^4 + 12 c^2 + 1,$$
then
$$ \color{magenta}{ a_{n+2} = (4 c^2 + 2) a_{n+1} - a_n, \; \; b_{n+2} = (4 c^2 + 2) b_{n+1} - b_n}. $$
I believe already that this doubly indexed family gives all solutions.
a b c
38 17 2
117 37 3
268 65 4
515 101 5
682 305 2
882 145 6
1393 197 7
2072 257 8
2943 325 9
4030 401 10
4443 1405 3
5357 485 11
6948 577 12
8827 677 13
11018 785 14
a b c
| {
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inverse trigonometric equations and inequalities Sorry for the list, wasn't sure if i should post each question in a separate post, but they are all under the same section and chapter. I don't expect for all questions to be answered, but any help on any of them is appreciated, thanks. I made sure all arithmetic calculations were correct. I didn't finish the problems completely, but i must be on the right track first to finish them (solve for x). ill be editing the post along as i find my own answers
(for all questions, find all solutions, solve for x)
1.
$$2sin^2(x) + (sin(2x)) = 0$$
use identity, sin(2x) = 2sin(x)cos(x)
$$2sin^2(x) + (2sin(x)cos(x)) = 0$$
factor
$$(2sin(x))(sin(x)+cos(x)) = 0$$
$$sin(x) = 0 $$
$$sin(x) = -cos(x)$$
$$\frac{sin(x)}{sin(x)} = -\frac{cos(x)}{sin(x)}$$
$$cot(x) = -1$$
sin(x)=0, cot(x)=-1 ... correct?
2.
$$6cos^2(\frac{x}{2})-7cos(\frac{x}{2})+2 = 0$$
cant factor, use quadratic formula ?
$$A= 6, B = -7, C = 2$$
$$\frac{7\pm \sqrt{1}}{12}$$
$$cos(\frac{x}{2}) = \frac{1}{2}$$
$$cos(\frac{x}{2}) = \frac{2}{3}$$
correct?
3.
$$ cos^2(x)-0.2sin(x) = 0.9$$
use identity
$$(1-sin^2(x))-\frac{1}{5}sin(x) = \frac{9}{10}$$
$$-sin^2(x)-\frac{1}{5}sin(x) = \frac{9}{10} - \frac{10}{10}$$
$$-sin^2(x)-\frac{1}{5}sin(x) + \frac{1}{10} = 0$$
negate to factor
$$sin^2(x)+\frac{1}{5}sin(x) - \frac{1}{10} = 0$$
cant factor, use quadratic formula
$$A=1,B=\frac{2}{10},C=-\frac{1}{10}$$
$$sin(x) = \frac{-\frac{2}{10}\pm \sqrt{\frac{4}{100}+(\frac{4}{10})}}{2}$$
$$sin(x) = \frac{-\frac{2}{10}\pm \sqrt{\frac{11}{25}}}{2}$$
$$sin(x) = 0.231662479$$ $$sin(x) = -0.431662479$$
correct?
4.
$$4sin^2(2x) = sin(2x) + 3$$
use identities
$$(4)(2sin(x)cos(x))^2 = 2sin(x)cos(x) + 3$$
$$16sin^2(x)cos^2(x) - 2sin(x)cos(x) - 3 = 0$$
factor-able
$$(2sin(x)cos(x)-1)(8sin(x)cos(x)+3)$$
i wont have sin(x) or cos(x) by itself unless i divide
$$sin(x) = \frac{1}{2cos(x)}$$
$$sin(x) = \frac{-3}{8cos(x)}$$
incorrect ... ?
5.
$$cos(x)csc(x) = cot^2(x)$$
identities
$$\cos(x)\frac{1}{sin(x)} = cot^2(x)$$
$$cot(x) = cot^2(x)$$
$$1 = cot(x)$$
correct?
| (Partial answers)
General comment: It seems that the questions ask for $x$, perhaps in some interval like $[0,2\pi)$, and perhaps without restriction. Nowhere have you mentioned what the possible values of $x$ are.
1) We have the possibility $\sin x=0$, giving $x=n\pi$, where $n$ ranges over the integers. For the possibility $\sin x+\cos x=0$, which you left undealt with, transform perhaps to $\tan x=-1$. That has solutions $x=\frac{3\pi}{4}$ and $x=-\frac{3\pi}{4}$. All solutions are given by $x=2n\pi \pm \frac{3\pi}{4}$.
2) Not that it matters much, but one can factor. You have obtained the right values for $\cos(x/2)$. That in principle tells you all possible values of $x/2$, and then of $x$. For $\cos(x/2)$, there are the solutions $x/2=2n\pi\pm \frac{\pi}{3}$. For $\cos(x/2)=\frac{2}{3}$, the expressions for $x/2$ are not nice.
4) You went to too much trouble. We have $4\sin^2(2x)-\sin(2x)-3=0$, already a quadratic. Factor as $(4\sin(2x)+3)(\sin(2x)-1)=0$.
But one can finish from where you got to, using $2\sin x\cos x=\sin(2x)$.
5) Don't forget about $\cot x=0$.
| {
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Solve the integral: $ \iint_D (x^2 \tan(x) + y^3 + 4) dxdy$ Solve the following integral
$$ \iint_{D} (x^2 \tan(x) + y^3 + 4)dxdy$$
where $D = \{ (x,y): x^2 + y^2 \leq 2\}$
I thought that a polar coordinates transformation would work but it takes an horrible form. Any suggestion to solve it?
Thanks!
| Being odd function,
$$\int_{-\sqrt{2}}^{\sqrt 2 } \int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} x^2 \tan(x) \, dydx = \int_{-\sqrt{2}}^{\sqrt 2 } \int_{-\sqrt{2-y^2}}^{\sqrt{2-y^2}} x^2 \tan(x) \, dxdy = 0$$
rest is easy.
EDIT:: the integral consist of three parts. The first part which I showed above is zero by symmetry of bounds for odd function.
$$\iint_{D} (x^2 \tan(x) + y^3 + 4)dxdy$$
The second part is also zero since it is odd function of $y$
$$\int_D y^3 \, dx dy = \int_{-\sqrt{2}}^{\sqrt 2 } \int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} y^3 \, dy dx = 0$$
The third part is
$$\int_D 4 \, dx dy = 4 \int_{-\sqrt{2}}^{\sqrt 2 } \int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} 1 \, dx dy = 4 \cdot \pi 2 = 8 \pi $$
The answer is $8 \pi $ you can check it here.
| {
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"timestamp": "2023-03-29T00:00:00",
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A tricky question on sequences of functions. Find
$$\lim_{n \to \infty}\left[
{1 \over x + 1} + {2x \over \left(x + 1\right)\left(x + 2\right)}
+{3x^{2} \over \left(x + 1\right)\left(x + 2\right)\left(x + 3\right)}
+ \cdots
+ {nx^{n-1} \over
\left(x + 1\right)\left(x + 2\right)\ldots\left(x + n\right)}\right]$$
| HINT:
For integer $r\ge1$,
$$\frac{rx^{r-1}}{(x+1)(x+2)\cdots(x+r-1)(x+r)}+\frac{x^r}{(x+1)(x+2)\cdots(x+r-1)(x+r)}=\frac{x^{r-1}}{(x+1)(x+2)\cdots(x+r-1)}$$
$$\iff \frac{rx^{r-1}}{(x+1)(x+2)\cdots(x+r-1)(x+r)}$$
$$=\frac{x^{r-1}}{(x+1)(x+2)\cdots(x+r-1)}-\frac{x^r}{(x+1)(x+2)\cdots(x+r-1)(x+r)} $$
Set $r=1,2,3,c\dots, n-1,n$ to add to recognize the Telescoping Series
Finally set $n\to\infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/791165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16\sqrt 2}$ This integral below
$$
I:=\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16 \sqrt 2}
$$
is what I am trying to prove. Thanks.
We can not expand the denominator as a series since the domain of integration is for $x\in [0,\infty)$. Next I wrote
$$
I=\int_0^\infty \log^2 x \frac{1+x^4-x^4+x^2}{1+x^4}dx=\int_0^\infty \log^2x \left(\frac{1+x^4}{1+x^4}+\frac{x^2-x^4}{1+x^4}\right)dx=\\
\int_0^\infty \log^2 x \, dx+\int_0^\infty \log^2 x \frac{x^2}{1+x^4}dx-\int_0^\infty \log^2 x \frac{x^4}{1+x^4}dx,
$$
however only the middle integral is convergent. I am not sure how to go about solving this problem. Thank you
| The integrand is invariant under inversion:
$$\log^2(1/x){1+(1/x)^2\over1+(1/x)^4}d(1/x)=-\log^2x{1+x^2\over1+x^4}dx$$
Thus
$$\begin{align}
\int_0^\infty \log^2x{1+x^2\over1+x^4}dx&=\int_0^1 \log^2x{1+x^2\over1+x^4}dx+\int_1^\infty \log^2x{1+x^2\over1+x^4}dx\\
&=\int_0^1 \log^2x{1+x^2\over1+x^4}dx-\int_1^0 \log^2x{1+x^2\over1+x^4}dx\\
&=2\int_0^1 \log^2x{1+x^2\over1+x^4}dx
\end{align}$$
Maye now you can expand things as a power series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 6,
"answer_id": 1
} |
Verifying Touchard's Identity $$C_{n+1} = \sum_{k=0}^{\lfloor n/2\rfloor}{n\choose 2k}\cdot C_k\cdot 2^{n-2k}$$
where $C_n$ are the Catalan numbers.
I think we start by diving both sides by $2^n$, but unsure of where to go from there
| Here is a proof using the method by Wilf from
Generatingfunctionology.
Suppose we are trying to show that
$$C_{n+1} = 2^n \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} C_k 2^{-2k}$$
where $C_n = \frac{1}{n+1}{2n\choose n}$ is the $n$-th Catalan number.
Recall the Catalan number recurrence
$$C_{n+1} = \sum_{k=0}^n C_k C_{n-k}$$
which implies for the generating function $C(z)$ of these numbers that
$$\frac{C(z)-1}{z} = C(z)^2$$
which has solutions $$C_{1,2}(z) = \frac{1\pm \sqrt{1-4z}}{2z}$$
of which the second one is analytic at zero so that
$$C(z) = \frac{1- \sqrt{1-4z}}{2z}.$$
Following Wilf we introduce the generating function $D(z)$ where
$$D(z) = \sum_{n\ge 0} z^n 2^n
\sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} C_k 2^{-2k}
= \sum_{k\ge 0} C_k 2^{-2k}
\sum_{n\ge 2k} {n\choose 2k} z^n 2^n
\\ = \sum_{k\ge 0} C_k 2^{-2k}
\sum_{n\ge 0} {n+2k\choose 2k} z^{n+2k} 2^{n+2k}
= \sum_{k\ge 0} C_k z^{2k}
\sum_{n\ge 0} {n+2k\choose 2k} z^n 2^n
\\ = \sum_{k\ge 0} C_k z^{2k} \frac{1}{(1-2z)^{2k+1}}.$$
This gives that
$$D(z) = \frac{1}{1-2z}
\frac{1-\sqrt{1-4z^2/(1-2z)^2}}{2z^2/(1-2z)^2}
\\ = \frac{(1-2z)^2}{1-2z}
\frac{1-\sqrt{1-4z^2/(1-2z)^2}}{2z^2}
= (1-2z)
\frac{1-\sqrt{1-4z^2/(1-2z)^2}}{2z^2}
\\= \frac{1-2z-\sqrt{(1-2z)^2-4z^2}}{2z^2}
= \frac{1-2z-\sqrt{1-4z}}{2z^2}.$$
On the other hand
$$\sum_{n\ge 0} C_{n+1} z^n
= \frac{C(z)-1}{z}
= \frac{1}{z} \left(\frac{1- \sqrt{1-4z}}{2z} -1 \right)
\\ = \frac{1}{z} \left(\frac{1- 2z -\sqrt{1-4z}}{2z}\right)
= \frac{1- 2z -\sqrt{1-4z}}{2z^2}.$$
We have equality, QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Consider a quadratic equation $az^2+bz+c=0$ where a,b,c are complex numbers. Prove that the equation has one purely imaginary root is given ... Problem :
Consider a quadratic equation $az^2+bz+c=0$ where a,b,c are complex numbers. Prove that the condition such that the equation has one purely imaginary root is given by $(b\overline{c}+c\overline{b})(a\overline{b}+\overline{a}b)+(c\overline{a}-a\overline{c})^2=0$
I am not getting any clue on this how to proceed such equation in complex numbers. Please suggest thanks a lot.
| Let $r$ be the purely imaginary root, so that $\overline{r} = -r$. We have these conjugate equations:
$$a r^2 + b r^1 + c = 0 \qquad \text{and} \qquad \overline{a} r^2 - \overline{b} r^1 + \overline{c} = 0$$
Here's a neat trick: Move the exponents on $r$ to subscripts ...
$$a r_2 + b r_1 + c = 0 \qquad \text{and} \qquad \overline{a}r_2 - \overline{b} r_1 + \overline{c} = 0$$
... and solve the linear system for them ...
$$
r_1 = \frac{a\overline{c}-\overline{a}c}{a\overline{b}+\overline{a}b} \qquad\qquad
r_2 = -\frac{b\overline{c}+\overline{b}c}{a\overline{b}+\overline{a}b}
$$
(Dealing with the case of $a\overline{b}+\overline{a}b=0$ is left as an exercise to the reader.) Now, return the subscripts to their positions as exponents ...
$$
r^1 = \frac{a\overline{c}-\overline{a}c}{a\overline{b}+\overline{a}b} \qquad\qquad
r^2 = -\frac{b\overline{c}+\overline{b}c}{a\overline{b}+\overline{a}b}
$$
... and simply observe the deep algebraic truth, $r^2 = (r^1)^2$ ...
$$\left(\;\frac{a\overline{c} -\overline{a}c}{a\overline{b}+\overline{a}b}\;\right)^2 = -\frac{b\overline{c}+\overline{b}c}{a\overline{b}+\overline{a}b}$$
Squaring and clearing fractions, we have
$$(a\overline{c} -\overline{a}c)^2 = -(a\overline{b}+\overline{a}b)( b\overline{c}+\overline{b}c)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int\limits_0^\pi \frac{x}{1+\sin^2x} \ dx$ How can one evaluate
$$\int_0^\pi \frac{x}{1+\sin^2x} \ dx\ ?$$
| let $x = -u +\dfrac{\pi}{2}$, then $\sin^2x = \sin^2(-u +\frac{\pi}{2}) = \cos^2u$, and also $dx = -du$. So:
$I = \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{\frac{\pi}{2} - u}{1 + \cos^2u}du = \pi\cdot \displaystyle \int_{0}^\frac{\pi}{2} \dfrac{1}{1 + \cos^2u}du$.
At this point you can introduce $t = \tan\left(\dfrac{u}{2}\right)$, and continue the fraction decomposition to finish it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
determinant inequality $\det(A^2+AB+B^2)\geq\det(AB-BA)$
$A,B$ are two $2\times 2$ real matrices, then
$$\det(A^2+AB+B^2)\geq\det(AB-BA)$$
The inequality is equivalent to the following problem: Let $X=A+\dfrac{B}{2},Y=-\dfrac{B}{2}$
$$\det[(X-Y)(X+Y)-2(X^2+Y^2)]≥4\det(XY-YX)$$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=353&t=588819
| For any two $2\times2$ matrices $A$ and $B$, the following identity holds:
$\renewcommand{\tr}{\operatorname{tr}}$
$\renewcommand{\adj}{\operatorname{adj}}$
$$
\det(X+Y) \equiv \det(X) + \det(Y) + \tr(X\adj(Y)).\tag{$\ast$}
$$
Therefore,
\begin{align}
\det(AB-BA)
&=2\det(AB) + \tr(AB\adj(-BA))\\
&=2\det(AB) - \tr(AB\adj(A)\adj(B)).
\end{align}
Write $t=\tr(A\adj(B))=\tr(B\adj(A))$. Then
\begin{align}
&\det(A^2+AB+B^2)\\
=&\det((A+B)^2 - BA)\\
=&\det((A+B)^2) + \det(-BA) + \tr((A+B)^2\adj(-BA))\quad\text{ by } (\ast)\\
=&\det(A+B)^2 + \det(AB) - \tr((A^2+B^2+AB+BA) \adj(A)\adj(B))\\
=&\det(A+B)^2 + \det(AB) - (\det(A)+\det(B))t - \tr(AB \adj(A)\adj(B)) - 2\det(AB)\\
=&\left(\det(A)+\det(B)+t\right)^2 - 3\det(AB) - (\det(A)+\det(B))t + \det(AB-BA)\\
=&\left(t + \frac{\det(A)+\det(B)}2\right)^2 + \frac34\left(\det(A)-\det(B)\right)^2 + \det(AB-BA)\\
\ge&\det(AB-BA).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_0^\infty\!\!\int_0^\infty\!\!\int_0^\infty\!\frac{(xyz)^{-1/7}(yz)^{-1/7}z^{-1/7}}{(x+1)(y+1)(z+1)}dx\,dy\,dz$ I am looking for guidance evaluating the following integral. $$\int_0^\infty\!\!\!\int_0^\infty\!\!\int_0^\infty\!\!\dfrac{(xyz)^{-1/7}(yz)^{-1/7}z^{-1/7}}{(x+1)(y+1)(z+1)}\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$
Its value can be found easily with a run through Mathematica (I will not post the answer) but there is apparently a common method to computing integrals of this type.
This problem be reduced to evaluating integrals of the type $$\int_0^\infty \dfrac{x^{-a}}{x+1}\mathrm{d}x.$$
I imagine there is some sort of differentiation under the integral sign that can be used but I am at loss.
| $$\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}\frac{(xyz)^{-1/7}(yz)^{-1/7}(z)^{-1/7}}{(x+1)(y+1)(z+1)}dxdydz$$
$$
I = \int\limits_0^{ + \infty } {\int\limits_0^{ + \infty } {\int\limits_0^{ + \infty } {\frac{{\left( {xyz} \right)^{ - \frac{1}
{7}} \left( {yz} \right)^{ - \frac{1}
{7}} z^{ - \frac{1}
{7}} }}
{{\left( {x + 1} \right)\left( {y + 1} \right)\left( {z + 1} \right)}}dxdydz} } } = \left( {\int\limits_0^{ + \infty } {\frac{{x^{1 - \frac{1}
{7} - 1} }}
{{\left( {x + 1} \right)^{1 - \frac{1}
{7} + \frac{1}
{7}} }}dx} } \right)\left( {\int\limits_0^{ + \infty } {\frac{{y^{1 - \frac{2}
{7} - 1} }}
{{\left( {y + 1} \right)^{1 - \frac{2}
{7} + \frac{2}
{7}} }}dy} } \right)\left( {\int\limits_0^{ + \infty } {\frac{{z^{1 - \frac{3}
{7} - 1} }}
{{\left( {z + 1} \right)^{1 - \frac{3}
{7} + \frac{3}
{7}} }}dz} } \right)
$$
now use
$$
{\rm B}\left( {x,y} \right) = \int\limits_0^{ + \infty } {\frac{{t^{x - 1} }}
{{\left( {1 + t} \right)^{x + y} }}dt}
$$
then
$$
= {\rm B}\left( {1 - \frac{1}
{7},\frac{1}
{7}} \right){\rm B}\left( {1 - \frac{2}
{7},\frac{2}
{7}} \right){\rm B}\left( {1 - \frac{3}
{7},\frac{3}
{7}} \right) = {\rm B}\left( {\frac{6}
{7},\frac{1}
{7}} \right){\rm B}\left( {\frac{5}
{7},\frac{2}
{7}} \right){\rm B}\left( {\frac{4}
{7},\frac{3}
{7}} \right)
$$
$$
= \frac{{\Gamma \left( {\frac{6}
{7}} \right)\Gamma \left( {\frac{1}
{7}} \right)}}
{{\Gamma \left( {\frac{6}
{7} + \frac{1}
{7}} \right)}} \cdot \frac{{\Gamma \left( {\frac{5}
{7}} \right)\Gamma \left( {\frac{2}
{7}} \right)}}
{{\Gamma \left( {\frac{5}
{7} + \frac{2}
{7}} \right)}} \cdot \frac{{\Gamma \left( {\frac{4}
{7}} \right)\Gamma \left( {\frac{3}
{7}} \right)}}
{{\Gamma \left( {\frac{4}
{7} + \frac{3}
{7}} \right)}} = \Gamma \left( {\frac{6}
{7}} \right)\Gamma \left( {\frac{1}
{7}} \right) \cdot \Gamma \left( {\frac{5}
{7}} \right)\Gamma \left( {\frac{2}
{7}} \right) \cdot \Gamma \left( {\frac{4}
{7}} \right)\Gamma \left( {\frac{3}
{7}} \right)
$$
$$
= \Gamma \left( {1 - \frac{1}
{7}} \right)\Gamma \left( {\frac{1}
{7}} \right) \cdot \Gamma \left( {1 - \frac{2}
{7}} \right)\Gamma \left( {\frac{2}
{7}} \right) \cdot \Gamma \left( {1 - \frac{3}
{7}} \right)\Gamma \left( {\frac{3}
{7}} \right)
$$
$$
= \frac{\pi }
{{\sin \left( {\frac{\pi }
{7}} \right)}} \cdot \frac{\pi }
{{\sin \left( {\frac{{2\pi }}
{7}} \right)}} \cdot \frac{\pi }
{{\sin \left( {\frac{{3\pi }}
{7}} \right)}} = \frac{{8\sqrt 7 }}
{7}\pi ^3 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/797955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 3
} |
Sum of Residues Modulo $p^2$. Let $p$ be an odd prime. Prove that
$$ \sum_{k = 1}^{p-1} k^{2p-1} \equiv \frac{p(p+1)}2 \pmod{p^2}$$
| @user8268 (currently deleted) answer has the right idea, though there were algebraic errors. The idea is to group for $1 \leq k \leq \frac{p-1}{2}$
\begin{align}
k^{2p-1}+(p-k)^{2p-1} \equiv k^{2p-1}+[(2p-1)pk^{2p-2}-k^{2p-1}] &\equiv p(2p-1)k^{2p-2} \pmod{p^2} \\
&\equiv p(2p-1) \pmod{p^2}
\end{align}
where in the last step we have used $p \mid k^{2p-2}-1$ by Fermat's little theorem.
Thus
\begin{align}
\sum_{k=1}^{p-1}{k^{2p-1}}=\sum_{k=1}^{\frac{p-1}{2}}{\left(k^{2p-1}+(p-k)^{2p-1}\right)} & \equiv \frac{p-1}{2}p(2p-1) \pmod{p^2} \\
& \equiv -\frac{p(p-1)}{2} \pmod{p^2}\\
&\equiv \frac{p(p+1)}{2} \pmod{p^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How can I solve these pde's? Three different problem I got:
1.. $xu_x+2x^2u_y-u=x^2e^x$ and $u(x,x^2+x)=xe^x+x^2$
2.. $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0, \quad x\neq y$
3.. $(y+xu)u_x+(x+yu)u_y=u^2-1$
Couldnt even start. Could you at least give me a hint? thanks.
edit1:
at 2. from $B^2-AC=\frac {(x-y)^2}4>0 $ so the equation is hyperbolic. so we can write its cononical forms. is it true?
edit2: for the first one :
by taking $\eta=x^2-y$ and $\xi=x$ I found $u(x,y)=xe^x+x(-x+1-e^{-x})$
for the second one :
I took $\xi=\frac{x^2-y^2}{2}$ and $\eta=x-y$ and found $u(x,y)=\frac {f^*(\frac{x^2-y^2}{2})}{x-y}+g(x-y)$
edit 3:----------------------
1-) $\frac ba=\frac {2x^2}{x}=2x=\frac {dy}{dx}$
$x^2-y=c$, $\eta =x^2-y$ and $\xi =x$
$u_x=w_{\xi}+w_{\eta}2x$ and $ u_y=-w_{\eta}$
$\xi(w_{\xi}+w_{\eta}2\xi)-2\xi ^2w_{\xi}-w=\xi^2e^\xi $
$w_{\xi}-\frac w \xi=e^\xi$ multiplying by $\frac 1\xi$
$\frac{d(w\frac 1\xi)}{d\xi}=e^\xi$
so we got $w=\xi e^\xi+g(\eta)$
$u(x,y)=xe^x+g(x^2-y)$
and $u(x,x^2+x)=xe^x+x^2 =xe^x+g(x^2-x^2-x)$
$g(-x)=x^2$, $g(x)=x^2$ so the $u(x,y)=xe^x+x^2$
//--------------------------------------------
3)$\frac {\frac{dx}{dt}}{y+ux}=\frac {\frac{dy}{dt}}{x+yu}=\frac {\frac{du}{dt}}{u^2-1}$
$\frac {\frac{dx}{dt}-\frac{dy}{dt}}{y+ux-x-yu}=\frac {\frac{du}{dt}}{u^2-1}$
$\frac {\frac{dx}{dt}-\frac{dy}{dt}}{(1-u)(y-x)}=\frac {\frac{du}{dt}}{u^2-1}$
$\frac {\frac{d(x-y)}{dt}}{y-x}=\frac {\frac{du}{dt}}{u+1}$
$ln(u+1)+ln(x-y)=c_1$------------------- (1) we got one independent solution
we need another
$\frac {\frac{dx}{dt}+\frac{dy}{dt}}{y+ux+x+yu}=\frac {\frac{du}{dt}}{u^2-1}$
$\frac {\frac{dx}{dt}+\frac{dy}{dt}}{(1+u)(x+y)}=\frac {\frac{du}{dt}}{u^2-1}$
$\frac {\frac{d(x+y)}{dt}}{x+y}=\frac {\frac{du}{dt}}{u-1}$
$ln(u-1)-ln(x+y)=c_2$----------------------(2)
from (1) and (2) which form is $h(x,y,u)=c_1$ and
$j(x,y,u)=c_2$
general solution would be in $j(x,y,u)=F(h(x,y,u))$ Where F is an arbitrary.
so $ln(u-1)-ln(x+y)=F(ln(u+1)+ln(x-y))$
| 1..
$xu_x+2x^2u_y-u=x^2e^x$
$u_x+2xu_y=\dfrac{u}{x}+xe^x$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dy}{dt}=2x=2t$ , letting $y(0)=y_0$ , we have $y=t^2+y_0=x^2+y_0$
$\dfrac{du}{dt}=\dfrac{u}{x}+xe^x=\dfrac{u}{t}+te^t$ , we have $u=te^t+tf(y_0)=xe^x+xf(y-x^2)$
$u(x,x^2+x)=xe^x+x^2$ :
$xe^x+xf(x)=xe^x+x^2$
$f(x)=x$
$\therefore u(x,y)=xe^x+x(y-x^2)=xe^x-x^3+xy$
2.. Solving $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0$
3.. :How to find the general solution of $(y+ux)u_x+(x+yu)u_y=u^2-1$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/804762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $2^n(\cos^n(\frac{2\pi}{9})+\cos^n(\frac{4\pi}{9})+\cos^n(\frac{8\pi}{9}))\in\mathbb{Z}$
*
*$a_n=2^n\left[\cos^n\left(\dfrac{2\pi}{9}\right)+\cos^n\left(\dfrac{4\pi}{9}\right)+\cos^n\left(\dfrac{8\pi}{9}\right)\right]$. Show that $a_n\in\mathbb{Z}$ for all $n\in\mathbb{Z}$.
*Find the last digit in $a_{10^6}$.
My progress:
*
*I believe it can be solved by expressing $a_n$ as a linear recurrence by using the cubic equation $x^3-3x+1=0$ which has the roots $x_1=2\cos\left(\dfrac{2\pi}{9}\right),x_2=2\cos\left(\dfrac{4\pi}{9}\right),x_3=2\cos\left(\dfrac{8\pi}{9}\right)$. I'm just not really sure how to connect $a_n$ to this cubic equation.
*I guess that if you could find a linear recurrence to $a_n$, the last digit would be cyclic and could be found that way. I was hoping to solve it without using computer work, but maybe that's tough.
Sorry if my $\TeX$ writing isn't correct. I'm still learning. Thanks!
| The observation on the zeros of $x^3 - 3 x + 1$ means that:
$$
1 - \frac{12}{z} + \frac{8}{z^3}
$$
has $2 \cos \frac{2 \pi}{9}, 2 \cos \frac{4 \pi}{9}, 2 \cos \frac{8 \pi}{9}$ as inverses of its zeros,
which means that the recurrence:
$$
a_{n + 3} - 12 a_{n + 2} + 8 a_n = 0
$$
has general solution:
$$
a_n
= c_1 2^n \cos^n \frac{2 \pi}{9}
+ c_2 2^n \cos^n \frac{4 \pi}{9}
+ c_3 2^n \cos^n \frac{8 \pi}{9}
$$
From the recurrence it is obvious that $a_n \in \mathbb{Z}$ for all $n$ if $a_0, a_1, a_2 \in \mathbb{Z}$. Would need to check those three cases of the above... in any case, $a_0 = 3$ is an integer. One down, two to go.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/805976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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real solution of eqn. in $\sin x+2\sin 2x-\sin 3x = 3,$ where $x\in (0,\pi)$. The no. of real solution of the equation $\sin x+2\sin 2x-\sin 3x = 3,$ where $x\in (0,\pi)$.
$\bf{My\; Try::}$ Given $\left(\sin x-\sin 3x\right)+2\sin 2x = 3$
$\Rightarrow -2\cos 2x\cdot \sin x+2\sin 2x = 3\Rightarrow -2\cos 2x\cdot \sin x+4\sin x\cdot \cos x = 3$
$\Rightarrow 2\sin x\cdot \left(-\cos 2x+2\cos x\right)=3$
Now I did not understand how can i solve it.
Help me
Thanks
| $$\begin{cases}\sin 2x=2\sin x\cos x\\{}\\\sin 3x=\sin2x\cos x+\sin x\cos2x=2\sin x\cos^2x+\sin x(1-2\sin^2x)\end{cases}$$
Thus we get
$$0=\sin x+4\sin x\cos x-2\sin x\cos^2x-\sin x+2\sin^3x$$
Divide al through by $\;\sin x\;$ (why can we?):
$$4\cos x-2\cos^2x+2\sin^2x=0\iff2\cos x-\cos^2x+1-\cos^2x=0\iff$$
$$2\cos^2x-2\cos x-1=0\iff \ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometrical Solve There are 2 different values of $ \ \theta \ $. They are $ \ a \ $ and $ \ b \ $, such that $ \ 0 \ < \ a,b \ < \ 360^\circ \ $.
If $ \ \sin(\theta+\phi) = \frac{1}{2} \sin2\phi \ $ , prove that $$ \ \frac{ \sin a \ + \ \sin b}{\cos a \ + \ \cos b} \ = \ \cot\phi \ . $$
| We have $\displaystyle\sin\theta\cos\phi=\sin\phi(\cos\phi-\cos\theta)$
Squaring we get $\displaystyle\sin^2\theta\cos^2\phi=\sin^2\phi(\cos\phi-\cos\theta)^2$
$\displaystyle\implies \sin^2\phi(\cos^2\phi+\cos^2\theta-2\cos\phi\cos\theta)=(1-\cos^2\theta)\cos^2\phi$
$\displaystyle\iff \cos^2\theta-2\cos\phi\sin^2\phi\cos\theta+\sin^2\phi\cos^2\phi-\cos^2\phi=0$
whose roots are $\displaystyle\cos a,\cos b\implies \cos a+\cos b=2\cos\phi\sin^2\phi$
Similarly starting with, $\displaystyle\cos\theta\sin\phi=\cos\phi(\sin\phi-\sin\theta),$
we shall find $\displaystyle\sin a+\sin b=2\cos^2\phi\sin\phi$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: for $\forall x\ne 0, \cos x < 1 - {x^2\over 2} + {x^4\over 24}$
Prove: for $\forall x\ne 0, \cos x < 1 - {x^2\over 2} + {x^4\over 24}$
What I did:
We can prove:
$${\cos x -1 + {x^2\over 2} \over {x^4\over 24}} < 1$$
Lets define:
$f(x) = \cos x -1 + {x^2\over 2}$ and $g(x)= {x^4\over 24}$
By LMVT:
$${{f(x) - f(0)} \over {g(x) - f(0)}} = {f'(y)\over g'(y)} = {{-\sin y + y} \over {4x^3\over 24}} = {{-\sin y + y} \over {x^3\over 6}}\text{ where }y\in(0,x)$$
I tried to show the last expression is smaller than $1$, but without success.
What's the trick? Maybe the $f(x), g(x)$ are wrong?
| This solution is way too long but can be tightened up and only uses Calc I methods.
First, the only solution of $\sin x = x$ is $x = 0$.
If there existed a second point $x$ with $\sin x = x$ the mean value theorem would provide a point $c$ in between $0$ and $x$ with $\cos c = 1$. This means $c$ is a multiple of $2\pi$, so that $|x| \ge 2\pi$. Since $|\sin x| = 1$ this is impossible.
Second, the only solution of $\cos x = 1 - \dfrac{x^2}{2}$ is $x = 0$.
Define $f(x) = 1 - \dfrac{x^2}{2} - \cos x$. If there existed a second point with $f(x) = 0$, the mean value theorem would provide a point $c$ in between $0$ and $c$ with $f'(c) = 0$. This point satisfies $\sin c = c$, which is impossible since $c \not = 0$. This means that $f$ has no zeroes other than $x = 0$.
Third, $1 - \dfrac{x^2}{2} < \cos x$ for all $x \not= 0$.
Define $f$ as above. Since the only zero of $f$ is at $x=0$, it suffices to show that $f(x)$ is negative in a neighborhood of $0$. According to the intermediate value theorem, $f$ can then never be positive. You can use the first derivative test to show that $f$ has a local maximum at $x = 0$ and is thus negative in a neighborhood of $0$.
Fourth, the only solution of $\sin x = x - \dfrac{x^3}{3}$ is $x=0$.
If there existed a point $x \not= 0$ satisfying this, the mean value theorem would provide a point $c$ in between $0$ and $x$ with $\cos c = 1 - \dfrac{c^2}{2}$, contrary to the above result.
Finally, $\cos x < 1 - \dfrac{x^2}{2} - \dfrac{x^4}{24}$ for all $x \not= 0$.
Define $g(x) = \cos x - 1 + \dfrac{x^2}{2} - \dfrac{x^4}{24}$. Argue as above. Show that $g$ has only one zero, and that $g$ must be negative in a neighborhood of $x=0$. Then you're done!
| {
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"timestamp": "2023-03-29T00:00:00",
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An Inequality Problem $1 \le \frac{a}{1-ab}+\frac{b}{1-bc}+\frac{c}{1-ac} \le \frac{3\sqrt{3}}{2}$ If $a,b,c>0$, are positive real numbers such that $a^2+b^2+c^2=1$ then, the following Inequalities hold:
$\displaystyle 1 \le \frac{a}{1-ab}+\frac{b}{1-bc}+\frac{c}{1-ac} \le \frac{3\sqrt{3}}{2}$
$\displaystyle 1 \le \frac{a}{1+ab}+\frac{b}{1+bc}+\frac{c}{1+ac} \le \frac{3\sqrt{3}}{4}$
Homogenizing the first inequality as,
$\displaystyle \frac{a}{a^2+b^2+c^2-ab}+\frac{b}{a^2+b^2+c^2-bc}+\frac{c}{a^2+b^2+c^2-ac} \le \frac{3\sqrt{3}}{2\sqrt{a^2+b^2+c^2}}$
and noting the cyclic symmtery I tried Rearrangement Inequalities Inequalities.
Using $a^2+b^2+c^2-ab = \frac{1}{4}(a+b)^2+\frac{3}{4}(a-b)^2+c^2 \ge \frac{1}{4}(a+b)^2+c^2 \ge (a+b)c$
$\displaystyle \sum_{cyc} \frac{a}{a^2+b^2+c^2-ab} \le \sum_{cyc} \frac{a}{ac+bc}$ might help, but I couldn't get anywhere with it.
Thank you!
| Please consider the following as a comment, I'm not able to say if the Holder equality is valid in that case.
RHS for the first inequality :
$$1-ab=1+\frac{(a-b)^2-a^2-b^2}{2}=\frac12+\frac{(a-b)^2+c^2}{2}\geq \frac{1+c^2}{2}$$
so
$$\sum_{cyc} \frac{a}{1-ab}\leq \sum_{cyc} \frac{a}{\frac{1+c^2}{2}}$$
We then use the Hölder's inequality :
$$\sum_{cyc} \frac{a}{1-ab}\leq \left(\sum_{cyc} a^2\right)^{\frac12}\left(\sum_{cyc} \frac{1+c^2}{2}\right)^{-1}\left(\sum_{cyc} 1\right)^{\frac32}=\frac{3\sqrt{3}}{2}$$
LHS for the first inequality :
$$1-ab\leq 1$$
so
$$\sum_{cyc} \frac{a}{1-ab}\geq \sum_{cyc} a$$
Or :
$$1=\sum_{cyc} a^2\leq \left(\sum_{cyc} a\right)^{2}$$
so
$$1\leq \sum_{cyc} \frac{a}{1-ab}$$
| {
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"url": "https://math.stackexchange.com/questions/815534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to prove that $ \sum_{n=0}^\infty \frac{1}{(2n+1)^2} + \sum_{k=1}^\infty \frac{1}{(2k)^2}=\frac{4}{3} \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$ How to prove
$$ \sum_{n=0}^\infty \frac{1}{(2n+1)^2} + \sum_{k=1}^\infty \frac{1}{(2k)^2}=\frac{4}{3} \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$
| Consider the following series:
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n^{2}} &= \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \cdots \\
&= \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} + \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}}
\end{align}
which leads to
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} = \frac{1}{4} \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} + \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}}
\end{align}
or
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} = \frac{1}{3} \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}}.
\end{align}
Now,
\begin{align}
\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} + \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} = \frac{4}{3} \ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}}
\end{align}
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Is there a proof that $ x^2 > (x-y)(x+y)$? I noticed that when I did the math, $5^2 > 4 \cdot 6$, and $10^2$ is greater than $9 \cdot 11$, etc. I looked for a proof of this and couldn't find one. Assuming $x$ and $y$ are real numbers and $y ≠ 0$.
|
For all $y > 0$, we have $y^2 > 0$ for all $y \not= 0$,
\begin{align}
x^2 &> x^2-y^2 =(x+y)(x-y)
\end{align}
I see that you also did some sample arithmetic with $x^2 \ge (x+y)(x-y)$. But try also testing $$x^2 > x^2-y^2$$ as well. E.g. $10^2 > 10^2-1 ^2$ or $5^2 \ge 5^2-(-9)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Simplifying $\sqrt[4]{161-72 \sqrt{5}}$ $$\sqrt[4]{161-72 \sqrt{5}}$$
I tried to solve this as follows:
the resultant will be in the form of $a+b\sqrt{5}$ since 5 is a prime and has no other factors other than 1 and itself. Taking this expression to the 4th power gives $a^4+4 \sqrt{5} a^3 b+30 a^2 b^2+20 \sqrt{5} a b^3+25 b^4$. The integer parts of this must be equal to $161$ and the coeffecients of the roots must add to $-72$. You thus get the simultaneous system:
$$a^4+30 a^2 b^2+25 b^4=161$$
$$4 a^3 b+20 a b^3=-72$$
In an attempt to solve this, I first tried to factor stuff and rewrite it as:
$$\left(a^2+5 b^2\right)^2+10 (a b)^2=161$$
$$4 a b \left(a^2+5 b^2\right)=-72$$
Then letting $p = a^2 + 5b^2$ and $q = ab$ you get
$$4 p q=-72$$
$$p^2+10 q^2=161$$
However, solving this yields messy roots. Am I going on the right path?
| I think this problem is a lot easier than the other answers would have one believe. Since
$$161^2-5\cdot72^2=(161+72\sqrt5)(161-72\sqrt5)=1$$
We can see that $161+72\sqrt5$ is a unit in the ring of integers of $\mathbb{Q}(\sqrt5)$ Thus it must be $(\pm1)$ times a power of the fundamental unit, $\phi=\frac{1+\sqrt5}2$:
$$161+72\sqrt5=\phi^n$$
Solving for $n$ we have
$$n=\frac{\ln(161+72\sqrt5)}{\ln\left(\frac{1+\sqrt5}2\right)}=12$$
Thus
$$(161+72\sqrt5)^{1/4}=\phi^3=2+\sqrt5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 2
} |
Question on how to find this particular limit..
Find the following limit $$\lim_{x \to 0^-}\left(\dfrac{\cos(x)}{\sin(x)}-\dfrac{1}{x}\right)$$
I checked online and the correct answer is $-\infty$ but I am not sure how to get to it via LHopital??
My Attempt
As it is a $0/0$ type limit I tried using LHopital's;
$$\lim_{x \to 0^-}\left(\dfrac{\cos(x)}{\sin(x)}-\dfrac{1}{x}\right)=\left(\dfrac{x\cos(x)-\sin(x)}{x\sin(x)}\right)$$
and differentiating we get;
$$\lim_{x \to 0^-}\left(\dfrac{-x\sin(x)}{x\cos(x)+\sin(x)}\right)$$
This is still a $0/0$ type limit... differentiating further doesn't help either...
Any help or advice would be much appreciated
| Bully for Taylor:
$$\begin{align}
\frac{\cos x}{\sin x} - \frac{1}{x} &= \frac{x\cos x - \sin x}{x\sin x}\\
&= \frac{x\left(1-\frac{x^2}{2} + O(x^4)\right)-\left(x - \frac{x^3}{6} + O(x^5)\right)}{x\left(x - \frac{x^3}{6} + O(x^5)\right)}\\
&= \frac{-\frac{x^3}{3} + O(x^5)}{x^2\left(1-O(x^2)\right)}\\
&= -\frac{x}{3}\frac{1+O(x^2)}{1-O(x^2)},
\end{align}$$
and we find the limit for $x\to 0$ is $0$, not $-\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Fractions in Questions and Answers
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