Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proving convergence of series $\sum_{k=1}^n {\sqrt{k^3+1}-\sqrt {k^3-1}}$ How to prove convergence of the following?
$$\sum_{k=1}^n {\sqrt{k^3+1}-\sqrt {k^3-1}}$$
Thanks!
| hint:$$\sum_{k=1}^n (\sqrt{k^3+1}-\sqrt {k^3-1})=\\\sum_{k=1}^n (\sqrt{k^3+1}-\sqrt {k^3-1})\frac{\sqrt{k^3+1}+\sqrt {k^3-1}}{\sqrt{k^3+1}+\sqrt {k^3-1}}
=\\\sum_{k=1}^n \frac{2}{\sqrt{k^3+1}+\sqrt {k^3-1}}\sim \sum_{k=1}^n \frac{2}{2\sqrt{k^3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit as $x$ approaches infinity of a ratio of exponential functions. The problem appears as follows:
Use algebra to solve the following:
$$\lim_{x\to \infty} \frac{2^{3x+2}}{3^{x+3}}$$
The result is infinity which makes intuitive sense but I can’t get it to yield algebraically.
Attempt:
$$\frac{2^{3x+2}}{3^{x+3}} = \f... | Since
\begin{align*}
\lim_{x\to \infty} \frac{2^{3x+2}}{3^{x+3}}
&= \lim_{x\to \infty} \frac{2^2 (2^3)^x}{3^3 3^x} \\
&= \lim_{x\to \infty} \frac{2^2}{3^3}\left( \frac{8}{3} \right)^x \\
&= \frac{2^2}{3^3} \lim_{x\to \infty} \left( \frac{8}{3} \right)^x \\
&> \frac{2^2}{3^3} \lim_{x\to \infty} 2^x, \\
\end{align*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Bilinear Transformation I'm studying about digital control systems and I was given this transformation $$s=\frac{2}{T}\frac{z-1}{z+1}$$ But I don't like just accepting things so I looked it up to find the proof and I found this on wikipedia
I'm aware of the taylor and maclaurin series but I can't quite get what's happ... | From the power series of the logarithm
$$\log(1+x) = - \sum_{n>0} \frac{(-x)^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\ldots$$
(when $\vert x \vert < 1)$, we can immediately deduce
$$\log(1-x) = - \sum_{n>0} \frac{x^n}{n}=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}-\ldots,$$
and su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving BCH formula
I managed to solve a) by differentiation. But, I am stuck at b). I can't see why change from $B$ to $e^B$ puts the series of a) in the exponential function. Also, I think I have to use a) and b) to show c). However I can't find a way to do so. Could anyone please explain to me?
| Ok here we go, using the identities given we clearly have
$$
\begin{align}
e^{sA}e^{sB} &= e^{sA} e^{sB} e^{-sA} e^{sA} \\
&= \exp\left(sB + s^{2}[A,B] + \frac{s^3}{2}[A,[A,B]] + \mathcal{O}(s^4) \right)e^{sA} \tag{1}
\end{align}
$$
and we want to show that after expansion this agrees, up to terms of $\mathcal{O}(s^4)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2433158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}=4x^2.$ I just need help to finetune this solution. Only having a correct answer is not sufficient to comb home 5/5 points on a problem like this. Thoughts on improvements on stringency? Is there any logical fallacy or ambiguity? Any input is very welc... | Taking derivative of LHS we have $$\frac{d}{dx}\left(\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}\right)=0$$
Therefore $$\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}=\text{ constant }$$
Calculation for $x=\dfrac{1}{\sqrt3}$ gives
$$\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Finding the Equations of Angle Bisectors The question is as follows:
Let $K(5, 12)$, $L(14, 0)$ and $M (0,0)$. The line $x + 2y = 14$ bisects angle $MLK$. Find equations for the bisectors of the angles $KML$ and $MKL$.
Any help will be truly appreciated!
| Let $KD$ be bisector of $\Delta KLM$.
Thus, since $$\frac{LD}{DM}=\frac{KL}{KM}=\frac{\sqrt{9^2+12^2}}{\sqrt{5^2+12^2}}=\frac{15}{13},$$
we obtain
$$D\left(\frac{13\cdot14+0}{15+13},0\right)$$ or
$$D(6.5,0).$$
Thus, $$m_{KD}=\frac{12-0}{5-6.5}=-8$$ and for the equation of $KD$ we obtain:
$$y-12=-8(x-5)$$ or
$$y=-8x+52.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding sum of the series $\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$ Find the sum: $$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$$
My method:
I tried to split it into partial fractions like: $\dfrac{A}{r}, \dfrac{B}{r+d}$ etc. Using this method, we have 4 equations in $A,B,C,D$! And solving them takes much time. I... | (Modified answer)
Define $r^{\overline{m}(d)}$ as the $d$-tuple rising factorial, i.e.
$r^{\overline{m}(d)}=r(r+d)(r+2d)\cdots(r+(m-1)d) $
and $r^{-\overline{m}(d)}=\frac 1{r^{\overline{m}(d)}}$
, e.g. $r^{\overline{4}(3)}=r(r+3)(r+6)(r+9)$.
$$\begin{align}
S(m,d)&=\sum_{r=1}^n\frac 1{r(r+d)(r+2d)\cdots (r+(m-1)d)}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Proving $\int_{0}^\pi \frac{2\cos 2\theta + \cos 3\theta}{5+4\cos\theta} = \frac{\pi}{8}$ Given that $$\int_{|z|=1|}\frac{z^2}{2z+1} dz = \frac{i\pi}{4}$$,
show $$\int_{0}^\pi \frac{2\cos 2 \theta + \cos 3\theta}{5+4\cos\theta} = \frac{\pi}{8}$$.
I saw the bounds of the latter integral and thought that I should try a... | Note that
$$2\int^\pi_0 \cos(n\theta)f(\cos \theta ) d\theta = \int^{\pi}_{-\pi} \cos(n\theta)f(\cos\theta ) d \theta $$
Now you can use that $\gamma(t) = e^{i t}$ where $t \in [-\pi , \pi )$ to prove
$$\Re \oint_{|z|=1} z^n f\left(\frac{z+z^{-1}}{2} \right) \frac{dz}{iz} = 2\int^\pi_0 \cos(n\theta)f(\cos \theta ) d\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is $\mathbb Q[\sqrt{11} + 10^\frac{1}{3}]$ equal to $\mathbb Q[\sqrt{11}, 10^\frac{1}{3}]$?
Is $\mathbb Q[\sqrt{11} + 10^\frac{1}{3}]$ equal to $\mathbb Q[\sqrt{11}, 10^\frac{1}{3}]$?
It is clear that $\sqrt{11} + 10^\frac{1}{3} \in \mathbb Q[\sqrt{11}, 10^\frac{1}{3}]$ and that $\mathbb Q \subset \mathbb Q[\sqrt{11}... | Let $\newcommand{\al}{\alpha}\al=\sqrt{11}+10^{1/3}$. Then
$$(\al-\sqrt{11})^3=10.$$
So
$$\al^3+33\al-10=(3\al^2+11)\sqrt{11}$$
and then $\sqrt{11}\in\Bbb Q(\al)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ The roots of the equation $2x^2-3x+6=0$ are α and β. Find a quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$.
The answer is $4x^2+5x+4=0$
I don't know how to get to the answe... | $$
\left( x - \frac \alpha \beta \right)\left( x - \frac\beta\alpha\right) = 0
$$
$$
(\beta x-\alpha)(\alpha x-\beta) = 0
$$
$$
\alpha\beta x^2 - (\alpha^2+\beta^2) x + \alpha \beta = 0 \tag 1
$$
If $2x^2-3x+6=0$ then $x = \dfrac{3 \pm \sqrt{-39}} 4.$ So
$$
\alpha\beta = \frac{3+\sqrt{-39}} 4 \cdot \frac{3-\sqrt{-39}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$ If α and β are the roots of the equation $3x^2+5x+4=0$, find the value of $α^3+β^3$
How can I factorize the expression to use the rule of sum and product of roots?
The answer is $\frac{55}{27}$
| Rename $a=\alpha$ and $b=\beta$
Since $a$ is a solution of $3x^2+5x+4=0$ we have $3a^2+5a+4=0$ and thus $a^2={1\over 3}(-5a-4)$, so:
$$ 3a^3 = -5a^2-4a = -{5\over 3}(-5a-4) -4a = {13\over 3}a+{20\over 3}$$
the same is true for $b$, so we have
$$9a^3+9b^3 = 13(a+b)+40 = 13{-5\over 3 }+40 = {55\over 3}$$
and thus conclus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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How to find minimum and maximum value of x, if x+y+z=4 and $x^2 + y^2 + z^2 = 6$? I just know that putting y=z, we will get 2 values of x. One will be the minimum and one will be the maximum. What is the logic behind it?
| The second condition gives
$$x^2+(y+z)^2-2yz=6$$ or
$$x^2+(4-x)^2-2yz=6$$ or
$$yz=x^2-4x+5,$$
which with $y+z=4-x$ gives
$$(4-x)^2-4(x^2-4x+5)\geq0$$ or
$$3x^2-8x+4\leq0$$ or
$$\frac{2}{3}\leq x\leq2.$$
The equality occurs for $y=z$ because $y$ and $z$ are roots of the equation
$$t^2-(4-x)t+x^2-4x+5=0$$ and the equalit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
for $x^2+y^2=a^2$ show that $y''=-(a^2/y^3)$ For $x^2+y^2=a^2$ show that $y''=-(a^2/y^3)$
I got that
$y^2=a^2-x^2$
$y'=-x/y$
$y''=(-1-y'^2)/y$
But then I get stuck.
| I would prefer to use parametric equation for these type of problems
$x=a\cos \alpha$
$y=a\sin \alpha$
$\frac{dy}{d\alpha}=a \cos \alpha$
$\frac{dx}{d\alpha}=-a \sin \alpha$
$\frac{dy}{dx}=- \cot \alpha$
$\frac{{d}^2y}{d{x}^2}=-\frac{d}{d\alpha} \cot \alpha \cdot\frac{d\alpha}{dx}$
$\frac{{d}^2y}{d{x}^2}=\frac{ (\opera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
triple integration question . limit problem(substitution) I have to evaluate the following integral-
$$\iiint x^2y \,dx \,dy \,dz$$ over the region bounded by $ \frac{x^2}1+\frac{y^2}4+\frac{z^2}9=1.$
I take the limits as -
for x: $-1$ to $1$
for y: $-2(1-x^2)^{\frac 12})$ to $2((1-x^2)^{\frac 12})$
for z: $-3((1-x^2-(... | $$K=\iiint_E x^2y \,\mathrm dx \,\mathrm dy \,\mathrm dz$$ where $E$ is the ellipsoid $E=\left\{ (x,y,z)\in\Bbb R^3: \frac{x^2}1+\frac{y^2}4+\frac{z^2}9=1\right\}$
So we have
$$
|z|\le 3,\quad |x|\le \sqrt{1-\frac{z^2}{9}},\quad |y|\le 2 \sqrt{1-\frac{x^2}{1}-\frac{z^2}{9}}
$$
and
$$
\begin{align}
K&={\Large\int}_{-3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Number of ways of arranging ten girls and three boys if the boys separate the girls into groups of sizes $3, 3, 2, 2$ Ten girls are to be divided into groups of sizes $3,3,2,2$. Also, there are $3$ boys. Number of ways of linear sitting arrangement such that between any two groups of girls, there is exactly one boy (no... | Your solution is correct.
Here is another way to think about the problem:
We will seat the girls in two blocks of three seats and two blocks of two seats. There are $\binom{4}{2}$ ways to choose the positions of the blocks of three seats. Once the blocks of seats have been arranged, there are $\binom{10}{3}$ ways to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find all n for which $2^n + 3^n$ is divisible by $7$? I am trying to do this with mods. I know that:
$2^{3k+1} \equiv 1 \pmod 7$, $2^{3k+2} \equiv 4 \pmod 7$, $2^{3k} \equiv 6 \pmod 7$ and $3^{3k+1} \equiv 3 \pmod 7$, $3^{3k+2} \equiv 2 \pmod 7$, $3^{3k} \equiv 6 \pmod 7$, so I thought that the answer would be when $... | Note that $3^6\equiv 1$, not $6$. You actually need to do this with exponents $6k, 6k+1, 6k+2$ and so on. Fermat's little theorem says that that's enough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Fill out a group table with 6 elements Let $G=\{0,1,2,3,4,5\}$ be a group whose table is partially shown below:
\begin{array}{ c| c | c | c | c |c|c|}
* & 0& 1 & 2 & 3& 4 & 5\\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
1 & 1 & 2 & 0 & 4 & & \\
... | I am guessing it is isomorphic to the dihedral group on $6 $ elements. You have gone wrong in the last "quater" of the grid.
\begin{array}{ c| c | c | c | c |c|c|}
* & 0& 1 & 2 & 3& 4 & 5\\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
1 & 1 & 2 & 0 & 4 & 5 & 3 \\
\hline
2 & 2 & 0 & 1 & 5 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Conditional probability of multivariate gaussian I'm unsure regarding my (partial) solution/approach to the below problem. Any help/guidance regarding approach would be much appreciated.
Let $\mathbf{X} = (X_1, X_2)' \in N(\mu, \Lambda ) $ , where
$$\begin{align}
\mu &= \begin{pmatrix}
1 \\
1
... | From where you are, simply use the relation $(a=3)$
$$
Y_1 | Y_2=a
\sim
\mathcal{N}
\left( \mu_1+
\rho\frac{\sigma_1}{\sigma_2} (a-\mu_2),
(1-\rho^2) \sigma_1^2
\right)
$$
and then compute the probability as
$$
1-\Phi_{Y1|Y_2}(2)
$$
using the cdf of $Y_1|Y_2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Show that $(a+2b+\frac{2}{a+1})(b+2a+\frac{2}{b+1})\geqslant16$ if $ ab\geqslant1$ $a$, $b$ are positive real numbers.
Show that $$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)\geqslant16$$
if
$$ ab\geqslant1$$
Should I use AM-GM inequality? If yes, where?
| Let $a+b=2u$.
Thus, by AM-GM $u\geq1$ and by C-S and AM-GM we obtain:
$$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)=$$
$$=\left(a+b+b+\frac{2}{a+1}\right)\left(a+b+a+\frac{2}{b+1}\right)\geq$$
$$\geq\left(a+b+\sqrt{ab}+\frac{2}{\sqrt{(a+1)(b+1)}}\right)^2\geq$$
$$\geq\left(2u+1+\frac{4}{a+1+b+1}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Prove that: $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\ge\frac{2}{3}$ I've got three inequalities:
$\forall n\in\mathbb N:$
$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \ge\frac{1}{2}$$
$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \ge\frac{7}{12}$$
$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\ge\frac{2}{3}... | We know that $$\color{red}{ \frac{x}{x+1}\le\ln(x+1)\le x~~~\forall ~x>0\tag{1}\label{eq}}$$
taking $x=\frac{1}{n+k}~~0\le k\le n$ this lead to $$\ln\left(\frac{1}{n+k}+1\right)\le\frac{1}{n+k}~~~\forall ~~0\le k\le n $$
that is
\begin{split}
\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} &=&\sum_{k=0}^{n} \frac{1}{n+k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Deductions from combined equation of two circles
Given combined equation of two circles
$$x^4+y^4+2x^2y^2-4x^3-4xy^2+2x^2+2y^2+4x-3=0$$
In order to find the distance between the centres and the number of common tangents to the circles,
one way would be to factorize the equation as
$$(x^2+y^2-1)(x^2+y^2-4x+3)=0$$
Howe... | Assuming the general equation of a circle as
$$x^2+y^2+2gx+2fy+c=0$$
For a system of two circles, the combined equation will be
$$(x^2+y^2+2g_1x+2f_1y+c_1)(x^2+y^2+2g_2x+2f_2y+c_2)=0$$
$$\Rightarrow x^4+2x^2y^2+y^4+2(g_1+g_2)x^3+2(f_1+f_2)x^2y+2(g_1+g_2)xy^2+2(f_1+f_2)y^3+(4g_1g_2+c_1+c_2)x^2+(4f_1f_2+c_1+c_2)y^2+4(g_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Area of region enclosed by the locus of a complex number Find the area of region enclosed by the locus of $z$ given by $\arg(z-i) - \arg(z+i)= \frac{2\pi}{3}$ and imaginary axis (where $i= \sqrt {-1}$)
What I did was I put $$\tan (\alpha) =z-i$$ and $$\tan (\beta) = z+i$$ and solving for $\tan(\alpha-\beta) $ I got the... | Let $z=x+yi$ where $x,y\in\mathbb R$.
We have, using this,
$$\begin{align}&\text{Arg}(x+(y-1)i)-\text{Arg}(x+(y+1)i)\\\\&=\begin{cases}\arctan\left(\frac{y-1}{x}\right)-\arctan\left(\frac{y+1}{x}\right)&\text{if$\ x\gt 0\ \text{or}\ (x\lt 0\ \text{and}\ y\ge 1)\ \text{or}\ (x\lt 0\ \text{and}\ y\lt -1)$}\\\\\arctan\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square. Problem: For what pair of positive integers $(a,b)$ is $3^a + 7^b$ a perfect square.
First obviously $(1,0)$ works since $4$ is a perfect square, $(0,0)$ does not work, and $(0,1)$ does not work, so we can exclude cases where $a$ or $b$ are zero for the... | The technique I mentioned is working. One of the cases is
$$ 1 + 2 \cdot 3^c = 7^d, $$
where we think that $c=d=1$ gives the largest such answer. Subtract $7$ from both sides,
$$ 2 \cdot 3^c - 6 = 7^d - 7. $$ Let $y+1 = c,$ $x+1 = d,$ for
$$ 6 \cdot 3^y - 6 = 7 \left( 7^x -1 \right), $$
$$ 6 \left( 3^y -1 \right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How can we see $ax^2-bx-\bar b x+c,a>0,c \ge 0 $ attains minimum when $x=\frac{\Re b}{a}$ How can we see $ax^2-bx-\bar b x+c, a>0,c\ge 0$ attains minimum (just minimize it over the set of all possible $x$ s.t. the quadratic function is real) when $x=\frac{\Re b}{a}$ where $\Re b$ is the real part of $b$? I know that $... | Let $\beta = \text{Real}(b)$. Notice
$$
\begin{aligned}
ax^2 - bx - \bar{b}x + c &= ax^2 - (b + \bar{b}) x + c \\
&= ax^2 - 2 \beta x + c \\
&= a\left( x^2 - 2 \frac{\beta}{a} x\right) + c \\
&= a \left( x - \frac{\beta}{a} \right)^2 + c - \frac{\beta^2}{a}
\end{aligned}
$$
where the last equality follows by completing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find $\mathbf{u} \cdot \mathbf{v}$, where $\mathbf{u} = \mathbf{e}^1 - \mathbf{e}^2 + \mathbf{e}^3 - \mathbf{e}^4$. Please note that Einstein summation notation is used.
Take $B = \{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e_3}, \mathbf{e}_4 \}$ as a basis for $\mathbb{R}^4$, where $\mathbf{e}_1 = (1, 0, 0, 0), \mathbf{e}_... | You made a mistake calculating $\mathbf{u}$ in the standard basis. It should be
$$ \mathbf{u} = (1,-\color{red}{2}, 2, -2) $$
If you carry out the computation the same way but with the $-2$ you'll get $\mathbf{u} \cdot \mathbf{v} = -5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Locus problem need help
My try
Let $(x,y)$ place on the demanded graph
$\sqrt {x^{2}+y^{2}}-1=\sqrt {x^{2}+(y+3)^{2}}$
$-2\sqrt {x^{2}+y^{2}}= 6y+8$
$-\sqrt {x^{2}+y^{2}}=3y+4$
$x^{2}+y^{2}= 16+9y^2+24y$
What should I do now?
| Let $P(h,k)$ be the point which is equidistant (call this distance $d$) from the circle and the line. Then the distance of $P$ from the center of the circle (which in this case is $(0,0)$) should be $d+1$ and since $d$ is also the distance (perpendicular distance) from the horizontal line $y=-3$, so $d=k+3$.
Thus we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find all functions $f(m)[(f(n))^2-1]=f(n)[f(m+n)-f(m-n)]$ Find all functions $f : \mathbb{N} \to \mathbb{N}$ satisfying
$$f(m)[(f(n))^2-1]=f(n)[f(m+n)-f(m-n)]$$ $\forall m>n$.
Attempted work :
Let $P(m,n)$ denote $f(m)[(f(n))^2-1]=f(n)[f(m+n)-f(m-n)], \forall m>n$.
Case 1 : $f(n)=1, \forall n \in \mathbb{N}$.
Case 2 ... | When $f(n)=a^n$ then $f$ verifies the functional equation.
$f(n)(f(m+n)-f(m-n))=a^n(a^{m+n}-a^{m-n})=a^m(a^{2n}-1)=f(m)(f(n)^2-1)$.
So let's set $f(1)=a\in\mathbb N^*$ and see if we can bring something by induction on $f(n)=a^n$.
$f(1)(f(m+1)-f(m-1))=f(m)(f(1)^2-1)\\\implies a(f(m+1)-a^{m-1})=a^m(a^2-1)\\\implies af(m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Highest common factor of divisors of a number
Given that a number is N=2910600 . Find total number of ways in which given number can be split into two factors such that their highest common factor is a prime number .
my attempt
$N=2^3 \cdot 3^3 \cdot 5^2 \cdot 7^2 \cdot 11$
Now, one of the number would have only the ... | Suppose that $n=ab$ with $\gcd(a,b)=p$ for some prime $p$. So $a=a'p$ and $b=b'p$ for some natural numbers $a'$ and $b'$. Hence $N=a'b'p^2$ with $\gcd(a',b')=1$.
So in order to construct all possible pairs $(a,b)$, we pick a prime $p$ and divide $N$ by $p^2$. Then we split the quotient in two coprime factors, $(a',b')$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do we minimize $\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right)$? Find the minimum value of the following function, where a and b are real
numbers.
\begin{align} f(x)&=\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right) \end{align}
Note: The solution should not contain ... | Hint:
$$(y-2c)(y-2d)=(y-(c+d))^2+4cd-(c+d)^2\ge?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to compute $\lim_{x\to 0}\frac{\sin(x^2+x)-x}{x^2}$ without L'Hospital's Rule or series? I came across this problem in an examination for beginners in Calculus:
Compute $\displaystyle \lim_{x\to 0}\frac{\sin(x^2+x)-x}{x^2}$.
It is easy to show that the limit is $1$ by using series or L'Hospital's Rule. But the e... | $$\begin{align} L = &\lim_{x\to 0}\frac{\sin(x^2+x)-x}{x^2} &\\ = &\lim_{x\to 0}\frac{\sin x^2 \cos x + \sin x \cos x^2-x}{x^2} &\\= &1+ \lim_{x\to 0}\frac{ \sin x \cos x^2-x}{x^2}&\\= &1+ \lim_{x\to 0}\frac{ (\sin x - x + x) \cos x^2-x}{x^2}&\\= &1+ \lim_{x\to 0}\frac{ (\sin x - x)\cos x^2 + x(\cos x^2 -1)}{x^2} &\\=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving that $\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$ are different positive real numbers.
First, I tried to simplify the proof statement but I got an even mo... | Use the rearrangement inequality. Assume without loss of generality $a\geq b\geq c>0$. Then we have $ab\geq ac\geq bc\,$ and $\,1/c^3\geq 1/b^3\geq 1/a^3$. Therefore the sorted sum-product
$$\frac{ab}{c^3}+\frac{ac}{b^3}+\frac{bc}{a^3}\geq\frac{bc}{c^3}+\frac{ab}{b^3}+\frac{ac}{a^3}=\frac{b}{c^2}+\frac{a}{b^2}+\frac{c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Proving equivalence of norms in $\mathbb{R}^2$
Let $\lVert \cdot \rVert_*:\mathbb{R}^2 \to \mathbb{R}, (x,y) \rightarrow \sqrt{x^2+2xy+3y^2}$ be a norm.
How can I find to constants $k,K \in \mathbb{R}^{>0}$ so that the following equivalence is given:$$k\lVert (x,y) \rVert_2 \leq \Vert (x,y) \rVert_* \leq K\Vert (x,y) ... | (How to make simple things difficult...)
By Young's inequality, for every $a>0$ you have that
$$
2|xy| = 2(a|x|)(|y|/a) \leq a^2 x^2 + \frac{y^2}{a^2}
$$
hence, for every $a, b > 0$,
$$
- a^2 x^2 - \frac{y^2}{a^2} \leq 2xy \leq b^2 x^2 + \frac{y^2}{b^2}\,.
$$
These inequalities give
$$
(1-a^2) x^2 + \left(3 - \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Compute $\lim_{\theta\rightarrow 0}\frac{\sin{(\tan{\theta})}-\sin{(\sin{\theta})}}{\tan{(\tan{\theta})}-\tan{(\sin{\theta})}}$ I rewrote it by writing the tan as sin/cos and cross multiplying:
$$\frac{\sin{(\tan{\theta})}-\sin{(\sin{\theta})}}{\tan{(\tan{\theta})}-\tan{(\sin{\theta})}}= \frac{\sin{(\tan{\theta})}-\sin... | Let $x = \sin \theta$. Then
$$
\lim_{\theta\rightarrow 0}\frac{\sin{(\tan{\theta})}-\sin{(\sin{\theta})}}{\tan{(\tan{\theta})}-\tan{(\sin{\theta})}}\\
= \lim_{x\rightarrow 0}\frac{\sin{(\frac{x}{\sqrt{1 - x^2}})}-\sin{(x)}}{\tan{(\frac{x}{\sqrt{1 - x^2}})}-\tan{(x)}}\\
$$
Further, by trig theorems (e.g. in wikipedia) ,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Integrate $\arctan{\sqrt{\frac{1+x}{1-x}}}$ I use partial integration by letting $f(x)=1$ and $g(x)=\arctan{\sqrt{\frac{1+x}{1-x}}}.$ Using the formula:
$$\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx,$$
I get
$$\int1\cdot\arctan{\sqrt{\frac{1+x}{1-x}}}dx=x\arctan{\sqrt{\frac{1+x}{1-x}}}-\int\underbrace{x\left(\arctan{\sq... | By letting $x=\frac{z-1}{z+1}$ the given integral boils down to
$$ 2\int \frac{\arctan\sqrt{z}}{(z+1)^2}\,dz\stackrel{z\mapsto u^2}{=}4\int\frac{u\arctan u}{(u^2+1)^2}\,du\stackrel{\text{IBP}}{=}\frac{u-(1-u^2)\arctan u}{u^2+1}. $$
Your method leads to the same solution by noticing that
$$\frac{d}{dx}\,\arctan\sqrt{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Solve for real $x$ if $(x^2+2)^2+8x^2=6x(x^2+2)$ Question:
Solve for real $x$ if $(x^2+2)^2+8x^2=6x(x^2+2)$
My attempts:
*
*Here's the expanded form:$$x^4-6x^3+12x^2-12x+4=0$$
*I've plugged this into several online "math problem solving" websites, all claim that "solution could not be determined algebraically, he... | By your hint we obtain that it's $$x^4-6x^3+12x^2-12x+4=0$$ or
$$x^4-2x^3+2x^2-4x^3+8x^2-8x+2x^2-4x+4=0$$ or
$$(x^2-2x+2)(x^2-4x+2)=0,$$ which gives the answer:
$$\{1+i,1-i,2+\sqrt2,2-\sqrt2\}.$$
If we don't know the answer then we can get the following.
For all real $k$ we have:
$$x^4-6x^3+12x^2-12x+4=$$
$$=(x^2-3x+k)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding value of determinant, Where $A+B+C+P+Q+R=0$ Finding value of Determinant $$\begin{vmatrix}
\tan (A+P) & \tan(B+P) & \tan(C+P)\\\\
\tan (A+Q)& \tan (B+Q) & \tan (C+Q)\\\\
\tan (A+R)&\tan (B+R) & \tan(C+R)
\end{vmatrix}$$ for all values of $A,B,C,P,Q,R$ Where $A+B+C+P+Q+R=0$
I did not understand how to st... | We have that if $X+Y+Z = 0$, then $\tan X+ \tan Y+\tan Z = \tan X \tan Y \tan Z$.
Hence for example $$\tan (A+P) \tan (B+Q) \tan (C+R) = \tan (A+P) +\tan (B+Q) +\tan (C+R)$$
Hence when you expand from R1, the determinant expands as
$$\begin{align} & \left(\tan (B+Q) +\tan (C+R)-\tan (C+Q) - \tan (C+R)\right)\\
+ & \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Prove that $x^2 + 4xy + 4y^2 + 3x + 6y - 2 = 0$ is the equation of two straight lines
Prove that the equation $x^2 + 4xy + 4y^2 + 3x + 6y - 2 = 0$ is of two straight lines. Find each line's equation, and find the angle between the two lines.
My Attempt:
Factorize $4y^2 + 6y - 2$; Can't be factorized without quadratic... | The given equation can be written as
$$ (x+2y)^2+3(x+2y)-2 = 0 $$
or as
$$ (x+2y)(x+2y+3) = 2 $$
and the only solutions of $z(z+3)=2$ are given by $z=\frac{-3\pm\sqrt{17}}{2}$, so our locus is given by the union of the parallel lines $x+2y=\frac{-3+\sqrt{17}}{2}$ and $x+2y=\frac{-3-\sqrt{17}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Center of mass of triangle created by parabola and two lines The parabola $y^2 = 3x$ is given.
Two perpendicular straight lines are drawn from the origin point, intersecting the parabola in points P and Q.
Find equation (in cartesian form) of the set of centers of mass of all triangles OPQ (where O is the origin point... | A generic line through the origin has equation $r:y=mx$ and we have $r^{\perp}:y=-\frac{1}{m}x$
The intersection of $r$ and $r^{\perp}$ with the parabola $y^2=3x$ are respectively $P\left(\frac{3}{m^2};\;\frac{3}{m}\right)$ and $Q(3m^2,\;-3m)$
The solution of the systems
$
\left\{
\begin{array}{l}
y^2=3x \\
y=mx \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Convergence of $\sum_n \frac{(-1)^n n^{\alpha}\sin{\frac{1}{n^{\alpha}}}}{n^{\beta} + (-1)^n}$
Discuss the convergence/divergence of the series of general term :$$\frac{(-1)^n n^{\alpha}\sin{\frac{1}{n^{\alpha}}}}{n^{\beta} + (-1)^n}\:\:\alpha,\beta >0$$
What I've done so far :
$$\begin{align}
\left\lvert\frac{(-1)^n... | Simply note that since $\alpha >0$ and $\beta >0$,
$$\frac{(-1)^n n^{\alpha}\sin{\frac{1}{n^{\alpha}}}}{n^{\beta} + (-1)^n} = \frac{(-1)^n}{n^\beta}(1+o(1))\left(1- \frac{(-1)^n}{n^\beta} +o\left( \frac{1}{n^\beta}\right) \right)$$
hence
$$\frac{(-1)^n n^{\alpha}\sin{\frac{1}{n^{\alpha}}}}{n^{\beta} + (-1)^n} = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that $\binom{ap}{p} \equiv a \pmod{ap}$
Let $N = ap$ be a composite positive integer, where $a > 1$ is a positive integer and $p$ is a prime. Prove that $$\binom{ap}{p} \equiv a \pmod{ap}.$$
We have $$\binom{ap}{p} = \dfrac{ap(ap-1) \cdots (ap-(p-1))}{p!} = \dfrac{a(ap-1)(ap-2) \cdots (ap-(p-1))}{(p-1)!}.$$ If ... | The proof is for $p>2$! Considering
$$\binom{ap}{p}=\frac{ap}{p}\binom{ap-1}{p-1} \tag{1}$$
and
$$ap-1 \equiv -1 \pmod{p}$$
$$ap-2 \equiv -2 \pmod{p}$$
$$...$$
$$ap-p+1 \equiv -p+1 \pmod{p}$$
we obtain
$$(ap-1)(ap-2)...(ap-p+1) \equiv (-1)^{p-1}(p-1)! \equiv (p-1)! \pmod{p} \tag{2}$$
But $$\binom{ap-1}{p-1}=Q \in \mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find angle in rad from sin I can't solve this and I have a test tomorrow, can you help me please:
$\alpha - \beta = \frac{3}{5} + \pi$
$\cos \alpha = \frac{1}{3}$
Determine $\sin\alpha - \beta$
That's the problem. Here's what I know:
$\sin \alpha = (\sqrt 24) / 5$
How do I progress? Thanks for the help.
| $$\alpha - \beta = \frac{3}{5} + \pi$$
$$\cos \alpha = \frac{1}{3}$$
so
$$\sin\alpha=\pm\frac{2\sqrt{2}}{3}$$
So the answer is
$$\pm\frac{2\sqrt{2}}{3}-\beta$$
and because of
$$\alpha - \beta = \frac{3}{5} + \pi$$
it is
$$\beta=\alpha-\pi-\frac{3}{5}$$
substituting to
$$\pm\frac{2\sqrt{2}}{3}-\beta$$
;
$$\pm\frac{2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2487627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$f(x)=\sqrt{\dfrac{x-\sqrt {20}}{x-3}}$, $g(x)=\sqrt{x^2-4x-12}$, find the domain of $f(g(x))$
$f(x)=\sqrt{\dfrac{x-\sqrt {20}}{x-3}}$, $g(x)=\sqrt{x^2-4x-12}$, find the domain of $f(g(x))$.
For some reason, I cannot get the correct answer. Here is what I tried.
$D_g:$
$$x^2-4x-12\ge0$$
$$(x-6)(x+2)\ge0$$
$$\boxed{(-... | The domain of $f$ is $(-\infty,3)\cup[\sqrt{20},\infty)$ (you seem to believe instead that $3>\sqrt{20}$).
Thus we need
$$
\begin{cases}
x^2-4x-12\ge 0 \\[4px]
\sqrt{x^2-4x-12}<3
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
x^2-4x-12\ge 0 \\[4px]
\sqrt{x^2-4x-12}\ge\sqrt{20}
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$ I'm trying to prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$
Suposse that $\sqrt{3}=a+b\sqrt{2}$
$\begin{align*}
\sqrt{3}&=a+b\sqrt{2}\\
3&=(a+b\sqrt{2})^2\\
3&=a^2+2\sqrt{2}ab+b^2\\
(3-a^2-12b^2)^2&=(2\sqrt{2}ab)^2\\
9-6a^2-12b^2+4a^2b^2+a^4+4b^4&=8a^2b^2
\end{align*}$
... | I would start the same:
$$\sqrt3 = a + b\sqrt{2},\ \ a,b\in\mathbb{Q}$$
then square both to get
$$3 = a^2 + 2ab\sqrt2 + b^2$$
just like you and then rearrange to get
$$\frac{3 - a^2 - b^2}{2ab} = \sqrt{2}$$
meaning that $\sqrt2$ would have to be rational, which we know is not true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2489947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}$ is rational if and only if $a^2-b$ and $\frac{1}{2}(a+\sqrt{a^2-b})$ are square. Let $a,b \in \mathbb{R}$ such that $a^2\geq b$. Prove that $\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}$ is rational if and only if $a^2-b$ and $\frac{1}{2}(a+\sqrt{a^2-b})$ are square.
$\Rightarrow... | Assuming "square" means square of a rational (otherwise any positive real is the square of a real number), it's not true if $a$ and $b$ are not assumed rational. For example, with $0 < b < 1/4$ and $a = b + 1/4$ we have $\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}} = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Cubed exponent equation
$$\left(2 · 3^x\right)^3 + \left(9^x − 3\right)^3 = \left(9^x + 2 · 3^x − 3\right)^3$$ Solve the equation
I got the answer to the problem, in which I evaluated the whole expression (which was quite hard) which was $0$ and $1/2$, is there a way to solve this problem in a less tedious way?
| Note that the entire equation is of the form
$$
a^3 + b^3 = (a+b)^3
$$
with $a = 2\cdot 3^x$ and $b = 9^x - 3$. This reduces to
$$
0 = 3a^2b + 3ab^2\\
0 = 3ab(a+b)
$$
which has the three solutions $a = 0$, $b = 0$, or $a+b = 0$. Clearly, $a = 0$ can't happen. $b = 0$ means $9^x = 3$, which has the solution $x = \frac12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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Computing: $ \lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2} $ $$
\lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}
$$
The answer is 0. Cannot seem to understand how the answer is 0.I know that the first part is 0 but I'm confused on how to deal with the natural log?
Why is squeeze theorem not a g... | For all norm we have $|x|, |y|\le \|(x,y)\|$
then we get $$|x^5+ y^5|\le 2\|(x,y)\|^5$$
Since $\lim_{x\to 0}x\ln x = 0$ and $\ln(x^2+y^2) = 2\ln\|(x,y)\| $ we obtain,
$$\left|\lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}\right| \le \lim_{\|(x,y)\|\to 0} 4\|(x,y)\|\left|\ln\|(x,y)\|\right| =0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Questions concerning smallest fraction between two given fractions. I recently encountered this on a practice test
Find the smallest positive integer $n$ such that there exists an integer $m$ satisfying $0.33 < \frac{m}{n} < \frac{1}{3}$
My answer:
$$\begin{align}0.33 < \frac{m}{n} < \frac{1}{3}&\implies(\frac{33}{1... |
Can we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\frac{a}{b}<\frac{m}{n}<\frac{c}{d}$).
This answer uses your method.
In the following, $a,b,c,d,m,n$ are positive integers.
$\frac ab\lt \frac mn\lt \frac cd$ is equivalent t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2494774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Expected number of distinct integers in a random multiset
Consider a random multiset $X_n$ of size $n$ sampled uniformly from the integers $\{1,\dots,n\}$. What is the expected number of distinct integers in $X_n$?
To clarify, each multiset of size $n$ should occur with equal probability.
From computer simulations ... | We get from first principles the following bivariate generating
function with $A_q$ marking the value $q$ being sampled and $u$
marking different types being seen:
$$G(z, u) =
\prod_{q=1}^n \left(1+uzA_q+uz^2A_q^2+uz^3A_q^3+\cdots\right)
\\ = \prod_{q=1}^n \left(1+u\frac{zA_q}{1-zA_q}\right).$$
Extracting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a first order differential equation in terms of Lambert W-function I am having great difficulty solving the following equation
$$\ \frac{ax}{(bx^2 + c)} = \frac{dx}{dt} $$
I have re-edited the question. Any help is appreciated. Thank you.
| For the equation
$$\frac{ax}{(bx^2 + c)} = \frac{dx}{dt}$$
it can be seen that
\begin{align}
\frac{dx}{dt} &= \frac{a}{2 b} \, \frac{2 b x}{b x^{2} + c} = \frac{a}{2 b} \, \frac{d}{dt} \, \ln(b x^{2} + c )
\end{align}
which yields
$$x(t) = \frac{a}{2 b} \, \ln(b x^{2} + c) + c_{0}.$$
Solving for $x(t)$ requires much s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve summation of $\sum_{j=0}^{n-2}2^j (n-j)$ Question
While Solving a recursive equation , i am stuck at this summation and unable to move forward.Summation is
$$\sum_{j=0}^{n-2}2^j (n-j)$$
My Approach
$$\sum_{j=0}^{n-2}2^j (n-j) = \sum_{j=0}^{n-2}2^j \times n-\sum_{j=0}^{n-2}
2^{j} \times j$$
$$=n \times (2^{n... | In general:
\begin{align}
\sum_{k=0}^{n} x^{k} &= \frac{1 - x^{n+1}}{1-x} \\
\sum_{k=0}^{n} k \, x^{k} &= \frac{x \, (1 - (n+1) x^{n} + n x^{n+1})}{(1-x)^{2}}
\end{align}
Now
\begin{align}
\sum_{k=0}^{n} k \, \frac{1}{x^{k}} &= \frac{n - (n+1) x + x^{n+1}}{x^{n} \, (1-x)^{2}} \\
\sum_{k=0}^{n} k \, x^{n-k} &= \frac{n -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How prove this $\sum_{k=2}^{n}\frac{1}{3^k-1}<\frac{1}{5}$ show that $$\sum_{k=2}^{n}\frac{1}{3^k-1}<\dfrac{1}{5}\tag1$$
I try to use this well known: if $a>b>0,c>0$,then we have
$$\dfrac{b}{a}<\dfrac{b+c}{a+c}$$
$$\dfrac{1}{3^k-1}<\dfrac{1+1}{3^k-1+1}=\dfrac{2}{3^k}$$
so we have
$$\sum_{k=2}^{n}\dfrac{1}{3^... | Note that for $n\ge2$,
$$\begin{align}\sum_{k=2}^n\frac{1}{3^k-1}&=\frac{1}{8}\end{align}+\sum_{k=3}^n\frac{1}{3^k-1}<\frac{1}{8}+\sum_{k=3}^n\frac{1}{3^k-3}=\frac{1}{8}+\frac{1}{3}\sum_{k=2}^{n-1}\frac{1}{3^k-1}$$
so you can use induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove that: $1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$ Prove that:
$$1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$$
I know only this method:
$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4... | $$\sum_{k= 1}^{200}(-1)^{k+1}k^{-1}=1-\frac 12+\frac 13-\frac 14+\cdots+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}=$$
$$\color{blue}{\left(1+\frac 13+\frac 15 +\dots+ \frac 1{199}\right)-\left(\frac 12+ \frac 14+\frac 16\cdots+\frac {1}{200}\right)}=\color{red}{\frac{1}{101}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Ramanujan's series $1+\sum_{n=1}^{\infty}(8n+1)\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}$ Ramanujan gave the following series evaluation $$1+9\left(\frac{1}{4}\right)^{4}+17\left(\frac{1\cdot 5}{4\cdot 8}\right)^{4}+25\left(\frac{1\cdot 5\cdot 9}{4\cdot 8\cdot 12}\right)^{4}+\cdots=\dfrac{2\sqr... | (Too long for a comment. But this factoid might be useful.) If I remember correctly, a pair of series in that letter was,
$$U_1 = 1-5\left(\frac{1}{2}\right)^{3}+9\left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}-13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3}+\cdots=\dfrac{2}{\pi}$$
$$V_1= 1+9\left(\frac{1}{4}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Help solving variable separable ODE: $y' = \frac{1}{2} a y^2 + b y - 1$ with $y(0)=0$ I am studying for an exam about ODEs and I am struggling with one of the past exam questions. The past exam shows one exercise which asks us to solve:
$$y' = \frac{1}{2} a y^2 + b y - 1$$with $y(0)=0$
The solution is given as
$$y(x) =... | This is a Riccati equation. Let us introduce the change of variable $z = y+\alpha$. The differential equation $y' = \frac{1}{2}ay^2 + by - 1$ rewrites as
\begin{aligned}
z' &= \frac{a}{2}(z^2-2\alpha z + \alpha^2) + b(z-\alpha) - 1\\
&= \frac{a}{2}z^2 + (b-a\alpha)z + \frac{a}{2}\alpha^2 - b\alpha - 1 \, .
\end{aligned... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Distance between points and lines Let P=(3,1,4), Q=(1,2,5), R=(0,0,1) and S=(4,3,6). Let A be the line passing through P and Q, and let B be the line passing through R and S.
a) What is the distance between R and A?
b) What is the distance between A and B?
So I tried to use the distance between points formula but it d... | The answers are:
a) $\frac{\sqrt{165}}{3}$
b) $\frac{\sqrt3}{3}$
The lines can be represented as follows:
\begin{align}
a(x) &= a_0 + u_a x, \\
b(x) &= b_0 + u_b y.
\end{align}
Here, $a_0$ can be $P$ or $Q$ (or any linear combination) and $b_0$ can be $R$ or $S$. I choose for $a_0=P$ and $b_0=R$. The unit vector $u_a$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\varepsilon$-$\delta$ definition of a limit [Calculus] $$\lim_{x\to 2} x^2 - 8x + 8= -4 $$
Now to express $ f(x) - l $ it as $x - x_0$ , we write $ x = (x - x_0) + x_0$.
So \begin{align}
f(x) - l &= x^2 - 8x + 8 + 4\\
&= (x - 2 + 2)^2 - 8(x - 2 + 2)+ 12\\
&= (x - 2)^2 +4(x-2) + 4 - 8(x - 2) - 16+ 12 \\
&= (x - 2)^2 - ... | Let $\varepsilon >0$. For $|x - 2| < \delta = \sqrt{4+\varepsilon} - 2$ we have:
\begin{align}
\left|x^2-8x+8+4\right| &= \left|x^2 - 8x + 12\right|\\
&= \left|(x-2)^2 - 4(x-2)\right|\\
&= \left|x-2\right|\left|x-2-4\right|\\
&\le |x-2|\big(|x - 2| + 4\big)\\
&< \delta(\delta + 4)\\
&=\varepsilon
\end{align}
Therefore
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Divergence of reciprocal of primes, Euler On Wikipedia at link currently is:
\begin{align}
\ln \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \ln\left( \prod_p \frac{1}{1-p^{-1}}\right)
= -\sum_p \ln \left( 1-\frac{1}{p}\right) \\
& {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \righ... | You can in fact do better and show that $K < 0.43$.
$$
\sum_{p}\frac{1}{2p^2} + \sum_{p}\frac{1}{3p
^3} + \sum_{p}\frac{1}{4p^4} + \cdots = \sum_{n=1}^{\infty}\sum_{p}\frac{1}{np^n}
$$
$$
< \sum_{n=1}^{\infty}\sum_{k=2}^{\infty}\frac{1}{nk^n} = \sum_{n=2}^{\infty}\frac{\zeta(n) - 1}{n} = 1 - \gamma \approx 0.422785... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why Cholesky Decomposition is Only Applicable to Positive-Definite Matirx The textbook I uses says that only when matrix $A \in \mathbb{R}^{n\times n}$ is positive-definite the Cholesky decomposition of $A$ is unique but it does not provide any proof.
I did Cholesky decomposition to a positive-semidefinite matrix
$
A ... | Let's try to compute (all the possible) Cholesky decompositions of a $2 \times 2$ real positive-semidefinite matrix
$$ \begin{pmatrix} d & e \\ e & f \end{pmatrix}. $$
We want
$$ \begin{pmatrix} a & 0 \\ b & c \end{pmatrix} \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} = \begin{pmatrix} a^2 & ab \\ ab & b^2 + c^2 \end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2508589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Determine galois group $Gal \Big(\frac{\mathbb{Q}(\sqrt[3]{3},\sqrt{-3})}{\mathbb{Q}}\Big)$ I've had some hard time determining Galois group $Gal \Big(\frac{\mathbb{Q}(\sqrt[3]{3},\sqrt{-3})}{\mathbb{Q}}\Big)$ because I didn't know exactly how to compute the order of the elements. See here for the computation of the o... | Consider the automorphism $\sigma$ such that $\sigma(\sqrt[3]{3}) = \omega \sqrt[3]{3}$ and $\sigma(\sqrt{-3}) = \sqrt{-3}$ and note that by your expression for $\omega$, then $\sigma(\omega) = \omega$:
$$
\sigma(\omega) = \sigma \left(\frac{-1 + \sqrt{-3}}{2}\right) = \frac{-1 + \sigma(\sqrt{-3})}{2} = \frac{-1 + \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2510098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to expand $(x^{n-1}+\cdots+x+1)^2$ (nicely) sorry if this is a basic question but I am trying to show the following expansion holds over $\mathbb{Z}$:
$(x^{n-1}+\cdots+x+1)^2=x^{2n-2}+2x^{2n-3}+\cdots+(n-1)x^n+nx^{n-1}+(n-1)x^{n-2}+\cdots+2x+1$.
Now I can show this in by sheer brute force, but it wasn't nice and ce... | Induction is the best way to do it:
Base case: $(x+1)^2 = x^2 + 2x + 1$.
Assume $(x^{n-1} + ... + 1)^2= x^{2n-2} + 2 x^{2n-3} + 3 x^{2n-4} + .... +nx^{n-1} + (n-1)x^{n-2} +.....+3x^2 + 2x + 1$
Then $(x^{n} + ... + 1)^2= [ x^{n} + (x^{n-1} + ... 1)]^2=$
$x^{2n} + 2x^n(x^{n-1} + ... 1) + (x^{n-1} + ... 1)^2] =$
$x^{2n} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2511494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Integral of a power combined with a Gaussian hypergeometric function I think the following is true for $k \ge 3$,
$$
\int_0^{\infty } (w+1)^{\frac{2}{k}-2} \, _2F_1\left(\frac{2}{k},\frac{1}{k};1+\frac{1}{k};-w\right) \, dw
=
\frac{\pi \cot \left(\frac{\pi }{k}\right)}{k-2}
.
$$
I have checked Table of Integrals, Seri... | Expanding on my comment, we can alternatively use Euler's integral representation of the hypergeometric function, along with a well-known integral representation of the digamma function.
We also need the reflection formula for the digamma function.
$$\begin{align} I(k) &= \int_{0}^{\infty}(1+w)^{2/k-2} \, _2F_1 \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve for $a,b,c,d \in \Bbb R$, given that $a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$ Today, I came across an equation in practice mock-test of my coaching institute, aiming for engineering entrance examination (The course for the test wasn't topic-specific, it was a test of complete high school mathematics). It was havi... | Your idea of turning the right side into a sum of perfect squares is a good one. Observing that
$$
(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2=2(a^2+b^2+c^2+d^2-ab-bc-cd-da)
$$
we multiply
$$
a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac{2}{5}=0
$$
by $2$ and rewrite the result as
$$
(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2+2ad-2d+\frac{4}{5}=0.
$$
We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Evaluating $\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}$ Evaluate : $$\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}$$
I tried using logarithmic functions, but cannot evaluate this please help me
| One could make use of the digamma function and obtain:
\begin{align}
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{x \, k + 1} &= \frac{1}{2 x} \, \left[ \psi\left(\frac{x+1}{2 x}\right) - \psi\left(\frac{1}{2 x}\right) \right] \\
\psi\left(\frac{1}{6}\right) &= - \gamma - \frac{\sqrt{3} \, \pi}{2} - \frac{3}{2} \, \ln(3) - 2 \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve $A^2=B$ where $B$ is the $3\times3$ matrix whose only nonzero entry is the top right entry Find all the matrices $A$ such that $$A^2= \left( \begin {array}{ccc} 0&0&1\\ 0&0&0 \\ 0&0&0\end {array} \right) $$ where $A$ is a $3\times 3$ matrix.
$A= \left( \begin {array}{ccc} 0&1&1\\ 0&0&1 \\ 0&0&0\end {array} \ri... | I don't know about all solutions, but given one solution, you can generate an infinity many other solutions: Let
$$
P=
\begin{pmatrix}
1 & x & y\\
0 & 1 & z \\
0 & 0 & 1\\
\end{pmatrix}
$$
Note that
$$
P^{-1}=
\begin{pmatrix}
1 & -x & xz-y\\
0 & 1 & -z \\
0 & 0 & 1\\
\end{pmatrix}
$$
If
$$
A^2=
\begin{pmatrix}
0 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2516554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
proving by multiplying Let (*) be $ \left(1+\frac{1}{n}\right)\left(1-\frac{1}{n^2}\right)^n < 1, ∀n∈ℕ$
I have tried many ways to get to (*)
We have $\left(1-\frac{1}{n^2}\right)^n <1$ and $-\left(1+\frac{1}{n}\right) < -1 <1$
How can you infer (*) from these two last inequalities?
| Proof: We know that $a_n=(1+\frac{x}{n})^n$ is a strictly increasing sequence for $x\geq -1$ and it is bounded above to $e^x$.Letting $x=\frac{-1}{n}$, we get: $$(1-\frac{1}{n^2})^n < e^{-1/n}$$
Therefore,
$$(1+\frac{1}{n})(1-\frac{1}{n^2})^n < (1+\frac{1}{n})\times e^{-1/n}$$
But, it is a well-known fact that for all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1} n^{n^2}}$ Woflram gives $\frac{1}{e}$ as the limit, but I failed to obtain it. Please help.
$$\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1}n^{n^2}}$$
| $$
\begin{align}
&\frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1}n^{n^2}}\\[6pt]
&=\frac{\left(\color{#C00}{1+\frac1n}\right)^{2n^2+2n+1}}{\left(\color{#090}{1+\frac2n}\right)^{n^2+2n+1}}\\[3pt]
&=\left(\frac{\color{#C00}{1+\frac2n+\frac1{n^2}}}{\color{#090}{1+\frac2n}}\right)^{n^2+2n}\left(\color{#C00}{1+\frac1n}\right)^{-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that $\frac{a}{a+1}+\frac{b}{b+1}\geq\frac{a+b}{1+a+b}\geq\frac{c}{1+c}$ whenever $c\leq a+b$ I found a post on ME here.
It sais:
Try to show that $$\frac{a}{a+1}+\frac{b}{b+1}\geq\frac{a+b}{1+a+b}\geq\frac{c}{1+c}$$ whenever $c\leq a+b$
I wasn't able to solve this. How does it work?
| We know that,
$$\frac{a}{1+a} \ge \frac{a}{1+a+b} $$
and
$$\frac{b}{1+b} \ge \frac{b}{1+a+b} $$
Adding the above two equations.
$$\frac{a}{1+a} + \frac{b}{1+b} \ge \frac{a}{1+a+b} + \frac{b}{1+a+b} $$
$$\frac{a}{1+a} + \frac{b}{1+b}\ge \frac{a+b}{1+a+b} $$
Now assume $a+b = k$
Given
\begin{align}
& a+b \ge c
\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Fourier Motzkin elimination with positive coefficients only How can we use Fourier-Motzkin elimination on system of inequalities with positive coefficients preceding each variable $x_1$ to $x_2$. Obviously, in this case we will only have an upper bound as a solution but how do we find it.
For example:
$5x_1 + 3x_2 \le... | Rewrite your two inequalities as
$$x_1 \le \frac{8}{5}- \frac{3}{5}x_2 \quad \textrm{and} \quad x_1 \le \frac{15}{2} -\frac{5}{2}x_2$$
Then
$$x_1 \le \min\{\frac{8}{5}- \frac{3}{5}x_2, \frac{15}{2} -\frac{5}{2}x_2 \}$$
This corresponds to the region of $\mathbb{R}^2$ below the graph of the function
$$x_1 = f(x_2) = \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2527361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Taylor inside an integral I know the following integral should be:
$$ \int_{0}^{1} \frac{dx}{\sqrt{1-x^2-\epsilon(1-x^3)}} \approx \pi/2 + \epsilon$$
for $\epsilon$ small. What I do is set $-\epsilon(1-x^3)=y$ and then since $y$ it's always greatly smaller than $(1-x^2)$ I do Taylor expansion around $y=0$:
$$ \frac{dx... | Using Taylor series you will end up having to justify evaluation at the improper bound of $1$, which will require further details that neither of the other answers have addressed. Instead, you could just note that
$$\frac{1-x^3}{1-x^2} = \frac{(1-x)(1+x+x^2)}{(1-x)(1+x)} = 1 + \frac{x^2}{1+x} \in [1,\tfrac32)$$ for $x ... | {
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"url": "https://math.stackexchange.com/questions/2529544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate the integral $\int_0^{\pi}\frac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx$?
Evaluate the integral $$\int_0^{\pi}\frac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx.$$
I have no idea.
$$\int_0^{\pi}\dfrac{a^n\sin^2x+b^n\cos^2x}{a^{2n}\sin^2x+b^{2n}\cos^2x}dx=\int_0^{\pi/2}\dfrac{a^... | A complex analysis approach. I assume that $a>b>0$ (the general case is left to the reader).
Let $A=a^n$, $B=b^n$, and $t=2x$, then, by the half-angle formulas, the given integral is
$$\begin{align}
I&=\frac{1}{2}\int_0^{2\pi}\frac{A(1-\cos(t))+B(1+\cos(t))}{A^2(1-\cos(t))+B^2(1+\cos(t))}dt
=\frac{1}{2}\int_0^{2\pi}\f... | {
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"source": "stackexchange",
"question_score": "5",
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Finding the trace of $T^4$ and $T^2$?
Q. Let $V$ denote the (complex) vector space of complex polynomials of degree at most $9$ and consider the linear operator $T:V \to V$ defined by $$T(a_0+a_1 x+a_2 x^2+\cdots+a_9 x^9)=a_0 +(a_2 x +a_1 x^2)+(a_4 x^3+ a_5 x^4 + a_3 x^5)+(a_7 x^6 + a_8 x^7 + a_9 x^8 + a_6 x^9).$$
(a... | Notice that $T$ just acts as $x^i \mapsto x^{\sigma(i)}$, where $\sigma$ is a permutation on $\{0, \ldots, 9\}$ given as:
$$\sigma = (1\,2)(3\,5\,4)(6\,9\,8\,7)$$
So $T^2$ acts as $x^i \mapsto x^{\sigma^2(i)}$, where $$\sigma^2 = \sigma \circ \sigma = (1\,2)^2(3\,5\,4)^2(6\,9\,8\,7)^2 = (3\,5\,4)^2(6\,9\,8\,7)^2$$
so
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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prove that $\displaystyle A\cdot B= 2^{-6}\bigg(\csc \frac{\pi}{22}-1\bigg)$
If $\displaystyle A= \prod^{5}_{k=1}\cos \left(\frac{k\pi}{11}\right)$ and $\displaystyle B= \sum^{5}_{k=1}\cos \left(\frac{k\pi}{11}\right)$ then show that $\displaystyle A\cdot B= 2^{-6}\bigg(\csc \frac{\pi}{22}-1\bigg)$
Attempt: $\display... | COMMENT.- This is not properly a hint but a (nice) remark that I can not avoid mentioning. Given the value $x=\dfrac{\pi}{11}$ there are triangles $\triangle{ABC}$ like that of the attached figure intimately linked for which there must be a counterpart of the proposed problem in angles with the sides and bisectors $BB'... | {
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"source": "stackexchange",
"question_score": "3",
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Sphere with smallest radius
Find the smallest sphere which touches the lines $\frac{x-5}{2}=\frac{y-2}{-1}=\frac{z-5}{-1}$ and $\frac{x+4}{-3}=\frac{y+5}{-6}=\frac{z-4}{4}$
The general equation of the sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$. Its centre is $(-u,-v,-w)$ and radius is $\sqrt(u^2+v^2+w^2-d)$. Since the g... | This video nicely explains a different approach based on geometric reasoning:
The two lines $L_1$ and $L_2$ are $<2t_1+5, 2-t_1,5-t_1>$ and $<-4-3t_2, -5-6t_2, 4+4t_2>$
The vector $\vec{P}$ between the two touchpoints $P_1$ and $P_2$ must be perpendicular to both direction vectors of the lines $L_1$ and $L_2$.
$$\begin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can this definite integral involving series be solved without a calculator? I got this question today but I can't see if there is any good way to solve it by hand.
Evaluate the definite integral
$$\int_2^{12}\frac{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}{\sqrt{x\sqrt{x\sqrt{x}...}}}\,\mathrm{d}x$$
where the series in the numer... | As you did, let $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+y}$. Clearly $y\ge0$ and $y$ satisfies
$$ y^2-y-x=0 $$
from which you have
$$ y=\frac{1\pm\sqrt{4x+1}}{2}. $$
Since $x\in[2,12]$ and $y\ge0$, you must choose "$+$". Since if you choose "$-$", then
$$ y=\frac{1-\sqrt{4x+1}}{2}<0. $$
Now, under $t=\sqrt{1+4x}$
\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Getting x in terms of y I have this equation:
$$\dfrac{x}{y} = \dfrac{y-x}{x}$$
How would I separate $x$ and $y$ in $x^2+xy-y^2=0$ ?
| $$\frac{x}{y}=\frac{y-x}{x}$$
$$x^2=y(y-x)$$
$$x^2=y^2-yx$$
$$x^2+yx=y^2$$
$$x^2+yx+\left(\frac{y}{2}\right)^2=y^2+\left(\frac{y}{2}\right)^2$$
$$\left(x+\frac{y}{2}\right)^2=\frac{5}{4}y^2$$
$$x+\frac{y}{2}=\pm\sqrt{5}\frac{y}{2}$$
$$x=\frac{y}{2}(-1\pm\sqrt{5})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Find real solutions in $x$,$y$ for the system $\sqrt{x-y}+\sqrt{x+y}=a$ and $\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2.$
Find all real solutions in $x$ and $y$, given $a$, to the system:
$$\left\{
\begin{array}{l}
\sqrt{x-y}+\sqrt{x+y}=a \\
\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2 \\
\end{array}
\right.
$$
From a math olympiad... | Square the first
$x-y+2\sqrt{x§2-y^2}+x+y=a^2$
$\sqrt{x^2-y^2}=\dfrac{a^2-2x}{2}$
Plug into the second
$\sqrt{x^2+y^2}=\dfrac{a^2-2x}{2}+a^2=\dfrac{3a^2-2x}{2}$
Square again
$x^2+y^2=\dfrac{9a^4-12a^2x+4x^2}{4}$
$x^2-y^2=\dfrac{a^4-4a^2x+4x^2}{4}$
Adding the last two equations we get
$2x^2=\dfrac{10a^4-16a^2x+8x^2}{4}\... | {
"language": "en",
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If $p$ is prime, $p\ne3$ then $p^2+2$ is composite I'm trying to prove that if $p$ is prime,$p\ne3$ then $p^2+2$ is composite. Here's my attempt:
Every number $p$ can be put in the form $3k+r, 0\le r \lt 3$, with $k$ an integer. When $r=0$, the number is a multiple of 3, so that leaves us with the forms $3k+1$ and $3k+... | Since $(p-1)p(p+1)$ is product of three consecutive numbers it must be divisible by $3$. Since $p\ne 3$ we see that $3$ divides $p^2-1$. So $p^2+2$ is also divisible by $3$ and since it is more than $3$ it is composite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Evaluating $\int_0^{\infty} \frac{\tan^{-1} x}{x(1+x^2)} dx$ The question is to evaluate $$\int_0^{\infty} \frac{\tan^{-1} x}{x(1+x^2)} dx$$
I used the substitution $x=\tan a$ then the given integral becomes $\int_0^{\pi/2} \frac{\tan^{-1}(\tan a)}{\tan a} da$
Now $\tan^{-1} (\tan a)=a \forall a \in [0,\pi /2]$ so that... | Instead of using Feynman's trick and differentiating under the integral sign as @Robert Israel did, in a very similar manner one may evaluate the integral by converting it to a double integral first.
Noting that
$$\int^1_0 \frac{dt}{1 + x^2 t^2} = \frac{\tan^{-1} x}{x},$$
the integral may be rewritten as
$$I = \int^\in... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
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$\tan(x) = 3$. Find the value of $\sin(x)$ I’m trying to figure out the value for $\sin(x)$ when $\tan(x) = 3$. The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm doing wrong?
1.) $\tan(x) = 3$, then $\frac{\sin(x)}{\cos(x)} = 3$.
2.) Then $\cos(x) = ... | In general,
if
$\tan(x) = a$
then
$a^2
=\tan^2(x)
=\dfrac{\sin^2(x)}{\cos^2(x)}
=\dfrac{\sin^2(x)}{1-\sin^2(x)}
$
so
$\dfrac1{a^2}
=\dfrac{1-\sin^2(x)}{\sin^2(x)}
=\dfrac{1}{\sin^2(x)}-1
$
so
$\dfrac{1}{\sin^2(x)}
=\dfrac1{a^2}+1
=\dfrac{1+a^2}{a^2}
$
so
$\sin^2(x)
=\dfrac{a^2}{1+a^2}
$
or
$\sin(x)
=\dfrac{a}{\sqrt{1+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546705",
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"source": "stackexchange",
"question_score": "1",
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How do you factorize a polynom in $\mathbb{Z}_2$? How can you efficiently factorize a polynom (in $\mathbb{Z}_2$) ?
Example in this answer:
$$x^{16}-x=x (x + 1) (x^2 + x + 1) (x^4 + x + 1) (x^4 + x^3 + 1) (x^4 + x^3 + x^2 + x + 1)$$
How do you do this by hand?
| Start with the factorization of $x^{15}-1$ into cyclotomic polynomials:
$$
x^{16}-x = x(x^{15}-1)=x \Phi_1(x) \Phi_3(x) \Phi_5(x) \Phi_{15}(x)
$$
which gives
$$
x^{16}-x = x(x - 1)(x^2 + x + 1) (x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)
$$
We then need to further factor these mod 2. The last fact... | {
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"source": "stackexchange",
"question_score": "4",
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On the asymptotics of $a_n=a_{n-1}^k+a_{n-2}^k$ where $k>1$ and $a_0=0, a_1=1$ Consider the sequence defined by $a_0=0,a_1=1, a_n=a_{n-1}^k+a_{n-2}^k$, where $k$ is a fixed integer larger than $1$.
One finds $a_n\sim a_{n-1}^k$ and thus $a_n\sim \alpha_k^{k^n}$ where $\alpha_k$ is a constant which depends on $k$.
It ... | Misha showed that the limit in the OP is $1$. I'll prove in particular $\lim\limits_{k\to\infty} \alpha_k^{k^3}=\lim\limits_{n\to\infty}a_n^{1/k^{n-3}}=2$.
On one hand, by induction it's easy to prove $a_n>2^{k^{n-3}}$, and it does come to mind when looking at the first terms. On the other hand, we find $a_n<2^{k^{n-3}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Integer part of $\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$ Find the value of the following infinite series:
$$\left\lfloor\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}\right\rfloor$$
Now, my doubt is whether it's $2$ or it's $3$. I'm not sure if it just converges to $3$ but not actually reaches it or if it compl... | set
$$x =\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$$
then
$$x^3 = 24+\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$$
$$x^3 = 24 + x$$
Shooting for the rational root theorem, consider the divisors of $24$. 2 does not satisfy the equation, clearly as $8 \neq 26$. However, $3^3 = 24 + 3$, so
$$3 =\sqrt[3]{24+\sqrt[3]... | {
"language": "en",
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If $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$; $a,b,$ are positive integers. Is it true: $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$?
Somehow I can't find counterexample nor to prove it. I try to write it $a=gx$ and $b=gy$ where $g=\gcd(a,b)$ but didn't help. It seems that there is no $a\ne b$ such that $a+b\mid a^4+b^4$. O... | A little handwavy but enough to get a counter example.
It's easy to verify $a+b|a^{2k} - b^{2k}$ (by noting $\frac {a^m - b^m}{a-b} = a^{m-1}+ a^{m-2}b+ ... + ab^{m-2} + b^{m-1}$ and replacing $b$ with $-b$ and noting $(-b)^{2k} = b^{2k}$.)
$a + b|a^4 - b^4$ and $a+b|a^2 - b^2$
so if $a+b|a^4 + b^4$ then $a+b|(a^4+b^4)... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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A matrix raised to a high power ($87$) So, I have this matrix: $$\pmatrix {0&0&0&-1\\1&0&0&0\\0&1&0&0\\0&0&1&0}^{87}$$
My teacher never discussed eigenvalues. So, I do not know what they are and there must be another way to do this (without multiplying the matrix $87$ times).
Thanks for your help.
| It is easy to work with this sparse matrix, because its powers remain sparse.
So
$$A^2 = \pmatrix{
0&0&-1&0\\
0&0&0&-1\\
1&0&0&0\\
0&1&0&0}
$$
$$A^4 = (A^2)^2 = \pmatrix{
-1&0&0&0\\
0&-1&0&0\\
0&0&-1&0\\
0&0&0&-1} = -I
$$
$$A^8 = (-I)^2 = I$$
So for integer $n$, $A^{8n} = I$. Then $A^{80} = I$, and $A^{84} = -I$. So
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Ratio of an inscribed circle's tangent to original square In the diagram, the circle is inscribed within square $PQRS,$ $\overline{UT}$ is tangent to the circle, and $RU$ is $\frac{1}{4}$ of $RS.$ What is $\frac{RT}{RQ}$?
| Circle has equation $x^2+y^2=4$
$U(1;\;-2)$
The equation of a general line passing through $U$ is
$mx-y-m-2=0$
This line is tangent to the circle if its distance from the center is equal to the radius, that is
$$\frac{|-m-2|}{\sqrt{m^2+1}}=2\to m^2+4m+4=4m^2+4\to m_1=0;\;m_2=\frac{4}{3}$$
One tangent is the line contai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximum and minimum value of a function.
Let us consider a function $f(x,y)=4x^2-xy+4y^2+x^3y+xy^3-4$. Then find the maximum and minimum value of $f$.
My attempt. $f_x=0$ implies $8x-y+3x^2y+y^3=0$ and $f_y=0$ implies $-x+8y+x^3+3xy²=0$ and $f_{xy}=3x^2+3y^2-1$. Now $f_x+f_y=0$ implies $(x+y)((x+y)^2+7)=0$ implies $x... | Note that $f(x,y)=(xy+4)(x^2+y^2-1)$ and therefore
$$\lim_{x\to+\infty}f(x,x)=\lim_{x\to+\infty}(x^2+4)(2x^2-1)=+\infty$$
and
$$\lim_{x\to+\infty}f(x,-x)=\lim_{x\to+\infty}(-x^2+4)(2x^2-1)=-\infty.$$
What may we conclude?
P.S. The critical point $(0,0)$ is a local minimum.
| {
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"timestamp": "2023-03-29T00:00:00",
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In the sequence $1,4,11,26$… each term is $2⋅(n-1)^{th}$ term $+ n$. What is the $n^{th}$ term? I readily see that it is $2^{n+1} - (n+2)$ but how can I deduce the $n^{th}$ term from the given pattern i.e. $2⋅(n−1)^{th}$ term $+n$ .
| So we have $a_0=0$ and $a_n=2a_{n-1}+n$ for $n\ge 1$. Let $A(x)=\sum_{n=0}^{\infty}{a_n x^n}$. Then multiplying both sides of the recurrence by $x^n$ and summing over $n\ge 1$, we get
$$
\begin{split}
A(x)=A(x)-a_0&=2xA(x)+\sum_{n\ge 1}{n x^n}\\
&=2xA(x)+x\left(\sum_{n\ge 0}{x^n}\right)'\\
&=2xA(x)+x\left(\frac{1}{1-x}... | {
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"url": "https://math.stackexchange.com/questions/2560866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate the limit $\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}$ which involves rational functions?
Find $$\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}.$$
I have tried rationalizing but there is no pattern that I can observe.
Edit:
So we forget about the $x$ that is multiplied to both the functions a... | This limit would be equal to $1/2$. See my demo :
$$
\lim_{x \to \infty} {x\sqrt{x^2+1}-x (x^3+1)^{1/3}} =
\lim_{x \to \infty} { x (\sqrt{x^2+1}-(x^3+1)^{1/3})}
$$
Now, let us see how much is each member, by Laurent-Taylor expansion :
$$
\sqrt{x^2+1} = x+ {1 \over {2x}} - {1 \over {8x^3}}+{1 \over {16x^5}}-...
$$
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find the value of $9\tan{10^{\circ}}+2\tan{20^{\circ}}+4\tan{40^{\circ}}-\tan{80^{\circ}}$
Find the value of
$$9\tan{10^{\circ}}+2\tan{20^{\circ}}+4\tan{40^{\circ}}-\tan{80^{\circ}}$$
It seem the answer is $0$, see http://www.wolframalpha.com/input/?i=9tan(10)%2B2tan(20)%2B4tan(40)-tan(80).
| with the help of $$\tan \theta = \cot \theta-2\cot 2\theta$$
Replace $\theta \rightarrow 2\theta$ , we have $$2\tan 2\theta =2\cot 2\theta -4\cot(4\theta)$$
similarly $$4\tan(4\theta) = 4\cot(4\theta)-4\cot(8\theta)$$
So $$\tan \theta+2\tan 2\theta+2\tan 4\theta = -8\cot(8\theta)$$
put $\theta = 10^\circ$
$$\tan 10^\ci... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Applicability of L'Hôpital rule on infinite sum. $$\lim_{n \to \infty} \left( \frac{n}{n^2+1} + \frac{n}{n^2+2} + \frac{n}{n^2+3} + \space ... \space + \frac{n}{n^2+n}\right) $$
As $n$ is not $\infty$ but tends to $\infty$ I can split the limit of sums into sum of limits. i.e.
$$\lim_{n \to \infty} \frac{n}{n^2+1} +\... | Notice that if you allow splitting the limit into sum of limits you can arrange it to be any real number.
Indeed, take any $x \in \mathbb{R}$. For $k \in \{1, \ldots, n\}$ we have:
$$\displaystyle\lim_{n \to \infty} \frac{n}{n^2+k} = 0 = \lim_{n\to\infty} \frac{x}{n}$$
So we would obtain:
\begin{align}
\lim_{n \to \inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Looking for a proof of an interesting identity Working on a problem I have encountered an interesting identity:
$$
\sum_{k=0}^\infty \left(\frac{x}{2}\right)^{n+2k}\binom{n+2k}{k}
=\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{x}\right)^n,
$$
where $n$ is a non-negative integer number and $x$ is a real number with ... | Extracting coefficients on the RHS we get the integral
(coefficient on $x^{n+2k}$)
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+2k+1}}
\frac{1}{\sqrt{1-z^2}}
\left(\frac{1-\sqrt{1-z^2}}{z}\right)^n
\; dz.$$
Now we put $(1-\sqrt{1-z^2})/z = w$ so that $z = 2w/(1+w^2).$ This has
$w = \frac{1}{2} z + \cdots$ so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Find the spectral decomposition of $A$ $$
A= \begin{pmatrix} -3 & 4\\ 4 & 3
\end{pmatrix}
$$
So i am assuming that i must find the evalues and evectors of this matrix first, and that is exactly what i did.
The evalues are $5$ and $-5$, and the evectors are $(2,1)^T$ and $(1,-2)^T$
Now the spectral decomposition of $A... | \begin{align}
\begin{bmatrix} -3 & 4 \\ 4 & 3\end{bmatrix}\begin{bmatrix} 2 \\ 1\end{bmatrix}= \begin{bmatrix} -2 \\ 11\end{bmatrix}
\end{align}
The eigenvector is not correct. The correct eigenvecor should be $\begin{bmatrix} 1 & 2\end{bmatrix}^T$ since
\begin{align}
\begin{bmatrix} -3 & 4 \\ 4 & 3\end{bmatrix}\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find floor of sum $\sum_{k=1}^{80} k^{-1/2}$ We have to find the floor $\lfloor S \rfloor$ of the following sum:
$$S = \sum_{k=1}^{80}\frac{1}{\sqrt k}$$
What I did was to find a approximate series that this series is near to. Let that series have general term $T_k$ and original series may have general term $a_k$. We ... | We have that $\sqrt{n+a+1}-\sqrt{n+a}$ behaves like $\frac{1}{2\sqrt{n}}$ for large values of $n$, and by picking $a=-\frac{1}{2}$ we get a telescopic term providing an accurate approximations of $\frac{1}{\sqrt{n}}$ for any $n\geq 1$:
$$ 2\sqrt{n+\tfrac{1}{2}}-2\sqrt{n-\tfrac{1}{2}} =\frac{1}{\sqrt{n}}+E(n),\qquad \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Calculate marginal distribution, $P(X=0|Y > 0)$, expected value and variance. Let $X$ be a random variable with values $0$ and $1$.
Let $Y$ be a random variable with values in $\mathbb{N_0}$.
Let $ p \in (0,1)$ and $ P(X=0, Y=n) = p \cdot \frac{e^{-1}}{n!} $ and $ P(X=1, Y=n) = (1-p) \cdot \frac{2^{n}e^{-2}}{n!} $.
... | First, you can compute $P(X=0)$ using the law of total probability.
\begin{align*}
P(X=0) &= \sum_{n}P(X=0,Y=n) \\
&= p \sum_{n}{\frac{\exp(-1)}{n!}}
= p
\end{align*}
Next, note that
\begin{align*}
P(Y=n|X=0) &= \frac{P(X=0,Y=n)}{P(X=0)} \\
&= \frac{\exp(-1)}{n!}
\end{align*}
That is $Y|X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $3^n - 4(2^n) + (-1)^n + 6 \equiv 0 \mod 24 $ Is it possible to prove that $3^n - 4(2^n) + (-1)^n + 6 \equiv 0 \mod 24 $ for $n \geq 1 $ . I know that it is true because $ \frac{3^n - 4(2^n) + (-1)^n + 6}{24}$ represents the number of ways to uniquely $4$-colour an n-cycle , excluding permutations of colours... | Let $x_n = 3^n - 4(2^n) + (-1)^n$. By expanding $(x-3)(x-2)(x+1)=x^3 - 4 x^2 + x + 6$, we get
$$
x_n= 4x_{n-1}-x_{n-2}-6x_{n-3} \quad\mbox{ for } n\ge 4
$$
Let $y_n = x_n + 6$. Then
$$
y_n= 4y_{n-1}-y_{n-2}-6y_{n-3} + 24
$$
and the claim follows immediately by induction after checking it for $y_1, y_2, y_3$, which are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Calculating $\int_0^\pi \log(1-2a\cos (x)+a^2)\,dx$ I need to calculate this integral using Riemann sum.
$$\int_0 ^\pi \log(1-2\alpha \cos (x) +\alpha^2){\rm d}x$$
a). For $|\alpha|<1$; b). For $|\alpha| > 1$.
I know one way of computing this using substitutions and symmetries, but it is necessary to do with Riemann su... | Here is an elementary real method.
Let
$$I(a) = \int^\pi_0 \ln (1 - 2a \cos x + a^2) \, dx, \quad a \in \mathbb{R}.$$
To find the value of $I(a)$ a number of properties for the function $I(a)$ will be estiblished.
Firstly, note that $I(0) = 0$. Also $I(a)$ is even since
\begin{align*}
I(-a) &= \int^\pi_0 \ln (1 + 2a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
find the $\sum\limits_{k=0}^n \frac{(-1)^k}{k! (2k+1)} \frac{1}{(n-k)!}$ I'm stuck on computing the sum of
\begin{align*}
\sum\limits_{k=0}^n \frac{(-1)^k}{k! (2k+1)} \frac{1}{(n-k)!}
\end{align*}
I tried some manipulations which include
\begin{align*}
\frac{1}{n!} \binom{n}{k} = \frac{1}{k! (n-k)!}
\end{align*}
but st... | Use
\begin{eqnarray*}
\int_{0}^{1} x^{2k} dx =\frac{1}{2k+1}.
\end{eqnarray*}
interchange the order the integral and plum
\begin{eqnarray*}
\sum_{k=0}^{n} \frac{(-1)^k}{(2k+1) k! (n-k)!} &=& \frac{1}{n!} \int_{0}^{1} \sum_{k=0}^{n}x^{2k} \binom{n}{k} \\
&=& \frac{1}{n!} \int_{0}^{1} (1-x^2)^n dx \\
\end{eqnarray*}
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Trigonometric equation: $\ln(\sin x + \cos x)^{1+\sin 2x}= 2$
$\ln(\sin x + \cos x)^{1+\sin 2x}= 2$
I am unable to solve it. I tried this way:
$(\sin x + \cos x)^{1+\sin 2x}= e^2$
I know that:
$\sin x + \cos x \le \sqrt2 $
$1+ \sin 2x \le 2 $
I don't know how to utilise this idea in my solution.
| Let $X = \sin x + \cos x$. Since $1+\sin 2x = (\sin x + \cos x)^2$, we have
$$ X^2 \ln X = 2. $$
Then
\begin{align}
X^2 \ln X^2 = 4 \implies Y \ln Y = 4
\end{align}
where $Y = X^2$.
It follows from here that $Y = e^{W(4)}$.
Thus $X = e^{W(4)/2}$.
Since
$$X = \sin x + \cos x = \sqrt{2}\sin(x + \pi/4)$$
we conclude tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
On the behaviour of $\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n$ I have to find the limit : (let $k\in \mathbb{R}$)
$$\lim_{n\to \infty}n^k \left(\Big(1+\frac{1}{n+1}\Big)^{n+1}-\Big(1+\frac{1}{n}\Big)^n \right)=?$$
My Try :
$$\lim_{n\to \infty}\frac{n^k}{\Big(1+\frac{1}{n}\Big)^n} \left(\frac{\B... | Using only the Binomial Theorem and Bernoulli's Inequality:
$$
\begin{align}
\hspace{-1cm}\left(1+\frac1{n+1}\right)^{n+1}\!\!-\left(1+\frac1n\right)^n
&=\left(\frac{n+2}{n+1}\right)^{n+1}-\left(\frac{n+1}n\right)^n\tag{1a}\\
&=\color{#C00}{\left(\frac{n+1}n-\frac1{(n+1)n}\right)^{n+1}}-\color{#090}{\left(\frac{n+1}n\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Calculate the determinant of $A-5I$ Question
Let $ A =
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
6 & 7 & 8 & 9 & 10 \\
11 & 12 & 13 & 14 & 15 \\
16 & 17 & 18 & 19 & 20 \\
21& 22 & 23 & 24 & 25
\end{bmatrix}
$.
Calculate the determinant of $A-5I$.
My approach
the nullity of $A$ is $3$, so the algebraic multiplicity of $\lam... | To determine $\det(A - 5I)$, it suffices to compute the eigenvalues of $A$. To that end, we note that $A$ has rank $2$, which means that it has at most $2$ non-zero eigenvalues.
Let $\lambda_1,\lambda_2$ denote these eigenvalues. We note that
$$
\lambda_1 + \lambda_2 = \operatorname{tr}(A) = 65\\
\lambda_1^2 + \lamb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
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Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.