Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find $K$ in $2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$? The problem is as follows:
Which value should have $K$ in the expression from below so it makes an identity?
$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&11\\
2.&12\\
3.&14\\
4.&15\\
\end{array}$
I'm confused exactly how to get that particular value for $K$.
The only thing which I was able to spot is that: (Using manipulations on the left side of the equation)
$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$
$2\csc 30 ^\circ (2\cos 7x \cos x)(2\cos 4x \cos 2x)-1$
$4 (2\cos 7x \cdot \cos 2x)(2 \cos x \cdot 4x)-1$
$4(\cos 9x + \cos 5x)(\cos 5x+\cos 3x)-1$
$4(\cos 9x \cos 5x +\cos 9x \cos 3x + \cos 5x \cos 5x + \cos 5x \cos 3x)-1$
$2(\cos 14x + \cos 4x + \cos 11x + \cos 6x + 1+\cos 10x + \cos 8x + \cos 2 x) - 1$
However at the point I reached here then I got stuck it seems that this might not be the intended strategy. What could be done here to solve this problem?. Is there any trick or what?.
Could someone help me with an orderly method to solve this without much fuss?.
| Let $x = \pi/4$. Then $\cos 2x = 0, \cos 6x = 0$, and $\cos 8x = 1$, which gives:
$$0 - 1 = \sin(K\pi/4) \cdot \csc \pi/4 \Rightarrow\sin(K\pi/4) = -\frac{1}{\sqrt2}$$
$$\Rightarrow K \pi/4 + 2\pi k= 5\pi/4, 7\pi/4$$
which only $K = 15$ satisfies from the given options.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find $\int \frac 1 {(x^2 +a^2)^2} dx $. I am trying to integrate $\int \frac 1 {(x^2+a^2)^2} \ dx$. The only thing that I can think to try is substitution, $u=x^2+a^2$ so that $\frac{du}{dx}=2x \Rightarrow du = 2x\ dx = 2\sqrt{u-a^2}\ du$ and then the integral becomes
$$ \int \frac{1}{u^2} (2\sqrt{u-a^2} \ du) $$
which does not seem productive.
I could try perhaps to separate with integration by parts, setting $u = \frac{1}{x^2 +a^2}$ and $dv = \frac{dx}{x^2+a^2}$. Then we obtain $\frac{du}{dx} = -\frac{1}{(x^2+a^2)^2}(2x)$ and $v = \frac 1 a \tan^{-1}(x/a)$. Then the integral becomes
$$ \begin{align*}
uv - \int v \ du &= \left(\frac{1}{x^2 +a^2}\right)\left(\frac 1 a \tan^{-1}(x/a)\right) - \int \left(\frac 1 a \tan^{-1}(x/a)\right) \left( -\frac{1}{(x^2+a^2)^2}(2x) \right) \ du
\end{align*}$$
but this also looks like it's headed nowhere good. Advice?
| hint
Write the integral as
$$\frac{1}{a^2}\int \frac{a^2+x^2-x^2}{(a^2+x^2)^2}dx$$
$$\frac{1}{a^2}\Bigl(\int\frac{dx}{a^2+x^2}-\frac 12\int x\frac{2xdx}{(a^2+x^2)^2}
\Bigr)$$
Put $ x=at $ for the first integral and use by parts for the second.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Finding the particular solution of a differential equation using at least three different methods. Find the particular solution $(x^2+6y^2)dx-4xydy=0$; when $x=1$, $y=1$ using at least three different methods.
I have done the first two. Can somebody help me with the third method.
Method 1: Homogenous Equation
Let $y=vx; dy=vdx+xdv$
$(x^2+6x^2v^2)dx-4x(vx)(vdx+xdv)=0$
$(x^2+2x^2v^2)dx-4x^3vdv=0$
$(1+2v^2)dx-4xvdv=0$
$\int\frac{dx}{4x}-\int\frac{v}{1+2v^2}dv=0$
$\ln{x}-\ln{(2v^2+1)}=C$
$\ln{(\frac{x}{2v^2+1})}=\ln{C}$
$\frac{x}{2v^2+1}=\frac{1}{C}$
$C=3$
$3x=2v^2+1$
$3x^3=2y^2+x^2$
$2y^2=x^2(3x-1)$
The particular solution by method 1 is $2y^2=x^2(3x-1)$.
Method 2: Bernoulli Equation
$2y\frac{dy}{dx}-\frac{x^2+6y^2}{2x}=0$
$2y\frac{dy}{dx}-\frac{3y^2}{x}=\frac{x}{2}$
Let $v=y^2; dv=2ydy$
$\frac{dv}{dx}-\frac{3v}{x}=\frac{x}{2}$
$P(x)=-3x^{-1}$; I.F.$=e^{-3\int x^{-1}dx}=x^{-3}$
$vx^-3=\frac{1}{2}\int\frac{dx}{x^2}$
$2vx^{-3}=-x^{-1}+C^{-1}$
$2y^2x^{-3}+x^{-1}=C^{-1}$
$C=\frac{1}{3}$
$2y^2+x^2=3x^3$
$2y^2=x^2(3x-1)$
The particular solution by method 2 is also $2y^2=x^2(3x-1)$.
| $$(x^2+6y^2)dx-4xydy=0$$
$$(x^2+6y^2)dx-2xdy^2=0$$
$$x^2dx+2(3y^2dx-xdy^2)=0$$
Multiply by $x^2$:
$$x^4dx+2(3x^2y^2dx-x^3dy^2)=0$$
$$x^4dx+2(y^2dx^3-x^3dy^2)=0$$
Divide by $x^6$:
$$\dfrac {dx}{x^2}-2d\left (\dfrac {y^2}{x^3}\right)=0$$
Integrate.
$$\dfrac {1}{x}+2\left (\dfrac {y^2}{x^3}\right)=C$$
$$y(1)=1 \implies C=3$$
Therefore:
$$x^2+2y^2=3x^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
calculation $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{1}{k}\sin\frac{k\pi}{n+1}$. calculation$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{1}{k}\sin\frac{k\pi}{n+1}.$$
I try to use $|\sin x-x|\leqslant\frac{x^2}{2}$, but I don't feel like it's going to work. Is there any other way?
| Let $ n\in\mathbb{N}^{*} $, we have the following : \begin{aligned}\sum_{k=1}^{n}{\frac{1}{k}\sin{\left(\frac{k\pi}{n+1}\right)}}&=\frac{2}{n+1}\int_{0}^{\frac{\pi}{2}}{\sum_{k=1}^{n}{\cos{\left(\frac{2kx}{n+1}\right)}}\,\mathrm{d}x}\\ &=\frac{2}{n+1}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}\sin{\left(\frac{nx}{n+1}\right)}}{\sin{\left(\frac{x}{n+1}\right)}}\,\mathrm{d}x}\end{aligned}
For all $ x\in\left(0,\frac{\pi}{2}\right] $ : \begin{aligned}\left\vert\frac{\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)\sin{\left(\frac{x}{n+1}\right)}}-\frac{\sin{x}}{x}\right\vert&=\left\vert\frac{\left(x-\left(n+1\right)\sin{\left(\frac{x}{n+1}\right)}\right)\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)x\sin{\left(\frac{x}{n+1}\right)}}+\frac{\sin{\left(\frac{nx}{n+1}\right)}-\sin{x}}{x}\right\vert\\ &\leq\frac{\sin{\left(\frac{nx}{n+1}\right)}}{x\sin{\left(\frac{x}{n+1}\right)}}\left\vert\frac{x}{n+1}-\sin{\left(\frac{x}{n+1}\right)}\right\vert+\frac{\left\vert\sin{x}-\sin{\left(\frac{nx}{n+1}\right)}\right\vert}{x}\\ &\leq\frac{x^{2}\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)^{3}\sin{\left(\frac{x}{n+1}\right)}}+\frac{2\sin{\left(\frac{x}{2\left(n+1\right)}\right)}\cos{\left(\frac{\left(2n+1\right)x}{2\left(n+1\right)}\right)}}{x}\end{aligned}
Combining the inequalities : $ \left(\forall y\in\left[0,\frac{\pi}{2}\right]\right),\ \frac{2}{\pi}x\leq\sin{x}\leq x $, $ \left(\forall y\in\mathbb{R}\right),\ \cos{y}\leq 1 $ and $ \left(\forall y\in\left[0,\frac{\pi}{2}\right]\right),\ \sin{y}\leq 1 $, we get : \begin{aligned}\left\vert\frac{\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)\sin{\left(\frac{x}{n+1}\right)}}-\frac{\sin{x}}{x}\right\vert\leq\frac{x^{2}\times 1}{\left(n+1\right)^{3}\times\frac{2x}{\pi\left(n+1\right)}}+\frac{2\times\frac{x}{2\left(n+1\right)}\times 1}{x}&=\frac{\pi x}{2\left(n+1\right)^{2}}+\frac{1}{n+1}\\ &\leq\frac{n+1+\frac{\pi^{2}}{4}}{\left(n+1\right)^{2}}\end{aligned}
Thus : $$ \frac{1}{n+1}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}\sin{\left(\frac{nx}{n+1}\right)}}{\sin{\left(\frac{x}{n+1}\right)}}\,\mathrm{d}x}\underset{n\to +\infty}{\longrightarrow}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}\sin{x}}{x}\,\mathrm{d}x}=\frac{1}{2}\int_{0}^{\pi}{\frac{\sin{x}}{x}\,\mathrm{d}x} $$
Which means : $$ \lim_{n\to +\infty}{\sum_{k=1}^{n}{\frac{1}{k}\sin{\left(\frac{k\pi}{n+1}\right)}}}=\mathrm{Si}\left(\pi\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do you solve $5\sin^2(x)+8\sin x \cos x-3=0$? I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into $\sin 2x$ but then I have $\sin^2(x)$ and $\sin 2 x$ together.
| Rewrite it as $5\sin^2(x)+8\sin x \cos x-3(\sin^2 x + \cos^2 x)$, which gives $2 \sin^2 x + 8 \sin x \cos x - 3 \cos^2 x$, a quadratic in $\sin x$.
Using the quadratic formula, the roots of $2u^2 + 8u - 3$ are $-2 - \sqrt{11/2}$ and $-2 + \sqrt{11/2}$. Hence this is equivalent to $(u + 2 + \sqrt{11/2})(u + 2 - \sqrt{11/2}) = 0$, so the original factorises as $(\sin x + (2 + \sqrt{11/2})\cos x)(\sin x + (2 - \sqrt{11/2})\cos x) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24]=(z+x-7)[y(z+x)+24]$, find $x^2+y^2+z^2$
Let x, y, z be pairwise distinct real numbers, if $$(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24] $$ $$=(z+x-7)[y(z+x)+24]$$, find $x^2+y^2+z^2$
I've tried many ways but couldn't find a working way to solve it. I tried letting $(x+y-7)[z(x+y)+24] = k$ and $t = x + y + z$ but none of these gives useful transformation as far as I can see. Could somebody shed some lights on this? Thanks in advance.
| I write $u = y + z$, $v = z + x$, $w = x + y$ and $s = (u + v + w) / 2 (= x + y + z)$.
The equation becomes $$(u - 7)(us - u^2 + 24) = (v - 7)(vs - v^2 + 24) = (w - 7)(ws - w^2 + 24).$$
Taking the difference of the first and the second term, we get $$(u - v)((u + v - 7)s - (u^2 + uv + v^2 - 7(u + v) - 24)) = 0.$$
Since $x, y, z$ are all different, we know that $u, v, w$ are all different. This leads to $$(u + v - 7)s = u^2 + uv + v^2 - 7(u + v) - 24 \tag{1}$$ and symmetrically $$(v + w - 7)s = v^2 + vw + w^2 - 7(v + w) - 24.$$ Taking the difference of the above two identities, we get $$(u - w)s = (u - w)(u + w + v - 7)$$ which leads to $$s = 2s - 7$$ and thus $s = 7$.
Now (1) becomes $$u^2 + uv + v^2 - 14(u + v) + 25 = 0. \tag{2}$$ Taking the symmetric sum of (2), we get $$2\sum u^2 + \sum uv - 56s + 75 = 0. \tag{3}$$
If we write $t = \sum x^2$, then we have $\sum xy = (s^2 - t)/2 = (49 - t)/2$. It follows that $$\sum u^2 = \sum (y^2 + 2yz + z^2) = 2\sum x^2 + 2\sum xy = 49 + t$$ and $$\sum uv = ((2s)^2 - \sum u^2) / 2 = (147 - t) / 2.$$
Thus the equation (3) reads $$2(49 + t) + (147 - t) / 2 = 317$$ and we deduce that $t = 97$.
Finally, it is possible to give examples of $x, y, z$ satisfying these equalities.
E.g. $x = 3, y = 2 + 2\sqrt{10}, z = 2 - 2\sqrt{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4048713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Equality of two elliptic integrals by wolfram 1 and wolfram 2
It's true that $\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{2}{\sqrt{1-\sin^2\theta\sin^2\phi}} d\phi d\theta =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin\theta\sin\phi}} d\phi d\theta$ ?
If yes, I need a way to prove the equality of the following two integrals
I tried everything but I am unable to convert into a standard form so How do I solve this problem.
Addition 1: For the second integral
$\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin\theta\sin\phi}} d\phi d\theta =\bigg(\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin\theta}}d\theta\bigg)^2 $ and we can use that $2 \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx = \frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)} = \frac{\Gamma \left( \frac{1}{4}\right)^2}{\sqrt{2\pi}}$
Addition 2: For the first integral
Let $\displaystyle K(k)=\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1-k^2\sin^2 t}}dt$ ( Complete Elliptic Integral of the First Kind). we know that $ \displaystyle K(k)=\frac{\pi}{2}\sum_{n=0}^\infty \left(\frac{(2n)!}{2^{2n}(n!)^2}\right)^2k^{2n}$. Then it's not difficult de show that $\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^2\theta\sin^2\phi}} d\phi d\theta=(\pi /2 )^2 \sum _{n=0}^{\infty }(\frac{(2n)!}{4^n(n!)^2})^3$.
The egality of the two integrals hold if we can calculte $$\sum _{n=0}^{\infty }(\frac{(2n)!}{4^n(n!)^2})^3$$
Wolfram gives
Addition 3:
This link gives
| I wish i have found out the following solution (thanks go to Etanche and Jandri).
\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^2(\theta)\sin^2 \varphi}}d\varphi d\theta\\
&\overset{z\left(\varphi\right)=\arcsin\left(\sin(\theta)\sin \varphi\right)}=\int_0^{\frac{\pi}{2}} \left(\int_0^ \theta\frac{1}{\sqrt{\sin(\theta-z)\sin(\theta+ z)}}dz\right)d\theta\tag1\\
&=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du \tag2\\
&=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du+\underbrace{\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{\frac{\pi}{2}}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du}_{w=\pi-v}\\
&=\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\
&=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv-\int_0^{\frac{\pi}{2}} \left(\int_{0}^{u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\
&=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv-\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin u}}\left(\int_0^{u}\frac{1}{\sqrt{\sin v}}dv\right)du\\
&=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv-\frac{1}{2}\left(\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin v}}dv\right)^2\tag3\\
&=\boxed{\frac{1}{2}\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv}
\end{align}
$(1)$: $\displaystyle dz=\dfrac{\sqrt{\sin^2\theta-\sin^2 z}}{\sqrt{1-\sin^2 z}}d\varphi$, $\sin^2 a-\sin^2 b=\sin(a-b)\sin(a+b)$
$(2)$: Perform the change of variable $u=\theta-z,v=\theta+z$
$(3)$: $\displaystyle\int_0^{\frac{\pi}{2}} f(x)f^\prime(x)dx=\frac{1}{2}\left(f^2\left(\frac{\pi}{2}\right)-f^2(0)\right)$ and $\displaystyle f(x)=\int_0^{x}\frac{1}{\sqrt{\sin u}}du$
(Edit: problem fixed again)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Prove $\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}≥\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Question : Prove $$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
$(a, b, c \in \mathbb{R}^+)$
I tried to solve it like this :
$$\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \; (\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2})$$
Am I doing this right? How can I finish this problem?
| There are million ways to prove this but the most immediate one would be the Rearrangement Inequality.
No matter the order of $a,b,c,$ the triplets $\left(\dfrac{1}{a^3}, \dfrac{1}{b^3}, \dfrac{1}{c^3}\right)$ and $\left(a^2, b^2, c^2\right)$ have the reverse arrangement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
What is the domain of k? $2|x-2|-3|x-3|=k$ What is the domain of k?
$2|x-2|-3|x-3|=k$
Using the definition of absolute function, I obtain:
$|x-2|= \begin{cases}
x-2, & x\ge2 \\
2-x, & x<2\\
\end{cases}\\$
$|x-3|= \begin{cases}
x-3, & x\ge3 \\
3-x, & x<3 \\
\end{cases} \\$
For $x<2$
$x-5=k$
For $2\le x<3$
$5x-13=k$
For $x \ge 3$
$-x+5=k$
But I need to find the domain of k.
| From the $3$ rules that you worked out, the graph of $2|x-2|-3|x-3|$ increases on $(-\infty, 3]$ and then decreases from $[3, \infty)$.
Hence the possible values of $k$ are $(-\infty, f(3)) = (-\infty, 2]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find sum of series $S = 1 + 3x + 5x^2 + ... + (2n + 1)x^n$, using $(1-x)S$ I study maths as a hobby. I am stuck on the following problem, which is an A level question from the Oxford board set probably in the '70s:
If $S = 1 + 3x + 5x^2 + ... + (2n + 1)x^n$, prove, by considering $(1 - x)S$ or otherwise, that if $x \ne 1$,
$$S = \frac{1 + x - (2n + 3)x^{n + 1} + (2n + 1)x^{n + 2}}{(1 - x^2)}$$
Firstly, I am puzzled that the nth term of S is given as $(2n + 1)x^n$.
For example, its 3rd term, $5x^2$ should surely be $7x^3$. Or does the numbering of n start at $n = 0$?
However,
$S = 1 + 3x + 5x^2 + ... + (2n + 1)x^n$
$Sx = x + 3x^2 + 5x^3 + ... + (2n + 1)x^{n + 1}$
$(1 - x)S = 1 + 2x + 2x^2 + ...+ (2n +1)x^{n+1}$
But I don't see how to get from this to the needed proof.
| The numbering starts at $n = 0$.
Your evaluation is not quite complete. Instead,
$$\begin{align}
S &= 1 + 3x + 5x^2 + 7x^3 + \cdots + \hphantom{(\vphantom{} - 1)}(2n + 1)x^n \\
xS &= \hphantom{1 + 3}x + 3x^2 + 5x^3 + \cdots + (2(n-1)+1) x^n + (2n + 1)x^{n+1} \\
(1-x)S &= 1 + 2x + 2x^2 + 2x^3 + \cdots + \hphantom{((n-1)+1)}2x^n \color{red}{- (2n+1)x^{n+1}}.
\end{align}$$
Therefore, $$(1-x)S + 1 + (2n+1)x^{n+1} = 2 + 2x + 2x^3 + \cdots + 2x^n = 2(1 + x + x^2 + \cdots + x^n).$$ Since we know that the sum of a finite geometric series with common ratio $x$ is $$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1} - 1}{x - 1},$$ it follows that $$(1-x)S = \frac{2(x^{n+1} - 1)}{x-1} - 1 - (2n+1)x^{n+1},$$ and dividing by $1-x$ and simplifying gives the desired value of $S$.
For the OP's benefit:
$$\begin{align}
S &= \frac{1}{1-x} \left( \frac{2(x^{n+1} - 1)}{x-1} - 1 - (2n+1)x^{n+1} \right) \\
&= -\frac{2(x^{n+1} - 1) - (x-1) - (2n+1)x^{n+1}(x-1)}{(x-1)^2} \\
&= -\frac{2x^{n+1} - 2 - x + 1 - (2n+1) x^{n+2} + (2n+1) x^{n+1}}{(x-1)^2} \\
&= \frac{(2n+1) x^{n+2} - (2n+3)x^{n+1} + x + 1}{(x-1)^2}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why do this series converge to 2 different values? Firstly, I consider the Fourier series of the $2\pi$-periodic function $f=\cos{\frac{3}{2}\pi}$ if $-\pi\leq x\leq \pi.$ And I get $$\cos\frac{3}{2}t \sim -\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nt.$$ Since I find that $f\in C^2(T),$ i.e. $f,f’$ and $f’'$ are continuous on $T,$ then $$\cos\frac{3}{2}t = -\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nt \text{ (converge uniformly)}.$$
Let $x=\frac{\pi}{2},$ then
\begin{align}
\cos\frac{3}{2}(\frac{\pi}{2})&=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nx=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos n\frac{\pi}{2}\notag\\
&=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{2n+1} }{(\frac{9}{4}-(2n)^2)\pi}(-1)^n=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi}.\notag
\end{align}
Let $x=\frac{3}{2}\pi,$ then
\begin{align}
\cos\frac{3}{2}(\frac{3}{2}\pi)&=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos nx=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-n^2)\pi} \cos n\frac{3}{2}\pi\notag\\
&=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{2n+1} }{(\frac{9}{4}-(2n)^2)\pi}(-1)^n=-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi}.\notag
\end{align}
I find that these two same series $$-\frac{2}{3\pi}+\sum_{n=1}^\infty \frac{3(-1)^{n+1} }{(\frac{9}{4}-(2n)^2)\pi} $$ converge to 2 different values, which are $$\cos\frac{3}{2}(\frac{\pi}{2}) \text{ and } \cos\frac{3}{2}(\frac{3}{2}\pi).$$ Did I make any mistake? Thanks~
| No, but your function is not $cos(\frac{3x}{2})$ as such, it is a periodic extension of that defined in $-\pi$ to $\pi$. If we call this function $f(x)$ then you can see in the graph below that $f(\frac{\pi}{2})$ and $f(\frac{3 \pi}{2})$ are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute close-form of $\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}$ I came across the improper trigonometric integral recently shared in a Chinese web forum
\begin{align}
\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}\\
\end{align}
which amuses me because of its appearance. What is more amusing is the claim, without providing the proof, that is has the close-form result
\begin{align}
\frac{\sqrt{8\sqrt2+2\sqrt7}}{7^{3/4} }\tanh^{-1} \frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 +\sqrt7}
+\frac{\sqrt{8\sqrt2-2\sqrt7}}{7^{3/4} }\tan^{-1}\frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 - \sqrt7}
\end{align}
I was able to verify it numerically; yet rather curious in how it could ever be derived. Based on my knowledge and experience, I do not assume it would be easy.
I was able to find an unsolved post here from a long time ago, which only reexpresses the corresponding indefinite integral via the tangent half-angle substitution as
$$\int\frac{2t(1-t)}
{2 t^4-3 t^3+3 t^2+t+1}dt$$
but gives up due to the complexity in partial fractionalization. I am doubtful that this approach would lead to the explicit expression claimed above. I am interested in any suitable methods producing the close-form.
|
Evaluate
$$\int_0^{\pi / 2} \frac{dx}{\sin x + \cos x + \tan x + \sec x + \csc x + \cot x} .$$
As in the linked question, applying the classical Weierstrass substitution, $$x = 2 \arctan t, \qquad dx = \frac{2 \,dt}{t^2 + 1},$$ transforms the integral to
$$
\int_0^1 \frac{2 t (1 - t)}{2 t^4 - 3 t^3 + 3 t^2 + t + 1}.
$$
One can show that the Galois group of the polynomial $$f(t) := 2 t^4 - 3 t^2 + 3 t^2 + t + 1$$ in the denominator of the integrand
is $D_8$, suggesting applying a transformation that takes advantage of that symmetry.*
The following shows one way of doing so, in particular transforming the integrand to one to which the Method of Partial Fractions can be applied somewhat more easily than the one in $t$. I don't know whether the particular substitution used here is in any sense the nicest choice.
Let $\alpha := \cot \frac{3 \pi}{8} = \sqrt{2} - 1$. Applying to the original integral the substitution
$$x = \frac{\pi}{4} - 2 \arctan (\alpha u), \qquad dx = -\frac{2 \alpha \,du}{\alpha^2 u^2 + 1} ,$$
simplifying considerably, and taking advantage of the evenness of the integration in $u$ gives that the integral is equal to
$$\frac{2 (8 + 5 \sqrt{2})}{7} \int_0^1 \frac{1 - u^2}{u^4 + \beta^4} du , \qquad \textrm{where } \beta := \frac{\sqrt{5 + 4 \sqrt{2}}}{\sqrt[4]{7}} .$$ Remark 1 The substitution $x \rightsquigarrow u$ is related to the Weierstrass substitution, $x \rightsquigarrow t$ by the linear fractional transformation $$t = \frac{\alpha (1 - u)}{\alpha^2 u + 1} .$$
Remark 2 Perhaps despite appearances the substitution $x \rightsquigarrow u$ is not particularly clever: It is the composition of a translation that centers the domain of integration on the origin, the classical Weierstrass substitution, and a dilation to make the coefficients and limits in the resulting integrand nicer.
Over $\Bbb Q(\beta) = \Bbb Q(\beta, \sqrt{2})$ (or just $\Bbb R$), the denominator of the integrand in $u$ factors into irreducible polynomials as
$$u^4 + \beta^4 = (u^2 + \sqrt{2} \beta u + \beta^2) (u^2 - \sqrt{2} \beta u + \beta^2),$$ so the rest of the computation can be handled with standard techniques: Applying the Method of Partial Fractions gives the decomposition
$$\frac{1 - u^2}{u^4 + \beta^4} = \frac{A u + B}{u^2 + \sqrt{2} \beta u + \beta^2} + \frac{C u + D}{u^2 - \sqrt{2} \beta u + \beta^2} .$$
Evenness of the integrand implies that $C = -A, D = B$, reducing the equation in the unknown coefficients to a $2 \times 2$ system in $A, B$, and some elementary algebra gives
$$A = -\frac{\beta^2 + 1}{2 \sqrt{2} \beta^3}, \qquad B = -\frac{1}{\beta^2} .$$
Separating each summand into the sum of a scalar multiple of $$\frac{u}{u^2 \pm \sqrt{2} \beta u + \beta^2} \qquad \textrm{and one of} \qquad \frac{1}{u^2 \pm \sqrt{2} \beta u + \beta^2} ,$$ and integrating respectively gives antiderivatives
$$\frac{1}{2} \log \left[\sqrt{2} \beta u \pm (\beta^2 + u^2)\right] \qquad \textrm{and} \qquad \arctan \left(1 \pm \frac{\sqrt{2} u}{\beta}\right) .$$ We can combine the two $\log$ terms with the identity $\operatorname{artanh} x = \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)$, yielding
$$\operatorname{artanh} \left(\frac{u^2 + \beta^2}{\sqrt{2} \beta u}\right),$$
and we can combine the two $\arctan$ terms with the arctan sum identity, $\arctan x + \arctan y = \arctan \left(\frac{x + y}{1 - xy}\right)$, yielding
$$
\arctan \left(\frac{\beta^2}{u^2}\right) .
$$
Putting this all together gives essentially the given exact form, and a numerical evaluation suggests that the expressions indeed agree, with value $0.1760244214\ldots$.
*The discriminant of the irreducible quartic $f$ is $\Delta := 2^5 7^3$; now, (a) $f$ remains irreducible over $\Bbb Q(\sqrt{\Delta}) = \Bbb Q(\sqrt{14})$, and (b) the cubic resolvent of $f$ factors over $\Bbb Q$ as a product of an irreducible quartic and a linear polynomial (up to a scalar multiple, $(2 x − 1) (4 x^2 − 4 x − 13)$), and these two facts together imply that $\operatorname{Gal}(f) \cong D_8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 3
} |
Does $\int_0^{2\pi} \sqrt{(1-a\cos x)^2 + b} \mathrm{d}x$ evaluate in terms of elliptic integrals? Consider the integral
$$\int_0^{2\pi} \mathrm{d}x \sqrt{(1-a\cos x)^2 + b}$$
with $a,b>0$. I believe I can somehow evaluate this integral by incomplete elliptic integrals. This would be obvious, for example, if $a=1$. But, in the general case, this integral appears rather challenging. Mathematica, and SymPy have not been useful. And the Weierstrass substitution returns polynomials of 4th degree.
| Here is a step-by-step solution where each step is not too hard.
Step 1: Instead of doing a Weierstrass substitution, substitute $t=\cos x$ and take factors out to get
$$I=2a\int_{-1}^1\frac{\sqrt{(t-1/a)^2+b/a^2}}{\sqrt{1-t^2}}\,dt$$
The roots of the upper quadratic are always complex since $a,b>0$: $c=\frac{1+\sqrt{-b}}a$ and $c^*$.
$$I=2a\int_{-1}^1\frac{\sqrt{(t-c)(t-c^*)}}{\sqrt{1-t^2}}\,dt$$
Step 2: Perform a linear fractional transformation that keeps the integrand's poles at $\pm1$ but shifts the zeros at $c$ and $c^*$ to the imaginary axis, so as to leave only even powers of the polynomial under root. This is the most involved part. Define$\newcommand{Re}{\operatorname{Re}}$
$$A=\frac{1+|c|^2+|c^2-1|}{2\Re(c)}>1$$
$$A_1=A^2-2A\Re(c)+|c|^2>0\qquad A_2=A^2|c|^2-2A\Re(c)+1>0\qquad B=\frac{A_2}{A_1}>0$$
The actual substitution is $u=\frac{At+1}{t+A}$. After taking out more factors
$$I=2a\sqrt{A_1(A^2-1)}\int_{-1}^1\frac1{(u+A)^2}\sqrt{\frac{u^2+B}{1-u^2}}\,du$$
Define $g=2a\sqrt{A_1(A^2-1)}$ and move on.
Step 3: Multiply top and bottom by $(u-A)^2\sqrt{u^2+B}$ to get
$$I=g\int_{-1}^1\frac{(u-A)^2(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du$$
$$=g\left(\int_{-1}^1\frac{(u^2+A^2)(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du-A\int_{-1}^1\frac{(2u)(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du\right)$$
The two integrands are even and odd about zero respectively, so
$$I=2g\int_0^1\frac{(u^2+A^2)(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du$$
We perform a partial fraction decomposition of the rational part of the integrand:
$$I=2g\int_0^1\left(1-\frac{3A^2+B}{A^2-u^2}+\frac{2A^2(A^2+B)}{(A^2-u^2)^2}\right)\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du$$
Step 4: Finally we use elliptic integrals (all arguments here follow Mathematica/mpmath conventions). Byrd and Friedman 213.11 gives$\newcommand{sn}{\operatorname{sn}}$
$$I=\frac{2g}{\sqrt{1+B}}\left(V_0-\frac{3A^2+B}{A^2-1}V_1+\frac{2A^2(A^2+B)}{(A^2-1)^2}V_2\right)$$
where
$$V_k=\int_0^{K(m)}\frac1{(1-n\sn^2u)^k}\,du\qquad m=\frac1{B+1}\qquad n=\frac1{1-A^2}$$
Step 5: B&F 336.00, .01, .02 gives solutions for the $V_k$.
$$V_0=K(m)\qquad V_1=\Pi(n,m)$$
$$V_2=\frac{nE(m) + (m-n)K(m) + (2nm+2n-n^2-3m)\Pi(n,m)}{2(n-1)(m-n)}$$
Putting everything together and simplifying a lot we get the final answer.
$$I=4a\sqrt{-n(A_1+A_2)}(E(m)-K(m)+(1-n)\Pi(n,m))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Problem 6.16 Ordinary Differential Equations and Dynamical Systems Teschl While reading the book Ordinary Differential Equations and Dynamical Systems by Gerald Teschl, I got stuck at Problem 6.16. It states:
Consider the system
\begin{eqnarray*}
\dot{x} = x-y-x(x^2+y^2)+\dfrac{xy}{\sqrt{x^2+y^2}} \\
\dot{y} = x+y-y(x^2+y^2)-\dfrac{x^2}{\sqrt{x^2+y^2}}.
\end{eqnarray*}
Show that $(1,0)$ is not stable even though $\lim\limits_{t \to +\infty} |\phi(t,x) - x_0| = 0$.
I was able to show the second part, but i could not show that $(1,0)$ is not stable. By plotting the system on mathlab I could see that it isn't stable, but how do I show it using the definition of stability?
| Hint.
From
\begin{eqnarray*}
\dot{x} = x-y-x(x^2+y^2)+\dfrac{xy}{\sqrt{x^2+y^2}} \\
\dot{y} = x+y-y(x^2+y^2)-\dfrac{x^2}{\sqrt{x^2+y^2}}.
\end{eqnarray*}
we have
\begin{eqnarray*}
x\dot{x} = x^2-xy-x^2(x^2+y^2)+\dfrac{x^2y}{\sqrt{x^2+y^2}} \\
y\dot{y} = xy+y^2-y^2(x^2+y^2)-\dfrac{x^2y}{\sqrt{x^2+y^2}}.
\end{eqnarray*}
and after addition
$$
(x^2+y^2)'=(x^2+y^2)(1-(x^2+y^2))
$$
or making $r=\sqrt{x^2+y^2}$
$$
\dot r = r(1-r^2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4061037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
On the asymptotic behavior of Elliptic integral near $k=1$ I was trying to prove that as $k\to1^-$,
$$
\int_0^1{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}}\sim\frac12\log{1\over1-k}
$$
Since the original integral is difficult to work with, I split it:
$$
\int_0^1{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}}=\int_0^k+\int_k^1
$$
in which it is evident that
$$
\begin{aligned}
\int_0^k{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}}
&=\int_0^k{\mathrm dx\over1-x^2}\cdot{\sqrt{1-x^2}\over\sqrt{1-k^2x^2}} \\
&=\frac12\int_0^k{\mathrm dx\over1-x}\cdot{\sqrt{1-x^2}\over\sqrt{1-k^2x^2}}+\mathcal O(1) \\
&=\frac12\int_0^k{\mathrm dx\over1-x}+\mathcal O\left(\int_0^k{1-k\over1-x}\mathrm dx\right)+\mathcal O(1) \\
&=\frac12\log{1\over1-k}+\mathcal O(1)
\end{aligned}
$$
I wonder how to show that $\int_k^1$ converges to zero as $k\to1^-$ so that I can complete the proof.
P.S. Alternative methods are welcome since I have been stuck on this for so much time :D
| If $0<k<x<1$, then
\begin{align*}
\sqrt {1 - x^2 } \sqrt {1 - k^2 x^2 } = \sqrt {1 - x} \sqrt {1 + x} \sqrt {1 - kx} \sqrt {1 + kx} \ge \sqrt {1 - x} \sqrt {1 - k} .
\end{align*}
Hence,
$$
0 \leq \int_k^1 {\frac{{dx}}{{\sqrt {1 - x^2 } \sqrt {1 - k^2 x^2 } }}} \le \frac{1}{{\sqrt {1 - k} }}\int_k^1 {\frac{{dx}}{{\sqrt {1 - x} }}} = 2.
$$
This bound together with your estimate for the integral between $0$ and $k$ imply the desired asymptotics.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Can you solve $\int 1/(x^2+1)^2\, dx$ I know that the integral looks like like the anti-derivative of $\arctan$, but i don't know how to use this fact so I tried to use a fraction decomposition to: $$\frac{1}{(x^2+1)^2}$$
to solve the anti-derivative : $$\int \frac{1}{(x^2+1)^2}\, dx$$
so :$$\frac{1}{(x^2+1)^2}=\frac{A}{x^2+1}+\frac{B}{x^2+1}\iff \frac{1}{x^2+1}=A+B$$
$$1+0x^2=(A+B)x^2+(A+B)1$$
Thus I got this system of equations:
$$\cases{A+B=1 \\ A+B=0}$$ Which is impossible. I also tried to use substitution but it didn't help me.
| Integrate as follows
\begin{align}
\int {\dfrac{1}{(x^2+1)^2}}\,dx
&= \int {\dfrac{1}{x^2+1}}\,dx - \int {\dfrac{x^2}{(x^2+1)^2}}\,dx\\
&= \int {\dfrac{1}{x^2+1}}\,dx + \frac1{2}\int x\>d\left({\dfrac{1}{x^2+1}}\right)\\
&= \frac1{2} \frac x{x^2+1} + \frac1{2}\int {\dfrac{1}{x^2+1}}\,dx\\
&= \frac1{2} \frac x{x^2+1} + \frac1{2}\tan^{-1} x+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4065559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
why is the gradient vector to rotated ellipse wrong? A formula for the rotated ellipse can be found by taking the standard ellipse
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad\qquad\qquad\qquad {\bf{x}}(t) = (a\cos t, b\sin t) \qquad \text{ for } t\in [0, 2\pi]
$$
and rotating it counter-clockwise using a rotation matrix
$$
R{\bf{x}}(t) =
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
a\cos t \\
b\sin t
\end{pmatrix}
=
\begin{pmatrix}
a\cos t\cos\theta - b\sin t\sin \theta\\
a\cos t\sin \theta + b \sin t \cos \theta
\end{pmatrix}
=
{\bf{x}}_r(t)
$$
I can easily find the gradient of the standard ellipse (non-rotated) by taking the derivative of the first formula.
$$
g =
\begin{pmatrix}
\frac{2x}{a^2}, &
\frac{2y}{b^2}
\end{pmatrix}^\top
$$
Now I tried to methods to compute the gradient at a point on the rotated ellipse.
*
*Method 1: I take the gradient of
$$
\frac{(x\cos\theta - y\sin\theta)^2}{a^2} + \frac{(x\sin\theta + y\cos\theta)^2}{b^2} = 1
$$
obtaining
$$
g_{r1} =
\begin{pmatrix}
\frac{2(x\cos\theta - y\sin\theta)\cos\theta}{a^2} + \frac{2(x\sin\theta + y\cos\theta)\sin\theta}{b^2} \\
\frac{-2(x\cos\theta - y\sin\theta)\sin\theta}{a^2} + \frac{2(x\sin\theta + y\cos\theta)\cos\theta}{b^2}
\end{pmatrix}
$$
*Method 2: Rotating the gradient of the non-rotated ellipse counterclockwise
$$
g_{r2} = \begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
\frac{2x}{a^2} \\
\frac{2y}{b^2}
\end{pmatrix}
=
\begin{pmatrix}
\frac{2x\cos\theta}{a^2} - \frac{2y\sin\theta}{b^2}\\
\frac{2x\sin\theta}{a^2} + \frac{2y\cos\theta}{b^2}
\end{pmatrix}
$$
To me, method 1 should work while method 2 should be wrong. However, as you can see in the plot below, $g_{r1}$ seems to actually be totally wrong while $g_{r2}$ is right. How come $g_{r2}$ seems to be right but not $g_{r1}$?
| Method 1 is wrong because the equation
$$
\frac{(x\cos\theta - y\sin\theta)^2}{a^2} + \frac{(x\sin\theta + y\cos\theta)^2}{b^2} = 1
$$
expresses the fact that the result of rotating the point $(x,y)$ belongs to the ellipse $x^2/a^2+y^2/b^2=1$, but this is not what you need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Uniform Convergence of $f_n(x) = \frac{n(1-2x) + \cos x -x\sin^2 x}{2n + \sin^2 x}$ [proof verification] As the title suggests, I'm trying to show that $f_n(x) = \frac{n(1-2x) + \cos x -x\sin^2 x}{2n + \sin^2 x}$ converges uniformly. I tried showing that $f_n(x)$ converges uniformly to the pointwise limit $\frac{1-2x}{2}$ but I couldn't get it to work out, so I've tried to show it's uniformly Cauchy like so:
Let $f_n(x) = \frac{n(1-2x) + \cos x -x\sin^2 x}{2n + \sin^2 x}$.
Let $m, n \in \mathbb{N}$ such that $m > n$. Consider
$$|f_n(x) - f_m(x)| = |\frac{(m-n)(2\cos(x) - \sin^2(x))}{(2m+\sin^2(x))(2m+\sin^2(x))}| \leq \frac{|m-n|(2)}{4mn} = \frac{|n - m|}{2mn} < \frac{n}{2n^2} < \frac{1}{2n} < \epsilon$$
So let $\epsilon > 0$.
Choose $N \in \mathbb{N}$ such that $N > \frac{1}{2\epsilon}$.
Then for all $m > n > N$,
$$|f_n(x) - f_m(x)| < \epsilon$$
So $f_n$ is uniformly Cauchy, so it must converge uniformly.
But after checking this result graphically I think that it's wrong, and I'm not really sure what else to do.
Any help would be greatly appreciated.
| The convergence is uniform for all $x \in \mathbb{R}$.
The pointwise limit is clearly $f(x) =(1-2x)/2$ and proceeding directly,
$$|f_n(x) - f(x) | = \left|\frac{\frac{1-2x}{2} + \frac{\cos x}{2n}- \frac{x\sin^2x}{2n}}{1 + \frac{\sin^2x}{2n}}- \frac{1-2x}{2} \right| \\ = \frac{\left|\frac{1-2x}{2} + \frac{\cos x}{2n}- \frac{x\sin^2x}{2n}- \frac{1-2x}{2}- \frac{\sin^2x}{2n}\frac{1}{2} + \frac{\sin^2 x}{2n} \frac{2x}{2}\right|}{1 + \frac{\sin^2x}{2n}} \\ \leqslant \left|\frac{\cos x}{2n}- \frac{\sin^2x}{4n} \right|\leqslant \frac{3}{4n}\underset{n \to \infty}\longrightarrow 0$$
Checking your approach, with $m >n$, we have
$$f_m(x) - f_n(x) \\=\frac{m(1-2x)2n + m(1-2x)\sin^2x + (\cos x - x\sin^2x)2n + (\cos x - x\sin^2x)\sin^2 x}{(2m+\sin^2x)(2n+ \sin^2 x)} \\ -\frac{n(1-2x)2m + n(1-2x)\sin^2x + (\cos x - x\sin^2x)2m + (\cos x - x\sin^2x)\sin^2 x}{(2m+\sin^2x)(2n+ \sin^2 x)} \\ = \frac{(m-n)(1-2x)\sin^2 x +2(n-m)(\cos x - x \sin^2 x)}{(2m+\sin^2x)(2n+ \sin^2 x)}\\ = \frac{(m-n)(\sin^2 x -2\cos x)}{(2m+\sin^2x)(2n+ \sin^2 x)},$$
and, thus,
$$|f_m(x) - f_n(x)| = \frac{(m-n)|\sin^2 x -2\cos x|}{(2m+\sin^2x)(2n+ \sin^2 x)}\leqslant \frac{3(m-n)}{4mn} = \frac{3}{4n} - \frac{3}{4m} < \frac{3}{4n}$$
For all $m > n > N(\epsilon)$ where $N(\epsilon) > \frac{4}{3\epsilon}$ we have $|f_m(x) - f_n(x)| < \epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating this Riemann sum $$\lim_{x\rightarrow\infty}\sum_{k=1}^{n}\frac{6(k+1)^2}{n^3}\sqrt{1+\frac{2(k+1)^3}{n^3}}$$
How do we determine if it is a Right/Left or midpoint Riemann Sum?
How do we find the values a and b?
Is $$f(x)=\frac{3x^2}{2}\sqrt{1+\frac{x^3}{4}}$$ possible? if $$x=\frac{2(k-1)}{n}$$
But then again how do I find the interval for which I have to integrate?
| $$
\sum_{k=1}^{n}\frac{6\left(k+1\right)^2}{n^3}\sqrt{1+\frac{2\left(k+1\right)^3}{n^3}}=\sum_{k=2}^{n+1}6\frac{k^2}{n^3}\sqrt{1+2\frac{k^3}{n^3}}
$$
You need to have only an expression with $\frac{k}{n}$ to idenfity with $f\left(\frac{k}{n}\right)$. So you write it as
$$
\frac{1}{n}\sum_{k=2}^{n+1}6\frac{k^2}{n^2}\sqrt{1+2\frac{k^3}{n^3}}
$$
Can you identify $a$, $b$ and $f$ now ? (notice also that's it starts at $k=2$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of possible solutions to $2 \le |x-1||y+3| \le 5$, where $x$ and $y$ are negative integers
If $$2 \le |x-1||y+3| \le 5$$ and both $x$ and $y$ are negative integers, find the number of possible combinations of $x$ and $y$ .
Below is my solution approach :-
As $x$ is a negative integer, hence $|x-1|$ in the $2 \le |x-1||y+3| \le 5$ will be come $-(x-1)$ or $(1-x)$ and the main equation will transform into $2 \le (1-x)|y+3| \le 5$.
$1st$ case when $y+3 \ge 0 \Rightarrow y \ge -3 \Rightarrow y \in \{{-3,-2,-1}\} $ as $y$ is a negative integer :
For $y=-3$ we can see that there is no valid solution for $x$ as $|y+3|$ part will become $0$, hence this case is invalid.
For $y=-2$ we get the solution set for $x$ to be $x \in \{{-4,-3,-2,-1}\} $ and total number of solutions possible in this case is $4$.
For $y=-1$ we get the solution set for $x$ to be $x \in \{{-1}\} $ and total number of solutions possible in this case is $1$.
So for this $1st$ case when $y+3 \ge 0$ we have in total 5 solutions.
$2nd$ case when $y+3 \lt 0 \Rightarrow y \lt -3$ and in this case $y$ will have infinite values and I am not able to proceed from here.
The answer for the total number of solutions provided is $10$ and you can see that I've been able to find out the $5$ solutions in my $1st$ case. Can someone please guide or help me about how to proceed in the 2nd case?
Thanks in advance !
| It is not true that $y$ can take infinitely many numbers.
We have since $x \le -1$, $|x-1|\ge 2, $
$$\frac1{|x-1|}\le \frac12$$
we must have
$$|y+3| \le \frac{5}{|x+1|}\le \frac52$$
If $y< -3$, then we have $$-y-3 \le \frac52 $$
$$-\frac{11}2 \le y $$
Hence $-5 \le y < -3$, that is $-5 \le y \le -4$.
For $y=-4=-3-1$, by symmetry, the same $x$ for which $y=-3+1=-2$ satisfies the inequality would work, there are $4$ of them.
For $y=-5=-3-2$, by symmetry, the same $x$ for which $y=-3+2=-1$ satisfies the inequlaity would work, there is $1$ of them.
Hence in total, there are $10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\prod_{k=1}^m \tan \frac{k\pi}{2m+1}$
How to evaluate this?
$$\prod_{k=1}^m \tan \frac{k\pi}{2m+1}$$
My work
I couldn't figure out a method to solve this product.
I thought that this identity could help.
$$\frac{e^{i\theta}-1}{e^{i\theta}+1}=i\tan \frac{\theta}{2}$$
By supposing $z=e^{\frac{2k\pi}{2m+1}i}$, then,
$$i\tan \frac{k\pi}{2m+1}=\frac{z-1}{z+1}$$
So, the product
$$\displaystyle\prod_{k=1}^m \tan\frac{k\pi}{2m+1}=\displaystyle\prod_{k=1}^m \frac{z-1}{i(z+1)}=\displaystyle\prod_{k=1}^m \frac{e^{\frac{2k\pi}{2m+1}i}-1}{i(e^{\frac{2k\pi}{2m+1}i}+1)}$$
which is getting more complicated.
Answer is $\sqrt{2m+1}$. Any help is appreciated.
| Let
$$f(x) = x^{n-1} + x^{n-2} + x^{n-3} \>\cdots \> +\>x +1
= \prod_{k=1}^{n-1}(x - e^{i\frac{2\pi k}n})
$$
and set $n=2m+1$ to evaluate
$$\frac{f(1)}{f(-1)}= 2m+1
= \prod_{k=1}^{2m}\frac{1 - e^{i\frac{2\pi k}{2m+1}}}{1 + e^{i\frac{2\pi k}{2m+1}}}
= i^{2m} \prod_{k=1}^{2m}\tan \frac{k\pi}{2m+1}
= \prod_{k=1}^{m}\tan^2\frac{k\pi}{2m+1}
$$
Thus
$$\prod_{k=1}^m \tan \frac{k\pi}{2m+1}=\sqrt{2m+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Domain of $\frac{x^2-3x-10}{x+2}\div\frac{x^2-25}{x}$ I saw a question asked to simplify
$\dfrac{x^2-3x-10}{x+2}\div\dfrac{x^2-25}{x}$
It is not so hard to do : $$\dfrac{(x-5)(x+2)}{x+2}\times\dfrac{x}{(x-5)(x+5)}=\dfrac x{x+5}$$
But my question is: should we have necessary all the conditions $x\ne-2$ , $x\ne5$ , $x\ne-5$ and also $x\ne0$ (because in former fraction we had $x$ in denominator) ? or we should have some of them?
| $\dfrac{x^2-3x-10}{x+2}\div\dfrac{x^2-25}{x}$ is our expression, so we can clearly see that $x \ne 0,-2. $ Also, we cannot have $x=5,-5$ because that would make the $2$nd fraction $0$, and we can't divide by zero. Thus, our restrictions are $x \ne -5,-2,0,5$. These restrictions must be kept throughout the whole process, or else our end product will be different from the start product. For example, $x^2$ and $x^2, x\ne 0$ are different graphs because the $2$nd one has a hole in the middle.
A more thorough explanation:
$$\dfrac{x^2-3x-10}{x+2}\div\dfrac{x^2-25}{x} = {\dfrac{x^2-3x-10}{x+2}\over\dfrac{x^2-25}{x}}$$
If we multiply the top and bottom by $x(x+2)$, you have to note that $\frac aa = 1$ is only true if $a \ne 0$, so we have to keep the restriction that $x \ne -2,0$
Thus, we get $$ {(x^2-3x-10)(x)\over(x^2-25)(x+2)}$$
Which becomes $$(x-5)(x+2)(x) \over (x-5)(x+5)(x+2)$$
When we simplify the ${(x-5)(x+2) \over (x-5)(x+2)}=1$ part, we have to note again that $\frac aa = 1 $ only when $a \ne 0$, so we have to keep the restriction that $x \ne -2,5$. Thus, we have finally simplified it down to $\frac{x}{x+5}$, which can't have $x=-5$. Thus, we have simplified, keeping all the restrictions that $x \ne 0,-2,-5,-5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$ \int\limits_{-\infty}^{+\infty} \frac{2}{(x-1)\cdot(x^2-6x+10)}\,\mathrm{d}\,x$ how to approach such types of problems as improper integral?
$$ \int\limits_{-\infty}^{+\infty} \frac{2}{(x-1)\cdot(x^2-6x+10)}\,\mathrm{d}\,x$$
is it also solve by complex theory integration?
| Your integral is$$\int_{-\infty}^\infty\left(\tfrac{2}{5(x-1)}-\tfrac{2(x-5)}{5(x^2-6x+10)}\right)dx=\left[\tfrac15\ln\tfrac{(x-1)^2}{x^2-6x+10}+\tfrac45\arctan(x-3)\right]_{-\infty}^\infty.$$As $x\to\pm\infty$, the numerator and denominator of $\tfrac{(x-1)^2}{x^2-6x+10}$ are both asymptotic to $x^2$, so the logarithm $\to\ln1=0$, and the integral is $\tfrac45(\tfrac{\pi}{2}-\tfrac{-\pi}{2})=\tfrac{4\pi}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Cambridge Admissions Exam: Where is the mistake? I am trying to solve the following problem.
My work:
Note: I am having only issues with (ii) but I have included my work for (i).
(i) I got the velocity components as
$$ v_x=v\cos \theta \tag{Horizontal Velocity} $$
$$v_y=v\sin\theta - tg \tag{Vertical Velocity}$$
We have $$ \tan \phi= \frac{v_y}{v_x}=\tan \theta -\frac{tg}{v \cos \theta}.$$
From the velocity components, we get the displacement components to be
$$s_x=vt\cos \theta \tag{Horizontal Displacement}$$
$$s_y=vt\sin \theta - t^2\frac{g}{2} \tag{Vertical Displacement}$$
By definition,
$$\tan \alpha = \frac{s_y}{s_x}=\tan \theta - \frac{tg}{2v\cos \theta} \tag{$*$}$$
Equating the two expressions through $\frac{tg}{v\cos\theta}$ gives
$$\tan \theta + \tan \phi = 2\tan \alpha$$
as desired.
(ii) Here is where I start to have issues. From $(*)$ we have
$$\tan \alpha - \tan \theta = - \pm \frac{tg}{2v} \sqrt{1+\tan^2 \theta}$$
$$\implies 4\tan^2 \alpha + 4\tan^2 \theta-8\tan\theta \tan \alpha= \frac{t^2g^2}{v^2}(1+\tan^2\theta)$$
$$\iff (4-\frac{t^2g^2}{v^2})\tan^2\theta -8\tan \theta \tan \alpha +4\tan^2 \alpha -\frac{t^2g^2}{v^2}=0$$
If $\tan\theta$ and $\tan \theta'$ are the solutions for $\tan \theta$ in this equation, then
$$\tan \theta + \tan \theta'= \frac{8\tan \alpha}{4-\frac{t^2g^2}{v^2}}$$
$$ \tan \theta \tan \theta' = \frac{4\tan^2\alpha - \frac{t^2g^2}{v^2}}{4-\frac{t^2g^2}{v^2}}$$
and thus $$\tan(\theta + \theta')=\frac{\tan \theta + \tan \theta'}{1-\tan \theta \tan \theta'}=\frac{\frac{8\tan \alpha}{4-\frac{t^2g^2}{v^2}}}{1-\frac{4\tan^2\alpha - \frac{t^2g^2}{v^2}}{4-\frac{t^2g^2}{v^2}}}= \frac{\frac{8\tan \alpha}{4-\frac{t^2g^2}{v^2}}}{\frac{4-4\tan^2 \alpha}{4-\frac{t^2g^2}{v^2}}}=\tan{2\alpha}$$
which is wrong as we require $\tan(\theta + \theta')=-\cot \alpha$.
Does anybody know where my mistake is or why is this method wrong?
| Please note that you are assuming that the time taken at both angles to reach $Q$ is same which is not correct assumption. We only know that the initial speed is same. So you should rather eliminate $t$ and write the projectile motion in terms of $x, y, u$ and $\theta$ where $x$ and $y$ are horizontal and vertical distance from $P$ to $Q$ respectively. $u$ is the initial speed.
$\displaystyle y = \tan\theta \cdot x - \frac{g}{2 \cdot u^2 \cdot \cos^2\theta} \cdot x^2 \ $
Rewriting in terms of $\tan \theta$,
$\displaystyle y = \tan\theta \cdot x - \frac{g \cdot (1+ \tan^2\theta)}{2 \cdot u^2} \cdot x^2$
$\displaystyle \frac{g \cdot x^2}{2 \cdot u^2} \cdot \tan^2\theta - x \cdot \tan \theta + (y + \frac{g \cdot x^2}{2 \cdot u^2}) = 0$
$\tan \theta + \tan \theta' = \displaystyle \frac{2 \cdot u^2}{g \cdot x}$
$\tan \theta \tan \theta' = \displaystyle 1 + \frac{2 \cdot u^2 \cdot y}{g \cdot x^2}$
Now using $ \ \tan (\theta+\theta') = \displaystyle \frac{\tan \theta + \tan \theta'} {1 - \tan \theta \tan \theta'}$, we get
$\tan (\theta + \theta') = \displaystyle - \frac{x}{y} = - \cot \alpha$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6+\sqrt{14})$ Sifting through my old Galois Theory notes, I found a proof that $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6 + \sqrt{14})$, but my proof was chicken-scratch so I can no longer decipher it.
Can somebody come up with a way of showing this?
| A Galois-theoretic approach would analyse the conjugates. We know that the conjugates of $\alpha := \sqrt{5+\sqrt{21}}$ are $\pm\sqrt{5 \pm \sqrt{21}}$ and the conjugates of $\frac12(\sqrt6 + \sqrt{14})$ are $\frac12(\pm \sqrt6 \pm \sqrt{14})$, so the interesting conjugate to analyse would be $\beta := \sqrt{5-\sqrt{21}}$ (the other two are $-\alpha$ and $-\beta$).
We compute:
$$\begin{array}{rcl}
(\alpha + \beta)^2
&=& \alpha^2 + \beta^2 + 2\alpha\beta \\
&=& (5+\sqrt{21}) + (5-\sqrt{21}) + 2\sqrt{(5+\sqrt{21})(5-\sqrt{21})} \\
&=& 10 + 2\sqrt{5^2 - 21} \\
&=& 14
\end{array}$$
So $\alpha + \beta = \sqrt{14}$ because it is positive.
We also compute:
$$\begin{array}{rcl}
(\alpha-\beta)^2
&=& \alpha^2+\beta^2-2\alpha\beta \\
&=& 10 - 2\sqrt{5^2-21} \\
&=& 6
\end{array}$$
So $\alpha - \beta = \sqrt6$ because $\alpha > \beta$.
Therefore, $\alpha = \frac12[(\alpha+\beta)+(\alpha-\beta)] = \frac12(\sqrt6+\sqrt{14})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Bound for Gamma Function Expression In Sobolev's The Efimov Effect. Discrete Spectrum Asymptotics, the following inequality is stated on page 112 $$\int_{1}^{+\infty}\frac{x^{-(n+\frac{1}{2})}}{(x^2-1)^{\frac{3}{4}}}dx=\frac{\Gamma(\frac{n}{2}+\frac{1}{2})\Gamma(\frac{1}{4})}{2\Gamma(\frac{n}{2}+\frac{3}{4})}\le\frac{8}{3^{\frac{3}{4}}(2n+1)^{\frac{1}{4}}},$$ where $n$ is a natural number. Using Gautschi's inequality I was able to prove the looser bound $$\le\frac{\Gamma(\frac{1}{4})\sqrt{10}}{2(2n+1)^{\frac{1}{4}}},$$ but I can't get that one. Can anybody help me to find that out?
| We have
$$
\frac{{\Gamma \left( {\frac{n}{2} + \frac{1}{2}} \right)\Gamma \left( {\frac{1}{4}} \right)}}{{\Gamma \left( {\frac{n}{2} + \frac{3}{4}} \right)}} = \int_0^1 {t^{\frac{n}{2} - \frac{1}{2}} (1 - t)^{ - \frac{3}{4}} dt} = \int_0^{ + \infty } {e^{ - s\left( {\frac{n}{2} + \frac{1}{8}} \right)} s^{ - \frac{3}{4}} \left( {\frac{{s/2}}{{\sinh (s/2)}}} \right)^{\frac{3}{4}} ds}
$$
for all $n\geq 0$. Thus,
$$
\frac{{\Gamma \left( {\frac{n}{2} + \frac{1}{2}} \right)\Gamma \left( {\frac{1}{4}} \right)}}{{\Gamma \left( {\frac{n}{2} + \frac{3}{4}} \right)}} \le \int_0^{ + \infty } {e^{ - s\left( {\frac{n}{2} + \frac{1}{8}} \right)} s^{ - \frac{3}{4}} ds} = \frac{{\sqrt 2 \Gamma \left( {\frac{1}{4}} \right)}}{{\left( {2n + \frac{1}{2}} \right)^{1/4} }}
$$
for all $n\geq 0$. Finally,
$$
\frac{{\Gamma \left( {\frac{n}{2} + \frac{1}{2}} \right)\Gamma \left( {\frac{1}{4}} \right)}}{{2\Gamma \left( {\frac{n}{2} + \frac{3}{4}} \right)}} \le \frac{8}{{3^{3/4} (2n + 1)^{1/4} }}\frac{{\sqrt 2 3^{3/4} \Gamma \left( {\frac{1}{4}} \right)(2n + 1)^{1/4} }}{{16\left( {2n + \frac{1}{2}} \right)^{1/4} }} < \frac{8}{{3^{3/4} (2n + 1)^{1/4} }}
$$
for all $n\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Number of digits in $8^n$ in base $6$ Let $d(n)$ be the number of digits when $8^n$ is written in base $6$. Find the closed form expression for $d(n)$.I tested for few numbers:
$$8^1=12$$
$$8^2=144$$
$$8^3=2212$$
So i felt that $d(n)=n$. But to formally prove it, i tried with induction.
Let $d(n)=n$, then we have
$$8^n=a_0a_1a_2....a_{n-1}, \: a_i \in \left\{0,1,2,3,4,5\right\}, \:a_0 \ne 0$$
Now we have
$$8^{n+1}=8^n \times 12$$
I am stuck here?
| $d(n) = n+1$ only works for small values of $n$. Note the leading digit is creeping up eventually you will have an $8^k = 5xxxx...x$ with $k+1$ digits but $8^{k+1} = 1yxxxx....x$ with $k + 3$ digits.
Indeed $8^6 = 5341344$ with $7$ digits and $8^7 = 112541012$ with $9$ digits.
=====
You can think backwards. If $8^n$ has $K$ digits then $8^n \ge {1\underbrace{000.....0}_K}_{base\ 6} =6^{K-1}$ and $8^n \le {\underbrace{5555.....5}_K}_{base\ 6}=6^{K}-1$ so $6^{K-1} \le 8^n < 6^K$.
So solve for $K$
$\log_6 6^{K-1} \le \log_68^n < \log_6 6^K$
$K-1 \le n\cdot \log_6 8 < K$
$K-1 = \lfloor n\cdot \log_6 8 \rfloor$
$K = \lfloor n\cdot \log_6 8 \rfloor
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why is the above matrix singular? Let
$$A=\begin{bmatrix}1&2&3\\3&4&5\\5&6&7\end{bmatrix}$$
with $\det(A) = 0$. We know if $\det(A) = 0$, then $A$ is singular. Also, we know if $A$ is singular, columns and rows are linearly dependent. Why are the columns or rows linearly dependent?
| We see that in every row the elements increase by 1 when going from column 1 to column 2 to column 3. In other words:
$$
\begin{pmatrix}2\\4\\6\end{pmatrix}-\begin{pmatrix}1\\3\\5\end{pmatrix}=
\begin{pmatrix}1\\1\\1\end{pmatrix}=
\begin{pmatrix}3\\5\\7\end{pmatrix}-\begin{pmatrix}2\\4\\6\end{pmatrix}
$$
Rearranging, you will get:
$$
2\begin{pmatrix}2\\4\\6\end{pmatrix}=
\begin{pmatrix}1\\3\\5\end{pmatrix}+\begin{pmatrix}3\\5\\7\end{pmatrix}
$$
which is clearly a linear relation between the columns. Thus, columns are not linearly independent
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to integrate definite integral of form $\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx$ During WKB approximation maths I typically end up deducing integrals that need to be calculated of the form:
$$\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx$$
where a is a constant.
I know the following standard integral
$$\int_0^{1} \sqrt{1-\sqrt{y}} \space dy = \frac{8}{15}$$
My question is how do I Use the standard integral above to deduce that
$$\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx = \frac{8 a^{\frac{5}{2}}}{15}$$
| Letting $\sqrt x=a \sin^2 \theta$ transforms the integral into
$$
\begin{aligned}
I &=4\int_0^{\frac{\pi}{2}} \sqrt{a-a \sin ^2 \theta} a^2 \sin ^3 \theta \cos \theta d \theta \\
&=4 a^{\frac{5}{2}}\int_0^{\frac{\pi}{2}} \sin ^3 \theta \cos ^2 \theta d \theta \\
&=-4 a^{\frac{5}{2}}\int_0^{\frac{\pi}{2}}\left(1-\cos ^2 \theta\right) \cos ^2 \theta d(\cos \theta) \\
&=-4 a^{\frac{5}{2}}\left[\frac{\cos ^2 \theta}{3}-\frac{\cos ^5 \theta}{5}\right]_0^{\frac{\pi}{2}} \\
&=\frac{8 a^{\frac{5}{2}}}{15}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\lim\limits_{(x,y) \to (0,0)} \frac{xy(y^2-x^2)}{x^2+y^2}=0$ Prove $\lim\limits_{(x,y) \to (0,0)} \frac{xy(y^2-x^2)}{x^2+y^2}=0$
My attempt: For all $\epsilon > 0$ there is a $\delta > 0$ such that $\left | \frac{xy(y^2-x^2)}{x^2+y^2} \right| < \epsilon \rightarrow \sqrt{x^2+y^2} < \delta$.
Note important facts: $ (1) \ \sqrt{x^2} < \sqrt{x^2 + y^2}$ and $ (2) \ x^2 < x^2 + y^2$
$$\left | \frac{xy(y^2-x^2)}{x^2+y^2}\right | = \left | \frac{\sqrt{x^2}\sqrt{y^2}(y^2-x^2)}{x^2+y^2} \right| < \left| \frac{\sqrt{x^2 + y^2}\sqrt{x^2 + y^2}(y^2-x^2)}{x^2+y^2}\right | = |y^2-x^2| $$
From $(2)$ we can show that :
$$|y^2-x^2| < |x^2+y^2-x^2| = |y^2| < |x^2 + y^2| < \delta^2 $$
Hence let $\epsilon = \delta^2$. Does this relationship work?
| Your last step is not correct as for example take $|2^2-5^2|\lt |5^2+2^2-5^2|$, which is clearly not true.
You may proceed like this also:
For any $\epsilon\gt 0$,
$\begin{align}|\frac{xy(y^2-x^2)}{x^2+y^2} |& = |\frac{\sqrt{x^2}\sqrt{y^2}(y^2-x^2)}{x^2+y^2} | \\& \lt | \frac{\sqrt{x^2 + y^2}\sqrt{x^2 + y^2}(y^2+x^2)}{x^2+y^2} | = y^2+x^2\lt \epsilon \end{align}$
Now choose $\delta =\sqrt \epsilon$ to prove your claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Fractional derivative with linear algebra I have this question in linear algebra that asks us to represent $\frac{d}{dx}\sin x$ and $\frac{d}{dx}\cos x$ as a matrix in the basis of $\binom{1}{0}=\sin x, \ \binom{0}{1}=\cos x$ using eigenvectors and eigenvalues. So, I guessed $\left[\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}\right]_B$, s.t.
$$
\frac{d}{dx}\sin x=\left[\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}\right]\cdot\binom{1}{0}=\binom{0}{1}=\cos x \\
\frac{d}{dx}\cos x=\left[\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}\right]\cdot\binom{0}{1}=\binom{-1}{0}=-\sin x
$$
The question then asks me to do it for the square root of a derivative operator, like so $\sqrt{\frac{d}{dx}}$. So, I found the eigenvectors and values.$$
PDP^{-1}=\left[\begin{matrix} 1 & 1 \\ i & -i \end{matrix}\right]\cdot\left[\begin{matrix} i & 0 \\ 0 & -i \end{matrix}\right]\cdot\frac{1}{2}\left[\begin{matrix} 1 & i \\ 1 & -i \end{matrix}\right] \\
PD^{\frac{1}{2}}P^{-1}=\left[\begin{matrix} 1 & 1 \\ i & -i \end{matrix}\right]\cdot\left[\begin{matrix} i & 0 \\ 0 & -i \end{matrix}\right]^{\frac{1}{2}}\cdot\frac{1}{2}\left[\begin{matrix} 1 & i \\ 1 & -i \end{matrix}\right] \\
=\left[\begin{matrix} 1 & 1 \\ i & -i \end{matrix}\right]\cdot\left[\begin{matrix} e^{i\frac{\pi}{4}} & 0 \\ 0 & e^{-i\frac{\pi}{4}} \end{matrix}\right]\cdot\frac{1}{2}\left[\begin{matrix} 1 & i \\ 1 & -i \end{matrix}\right] \\
=\frac{1}{2}\left[\begin{matrix} 1 & 1 \\ i & -i \end{matrix}\right]\cdot\left[\begin{matrix} e^{i\frac{\pi}{4}} & ie^{i\frac{\pi}{4}} \\ e^{-i\frac{\pi}{4}}& -ie^{-i\frac{\pi}{4}}\end{matrix}\right] \\
=\frac{1}{2}\cdot\left[\begin{matrix}e^{i\frac{\pi}{4}}+e^{-i\frac{\pi}{4}} & i(e^{i\frac{\pi}{4}}-e^{-i\frac{\pi}{4}}) \\ i(e^{i\frac{\pi}{4}}-e^{-i\frac{\pi}{4}}) & i^2(e^{i\frac{\pi}{4}}+e^{-i\frac{\pi}{4}})\end{matrix}\right] \\
=\frac{1}{2}\cdot2\cdot\left[\begin{matrix} \cos{\frac{\pi}{4}} & -\sin{\frac{\pi}{4}} \\ -\sin{\frac{\pi}{4}} & -\cos{\frac{\pi}{4}}\end{matrix}\right] \\
=\frac{1}{\sqrt{2}}\cdot\left[\begin{matrix}1 & -1 \\ -1 & -1\end{matrix}\right]
$$
However, when I tested $\sin x$, I got:$$
\frac{1}{\sqrt{2}}\cdot\left[\begin{matrix}1 & -1 \\ -1 & -1\end{matrix}\right]\cdot\binom{1}{0}=\frac{1}{\sqrt{2}}\cdot\binom{1}{-1} \\
\frac{1}{\sqrt{2}}\cdot\left[\begin{matrix}1 & -1 \\ -1 & -1\end{matrix}\right]\cdot\frac{1}{\sqrt{2}}\cdot\binom{1}{-1}=\binom{1}{0}=\sin x
$$ which is a weird result because I expected $\cos x$. I redid my working again and got the same result. What did I do wrong here? Could someone help me find that error please? Thank you very much for your help.
| Your calculation is very wrong. E.g. $P^{-1}$ is not $\frac12\pmatrix{1&i\\ 1&-i}$ but $\frac12\pmatrix{1&-i\\ 1&i}$ and the eigendecomposition of your matrix isn't $P\pmatrix{i\\ &-i}P^{-1}$ but $P\pmatrix{-i\\ &i}P^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Computing $\int \frac{-\mathrm{d}x}{\sqrt{ax^2+bx+c}}$ First I complete the square $$ax^2+bx+c = a\left(\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}\right)$$
and let $d=\frac{4ac-b^2}{4a}$. Then $$\int \frac{-\mathrm{d}x}{\sqrt{ax^2+bx+c}}=\frac{1}{\sqrt{a}}\int \frac{-\mathrm{d}x}{\sqrt{\left(x^2+\frac{b}{2a}\right)^2+\frac{d}{a}}}$$
$$=\frac{1}{\sqrt{a}}\int \frac{-\mathrm{d}x}{\sqrt{\left(\frac{a}{d}\right)\left(x^2+\frac{b}{2a}\right)^2+1}}$$
Let $u=\sqrt{\frac{a}{d}}\left(x+\frac{b}{2a}\right)$. Then $\mathrm{d}u=\sqrt{\frac{a}{d}}\ \mathrm{d}x$ and the integral becomes
$$\frac{\sqrt{d}}{a}\int\frac{-\mathrm{d}u}{\sqrt{u^2+1}}$$
But I don't think this is right. Ultimately I'm looking for an answer in terms of inverse trigonometric functions, hopefully, $\arccos u$ depending on $a,b,c$. What am I doing wrong here?
| $\frac{\sqrt{d}}{a} \int \frac{-du}{\sqrt{u^{2}+1}}$.
Currently I can see only solution $\sqrt{u^{2}+1} =t - u$
$u=\frac{t^{2} - 1}{2t}. \sqrt{u^{2}+1} = \frac{t^{2}+1}{2t}, du=\frac{t^{2} + 1}{2t^{2}}$
$\int \frac{-du}{\sqrt{u^{2}+1}} = -\int \frac{dt}{t} = -ln|t|+C = -ln|u+\sqrt{u^2+1}| + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4090314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Clarification on partial fraction expansion I would like to use the cover up method for the following equation.
$$\frac{1}{x^2(x+1.79)}$$
and it breaks down into
$$\frac{A}{x}\quad\frac{B}{x^2}\quad\frac{C}{(x + 1.79)}$$
I realize that you cover up $(x+1.79)$ to get $x = -1.79$ but what do you do for the other variables?
Thanks for your help.
| Here it is another way to approach for the sake of curiosity.
On one hand, we have that
\begin{align*}
\frac{1}{x^{2}(x+k)} & = \frac{1}{k}\left(\frac{k}{x^{2}(x+k)}\right)\\\\
& = \frac{1}{k}\left(\frac{(x + k) - x}{x^{2}(x+k)}\right)\\\\
& = \frac{1}{kx^{2}} - \frac{1}{kx(x+k)}
\end{align*}
On the other hand, we have that
\begin{align*}
\frac{1}{x(x+k)} & = \frac{1}{k}\left(\frac{k}{x(x+k)}\right)\\\\
& = \frac{1}{k}\left(\frac{(x + k) - x}{x(x+k)}\right)\\\\
& = \frac{1}{kx} - \frac{1}{k(x+k)}
\end{align*}
Gathering both results, it results that
\begin{align*}
\frac{1}{x^{2}(x+k)} = \frac{1}{kx^{2}} - \frac{1}{k^{2}x} + \frac{1}{k^{2}(x+k)}
\end{align*}
At your case, $k = 1.79$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4091789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding Taylor Series from existing Series If the Taylor Series of $\ln(x)$ is known:
$$\ln(x) = (x-1) -\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\frac{1}{5}(x-1)^5-...$$
Can one find the Taylor series of
$$f(x)= \frac{x}{1-x^2}$$
by manipulating the Taylor series of ln(x)?
| As an alternate without using $\ln x$. Make use of the geometric series:
$$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n \implies \frac{1}{1-x^2} = \sum_{n=0}^\infty x^{2n} \implies \frac{x}{1-x^2} = \sum_{n=0}^\infty x^{2n+1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4094977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find $\iiint_D z^{2}dV$ where $D$ is restricted to the given surfaces Find the following triple integral where $D$ is restricted to the the surfaces $x^2+y^2+z^2=R^2$ and $x^2+y^2+z^2=2Rz$
$$\iiint_D z^{2}dV$$
The surface $x^2+y^2+z^2=R^2$ represents a sphere centered at the origin and $x^2+y^2+z^2=2Rz$ can be written as $x^2+y^2+(z-R)^2=R^2$ which is a sphere centered at $(0,0,R)$, I used geogebra to see the graph of the surfaces, though I'm unable to determine the limits of the integration.
| I'm going to set the integral up in spherical coordinates, which are a good choice here because the domain is bounded by spheres.
First, let's look at the bounds on the radius, $\rho$. Let's rewrite the equations of the spheres using spherical coordinates. The equation for the sphere centered at the origin should be $\rho^2 = R^2 \to \rho = R$, and the equation for the off-centered sphere should be $\rho^2 = 2R\rho\cos\phi \to \rho = 2R\cos\phi$.
Now, if you imagine looking from the origin outwards, in the region of interest you should see that if we look straight up then we're bounded by the sphere centered at the origin, then as we start looking further down our upper bound gets switched to the sphere centered at $(0, 0, R)$. So, we're going to have to split our domain at the intersection.
To get information about the intersection, we can set our two equations for our spheres equal to get $R = 2R\cos\phi \to \cos\phi = \frac{1}{2}$, and because $0 \leq \phi \leq \pi$, this gives us $\phi = \frac{\pi}{3}$ at the intersection. So, for $0 \leq \phi \leq \frac{\pi}{3}$, we should have $0 \leq \rho \leq R$, and for $\frac{\pi}{3} \leq \phi \leq \frac{\pi}{2}$, we should have $0 \leq \rho \leq 2R\cos\phi$.
Finally, we need bounds for the polar angle, $\theta$. This is the easiest one here because we just need to go through one full rotation: $0 \leq \theta \leq 2\pi$.
Now we can write out our iterated integral, remembering to include the Jacobian. ($dV = \rho^2 \sin\phi d\rho d\phi d\theta)$
$$\iiint_D z^2 dV = \int_0^{2\pi} \int_0^{\frac{\pi}{3}} \int_{0}^{R} \rho^4 \cos^2\phi \sin\phi d\rho d\phi d\theta + \int_0^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_{0}^{2R\cos{\phi}} \rho^4 \cos^2\phi \sin\phi d\rho d\phi d\theta $$
And from here we integrate just like a normal multiple integral.
$$\int_0^{2\pi} \int_0^{\frac{\pi}{3}} \int_{0}^{R} \rho^4 \cos^2\phi \sin\phi d\rho d\phi d\theta = \int_0^{2\pi} \int_0^{\frac{\pi}{3}} \dfrac{\rho^5}{5}\Bigg|_{0}^R \cos^2\phi \sin\phi d\phi d\theta$$
$$= \frac{R^5}{5}\int_0^{2\pi} \int_0^{\frac{\pi}{3}} \cos^2 \phi \sin\phi d\phi d\theta$$
$$u = \cos\phi, du = -\sin\phi d\phi$$
$$\frac{R^5}{5}\int_0^{2\pi} \int_{\frac{1}{2}}^{1} u^2 du d\theta = \frac{R^5}{5}\int_0^{2\pi} \frac{1}{3}u^3\Bigg|_{\frac{1}{2}}^{1} du d\theta = \frac{7\pi R^5}{60}$$
$$\int_0^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_{0}^{2R\cos{\phi}} \rho^4 \cos^2\phi \sin\phi d\rho d\phi d\theta = \int_0^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\dfrac{\rho^5}{5}\Bigg|_{0}^{2R\cos\phi} \cos^2\phi \sin\phi d\phi d\theta$$
$$= \frac{32R^5}{5}\int_0^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cos^7 \phi \sin\phi d\phi d\theta = \frac{32R^5}{5}\int_0^{2\pi} \int_{0}^{\frac{1}{2}} u^7 du d\theta = \frac{\pi R^5}{160}$$
$$\iiint_D z^2 dV = \frac{7\pi R^5}{60} + \frac{\pi R^5}{160} = \frac{59\pi R^5}{480}$$
Hope this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine the ratio between the base area of the trunk $ T_2 $ and the trunk $ T_1 $ Two cone trunks $ T_1 $ and $ T_2 $ have a common base of radius equal to $ 8 $ cm, the other bases being concentric circles. Knowing that the radius of the larger base of $ T_1 $ is equal to $ 15 $ cm, and the volume of $ T_1 $ is three times the volume of $ T_2 $, determine the ratio between the base area of the trunk $ T_2 $ and the trunk $ T_1 $, this ratio being between non-common bases
Can anyone give me a tip?
| Volume of trunk $T_1$ is:
$V_1=\frac 12 h(8^2\times 3,14+15^2\times 3,14=A_1)$
And that of trunk $T_2$ is:
$V_2=\frac 12 h(8^2\times 3.14+ 3.14 \times R^2=A_2)$
where $h$ is the height of trunks and $A_1$ and $A_2$ are the areas of trunks $T_1$ and $T_2$ on the top.
$V_1=3V_2$
$\frac 12 h(8^2\times 3,14+15^2\times 3,14=A_1)=\frac 32 h(8^2\times 3.14+ 3.14 \times R^2=A_2)$
$\rightarrow 2\times 8^2\times 3.14+3A_2=A_1 $
$2\times 8^2\times 3.14\approx 402$
$A_1=15^2\times 3.14\approx 706$
so we have:
$ 402+3A_2=(706=A_1)$
dividing both sides by $706$ we get:
$\frac{3A_2}{A_1}=1-\frac{402}{706}=\frac {304}{706} \Rightarrow \frac {A_2}{A_1}=\frac{304}{3\times 706}\approx 0.144=\frac{144}{1000}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Taylor Series $\sin(3x)$ centered at $a=\frac{\pi}{6}$ Using a table I got
\begin{array}{c|c|c}
n & f^{n}(x) & f^{n}\left(\frac{\pi}{6}\right) \\
\hline
0 & \sin(3x) & 1\\
1 & 3\cos(3x) & 0\\
2 & -9\sin(3x) & -9\\
3 & -27\cos(3x) & 0\\
4 & 81\sin(3x) & 81
\end{array}
From this table, I expanded $\sin(3x)$ centered at $a=\frac{\pi}{6}$, I get
\begin{align}
\require{cancel}
f(x) &= \frac{1}{0!}\left(x-\frac{\pi}{6}\right)^0+\frac{0}{1!}\left(x-\frac{\pi}{6}\right)^1-\frac{9}{2!}\left(x-\frac{\pi}{6}\right)^2+\frac{0}{3!}\left(x-\frac{\pi}{6}\right)^3+\frac{81}{4!}\left(x-\frac{\pi}{6}\right)^4+\cdots \\
&= \frac{1}{0!}\left(x-\frac{\pi}{6}\right)^0+\cancel{\frac{0}{1!}\left(x-\frac{\pi}{6}\right)^1}-\frac{9}{2!}\left(x-\frac{\pi}{6}\right)^2+\cancel{\frac{0}{3!}\left(x-\frac{\pi}{6}\right)^3}+\frac{81}{4!}\left(x-\frac{\pi}{6}\right)^4+\cdots \\
&= \frac{1}{0!}\left(x-\frac{\pi}{6}\right)^0-\frac{3^2}{2!}\left(x-\frac{\pi}{6}\right)^2+\frac{3^4}{4!}\left(x-\frac{\pi}{6}\right)^4-\frac{3^6}{6!}\left(x-\frac{\pi}{6}\right)^6
\end{align}
Is the following Taylor Series correct for the function $f(x)=\sin(3x)$ centered at $a=\frac{\pi}{6}$?
\begin{align}
f(x) &=\sum_{n=1}^{\infty}(-1)^n\frac{3^{2n-2}}{(2n-2)!}\left(x-\frac{\pi}{6}\right)^{2n-2}
\end{align}
| That is correct, but it is simpler to do it as follows:\begin{align}\sin(3x)&=\sin\left(3\left(x-\frac\pi6\right)+\frac\pi2\right)\\&=\cos\left(3\left(x-\frac\pi6\right)\right)\end{align}and now, since$$\cos(x)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n},$$you have\begin{align}\sin(3x)&=\cos\left(3\left(x-\frac\pi6\right)\right)\\&=\sum_{n=0}^\infty\frac{(-1)^n3^n}{(2n)!}\left(x-\frac\pi6\right)^{2n}.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Inequality $\Gamma\left(\sin\left(\frac{1}{x}\right)\right)-\Gamma\left(\sin^2\left(\frac{1}{x}\right)\right)-x+x^2+\frac{1}{3}\geq 0$ Problem found with the help of Desmos and my imagination .
Let $x\geq 1$ then we have :
$$\Gamma\left(\sin\left(\frac{1}{x}\right)\right)-\Gamma\left(\sin^2\left(\frac{1}{x}\right)\right)-x+x^2+\frac{1}{3}\geq 0$$
Some informations :
We have the limits :
$$\lim_{x\to \infty}\Gamma\left(\sin\left(\frac{1}{x}\right)\right)-x=-\gamma$$
And :
$$\lim_{x\to \infty}\Gamma\left(\sin^2\left(\frac{1}{x}\right)\right)-x^2=-\gamma+\frac{1}{3}$$
It doesn't work with higher degree like $3$.It seems to be a little mysterious .
The derivative is very complicated and I don't expose it here .
How to show the inequality ?
| Some thoughts
Let $f(v) = \Gamma(v) - \frac{1}{v}$.
Fact 1: $f'(v) > 0$ for all $v$ in $(0, 1)$.
Now, by Fact 1, we have
$$\Gamma(\sin(1/x)) - \frac{1}{\sin (1/x)}
\ge \Gamma(\sin^2(1/x)) - \frac{1}{\sin^2(1/x)}.$$
Thus, it suffices to prove that
$$\frac{1}{\sin (1/x)} - \frac{1}{\sin^2(1/x)} - x + x^2 + \frac{1}{3}\ge 0$$
or
$$- \left(\frac{1}{\sin (1/x)} - \frac{1}{2}\right)^2 - x + x^2 + \frac{7}{12} \ge 0.$$
Fact 2: $1/2 < \frac{1}{\sin (1/x)} \le \frac{60 x^3 + 3x}{60x^2 - 7}$ for all $x \ge 1$.
By Fact 2, it suffices to prove that
$$- \left(\frac{60 x^3 + 3x}{60x^2 - 7} - \frac{1}{2}\right)^2 - x + x^2 + \frac{7}{12} \ge 0$$
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
polynomial with complex roots
As I understans - this polynomial has 4 different roots. We are given $z_1=3-2i$, so $z_2=3+2i$. So we already have 2 of 4 roots. I expressed the polynomial as $(x^2+2)(x^2-6x+13)$. I solved each of these equations separately and got $x_{1,2}=\pm i$ for $(x^2+2)$ and $x_{3,4}=\pm 2i$ for $(x^2-6x+13)$
So there are 6 roots in total. What am I doing wrong here?
Thanks in advance.
| I think you are miscalculating the roots of the polynomial.
Indeed, since $3 - 2i$ is a root and the coefficients of $p(z)$ are real, $3 + 2i$ is also a root.
Taking advantage of such fact, we can conclude that $z^{2} - 6z + 13$ is a factor of $p(z)$.
Hence we have that
\begin{align*}
z^{4} - 6z^{3} + 15z^{2} - 12z + 26 & = (z^{4} - 6z^{3} + 13z^{2}) + (2z^{2} - 12z + 26)\\\\
& = z^{2}(z^{2} - 6z + 13) + 2(z^{2} - 6z + 13)\\\\
& = (z^{2} + 2)(z^{2} - 6z + 13)
\end{align*}
whence it results the other roots are given by $\pm i\sqrt{2}$.
Hopefully this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx$ I want to evaluate $\displaystyle \int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx$ but it's quite difficult.
I have tried to rewrite the integral as
$$\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx=\frac{\pi }{2}\int _0^{\frac{\pi }{2}}\tan \left(x\right)\ln ^2\left(\sin \left(x\right)\right)\:dx-\int _0^{\frac{\pi }{2}}x\tan \left(x\right)\ln ^2\left(\sin \left(x\right)\right)\:dx$$
I've also tried to integrate by parts in multiple ways yet I cant go forth with this integral, I also tried using the substitution $t=\tan{\frac{x}{2}}$ but cant get anything to work, I'll appreciate any sort of help.
I also tried using the classical expansion
$$\ln \left(\cos \left(x\right)\right)=-\ln \left(2\right)-\sum _{n=1}^{\infty }\frac{\left(-1\right)^n\cos \left(2nx\right)}{n},\:-\frac{\pi }{2}<x<\frac{\pi }{2}$$
But it only gets worse.
| My approach.
$$\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx$$
$$=\frac{1}{4}\int _0^{\infty }\frac{\arctan \left(x\right)\ln ^2\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx=\frac{\pi }{8}\int _0^{\infty }\frac{\ln ^2\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx$$
$$-\frac{1}{4}\int _0^{\infty }\frac{\arctan \left(\frac{1}{x}\right)\ln ^2\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx$$
$$\int _0^{\infty }\frac{\arctan \left(\frac{1}{x}\right)\ln ^2\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx=\frac{\pi }{2}\int _0^{\infty }\frac{x\ln ^2\left(\frac{x^2}{1+x^2}\right)}{1+x^2}\:dx-\int _0^{\infty }\frac{x\arctan \left(\frac{1}{x}\right)\ln ^2\left(\frac{x^2}{1+x^2}\right)}{1+x^2}\:dx$$
$$=\frac{\pi }{2}\int _0^{\infty }\frac{x\ln ^2\left(\frac{x^2}{1+x^2}\right)}{1+x^2}\:dx-\frac{4}{3}\int _0^{\infty }\frac{x\arctan ^3\left(\frac{1}{x}\right)}{1+x^2}\:dx-\frac{4}{3}\operatorname{\mathfrak{I}} \left\{\int _0^{\infty }\frac{x\ln ^3\left(\frac{x}{x-i}\right)}{1+x^2}\:dx\right\}$$
$$=\frac{\pi }{4}\int _0^1\frac{\ln ^2\left(x\right)}{1-x}\:dx+4\int _0^{\frac{\pi }{2}}x^2\ln \left(\sin \left(x\right)\right)\:dx-\frac{4}{3}\operatorname{\mathfrak{I}} \left\{3\operatorname{Li}_4\left(2\right)+i\pi \ln ^3\left(2\right)-6\zeta \left(4\right)\right\}$$
$$=\frac{5\pi }{4}\zeta \left(3\right)-\frac{\pi ^3}{6}\ln \left(2\right)-\frac{2\pi }{3}\ln ^3\left(2\right)$$
Thus.
$$\frac{1}{4}\int _0^{\infty }\frac{\arctan \left(x\right)\ln ^2\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx=\frac{\pi }{8}\zeta \left(3\right)-\frac{1}{4}\left(\frac{5\pi }{4}\zeta \left(3\right)-\frac{\pi ^3}{6}\ln \left(2\right)-\frac{2\pi }{3}\ln ^3\left(2\right)\right)$$
Therefore.
$$\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx=-\frac{3\pi }{16}\zeta \left(3\right)+\frac{\pi ^3}{24}\ln \left(2\right)+\frac{\pi }{6}\ln ^3\left(2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4113352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
} |
How to prove that the following is Cauchy? So I have to prove that the following is a Cauchy sequence by definition:
$|u_{n+2}-u_{n+1}|\leq \frac{1}{2}|u_{n+1}-u_n|$.
I know that, $$|u_{n+2}-u_{n+1}|\leq \frac{1}{2^n} |u_2-u_1|$$
How do I proceed further?
| For $m \ge n \ge 2$, note that
\begin{align}
|u_m - u_n| &= |u_m - u_{m - 1} + u_{m - 1} - \cdots + u_{n + 1} - u_n| \\
& \le |u_m - u_{m - 1}| + \cdots + |u_{n + 1} - u_n| \\
& \le \frac{1}{2^{m-2}}|u_2 - u_1| + \cdots + \frac{1}{2^{n-1}}|u_2 - u_1| \\
& = |u_2 - u_1|\left(\frac{1}{2^{n-1}} + \cdots + \frac{1}{2^{m - 2}}\right) \\
& \le |u_2 - u_1|\left(\frac{1}{2^{n-1}} + \cdots \right) \\
& = \frac{|u_2 - u_1|}{2^{n - 2}}.
\end{align}
Can you conclude now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximize $x + y$ with constraint $y \cdot x^2=15$. As the title says, I am asked to maximize the sum of two numbers $x + y$ when $x^2 \cdot y = 15$.
The thing is... I tried substitution but got nowhere, it seems $h(x) = x + \frac{15}{x^2}$ only has a minimum value, not a maximum.
I've been starting to think there is an error in the excercise. I tried Lagrange multiplicators next, having:
\begin{align}
f(x,y) &= x+y\\
g(x,y) &= x^2\cdot y - 15 = 0
\end{align}
Then:
\begin{align}
\nabla f = \lambda \nabla g
\end{align}
that gives:
\begin{align}
2xy &= 1\\
2x^2 &= 1\\
x^2\cdot y &= 15
\end{align}
then we find
\begin{align}
2xy &= 2x^2\\
x(2y - x) &= 0\\
x = 0 &\lor x = 2y
\end{align}
Now, from $x^2 \cdot y = 15$ we know $x,y>0$, then we only have one solution $x=2y$, applying that to $g(x)$ we get $x^2 \cdot \frac{x}{2} = 15 \implies x = \sqrt[3]{30}, y = \frac{\sqrt[3]{30}}{2}$.
Then the only point that I can use from Lagrange is $(\sqrt[3]{30}, \frac{\sqrt[3]{30}}{2})$. But what guarantee do I have that said point is the maximum of the sum of $x + y$?. Doesn't Lagrange only find extremes? How do I check this result?
As I said earlier, if I graph $h(x) = x + \frac{15}{x^2}$ in $R^2$ I find that $x=\sqrt[3]{30}$ is the value in the $x$ axis where the minimum value of $h(x)$ is, so this only feeds my suspicions about the problem having an error and the point needed being actually a minimum instead of a maximum.
Can anyone shed some light into this?
| Since $x^2y=15$, we have $y\gt0$. To maximize $x+y$, we want $x\gt0$ as well. However, for $x\gt0$, we have the reverse inequality
$$
\frac{x+y}3=\overbrace{\frac{2\cdot\color{#C00}{\frac x2}+\color{#090}{y}}3\ge\left(\left(\color{#C00}{\frac x2}\right)^2\color{#090}{y}\right)^{1/3}}^\text{AM-GM}=\left(\frac{15}4\right)^{1/3}
$$
But as Alan says, we have $y=\frac{15}{x^2}$, so
$$
x+y=x+\frac{15}{x^2}
$$
which can be made as large as we wish by making $x$ very small or very big.
If we consider $x\lt0$, then $x+\frac{15}{x^2}:(-\infty,0)\to(-\infty,\infty)$, so there is no minimum or maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Factorizing $\sum_{abc} a(b^3-c^3)$
Factor
$$\sum_{abc} a(b^3-c^3)$$
The solution given is as follows:
If we consider it as $f(a)$ then when $a=b$, $f(b)= 0$; thus, $b$ is a factor of $f(a)$. And since the equation is cyclic, other two factors are also cyclic. So the factors are:
$$(a-b)(b-c)(c-a)$$
Now, my first doubt is why are we ignoring the other factors. For example, for $f(a)$, $a=b$ and $a=c$, $f(b)=f(c)=0$; thus, for $f(a)$ the factors should be $(a-b)(a-c)$.
Thus the overall factor of our polynomial then comes out to be:
$$-(a-b)^2(b-c)^2(c-a)^2$$
Why is this not considered?
Besides this doubt, the book further says, since our polynomial is in a degree $4$ and the factors are of degree $3$, we need an extra first degree cyclic expression which will be another factor. Which is $m(a+b+c)$, with $m$ being a constant.
The conclusion is that the factors of the given polynomial are
$$(a-b)(b-c)(c-a)m(a+b+c)$$
How did we guessed this extra term just by knowing that our degree was off?
| A different point of view: looking for a determinantal common expression, a method that often works.
If one knows Vandermonde determinants like this one
$$\det \begin{pmatrix}
1 &1 &1 \\
a &b &c \\
a^2 &b^2 &c^2\end{pmatrix}=(a-b)(b-c)(c-a) \\$$
the similitude is evident ; you just have to modify the last line:
$$\det \begin{pmatrix}
1 &1 &1 \\
a &b &c \\
a^3 &b^3 &c^3\end{pmatrix} \\$$
And now, it remains to expand this determinant in two different ways:
*
*with respect to the second line for the initial expression
*by using $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$, giving:
$$\det \begin{pmatrix}
1 &0 &0 \\
a &b-a &c-a \\
a^3 &b^3-a^3 &c^3-a^3\end{pmatrix} =\det \begin{pmatrix}
1 &0 &0 \\
a &(b-a) &(c-a) \\
a^3 &(b-a)(b^2+ba+a^2) &(c-a)(c^2+ca+a^2)\end{pmatrix} $$
$$=(b-a)(c-a)(c^2+ca+a^2-b^2-ba-a^2) =(b-a)(c-a)(c-b)(c+b+a)$$
as desired.
One can find this expansion as well here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving for the derivative of absolute value, gone wrong The absolute value $|x|$ can be represented as $\sqrt{x^{2}}$, as per this question.
Let $f(x) = |x| = \sqrt{x^{2}}$. Solving for $f'(x)$, let $u = x^{2}$. Then,
\begin{align*}\frac{df}{dx} &= \frac{df}{du}\cdot\frac{du}{dx} \\ &= \frac{d}{du}(\sqrt{u})\cdot\frac{d}{dx}(x^{2}) \\ &=\frac{1}{2\sqrt{u}}\cdot2x \\ &= \frac{x}{\sqrt{x^{2}}} \\ &= \frac{x}{|x|}\end{align*}
We can also see that $\sqrt{x^{2}} = \left(\sqrt{x}\right)^{2}$. Then, let $u = \sqrt{x}$. Solving for $f'(x)$,
\begin{align*}\frac{df}{dx} &= \frac{df}{du}\cdot\frac{du}{dx} \\ &= \frac{d}{du}(u^{2})\cdot\frac{d}{dx}(\sqrt{x}) \\ &= 2u\cdot \frac{1}{2\sqrt{x}} \\ &= \frac{\sqrt{x}}{\sqrt{x}} \\ &= 1\end{align*}
I think the problem here is by letting $u = \sqrt{x}$. What seems to be the problem?
| Consider real valued functions $f(x)=\sqrt { x^2}$ and $g(x)=(\sqrt{ x})^2$. $f$ and $g$ have different domains. Domain of $f=\mathbb R$ and that of $g$ is $[0,\infty)$ and hence $f$ and $g$ are not the same.
We'll have $f(x)=g(x)$ if $x\in [0,\infty)$.
Therefore in second part, you have unknowingly assumed that $x\gt 0$ and therefore derivative is $1$ which tallies with your result in first part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How many different ways to distribute 30 different books to 3 people, such that the number of books given to each of them create an arithmetic series?
How many different ways to distribute $30$ different books to $3$ people, such that the number of books given to each of them create an arithmetic series?
So, I thought one of them should have exactly $10$ books, and the others can partition rest of the books however they like.
Example: $1 - 10 - 19$
Choose $10$ books, $\binom{30}{10}$, any of them can have exactly $10$ books, so multiply it by $3$, also others can partition the books however they like, which is $2^{20}$.
In total $\binom{30}{10} \times 3 \times 2^{20}$.
Then, I guess I should be done here. Except, when all of them have $10$ books, so $10-10-10 $ right? Therefore I should discard $2 \times \binom{30}{10} \times \binom{20}{10}$.
In conclusion, $\ 3 \times 2^{20} \times \binom{30}{10} - (2 \times \binom{30}{10} \times \binom{20}{10})$
Is my answer true? If you have an another approach, can you share? Thank you in advance.
| It is better to treat the $10-10-10$ case completely separately from the rest. The number of ways out of $2^{20}$ for the other two people to get $10$ books is $\binom{20}{10}$. The number of ways to distribute the books equally is the multinomial coefficient $\binom{30}{10,10,10}=\binom{30}{20}\binom{20}{10}$. Thus the answer is $3\binom{30}{10}\left(2^{20}-\binom{20}{10}\right)+\binom{30}{10,10,10}$, which matches your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4124335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Problem on finding the maxima of a function I need to find the maximum value of $f(x) = (2\sin A + 3\cos A + 4)^2 * (6-2\sin A-3\cos A)^3$ as $A$ ranges over all real numbers.
For this I performed the derivative tests by plugging in $2\sin A + 3\cos A = t$, which yielded me the value of $A = \tan^{-1}[-2/3]$.
Here's where I am stuck. As A lies in the second or fourth quadrant, I am not sure as to what signs I should pick for $\sin A$ and $\cos A$. Calculation gave me the maximum value to be $3456$ at $\sin A = \sqrt{9/13}$
and $\cos A = \sqrt{4/13}$, after taking the positive value of sinA and the negative value of cosA... but why?
Please help me out with this problem. Thanks in advance.
| For the ease of calculation, let $t=6-(2\sin A+3\cos A)$
$$\implies6-\sqrt{2^2+3^2}\le t<6+\sqrt{2^2+3^2}$$
$$g(t)=(10-t)^2\cdot t^3$$
Using AM-GM inequality for $(10-t,t>0\iff0<t<10$
$$\dfrac{10}5=\dfrac{2\cdot\dfrac{10-t}2+3\cdot\dfrac t3}{2+3}\ge\sqrt[2+3]{\left(\dfrac{10-t}2\right)^2\left(\dfrac t3\right)^3}$$
the maximum of the RHs will be attained if $$\dfrac{10-t}2=\dfrac t3\iff30-3t=2t\iff t=6$$
$$\iff2\sin A+3\cos A=0$$
$$\iff\dfrac{\sin A}3=\dfrac{\cos A}{-2}=\pm\sqrt{\dfrac{\sin^2A+\cos^2A}{(3)^2+(-2)^2}}$$
So, we need $\sin A=\dfrac{3b}{\sqrt{13}},\cos A=\dfrac{-2b}{\sqrt{13}}$ where $b=\pm1$
So, we only need opposite signs of $\sin A,\cos A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4125160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find out intersections of two absolute value functions? If we are asked to find out common points where two equations intersect each other, for example let us assume that two equations are $Y=12-x^2$ and $Y=4x$, we can solve it in the following way:
\begin{align*}4x&= 12-x^2\\
x^2+4x-12&=0\\
x^2+6x-2x-12&=0\\
x(x+6)-2(x+12)&=0\\
(x-2)(x+6)&=0
\end{align*}
So we have now two roots of $x$, $2$ and $-6$.
Plugging these values in one of the above equations we get two values for $y$, $8$ and $-24$.
So in $(2,8)$ and $(-6,-24)$ the above two lines intersect each other.
But what will we do if we are asked to find out the points where another two equations, $y=|x|$ and $y=|x^2-4|$ meet?
Is there any methodical approach which help find out the common points?
| Here's one way to approach it:
We can rewrite the absolute value functions in piecewise notation:
$$y = |x| = \begin{cases} -x, & x\le 0\\
x, &x>0
\end{cases}$$
and
$$y = |x^2 - 4| = |(x-2)(x+2)| = \begin{cases} x^2 - 4, &x\in (-\infty,-2)\cup(2,\infty)\\4 - x^2, &x\in [-2,2]
\end{cases}.$$
So, looking at overlapping domains, you need to find:
*
*Intersection points of $y = -x$ and $y = x^2 - 4$ for $x \le -2$.
*Intersection points of $y = -x$ and $y = 4-x^2$ for $x \in (-2,0)$.
*Intersection points for $y = x$ and $y = 4-x^2$ for $x\in [0,2)$.
*Intersection points for $y = x$ and $y = x^2 - 4$ for $x \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_0^1 \frac{f(x)}{\sqrt{1+x^2}}dx$
Let $f(x)$ be continuous on $[0;1]$, with $f(0) = 0; f(1) = 1$ and
$$\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx = \dfrac{1}{\ln\left(1+\sqrt{2}\right)}$$
Find ${\displaystyle \int_0^1} \dfrac{f(x)}{\sqrt{1+x^2}} \,dx$
*
*Attempt:
I tried to use Cauchy-Scharwz as below:
$$\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx \cdot \int_0^1 \dfrac{f(x)}{\sqrt{1+x^2}} \,dx \geq \left(\int_0^1 \sqrt{[f'(x)]^2 \sqrt{1+x^2}} \cdot \sqrt{\dfrac{f(x)}{\sqrt{1+x^2}}} \,dx\right)^2$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left(\int_0^1 f'(x) \sqrt{f(x)} \,dx\right)^2$$
I was able to find ${\displaystyle \int_0^1} f'(x) \sqrt{f(x)} \,dx = \dfrac{2}{3}$, but the problem is I can't show if the equality is happen or not, so my attempt isn't helpful at all.
Is there a better way to approach this?
| From the @VIVID answer, we can then find $f(x)$ and solve for the original integral
As he shows:
$$1 = \int_0^1 \frac{1}{\sqrt{1+x^2}}dx\cdot\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx \geq \left(\int_0^1 f'(x) dx\right)^2 = 1$$
The equality happens when:
$$[f'(x)]^2 \sqrt{1+x^2} = \frac{k}{\sqrt{1+x^2}} \,\,\text{(k is a constant)}$$
$$\implies f'(x) = \sqrt{k} \cdot \frac{1}{\sqrt{1+x^2}} \,\,\,\text{ or }\,\,\, f(x) = \sqrt{k} \,\ln(|\sqrt{x^2 + 1} + x|) + C$$
From the conditions $f(0) = 0; f(1) = 1$, you can find $k$ and $C$. Hence the function will be:
$$f(x) = \frac{1}{\ln(1 + \sqrt{2})} \cdot \ln(|\sqrt{x^2 + 1} + x|)$$
Now solve for the integral:
$$\int_0^1 \frac{f(x)}{\sqrt{1+x^2}} \, dx = \frac{1}{\ln(1 + \sqrt{2})} \cdot \int_0^1 \frac{\ln(|\sqrt{x^2 + 1} + x|)}{\sqrt{1+x^2}} \, dx$$
Let $u = \ln(|\sqrt{x^2 + 1} + x|)$ then the integral become:
$$\frac{1}{\ln(1 + \sqrt{2})} \cdot \int_0^{\ln(1 + \sqrt{2})} u \,du = \color{red}{\frac{1}{2} \cdot \ln(1 + \sqrt{2})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
$2$ circles internally tangent to a third one with given values of radii and length of 2 chords. Find the value of a certain product of lengths The line AB goes through P. The radius of the red circle is 81, the large black circle is 99, and the blue circle 18. The line XY is tangent to the red and blue circles.
Find NX*NA.
I made an attempt via finding many side lengths (Found that the distance between O_2 and X is 54, and that ABN and YXN are similar. However, this didn't lead to anywhere. I also found the ratio between AX and XN is 9/2.
I've also found that sin(ANB) is 2/3.
|
Edit: Adding a solution using similarity of triangles and Pythagoras theorem
$\small O_2X^2 = XT^2 + O_2T^2 = XT^2 + O_1O_2^2 - (O_1X - XT)^2= 18^2 + 81^2 - 63^2$
$\implies \small O_2X = 54$.
As $O_1O = O_2P = 18, O_1P = OO_2 = 81, $ then we must have $ \small \ O_1P \parallel OB$.
Now you can show $ \small \angle AO_1P = \angle OO_1O_2$.
If $\small AU \perp O_1P \text { and } O_2W \perp AO, \triangle O_1 O_2 W \cong \triangle O_1 A U$.
$\small AU^2 = AO_1^2 - O_1U^2 = 81^2 - 9^2, AP^2 = AU^2 + PU^2 = 81^2 - 9^2 + (81-9)^2$
$ \displaystyle \small AU = 36 \sqrt{5}, AP = 108, AV = \frac{11}{9} AU = 44 \sqrt5, OV = 11$
$\small AO_2^2 = AV^2 + VO_2^2 \implies AO_2 = 54 \sqrt5$
Now note that $ \displaystyle \small \angle O_2AX = \frac{1}{2} \angle O_2O_1X \implies \triangle O_2AS \sim \triangle O_2O_1M$
That leads to $ \small O_2S = 18 \sqrt5$ and applying Pythagoras theorem gives you $\small AS$ and $\small SX$. Adding them,
$\small AX = 36 (\sqrt{10}+1)$
As $ \small \ O_1P \parallel OB, \angle AXP = \angle ANB \implies PX \parallel BN$
$ \therefore \displaystyle \small \frac{AN}{AX} = \frac{AB}{AP} = \frac{OB}{O_1P} = \frac{99}{81} = \frac{11}{9} \implies
AN = \frac{11}{9} \cdot AX, XN = \frac{2}{9} \cdot AX$
$ \displaystyle \small NX \cdot NA = \frac{2}{9} \cdot \frac{11}{9} \cdot AX^2 = 352 (11 + 2 \sqrt{10})$
Solution using trigonometry:
For calculation of $ \small AP$ and $ \small O_2X$ and establishing $ \small PX \parallel BN$, please see previous solution.
If $ \displaystyle \small \angle PAO_2 = \alpha$, $\sin \alpha = \sin (\frac{1}{2} \angle PO_1O_2) = \frac{9}{81} = \frac{1}{9}$
If $ \displaystyle \small \angle O_2 A X = \beta$, $\sin \beta = \sin (\frac{1}{2} \angle O_2 O_1 X) = \frac{27}{81} = \frac{1}{3}$
If $ \small \angle AXP = \gamma \text { and } R \text { is circumradius of } \triangle APX, R = 81$.
$ \displaystyle \small 2R \sin \gamma = AP \implies \sin \gamma = \frac{2}{3}$
$ \small AX = 2 R \sin \angle APX = 2 R \sin (180^0 - (\alpha + \beta) - \gamma) \small = 162 \sin (\alpha+\beta+\gamma)$
$ \displaystyle \small \sin (\alpha + \beta) = \frac{1}{9} \cdot \frac{2\sqrt2}{3} + \frac{1}{3} \cdot \frac{4\sqrt5}{9} = \frac{4\sqrt5+2\sqrt2}{27}$
$ \displaystyle \small \cos(\alpha + \beta) = \frac{8\sqrt{10}-1}{27}, \cos \gamma = \frac{\sqrt5}{3}$
$ \displaystyle \small \sin(\alpha+\beta+\gamma) = \frac{2 (\sqrt{10}+1)}{9}$
So $ \small AX = 36 (\sqrt{10}+1)$
From here, follow the same steps as in previous solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In $\triangle{ABC}$, $\angle ABC=45^ \circ$. $X$ is a point on $BC$ such that $BX=\frac{1}{3}BC$ and $\angle AXC=60^ \circ$. Find $\angle ACB$. Problem
In $\triangle{ABC}$, $\angle ABC=45^ \circ$. $X$ is a point on $BC$ such that $BX=\frac{1}{3}BC$ and $\angle AXC=60^ \circ$. Find $\angle ACB$.
The problem looks easy. Though I couldn't solve it in an efficient way. Finally I solved it using trigonometry.
Trig solution
Let $BX = a$ units, then $BC = 3a$ and $XC = 3a-a= 2a$ units. $\angle AXC =60^ \circ$ and $\angle ABC= 45^ \circ$, then $\angle BAX= 60^ \circ -45^ \circ = 15^ \circ$.
Applying sine rule in $\triangle ABX$,
$$\frac{BX}{\sin \angle BAX} = \frac{AX}{\sin \angle ABC}$$
$$\implies \frac{a}{\sin 15°} = \frac{AX}{\sin 45°} \tag{1}$$
In $∆AXC$ , let $\angle ACB = \theta$, then $\angle XAC = (120 - \theta)$ and by sine rule,
$$\frac{XC}{\sin \angle XAC} = \frac{AX}{\sin \angle ACB}$$
$$\implies \frac{2a}{\sin (120 - \theta)} = \frac{AX}{\sin \theta} \tag{2}$$
Dividing $(1)$ by $(2)$,
$$\frac{\sin (120-\theta)}{2\sin 15^ \circ}= \frac{\sin \theta}{\sin 45°}$$
$$\implies 2\sin 15°\cdot\sin \theta = \sin 45°\cdot\sin (120-\theta)$$
$$\implies \frac{\sqrt{3}–1}{\sqrt 2}.\sin \theta = \frac{1}{\sqrt 2}.(\sin120°.\cos \theta - \cos 120°.\sin \theta).$$
$$\implies (\sqrt{3}–1).\sin \theta = \frac{\sqrt 3}{2}.\cos \theta + \frac{1}{2}.\sin \theta$$
$$\implies \tan \theta= 2+\sqrt 2$$
$$\implies \theta=75^ \circ$$
Thus, $\angle ACB = 75°$.
This solution is impossible without knowing the values of $\sin 15^ \circ$ and $\tan 75^ \circ$. And I find trigonometry boring. So, can this problem be solved in some other ways?
|
Draw a perp from $C$ to $AX$. Now as $ \displaystyle \angle AXC = 60^0, OX = \frac{CX}{2} = BX$
As $\angle AXB = 120^0$, $\angle XOB = \angle OBX = 30^0$. Also, $\angle OCX = 30^0$.
So $OB = OC$.
Now $\angle ABO = \angle XAB = 15^0 \implies OA = OB$
$OA = OB = OC$ and hence $O$ is circumcenter of the $\triangle ABC$.
As $\angle AOB = 150^0$, $\angle ACB = 75^0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4131869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to find the critical points of $f(x,y,z) = x^4 + y^4 + z^4 - 4xyz$? So I'm trying to find the critical points of this function $f(x,y,z) = x^4 + y^4 + z^4 - 4xyz$, to do that I try to find the points where the gradient of f is equal to $(0,0)$, though I can't solve the systems of equations, is this even the right way to do it?
Here's the system of equations that I'm trying to solve:
$4x^3-4yz = 0$
$4y^3-4xz = 0$
$4z^3-4xy = 0$
I first do this:
$x^3 = yz$
$y^3 = xz$
$z^3 = xy$
Then this:
$y = \dfrac{x^3}{z}$
$z = \dfrac{y^3}{x}$
$x = \dfrac{z^3}{y}$
But then I just go in circles by replacing each variable, I'm probably doing something wrong but I can't see it...
| Once I have
$x^3=yz$
$y^3=xz$ and
$z^3=xy$
I would notice that dividing the first equation by the second eliminates z: $\frac{x^3}{y^3}= \frac{y}{x}$ so that $x^4= y^4$. That has the two solution y= x and y= -x.
If y= x, the three equations are
$x^3= xz$ so $x^2= z$
$x^3= xz$ again, and
$z^3= x^2$.
Since $z= x^2$, $x^6= x^2$.
$x^6- x^2= x^2(x^4- 1)= x^2(x^2- 1)(x^2+ 1)= 0.
there are three solutions.
1)x= 0, y= 0, z= 0.
2)x= 1, y= 1, z= 1.
3)x= -1, y= -1, z= -1.
Now, if y= -x, then the three equations become
$x^3= -xz$ so that $x^2= -z$
$-x^3= xz$ again and
$z^3= -x^2$.
Since $z= -x^2$, $z^3= -x^6= -x^2$
$x^6- x^2= x^2(x^4- 1)= x^2(x^2- 1)(x^2+ 1)= 0$
That gives the same x values as before but now y= -x and $z= -x^2$. We have
*
*x= 0, y= 0, z= 0
*x= 1, y= -1, z= -1
*x= -1, y= 1, z= -1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
I tried to use Newton's binomial formula as:
$$
(1 + x + x^2)^6 = \sum_{k = 0}^{6}\left( \binom{6}{k}(1 + x)^{n-k} x^{2k}\right)
$$
and that's all I can think of, other then just to compute it.
| We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{[x^6]}&\color{blue}{((1+x)+x^2)^6}\\
&=[x^6]\sum_{k=0}^6\binom{6}{k}x^{2k}(1+x)^{6-k}\tag{2}\\
&=\sum_{k=0}^3\binom{6}{k}[x^{6-2k}](1+x)^{6k-k}\tag{3}\\
&=\sum_{k=0}^3\binom{6}{k}\binom{6-k}{6-2k}\tag{4}\\
&=\binom{6}{0}\binom{6}{6}+\binom{6}{1}\binom{5}{4}+\binom{6}{2}\binom{4}{2}+\binom{6}{3}\binom{3}{0}\\
&=1\cdot1+6\cdot 5+15\cdot 6+20\cdot 1\\
&\,\,\color{blue}{=141}
\end{align*}
Comment:
*
*In (2) we use (1) and apply the binomial theorem.
*In (3) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$. We also set the upper limit to $3$ since other indices do not contribute.
*In (4) we select the coefficient of $x^{6-2k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 4
} |
Evaluation of $\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=2-2\ln(2)$ I came across the following statements
$$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \tag{2}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(4 n+1)}=4-\frac{\pi}{2}-3 \ln 2 \qquad \tag{3}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(6 n+1)}=6-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2 \qquad \tag{4}$$
The (1) by partial fractions
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{2}{2n+1}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}$$
Recall the Digamma function
$$\psi(x+1)=\gamma+\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+x}$$
Therefore
$$\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}=\psi(1+\frac{1}{2})-\gamma$$
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\psi\left(\frac{3}{2}\right)-\gamma$$
In the same token we can derive the relation for the other three ralations. My Question is: can we calculate the values of the digamma function for those values without resorting in the Gauss´s Digamma formula?
$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln (2 m)-\frac{\pi}{2} \cot \left(\frac{r \pi}{m}\right)+2 \sum_{n=1}^{\left\lfloor\frac{m-1}{2}\right\rfloor} \cos \left(\frac{2 \pi n r}{m}\right) \ln \sin \left(\frac{\pi n}{m}\right)$$
I tried this approach also, but I think the resulting integral is divergent $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}x^{2n}dx=\int_{0}^{1}\sum_{n=1}^{\infty} \frac{x^{2n}}{n}=-\int_{0}^{1}\ln(1-x^2)dx $$
| $\def\o{\omega_k}$
$$
\sum_{n = 1}^\infty {\frac{1}{{n(2n + 1)}}} = 2\sum_{n = 1}^\infty\left(\frac1{2n}-\frac1{2n+1}\right)=2\left(1-\sum_{n = 1}^\infty\frac{(-1)^n}n\right)=2-2\log2.
$$
Observe the additional factor $2$ comparing with a typo in your question.
Inspired by this result we can elaborate the general solution:
We note that:
$$S_k=\sum_{n=1}^\infty\frac1{kn(kn+1)}=\sum_{n = 1}^\infty\left(\frac1{kn}-\frac1{kn+1}\right)=1-\frac1k
\sum_{j=1}^k(1-\o^{-j})\log\left(1-\o^j\right),\tag1$$
where
$\o=e^{\frac{2\pi i}k}$
is the $k$-th primitive root of unity.
To verify $(1)$ observe:
$$
\frac1k\sum_{j=1}^k(\o^j)^p=\mathbb{1}_{p\equiv0\pmod k}.
$$
Of course it is of interest to represent the solution (1) in terms of real parameters used in the question. This can be done as follows. Observe that we can omit in the sum the term $j=k$ and combine the complex conjugated terms $j$ and $k-j$ to obtain:
$$
S_k=1-\frac1k\left\{\underbrace{\left[1+(-1)^k\right]}_{s_k}\log2+
\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}
\left[(1-\o^{-j})\log\left(1-\o^j\right)+(1-\o^{j})\log\left(1-\o^{-j}\right)\right]\right\},\\
$$
where we singled out the unpaired term $2\log2$ for $j=\frac k2$ in the case of even $k$.
The sum can be rewritten as:
$$\begin{align}
&\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}\left\{\left(1-\cos\frac{2\pi j}k\right)\log\left(\left[1-e^{\frac{2\pi j}ki}\right]\left[1-e^{-\frac{2\pi j}ki}\right]\right)+i\sin\frac{2\pi j}k\log\left(
\frac{1-e^{\frac{2\pi j}ki}}{1-e^{-\frac{2\pi j}ki}}\right) \right\}\\
&=\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}\left\{\left(1-\cos\frac{2\pi j}k\right)\log\left(2-2\cos\frac{2\pi j}k\right)+\pi\left(1-\frac{2j}k\right)\sin\frac{2\pi j}k \right\},\\
\end{align}$$
so that finally:
$$\begin{align}
S_k&=1-\frac1k\left\{s_k\log2+\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}
\left[4\sin^2\frac{\pi j}k \log\left(2\sin\frac{\pi j}k\right)+
\pi\left(1-\frac{2j}k\right)\sin\frac{2\pi j}k\right]\right\}\\
&=1-\log2-\frac1k\left\{4\sum_{j=1}^{\left\lfloor\frac {k-1}2\right\rfloor}\sin^2\frac{\pi j}k \log\left(\sin\frac{\pi j}k\right)+
\frac\pi2\cot\frac\pi k\right\}.\tag2
\end{align}$$
The derivation of the last line is given below.
Appendix
We are going to prove:
$$Z_k=\sum_{0<j<\frac k2}
\left(1-\frac{2j}k\right)\sin\frac{2\pi j}k=\frac12\cot\frac\pi k.$$
First we observe that due to symmetry:
$$
Z_k=\frac12\sum_{0\le j<k}\left(1-\frac{2j}k\right)\sin\frac{2\pi j}k.
$$
Next:
$$\begin{align}
4Z_k\sin\frac\pi k&=\sum_{0\le j<k}\left(1-\frac{2j}k\right)
\left(\cos\frac{\pi(2j-1)}k-\cos\frac{\pi(2j+1)}k\right)\\
&=\sum_{0\le j<k}\left(1-\frac{2j}k\right)\cos\frac{\pi(2j-1)}k
-\sum_{1\le j\le k}\left(1+\frac2k-\frac{2j}k\right)\cos\frac{\pi(2j-1)}k\\
&=\underbrace{\cos\frac{\pi}k}_{j=0}
+\underbrace{\cos\frac{\pi}k}_{j=k}
-\frac2k\underbrace{\sum_{1\le j\le k}\cos\frac{\pi(2j-1)}k}_{=0}\\
&=2\cos\frac{\pi}k.
\end{align}$$
The proof of the another identity used in $(2)$:
$$\sum_{0<j<k}\sin^2\frac{\pi j}k=\frac k2$$
valid for $k\ge2$ is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
How to find the unit digit in the product of exponents?
What is the unit digit in the product $3547^{153}\times251^{72}$?
I know that the required digit is equal to the unit digit in $7^{153}\times1^{72}$. And I also know that $1^{72}=1$, as $1$ is the multiplicative identity. But I am having trouble evaluating $7^{153}$.
Similarity,
What is the unit digit in $264^{102}+264^{103}$?
$$264^{102}+264^{103}=264^{102}×(1+264)=264^{102}\times265=\cdots$$
| $251^{72} \equiv 1 \pmod{10}$
$3547^{153} \equiv 7^{153} \pmod{10}$
Cyclic sequence of $7^{n}$ is as follows.
$7^{1} = 7$
$7^{2} = 49$
$7^{3} = 343$
$7^{4} = 2401$
Lets divide $153$ by $4$ and the remainder is $1$.
Thus, the unit digit of $7^{153}$ is equal to the unit digit of $7^{1} = 7.$
Hence the unit digit of $3547^{153}\times251^{72}$ is equal to $7.$
$264^{102} \equiv 4^{102} \pmod{10}$
Cyclic sequence of $4^{n}$ is as follows.
$4^{1} = 4$
$4^{2} = 16$
Lets divide $102$ by $2$ and the remainder is $0$.
Thus, the unit digit of $4^{102}$ is equal to the unit digit of $4^{2} = 6.$
Hence the unit digit of $264^{102}+264^{103}$ is equal to $0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4146309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Compute $\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$ Compute the given line integral:
*
*$$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$$
Let $P=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}},Q=\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}, R=\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}$ then $$\color{blue}{\frac{\partial P}{\partial y}}=\frac{-xy}{(x^2+y^2+z^2)^{3/2}},\color{green}{\frac{\partial P}{\partial z}}=\frac{-xz}{(x^2+y^2+z^2)^{3/2}}$$$$\color{blue}{\frac{\partial Q}{\partial x}}=\frac{-xy}{(x^2+y^2+z^2)^{3/2}},\color{red}{\frac{\partial Q}{\partial z}}=\frac{-yz}{(x^2+y^2+z^2)^{3/2}}$$$$\color{green}{\frac{\partial R}{\partial x}}=\frac{-xz}{(x^2+y^2+z^2)^{3/2}},\color{red}{\frac{\partial R}{\partial y}}=\frac{-yz}{(x^2+y^2+z^2)^{3/2}}$$
So the vector field is conservative which means there is $f$ such that $\vec \nabla f=F$, I don't know how to continue.
| Compute the given line integral:
$$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$$
This can be made a lot simpler by change of coordinates into spherical. Recall that:
$$ \rho= \sqrt{x^2 +y^2 +z^2}$$
This leads to:
$$ d \rho = \frac{x dx + y dy + zdz}{\sqrt{x^2 + y^2 +z^2}}$$
Hence, our integral just becomes to evaluate:
$$ \rho(3,4,5) - \rho(0,0,0)=5\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
value of a $\frac{ p_t(z)}{z^m}$ Fundamental Theorem of Algebra by using Algebraic topology
In wikipidea it is written that
let $p(z)$ be a polynomial in $\mathbb{C}$ ;
$p(z)=z^m +a_1z^{m-1} +\cdots+ a_m$
Define a homotopy: $p_t(z)=tp(z) + (1-t)z^m$
then
$$\frac{ p_t(z)}{z^m} =1 + t \left(\frac{a_1}{z} +\cdots+\frac{a_m}{z}\right)$$
My doubt :Im not getting why $\frac{p_t(z)}{z^m} =1 + t (\frac{a_1}{z} +\cdots+\frac{a_m}{z})\text{?}$
My thinking :
\begin{align}
& \frac{ p_t(z)}{z^m} =\frac{tp(z)}{z^m} + \frac{(1-t)z^m}{z^m} \\[8pt]
= {} & \frac{tp(z)}{z^m}+(1-t)\neq1 + t \left(\frac{a_1}{z} +\cdots+\frac{a_m}{z}\right)
\end{align}
| $p_t(z) = tp(z) + (1-t)z^m = z^m + t(p(z) - z^m)$ so $$\frac{p_t(z)}{z^m} = 1 + t(\frac{p(z)}{z^m} - 1) = 1 + t(1 + \frac{a_1}{z} + \frac{a_2}{z^2} + \cdots + \frac{a_m}{z^{m-1}} - 1) = \\ 1+t(\frac{a_1}{z} + \frac{a_2}{z^2} + \cdots + \frac{a_m}{z^{m-1}})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The equation $\tan x = \tan 2x \tan 4x \tan 8x$ In the question we have the equality
$$\tan 6^{\circ} \tan 42^{\circ} = \tan 12^{\circ} \tan 24^{\circ}$$
which is equivalent to
$$ \tan 6^{\circ} = \tan 12^{\circ} \tan 24^{\circ} \tan 48^{\circ}$$
This means that the equation
$$\tan x = \tan 2x \tan 4x \tan 8x$$
has the solution $x =6^{\circ} = \frac{\pi}{30}$. How to find all the solution of this equation?
| Factorize the equation as follows
\begin{align}
& \tan x-\tan 2x \tan 4x \tan 8x \\
=& \frac{\sin x \cos 2x \cos 4x \cos 8x-\cos x \sin 2x \sin 4x \sin 8x }
{\cos x \cos 2x \cos 4x \cos 8x }\\
=& \frac{\sin x [\cos 2x \cos 4x \cos 8x-(1+\cos 2x)\sin 4x \sin 8x ]}
{\cos x \cos 2x \cos 4x \cos 8x }\\
=& \frac{\tan x (\cos 12x -2\sin 2x \sin 8x )}
{\cos 4x \cos 8x }\\
=& \frac{\tan x }{\cos8x} \left( 4 \cos^2 4x-3 -4\sin 2x \sin 4x\right)\\
=& \frac{\tan x }{\cos8x} \left(16\cos^4 2x +8\cos^3 2x -16\cos^2 2x-8\cos 2x +1 \right)\\
=& \frac{16\tan x }{\cos8x} \left(\cos^2(2x-\frac{2\pi}3) - \frac12 \cos(2x-\frac{2\pi}3) -\frac14\right)\\
&\hspace{15mm}\times \left(\cos^2(2x+\frac{2\pi}3) - \frac12 \cos(2x+\frac{2\pi}3) -\frac14\right)\\
=& \frac{16\tan x }{\cos8x} \left(\cos(2x-\frac{2\pi}3) - \cos\frac\pi5\right)\left(\cos(2x-\frac{2\pi}3) - \cos\frac{3\pi}5\right)\\
&\hspace{15mm}\times \left(\cos(2x+\frac{2\pi}3) - \cos\frac\pi5\right)\left(\cos(2x+\frac{2\pi}3) - \cos\frac{3\pi}5\right)\\
\end{align}
where $ \cos\frac{\pi}5 +\cos\frac{3\pi}5 =\frac12$ and $ \cos\frac{\pi}5\cos\frac{3\pi}5 =-\frac14$ are recognized. Then, the full set of the solutions are $x=n\pi,\>n\pi \pm \frac{\pi}3\pm \frac\pi{10}$ and $n\pi \pm \frac{\pi}3\pm \frac{3\pi}{10}
$, or
$$x-n\pi=0, \>\pm \frac\pi{30}, \>\pm \frac{7\pi}{30} , \>\pm \frac{13\pi}{30} , \>\pm \frac{19\pi}{30}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int_0^\infty \frac{dx}{(x+\sqrt{1+x^2})^2}$ How to evaluate this definite integral from MIT Integration Bee 2006?
$$\int_0^\infty \frac{dx}{(x+\sqrt{1+x^2})^2}.$$
So far, I have shown that the indefinite integral is
$$\frac{2x^3 + 3x - 2(1+x^2)^{3/2}}{3}.$$
At $x = 0$, the expression above equals $-\dfrac{2}{3}.$
Using WolframAlpha, I also know that the definite integral equals $\dfrac{2}{3}$.
So the only thing left to show is $$\lim_{x\rightarrow \infty} \frac{2x^3 + 3x - 2(1+x^2)^{3/2}}{3}=0.$$
I'm not sure how to calculate this limit.
| It is easy to evaluate that definite integral by successive substitutions. I don't see why you evaluate the indefinite one first.
\begin{align}\int_0^\infty \frac{\mathrm dx}{(x+\sqrt{1+x^2})^2}&=\int_0^{\pi/2} \frac{\sec^2 u}{(\tan u+\sec u)^2}\mathrm du \text{ ,via $x=\tan u$}\\&=\int_0^{\pi/2}\frac{\mathrm du}{(\sin u+1)^2}\\&=2\int_0^1 \frac{1+t^2}{(1+t)^4}\mathrm dt\text{ ,via $t=\tan \frac{u}{2}$}\\&=2\int_1^2 \frac{(w-1)^2+1}{w^4}\mathrm dw\text{ ,via $w=t+1$}\\&=2\int_1^2 \left(\frac{1}{w^2}-\frac{2}{w^3}+\frac{2}{w^4}\right)\mathrm dw\\&=\frac{2}{3}\end{align}
Footnote
Weierstrass substitution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Finding coefficient of $x^{111}$ in $P=(1+x)+2(1+x)^2+3(1+x)^3+\ldots+1000(1+x)^{1000}$
Find the coefficient of $x^{111}$ in $P=(1+x)+2(1+x)^2+3(1+x)^3+\ldots+1000(1+x)^{1000}$.
Here is what I did. It is an AGP. So we can write it as follows.
$$\begin{aligned}P&=(1+x)\frac{\mathrm d}{\mathrm dx}\sum_{r=1}^{1000}(1+x)^r=\sum_{r=1}^{1000}r(1+x)^{r}\\ &= (1+x)\frac{\mathrm d}{\mathrm dx}\left[\frac{1+x}{x}\left((1+x)^{1000}-1\right)\right]\\ &= (1+x)\left[\frac{1001(1+x)^{1000}}{x}-\frac{(1+x)^{1001}}{x^2}+\frac{1}{x^2}\right]\\ &= \frac{1001}{x}\cdot(1+x)^{1001}-\frac{1}{x^2}\cdot(1+x)^{1002}+\frac{1+x}{x^2}\end{aligned}$$
So as per this expression, the coefficient of $x^{111}$ should be $1001\cdot {1001 \choose 112}-{1002\choose 113}$. But the given answer is $1000\cdot{1001\choose 112}-{1001\choose 113}$. Can somebody spot where the mistake is? Thanks.
| You made a mistake when taking the derivative, it should be $$(1+x)\left[\frac{1000(1+x)^{1000}}{x}-\frac{(1+x)^{1000}}{x^2}+\frac{1}{x^2}\right].$$ Once this is fixed, you will get the right answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
let $m,n \in N$ such that $2m^2+m=3n^2+n$, Find all integral solution of $2m^2+m=3n^2+n$?
let $m,n \in N$ such that $2m^2+m=3n^2+n$. Find all integral solution of $2m^2+m=3n^2+n$ ?
My work:
$$2m^2-2n^2+m-n=n^2$$
$$\implies 2(m-n)(m+n)+(m-n)=n^2$$
$$\implies(2m+2n+1)(m-n)=n^2$$
Case $1$
$2m+2n+1=n^2$ and $m-n=1$
Now putting $m=n+1$ in $2m+2n+1=n^2$
which lead me to $2(n+1)+2n+1=n^2$
$$\implies n^2-4n-3=0$$
above equation give me irrational roots.
Case 2
$2m+2n+1=n$ and $m-n=n$
Now putting $m=2n$ in $2m+2n+1=n$
which leads me to $2(2n)+2n+1=n$
$\implies n=\frac{-1}{5}$
which is not a natural number.
I am not getting any solution to the given equations. Is my approach correct?
Also is it necessary that $(m-n)$ and $(2m+2n+1)$ must be prefect square
Please Suggest a solution without using Modular Arithmetic
| $$\begin{align*}
2m^2+m&=3n^2+n \\
\frac{1}{8}\left((4m+1)^2-1\right)&=\frac{1}{12}\left((6n+1)^2-1\right)\\
3(4m+1)^2-3&=2(6n+1)^2-2
\end{align*}$$
Let $x=(4m+1)^2$ and $y=(6n+1)^2$, we have,
$$3x^2-2y^2=1$$
The above equation has been solved here
The general solution is obtained from the recurrences,
$$x_{n+1}=5x_n+4y_n,\;y_{n+1}=6x_n+5y_n,\;(x_1,y_1)=(1,1). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$n^3+n+1$ is perfect square find all positive integers $n$ for which $n^3+n+1$ is a perfect square
$n^3+n+1$ is odd so $8|n^3+n$ so $8|n(n^2+1)$ if $2|n^2+1$ ,$v_2(n^2+1)=1$ so $n$ must be even, contradiction. So $8\mid n.$
Also by checking $\bmod7$ we understand that $n=0\pmod7$ or $n=2\pmod7$ so $56\mid n$ or $n=16\pmod{56}$ also $72$ is an answer!
Another approach: add $n^2$ to both sides: $(n+1)(n^2+1)=a^2+n^2$ It's obvios that $\gcd(a,n)=1$ so by vitek lemma all prime divisors of $a^2+n^2$ are $4k+1$ so the same for prime divisors of $n+1$.
Another approach: let $[\sqrt{n^3}]=c$
So $n^3<(c+1)^2$ so $n+1<c$ so $n^3+n+1<(c+2)^2$ so if $n^3+n+1=a^2$ $a$ is less than or equal to $c+1$ on the other hand it's obvious that $a>c$ so $a=c+1$
| The solutions correspond to the integral points on the elliptic curve
$$
y^2=x^3+x+1,
$$
which are given by
$$
(0,\pm 1),\;(72,\pm 611).
$$
See for example LMFDB.
It is easy to obtain the points $(72,\pm 611)$ starting from $(0,1)$, see
Find an affine point on $E:y^2=x^3+x+1$
but it requires more on elliptic curves to show that there are no other integer points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
trigonometrical inequality $\tan{A} + \tan{B} + \tan{C} = \tan{A}\tan{B}\tan{C}$, Hint to prove $S2\neq 1$? I was studying conditional identities for triangle in trigonometry where I had to prove that
$\tan{A} + \tan{B} + \tan{C} = \tan{A}\tan{B}\tan{C}$
So I started with $\tan({A+B+C)}=\frac{S1-S3}{1-S2}$ where $S1=\sum_{cyc}\tan A$, $S2=\sum_{cyc}\tan A.\tan B$ and $S3=\prod_{cyc}\tan A$
Now since $A+B+C=\pi$ and since $\tan\pi=0$, therefore $\frac{S1-S3}{1-S2}=0$ and hence $S1=S3$ but I got stuck when I realized that another condition, $1-S2\neq0$ should also be true. Now I can't prove this.
I'm not having any idea how to start to prove $S2\neq1$ for a triangle. please give me a hint.
| $~~~~\tan A\tan B+\tan B\tan C+\tan A\tan C\\=\tan A \cdot [\tan (B+C) \cdot (1-\tan B ~tan C)]+\tan B\tan C\\
=-(\tan A)^2 \cdot (1-\tan B ~tan C)+\tan B\tan C\\
=(\tan^2 A+1)(\tan B \tan C)-\tan^2 A$
Assume there exist solution for $~\tan A\tan B+\tan B\tan C+\tan A\tan C=1$
$⇒ (\tan^2 A+1)(\tan B \tan C)=\tan^2 A+1$
$⇒\tan B \tan C=1$
$~~~~~\tan A=-\tan (B+C)=\frac{\tan B+\tan C}{1-\tan B \tan C}~~$ which is not defined $($as denominator $= 0)$.
Hence $~\tan A\tan B+\tan B\tan C+\tan A\tan C \ne 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proof of Mergesort $N$ elements with $N \log N + O(N)$ comparisons In the book "An introduction to the Analysis of Algorithms", written by Robert Sedgewick and Philippe Flajolet during the proof of the
Theorem 1.1: (Mergesort) To sort an array of N elements, Mergesort uses $N \log N + O(N)$ comparions.
It contains the following part:
To get an indication for the nature of the solution to this recurrence, we consider the case when N is a power of 2:
$C_{2^n} = 2C_{2^n-1} + 2^n$ for $n \geq 1$ with $C_1 = 0$.
Dividing both sides of this equation by $2^n$, we find that
$$
\frac{C_{2^n}}{2^n} = \frac{C_{2^n-1}}{2^{n-1}} + 1 = \frac{C_{2^n-2}}{2^{n-2}} + 2 = \frac{C_{2^n-3}}{2^{n-3}} + 3 = ... = \frac{C_{2^0}}{2^0} + n = n.
$$
I can't understand the trick behind incrementing the right side of the term (+1, then +2, then +3).
Why does the change of the left side lead to the increment on the right side?
Any help is kindly appreciated.
| You basically use the fact $C_{2^n}=2C_{2^{n-1}}+2^n$ for all $n\geq 1$ recursively (I think you meant $2^{n-1}$ as index and not $2^n-1$). Dividing everything by $2^n$ yields
\begin{align}
\frac{C_{2^n}}{2^n} &= \frac{2C_{2^{n-1}}+2^n}{2^n} = \frac{2C_{2^{n-1}}}{2^n} + \frac{2^n}{2^n} = \frac{C_{2^{n-1}}}{2^{n-1}}+ 1 \\
&=\frac{2C_{2^{n-2}}+2^{n-1}}{2^{n-1}}+ 1 =\frac{2C_{2^{n-2}}}{2^{n-1}}+ 1+1 =\frac{C_{2^{n-2}}}{2^{n-2}}+ 2
\end{align}
I hope you see the recursive pattern now. If you go on until the exponent in the index has been reduced to $0$ you get:
$$
\frac{C_{2^0}}{2^0} + n = \frac{0}{1} + n = n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4159209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How many polynomials $P(x)$ of degree not exceeding 3 are there such that each coefficient is a non-negative integer less than 10, and that $P(-1)=-9$
How many polynomials $P(x)$ of degree not exceeding 3 are there such that each coefficient is a non-negative integer less than 10, and that $P(-1)=-9$?
My attempt:There would be one linear polynomial i.e. $9x=0$
Edit:Let the quadratic equations be of the form $ax^2+bx+c=0$ then the required condition would be $a+c-b=-9$ but minimum possible value of $a+cl-b$ is $-8$ so there would be no such quadratic equations.
Now, let the cubic polynomial be of the form $ax^3+bx^2+cx+d$ then we need such cubics which satisfy this condition:
$-(a+c)+b+d=9\\i.e. b+d=9+a+c$
Now $a$ can be anything between 1 and 9 while $\text{b,c and d}$ can be anything between 0 to 9.
Now, maximum value of $b+d$ would be 18. So,the above condition would reduce to $18-9=9=a+d$ with the condition that both $\text{a and b}$ are non negative integers less than 10 and $a$ can not be equal to zero. Using, combinatorics it can be easily found that number of solutions of this equation is $\binom{9}1=9$.
Second largest value of $b+d$ would be 17.Thus, giving the condition $a+c=8$ which has 8 solutions. Besides,$b+d=17$ can be written in $2$ ways. So, total number of solutions is equal to $8.2=16$.
Continuing further similarly we get,
$a+c=7=> \text{Number of solutions}=7.3=21\\ a+c=6=> \text{Number of solutions}=6.4=24\\a+c=5=> \text{Number of solutions}=5.5=25\\a+c=4=> \text{Number of solutions}=4.6=24\\a+c=3=> \text{Number of solutions}=3.7=21\\a+c=2=> \text{Number of solutions}=2.8=16\\a+c=1=> \text{Number of solutions}=1.9=9$
Adding all these should give answer. But their sum is 166 which is the wrong answer.
Correct answer: $\binom{12}3=220$
What am I doing wrong?
| $$(b-a)+(d-c)=9$$
Imagine a line with 9 red balls and 3 green balls. Consider the following example,
The number of red balls from the start till the first green ball gives you the value for $a$.
The number of red balls from the start till the second green ball gives you the value for $b-a$.
The number of red balls from the second green ball till the third green ball gives you the value for $c$.
The number of red balls from the second green ball till the end gives you the value for $d-c$.
This has $\binom{12}3=220$ solutions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4162694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\int _0^1\frac{12\arctan ^2 x\ln (\frac{(1-x)^2}{1+x^2})-\ln ^3(\frac{(1-x)^2}{1+x^2})}{x}\:dx$ I want to find and prove that: $$\int _0^1\frac{12\arctan ^2\left(x\right)\ln \left(\frac{\left(1-x\right)^2}{1+x^2}\right)-\ln ^3\left(\frac{\left(1-x\right)^2}{1+x^2}\right)}{x}\:dx=\frac{9 \pi ^4}{16}$$
I tried to split the integral:
$$=12\int _0^1\frac{\arctan ^2\left(x\right)\ln \left(\frac{\left(1-x\right)^2}{1+x^2}\right)}{x}\:dx+12\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x^2\right)}{x}\:dx$$
$$-6\int _0^1\frac{\ln \left(1-x\right)\ln ^2\left(1+x^2\right)}{x}\:dx+\int _0^1\frac{\ln ^3\left(1+x^2\right)}{x}\:dx-8\int _0^1\frac{\ln ^3\left(1-x\right)}{x}\:dx$$
But the first $3$ integrals are very tough, expanding some terms yield very complicated series.
I also attempted to integrate by parts but it was rather messy and I'd prefer not to put it here. I have faith that there must be a trick for finding this one because of its closed forms simplicity.
Note $2$: I found this integral in a certain mathematics group, I've ask for hints but received none.
| Integrals of such form (i.e. combination of $\arctan x, \log x, \log(1\pm x), \log(1+x^2$)) have systematic way of calculation. If you randomly write down such an integral, there is likely no slick way.
However, for this particular one, there is something behind. Writing $a=\arctan x, b=\log(\frac{(1-x)^2}{1+x^2})$, I will simultaneously prove OP's question as well as the following generalizations
$$\begin{aligned}I_3 &= \color{red}{\int_0^1 \frac{12a^2b-b^3}{x} dx = \frac{9\pi^4}{16}} \\
I_5 &= \int_0^1 \frac{-80 a^4 b+40 a^2 b^3-b^5}{x} dx = \frac{33 \pi ^6}{8} \\
I_7 &= \int_0^1 \frac{448 a^6 b-560 a^4 b^3+84 a^2 b^5-b^7}{x} dx = \frac{2193 \pi ^8}{32} \end{aligned}$$
Here numerator of $I_n$ is real part of $(2ai-b)^n$. Write
$$I_n=\int_0^1 \frac{\Re[(2ai-b)^n]}{x}dx = -2^n \int_0^1 \frac{\Re[(\log(1-x)-\log(1+ix))^n]}{x}dx $$
then for $n$ odd, $\frac{I_n}{\pi^{n+1}} \in \mathbb{Q}$. More precisely, if $n=2m+1$,
$$\tag{*}I_n = 2^{2n-2}3\pi^{n+1}(-1)^{m+1} \sum_{r=0}^m \binom{n}{2r+1}
\left(\frac{3}{8}\right)^{2r}
\frac{1}{n-2r}B_{n-2r}\left(\frac{3}{8}\right)$$
here $B_n(x)$ is Bernoulli polynomial, from which above cited values of $I_n$ follows.
Proof of $(*)$:
Integrate $\frac{(\log(1-z)-\log(1+iz))^n}{z}$ around quarter circle in the first quadrant, the integral from $0$ to $i$ is
$$\begin{aligned}2^n \int_0^i \frac{(\log(1-x)-\log(1+ix))^n}{x}dx &= 2^n \int_0^1 \frac{(\log(1-ix)-\log(1-x))^n}{x}dx \\
&= -2^n \int_0^1 \frac{(\log(1-x)-\log(1-ix))^n}{x}dx \end{aligned}$$
so the real part of $\int_0^i$ is $-I_n$. Hence, if $C$ is the quarter circle,
$$\begin{aligned} 2I_n &= 2^n\Re \int_C \frac{(\log(1-z)-\log(1+iz))^n}{z}dz \\
&= -2^n\Im \int_0^{\pi/2} (\log(1-e^{ix})-\log(1+ie^{ix}))^n dx
\end{aligned}$$
Now $$\log(1-e^{ix})-\log(1+ie^{ix}) = \log\left(\frac{\sin(x/2)}{\sin(\pi/4-x/2)}\right)-\frac{3i\pi}{4}$$
Hence if we write $n=2m+1$,
$$I_n = 2^n \sum_{r=0}^m \binom{2m+1}{2r+1} (-1)^r\left(\frac{3\pi}{4}\right)^{2r+1}\int_0^{\pi/4} \log^{2(m-r)}\left(\frac{\sin x}{\sin(\pi/4-x)}\right) dx$$
As a lesser known fact, integrals like those in the summand are always rational multiple of $\pi^{2(m-r)+1}$:
$$\int_0^{\pi/4}\log^{2n}\left(\frac{\sin x}{\sin(\pi/4-x)}\right) dx = \pi^{2n+1}\frac{(-1)^{n+1}2^{2n+1}}{2n+1}B_{2n+1}\left(\frac{3}{8}\right)$$
QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Find the number of dimensions and a basis $W$ of a generating set of $U$ Given the system of vectors $U= \left \{ \left ( 1, 2, 3 \right ), \left ( 0, -1, -2 \right ), \left ( 2, 3, 4 \right ), \left ( 1, 0, -1 \right ) \right \}.$ Find the number of dimensions and a basis $W$ of a generating set of $U.$
We use the rows reduction
$$\begin{pmatrix} 1 & 0 & 2 & 1\\ 2 & -1 & 3 & 0\\ 3 & -2 & 4 & -1 \end{pmatrix}\xrightarrow{2R_{1}- R_{2}}\begin{pmatrix} 1 & 0 & 2 & 1\\ 0 & 1 & 1 & 2\\ 3 & -2 & 4 & -1 \end{pmatrix}\xrightarrow{3R_{1}- 2R_{2}- R_{3}}\begin{pmatrix} 1 & 0 & 2 & 1\\ 0 & 1 & 1 & 2\\ 0 & 0 & 0 & 0 \end{pmatrix}$$
then $\operatorname{rank}= 2.$
I don't have a way of thinking to continue. What is the relationship between $W$ and $U,$ I need to the help ?!
Edit. Find the value of $k$ so that $u= \left ( 2, 3, k^{2}+ 1 \right )$ is a linear combination of $W,$ and what is $\left [ u \right ]_{W}\!.$
| Actually, you should work with the matrix$$\begin{bmatrix}1&2&3\\0&-1&-2\\2&3&4\\1&0&-1\end{bmatrix}$$instead. Note that if you add to the third row the first row times $-2$ and you add to the fourth row the first row times $-1$, then you get$$\begin{bmatrix}1&2&3\\0&-1&-2\\0&-1&-2\\0&-2&-4\end{bmatrix}.$$Now, if you add to the third row the second row times $-1$ and if you add to the fourth row the second row times $-2$, then you get$$\begin{bmatrix}1&2&3\\0&-1&-2\\0&0&0\\0&0&0\end{bmatrix}.$$So, a basis of the space spanned by $U$ is $\{(1,2,3),(0,-1,-2)\}$ and its dimension is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $(b + c) , (c + a) , (a + b)$ are in H.P. then find the relation between $\dfrac{a}{b+c} , \dfrac{b}{c+a} , \dfrac{c}{a+b}$ .
If $(b + c) , (c + a) , (a + b)$ are in H.P. then $\dfrac{a}{b+c} , \dfrac{b}{c+a} , \dfrac{c}{a+b}$ are in $(i)$ A.P. $(ii)$ G.P. $(iii)$ H.P. $(iv)$ None of These.
What I Tried:- I have that $(b+c),(c+a),(a+b)$ are in H.P. .
$\implies \dfrac{1}{b+c}, \dfrac{1}{c+a}, \dfrac{1}{a+b}$ are in A.P. .
$\implies \dfrac{1}{c+a} - \dfrac{1}{b+c} = \dfrac{1}{a+b} - \dfrac{1}{c+a}.$
$\implies \dfrac{(b + c) - (c + a)}{(c + a)(b+c)} = \dfrac{(c+a) - (a+b)}{(a+b)(c+a)}.$
$\implies \dfrac{(b - a)}{(b + c)} = \dfrac{(c - b)}{(b + a)}.$
$\implies (b^2 - a^2) = (c^2 - b^2).$
$\implies 2b^2 = a^2 + c^2.$
So I was able to prove that $a^2 , b^2 , c^2$ are in A.P. , but I am not sure how to show the claim in the question.
Can anyone help me? Thank You.
| Hint 1:
Add $1$ to each term as mentioned in the comments by @SathvikAcharya
Hint 2:
If $p,q,r$ is an arithmetic progression, $\lambda p,\lambda q,\lambda r$ is also an arithmetic progression where $\lambda$ is a real number
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A discussion about exponential diophantine equations and pythagorean triples
Do all pythagorean triples $(a,b,c)$ have the identity that the three
(exponential diophantine) equations
\begin{equation} a^x+b^y=c^z \end{equation} \begin{equation}
b^y+c^z=a^x \end{equation} \begin{equation} a^x+c^z=b^y \end{equation}
all have no more than one positive integer solution $(x,y,z)$?
This question is inspired by three exponential diophantine equations (1)$3^x+4^y=5^z$,(2) $4^y+5^z=3^x$, and (3) $3^x+5^z=4^y$.
Equations (1)(2)(3) all have no more than $1$ unique integer solution. Here, I'll give a proof for equation (3).
Find all positive integer solutions $(x,y,z)$ of the equation $3^x+5^z=4^y$.
$\because 3^x+5^z=4^y$, $\therefore (-1)^x+1^z\equiv 0\;\; (\operatorname{mod} 4)$. Also, since $x,y,z$ are all positive integers, so $x$ must be a multiple of $2$. Let $x=2a$. Then $3^{2a}+5^z=2^{2y}$, so we can get $$5^z=\left(2^y+3^a\right)\left(2^y-3^a\right).$$
Suppose $2^y+3^a=5^m, 2^y-3^a=5^n\;\;(m,n\in\mathbb{N}, m>n)$, subtracting the two equations, we get $2\cdot3^a=5^m-5^n$. If $n\geqslant1$, then $5\mid 5^m-5^n$, so $5\mid 2\cdot3^a$, which is a contradiction. Therefore, $n=0$.
We have $2^y=3^a+1$. For $a=1$, we get $y=2$. For $a\geqslant 2$, writing the equation in modulo $4$, we get $0\equiv (-1)^a+1$, so $a$ must be odd. Let $a=2k+1$, then $3^{2k+1}+1=2^a$. i.e. $3\cdot 9^k+1=2^y$. Writing the equation in modulo $8$, we get $3\cdot 1^k+1\equiv 0$. Which is another contradiction, so $(a,y)=(1,2)$, which leads us to $x=2a=2$.
Plugging in $x=2, y=2$ into the original equation (equation (3)), we get $5^z=7$. Therefore, $z$ is not an integer, so the exponential
diophantine equation $3^x+5^z=4^y$ doesn't have any solutions.
I'm curious that is the above statement (at the beginning of the question) true for all pythagorean triples. e.g. Does the exponential
diophantine equation $20^x+29^z=21^y$ have no more than one unique solution? Is there a general way to tackle with this problem?
Note:
*
*Again, I hope my question is not a duplication.
*$\mathbb{N}$ is the set of positive integers.
| An argument mod $3$ shows that $z$ is even then let $z=2{z_1}.$
Hence $3^x = 2^{2y} - 5^{2z_1} = ( 2^y + 5^{z_1})(2^y - 5^{z_1})$
$\bullet\ 2^y - 5^{z_1}\ne 1$
$2^y + 5^{z_1} \pmod 3 \implies z_1 \not\equiv y \pmod 3$
$2^y - 5^{z_1} \pmod 3 \implies z_1 \equiv y \pmod 3$
This is a contradiction.
$\bullet\ 2^y - 5^{z_1} = 1$
According to Catalan's conjecture, $2^y - 5^{z_1} = 1$ has no positive integer solutions.
Hence $3^x+5^z=4^y$ has no positive integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\lim_{x\to 1}\frac{x^2-ax+b}{x-1} = 5$ then find $a+b$
If
$$\lim_{x\to 1}\frac{x^2-ax+b}{x-1} = 5$$
then find $a+b$
The solution which I have says that $x-1$ tends to $0$. Therefore numerator $x^2 - ax + b$ must tend to $0$ because the limit exists.
My query is: why it is written, that limit exists so numerator of tend to $0$.
| $$\frac{x^2-ax+b}{x-1}=x-a+1+\frac{b-a+1}{x-1}$$ and for the limit to exist, the numerator of the fraction must be zero.
Hence
$$\begin{cases}b-a+1=0,\\1-a+1=5.\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$ I am a student in Germany, and I prepare for Math Olympiad by solving math problems. I have been solving the following question, which took about 4 hours to solve.
Prove the following inequality without using calculator:
$$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$
Can you check my proof? It would be really grateful.
First, we can define function $f(x)$ as following:
$$f(x) = \sqrt[3]{x+1} - \sqrt[3]{x}\space(x > 0)$$
$$f(3) = \sqrt[3]{3+1} - \sqrt[3]{3} = \sqrt[3]{4} - \sqrt[3]{3}$$
$$f(2) = \sqrt[3]{2+1} - \sqrt[3]{2} = \sqrt[3]{3} - \sqrt[3]{2}$$
Then, we will differentiate $f(x)$ to check whether $f(x)$ is a decreasing function or not. $f '(x)$ must be a falling function if $f '(x)$ < 0.
$$f'(x) = \frac{1}{3\sqrt[3]{(x+1)^2}} - \frac{1}{3\sqrt[3]{x^2}}$$
Since the minuend is smaller than the subtrahend (minuend has a bigger denominator than the denominator of subtrahend), we can say $f '(x)$ is less than 0 which makes $f(x)$ a decreasing function. Falling function means that $f(a) > f(a+1)$. Substitute $a=2$ and we get:
$$f(2) > f(3)$$
$$\sqrt[3]{3} - \sqrt[3]{2} > \sqrt[3]{4} - \sqrt[3]{3}$$
$$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$
Thank you for reading this text, but it would be more grateful if you check my solution, and comment my solution.
I wish you a beautiful day, and stay safe.
| The standard difference quotient for approximating $f''(a)$ is
$$ \frac{f(a-h) - 2 f(a) + f(a+h)}{h^2} $$
Here $f(x) = \sqrt[3]x $ is infinitely differentiable, while the second and fourth derivatives are negative for $0 < h < a.$ Indeed
$$ f(a-h) - 2 f(a) + f(a+h) = f''(a) h^2 + \frac{h^3}{6} \left( f'''( \xi) - f'''(\eta) \right) $$
where $ a-h < \eta < a < \xi < a+h .$ This is just the Taylor series with remainder. As the fourth derivative is negative we see $ f'''( \xi) - f'''(\eta) < 0.$ Thus we find,
$$ f(a-h) - 2 f(a) + f(a+h) <0$$
Her $f$ is the cube root, $a=3$ and $h=1$ so
$$ \sqrt[3]2 - 2\sqrt[3]3 + \sqrt[3]4 < 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Cubic equation problem $\frac{x^3}{3}-x=k$ The cubic function $\frac{{{x^3}}}{3} - x = k$ has three different roots $\alpha,\beta,\gamma$ about the real number k. Let's call the minimum value of $|\alpha|+|\beta|+|\gamma|$ as $m$. FInd the value of $m^2$.
My approach is as follow
$\frac{{{x^3}}}{3} - x = k \Rightarrow f\left( x \right) = \frac{{{x^3}}}{3} - x - k$
$f'\left( x \right) = {x^2} - 1 = 0$
Hence $x = \pm 1$
$f\left( 1 \right) = \frac{1}{3} - 1 - k = - \left( {k + \frac{2}{3}} \right)\& f\left( { - 1} \right) = - \frac{1}{3} + 1 - k = - \left( {k - \frac{2}{3}} \right)$
For real roots $f\left( 1 \right)f\left( { - 1} \right) < 0$
Therefore $ - \left( {k + \frac{2}{3}} \right) \times \left( { - \left( {k - \frac{2}{3}} \right)} \right) < 0$
$k \in \left( { - \frac{2}{3},\frac{2}{3}} \right)$
We know that $\alpha+\beta+\gamma=0$
But how we will find the minimum values of the sum of the modulus of the roots.
| Hint $\;-\;$ using $\,\alpha\beta+\beta\gamma+\gamma\alpha=-3\,$:
$$
\require{cancel}
\begin{align}
m^2 &= \left(|\alpha|+|\beta|+|\gamma|\right)^2
\\ &= \alpha^2+\beta^2+\gamma^2 + 2\left(|\alpha\beta|+|\beta\gamma|+|\gamma\alpha|\right)
\\ &= \cancel{\left(\alpha+\beta+\gamma\right)^2} - 2(\alpha\beta+\beta\gamma+\gamma\alpha)+ 2\left(|\alpha\beta|+|\beta\gamma|+|\gamma\alpha|\right)
\\ &= 6 + 2\left(|\alpha\beta|+|\beta\gamma|+|\gamma\alpha|\right)
\\ &\ge 6 + 2\,|\alpha\beta+\beta\gamma+\gamma\alpha|
\\ &= 12
\end{align}
$$
This gives a lower bound on $m^2\,$. To prove it's an actual minimum, it is enough to find a $k$ such that $|\alpha|+|\beta|+|\gamma| = 2 \sqrt{3}\,$, which turns out not to be too hard.
Note: the above assumes the roots are real (per OP's comments), in order for $\,|\alpha|^2=\alpha^2\,$ to hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim\limits_{(x,y)\to(1,-1)}{\frac{e^{x-y}\tan{(x+y)}}{x^2-y^2}}$ I need to find $\lim\limits_{(x,y)\to(1,-1)}{\frac{e^{x-y}\tan{(x+y)}}{x^2-y^2}}$. I came up with this:
Since $x^2-y^2$ changes its sign, I can'y apply $\tan{(x+y)}\sim(x+y)$ just yet. So,$\lim\limits_{(x,y)\to(1,-1)}{|\frac{e^{x-y}\tan{(x+y)}}{x^2-y^2}|}=\lim\limits_{(x,y)\to(1,-1)}{\frac{e^{x-y}|\tan{(x+y)}|}{|x^2-y^2|}}=\lim\limits_{(x,y)\to(1,-1)}{\frac{e^{x-y}|x+y|}{|x-y|.|x+y|}}=\lim\limits_{(x,y)\to(1,-1)}{\frac{e^{x-y}}{|x-y|}}=\frac{e^2}{2}$
Is this correct and is there a better way to approach the problem?
| What you did was to compute the limit of $\left|\frac{e^{x-y}\tan(x+y)}{x^2-y^2}\right|$, not of $\frac{e^{x-y}\tan(x+y)}{x^2-y^2}$.
You have$$\lim_{(x,y)\to(1,-1)}\frac{e^{x-y}\tan(x+y)}{x^2-y^2}=\lim_{(x,y)\to(1,-1)}\frac{e^{x-y}}{x-y}\frac{\tan(x+y)}{x+y}=\frac{e^2}2,$$since$$\lim_{(x,y)\to(1,-1)}\frac{\tan(x+y)}{x+y}=\lim_{h\to0}\frac{\tan h}h=\tan'(0)=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}$?
Evaluate $$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}.$$
I solved the problem with the Taylor series expansion of $\cos x$. Here is my solution:
$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1-\{1-\frac{(7x)^2}{2!}+\frac{(7x)^4}{4!}-\frac{(7x)^6}{6!}+\dots\}}{3x^2}\\ =\lim\limits_{x\to 0}\frac{x^2(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1}{3}(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)\\ =\frac {49}{6}$
Can this be solved without using the Taylor series?
| $$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}=\lim\limits_{x\to 0}\frac{1-(1-2\sin^2\frac{7x}{2})}{3x^2}=\lim\limits_{x\to 0}\frac{2\sin^2\frac{7x}{2}}{3x^2}=\frac{49}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Sum with Bernoulli poly. $\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n+1}{2k} B_{2k}(1/4) $ I have a roundabout proof for
$$ \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n+1}{2k} B_{2k}(1/2)= \frac{n+1}{2^n} $$
where $B_{2k}(x)$ are the Bernoulli polynomials,$B_0(x)=1, B_2(x)=x^2-x+1/6,$ etc.
Changing the argument of the Bernoulli polynomial and observing patterns (aided by
OEIS A167205) I have a relationship which has
been checked to high values of positive integer $n,$
$$ \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n+1}{2k} B_{2k}(1/4)= \frac{n+1}{2^{2n \ - \ (1 \ - \ (-1)^n)/2)}} \ \frac{3^n+1}{3- \ (-1)^n }$$
How can this equation be proved?
| We seek to show that with $B_{2k}(x)$ a Bernoulli polynomial we have
$$\sum_{k=0}^{\lfloor n/2 \rfloor}
{n+1\choose 2k} B_{2k}(1/4) =
\frac{1}{2^{2n+1}} (n+1) (1+3^n).$$
We will use the EGF of the Bernoulli polynomials which is
$$\sum_{n\ge 0} B_n(x) \frac{z^n}{n!} =
\frac{z \exp(zx)}{\exp(z)-1}.$$
The LHS of the proposed identity is
$$(n+1) \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} B_{2k}(1/4) \frac{1}{n+1-2k}$$
so we really only have to prove that
$$\sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} B_{2k}(1/4) \frac{1}{n+1-2k}
= \frac{1}{2^{2n+1}} (1+3^n).$$
We get two pieces for the LHS, the first call it A is
$$\frac{1}{2} \sum_{k=0}^{n}
{n\choose k} B_{k}(1/4) \frac{1}{n+1-k}$$
and the second call it B
$$\frac{1}{2} \sum_{k=0}^{n}
{n\choose k} (-1)^k B_{k}(1/4) \frac{1}{n+1-k}.$$
What we have here is a convolution of two exponential generating
functions. Recall that when we multiply two exponential generating
functions of the sequences $\{p_n\}$ and $\{q_n\}$ we get that
$$ P(z) Q(z) = \sum_{n\ge 0} p_n \frac{z^n}{n!}
\sum_{n\ge 0} q_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} p_k q_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} p_k q_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} p_k q_{n-k}\right)\frac{z^n}{n!}$$
Observe that
$$\sum_{p\ge 0} \frac{1}{p+1} \frac{z^p}{p!} =
\frac{\exp(z)-1}{z}.$$
Therefore the EGF of piece A is
$$\frac{1}{2} \exp(z/4)$$
and the EGF of piece B is
$$\frac{1}{2} \frac{-z \exp(-z/4)}{\exp(-z)-1}
\frac{\exp(z)-1}{z}
= \frac{1}{2} \frac{-z \exp(3z/4)}{1-\exp(z)}
\frac{\exp(z)-1}{z}
\\ = \frac{1}{2} \exp(3z/4).$$
Extracting the coefficient on $[z^n]$ (EGF not OGF) on A and B
we find
$$\frac{1}{2} n! [z^n] (\exp(z/4) + \exp(3z/4)) =
\frac{1}{2^{2n+1}} (1+3^n)$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4199582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Understanding matrix multiplication for visualizing what is happening under the hood Take the case of this matrix multiplication:
$$
A x=
\begin{pmatrix}
1 & -1 & 2\\
0 & -3 & 1\\
\end{pmatrix}
\begin{pmatrix}
2 \\ 1 \\ 0
\end{pmatrix}
$$
The answer of which is
$
\begin{pmatrix}
1 \\
-3
\end{pmatrix}
$.
Source: https://mathinsight.org/matrix_vector_multiplication
I understand there are three components in $A$ and $x.$ So how can matrix multiplication have two (not sure if the component will be the right term) terms as part of the answer leading to matrix multiplication? What is the way to visualize the result? I think with three components, the matrix multiplication should have the result in three parts. I know I am missing something.
| We look at the problem by considering two aspects. We start with matrix multiplication and we will see that multiplication of a matrix with a vector can be seen as special case of it. Then we look at how the elements of the product matrix in the current problem are calculated.
Matrix multiplication:
*
*An $(m\times \color{blue}{n})$-matrix $A$ consisting of $m$ rows and $\color{blue}{n\ \mathrm{columns}}$ multiplied with an $(\color{blue}{n} \times q)$-matrix $B$ consisting of $\color{blue}{n\ \mathrm{rows}}$ and $q$ columns gives an $(m\times q)$-matrix $A\cdot B$.
*A vector $x$ of dimension $n$ can be seen as $(\color{blue}{n}\times 1)$ matrix. Multiplication of an $(m\times \color{blue}{n})$-matrix $A$ with $x$ gives $Ax$ which is consequently an $(m\times 1)$ matrix.
*Current problem:
\begin{align*}
A=\begin{pmatrix}
1 & -1 & 2\\
0 & -3 & 1\\
\end{pmatrix}
\end{align*}
is a $(2\times 3)$ matrix. The vector $x=\begin{pmatrix}
2 \\
1\\
0
\end{pmatrix}$
is a $(3\times 1)$-matrix and we therefore obtain $Ax$, which is a $(2\times 1)$-matrix.
Elements of the matrix product:
*
*With the settings from above we have $A\cdot B=C=\left(c_{i,j}\right)_{1\leq i\leq m,1\leq j\leq q}$. The element $c_{i,j}$ denotes the element in the $i$-th row and $j$-th column of $C$. We calculate $c_{i,j}$ by multiplying the elements of the $i$-th row and $j$-th column elementwise and adding them up.
We demonstrate this with the current problem. We consider
\begin{align*}
\color{blue}{Ax}
&=\begin{pmatrix}
1 & -1 & 2\\
0 & -3 & 1\\
\end{pmatrix}
\begin{pmatrix}
2 \\
1 \\
0
\end{pmatrix}\\
\end{align*}
and we know from above that $Ax=(c_{i,j})_{1\leq i\leq 2, 1\leq j\leq 1}$ gives a $(2,1)$-matrix with elements $c_{1,1}$ and $c_{2,1}$.
*
*Element $c_{1,1}$: Multiplication of first row with first column gives
\begin{align*}
\left(\color{blue}{c_{1,1}}\right)&=
\begin{pmatrix}
1 & -1 & 2\\
\end{pmatrix}
\begin{pmatrix}
2 \\
1 \\
0
\end{pmatrix}\\
&=\left(1\cdot 2+(-1)\cdot 1+2\cdot 0\right)\\
&=(\color{blue}{1})\\
\end{align*}
*Element $c_{2,1}$: Multiplication of second row with first column gives
\begin{align*}
\left(\color{blue}{c_{2,1}}\right)&=
\begin{pmatrix}
0 & -3 & 1\\
\end{pmatrix}
\begin{pmatrix}
2 \\
1 \\
0
\end{pmatrix}\\
&=\left(0\cdot 2+(-3)\cdot 1+1\cdot 0\right)\\
&=(\color{blue}{-3})\\
\end{align*}
We conclude
\begin{align*}
\color{blue}{Ax}=\begin{pmatrix}
1 & -1 & 2\\
0 & -3 & 1\\
\end{pmatrix}
\begin{pmatrix}
2 \\
1 \\
0
\end{pmatrix}=
\begin{pmatrix}
c_{1,1} \\
c_{2,1}
\end{pmatrix}\color{blue}{=\begin{pmatrix}
1 \\
-3
\end{pmatrix}}\tag{1}
\end{align*}
Note, with some experience this matrix calculation can be reduced to (1) without doing any intermediate steps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find matrices that satisfy subspaces conditions? I'm having a hard time trying to learn some linear algebra wizardry. Can someone explain to me how to solve this problem:
P4.18 Find 2x3 matrices A,B, and C, that satisfy the conditions.
P4.18 Find $2 \times 3$ matrices $A,B,C$ that satisfy the conditions.
*
*$R(A) = \operatorname{span}((1,2,3))$ and $C(A) = \operatorname{span}((1,2)^T)$,
*$N(B) = \operatorname{span}((1,1,1)^T)$,
*$N(C^T) = \operatorname{span}((1,3))$.
Note: when I write $R(A)$, $N(B)$, $N(C^T)$, I mean rowspace of A, Null-space of B and left-null-space of C.
Thank you considering answering this question!! The problem is from the book No Bullshit Linear Algebra by Ivan Savov, on page 255. The suggested solution is:
a)
1, 2, 3
2, 4, 6
b)
1, -1, 0
1, 0, -1
c)
3, 6, 9
-1, -2, -3
Other answers are possible.
| I’ll give this a shot:
Let’s say $A$ looks like $\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$, then $R(A)= K_1\cdot \begin{bmatrix} a_{11} & a_{12} & a_{13}\end{bmatrix} + K_2\cdot \begin{bmatrix} a_{21} & a_{22} & a_{23} \end{bmatrix}=K\cdot\begin{bmatrix} 1&2&3\end{bmatrix} = span\left\{\begin{bmatrix}1&2&3\end{bmatrix}\right\}$
for some constants $K,K_1,K_2$
and since there is only one vector in the span, we have a basis for the row-space of $A$ and the rows of $A$ must be linearly dependent and hence multiples of each other:
$k\cdot\begin{bmatrix} a_{11} & a_{12} & a_{13}\end{bmatrix}= \begin{bmatrix} a_{21} & a_{22} & a_{23}\end{bmatrix}$
for some constant $k$. Now every column vector of $A$ looks like $a_{1j}\cdot\begin{bmatrix}1 \\k\end{bmatrix}$ for $j=1,2,3$.
$C(A)= a_{11}\cdot\begin{bmatrix}1 \\k\end{bmatrix}+ a_{12}\cdot\begin{bmatrix}1 \\k\end{bmatrix}+ a_{13}\cdot\begin{bmatrix}1 \\k\end{bmatrix}=\mathcal{K}\cdot\begin{bmatrix}1 \\k\end{bmatrix}=span\left\{\begin{bmatrix}1 \\k\end{bmatrix}\right\} $
and the second clue gives us that $C(A)= span\left\{\begin{bmatrix}1 \\2\end{bmatrix}\right\}$ and $k=2$.
$A=\begin{bmatrix} 1&2&3\\2&4&6\end{bmatrix}$ seems to work fine (letting $[a_{11} \: a_{12} \: a_{13} ]= [1\:2\:3]$).
On the other hand, the null-space is the solution space of the homogeneous system(essentially add a column of zeros, and solve the system)
The solution space of $\left[\begin{array}{ccc|c}b_{11}&b_{12}&b_{13}&0\\b_{21}&b_{22}&b_{23}&0\end{array}\right]$ is said to look like $t\cdot \begin{bmatrix}1\\1\\1\end{bmatrix}$.
An easy $2\times 3$ matrix which would have this could be
$B=\begin{bmatrix}1&0&-1\\0&1&-1\end{bmatrix}$ which is similar to the answer provided only the rows are flipped.
To see why, one can imagine solving the system of linear equations parametrically:
$\left\{\begin{array}{c}x-z=0\\y-z=0\end{array}\right.$ one can bring $z$ over to the right and let $z=t$
$\left\{\begin{array}{c}x=z=t\\y=z=t\end{array}\right.$ so the general solution looks like $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}t\\t\\t\end{bmatrix}=t\cdot\begin{bmatrix}1\\1\\1\end{bmatrix}$.
Similarly for $C=\begin{bmatrix}3&6&9\\-1&-2&-3\end{bmatrix}$ is an acceptable answer because $C^T = \begin{bmatrix}3&-1\\6&-2\\9&-3\end{bmatrix}$ and the system of equations
$\left\{\begin{array}{c}3x-y=0\\6x-2y=0\\9x-3y=0\end{array}\right.\Longrightarrow \left\{\begin{array}{c}3x=y\\6x=2y\\9x=3y\end{array}\right. \Longrightarrow 3x=y$
the general solution has form $\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\3x\end{bmatrix}=x\cdot\begin{bmatrix}1\\3\end{bmatrix}=span\left\{\begin{bmatrix}1\\3\end{bmatrix}\right\}$
I think there a $^T$ missing in #3.
Hope this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
if $S_n = (3 + \sqrt{5})^n + (3 - \sqrt {5})^n$ then show that $S_n$ is an integer by induction this is all that i have tried:
let $n=1$ so equation gives $(3 + \sqrt {5}) + (3 - \sqrt {5}) = 6$ which is an integer
so it is true for $n=1$
now let it be true for $k \ge n$ then we have:
$(3 + \sqrt {5})^k + (3 - \sqrt {5})^k$ is an integer
for k+1: $(3 + \sqrt {5})^k(3 + \sqrt {5}) + (3 - \sqrt {5})^k(3 - \sqrt {5})$
$= 3((3 + \sqrt {5})^k + (3 - \sqrt {5})^k) + \sqrt {5}((3 + \sqrt {5})^k - (3 - \sqrt {5})^k)$
$= (integer) + \sqrt {5}((3 + \sqrt {5})^k - (3 - \sqrt {5})^k)$
from here I started facing a problem.
I wasn't able to prove $\sqrt {5}((3 + \sqrt {5})^k - (3 - \sqrt {5})^k)$ an integer.
can someone please help me out?
(ONLY USING INDUCTION!)
| Let's do a double induction.
To show: $(3+\sqrt5)^k+(3-\sqrt5)^k$ and $\sqrt5(3+\sqrt5)^k-\sqrt5(3-\sqrt5)^k$ are integers.
Base Case: trivial
Inductive Case: You've shown why $(3+\sqrt5)^k+(3-\sqrt5)^k$ is an integer. And, $$(3+\sqrt5+3-\sqrt5)\cdot((3+\sqrt5)^k+(3-\sqrt5)^k)=(3+\sqrt5)^{k+1}+(3-\sqrt5)^{k+1}+3((3+\sqrt5)^k+(3-\sqrt5)^k)+\sqrt5(3+\sqrt5)^k-\sqrt5(3-\sqrt5)^k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to prove that the limit of $f(x,y)=\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}$ is $0$ as $(x,y)$ approaches $(0,0)$? I need to show that $\displaystyle\lim_{(x,y)\rightarrow(0,0)}\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}$ exists. I know that it is equal to zero. Until now, all I know how to do is to prove using the $\epsilon-\delta$ definition. I tried to prove it by the following way:
Let $\delta>0$ be a real number such that $0<\sqrt{x^2+y^2}<\delta$. We have $0< x^2+y^2<\delta^2$.
We know that $\left\vert x^2\sin(y)+y^2\sin(x)\right\vert\leq\left\vert x^2\sin(y)\right\vert+\left\vert y^2\sin(x)\right\vert=x^2\left\vert\sin(y)\right\vert+y^2\left\vert\sin(x)\right\vert\leq x^2+y^2$.
Thus
$$0<\left\vert x^2\sin(y)+y^2\sin(x)\right\vert\leq x^2+y^2<\delta^2$$
and dividing the inequality by $x^2+y^2>0$ we have
$$ 0<\dfrac{\left\vert x^2\sin(y)+y^2\sin(x)\right\vert}{x^2+y^2}<1<\dfrac{\delta^2}{x^2+y^2}. $$
I thought that this could be useful because $\dfrac{\left\vert x^2\sin(y)+y^2\sin(x)\right\vert}{x^2+y^2}=\left\vert\dfrac{x^2\sin(y)+y^2\sin(x)}{x^2+y^2}\right\vert$, but I don't know how to proceed.
| Recall that $|\sin t|\leq |t|$ for any $t\in\mathbb{R}$. For any $x,y\in\mathbb{R},$
we have that
\begin{eqnarray*}
& & \left|x^{2}\sin y+y^{2}\sin x\right|\\
& \leq & \left|x^{2}\sin y\right|+\left|y^{2}\sin x\right|\\
& \leq & |x^{2}y|+|y^{2}x|.
\end{eqnarray*}
Now, let $\varepsilon>0$ be given. Define $\delta=\frac{1}{2}\varepsilon$.
Let $(x,y)\in\mathbb{R}$ that satisfies $0<\sqrt{x^{2}+y^{2}}<\delta$.
We have estimation:
\begin{eqnarray*}
& & \left|\frac{x^{2}\sin y+y^{2}\sin x}{x^{2}+y^{2}}\right|\\
& \leq & \left|\frac{x^{2}}{x^{2}+y^{2}}\right||y|+\left|\frac{y^{2}}{x^{2}+y^{2}}\right||x|\\
& \leq & |y|+|x|\\
& \leq & \sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+y^{2}}\\
& < & 2\delta\\
& = & \varepsilon.
\end{eqnarray*}
This shows that $\lim_{(x,y)\rightarrow(0,0)}\frac{x^{2}\sin y+y^{2}\sin x}{x^{2}+y^{2}}=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4205040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Is it true that $\forall x>3, \exists yx$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 ?$
Is it true that $\forall x>3, \exists y<x$ and $\exists z>x$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 ?$
Here is my proof but I am stuck in the middle. And I am also wondering if I am on the right track?
Let $y = x - a$ for some real number $a$ that $a>0$, and $z = x + b$ for some real number $b$ that $b>0$.
Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 \Rightarrow \frac{1}{x}+\frac{1}{x-a}+\frac{1}{x+b}=1.$
After combining similar terms, I got:
$a = \frac{x(x^2+bx-3x-2b)}{x^2+(b-2)x-b}$, and $b = \frac{x(x^2-ax-3x+2a)}{-x^2+(a+2)x-a}$
For $a$ exist, $x^2+(b-2)x-b \neq 0$. And for $b$ exist, $-x^2+(a+2)x-a \neq 0$.
Let $f(x) = x^2+(b-2)x-b $ and $g(x) = -x^2+(a+2)x-a $.
By the formula of quadratic function: $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$:
$f(x)\neq 0$ $\Rightarrow (b-2)^2+4b < 0$ ,and $g(x)\neq 0\Rightarrow(a+2)^2-4a<0$.
Since $(b-2)^2+4b \geq 0$,and $(a+2)^2-4a\geq0$ , no matter how we choose $a$ and $b$, $f(x)=0$ and $g(x)=0$ always have a solution.
Then I stuck at here, I understand that I need to prove $f(x)\neq0$ and $g(x)\neq0$ for all $x >3$ and $a,b$ have to be positive to make the statement true. Otherwise, it's false. Really hope to get some hints/help!! Thank you!
| For every $n\in N \gt 2$, we show that there exists $n$ pairwise distinct natural numbers whose sum of inverses equals $1$.
Proof by Induction.
Making induction on $n\geq 3$, the initial case $n=3$ following from
$\frac {1}{2} + \frac {1}{3}+\frac {1}{6}=1$.
We now assume by Induction hypothesis, that, for a certain natural number $k\geq 3 $ there exists natural numbers $x_1\lt x_2\lt ...\lt x_k$ such that
$\frac {1}{x_1} + \frac {1}{x_2}+...+\frac {1}{x_k}=1$.
Multiplying through the equality above by $\frac {1}{2}$ and adding $\frac {1}{2}$ on both sides of the resulting equality , we obtain
$\frac {1}{2}+\frac {1}{2x_1} + \frac {1}{2x_2}+...+\frac {1}{2x_k}=1$.
Since $2\lt 2x_1\lt 2x_2\lt ...\lt 2x_k$, we have $k+1$ pairwise distinct natural numbers whose sum of inverses equals $1$ and this completes the induction step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Show that the sum of the squares of 3,4,5 and 6 consecutive numbers can not be a square $\textbf{Edit:}$
Thank you so far for the answers. I still do not understand how to prove that the sum of 6 consecutive squares is not a square.
I've tried ${6d^2+30d+55 \ne n^2 \Rightarrow 6(d^2 + 5d +9)+1 \ne n^2}$ and would somehow like to elaborate this, but (yet again) I get stuck.
Is is correct to say that ${6(d^2 + 5d +9)+1}$ is on the form
${6k+1}$? Because then I could possibly stop there, having
${6k+1 \ne n^2}$
Thanks!
${\rule{8cm}{0.4pt}}$
\begin{equation}
\end{equation}
I've gotten this far:
$\textbf{Sum of 3 consecutive squares}$
Rewrite of problem:
\begin{equation}
\begin{aligned}
(d-1)^2+d^2+(d+1)^2 &\ne n^2\\
3d^2+2 &\ne n^2 \\
\label{p1_1}
\end{aligned}
\end{equation}
Assume ${n=d\cdot3+r}$, where ${r=0,1}$ or $2$.
Then,
\begin{equation}
n^2 \equiv_3 r^2 \equiv_3 0 \vee 1
\end{equation}
Conclusion:
Since ${r=2}$ in ${3d^2+2}$ and ${r=0 \vee 1}$ in ${n^2 \equiv_3 r^2}$ we have shown that
${3d^2+2 \ne n^2}$.
Question: Is my conclusion and reasoning valid and "good enough"?
$\textbf{Sum of 4 consecutive squares}$
Rewrite of problem:
\begin{equation}
\begin{aligned}
d^2+(d+1)^2 + (d+2)^2 + (d+3)^2 &\ne n^2\\
4(d^2+3d+3)+2&\ne n^2\\
\label{p1_b}
\end{aligned}
\end{equation}
${\Rightarrow} {r=2}$.
Assume ${n=d\cdot4+r}$, where ${r=0,1,2}$ or $3$.
Then,
\begin{equation}
n^2 \equiv_4 r^2 \equiv_4 0 \vee 1
\end{equation}
Conclusion:
Same reasoning as in conclusion of sum of 3 consecutive squares.
$\textbf{Sum of 5 consecutive squares}$
\begin{equation}
\begin{aligned}
(d-2)^2+(d-1)^2+d^2+(d+1)^2+(d+2)^2 &\ne n^2\\
5d^2+10 &\ne n^2\\
5(d^2+2) &\ne n^2\\
\label{p1_c}
\end{aligned}
\end{equation}
${\Rightarrow r=2 \vee 10}$ (is this correct?)
${n^2 \equiv_5 r^2 \equiv_5 0 \vee \pm 1}$
Conclusion:
Same reasoning as in conclusion of sum of 3 consecutive squares.
$\textbf{Sum of 6 consecutive squares}$
\begin{equation}
\begin{aligned}
d^2+(d+1)^2+(d+2)^2+(d+3)^2+(d+4)^2+(d+5)^2 &\ne n^2\\
6d^2+30d+55&\ne n^2\\
\label{p1_d}
\end{aligned}
\end{equation}
Here I'm stuck.
I feel pretty unsure about all my reasonings and would therefore appreciate your help. Please, feel free to be verbose in your answers.
Thank you!
| Of six consecutive numbers, three are even and three are odd. The square of the even numbers are congruent to $0$ mod $4$; the squares of the odds are congruent to $1$ mod $4$. Consequently the sum of six consecutive squares is congruent to $3$ mod $4$, which cannot be a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Function Analysis
I tried to differentiate the function and set the value at =0 in order to find x in terms of a and b. Then, I also tried to plug the value of x into in and set the value of y into 1 and 4. At the end, I was struggling since it became more tedious. So anyone can suggest a nicer way to solve this? Thanks in advance
| I found a solution which is a little bit tedious but less than some of the others.
The equation $f(x)=1$ is equivalent to:
\begin{align*}
ax^2 + 4x + b &= x^2 + 2
\\\iff (a-1)x^2 + 4x + (b-2) &= 0
\end{align*}
This is a quadratic equation in $x$. If $f$ has a minimum at $x$ too, then we can assume the equation has one double solution. Therefore the discriminant of the quadratic polynomial must be zero:
\begin{align*}
0 &= 4^2 - 4(a-1)(b-2) = 4(2a+b + 2 - ab)
\end{align*}
In a similar way, the equation $f(x) = 4$ gives
\begin{align*}
ax^2 + 4x + b &= 4(x^2 + 2)
\\\iff (a-4)x^2 + 4x + (b-8) &= 0
\end{align*}
Since $4$ is the maximum value, we expect a double solution, so the discriminant is zero as well:
\begin{align*}
0 &= 4^2 - 4(a-4)(b-8) = 4(8a+4b - 28 - ab)
\end{align*}
Putting these together, we have two equations in the two unknowns $a$ and $b$:
\begin{align*}
2a+b - ab &= -2 \tag{1}
\\ 8a + 4b - ab &=28 \tag{2}
\end{align*}
Subtracting ($1$) from ($2$) gives
$$
6a+3b = 30 \implies 2a+b = 10 \tag{3}
$$
Subtracting $4$ times ($1$) from ($2$) gives
$$
3ab = 36 \implies ab = 12 \tag{4}
$$
We can substitute ($3$) into ($4$) to get
\begin{align*}
a(10-2a) = 12 \implies
a^2 - 5a + 6 = 0
\end{align*}
The solutions are $a = 2$ (yielding $b=6$) and $a=3$ (yielding $b=4$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is $d$ a multiple of $4$ in the equation $(x+a)(x+b)(x+c)+d=(x+w)(x+y)(x+z)$ where $a$, $b$, $c$, $d$, $w$, $y$, and $z$ are integers? I tried to find the value of integers $a$, $b$, $c$, $d$, $w$, $y$, and $z$ so that the polynomial
$(x+a)(x+b)(x+c)+d$ and $(x+w)(x+y)(x+z)$ are equal.
By using brute-force approach with Microsoft Excel, I find some solutions like:
$(x-13)(x-12)(x-8) - 12 = (x-9)(x-10)(x-14)$
$(x-13)(x-10)(x-10) + 4 = (x-12)(x-9)(x-12)$
$(x+6)(x+7)(x+11) - 12 = (x+9)(x+10)(x+5)$
$(x-7)(x+3)(x-13) - 288 = (x-1)(x-15)(x-1)$
and so on.
To my surprise, all the values of $d$ that I obtained are multiples of $4$.
Is it true that $d$ is always a multiple of $4$? If so, why?
I tried to use Wolfram Alpha to find out why this happens but it gave the solutions $w$, $y$, and $z$ in terms of $a$, $b$, $c$, and $d$ and the solutions are very long and complicated.
Note: I have also tried to find the integers
$a$, $b$, $c$, $d$, $m$, $v$, $w$, $y$, and $z$ that solve the equation $(x+a)(x+b)(x+c)(x+d)+m =(x+v)(x+w)(x+y)(x+z)$ and found that $m$ is also a multiple of $4$. Maybe there is some generalization to what I found.
| Indeed, every possible integral $d$ must be a multiple of $4$. To see this, first note that, for the identity to hold, we must have:
$$a+b+c=w+y+z$$ and $$ab+bc+ac=wy+yz+zx$$
This means that, $$a+b+c \equiv w+ y+z \mod 4 {\tag 1}$$ and $$ab+bc +ac \equiv wz+zy+wy \mod 4 {\tag 2}$$
Thus, modulo $4$, only possible unordered triplets for $\{(a,b,c), (w,y,z)\}$ satisfying both conditions $(1)$ and $(2)$ are:
If $a+b+c \equiv 0 \mod 4$-
$$\{(0,0,0), (0,2,2)\}$$
If $a+b+c \equiv 1 \mod 4$-
None
If $a+b+c \equiv 2 \mod 4$-
$$\{(0,3,3), (1,1,0)\}$$
$$\{(0,2,0),(2,2,2)\}$$
If $a+b+c \equiv 3 \mod 4$-
$$\{(1,1,1),(3,1,3)\}$$
$$\{(0,3,0),(3,2,2)\}$$
These pairs are those apart from the ones where $(a,b,c)=(w,y,z)$ which are of course satisfying both conditions.
For each of the listed above pairs, check that $d=wyz-abc \equiv 0 \mod 4$.
Hence $4|d$ is always true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4211880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
In triangle $ABC$, $AP$ is the angle bisector In triangle $ABC$, $AP$ is the angle bisector of $\measuredangle A$. If $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on $AC$, find $AB$ and $AC$.
$AP$ is the angle bisector of $\measuredangle BAC$ which means that $\measuredangle BAP=\measuredangle CAP=\alpha$. On the other hand, $OA=OP=R_{\triangle ABP}=R$, so $\measuredangle PAO=\measuredangle APO=\alpha$. This makes $OP\parallel AB$. I am trying to figure out the best way to use that parallelity.
The most straightforward think we can note is $\triangle OPC\sim\triangle ABC$. I don't see if this can be helpful, though.
Another true thing is $\dfrac{BP}{PC}=\dfrac{AB}{AC}=\dfrac{16}{20}=\dfrac{4}{5}$.
| Let's say the radius of the circle is $r$. Call the intersection of $OC$ with the circle, $D\neq A$.
Using similarity, we have
$$\frac{r+DC}{r}=\frac{20}{16}$$
$$4(r+DC)=5r$$
$$DC=\frac{r}{4}$$
Evaluating the power of point $C$, we get that
$$CD\cdot AC=20\cdot 36$$
$$\frac{r}{4}\cdot \frac{9r}{4}=20\cdot 36$$
$$r=16\sqrt{5}$$
$$AC=\frac{9r}{4}=\boxed{36\sqrt{5}}$$
Using angle bisector theorem, we have
$$16\cdot AC=20\cdot AB$$
$$AB=\frac{4}{5}\cdot 36\sqrt{5}$$
$$AB=\boxed{\frac{144\sqrt{5}}{5}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
$x_1$ , $x_2$ are roots of $x=5-x^2$. Find the equation with roots $\frac1{(x_1+1)^3}$ and $\frac1{(x_2+1)^3}$.
Suppose $x_1$ , $x_2$ are the roots of the equation $x=5-x^2$.Then
$\dfrac1{(x_1+1)^3}$ and $\dfrac1{(x_2+1)^3}$ are roots of which
equation?
$1)125x^2+16x=1$
$2)125x^2=16x+1$
$3)125x^2=12x+1$
$4)125x^2+12x=1$
I solved this problem with the following approache,
I've denoted the roots of the original equation by $\alpha$ and $\beta$ rather than $x_1$, $x_2$ ,
$S=\alpha+\beta=-1$ and $P=\alpha\beta=-5$. We find $S'$ and $P'$ of the new equation,
$$P'=\dfrac1{[(\alpha+1)(\beta+1)]^3}=\dfrac1{(S+P+1)^3}=-\frac1{125}$$
$$S'=\dfrac{(\beta+1)^3+(\alpha+1)^3}{[(\alpha+1)(\beta+1)]^3}=-\frac1{125}\times\left((\alpha^3+\beta^3)+3(\alpha^2+\beta^2)+3(\alpha+\beta)+2\right)$$
$$=-\dfrac{(S^3-3PS)+3(S^2-2P)+3S+2}{125}=-\dfrac{-16+33-3+2}{125}=-\frac{16}{125}$$
Hence the new equation is $125x^2+16x-1=0$. And the answer is first choice.
This was a problem from a timed exam. So can you solve it with other approaches (preferably quicker one) ?
| I guess I actually "went there" by looking at the roots themselves: for $ \ x^2 + x - 5 \ = \ 0 \ \ , $ we have $ \ x_{1,2} \ = \ -\frac12 \ \pm \ \frac{\sqrt{21}}{2} \ \ . $ For the equation choices, all of them are of the form $ \ 125·x^2 \ \pm \ \begin{array}{c} 12 \\ 16 \end{array}·x \ - \ 1 \ = \ 0 \ \ , $ so the middle term is chiefly what we are going to want to distinguish. We will concentrate on the monic polynomial $ \ x^2 \ \pm \ \frac{12? \ 16?}{125} ·x \ - \ \frac{1}{125} \ \ $ and see what the transformed zeroes prove to be.
For numbers of the form $ \ x_{1,2} \ = \ -\frac12 \ \pm \ \frac{\sqrt{D}}{2} \ \ , $ we find
$$ x_{1,2} + 1 \ \ = \ \ +\frac12 \ \pm \ \frac{\sqrt{D}}{2} \ \ \Rightarrow \ \ \frac{1}{x_{1,2} + 1} \ \ = \ \ \frac{2}{1 \ \pm \ \sqrt{D}} \ \ = \ \ \frac{2·(1 \ \mp \ \sqrt{D})}{1 \ - \ D} $$
$$ \Rightarrow \ \ x'_{1,2} \ \ = \ \ \left( \frac{1}{x_{1,2} + 1} \right)^3 \ \ = \ \ \frac{8·(1 \ \mp \ \sqrt{D})^3}{(1 \ - \ D)^3} \ \ . $$
We pretty definitely don't want to have to multiply these expressions out, but that won't be necessary. The middle coefficient of the new monic polynomial is $ \ -(x'_1 \ + \ x'_2) \ = \ \frac{8}{(1 \ - \ D)^3} \ · \ [ \ (1 \ + \ \sqrt{D})^3 \ + \ (1 \ - \ \sqrt{D})^3 \ ] \ \ . $ We don't even really need to carry out the binomial expansions of the terms in brackets, since it is clear that the terms with odd powers of $ \ \sqrt{D} \ $ will "cancel" in the sum (we do need to know the binomial coefficients). The middle coefficient of the monic polynomial is therefore
$$ -(x'_1 \ + \ x'_2) \ \ = \ \ \frac{8}{(1 \ - \ D)^3} \ · \ 2 · (1 \ + \ 3D ) \ \ ; $$
with $ \ D \ = \ 21 \ \ $ for our problem, we find
$$ -(x'_1 \ + \ x'_2) \ \ = \ \ \frac{8 \ · \ 2 \ · \ (1 \ + \ 3·21 )}{(1 \ - \ 21)^3} \ \ = \ \ \frac{16 \ · \ 64}{ - 20^3} \ \ = \ \ -\frac{4^2 \ · \ 4^3}{ 4^3 · 5^3} \ \ = \ \ -\frac{16}{125} \ \ . $$
[With just a bit more -- but unrequired -- effort, we can confirm that
$$ x'_1 \ · \ x'_2 \ \ = \ \ \frac{8^2 \ · \ [ \ (1 \ + \ \sqrt{D}) \ · \ (1 \ + \ \sqrt{D}) \ ]^3}{(1 \ - \ D)^6} \ \ = \ \ \frac{4^3 \ · \ \ (1 \ - \ D)^3}{(1 \ - \ D)^6}\ \ = \ \ \frac{4^3 }{(1 \ - \ D)^3} $$ $$ = \ \ \frac{ 4^3}{ - 4^3 · 5^3} \ \ = \ \ -\frac{1}{125} \ \ . \ ] $$
Hence, the transformed quadratic equation is $ \ 125·\left(x^2 \ + \ \frac{16}{125}·x \ - \ \frac{1}{125} \right) \ = \ 0 \ \ , $ which is choice $ \ \mathbf{(1)} \ \ . $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Expressing $f(x) = \frac{x-1}{x+1}$ as a sum of an even and odd function I am trying to write $$f(x) = \frac{1-x}{x+1}$$ as a sum of an even and odd function. One solution, though messy, is to use the following derivation:
We work backwards. Let $f = f_e + f_0$, where $f_e$ is even and $f_0$ odd. Then $f(x) = f_e (x) + f_0 (x)$ for all $x$ in the domain of $f$. But then
$$
f(-x) = f_e (-x) + f_0 (-x) = f_e (x) - f_0 (x).
$$
Adding, we eliminate $f_0 (x)$:
$$
f(x) + f(-x) = 2f_e (x),
$$
so
$$
f_e (x) = \frac{f(x) + f(-x)}{2}.
$$
Now, $f_0 = f(x) - f_e$. Then:
$$
f_0 (x) = f(x) - \frac{f(x) + f(-x)}{2} = \frac{2f(x) - f(x) - f(-x)}{2} = \frac{f(x) -f(-x)}{2}.
$$
One easily checks that $f_e$ is even and $f_0$ odd.
Using this derivation, I get:
\begin{align*}
\frac{1-x}{1+x} = \frac{1}{2}\left(\frac{1-x}{1+x} + \frac{1 + x}{1 - x} \right) + \frac{1}{2} \left(\frac{1-x}{1+x} - \frac{1+x}{1-x} \right).
\end{align*}
Assuming I haven't made an algebra mistake, this seems to work, but it's not elegant. I did some research into this and found another proposed solution, but I believe it to be faulty. First, I'll just present it.
\begin{align*}
\frac{1-x}{1+x} & = \frac{1-x}{1+x} \cdot \frac{1-x}{1-x} \\
& = \frac{1 - 2x + x^2}{1 - x^2} \\
& = \frac{x^2 + 1}{1 - x^2} + \frac{2x}{1 - x^2}.
\end{align*}
One then checks that the first function is even and the second is odd.
My problem with this solution is that it is not valid for every $x$ in the domain of $f$. By definition, the domain of $f$ is $\mathbb{R} \setminus \{-1\}$. But this solution, in multiplying by $\frac{1-x}{1-x}$, presupposes that $x \neq 1$, which does not seem to me to be allowed.
I'm back at square one, then, because the solution I found does not seem in any way elegant or natural, and I'm assuming there is some kind of a trick that I am missing. I tried polynomial long division, partial fractions, and so forth, and nothing, other than the above derivation, brought me any progress.
| In both cases, your solutions are undefined when $x=\pm1$. And that's natural. After all, $f(x)$ is undefined when $x=-1$ and, if you express $f(x)$ as $f_o(x)+f_e(x)$, with $f_o$ odd and $f_e$ even, then it is natural that at least one of the functions $f_o$ and $f_e$ is undefined when $x=-1$. But then, since $f_o(-x)=-f_o(x)$ and $f_e(-x)=f_e(x)$, if $f_o(x)$ is undefined when $x=-1$, then it is also undefined when $x=1$, and the same thing applies to $f_e(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Area of Cyclic Quadrilateral
*
*Let $ABCD$ be a quadrilateral inscribed in a circle with diameter $AC$, and let $E$ be the foot of perpendicular from $D$ onto $AB$. If $AD=DC$ and the area of quadrilateral $ABCD$ is $24$, find $DE$.
Here's what I did:
Assuming the radius of the circle to be $r$, $AD=DC=a$, $AB=b$ and $BC=c$, I found out that:
• $a^2=2r^2$, and $b^2+ c^2 = 4r^2$, using $Pythagoras$
•Next $ $ $[ABCD]= \frac {a²}{2} + \frac {bc}{2} $.
Substituting the values, I got $b+c=4\sqrt6$.
•Finally, we have $$[ABCD] = \frac {(b+c)(b+c)(2a+b-c)(2a+c-b)}{16} \\... using \ Brahmagupta \ Formula$$
Substituting values, I got $bc=24$ $\Rightarrow$ $ b=c=2\sqrt6$.
I am stuck here.
What should I do next?
Thanks.
|
If $r$ is the radius of the circle, by Ptolemy Theorem,
$AC \cdot BD = AB \cdot CD + BC \cdot AD$
$2r \cdot BD = (AB + BC) \cdot AD \tag1$
As $AD = DC$, $\angle ABD = \angle CBD = 45^ \circ$
So, $\angle AOD = 90^ \circ \implies AD = r \sqrt2$
Plugging into $(1)$,
$BD = \cfrac{1}{\sqrt2} (AB + BC) \tag2$
Area of quadrilateral is,
$\cfrac{1}{2} (AB \cdot BD \sin 45^ \circ + BC \cdot BD \sin 45^ \circ) = 24$
$(AB+BC) \cdot BD = 48 \sqrt2 \tag3$
From $(2)$ and $(3)$, we easily see that $BD = 4 \sqrt3$ and then $DE = BD \sin 45^0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove $\sum_{cyc}\frac{xy+1}{(x+y)^2}\geq 3$ when $x^2+y^2+z^2+(x+y+z)^2\leq 4$.
Let $x,y,z\in \Bbb{R}^+$ such that $x^2+y^2+z^2+(x+y+z)^2\leq 4$. Prove that $$\sum_{cyc}\frac{xy+1}{(x+y)^2}\geq 3.$$
As there are three fractions in the left side and a single term in the right side, I thought Cauchy-Schwarz might be of help. But from Cauchy-Schwarz, I got $$\sum_{cyc}\frac{xy+1}{(x+y)^2}\geq \frac{\left(\sum_{cyc}\sqrt{xy+1}\right)^2}{x^2+y^2+z^2+(x+y+z)^2}.$$
Then we have to prove that $$\left(\sum_{cyc}\sqrt{xy+1}\right)^2\geq12.$$
But this doesn't seem to work. So, how to solve the problem?
| $$\sum_{cyc}\frac{xy+1}{(x+y)^2}-3\geq\sum_{cyc}\frac{2xy+\sum\limits_{cyc}(x^2+xy)}{2(x+y)^2}-3=\frac{\sum\limits_{cyc}(x^2+3xy)\sum\limits_{cyc}(x^4-x^2y^2)}{2\prod\limits_{cyc}(x+y)^2}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Analytically minimize $x^2 + y^2$ constrained to $(x+1)^3 =- y^2$ I have the following problem:
$$\text{min} \ x^2 + y^2$$
$$s.t. \ (x+1)^3 =- y^2$$
What I did was substituting, so I got the function $f(x) = x^2 - (x+1)^3 $ but I don't know how to get analytically to the minimum. Graphing the original problem gives $x = -1$ and $y=0$ as the solution but I can't get there using my new function, I did the first derivative equal to $0$ but that has no real solutions. I did Lagrange method on the original problem but it was nonsense too. Thanks for the help.
| Believe in Algebra
If $x+1>0$, then we get a contradiction: $y^2<0$.
This immediately implies, $x+1≤0\implies x≤-1$ must be.
Then, we can define the function $f(x)$, such that $$f(x)=x^2-(x+1)^3,~x≤-1.$$
Method $-1$
$$\begin{align}x≤-1\implies &\begin{cases}x^2≥1\\ (x+1)^3≤0\end{cases}\\\\
\implies &\begin{cases}x^2≥1\\ -(x+1)^3≥0
\end{cases}\\\\ \implies &x^2-(x+1)^3≥1.\end{align}$$
Finally, we conclude that,
$$\begin{align}&\min \left\{x^2-(x+1)^3\mid x≤-1\right\}=1,\\
&~~~~~~~~~~~~\text{at}~|x|=1,~x+1=0 \\ \\
\implies &\min \left\{x^2-(x+1)^3\mid x≤-1\right\}=1,\\
&~~~~~~~~~~~~~~~~~~~~~\text{at}~x=-1\\\ \end{align}$$
This means,
$$\min \left\{x^2+y^2 \mid (x+1)^3=-y^2\right\}=1,\\
\text{at}~x=-1,~y=0.$$
Method $-2$
$$\begin{align}x^2-(x+1)^3&=(x+1)^2-2x-1-(x+1)^3\\&=(x+1)^2-2(x+1)+1-(x+1)^3\\
&=(x+1)^2-(x+1)^3-2(x+1)+1\\
&=z^3+z^2+2z+1≥1,~z=-(x+1)≥0.\\ \end{align}$$
Method $-3\text{a}$
$$\begin{align}x^2-(x+1)^3&=x^2-1-(x+1)^3+1=\\
&=(x-1)(x+1)-(x+1)^3+1\\
&=(x+1)(x-1-x^2-2x-1)+1\\
&=\underbrace{-(x+1)}_{≥0}\underbrace{(x^2+x+2)}_{>0}+1≥1.\end{align}$$
Method $-3\text{b}$
$$\begin{align}x^2-(x+1)^3&=-(x+1)(x^2+x+2)+1\\
&=-(x+1)\left((x+1)^2-(x+1)+2)\right)+1\\
&=-(x+1)^3+(x+1)^2-2(x+1)+1\\
&=z^3+z^2+2z+1≥1,~z=-(x+1)≥0.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 4
} |
If $x^3+y^2$ and $x^2+y^3$ are integers, show that $x$, $y$ are integers
Given rational numbers $x$, $y$ such that $x^3+y^2$ and $x^2+y^3$ are integers, prove that $x$, $y$ are integers.
For this problem I don't even know how to start tackling it. I tried various ways:
*
*Letting $x = \frac{a}{b}$, $y = \frac{c}{d}$, which only makes it more complicated
*Doing operations on them: $x^3+y^2+x^2+y^3=x^2(x+1)+y^2(y+1)$, which I have no idea how to continue;
$(x^3+y^2)(x^2+y^3)=x^5+y^5+x^3y^3+x^2y^2$, and it's also probably too complex to break down
$x^3+y^2-x^2-y^3=(x-y)(x^2+xy+y^2)+(y-x)(x+y)$ seems more plausible, but I also can't do anything with this.
I would really appreciate any way of tackling this problem, because I have spent a while thinking about this problem.
EDIT: I have found a solution to this problem (this was edited 2 days after this was posted, so there were also answers before this):
Suppose $x = \frac{a}{b}$, $y = \frac{c}{d}$ in which $a$, $b$, $c$, $d$
are integers and $\gcd(a,b) = \gcd(c,d) = 1$. Then we have
$x^2+y^3=\frac{a^2}{b^2}+\frac{c^3}{d^3}=\frac{a^2d^3+b^2c^3}{b^2d^3}$,
so:
$b^2|a^2d^3+b^2c^3$, which means $b^2|a^2d^3$, and $\gcd(a,b)=1$ so
$b^2|d^3$, and
$d^3|a^2d^3+b^2c^3$, which means $d^3|b^2$, so $b^2=d^3$.
Doing the same to $x^3+y^2$, we get $b^3=d^2$. From that we get
$b^5=d^5$, so $b=d$. Substituting to $b^2=d^3$, we get $b^2=b^3$,
hence $b=d=1$, which implies that $x$ and $y$ are integers.
| Valuations to the rescue.
Hint: Assume that at least one of $x,y$ is not an integer. Let $p$ be a prime factor appearing in one of the denominators. Without loss of generality we can assume that $p$ appears in the denominator of $x$ to at least as high a power than in the denominator of $y$ (otherwise swap their roles in what follows). Show that this implies that $p^3$ is a factor of the denominator of $x^3+y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 3,
"answer_id": 2
} |
Solving the eigenvalue from a set of coupled second order differential equation numerically I met a problem in solving a set of coupled differential equation. Given a set of equation as shown in below:
\begin{align*}
A_1\psi_1(z)+A_2\frac{d^2\psi_1(z)}{dz^2}+A_3\frac{d\psi_2(z)}{dz}&=\lambda\psi_1(z)\\
A_4\psi_2(z)+A_5\frac{d^2\psi_2(z)}{dz^2}+A_3\frac{d\psi_1(z)}{dz}&=\lambda\psi_2(z)
\end{align*}
with the following 4 boundary conditions:
\begin{align*}
\psi_1(0)=\psi_1(d)&=0\\
\psi_2(0)=\psi_2(d)&=0
\end{align*}
where $A_{i}$ is a constant coefficient.
I've been stuck in this question for a long time. As the coupling term $A_{3}$ present in the problem, we cannot simply solve it analytically. Without the boundary condition on the first order derivative, I was unable to determine the value of $\lambda$. Is the boundary value sufficient to tackle the problem? Is there any numerical method to solve this type of eigenvalue problem?
| Hint.
Using the Laplace transform we have
$$
\cases{
(A_1+s^2A_2-\lambda)\Psi_1(s)+A_3 s \Psi_2(s) = A_2(s\psi_1(0)+\dot\psi_1(0))+A_3\psi_2(0)\\
(A_4+s^2A_5-\lambda)\Psi_2(s)+A_3 s \Psi_1(s) = A_5(s\psi_2(0)+\dot\psi_2(0))+A_3\psi_1(0)\\
}
$$
and then
$$
\cases{
\Psi_1(s) = \frac{A_3 (A_3 \psi_1(0) s-A_$ \psi_2(0)+A_5 \dot\psi_2(0)s+\lambda \psi_2(0))-A_2 (\dot\psi_1(0)+\psi_1(0) s)
\left(A_4+A_5 s^2-\lambda \right)}{\left(A_1+A_2s^2-\lambda \right) \left(\lambda-A_4-A_5 s^2 \right)+A_3^2s^2}\\
\Psi_2(s) = \frac{\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}{\left(A_1+A_2s^2-\lambda \right) \left(\lambda-A_4-A_5 s^2 \right)+A_3^2s^2}
}
$$
so the eigenfunctions are linked to the denominator roots or
$$
\left(A_1+A_2s^2-\lambda \right) \left(\lambda-A_4-A_5 s^2 \right)+A_3^2
s^2 = 0
$$
NOTE
Assuming numerical values to $A_k = 1$ and after Laplace inversion we obtain the following set of conditions
$$
\left(
\begin{array}{cccc}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\frac{2 \cosh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & -\frac{2 \sinh \left(\frac{L}{2}\right)
\sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \cosh \left(\frac{L}{2}\right) \cosh \left(\frac{1}{2} L \sqrt{4 \lambda
-3}\right)-\frac{\sinh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \frac{\cosh
\left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}}-\sinh \left(\frac{L}{2}\right) \cosh \left(\frac{1}{2}
L \sqrt{4 \lambda -3}\right) \\
-\frac{2 \sinh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \frac{2 \cosh \left(\frac{L}{2}\right)
\sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \frac{\cosh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4
\lambda -3}\right)}{\sqrt{4 \lambda -3}}-\sinh \left(\frac{L}{2}\right) \cosh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right) & \cosh
\left(\frac{L}{2}\right) \cosh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)-\frac{\sinh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4
\lambda -3}\right)}{\sqrt{4 \lambda -3}} \\
\end{array}
\right)\left(\begin{array}{c}\dot\psi_1(0)\\ \dot\psi_2(0)\\ \psi_1(0)\\ \psi_2(0)\end{array}\right) = 0
$$
and
$$
\det(M) = \frac{2 \left(\cosh \left(L \sqrt{4 \lambda -3}\right)-1\right)}{4 \lambda -3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Two methods of finding out $\frac{d^2y}{dx^2}$ when $\frac{dy}{dt} = 4t-2$ and $\frac{dx}{dt} = 2t + 3$ Find $\frac{d^2y}{dx^2}$ given that $\frac{dy}{dt} = 4t-2$ and $\frac{dx}{dt} = 2t + 3$ at $t=2$
I tried two methods for this question, both of which give me different answers. Which one is wrong and why so?
Method $1$: $\frac{dy}{dt} = 4t-2$ hence $\frac{d^2y}{dt^2} = 4$, similarly $\frac{d^2x}{dt^2} = 2$ therefore, $\frac{d^2y}{dx^2} = 2$
Method $2$: $\frac{dy}{dx} = \frac{4t-2}{2t+3}$ therefore $\frac{d^2y}{dx^2} = \frac{(2t+3)4 - (4t-2)2}{(2t+3)^2}.\frac{dt}{dx} = \frac{16}{(2t+3)^2}.\frac{1}{2t+3} = \frac{16}{7^3}$
I can't understand why one method is wrong while the other is right, however they have very different answers. What am I doing wrong?
| Method #2 is the correct one, using the formula...
$$\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} $$
which is valid.
Method #1 is using a formula ...
$$\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} \frac{d^2t}{dx^2} \tag{???}$$
which is not valid. To see this consider the case where $t=x$,for which
$\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2}$, but $(???)$ would give $\frac{d^2y}{dx^2} = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how to prove $\sum_{cyc}(1-a_{1}+a_{1}a_{2})^2\ge\frac{n}{2}$ let $a_{i}\in [0,1]$,prove or disprove
$$f_{n}=(1-a_{1}+a_{1}a_{2})^2+(1-a_{2}+a_{2}a_{3})^2+\cdots+(1-a_{n}+a_{n}a_{1})^2\ge\dfrac{n}{2}$$
I can only prove $n=3$.
$$f_{3}=\sum_{cyc}[1-a_{1}(1-a_{2})]^2=3-2\sum_{cyc}a_{1}(1-a_{2})+\sum_{cyc}a^2_{1}(1-a_{2})^2$$
let $$t=a_{1}(1-a_{2})+a_{2}(1-a_{3})+a_{3}(1-a_{1})=1-(1-a_{1})(1-a_{2})(1-a_{3})-a_{1}a_{2}a_{3}\le 1$$
and
$$\sum_{cyc}a^2_{1}(1-a_{2})^2=(\sum_{cyc}a_{1}(1-a_{2}))^2-2\sum_{cyc}a_{1}a_{2}(1-a_{1})(1-a_{3})\ge t^2-\dfrac{1}{4}\sum_{cyc}a_{2}(1-a_{3})=t^2-\dfrac{1}{4}t$$
so we have
$$f_{3}\ge 3-2t+t^2-\dfrac{1}{4}t\ge\dfrac{3}{2}$$
But for $n$ I can't,maybe use induction to prove it?
| I would say that you can not prove it using induction.
Firstly, let us rewrite $f(n)$ as $$f_{n}=(1+a_{1}(a_{2}-1))^2+(1+a_{2}(a_{3}-1))^2+\cdots+(1+a_{n}(a_{1}-1))^2$$
By inductive hypothesis, $$(1+a_{1}(a_{2}-1))^2+(1+a_{2}(a_{3}-1))^2+\cdots+(1+a_{n}(a_{1}-1))^2\ge\dfrac{n}{2}$$
Therefore, we have that $$(1+a_{1}(a_{2}-1))^2+(1+a_{2}(a_{3}-1))^2+\cdots+(1+a_{n}(a_{n+1}-1))^2+(1+a_{n+1}(a_{1}-1))^2\geq$$
$$\dfrac{n}{2}-(1+a_{n}(a_{1}-1))^2+(1+a_{n}(a_{n+1}-1))^2+(1+a_{n+1}(a_{1}-1))^2$$
We have to prove that $$(1+a_{n}(a_{n+1}-1))^2+(1+a_{n+1}(a_{1}-1))^2-(1+a_{n}(a_{1}-1))^2\geq\frac{1}{2}$$
Unfortunately, this inequality is false (take for instance $a_n=0$, $a_{n+1}=1$, and $a_1$ whatever value you want lesser than $\frac{3}{4}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Integration by Substitution with $~ t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right) ~$ The angle was halved I've been encountering the problem of the below integration.
$$ \int_{0 }^{2\pi } \frac{ d\theta }{ R+ r \cdot \cos^{}\left(\theta_{} \right) } \tag{1} $$
The official description states that the above integration formula can be calculated using substitution of integration.
$$ t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \therefore ~~ \int_{0 }^{2\pi } \frac{ d\theta }{ R+ r \cdot \cos^{}\left(\theta_{} \right) } = \frac{ 2\pi }{ \sqrt{ R ^{2} -r ^{2} } } $$
Currently I can't derive the above RHS.
I think firstly find out the form of result of calculations of indefinite integral of eqn1 is wiser way.
What I tried so far are as below.
$$ t= \tan^{}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \frac{ \theta_{} }{ 2 }= \tan^{-1} \left( t \right) ~~ \leftarrow~~ \text{Thought that this approach won't work} $$
$$ \frac{ dt }{ d\theta } = \sec^{2}\left( \frac{ \theta_{} }{ 2 } \right) \cdot \frac{1}{2} $$
$$ \frac{ 2dt }{ d\theta } = \sec^{2}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \frac{ d\theta }{ 2 dt } =\sec^{-2}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \frac{ d\theta }{ 2 dt } = \left( \sec^{}\left( \frac{ \theta_{} }{ 2 } \right) \right)^{-2} $$
$$ \frac{ d\theta }{ 2 dt } = \left( \cos^{-1}\left( \frac{ \theta_{} }{ 2 } \right) \right)^{-2} $$
$$ \frac{ d\theta }{ 2 dt } = \cos^{2}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ d\theta = 2 dt \cdot \cos^{2}\left(\frac{ \theta_{} }{ 2 } \right) $$
First things to first, the equation1 has $~ \cos^{}\left(\theta_{} \right) ~$ however how can I handle $~ t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right) ~$ ??
| The goal with the $t=\tan\frac{\theta}{2}$ substitution is to find all trig functions in terms of $t$. Of course with any integral substitution, the first thing is to find $\frac{dt}{d\theta}$.
$$\frac{dt}{d\theta}=\frac{1}{2}\sec^2\frac{\theta}{2}$$
$$\frac{dt}{d\theta}=\frac{t^2+1}{2}$$
Since we also have a $\cos\theta$ term in our integrand, we need to find $\cos\theta$ in terms of $t$. We start with
$$\cos\theta=2\cos^2\frac{\theta}{2}-1$$
$$\cos\theta=\frac{2}{\sec^2\frac{\theta}{2}}-1$$
$$\cos\theta=\frac{2}{t^2+1}-1$$
$$\cos\theta=\frac{1-t^2}{t^2+1}$$
As for the bounds of the resulting integral, we can split the bounds in half, apply the substitution, and then merge the two bounds. This gives us a bounds of $t\in (-\infty,\infty)$, providing that both halves are finite. Our integral is now
$$\int_{-\infty}^\infty \frac{2}{t^2+1}\cdot\frac{1}{R+\frac{1-t^2}{t^2+1}r}\, dt$$
$$=\int_{-\infty}^\infty 2\cdot\frac{1}{(R-r)t^2+(R+r)}\, dt$$
$$=\frac{2}{R+r}\int_{-\infty}^\infty \frac{dt}{\frac{R-r}{R+r}t^2+1}$$
$$=\frac{2}{\sqrt{R^2-r^2}}\left[\tan^{-1} \left(\sqrt{\frac{R-r}{R+r}}t\right)\right]_{-\infty}^\infty$$
As long as $R>r$, then we can see that
$$\lim_{t\to\infty} \tan^{-1} \left(\sqrt{\frac{R-r}{R+r}}t\right)=\frac{\pi}{2}$$
$$\lim_{t\to -\infty} \tan^{-1} \left(\sqrt{\frac{R-r}{R+r}}t\right)=-\frac{\pi}{2}$$
Hence, our integral evaluates to
$$=\boxed{\frac{2\pi}{\sqrt{R^2-r^2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Calculate the volume between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$. Calculate the volume between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$.
Attempt
We project on the $xy$ plane the intersection between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$, which is the circle $x^2+y^2=1, z=1$.
We can conclude that the region between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$ can be described by
$$-1\leq x\leq 1, -\sqrt{1-x^2}\leq y \leq \sqrt{1-x^2}, x^2+y^2\leq z \leq \sqrt{x^2+y^2}$$
The volume is given by
$$V=\int \int \int_W dxdydz=\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{x^2+y^2}^{\sqrt{x^2+y^2}} dxdydz$$
When I try to solve this, I get a difficult expression and cannot calculate it. So I think, everything I have done is wrong.
| As mentioned in comments, this is easier in cylindrical coordinates.
$x = r \cos\theta, y = r \sin\theta, z = z$
Paraboloid surface is $z = x^2 + y^2 = r^2$ and surface of the cone is $z = \sqrt{x^2+y^2} = r$
i) going in the order $dr$ first,
$z \leq r \leq \sqrt z, 0 \leq z \leq 1, 0 \leq \theta \leq 2\pi$
$\displaystyle \int_0^{2\pi} \left[\int_0^1 \left[\int_z^{\sqrt z} r \ dr \right] \ dz \right]\ d\theta $
ii) going in the order $dz$ first,
you can set up the integral using bounds,
$r^2 \leq z \leq r, 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi \ $.
Both integrals are straightforward to evaluate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Are there positive integers $x$, $y$ and prime numbers $p$ so that $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{p}$ I have a solution for this, but I'm not really sure about that:
We have: $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{p}$, multiplying both sides by $\sqrt[3]{x^2}-\sqrt[3]{xy}+\sqrt[3]{y^2}$ we get $x+y=\sqrt[3]{p}(\sqrt[3]{x^2}-\sqrt[3]{xy}+\sqrt[3]{y^2})$ so $\sqrt[3]{px^2}-\sqrt[3]{pxy}+\sqrt[3]{py^2}$ is an integer.
Suppose $\sqrt[3]{px^2}$, $\sqrt[3]{pxy}$, and $\sqrt[3]{py^2}$ are all integers, we can easily see that $p|x$ and $p|y$.
Let $x^2=p^2a^3$, $y^2=p^2b^3$ where $a$ and $b$ are positive integers. Then we can see that $a$ and $b$ also have to be perfect squares since $x=p\sqrt{a^3}$ and $y=p\sqrt{b^3}$ are integers. Since $\gcd(2,3)=1$, we can let $x^2=p^2a'^6$, $y^2=p^2b'^6$, or $x=pa'^3$, $y=pb'^3$ where $a'$ and $b'$ are integers. Subbing that to the original equation we get $\sqrt[3]{pa'^3}+\sqrt[3]{pb'^3}=\sqrt[3]{p}$, so $\sqrt[3]{p}(a'+b')=\sqrt[3]{p}$. But $a'+b'>1$, so $\sqrt[3]{p}(a'+b')>\sqrt[3]{p}$, hence a contradiction.
I can see I made a lot of assumptions here, like "Let $x^2=p^2a^3$, $y^2=p^2b^3$", or assuming all $\sqrt[3]{px^2}$, $\sqrt[3]{pxy}$, and $\sqrt[3]{py^2}$ are integers. Are there any problems with my work or is it good to go? And moreover, do you have a better solution than this? I appreciate your time and effort for this and thanks a lot in advance.
| Without using a specific formula, you can also construct a solution as follows:
$$\begin{align}&\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{p},\thinspace x,y\in\mathbb Z^{+}; p\in\mathbb P\\
\implies &x+y+3\sqrt [3]{xy}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)=p\\
\implies &x+y+3\sqrt[3]{pxy}=p\\
\implies &27 pxy=(p-x-y)^3 \\
\implies &p\mid (p-x-y)^3\\
\implies &p\mid p-x-y \\
\implies &p≤p-x-y ,\thinspace p>x+y≥2 \\
\implies &x+y≤0,~\text{A contradiction.}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$ Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms
My Attempt:
The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac{1}{2(4i+1)}-\frac{1}{4i+2}+\frac{1}{2(4i+3)}\right)=\frac{1}{2}\sum_{i=0}^\infty \int_0^1 \left(x^{4i}-2x^{4i+1}+x^{4i+2}\right) \, dx$$
So,$$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4} + \frac{x^2}{1-x^4}\right)dx=\frac{1}{2} \int_0^1 \left(\frac{1+x^2}{1-x^4}-\frac{2x}{1-x^4}\right)\,dx = \frac{1}{2} \int_0^1 \left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)\,dx$$
$$=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\int_{0}^{1}\frac{2x}{1-x^4}dx=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\frac{1}{2}\int_{0}^{1}\frac{1}{1-y^2}dy=0(y=x^2)$$
which is obviously absurd since all terms of $S$ are positive.
But if I do like this then I am able to get the answer, $$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4}+\frac{x^2}{1-x^4}\right)dx=\frac{1}{2}\int_{0}^{1}\frac{(1-x)^2}{1-x^4}dx=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1+x}-\frac{x}{1+x^2}\right)dx=\frac{\ln2}{4}$$
What is wrong with the previous approach
| Find the limit
$$\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$$
The lim can be writen as:
$$S=\lim_{n\rightarrow\infty}\sum_{k=0}^n\frac{1}{(4k+1)(4k+2)(4h+3)}$$
$$=\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{4k+2}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right)=\frac{1}{2}\sum_{k=0}^{\infty}\left(\frac{1}{(4k+1)(4k+2)}-\frac{1}{(4k+3)(4k+3)}\right)$$
$$=\frac{1}{2}\sum_{k=0}^{\infty}\left\{\left(\frac{1}{4k+1}-\frac{1}{4k+2}\right)-\left(\frac{1}{4k+2}-\frac{1}{4k+3}\right)\right\}=\frac{1}{2}(S_1-S_2)$$
Now we will use the special function (digamma function):
$$\Psi(x)=\frac{d}{dx}\ln\Gamma(x)$$
and the identity
$$\sum_{k=0}^{\infty}\left(\frac{1}{k+y}-\frac{1}{k+x}\right)=\Psi(x)-\Psi(y)$$
With $y=\frac{1}{4},\quad x=\frac{2}{4}$:
$$\sum_{k=0}^{\infty}\left(\frac{4}{4k+1}-\frac{4}{2k+2}\right)=4\sum_{k=0}^{\infty}\left(\frac{1}{4k+1}-\frac{1}{2k+2}\right)=\left(\Psi\left(\frac{2}{4}\right)-\Psi\left(\frac{1}{4}\right)\right)\rightarrow$$
$$S_1=\frac{1}{4}\left(\Psi\left(\frac{2}{4}\right)-\Psi\left(\frac{1}{4}\right)\right)$$
With $y=\frac{2}{4},\quad x=\frac{3}{4}$:
$$\sum_{k=0}^{\infty}\left(\frac{4}{4k+2}-\frac{4}{2k+3}\right)=4\sum_{k=0}^{\infty}\left(\frac{1}{4k+2}-\frac{1}{2k+3}\right)=\left(\Psi\left(\frac{3}{4}\right)-\Psi\left(\frac{2}{4}\right)\right)\rightarrow$$
$$S_2=\frac{1}{4}\left(\Psi\left(\frac{3}{4}\right)-\Psi\left(\frac{2}{4}\right)\right)$$
Hence
$$S=-\frac{1}{8}\left(\Psi\left(\frac{1}{4}\right)-2\Psi\left(\frac{1}{2}\right)+\Psi\left(\frac{3}{4}\right)\right)$$
The particular values of $\Psi(x)$ are:
$\psi\left(\frac{1}{2}\right)=-C-2\ln2$
$\psi\left(\frac{1}{4}\right)=-C-\frac{\pi}{2}-3\ln2$
$\psi\left(\frac{3}{4}\right)=-C+\frac{\pi}{2}-3\ln2$
Whwre $C$ is the Euler's constante.
Replacing in $S$ we obtine
$$S=\frac{\ln2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
Prove that: $a^{12}+b^{12}+c^{12} \geq \frac{2049}{8}\cdot a^4b^4c^4 $ Let $a,b,c$ be real and $a+b+c=0.$
Prove that:$$a^{12}+b^{12}+c^{12} \geq \frac{2049}{8}\cdot a^4b^4c^4$$
I know we can make the following:
Let $bc\geq0$ thus we need to prove
$$b^{12}+c^{12}+(b+c)^{12}\geq \frac{2049}{8} b^4c^4(b+c)^4$$or
$$\frac{(2b+c)^2(b+2c)^2(b-c)^2(4b^6+12b^5c+99b^4c^2+178b^3c^3+99b^2c^4+12bc^5+4c^6)}{8}\geq 0\text{,}$$which is true.
But I'm trying to find another solution. Can be?
| Alternative proof:
If $abc = 0$, the desired inequality is clearly true.
If $abc \ne 0$, WLOG, assume that $abc = -1$.
WLOG, assume further that $a, b > 0$ and $c < 0$.
We need to prove that $a^{12} + b^{12} + c^{12} \ge \frac{2049}{8}$.
Let $q = ab + bc + ca$. It is easy to obtain
$a^{12} + b^{12} + c^{12} = 2q^6-24q^3+3$. The details are given at the end.
Thus, it suffices to prove that
$$2q^6-24q^3+3 \ge \frac{2049}{8}$$
or (since $q < 0$)
$$q^3 \le - 27/4.$$
We have $a^3 + qa + 1 = a^3 + (ab + bc + ca)a + 1 = a^2(a + b + c) + abc + 1 = 0$.
Thus, we have
$q = - a^2 - \frac{1}{a} = - (a^2 + \frac{1}{2a} + \frac{1}{2a})
\le -3 \sqrt[3]{a^2 \cdot \frac{1}{2a} \cdot \frac{1}{2a}} = - 3\sqrt[3]{1/4}$.
Thus, we have $q^3 \le - 27/4$.
We are done.
The details of $a^{12} + b^{12} + c^{12} = 2q^6-24q^3+3$:
From $a + b + c = 0$ and $abc = -1$, it is easy to verify that
\begin{align*}
a^3 + qa + 1 &= 0, \\
b^3 + qb + 1 &= 0, \\
c^3 + qc + 1 &= 0.
\end{align*}
We have
\begin{align*}
&a^2 + b^2 + c^2 = (a + b + c)^2 - 2q = -2q, \\
&a^3 + b^3 + c^3 = -q(a + b + c) - 3 = - 3, \\
&a^4 + b^4 + c^4 = -q(a^2 + b^2 + c^2) - (a + b + c) = 2q^2.
\end{align*}
Then, we have
\begin{align*}
&a^{12} + b^{12} + c^{12}\\
=\,& (-qa - 1)^4 + (-qb - 1)^4 + (-qc - 1)^4\\
=\, & q^4(a^4 + b^4 + c^4) + 4q^3(a^3 + b^3 + c^3) + 6q^2(a^2 + b^2 + c^2) + 4q(a + b + c) + 3\\
=\, & 2q^6-24q^3+3.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $x^4 + ax^3 + 3x^2 +bx +1\geq0$, $\forall x\in \mathbb{R}$, find maximum of $a^2+b^2$
Given that $x^4 + ax^3 + 3x^2 +bx +1$ is always greater than equal to $0$ for all $x$ belongs to $\mathbb R$, find $\max(a^2 + b^2)$.
What I did was to show that that above expression is equivalent to $$[x(x+a/2)]^2 + [(12-a^2)(x + 2b/(12-a^2))^2]/4 + (12-a^2 - b^2) /(12-a^2)\geq 0,$$
from this one case is that if $ 12-a^2\geq 0$, then the expression trivially is greater/equal zero, which given max of $a^2+b^2$ to be $12$, but what about if $12-a^2<=0$, then is max of $a^2+ b^2$ greater than 12 possible and what is the exact maximum ? Conclusion: there are two such $f(x)$ for which maxima is attained those are $f(x) = x^4 -2√5x^3 + 3x^2+2√5x +1$ and other as said by achille hui.
|
I have started using MathJax but am not fluent in using it so posting the solution like this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
I don't agree this evaluation of the volume of $E=\{(x,y,z) \in \mathbb{R}^3 \ | \ 4 \leq x^2+y^2+z^2 \leq 9, x^2+y^2 \geq 1\}$ I've found the following solution of the problem "Evaluate the volume of $E=\{(x,y,z) \in \mathbb{R}^3 \ | \ 4 \leq x^2+y^2+z^2 \leq 9, x^2+y^2 \geq 1\}$.", but I don't agree with that from some point to the end, can someone help me understand if I'm wrong or right?
The set is even in all the three coordinates $x=r \cos t,\, y=r \sin t,\, z=z$ with $r \geq 0$ and $0 \leq t \leq 2\pi$, the volume is given by $8$ times the volume of the set
$F=\{\{(x,y,z) \in \mathbb{R}^3 \ | \ 4 \leq x^2+y^2+z^2 \leq 9, x^2+y^2 \geq 1, x \geq0,y\geq0,z\geq0\}$.
Using cylindrical coordinates, we get that $F$ becomes the set
$$G=\{(r,t,z) \in \mathbb{R}^3 \ | \ 4 \leq r^2+z^2 \leq 9,r^2\geq1, r\cos t \geq 0, r \sin t \geq0,z\geq0\}$$
$$=\{(r,t,z) \in \mathbb{R}^3 \ | \ 4-r^2 \leq z^2 \leq 9-z^2,r\geq1, 0 \leq t \leq \pi/2 ,z \geq 0\}$$
Now it is $4-r^2 \leq z^2 \leq 9-r^2 \iff \sqrt{4-r^2} \leq z \leq \sqrt{9-r^2}$ with the conditions $4-r^2 \geq 0$ and $9-r^2 \geq 0$; those two conditions together and the fact that $r \geq 1$ lead to $1 \leq r \leq 2$; since it must be $z \geq 0$ and $z \geq \sqrt{4-r^2}$, being the root always nonnegative it is $z \geq \sqrt{4-r^2}$. So in my opinion it should be
$$\text{Volume}(E)=8\iiint_G rdrdtdz=8\int_0^{\pi/2}\int_1^2 \int_{\sqrt{4-r^2}}^{\sqrt{9-r^2}}rdzdrdt$$
However the solution uses my same reasoning except when it must be discussed the interval of integration for $z$: it says that we must split the integral in the sum of these two integrals and so according to the solution it should be
$$\text{Volume}(E)=8\int_0^{\pi/2}\int_1^2 \int_{\sqrt{4-r^2}}^{\sqrt{9-r^2}}rdzdrdt+8\int_0^{\pi/2}\int_2^3 \int_{0}^{\sqrt{9-r^2}}rdzdrdt$$
Because it says that the conditions $4 \leq r^2+z^2 \leq 9,z\geq 0$ are equivalent to the system
$$\begin{cases} z \leq -\sqrt{4-r^2} \vee z \geq \sqrt{4-r^2} \\ -\sqrt{9-r^2} \leq z \leq \sqrt{9-r^2} \\ z \geq 0\end{cases} \iff \begin{cases} z \geq \sqrt{4-r^2} \\ 0\ \leq z \leq \sqrt{9-r^2}\end{cases}$$
And it says that $\sqrt{4-r^2} \leq z \implies 4-r^2 \geq 0$ and $\sqrt{9-r^2} \geq z \implies 9-r^2 \geq 0$, so it has to split the integral in the sum of two integrals that have different intervals on $r$ because of those implications. But I don't understand this latter statement, because it is not like I'm having two different conditions from above or two different conditions from below on the variable $z$ that can vary according to the interval where $r$ belongs to, because it is always $\sqrt{4-r^2} \geq 0$ and so the two conditions $z \geq 0$ and $z \geq \sqrt{4-r^2}$ are basically the only one condition $z \geq \sqrt{4-r^2}$. Am I wrong or am I right? Thanks.
| In cylindrical coordinates you get the set of conditions
$$
4\leqslant r^2+z^2\leqslant 9,\quad r^2\geqslant 1,\quad r\cos t\geqslant 0,\quad r\sin t\geqslant 0,\quad z\geqslant 0\tag1
$$
However these inequalities can be simplified a lot because in cylindrical coordinates $r\geqslant 0$ ever, so we get
$$
4\leqslant r^2+z^2\leqslant 9,\quad r\geqslant 1,\quad \cos t\geqslant 0,\quad \sin t\geqslant 0,\quad z\geqslant 0\tag2
$$
And so
$$
4\leqslant r^2+z^2\leqslant 9,\quad z\geqslant 0,\quad r\geqslant 1,\quad 0\leqslant t\leqslant \frac{\pi }{2}\tag3
$$
Now, the inequality $4\leqslant z^2+r^2\leqslant 9$ is satisfied for some $z$ if and only if $r\leqslant 3$, and together with the condition $r\geqslant 1$ we have that at most the range of $r$ is $[1,3]$. When $r\in[1,2]$ we have that the first inequality in (3) gives $z\in[\sqrt{4-r^2},\sqrt{9-r^2}]$ (because $z\geqslant 0$), and when $r\in[2,3]$ the inequality $4\leqslant r^2+z^2$ is trivially true for all values of $z$ so we only must care about the inequality $r^2+z^2\leqslant 9$, then in this case the range of $z$ is $[0,\sqrt{9-r^2}]$ (again, because $z\geqslant 0$).
These conditions are exactly the same of the given answer of your book.
I will add a different way to see it using set theory: let the conditions
$$
4-r^2\leqslant z^2\leqslant 9-r^2,\quad z\geqslant 0,\quad r\geqslant 1
$$
Then using intervals the conditions are equivalent to
$$
z^2\in [4-r^2,9-r^2]\cap [0,\infty )=[\max\{0,4-r^2\},9-r^2]
$$
and
$$
\max\{0,4-r^2\}=\begin{cases}
0,& r>2\\
4-r^2,&r\in[1,2]
\end{cases}
$$
Finally note when $r>3$ then the interval $[\max\{0,4-r^2\},9-r^2]$ is the empty set.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the length of DQ and the angle QDB on the triangle Find the length of DQ and the angle QDB.
To find the length of DQ, I use cosine rule:
\begin{align}
DQ^2&=QB^2+BD^2-2\cdot QB\cdot BD \cos(\angle QBD)\\
&=9+32-24\sqrt{2}\cos(\angle QBD)\\
&=41-24\sqrt{2}\cos(\angle QBD).
\end{align}
I don't know the angle $\angle QBD$. Now I try to find it using sine rule.
Consider that $\angle TBD =\angle QBD$.
\begin{align}
\dfrac{TD}{\sin(\angle TBD)}=\dfrac{DB}{\sin(\angle DTB)}.
\end{align}
We can see that to find $\angle QBD=\angle TBD$, need the $\angle DTB$. But we don't know the value $\angle DTB$. I'm get stuck here. I can't compute the length DQ and the $\angle QDB$.
Anyone know how to find it?
| Using Stewart Theorem
$$DQ^2=\frac{TD^2\times BQ+DB^2\times TQ}{BT}-BQ\times TQ$$
$$DQ^2=\frac{25 \times 3 +32\times 2}{5}-3\times 2$$
$$DQ^2=\frac {109}{5}$$
$$DQ=\sqrt{\frac {109}{5}}$$
Using Cosine Theorem
$$\cos\angle QDB = \frac{DQ^2+DB^2-QB^2}{2\times DQ \times DB}$$
$$\cos\angle QDB = \frac{\frac{109}{5}+32-9}{2\times \sqrt\frac{109}{5}\times 4\sqrt2}$$
$$\cos\angle QDB =\frac{14\sqrt{1090}}{545}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.