Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
prove that $\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Assume: $a,b,c >0$ prove that :
$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
| HINT: One may start proving the inequality:
$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\ \frac{9}{a+b+c}+ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
that is easy to prove. I met some time ago this inequality and have just remembered now. Of course, it's easy only if you met it before, otherwise it's rather hard to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Prove that: $\frac1{20}\le \int_{1}^{\sqrt 2} \frac{\ln x}{\ln^2x+1} dx$ I'm interested in proving the following integral inequality:
$$\frac1{20}\le \int_{1}^{\sqrt 2} \frac{\ln x}{\ln^2x+1} dx$$
According to W|A the result of this integral isn't pretty nice, and involves the exponential integral.
| $$\begin{array}{c l} \int_1^{\sqrt2} \frac{\log x}{1+(\log x)^2}dx & =\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}e^udu \\ & \ge\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}du \\ & = \left[\frac{1}{2}\log(1+u^2)\right]_0^{\frac{1}{2}\log 2} \\ & =\frac{1}{2}\log\left(1+\frac{(\log 2)^2}{4}\right) \\ & \approx 0.056714899 \\ & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Computing the derivative from the definition Using the limit definition of the derivative which I know is:
$$f'(x)=\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{h}\right)$$
I am trying to solve this problem
$$f(x)= \frac{x}{x+2} $$
How do I go about properly solving this, I seemed to get
$$\frac{x}{x+2}\ $$
as my answ... | First. The derivative function of $f(x)$ is not $\frac{f(x+h)-f(x)}{h}$. (That would be a function of two variables, $x$ and $h$). Rather, the derivative function of $f(x)$ is
$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$
The limit is important. Without the limit, it's just plain wrong.
Second. I don't know how you "got back... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Integration Example How can i find the integration of this example
$$\int \frac{\sin x}{\sin x - \cos x } dx$$
I tried first add cos and then substracting cos but then what about $$\int \frac{\cos x}{\sin x - \cos x } dx\ ?$$
| Let $f(a) = \int \frac{\sin(ax)}{\sin(x) - \cos(x)}dx$.
Differentiating throughout by a, we get
$$f'(a) = a\int \frac{\cos(ax)}{\sin(x) - \cos(x)}dx$$
Therefore, $$af(a) - f'(a) = a\int \frac{\sin(ax) - \cos(ax)}{\sin(x) - \cos(x)}dx$$
Substituting $a = 1$, we get
$$f(1) - f'(1) = x + C_1$$
Also,
$$af(a) + f'(a) = a\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/162783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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To find the square root of a polynomial My question is:
Find the value of $k$ such that $$4x^6 - 24x^5 + 20x^4 + 68x^3 -44x^2 - 40x + k$$ is a perfect square.
hey all i have made an edit. Sorry for the inconvenience.
Any help to solve this question would be greatly appreciated.
| If the goal is to find $k$ so that the polynomial is a perfect square, start by noting that it must be the square of a cubic polynomial:
$$4x^6 - 24x^5 + 20x^4 + 68x^3 -44x^2 - 40x + k=(ax^3+bx^2+cx+d)^2\;.$$
Clearly this immediately require that $a=2$. Now the square of $(2x^3+bx^2+cx+d)^2$ is
$$4x^6+4bx^5+(4c+b^2)x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/163219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Will the following expression be irrational, rational or integer? Will the following expression be irrational, rational or integer?
$$\sqrt[3]{\sqrt a +b} - \sqrt[3]{\sqrt a -b}$$
where $a$ = $52$ and $b$ = $5$ .
By intuition, I think this will be an integer.
| Let's use the identity $(\alpha + \beta)^3 = \alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)$
Set $\alpha =\sqrt[3]{b+\sqrt{a}} \text{ and } \beta= \sqrt[3]{b - \sqrt{a}} \text { and } \alpha+\beta=x$
We know that $\alpha\beta = \sqrt[3]{b+\sqrt{a}} \times \sqrt[3]{b - \sqrt{a}} = \sqrt[3]{b^2-a} = \sqrt[3]{25-52} = -3$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/163440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Compute $\sum_0^{n-1}2^i11^{n-i-1}\bmod10^9$ when $n=13^{17}$ Given the following function $f$
$f(1)=1$
$f(n)=11\cdot f(n-1)+2^{n-1}$
I would like to compute $f(13^{17})\mod 10^9$ and ended up using the following :
$f(n)=\sum_{i=0}^{n-1}({11^{n-(i+1)}\cdot 2^i})$
though I am able to quickly compute a single term usin... | Clearly $f(n)$ will be of the form $a \cdot 11^n+b \cdot 2^n$ where $a,b$ are integers.
Given $f(1)=1$, but $f(1)=11a+2b$.
Similarly $f(2)=11f(1)+2=13$, but $f(2)=121a+4b$.
Solving for $a,b$, we get $f(n)= \frac{11^n-2^n}9$.
$f(13^{17})=\frac{11^{13^{17}}-2^{13^{17}}}9$.
$11^{13^{17}} = (10+1)^{13^{17}}=1+{13^{17}} \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given that $\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(xy)=11/12π$, prove that when $x=1, dy/dx=-1-\sqrt{3}/2$ Given that $x$ and $y$ satisfy the equation:
$$\arctan(x)+\arctan(y)+\arctan(xy)=11/12π$$
Prove that, when $x=1, dy/dx=-1-\sqrt{3}/2$.
I tried to differentiate both sides:
$$1/(1+x^2)+y/(1+y^2)+(y+x\,dy/dx)/(1+(xy)^2... | $x=1\implies \arctan 1+2\arctan y=11\pi/12=\arctan y=\pi/3\implies y=\sqrt 3$. Thus, taking derivative on both sides gives,
$$
\begin{align}
& \frac{1}{1+x^2}+\frac{1}{1+y^2}\frac{dy}{dx}+\frac{1}{1+x^2y^2}\left(y+x\frac{dy}{dx}\right)=0 \\[10pt]
& \implies \frac{1}{2}+\frac{1}{4}\left(\frac{dy}{dx}\right)+\frac{1}{4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Equation of right circular cylinder with radius of the base as 2 units. Obtain the equation of right circular cylinder with radius of the base as 2 units. Its axis passes through $(1, 2, 3)$ and direction cosines are given as $(2, -3, 6)$
I got $45x^2+40y^2+13z^2+12xy-36yz-24zx-42x-280y-126z+294 = 0$
| Put
$$
\begin{gathered}
A = \left( {1,2,3} \right) \hfill \\
X = \left( {x,y,z} \right) \hfill \\
\mathbf{n} = \frac{1}
{7}\left( {2, - 3,6} \right) \hfill \\
\end{gathered}
$$
Then the required cylinder is the lieu of points such that:
$$
2 = \left| {\mathop {AX}\limits^ \to \times \mathbf{n}} \right|
$$
tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/168236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Can $a^2+b^2+2ac$ be a perfect square if $c\neq \pm b$?
Can $a^2+b^2+2ac$ be a perfect square if $c\neq \pm b$?
$a,b,c \in \mathbb{Z}$.
I have tried some manipulations but still came up with nothing. Please help.
Actual context of the question is:
Let say I have an quadratic equation $x^2+2xf(y)+25$ that I have to... | A small manipulation changes the problem into a more familiar one. We are interested in the Diophantine equation $a^2+b^2+2ac=y^2$. Complete the square. So our equation is equivalent to $(a+c)^2+b^2-c^2=y^2$. Write $x$ for $a+c$. Our equation becomes
$$x^2+b^2=y^2+c^2.\tag{$1$}$$
In order to get rid of trivial soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/168619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
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Inequality $\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\cdots+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$ Let:$a,b,c,d>0$ be real numbers ,how to prove that :
$$\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\frac{b^2+c^2+d^2}{b^5+c^5+d^5}+\frac{c^2+d^2+a^2}{c^5+d^5+a^5}+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$$.
Ed... | Also, by Muirhead(or Chebyshov) and AM-GM we obtain:
$$\sum_{cyc}\frac{a^2+b^2+c^2}{a^5+b^5+c^5}\leq\sum_{cyc}\frac{3}{a^3+b^3+c^3}\leq\sum_{cyc}\frac{1}{abc}=\frac{a+b+c+d}{abcd}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Analytic expression for the primitive of square root of a quadratic Can an analytic expression be given for
$$\int \sqrt{ax^2 + bx +c} \, dx$$
I think substitution doesn't work in this case (I need to compute the integral $\int_0^t \ldots$).
| To deal with the integral, we first use method of completing square and then the well-know result
$$
\int \sqrt{x^2-a^2} d x=\frac{1}{2}\left[x \sqrt{x^2-a^2}-\ln \left|x+\sqrt{x^2-a^2}\right| \right ]+C
$$
Case 1: $a>0$ and $b^2>4ac$
$$
\begin{aligned}
I &=\int \sqrt{a x^2+b x+c} d x\\&=\int \sqrt{\left(\sqrt{a} x+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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The minimum value of $(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)$ if $x+y+z=1$ $x, y, z$ are three distinct positive reals such that $x+y+z=1$, then the minimum possible value of $(\frac{1}{x}-1) (\frac{1}{y}-1) (\frac{1}{z}-1)$ is ?
The options are: $1,4,8$ or $16$
Approach: $$\begin{align*}
\left(\frac{1}{x} -1\... | If we put no constraint on $x$, $y$, and $z$ apart from $x$, $y$, $z$ positive and $x+y+z=1$, then indeed your calculation, and the one by Patrick Da Silva, show that the minimum value is $8$, attained at $x=y=z=\frac{1}{3}$.
However, the problem specifies that $x$, $y$ and $z$ are distinct real numbers. If we take th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Factorise the determinant $\det\Bigl(\begin{smallmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{smallmatrix}\Bigr)$ Factorise the determinant $\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}$.
My textbook only provides two simple examples.
Really have no idea how ... | Regard $b$ and $c$ as constants, and the determinant as a polynomial in $a$. Then find ways of making the determinant equal to 0, and by the Factor Theorem you'll get a factor of the determinant.
Obvious choices: set $a=b$ or $a=c$ and you'll have two identical rows, so $(a-b)$ and $(a-c)$ are factors.
We can clearly s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Inequality for cosines Is the following inequality in a triangle known?
$$4(\cos A + \cos B + \cos C) \le 3 + \cos \left(\frac{B-C}{2}\right) + \cos \left(\frac{C-A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$$
It looks correct to me but I would appreciate if someone confirm it.
| Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, we need to prove that
$$2\sum_{cyc}\frac{a^2+b^2-c^2}{ab}\leq3+\sum_{cyc}\left(\sqrt{\frac{(a+b+c)^2(b+c-a)(a+c-b)}{16c^2ab}}+\sqrt{\frac{(a+b-c)^2(b+c-a)(a+c-b)}{16c^2ab}}\right)$$ or
$$2\sum_{cyc}c(a^2+b^2-c^2)\leq3abc+\sum_{cyc}\frac{a+b}{2}\sqrt{ab(a+c-b)(b+c-a)}$$ or
$$4\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 3
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How to prove that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer As the title says, I'm trying to show that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer.
I suppose there's probably some heavy duty classification theorems that give one line proofs to this but I don't have any of that ... | If we start with $\sqrt{-3}x + 1 = 10^{2/3}$, we get
$$(\sqrt{-3}x+1)^3 = 100$$
hence
$$100 = -3\sqrt{-3}x^3 - 9x^2 + 3\sqrt{-3}x + 1.$$
Therefore,
$$3\sqrt{-3}x^3 + 9x^2 - 3\sqrt{-3}x + 99 = 0.$$
Dividing through by $3\sqrt{-3}$ we obtain
$$x^3 + \frac{3}{\sqrt{-3}}x^2 - x + \frac{33}{\sqrt{-3}}=0$$
and rationalizin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/175206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
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Evaluating $\int ^\frac{\pi}{2}_{0} \sin\left(2x+\frac{\pi}{4}\right)\ dx$
Find the exact value of the following definite integral:
$$\int ^\frac{\pi}{2}_{0} \sin\left(2x+\frac{\pi}{4}\right)\:dx=\left[-\frac{1}{2}(2x+\frac{\pi}{4})\right]^\frac{\pi}{2}_{0}$$
$$=-\frac{1}{2}\left(2\frac{\pi}{2}+\frac{\pi}{4}\rig... | You have forget $\cos$ after you have done antiderivative.
Solve-
$$\int^\frac{\pi}{2}_{0}\sin\left(2x+\frac{\pi}{4}\right)\:dx=\left[-\frac{1}{2}\cos\left(2x+\frac{\pi}{4}\right)\right]^\frac{\pi}{2}_{0}$$
$$=-\frac{1}{2}\cos\left(2\frac{\pi}{2}+\frac{\pi}{4}\right)+\frac{1}{2}\cos\left(2\cdot 0+\frac{\pi}{4}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/175815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Poisson summation formula (in general) Define Poisson kernel as $$ P_r ( \theta) := \frac{1}{2\pi} \frac{1-r^2}{1- 2r \cos \theta + r^2} $$ Then I want to prove the Poisson summation formula which is $$ P_r (2\pi x) = \sum_{n=-\infty}^\infty P_y (x+n)\;\;\;\;\text{(here $r = e^{-2 \pi y} $}) $$
| We will use the result that
$$\sum_{n=-\infty}^{\infty} \frac{y}{(x+n)^2+y^2}= \frac{1}{2} \frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}} = P_y(2\pi\,x)\,, $$
Recalling Poisson formula in the upper half plane for $y>0$,
$$ P_{y}(x) = \frac{y}{x^2+y^2}\,. $$
We construct the sum
$$ \sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Indefinite Integral of $\sqrt{\sin x}$ $$\int \sqrt{\sin x} ~dx.$$
Does there exist a simple antiderivative of $\sqrt{\sin x}$? How do I integrate it?
| Since $\sqrt{\sin(x)} = \sqrt{1 - 2 \sin^2\left(\frac{\pi}{4} -\frac{x}{2}\right)}$, this matches with the elliptic integral of the second kind:
$$\begin{align*}
\int \sqrt{\sin(x)} \, \mathrm{d} x &\stackrel{u = \frac{\pi}{4}-\frac{x}{2}}{=} -2 \int \sqrt{1-2 \sin^2(u)} \,\mathrm{d} u\\ &= -2 E\left(u\mid 2\right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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hints on solving DE How to solve this DE?
$$ {dx \over x} = {dy \over y} = {dz \over z - a \sqrt{x^2+y^2+z^2}}$$
From the first part, I get $y = c_1x$. How to find the other solution?
The answer according to answer sheet is $ z + \sqrt{x^2 + y^2 + z^2} = c_2$. Thank you for help.
| Let
\begin{equation}
{\frac{dx}{x}} = {\frac{dy}{y}} = {\frac{dz}{z - a \sqrt{x^2+y^2+z^2}}} = K
\end{equation}
\begin{equation}
{\frac{2xdx}{2x^{2}}} = {\frac{2ydy}{y^{2}}} = {\frac{2zdz}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}}} = K
\end{equation}
Then
\begin{equation}
\frac{dx^{2}}{2x^{2}} = \frac{dy^{2}}{2y^{2}} = \frac{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The probability of one Gaussian larger than another. For two Gaussian-distributed variables, $ Pr(X=x) = \frac{1}{\sqrt{2\pi}\sigma_0}e^{-\frac{(x-x_0)^2}{2\sigma_0^2}}$ and $ Pr(Y=y) = \frac{1}{\sqrt{2\pi}\sigma_1}e^{-\frac{(x-x_1)^2}{2\sigma_1^2}}$. What is probability of the case X > Y?
| I assume that $X$ and $Y$ are independent. Let $Z=X-Y$ then $Z\sim\cal{N}(x_0-y_0,\sigma_0^2+\sigma_1^2)$. Accordingly
$$P(Z>0)=\int_0^\infty\frac{1}{\sqrt{2\pi(\sigma_0^2+\sigma_1^2)}}\exp\left(\frac{-(z-x_0+y_0)^2}{2(\sigma_0^2+\sigma_1^2)}\right)\mathrm{d}z$$
if we use the complementary error function
$$\operatornam... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Evaluating $\int_1^3\frac{\ln(x+2)}{x^2+2x+15} \ dx$ Could you please give me a hint on how to compute:
$$
\int_1^3\frac{\ln(x+2)}{x^2+2x+15}dx
$$
Thank you for your help
| It is not really a simple integral (even if nothing special happens in the range $(1,3)$). Sasha gave a fine approximation (+1) let's provide the dilogarithm answer...
Let's start by factoring the denominator $\ x^2+2x+15$ :
The reduced discriminant is $\Delta=1-1\cdot 15=-14\ $ so that that it will have two complex co... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim\limits_{n\to \infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n}$ Show that $$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n}\right)=\log 6$$ Here I need to use the definition of integral but I faced problem in range . Please help.
| Maybe it is intended that you mention Riemann sums explicitly.
Rewrite our sum as
$$\frac{1}{n}\left(\frac{1}{1+\frac{1}{n}} + \frac{1}{1+\frac{2}{n}} + \frac{1}{1+\frac{3}{n}}+\cdots +\frac{1}{1+\frac{5n}{n}} \right).$$
We recognize this as a (right) Riemann sum for the integral
$$\int_0^5 \frac{dx}{1+x},$$
which ha... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding positive integer solutions to $n = ax^2 +by^2 - cxy$ How can I find the positive integer solutions to $x$ and $y$, given that $n$, $a$, $b$ and $c$ are all positive integers, in an equation of the form:
$$n = ax^2 + by^2 - cxy.$$
Specifically, I want to find the positive integer solutions to the following equat... | $ax^2+by^2-cxy=n$
Expressing as a quadratic equation of x, $ax^2-x(cy)+by^2-n=0$
As x is positive integer, $=>(cy)^2-4.a(by^2-n)=(c^2-4ab)y^2 +4an$ must be perfect square.
$c=16, a=3, b=20 =>16y^2+12n=d^2(say)$=>d is even=2e(say)
$=>4y^2+3n=e^2$=>e is odd $>e^2≡1(mod\ 8)$
$=>e^2-4y^2 \equiv 1 \pmod4$ => $3n \equiv 1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
| Since $(2,0,0)\succ\left(\frac{4}{3},\frac{1}{3},\frac{1}{3}\right)$, by Murhead we obtain:
$$a^2+b^2+c^2\geq\sum_{cyc}a^{\frac{a}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}=a+b+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 7
} |
Evaluating the elliptic integral $\int_{-\pi}^\pi\frac{dx}{\sqrt{(t-2\cos x)^2-4}}$ I have the following integral,
$$I(t)=\int_{-\pi}^\pi\frac{dx}{\sqrt{(t-2\cos x)^2-4}},$$
where $t>4$ is a real parameter. I know from messing around numerically and playing with Mathematica that
$$I(t)=\frac{4}{t}K\left(\frac{16}{t^2}\... | Since $\cos(-x) = \cos(x)$ we can reduce the integration range to $(0,\pi)$, and then do the change of variable $u = \cos(x)$:
$$
\int\limits_{-\pi}^\pi \frac{\mathrm{d} x}{\sqrt{(t-2 \cos(x))^2-4}} = \int_0^\pi \frac{\mathrm{d} x}{\sqrt{\left(\frac{t}{2} - \cos(x)\right)^2 -1}} = \int_{-1}^1 \frac{2 \mathrm{d} u}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/182229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Finite summation
Possible Duplicate:
why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$
What is the proof without induction for :
$(1)$ $\sum_{i=1}^n\ i^2= \frac{n(n+1)(2n+1)}{6}$
$(2)$ $\sum_{i=1}^n\ i^3=\frac{n^2(n+1)^2}{4}$
| There are probably many ways. For example, if you already know that $\sum_{i=1}^n i = \frac{n(n+1)}{2}$, it follows $$3\sum_{i=1}^n i^2 + 3\sum_{i=1}^n i + n = \sum_{i=1}^n (3i^2 + 3i + 1) = \sum_{i=1}^n ((i+1)^3 - i^3) = (n+1)^3 - 1$$ since the last sum telescopes, so $$\sum_{i=1}^n i^2 = \frac{1}{3}((n+1)^3 - 1 - 3\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I solve a Continued Fraction of solution to quadratic equation? I know that it is possible to make a CF (continued fraction) for every number that is a solution of a quadratic equation but I don't know how.
The number I'd like to write as a CF is:
$$\frac{1 - \sqrt 5}{2}$$
How do I tackle this kind of problem?
| Suppose $x$ is a root of $p(z) = z^2 - b z - c$. Then, diving $p(z)$ over $z$ and solving that for $z$ gets us
$$
z = b+ \frac{c}{z}
$$
Iterating:
$$
z = b + \cfrac{c}{b + \cfrac{c}{z}} = \cfrac{b}{c + \cfrac{c}{b+ \ddots}}
$$
Since $\frac{1-\sqrt{5}}{2}$ is a root of $z^2 - z -1$ we have:
$$
\frac{1-\sqrt{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed? Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?
one of the digits which can be formed is $4444$
$4$ digit numbers greater than $3000$, which consists of only $2's$ ... | You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.
First place can have either $3$ or $4$. So we have two choices. Lets analyse it.
If first place is $3$ - so we have to choose 3 digits from $(2,2,3, 3,4,4,4,4)$. So any place can have either 2 or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the... | *
*$n=2k$:
$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16\\=(4k^2-1)(4k^2+5)+16\\=16k'^4+16k''^2+11\\=16k+11$$
Which is not $16k$.
*
*$n=2k+1$:
$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16\\=(4k^2+4k)(4k^2+4k+6)+16\\=8\underbrace{k(k+1)}
_{2k}(2k^2+2k+3)+16$$
Which is $16k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Prove that $\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$ let $ABC$ be an acute triangle with all angles greater than $45^o$
Prove that
$$\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$$
I let $\tan A=a$, $\tan B=b$, $\tan C=c$ with $a+b+c=abc$ then the inequali... | Putting $D=A-45^{\circ}$ etc, $D+E+F=(A+B+C-135^{\circ})=45^{\circ}$
$1+\tan A=1+\tan(D+45^{\circ})=\frac{2}{1-\tan D}$ appyling $\tan(A+B)=\frac{\tan A+ \tan B}{1-\tan A \tan B}$
So, $\frac{2}{1 + \tan A}=1-\tan D$
Now the problem reduces to minimize $\tan D+\tan E+ \tan F$ where $D+E+F=45^{\circ}$ and $D,E,F>0$
If we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/187938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
I know that, $S_{2n}+4S_{n}=n(2n+1)^2$. Is there a way to find $S_{2n}$ or $S_{n}$ by some mathematical process with just this one expression? $S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.
when, ... | Let $S_n=an^3+bn^2+cn+d$ where $a,b,c,d$ are rational numbers.
So, $S_{2n}+4S_n=n^3 12a+n^2 8b + n 6c+5d$
$\implies n^3 12a+n^2 8b + n 6c+5d= n(2n+1)^2=4n^3+4n^2+n$
Comparing the coefficients of the different powers on $n$,
$12a=4,8b=4,6c=1,d=0$
So, $6S_n=2n^3+3n^2+n=n(n+1)(2n+1)\implies S_n=\frac{n(n+1)(2n+1)}{6}$
Als... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Limit of equation as x tends to -1 I was given the following expression and had to find the limit as: $$ x \rightarrow 1, x \rightarrow - 1, x \rightarrow \infty $$
$$ \lim_{x \to -1} \frac{x^2 +3x +2}{x^2 -1} = \lim_{x \to -1} \frac{\frac{x^2}{x^2} + \frac{3x}{x^2} + \frac{2}{x^2}}{\frac{x^2}{x^2} - \frac{1}{x^2}} = \... | $$
\require{cancel}
\begin{equation*}
\lim \frac{x^2 +3x +2}{x^2 -1}= \lim \frac{\cancel{(x+1)}(x+2)}{\cancel{(x+1)}(x-1)}= \lim \frac{x+2}{x-1}=
\begin{cases} -\frac 12 & \text{if $x \to -1$,}
\\
+\infty &\text{if $x \to 1$.}
\end{cases}
\end{equation*}$$
When $x \to \infty$ you should consider the terms with the bigg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$ \sum_{k=1}^{\infty} \ln{\left(1 + \frac{1}{4 k^2}\right)}$ Computing this sum Compute the limit:
$$ \sum_{k=1}^{\infty} \ln{\left(1 + \frac{1}{4 k^2}\right)}$$
My teacher says it can be solved by only using high school knowledge, but I don't see how. What did I try? Well, I thought of Riemann sums but I see no way to... | As an alternative to @Seiros solution, albeit far less elegant is to use $$\log\left(1+\frac{1}{4k^2}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \frac{1}{4^{n+1} k^{2n+2}}$$
Then
$$\begin{eqnarray}
\sum_{k=1}^\infty \log\left(1+\frac{1}{4k^2}\right) &=& \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \frac{\zeta(2n+2)}{4^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Proving the inequality $a^2+b^2+c^2+ab+bc+ca\ge6$ Given that $a$, $b$, $c$ are non-negative real numbers such that $a+b+c=3$, how can we prove that:
$a^2+b^2+c^2+ab+bc+ca\ge6$
| $$ a^2 + b^2 + c^2 + ab + bc + ac = (a+b+c)^2 - (ab + bc + ac) = 9 - (ab + bc + ac)$$
Now it remains to show that max value of $(ab + bc + ac)$ is $3$. For that, we know the AM-GM equality ( for $a,b, c >0$ ) that $3(a^2 + b^2 + c^2) \geq (a+ b + c)^2 \geq 3(ab +bc +ac)$. From the last two part we have $(a+ b + c)^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
What is the right approach to use for factoring a rational inequality Here is an example:
$\frac{x^2 - 3x + 2}{x + 1} -5 > 0$
My approach would be to factor, find the undefined areas and the zeros, and then pick some points in the intervals left to see what I find. I'm not really sure if that's the right way to go, and... | This is not a general approach, but in this example $x=-1$ is 'special', so split the consideration into $I_- = (-\infty,-1)$ and $I_+ = (-1, +\infty)$.
On $I_+$, we have $x^2 - 3x + 2 - 5 (x + 1) = x^2-8x-3 > 0$. The factors are $4 \pm \sqrt{19}$, hence if $x \in I_+$, then $\frac{x^2 - 3x + 2}{x + 1} -5 > 0$ iff $x >... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Inequality. $\frac{a}{(b+c)^4}+\frac{b}{(c+a)^4}+\frac{c}{(a+b)^4} \geq \frac{3}{2(a+b)(b+c)(c+a)}$
For $a,b,c >0$ prove that :
$$\frac{a}{(b+c)^4}+\frac{b}{(c+a)^4}+\frac{c}{(a+b)^4} \geq \frac{3}{2(a+b)(b+c)(c+a)}.$$
I don't know how should I start. It is difficult for me because the denominators has the power equa... | The inequality can be written
$$
X := \sum_{cyc} \frac {a(a + b)(a + c)}{(b + c)^3} \geq \frac 3 2
$$
Let's suppose $a \leq b \leq c$. We have
$$
\frac 1 {b + c} \leq \frac 1 {a + c} \leq \frac 1 {a + b} \\
\frac {a(a + b)(a + c)} {(b + c)^2} \leq
\frac {b(b + c)(b + a)} {(c + a)^2} \leq
\frac {c(c + a)(c + b)} {(a + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate this limit: $\lim\limits_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$? I have difficulties in evaluating $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$$
Could you give me a hint how to start solving this? (I know the result is $3$)
Thanks a lot !
| For convenience, change variables to $t = x - 1$, so you're looking at
$\dfrac{\sqrt{4+t}-2}{(8+t)^{1/3}-2}$.
$$ \sqrt{4+t} = 2 \sqrt{1+\frac{t}{2}} = 2 \left(1 + \frac{t}{4} + \ldots\right) $$
$$ (8+t)^{1/3} = 2 \left(1 + \frac{t}{8}\right)^{1/3} = 2 \left(1 + \frac{t}{24} + \ldots\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/201782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$.
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$
I square the both sides and get $(5+2\sqrt{6})^{x}+(5-2\sqrt{6})^{x}=98$. But I don't know how to carry on. Please help. Thank you.
| Let $t=( \sqrt{5 + 2\sqrt{6}})^{x}\implies (\sqrt{5-2\sqrt{6}})^{x}=\frac{1}{t} $
Thus the equation becomes,
$ t+\frac{1}{t}= 10\implies t^2-10t+1=0$ which is a quadratic equation and have roots $t=5+2\sqrt 6,5-2\sqrt 6$
If $t=5+2\sqrt 6\implies ( \sqrt{5 + 2\sqrt{6}})^{x}=5+2\sqrt 6\implies (5+2\sqrt 6)^{1-x/2}=1\impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/202078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Finding domains? I need to find what $(f + g)(x)$ would equal here..
$$f(x) = \sqrt{25 − x^2},\quad g(x) = \sqrt{x^2 − 4}$$
Am I supposed to just add $\sqrt{25 − x^2}$ and $\sqrt{x^2 − 4}$? And if so, how? And how would I then determine the domain?
| If $f(x)=\sqrt{25-x^2}$ and $g(x)=\sqrt{x^2-4}$, then $(f+g)(x)=\sqrt{25-x^2}+\sqrt{x^2-4}$, so yes: to find the value of $f+g$ at any point $x$, just add the values of $f(x)$ and $g(x)$.
In the context of this question the domain of $f+g$ is the set of all real numbers $x$ for which $(f+g)(x)$ can be calculated as a r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/202505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to derive the equation for x in a quadratic equation?
Possible Duplicate:
Why can ALL quadratic equations be solved by the quadratic formula?
How to derive this:
$x = \frac{-b + {\sqrt{b^2 + 4ac}}}{2a}$
From this:
$ax^2 + bx + c = 0$
I know this may be a little elementary :)
| from $ax^2 + bx + c = 0 \Leftrightarrow 4a^2x^2 + 4abx + b^2 = b^2 - 4ac$.
Hence $(2ax + b)^2 = b^2 - 4ac$ then $2ax + b = \pm \sqrt{b^2 - 4ac}$.
from this, we have $ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/203195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Help? I cannot do the integration by parts correctly?? Can someone please show me how to integrate
$$\int_0^\infty\frac4{\pi b^2}x^2e^{-x^2/b^2}dx\;?$$
please show steps how to integrate this problem. This is what i have so far.
$$\frac4{\pi b^2}\int_0^\infty x^2 e^{-x^2/b^2}dx\;.$$
Then i take the property I^2 = $$\i... | $$\begin{align*}u=&x&u'=&1\\v'=&xe^{-x^2/b^2}&v=&-\frac{b^2}{2}e^{-x^2/b^2}\end{align*}\;\;\;\;\;\;\Longrightarrow$$
$$\Longrightarrow \int_0^\infty x^2e^{-x^2/b^2}dx=\left.-\frac{b^2x}{2}e^{-x^2/b^2}\right|_0^\infty+\frac{b^2}{2}\int^\infty_0e^{-x^2/b^2}dx=\frac{b^3}{2}\sqrt\frac{\pi}{2}$$
And thus your integral equal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/204740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How do you find a polynomial with integer coefficients with this sum of radicals as a root? Let $$x=\sqrt{a + \sqrt{b}} + \sqrt{c + \sqrt{d}}$$
How do you find the polynomial with this value as a root? Where a, b, c, and d are integers.
| For the sake of defending my honor, here's most of the solution I alluded to involving squaring. It is somewhat messy, and the solution alluded to by Patrick Da Silva is nicer.
First square:
$$x^2 = a + \sqrt{b} + 2 \sqrt{(a + \sqrt{b})(c + \sqrt{d})} + c + \sqrt{d}.$$
Second square:
$$(x^2 - a - \sqrt{b} - c - \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/208487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ I am quite new to generating functions concept and I am really finding it difficult to know how to approach problems like this. I need to find the sum of $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ using generating functions. How do I proceed about it?
| Note that if $A(z) = \sum_{n \ge 0} a_n z^n$, then
$$
z \frac{\mathrm{d}}{\mathrm{d} z} A(z)
= \sum_{n \ge 0} n a_n z^n
$$
and also:
$$
\frac{A(z)}{1 - z}
= \sum_{n \ge 0} \left( \sum_{0 \le k \le n} a_k \right) z^n
$$
Starting with:
$$
\sum_{n \ge 0} z^n
= \frac{1}{1 - z}
$$
the generating function for the sum y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/209268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 3
} |
how to find the series $x + x^{1 + \frac{1}{2}} + x^{1 + \frac{1}{2}+ \frac{1}{3}} +...$ is convergent. The series $x + x^{1 + \frac{1}{2}} + x^{1 + \frac{1}{2}+ \frac{1}{3}} +...$
is convergent if
(A) $x>e$
(B) $x<e $
(C) $x<1/e$
(D) $x>1/e$
I think the answer is C, but I could not determine the condition... how to... | We have
\begin{align*}
x^{\sum_{i=1}^n \frac 1i} &\le x^{\log n + 1}\\
&= x \cdot \exp(\log n\cdot \log x)\\
&= x \cdot n^{\log x}
\end{align*}
and
\begin{align*}
x^{\sum_{i=1}^n \frac 1i} &\ge x^{\log n}\\
&= \exp(\log n\cdot \log x)\\
&= n^{\log x}
\end{align*}
and hence
\[ \sum_{n=1}^\infty x^{\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/209711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Common tangent to two circles
Find the equations of the common tangents to the 2 circles:
$$(x - 2)^2 + y^2 = 9$$
and
$$(x - 5)^2 + (y - 4)^2 = 4.$$
I've tried to set the equation to be $y = ax+b$, substitute this into the 2 equations and set the discriminant to zero, we then get a simultaneous quadratic equation... | Like in my other answer, the equation of any tangent of $$(x-2)^2+y^2=3^2$$ will be $$x\cos t+y\sin t-(2\cos t+3)=0$$
Now if this has to be tangent of $(x-5)^2+(y-4)^2=4$ if the radius $=$ the distance from the center$(5,4)$
$$\dfrac{|5\cos t+4\sin t-(2\cos t+3)|}{\sqrt{\cos^2t+\sin^2t}}=2$$
$$\implies4\sin t+3\cos t-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/211538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Frequency of a trigonometric function - Where is my mistake? I need to find the frequency of the following trigonometric function.$$y=\sin^4(x)+\cos^4(x)$$
The "answers" section says the answer is: $$F_y=\frac{\pi}{2}$$
This is what i did:
Finding $\sin(x)^4$ frequency (I'll call it F1):
$$\cos(2x)=1-\sin^2(x)$$
$$\sin... | You might consider using Euler's formula, which can be used to obtain
$$ \cos(x) = \frac{e^{ix} + e^{-ix}}{2} $$ and
$$ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}. $$
Put $z = e^{ix}$, so that
$$\cos(x) =\frac{z+1/z}{2}$$
and
$$\sin(x) = \frac{z-1/z}{2i}$$
This gives
$$\cos(x)^4+\sin(x)^4 =
\left(\frac{z+1/z}{2}\right)^4 +... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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When is $1^5 + 2^5 + \ldots + n^5$ a square? When is $1^5 + 2^5 + \ldots + n^5$ a square? I found that this happens sometimes: $n=13$ gives $1001^2$, $n=133$ gives $9712992^2$ and $n=1321$ gives $942162299^2$.
I feel that the identity$$\displaystyle\sum_{i=1}^n i^5 = \tfrac{1}{12}[2n^6+6n^5+5n^4-n^2]$$ will be useful, ... | This is OEIS A031138, which lists some more and says $a(n) =11\cdot(a(n-1)-a(n-2)) + a(n-3) \\ a(n)=-1/2+((3-\sqrt 6)/4)\cdot(5+2\sqrt 6)^n+((3+\sqrt 6)/4)\cdot(5-2\sqrt 6)^n$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
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Expected Number of Successes in a Sample
$200$ calculators are ordered and of those $200$, $20$ are broken. $10$ calculators are selected at random. Calculate the expected value of broken calculators in the selection.
Solution:
Chance of broken calculator: $\dfrac{1}{10}$.
Do I need to calculate the odds of $0$ - $1... | Your shorter formula doesn't work. Suppose the question asked you to select two calculators instead of ten. To compute the expected value "by hand", there are three cases to consider (we don't care about the case when no calculators are broken).
First Calculator Broken: This event happens with probability $(1/10)(9/10)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to eliminate these extra solutions? (finding the gcd of two expressions)
Prove that for any integer $n$, $\gcd (3n^2+5n+7, n^2+1)=1$ or $41$.
The following answer is convoluted because I've intentionally created excess solutions. However, I can't figure out how to eliminate them! Anyone?
Let $$d=\gcd (3n^2+5n+7, ... | Suppose that
$$
(3n^2+5n+7,n^2+1)=(5n+4,n^2+1)\ne1\tag{1}
$$
then either
$$
(5n+4,n+i)=(4-5i,n+i)\ne1\tag{2}
$$
or
$$
(5n+4,n-i)=(4+5i,n-i)\ne1\tag{3}
$$
Since $4-5i$ is a Gaussian prime, $(2)\Rightarrow4-5i\,|\,n+i$. That is,
$$
\frac{n+i}{4-5i}=\frac{(4n-5)+(5n+4)i}{41}\in\mathbb{Z}[i]\tag{4}
$$
which is true if and ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$\cos^n x-\sin^n x=1$ For $0 < x < 2\pi$ and positive even $n$, the only solution for $\cos^n x-\sin^n x=1$ is $\pi$.
The argument is simple as $0\le\cos^n x, \sin^n x\le1$ and hence $\cos^n x-\sin^n x=1$ iff $\cos^n x=1$ and $\sin^n x=0$.
My question is that any nice argument to show the following statement?
'For $0 <... | We leave the case $n = 1$ and $n = 2$ separately, and assume $n \geq 3$ from now on.
Observe that if $|r| \leq 1$, then $|r^n| \leq r^2$ with equality if and only if $r = 0$ or $|r| = 1$. Then it follows that
$$1 = \left|\cos^n x - \sin^n x\right| \leq \left|\cos^n x\right| + \left|\sin^n x\right| \leq \cos^2 x + \sin^... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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What is the number of trailing zeros in a factorial in base ‘b’? I know the formula to calculate this, but I don't understand the reasoning behind it:
For example, the number of trailing zeros in $100!$ in base $16$:
$16=2^4$,
We have:
$\frac{100}{2}+\frac{100}{4}+\frac{100}{8}+\frac{100}{16}+\frac{100}{32}+\frac{100}{... | Suppose that $b=p^m$, where $p$ is prime; then $z_b(n)$, the number of trailing zeroes of $n!$ in base $b$, is
$$z_b(n)=\left\lfloor\frac1m\sum_{k\ge 1}\left\lfloor\frac{n}{p^k}\right\rfloor\right\rfloor\;.\tag{1}$$
That may look like an infinite summation, but once $p^k>n$, $\left\lfloor\frac{n}{p^k}\right\rfloor=0$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/226868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Question involving some roots of the equation $31z^{15} - z^{10} + 32 = 0 $ Consider the equation $$31z^{15} - z^{10} + 32 = 0.$$ What would be the sum of all those roots of the equation whose real part is positive? Only trivially trying to solve the equation I find not helpful. Even factorizing we get , by putting $z^... | $31z^{15}-z^{10}+32=0\implies 31(z^{15}+1)-(z^{10}-1)=0$
$31(z^5+1)(z^{10}-z^5+1)-(z^5+1)(z^5-1)=0$
$(z^5+1)\{31(z^{10}-z^5+1)-(z^5-1) \}=0 $
$(z^5+1)\{31z^{10}-32z^5+32 \}=0 $
If $z^5+1=0, z^5=-1=e^{(2m+1)\pi i} $ where $m$ is any integer.
So, $z=e^{\frac{(2m+1)\pi i}5}$ where any $5$ in-congruent values of $m\pmod 5$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/227324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding Lebesgue Integral of $\frac{1}{\sqrt{x}}$ over $(0,1]$ How do I rigorously discover what
$$
\int_{(0,1]} \frac{1}{x^{1/2}} = \underset{0 \le \phi \le \frac{1}{\sqrt{x}}}{\sup} \int_{(0,1]} \phi
$$
(for $\phi$ a simple function) is? Note that I have already shown in another exercise that such an integral exists... | Here is a brute force approach using the dominated convergence theorem (DCT). The idea is to create a sequence of simple functions $s_n$ such that $s_n(x) \leq \frac{1}{\sqrt{x}}$, and $s_n(x) \to \frac{1}{\sqrt{x}}$. The DCT shows that $\int s_n \to \int \frac{1}{\sqrt{x}}$.
Let $s_n = \sum_{k=n}^{n^2-1} \frac{k}{n} 1... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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How to find solutions of $x^2-3y^2=-2$? According to MathWorld,
Pentagonal Triangular Number: A number which is simultaneously a pentagonal number $P_n$ and triangular number $T_m$. Such numbers exist when
$$\frac{1}{2}n(3n-1)=\frac{1}{2}m(m+1).$$
Completing the square gives
$$(6n-1)^2-3(2m+1)^2=-2.$$
Substitu... | Here's another approach.
$$X^2-AY^2=B\tag1$$
$$x^2-Ay^2=1\tag2$$
If we know fundamental solution $(a,b/A)$ for $(2)$ and “trivial” solutions $(t,v)$ for $(1)$ then:
$$X_n = \sum_{k=0}^{n}\frac{a^{n-k}b^k\displaystyle\binom{n}{k}\left(\left(\left\lceil\frac{k}{2}\right\rceil -\left\lfloor\frac{k}{2}\right\rfloor\right)v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/228356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 5
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As shown in the figure: Prove that $a\,+\,b\,+\,c=d$ Geometry: Auxiliary Lines
As shown in the figure: Prove that $a\,+\,b\,+\,c=d$
|
Using sine law of triangle, $$\frac e{\sin(144^{\circ}-x)}=\frac g{\sin 36^{\circ}}$$ and $$\frac e{\sin(138^{\circ}-x)}=\frac {g+f}{\sin 42^{\circ}}$$
So, $$f=e\left(\frac{\sin 42^{\circ}}{\sin(138^{\circ}-x)}-\frac{\sin 36^{\circ}}{\sin(144^{\circ}-x)}\right)$$
$$=e\left(\frac{\sin 42^{\circ}\sin(144^{\circ}-x)-\sin... | {
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"source": "stackexchange",
"question_score": "6",
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Understanding simplification of an algebra problem I have the problem $x^3 - 2x^2$. My book tells me that this problem is simplified to $x^3 (1 -(\frac{2}{x}))$. How does that work?
This step of my book I am in about the "end behavior" of trying to graph a polynomial function. Apparently once you get $x^3 (1 -(\frac{2}... | This is only valid when $x \neq 0$. It follows from the distributive rule. If $x \neq 0$, then
\begin{align*}
x^3 \left(1 - \frac{2}{x} \right) &= x^3 \cdot 1 - x^3 \cdot \frac{2}{x} \quad \text{(by distributive rule)}\\
&= x^3 - 2x^2.
\end{align*}
As Cameron Buie explained nicely, $x^3$ and $x^3(1 - \frac{2}{x})$ ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/231279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate $3^{45357} \mod 5$? I wrote some code, here is what it gives:
\begin{align*}
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
3^7 \mod 5 = 2 \\\\
3^8 \mod 5 = 1 \\
3^9 \mod 5 = 3 \\
3^{10} \mod 5 = 4 \\
3^{11} \mod 5 = 2 \... | Using Fermat's Little theorem,
$a^{p-1}\equiv1\pmod p$ if $(a,p)=1$
$\implies (a^{p-1})^d\equiv1^d\pmod p \equiv1 \pmod p$
if $b=c+(p-1)d$ i.e., $b\equiv c\pmod {p-1}$ where $a,b,c,d$ are non-negative integers and $p$ is prime,
$ a^b=a^{c+(p-1)d}=a^c\cdot(a^{p-1})^d\equiv a^c\pmod p$
Here, $3^{5-1}\equiv 1\pmod 5$ as... | {
"language": "en",
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"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Integration questions on $\int \frac{x^4\left ( 1-x \right )^4}{1+x^2} \, dx$ and $\int \frac{x^4}{x^4+5x^2+4} \, dx$ I would appreciate any hints on how to solve the following integration problems, they are my homework questions btw:
$$\int \frac{x^4\left ( 1-x \right )^4}{1+x^2} \, dx$$
$$\int \frac{x^4}{x^4+5x^2+4} ... | Here are a couple of off-hand suggestions. There are undoubtedly more efficient ways.
For the first integral, you could multiply out the top, and use polynomial long division. You will get something that has the shape $P(x)+\frac{Ax}{1+x^2}+\frac{B}{1+x^2}$.
Integrating the polynomial will be easy. For $\frac{Ax}{1+x^2... | {
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"answer_count": 1,
"answer_id": 0
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Interpolation of a function Given the function
$$f (x) = x\bigg(x − {1\over4}\bigg)\bigg(x − {1\over2}\bigg)$$
How can I interpolate $f(x)$ with $p(x) = a_0T_0(x) + a_1T_1(x) + a_2T_2(x) + a_3T_3(x)$ to show that
$$a_0 = -{3\over8},\ \ a_1 = {7\over8},\ \ a_2 = −{3\over8},\ \ a_3 = {1\over4}$$
| First, note that, the Chebyshev polynomials have the property
$$\int_{-1}^1 T_n(x)T_m(x)\,\frac{dx}{\sqrt{1-x^2}}= \begin{cases} 0 &: n\ne m \\ \pi &: n=m=0\\ \pi/2 &: n=m\ne 0 \end{cases}\rightarrow (*)\,.$$
The first four polynomials are $ T_0(x)=1, T_1(x)=x, T_2=2x^2-1, T_3=4 x^3-3x.$
Now, we have
$$ f(x) \sim a_0T_... | {
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"source": "stackexchange",
"question_score": "1",
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Parameter values that make function values side lengths of a triangle I have been trying to solve the following problem for more than a week without any success.
Given the function:
$$f(x)=\frac{x^2+mx+4}{x^2+x+4}$$
Find all possible values of the parameter $m$ such that for any three numbers $a,b,c$ the corresponding... | If $f(a), f(b), f(c)$ are sides of a triangle if they satisfy the triangle inequalities
$$ f(a) + f(b) > f(c), \quad f(a) + f(c) > f(b), \quad f(b) + f(c) > f(a).$$
If we require that this holds for every $a,b,c$, then by symmetry we only need to prove that, for all $a,b,c$, $f(a) + f(b) > f(c)$. Writing out this inequ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Prove via induction $\frac{(n+1)^2}{2^n}\le\frac{9}{4}$ How do you prove in induction that:
$$\frac{(n+1)^2}{2^n}\le\frac{9}{4}$$
This is what I keep getting:
Checking for $n=1$ we get $2\le\frac{9}{4}$.
Assuming it's true for $n$ and checking for $n+1$ I get this:
$$\frac{(n+2)^2}{2^{n+1}}=\frac{2(n+1)^2-n^2+2}{2\time... | The simple answer is to check $n=2$ too, where you have
$$\frac{(n+1)^2}{2^n}= \frac{(2+1)^2}{2^2} = \frac{9}{4} \le\frac{9}{4}$$ and then do the induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$ $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$$
I tried to say we can erase the $1$ from the equation, as it's a constant. But I don't know how to do the rest without running into this mistake... | My answer here
(Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $)
shows that
$\sqrt[a]{n^a+n^{a-c}}
=n+\dfrac{1}{an^{c-1}}+O(n^{-(2c-1)})
$.
If $a=3, c=1$,
$\sqrt[3]{n^3+n^{2}}
=n+\dfrac{1}{3}+O(n^{-1})
$.
If $a=3, c=3$,
$\sqrt[3]{n^3+1}
=n+\dfrac{1}{3n^2}+O(n^{-5})
=n+O(n^{-2})
$.
Their... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/236901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determining the number $N$ Let $1 = d_1 < d_2 <\cdots< d_k = N$ be all the divisors of $N$ arranged in increasing order. Given that $N=d_1^2+d_2^2+d_3^2+d_4^2$, determine $N$. The divisors include $N$. It seems that $130$ is an answer. Is there another possible answer for $N$?
| $N$ is even (if not then all $d_i$ are odd, making $\sum_{i=1}^4 d_i^2$ even). Therefore $d_1=1$ and $d_2=2$, and at exactly one of $d_3$ and $d_4$ is even.
Suppose that $4 \mid n$. Then one of $d_3, d_4$ is $4$ and the other is an odd prime $p$. Since $N=21+p^2$ and $p \mid N$, we have $p \mid 21$. But $4 \nmid 21... | {
"language": "en",
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"source": "stackexchange",
"question_score": "9",
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solving the equation $x^4-5x^3+11x^2-13x+6=0$ by given condition 1.(a) solve the equation $x^4-5x^3+11x^2-13x+6=0$ , given that two of its roots $p$ & $q$ are connected by the relation $3p+2q=7$
(b) solve the equation $x^4-5x^3+11x^2-13x+6=0$ which has two roots whose difference is $1$
did I need to solve these problem... | Let $$f(x) = x^4 - 5x^3 + 11x^2 - 13x + 6$$ We get that $$f(1) = 1 - 5 + 11 - 13 + 6 = 0$$ and $$f(2) = 16 - 5 \times 8 + 11 \times 4 - 13 \times 2 + 6 = 16 - 40 + 44 - 26 + 6 = 0$$
Hence, we have that $$f(x) = (x-1)(x-2)(x^2 + ax + b)$$ Plugging in $x=0$, we get that
$$f(0) = 2b = 6 \implies b = 3.$$ Plugging in $x=3$... | {
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"source": "stackexchange",
"question_score": "1",
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Find the maxium $\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}$ 1) $a, b, c$ are triangle edges's length such that $abc = 1$. Find max:
$$\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}$$
My idea: $$\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}=\f... | Hint: For your second question, subtracting the equations in pairs suggests the substitution $$a=x+2,\quad b=y+2, \quad c=z+2.$$
Try this in the original set of equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/243077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this series convergent or divergent? Kindly asking, what can I do about series
$$
\left(\frac{1}{3}\right)^2+\left(\frac{1\times 4}{3\times 6}\right)^2+\left(\frac{1\times 4\times 7}{3\times 6\times 9}\right)^2+...+\left(\frac{1\times 4\times 7\times...\times (3n-2)}{3\times 6\times 9\times...\times3n}\right)^2+...$... | $$\frac{1\times 4\times 7\times\cdots\times (3n-2)}{3\times 6\times 9\times\cdots\times3n}\leqslant\frac1{(n+1)^{2/3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/245925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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An algebra problem related to trigonometry Let $\alpha $ be an acute angle such that $\sin \alpha = \dfrac{3 \sqrt 3}{14}$.
Prove that
$$\frac{2\cdot7^{n}}{\sqrt 3}\sin \left(n\alpha + \dfrac{\pi}{3} \right) \in \mathbb{Z} \qquad\forall n>0. $$
| This is equivalent to saying that for all $n>0$
$$
\frac{2\cdot7^{n}}{\sqrt{3}}\sin\left(n\alpha+\frac\pi3\right)\in\mathbb{Z}
$$
Note that $\cos(\alpha)=\frac{13}{14}$, so that
$$
e^{i\alpha}=\frac{13}{14}+i\frac{3\sqrt3}{14}
$$
therefore
$$
\begin{align}
e^{i(n\alpha+\pi/3)}
&=\left(\frac{13}{14}+i\frac{3\sqrt3}{14}\... | {
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"source": "stackexchange",
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How to solve second degree recurrence relation? For first degree recurence relation it is as simple as $f(n)=a^n\cdot f(0)+b\dfrac{a^n-1}{a-1}$.
But how do you solve second degree?
For example
$$f(n)=\begin{cases}
1,&\text{for }n=1\\
2,&\text{for }n=2\\
-3f(n-1)+4f(n-2),&\text{for }n>2\;.
\end{cases}$$
I tried googling... | Or use generating functions. Rewite the recurrence as:
$$
f(n + 2) = - 3 f(n + 1) + 4 f(n) \qquad f(1) = 1, f(2) = 2
$$
Define:
$$
F(z) = \sum_{n \ge 0} f(n + 1) z^n
$$
By the properties of ordinary generating functions (see e.g. Wilf's "generatingfunctionology"):
$$
\frac{F(z) - f(1) - f(2) z}{z^2} = - 3 \frac{F(z) - ... | {
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Confused about an integral from MIT integration bee 2012 One of the integrals is:
$$\int \frac{\mathrm{d}x}{2+2\sin x + \cos x}\, \mathrm{d}x $$
How can there be two $\mathrm{d}x$?
MIT's Integration Bee
| The extra $dx$ is definitely a typo. To solve this integral we have to substitute $u=\tan\frac{x}{2}$. Then, $\sin x=\frac{2u}{1+u^2}$, $\cos x=\frac{1-u^2}{1+u^2}$ and $dx=\frac{du}{1+u^2}$. Thus,
$$\int\frac{dx}{2+2\sin x+\cos x}=\int\frac{du}{2+2u^2+4u+1-u^2}=\int\frac{du}{u^2+4u+3}=\int\frac{du}{(u-1)(u-3)}=\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/255508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
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Exercise about MacLaurin's polynomial and small-o In class the professor wrote the following limit:
$\lim_{x\to 0} \frac{\sinh^2 (x) -x^2}{x^4}$
So he "expanded" (sorry for my English) the MacLaurin's formula for $\sinh x$ up to the 3rd power, and got: $x + \frac{x^3}{3!} + o(x^4)$
When he squared the MacLaurin's poly... | The $o(x^7)$ and $o(x^8)$ are "absorbed" into the $o(x^5)$.
Suppose for example that $f(x)=o(x^7)$. Formally, what that means is that
$$\lim_{x\to 0} \frac{f(x)}{x^7}=0.$$
We show that $f(x)=o(x^5)$. Suppose that $f(x)=o(x^7)$. Then
$$\lim_{x\to 0}\frac{f(x)}{x^5}=\lim_{x\to 0}x^2\frac{f(x)}{x^7}=(0)(0)=0.$$
Informal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/256089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a matrix $B$ such that $B^2=A$ $A= \begin{pmatrix} 1 & 2+3i \\ 2-3i & -1 \end{pmatrix}$
What is matrix $B$ such that $B^2=A$?
Its eigenvalues are $\sqrt{14}, -\sqrt{14}$
and I tried to use formula $B=U\sqrt{\lambda}U^{H}$ where $U$ is unitary matrix.
But then $\sqrt{ }$ of $-\sqrt{14}$ is not possible.
How can... | The characteristic polynomial of $A$ is $\lambda^2 - 14$, so $A^2 - 14 I = 0$.
Thus $(a A + bI)^2 = (14 a^2 + b^2) I + 2 a b A$. To make this equal to $A$, we want
$2ab = 1$ and $14 a^2 + b^2 = 0$. Thus $b = 1/(2a)$ and $14 a^2 = -1/(4a^2)$, or
$a^4 = -1/56$. Take $a$ to be any of the four fourth roots of $-1/56$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/256757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
| Hint $\,\ c\mid a\!+\!b,\overbrace{a^2\!-\!ab\!+\!b^2}^{\large f(b)}\Rightarrow\,c\mid 3a^2,3ab,3b^2\Rightarrow\,c\mid(3a^2,3ab,3b^2)=3(a,b)^2 = 3 $
by $\,\ {\rm mod}\ \ a\!+\!b\!:\ \ \color{#c00}{b\equiv -a}\,\Rightarrow\, f(\color{#c00}b)\equiv f(\color{#c00}{-a})\equiv 3a^2\equiv 3a(-b)\equiv 3(-b)(-b) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/257392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that if $a^x=b^y=(ab)^{xy}$, then $x+y=1$ using logarithms
Prove that if $a^x=b^y=(ab)^{xy}$, then $x+y=1$.
How do I use logarithms to approach this problem?
| Using logarithms:
Since $a^x = b^y$,
$$
\log a^x = \log b^y \quad \Rightarrow \quad x \log a = y \log b
\quad \Rightarrow \quad \log a = \frac{y}{x} \log b
$$
Then, since $b^y = (ab)^{xy}$,
$$
\log b^y = \log (ab)^{xy}
\quad \Rightarrow \quad
y \log b = xy \log (ab) = xy \left( \log a + \log b\right)
$$
Let's... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/258332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove that $\int_0^{+\infty} \frac{\ln x}{a^2+x^2} dx = \frac{\pi\ln a}{2a}$ Is that true $$\int_0^{+\infty} \cfrac{\ln x}{a^2+x^2} dx = \cfrac{\pi\ln a}{2a},$$ where $a>0$ ? And how to compute it?
| Let
$$
\mathcal{I}(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}\ dx.
$$
Using substitution $u=\dfrac{a^2}{x}\;\Rightarrow\;x=\dfrac{a^2}{u}\;\Rightarrow\;dx=-\dfrac{a^2}{u^2}\ du$ yields
\begin{align}
\mathcal{I}(a)&=\int_0^\infty\frac{\ln \left(\dfrac{a^2}{u}\right)}{\left(\dfrac{a^2}{u}\right)^2+a^2}\cdot \dfrac{a^2}{u^2}\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/260621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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Finding the number of points on the straight line joining $(-4,11)$ and $(16,-1)$ Find the number of points on the straight line which joins $(-4,11)$ and $(16,-1)$ whose coordinates are positive integers.
a) $1$
b) $2$
c) $3$
d) $4$
| The slope of this line is $$\frac{-1-11}{16-(-4)}=\frac{-12}{20}=-\frac35\;.$$ Thus, the equation of the line is $y-11=-\frac35(x-(-4))=-\frac35x-\frac{12}5=-\frac15(3x+12)$. Clearly $y$ is an integer precisely when $3x+12$ is a multiple of $5$, which is the case precisely when $x+4$ is a multiple of $5$. These values ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/262548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help with Inequality Given that $x, y, z$ are nonnegative real numbers such that :
$$x^2 + y^2 + z^2 + xyz = 4$$
Prove that
$0 ≤ xy + yz + zx − xyz ≤ 2$
| Let $x=\frac{2a}{\sqrt{(a+b)(a+c)}}$ and $y=\frac{2b}{\sqrt{(a+b)(b+c)}}$, where $a$, $b$ and $c$ be positives.
Hence, $z=\frac{2c}{\sqrt{(a+c)(b+c)}}$ and the left inequality it's
$$4\sum_{cyc}\frac{ab}{(a+b)\sqrt{(a+c)(b+c)}}\geq\frac{8abc}{(a+b)(a+c)(b+c)}$$ or
$$\sum_{cyc}ab\sqrt{(a+c)(b+c)}\geq2abc,$$
which is tru... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/262608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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limits calculus I am having trouble understanding part of the solution to this simple problem.
$\lim_{x \to 2} (x^2 + 3x) = 10$
Solution:
Let $\epsilon > 0$
$| x - 2 | < \delta$ and $| x^2 +3x -10 | < \epsilon$
since $x^2 +3x -10 = (x - 2)^2 + 7x -14 = (x - 2)^2 + 7x -14 = ( x -2 )^2 +7(x-2)$
$|(x-2)^2 +7(x-2)| \leq |... | We have been challenged with an $\epsilon$, perhaps $\epsilon=1/1000$. We want to come up with a $\delta$ such that if $|x-2|\lt \delta$, then for sure $|x^2+3x-10|\lt \epsilon$.
Suppose that after some calculation, we announce that $\delta=\epsilon/8$ does the job. Then triumphantly the challenger could say that she ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $m,n\in\mathbb{N}$ and $f(m) \mid f(n)$ for the $f$ given below, must $m=n$? If for two natural number $m$ and $n$, $(2^{2m+1}-1)2^{4m-2}(2^{2m+1}+2^{m+1}+1)\mid(2^{2n+1}-1)2^{4n-2}(2^{2n+1}+2^{n+1}+1)$, then $m=n$?
| False as stated. The divisibility also holds for example, when $n=1$ and $m=13$.
When $m=1$, the left hand side is
$$
(2^3-1)2^2(2^3+2^2+1)=7\cdot4\cdot13.
$$
When $n=13$ we obviously have that
$$
(2^3-1)\mid(2^{27}-1)\qquad\text{and}\qquad2^2\mid2^{4\cdot13-2}.
$$
The last factor is a bit trickier, but from Little Fe... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help me solve this olympiad challenge? Given:
$$p(x) = x^4 - 5773x^3 - 46464x^2 - 5773x + 46$$
What is the sum of all arctan of all the roots of $p(x)$?
| Let $p_1,p_2,p_3,p_4$ be the roots of $p(x)=0$
So, we need $\arctan p_1+\arctan p_2+\arctan p_3+\arctan p_4$
Using Vieta's Formulae,
$\sum p_i=p_1+p_2+p_3+p_4=\frac{5773}1$
$\sum p_ip_j=p_1p_2+p_1p_3+p_1p_4+p_2p_3+p_2p_4+p_3p_4=\frac{46464}1$
$\sum p_ip_jp_k=p_1p_2p_3+p_1p_2p_4+p_1p_3p_4+p_2p_3p_4=\frac{5773}1$
$p_1p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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A relationship between matrices, bernoulli polynomials, and binomial coefficients We define the following polynomials, for $n≥0$:
$$P_n(x)=(x+1)^{n+1}-x^{n+1}=\sum_{k=0}^{n}{\binom{n+1}{k}x^k}$$
For $n=0,1,2,3$ this gives us,
$$P_0(x)=1\enspace P_1(x)=2x+1\enspace P_2(x)=3x^2+3x+1\enspace P_3(x)=4x^3+6x^2+4x+1$$
We the... | Here's another way to look at this. The Bernoulli polynomials can be defined by the property
$$\int_x^{x+1} B_n(u) \, du = x^n.$$
So if we let $T$ be the operator from the set of polynomials to itself given by $(Tf)(x) = \int_x^{x+1} f(u) \, du$, then we have $(TB_n)(x) = x^n$. The operator $T$ sends $x^n$ to $$\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Find the acute angle $x$ for $\tan x = \tan(x+10^\circ)\tan(x+20^\circ)\tan(x+30^\circ)$. How to solve the following equation?
$$\tan x= \tan(x+10^\circ)\tan(x+20^\circ)\tan(x+30^\circ)$$
| $$\frac{\sin x\cos(x+10)}{\cos x\sin(x+10)}=\frac{\sin(x+20)\sin(x+30)}{\cos(x+20)\cos(x+30)}$$
Applying componendo and dividendo,
$$\frac{\cos x\sin(x+10)+\sin x\cos(x+10)}{\cos x\sin(x+10)-\sin x\cos(x+10)}
=\frac{\cos(x+20)\cos(x+30)+\sin(x+20)\cos(x+30)}{\cos(x+20)\cos(x+30)-\sin(x+20)\cos(x+30)}$$
$$\frac{\sin(2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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please solve a 2013 th derivative question? $ f(x) = 6x^7\sin^2(x^{1000}) e^{x^2} $
Find $ f^{(2013)}(0) $
A math forum friend suggest me to use big O symbol, however have no idea what that is, so how does that helping?
| Note that,
$$ 6\,x^{7} \sin\left(x^{1000}\right)\sin\left(x^{1000}\right)e^{x^2} $$
$$ = 6\,x^{7} \left( x^{1000}-\frac{x^{3000}}{3!}+\dots \right)\left( x^{1000}-\frac{x^{3000}}{3!}+\dots \right)\left(1+\frac{x^2}{1!}+\frac{x^4}{4!}+\dots\right) $$
$$ = 6x^7x^{2000}\left( 1-\frac{x^{2000}}{3!} +\dots\right)^2\left(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/267846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 2,
"answer_id": 0
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$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$
Find $$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$$
I know that $\lim_{x \to 0} \frac{\sin x}{x}=1$ But here $\sin \left(\frac{1}{3 \sqrt{x}}\right)$ is given when $x \to 0$. Need help.
| If we let $1/x=y$
$$\lim_{y \to \infty}\frac{\sin \left(\frac{\sqrt{y}}{3 }\right)}{|y|\sqrt{7}}=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Check convergence - look complicated $$\sum_{n=2}^\infty(\cos((n^{3}+\sqrt{n}+7)^{\frac{1}{3}})-\cos((n^{3}-2\sqrt{n}+3)^{\frac{1}{3}}))$$
Check convergence.
Please verify my answer below.
$$\sum_{n=2}^\infty(\cos((n^{3}+\sqrt{n}+7)^{\frac{1}{3}})-\cos((n^{3}-2\sqrt{n}+3)^{\frac{1}{3}}))=$$
$$=\sum_{n=2}^\infty-2\sin\... | Let me suggest some simplifications. The $n$th term is the difference of two cosines and the function cosine is $1$-Lipschitz hence each term is at most the difference of the two cube roots. Furthermore, the derivative of the cube root function is decreasing hence for every $a\gt b\gt0$,
$$
a^{1/3}-b^{1/3}=\int_b^a\fra... | {
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If $r = \min \left\{ a , \frac{b}{ a+ a^2 b^3} \right\}$ , find $ r_{\max}$ Studying differential equations I came cross through this:
Let $ \displaystyle{ r = \min \left\{ a , \frac{b}{ a+ a^2 b^3} \right\} } $, where $ a,b >0$. Find $ r_{ \max} $.
Here is what I did:
Fix $ a> 0$ and define $ \displaystyle{ g(b) = \fr... | It looks correct to me. The reason that is correct is because you're finding when the two values are equal. If they aren't equal, then since the real numbers are totally ordered, we know that one will be less than the other.
Solving for $a$, we have $$a^{7/3}=\frac{2^{2/3}}{3}.$$ Taking the $3/7$ root of each side yiel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Use Riemann Sums to Find Area bounded by Curve How can I find the area bound by $\;x=0,\, x=1,\;$ the $\;x$-axis ($y = 0$) and $\;y=x^2+2x\;$ using Riemann sums?
I want to use the right-hand sum. Haven't really found any good resources online to explain the estimation of areas bounded by curves, hoping anyone here can... | Right Riemann Sums place the right corner of the rectangles on the curve.
Right Riemann Sums are an overestimation of area because of all the extra space that is not under the curve that is still calculated in the area because it is inside the rectangles.
For your problem, we have:
$f(x) = x^{2} + 2x$
With: $a = 0, b =... | {
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"url": "https://math.stackexchange.com/questions/281450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of strings of length n formed by $\{0,1,2\}$ such that $1$ and $2$ do not occur successively. What is the number of ways of forming a string of length $n$ from the set $\{0,1,2\}$ such that $1$ and $2$ do not occur successively.
| There is an interesting related problem that has a solution using inclusion-exclusion. This is the restriction of the inadmissible words to words where $1$ and $2$ occur in order. This requires that we count the number of $n$-strings having $k$ instances of $1$ followed by $2$ ocurring successively with $1\le k \le \lf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
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Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
| Let $u:=a^2, v:=b^2, w:=c^2$, we have $u+v+w=3$.
Consider the function
$$f(x)=\ln (1+x^{\frac{1}{2}}+x^{\frac{3}{2}}),\ 0<x\leq 3$$
it's easy to compute that $f''(x)<0$.
by Jensen's inequality, we have
$$\sum\ln (1+u^{\frac{1}{2}}+u^{\frac{3}{2}})\leq3f(\dfrac{\sum u}{3})=3\ln 3$$
that is
$$\prod(a^3+a+1) \leq 27$$
Whe... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
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"answer_id": 5
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Prove $6 \nmid [\left( \sqrt[3]{28} - 3 \right)^{-n}]$ Prove that: $$6 \not\left|\ \left\lfloor\frac 1 {(\sqrt[3]{28} - 3)^{n}}\right\rfloor \ (n \in Z^+)\right.$$
($\lfloor x\rfloor$ = largest integer not exceeding $x$)
I am very bad as English and number theory, please help me
| If we set $\eta=\sqrt[3]{28}$ and $\omega=\dfrac1{\eta-3}=\dfrac{\eta^3-27}{\eta-3}=\eta^2+3\eta+9$, then, working $\bmod\ \eta^3-28$:
$$
\begin{align}
\omega^0&=1\\
\omega^1&=9+3\eta+\eta^2\\
\omega^2&=249+82\eta+27\eta^2\\
\omega^3&=6805+2241\eta+738\eta^2
\end{align}\tag{1}
$$
Solving the linear equations involved y... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Improper integral with log and absolute value: $\int^{\infty}_{0} \frac{\log |\tan x|}{1+x^{2}} \, dx$ How do you show that $$ \int^{\infty}_{0} \frac{\log |\tan x|}{1+x^{2}} \, dx = \frac{\pi}{2} \log (\tanh 1)\, ?$$
I know that the integral converges since $\log |\tan x| = \frac{1}{2} \log(\tan^{2}x)$, and $\log (\ta... | How about:
$$
\int_0^\infty \frac{\log | \tan x |}{1+x^2} \mathrm{d} x = \frac{1}{2} \int_0^\infty \frac{\log \tan^2 x }{1+x^2} \mathrm{d} x = \frac{1}{4} \int_{-\infty}^\infty \frac{\log \tan^2 x }{1+x^2} \mathrm{d} x \tag{$\ast$}
$$
Writing the integral over $\mathbb{R}$ as the average of integrals over $\mathbb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/285960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Evaluate integral with trigonometric functions in denominator $$ \int \frac {1}{\sin x + \cos x} dx$$
How would I go about solving this?
| The easiest way is to proceed as @AymanHourieh has suggested. Another approach is using Weierstrass substitution. Let $t = \tan(x/2)$. Note that
$$\sin(x) = \dfrac{2 \tan(x/2)}{1+\tan^2(x/2)} = \dfrac{2t}{1+t^2}; \,\,\,\,\, \cos(x) = \dfrac{1-\tan^2(x/2)}{1+\tan^2(x/2)} = \dfrac{1-t^2}{1+t^2}$$
$$dt = \dfrac{\sec^2(x/2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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to compute $\int_{2}^{3}\sqrt{1+\frac{1}{x^{2}}} \: \: dx$ Help me please to compute: $\int_{2}^{3}\sqrt{1+\frac{1}{x^{2}}} \: \: dx$
Thanks a lot!
| Hint: let $x=\tan{\theta}$, $dx=\sec^2{\theta} d \theta$:
$$\begin{align} \int_{2}^{3} dx \: \sqrt{1+\frac{1}{x^{2}}} &= \int_{\arctan{2}}^{\arctan{3}} \frac{d \theta}{\sin{\theta} \cos^2{\theta}} \\ &= \int_{\arctan{2}}^{\arctan{3}} \frac{d \theta \, \sin{\theta}}{\sin^2{\theta} \cos^2{\theta}} \end{align} $$
Now let... | {
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Solving $\;5^{2x}-4\cdot 5^x=12$ I need to solve $\quad\displaystyle 5^{2x}-4\cdot 5^x=12$.
I've only gotten this far: $\quad \displaystyle 5^{2x}-20^x=12.$
I don't know what to do next.
Thanks in advance!
| Please note that $4\cdot5^x = 4(5^x)\neq 20^x$
If we let $y=5^x$, then $$(5^x)^2-4(5^x) =12 \;\implies \;y^2 - 4y = 12 \;\implies y^2 - 4y - 12 = (y-6)(y+2) = 0$$
$\implies \; y = 6\;$ or $\;y = -2$
So $y = 5^x = 6\; $ or $\; y = 5^x = -2$
Can you take it from here?
You can omit $\;y= 5^x = -2\;$ as a solution if we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/288996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Diff eq. transformation polar coordinates I have $(x',y')=(x-y-x(x^2+y^2)+\frac{xy}{\sqrt{x^2+y^2}},x+y-y(x^2+y^2)-\frac{x^2}{\sqrt{x^2+y^2}} )$
Now I want to use polar coordinates $(x,y)=(r\cos(t),r\sin(t))$ to get $(r',t')=(r(1-r^2),2\sin(\frac{t}{2})^2)$
I do not see this relation. When I put $x=\cos t$, $y=\sin t$ ... | Using $x=r \cos{t}$, $y=r \sin{t}$:
$$x'=(\cos{t}) r' - r (\sin{t}) \, t'$$
$$y'=(\sin{t}) r' + r (\cos{t}) \, t'$$
So we get
$$\left ( \begin{array}\\ \cos{t} & -r \sin{t} \\ \sin{t} & r \cos{t} \end{array} \right ) \left ( \begin{array}\\ r' \\ t' \end{array} \right ) = \left ( \begin{array}\\ r \cos{t} - r \sin{t} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/290209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
First week in Linear Algebra, need some help on this simple problem If you have
$5x + 2y + z = 0$
$2x + y = 0$ and you're asked to solve using back-substitution how would you go about doing it?
Initially I thought just simply the following:
$x + \frac{2}{5} y + \frac{1}{5} z = 0$ (divide by $5$)
$2x + y ... | Of course the accepted answer is completing one, but look at this one. Maybe it looks fine either. We have: $$
\left\{
\begin{array}{ll}
5x+2y+z=0 \\
2x+y=0 &
\end{array}
\right.$$
Now multiply the second equation by $-2$:
$$ \left\{
\begin{array}{l1}
5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/291190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How is this linear 2nd-order ODE solved? In this article, the authors present the inhomogeneous equation
$$\ddot{\phi}_2 + \phi_2 + g_2\phi_1^2 + \omega_1\ddot{\phi}_1 = 0,\tag{11}$$
where
$$ \phi_1 = p_1 \cos(\tau + \alpha), \tag{13}$$
followed by its solution
$$\phi_2 = p_2\cos(\tau + \alpha) + q_2\sin(\tau + \alpha)... | Particular solution can be written in the form\begin{align}
\phi_2 = p_2 \cos(\tau + \alpha) + q_2 \sin(\tau + \alpha) + C \cos(2(\tau + \alpha)) + D\sin(2(\tau + \alpha)) + E.
\end{align}
Now after doing derivative, we get,
$$\dot{\phi_2}= -p_2 sin(\tau +\alpha)+q_2cos(\tau+ \alpha)-2Csin(2\tau+2\alpha)+2Dcos(2\tau+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/292406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Is this function convex or non-convex? Let
$$f(a,b,c,d) = \frac{(a-b) \cdot (c-d)}{\sqrt{(a-b)^2+(c-d)^2}}$$
where $a,b,c,d$ are variables. Is this function convex or non-convex?
| We have
$$\frac{\partial f}{\partial a} = \frac{c-d}{\sqrt{(a-b)^2+(c-d)^2}}-\frac{(a-b)^2(c-d)}{\left[(a-b)^2+(c-d)^2\right]^{3/2}}=\frac{(c-d)^3}{\left[(a-b)^2+(c-d)^2\right]^{3/2}}$$
$$\frac{\partial^2f}{\partial a^2} = -\frac{3(a-b)(c-d)^3}{\left[(a-b)^2+(c-d)^2\right]^{5/2}},$$
which is negative when $a>b$ and $c>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/293837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
A simplification regarding matrices Let $$A = \left( \begin{array}{cc}
0 & -1 \\
1 & 0 \\
\end{array} \right )$$
What can I say about the matrix $A^k$ for arbitrary natural $k$. Is there some way to express $A^k$ in terms of $A$. I have observed that $A^4 = I$.
EDIT: I want a better form of $A^k$ in order to fin... | We can diagonalize $A$ by calculating the Jordan Normal Form, using it's eigenvalues, $\lambda_{1, 2} = \pm i$, where $i$ is the imaginary unit for this matrix as:
$$A = \left( \begin{array}{cc}
0 & -1 \\
1 & 0 \\
\end{array} \right ) = S \cdot J \cdot S^{-1} = \left( \begin{array}{cc}
-i & \ i
\\ 1 & 1 \\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/295486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Sum of combinations of n taken k where k is from n to (n/2)+1 I wonder if there's a formula for obtaining the sum of $n\choose k$'s where $k$ is from $n$ to $\frac{n}{2}+1$. I found out that in odd numbers, it is $2^{n-1}$ (powerset divided by $2$).
*
*1 = 1
*3 = 4
*5 = 16
*7 = 64
However, It is not the case ... | Recall that
$$\sum_{k=0}^n \dbinom{n}k = 2^n$$
Also, recall that
$$\dbinom{n}k = \dbinom{n}{n-k}$$
Hence, for odd $n$, we have
\begin{align}
2^n & = \sum_{k=0}^n \dbinom{n}k\\
& = \sum_{k=0}^{(n-1)/2} \dbinom{n}k + \sum_{k=(n+1)/2}^n \dbinom{n}k\\
& = \sum_{k=0}^{(n-1)/2} \dbinom{n}{n-k} + \sum_{k=(n+1)/2}^n \dbinom{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/299273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
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