Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate $\lim\limits_{x\to\ 0}(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)})$
Evaluate $$\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)}\right)$$
It is easy with L'Hospital's rule, but takes too much time to calculate derivatives. What are some shorter methods?
| The problem as corrected by the OP is now more symmetric. We can consider cases characterized as
$$
L = \lim_{x \to 0} \frac{1}{f(x)g^{-1}(x)}-\frac{1}{f^{-1}(x)g(x)}
$$
where smooth $f(x) = x + \varepsilon_f x^m + o(x^{m+1}), g(x) = x + \varepsilon_g x^n + o(x^{n+1})$ with integers $m, n > 1$. Then $f^{-1}(x) = x - \varepsilon_f x^m + o(x^{m+1}), g^{-1}(x) = x - \varepsilon_g x^n + o(x^{n+1})$, and
$$
\begin{align}
L & = \lim_{x \to 0} \frac{1}{(x+\varepsilon_fx^m)(x-\varepsilon_gx^n)} -
\frac{1}{(x-\varepsilon_fx^m)(x+\varepsilon_gx^n)} \\
& = \lim_{x \to 0} \frac{2\varepsilon_gx^{n+1}-2\varepsilon_fx^{m+1}}
{x^4+o(x^6)} \\
& = \lim_{x \to 0} 2\varepsilon_gx^{n-3}-2\varepsilon_fx^{m-3}
\end{align}
$$
In this case, we have $f(x) = \sin x, g(x) = \tan x$, so $\varepsilon_f = -1/6, \varepsilon_g = 1/3, m = n = 3$, and then
$$
L = \lim_{x \to 0} 2 \left(\frac{1}{3}\right) - 2 \left(-\frac{1}{6}\right) = 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Prove an iff condition for the existence of LU decomposition I'm asked to prove that the matrix
$$A=\begin{pmatrix}
a & b & 0\\
c & d & e\\
0 & 1 & g
\end{pmatrix}$$
has a $LU$ decomp iff $a\not=0$ and $ad\not=bc$. However, I do not believe this to be true since we have
$$\frac{1}{2}\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
1 & 1 & 1
\end{pmatrix}\begin{pmatrix}
0 & 1 & 0\\
0 & 1 & 1\\
0 & 0 & 1
\end{pmatrix}=\begin{pmatrix}
0 & \frac{1}{2} & 0\\
0 & 1 & \frac{1}{2}\\
0 & 1 & 1
\end{pmatrix}$$
Where the LHS is a $LU$ decomp of the RHS, and the RHS has the form mentioned in the question, with $a=ad=bc=0$, contradicting the iff condition.
So what's going on here? Is this question just wrong, or am I not understanding something?
Thanks
| Given: $A=\begin{bmatrix}
a &b &0 \\
c&d &e \\
0 &1 &g
\end{bmatrix}$, we use the fact the the LU decomposition is possible if no row exchanges are required.
$\Rightarrow :$ Assuming $A$ has an $LU$ decomposition, we start by doing an elimination step. If $a\ne 0$, then the multipliers are $m_{21}=\frac{c}{a}$ and $m_{31}=0$. Hence the elimination produces
$$A^{(1)}=\begin{bmatrix}
a &b &0 \\
0&d-\frac{bc}{a} &e \\
0 &1 &g
\end{bmatrix}$$
A row exchange will not be required if $d\ne \frac{bc}{a}$, or $ad\ne bc$. If this is the case, we proceed with another elimination step with $m_{32}=\frac{1}{d-\frac{bc}{a}}$, from which we obtain
$$A^{(2)}=\begin{bmatrix}
a &b &0 \\
0&d-\frac{bc}{a} &e \\
0 &0 &g-\frac{e}{d-\frac{bc}{a}}
\end{bmatrix}$$
Therefore the decomposition becomes
$$A=LU=\begin{bmatrix}
1 &0 &0 \\
-\frac{c}{a} &1 &0 \\
0 &-\frac{1}{d-\frac{bc}{a}} & 1
\end{bmatrix} =\begin{bmatrix}
a &b &0 \\
0&d-\frac{bc}{a} &e \\
0 &0 &g-\frac{e}{d-\frac{bc}{a}}
\end{bmatrix}$$
$\Leftarrow$: Let us prove by contrapositive. Assume $a=0$ or $ad=bc$.
If $a=0$, then $A$ will be
$$A=\begin{bmatrix}
0 &b &0 \\
c&d &e \\
0 &1 &g
\end{bmatrix}$$
We see that the first pivot is zero, and hence row exchanges are required and the LU decomposition is not possible.
If $ad=bc$, then a similar step of elimination (see above) produces
$$A^{(1)}=\begin{bmatrix}
a &b &0 \\
0&\frac{1}{a}0=0 &e \\
0 &1 &g
\end{bmatrix}$$
which shows a zero second pivot, and hence requires a row exchange.
Therefore we conclude that the $LU$ decomposition exists iff $a\ne 0$ and $ad \ne bc$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $f(x) = \frac{\sin^{-1} x}{\sqrt{1- x ^2}}$, then evaluate $(1-x^2)f''(x) - xf(x)$
$f(x) = \dfrac{\sin^{-1} x}{\sqrt{1- x ^2}}$
Differentiating the given function, we get
$f'(x) = \dfrac{1 + \dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}}{1-x^2}$
which can also be written as
$f'(x) = \dfrac{1 + xf(x)}{1-x^2}$
Differentiating this one more time
$f''(x) = \dfrac{(1-x^2)[xf'(x) + f(x)] - [1 + xf(x)](-2x)}{(1-x^2)^2}$
Now the next obvious step is to substitute for $f'(x)$ since we have no $f'(x)$ term in the question.
Which further simplify the equation
$f''(x) = \dfrac{3x[1 + xf(x)] + (1-x^2)f(x)}{(1-x^2)^2}$
Head scratch, I have no idea how to proceed. Possibilities are I have made a mistake above, if so please point out and help me take this ahead.
| Your question has probably a typo, here's how.
Given $$f(x)=\frac{\sin^{-1}x}{\sqrt{1-x^2}}$$
Using quotient rule,
$$f'(x)=\frac{1+\frac{x\sin^{-1}x}{\sqrt{1-x^2}}}{1-x^2}$$
Re arranging this we get,
$$(1-x^2)f'(x)=1+x\times \frac{\sin^{-1}x}{\sqrt{1-x^2}}$$
$$(1-x^2)f'(x)=1+xf(x)$$
We get finally, $$(1-x^2)f\color{red}{'}(x)-xf(x)=\color{blue}{1}$$
This is the statement, which gives you the same result as mentioned in comments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluate $\int\frac{d\theta}{1+x\sin^2(\theta)}$ We have to evaluate $$\int\frac{d\theta}{1+x\sin^2(\theta)}$$
Here is all my steps:
$$\tan\frac{\theta}{2}=\omega\Rightarrow \sin(\theta)=\frac{2\omega}{1+\omega^2}\Rightarrow 1+x\sin^2{(\theta)}=1+\frac{x 4\omega^2}{(\omega^2+1)^2}$$
Therefore: $$\int\frac{d\theta}{1+x\sin^2(\theta)}=2\int\frac{\omega^2+1\:\ d \omega}{(\omega^2+1)^2+x(2\omega)^2}=2\left(\int\frac{\omega^2\:\ d \omega}{(\omega^2+1)^2+x(2\omega^2)}+\int\frac{d\omega}{(\omega^2+1)^2+x(2\omega)^2}\right)$$
I don't know if I'm wrong somewhere but I don't have ideea how can I continue...
| Instead of the half-angle substitution I would try using
$$\frac{1}{1+x\sin^2\theta} = \frac{1}{(1+x)\sin^2\theta + \cos^2\theta}= \frac{\sec^2\theta}{(1+x)\tan^2\theta + 1}.$$
Now substitute $u=\sqrt{|1+x|}\tan \theta$ (I don't see a way to avoid splitting the cases $x>-1$ and $x<-1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Calculus differentiable gra Hey guys so stuck on a calculus question. So far all I know is that $d$ should equal $4$. I then got the derivative of the a b c d function to $3ax^2 + 2bx + c$ , subbed in $(0,4)$ to get $c$, which was also $4$. Just wondering if I'm on the right track and where I should go from here.
A section of road, represented by the line $y = x + 4$ when $x \leq 0$, is to
be smoothly connected to another section of road, represented by $y = 4
> –x$ when $x \geq 4$ , by means of a curved section of road, represented by a
cubic curve $y = ax^3+ bx^2+ cx + d$ . Find $a, b, c$ and $d$ such that the
function $f(x)$ is everywhere differentiable (and therefore everywhere
continuous), where
$$f(x) = \begin{cases}
x+4 &x\leq 0 \\
ax^3 +bx^2 +cx +d &0<x<4 \\
4-x &x\geq 4
\end{cases}$$
| you have $$ f(x) = \begin{cases}
x+4 &x\leq 0 \\
ax^3 +bx^2 +cx +d &0<x<4 \\
4-x &x\geq 4
\end{cases}$$
making $f$ continuous at $x=0$ and at $x = 4$ requires $$d = 4,\quad 64a+16b+4c+d= 0\tag 1$$
we also have $$f'(x) = \begin{cases}
1 &x < 0 \\
3ax^2 +2bx +c & 0<x<4 \\
-1 &x\geq 4
\end{cases}$$
making $f'$ continuous at $x=0$ and at $x = 4$ requires $$c = 1,\quad 48a+8b+c= -1\tag 2$$
you can solve $(1)$ and $(2)$ to find all the constants $a, b, c, d.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
An unusual limit involving $e$ From MathWorld I have the following quote:
$e$ is given by the unusual limit $$\lim_{n \to \infty}\left(\frac{(n + 1)^{n + 1}}{n^{n}} - \frac{n^{n}}{(n - 1)^{n - 1}}\right) = e\tag{1}$$ Now if we put $$a_{n} = \frac{(n + 1)^{n + 1}}{n^{n}}$$ then the above result says that $$\lim_{n \to \infty}(a_{n} - a_{n - 1}) = e$$ Now we know that
Theorem: If $a_{n} - a_{n - 1} \to L$ then $a_{n}/n \to L$.
Using the example here we see that $$\frac{a_{n}}{n} = \frac{n + 1}{n}\left(1 + \frac{1}{n}\right)^{n} \to e$$ I am thinking of some partial converse to the standard theorem above which can get us from limit of $a_{n}/n$ to limit of $a_{n} - a_{n - 1}$.
Another option to prove $(1)$ is to make use of real variable theory. Thus we can put $x = 1/n$ and deal with function $f(x) = \left(1 + \dfrac{1}{x}\right)(1 + x)^{1/x}$ and use Taylor expansion $$(1 + x)^{1/x} = e - \frac{ex}{2} + \frac{11e}{24}x^{2} + \cdots$$ and $$g(x) = \left(\frac{1}{x} - 1\right)(1 - x)^{-1/x}$$ and calculate the limit of $f(x) - g(x)$ as $x \to 0$. This way we see that $$f(x) = \frac{e}{2} + \frac{e}{x} + o(1), g(x) = -\frac{e}{2} + \frac{e}{x} + o(1)$$ and clearly $f(x) - g(x) \to e$ as $x \to 0$.
Is there a proof without using real variable theory and just dealing with theorems on sequences which establishes the result $(1)$?
| For $n > 1$, we can write
$$a_n - a_{n-1} = \biggl(1+\frac{1}{n}\biggr)^n\cdot \biggl(n+1 - (n-1)\frac{n^{2n}}{(n^2-1)^n}\biggr).\tag{1}$$
Bernoulli's inequality says on the one hand that
$$\frac{n^{2n}}{(n^2-1)^n} = \biggl(1 + \frac{1}{n^2-1}\biggr)^n \geqslant 1 + \frac{n}{n^2-1},$$
so
$$n+1 - (n-1)\frac{n^{2n}}{(n^2-1)^n} \leqslant 2 - \frac{n}{n+1} = 1+ \frac{1}{n+1},$$
and on the other hand it says
$$\frac{n^{2n}}{(n^2-1)^n} = \frac{1}{\Bigl(1 - \frac{1}{n^2}\Bigr)^n} \leqslant \frac{1}{1-\frac{1}{n}} = \frac{n}{n-1},$$
whence
$$n+1 - (n-1)\frac{n^{2n}}{(n^2-1)^n} \geqslant (n+1) - n = 1.$$
Since the first factor in $(1)$ converges to $e$, and the second to $1$, as we just saw, it follows that $a_n - a_{n-1} \to e.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
Find $\lim\limits_{n\to\infty}y_n$ if $y_1=\frac{x}{2},y_n=\frac{x}{2}+\frac{y^2_{n-1}}{2},0\le x \le 1,n=2,3,...$ Is it a good approach to use induction? If $0\le x \le 1$ then $0\le y_1 \le \frac{5}{8}$. Suppose that $$0\le y_n \le \frac{5}{8}$$ and prove $$0\le y_{n+1} \le \frac{5}{8}$$ If $$y_{n+1}=\frac{x}{2}+\frac{y^2_n}{2}$$ what should be the induction step?
| Since the only possible limits are $1\pm\sqrt{1-x}$, let $z_n=1-y_n$, find the recursion for $z_n$, and check the consequences of $z_n^2<1-x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral using contour integration Here is the integral I want to evaluate:
$$\int_{0}^{2\pi} \frac{dx}{a+b \cos x }, \quad a>b >0$$
Apparently there are limitations as to what values $a, b$ are supposed to take but let us not concern about this. Since using the sub $u =\tan \frac{x}{2}$ (the Weiersstrass sub) results that the integral is $0$ (as it should, since it is not $1-1$ function in this interval) I got down down the way of contour integration. Hence:
$$\begin{aligned}
\int_{0}^{2\pi}\frac{dx}{a+ b\cos x} &\overset{x=i \ln u}{=\! =\! =\!} \oint \limits_{|z|=1} \frac{dz}{iz \left [ a + \frac{b}{2}\left ( z+z^{-1} \right ) \right ]} \\
&= \frac{1}{i} \oint \limits_{|z|=1} \frac{dz}{za + \frac{bz^2}{2}+\frac{b}{2}}\\
&=\frac{2}{i} \oint \limits_{|z|=1} \frac{dz}{bz^2 +b +2za} \\
&= \frac{2}{i} 2\pi i \sum_\text{residues} f(z)\\
&= \frac{4\pi}{1+ \sqrt{1-8ab}+2a}
\end{aligned}$$
However, judging by intuition this must not be the result. Because this one restricts the integral too much. What i mean is, that for $a=6, b=3$ we have that:
$$\int_0^{2\pi} \frac{dx}{6+3\cos x}= \frac{2\pi}{3\sqrt{3}}$$
My formula cannot derive the result because then radical would be negative. What am I doing wrong here?
| You have a pole at $z=\dfrac{\sqrt{a^2-b^2}-a}b$ and one at $z=-\dfrac{\sqrt{a^2-b^2}+a}b$. The pole at $z=\dfrac{\sqrt{a^2-b^2}-a}b$ has is simple and has residue $\dfrac{1}{2\sqrt{a^2-b^2}}$. The other pole is outside the circle and therefore does not matter.
The final answer is therefore ${4\pi}\dfrac{1}{2\sqrt{a^2-b^2}}=\dfrac{2\pi}{\sqrt{a^2-b^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
inequalities with fraction problem x $\frac{1}{x} > \frac{2x} {x^2 +2}$ solving this inequalities:
My long solution (wrong) :
multiplying $(x^2 + 2)^2 (x)^2 \dots$ (multiplying square of each denominator, getting rid of the > or < 0)
$x (x^2 +2)^2 > 2x(x^2) ( x^2 +2)$
$x(x^4+4x^2 +4)>2x^2(x^2+2)$
$x^5+4x^3+4x>2x^4+4x^2$
$x^5+4x^3+4x-2x^4-4x^2>0$
$x ( x^4 + 4x^2 + 4 +2x^3 -4x) >0 $
$x>0$ or ^ ...
This solving method doesn't look right
| We need $$\dfrac1x-\dfrac{2x}{x^2+2}>0$$
$$\iff\dfrac{x^2+2-2x^2}{x(x^2+2)}>0$$
As $x^2+2>0$ for real $x,$ multiplying both sides by $x^2(x^2+2)$
$$\iff x(2-x^2)>0\iff\{x-(-\sqrt2)\}\cdot x\cdot (x-\sqrt2)<0$$
Clearly, we need to check for $(-\infty,-\sqrt2);(-\sqrt2,\sqrt2);(\sqrt2,\infty)$
As the product of three terms is negative, we need exactly one $(0<x<\sqrt2)$ or all three to be negative $(x<-\sqrt2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Show that $n^n<(n!)^2$ I want to show that $\lim\limits_{n \to \infty}\frac{n^n}{(n!)^2}=0$
But I have absolutely no idea besides that $\frac{n^n}{(n!)^2}=\frac{n}{1}\cdot \frac{n}{2}\cdot ...\cdot \frac{n}{(n-1)^2}\cdot \frac{n}{n^2}$
Help me please.
| Let's check ratio of $a_n$ and $a_{n+1}$:
$$a_n = \frac{n^n}{n!^2}$$
$$a_{n+1} = \frac{(n+1)^{n+1}}{(n+1)!^2}$$
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}}{(n+1)!^2} : \frac{n^n}{n!^2} =
\frac{(n+1)^{n+1}}{n^n}\frac{n!^2}{(n+1)!^2} = \left(1+\frac{1}{n}\right)^n \frac{n!^2(n+1)}{n!^2(n+1)^2} = \frac{1}{n+1}\left(1+\frac{1}{n}\right)^n \sim\frac en,
$$
hence
$$a_n \sim \frac{e^n}{n!}\to 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Asymptotic expansion computation I found a paper in which the following expression
$$\log\left(1 - \frac{1}{x}\right) + \frac{1}{x} + \frac{A}{x^2}+\epsilon x = 0$$
Where $\epsilon$ is a constant of order $10^{-2}$ to $10^{-5}$, $A$ is a constant of order unity, is approximated for $\frac{1}{x}<<1$ as
$$\epsilon x^4 + \left(A-\frac{1}{2}\right) x - \frac{1}{3} + O\left(\frac{1}{x}\right) = 0$$
I am unable to derive this result.
Going back to the first equation, I denote $\frac{1}{x} = y$.
Then by using the Taylor expansion
$$\log(1-y) = y -\frac{1}{2}y^2 +\frac{2}{3}y^3 +...$$
And substitute I obtain
$$2y +\left(a-\frac{1}{2}\right)y^2 + \frac{2}{3}Y^3 + \frac{\epsilon}{y} +O(x^4)= 0$$
And it is not even close. Any hint on my error would be very appreciated, thanks
| $$\log\left(1 - \frac{1}{x}\right) + \frac{1}{x} + \frac{A}{x^2}+\epsilon x =-\frac1x-\frac1{2x^2}-\frac1{3x^3}+O\left(\frac1{x^4}\right)+ \frac{1}{x} + \frac{A}{x^2}+\epsilon x=0$$
then
$$x^3\left(\left(A-\frac12\right)\frac1{x^2}-\frac1{3x^3}+O\left(\frac1{x^4}\right)+ \frac{A}{x^2}+\epsilon x\right)=(A-\frac12)x-\frac13+O\left(\frac1{x}\right)+Ax+\epsilon x^4=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find out all solution of trigonometric equation $\tan \theta = - \frac{\sqrt 3}{3}$ $\tan \theta = - \frac{\sqrt 3}3$
I thought it was $2\pi\over3$ and $5\pi\over3$ but I was wrong
please help
| For any $a\in\Bbb R$,
$$\tan\theta=a\iff \theta=\arctan a+ n\pi,\, n\in\Bbb Z$$
$\arctan x$ is the unique number $y$ in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ such that $\tan y=x$.
In this case, $$\tan\theta=-\frac{\sqrt{3}}{3}\iff \theta=-\frac{\pi}{6}+n\pi,\, n\in\Bbb Z,$$
because $\arctan\left(-\frac{\sqrt{3}}{3}\right)=-\frac{\pi}{6}$, which is in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ and
$$\tan\left(-\frac{\pi}{6}\right)=\frac{\sin\left(-\frac{\pi}{6}\right)}{\cos\left(-\frac{\pi}{6}\right)}=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$$
There are infinitely many solutions for $\theta$, all generated by above formula.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the maximum of an expression in three variables I am trying to find the max of $\frac{(a)(a-1)(9-a)}{2} + \frac{(b)(b-1)(9-b)}{2} + \frac{(c)(c-1)(9-c)}{2}$ subject to $a+b+c=9$ over nonnegative integers. I originally tried to see if something could work with Jensen's or calculus, and I have a conjecture that I haven't been able to prove that the maximum occurs when $a=5, b=4, c=0$ by simply plugging in an initial guess $a=b=c=3$ and then trying $a=4, b=3, c=2$ and then eventually $a=4, b=4, c=1$. Then because the function $f(x) = \frac{(x)(x-1)(9-x)}{2}$ is increasing when $x<6$ and $f(0)=f(1)=0$ we know that $(5,4,0) > (4,4,1)$.
Is there a more rigorous approach to this?
| Without loss of generality $a \geq b \geq c$. So $c \leq 3$ and if you write down a list of all the values of $\frac{x(x-1)(9-x)}{2}$ for $0 \leq x \leq 9$ it's easy to check which pair $(a,b)$ maximises the total for each of the four possessible values of $c$, then see which of those is the greatest.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving a three variable equation I have three given values, suppose a=1.86, b=2.6 and c=4.2. Now I have to figure out x,y,z such that
*
*$x\gt 0,y\gt 0$ and $z\gt 0$
*$x+y+z=1$
*$a*x\gt 1, b*y\gt 1$ and $cz\gt 1$
I need a generalized solution steps for this to implement in programming.
Thanks.
| $\left\{ {\begin{array}{*{20}{l}}
{x > \frac{1}{a} \Rightarrow x = \frac{1}{a} + X}\\
{y > \frac{1}{b} \Rightarrow y = \frac{1}{b} + Y}\\
{z > \frac{1}{c} \Rightarrow z = \frac{1}{c} + Z}
\end{array}} \right.$ where $0 < X,Y,Z$
$ \Rightarrow \frac{1}{a} + X + \frac{1}{b} + Y + \frac{1}{c} + Z = 1 \Rightarrow X + Y + Z = 1 - \frac{1}{a} - \frac{1}{b} - \frac{1}{c}$ $\hspace{.3cm}$(*)
(*) is an equation of surface where cut each axis in $1 - \frac{1}{a} - \frac{1}{b} - \frac{1}{c}$.
Your solution is all the point on this surface which that lie on the first quadrant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Differential equation $y\cdot y'' - (y')^2 - 1 = 0$ I'm trying to solve this equation but at the end I'm stuck and can't reache the answer. I use the substitutions $y'=p$ and $y'' = p'\cdot p$:
$$y \cdot p'p - p^2 - 1 = 0 \implies y\cdot p \frac {dp}{dy} - (p^2 + 1) = 0 \implies \int \frac {p}{p^2 + 1}dp = \int \frac {dy}{y} \implies \frac{1}{2} \ln \left|C_1(p^2 + 1) \right | = \ln y \implies y = C_1 \sqrt{p^2+1} \implies p = \sqrt {C_1y^2 - 1} $$
Next step:
$$ y' = \sqrt {C_1 y^2 - 1} \implies \frac{dy}{dx} = \sqrt {C_1 y^2 - 1} \implies \int \frac {dy}{C_1 \sqrt{ y^2 - \frac {1}{C_1^2} }} = \int dx \implies \\ C_1 \ln \left |y + \sqrt{y^2 - C^2_1} \right | = x + C_2 \implies y + \sqrt{y^2 - C_1^2} = \exp {\frac {x+C_2}{C_1}} $$
Here I don't know how to go on. The answer should be $ y = \frac {C_1}{2} \left ( \exp(\frac{x+C_2}{c_1}) + \exp(-\frac{x+C_2}{c_1}) \right )$
| Maybe your $C_1$ and $C_2$ are different to the answer's. But let solve your equation:
$$
y+\sqrt{y^2-C_1^2}=\exp(\frac{x+C_2}{C_1})\\
\sqrt{y^2-C_1^2}=\exp(\frac{x+C_2}{C_1})-y\\
y^2-C_1^2=\exp(2\frac{x+C_2}{C_1})-2y\exp(\frac{x+C_2}{C_1})+y^2\\
y=\frac{1}{2}(\exp(\frac{x+C_2}{C_1})+C_1^2\exp(-\frac{x+C_2}{C_1}))
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $x^8 \equiv 3 \pmod {13}$ I need to find all solutions to $x^8 \equiv 3 \pmod {13}$.
What I've tried: I know $2$ is a primitive root modulo $13$.
So it is equivalent to solve
$2^{8t} \equiv 2^4 \pmod {13}$
Then I get $t = 2 + 3k$.
I think I'm wrong... and if not, what are the final solutions??
| No, your answer is correct, $t = 2,5,8,11$ give respectively $x = 4,6,9,7$.
For the sake of Google, I will write out the method OP probably used, which is different from the one used in the other answer.
Knowing that $2$ is a primitive root modulo $13$ and that $3 \equiv 2^4 \pmod{13}$, we let $x = 2^t$ and we wish to find all values of $t$ such that
$$\left(2^t\right)^8 \equiv 2^{8t} \equiv 2^4 \pmod{13}.$$
We use the fact that if $a$ is a primitive root modulo $p$, then $a^x \equiv a^y \pmod p$ if and only if $x \equiv y \pmod{(p-1)}$, so we wish to solve
$$8t \equiv 4 \pmod{12}.$$
The method to solve congruences of the form $ax \equiv b \pmod n$ is well-known (see for example these notes) and we find that the solutions are $t \equiv 2 \pmod 3$, which yields $t = 2,5,8,11$. In turn, those values yield $x = 4,6,9,7$ (which matches the other answer).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why does $\int_0^{2\pi} (1+2\cos(x))/(5+4\cos(x))\,dx$ vanish? The standard substitution $y=\tan(x/2)$ shows that
$$ \int_0^{2\pi} \frac{1+2\cos(x)}{5+4\cos(x)}\,dx = 0. $$
What is the "real explanation" for this fact? My guess is that the "book proof" involves contour integration; is this correct? Is there an elegant "calculus proof" avoiding technical computations?
Thanks!
| Consider the more general integral, where $d \neq 0$,
\begin{align}\tag{1}
I = \int_{0}^{2 \pi} \frac{a + b \cos x}{c + d \cos x} \, dx.
\end{align}
Now by splitting the integral into two regions, $[0, \pi)$ and $(\pi, 2 \pi]$, then in the second integral let the variable be shifted by $\pi$. This is seen by:
\begin{align}
I &= \int_{0}^{\pi} \frac{a + b \cos x}{c + d \cos x} \, dx + \int_{\pi}^{2 \pi} \frac{a + b \cos x}{c + d \cos x} \, dx \\
&= \int_{0}^{\pi} \frac{a + b \cos x}{c + d \cos x} \, dx + \int_{0}^{\pi} \frac{a - b \cos x}{c - d \cos x} \, dx. \tag{2}
\end{align}
The general form of the integrals is
\begin{align}
\int_{0}^{\pi} \frac{a + b \cos x}{c + d \cos x} \, dx &= \left[ \frac{b x}{d} + \frac{2(bc - ad)}{d \sqrt{d^2 - c^2}} \, \tanh^{-1}\left( \frac{(c-d)}{\sqrt{d^2 - c^2}} \, \tan\left(\frac{x}{2}\right) \right) \right]_{0}^{\pi} \\
&= \frac{b \pi}{d} - \frac{i \, \pi (bc-ad)}{d \sqrt{d^2 - c^2}}. \tag{3}
\end{align}
and, upon letting $(b,d) \to (-b , -d)$,
\begin{align}
\int_{0}^{\pi} \frac{a - b \cos x}{c - d \cos x} \, dx = \frac{b \pi}{d} - \frac{i \, \pi(bc-ad)}{d \sqrt{d^2 - c^2}}. \tag{4}
\end{align}
Using (3) and (4) in (2) yields
\begin{align} \tag{5}
I = \frac{2 \pi }{d} \, \left( b - \frac{i \, (bc-ad)}{\sqrt{d^2 - c^2}} \right).
\end{align}
From (1) and (5) the integral is
\begin{align}\tag{6}
\int_{0}^{2 \pi} \frac{a + b \cos x}{c + d \cos x} \, dx = \frac{2 \pi }{d} \, \left( b - \frac{i \, (bc-ad)}{\sqrt{d^2 - c^2}} \right).
\end{align}
In order to remove the factor of $i$ it is required that $c > d$, $d \neq 0$. This leads to
\begin{align}\tag{7}
\int_{0}^{2 \pi} \frac{a + b \cos x}{c + d \cos x} \, dx = \frac{2 \pi }{d} \, \left( b - \frac{bc-ad}{\sqrt{c^2 - d^2}} \right).
\end{align}
The problem asked has the values $(a,b,c,d) = (1,2,5,4)$ for which it is quickly determined that
\begin{align}\tag{6}
\int_{0}^{2 \pi} \frac{1 + 2 \cos x}{5 + 4 \cos x} \, dx = 0.
\end{align}
Another example is $(a,b,c,d) = (\beta^{2}, 1, 6, 4)$, where $2 \beta = 1 - \sqrt{5}$, which yields the result
\begin{align}
\int_{0}^{2 \pi} \frac{\beta^{2} + \cos x}{6 + 4 \cos x} \, dx = 0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
How exactly do we do Gauss elimination?
This is a matrix:
$$\begin{bmatrix}
1 & 1 & 1\\
1 & 2 & 3\\
1 & 3 & k
\end{bmatrix}\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=
\begin{bmatrix}
3\\
6\\
4+k
\end{bmatrix}$$
Find $k$ so that it has no unique solution. Solve the equations for this value of $k$.
I found $k = 5$ by using $\text{determinant}=0$.
Then I tried Gauss elimination, from what I understand from my lecturer, I use the 3 rules of Gauss elimination to solve, I just randomly use whatever I need to achieve echelon form. I have come to this:
$$\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 3\\
0 & 0 & 1 & 0\\
\end{bmatrix}$$
The answer is
$$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\3\\0\end{bmatrix}+t\cdot\begin{bmatrix}1\\-2\\1\end{bmatrix}$$
Where does the $t$ come from? Does it involve eigenvalues?
| You did not row reduce correctly. Since we are told that there are infinitely many solutions, we know that the final result should have free variables, so there should be at most two pivots. We should get:
$$
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
1 & 2 & 3 & 6 \\
1 & 3 & 5 & 9 \\
\end{array}\right]
\sim
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 3 \\
0 & 2 & 4 & 6 \\
\end{array}\right]
\sim
\left[\begin{array}{ccc|c}
1 & 0 & -1 & 0 \\
0 & 1 & 2 & 3 \\
0 & 0 & 0 & 0 \\
\end{array}\right]
$$
Translating back to a system, we have:
\begin{cases}
x - z = 0 \\
y + 2z = 3
\end{cases}
Taking $z = t$ and solving for $x$ and $y$ in terms of $t$, we conclude that the general solution is:
$$
\begin{bmatrix}x \\ y \\ z\end{bmatrix}
=
\begin{bmatrix}t \\ 3 - 2t \\ t\end{bmatrix}
=
\begin{bmatrix}0 \\ 3 \\ 0\end{bmatrix}
+
t\begin{bmatrix}1 \\ -2 \\ 1\end{bmatrix}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Value of $\int_{|z+1|=2} \frac{z^2}{4-z^2}dz$ $\int_{|z+1|=2} \frac{z^2}{4-z^2}dz=-2\pi i$.
Am I correct, I used cauchy integral formula
| Since $z^2-4=(z-2)(z+2)$ and $z=-2$ is the only zero inside $|z+1|=2$. Assuming $|z+1|=2$ is positively oriented, Cauchy's Integral Formula gives
$$
\int_{|z+1|=2}\frac{z^2}{4-z^2}dz =- \int_{|z+1|=2}\frac{z^2}{z^2-4}dz = -\int_{|z+1|=2}\frac{z^2}{(z-2)(z+2)}dz =- 2 \pi i \left( \frac{z^2}{z-2} \right)_{z=-2} = 2\pi i
$$
Hence your result is not the correct one! If $|z+1|=2$ is negatively oriented, then you have the correct result!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$ Determine the Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$.
$f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}} = \arcsin(x)\frac{1}{\sqrt{1-x^2}}$
It is known:
(1.) $\arcsin(x) = \sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
(2.) $\frac{d}{dx} (\arcsin(x)) = \frac{1}{\sqrt{1-x^2}} = \frac{d}{dx}(\sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}) = \sum_{n=0}^\infty\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}$
(3.) In third step I multiplied these two series, but I am not sure whether it is correct:
$\arcsin(x)\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
EDIT:
How did you get this sum $\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}$ ?
And, in case I have to determine the product of the (some other) series, which one of these should I write?
(a) $\sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
or
(b) $ = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m-1)!!x^{2m+1}}{2^mm!(2m+1)})\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}$
| We have, by the extended binomial theorem:
$$ \frac{1}{\sqrt{1-x^2}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n} x^{2n},\tag{1} $$
hence by integrating termwise:
$$ \arcsin x=\sum_{n\geq 0}\frac{1}{(2n+1)4^n}\binom{2n}{n} x^{2n+1}=x\cdot\phantom{}_2 F_1\left(\frac{1}{2},\frac{1}{2};\frac{3}{2};z^2\right)\tag{2} $$
Now we may notice that:
$$ \sum_{a+b=N}\frac{1}{(2a+1)4^a}\binom{2a}{a}\frac{1}{(2b+1)4^b}\binom{2b}{b} = \frac{4^N}{(N+1)(2N+1)\binom{2N}{N}}\tag{3}$$
We may find the Taylor series of $\arcsin^2(z)$ by exploiting the Catalan's convolution formula, the hypergeometric identity:
$$ \phantom{}_2 F_1\left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};z\right)^2 = \phantom{}_3 F_2\left(a,b,\frac{a+b}{2};\frac{a+b+1}{2},a+b;z\right)\tag{4}$$
the Lagrange's inversion theorem or the rather simple technique shown in this answer. So we have:
$$\arcsin^2(x) = \sum_{n\geq 0}\frac{4^n}{(n+1)(2n+1)\binom{2n}{n}}x^{2n+2}\tag{5}$$
and by differentiating $(5)$:
$$ \frac{\arcsin x}{\sqrt{1-x^2}} = \sum_{n\geq 0}\frac{4^n}{(2n+1)\binom{2n}{n}}x^{2n+1}. \tag{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Finding value of an expression If $x^2-3x-1=0$ then find the value of $(x^6+1)/x^3$ I tried to solve the quadratic but it became too complicated any way of doing this without a calculator
| Since $x^2=3x+1$, we have $x^3=3x^2+x=10x+3$.
In the same way, $\frac{1}{x^2}=1-\frac{3}{x}$ gives $\frac{1}{x^3}=\frac{1}{x}-\frac{3}{x^2}=\frac{10}{x}-3$, hence:
$$\frac{x^6+1}{x^3} = x^3+\frac{1}{x^3} = 10\left(x+\frac{1}{x}\right) = \pm 10\sqrt{13},$$
because $x+\frac{1}{x}$ is $\pm\sqrt{\Delta}$, since the product of the roots of $x^2-3x-1$ is $-1$ by Vieta's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rank-reducibility of Latin squares Consider the following Latin square with rank $N= 9$:
$$
\begin{bmatrix}
5 & 3 & 1 & 2 & 4 & 7 & 6 & 8 & 9 \\
3 & 7 & 9 & 6 & 8 & 4 & 5 & 2 & 1 \\
8 & 5 & 4 & 9 & 1 & 2 & 7 & 3 & 6 \\
9 & 2 & 3 & 8 & 7 & 6 & 1 & 4 & 5 \\
7 & 4 & 5 & 3 & 9 & 1 & 2 & 6 & 8 \\
6 & 1 & 2 & 5 & 3 & 9 & 8 & 7 & 4 \\
2 & 9 & 6 & 7 & 5 & 8 & 4 & 1 & 3 \\
4 & 8 & 7 & 1 & 6 & 5 & 3 & 9 & 2 \\
1 & 6 & 8 & 4 & 2 & 3 & 9 & 5 & 7 \\
\end{bmatrix}$$
This LS is rank-reducible, as it has the property that
we can obtain a Latin square $L'$ with rank $N'=8$ directly from it.
The value $9$ in $L_{4,1}$ is the key to this reduction. For every row $R$, if $L_{R,C} = 9$ then
$L_{R,1} = L_{4,C}$.
This allows us to remove row 4 and column 1, change each $L_{R,C} = 9$ to $L_{R,1}$ in the remaining
rows, and the result will be a Latin square of rank $N=8$:
$$
\begin{bmatrix}
3 & 1 & 2 & 4 & 7 & 6 & 8 & 5 \\
7 & 3 & 6 & 8 & 4 & 5 & 2 & 1 \\
5 & 4 & 8 & 1 & 2 & 7 & 3 & 6 \\
4 & 5 & 3 & 7 & 1 & 2 & 6 & 8 \\
1 & 2 & 5 & 3 & 6 & 8 & 7 & 4 \\
2 & 6 & 7 & 5 & 8 & 4 & 1 & 3 \\
8 & 7 & 1 & 6 & 5 & 3 & 4 & 2 \\
6 & 8 & 4 & 2 & 3 & 1 & 5 & 7 \\
\end{bmatrix}$$
If the pivot value in $V=L_{4,1}$ was any other value than 9, we would just swap the values $V$ and $9$ throughout the square before
proceeding with the reduction.
I wonder, then, are there any other cases which would allow a similarly direct construction of a reduced-rank Latin square?
| You're essentially doing transversal prolongation in reverse.
A transversal is a set of entries with one representative from each row, each column, and each symbol. We can extend a Latin square of order $n$ with a transversal to a Latin square of order $n+1$ by (a) replacing the transversal with a new symbol, (b) adding the corresponding symbols to the new margins, and (c) placing the new symbol in the intersection of the new margins.
Here's a simple example:
$\begin{bmatrix}
3 & 1 & \color{red}{\mathbf 2} \\
\color{red}{\mathbf 1} & 2 & 3 \\
2 & \color{red}{\mathbf 3} & 1 \\
\end{bmatrix}
\longmapsto
\begin{bmatrix}
3 & 1 & \color{blue}{\mathbf \infty} & \color{red}{\mathbf 2} \\
\color{blue}{\mathbf \infty} & 2 & 3 & \color{red}{\mathbf 1} \\
2 & \color{blue}{\mathbf \infty} & 1 & \color{red}{\mathbf 3} \\
\color{red}{\mathbf 1} & \color{red}{\mathbf 3} & \color{red}{\mathbf 2} & \color{blue}{\mathbf \infty} \\
\end{bmatrix}
$
In your example, you've started off with a prolonged Latin square (with its rows and columns permuted), and have discovered the Latin square it was prolonged from.
This construction generalizes to $k$-plexes ($k$ representatives of each row, column, and symbol) where $k \leqslant n/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int{ \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}}dx$ using Pascal inversion (Note: I apreciate very much who marked this as a duplicate but I would like an answer for why my proof is wrong)
This is my solution, I have no clue why it failed. Let's start:
define
$$I_n(m) = \int_{0}^{x} \frac{t^m}{1 + t + t^2/2 + ... + t^n/n!}\ dt$$
so it should be true that
$$\sum_{m=0}^{n}\frac{I_n(m)}{m!} = x$$
Then I use Pascal inversion:
$$\sum_{m=0}^n \frac{n! I_n(m)}{m!} = n!x$$
$$\sum_{m=0}^n {n\choose m} B_n(m) = n!x$$
where $B_n(m) = (n-m)!I_n(m)$
by Pascal's formula:
$$I_n(n) = (-1)^nxn! \sum_{m=0}^{n} \frac{(-1)^m}{(n-m)!}$$
what did I do wrong ????
| You may observe that
$$
\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'=1 + x + \frac{x^2}{2} + \cdots + \frac{x^{n-1}}{(n-1)!}
$$ giving
$$
\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)-\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'=\frac{x^n}{n!}
$$ and
$$
\begin{align}
\int \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}dx&=n!\int\frac{\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)-\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}\:dx\\\\
&=n!\int dx-n!\int\frac{\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'}{\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)}\:dx.
\end{align}
$$ Thus
$$
\int \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}dx=n!\:x-n!\ln \left| 1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right|+C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
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Trigonometry equation maximum Given the equation: $\cos x + \sqrt3 \sin x = a^2$ find the maximum value for $a$ for which the equation has solutions and for this case solve the equation, $a \in \mathbb{R}$.
I'm guessing I need to find the maximum for the function and for this I have to differentiate it and solve it when it's derivative is $0$? At a first glance I'd say the maximum for $a$ is $\sqrt2$ when $x=\frac{\pi}{3}$, but how do I go around proving this?
| Since \begin{align}\cos x+\sqrt 3\sin x &=2\left( \frac 12\cos x+\frac {\sqrt 3}2\sin x \right) \\
& =2 \sin\left(x + \frac{\pi}{6}\right) = a^2\end{align}
Since $\sin x$ oscillates between $-1$ and $1$, we need $\displaystyle -1 \leq \frac{a^2}{2} \leq 1$. The maximum value of $a$ is easily seen to be $a = \sqrt{2}$ since $\displaystyle \frac{a^2}{2}\leq 1 \implies a^2 \leq 2$.
An alternative way (but much less elegant, imo) is to differentiate and set the derivative equal to zero, as shown below:
$$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\cos x + \sqrt{3} \sin x\right) = \sqrt{3}\cos x - \sin x.$$
The derivative first vanishes at $x = \frac{\pi}{3}$ and has the general solution of $x = n\pi - \frac{2\pi}{3}$. When $x = \frac{\pi}{3}$, you need only check that this is a maximum point (hint: look at the second derivative) then substitute it into the original equation to deduce that $a^2 = 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x^2-|5x-3|-x<2,\ \ x\in \mathbb{R} $
Solve $x^2-|5x-3|-x<2,\ \ x\in \mathbb{R} $
I tried $x^2-|5x-3|-x<2$ ,
case $1$ , $x^2-(5x-3)-x<2,\ x\geq 0 \\
x^2-6x+1<0 \\
3-2\sqrt2 < 3+2\sqrt2 \\
0.17<x<5.8\\
$
$x^2-(5x-3)-x<2$ ,
case $2$ , $x^2+(5x-3)-x<2,\ x< 0 \\
x^2+4x-5<0 \\
-5 < x< 1\\
$
The region common is $3-2\sqrt2<x<1$
But the book gives answer $-5<x<3+2\sqrt2$ . I am confused.
| Case 1 is not for $x>0$ but for $5x-3>0\implies x>\frac{3}{5}$
So for case 1 you have $\frac{3}{5}<x<3+\sqrt{2}$ (since $\frac{3}{5}>3-\sqrt{2}$
For case 2 you have $x<\frac{3}{5}$, so $-5<x<\frac{3}{5}$
So the general solution is $-5<x<3+\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Can anyone prove the identity $\sum_{m=-\infty}^\infty (z+\pi m)^{-2} = (\sin z)^ {-2} $ I came across this identity in a paper on elliptic curves, and the proof wasn't provided. It really irked me, and I couldn't find an explanation anywhere else. Can anyone shed some light?
$$\sum_{m=-\infty}^\infty (z+\pi m)^{-2} = (\sin z)^ {-2} $$
| Following the suggestion by @Lucian, we use the Euler's infinite product representation of the sine function and write
$$\sin z=z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2\pi^2}\right)\tag 1$$
Next, taking the logarithmic derivative of $(1)$ yields
$$\cot z=\frac1z-2z\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2-z^2} \tag 2$$
And then, taking the derivative of $(2)$ and multiplying by $-1$ yields
$$\begin{align}
\csc^2z&=\frac{1}{z^2}+2\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2-z^2} +4z^2\sum_{n=1}^{\infty}\frac{1}{(n^2\pi^2-z^2)^2} \\\\
&=\frac{1}{z^2}+2\sum_{n=1}^{\infty}\frac{(n^2\pi^2+z^2)}{(n^2\pi^2-z^2)^2}\\\\
&=\frac{1}{z^2}+\sum_{n=1}^{\infty}\left(\frac{1}{(n\pi+z)^2}+\frac{1}{(n\pi-z)^2}\right)\\\\
&=\frac{1}{z^2}+\sum_{n=1}^{\infty}\frac{1}{(n\pi+z)^2}+\sum_{n=-\infty}^{-1}\frac{1}{(n\pi+z)^2}\\\\
&=\sum_{n=-\infty}^{\infty}\frac{1}{(n\pi+z)^2}
\end{align}$$
as was to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find the value of $x$ such that $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$ Find the value of $x$, $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$$
Help guys please, I have tried and I got, $x=-2, x=1$, and I think it's wrong
| This should be a comment but then too long ...
Picking up on the solution given by Jan Erlang:
$$\left(x^4-8 x^2+10\right)^2 \left(x^4-8 x^2+14\right)^2+x-4=\left(x^2-x-3\right) \left(x^2+x-4\right) \left(x^{12}-24
x^{10}-x^9+228 x^8+16 x^7-1087 x^6-88 x^5+2720 x^4+191 x^3-3380
x^2-136 x+1633\right)$$
Now $x^2-x-3=0$ yields: $\frac{1}{2} \left(1\pm\sqrt{13}\right)$ and $x^2+x-4=0$ yields $\frac{1}{2} \left(-1\pm\sqrt{17}\right)$ (where only positive roots are acceptable). Further $$x^{12}-24
x^{10}-x^9+228 x^8+16 x^7-1087 x^6-88 x^5+2720 x^4+191 x^3-3380
x^2-136 x+1633=0$$ yields $6$ more positive roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrating $\sqrt{1-x^2}$ without using trigonometry I am a beginning calculus student. Tonight I had a thought. Maybe I could calculate $\pi$ using integration, but no trig.
The problem is that I don't really know where to start. I thought perhaps I could use the Chain Rule for derivatives.
$~~~~~~~~~$
| Faster converging series
$\sqrt{1-x^2} = 1 - \frac{1}{1!} \frac{1}{2} x^2 - \frac{1}{2!} \frac{1}{2} \frac{1}{2} x^4 - \frac{1}{3!} \frac{1}{2} \frac{1}{2} \frac{3}{2} x^6 + \cdots$ [by Binomial theorem]
$\ = 1 - \sum_{k=1}^\infty \frac{(2k-3)!!}{2^k k!} x^{2k}$ [where $n!!$ is the double-factorial of $n$]
$\int \sqrt{1-x^2}\ dx = x - \sum_{k=1}^\infty \frac{(2k-3)!!}{2^k k! (2k+1)} x^{2k+1}$ [which converges faster when $x = \frac{1}{2}$ than $x = 1$]
$π = 12 \cdot \left( \int_{0}^{\frac{1}{2}} \sqrt{1-x^2}\ dx - \frac{\sqrt{3}}{8} \right)$ [the integral is $\frac{1}{12}$ of the pie plus a triangle]
$\ = 12 \cdot \left( \frac{1}{2} - \sum_{k=1}^\infty \frac{(2k-3)!!}{2^{3k+1} k! (2k+1)} \right ) - \frac{3\sqrt{3}}{2} = 6 - \frac{3\sqrt{3}}{2} - 12 \sum_{k=1}^\infty \frac{(2k-3)!!}{2^{3k+1} k! (2k+1)}$
Notes
Binomial theorem for non-integer powers can be proven by using Taylor series of $x \mapsto (1+x)^r$ on $(0,1)$. But that requires knowledge of the derivatives, for which the simplest proof is by product rule and chain rule for integer powers, and by implicit differentiation for extending to rational powers, followed by limits for extending to real powers.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Computing $\sum _{k=1}^{\infty } \frac{\Gamma \left(\frac{k}{2}+1\right)}{k^2 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}$ in closed form What tools other than beta function you might like to use here?
$$\sum _{k=1}^{\infty } \frac{\displaystyle \Gamma \left(\frac{k}{2}+1\right)}{\displaystyle k^2 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}\approx 1.27541$$
Supplementary question: calculating
$$\sum _{k=1}^{\infty } \frac{\displaystyle \Gamma \left(\frac{k}{2}+1\right)}{\displaystyle k^3 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}\approx 1.02593$$
Is there a way to generalize it and get such a calculation?
$$\sum _{k=1}^{\infty } \frac{\displaystyle \Gamma \left(\frac{k}{2}+1\right)}{\displaystyle k^n \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}$$
| Writing $\dfrac{\Gamma\left(\frac{k}{2}+1\right)}{\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}$ as an integral and exchanging summation and integration, we get
\begin{align}
\sum^\infty_{k=1}\frac{\Gamma\left(\frac{k}{2}+1\right)}{k^2\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}
&=\frac{2}{\sqrt{\pi}}\int^1_0\frac{x{\rm Li}_2(x)}{\sqrt{1-x^2}}\ {\rm d}x\tag1\\
&=-\frac{2}{\sqrt{\pi}}\int^1_0\frac{\sqrt{1-x^2}\ln(1-x)}{x}\ {\rm d}x\tag2\\
&=-\frac{1}{\sqrt{\pi}}\int^\pi_0\frac{\cos^2{x}\ln(1-\sin{x})}{\sin{x}}\ {\rm d}x\tag3\\
&=-\frac{1}{\sqrt{\pi}}\left[\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n}\mathcal{A}_n+2\sum^\infty_{n=0}\frac{(-1)^{n-1}}{2n+1}\mathcal{B}_n\right]\tag4\\
\end{align}
where
\begin{align}
\mathcal{A}_n
=\int^\pi_0\frac{\cos^2{x}}{\sin{x}}(\cos(2nx)-1)\ {\rm d}x
\ \ \ , \ \ \ \mathcal{B}_n
=\int^\pi_0\frac{\cos^2{x}}{\sin{x}}\sin((2n+1)x)\ {\rm d}x\\
\end{align}
Using simple trigonometric identities, it is not hard to see that, for $n\in\mathbb{N}$,
\begin{align}
\mathcal{A}_n-\mathcal{A}_{n-1}&=-\ \frac{1}{2n-3}-\frac{2}{2n-1}-\frac{1}{2n+1}\tag5\\
\mathcal{B}_n-\mathcal{B}_{n-1}&=0\tag6
\end{align}
Thus we may obtain the closed forms for both sequences.
\begin{align}
\mathcal{A}_n=2H_n-4H_{2n}+\frac{1}{2n-1}-\frac{1}{2n+1}+2\ \ \ , \ \ \ \mathcal{B}_n=\pi-\frac{\pi}{2}\delta_{n0}\tag7
\end{align}
Using the Taylor series for $\ln(1-x)$ and $\arctan{x}$, as well as the well-known identities
\begin{align}
\sum^\infty_{n=1}\frac{(-1)^{n-1}H_n}{n}&=\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}\tag8\\
\sum^\infty_{n=1}\frac{(-1)^{n-1}H_{2n}}{n}&=\frac{5\pi^2}{48}-\frac{1}{4}\ln^2{2}\tag9
\end{align}
gives us
\begin{align}
\sum^\infty_{k=1}\frac{\Gamma\left(\frac{k}{2}+1\right)}{k^2\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}
&=-\frac{1}{\sqrt{\pi}}\left[2\left(\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}\right)-4\left(\frac{5\pi^2}{48}-\frac{1}{4}\ln^2{2}\right)+2-2\pi\left(\frac{\pi}{4}\right)+2\left(\frac{\pi}{2}\right)\right]\\
&=\frac{3\pi^2-4\pi-8}{4\sqrt{\pi}}\tag{10}
\end{align}
as the closed form.
Explanation:
$(1)$: Write $\displaystyle\frac{\Gamma\left(\frac{k}{2}+1\right)}{\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}=\frac{2}{\sqrt{\pi}}\int^1_0\frac{x^{k+1}}{\sqrt{1-x^2}}\ {\rm d}x$.
$(2)$: Integrated by parts.
$(3)$: Substitute $x\mapsto\sin{x}$ then $x\mapsto\pi-x$.
$(4)$: Use the fact that $\ln(2-2\sin{x})=2\mathrm{Re}\ln(1+ie^{ix})$ then expand the $\mathrm{RHS}$.
$(5)$: Write $\displaystyle\mathcal{A}_n-\mathcal{A}_{n-1}=-\int^\pi_0(1+\cos(2x))\sin((2n-1)x)\ {\rm d}x$.
$(6)$: Write $\displaystyle\mathcal{B}_n-\mathcal{B}_{n-1}=\int^\pi_0(1+\cos(2x))\cos(2nx)\ {\rm d}x$.
$(7)$: Sum up $(5)$ and $(6)$.
$(8),(9)$: Let $z=-1$, $z=i$ in $\displaystyle\sum^\infty_{n=1}\frac{H_n}{n}z^n={\rm Li}_2(z)+\frac{1}{2}\ln^2(1-z)$.
$(10)$: Apply $(7)$, $(8)$, $(9)$ to $(4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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} |
Calculate Infinite Limit I'm trying to calculate the limit and when I get to the last step I plug in infinity for $\frac 8x$ and that divided by -4 I get - infinity for my answer but the book says 0. Where did I go wrong?
$$
\frac {8x^3-x^2}{7+11x-4x^4}
$$
Divide everything by $x^4$
$$
\frac {\frac{8x^3}{x^4}-\frac{x^2}{x^4}}{\frac{7}{x^4}+\frac{11x}{x^4}-\frac{4x^4}{x^4}}
$$
Results
$$
\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4} = \infty
$$
| Suppose that $n,m\in \mathbb{N}$:
$$f(x)=\dfrac{a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}}{b_{m}x^m+a_{m-1}x^{m-1}+\dots+b_{1}x+b_{0}}$$
We have three case:
Case $1$:
$$n< m\Longrightarrow \lim_{x\rightarrow\infty}f(x)=0$$
Case $2$:
$$m<n\Longrightarrow \lim_{x\rightarrow\infty}f(x)=\infty$$
Case $3$:
$$n=m\Longrightarrow \lim_{x\rightarrow\infty}f(x)=\dfrac{a_{n}}{b_{m}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there anyway to show $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$ other than taking derivatives? The purpose is to show $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$ for any $x,y\in\Bbb{R}$.
Taking partial derivatives with respect to $x,y$ respectively $\frac{{\partial \frac{{\cos x - \cos y}}{{x - y}}}}{{\partial x}} = 0$, $\frac{{\partial \frac{{\cos x - \cos y}}{{x - y}}}}{{\partial y}} = 0$ gives $\sin x={ - \frac{{\cos x - \cos y}}{{x - y}}}$ and $\sin y={ - \frac{{\cos x - \cos y}}{{x - y}}}$. Plugging them back and we have the desired result $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$.
What I want to ask is, is there more decent way to show the inequity? Taking derivatives looks clumsy. Hope someone can help. Thank you!
| Using the difference to product identity $\cos x - \cos y = -2\sin\dfrac{x-y}{2}\sin\dfrac{x+y}{2}$ along with the inequality $|\sin \theta| \le |\theta|$ gives:
$\left|\dfrac{\cos x - \cos y}{x-y}\right| = \left|\dfrac{-2\sin\frac{x-y}{2}\sin\frac{x+y}{2}}{x-y}\right| = \dfrac{2\left|\sin\frac{x-y}{2}\right|\left|\sin\frac{x+y}{2}\right|}{|x-y|} \le \dfrac{2\left|\frac{x-y}{2}\right| \cdot 1}{|x-y|} = 1$.
Note that both $\cos x - \cos y = -2\sin\dfrac{x-y}{2}\sin\dfrac{x+y}{2}$ and $|\sin \theta| \le |\theta|$ can be proven without calculus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If eigenvalues are $\pm 1$, then matrix is orthogonal If eigenvalues of a matrix are $1$, or $-1$, or both of them, does that necessarily means that the matrix is orthogonal?
The only thing I'm certain about is that the matrix is non-singular, due to absence of zero among the eigenvalues.
| The eigenvectors with eigenvalue $1$ have to be orthogonal to the eigenvectors with eigenvalue $-1$ in order for the matrix to be orthogonal. A $2\times2$ counterexample using this idea is not difficult to construct. For example, let $M$ be a linear operator on $\mathbb{R}^{2}$ for which
$$
M\left[\begin{array}{c}1 \\ 1\end{array}\right]=\left[\begin{array}{c}1 \\ 1\end{array}\right] \\
M\left[\begin{array}{c}0 \\ 1\end{array}\right]=-\left[\begin{array}{c}0 \\ 1\end{array}\right].
$$
Then $M$ has eigenvalues $1$ and $-1$, but the eigenvectors corresponding to these eigenvalues are not orthgonal. To find $M$,
\begin{align}
M\left[\begin{array}{c}x \\ y\end{array}\right]
& = M\left(x\left[\begin{array}{c}1 \\ 1\end{array}\right]+(y-x)\left[\begin{array}{c}0 \\ 1\end{array}\right]\right) \\
& = x\left[\begin{array}{c}1 \\ 1\end{array}\right]+(x-y)\left[\begin{array}{c}0 \\ 1\end{array}\right] \\
& = \left[\begin{array}{cc}1 & 0\\2 & -1\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]
\end{align}
So a simple counterexample is
$$
M = \left[\begin{array}{cc}1 & 0 \\ 2 & -1\end{array}\right].
$$
It is easy to check that $M$ has the correct eigenvectors with eigenvalues $\pm 1$. And $M$ is not orthogonal:
$$
M^{\perp}M = \left[\begin{array}{cc}1 & 2 \\ 0 & -1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 2 & -1\end{array}\right]
= \left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]\ne I.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}$ for all $n\in\mathbb{N}$.
Show that $$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}\leq 2^{n}\qquad (n\in \mathbb{N}).$$
I want to show the last step, that is, the inductive step. Assume that this equation is true for some $n=k$. Note that
$$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}=\prod_{i=1}^{n}\left [ 2-\frac{1}{i} \right ].$$
For the case $n=k+1$ would be
$$\prod_{i=1}^{k+1}\left [ 2-\frac{1}{i} \right ]=\prod_{i=1}^{k}\left [ 2-\frac{1}{i} \right ]\left [ 2-\frac{1}{k+1} \right ]\leq 2^{k}\left [ 2-\frac{1}{k+1} \right ]=2^{k+1}-\frac{2^{k}}{k+1}.$$
I have to show that the last term is $\leq 2^{k+1}$. But it's not possible to show that $2^{k}/(k+1)\geq 0$.
| Another approach:
$$\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{1\cdot 2\cdot 3\cdot \ldots \cdot n}< \frac{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}{1\cdot 2\cdot 3\cdot \ldots \cdot n}=\frac{2^n n!}{n!}=2^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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When are we permitted to multiply or divide both sides of an equation by a variable? As it is said in the mathematics books (at least the one I have), we are not permitted to divide or multiply both sides of an equation by a variable, because it is possible to lose some answers. For example, in the following equation
$$x^2=x$$
if we divide both sides by $x$, we would have $x = 1$, but the original equation has two answers, $0$ and $1$, and we've lost $x = 0$ by dividing it by a variable.
But in the same book, in order to solve a rational equation, the author multiplies both sides of the equation by $x(x-2)$, solving the equation:
$$\begin{align*}
\frac {x+2}{x-2} - \frac1x &= \frac{2}{x(x-2)}\\\\
x(x-2)\frac {x+2}{x-2} - x(x-2)\frac 1x &= x(x-2)\frac{2}{x(x-2)}\\\\
x(x+2)-(x-2)&=2\\\\
x^2+x&=0\\\\
x(x+1)&=0\\\\
x&=0\\
x&=-1
\end{align*}$$
Can someone please explain when we are allowed to do this and when we are not? I got a bit confused!
| When dividing by any quantity,
or when canceling out two quantities in a ratio
(for example, canceling $x$ and $x$ to find that $\frac xx=1$),
you need to be aware of what assumptions you have to make so
that the division or canceling makes sense,
and remember that those assumptions apply to any results you get.
For example, with
$$x^2 = x,$$
$x \neq 0$ then you can divide by $x$.
One way to keep track of your assumptions is to work out different
"cases" of the solution:
Case $x = 0$: Then the equation becomes $0^2 = 0$, which is true,
so $x = 0$ is a solution.
Case $x \neq 0$: Then since $x \neq 0$, you can divide by $x$,
so $x^2 = x$ implies $x = 1$. This is consistent with the assumptions
(namely that $x\neq 0$), so $x = 1$ is a solution.
If you look at all possible cases (and here, since $x$ either is or isn't zero,
we have already covered all possible cases),
the complete solution set consists of all solutions you find in all those cases;
that is, for this problem the solution set is $x=\{0,1\}$
(in other words, $x=0$ or $x=1$).
Now consider the example from the book:
$$\frac {x+2}{x-2} - \frac1x = \frac{2}{x(x-2)}.$$
Two equal quantities multiplied by the same quantity
are two equal quantities (even if we multiply both by zero!),
so we know that
$$x(x-2)\frac {x+2}{x-2} - x(x-2)\frac 1x = x(x-2)\frac{2}{x(x-2)}.$$
The tricky part is what comes next. It looks like the multiplier $x-2$
on the right-hand side should cancel the divisor $x-2$,
that is, $\frac {x-2}{x-2} = 1.$ But this is true only if $x-2 \neq 0$;
if $x - 2 = 0$ then $\frac {x-2}{x-2} = \frac 00,$ which is undefined.
Similarly, we can only cancel $x$ and $x$ on the right-hand side if $x \neq 0$.
So again we have two cases:
Case $x - 2 = 0$ or $x = 0$: In this case, the term on the right-hand
side of the equation evaluates to $\frac 20$, which is undefined,
so there are no solutions in this case.
Case $x - 2 \neq 0$ and $x \neq 0$:
In this case we can cancel $x-2$ with $x-2$ and cancel $x$ with $x$,
so we have
$$\begin{eqnarray}
x(x+2) - (x-2) &=& 2.\\
x^2 + x &=& 0.\\
x(x+1) &=& 0.
\end{eqnarray}$$
Now, remembering that we are still working out the case where
$x - 2 \neq 0$ and $x \neq 0$, we can divide both sides by $x$:
$$\begin{eqnarray}
x+1 &=& 0. \\
x &=& -1.
\end{eqnarray}$$
So $x = -1$ is the only solution in this case.
(Alternatively, if we used $x(x+1)=0$ to conclude that "$x=0$ or $x+1=0$",
we would still be working under the assumption that $x\neq 0$,
and from those facts we could conclude simply that $x+1=0$.)
Since there were no solutions from the other case,
$x = -1$ is altogether the only solution of the equation.
Note that this method did not identify $x=0$ as a solution.
That's because I do not accept that it is a solution:
if you set $x=0$ in $\frac {x+2}{x-2} - \frac1x = \frac{2}{x(x-2)},$
you get a term of $\frac 10$ on the left and you get $\frac 20$
on the right, and both of those are undefined.
I hope that (eventually) the book agrees with this.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limits. Finding positive value Find a real number k such that the limit $$\lim_{n\to\infty}\ \left(\frac{1^4 + 2^4 + 3^4 +....+ n^4}{n^k}\right)$$ has as positive value.
If I am not mistaken every even $k$ can be the answer. But the answer is 5.
| First note that
$$\sum\limits_{i=1}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
$$=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}$$
So now we have
$$\lim\limits_{n\to\infty} \left(\frac{1^4 + 2^4 + 3^4 +\cdots + n^4}{n^k}\right)$$
$$=\lim\limits_{n\to\infty} \left(\frac{\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}}{n^k}\right)$$
$$=\lim\limits_{n\to\infty} \left(\frac{n^5}{5n^k}+\frac{n^4}{2n^k}+\frac{n^3}{3n^k}-\frac{n}{30n^k}\right)$$
If $k\lt 5$, then
$$\lim\limits_{n\to\infty} \left(\frac{n^5}{5n^k}+\frac{n^4}{2n^k}+\frac{n^3}{3n^k}-\frac{n}{30n^k}\right)=\infty$$
If $k\gt 5$, then
$$\lim\limits_{n\to\infty} \left(\frac{n^5}{5n^k}+\frac{n^4}{2n^k}+\frac{n^3}{3n^k}-\frac{n}{30n^k}\right)=0$$
If $k=5$, then
$$\lim\limits_{n\to\infty} \left(\frac{n^5}{5n^5}+\frac{n^4}{2n^5}+\frac{n^3}{3n^5}-\frac{n}{30n^5}\right)$$
$$=\lim\limits_{n\to\infty} \left(\frac{1}{5}+\frac{1}{2n}+\frac{1}{3n^2}-\frac{1}{30n^4}\right)=\frac15$$
Therefore the answer is $k=5$ because zero is not positive nor negative and infinity is not quantifiable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this inequality: $\frac n2 \le \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac1{2^n - 1} \le n$ $\dfrac{n}{2} \le \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^n - 1} \le n $
I've Tried for hours but didn't got any striking idea. I don't have any efforts to show rather than induction. Please Try If you 've non-inductive proof.
| It's a standard trick. Let
$$
H_n = 1 + \frac12 +\ldots+ \frac1n.
$$
Look on $H_{2^n-1}$:
$$
n=1: H_1 = 1=1\\
n=2: H_3 = 1 + \frac12 + \frac13 < 1 + \frac12 + \frac12 = 1 + 1=2\\
n=3: H_7 = 1 + \frac12 + \frac13 + \left(\frac14+\frac15+\frac16+\frac17\right) < 1 + \frac{2}{2} + \left(\frac14+\frac14+\frac14+\frac14\right)=1 + 2 =3
$$
So, $H_{2^n-1} < n$. Analogously,
$$
n=1: H_1 = 1=1\\
n=2: H_3 = 1 + \frac12 + \frac13 > 1 + \frac14 + \frac14 = 1 + \frac{1}{2}\\
n=3: H_7 = 1 + \frac12 + \frac13 + \left(\frac14+\frac15+\frac16+\frac17\right) > 1 + \frac{1}{2} + \left(\frac18+\frac18+\frac18+\frac18\right)=1 + \frac{2}{2},
$$
or in general
$$
H_{2^n-1}>1 + \frac{n-1}{2} = \frac{n+1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Solve the trigonometric equation: $\sin {3x} = 4 \sin^2 x$ Solve the equation $\sin{3x} = 4 \sin^2 x$.
I tried to change the $\sin{3x}$ to $3\sin x\cos x$ then solve it, but I could not find the correct answer.
| Notice, $$\sin 3x=4\sin^2 x$$ $$\implies 3\sin x-4\sin^3x=4\sin^2 x$$ $$\implies 4\sin^3x+4\sin^2 x-3\sin x=0$$$$\implies \sin x(4\sin^2x+4\sin x-3)=0$$
$$\implies \sin x(2\sin x-1)(2\sin x+3)=0$$ As you have not mentioned any condition of the unknown value $x$ hence writing the general solution for $x$ as follows $$\implies \color{blue}{\sin x=0}\iff \color{blue}{x=n\pi} $$ $$\implies \color{blue}{2\sin x-1=0\ \text{or}\ \sin x=\frac{1}{2}\ or \ \sin x=\sin\frac{\pi}{6}}\iff \color{blue}{x=2n\pi+\frac{\pi}{6}} $$
$$or\ \color{blue}{\sin x=\sin\frac{5\pi}{6}}\iff \color{blue}{x=2n\pi+\frac{5\pi}{6}} $$
$$\implies \color{blue}{2\sin x+3=0\ \text{or}\ \sin x=\frac{-3}{2}}\iff \color{blue}{\sin x\neq \frac{-3}{2}} $$ Now, the complete general solution can be written as $$\color{blue}{x=[n\pi]\cup\left[2n\pi+\frac{\pi}{6}\right]\cup\left[2n\pi+\frac{5\pi}{6}\right]}$$
Where $\color{blue}{\text{n is any integer}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the matrix $\mathbf{A}$ if $A\binom{7}{-1} = \binom{6}{2}.$ Find the $2\times2$ matrix $A$ where $A^2=A$ and
$$A\begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$$
I tried plugging in: $A= \begin{pmatrix}a&b\\c&d\end{pmatrix}$ but that became messy very quickly. I got the equations:
$7a-b = 6$
$7c-d = 2$
$a^2+bc = a$
$ab+bd = b$
$ac + cd = c$
$bc + d^2 = d$ from trying that method. What should I do?
| Hint: Since you have $7a-b=6$ and $7c-d=2$ then $b=7a-6$ and $d=7c-2.$ Moreover, $A^2=A$ then $\det(A)^2=\det(A)$ which implies $\det(A)=0$ or $\det(A)=1$ then $ad=bc$ or $ad-bc=1.$ If $ad=bc$ then you get $a(7c-2)=c(7a-6)$ then $-2a=-6c$ so the coefficients of the matrix $A$ are of the form $a=3c,$ $b=7a-6=21c-6 $ and $d=7c-2.$ If $ad-bc=1$ then $-2a+6c=1$ so the coefficients of the matrix $A$ are of the form $c=\frac{2a+1}{6}$ $b=7a-6$ and $d=7c-2=\frac{14a-5}{6}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solving $6 \cos x - 5 \sin x = 8$ My attempt:
Using the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi)$$
$$
\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8
$$
$$
\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}
$$
And then assume:
$$
\frac{6}{\sqrt{51}}= \cos \psi ; \frac{5}{\sqrt{51}}= \sin\psi ;
$$
$$
\cos \psi \cos x - \sin \psi \sin x = \cos (x+ \psi) = \cos(x + \arccos ( \frac{6}{\sqrt{51}}))
$$
$$
x + \arccos\left(\frac{6}{\sqrt{51}}\right) = \arcsin\left( \frac{8}{\sqrt{51}}\right)
$$
$$
x \approx 12^\circ
$$
But answer is:
$$
-\frac{\pi}{4} + (-1)^n \frac{\pi}{4} + \pi n , n\in\Bbb Z
$$
| $6^2 + 5^2 = 61 < 8^2$.
So your attempted answer is ok except for a digit and the fact that you get
$$
\cos(\text{something}) = \frac 8 {\sqrt{61}} >1.
$$
A cosine of a complex number can be bigger than $1$, but a cosine of a real number cannot.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Baby Rudin claim: $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...$ converges This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence of rearrangements.
Let $\{s_n\}$ be the sequence of partials sums of the series then for $n \ge 0$
$s_{3(n+1)} = \sum ^n _ {k=0} \frac{1}{4k+1} + \frac{1}{4k+3} - \frac{2}{4k+4}$
We can view it as the sequence(on $n$) of partials sums of
$\sum_0 a_n = \sum_0 \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{2}{4n+4}$
Where $|a_n| = a_n = \frac{1}{4n+4}\{\frac{3}{4n+1}+\frac{1}{4n+3}\} \le \frac{1}{4n^2}$.
By the comparison test $s_{3(n+1)}$ converges to some real $\alpha$.
But $s_{3(n+1)+1} = s_{3(n+1)}+ \frac{1}{4n+5}$ and $s_{3(n+1)+2} = s_{3(n+1)}+ \frac{1}{4n+5}+\frac{1}{4n+7} $hence we have a partition of $\{s_n\}$ into subsequences which tend to $\alpha$ and this implies $s_n \rightarrow \alpha$.
Is my proof correct? Any alternative solutions are appreciated.
| The following argument shows that the series converges, and gives its sum:
$\hspace{.3 in}1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots=\ln 2$ $\;\;\;$so
$\hspace{.27 in}\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots=\frac{1}{2}\ln 2$. $\;\;\;$Inserting zeros, we get
$\hspace{.26 in}0+\frac{1}{2}+0-\frac{1}{4}+0+\frac{1}{6}+0-\frac{1}{8}+0+\frac{1}{10}+\cdots=\frac{1}{2}\ln 2$.
Adding this to the original series gives
$\hspace{.26 in}1+0+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+0+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+0+\cdots=\frac{3}{2}\ln 2$, $\;\;$ so
$\hspace{.25 in} 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\cdots=\frac{3}{2}\ln 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How to integrate this : $ \int \cot(5x) \tan(2x) \mathrm{d}x$ Methods to integrate this integral:
$$\int \cot(5x) \tan(2x) \mathrm{d}x$$
I have tried several methods, step by step, and they have led to invalid results. Helpful hints or processes are welcome.
I did this further :
$$ \cot(5x) = \frac{1}{\tan(2x+ 3x)}$$
$$= \frac{1}{\frac{\tan 2x +\tan 3x}{1-\tan 2x\tan 3x}}$$
Simplifying it further by breaking $\tan(3x) = \tan(2x +x)$ again we get
$$= \frac{1-\tan^2 x -2\tan x \tan 2x}{2\tan 2x -\tan^2 2x . \tan x +\tan x}$$
now the integral is $$ = \frac{1-\tan^2 x -2\tan x \tan 2x}{2\tan 2x -\tan^2 2x . \tan x +\tan x} \tan 2x $$
Can we do something with this further thanks..
| Letting $\tan x=z$, the integral can be rewritten as $$I=\int \frac{2z(1-10z^2+5z^4)}{(1-z^4)(5z-10z^3+z^5)}dz\\=2\int\frac{1-10z^2+5z^4}{(1-z^4)(5-10z^2+z^4)}dz\\=2\int \frac{1}{1-z^4}dz-8\int \frac{1}{5-10z^2+z^4}dz\\=2J-8K$$ where $J$ can be easily evaluated. As for $K$, it can be written as $$\int \frac{1}{(z^2-5-2\sqrt{5})(z^2-5+2\sqrt{5})}dz$$ which can then be evaluated without much effort. I admit that the final expressions will be ugly, but the evaluation process is straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is there a way to show the following equality without induction I wanted to show the following equality without using induction:
$$
\sum_{k=2}^n \frac{1}{k(k-1)} = \frac{n-1}{n}
$$
Any hint on how to do it?
| Notice, $$\sum_{k=2}^{n}\frac{1}{k(k-1)}$$ $$=\sum_{k=2}^{n}\left(\frac{1}{k-1}-\frac{1}{k}\right)$$ $$=\left(\frac{1}{2-1}-\frac{1}{2}\right)+\left(\frac{1}{3-1}-\frac{1}{3}\right)+\left(\frac{1}{4-1}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{(n-1)-1}-\frac{1}{n-1}\right)+\left(\frac{1}{n-1}-\frac{1}{n}\right)$$ $$=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n-1}-\frac{1}{n}\right)$$ $$=\left(\frac{1}{1}-\frac{1}{n}\right)$$ $$=\frac{n-1}{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Confusion about the integral $\int dT/(1-T^2)$ From some reference on the internet we have the following real valued function and its derivative:
$$
M(T) = \frac{\sqrt{1-T^2}}{1+T}
\quad \Longrightarrow \quad \frac{dM}{dT} = - M/(1-T^2)
$$
The reverse of differentiation is integration:
$$
\frac{dM}{dT} = - M/(1-T^2) \quad \Longrightarrow \quad \int\frac{dM}{M} = - \int \frac{dT}{1-T^2}
= \int \frac{dT}{T^2-1}\quad \Longrightarrow \\
\int\frac{dM}{M} = \frac{1}{2} \left[\, \int \frac{dT}{T-1} -\int \frac{dT}{T+1}\,\right]
\quad \Longrightarrow \\ \ln|M| = \frac{1}{2}\left[\,\ln|T-1|-\ln|T+1|\,\right] =
\ln\left(\sqrt{\left|\frac{T-1}{T+1}\right|}\right) \quad \Longrightarrow \\
M = \sqrt{\frac{1-T}{1+T}} = \frac{\sqrt{1-T^2}}{1+T}
$$
Here is where the confusion starts, because we also could have proceeded in the following way:
$$
\frac{dM}{dT} = - M/(1-T^2) \quad \Longrightarrow \quad \int\frac{dM}{M} = - \int \frac{dT}{1-T^2}
\quad \Longrightarrow \\
\int\frac{dM}{M} = - \frac{1}{2} \left[\, \int \frac{dT}{1-T} + \int \frac{dT}{1+T}\,\right]
\quad \Longrightarrow \\ \ln|M| = - \frac{1}{2}\left[\,\ln|1-T|+\ln|1+T|\,\right] =
\ln\left(\frac{1}{\sqrt{\left|1-T^2\right|}}\right) \quad \Longrightarrow \\
M = \frac{1}{\sqrt{1-T^2}}
$$
Which is clearly wrong. But .. where is the error?
| You need to pay attention to the constant of integration and don't ignore absolute value signs.
So you would have the same problem if you tried to do the same thing with this equaltiy $\frac{1}{1-x}=-\frac{1}{x-1}$
$\int \frac{1}{1-x}=-\int\frac{1}{x-1}$
$\log|1-x|=\log|x-1|+c_1$ Here you missed out the constant, also you are missing a minus sign
$|1-x|= c_2|x-1|$ In this step you missed out the absolute value signs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
| Let $X \sim G(1-x)$, a geometric random variable with success probability $1-x$. We have
$$
\mathbb{E}[X] = \sum_{n=1}^\infty nx^{n-1}(1-x).
$$
On the other hand, we know that $\mathbb{E}[X] = 1/(1-x)$, and we deduce the formula.
We can argue that $\mathbb{E}[X] = 1/(1-x)$ in many ways. One way is to consider $N$ different trials with success probability $1-x$. The number of successful trials is roughly $(1-x)N$, and so the average distance between successful ones (which is distributed according to $X$) is roughly $N/((1-x)N) = 1/(1-x)$. (This argument can be formalized.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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Is this problem wrongly built? Or is there a solution which I don't know how to arrive at? I was solving a Cauchy-Schwarz's inequality based problem. Given that $x^2+y^2+z^2=1$ I am supposed to show that $x+y+z \le 6$. After struggling for a while I realised that I could solve this inequality had the condition been $x^2+y^2+z^2=12$. This was my solution:
$(x.1+y.1+z.1)^2 \le (x^2+y^2+z^2)(1^2+1^2+1^2)$
$\implies (x+y+z)^2 \le 12*3$
$\implies (x+y+z)^2 \le 12*3 \le 36$
Taking square root on both sides, I got
$(x+y+z) \le 6$
Now I am wondering whether I detected a typo in the problem ($x^2+y^2+z^2=12$, not $x^2+y^2+z^2=1$) or is there a real way to reach $(x+y+z) \le 6$ starting from $x^2+y^2+z^2=1$ based on Cauchy-Schwarz principles!
The main reason for this doubt is I am stuck on an extension of this problem in which I need to prove $x^3+y^3+z^3 \ge 24$
Any help is appreciated.
| It seems that there is no need for Cauchy inequality. Obviously $x\leq1$. (why?). Similarly $y$ and $z$ are. So $x+y+z\leq 3$. This inequality can be strengthen since x, y, z can't take value 1 at the same time.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$
Evaluate $$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$
I thought of rewriting this as $$\large{\int\dfrac{dx}{x^5(1+\frac{1}{x^4})^{3/4}}}$$ and substituting $$u^4=\left(1+\frac{1}{x^4}\right)\Rightarrow u=\left(1+\frac{1}{x^4}\right)^{1/4}$$ and subsequently I got $$du=\dfrac{1}{4}\left(1+\frac{1}{x^4}\right)^{-3/4}\times (-4x^{-5})dx$$
However, I can not think of how to proceed further. Any help would be truly appreciated. Many thanks in advance!
| Let, $1+\frac{1}{x^4}=t \implies \frac{-4dx}{x^5}=dt$$$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$ $$=\int \frac{dx}{x^5\left(1+\frac{1}{x^4}\right)^{3/4}}$$ $$=\frac{-1}{4}\int\frac{dt}{\left(t\right)^{3/4}}$$ $$=\frac{-1}{4}\int (t)^{-3/4}dt$$ $$=\frac{-1}{4}\frac{t^{1/4}}{1/4}+C$$ $$=-\left(1+\frac{1}{x^4}\right)^{1/4}+C$$$$=-\frac{(1+x^4)^{1/4}}{x}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Using Lagrange's Method in Finding Extreme Values of $x^2 + y^2 + z^2$ for $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ (New to This Method) Did I do this hw question correctly (at least in theory, I do not expect anyone to check my algebra work)? In particular, did I solve for lambda and plug lambda back into my equations for x,y, and z correctly, or is there a better way? Something does not feel right about it. Thanks!
Apply Lagrange's method in finding the extreme values $x^2 + y^2 + z^2$ subject to the constraint $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, where $a > b > c > 0$.
Here is my work thus far:
Let $f(x,y,z) = x^2 + y^2 + z^2$, and let $g(x,y,z) - k = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} - 1$.
Then let $u=f(x,y,z) + \lambda(g(x,y,z) - k)$. This gives us: $$u=(x^2 + y^2 + z^2) + \lambda\frac{x^2}{a^2} + \lambda\frac{y^2}{b^2} + \lambda\frac{z^2}{c^2} - \lambda$$
Taking the partial of u with respect to each variable (including lambda) and setting it equal to zero, we get:
$u_x = 2x + \frac{2\lambda x}{a^2} = 0$
$u_y = 2y + \frac{2\lambda y}{b^2} = 0$
$u_z = 2z + \frac{2\lambda z}{c^2} = 0$
$u_\lambda = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
Solving for x, y, and z, we get $x = -\frac{\lambda}{a^2},\ y = -\frac{\lambda}{b^2},\ z = -\frac{\lambda}{c^2}$, which we can plug into our partial, $u_\lambda$, for x, y, and z. In our partial, we get $$\frac{\lambda^2}{a^6} + \frac{\lambda^2}{b^6} + \frac{\lambda^2}{c^6} = 1$$
Solving for $\lambda$ gives us the difficult solution of $$\lambda = \pm \frac{a^3b^3c^3}{\sqrt{a^6b^6 + a^6c^6 + b^6c^6}}$$
Since $a,\ b,\ c>0$, the plugging in the positive solution for $\lambda$ will give us our max and vice versa for the min (if the function were odd, see below). Plugging $\lambda$ into our x, y, and z equations yield:
$x = \pm \frac{ab^3c^3}{\sqrt{a^6b^6 + a^6c^6 + b^6c^6}}$
$y = \pm \frac{a^3bc^3}{\sqrt{a^6b^6 + a^6c^6 + b^6c^6}}$
$z = \pm \frac{a^3b^3c}{\sqrt{a^6b^6 + a^6c^6 + b^6c^6}}$
Since f is even (each variable is raised to the second power), then either the positive or negative solution will give a positive result.
| To comment on your solution I think you have a mistake when you got $x=-\frac\lambda{a^2}$ from $2x+\frac{2\lambda x}{a^2}=0$. (I will add link to the original version, in case the post will be subsequently changed.)
In fact, you can rewrite this equation as
$$2x+\frac{2\lambda x}{a^2}=2x\left(1+\frac\lambda{a^2}\right)=0,$$
which means that either $x=0$ or $\lambda = -a^2$.
You can make similar argument for the other two equations.
I just want to point out that it is relatively easy at least to see whether the final result is correct.
Manipulating inequalities. You get
$$\frac{x^2+y^2+z^2}{a^2}= \frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{a^2} \le \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
which implies
$$x^2+y^2+z^2\le a^2.$$
Since this value is attained for $x=a$, $y=z=0$, you see that the maximum is $a^2$. You can find the minimal value in a similar fashion.
Geometric interpretation. We know that
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
is the equation of ellipsoid with principal axes of lengths $a$, $b$, $c$. The question is asking about the closest and the furthest point of the ellipsoid from the origin. Any such point must lie on some of the axes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integration validity of $\int\frac{1}{\sqrt{a^2 + x^2}}\,dx$ I'm just wondering if the following integration is valid.
\begin{array}{l}
\int {\frac{1}{{\sqrt {{a^2} + {x^2}} }}} dx\\
{\rm{Let }}{u^2} = {a^2} + {x^2}\\
2udu = 2xdx\\
\frac{{du}}{x} = \frac{{dx}}{u}\\
{\rm{Let }}\frac{{du}}{x} = \frac{{dx}}{u} = A\\
du = Ax\\
dx = Au\\
\frac{{du + dx}}{{x + u}} = \frac{{Ax + Au}}{{x + u}} = A = \frac{{dx}}{u}\\
\int {\frac{1}{{\sqrt {{a^2} + {x^2}} }}} dx\\
= \int {\frac{1}{{\sqrt {{u^2}} }}} dx\\
= \int {\frac{1}{u}} dx\\
= \int {\frac{{du + dx}}{{x + u}}} \\
= \int {\frac{{d\left( {u + x} \right)}}{{x + u}}} \\
= \ln \left| {x + u} \right| + C\\
= \ln \left| {x + \sqrt {{a^2} + {x^2}} } \right| + C
\end{array}
| Even another view on this technique could be provided via Euler substitutions. What there appeared, especially in final crucial step, was $x + u$ in the denominator, so why from the beggining assume the substitution :
$$\sqrt{a^2 + x^2} = u - x$$
Squaring both sides :
$$a^2 = u^2-2ux$$ and taking differential
$$u\mathrm{d}x=(u-x)\mathrm{d}u$$ hence
$$I = \int \frac{\mathrm{d}x}{\sqrt{a^2+x^2}} = \int \frac{(u-x)\mathrm{d}u}{(u-x)u} = \int \frac{\mathrm{d}u}{u} = \ln |u| +C = \ln\left(x+\sqrt{a^2+x^2}\right) +C $$
This approach is therefore equivallent, so what may appear "magical" is the taking multiple differentials, but this can be avoided : eliminating variable $x$ we get :
$$x = \frac{u^2-a^2}{2u}$$
Ergo $$ \sqrt{a^2+x^2} = \sqrt{a^2+\frac{(u^2-a^2)^2}{4u^2}}=\frac{u^2+a^2}{2u} $$ and for $\mathrm{d}x$ :
$$\mathrm{d}x = \mathrm{d}\left(\frac{u^2-a^2}{2u}\right)=\frac{u^2+a^2}{2u^2}\mathrm{d}u$$
Therefore
$$I = \int \frac{\mathrm{d}x}{\sqrt{a^2+x^2}} = \int \frac{\frac{u^2+a^2}{2u^2}\mathrm{d}u}{\frac{u^2+a^2}{2u}} =\int \frac{\mathrm{d}u}{u} = \ln |u| +C = \ln\left(x+\sqrt{a^2+x^2}\right) +C $$
And I was just curious what king of "magic" would arise from the third Euler substitution, clearly different than the first two, which are here somewhat equivalent, and yes here it is :
Let $$\sqrt{a^2 + x^2} = tx$$
Or $$a^2 + x^2 = t^2x^2$$
Taking differential$$2x\mathrm{d}x = 2tx^2\mathrm{d}t+2xt^2\mathrm{d}x$$
Rearanging : $$\frac{\mathrm{d}x}{tx} = \frac{\mathrm{d}t}{1-t^2}$$
Ergo
$$I=\int\frac{\mathrm{d}x}{\sqrt{a^2+x^2}}=\int\frac{\mathrm{d}x}{tx}=\int\frac{\mathrm{d}t}{1-t^2}=\frac{1}{2}\ln\left|\frac{1+t}{1-t}\right|+C=\frac{1}{2}\ln\left|\frac{x+\sqrt{a^2+x^2}}{x-\sqrt{a^2+x^2}}\right|+C$$
Indeed, after rationalisation the denominator in $\ln$ :
$$\frac{1}{2}\ln\left|\frac{x+\sqrt{a^2+x^2}}{x-\sqrt{a^2+x^2}}\right| + C = \frac{1}{2}\ln\left|\frac{\left(x+\sqrt{a^2+x^2}\right)^2}{x^2-(a^2+x^2)}\right| +C = \ln\left(x+\sqrt{a^2+x^2}\right) + C' $$
Note : Substitution $\sqrt{a^2+x^2}=a+tx$ is also possible ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Separating variables by substitution in a homogenous ODE I am brand new to ODE's, and have been having difficulties with this practice problem. Find a 1-parameter solution to the homogenous ODE:$$2xy \, dx+(x^2+y^2) \, dy = 0$$assuming the coefficient of $dy \ne 0$
The textbook would like me to use the subsitution $x = yu$ and $dx=y \, du + u \, dy,\ y \ne 0$
Rewriting the equation with the subsititution:
$$2uy^2(y \, du + u \, dy)+(x^2+y^2) \, dy = 0$$
divide by $y^2$
$$2u(y \, du + u \, dy) + (u^2+1 ) \, dy=0$$
but after further simplification I end up getting ${dy \over y}$ which would mean I would get a logarithm after integrating, and the answer is given as $$3x^2y+y^3 = c$$
Could I get some help/hints as to how this answer was obtained?
| The variables $y$ and $u$ can be separated from each other:
$$
2u(y \, du + u \, dy) + (u^2+1 ) \, dy=0
$$
$$
2u\left( du + u \, \frac{dy} y \right) + (u^2+1 ) \, \frac{dy} y = 0
$$
$$
2u\left( \frac{du} u + \frac{dy} y \right) + \left( u + \frac 1 u \right) \, \frac{dy} y = 0
$$
$$
2 \left( \frac{du} u + \frac{dy} y \right) + \left( 1 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
2 \frac{du} u + 2 \frac{dy} y + \left( 1 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
2 \frac{du} u + \left( 3 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
\left( 3 + \frac 1 {u^2} \right) \, \frac{dy} y = -2 \frac{du} u
$$
$$
\frac{dy} y = \frac{-2 \, du/u}{3 + \frac 1 {u^2}}
$$
$$
\frac{dy} y = \frac{-2 u\, du}{3u^2 + 1}
$$
$$
\log|y| = - \frac 1 3 \log|3u^2 + 1| + \text{constant}
$$
$$
y = (3u^2+1)^{-1/3}\cdot\text{constant}
$$
$$
y^3 = \frac{\text{constant}}{3u^2+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding indefinite integral $\int \frac{x^2}{(x \sin x + \cos x)^2} dx $ I need hint in finding the integral of
$$\int \frac{x^2}{(x \sin x + \cos x)^2} dx $$
I tried dividing the term by $x^2\cos^2x$ and then substituting $\tan x$.
| From the formula
$$
D\frac fg=\frac{f'g-fg'}{g^2},
$$
we might guess that we have a quotient $f/g$ with $g=x\sin x+\cos x$, but what is then $f$?
First, we note that
$$
g'=x\cos x.
$$
Well, with the trig-one, we can write
$$
\begin{aligned}
\frac{x^2}{(x\sin x+\cos x)^2}
&=\frac{x^2\sin^2x+x^2\cos^2x}{(x\sin x+\cos x)^2}
=\frac{x\sin x(x\sin x+\cos x)-x\cos x(\sin x-x\cos x)}{(x\sin x+\cos x)^2}\\
&=\frac{x\sin x\cdot g-g'\cdot (\sin x-x\cos x)}{(x\sin x+\cos x)^2}.
\end{aligned}
$$
By pattern matching, $f=\sin x-x\cos x$ (note that, then $f'=x\sin x$).
We conclude that our guess was lucky, and that
$$
\int \frac{x^2}{(x\sin x+\cos x)^2}\,dx = \frac{\sin x-x\cos x}{x\sin x+\cos x}+C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Show that any 2D vectors can be expressed in the form...
(a) Show that any 2D vector can be expressed in the form
$s \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix},$
where $s$ and $t$ are real numbers.
(b) Let $u$ and $v$ be non-zero vectors. Show that any 2D vector can be expressed in the form
$s u + t v$
where $s$ and $t$ are real numbers, if and only if of the vectors $u$ and $v$, one vector is not a scalar multiple of the other vector.
No idea what to do to start these problems. Hints appreciated.
| (a) Let $\begin{pmatrix} x\\ y\end{pmatrix}$ a 2D vector. Finding $s$ et $t$ such that $$s \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix}=\begin{pmatrix} x\\ y\end{pmatrix}$$
is the same thing as solving:
$$\begin{cases}3s+2t=x\\
-s+7t=y\end{cases}$$
where $s$ and $t$ are the unknowns.
(b) Now if you replace $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 7 \end{pmatrix}$ by $u$ and $v$, what are the conditions on the previous system for it to admit a (unique) solution?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the least $N$ so there is no square
Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000 \cdot N$ contains no square of an integer.
Let $x^2$ appear before $1000N$ so:
$(x+1)^2 - x^2 > 1000 \implies x \ge 500$
Let $A = [1000N, 1000(N+1)]$
So I let $x=500$ then, $x^2 = 250000$, obviously this is impossible since the set has a square already.
So now I need to set some number, $k^2 \ge 501^2 \equiv 0 \pmod{1000}$
So I need to solve the quadratic residue:
$k^2 \equiv 0 \pmod{1000}$ with constraints for $k$.
Lets see: $500^2 = 250000$ and $501^2 = 251001$ and $502^2 = 252004$
$501^2 = 500^2 + 1000 + \sum_{k=1}^{1} 2k - 1$
$502^2 = 500^2 + 2(1000) + \sum_{k=1}^{2} 2k - 1$
The next $n$th square after $500$ is generalized:
$p = 500^2 + n(1000) + \sum_{k=1}^{n} 2k - 1$
Take $p \pmod{1000}$
$p \equiv n + \sum_{k=1}^{n} 2k - 1 \pmod{1000}$
$p \equiv n + n^2 \pmod{1000}$
$n(n+1) \equiv 0 \pmod{1000}$
$n \equiv 0$ and $n \equiv -1 \pmod{1000}$
But this method doesnt work properly. The final answer is: $N = 282$.
Please some hints only.
| To have $(500+a)^2<1000N$ and $(500+a+1)^2\ge 1000(N+1)$ we need
$$500^2+1000a+a^2<1000N,\qquad 500^2+1000(a+1)+(a+1)^2\ge 1000(N+1) $$
and hence
$$ 250 +a+1+\left\lfloor\frac{(a+1)^2}{1000}\right\rfloor\ge N +1>N>250+a+\left\lfloor\frac {a^2}{1000}\right\rfloor$$
Therefore look for small $a$ with $\left\lfloor\frac{(a+1)^2}{1000}\right\rfloor>\left\lfloor\frac {\strut a^2}{1000}\right\rfloor$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the sum of the series below Find the sum
$$(1\cdot2)+(1\cdot3)+(1\cdot4)+\cdots+(1\cdot2015)+(2\cdot3)+(2\cdot4)+\cdots+(2\cdot2015)+\cdots+(2014\cdot2015)$$
What I have tried...
We are looking for $$S(n)=\sum_{i=1}^{n-1}\sum_{j=i+1}^nij$$ when $n=2015$
$$S(n)=\sum_{1\le i<j\le n}ij$$ $$=\frac 12\left((\sum_{i=1}^n i)^2-\sum_{i=1}^ni^2\right)$$
$$S(n)=\frac 12\left(\frac{n^2(n+1)^2}4-\frac{n(n+1)(2n+1)}6\right)$$
$$=\frac{(n-1)n(n+1)(3n+2)}{24}$$
And so $$\boxed{S(2015)=\frac{2014\times 2015\times 2016\times 6047}{24}}$$
Is the solution wrong or not?
| Summing along diagonals gives a nice symmetric derivation of the solution, using binomial coefficients and minimizing messy manipulation of fractions.
$$\begin{align}
S(n)&=\sum_{i=1}^{n-1}\sum_{j=i+1}^nij
\color{lightgray}{=\sum_{1\le i<j\le n} ij}\\\\
&=\begin{cases}\begin{matrix}
\color{orange}{1\cdot 2} &+1\cdot 3 &+1\cdot 4 &+\cdots &+\color{purple}{1\cdot (n-2)} &\color{blue}{+1\cdot (n-1)} &\color{green}{+1\cdot n}&\\
&\color{orange}{+2\cdot 3} &+2\cdot 4 &+\cdots &+2\cdot (n-2) &\color{purple}{+2\cdot (n-1)} &\color{blue}{+2\cdot n}&\\
& &\color{orange}{+3\cdot 4} &+\cdots &+3\cdot (n-2) &+3\cdot (n-1) &\color{purple}{+3\cdot n}& \\
& & &\ddots & & &\vdots &\\
& & & & & &\color{orange}{+(n-1)\cdot n}&
\end{matrix}\end{cases}\\\\
&\color{lightgray}{=\sum_{r=1}^{n-1}\sum_{i=1}^{n-r}i(i+r)
\qquad\qquad\text{using}\ r=j-i}\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\sum_{i=1}^mi(n-m+i)
\qquad\text{using}\ m=n-r}\\\\
&=\begin{cases}\begin{array}{r}
\color{green}{1\cdot n}&\\
\color{blue}{+1\cdot (n-1)+2\cdot n}&\\
\color{purple}{+1\cdot (n-2)+2\cdot (n-1)+3\cdot n}&\\
\vdots\\
\color{orange}{+1\cdot 2+2\cdot3 +3\cdot 4+\cdots +(n-3)\cdot (n-2)+(n-2)\cdot (n-1)+(n-1)\cdot n}\end{array}\end{cases}\\\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\sum_{i=1}^m ni-i(m-i)}\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\sum_{m=1}^m\sum_{k=i+1}^m i\right]}\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\sum_{k=2}^m\sum_{i=1}^{k-1}i\right]}\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\sum_{k=2}^m\binom k2\right]}\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\binom {m+1}3\right]}\\
&=\begin{cases}\begin{array}{l}\quad
\color{Green}{n}&\\
\color{blue}{+3n-1}\\
\color{purple}{+6n-4}&\\
+10n-10&\\
\vdots\\
+\binom {m+1}2n-\binom {m+1}3&\\
\vdots \\
\color{orange}{+\binom n2 n-\binom n3}\end{array}\end{cases}\\\\
&=\sum_{m=1}^{n-1}\binom {m+1}2 n-\binom {m+1}3\\\\
&=\binom {n+1}3n-\binom {n+1}4\\\\
&=n\binom {n+1}3-\frac{n-2}4\binom{n+1}3\\\\
&=\frac{3n+2}4\binom{n+1}3\qquad\blacksquare\\\\
S(2015)&=\frac {6047}4\binom{2016}3=2\;061\;359\;653\;080\qquad\blacksquare
\end{align}$$
Note
The summation can also be computed by summing columns instead of rows (i.e. swapping $i$ and $j$) as follows:
$$\begin{align}
S(n)
&=\sum_{i=1}^{n-1}\sum_{j=i+1}^n ij\\
&=\sum_{1\le i<j\le n}ij\\
&=\sum_{j=2}^n\sum_{i=1}^{j-1}ij\\
&=\sum_{j=1}^n j\sum_{i=1}^{j-1}\binom i1\\
&=\sum_{j=1}^n j\binom j2\\
&=\sum_{j=1}^n (j+1)\binom j2-\binom j2\\
&=\sum_{j=1}^n 3\binom {j+1}3-\binom j2\\
&=3\binom {n+2}4-\binom {n+1}3\\
&=\frac {3(n+2)}4\binom {n+1}3-\binom {n+1}3\\
&=\frac {3n+2}4\binom {n+1}3\qquad\blacksquare
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Finding the kernel of a linear map Our exercise is to find all solutions to the equation $Ax = 0$, among others for the following matrix
$$A =\begin{pmatrix} 6 & 3 & -9 \\ 2 & 1 & -3 \\ -4 & -2 & 6 \end{pmatrix}.$$
This amounts to finding the kernel, and obviously, the rows of the matrix are multiples of each other, so we can reduce the equations to:
$$6x_1 + 3x_2 - 9x_3 = 0.$$
Choosing $x_3 = 0$, $x_1 = 1$ and $x_2 = -2$ would fulfill the equation in my opinion.
However, the official solution is described as the following set:
$$\lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}$$
with $\mu, \lambda \in$ R.
My questions now:
Where does the vector (and its multiples) $\lambda \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$ come from?
Also, since after the reduction we are left with only one row, the dimension of the kernel should accordingly be $1$ - however $\left[\begin{smallmatrix} 0 \\ 3 \\ 1 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 1 \\ -2 \\ 0 \end{smallmatrix}\right]$ (from $x_3 = 0$, $x_1 = 1$ and $x_2 = -2$) seem to be two different basis vectors - what have I done or understood wrongly?
Many thanks
| First, note that
$$\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} =
\begin{pmatrix} 1\\1\\1\end{pmatrix} - \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}, $$
So your missing solution
$\left[\begin{smallmatrix} 1 \\ -2 \\ 0 \end{smallmatrix}\right] $
is, in fact, a special case of the general solution
$ \lambda \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right] + \mu \left[\begin{smallmatrix} 0 \\ 3 \\ 1 \end{smallmatrix}\right] $ with $\lambda =1$ and $\mu = -1$.
Second, the general solution can be obtained from reduced equation $6x_1 + 3x_2 - 9x_3 = 0$ by imposing certain assumptions:
*
*For example, if we assume $x_1 = 0$, we will get the following
$$
\begin{cases} x_1 = 0 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases}
\implies \begin{cases}x_1 = 0 \\ 3x_2 - 9x_3 = 0 \end{cases}
\implies \begin{cases}x_1 = 0 \\ x_2 = 3 x_3\end{cases}
$$
Assuming parametrization $x_3 = \mu$, we get
$$
\begin{cases} x_1 = 0 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases}
\implies
\begin{cases}x_1=0 \\ x_2 = 3 x_3 \\ x_3 = \mu & - \operatorname{parameter}\end{cases}
\iff
\begin{cases}x_1 = 0 \\ x_2 = 3\mu \\ x_3 = \mu \end{cases}
$$
We can rewrite this solution in the vector form:
$$
\begin{cases}x_1 = 0 \\ x_2 = 3\mu \\ x_3 = \mu \end{cases}
\iff
x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}
= \begin{bmatrix} 0 \\ 3\mu \\ \mu \end{bmatrix}
\iff
x
= \begin{bmatrix} 0 \\ 3 \\ 1\end{bmatrix} \mu
$$
*Similarly, imposing condition $x_1 = x_2$, we write
$$
\begin{cases} x_1 = x_2 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases}
\implies \begin{cases}x_1 = x_2 \\ 6x_1 + 3x_2 - 9x_3 = 0 \end{cases}
\implies \begin{cases}x_1 = x_2 \\ x_1 = x_3\end{cases}
$$
Assuming parametrization $x_1 = \lambda$, we get
$$
\begin{cases}
x_1 = \lambda & -\operatorname{parameter}\\ x_2 = x_1 \\ x_3 = x_1
\end{cases}
\implies
\begin{cases} x_1 = \lambda \\ x_2 = \lambda \\ x_3 = \lambda \end{cases}
$$
The vector form of the solution then looks like
$$
\begin{cases} x_1 = \lambda \\ x_2 = \lambda \\ x_3 = \lambda \end{cases}
\implies
x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}
= \begin{bmatrix} \lambda \\ \lambda \\ \lambda \end{bmatrix}
\iff
x= \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix} \lambda
$$
Third, Since parametrized solutions $x = x(\lambda)$ and $x = x(\mu)$ are linearly independent, and since the equation $ 6x_1 + 3x_2 - 9x_3 = 0$ is linear, the general solution can be written as the sum of two independent parametrized particular solutions, i.e.
$$
x = x_\lambda + x_\mu = \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix} \lambda + \begin{bmatrix} 0 \\ 3 \\ 1\end{bmatrix} \mu
$$
Finally, Note that we could have imposed different assumptions on $x$, and that would result in different parametrized solutions.
However, as long as these solutions will be linearly independent, the will still span the same two-dimentional space of solutions which we have now.
Thus, the kernel of $A$, which is also a space of all solutions of the system $Ax = 0$, is the linear span of vectors $\left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 0 \\ 3 \\ 1 \end{smallmatrix}\right]$:
$$
\ker A
= \operatorname{span} \left(\; \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3 \\ 1\end{bmatrix} \;\right)
= \left\{\; \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3 \\ 1\end{bmatrix} \; \right\}
= \left\{\ \vec v = \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix} \lambda + \begin{bmatrix} 0 \\ 3 \\ 1\end{bmatrix} \mu
\mathrel{\bigg|}
\lambda, \mu \in \mathbb R\ \right\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
Show that for any two positive real numbers $a$ and $b$, $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
My attempt: $(\sqrt a-\sqrt b)^2\geq0\\\frac{a+b}{2}\geq \sqrt{ab}$
$(a-b)^2\geq0\\a^2+b^2\geq2ab \\\sqrt{\frac{a^2+b^2}{2}}\geq\sqrt{ab}$
I do not know how to link them together. Appreciate any tips.
| By setting:
$$ M_p(a,b)=\lim_{q\to p^+}\left(\frac{a^q+b^q}{2}\right)^{\frac{1}{q}}$$
we have to prove: $$2M_1\geq M_0+M_2, $$
but $M_0^2+M_2^2 = 2M_1^2$, hence the previous line follows from Cauchy-Schwarz inequality:
$$ M_0+M_2 \leq \sqrt{2}\sqrt{M_0^2+M_2^2} = 2M_1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means that we can set partial derivatives of $f$ equal to $0$ and try to solve for $x$ and $y$
$\nabla f = (1+2x, 1+2y) = (0,0) \implies (x,y) = (\frac{-1}{2}, \frac{-1}{2})$
So we get the minimal value $f(\frac{-1}{2}, \frac{-1}{2}) = \frac{-1}{2} + \frac{-1}{2} + (\frac{-1}{2})^2 \frac{-1}{2})^2 = -\frac{1}{2}$
But how about the maximal value? How does $x^2 + y^2 = 1$ restrict $f$?
| $x^2 + y^2 =1$. So $f(x,y) = x + y + 1$
Now consider $x + y = a ,$ Then $y = x - a$
$x^2 + y^2 = 1$ So $x^2 + (x-a)^2 = 1$
$2 x^2 -2 ax + a^2 -1 = 0$
The determinant must be non negative : $4^2 - 8 a^2 + 8 >= 0 : a^2 <=2$
Which means maximum of $a$ is$\displaystyle \sqrt{2}$ and minimum is $\displaystyle -\sqrt{2}$
So $f(x,y)_{\bf{Min.}} = 1-\sqrt{2}$ and $f(x,y)_{\bf{Max.}} = 1+\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $ \ \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} \ = \ 0 \ $ has a solution in $ \ (-1,1) $ If $a$ and $b$ are positive numbers, prove that the equation
$$\frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} = 0$$
has at least one solution in the interval $ \ (-1,1) \ $ .
The question is from the exercises section of a textbook chapter on limits/continuity.
I've been stumped on this one for a couple of days. I've been trying to calculate $\lim _{x \to -1}$ and $\lim _{x \to 1}$ and then show the function is continuous to show a root must lie in the interval. Factorising the denominators gives...
$$\frac{a}{(x+1)(x^2+x-1)} + \frac{b}{(x-1)(x^2+x+2)} = 0$$
So of course $x = 1$ and $x = -1$ are undefined and so the limits will be one-sided. Playing around with equation I haven't been able to find an equivalent function across $x \neq -1, x \neq 1$.
The only thing I have been able to show is
$$\frac{a}{b} = - \frac{(x+1)(x^2+x-1)}{(x-1)(x^2+x+2)}$$
and so
$$\lim _{x \to -1} \frac{a}{b} = 0, \lim _{x \to 1} \frac{b}{a} = 0$$
but I'm not sure if this is significant or I'm overthinking things. Could anyone point me in the right direction?
| Proof:
Let $f(x)=\frac{a}{x^{3}+2 x^{2}-1}+\frac{b}{x^{3}+x-2} .$ We want to prove that $f$ has at least one solution in the interval
$(-1,1) .$ It suffices to prove that $f(x)=0 .$ By factorization, $f(x)=\frac{a}{(x+1)(x-a)(x-b)}+$$\frac{b}{(x-1)\left(x^{2}+x+2\right)} .$ We denote $\frac{-1+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}$ as $a, b$ respectively. We find $x$ such that $f$ is discontinuous. Clearly, $x=\pm 1, a, b .$ However, since the given interval is $(-1,1)$ then $f$ is discontinuous at $x=a .$ Now, suppose $r, s$ is in $(a, 1)$ with $r>a$ and $s<1$ Clearly, $f$ is continuous on $[r, s] .$ Note that, $\lim _{x \rightarrow a^{+}} f(x)=\infty$ and $\lim _{x \rightarrow 1^{-}} f(x)=-\infty .$ Hence, by the Intermediate Value Theorem, since $f$ is continuous on the closed interval $[r, s]$ and previously we have shown that $-\infty<f(x)<\infty$ then there exist a $c \in(r, s)$ such that $f(c)=0$. Hence,
$f(x)$ has at least one solution in $(-1,1)$ namely, $x=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}-\sqrt{2(2-x)(1-x^2)}-4\sqrt{1-x}(2-\sqrt{4-x^2})}{x\sqrt{1-x}(2-\sqrt{4-x^2})}$ I can't find what is wrong when using L'Hospital's rule on this limit.
Derivative in denominator is $$\frac{-5x^3+4x^2-\sqrt{4-x^2}(6x-4)+12x-8}{2\sqrt{(1-x)(4-x^2)}}$$
and in numerator is $\frac{2\sqrt{2(4-x^2)(1+x)}(2-\sqrt{4-x^2})-4x\sqrt{2(1-x^2)(1-x)}-\sqrt{2+x}(x^2-2x(2-x)-1)-\sqrt{(2+x)(1-x^2)}}{\sqrt{2(4-x^2)(1-x^2)}}$
This gives undefined statement $\frac{0}{0}$.
I tried applying L'Hospital's rule again, but that again gives $\frac{0}{0}$
Limit should be $L=1$
| using Bernoulli
\begin{align}
x \to 0 \hspace{10mm} &
{\color{Red}{(1+ax)^n \approx 1+anx} } \\
& \sqrt{1-x^2} = (1-x^2)^{\frac{1}{2}} \approx 1-\frac{1}{2}x^2 \\
& \sqrt{4-x^2}=\sqrt{4(1-\frac{x^2}{4}})=2(1-\frac{x^2}{4})^{\frac{1}{2}} \approx 2(1-\frac{1}{2} \frac{x^2}{4})=2-\frac{x^2}{4}
\end{align}
so by putting them in limit :
\begin{align}
L &= \lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-\sqrt{1-x^2})-4\sqrt{1-x}(2-\sqrt{4-x^2})}{x\sqrt{1-x}(2-\sqrt{4-x^2})} \\
&= \lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-(1-\frac{1}{2}x^2))-4\sqrt{1-x}(2-(2-\frac{x^2}{4}))}{x\sqrt{1-x}(2-(2-\frac{x^2}{4}))} \\
&= \lim_{x \to 0} \frac{\sqrt{2(2-x)} (\frac{1}{2}x^2)-4\sqrt{1-x}(\frac{x^2}{4})}{x\sqrt{1-x}(\frac{x^2}{4})} \\
&= \lim_{x \to 0} \frac{x^2(\frac{1}{2}\sqrt{2(2-x)}-\sqrt{1-x})}{\frac{x^3}{4}\sqrt{1-x}} \\
&= \lim_{x \to 0} \frac{(\sqrt{\frac{2(2-x)}{4}}-\sqrt{1-x})}{\frac{x}{4}\sqrt{1-x}} =
\lim_{x \to 0} \frac{(\sqrt{1-\frac{x}{2}}-\sqrt{1-x})}{\frac{x}{4}\sqrt{1-x}}
\end{align}
use bernoulli again
\begin{align}
L = \lim_{x \to 0} \frac{(1-\frac{x}{4})-(1-\frac{x}{2})}{\frac{x}{4}\sqrt{1-x}} = \lim_{x \to 0} \frac{\frac{x}{4}}{\frac{x}{4}\sqrt{1-x}}
\end{align}
simplify $\frac{x}{4}$
$$\lim_{x \to 0} \frac{1}{\sqrt{1-x}} = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the value of $(x^{2}+y^{2})$ If $x^{3}-3xy^{2}=10$, $y^{3}-3x^{2}y=30$
Find $$(x^{2}+y^{2})$$
I tried but I got nothing, any help please?
| $$(x+iy)^3 = (x^3-3xy^2) + i(3x^2y-y^3) $$
gives $(x+iy)^3 = 10-30 i $. If we multiply by the conjugate (i.e. take norms):
$$ (x^2+y^2)^3 = 10^2+30^2 = 1000 $$
we immediately get $x^2+y^2=\color{red}{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Complex number and exponent If $z = -1+ i\sqrt{3}$
Is it possible that to prove by using induction $z^{2n}+2^n\cdot z^n+2^{2n}=0$ if $n$ is not multiple of $3$.
I know other way of proving it.
| Notice, $$z=-1+i\sqrt{3}=2\left(-\frac{1}{2}+i\frac{\sqrt 3}{2}\right)$$
$$=2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)$$
Now, we have $$z^{2n}+2^nz^n+2^{2n}=\left(2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)\right)^{2n}+2^n\left(2\left(\cos\frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)\right)^n+2^{2n}$$ $$=2^{2n}\left(\cos\frac{4n\pi}{3}+i\sin \frac{4n\pi}{3}\right)+2^{2n}\left(\cos\frac{2n\pi}{3}+i\sin \frac{2n\pi}{3}\right)+2^{2n}$$ $$=2^{2n}\left[\cos\frac{4n\pi}{3}+i\sin \frac{4n\pi}{3}+\cos\frac{2n\pi}{3}+i\sin \frac{2n\pi}{3}+1\right]$$
$$=2^{2n}\left[\left(\cos\frac{4n\pi}{3}+\cos\frac{2n\pi}{3}\right)+i\left(\sin \frac{4n\pi}{3}+\sin \frac{2n\pi}{3}\right)+1\right]$$
$$=2^{2n}\left[2\cos(n\pi)\cos\left(\frac{n\pi}{3}\right)+2i\sin(n\pi)\cos\left(\frac{n\pi}{3}\right)+1\right]$$
$$=2^{2n}\left[2\cos(n\pi)\cos\left(\frac{n\pi}{3}\right)+1\right]$$
$\forall\quad n\neq 3m$ where $m$ is an integer.
step 1: setting $n=1$ in the above expression, we get $$2^{2}\left[2\cos(\pi)\cos\left(\frac{\pi}{3}\right)+1\right] $$ $$=4\left[2(-1)\frac{1}{2}+1\right]=0$$ Hence, the given equality is true for $n=1$
*Let's assume that it holds for $n=k$ then we get
$$2^{2k}\left[2\cos(k\pi)\cos\left(\frac{k\pi}{3}\right)+1\right]=0$$
$$\iff 2\cos(k\pi)\cos\left(\frac{k\pi}{3}\right)=-\frac{1}{2}$$
3. Substituting $n=k+1$ in the above equality, we get
$$=2^{2(k+1)}\left[2\cos((k+1)\pi)\cos\left(\frac{(k+1)\pi}{3}\right)+1\right]$$
$$=4\cdot2^{2k}\left[\cos\left(k\pi+\pi\right)\cos \left(\frac{k\pi}{3}+\frac{\pi}{3}\right)+1\right]$$
$$=4\cdot2^{2k}\left[-\cos(k\pi)\left\{\cos\frac{k\pi}{3}\cos\frac{\pi}{3}-\sin\frac{k\pi}{3}\sin\frac{\pi}{3}\right\}+1\right]$$
$$=4\cdot2^{2k}\left[-\cos(k\pi)\left\{\cos\frac{k\pi}{3}\frac{1}{2}-\sin\frac{k\pi}{3}\frac{\sqrt 3}{2}\right\}+1\right]$$
$$=4\cdot2^{2k}\left[-\frac{1}{2}\cos(k\pi)\cos\frac{k\pi}{3}+\frac{\sqrt 3}{2}\cos (k\pi)\sin\frac{k\pi}{3}+1\right]$$ substituting value from (2)
$$=4\cdot2^{2k}\left[-\frac{1}{2}\left(-\frac{1}{2}\right)+\frac{\sqrt 3}{2}\cos (k\pi)\sin\frac{k\pi}{3}+1\right]$$ $$=4\cdot2^{2k}\left[\frac{\sqrt 3}{2}\cos (k\pi)\sin\frac{k\pi}{3}+\frac{5}{4}\right]\neq 0$$ Hence, we find from (1), (2) & (3) that the given equality doesn't hold for all positive integers $n\geq 2$
| {
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"url": "https://math.stackexchange.com/questions/1386316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compare $A=\frac{1.0\,000\,004}{(1.0\,000\,006)^2}$ and $ B=\frac{(0.9\,999\,995)^2}{0.9\,999\,998}$ My work:
*
*$1.0\,000\,004 = 1+\frac{4}{10^7}=1+\frac{1}{125\cdot 10^6}$
*$ (1.0\,000\,006)^2=(1+\frac{6}{10^7})^2=(1+\frac{3}{5\cdot 10^6})^2$
*$ (0.9\,999\,995)^2=(\frac{9\,999\,995}{10^7})^2=(\frac{1\,999\,999}{2\cdot10^6})^2$
*$ 0.9\,999\,998=\frac{9.999\,998}{10^7}=\frac{4\,999\,999}{5\cdot10^6}$
The two fractions become:
$A=\dfrac{1+\frac{1}{125\cdot 10^6}}{(1+\frac{3}{5\cdot 10^6})^2}$ and $B=\dfrac{(\frac{1\,999\,999}{2\cdot10^6})^2}{\frac{4\,999\,999}{5\cdot10^6}}$
At this stage, I don't know how to continue.
Any help is appreciated. Thank you.
| $$A=\frac {10^7\times (10^7+4)}{(10^7+6)^2}$$
$$B=\frac {(10^7-5)^2}{10^7\times (10^7-2)}$$
$$\frac AB=\frac {10^{14}(10^7+4)(10^7-2)}{(10^7+6)^2(10^7-5)^2}$$
The numerator is $10^{28}+2\cdot10^{21}-8\cdot10^{14}$
The denominator is $10^{28}+2\cdot 10^{21}-59\cdot10^{14}-60\cdot10^7+900$
It is easy to see from this which is larger.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} ={a}(x-y)$ Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y)$. $a$ is a constant.
I have the final answer, which is $$\frac{ \mathrm{d}y}{ \mathrm{d}x} = \sqrt{\frac{1-y^2}{1-x^2}}.$$
But I've only been able to get till $$\frac{ \mathrm{d}y}{ \mathrm{d}x} = \frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}.\sqrt{\frac{1-y^2}{1-x^2}}.$$
How can the two terms be cancelled off? Thanks.
| Given $$\sqrt{1-x^2}+\sqrt{1-y^2} = a(x-y)\;,$$ Where $a$ is constant.
Now Put $x=\sin \alpha$ and $y = \sin \beta$
So $$\sqrt{1-\sin^2 \alpha}+\sqrt{1-\sin^2 \beta} = a(\sin \alpha-\sin \beta)$$
So $$\displaystyle \cos \alpha+\cos \beta = a(\sin \alpha-\sin \beta)$$
So $$\displaystyle 2\cos \left(\frac{\alpha+\beta}{2}\right)\cdot \cos \left(\frac{\alpha-\beta}{2}\right) = 2a\cdot \cos \left(\frac{\alpha+\beta}{2}\right)\cdot \sin \left(\frac{\alpha-\beta}{2}\right)$$
So we get $$\displaystyle 2\cos \left(\frac{\alpha+\beta}{2}\right)\cdot \left[\cos \left(\frac{\alpha-\beta}{2}\right)-a\cdot \sin \left(\frac{\alpha-\beta}{2}\right)\right] = 0$$
So either $$\displaystyle \cos \left(\frac{\alpha+\beta}{2}\right) = 0\Rightarrow \alpha+\beta = 2n\pi\pm \pi,n\in \mathbb{Z}$$ or $$\displaystyle \tan \left(\frac{\alpha-\beta}{2}\right) = \frac{1}{a}$$
So $\displaystyle \tan \left(\frac{\alpha-\beta}{2}\right) = \frac{1}{a}\Rightarrow \left(\frac{\alpha-\beta}{2}\right) = \tan^{-1}\left(\frac{1}{a}\right)\Rightarrow \alpha-\beta = 2\tan^{-1}\left(\frac{1}{a}\right)$
so $$\displaystyle \sin^{-1}(x)-\sin^{-1}(y) = 2\tan^{-1}\left(\frac{1}{a}\right)$$
Now Differentitae both side w r to $x\;,$ We get $\displaystyle \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx} = 0$
so we get $\displaystyle \frac{dy}{dx} = \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} = \sqrt{\frac{1-y^2}{1-x^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ I was trying to integrate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ and by applying on what i saw on the formula of inverse trigonometric functions, there is formula like $\frac{1}{a}Arcsec\frac{u}{a}$ = $\int \frac{du}{u\sqrt(u^2-a^) }$ so my answer is $\frac{1}{a}Arcsec\frac{x}{a}$ but my friend said it should be something like $arcsin?$
| $$I=\int\dfrac1{x\sqrt{x^2-a^2}}dx=\int\dfrac x{x^2\sqrt{x^2-a^2}}dx$$
Let $\sqrt{x^2-a^2}=u\implies\dfrac x{\sqrt{x^2-a^2}}dx=du$ and $x^2=u^2+a^2$
$$I=\int\dfrac{du}{u^2+a^2}=\dfrac1a\arctan\dfrac ua+K$$
Now we can prove $$\arctan y=\arcsin\dfrac y{\sqrt{1+y^2}}\text{ OR } \arcsin z=\arctan\dfrac z{\sqrt{1-z^2}}$$
and use $\arcsin a+\arccos a=\dfrac\pi2$ for $|a|\le1$
and $\arccos p=\text{arcsec}\dfrac1p$ for $|p|\le1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Quadratic formula does not work If I put the equation: $5x^2-x-4 =0$ in the quadratic formula, than I get
$x = 1$ or $x = \frac{-4}{5}$
but the real zeros are: $x = -1$ or $x = \frac{4}{5}$
Can somebody explain me if the quadratic formula fails or me?
| Plug in the roots the quadratic formula gave you
$5(1)^2 - (1) - 4 $
$=5 - 5 $
$= 0$
And your other root
$5(- \frac{4}{5})^2 - (-\frac{4}{5}) - 4 $
$ = \frac{20}{5} -\frac{(16+4)}{5}$
$= 0$
And you can verify your roots work.
The quadratic formula is correct.
I recommend looking at the proof of the quadratic formula if you never have before to possibly help convince you that it does indeed work and more importantly how it works.
if
$ax^2 + bx + c = 0$
then
$ax^2 + bx = -c$
$a(x^2 + \frac{b}{a}x) = -c$
$a(x^2 + \frac{b}{a}x + [\frac{b}{2a}]^2 - [\frac{b}{2a}]^2) = -c$
$a(x^2 + \frac{b}{a}x + [\frac{b}{2a}]^2) - a[\frac{b}{2a}]^2$ = -c
$a(x^2 + \frac{b}{a}x + [\frac{b}{2a}]^2) = \frac{b}{4a}^2 -c$
$a(x + \frac{b}{2a})^2 = \frac{b^2-4ac}{4a}$
$(x + \frac{b}{2a})^2 = \frac{b^2-4ac}{4a^2}$
$(x + \frac{b}{2a}) = \pm\sqrt{\frac{b^2-4ac}{4a^2}} = \frac{\pm\sqrt{b^2-4ac}}{2a}$
$x=-\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
The quadratic formula is devised by observing and analyzing the relationship between the objects $a$,$b$, and $c$ in the form of $ax^2 + bx + c = 0$.
| {
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"url": "https://math.stackexchange.com/questions/1388309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the trigonometric equation has no solution Prove that the trigonometric equation $$\frac{\sin^3 x}{1-\sin x}+\frac{\cos^3 x}{1-\cos x}=-1$$ has no solution. I tried applying $T2's$ lemma to contradict but could only do so for the first and third quadrant values if $x$. There must be some good proof without the restrictions in the values of $x$. Thanks.
| $$\frac{\sin^3x}{1-\sin x}+\frac{\cos^3x}{1-\cos x}=-1\tag1$$
Let $f(x)$ be the LHS of $(1)$. Then, by AM-GM inequality, we have
$$\begin{align}f(x)&=\frac{(1-\sin x)(-\sin^2x-\sin x-1)+1}{1-\sin x}+\frac{(1-\cos x)(-\cos^2x-\cos x-1)+1}{1\cos x}\\&=-\sin^2x-\sin x-1+\frac{1}{1-\sin x}-\cos^2x-\cos x-1+\frac{1}{1-\cos x}\\&=-1-\sin x-1+\frac{1}{1-\sin x}-\cos x-1+\frac{1}{1-\cos x}\\&=(1-\sin x)+\frac{1}{1-\sin x}+(1-\cos x)+\frac{1}{1-\cos x}-5\\&\ge 2\sqrt{(1-\sin x)\cdot\frac{1}{1-\sin x}}+2\sqrt{(1-\cos x)\cdot\frac{1}{1-\cos x}}-5\\&=2+2-5\\&=-1\end{align}$$
Thus, we have $f(x)\ge -1$.
The equality is attained when $1-\sin x=\frac{1}{1-\sin x}$ and $1-\cos x=\frac{1}{1-\cos x}$, i.e. when $\sin x=\cos x=0$, but there is no such $x$. So, we have $f(x)\gt -1$.
Hence, $(1)$ has no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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using half identities to find exact value of each trigonometric expression a) $\cos{({105}^{°})}$
b) $\sin{(\frac{3\pi}{8})}$
c) $\cot{({67.5}^{°})}$
please do explain how you are able to get the answer as I'm still confused about this topic... Thank you
| Notice, $$\cos 105^\circ=\cos(90^\circ+15^\circ)=-\sin 15^\circ=-\sqrt{\frac{1-\cos 30^\circ}{2}}=-\sqrt{\frac{1-\frac{\sqrt 3}{2}}{2}}=-\sqrt{\frac{2-\sqrt 3}{4}}$$ $$=-\sqrt{\frac{4-2\sqrt 3}{8}}= -\frac{\sqrt 3-1}{2\sqrt 2}=\frac{1-\sqrt 3}{2\sqrt 2}$$ $$\implies \sin\frac{3\pi}{8}=\sin 67.5^\circ=\sin(90^\circ-22.5^\circ)=\cos 22.5^\circ=-\sqrt{\frac{1+\cos 45^\circ}{2}}=\sqrt{\frac{1+\frac{1}{\sqrt2}}{2}}=\sqrt{\frac{\sqrt 2+1}{2\sqrt 2}}=\sqrt{\frac{2+\sqrt 2}{4}}=\frac{\sqrt{2+\sqrt 2}}{2}$$
$$\implies \cot 67.5^\circ=\sqrt{\csc^267.5-1}=\sqrt{\frac{1}{\sin^2 67.5^\circ}-1}=\sqrt{\left(\frac{2}{\sqrt{2+\sqrt 2}}\right)^2-1}=\sqrt{\frac{2-\sqrt2}{2+\sqrt 2}}=\sqrt{\frac{(2-\sqrt2)^2}{4-2}}=\frac{2-\sqrt 2}{\sqrt2}=\sqrt 2-1$$
Thus, we have all the values as follows $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\cos 105^\circ=\frac{1-\sqrt3}{2\sqrt2}}}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\sin \frac{3\pi}{8}=\frac{\sqrt{2+\sqrt2}}{2}}}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\cot 67.5^\circ=\sqrt 2-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388757",
"timestamp": "2023-03-29T00:00:00",
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Differentiate $y= x(4-9x^4)^4$ So the textbook tells me to do the following question but I can't seem to get the right answer. (answer: $(4-9x^4)^3(4-45x^4)$
This is what I've done:
apply the product rule
$u = x$
$u' = 1$
$v = (4-9x^4)^4$
$v' = -144x^3 (4-9x^4)^3$
I used the product rule and got: $(4-9x^4)^3 (-153x^4+ 4)$
What am I doing wrong?
edit: the textbook has the wrong answer, thanks anyways :D
| $y = x\left(4-9x^4\right)^4$
Use the product rule, $\left(uv\right)' = u'v + uv'$, setting:
$u = x$, $v = \left(4-9x^4\right)^4$.
Then,
$u' = 1$
Using the chain rule on $v$, we get:
$v' = 4\left(4-9x^4\right)^3 \cdot \left(-36x^3\right) = -144x^3\left(4-9x^4\right)^3$
Now, $\left(uv\right)' = u'v + uv' = \left(4-9x^4\right)^4 + -144x^4\left(4-9x^4\right)^3 = \left(4-9x^4\right)^3\left(4-9x^4-144x^4\right)$.
The answer is therefore $\left(4-9x^4\right)^3\left(4-153x^4\right)$, so you were correct, assuming the problem was transcribed correctly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve this trigonometric equation $ \sin2x-\sqrt3\cos2x=2$ Solve equation: $$ \sin2x-\sqrt3\cos2x=2$$
I tried dividing both sides with $\cos2x$ but then I win $\frac{2}{\cos2x}$.
| Notice, we have $$\sin 2x-\sqrt 3\cos 2x=2$$ $$\frac{1}{2}\sin 2x-\frac{\sqrt 3}{2}\cos 2x=1$$ $$\sin 2x\cos\frac{\pi}{3}-\cos 2x\sin \frac{\pi}{3}=1$$
$$\sin\left(2x-\frac{\pi}{3}\right)=1=\sin \frac{\pi}{2}$$
Now, writing the general solution for $x$, we get $$2x-\frac{\pi}{3}=2n\pi+\frac{\pi}{2}$$
$$2x=2n\pi+\frac{\pi}{2}+\frac{\pi}{3}=2n\pi+\frac{5\pi}{6}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=n\pi+\frac{5\pi}{12}}}$$
Where, $n$ is any integer
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Explanation of a method to compute $\sum_{k \le n} k^2$ I was searching for methods to compute $\sum_{k\le n} k^2$. I stumbled across this (which is an answer provided by Gareth Rees to this question).
"..Represent $k^2$ in terms of falling powers (easy by inspection in this case, but you can use Stirling subset numbers to convert): $$ k^2 = k^{\underline 2} + k^{\underline 1}$$
Sums of falling powers are easy, just like integration of ordinary powers, except for the treatment of limits: $$ \sum_{k=1}^n k^{\underline 2} + k^{\underline 1} = \bigg({1\over 3}k^{\underline 3} + {1\over 2}k^{\underline 2}\bigg)\ \bigg|^{n+1}_0$$
And then convert back into ordinary powers (by expansion, or using signed Stirling cycle numbers): $$ {1\over 3}((n+1)^3 - 3(n+1)^2 + 2(n+1)) + {1\over 2}((n+1)^2 - (n+1))..."$$
Could someone explain what is going on in this answer and provide some useful and explanatory links?
Thank you.
| Instead of falling powers I might use Binomial Coefficients:
$$ k^2 = 2\, \frac{k(k-1)}{2} + k = 2\binom{k}{2} + \binom{k}{1}$$
Now if we add over all $k$ we can get other binomial coefficients:
$$ \sum_{k \leq n} \binom{k}{2} = \binom{n+1}{3} \text{ and } \sum_{k \leq n} \binom{k}{1} = \binom{n+1}{2}$$
This feels just like the forumla $\int x^k = \tfrac{1}{k+1} x^{k+1} + C$ and finally
$$ \sum_{k \leq n} k^2
= 2\sum_{k \leq n} \binom{n}{2} + \sum_{k \leq n} \binom{n}{1}
= 2\binom{n+1}{3} + \binom{n+1}{2}
= 2\cdot \frac{(n+1)n(n-1)}{6} + \frac{(n+1)n}{2}$$
This recovers the correct formula since:
$$ 2\cdot \frac{(n+1)n(n-1)}{6} + \frac{(n+1)n}{2} = \frac{n(n+1)(2n+1)}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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sequence is cauchy how to prove that the recursive sequence $a_0\ge 0$, $a_{n+1}=\frac{3(1+a_n)}{3+a_n}$ is a cauchy sequence? The sequence seems to be bounded and if the sequence is monotonic increasing (I still dont know if it is..), it is convergent, then the sequence must be cauchy. But how to prove with the definition of cauchy sequence if the sequence is cauchy? If I try to start with $|a_{n+1}-a_n|=...$ I don't do useful calculations.. Regards
| Define $f(x) = \dfrac{3 + 3x}{3 + x}, x \geq 0$.
It is easily seen that
$$f'(x) = \frac{3(3 + x) - (3 + 3x)}{(3 + x)^2} = \frac{6}{(3 + x)^2} \in \left(0, \frac{2}{3}\right]$$
for all $x \geq 0$.
It then follows by the mean value theorem that, for every positive integer $n$:
\begin{align*}
& \left|a_{n + 1} - a_n\right| = \left|f(a_n) - f(a_{n - 1})\right| \\
\leq & \frac{2}{3}\left|a_n - a_{n - 1}\right| \\
\leq & \cdots \\
\leq & \left(\frac{2}{3}\right)^{n}\left|a_1 - a_0\right|
\end{align*}
Therefore, for every positive integer $p$, we have
\begin{align*}
& \left|a_{n + p} - a_n\right| \leq \sum_{k = 1}^p \left|a_{n + k} - a_{n + k - 1}\right| \\
\leq & \sum_{k = 1}^p \left(\frac{2}{3}\right)^{n + k - 1}\left|a_1 - a_0\right| \\
\leq &\left|a_1 - a_0\right|\left(\frac{2}{3}\right)^n\sum_{k = 0}^\infty \left(\frac{2}{3}\right)^k \\
= & 3\left|a_1 - a_0\right|\left(\frac{2}{3}\right)^n \\
\to & 0
\end{align*}
as $n \to \infty$, showing that $\{a_n\}$ is a Cauchy sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $\displaystyle\lim_{x\to 0}\tfrac{\sqrt{x+25}-5} {\sqrt{x+16}-4}$ \begin{eqnarray}
\\&\lim_{x\to 0}\frac{\sqrt{x+25}-5} {\sqrt{x+16}-4}
\end{eqnarray}
Undefined limit
\begin{eqnarray}
\frac{0} {0}
\end{eqnarray}
| METHOD 1:
One quick way is to write
$$\sqrt{x+a^2}=|a|\sqrt{1+x/a^2}=|a|\,\left(1+\frac12 \left(\frac{x}{a^2}\right)+O\left(x\right)^2\right)$$
Then, we have
$$\frac{\sqrt{x+25}-5}{\sqrt{x+16}-4}=\frac45 +O(x)\to \frac45$$
METHOD 2:
Using L'Hospital's Rule gives
$$\lim_{x\to 0}\frac{\sqrt{x+25}-5}{\sqrt{x+16}-4}=\lim_{x\to 0}\frac{\frac{1}{2\sqrt{x+25}}}{\frac{1}{2\sqrt{x+16}}}=\frac45$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve trigonometric inequality $ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $ Solve this trigonometric inequality:
$$ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $$
My steps:
$$ \cos x \cos 2x - \sin x \sin 2x < - \sin 6x $$
$$ \cos 3x < \sin (-6x)$$
$$ \cos 3x < \cos (\frac{\pi}{2}+6x) $$
From this we get:
$$ 3x > \dfrac{\pi}{2}+6x+2k\pi$$
$$ -3x > \dfrac{\pi}{2}+2k\pi$$
$$ x < -\dfrac{\pi}{6}+ \dfrac{2k\pi}{3}$$
and
$$ 3x < -\dfrac{\pi}{2} - 6x + 2k\pi$$
$$ 9x < -\dfrac{\pi}{2} + 2k\pi$$
$$ x < - \dfrac{\pi}{18} + \dfrac{2k\pi}{9} $$
as you can see the solution is not correct
| $$\cos { \left( 3x \right) < } -\sin { \left( 6x \right) } \\ \cos { \left( 3x \right) < } -2\sin { \left( 3x \right) \cos { \left( 3x \right) } } \\ \cos { \left( 3x \right) } \left( 1+2\sin { \left( 3x \right) } \right) <0$$
$$1.\cos { \left( 3x \right) } >0\quad and\quad 1+2\sin { \left( 3x \right) } <0\\ 2.\cos { \left( 3x \right) } <0\quad and\quad 1+2\sin { \left( 3x \right) } >0\\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all integer values for $m$ such that $x_1,x_2\in\mathbb{Z}$ We have $(m+1)x^2-(2m+1)x-2m=0$ where $m\in\mathbb{R}\backslash\left\{-1\right\}$. We need to find all integer values for $m$ such that both roots of the equation are integers.
Here is all my steps:
*
*$x_{1,2}=\frac{(2m+1)\pm\sqrt{12m^2+12m+1}}{2(m+1)}\in\mathbb{Z}$
$\Rightarrow 2(m+1)\:|\:(2m+1)\pm\sqrt{12m^2+12m+1}$
$\Rightarrow 2(2m+2)+1\pm\sqrt{12(2m+2)^2+12(2m+2)+1}=0$
*$x_1=2(2m+2)+1+\sqrt{12(2m+2)^2+12(2m+2)+1}=0$
$\Rightarrow \sqrt{12(2m+2)^2+12(2m+2)+1}=-4m-5$
$\Rightarrow 12(2m+2)^2+12(2m+2)+1=(-4m-5)^2$
$\Rightarrow 12(4m^2+8m+4)+24m+25=16m^2-40m+25$
$\Rightarrow 48m^2+96m+48+24m+25=16m^2-40m+25$
$\Rightarrow 32m^2+160m+48=0$|:16
$\Rightarrow 2m^2+10m+3=0$ but $\sqrt{\Delta}\notin\mathbb{N}$
Now we have to verify for $x_2$ like above:
*
*$x_2=48m^2+96m+48+24m+25=16m^2+40m+25$
$\Rightarrow 32m^2+80m+48=0|:16$
$\Rightarrow 2m^2+5m+3=0 \Rightarrow \sqrt{\Delta}=1$
$\Rightarrow m_1=-1\in\mathbb{Z}, m_2\notin\mathbb{Z}$
But as the author says $m\in\mathbb{R}\backslash\left\{-1\right\}$
| I'm sorry, but I find as the only answers $m=0$ or $m=-2$ or $m=-1$ (though in the latter case there are not two solutions).
To see this, note that if $m=-1$, it' not a quadratic equation, and the solution is $x=-2$.
If $m\neq -1$, let $x_1, x_2$ be the roots; use Vieta's relations:
$$x_1x_2=-\frac{2m}{m+1},\quad x_1+x_2=\frac{2m+1}{m+1}.$$
If $m, x_1,x_2$ are integers, this implies $m+1$ divides both $2m$ and $2m+1$, hence it divides $1$, whence $m+1=\pm 1\iff m=0,-2$.
For $m=0$, the equation is $\;x^2-x=0$, which has roots $0,1$.
For $m=-2$, the equation is $\;-x^2+3x+4=0$, which has roots $-1,4$.
| {
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"url": "https://math.stackexchange.com/questions/1394220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find $\int \frac{5x^4+4x^5}{(x^5+x+1)^2}$ $\displaystyle \int \frac{5x^4+4x^5}{(x^5+x+1)^2}$
Since in the denominator of the integrand,$(x^5+x+1)^2$ is there.So the answer must be in the form $\displaystyle \frac{f(x)}{(x^5+x+1)}$, but i could not figure out what $f(x)$ i should suppose.I could not imagine any other way out.Please help...
| Let $$\displaystyle I = \int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \int\frac{\left[\left(5x^4+1\right)+(4x^5-1)\right]}{(x^5+x+1)^2}dx$$
So $$\displaystyle I = \underbrace{\int\frac{5x^4+1}{(x^5+x+1)^2}dx}_{J}+\underbrace{\int \frac{4x^5-1}{(x^5+x+1)^2}dx}_{K}$$
So for Calculation of
$$\displaystyle J= \int \frac{5x^4+1}{(x^5+x+1)^2}dx$$
Let $(x^5+x+1) = t\;,$ Then $(5x^4+1)dx = dt$
So Integral $$\displaystyle J = \int\frac{1}{t^2}dt =-\frac{1}{t} = -\left(\frac{1}{x^5+x+1}\right)$$
Now for Calculation of
$$\displaystyle K= \int\frac{4x^5-1}{(x^5+x+1)^2}dx = \int\frac{4x^5-1}{x^2\cdot \left(x^4+1+x^{-1}\right)^2}dx = \int \frac{\left(4x^3-x^{-2}\right)}{\left(x^4+1+x^{-1}\right)^2}dx$$
Now Put $\displaystyle \left(x^4+1+x^{-1}\right) = u\;,$ Then $\left(4x^3-x^{-2}\right)dx = du$
So Integral $$\displaystyle K = \int\frac{1}{u^2}du = -\frac{1}{u} = -\frac{1}{x^4+1+x^{-1}} = -\left(\frac{x}{x^5+x+1}\right)$$
So we get $$\displaystyle I = J+K = -\frac{1}{x^5+x+1}-\frac{x}{x^5+x+1} = -\left(\frac{x+1}{x^5+x+1}\right)+\mathcal{C}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Use calculus to find the area bounded by the circle $x^2+y^2-2x-2y-23=0$ and the pair of lines $x^2+2xy+y^2-7x-7y+12=0$ Use calculus to find the area bounded by the circle $x^2+y^2-2x-2y-23=0$ and the pair of lines $x^2+2xy+y^2-7x-7y+12=0$.
I tried to solve the two equations by subtracting them.
$2xy+35=5x+5y$,on further solving it is getting too complicated and not giving solutions.Hence could not move ahead.Please help me.
| The lines are $x+y=3$ and $x+y=4$. These are parallel lines. The equation of the circle is $(x-1)^2+(y-1)^2=25$. Hence you may rotate the lines so that they are parallel to the $x$-axis. You may also place the origin at the center of the circle. The distance between these lines is $1/\sqrt2$ and the distance between $x+y=3$ from the center of the circle is $1/\sqrt2$. Hence the area is $$S=2\int_{1/\sqrt2}^{2/\sqrt2}\sqrt{25-y^2}dy=-\frac{7}{2}+\sqrt{46}-25 \arcsin\left(\frac{1}{5 \sqrt{2}}\right)+25 \arcsin \left(\frac{\sqrt{2}}{5}\right)$$
Before rotation:
After rotation and centering:
| {
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"timestamp": "2023-03-29T00:00:00",
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$\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$
How can I show that $\arctan (x) + \arctan(1/x) =\frac{\pi}{2}$?
I tried to let $x = \tan(u)$. Then
$$ \arctan(\tan(u)) + \arctan(\tan(\frac{\pi}{2} - x)) = \frac{\pi}{2}$$
but it does not seem useful.
I'd appreciate most a proof that gives intuition and / or uses geometric
insight.
| $\arctan(a)+\arctan(b)
=\arctan(\frac{a+b}{1-ab})
$.
Therefore
$\arctan(x)+\arctan(1/x)
=\arctan(\frac{x+1/x}{1-1})
=\arctan(\frac{x+1/x}{0})
=\pi/2
$.
If this bothers you,
$\begin{array}\\
\arctan(x-c)+\arctan(1/x)
&=\arctan(\frac{x-c+1/x}{1-(x-c)/x)})\\
&=\arctan(\frac{x-c+1/x}{c/x})\\
&=\arctan(\frac{x^2-cx+1}{c})\\
&=\arctan(\frac{x^2+1}{c}-x)\\
&\to \pi/2
\quad\text{ as } c \to 0\\
\end{array}
$.
| {
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"url": "https://math.stackexchange.com/questions/1403539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why does the diophantine equation $x^2+x+1=7^y$ have no integer solutions? This following Problem is from Pell equation chapters exercise
Let $y>3$ positive integer numbers, show that following diophantine equation
$$x^2+x+1=7^y\tag{1}$$ has no integer solutions.
I tried write the equation
$$(2x+1)^2+3=4\cdot 7^y$$
if $y=2k$ then we have
$$(2\cdot 7^k+2x+1)(2\cdot 7^k-2x-1)=3$$
this case has no integer.
But $y$ is odd number, How to prove equation (1) has no integer solutions for $x,y (y>3)$? Any help would be appreciated.
| A sketch of my thoughts:
Let $\nu_7(n)=\max\{m\in\mathbb{N}:7^m\mid n\}$. A good idea may be to prove that if $\nu_7(x^2+x+1)=\nu_7(x^3-1)-\nu_7(x-1)=y>3$ then $x$ has to be large, say $x\geq 7^{y-1}$. In such a case, however, $x^2+x+1$ is too big to be just $7^y$ and it must have some other prime factor.
By this way, the problem boils down to finding (or, at least, lower-bounding) the two elements of order three in $G=\left(\mathbb{Z}_{/(7^y\mathbb{Z})}\right)^*$, that is a cyclic group with $o(G)=6\cdot 7^{y-1}$.
If $y=1$, that elements are $2$ and $4$. If $y=2$, that elements are $18$ and $30$. If $y=3$, that elements are $18$ and $18^2$. In general, we may compute such elements by solving $z^2+3\equiv 0\pmod{7^y}$. If $y=4$, such elements are $1047$ and $1353$. If $y=5$, such elements are $1353$ and $15453$.
We just need to find a pattern, or an alternative reason for which $4\cdot 7^y-3$ cannot be a square for $y>3$. Obviously if $y$ is even $4\cdot 7^y-3$ is too close to a square to be a square itself, so we may assume that $y$ is odd.
| {
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"url": "https://math.stackexchange.com/questions/1403641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$ For $a\geq2$,if the value of the definite integral $\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$.Find the value of $a$.
Substituting $x-\frac{1}{x}=t$ does not seem to work here,what is the good substitution to make it integrable?Thanks in advance.
| Let $$\displaystyle I = \int_{0}^{\infty}\frac{1}{a^2+\left(x-\frac{1}{x}\right)^2}dx = \int_{0}^{\infty}\frac{1}{x^2+\frac{1}{x^2}+(a^2-2)}dx\;,$$ Where $a^2-2 = k \geq 0$
So $$\displaystyle I = \int_{0}^{\infty}\frac{x^2}{x^4+kx^2+1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{(x^2+1)+(x^2-1)}{x^4+kx^2+1}dx$$
So we get $$\displaystyle I = \underbrace{\int_{0}^{\infty}\frac{x^2+1}{x^4+kx^2+1}dx}_{I}+\underbrace{\int_{0}^{\infty}\frac{x^2-1}{x^4+kx^2+1}dx}_{J}$$
Now $$\displaystyle I = \int_{0}^{\infty}\frac{x^2+1}{x^4+kx^2+1}dx = \int_{0}^{\infty}\frac{1+x^{-2}}{x^2+x^{-2}+k} = \int_{0}^{\infty}\frac{(1+x^{-2})}{\left(x-\frac{1}{x}\right)^2+\left(\sqrt{k+2}\right)^2}dx$$
Now Substute $$\displaystyle \left(x-\frac{1}{x}\right) = t\;,$$ Then $\left(1+\frac{1}{x^2}\right)dx = dt$ and Changing Limit, We get
and also Put $a^2-2 = k\Rightarrow a^2=k+2.$
$$\displaystyle J = \int_{-\infty}^{\infty}\frac{1}{t^2+a^2}dt = 2\int_{0}^{\infty}\frac{1}{t^2+a^2}dt = \frac{\pi}{2a}$$
Similarly for $$\displaystyle J = \int_{0}^{\infty}\frac{x^2-1}{x^4+kx^2+1}dx = $$
Put $\displaystyle x= \frac{1}{u}\;,$ Then $\displaystyle dx = -\frac{1}{u^2}du$ and Changing Limit, We get
$$\displaystyle K = -\int_{\infty}^{0}\frac{1-u^2}{u^4+ku^2+1}du = \int_{0}^{\infty}\frac{1-u^2}{u^4+ku^2+1}du = -\int_{0}^{\infty}\frac{u^2-1}{u^4+ku^2+1}du $$
so we get $$\displaystyle K = -\int_{0}^{\infty}\frac{x^2-1}{x^4+kx^2+1}dx=-K $$
above we have used the formulae
$\displaystyle \int_{a}^{b} f(x)dx = -\int_{b}^{a} f(x)dx $ and $\displaystyle \int_{a}^{b} f(t)dt = \int_{a}^{b} f(x)dx $
So we get $K=0$
So $$\displaystyle I = J+K = \frac{\pi}{2a}+0 = \frac{\pi}{5050}$$
so we get $\displaystyle a= 2525$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$ $(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$ Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$
$(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$
I tried to prove it but my answer is not correct.
For first part,As $0\leq x\leq2\Rightarrow 2\leq x^2+2\leq6\Rightarrow\frac{1}{6}\leq\frac{1}{x^2+2}\leq\frac{1}{2}\Rightarrow\frac{1}{6}\int_{0}^{2}1dx\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{1}{2}\int_{0}^{2}1dx$$\Rightarrow\frac{1}{3}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq1$
In the second part,$0\leq x\leq2\Rightarrow -2<x^2-x<4\Rightarrow e^{-2}<e^{x^2-x}<e^4\Rightarrow2e^{-2}<\int_{0}^{2}e^{x^2-x}dx<2e^4$
Where have i gone wrong?What is the correct method to solve it.
| For the second problem, we note that
$$-\frac14\le x^2-x\le2$$for $x\in[0,2]$. Note that we can easily find the minimum by taking the derivative and setting it to $0$, while we merely check the endpoints to find the maximum.
Inasmuch as the exponential function is monotonically increasing, then
$$e^{-\frac14}\le e^{x^2-x}<e^2 \tag 1$$
Then, we see immediately that
$$2\,e^{-\frac14}\le \int_0^2 e^{x^2-x}\,dx \le 2\,e^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Rationalization of a fraction Can someone please explain how i can rationalize the fraction $\frac{2 +\sqrt{3}}{(2-\sqrt{3})^3}$ so that i obtain an answer that has an exponent of 6. I basically need to compare this value with $\dfrac{(10)^6}{(3)^6}$ to see which is greater. Thanks in advance for helping out!
| \begin{align*}
\frac{2+\sqrt{3}}{(2-\sqrt{3})^3}&=\frac{2+\sqrt{3}}{(2-\sqrt{3})^3}\cdot\frac{(2+\sqrt{3})^3}{(2+\sqrt{3})^3}\\
&=\frac{(2+\sqrt{3})^4}{(2^2-3)^3}\\
&=\frac{(2+\sqrt{3})^4}{1}\\
&=(2+\sqrt{3})^4
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are the Eigenvectors in the following matrix? I have the matrix A:
\begin{bmatrix}
4 & 2 & 2\\
2 & 4 & 2\\
2 & 2 & 4\\
\end{bmatrix}
I found $\lambda I_n - A$ to be:
\begin{bmatrix}
(\lambda -4) & -2 & -2\\
-2 & (\lambda -4) & -2\\
-2 & -2 & (\lambda -4)\\
\end{bmatrix}
Applying Sarrus' Rule, and with help in a previous question, I determined the Eigenvalues to be $\lambda = 8$ or $2$ from the polynomial $\lambda^3 -12\lambda^2 +36\lambda -32 = 0$
If I now get the matrix
\begin{bmatrix}
-2 & -2 & -2\\
-2 & -2 & -2\\
-2 & -2 & -2\\
\end{bmatrix} when verifying $\lambda = 2$, does this mean that $\lambda = 2$ is not a valid Eigenvalue? Or does it mean that $(\lambda I_n -A)v = 0$
Vector v = 1? And creates an Eigenspace from this?
Checking for $\lambda = 8$ gives me:
\begin{bmatrix}
6 & -2 & -2\\
-2 & 6 & -2\\
-2 & -2 & 6\\
\end{bmatrix}
How do I get the Eigenvectors for this?
| An easier way to check for eigenvalues is to factorize
$$
x^3 - 12x^2 + 36x -32 = \left(x-8\right)\left(x-2\right)^2
$$
Then to get the eigenvectors, input the eigenvalues one by one to
$$
(\lambda I - A) \cdot \begin{bmatrix}S_1\\S_2\\S_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}
$$
and solve for $S_1$, $S_2$ and $S_3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$ The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$,(where $[]$ denotes the greatest integer function) belongs to the interval $(a,\frac{b}{c}]$,where $a,b,c\in N$ and $\frac{b}{c}$is in its lowest form.Find the value of $a+b+c+abc$.
I tried to solve this question.Let $\left[\frac{3}{x}\right]=t$,therefore $\left[\frac{4}{x}\right]=5-t$.
$t\leq \frac{3}{x}<t+1$......(i) and $5-t\leq \frac{4}{x}<6-t$.....(ii)
Add (i) and (ii),we get $5\leq\frac{3}{x}+\frac{4}{x}<7$
$5\leq\frac{7}{x}<7$
Then i solved $5\leq\frac{7}{x}$ and $\frac{7}{x}<7$
and took the intersection of the two solution sets and got $x\in (1,\frac{7}{5}]$,so my $a+b+c+abc=48$ but the answer is given to be 20.What is wrong in my method.Please help me.
| Here are some hints.
Note that as $x$ increases, the sum can get no larger - it is decreasing but not strictly so, because it is sometimes constant. Also with $x=1$ the sum is equal to $7$ so we must have $x \gt 1$ for a sum as low as $5$.
Now if $x\gt 1$ we have immediately that $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]\le 2+3=5$ so we must have equality with $\left[\frac{3}{x}\right]=2$ and $\left[\frac{4}{x}\right]=3$
You should be able to complete things from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trying to solve the trig equation $\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$ The equation is
$$\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$$
My solution goes like this
$$
\begin{cases}
3+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x) \\
\frac{\sin(x)}{\sqrt 3}+3\cos(x) \ge 0
\end{cases}
$$
$$3(\sin^2(x)+\cos^2(x))+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x)$$
$$2\cos^2(x)+\frac{\sin^2(x)}{3}-3\sin^2(x)+\frac{6}{\sqrt 3}\sin(x)\cos(x)=0$$
I multiply by 3 and divide by $\cos^2(x)$:
$$8\tan^2(x)-6\sqrt{3}\tan(x)-6=0$$
Let $t=\tan(x)$, then
$$4t^2-3\sqrt{3}t-3=0$$
$$t_1=\frac{7\sqrt 3}{2}$$
$$t_2=\frac{11\sqrt 3}{4}$$
The solutions for $x$ would be arc-tangents of these values.
But the textbook's answer is
$$\color{green}{x_1=\frac{\pi}{3}+2\pi n; x_2=-\arctan\left(\frac{\sqrt 3}{4}\right)+2\pi n}$$
Where did I make a mistake?
P.S. From the textbook
| You made a mistake in factorising the following equation:
$$ 4t^2 -3\sqrt3 t -3 = 0$$
$$ t = \frac{3\sqrt 3 \pm \sqrt{75}}{8}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Finding the maximum of a function on $ \Bbb{S}^{7} $. I'm trying to find the maximum of the function
$$2 a^2 h+\sqrt{3} a d f+\sqrt{3} a e g+2 b^2 h-\sqrt{3} b d g+\sqrt{3} b e f\\+2 c^2
h+\sqrt{3} c d^2+\sqrt{3} c e^2-\sqrt{3} c f^2-\sqrt{3} c g^2\\-d^2 h-e^2 h-f^2
h-g^2 h-2 h^3$$
on the sphere $$a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2=1 \text{.}$$
I suspect it is $2$, but I can't come up with a proof. Lagrange multipliers don't work, since the resulting equations don't have an analytic solution.
The value $2$ is attained for $h=-1$.
My current idea is that terms of the form $x^2y$ can't be bigger than $\frac{2}{3\sqrt{3}}$ and terms of the form $xyz$ can't be bigger than $\frac{1}{3\sqrt{3}}$, but I'm not sure this helps.
I would also be satisfied with a slightly higher bound than $2$.
Additional information:
The function is proportional to $\sum_{ijk=1}^8 d_{ijk}p^i p^j p^k$ (the constraint is just $\sum_i (p^i)^2 = 1$), where $ \tau_i\tau_j = -\frac{2}{3}\delta_{ij} +(f_{ijk}-i\, d_{ijk})\tau_k$ holds for the generators $\tau_i$ of $SU(3)$ with $[\tau_i,\tau_j]=f_{ijk} \tau_k$ and the normalization condition $\mathrm{Tr}(\tau_i\tau_j) = -2\delta_{ij}$ is imposed.
| (Update: I just realize that this is very similar to @Michael 's earlier solution.)
Your expression ($=:\Phi_0$) can be simplified using the following measures:
$$a=r\cos\alpha,\quad b=r\sin\alpha,\quad d=s\cos\beta,\quad e=s\sin\beta,\quad f=t\cos\gamma,\quad g=t\sin\gamma\ .$$
After some computation this results in
$$\Phi_1=2c^2h+\sqrt{3}c(s^2-t^2)-h(2h^2-2r^2+s^2+t^2)+\sqrt{3}rst\cos(\alpha-\beta+\gamma)\ .$$
Therefore we are left with the task of maximizing
$$\Phi_2:=2c^2h+\sqrt{3}c(s^2-t^2)-h(2h^2-2r^2+s^2+t^2)+\sqrt{3}rst$$
under the constraints
$$r\geq0, \quad s\geq0,\quad t\geq0,\quad r^2+s^2+t^2+c^2+h^2=1\ .$$
Using the main constraint we can rewrite $\Phi_2$ as
$$\Phi_3=-h^3+h\bigl(-1+3(c^2+r^2)\bigr)+\sqrt{3}\bigl(rst+c(s^2-t^2)\bigr)\ .$$
This suggests writing
$$c=x\cos\phi,\quad r=x\sin\phi\ ,$$
which then leads to
$$\Phi_4=-h(1+h^2-3x^2)+\sqrt{3}x\bigl((s^2-t^2) \cos\phi+st\sin\phi\bigr)\ ,\tag{1}$$
whereby the constraints now are
$$x\geq0, \quad s\geq0,\quad t\geq0,\quad x^2+s^2+t^2+h^2=1\ .$$
Maximizing $(1)$ over $\phi$ brings us to
$$\Phi_5=-h(1+h^2-3x^2)+\sqrt{3}x\sqrt{(s^2-t^2)^2+s^2 t^2}\ .$$
For given sum $s^2+t^2$ the square root on the right hand side is maximal when one of the variables is $=0$ (check this!), so that we arrive at
$$\Phi_6=-h(1+h^2-3x^2)+\sqrt{3}x s^2\tag{2}$$
under the constraints
$$x\geq0,\quad s\geq0,\quad x^2+s^2+h^2=1\ .$$
Eliminating $s^2$ from $(2)$ using the constraint leaves us with
$$\Phi_7=-h(1+h^2-3x^2)+\sqrt{3}x(1-h^2-x^2)\qquad(-1\leq h\leq 1, \ 0\leq x\leq\sqrt{1-h^2})\ .$$
This could be analyzed further, but a 3D plot of $\Phi_7$ as a function of $h$ and $x$ reveils that $\Phi_7$ indeed takes the maximal value $2$ when $h=-1$.
| {
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"url": "https://math.stackexchange.com/questions/1413299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Summation of the reciprocals of the product of consecutive integers It is well known that there is a closed formula for:
$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{(n)(n + 1)}$$
And likewise for:
$$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \cdots + \frac{1}{(n)(n + 1)(n+2)}$$
I am wondering if there is a closed formula for:
$$f(n, k) = \sum_{i=1}^n \frac{1}{\prod_{j=0}^k (i + j)}$$
Note that putting $k = 1$ and $k = 2$ in the above function yields the above two series.
| Notice
$$\begin{align}
\frac{1}{\prod\limits_{j=0}^{k}(i+j)} = \frac{1}{k}\left(\frac{(i+k)-i}{\prod\limits_{j=0}^{k}(i+j)}\right)
&= \frac{1}{k}\left[\frac{1}{\prod\limits_{j=0}^{k-1}(i+j)}-\frac{1}{\prod\limits_{j=1}^{k}(i+j)}\right]\\
&= \frac{1}{k}\left[\frac{1}{\prod\limits_{j=0}^{k-1}(i+j)}-\frac{1}{\prod\limits_{j=0}^{k-1}((i+1)+j)}\right]\end{align}$$
The sum can be recasted as a telesoping one. This leads to
$$\sum_{i=1}^n\frac{1}{\prod\limits_{j=0}^{k}(i+j)}
= \frac{1}{k}\left[\frac{1}{\prod\limits_{j=0}^{k-1}(1+j)}-\frac{1}{\prod\limits_{j=0}^{k-1}((n+1)+j)}\right] = \frac{1}{k}\left[\frac{1}{k!} - \frac{n!}{(n+k)!}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How do you factor $\frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}=$? \begin{align}
& \frac{2x^2-x-1}{x^2-9} \cdot \frac{x+3}{2x+1}= \frac{2x^2-x-1}{(x-3)(x+3)} \cdot \frac{x+3}{2x+1} \\[10pt]
= {} & \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)} \cdot \frac{1}{2x+1}= \frac{2x^2-x-1}{(x-3)(2x+1)}
\end{align}
Then I use quadratic formula on numerator to factor it :
$a=2,b=-1,c=-1$
$$=\frac{2(x+2)(x-\frac{5}{2})}{(x-3)(2x+1)}$$
But apparently this can be factored further. What else can I do?
| HINT:
$2x^2-x-1=(x-1)(2x+1)$
$x^2-9=(x+3)(x-3)$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Evaluation of $ \int\frac{(1+x)\left[(1-x+x^2)\cdot (1+x+x^2)+x^2\right]}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$
Evaluation of $\displaystyle \int\frac{(1+x)\left[(1-x+x^2)\cdot (1+x+x^2)+x^2\right]}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$
$\bf{My\; Try::}$ We can write $$1+2x+3x^2+4x^3+3x^4+2x^5+x^6 $$
$$= (1+x)+(x+x^2)+2(x^2+x^3)+2(x^3+x^4)+(x^4+x^5)+(x^5+x^6)$$
$$ = (1+x)\left[1+x+2x^2+2x^3+x^4+x^5\right]$$
So Integral $$\displaystyle I = \int\frac{[(1-x+x^2)\cdot (1+x+x^2)+x^2]}{1+x+2x^2+2x^3+x^4+x^5}dx$$
So $$\displaystyle I = \int\frac{(1+x^2)^2}{1+x+2x^2+2x^3+x^4+x^5}dx$$
Now How can I solve it, Help me
Thanks.
| Well, you're really almost done. All you have to do is realize that the denominator factors more: $$1+x+2x^2+2x^3+x^4+x^5=(1+x)(1+x^2)^2$$
This of course gives you
$$I=\int\frac{1}{1+x}\,dx=\ln|1+x|+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sin x+2x\ge\frac{3x(x+1)}{\pi}$ for all $x\in [0,\frac{\pi}{2}]$ Question: Prove that $\sin x+2x\ge\frac{3x(x+1)}{\pi}$ for all $x\in [0,\frac{\pi}{2}]$
How did $\pi$ come in the expression?
This is how I tried to solve.
$\sin x\le x$ so $\sin 2x\le 2x$
$\sin x+2x\le 3x$.
Applying AM GM inequality: $\sin x+2x\ge 2\sqrt{2x\sin x}$
$3x\ge 2\sqrt{2x\sin x}$
This how far I can go.
| Just expanding mastrok's answer, over $I=\left(0,\frac{\pi}{2}\right)$ we have:
$$ \frac{d^2}{dx^2}\left(\sin x+2x\right) = -\sin x<0 \tag{1}$$
Hence $f(x)=\sin(x)+2x$ is a concave function, and since:
$$ \frac{d^2}{dx^2}\left(\frac{3}{\pi}x(x+1)\right)=\frac{6}{\pi}>0\tag{2}$$
$g(x)=\frac{3}{\pi}x(x+1)$ is a convex function. We have $f(0)=g(0)=0$ and since $\pi>2$:
$$ f\left(\frac{\pi}{2}\right)=\pi + 1>\frac{3}{2}+\frac{3\pi}{4}=g\left(\frac{\pi}{2}\right)\tag{3}$$
proving that:
$$ \forall x\in I,\qquad f(x)>\left(2+\frac{2}{\pi}\right)x>\left(\frac{3}{2}+\frac{3}{\pi}\right)x>g(x).\tag{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of a trigonometric rational expression How to evaluate the limit of this expression?
$$\lim_{x\to\infty} \frac{\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}$$
I managed to simplify the denominator into a sinus form by the Pythagorean formula, and also modified the argument of the sinus in the numerator by dividing with a conjugate which I think is necessary, into this final form:
$$\lim_{x\to\infty} \frac{\sin^2\left(\frac{1}{\sqrt{x+1}+\sqrt{x}}\right)}{\sin^2\frac{1}{x}}$$
And that's where I got stuck. So far I've tried to come up with some solution to eliminate the sinus from the denominator, without success. I'm sure the solution is pretty obvious but I don't quite know how to modify the trigonometric functions in this fashion, so any help would be most appreciated.
| Observe
\begin{align}
\lim_{x\to\infty}\frac{\sin^2\left(\sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}&=\lim_{x\to\infty}\frac{\sin^2\left(\sqrt{x+1}-\sqrt{x}\right)}{\sin^2\frac{1}{x}}\\
&=\lim_{x\to\infty}\frac{\left(\sqrt{x+1}-\sqrt{x}\right)^2\frac{\sin^2\left(\sqrt{x+1}-\sqrt{x}\right)}{(\sqrt{x+1}-\sqrt{x})^2}}{\frac{1}{x^2}\frac{\sin^2(1/x)}{1/x^2}}
\end{align}
Now, setting $t=\dfrac{1}{\sqrt{x+1}+\sqrt{x}}$ we note $x\to\infty$ implies $t\to 0^+$, and since $\displaystyle{\lim_{t\to 0}\frac{\sin t}{t}=1}$
$$\lim_{x\to\infty}\frac{\sin^2(\sqrt{x+1}-\sqrt{x})}{(\sqrt{x+1}-\sqrt{x})^2}=\left[\lim_{x\to \infty}\frac{\sin \frac{1}{\sqrt{x+1}+\sqrt{x}}}{\frac{1}{\sqrt{x+1}+\sqrt{x}}}\right]^2=\left(\lim_{t\to 0^+}\frac{\sin t}{t}\right)^2=(1)^2=1$$
Similarly, putting $t=1/x$ we have $t\to0^+$ as $x\to\infty$, and
$$\lim_{x\to\infty}\frac{\sin^2(1/x)}{1/x^2}=\left(\lim_{t\to 0^+}\frac{\sin t}{t}\right)^2=1^2=1$$
Then,
\begin{align}
\lim_{x\to\infty}\frac{\sin^2\left(\sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}&=\left(\lim_{x\to \infty}\frac{\left(\sqrt{x+1}-\sqrt{x}\right)^2}{\frac{1}{x^2}}\right)\lim_{x\to\infty}\frac{\frac{\sin^2\left(\sqrt{x+1}-\sqrt{x}\right)}{(\sqrt{x+1}-\sqrt{x})^2}}{\frac{\sin^2(1/x)}{1/x^2}}\\[5pt]
&=\left(\lim_{x\to\infty}x^2\left(\frac{1}{\sqrt{x+1}+\sqrt{x}}\right)^2\right)\cdot\frac{1}{1}\\
&=\lim_{x\to\infty}\left(\frac{x}{\sqrt{x+1}+\sqrt{x}}\right)^2
\end{align}
You can determine the last limit.
| {
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Evaluation of $\int\frac{\sin 2x}{(3+4\cos x)^3}\,dx$
Evaluation of $$\int\frac{\sin 2x}{(3+4\cos x)^3}\,dx$$
$\bf{My\; Try::}$ Let $$\displaystyle \int\frac{\sin 2x}{(3+4\cos x)^3}dx = 2\int\frac{\sin x\cos x}{(3+4\cos x)^3}dx$$
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^3 x.$
So we get $$\displaystyle I = 2\int\frac{\sec x\cdot \tan x}{(3\sec x+4)^3}dx$$
Now Put $\displaystyle (3\sec x+4) = t\;,$ Then $3\sec x\cdot \tan xdx = dt$
So we get $$\displaystyle I = \frac{2}{3}\int \frac{1}{t^3}dt = -\frac{1}{3t^2}+\mathcal{C} = -\frac{1}{3(3\sec x+4)^2}+\mathcal{C}$$
But answer is http://www.wolframalpha.com/input/?i=INTEGRATION+OF+%28sin+2x%29%2F%283%2B4cos+x%29%5E3
Where I gave Done Wrong,
Thanks
| Hint:
By inspection,
$$\int\frac{\sin x\cos x}{(3+4\cos x)^3}dx=-\int \frac{t}{(3+4t)^3}dt$$
which can be decomposed with
$$4t=(3+4t)-3,$$ to yield elementary functions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a good strategy to compute$f(x) = e^x −\cos x − \sin x$ for $x$ near $0$ Find a good strategy to compute $f(x) = e^x − \cos x − \sin x$ for $x$ near 0.
In five-decimal-digit arithmetic, compute $f(0.1)$ using the straightforward method
and your better strategy, and compare the difference. (Hint: The computer can only
store numbers with five decimal digits at any step.)
so I know using the straightforward strategy that:
$f(0.1) = e^{(0.1)} - \cos(0.1) - \sin(0.1)$ (since the computer can only store 5 decimal digits at any step ->
$= 1.10517 - 0.99500 - 0.09998$
$= 0.01019 $
so I think my teacher mentioned something in class saying we should use either taylor or macclaurin series? Can someone walk me through how exactly I would do that for my "better" strategy?
| With Taylor series you have
$$\begin{align} f(x) &= \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \cdots \right)\\
&-\left(1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots\right)\\
&- \left(x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots\right)\end{align}$$
Now, notice that the terms after $x^5$ were deliberately left out because they won't be significant if your computer can store only $5$ decimal places, as all these terms would be at least in the $6$th decimal place ($0.1^6, 0.1^7, \cdots$).
So you may as well leave those terms out and approximate $f$ as
$$$$\begin{align} f(x) &\approx \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} \right)\\
&-\left(1 - \frac{x^2}{2} + \frac{x^4}{24} \right)\\
&- \left(x - \frac{x^3}{6} + \frac{x^5}{120} \right)\end{align}$$$$
| {
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"timestamp": "2023-03-29T00:00:00",
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To find maximum value when sum is fixed Given that $a,b,c,d,e$ and $f$ are six non-negative real numbers such that
$a+b+c+d+e+f=1$.
Find the maximum value of
$ab+bc+cd+de+ef$.
My Approach
After lot of thinking I reduced the problem to following:
Given that $a,b,c,d,e$ and $f$ are six non-negative real numbers such that
$a+b+c=1$.
Find the maximum value of
$ab+bc$.
which is quite easy to solve:
i.e. $ab+bc=b(a+c)=b(1-b)=\frac{1}{4}-\left(b-\frac{1}{2}\right)^2\leq\frac{1}{4}$.
Not able to extrapolate to the given problem in hand.
Is there a Calculus approach.
| Let $0\leq a \leq b \leq c \leq d \leq f \leq e\leq 1$
$$a+b+c+d+f+e=1 \rightarrow (a+b+c+d+f+e-1)^2=0$$
$$\rightarrow a^2+b^2+c^2+d^2+f^2+(e-1)^2+....=0$$
so we have $a^2=b^2=c^2=d^2=f^2=(e-1)^2=0$ i.e., $a=b=c=d=f=0$ and $e=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Forming a committee from $4$ gentlemen and $4$ ladies with certain conditions
From $4$ gentlemen and $4$ ladies a committee of $5$ is to be formed .
If the committee consists of $1$ president, $1$ vice president and $3$ secretaries.
What will be the number of ways of selecting the committee with at least $3$
women such that at least one women holds the post of president or vice-president?
I tried
Case $1$
$3W2M$
$\dbinom{4}{1}\times \dbinom{4}{1} \times \dbinom{3}{2} \times \dbinom{3}{1}=144 $
$+$
$\dbinom{4}{1}\times \dbinom{3}{1} \times \dbinom{2}{1} \times \dbinom{4}{2}=144 $
Case $2$
$4W1M$
$\dbinom{4}{1}\times \dbinom{3}{1} \times \dbinom{2}{2} \times \dbinom{3}{1}=36 $
$+$
$\dbinom{4}{1}\times \dbinom{4}{1} \times \dbinom{3}{3}=16 $
Total ways=$338$.
But the answer given is $512$.
I look for a short and simple way.
I have studied maths up to $12$th grade.
| Count the number of committees with $\color\red{3\text{ women}}$ and $\color\green{2\text{ men}}$:
$$\color\red{\binom43}\cdot\color\green{\binom42}\cdot\binom51\cdot\binom41\cdot\binom33=480$$
Subtract the number of committees where all $3$ women are secretaries:
$$\color\red{\binom43}\cdot\color\green{\binom42}\cdot\color\green{\binom21}\cdot\color\green{\binom11}\cdot\color\red{\binom33}=48$$
Add the number of committees with $\color\red{4\text{ women}}$ and $\color\green{1\text{ man}}$:
$$\color\red{\binom44}\cdot\color\green{\binom41}\cdot\binom51\cdot\binom41\cdot\binom33=80$$
Hence the number of committees is $480-48+80=512$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $\sin\theta+\cos\theta=1$ prove that $\cos\theta-\sin\theta=\pm1$ So my work,
Squaring both sides $$(\sin\theta+\cos\theta)^2=1$$
$$1+2\sin\theta\cos\theta=1\ \ \ \ \ \text{-------(i)}$$
$$\sin\theta\cos\theta=0 \ \ \ \ \ \text{------(ii)}$$
So reverting back to $(i)$,
$$\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-4\sin\theta\cos\theta=1-4\sin\theta\cos\theta$$
$$(\cos\theta-\sin\theta)^2=1-4\sin\theta\cos\theta$$
$$\cos\theta-\sin\theta=\pm1$$
But my teacher says that there is a shorter solution than that, so please can someone help me find that?
| Let $x = \cos\theta$ and $y = \sin\theta$. Then $(x,y)$ is a point on
the unit circle. But the equation
$$\sin \theta + \cos \theta = 1$$
says that $x + y = 1$. So $(x,y)$ must be on the line given by $x + y = 1$,
that is, the line that intersects the unit circle at $(1,0)$ and $(0,1)$.
In fact, since $(x,y)$ is on that circle and on that line,
it must be one of those two points.
Case 1: $(x,y) = (0,1)$.
\begin{align}
\cos \theta &= 0 \\ \sin \theta &= 1 \\ \cos\theta - \sin\theta &= -1
\end{align}
Case 2: $(x,y) = (1,0)$
\begin{align}
\cos \theta &= 1 \\ \sin \theta &= 0 \\ \cos\theta - \sin\theta &= 1
\end{align}
And those are the only two possible cases that can occur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Minimum of $\frac{1}{x+y}+\frac{1}{x+z}-\frac{1}{x+y+z}$ for $0\leq x+y,y+z,z+x\leq 1$ Let $0\leq x,y,z,x+y,y+z,z+x\leq 1$. What is the minimum of $$F(x,y,z)=\frac{1}{x+y}+\frac{1}{x+z}-\frac{1}{x+y+z}?$$
We have $F(1/2,1/2,1/2)=4/3$. Since the constraints are on $x+y,y+z,z+x$ instead of $x,y,z$, taking partial derivative with respect to $x,y,z$ doesn't help much. We cannot change one variable without affecting the others.
| Hint: Do the substitution $ a= x+y, b = y + z, c = z + x $, and we have $ 0 \leq a, b, c, \leq 1 $ and we want to minimize
$$ \frac{1}{a} + \frac{1}{c} - \frac{2}{a+b+c}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How can I solve this recursive function $f(n) = f(f(n+1))$? I am trying to solve this:
$$
f(n) =
\begin{cases} n - 1,& n > 5\\
f(f(n+1)),& n\leqslant 5
\end{cases}
$$
What is the technical name of this kind of function ? --> $f(f(n+1))$
A function within another function?
\begin{align}
f(1) &= f(f(1+1)) = f(f(2)) \\
f(2) &= f(f(2+1)) = f(f(3))\\
f(3) &= f(f(3+1)) = f(f(4))\\
f(4) &= f(f(4+1)) = f(f(5))\\
f(5) &= 5 - 1 = 4\\
f(6) &= 6 - 1 = 5\\
f(7) &= 7 - 1 = 6
\end{align}
I can only find the answer when $n > 5$. When $n\leqslant5$, I can only see $f(f(k))$. How to find the answers of $f(1)$, $f(2)$, $f(3)$ and $f(4)$ ? Thank you very much.
| An example will make it understand better
$ {let}$ $ n = 5$ now $ f(n)=f(f(n=1))$
so, $f(5)=f(f(5+1))$
$=f(f(6)) = f(5)$
$f(5)=f(f(5+1))$
$=f(f(6)) = f(5)$
$f(5)=f(f(5+1))$
$=f(f(6)) = f(5)$
..............
and it goes on so the function is not defined at 5
Now let $n=4$
so, $f(3)=f(f(3+1))$
$=f(f(4))$ now $f(4)$ will be $f(f(5))$
so, $f(3)=f(f(4))=f(f(f(5)))$
which again is not defined,
so we can say that the function is of indeterminate form for $n\leqslant5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Express $y= x^{3} - x^{2} - 5x - 3$ in its fully factorised form Don't know how to do this, please help. I have never done factorising cubic polynomials and don't know how to go about this
| $$x^3-x^2-5x-3=x^3+2x^2-3x^2-6x+x-3$$
$$x^3+2x^2+x-3x^2-6x-3=0$$
$$x(x^2+2x+1)-3(x^2+2x+1)=0$$
$$(x^2+2x+1)(x-3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
First order differential equation involving inverses My question is to find the solutions to the following
$\frac{df(x)}{dx} = f^{-1} (x)$
where $f^{-1} (x)$ refers to the inverse of the function f. The domain really isn't important, though I am interested in either (-inf, inf) or (0, inf), so if any solutions are known for more restricted domains then they are welcome.
I cannot find any material relating to this type of question in any of my calculus and differential equations textbooks and references; it seems quite unorthodox. Any material which covers this type of diff equation would be wlecome
| Here is a partial solution using power series composition.
To solve
$$
f(f'(x))=x\tag{1}
$$
take the derivative of $(1)$
$$
f'(f'(x))\,f''(x)=1\tag{2}
$$
If we can find a point $a\gt0$ so that $f'(a)=a$, we can expand $f'(a+x)$ around $x=0$. Let
$$
g(x)=f'(a+x)-a\tag{3}
$$
Applying $f'(a+x)=a+g(x)$ twice gives
$$
\begin{align}
f'(f'(a+x))
&=f'(a+g(x))\\
&=a+g(g(x))\tag{4}
\end{align}
$$
Differentiating $(3)$ yields
$$
g'(x)=f''(a+x)\tag{5}
$$
Plugging $(4)$ and $(5)$ into $(2)$, we get
$$
g'(x)=\frac1{a+g(g(x))}\tag{6}
$$
Iterating $(6)$ gives
$$
\begin{align}
g(x)
&=\frac xa-\frac{x^2}{2a^4}+\frac{(1+3a)x^3}{6a^8}-\frac{(1+3a+10a^2+15a^3)x^4}{24a^{13}}\\
&+\frac{\left(1+3a+10a^2+30a^3+55a^4+105a^5+105a^6\right)x^5}{120a^{19}}+\dots\tag{7}
\end{align}
$$
Therefore,
$$
\begin{align}
f(x)
&=a+a(x-a)+\frac{(x-a)^2}{2a}-\frac{(x-a)^3}{6a^4}+\frac{(1+3a)(x-a)^4}{24a^8}\\[6pt]
&-\frac{(1+3a+10a^2+15a^3)(x-a)^5}{120a^{13}}\\
&+\frac{\left(1+3a+10a^2+30a^3+55a^4+105a^5+105a^6\right)(x-a)^6}{720a^{19}}+\dots\tag{8}
\end{align}
$$
Comparison with Julián Aguirre's Answer
$$
\begin{align}
f(x)
&=\phi^{-1/\phi}x^\phi\\
&=\phi+\phi(x-\phi)+\frac{\phi-1}2(x-\phi)^2-\frac{5-3\phi}6(x-\phi)^3+\frac{18\phi-29}{24}(x-\phi)^4\\
&-\frac{217-134\phi}{120}(x-\phi)^5+\frac{1219\phi-1972}{720}(x-\phi)^6+\dots\tag{9}
\end{align}
$$
which matches $(8)$ with $a=\phi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Find $n$, where its factorial is a product of factorials I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.
I have tried simplifying as follows:
$$\begin{array}{}
3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\
(3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^4 \cdot 5^2 \cdot 4^2 \cdot 7 &= n! \\
\end{array}$$
I really didn't see this helping me.
I then tried $6 \cdot 6 \cdot 6 \cdot 6 \cdot 25 \cdot 16 \cdot 7$, but $25$ only has $5$ as a double factor.
Any ideas?
| You already have some clever answers; here's less clever approach. You want
$$
3!\cdot5!\cdot7!=n!=7!\cdot8\cdot9\cdots\cdot n,
$$
so, cancelling 7! and computing that $3!\cdot 5!=6\cdot120=720$, you want
$$
720=8\cdot9\cdots\cdot n.
$$
Now start dividing both sides by 8, then 9, etc. until you get the answer. Dividing by 8 gives you $90=9\cdots n$, and then you either see the answer already or divide both sides by 9 to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 7,
"answer_id": 5
} |
Prove that $\sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$
Prove that $\displaystyle \sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$
$\bf{My\; Try::}$ We can write $$\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{n!}{r!\times (n-r)!}\times \frac{r!\times 3!}{(r+3)!} = \frac{3!\times n!}{(n-r)!\times (r+3)!}$$
So our sum is $$ 3!\sum^{n}_{r=0}(-1)^r\frac{n!}{(n-r)!\times (r+3)!} = \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\frac{(n+3)!}{(r+3)!\times (n-r)!}$$
So we get Sum $$ \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\binom{n+3}{r+3} = -\frac{3!}{(n+1)(n+2)}\sum^{n}_{r=0}(-1)^{r+3}\binom{n+3}{r+3}$$
$$ = -\frac{3!}{(n+1)(n+2)(n+3)}\left[-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]$$
$$=-\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]+\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}\right]$$
So we get $$ = \frac{3!}{(n+1)(n+2)(n+3)}\left[1-(n+3)+\frac{(n+3)(n+2)}{2}\right] = \frac{3!}{(n+1)(n+2)}\times \frac{n^2+3n+2}{2} = \frac{3!}{(n+1)(n+2)(n+3)}\times \frac{(n+1)(n+2)}{2} = \frac{3!}{2(n+3)}$$
Can we prove it Using Combinatorial way,bcz above method is very lengthy
If yes then plz explain here
Thanks
| Your computational argument can be written a bit more simply:
$$\begin{align*}
\sum_{r=0}^n(-1)^r\frac{\binom{n}r}{\binom{r+3}r}&=3!\sum_{r=0}^n(-1)^r\frac{n!}{(n-r)!(r+3)!}\\
&=\frac6{(n+3)(n+2)(n+1)}\sum_{r=0}^n(-1)^r\binom{n+3}{r+3}\tag{1}\\
&=\frac6{(n+3)(n+2)(n+1)}\sum_{r=3}^{n+3}(-1)^{r-3}\binom{n+3}r\\
&=\frac6{(n+3)(n+2)(n+1)}\sum_{r=0}^n\left(\sum_{r=0}^{n+3}(-1)^{r-3}\binom{n+3}r-\sum_{r=0}^2(-1)^{r-3}\binom{n+3}r\right)\\
&=\frac6{(n+3)(n+2)(n+1)}\sum_{r=0}^2(-1)^{r-2}\binom{n+3}r\\
&=\frac6{(n+3)(n+2)(n+1)}\left(\binom{n+3}0-\binom{n+3}1+\binom{n+3}2\right)\\
&=\frac6{(n+3)(n+2)(n+1)}\left(1-(n+3)+\frac12(n+3)(n+2)\right)\\
&=\frac{6\left(\frac12(n+3)-1\right)}{(n+3)(n+1)}\\
&=\frac{3n+9-6}{(n+1)(n+3)}\\
&=\frac3{n+3}\;.
\end{align*}$$
For a slightly neater computational argument we can start by taking advantage of $(1)$ to rewrite the identity as
$$\sum_{r=0}^n(-1)^r\binom{n+3}{r+3}=\binom{n+3}3\cdot\frac3{n+3}=\binom{n+2}2\;.\tag{2}$$
Now
$$\begin{align*}
\sum_{r=0}^n(-1)^r\binom{n+3}{r+3}&=\sum_{r=0}^n(-1)^r\left(\binom{n+2}{r+2}+\binom{n+2}{r+3}\right)\\
&=\sum_{r=0}^n(-1)^r\binom{n+2}{r+2}+\sum_{r=0}^n(-1)^r\binom{n+2}{r+3}\\
&=\sum_{r=0}^n(-1)^r\binom{n+2}{r+2}+\sum_{r=1}^{n+1}(-1)^{r-1}\binom{n+2}{r+2}\\
&=\binom{n+2}2+(-1)^n\binom{n+2}{n+3}\\
&=\binom{n+2}2\;.
\end{align*}$$
It’s fairly easy to see what’s going on here combinatorially, though what follows is too informal to be a true combinatorial argument. The righthand side of $(2)$ of course is the number of $2$-element subsets of $[n+2]$. The lefthand side of $(2)$ is
$$\binom{n+3}3-\binom{n+3}4+\binom{n+3}5-+\ldots\;.$$
The first term counts the $3$-element subsets of $[n+3]$; those that contain $n+3$ are in one-to-one correspondence with the $2$-element subsets of $[n+2]$, so we’d like to throw away the $3$-element subsets of $[n=3]$ that don’t contain $[n+3]$. These are in one-to-one correspondence with the $4$-element subsets of $[n+3]$ that do contain $n+3$, so as a next approximation we’ll throw away all $4$-element subsets of $[n+3]$, getting $\binom{n+3}3-\binom{n+3}4$.
Now, of course, we need to add back in the $4$-element subsets of $[n+3]$ that don’t contain $n+3$. These are in one-to-one correspondence with the $5$-element subsets of $[n+3]$ that do contain $n+3$, so as a next approximation we add back in all $5$-element subsets of $[n+3]$, getting $\binom{n+3}3-\binom{n+3}4+\binom{n+3}5$.
At this point it’s clear that we have an informal induction argument going, and a bit of thought shows that it ends in the right place.
| {
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"url": "https://math.stackexchange.com/questions/1441827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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