Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$ How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$
What's the approach to it?
Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education... | Consider
$$
I(c)=\int_{0}^{\frac{\pi}{2}} \ln \left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x+c\right) d x, \textrm{ where } c\geq 0.
$$
Using my post, we have
$$
I(c)=\pi \ln \left(\frac{\sqrt{a^{2}+c}+\sqrt{b^{2}+c}}{2}\right)
$$
Differentiating $I(c)$ w.r.t. $c$ yields
$$
I^{\prime}(c)=\frac{\pi}{2 \sqrt{\left(a^{2}+c\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Prove ${\frac {1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{ \frac {1+{c}^{3}}{1+c{a}^{2}}}\ge 3 $
Let $a,b,c \ge0$, prove the on equality: $${\frac
{1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{ \frac
{1+{c}^{3}}{1+c{a}^{2}}}\ge 3 $$
I tried: $$LHS = \sum\frac 1{1+ab^2}+\sum \frac {a^4}{a... | Without loss of generality, let $a\geqslant b\geqslant c\geqslant0$. The target inequality is equivalent to: $$\left(\frac{1+{a}^{3}}{1+a{b}^{2}}-1\right)+\left(\frac{1+{b}^{3}}{1+b{c}^{2}}-1\right)+\left(\frac{1+{c}^{3}}{1+c{a}^{2}}-1\right)\geqslant0$$
That is
\begin{align*}
a\frac{a^2-b^2}{1+ab^2}+b\frac{b^2-c^2}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $n \ge \sqrt{n+1}+\sqrt{n}$ (how) Can I show that:
$n \ge \sqrt{n+1}+\sqrt{n}$ ?
It should be true for all $n \ge 5$.
Tried it via induction:
*
*$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.
*$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$
Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1... | What I would try first if I wanted it to be true - no tricks:
$n \ge \sqrt{n + 1} + \sqrt{n} \iff \frac{n}{\sqrt{n}} \ge \frac{\sqrt{n+1}}{\sqrt{n}} + \frac{\sqrt{n}}{\sqrt{n}} \iff \sqrt{n} \ge \sqrt{1 + \frac{1}{n}} + 1 \Leftarrow n \ge 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/403090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Show that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$ cannot be a square when $x\neq3$ and $x\neq2$ respectively. As the title says, I need help showing that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$ cannot be a square when $x\neq3$ and $x\neq2$ respectively, where $x$ is a natural number.
In order to do this, I hav... | If $x > 3$ and $n = x^4+x^3+x^2+x+1$ is a square then $4n$ is also a square. However
$$ \begin{eqnarray}4(x^4+x^3+x^2+x+1) &=& (2x^2 + x)^2 + 3x^2 + 4x+4 \\ &=& (2x^2+x+1)^2 - (x+1)(x-3) \end{eqnarray}$$
which shows that
$$(2x^2 + x)^2 < 4n < (2x^2+x+1)^2.$$
Since $4n$ is between two consecutive squares it cannot be a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/403473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$ $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$.
I struggled on it, but I didn't find any pattern to solve it.
| Idea: You can unwrap like this:
$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}}<2$$
if
$$\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}<2^2-1$$
if
$$\sqrt{3+\cdots+\sqrt{2006}}<(2^2-1)^2-2$$
and so on, so we want to show
$$2006 < (((2^2-1)^2-2)^2-\cdots)^2-2005$$
might as well prove it by induction for all $n$ rather than just ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/404653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
| $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$
since priority of multiplication $\times$ or $\cdot$ is greater than addition $+$ so expression will be:
$1+1+1+1+1+1+1+1+1+1+1+0+1\implies 12$
This is the order of operation:
$1$ B:- Brackets first
$2$ O:- Orders (i.e. Powers and Square Roots, etc.)
$3$ DM:- Division and Multiplicat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/405543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Trigonometric identities using $\sin x$ and $\cos x$ definition as infinite series Can someone show the way to proof that $$\cos(x+y) = \cos x\cdot\cos y - \sin x\cdot\sin y$$ and $$\cos^2x+\sin^2 x = 1$$ using the definition of $\sin x$ and $\cos x$ with infinite series.
thanks...
| In both cases, you'll want to use the Cauchy product, and the binomial theorem will be useful (for at least the first one), too. I leave the second one to you.
For the first one, $$\begin{align}\sin x\cdot\sin y &= \left(\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}\right)\cdot\left(\sum_{n=0}^\infty\frac{(-1)^ny^{2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/405628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Prove inequality: $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)} + xyz$
Let $x,y,z\in \mathbb R^+$ prove that:
$$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge xyz + \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$
The inequality $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{C-S}{\g... | The inequality is homogeneous, so we may assume that $xyz=1$. Then if we expand left and right hand side, we see that we have to show that
$$\sqrt{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 3} \ge 1 + \sqrt[3]{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 2}.$$
Let
$$u = \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/406779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Equilibrium distribution of a Markov Chain Can anyone please tell me or help me with this question shown below?
A drunken chess grandmaster dials a long string of digits on a standard telephone keypad (laid out as shown below). It takes more than alcohol to make a grandmaster forget the rules of chess, so each digit he... | Hint: the transition matrix looks like this (1 is the first entry and 0 is the last in each row/column). I have assumed that if the grand master presses $5$ first, we will randomly pick the next digit uniformly.
$$\left[\begin{array}{cccccccccc}
0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0\\
0 & 0 & 0 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/407154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve integrals of type $ \int\frac{1}{(a+b\sin x)^4}dx$ and $\int\frac{1}{(a+b\cos x)^4}dx$ $$\displaystyle \int\frac{1}{(a+b\sin x)^4}dx,~~~~\text{and}~~~~\displaystyle \int\frac{1}{(a+b\cos x)^4}dx,$$
although i have tried using Trg. substution. but nothing get
| Another method is the following:
$$
\frac{1}{(a+b\cos x)^4} = -\frac{1}{6}\frac{\partial^3}{\partial a^3}\frac{1}{a+b\cos x} \\
\begin{align}
\int\frac{\text{d}x}{(a+b\cos x)^4} & = -\frac16\int\frac{\partial^3}{\partial a^3}\frac{\text{d}x}{a+b\cos x} \\
& = -\frac16\frac{\partial^3}{\partial a^3}\int\frac{\text{d}x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/410841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Exponential equations involving natural numbers at power "x" Find x : $$4^x+15^x=9^x+10^x(2^x-3^x)(2^x-3^x-5^x)$$
| Oops - I made a mistake:
$x=0$ is NOT a root
since
$3^0-2^0 = 0$, not $1$.
Please retract my points.
I'll write it as
$$4^x+15^x=9^x+10^x(3^x-2^x)(5^x+3^x-2^x)$$
We see that $x=0$ is a root,
since
both sides are $2$
(because all the $n^x$ are 1).
(No it's not, but I talked fast, didn't I?)
I'll submit this and then loo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/411394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Is there a contradiction is this exercise? The following exercise was a resolution to this problem
Let $\displaystyle\frac{2x+5}{(x-3)(x-7)}=\frac{A}{(x-7)}+\frac{B}{(x-3)}\space \forall \space x \in \mathbb{R}$. Find the values for $A$ and $B$
The propose resolution was:
In order to isolate $A$ on the right side, m... | I would do it in another way. the equalation is $$\frac {2x+5}{(x-3)(x-7)}=\frac A {x-3}+\frac B {x-7}$$. what we can do is to take the common domniator on the right side getting $$\frac {2x+5}{(x-3)(x-7)}=\frac {A(x-7)+B(x-3)} {(x-7)(x-3)}$$. since the numerator in both sides is condcuted from numbers and x, we can co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/411658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Maximize area of a triangle with fixed perimeter If perimeter of a triangle is $2d$, what is the length of sides so the triangle has maximal area?
I found some solution using circle and angles, but I think I have to use derivatives.
I need help.
| Let $a$, $b$ and $c$ be the sides of a triangle. The perimeter, $p=a+b+c$, is fixed and we want to find the values of $a$, $b$ and $c$ that give the triangle maximum area. Heron's formula says that the triangle's area is
$$A=\sqrt{s(s-a)(s-b)(a-c)}$$
where s is the semiperimeter $\frac{a+b+c}{2}=\frac{p}{2}.$
Because p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Estimating the nested radical with $1,2,3,4,\dots$ under square root signs I require some assistance in proving the following inequality:
$$\sqrt{1 + \sqrt{2 + \sqrt{5}}} < \sqrt{ 1 + \sqrt{2 + \sqrt{3 + \sqrt{4 + \cdots}}}} < \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{5}}}}$$
I'm not quite sure how to create a rigorous enoug... | I was just curious if this was a way it could be resolved:
$0 < \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}} < \sqrt{ 1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}$
This is true because each term in the middle radical is less than $1$.
$\implies 1 < \sqrt{1 + \sqrt{\frac{5}{2^4}+\sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Taylor Series Remainder Use Taylor's Theorem to estimate the error in approximating $\sinh 2x$ by $2x + 4/3x^3$ on the interval $[-0.5,0.5]$.
For this question, I use the Taylor's remainder formular, $$
R_n(x)= \frac{f^{(n+1)}(z)(x-a)^{n+1}}{(n+1)!}$$
and I get $R_4(0.5) = 0.012\,85$.
Is this correct?
| Generally, you want to use:
$$\displaystyle |R_{n}(x)| \le \frac{M_{n+1}}{(n+1)!}|x-a|^{n+1}$$
where, $\displaystyle R_{n}(x) = f(x) -T_n(x)$ is the remainder term and $T_n(x)$ is the Taylor polynomial of degree $n$ for $f(x)$, centered at $x = a$.
For this problem, we have an odd function, $f(x) = \sinh 2x$ on the in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Geometric Inequality Related To Median, Altitude For a triangle $ABC$, let $m_{a}$, $h_{a}$ be $A$-median, $A$-altitude.
Define $m_{b}$,$h_{b}$ and $m_{c}$,$h_{c}$ likewise.
Prove that $\dfrac{h_{a}}{m_{b}}+\dfrac{h_{b}}{m_{c}}+\dfrac{h_{c}}{m_{a}}\leq 3$
I have no solution.
| Let $a^2=x$, $b^2=y$ and $c^2=z$.
Thus, by C-S $$\sum_{cyc}\frac{h_a}{m_b}\leq\sqrt{\sum_{cyc}h_a^2\sum_{cyc}\frac{1}{m_b^2}}=\sqrt{\sum_{cyc}\frac{4S^2}{a^2}\sum_{cyc}\frac{4}{2a^2+2c^2-b^2}}=$$
$$=\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)\sum\limits_{cyc}\frac{1}{a^2}\sum_{cyc}\frac{1}{2a^2+2c^2-b^2}}=$$
$$=\sqrt{\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving there are no integer solutions for $3x^2=9+y^3$
Prove there are no $x,y\in\mathbb{Z}$ such that $3x^2=9+y^3$.
Initial proof
Let us assume there are $x,y\in\mathbb{Z}$ that satisfy the equation, which can be rewritten as $$3(x^2-3)=y^3.$$ So, $$3 \mid y \Rightarrow 3^3 \mid y^3 \Rightarrow 3^2 \mid x^2 - 3.$$ ... | Suppose there is a solution to $3x^2 = 9 + y^3$.
Then $3$ divides the RHS hence $3|y^3$, so $3|y$. Let $y = 3a$ for some integer $a$.
Then $3x^2 = 9 + 27a^3$. Hence $9 | 3x^2$ and so $3|x^2$, implying $3|x$. Let $x = 3b$ for some integer $b$.
Then $27b^2 = 9 + 27a^3$, and after cancelling $9$'s we get $3b^2 = 1 + 3a^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Examples of how to calculate $e^A$ I'm trying to learn the process to discover $e^A$
For example, if $A$ is diagonalizable is easy:
$$A =\begin{pmatrix}
5 & -6 \\
3 & -4 \\
\end{pmatrix}$$
Then we have the canonical form $$J_A =\begin{pmatrix}
2 & 0 \\
0 & -1 \\
\end{pmat... | You might want to have a look at Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
Other references can be found in The matrix exponential: Any good books?
We have:
$$A = \begin{bmatrix}0 & 1 \\-2 & -2\end{bmatrix}$$
We want to find the characteristic polynomial and eigenvalues by so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/420763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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If $x+{1\over x} = r $ then what is $x^3+{1\over x^3}$? If $$x+{1\over x} = r $$ then what is $$x^3+{1\over x^3}$$
Options:
$(a) 3,$
$(b) 3r,$
$(c)r,$
$(d) 0$
| Given, $x+1/x=r$
$x^3+\frac1{x^3}=\left(x+\frac1x\right)^3-3\cdot x\cdot\frac1x\left(x+\frac1x\right)$
which gives us, $x^3+\frac1{x^3}$ = $ r^3-3r$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/421995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
| Form an equation with $a,b,c$ as its roots.
We have $ab+bc+ca=-\frac{1}{2}$ from the given relation.
So the equation with $a,b,c$ as its roots is,
$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0$
$\Rightarrow x^3-\frac{1}{2}x-abc=0$
If all of $a,b,c$ are of same sign then we have $a=b=c=0$ in which case the inequality is trivially ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/425187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 5
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Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. Can anyone give me a hint to evaluate this integral?
$$\int_0^\infty \frac{dx}{1+x^4}$$
I know it will involve the gamma function, but how?
| let $I=\int \dfrac{dx}{1+x^4}$
I=$\dfrac {1}{2}\int \dfrac {x^2+1-(x^2-1)}{1+x^4} $
splitting the fraction,
I=$\dfrac {1}{2} (\int \dfrac {x^2+1}{1+x^4} - \int \dfrac {x^2-1}{1+x^4} )$
let $I_1$ =$ \int \frac {x^2+1}{1+x^4}$
Dividing the numerator and denominator by $x^2$,
$I_1$= $ \int \dfrac {1+\dfrac {1}{x^2} }{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 2
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Need help in proving that $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$ We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$
I have tried and it gets confusing.
| $$\frac{\sin\theta -\cos\theta +1}{\sin\theta +\cos\theta -1}= \frac{1}{\sec\theta - \tan\theta}$$
By taking $$\mbox{L.H.S ( Left hand side )} = \frac{\sin\theta -\cos\theta +1}{\sin\theta +\cos\theta -1} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Norm of a linear transformation Let $T:\mathbb R^2\to \mathbb R^2$ be given by the matrix $\begin{pmatrix}a&b\\ c& d\end{pmatrix}$. Let $u:=a^2+b^2+c^2+d^2+2(ad-bc)$ and $v:=a^2+b^2+c^2+d^2-2(ad-bc)$.
I need to show that $\mid\mid T\mid\mid=\frac 12 (\sqrt u+\sqrt v)$.
I tried using the definition $\mid\mid T\mid\mi... | You have to find largest eigenvalue of $T^tT$ and show that it is equal to $\frac{u+v+2\sqrt{uv}}{4}.$ You get into the following equations $$(x-a^2-c^2)(x-b^2-d^2)-(ab+cd)^2=0\\\implies x^2-(a^2+b^2+c^2+d^2)x+(ad-bc)^2=0\\\implies x_\max=\frac{a^2+b^2+c^2+d^2+\sqrt{(a^2+b^2+c^2+d^2)^2-4(ad-bc)^2}}{2}\\=\frac{u+v+2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/429485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Matrix to power $2012$ How to calculate $A^{2012}$?
$A = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$
How can one calculate this? It must be tricky or something, cause there was only 1 point for solving this.
| $$A = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$$
$$A^2=\left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]\cdot \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$$
$$A^2 = \left[\begin{array}{ccc}{9-2-4}&{-3+2}&{-6+2+2}\\{6-4}&{-2+2}&{-4+2}\\{6-2}&{-2+1}&{-4+2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/429971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 3
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A problem of AM>GM
If $n$ is a positive integer with $n> 1$, prove that
$$2^{n(n+1)}>(n+1)^{(n+1)}\cdot\left(\frac{n}{1}\right)^n\cdot\left(\frac{n-1}{2}\right)^{(n-1)}\cdots\left(\frac{2}{n-1}\right)^2\cdot\frac{1}{n}$$
For solving it I have considered the numbers $\displaystyle{n+1,\,n,\,\frac{n-1}{2},\,\ldots,\f... | Hint:
$$\left(\frac{n}{1}\right)^n\cdot\left(\frac{n-1}{2}\right)^{(n-1)}\cdots\left(\frac{2}{n-1}\right)^2\cdot\frac{1}{n} = \binom{n}{0} \binom{n}{1} \cdots \binom{n}{n}$$
where $\binom{n}{k}$ is the usual binomial coefficient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/431476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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If $a,b$ are roots for $x^2+3x+1=0$.Calculating $(\frac{a}{b+1})^2 +(\frac{b}{a+1})^2$ If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
| From the content of this thread follows,
$x^2 + 3x +1 = 0 \Leftrightarrow x^2 +2x+1=-x \Leftrightarrow \boxed{(x+1)^2 =-x} (*)$
$\frac{a^2}{(b+1)^2}+\frac{b^2}{(a+1)^2} \Leftrightarrow
\left(\frac{b^2}{b^2}\right)\frac{a^2}{(b+1)^2}+\left(\frac{a^2}{a^2}\right)\frac{b^2}{(a+1)^2} \\
\hspace{3.05cm}\Leftrightarrow \u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/431606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 5
} |
Prove that $(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}>2$ for $x > 0$
Let $x>0$. Show that
$$(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}>2.$$
Do you have any nice method?
My idea $F(x)=(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}$
then we hvae $F'(x)=\cdots$ But it's ugly.
can you have nice methods? Thank you
by this I have see thi... | First to prove $f(x)=x^{\frac{1}{x}}$ have a max $f(e)$.
Edit:2nd version
when $x<e-1, f(x+1)>f(x)$
when $x>e,f(x)>f(x+1)$
for $f(x+1)>f(x)$, it is trivial .
when $e-1\le x\le e$, :
$f(x) $ is mono increasing, so $f(x)^{-1}_{min}=f(e)$, $f(x+1)$ is mono decreasing, so $ f(x+1)_{min}=f(e+1)$, thus,$f(x+1)+f(x)^{-1} > (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to integrate these integrals $$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \cos x}$$
$$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \sin x}$$
It seems that substitutions make things worse:
$$\int \frac {dx}{1+ \cos x} ; t = 1 + \cos x; dt = -\sin x dx ; \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (t-1)^2} $$
$$ \Rightarrow
\int \frac ... | Use this simple trigonometry manipulation:
$$
\frac{1}{1+\cos x}=\frac{1}{1+\cos x}\cdot\frac{1-\cos x}{1-\cos x}=\frac{1-\cos x}{\sin^2 x}.
$$
Therefore
$$
\begin{align}
\int\frac{1}{1+\cos x}\;dx&=\int\frac{1-\cos x}{\sin^2 x}\;dx\\
&=\int\frac{1}{\sin^2 x}\;dx-\int\frac{\cos x}{\sin^2 x}\;dx\\
&=-\int\;d(\cot x)-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Expressing polynomial roots expression in terms of coefficients This is my first question on MSE. Apologies in advance for any textual or LaTeX errors.
I'm stuck with this problem:
Given $x^3 - bx^2 + cx - d = 0$ has roots $\alpha$, $\beta$, $\gamma$, find an expression in terms of $b$, $c$ and $d$ for:
(i) $\alpha... | Let $y=1+\alpha^3$
Again, $\alpha^3=b\alpha^2-c\alpha+d$
$$\implies y-1-d=b\alpha^2-c\alpha$$
Cubing we get $$ (y-1-d)^3=b^3\alpha^6-c^3\alpha^3-3bc\alpha^2\cdot \alpha(b\alpha^2-c\alpha)$$
$$(y-1-d)^3=b^3(y-1)^2-c^3(y-1)-3bc(y-1)(y-1-d)$$ as $\alpha^3=y-1,b\alpha^2-c\alpha=y-1-d$
Arrange as $y^3+By^2+Cy+D=0$ whose r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Confused about harmonic series and Euler product So Euler argued that
$$1 + \frac{1}{2} + \frac{1}{3} + \frac {1}{4} + \cdots = \frac {2 \cdot 3 \cdot 5 \cdot 7 \cdots} {1 \cdot 2 \cdot 4 \cdot 6 \cdots} $$
which you can rearrange to
$$ \left( \frac {1 \cdot 2 \cdot 4 \cdot 6 \cdots} {2 \cdot 3 \cdot 5 \cdot 7 \cdots} ... | $$\prod_1^r{p_n-1\over p_n}=\prod_1^r\left(1-{1\over p_n}\right)^{-1}=\prod_1^r\left(1+{1\over p_n}+{1\over p_n^2}+\cdots\right)=\sum{1\over m}$$ where the sum is over all $m$ divisible by no primes other than $p_1,\dots,p_n$. Formally, the limit as $r\to\infty$ gives the first displayed equation in the question. Rigor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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where this series converges Given the series $$\sum_{j=0}^{\infty}\frac{1}{6j^2-5j+1}$$
I am completely stuck and do not understand the answer from my book which is $\pi^2/36-1$. I need explanation and different approach how this result is gained. Thanks
| Rewrite your series as :
\begin{align}
\tag{1}\sum_{j=0}^{\infty}\frac{1}{6j^2-5j+1}&=\sum_{j=0}^{\infty}\frac{1}{(3j-1)(2j-1)}\\
&=\sum_{j=0}^{\infty}\frac2{2j-1}-\frac3{3j-1}\\
\tag{2}&=-2+3+\sum_{j=1}^{\infty}\frac1{j-\frac 12}-\frac1{j-\frac 13}\\
&=1-\psi\left(1-\frac 12\right)+\psi\left(1-\frac 13\right)\\
\tag{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How find this value $\frac{a^2+b^2-c^2}{2ab}+\frac{a^2+c^2-b^2}{2ac}+\frac{b^2+c^2-a^2}{2bc}$ let $a,b,c$ such that
$$\left(\dfrac{a^2+b^2-c^2}{2ab}\right)^2+\left(\dfrac{b^2+c^2-a^2}{2bc}\right)^2+\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2=3,$$
find the value
$$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^... | I think the answer must be $1$ if we impose a triangle inequality. The individual terms are all cosines of angles of a triangle. Thus, the sum of those angles must be $\pi$. But the sum of the squares of their cosines is $3$; therefore each cosine must be $\pm 1$. But the sum of the angles is, again, $\pi$, so that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If $\sin a+\sin b=2$, then show that $\sin(a+b)=0$ If $\sin a+\sin b=2$, then show that $\sin(a+b)=0$.
I have tried to solve this problem in the following way :
\begin{align}&\sin a + \sin b=2 \\
\Rightarrow &2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)=2\\
\Rightarrow &\sin\left(\frac{a+b}{2}\rig... | HINT:
As $-1\le \sin x\le 1$ for real $x$
$\implies -2\le \sin a+\sin b\le 2$
The equality occurs if $\sin a=\sin b=1$
$a=2n\pi+\frac\pi2, b=2m\pi+\frac\pi2,$ for some integer $m,n$
So, $a+b=2\pi(m+n)+\pi=\pi(2m+2n+1)$ and we know $\sin r\pi=0$ for integer $r$
Alternatively, if $\sin x=1,\cos x=\pm\sqrt{1-1^2}=0\impl... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given that $a^2(a+k)=b^2(b+k)=c^2(c+k)$, find the value of $1/a+1/b+1/c$
Given $$a^2(a+k)=b^2(b+k)=c^2(c+k)$$
find the value of $1/a+1/b+1/c$.
I tried to derive a relation from the equality but it did not help my cause.
| HINT:
Let $$a^2(a+k)=b^2(b+k)=c^2(c+k)=d$$
So $a,b,c$ are the roots of $x^3+x^2k-d=0$
Using Vieta's Formula, $a+b+c=-k, ab+bc+ca=0,abc=d$
$$\text{Now,}\frac1a+\frac1b+\frac1c=\frac{ab+bc+ca}{abc}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Factor Equations Please check my answer in factoring this equations:
Question 1. Factor $(x+1)^4+(x+3)^4-272$.
Solution: $$\begin{eqnarray}&=&(x+1)^4+(x+3)^4-272\\&=&(x+1)^4+(x+3)^4-272+16-16\\
&=&(x+1)^4+(x+3)^4-256-16\\
&=&\left[(x+1)^4-16\right]+\left[(x+3)^4-256\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\rig... | Let $w = x + 2$
\begin{align}
(x+1)^4+(x+3)^4-272
&= (w-1)^4+(w+1)^4-272\\
&= 2w^4 + 12w^2 - 270\\
&= 2(w^4 + 6w - 135)\\
&= 2(w^4 +6w^2 + 9 - 144)\\
&= 2(w^2 + 3)^2 - 12^2)\\
&= 2(w^2 + 15)(w^2 - 9)\\
&= 2(w^2 + 15)(w - 3)(w + 3)\\
&=2(x^2 + 4x + 19)(x - 1)(x + 5)\\
\end{align}
\begin{align}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Generating function of the number of integer partitions of $n$ into all distinct parts Let $p_d (n)$ denote the number of integer partitions of $n$ into all distinct parts. I am given the following equation, but I can't figure out why it holds:
$$\sum_{n \ge 0} p_d(n)x^n = \prod_{i \ge 1}(1+x^i).$$
P.S. This example is... | Consider first the finite product
$$\prod_{i = 1}^n (1+x^i).$$
If you expand it, you get
$$\prod_{i = 1}^n (1+x^i) = \sum_{\varepsilon \in \{0,\,1\}^n} x^{\sum_{i=1}^n \varepsilon_i \cdot i}.$$
The term $x^k$ occurs as many times in that sum, as there are ways to write $k$ as a sum of distinct positive integers $\leqsl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/440461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Continuity and Finding Values Find value of a,b,c such that F is continuous on the real number system:
$$
f(x) = \left\{
\begin{array}{lr}
-1 & : x\le-1\\
ax^2+bx+c & : |x|<1,x\ not\ equal\ to\ 0\\
0 &:x=0\\
1 &:x\ge1
\end{array}
\right.
$$
We went over the following solution in clas... | The reason they break it down into two cases at $x=-1$ is that the function is fundamentally different on $\{x<-1\}$ than it is on $\{x>-1\}$.
The way they simplify into $a-b+c+1=0$ is by simply plugging $x=-1$ into $ax^2 + bx + c - (-1)$.
The way we knew that $c=0$ is that as $x$ gets small, $ax^2 + bx + c$ will get a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Exponentials of a matrix I just was working with matrix exponentials for solving problems in control theory. Suppose $A $ is a square matrix. How can we interpret $A_1 = e^ {\textstyle-A\log(t) }$, where $\log$ is natural logarithm? Is there a formula for extending the scalar case of $e^ {\textstyle-a\log(t) }$ which g... | With the matrix exponential, more care is necessary. One approach is to multiply each item in the matrix $A$ by $\ln t$ and then find the diagonal matrix and take advantage of its structure.
Try working these two examples and see if you get them.
Example 1: Matrix is in diagonal form
$A = \begin{bmatrix}1 & 0\\0 & 1\e... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculus Implicit Differentiation I'm learning implicit differentiation and I've hit a snag with the following equation.
$$
f(x, y) = x + xy + y = 2
$$
$$
Dx(x) + Dx(xy) + Dx(y) = Dx(2)
$$
$$
1 + xy' + y + y' = 0
$$
$$
xy' + y' = -1 - y
$$
$$
y'(x + 1) = 1 + y
$$
$$
y' = \dfrac{(1 + y)}{(x + 1)}
$$
$$
y'' =... | Your mistake seems to originate when moving from this:
$$
xy' + y' = -1 - y
$$
...to this, where you "lost the sign":
$$
y'(x + 1) = 1 + y$$
We need $$y'(x + 1) = -(1 + y)$$
Let's back up: $$\begin{align} 1 + xy' + y + y' & = 0 \\ \\ xy' + y' & = -1 - y\\ \\ & = -(1 + y)\end{align}$$
Then we factor out $y'$ on the l... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Big $\Omega$ question! Prove $(n-1)(n-2)(n-3)$ is $\Omega(n^3)$ Problem
Prove $(n-1)(n-2)(n-3)$ is $\Omega(n^3)$.
Attempt @ Solution
*
*$f(n) = n^3(1-6/n+11/n^2-6/n^3)$
*$g(n) = n^3$
*Show that there exists a $C > 0$ and $n_0$ such that $f(n) \ge Cg(n)$ for all $n > n_0$.
*I tried plugging in different numbers fo... | You will not be able to show that $f(n)\gt g(n)$, because it is in fact smaller.
What I would suggest is that if $n\ge 6$, then $n-3\ge \frac{n}{2}$, as are $n-2$ and $n-1$. Thus for $n\ge 6$, we have $f(n)\ge \frac{1}{8}g(n)$. So we can take $C=\frac{1}{8}$.
And $C$ certainly does not have to be an integer. In our pa... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $2^n < \binom{2n}{n} < 2^{2n}$ Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction.
First part: $2^n < \binom{2n}{n}$. The base $n = 1$ is trivial. Assume inductively that some $k$ satisfies our statement. The indu... | Let $A$ be the set of all binary words containing $n$ bits, $B$ be the set of all binary words containing $2n$ bits, exactly $n$ of which are $1's$, $C$ be the set of all binary words containing $2n$ bits.
Then as $B$ is a proper subset of $C$ we have $\binom{2n}{n} <2^{2n}$.
We define a function $E : B \to A$ by $E(x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 5
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Interpreting the ; in a series This question is linked to this question.
So, suppose I set $n=5$. Given the following formula:
$$\frac{1}{n}, \dots , \frac{n-1}{n} $$
Am I suppose to get:
$$
\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \hspace{8.2cm}(1)
$$
Or
$$
\frac{1}{5}, \frac{1}{4}, \frac{1}{3}, \frac{2}... | The commas separate values for a single given value of $n$. The semi-colons separate for different values of $n$:
$$\underbrace{\frac12}_{n=1};\underbrace{\frac{1}{3}, \frac{2}{3}}_{n=3};\underbrace{\frac14,\frac{2}{4},\frac{3}{4}}_{n=4};\ldots$$
The intention was that this form a single set of unique rational numbers;... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$
Problem: If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$
Solution: $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $
Differenting both sides,... | Yes and to complete: we have
$$\int f'(x)(f(x))^{-2}\;dx=C\int\sin(2x)\;dx$$
where $C=b^2-a^2$ so
$$-\frac{1}{f(x)}=-\frac{C}{2}\cos(2x)+C'$$
and you can take $f(x)$ from it.
| {
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"url": "https://math.stackexchange.com/questions/451131",
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"source": "stackexchange",
"question_score": "6",
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If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$.
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
I know how to solve a problem like
"If $\cos\alpha = \frac{\sqrt{3}}... | If $\sin\alpha + \cos\alpha = 0.2$, then squaring both sides and simplifying will produce $2 \sin \alpha \cos \alpha = -0.96$.
It follows that $\sin^2 \alpha - 2 \sin(\alpha) \cos(\alpha) + \cos^2 \alpha = 1.96$, from which we conclude $\sin\alpha - \cos\alpha = \pm 1.4$
From
$\left\{
\begin{array}{l}
\sin\al... | {
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"source": "stackexchange",
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Need help calculating this determinant using induction This is the determinant of a matrix of ($n \times n$) that needs to be calculated:
\begin{pmatrix}
3 &2 &0 &0 &\cdots &0 &0 &0 &0\\
1 &3 &2 &0 &\cdots &0 &0 &0 &0\\
0 &1 &3 &2 &\cdots &0 &0 &0 &0\\
0 &0 &1 &3 &\cdots &0 &0 &0 &0\\
\vdots &\vdots &\vdots&\ddots &\dd... | You have a tridiagonal matrix.
It's determinant can be written as a recurrence relation:
$$\det A \stackrel{\textrm{def}}{=} f_n = a_nf_{n-1} + c_{n-1}b_{n-1}f_{n-2}.$$
Define $f_0 = 1, f_{-1} = 0$.
However, your $a_n, b_n, c_n$ values are constant, so
$$ \det A = 3f_{n-1}+ 2 f_{n-2}.$$
Thus, you have
$$\begin{align*}
... | {
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"source": "stackexchange",
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not both $2^n-1,2^n+1$ can be prime. I am trying to prove that not both integers $2^n-1,2^n+1$ can be prime for $n \not=2$. But I am not sure if my proof is correct or not:
Suppose both $2^n-1,2^n+1$ are prime, then $(2^n-1)(2^n+1)=4^n-1$ has precisely two prime factors. Now $4^n-1=(4-1)(4^{n-1}+4^{n-2}+ \cdots +1)=3A$... | If $n$ is even $=2m,2^n-1=2^{2m}-1=4^m-1$ is divisible by $4-1=3$ and $4^m-1>3$ if $m\ge1\iff n\ge2$
If $n$ is odd $=2m+1,2^n+1=2^{2m+1}+1$ is divisible by $2+1=3$ and $2^{2m+1}+1>3$ if $m\ge1\iff n\ge3$
alternatively,
$$(2^n-1)(2^n+1)=4^n-1$$ is divisible by $4-1=3$
So, at least one of $2^n-1,2^n+1$ is divisible by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 1
} |
If $\sin\theta+\sin\phi=a$ and $\cos\theta+ \cos\phi=b$, then find $\tan \frac{\theta-\phi}2$. I'm trying to solve this problem:
If $\sin\theta+\sin\phi=a$ and $\cos\theta+ \cos\phi=b$, then find $\tan \dfrac{\theta-\phi}2$.
So seeing $\dfrac{\theta-\phi}2$ in the argument of the tangent function, I first thought of ... | Method $1:$
Squaring & adding what you have derived $$4\cos^2\frac{\theta-\phi}2=a^2+b^2$$
$$\implies \sec^2\frac{\theta-\phi}2=\frac4{a^2+b^2}$$
$$\implies \tan^2\frac{\theta-\phi}2=\frac4{a^2+b^2}-1=\frac{4-a^2-b^2}{a^2+b^2}$$
Method $2:$
$$\text{As }\quad2\cos\frac{\theta+\phi}2\cos\frac{\theta-\phi}2=b,$$
$$\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What is the number of real solutions of the following? $ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $ What is the number of real solutions of the following?
$$ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $$
My solution:
$$ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $$
... | Separate the square roots and square both sides. If you isolate the term $\sqrt{x+3-4\sqrt{x-1}}$, after simplifying you will obtain:
$$\sqrt{x+8-6\sqrt{x-1}}=3-\sqrt{x-1}$$
Squaring both sides again gives:
$$x+8-6\sqrt{x-1}=9-6\sqrt{x-1}+x-1$$
Which holds for all valid $x$. It follows that the original equation is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all
| Let $$f(y) = \int_{0}^{\infty} \frac{\sin^3{yx}}{x^3} \mathrm{d}x$$
Then,
$$f'(y) = 3\int_{0}^{\infty} \frac{\sin^2{yx}\cos{yx}}{x^2} \mathrm{d}x = \frac{3}{4}\int_{0}^{\infty} \frac{\cos{yx} - \cos{3yx}}{x^2} \mathrm{d}x$$
$$f''(y) = \frac{3}{4}\int_{0}^{\infty} \frac{-\sin{yx} + 3\sin{3yx}}{x} \mathrm{d}x$$
Therefore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 4
} |
matrix representation of polynomial Here is a polynomial $p(x,y) = (ax + by)^2$, it can be written like this $$p(x,y) = \left(\left[ \begin{array}{cc}
a & b \\
\end{array} \right] \left[ \begin{array}{c}
x\\
y\\
\end{array} \right]\right)^2$$
and I know that it can also be written as something like $v^TMv$, here $v = [... | $$\left[ \begin{array}{cc}
a & b \\
\end{array} \right] \left[ \begin{array}{c}
x\\
y
\end{array} \right] = \left[ \begin{array}{cc}
x & y \\
\end{array} \right] \left[ \begin{array}{c}
a\\
b
\end{array} \right]$$
Then $M= \left[ \begin{array}{c}
a & b
\end{array}\right] \left[ \begin{array}{c}
a\\
b
\end{array}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/454007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If a,b,c are sides of a triangle, prove: $ \sqrt{a+b-c} + \sqrt{b+c-a} + \sqrt{c+a-b} \le \sqrt{a} + \sqrt{b} + \sqrt{c} $ I did substitute $a=x+y, b=x+z, c=y+z$ and I arrived at $\sqrt{2x} + \sqrt{2y} + \sqrt{2z} \le \sqrt{x+y} + \sqrt{x+z} + \sqrt{y+z}$.
However, after this, I tried various methods like AM-GM and Cau... | Let $a\geq b\geq c$ and $f(x)=\sqrt{x}$.
Hence, $a+b-c\geq a+c-b\geq b+c-a$, $(a+b-c,a+c-b,b+c-a)\succ(a,b,c)$ and since $f$ is a concave function, the starting inequality it's just Karamata.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$ While I don't doubt that this question is covered somewhere else I can't seem to find it, or anything close enough to which I can springboard. I however am trying to prove $$\frac{1}{1^2} +\frac{1}{2^2} + \cdo... | $$\sum_{n=1}^\infty{1\over n^2}<1+\sum_{n=2}^\infty{1\over n^2-{1\over4}}= 1+\sum_{n=2}^\infty\left({1\over n-{1\over2}}-{1\over n+{1\over2}}\right)=1+{2\over3}={5\over3}\ .$$
By the way: Since the left hand side is $={\pi^2\over6}$ you obtain from this the estimate $\pi^2<10$, which is not bad. In fact $\pi^2\doteq9.8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Find $a,b\in\mathbb{Z}^{+}$ such that $\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$ find positive intergers $a,b$ such that
$\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$
Here i tried plugging
$x^3=a,y^3=b$
$(x+y-1)^2=x^2+y^2+1+2(xy-x-y)=49+20\sqrt[3]{6} $
the right hand part is a square hence can be... | This one is interesting.
Note that not both $a$ and $b$ are perfect cubes. Why? because otherwise the lhs would be a perfect cube!
Now, let's tackle the case where one of them is a perfect cube, say $a$. In that case, $(\sqrt[3]{a} - 1)$ is an integer. By elementary property of surds, square of this integer must match ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
seeming ugly limit i want to compute the limit
$$\lim_{x \rightarrow 0} \frac{e^x-1-x-\frac{x^2}{2}-\frac{x^3}{6}-\frac{x^4}{24}-\frac{x^5}{120}-\frac{x^6}{720}}{x^7}$$
Instead of doing some messy calculation, I think if there is some ingenious way to compute this limit, but i don't know how to do. thank you so much.
| Apply the L'Hôpital rule $7$ times (until the derivatives kill the polynomial $1+x+\ldots+\frac{x^6}{6!}$) and we find
$$\lim_{x \rightarrow 0} \frac{e^x-1-x-\frac{x^2}{2}-\frac{x^3}{6}-\frac{x^4}{24}-\frac{x^5}{120}-\frac{x^6}{720}}{x^7}=\lim_{x \rightarrow 0}\frac{e^x}{7!}=\frac{1}{7!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How find the of minimum $ab+bc+\frac{\sqrt{2}}2ac$ for $a^2+b^2=4,b^2+c^2=8$ let $a,b,c$ are real numbers and such $a^2+b^2=4,b^2+c^2=8$, find the minimum of
$$ab+bc+\dfrac{\sqrt{2}}{2}ac$$
if this problem ask find the maximum
we can use $AM-GM$
$$ab+bc+\dfrac{\sqrt{2}}{2}ac=\dfrac{1}{Ax}Aa\cdot xb+\dfrac{1}{Bz}Bb\cd... | Looks like a job for Lagrange multipliers...
let $$F(a,b,c,\lambda,\mu) = ab + bc + \frac{\sqrt{2}}{2} ac + \lambda (a^2 + b^2 - 4) + \mu(b^2 + c^2 - 8)$$
and solve $\dfrac{\partial F}{\partial a} = \dfrac{\partial F}{\partial b} = \dfrac{\partial F}{\partial c} = \dfrac{\partial F}{\partial \lambda} = \dfrac{\partial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/457766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing invariant factors from Smith normal form The goal is to find the Jordan Canonical Form of the matrix
$$A=\begin{bmatrix}2&1&1&2\\0&2&0&1\\0&0&2&-1\\0&0&0&1\end{bmatrix}$$
Since the matrix is already upper-triangular, it's obvious that the eigenvalues are 2 and 1, where 1 has geometric and algebraic multiplic... | I'll answer your second question first.
Let
$$P(x) = \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&(x-2)^2&0\\0&0&0&(x-1)(x-2)\end{bmatrix}.$$
We need to fix the bottom right $2 \times 2$ principal submatrix. I have explained how to do that here, so I'll just present the results here:
\begin{align*}
\begin{bmatrix}1&0&0&0\\0&1&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Quadratic sum of Gauss integers Let $A$ be a set, define $nA=\{x\mid x=a_1+a_2+\cdots a_k,a_i\in A,1\leq k\leq n\}$.
Denote $G=\{z\mid z=(a+bI)^2,a,b\in \mathbb Z,I=\sqrt{-1}\},K=\{z\mid z=a+2bI,a,b\in \mathbb Z,I=\sqrt{-1}\}$
What's the smallest integer $n$ such that $K=nG$ (if there exist)?
Lagrange's four-square t... | (This is not a full answer, but too long for a comment)
Claim: $n\leq 8$.
Given $a+2bi\in\mathbb Z[i]$ for some $a,b\in\mathbb Z$. By Langrange's theorem, we have:
$$|a|=x_1^2+x_2^2+x_3^2+x_4^2$$
$$|b|=y_1^2+y_2^2+y_3^2+y_4^2$$
for some $x_i,y_i\in\mathbb Z$. Now, define
$$c_a=\begin{cases}1&a\geq 0\\i&a<0\end{cases}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/461823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove the inequality $\,\frac{1}{\sqrt{1}+ \sqrt{3}} +\frac{1}{\sqrt{5}+ \sqrt{7} }+\ldots+\frac{1}{\sqrt{9997}+\sqrt{9999}}\gt 24$
Prove the inequality $$\frac{1}{\sqrt{1}+ \sqrt{3}}
+\frac{1}{\sqrt{5}+ \sqrt{7} }+......... +\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
My work:
Rationalizing the denominator gives
$$\... | Note that $\frac{\sqrt{3}-1}{2}>\frac{\sqrt{5}-\sqrt{3}}{2}$, etc. because $\sqrt{\phantom{x}}$ is concave down. So twice your left-hand side is greater than a telescoping sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/462118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Factorize $(x+1)(x+2)(x+3)(x+6)- 3x^2$ I'm preparing for an exam and was solving a few sample questions when I got this question -
Factorize : $$(x+1)(x+2)(x+3)(x+6)- 3x^2$$
I don't really know where to start, but I expanded everything to get :
$$x^4 + 12x^3 + 44x^2 + 72x + 36$$
I used rational roots test and Descarte... | $(x+1)(x+2)(x+3)(x+6)-3x^2$
$\rightarrow(x+1)(x+6)(x+2)(x+3)-3x^2$
$\rightarrow(x^2+x+6x+6)(x^2+2x+3x+6)-3x^2$
$\rightarrow(x^2+7x+6)(x^2+5x+6)-3x^2$
$\rightarrow(x^2+6+7x)(x^2+6+5x)-3x^2$
Put :- $x^2+6=y$ eq($1$) . So :-
$(y+7x)(y+5x)-3x^2$
$\rightarrow y^2+7xy+5xy+35x^2-3x^2$
$\rightarrow y^2+12xy+32x^2$
$\rightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 4,
"answer_id": 3
} |
Given $\sec \theta + \tan \theta = 5$ , Find $\csc \theta + \cot \theta $. The question is to find the value of $ \csc \theta + \cot \theta $ if $\sec \theta + \tan \theta = 5$ .
Here is what I did :
$\sec \theta + \tan \theta = 5$
$\sec \theta = 5 - \tan \theta $
Squaring both sides ,
$$\sec^2 \theta = 25 + \... | Here is a simpler solution to this problem:
$$\left(\sec(\theta)+\tan(\theta) \right)\left(\sec(\theta)-\tan(\theta) \right)=\sec^2(\theta)-\tan^2(\theta)=1$$
Since $\sec(\theta)+\tan(\theta)=5$ you get $\sec(\theta)-\tan(\theta)=\frac{1}{5}$.
Adding and subtracting these two relations you get
$$2\sec(\theta)=5+\frac15... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Evalute $\lim_{x\to-\infty} \frac{\sqrt{x^2+4x^4}}{8+x^2}$ Having a hard time with this. So far I have:
$$ \frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \frac{x\sqrt{1+4x^2}}{8+x^2}$$
| This is a match-in-heaven for high-school level Non-Standard Analysis (see Keisler textbook).
Since $x$ goes to infinity, we can replace it by the "larger than any real number" $H = \frac {1}{\epsilon} $ where $\epsilon$ is the infinitesimal (smallest than any real number), and dispense with the $\lim$ altogether.
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Creating sequences from natural numbers How much different arithmetic sequences can you make from the numbers 1 to 51 ?
Note:
The sequence length has to be 3 numbers.
The difference between each 2 numbers is positive.
| There are several ways to count in an organized manner.. We describe three of them.
(Our favourite way is the first, and relatives.)
First way: We need to choose two numbers $a$ and $b$ to serve as the ends of our sequence. The numbers $a$ and $b$ determine a three-term increasing arithmetic sequence precisely if $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to ... | Consider the polynomial $$P(z):=z^6+z^5+z^4+z^3+z^2+z+1\,.$$ Let $t:=z+\dfrac{1}{z}$. Therefore, $P(z)=z^3\,Q(t)$, where
$$Q(t):=t^3+t^2-2t-1\,.$$
Let $\omega:=\exp(\text{i}\theta)$, where $\theta:=\dfrac{2\pi}{7}$. Then, $\omega$, $\omega^2$, $\omega^3$, $\omega^4$, $\omega^5$, and $\omega^6$ are all the roots of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/470614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
The asymptotic expansion for the weighted sum of divisors $\sum_{n\leq x} \frac{d(n)}{n}$ I am trying to solve a problem about the divisor function. Let us call $d(n)$ the classical divisor function, i.e. $d(n)=\sum_{d|n}$ is the number of divisors of the integer $n$. It is well known that the sum of $d(n)$ over all po... | I will do the case of the sum being restricted to even integers for the sake of completing the original problem as stated by the OP.
Start with $\Lambda(s)$ for $\sum_{n \; \text{odd}} d(n)/n^s$ which is
$$\Lambda(s) = \left(1-\frac{1}{2^s}\right)^2\zeta(s)^2.$$
It follows that for $\sum_{n \; \text{even}} d(n)/n^s$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/471326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
$f$ holomorphic from unit disc to itself. $f\left(\frac{1}{2}\right) = f\left(-\frac{1}{2}\right) = 0$. Show that $|f(0)| \le 1/3$. I'm studying for a qual exam. I cannot solve the following problem:
Let $f$ be holomorphic from the unit disc to itself. $f\left(\frac{1}{2}\right) = f\left(-\frac{1}{2}\right) = 0$. Show... | Consider the function
$$g(z) = f(z) \frac{1 - r^2 z^2}{r^2 - z^2}$$
where $r = \frac{1}{2}$, which is holomorphic on $\mathbb{D}$ since $f$ has roots at $\pm r$. When $|z| = 1$, $|f(z)| \leq 1$ and
$$|g(z)| \leq \left|\frac{1 - r^2 z^2}{r^2 - z^2} \right|$$
We have
$$16 |1 - r^2 z^2|^2 = |4 - z^2|^2 = (4 - z^2) (4 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/471608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How prove this $\frac{1}{4}<\sum_{n=1}^{\infty}g\left(\frac{1}{n}\right)<1$ let $x>0$, and such
$$(1+x^2)f'(x)+(1+x)f(x)=1,g'(x)=f(x),f(0)=g(0)=0$$
show that
$$\dfrac{1}{4}<\sum_{n=1}^{\infty}g\left(\dfrac{1}{n}\right)<1$$
my idea: we can find
$$f(x)=e^{-\int\dfrac{1+x}{1+x^2}dx}\left(\int\dfrac{1}{1+x^2}e^{\int\dfr... | Notice that
$$(1+x^2)f'(x)+(1+x)f(x)=1\tag1$$
implies
$$ \left(e^{\arctan x}\sqrt{1+x^2}f(x)\right)'=\frac{e^{\arctan x}}{\sqrt{1+x^2}},\tag2$$
and $f(0)=0$. Thus, for $x>0$, it holds that
\begin{align*}
g'(x)&=f(x)\\&=\frac{1}{e^{\arctan x}\sqrt{1+x^2}}\int_0^x \frac{e^{\arctan t}}{\sqrt{1+t^2}}{\rm d}t\\
&\le \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/471894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Taylor Series Expansion of $f(x) = 2/(1-x)$ centered at $x=3$. Give the interval of convergence. Find the Taylor Expansion for the function f(x) = 2/(1-x) centered at x = 3.
Give the interval of convergence for this series.
So if I remember correctly, we first take the first four or so derivatives.
$f(x) = 2/(1-x)$
... | The idea is to take $y=x-3$ so $x=y+3$ and then
$$f(x)=\frac{2}{1-x}=\frac{-2}{2+y}=\frac{-1}{1+(y/2)}=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{y}{2}\right)^n=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{x-3}{2}\right)^n$$
The ratio test gives $R=2$ so the interval of convergence centred at $3$ is $(1,5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/473265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use the identity $(r+1)^3-r^3\equiv3r^2+3r+1$ to find $\sum_{r=1}^nr(r+1)$ Use the identity $(r+1)^3-r^3\equiv3r^2+3r+1$ to find $$\sum_{r=1}^nr(r+1)$$
I can obtain $$\sum_{r=1}^n3r^2+3r+1=(n+1)^3-1$$
and I think the next step is
$$3\sum_{r=1}^nr(r^2+1)+\frac13=\left((n+1)^3-1\right)$$
But how do I deal with the const... | $$3r^2+3r=3r(r+1)$$ and $$n(n+1)(n+2)=\dfrac{1}{4}n(n+1)(n+2)[(n+3)-(n-1)]\\=\dfrac{1}{4}n(n+1)(n+2)(n+3)-\dfrac{1}{4}(n-1)n(n+1)(n+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
I need to calculate $x^{50}$ $x=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}$, I need to calculate $x^{50}$
Could anyone tell me how to proceed?
Thank you.
| This is a very elementary approach based on finding the general form. If we do $x^2$, we find $$x^2=\begin{pmatrix}1&0&0\\1&1&0\\1&0&1\end{pmatrix},~~x^4=\begin{pmatrix}1&0&0\\2&1&0\\2&0&1\end{pmatrix}$$ so I guess that we have $$x^{2k}=\begin{pmatrix}1&0&0\\k&1&0\\k&0&1\end{pmatrix}$$ An inductive approach adimits thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 8,
"answer_id": 2
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Can I get a hint on solving this recurrence relation? I am having trouble solving for a closed form of the following recurrence relation.
$$\begin{align*}
a_n &= \frac{n}{4} -\frac{1}{2}\sum_{k=1}^{n-1}a_k\\
a_1 &= \frac{1}{4}
\end{align*}$$
The first few values are $a_1=\frac{1}{4},a_2=\frac{3}{8},a_3=\frac{7}{16},a_4... | By inspection we determine a particular solution to $2a_n-a_{n-1}=1/2$ is given by $a_n=1/2$ trivially -- try an ansatz of the form $a_n=k$ and thus we get $k=2k-k=1/2$.
Considering the homogeneous case, $2a_n-a_{n-1}=0$, let $a_n=\lambda^n$ hence:$$2\lambda^n-\lambda^{n-1}=0\\2\lambda-1=0\\\lambda=\frac12$$... and so ... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Limit of $\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot[{(x+\sin{\frac{1}{x}})}^{\frac{1}{3}} -x^{\frac{1}{3}}]}$ I need to find the limit of
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
| $$Limit=\lim_{x\rightarrow\infty}x^2[(1+\frac{\sin \frac{1}{x}}{x})^\frac{1}{3}-1]=\lim_{x\rightarrow\infty}x^2\frac{\sin \frac{1}{x}}{3x}=\frac{1}{3}$$
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Show that $\int \limits_{0}^{\infty}\frac{x}{\sinh ax}dx=\left(\frac{\pi}{2a}\right)^2$ Show that
$$\int \limits_{0}^{\infty}\frac{x}{\sinh ax}dx=\left(\frac{\pi}{2a}\right)^2$$
using 2 ways: the first using contour integration and the second using real analysis.
| $$a>0:$$
$$\int_0^{\infty} \frac{x\,dx}{\sinh ax}=\frac{1}{a^2}\int_0^{\infty}\frac{x\,dx}{\sinh x}=\frac{2}{a^2}\int_0^{\infty} \left(\frac{x}{e^{x}}\right)\frac{dx}{1-e^{-2x}}=\frac{2}{a^2}\int_0^{\infty}x\sum_{k=0}^{\infty}e^{-(2k+1)x}\,dx$$
Now, since
$$\int_0^{\infty} xe^{-kx}\,dx=\frac{1}{k^2}$$
We have:
$$\int_... | {
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"question_score": "5",
"answer_count": 5,
"answer_id": 2
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Find the roots of the given equation : $2^{x+2}.3^{\frac{3x}{x-1}} =9$ - Logarithm problem Find the roots of the given equation :
$2^{x+2}.3^{\frac{3x}{x-1}} =9$
My working :
Taking log on both sides we get : $$\log (2^{x+2}.3^{\frac{3x}{x-1}}) =\log 3^2 \Rightarrow (x+2)(\log2) + \frac{3x}{x-1}\log 3 = 2\log 3$$
N... | we could argue that 9 is divisible by 3 and not 2
so wlog
$\large2^{x+2}3^{\frac{3x}{x-1}}=9$
$2^{x+2}$ should be one,so $x=-2$
also
$ \large 3^{\frac{3x}{x-1}}=3^2$
${\frac{3x}{x-1}}=2\implies x=-2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Nonlinear ordinary differential equation (Elsgolts) Please, help me to solve the following non-linear ODEs:
\begin{align}
\tag 1 y &= (y')^4 -(y')^3 -2 \\
\tag 2 y' &= \dfrac{y}{x+ y^3}
\end{align}
Thanks.
| $$ y = y'^4 - y'^3 - 2 $$
$$ y' = p \Rightarrow y = p^4 - p^3 - 2 $$
$$ \Rightarrow y' = p = p^2(4p-3)\frac{dp}{dx} $$
$$ \Rightarrow p(4p-3) \ dp = dx \Rightarrow \int 4p^2 - 3p \ dp = \int dx $$
$$ \Rightarrow \frac{4p^3}{3} - \frac{3p^2}{2} + c_1 = x $$
$$ \Rightarrow \left\{\begin{matrix}
y = p^4 - p^3 - 2 \\
\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculate $ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $ How evaluate the following limit?
$$ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $$
I cannot apply L'Hopital because $ f(x) = 3 -\sqrt{5 -x} \neq 0 $ at $x = 5$
| I think the problem is $$\lim_{x\to4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$$
HINT:
$$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\frac{3^2-(5+x)}{1^2-(5-x)}\cdot\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}=-\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}$$ if $x\ne4$
Now here $x\to4\implies x\ne4$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\forall x \in \Bbb R, 0 \lt \frac{1}{ x^2+6x+10} \le 1$ I am having trouble understanding the meaning of this pictorially.
Do I just have to multiply across the inequality by $x^2+6x+10$ since $x^2+6x+10 \gt 0$ for all real $x$, giving:
$0 \lt1 \le x^2+6x+10$, giving that $0 \lt 1 $ and $x^2+6x+10 \ge 1$? ... | Let $\displaystyle y=\frac1{x^2+6x+10}\ \ \ \ (1)$
On rearrangement, $x^2y+6yx+10y-1=0\ \ \ \ (2)$ which is a Quadratic Equation is $x$
As $x$ is real, the discriminant must be $\ge0$
$\implies (6y)^2-4y(10y-1)\ge0\iff y^2-y\le0\iff y(y-1)\le 0$
$\implies 0\le y\le1$
But if $y=0,(2)\implies -1=0$ for finite $x$ which... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Congruence equation - Find all possible $k$ Let $x,\ y,\ z > 1 \in \mathbb{Z_+}$ and $\gcd(x,y,z)=1$.
Find all possible $k$ such :
$$ \sum_{\mathrm{cyc}}{x}\equiv\sum_{\mathrm{cyc}}{x^2y}\equiv\sum_{\mathrm{cyc}}{xy^2}\equiv 3xyz\equiv 0\pmod{k} $$
PS: It's a part of bigger task (I reduce it by myself).
| First, if $p$ is a prime not equal to 3 that divides $k$, then we have $xyz \equiv 0 \pmod{p}$, so WLOG $x \equiv 0 \pmod{p}$. Then, since $ x + y + z \equiv 0 \pmod{p}$, hence $y = - z \pmod{p}$. Substituting this into the other equation, we get that $ y^3 \equiv 0 \pmod{p}$, so $y \equiv 0 \pmod{p}$. But this gives $... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why does this series converge? My question is:
Why does the series $$ \sum_{j,k=1}^\infty \frac{1}{j^4+k^4} $$ converge?
I tested the convergence with Mathematica and Octave, but I can't find an analytical proof. In fact, numerical computations suggest that the value of the series is $<1$.
One obvious thing to do wo... | Grouping the terms by $i = (j + k)$, we get
$$
\sum_{i = 2}^\infty\:\: \sum_{j, k \gt 0; j + k = i} \frac{1}{j^4 + k^4}
$$
we can estimate that the inner sum, $\sum_{j, k \gt 0; j + k = i} \frac{1}{j^4 + k^4} \leq\frac{i - 1}{(i/2)^4}$, since we have $j^4 + k^4 \geq \left(\frac{j + k}{2}\right)^4 = (i/2)^4$. So we need... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Evaluating an indefinite integral I need help evaluating the following indefinite integral explicitly
$$\int \frac{1}{1+t^{2^{-n}}} dt$$
I would appreciate any help
| $\frac{1}{1+t^{\frac{1}{2^n}}} = \frac{1}{t^{\frac{1}{2^n}}}\frac{1}{1+\frac{1}{t^{\frac{1}{2^n}}}} = \frac{1}{t^{\frac{1}{2^n}}} \sum_{k=0}^\infty (-1)^k ( \frac{1}{t^{\frac{1}{2^n}}})^k = \sum_{k=0}^\infty (-1)^k \frac{1}{t^{{\frac{1}{2^n}}(k+1)}}$, and this convergence is uniform for $t \ge 2$.
Hence
\begin{eqnarra... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding a non-recursive formula for a recursively defined sequence So I have a recursive definition for a sequence, which goes as follows:
$$s_0 = 1$$
$$s_1 = 2$$
$$s_n = 2s_{n-1} - s_{n-2} + 1$$
and I have to prove the following proposition: The $n$th term of the sequence defined above is $s_n = \frac{n(n+1)}{2} + 1$.... | At the end of this answer is a brief description of inverting finite difference operators. In the case here
$$
\begin{align}
\Delta^2s_k
&=s_n-2s_{n-1}+s_{n-2}\\
&=1
\end{align}
$$
Is a second order finite difference operator. Inverting the operator by summing twice, we get that the solution is of the form
$$
s_n=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integrate $\int\frac{1}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}dx$
Solve the indefinite integral
$$
I=\int\frac{1}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}\;dx
$$
My Attempt:
$$
\begin{align}
I&=\int\frac{1}{\sin x+\cos x+\frac{1}{\sin x \cos x}+\frac{\sin x +\cos x}{\sin x\cos x}}\;dx\\
\\
&=\int\frac{\sin x\co... | You want
$\begin{align}
\int\frac{dx}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}
&= \int\frac{dx}{\sin x+\cos x+\frac{\sin x}{\cos x}
+\frac{\cos x}{\sin x}+\frac1{\sin x}+\frac1{\cos x}}\\
&= \int\frac{\sin x \cos x\ dx}{\sin^2 x \cos x+\cos^2 x \sin x+\sin^2 x
+\cos^2 x+\cos x+\sin x}\\
&= \int\frac{\sin x \cos x\ d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/497127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 2
} |
Integrating $\int \cos^3(x)\cos(2x) \, dx$ How would Integrate the following?
$$\int \cos^3(x)\cos(2x) \, dx.$$
I did
$$\int \cos^3(x)(1-2\sin^2(x)) \, dx = 2\int \cos^3(x)-\cos^3x\sin^2x \, dx$$
But I find myself stuck....
| \begin{align}
\int \cos^3(x) \cos(2x) dx &= \int \cos^3(x)(1-2\sin^2(x)) dx \\
&= \int \cos^2(x) \cdot \cos(x) (1-2\sin^2(x)) dx \\
&= \int (1-\sin^2x)\cdot \cos(x) \cdot(1-2\sin^2(x)) dx \\
&= \int ( \cos(x)- 3 \sin^2x \cdot \cos(x) + 2 \sin^4x \cdot \cos(x) ) dx.
\end{align}
Now using inverse chain rule which is
$$\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Radius of convergence of product
Let $\sum_{i=0}^\infty a_nz^n$ and $\sum_{i=0}^\infty b_nz^n$ be power series, and define the product $\sum_{i=0}^\infty c_nz^n$ by $c_n=a_0b_n+a_1b_{n-1}+\ldots+a_nb_0$. Find an example where the first two series has radius of convergence $R$, while the third (the product) has radius ... | A simpler example: let
$$
f(z) = \frac{1+z}{1-z} = \frac{1}{1-z} + \frac{z}{1-z}.
$$
Note that the first term is just the formula for the geometric sum with first term 1, $$\frac{1}{1-z} = 1 + z + z^2 + z^3 + \cdots, \qquad |z| < 1,$$ and the second term is the formula for a geometric sum with first term equal to the ... | {
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"url": "https://math.stackexchange.com/questions/498090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$.
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$
By trying out different values of $n$, I found t... | The matrix
$$N=\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$$
is nilpotent with index 2 of nilpotency: $N^2=0$ so by the binomial formula we have
$$\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}^n=(I_2+N)^n=\sum_{k=0}^n {n\choose k}N^k={n\choose 0}I_2+{n\choose 1}N=I_2+nN=\begin{pmatrix} 1& n\\ 0 & 1 \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 3
} |
Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$
Problem:Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ where $a,b,c$ are distinct real numbers
Solution:$(x-a)^3+(x-b)^3+(x-c)^3=0$
$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-a^3-b^3-c^3=0$
By Descartes rule of sign,number of positive ... | Set $m=(a+b+c)/3$, $A=a-m$, $B=b-m$, $C=c-m$ and $x=y+m$. Then your equation becomes
$$
(y-A)^3 + (y-B)^3 + (y-C)^3 = 0
$$
and, since $A+B+C=0$, your expansion applies to give
$$
y^3+(A^2+B^2+C^2)-\frac{A^3+B^3+C^3}{3}=0
$$
which is a suppressed cubic, whose discriminant is
$$
\frac{1}{4}\biggl(-\frac{A^3+B^3+C^3}{3}\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500132",
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"source": "stackexchange",
"question_score": "2",
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How to prove: $a+b+c\le a^2+b^2+c^2$, if $abc=1$? Let $a,b,c \in \mathbb{R}$, and $abc=1$. What is the simple(st) way to prove inequality
$$
a+b+c \le a^2+b^2+c^2.
$$
(Of course, it can be generalized to $n$ variables).
| A).
If values $~a,b,c~$ are positive, then
denote $m = \dfrac{a+b+c}{3},~$ ($m\ge \sqrt[3]{abc}=1$, AM-GM);
$\left\{\begin{array}{l}
a=m+\alpha,\\
b=m+\beta, ~~~~~~(\alpha+\beta+\gamma=0).\\
c=m+\gamma.
\end{array}\right.~~~
$
Then
$$
a^2+b^2+c^2 ~=~ 3m^2 + \alpha^2+\beta^2+\gamma^2 ~\ge~ 3m ~=~ a+b+c.
$$
B). I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
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A question by Ramanujan about a relational expression of a triangle I found the following question in a book without any proof:
Question : Suppose that each length of three edges of a triangle $ABC$ are $BC=a, CA=b, AB=c$ respectively. If
$$\frac1a=\frac1b+\frac1c, \frac2a=\frac{1}{c-b}-\frac{1}{c+b},$$
then prove
$$... | $$\begin{align}\text{Partial answer:}\qquad\qquad\quad\frac1a=\ \frac1b+\frac1c\quad\iff\quad&a=\frac1{\frac1b+\frac1c}=\frac{bc}{b+c}\qquad\qquad(1)\\\\\\\\\\\frac2a=\frac1{c-b}-\frac1{c+b}\quad\iff\quad\frac2a=\frac{2b}{c^2-b^2}\quad\iff\quad&a=\frac{c^2-b^2}b\ \qquad\qquad\qquad\quad\ (2)\end{align}$$
$$\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to solve equation? How to solve this equation in the set of real numbers?
$$(x^{2}+3y^{2}-7)^{2} + \sqrt{3-xy-y^2}=0$$
I tried to solve $x^{2}+3y^{2}-7=0$ and $\sqrt{3-xy-y^2}$=0 for x. But it did not help.
| $(x^2 + 3y^2 - 7)^2=0$ and $3-xy-y^2=0$
$x^2 + 3y^2 - 7=0$ and $3-xy-y^2=0$
$x=\pm (7-3y^2)^2$... So you have two cases:
You put $x=(7-3y^2)$ in the equation $3-xy-y^2=0$ and you solve
You put $x=-(7-3y^2)$ in the equation $3-xy-y^2=0$ and you solve . . .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/501399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Approaching modular arithmetic problems I'm a little stumbled on two questions.
How do I approach a problem like $x*41 \equiv 1 \pmod{99}$.
And given $2$ modulo, $7x+9y \equiv 0 \pmod{31}$ and $2x−5y \equiv 2 \pmod{31}$ (solve for $x$ only)?
When I solve for $x$ for the latter, I got a fraction as the answer and I'm n... | Finding the solution to $$x \times 41 \equiv 1 \pmod {99}$$ is equivalent to asking for the multiplicative inverse of $41$ modulo $99$. Since $\gcd(99,41)=1$, we know $41$ actually has an inverse, and it can be found using the Extended Euclidean Algorithm:
\begin{align*}
99-2 \times 41 &= 17 \\
41-2 \times 17 &= 7 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is the smallest value of $x^2+y^2$ when $x+y=6$? If $ x+y=6 $ then what is the smallest possible value for $x^2+y^2$?
Please show me the working to show where I am going wrong!
Cheers
| We have
$$2(x^2+y^2)=(x+y)^2 +(x-y)^2=36+(x-y)^2.$$
But $36+(x-y)^2$ is smallest when $x=y$. Thus the minimum value of $2(x^2+y^2)$ is $36$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 4
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How to integrate a function that has no exact integration and cannot be expanded by a Taylor series Today I posted a question about the integral
$$\int\frac{(\cos{c x})^2}{a+bx}dx$$
for which a Taylor series can be built and the integral solved for the desired approximation.
Another term in the stiffness matrix has the... | Maple finds a closed form for your antiderivative:
$$ 1/4\,{\frac {x\sin \left( 2\,cx \right) }{bc}}+1/8\,{\frac {\cos
\left( 2\,cx \right) }{{c}^{2}b}}-1/4\,{\frac {a\sin \left( 2\,cx
\right) }{c{b}^{2}}}+1/2\,{a}^{2}{\it Si} \left( 2\,cx+2\,{\frac {ac}
{b}} \right) \sin \left( 2\,{\frac {ac}{b}} \right) {b}^{-3}\\+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/504792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do you factor this? $x^3 + x - 2$ How do you factor $x^3 + x - 2$?
Hint: Write it as $(x^3-x^2+x^2-x+2x-2)$ to get $(x-1)(x^2+x+2)$
Note the factored form here. Thanks!
| By inspection, we see that $1$ is a root of $x^3 + x - 2$, so it is divisible by $x - 1$; alternatively, the rational roots theorem would suggest this too.
Now $x^2 + x + 2 = x^2 + x + \frac{1}{4} + \frac{7}{4} = (x + \frac{1}{2})^2 + \frac{7}{4}$ has no real roots, and is irreducible. If you're factoring over $\Bbb{C}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ without using L'Hopital $$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$
I tried using $\lim\limits_{x\to0} \frac{\sin x}x=1$.
But it doesn't work :/
| $$
L=\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=\lim_{x\to0}\frac{x-\sin x}{x\cos x-\sin x}\cos x = \lim_{x\to0}\frac{2x-\sin2x}{2x\cos2x-\sin2x}\cos2x\\
= \lim_{x\to0}\frac{x-\cos x\sin x}{x(1-2\sin^2x)-\cos x\sin x}\cos2x=\lim_{x\to0}\frac{x-\cos x\sin x}{x-\cos x\sin x-2x\sin^2x}\cos2x
$$
Which, noting that $\lim_{x\to0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
How to solve inequalities with infinite terms Consider the following inequality:
$x + 2 < 1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} ... $ with $x>0$.
Is there a general way to solve such an inequality with infinite terms?
The best I can do is some conjectures:
For $x = 2$ the right hand side equals 2, so I kno... | If $|\frac1x|<1 \iff |x|>1$
$$1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} ... =\frac1{1-\frac1x}=\frac x{x-1}$$
So, we need $\displaystyle x+2<\frac x{x-1}$
Multiplying either sides by $(x-1)^2,$
$\displaystyle\iff (x+2)(x-1)^2< x(x-1)$
If $x>1,$ we have $(x+2)(x-1)<x\iff x^2-2<0\iff -\sqrt2<x<\sqrt2$
$\implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How prove this $a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$ let $a,b,c>0$, show that
$$a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$$
I know this
$$a+b+c\ge 3\sqrt[3]{abc}$$
so
$$\Longleftrightarrow 6\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$$
But this maybe not true?
| WLOG suppose that $a\geq b\geq c$.
First see that using weighted AG-inequality, we have:
$$
c+3\sqrt[3]{abc}\geq 4\sqrt c \sqrt[4]{ab} (*)
$$
Then we have:
$$
a+b+c+3\sqrt[3]{abc}- 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=\\
(\sqrt{a}+\sqrt{b}-\sqrt{c})^2-4\sqrt{ab}+3\sqrt[3]{abc}\\
\geq (2\sqrt[4]{ab}-\sqrt{c})^2-4\sqrt{ab}+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/510104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Let P be a moving point such that if $PA$ and $PB$ are two tangents drawn from $P$ to the circle $x^2+y^2=1 ( $ A ,B being the points of contact) ,... Problem :
Let $P$ be a moving point such that if $PA$ and $PB$ are two tangents drawn from $P$ to the circle $x^2+y^2=1$ ( $A$, $B$ being the points of contact) , then ... | First note that the quadrilaterial $PAOB$ is a kite, this follows from the fact that the tangent from one point to a circle are from a same length, and the $AO=OB=r$.
From the fact that $\angle AOB = 60^{\circ}$ and using $AO=OB$ we get that the $\triangle AOB$ is equilaterial. This leads to conclusion $AB=r=1$, which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/510171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
integral of $\int\frac{2\sin(2x)-\cos(x)}{6-\cos^2x -4\sin x}\mathrm{d}x$ so i got this problem $$\int\dfrac{2\sin 2x-\cos x}{6-\cos^2x -4\sin x}\mathrm{d}x$$
now this is what i tried $=\int\dfrac{4\sin(x)\cos(x)-\cos(x)}{6-(1-\sin^2x) -4\sin x}\mathrm{d}x=\int\dfrac{\cos(x)(4\sin(x)-1)}{5+\sin^2x-4\sin x}\mathrm{d}x$ ... | $\large Hint:$ $u \equiv \sin\left(x\right)$.
\begin{align}
&
\int
{2\sin\left(2x\right) - \cos\left(x\right)
\over
6 - \cos^{2}\left(x\right) - 4\sin{x}}\,{\rm d}x
=
\int
{4\sin\left(x\right) - 1
\over
\sin^{2}\left(x\right) - 4\sin{x} + 5}\,\cos\left(x\right)\,{\rm d}x
\\[3mm]&=
\int{4u - 1 \over u^{2} - 4u + 5}\,{\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Derivation of the quadratic equation So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
| $ax^2+bx+c=0 (a\neq0)$
$x^2+\frac{b}{a}x+\frac{c}{a}=0$
$x^2+\frac{b}{a}x=-\frac{c}{a}$
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}$
$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$
$(x+\frac{b}{2a})=\pm\frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}$
$\therefore x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Definite integrals: Evaluate the integral
Evaluate:
$$ \int ^{1/2}_{1/4} \frac{dx}{ \sqrt{x-x^2}}dx$$
can u help me with this?
What is meant by the dx in the numerator?
EDIT:
ANSWER AS GIVEN IN THE BOOK
$$ \int ^{1/2}_{1/4} \frac{dx}{ \sqrt{x-x^2}}dx$$
$$=\int ^{1/2}_{1/4} \frac{1}{\sqrt {-(x^2-x+\frac{1}{4}- \frac... | $$I=\int_{\frac14}^{\frac12}\frac{dx}{\sqrt{\left(\frac12\right)^2-\left(x-\frac12\right)^2}} =\int_{\frac14}^{\frac12}\frac{2dx}{\sqrt{1^2-(2x-1)^2}}$$
Put $2x-1=\sin y\implies 2dx=\cos y dy,$
When $x=\frac12,\sin y=0\implies y=0$ and $x=\frac14,\sin y=-\frac12\implies y=-\frac\pi6$
$$\implies I=\int_{-\frac\pi6}^0\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/516181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\frac{a+b+c}{x+y+z}$ given $a^2+b^2+c^2$, $x^2+y^2+z^2$ and $ax+by+cz$. We are given $a^2+b^2+c^2=m$, $x^2+y^2+z^2=n$ and $ax+by+cz=p$ where $m,n$ and $p$ are known constants. Also, $a,b,c,x,y,z$ are non-negative numbers.
The question asks to find the value of $\dfrac{a+b+c}{x+y+z}$.
I have thought a lot about th... | I don't think this can be solved. Think of $A=(a, b, c)$ and $X=(x, y, z)$ as vectors. Then we know $A\cdot A$, $X\cdot X$ and $A\cdot X$. So all we know is the lengths of $A$ and $X$, and the angle between them. Now let $C=(1,1,1)$. What we want is
$$\frac{A\cdot C}{X\cdot C}$$
But if we hold $A$ and $C$ fixed we can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/518225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
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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.