Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How can I attack $\int_{-\infty}^\infty \frac{x}{[(x-vt)^2+a^2]^2} dx$? I want to evaluate
$$\int_{-\infty}^\infty \frac{x}{[(x-vt)^2+a^2]^2} dx$$
I know that
$$\int \frac{1}{(\xi^2+a^2)^2} = \frac{\xi}{2a^2(\xi^2+a^2)}+\frac{1}{2a^3}\arctan\frac{\xi}{a}$$
Using integration by parts and choosing $a:= a$, $\xi := x-vt$... | Start by substituting $\xi=x-vt$ to get
$$\eqalign{\int_{-\infty}^\infty \frac{x}{((x-vt)^2+a^2)^2} dx
&=\int_{-\infty}^\infty \frac{\xi+vt}{(\xi^2+a^2)^2} dx\cr
&=\int_{-\infty}^\infty \frac{\xi}{(\xi^2+a^2)^2} dx
+vt\int_{-\infty}^\infty \frac{1}{(\xi^2+a^2)^2} dx\ .\cr}$$
Now the first integral is zero becau... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/774142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A closed form for the infinite series $\sum_{n=1}^\infty (-1)^{n+1}\arctan \left( \frac 1 n \right)$ It is known that $$\sum_{n=1}^{\infty} \arctan \left(\frac{1}{n^{2}} \right) = \frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right). $$
Can we also find a closed form for th... | Let $ \displaystyle S(a) = \sum_{n=1}^{\infty}(-1)^{n-1} \arctan \left(\frac{a}{n} \right)$.
Since $ \displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{n}{n^{2}+a^{2}}$ converges uniformly on $\mathbb{R}$,
$$ \begin{align} S'(a) &= \sum_{k=1}^{\infty} (-1)^{n-1} \frac{n}{a^{2}+n^{2}} \\ &= \frac{1}{2} \sum_{n=1}^{\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/776182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 3
} |
no. of positive integral solutions of ||x - 1| - 2| + x = 3 What are the no. of positive integral solutions of ||x - 1| - 2| + x = 3 ?
My effort
||x - 1| - 2| = 3 - x
|x - 1| - 2 = 3 - x OR |x - 1| - 2 = x - 3
|x - 1| = 5 - x OR |x - 1| = x - 1
x - 1 = 5 - x OR x - 1 = x - 5 OR x - 1 $\geq$ 0
2x = 6 OR x $\geq$ 1
... | You should start with expanding the innermost absolute value:
$$
\vert \vert x-1 \vert -2 \vert = \begin{cases} \vert x-3 \vert & x \geqslant 1 \\ \vert -1-x \vert & x < 1 \end{cases} = \begin{cases} x-3 & x \geqslant 1, x > 3 \\ 3-x & x \geqslant 1, x \leqslant 3 \\ 1+x & x < 1, x \geqslant -1 \\ -1-x & x < 1, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/778419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Equation of tangent perpendicular to line I've got the homework question which I cannot solve.
Find the equation of the tangents to $4x^2+y^2=72$ that are
perpendicular to the line $2y+x+3=0$.
What I have done so far:
I have found the gradient of the line which is $m_1 = -\frac12$.
Which means that the equation per... | Let the point of tangency be the point $(x,y)$ on our curve. Your calculation shows that $\frac{-4x}{y}=2$, that is, $y=-2x$.
Also, we have $4x^2+y^2=72$. Substituting $-2x$ for $y$ in this equation, we get $8x^2=72$, and therefore $x=\pm 3$. That gives the two points $(3,-6)$ and $(-3,6)$.
The line with slope $2$ th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/778779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to find PV $\int_0^\infty \frac{\log \cos^2 \alpha x}{\beta^2-x^2} \, \mathrm dx=\alpha \pi$ $$
I:=PV\int_0^\infty \frac{\log\left(\cos^2\left(\alpha x\right)\right)}{\beta^2-x^2} \, \mathrm dx=\alpha \pi,\qquad \alpha>0,\ \beta\in \mathbb{R}.$$
I am trying to solve this integral, I edited and added in Principle v... | We make use of the identity
$$ \sum_{n=-\infty}^{\infty} \frac{1}{a^{2} - (x + n\pi)^{2}} = \frac{\cot(x+a) - \cot(x-a)}{2a}, \quad a > 0 \text{ and } x \in \Bbb{R}. $$
Then for $\alpha, \beta > 0$ it follows that
\begin{align*}
I := \mathrm{PV}\int_{0}^{\infty} \frac{\log\cos^{2}(\alpha x)}{\beta^{2} - x^{2}}
&= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
"answer_id": 0
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Prove or disprove inequality $\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\le\frac{a^4+b^4+c^4}{2abc}$. Let $a$, $b$ and $c$ be real numbers greater than $0$. Prove inequality $$\displaystyle{\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\le\frac{a^4+b^4+c^4}{2abc}}.$$
| $2abc\cdot \dfrac{a^2}{b+c} = 2a^3\cdot \dfrac{bc}{b+c} \leq 2a^3\cdot \dfrac{b+c}{4} = \dfrac{a^3(b+c)}{2}$. Thus:
$2abc\cdot \dfrac{a^2}{b+c} + 2abc\cdot \dfrac{b^2}{c+a} + 2abc\cdot \dfrac{c^2}{a+b} \leq \dfrac{(a^3b + ab^3) + (a^3c + ac^3) + (b^3c + bc^3)}{2} \leq \dfrac{(a^4 + b^4) + (b^4 + c^4) + (c^4 + a^4)}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How prove the constant term of $\left(1+x+\frac{1}{x}\right)^p\equiv1\pmod {p^2}$ if $p>3$ is odd prime number,show that:
the constant term of $$\left(1+x+\dfrac{1}{x}\right)^p\equiv1\pmod {p^2}$$
My try: since
$$(1+x+\dfrac{1}{x})^p=\sum_{k=0}^{p}\binom{p}{k}\left(x+\dfrac{1}{x}\right)^k=\sum_{k=0}^{p}\binom{p}{k}\sum... | For $p=2$, $(1+x+\frac{1}{x})^2=x^2+2x+3+\frac{2}{x}+\frac{1}{x^2}$.
$3\equiv 1\ (mod\ 4)$ ???
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/781750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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$l^2+m^2=n^2$ $\implies$ $lm$ is always a multiple of 3 when $l,m,n,$ are positive integers. Let $l,m,n$ be any three positive integers such that $l^2+m^2=n^2$
Then prove that $lm$ is always a multiple of 3.
| We show that $3$ must divde $lm$, by showing that $3$ divides $l$ or $3$ divides $m$.
Any integer $x$ is either divisible by $3$, or is congruent to $1$ or $-1$ modulo $3$. And iff $x\equiv \pm 1\pmod{3}$, then $x^2\equiv 1 \pmod{3}$.
Thus if neither $l$ nor $m$ is divisible by $3$, then $l^2+m^2\equiv 2\pmod{3}$. It... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove that $|f(x)| \le \frac{3}{2}$ when $f(x)=ax^2+bx+c$ Suppose $f(x) = ax^2+bx+c$ where $x \in [-1,1]$. If $f(-1),f(0),f(1)\in [-1,1]$ show that $|f(x)| \le \frac{3}{2}$ $\forall x \in [-1,1]$.
This is how I tried:
$f(0)=c$
$f(1)=a+b+c$
$f(-1)=a-b+c$
Putting $f(0)=c$ we get $f(1)-f(0)=a+b$, $f(-1)-f(0)=a-b$. Sol... | As mentioned above, this has already been answered in Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$. with the better bound $5/4$. But it seems that your
proof is almost complete and slightly shorter than (or at least ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Help on an integral I can't figure out how to compute the following integral:
$$\int_{-\infty}^{+\infty}\frac{\mathrm{d}z}{\sqrt{(x^2+y^2+z^2)^3}}$$
I think I should do some substitution, but I didn't figure it out. Can you please give me a hint?
According to Mathematica the result should be
$$\left[\frac{z}{(x^2+y^2)\... | Let $z=\sqrt{x^2+y^2}\tan\theta$, then $dz=\sqrt{x^2+y^2}\sec^2\theta\ d\theta$.
$$
\begin{align}
\int_{-\infty}^{+\infty}\frac{dz}{\sqrt{(x^2+y^2+z^2)^3}}&=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\sqrt{x^2+y^2}\sec^2\theta\ d\theta}{\sqrt{((x^2+y^2)+(x^2+y^2)\tan^2\theta)^3}}\\
&=\int_{\Large-\frac\pi2}^{\Large... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/784092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve: $\tan2x=1$ Are there any errors in my work? Thanks in advance! (Sorry for the bad format. I'm still new to this)
$\tan2x=1$
$\frac{2\tan x}{1-\tan^2x}=1$
$2\tan x=1-\tan^2x$
$0=1-\tan^2x-2\tan x$
$0 =-\tan^2x-2\tan x +1$
$0=\tan^2x+2\tan x-1$
$\frac{-(2)\sqrt{2^2-4(1)(-1)}}{2(1)}$
$x=0.4142, x=2.4142$
$\tan^{-... | Your solution is not very rigorous after $0 = tan^2x + 2tanx - 1$. After that line, the
equation should be $tanx = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)}$.
Also, after that, it's $tanx = \sqrt{2} - 1 \approx 0.4142$ or $tanx = \sqrt{2} + 1 \approx 2.4142$.
Then, $x = tan^{-1}(0.4142) \approx 22.5, 202.5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/784445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Deriving $\sum\limits_{i=1}^n \frac{1}{i^2} = \frac{π^2}{6}-ψ^{(1)}(n+1)$. How is the partial sums formula of the $$\sum\limits_{i=1}^n \frac{1}{i^2} = \frac{π^2}{6}-ψ^{(1)}(n+1)$$
derived?
| It can be shown with the Hurwitz Zeta function (see especially formulas (1) and (2))
$$\sum\limits_{i=1}^n \frac{1}{i^2}
= \sum\limits_{i=1}^{\infty} \frac{1}{i^2} - \sum\limits_{i=n+1}^{\infty} \frac{1}{i^2} \\
= \sum\limits_{i=1}^{\infty} \frac{1}{i^2} - \sum\limits_{i=0}^{\infty} \frac{1}{(i+n+1)^2} \\
= \zeta(2)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/784778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometry problem cosine identity Let $\cos^6\theta = a_6\cos6\theta+a_5\cos5\theta+a_4\cos4\theta+a_3\cos3\theta+a_2\cos2\theta+a_1\cos\theta+a_0$. Then $a_0$ is
(A) $0$ (B) $\frac{1}{32}$ (C) $\frac{15}{32}$ (D) $\frac{10}{32}$
Any hints on how to approach this?
| Using \begin{align}&2\cos^2 x=\cos 2x +1, \\&4\cos^3x=\cos 3x+3\cos x, \\ \text{and}&2\cos a \cos b= \cos(a+b)+\cos(a-b),\end{align}
we obtain,
\begin{align}&\cos^6 x=(\cos^3 x)^2\\
\\=&\left(\dfrac{\cos 3x+3\cos x}{4}\right)^2\\
\\=&\dfrac1{16}(\cos^2 3x + 9\cos^2 x+6\cos x\cos 3x)\\
\\=&\dfrac{1}{16}\left(\dfrac{\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/786169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find minimum value of the expression x^2 +y^2 subject to conditions Find the values of $x,y$ for which $x^2 + y^2$ takes the minimum value where $(x+5)^2 +(y-12)^2 =14$.
Tried Cauchy-Schwarz and AM - GM , unable to do.
| Another way is triangle inequality (essentially think of the triangle between the origin, the centre of the circle and any point on the circle):
$$\sqrt{x^2+y^2} +\sqrt{(x+5)^2+(y-12)^2} \ge \sqrt{(-5)^2+(12)^2} \implies x^2+y^2 \ge \left(13 - \sqrt{14}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Maximium value of $(b-a)\Big(\dfrac 34-\dfrac{a+b}2-\dfrac{a^2+ab+b^2}3\Big)$ For $b>a$ what is the maximum possible value of $(b-a)\Big(\dfrac 34-\dfrac{a+b}2-\dfrac{a^2+ab+b^2}3\Big)$ ?
| For a fixed $a$, the expression is
$$\frac34(b-a)-\frac12(b^2-a^2)-\frac13(b^3-a^3)$$
which is a third degree polynomial.
Differentianting respect to $b$ gives:
$$\frac 34-b-b^2$$ whose roots are $-3/2$ and $1/2$.
The maximum is reached when $b=1/2$. If $a>1/2$, there is no maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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the series $\displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})}$ converges uniformly in $[-a,a]$ I have to show that the series $\displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})}$ converges uniformly in $[-a,a], a>0$.
$$$$
That's what I have tried:
$w_n=\frac{(-1)^n... | Hint You may use Dirichlet's criterion, since $\sin (xn^{-1})$ is eventually decreasing in $[0,M]$ and $1-\cos (xn^{-1})$ is eventually decreasing in $[0,M]$ too, and both go uniformly to zero, and $$\sum_{n\geqslant 1} (-1)^{n}n^{-1/2}$$ converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/790290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving an integral using Laplace transform and inverse Laplace transform I want to solve this integral equation using Laplace:
$$ Y(t) + 3{\int\limits_0^t Y(t)}\operatorname d\!t = 2cos(2t)$$
if $$ \mathcal{L}\{Y(t)\} = f(s)$$
then, $$ f(s) + 3 \frac{f(s)}{s} = \frac{2s}{s^2+4} $$
doing some operations I obtain
$$ f(... | Building on my comment, try the following:
$$f(s)=\frac{2s^2}{(s+3)(s^2+4)} = \frac{As+B}{s^2+4} + \frac{C}{s+3}$$
So $As(s+3) + B(s+3) + C(s^2+4) = 2s^2$, whence\
\begin{align*}
A+C &= 2\\
3A + B &= 0\\
3B+4C &= 0
\end{align*}
We can shortcut the solution a bit, as follows:
Plug in $s = -3$, we get: $13C = 18$, whence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/790385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Integral determination I am trying to figure out this integral:
$\int \frac{x}{(x^2+4)^6}dx$
Substitution:
$t = x^2+4$
$dt = 2xdx => dx=\frac{dt}{2x}$
Then:
$\int \frac{x}{(x^2+4)^6}dx = \int \frac{x}{t^6}\frac{dt}{2x} = \frac{1}{2} \frac{1}{t^6} + C = $ but how to continue now?
| $$\int \frac{xdx}{(x^2+4)^6}=\frac12 \int \frac{dt}{t^6}=\int t^{-6}dt=\frac{1}{2}\cdot \frac{t^{-5}}{-5}+c=\frac{-1}{10(x^2+4)^5}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/791947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to finish this integration? I'm working with the integral below, but not sure how to finish it...
$$\int \frac{3x^3}{\sqrt[3]{x^4+1}}\,dx = \int \frac{3x^3}{\sqrt[3]{A}}\cdot \frac{dA}{4x^3} = \frac{3}{4} \int \frac{dA}{\sqrt[3]{A}} = \frac{3}{4}\cdot\quad???$$
where $A=x^4+1$ and so $dA=4x^3\,dx$
| Assuming I interpreted your very low quality picture correctly...
You got a good start on the problem. That's the right substitution. Rewrite...
$$\frac{3}{4} \int \frac{dA}{\sqrt[3]{A}} = \frac{3}{4} \int A^{-1/3}\,dA$$
...and use the backwards power rule...
$$ = \frac{3}{4} \frac{A^{2/3}}{2/3} +C = \cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find norm of operator I have a linear functional $$A: L_2[0,2] \to \mathbb R, Ax = \int_0^2(t^2+2)x(t)dt$$
I need to find $C$, trying to measure $C$ and $||Ax||$ to find it, but how can I do it in this problem?
| Here I'll play the role of Robin to Batman, and emulate my mentor's work in a more elementary manner, as befits a protoge of the master!
First of all, note that $y(t) = t^2 +2 \in L^2[0, 2]$; since the inner product $\langle u, v \rangle$ on $L^2[0, 2]$ is given by
$\langle u, v \rangle = \int_0^2 u(t)v(t) dt, \tag{1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/792839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding determinant for matrix using upper triangle method Here is an example of a matrix, and I'm trying to evaluate its determinant:
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
2 & 5 & -2 & 9 \\
3 & 7 & 0 & 1 \\
\end{pmatrix}
$$
When applying first row operation i get:
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 ... | Different row-operations affect the determinant of the matrix differently. Adding a multiple of one row to another will not change the determinant. However, multiplying a row by some factor will lead to the determinant being multiplied by the same factor.
So, since you multiplied $R_4$ by the factor $- \frac 12$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/793217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Algebraic Solving Contest Problem The problem is as follows
If $x^2+x-1=0$, compute all possible values of $\frac{x^2}{x^4-1}$
This was a no-calculator 10 min for 2 problem format contest.
I started by using quadratic formula, but the answers I got were too ugly to be plugged into the equation before time ran out. I tr... | Let $a, b$ be the roots of $x^2+x-1=0$. Then $a+b = -1, ab = -1$. For simplicity, consider now the reciprocals of the values we wish to find, i.e. $\dfrac{a^4-1}{a^2} = k_1$ and $\dfrac{b^4-1}{b^2} = k_2$, then
$$k_1 + k_2 = a^2+b^2-\left(\frac1{a^2}+\frac1{b^2} \right), \quad k_1 k_2 = \frac{(a^4-1)(b^4-1)}{(ab)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/799896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Extrema of $x+y+z$ subject to $x^2 - y^2 = 1$ and $2x + z = 1$ using Lagrange Multipliers
Find the extrema of $x+y+z$ subject to $x^2 - y^2 = 1$ and $2x + z = 1$ using Lagrange multipliers.
So I set it up:
$$
1 = 2x\lambda_1 + 2\lambda_2 \\
1 = -2y\lambda_1 \\
1 = \lambda_2
$$
Plug in for $\lambda_2$:
$$
1 = 2x\lambd... | Consider the constraints: you can simplify them in
$$ y=\pm\sqrt{y^2+1}, \qquad z=\mp 2\sqrt{y^2+1}+1$$
for all $y\in\mathbb{R}$.
In this way you can convert the constrained optimization in two unconstrained optimizations: substitute the two solutions in the objective function and you will find
$$- \sqrt{y^2+1}+y+1, \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/800575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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probability with replacement A box contains 4 red, 3 black and 2 white cubes. A cube is randomly drawn and has its color noted. The cube is then replaced, together with 2 more of the same color. A second cube is then drawn. Find the probability that
1. The second cube is black
2. the second cube is not white
| Divide into cases. With probability $\frac{4}{9}$ a red is drawn. If that happens, the probability of a black next is $\frac{3}{11}$.
With probability $\frac{3}{9}$ a black is drawn. If that happens, the probability of a black next is $\frac{5}{11}$.
With probability $\frac{2}{9}$ a white is drawn. If that happens, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/801203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof for $\sin(x) > x - \frac{x^3}{3!}$ They are asking me to prove $$\sin(x) > x - \frac{x^3}{3!},\; \text{for} \, x \, \in \, \mathbb{R}_{+}^{*}.$$ I didn't understand how to approach this kind of problem so here is how I tried:
$\sin(x) + x -\frac{x^3}{6} > 0 \\$
then I computed the derivative of that function to d... | Take a decreasing sequence of positive real numbers $a_n$ such that $a_n\to 0$.
Now, consider the sequence $b_k=\sum_{n=1}^k (-1)^{n-1}a_n$. The alternating series criterion guarantee us that it converges to some $b$.
Note that $b_1=a_1$, $b_2=b_1-a_2\in(0,b_1)$, $b_3=b_2+a_3\in(b_2,b_1)$, etc. So the limit $b$ is less... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 9,
"answer_id": 0
} |
Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$ Hi how can we prove this integral below?
$$
I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}
$$
I tried to use
$$
I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx
$$
and now tried changing variables to $y=x(1-x)$ ... | Setting $x\mapsto\frac1x$, we obtain
$$
\int_0^1\frac{\ln^2x}{x^2-x+1}\ dx
=\frac12\int_0^\infty\frac{\ln^2x}{x^2-x+1}\ dx.\tag1
$$
Note that $1+x^3=(1+x)(x^2-x+1)$, hence
$$
\frac12\int_0^\infty\frac{\ln^2x}{x^2-x+1}\ dx=\frac12\int_0^\infty\left[\frac{\ln^2x}{1+x^3}+\frac{x\ln^2x}{1+x^3}\right]\ dx.\tag2
$$
Now consi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 8,
"answer_id": 1
} |
Calculus limits please demonstrate question? I have been trouble doing some more difficult limits
I have these two examples from a teacher and think if i can just be shown the working and answer i should be fine from there out. Thanks for your help.
$(a)\; \lim\limits_{x\to-2}\dfrac{3x^2+8x+4}{2x^2+x-6} \\ \; \\ (b)\;... | $$\lim\limits_{x\to-2}\dfrac{3x^2+8x+4}{2x^2+x-6} = \lim_{x\to -2} \dfrac{(3x+2)(x+2)}{(2x -3)(x+2)} = \lim_{x\to -2}\frac{3x+2}{2x-3} = \frac{-4}{-7}= \frac 47 $$
For the second limit, divide numerator and denominator by $x^4$, and evaluate.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A logarithmic integral $\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$ How to prove the following
$$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$
I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are ... | $\displaystyle J=\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$
Perform the change of variable $y=\sqrt{\dfrac{1-x}{1+x}}$,
$\displaystyle J=-4\int^1_0 \dfrac{\log(x)}{1-x^2}dx=-4\int_0^1 \left( \log x\times\sum_{n=0}^{+\infty}x^{2n}\right) dx=-4\sum_{n=0}^{+\infty}\left(\int_0^1 x^{2n}\log x dx\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 10,
"answer_id": 4
} |
Line of intersection of planes How to find the line of intersection of the planes
$x·\begin{pmatrix}
1\\
2\\
3\\
\end{pmatrix}=0
$
and $x=\lambda_1\begin{pmatrix}
2\\
1\\
2\\
\end{pmatrix}+\lambda_2 \begin{pmatrix}
1\\
0\\
-1\\
\end{pmatrix}$. All I can figure is x is orthagonal to $\begin{pmatrix}
1\\
2\\
3\\
\end{pma... | So each vector in the second plane looks like $\begin{bmatrix}2\lambda_1+\lambda_2\\ \lambda_1\\ 2\lambda_1-\lambda_2\end{bmatrix}$, and to be in the first plane this would have to be perpendicular to $[1,2,3]^\top$. This yields $2\lambda_1+\lambda_2+2\lambda_1+6\lambda_1-3\lambda_2=0$, hence $10\lambda_1-2\lambda_2=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/809468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of $\frac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}$ as $x \rightarrow 0$ Find $\lim_{x \to 0} \dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}$
I came across this limit a long time ago and could easily obtain a straightforward solution by fi... | I suppose and think that this is really related to the coincidental coefficients. We can show it building for each piece a Taylor expansion built at $x=0$. As results, we have
$$\tan ^{-1}\left(\sin ^{-1}(x)\right)=x-\frac{x^3}{6}+\frac{13 x^5}{120}-\frac{173 x^7}{5040}+O\left(x^{9}\right)$$
$$\sin ^{-1}\left(\tan ^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/809632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
} |
How come the function and the inverse of the function are the same? What is the inverse of the function:
$$f(x)=\frac{x+2}{5x-1}$$
?
Answer:
$$f^{-1}(x)=\frac{x+2}{5x-1}$$
Can one of you explain how the inverse is the same exact thing as the original equation?
| Let $f(x)=y$. So we have $y=\frac{x+2}{5x-1}$. Now "swap" the variables so we have
$$x = \frac{y'+2}{5y'-1}$$
where $y' = f^{-1}(x)$ (to distinguish $y'$ from $y$). Now let's solve for $y'$.
\begin{align}
x &= \frac{y'+2}{5y'-1} \\
x(5y'-1) &= y'+2 & \text{multiply both sides by } 5y'-1 \\
5xy' - x&=y'+2 & \text{distr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I find the integral? I want to find the integral $$\int_{0}^{1}{\sqrt[3]{2x^3-3x^2-x+1}}\,\mathrm{d}x.$$
I tried, $$\int_{0}^{1}{\sqrt[3]{(-1 + 2 x) (-1 - x + x^2)}}\,\mathrm{d}x.$$
Put $t =-1 - x + x^2$, then $\mathrm{d} t = (2x + 1)\mathrm{d}x.$ And now, I can not find the integral.
| Note that $2x^3-3x^2-x+1=2(x-\frac{1}{2})^3-\frac{5}{2}(x-\frac{1}{2})$. Hence the integrand is symmetric about $x=\frac{1}{2}$, and since you are integrating with limits also symmetric about $\frac{1}{2}$, the value of the integral is just $0$.
$$\int_{0}^{1}{\sqrt[3]{2x^3-3x^2-x+1}}\,\mathrm{d}x=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/812032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find sum of $\frac{1}{\sin\theta\cdot \sin2\theta} + \frac{1}{\sin2\theta\cdot \sin3\theta} + \cdots + \frac{1}{\sin n \theta \sin (n+1)\theta}$ $$\sum_{k=1}^n \frac{1}{\sin k\theta \sin (k+1)\theta} = \dfrac{1}{\sin\theta\cdot \sin2\theta} + \dfrac{1}{\sin2\theta\cdot \sin3\theta} + \cdots + \frac{1}{\sin n \theta \si... | Here is a hint: the sum telescopes, with some careful manipulation:
$$\sin \theta = \sin((k+1)\theta - k\theta) = \sin(k+1)\theta \cos k\theta - \sin k\theta \cos (k+1)\theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Help with limit equation The question is:
$$\lim_{x \to + \infty} \frac{3x^4 - 11x^3 + 42x}{-15x^4 + 3x^2 +101}$$
So far my answer:
$$\lim_{x \to + \infty} \frac{x^4}{x^4} \frac{3 - \frac{11x^3}{x^4}+\frac{42x}{x^4}}{-15 + \frac{3x^2}{x^4}+ \frac{101}{x^4}}$$
cancel the $x^4$ gives us:
$$\lim_{x \to + \infty} \frac{3 -... | Yes. It is right. When you have ratios of polynomials, however, there is an easy way to analyze $$\lim_{x \to +\infty} \frac{p(x)}{q(x)}$$
It is $+\infty$ or $-\infty$ if the degree of $p(x)$ is greater than the degree of $q(x)$, and the sign depends on the sign of the ratio of $p(x)$ and $q(x)$'s dominant coefficients... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does $\frac{(x^2 + x-6)}{x-2} = \frac{(x+3)(x-2)}{x-2}$? I'm not the best at algebra and would be grateful if someone could explain how you can get from,
$$\frac{x^2 + x-6}{x-2}$$
to,
$$\frac{(x+3)(x-2)}{x-2}$$
| Note that the denominator of $\frac{(x^2 + x-6)}{x-2}$ and
$\frac{(x+3)(x-2)}{x-2}$ is the same, so it remains to show that the numerators are the same; i.e. that $x^2+x-6\equiv(x+3)(x-2)$.
Now, $\color{green}{(x+3)(x-2)} \equiv x^2\underbrace{-2x+3x}_{\equiv \ +x}-6 \equiv \color{green}{x^2+x-6},$ as required.
To exp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/817424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to find the following sum? $\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} $ I want to calculate the sum with complex analysis (residue)
$$
1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$ $$
1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} -... | $1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - \frac{1}{{23}} + \frac{1}{{25}} - .... = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 1}}} - \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 7}}} $
\begin{array}{l}
1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - \frac{1}{{23}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/817911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
} |
limits of function without using L'Hopital's Rule $\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x+ 1 - x}} = 1$ Good morning.
I want to show that without L'Hopital's rule :
$\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x + 1 - x}} = 1$
I did the steps
$
\begin{array}{l}
\mathop {\lim... | by use of Taylor series when ${x \to 1}$ you'll get $\ln x=(x-1)-\frac{{(x - 1)^2}}{2x^2}$
$\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - (x - 1)+\frac{{(x - 1)^2}}{2x^2}}}{{{x(x - 1)}-\frac{{(x - 1)^2}}{2x} + 1 - x}}$
after factoring $(x-1)$ you'll have
$\mathop {\lim }\limits_{x \to 1} \frac{{\frac{{(x - 1)}}{2x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Factoring in the derivative of a rational function Given that
$$
f(x) = \frac{x}{1+x^2}
$$
I have to find
$$\frac{f(x) - f(a)}{x-a}$$
So some progressing shows that:
$$
\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} =
\frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} =
\frac{x+xa^2... | Factor:
$$x+xa^2−a−ax^2=x-a-ax^2+xa^2=(x-a)-ax(x-a)=(1-ax)(x-a)$$
Thus, $$\frac{f(x)-f(a)}{x-a}=\frac{\frac{(1-ax)(x-a)}{(1+x^2)(1+a^2)}}{(x-a)}=\boxed{\frac{1-ax}{(1+x^2)(1+a^2)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
This double integral $$ \int_0^1\int_0^1x^3y^2\sqrt{1+x^2+y^2}\hspace{1mm}dxdy$$
We have to compute this up to 4 decimal places
| If you just need an approximation of this integral, you could use numerical methods. Matlab or Wolfram's Double Integral Calculator, for example, can do it just fine :
$$ \int_{0}^{1} \int_{0}^{1} x^{3}y^{2}\sqrt{1+x^{2}+y^{2}} \, \mathrm{d}x \mathrm{d}y \simeq 0.125065 $$
But if you want an exact computation, this mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How prove this Rāmā ujan Aiya kār identity $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}=3$ show that
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}=3$$
I know
$$3=\sqrt{1+8}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+15}}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot6}}}=\cdots$$
| For your question answer is given in this
http://www.isibang.ac.in/~sury/ramanujanday.pdf
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sum\limits_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}$ isn't divisible by 5 I have no idea
Prove that for any $n$ natural number this sum $$\sum\limits_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}$$ isn't divisible by $5$.
$\begin{array}{l}
\left( {1 + x} \right)^{2n + 1} - \left( {1 - x} \right)^{2n + 1} = \sum\limits_{k = 0}^{2n... | We will use $\mathbb{Z}_5[\sqrt2]$
$$
\begin{align}
a_n
&=\sum_{k=0}^n\binom{2n+1}{2k+1}2^{3k}\\
&=\frac1{4\sqrt2}\left(\sum_{k=0}^{2n+1}\binom{2n+1}{k}\sqrt8^{\,k}-\sum_{k=0}^{2n+1}\binom{2n+1}{k}(-\sqrt8)^k\right)\\[3pt]
&=\frac{\left(1+2\sqrt2\right)^{2n+1}-\left(1-2\sqrt2\right)^{2n+1}}{4\sqrt2}\\[9pt]
&\equiv3(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/822230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Proving that $\sqrt[3] {2} ,\sqrt[3] {4},1$ are linearly independent over rationals I was trying to prove that $\sqrt[3] {2} ,\sqrt[3] {4}$ and $1$ are linearly independent using elementary knowledge of rational numbers. I also saw this which was in a way close to the question I was thinking about. But I could not com... | First show that $1$ and $\sqrt[3]2$ are linearly independent. (This should be relatively easy.)
Then in order for $1$, $\sqrt[3]2$ and $\sqrt[3]4$ to be linearly dependent we must have
$$\sqrt[3]{4}=a+b\sqrt[3]2$$
for some $a,b\in\mathbb Q$.
(Since $\sqrt[3]{4}$ is a linear combination of $1$ and $\sqrt[3]2$.)
If we mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
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Urn with balls, distribution of random variable From an urn containing $6$ balls numerated $1,\ldots,6$ we randomly choose one, then again and stop only when we picked the ball with number $1$ on it. Let $X$ be the greatest number that appeared on balls we already pulled out. What's the distribution of $X\ $? And $\mat... | Edit: Later on, we describe a fast solution. But for reasons of nostalgia, we keep our first slow way.
The slow way: We assume that the balls are removed one at a time and not replaced. Then the probabilities can be found with a careful examination of cases.
With probability $\frac{1}{6}$ we have $X=1$.
We have $X=2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Three Circles Meeting at One Point We have three triples of points on the plane, that is, $X=\{x_1, x_2, x_3\}$, $Y=\{y_1, y_2, y_3 \}$, and $Z=\{z_1, z_2, z_3\}$, where $x_i, y_i, z_i$ are points on the plane. I was wondering if there is a simple (or not-so-simple) algebraic relation satisfied by the coordinates of th... | Let $\triangle ABC$ have edge-lengths $a=|BC|$, $b=|CA|$, $c=|AB|$. Suppose circles of radius $u$, $v$, $w$ about respective points $A$, $B$, $C$ meet at a common point $P$.
Coordinatizing, we can write $A=(0,0)$, $B=(c,0)$, $C=(b\cos A, b\sin A)$, $P=(p,q)$, so that
$$\begin{align}
u^2 &= p^2 + q^2 &(1)\\
v^2 &= (p-c)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\dfrac{\sigma_1(n)}{n} = \sigma_{-1}(n)$ where $\sigma_x(n)$ is the sum of the $x$th powers of the positive divisors of $n$. I computed $\dfrac{\sigma_1(n)}{n}$ and $\sigma_{-1}(n)$ on a good hundred values of $n$, and they seem to always match.
For example:
$\dfrac{\sigma_1(6)}{6} = \dfrac{1 + 2 + 3 + 6}{6... | $\sigma_k(p^a)=\dfrac{1-p^{k(a+1)}}{1-p^k}=p^{ka}\dfrac{p^{-k(a+1)}-1}{p^{-k}-1}=p^{ka}\sigma_{-k}(p^a)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/833178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $\frac{\sin\theta}{1-\cos\theta} \equiv \csc\theta + \cot\theta$ This must be proved using elementary trigonometric identities.
I have not been able to come to any point which seems useful enough to include in this post.
| Multiply the LHS by $1+\cos\theta$ yields
\begin{align}
\frac{\sin\theta}{1-\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta}&=\frac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}\\
&=\frac{\sin\theta(1+\cos\theta)}{\sin^2\theta}\qquad;\qquad\color{red}{\cos^2\theta+\sin^2\theta=1}\\
&=\frac{1+\cos\theta}{\sin\theta}\\
&=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Give a delta-epsilon proof that $\lim_{x \to \frac{1}{2}} \frac{1}{x} = 2$, is my answer right? One of my first $\delta$-$\epsilon$ proofs, but I don't feel that I'm heading in the right direction. I want to proof that:
$$
\lim_{x \to \frac{1}{2}} \frac{1}{x} = 2
$$
By the limit definition, For every $\epsilon > 0$... | The one equation should be
\begin{equation}
\frac{|1-2x|}{|x|} = \frac{2|\frac{1}{2}-x|}{|x|}\ .
\end{equation}
So you have that
\begin{equation}
d(x,\frac{1}{2}) < \delta \Rightarrow d(\frac{1}{x},2) < \frac{2 \delta}{|x|}\ .
\end{equation}
Then, the second problem I see is that you take $2|x| < 2\delta +1$ and concl... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why all composite numbers have this property? Define $f(n)=\sum\limits_{A \in S} f_{1}(n,A),\ n>2,\ n \in \mathbb{Z}$, where $S$ is the power set of $\{\frac{1}{2},\cdots ,\frac{1}{n-1}\}$.
Define $\ f_1(n,\varnothing)=1,\ f_{1}(n,A)=(-1)^{\#A-2n\Sigma(A)}$, where $\#A$ is the size of the set $A$ and $\Sigma(A)$ is t... | Because $f_1(n,A) = (-1)^{\#A-2n\sum A} = \prod_{x\in A} (-1)^{1-2nx}$, the sum over the power set of any set $B$ factors:
$$
\sum_{A\subset B} f_1(n,A) = \prod_{x\in B} \big( 1 + (-1)^{1-2nx} \big).
$$
In particular,
$$
f(n) = \prod_{k=2}^{n-1} \big( 1 + (-1)^{1-2n/k} \big).
$$
If $n$ is composite, then there exists a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $
Show that
$$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $$
Indeed,
First let's show
$7\mid x \text{ and } 7\mid y \Longrightarrow 7\mid x^2+y^2 $
we've $7\mid x \implies 7\mid x^2$ the same for $7\mid y \implies 7\mid ... | Your example is just a special case of this theorem,
Theorem
If $p\equiv 3\pmod 4$ Then whenever $p\mid x^2+y^2$, we have $p\mid x$ and $p\mid y$.
Proof
Assume that $p\mid x^2+y^2$ with $p\not\mid x,y$ then
$$x^2+y^2\equiv 0\pmod p$$
$$x^2\equiv -y^2\pmod p$$
Which means $$x^{2\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}}y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
help: isosceles triangle circumscribing a circle of radius r Please help me show that: the equilateral triangle of altitute $3r$ is the isosceles triangle of least area circumscribing a circle of radius $r$.
Iassumed the following:
base = $2a$
height = $h$
radius of circle = $r$
Area = $\frac{1}{2}(2a)h = ah$
$tan(2\th... | This is a nice example of a problem where the "right" choice for the independent variable makes a solution reasonably tractable, whereas other choices will lead to rather rough going. I tried a straight-ahead Cartesian approach (which led to a quintic equation) and the trigonometric approach you attempted (which gave ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/846673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$. I know that, for $|x|\leq 1$, $e^x$ can be bounded as follows:
\begin{equation*}
1+x \leq e^x \leq 1+x+x^2
\end{equation*}
Likewise, I want some meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$.
The first thing that comes to my mind is $\sqrt{a... | Maybe..
$$
\sqrt{1+x} = 1 + \frac{1}{2} x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + O(x^4).
$$
So, for small enough $x$,
$$
\sqrt{1+x} > 1 + \frac{1}{2}x - \frac{1}{8}x^2
$$
hence
$$
\sqrt{a^2 + b} = a \sqrt{1 + (b/a^2)} > a\Big(1 + \frac{b}{2a^2} -\frac{b^2}{8a^4} \Big)
$$
when $a$ is much larger than $b$. So, in this cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/847153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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If $4k^3+6k^2+3k+l+1$ and $4l^3+6l^2+3l+k+1$ are powers of two, how to conclude $k=1, l=2$ It is given that
$$4k^3+6k^2+3k+l+1=2^m$$
and
$$4l^3+6l^2+3l+k+1=2^n$$
where $k,l$ are integers such that $1\leq k\leq l$.
How do we conclude that the only solution is $k=1$, $l=2$?
I tried subtracting the two equations to get:... | The following argument has a lot of cases, but at least works (I think) to show the only solution is $2,1$.
Let $p(x,y)=4x^3+6x^2+3x+y+1.$ [I changed notation; $1 \le k \le l$ will be here $1 \le y \le x$], and then one wants each of $p(x,y)$ and $p(y,x)$ to be powers of $2$. Since the variables are each $\ge 1$ these ... | {
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"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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What am I supposed to do here now? $\tan(\pi/8) = \sqrt{2} -1$ complex analysis
Find $\sqrt{1+i}$, and hence show $\tan(\pi/8) = \sqrt{2}-1$
Okay so I know that $\sqrt{1+i} = 2^{1/4}e^{i\pi/8}$ and I know $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$ and $\cos x = \frac{e^{ix} + e^{-ix}}{2}$
If i directly substitute those d... | If you let $$\sqrt{1+i}=x+iy$$, then $$(x^2-y^2)+i2xy=1+i$$. then $$x^2-y^2=1$$ and $$2xy=1$$. Hence $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=2$$. Hence $$x^2+y^2=\sqrt{2}$$. Now adding this two terms $$x^2=\frac{1+\sqrt{2}}{2}$$ and $$y^2=\frac{\sqrt{2}-1}{2}$$. So $$\sqrt{1+i} =\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/850056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Please check my solution of $\int \sin^6(x)\cos^3(x) dx$ $$\int \sin^6(x)\cos^3(x) dx = \int \sin^6(x)(1-\sin^2(x))\cos(x)dx$$
$$\int \sin^6(x)\cos(x)dx - \int\sin^8x\cos xdx$$
Now, $\cos xdx = d(\sin x)$
$$\int u^6du - \int u^8du = \frac{1}{7}u^7 - \frac{1}{9}u^9 + C$$
$$\frac{1}{7}\sin^7(x) - \frac{1}{9}\sin^9(x) + C... | Your expression is correct. Using the double-angle identity $\cos 2x=1-2\sin^2 x$, you can verify that the more awkward Alpha version is also correct. For $7(1-2\sin^2 x)+11=18-14\sin^2 x$ and $\frac{18}{126}=\frac{1}{7}$ and $\frac{14}{126}=\frac{1}{9}$.
Remark: Fairly often, with trigonometric functions, verificatio... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
how to derive the fact that the integral of $1/\sin^2(x) = -\cot (x)$ I know how that the integral of $\dfrac{1}{\sin^2(x)} = -\cot (x)$, but how does derive this fact? Can you use half-angle formula to do this integral?
| We can use the tangent half-angle substitution to get rid of any trigonometric expression during the integration.
Let $\sin x=\dfrac{2t}{1+t^2}$. Then $\cos x=\dfrac{1-t^2}{1+t^2}$ and $dx=\dfrac{2dt}{1+t^2}$, and we have
$$\int\frac{1}{\sin^2x}dx=\int\frac{(1+t^2)^2}{4t^2}\frac{2}{1+t^2}dt=\int\frac{1+t^2}{2t^2}dt=\in... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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If $a+\frac{1}{b}, b+\frac{1}{c}, c+\frac{1}{a}\in\mathbb{Z}$, find $a+b+c$. Let $a,b,c$ be positive rational numbers such that
$a+\frac{1}{b}, b+\frac{1}{c}, c+\frac{1}{a}$
are all integers.
Find all the possible values of $a+b+c$.
it would be too complicate to solve by quadratic equation(the discriminant is 6 degree... | Ok I worked it out
$$a+b+c=3, \frac{7}{2} ,\frac{25}{6}, \frac{23}{6}$$
Assume $$\frac{x}{y}+\frac{z}{w}=\frac{xw+yz}{yw}$$
is an integer with both fractions in reduced form,
then $y|w$ and $w|y$ so $y=w$ (they are positive).
This means that
$a,b,c$ must be of the form
$$\frac{p}{q}, \frac{q}{r}, \frac{r}{p}$$ resp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$
Evaluate the limit
$$
\lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right)
$$
My Attempt:
To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B ... | Hint
Since $x$ goes to infinity, let us write $$\sqrt[3]{x^3+x^2+1}=x\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}}$$ and let us define $y=\frac{1}{x}+\frac{1}{x^2}$. Now, look at the Taylor expansion of $$\sqrt[3]{1+y}=1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)$$ and replace $y$ by its definition and expand to get finally $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
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For the polynomial For the polynomial, -2 is a zero. $h(x)= x^3+8x^2+14x+4$. Express $h(x)$ as a product of linear factors.
Can someone please explain and help me solve?
| Since its a cubic equation, you are looking at $$(x+A).(x+B).(x+C)=x^3+8x^2+14x+4$$
Obviously, you have to multiply and work hard towards solving this and getting the answer.
Alternatively, what you can do is represent the same thing as
$$x^3+8x^2+12x+2x+4=0$$
Or, $$x(x^2+8x+12)+2(x+2)$$
Or, $$x(x+6)(x+2)+2(x+2)$$
Or, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/852355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Basic induction proof methods so we're looking to prove $P(n)$ that
$$1^2+2^3+\cdots+n^3 = (n(n+1)/2)^2$$
I know the basis step for $p(1)$ holds.
We're going to assume $P(k)$
$$1^3+2^3+\cdots+k^3=(k(k+1)/2)^2$$
And we're looking to prove $P(k+1)$
What I've discerned from the internet is that I should be looking to add... | Assuming $P(k)$, you add $(k+1)^3$ on both sides of
$$
1^3 + 2^3 + \ldots + k^3 = (k(k+1)/2)^2
$$
to get
\begin{align}
1^3 + 2^3 + \ldots + k^3 + (k+1)^3
& = (k(k+1)/2)^2 + (k+1)^3 \\
& = \frac 14 k^2(k+1)^2 + (k+1)^3 \\
& = \frac 14\left(k^4 + 2k^3 + k^2 + 4k^3 + 12k^2 + 12k + 4\right) \\
& = \frac 14\left(k^4 + 6k^3 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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functions satisfying $f(x - y) = f(x) f(y) - f(a - x) f(a + y)$ and $f(0)=1$ A real valued function $f$ satisfies the functional equation
$$f(x - y) = f(x) f(y) - f(a - x) f(a + y) \tag 1 \label 1$$
Where $a$ is a given constant and $f(0) = 1$. Prove that $f(2a - x) = -f(x)$, and find all functions which satisfy the gi... | Partial progress:
Let $P(x,y)$ be the property that $f(x-y) = f(x)f(y)-f(a-x)f(a+y)$.
Then, $P(0,0)$ gives $f(0) = f(0)f(0)-f(a)f(a)$ which yields $f(a) = 0$.
Also, $P(a,x)$ gives $f(a-x) = f(a)f(x)-f(0)f(a+x)$ which yields $f(a-x)=-f(a+x)$.
Then, $P(x,x)$ gives $f(0) = f(x)f(x)-f(a-x)f(a+x)$ which yields $f(x)^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/852861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Proving an expression is composite I am trying to prove that $ n^4 + 4^n $ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even.
This problem is given in a section where induction is introduced, but I am not quite sure how induction could be us... | Some interesting factorizations of a polynomial of type $x^4+\text{const}$:
$$ x^4+4=(x^2+2x+2)(x^2-2x-2) \tag{1}$$
$$ x^4+1=(x^2+\sqrt[]{2}x+1)(x^2-\sqrt[]{2}x+1) \tag{2}$$
So one can ask, how to select the coefficients $a,b,c,d$ in
$$(x^2+ax+b)(x^2+cx+d) \tag{3}$$
such that all coefficients of the resulting polynomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/853615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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I'm missing the right substitute $\sqrt3\cos x=1-\sin x$ Please show me how to solve the following equation for $x$.
I've tried multiple substitutes but can't seem to find the right one.
$$\sqrt3\cos x=1-\sin x$$
| You don't have to guess the right substitution. Every equation of the form
$$
a\cos x+b\sin x = c
$$
can be managed with the following procedure.
Divide both sides by $\sqrt{a^2+b^2}$, which is different from $0$ unless $a=b=0$, which would make it trivial.
Now we can observe that there is a unique $\varphi$ such that
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/854592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
what is a smart way to find $\int \frac{\arctan\left(x\right)}{x^{2}}\,{\rm d}x$ I tried integration by parts, which gets very lengthy due to partial fractions.
Is there an alternative
| Compute using the series expansion:
\begin{align*}
\int \frac{\tan^{-1} x}{x^2} dx &= \int \frac 1 {x^2} \left(x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots\right) dx \\
&= \int \frac 1 x - \frac{x}{3} + \frac{x^3}{5} - \cdots dx \\
&= \ln x - \frac{x^2}{2 \cdot 3} + \frac{x^4}{4 \cdot 5} - \frac{x^6}{6 \cdot 7} + \cdots
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/854997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$
Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?
M... | Note that $K = \dfrac{2 \cdot 2 \cdot 4 \cdot 4 \cdots 100 \cdot 100}{1 \cdot 2 \cdot 3 \cdot 4 \cdots 99 \cdot 100} = \dfrac{2^{100}(50!)^2}{100!} = \dfrac{2^{100}}{\dbinom{100}{50}}$
It can be shown that the Central Binomial Coefficent satisfies:
$\left(1-\dfrac{1}{8n}\right)\dfrac{2^{2n}}{\sqrt{\pi n}} \le \dbinom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/855990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
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Evaluate the sum ${n \choose 1} + 3{n \choose 3} +5{n \choose 5} + 7{n \choose 7}...$ in closed form How do I evaluate the sum:$${n \choose 1} + 3{n \choose 3} +5{n \choose 5} + 7{n \choose 7} ...$$in closed form?
I don't really know how to start and approach this question. Any help is greatly appreciated.
| Starting with
\begin{align}
(1+t)^{n} = \sum_{k=0}^{n} \binom{n}{k} t^{k}
\end{align}
it is seen that
\begin{align}
\frac{1}{2} \left[ (1+t)^{n} - (1-t)^{n} \right] = \sum_{k=1}^{[(n+1)/2]} \binom{n}{2k-1} t^{2k-1}.
\end{align}
Now differentiating both sides leads to
\begin{align}
\sum_{k=1}^{[(n+1)/2]} (2k-1) \binom{... | {
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"source": "stackexchange",
"question_score": "3",
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Proof Verification for Discrete Math Class Prove that $n^2$ is even iff $n$ is even.
I proved it like this:
Case I: $n$ is even
1) $n = 2a$ $(a\in Z)$
2) $n^2 = 4a^2 = 2(2a^2)$
3) $2a^2 = K$ $(K \in Z)$
4) $n^2 = 2K$
Case II: $n$ is odd
1) $n = 2a + 1$ $(a\in Z)$
2)$n^2 = 4a^2 + 4a +1 = 2(2a^2 + 2a) + 1$
3) $2a^2 + 2a... | You are attempting to prove $A \iff B$.
Your book proves $A \Rightarrow B$ and $B \Rightarrow A$.
You prove $A \Rightarrow B$ and $\lnot A \Rightarrow \lnot B$. However, $\lnot A \Rightarrow \lnot B$ is logically equivalent to $B \Rightarrow A$.
All of these things can be checked with truth tables. For example:
$\begi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find orthogonal projection of vector on a subspace? Well, I have this subspace: $V = \operatorname{span}\left\{ \begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix},\begin{pmatrix}1 \\3 \\4\end{pmatrix}\right\}$
and the vector $v = \begin{pmatrix}9 \\0 \\0\end{pmatrix}$
How can I find the orthog... | Hint:
1) Compute an orthonormal base $v_1,v_2$ of $V$ using Gram-Schmidt
2) Consider the projector $p_V(x) = \sum_{i=1}^2 \langle v_i,x\rangle v_i$
3) Compute $p_V(v)$
So what you did is wrong because $\begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix}$,$\begin{pmatrix}1 \\3 \\4\end{pmatrix}$ are not ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/857942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Calculate $\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$ I have homework questions to calculate infinity sum, and when I write it into wolfram, it knows to calculate partial sum...
So... How can I calculate this:
$$\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$$
| OK, I answer the question with the hint:
$$\sum_{k=1}^n \frac 1 {(k+1)(k+2)} = \sum_{k=1}^n \left(\frac 1 {k+1} - \frac 1 {k+2}\right) = \\ = \left( \frac 1 2 - \frac 1 3 \right) + \left( \frac 1 3 - \frac 1 4 \right) + \left( \frac 1 4 - \frac 1 5 \right) + \ldots + \left( \frac 1 {n+1} - \frac 1 {n+2} \right) = \\ = ... | {
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"url": "https://math.stackexchange.com/questions/858751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Test for convergence $\sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1})$ Test for convergence $$\sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1})$$
So far I attempted to use the ratio test, but I'm stuck on what to do after.
$$\dfrac{\dfrac{\sqrt{n+2}-\sqrt{n}}{n+1}}{\dfrac{\sqrt{n+1}-\sqrt{n-1}}{n}} = \dfrac{n(\... | First note that
$$
\frac{\sqrt{n+1}+\sqrt{n-1}}{\sqrt n} > \frac{\sqrt{n+1}}{\sqrt n} > 1.
$$
Therefore
$$
\sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1}) < \sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1}) \frac{\sqrt{n+1}+\sqrt{n-1}}{\sqrt n}= \sum_{n=1}^\infty \frac{2}{n^{3/2}}
$$
which converges; hence the ... | {
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"url": "https://math.stackexchange.com/questions/861510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How do you find the following limit as x approaches infinity? $\lim_{x\to \infty} \sqrt{x^2+9} - \sqrt{x^2-2}$
I have tried multiplying by the conjugate but the square roots are throwing me off and I'm not sure what to do next. How do you solve this?
| $$\sqrt{x^2+9}-\sqrt{x^2-2}=\frac{(\sqrt{x^2+9}-\sqrt{x^2-2})(\sqrt{x^2+9}+\sqrt{x^2-2})}{\sqrt{x^2+9}+\sqrt{x^2-2}}=\frac{x^2+9-x^2+2}{\sqrt{x^2+9}+\sqrt{x^2-2}}=\frac{11}{\sqrt{x^2+9}+\sqrt{x^2-2}}$$
$$\lim_{x \to \infty} \sqrt{x^2+9}-\sqrt{x^2-2}=\lim_{x \to \infty } \frac{11}{\sqrt{x^2+9}+\sqrt{x^2-2}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Use a proof by cases to show that $\lfloor n/2 \rfloor$ * $\lceil n/2 \rceil$ = $\lfloor \frac{n^2}{4} \rfloor$ for all integers $n$. Question
Use a proof by cases to show that $\lfloor n/2 \rfloor$ * $\lceil n/2 \rceil$ = $\lfloor \frac{n^2}{4} \rfloor$ for all integers $n$.
My Attempt:
I can only think of two cases, ... | Why make it so complicated:
Case 1: n is even. Let n = 2k.
$\lceil n/2 \rceil \times \lfloor n/2 \rfloor = k * k = \lfloor n^2/4 \rfloor$
Case 2: n is odd. Let n = 2k + 1.
$\lceil n/2 \rceil \times \lfloor n/2 \rfloor = k (k + 1) = \lfloor n^2/4 \rfloor$ as $n^2/4 = (4k^2 + 4k + 1)/4 = k(k+1) + 1/4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/864140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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I need help finding the critical values of this function. So $h(t)=t^{\frac{3}{4}}-7t^{\frac{1}{4}}$. So I need to set $h'(t)=0$. So for $h'(t)$ the fattest I've gotten to simplifying os $h'(t)=\frac{3}{4 \sqrt[4]{t}}-\frac{7}{4\sqrt[4]{t^3}}$ and that is as farthest as I can simplify. So i'm having a had time having $... | $h(t)=t^{\frac{3}{4}}-7t^{\frac{1}{4}}$
$\Rightarrow$ $0=h'(t)=\frac{3}{4}t^{-\frac{1}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=\frac{3}{4}t^{-\frac{3}{4}+\frac{2}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=\frac{3}{4}t^{-\frac{3}{4}}t^{\frac{2}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=t^{-\frac{3}{4}}(\frac{3}{4}t^{\frac{2}{4}}-\frac{7}{4})$
Thus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Simplify [1/(x-1) + 1/(x²-1)] / [x-2/(x+1)] Simplify: $$\frac{\frac{1}{x-1} + \frac{1}{x^2-1}}{x-\frac 2{x + 1}}$$
This is what I did.
Step 1: I expanded $x^2-1$ into: $(x-1)(x+1)$. And got: $\frac{x+1}{(x-1)(x+1)} + \frac{1}{(x-1)(x+1)}$
Step 2: I calculated it into: $\frac{x+2}{(x-1)(x+1)}$
Step 3: I multiplied $x-\... | First recall that
\[ \frac{a}{b} \pm \frac{c}{d} =\frac{ad \pm cb}{bd} \]
And
\[ \frac{\frac{a}{b}}{\frac{c}{d}} =\frac{ad}{bc} \]
Now just simplify, no fancy fractions needed:
\[
\frac{\frac{1}{x-1}+\frac{1}{x^2-1}}{x-\frac{2}{x+1}}
= \frac{\frac{(x^2-1)+(x-1)}{(x-1)(x^2-1)}}{\frac{x(x+1)-2}{x+1}}
= \frac{\frac{(x-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find $\lambda$ if $\int^{\infty}_0 \frac{\log(1+x^2)}{(1+x^2)}dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}dx$ Problem : If $\displaystyle\int^\infty_0 \frac{\log(1+x^2)}{(1+x^2)}\,dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}\,dx$ then find the value of $\lambda$.
I am not getting any clue how to proceed as if I ... | From Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$, you can obtain
$$\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx=\pi\ln 2. $$
Define
$$I(\alpha)=\int_0^{1}\frac{\ln(\alpha x+1)}{x^2+1}dx. $$
Then
\begin{eqnarray*}
I'(\alpha)&=&\int_0^{1}\frac{x}{(\alpha x+1)(x^2+1)}dx=\int_0^1\left(\frac{x+\alpha}{(\alpha^2+1)(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Number theory proofs regarding perfect squares How do you prove that $3n^2-1$ is never a perfect square
| Another approach. First establish that perfect squares are either $0$ or $1$ modulo $4$:
$$(2k)^2 = 4k^2 \equiv 0\pmod 4$$
$$(2k+1)^2 = 4k^2 + 4k + 1 \equiv 1 \pmod 4$$
If $n$ is even, $n = 2m$ and,
$$3n^2 - 1 = 12m^2 - 1 \equiv -1 \equiv 3 \pmod 4$$
If $n$ is odd, $n = 2m+1$ and,
$$3n^2 - 1 = 12m^2 + 12m + 3 - 1 \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I reduce radian fractions? For example, I need to know $\sin (19π/12)$.
I need to use the subtraction formula. How do I get $(\text{what}) - (\text{what}) = 19π/12$? I am stuck at what are the radians
Do I divde it by something? What is the process?
| $2\pi$ radians represents a full circle turn so
*
*$\sin(2\pi+\theta)=\sin(\theta)$ and
*$\sin(2\pi-\theta)=\sin(-\theta)=-\sin(\theta)$
So $\sin \left(\frac{19}{12}\pi\right) = \sin \left(2\pi -\frac{5}{12}\pi\right) = \sin \left(-\frac{5}{12}\pi\right) = -\sin \left(\frac{5}{12}\pi\right)$
which since $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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What subject does this fall under (differential curves maybe?) I found a question where I don't even know where it comes from (except, vaguely, a Calc 2 class--but in my past Calc 2 class I never saw anything like this):
Find the angle of rotation needed to eliminate $xy$ from the equation $2x^2 +2\sqrt{3}xy+4y^2 +8x ... | Say that $Q\colon \mathbb R^2\rightarrow \mathbb R$ is the function that maps $(x,y)$ to $2x^2 +2\sqrt{3}xy+4y^2 +8x +8\sqrt{3}y$. Note that the set of points $(x,y)$ so that $Q(x,y)=50$ is an ellipse (picture).
Let $e_1$ be $(1,0)$ and $e_2$ be $(0,1)$, the vectors of the standard basis for $\mathbb R^2$.
Then we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/871828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you get from this expression to the other? Briefly, I am following the solution to a problem, which I understand up to the point where this expression is reached:
$$u_x = \frac{vy + \sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)}}{1 + y^2}$$
The solution then states that this is equivalent to:
$$u_x = \frac{c^2 - v^2}{\s... | \begin{align}
u_x =& \frac{c^2 - v^2}{\sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)} - vy}\\\\
=& \frac{\left(\sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)} + vy\right)(c^2 - v^2)}{\left(\sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)} - vy\right)\left(\sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)} + vy\right)}\\\\
=&\frac{\left(\sqrt{(vy)^2 + (c^2 - v^2)(1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Logarithm Equality. $$\sqrt {\log_a(ax)^{\frac{1}{4}} + \log_x(ax)^{\frac{1}{4}}} + \sqrt {\log _a{(\frac{x}{a})^{\frac{1}{4}}} + \log_x (\frac{a}{x})^\frac{1}{4}} = a,$$ for $a>0$ and different than 1... I keep getting $a = 1$, but that cannot be. I use log identities to transform the above into $$\sqrt {\frac{1}{2}... | $$\sqrt {
\log_a\left((ax)^\frac 14\right)
+ \log_x\left((ax)^\frac 14\right)
} + \sqrt {
\log_a\left(\left(\frac xa\right)^\frac 14\right)
+ \log_x\left(\left(\frac ax \right)^\frac 14\right)
} = a$$
$$\sqrt {
\frac 14\log_a\left(ax\right)
+ \frac 14\log_x\left(ax\right)
} + \sqrt {
\frac 14\log_a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show that $(a+b)^p\le 2^p (a^p+b^p)$ If I may ask, how can we derive that $$(a+b)^p\le 2^p (a^p+b^p)$$ where $a,b,p\ge 0$ is an integer?
| In fact we can make the stronger claim:
$$(a+b)^p\le 2^{p-1} (a^p+b^p)$$
if $a,b \ge 0$. As Mohammad Khosravi points out, this is equivalent to
$$(x+1)^p\le 2^{p-1} (x^p+1)$$
if $x \ge 0$. We prove this by induction. The case $p=1$ is easy: $x+1 \le x+1$. So suppose we have established that
$$(x+1)^{p-1}\le 2^{p-2} (x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$
Find the closed form $$a_{n}$$
since
$$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$
so
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+... | Incomplete answer (maybe it can be used):
If $\left|z\right|<1$ then $\frac{1}{1-z}=1+z+z^{2}+\cdots$
Applying this for $z=x^{4},x^{3},x^{2}$ you find:
$$\frac{1}{\left(1-x^{4}\right)\left(1-x^{3}\right)\left(1-x^{2}\right)}=\left(1+x^{4}+x^{8}+\cdots\right)\left(1+x^{3}+x^{6}+\cdots\right)\left(1+x^{2}+x^{4}+\cdots\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Integral: $\int_0^{\pi} \frac{x}{x^2+\ln^2(2\sin x)}\,dx$ I am trying to solve the following by elementary methods:
$$\int_0^{\pi} \frac{x}{x^2+\ln^2(2\sin x)}\,dx$$
I wrote the integral as:
$$\Re\int_0^{\pi} \frac{dx}{x-i\ln(2\sin x)}$$
But I don't find this easier than the original integral. I have seen solutions wh... | Here is an approach without using contour integration (Cauchy's theorem).
I've found the following result.
Theorem. Let $a$ be any real number.
Then
$$
\begin{align}
\displaystyle \int_{0}^{\pi} \frac{x}{x^2+\ln^2(2 e^{a} \sin x)}\mathrm{d}x
& = \, \frac{2 \pi^2}{\pi^2+4a^2},\tag1\\\\
\int_{0}^{\pi} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 1,
"answer_id": 0
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When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$... When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$. What is the denominator of the fraction?
I tried,
Let the numerator of the fraction be $x$ and the denominator be $y$.
Accordingly, $$\frac{x+4}... | In general you are correct: to solve for two unknowns, you would usually need two equations. But in this case you are lucky, and the one equation gives a solution for the one unknown you are asked to find.
$\dfrac{x+4}{y}=\dfrac{x}{y}+\dfrac{2}{3}$ so multiplying both sides by $3y$ gives $3x+12 = 3x+2y$ making the den... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Solving recurrence relation: Product form Please help in finding the solution of this recursion.
$$f(n)=\frac{f(n-1) \cdot f(n-2)}{n},$$
where $ f(1)=1$ and $f(2)=2$.
| I believe the formula is, for $n\geq 3$,
$$f(n) = 2^{F_{n-1}}\prod_{k=3}^{n}k^{-F_{n-k+1}}$$
where $F_1,F_2,F_3,\ldots$ is the standard Fibonacci sequence whose first two terms are $1$.
You can see this by just writing out the first several terms. For example,
$$f(9) =\dfrac{2^{21}}{3^{13}4^8 5^5 6^3 7^2 8^1 9^1}
$$
Ad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$
Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rare... | Without loss of generality, we assume that $a+b+c=1$. Then,
\begin{align*}
\frac{a}{(b+c)^2} + \frac{b}{(a+c)^2} + \frac{c}{(b+a)^2} &= \frac{a^2}{a(b+c)^2} + \frac{b^2}{b(a+c)^2} + \frac{c^2}{c(b+a)^2}\\
&\ge \left(\frac{a}{b+c} + \frac{b}{a+c}+\frac{c}{b+a} \right)^2\\
&=\left(\frac{1}{b+c} + \frac{1}{a+c}+\frac{1}{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 5
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Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$
If $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$$
Additional info:We can use AM-GM and Cauchy inequaliti... | The next thing you can try is Cauchy again
$$\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)(b+c+a)\geq (a+b+c)^2.$$
So
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c.$$
Now you can eliminate $a+b+c$ from your first inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How to describe $G/U$? Let $G=SL_2(\mathbb{C})$ and let $U = \{\left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right): x \in \mathbb{C}\}$. We have an action of $U$ on $G$ by right multiplication. By definition, $G/U$ is the set of all $U$-orbits under this action. How to describe $G/U$ as a set?
Take $\left( \begin... | For any $a \ne 0,$ and $c \in \mathbb{C},$ the orbit of $\begin{pmatrix} a & 0 \\ c & a^{-1} \end{pmatrix}$ under $U$ gives you $$\Big\{\begin{pmatrix} a & ax \\ c & cx+a^{-1} \end{pmatrix}: \; x \in \mathbb{C}\Big\},$$ so these are all distinct orbits ($a$ and $c$ are uniquely determined by the left column) and run th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/884287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Simplifying from POS using boolean algeabra I have a boolean function, f expressed in the Product of Sum form.
$$f = (A+B+C)\cdot(A+B+ \overline C)\cdot(\overline A + \overline B + \overline C) $$
On simplification I get,
$$ f = ((A+B) + (C \cdot\overline C))\cdot (\overline A + \overline B + \overline C) $$
$$ = (... | Your answer is correct, while the given answer is incorrect (or there is a typo somewhere). Indeed, let $A = B = C = 1$. Then observe that:
\begin{align*}
f &= (A+B+C)\cdot(A+B+ \overline C)\cdot(\overline A + \overline B + \overline C) \\
&= (1 + 1 + 1) \cdot (1 + 1 + 0) \cdot (0 + 0 + 0) \\
&= 1 \cdot 1 \cdot 0 \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/884796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Recognizing the sequence 1/16, 1/8, 3/16, 1/4, 5/16, ...
What is the missing number?
$$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}, \ \ \ [?]$$
$$A. \frac{5}{4}\quad B. \frac{3}{4}\quad C. \frac{5}{8}\quad D. \frac{3}{8}$$
Spoiler: Answer is $D$, but I don't know why.
Thanks
| Another sequence-recognizing technique
is to look at the difference between consecutive terms.
In this case,
$\frac{1}{8}-\frac{1}{16}
= \frac{1}{16}
$,
$\frac{3}{16}-\frac{1}{8}
= \frac{1}{16}
$,
$\frac{1}{4}-\frac{3}{16}
= \frac{1}{16}
$,
and
$\frac{5}{16}-\frac{1}{4}
= \frac{1}{16}
$.
Since the difference between
co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/885137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If $f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$, then what is the value of $f'(1)$
Find $f'(1)$ if $$f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$$
My attempt at the question:
Let $(x-\dfrac{2}{x})$ be $g(x)$
Then $$f(g(x)) = \sqrt{x-1} $$
Differentiating with respect to x:
$$f'(g(x))\cdot g'(x) = \frac{1}{2\sqrt{x-1}} $... | Let us assume ${\left(x-\frac{2}{x}\right) = 1, \text{the part inside}}$
It is found that $x = 2$ or $x = -1$
Taking the positive value of $x$
Therefore $f(1) = \sqrt {x-1}$ and
Thus $$f'(1) = \frac {1}{2}\cdot(x-1)^{\frac{1}{2}-1},$$ $ \text{1 for the value inside the function}$
$$\Rightarrow \frac{1}{2}\cdot(x-1)^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Recurrence of the form $2f(n) = f(n+1)+f(n-1)+3$ Can anyone suggest a shortcut to solving recurrences of the form, for example:
$2f(n) = f(n+1)+f(n-1)+3$, with $f(1)=f(-1)=0$
Sure, the homogenous solution can be solved by looking at the characteristic polynomial $r^2-2x+1$, so that in general a solution for the homogen... | we can use the generating function $F(x)=\sum_{k=1}^{\infty} f(k)x^k$
the recurrence relation may be written:
$$2f(n+1) = f(n+2)+f(n)+3$$
so multiplying by $x^{n+2}$ we have
$$2xf(n+1)x^{n+1} = f(n+2)x^{n+2}+x^2f(n)x^n+3x^{n+2}$$
and summing for $n=0$ to $\infty$
$$
2x(F(x)-f(0)) = F(x)-xf(1)-f(0) + x^2F(x) +3x^2(1-x)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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Evaluating $ \int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx$ I am trying to evaluate the indefinite integral of
$$\int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx.$$
The first thing I did was the substitution rule:
$u=1+x^2$, so that $\displaystyle x \, dx=\frac{du}2$ and $1-x^2=2-u$. The integral then transforms to
$$\int \sqrt{\fr... | $\bf{My\; Solution::}$ Let $\displaystyle I = \int x\sqrt{\frac{1-x^2}{1+x^2}}dx\;,$ Now Substitute $x^2=t\;,$ Then $\displaystyle xdx = \frac{dt}{2}$
So $\displaystyle I = \frac{1}{2}\int \sqrt{\frac{1-t}{1+t}}dt = \frac{1}{2}\int \frac{\left(1-t\right)}{\sqrt{1-t^2}}dt =\frac{1}{2}\int\frac{1}{\sqrt{1-t^2}}dt-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
Solving the logarithimic inequality $\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$ I tried solving the logarithmic inequality:
$$\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$$
several times but keeping getting wrong answers.
| Put $u=\log_2\left(\dfrac x2 \right)=\log_2x-1$
Note that $\log_2(x^2)=2(u+1)$, and $\log_2\left(\dfrac 2x\right)=-u$.
Hence inequality becomes
$$ \begin{align}
u-\dfrac {2(u+1)}u \leq1\\
\dfrac{u^2-3u-2}u \leq0\\
\dfrac{(u-\alpha)(u-\beta)}u \leq 0\\
\end{align}$$
where $\alpha=\dfrac {3+\sqrt{17}}2,\ \beta=\dfrac {3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Solving awkward quadratic equation to obtain "nice" solution. I would like to solve the following quadratic equation to get a "nice" analytic solution for $\rho$.
$\rho^2(r\sin\theta-2nr^2)+\rho(2nr^3-2r^2\sin\theta-2\sin\theta+2nr)-2nr^2+3r\sin\theta=0$
where $r=1-\cos\theta$ (I cannot see how this could be used to si... | It is not a particularly simple form.
Solution:
$\rho = (R \pm S)/Q$
Where
$Q = \textrm{ sin}( \frac{\pi}{2 n} )^3( 32 n - 16 \textrm{ cot}(\frac{\pi}{2 n}))$
$R = -12 \textrm{ cos}(\frac{\pi}{2 n}) + 6 \textrm{ cos}(\frac{3 \pi}{2 n} ) -
2 \textrm{ cos}( \frac{5 \pi}{2 n}) +
2 n \left(14 \textrm{ sin}( \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then $|z|=$? If z is a complex number and $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then find the value of $|z|$ .
I tried to put $ \frac {z^2 + z+ 1} {z^2 -z +1} =k $ then solve for $z$ and tried to find |z|, but it gets messy and I am stuck.
The answer giv... | Proposition. $\frac{z^2+z+1}{z^2-z+1}$ is real if and only if $z\in\Bbb R$ or $|z|=1$.
I'll let you handle the $\Leftarrow$ direction. For $\Rightarrow$, write $k=\frac{z^2+z+1}{z^2-z+1}=\frac{(z+z^{-1})+1}{(z+z^{-1})-1}\implies z+z^{-1}=\frac{k+1}{k-1}$.
When is $z+z^{-1}$ real for nonreal $z$? (What's its imaginary p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving or disproving inequality $ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \ge x + y + z $
Given that $ x, y, z \in \mathbb{R}^{+}$, prove or disprove the
inequality
$$ \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} \ge x + y + z $$
I have rearranged the above to:
$$ x^2y(y - z) + y^2z(z - x) + z^2x(x - y) \ge 0 ... | Consider this inequality:
$$(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0$$
Expand and simplify the above expression to get:
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
Substitute $a = \dfrac{1}{x}, b = \dfrac{1}{y}$ and $c = \dfrac{1}{z}$:
$$\dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2} \ge \dfrac{1}{xy} + \dfrac{1}{yz} + \dfrac{1}{z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/891997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Fermat: Prove $a^4-b^4=c^2$ impossible Prove by infinite descent that there do not exist integers $a,b,c$ pairwise coprime such that $a^4-b^4=c^2$.
| Lemma. If $a,b,c$ are coprime and $a^2+b^2=c^2$, then
$$ \{a,b\}=\{2pq,p^2-q^2\},\quad c=p^2+q^2 $$
with $p,q$ coprime and not both odd.
This follows from the fact that we are looking for rational points on the curve $x^2+y^2=1$, that allows the parametrization $\left(\frac{2t}{1-t^2},\frac{1-t^2}{1+t^2}\right)$. Assu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/893544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Calculate $\int\frac{dx}{x\sqrt{x^2-2}}$. The exercise is:
Calculate:$$\int\frac{dx}{x\sqrt{x^2-2}}$$
My first approach was:
Let $z:=\sqrt{x^2-2}$ then $dx = dz \frac{\sqrt{x^2-2}}{x}$ and $x^2=z^2+2$ $$\int\frac{dx}{x\sqrt{x^2-2}} = \int\frac{1}{x^2}dz = \int\frac{1}{z^2+2}dz = \frac{1}{2}\int\frac{1}{(\frac{z}{\sq... | The mistake is that you are integrating over $z$,you can do a substitution $$u=\frac{z}{\sqrt{2}},dz=\sqrt{2}du\\\frac{1}{2}\int\frac{\sqrt{2}du}{u^2+1}=\frac{\sqrt{2}}{2}\arctan(u)=\frac{1}{\sqrt{2}}\arctan(\frac{\sqrt{x^2-2}}{\sqrt{2}})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/895602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Inequality involving a finite sum this is my first post here so pardon me if I make any mistakes.
I am required to prove the following, through mathematical induction or otherwise:
$$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$$
I tried using mathematical induction th... | $P(k+1)=P(k)+\dfrac{1}{\sqrt{k+1}}\leq 2\sqrt{k}+\dfrac{1}{\sqrt{k+1}}=\dfrac{2\sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}}$, in the other hand we have $2\sqrt{k}\sqrt{k+1}\leq k+k+1=2k+1$ (using $2ab\leq a^2+b^2$), now we get $2\sqrt{k}\sqrt{k+1}+1\leq 2(k+1)$, and finally $\dfrac{2\sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}}\leq 2\sqrt{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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Proof by Induction: $(1+x)^n \le 1+(2^n-1)x$ I have to prove the following by induction:
$$(1+x)^n \le 1+(2^n-1)x$$
for $n \ge 1$ and $0 \le x \le 1$.
I start by showing that it's true for $n=1$ and assume it is true for one $n$.
$$(1+x)^{n+1} = (1+x)^n(1+x)$$
by assumption:
$$\le (1+(2^n-1)x)(1+x)$$
$$= 1+(2^n-1)x+x+(... | I'm not exactly clear as to what your final sentence means. But, since $0\le x\le 1$, we know that $x^2\le x\cdot x\le 1\cdot x\le x$, so that
$$(1+x)^{n+1} \le 1 + (2^n-1)x + x + (2^n-1)x^2
\le 1 + (2^n-1)x + x + (2^n-1)x = 1+2(2^n-1)x + x = 1 + (2^{n+1}-1)x,$$
which is the statement for $n+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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