Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Determinants using elementary row operations Let matrix $A$ be defined as \begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \vdots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{pmatrix}
And $det(A)=d$
Let $B=$
\begin{pmatrix}
2a_{11} & 2a_{12} & \cdot... | $$\begin{pmatrix}
2a_{11} & 2a_{12} & \cdots & 2a_{1n} \\
2a_{21} & 2a_{22} & \cdots & 2a_{2n} \\
\vdots & \vdots & \vdots & \vdots \\
2a_{n1} & 2a_{n2} & \cdots & 2a_{nn}\\
\end{pmatrix} +
\begin{pmatrix}
3a_{11} & 3a_{12} & \cdots & 3a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \vdots & \vd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1314087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Expansion of Generating Functions If you roll $10$ dice, how many ways can you get a total sum of top faces of $25$?
I understand how to write the generating function of $(x+x^2+ \dots +x^6)^{10}$ and the fact that you need to find the coefficient of $x^{25}$, but how do you do this? The conventional binomial theorem o... | I should mention, this answer assumes the dice are distinct. (e.g. you have a red die, a blue die, a green die, etc... or a "first die" a "second die" a "third die", etc...)
We know that each result on the die is minimum 1, so let us ask a similar question instead to simplify the arithmetic involved: What is the coef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim\limits_{x\to\ 0}(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)})$
Evaluate $$\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)}\right)$$
It is easy with L'Hospital's rule, but takes too much time to calculate derivatives. What are some shorter methods?
| Here is a solution to the EDITED problem asked by OP. I will make use of
some standard limits only, without l'Hospital rule nor Taylor series.
First, some simple transformations are required:
\begin{eqnarray*}
\frac{1}{\sin x\arctan x}-\frac{1}{\tan x\arcsin x} &=&\frac{1}{\sin
x\arctan x}-\frac{\cos x}{\sin x\arcsin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
with this inequality $\ln{x}\ln{(1-x)}<\sqrt{x(1-x)}$
If $0<x<1$, show that
$$\ln{x}\ln{(1-x)}<\sqrt{x(1-x)}$$
use derivative it's not easy, such
$$
f(x)=(\ln{x}\ln{(1-x)})^2-x(1-x),
$$
$$
f'(x)=2x-1+\dfrac{2\ln{x}\ln^2{(1-x)}}{x}+\dfrac{2\ln^2{x}\ln{(1-x)}}{x-1}
$$
and we $f(x)=f(1-x)$,then we prove inequality ho... | Let
\begin{equation*}
f(x)=\ln x\ln (1-x)-\sqrt{x(1-x)},\ for\ 0<x\leq \frac{1}{2}.
\end{equation*}
\begin{eqnarray*}
f^{\prime }(x) &=&\frac{\ln (1-x)}{x}-\frac{\ln x}{1-x}+\frac{x-\frac{1}{2}}{%
\sqrt{x(1-x)}} \\
&=&\frac{(1-x)\ln (1-x)}{x(1-x)}-\frac{x\ln x}{x(1-x)}+\frac{(x-\frac{1}{2})%
\sqrt{x(1-x)}}{x(1-x)} \\
f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
If $f(x) = \frac{\sin^{-1} x}{\sqrt{1- x ^2}}$, then evaluate $(1-x^2)f''(x) - xf(x)$
$f(x) = \dfrac{\sin^{-1} x}{\sqrt{1- x ^2}}$
Differentiating the given function, we get
$f'(x) = \dfrac{1 + \dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}}{1-x^2}$
which can also be written as
$f'(x) = \dfrac{1 + xf(x)}{1-x^2}$
Differentiatin... | we will differentiate using the implicit definition of $y.$ we have $$y = \frac{\sin^{-1} x}{\sqrt {1- x^2}} \to y^2(1-x^2) = (\sin^{-1} x)^2 \tag 1$$ differentiating once, we have $$ -2xy^2 +2yy'(1-x^2) =\frac 2{\sqrt{1-x^2}}\sin^{-1} x = 2y$$ diving out by $y, $ you get $$1+xy =y'(1-x^2) \tag 2 $$ differentiating ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find this closed form $\sum_{k=1}^{n}\left(\lfloor a_{k}\rfloor +\lfloor a_{k}+\frac{1}{2}\rfloor \right)$ Let $\dfrac{1}{a_{k}}=\dfrac{1}{k^2}+\dfrac{1}{k^2+1}+\cdots+\dfrac{1}{(k+1)^2-1}$
I need some ideas to exploit for finding the closed form of
$$\sum_{k=1}^{n}\left(\lfloor a_{k}\rfloor +\lfloor a_{k}+\dfrac{1}{2}... | Let
$$b_{n}=\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+n-1}+\dfrac{1}{n^2+n}+\dfrac{1}{n^2+n+1}+\cdots+\dfrac{1}{n^2+2n}$$
so
$$\dfrac{2}{n+1}<\dfrac{1}{n}+\dfrac{1}{n+2}=\dfrac{n+1}{n^2+n}+\dfrac{n}{n^2+2n}<b_{n}<\dfrac{n}{n^2}+\dfrac{n+1}{n^2+n}=\dfrac{2}{n}$$
so we have
$$\dfrac{2}{k+1}<\dfrac{1}{a_{k}}<\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do you solve the summation of $2-4+8-16+32- \dots 2^{48}$? This is a summation problem but I can't seem to figure out how to solve this with the mix of subtraction and addition.
| $$S=2-4+8-16+32- \dots -2^{48} = (-1)(-2+4-8+16-32+ \dots +2^{48}) $$
$$ = (-1)\left((-2)^1+(-2)^2+(-2)^3+ \dots +(-2)^{48}\right)$$
which is a geometric series with first term $a=-2$ and common ratio $r=-2$ and $n=48$ terms. $S$ then becomes
$$S = (-1)\times \frac{a(1-r^{n+1})}{1-r} = -1\times \frac{-2(1-(-2)^{49})}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
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What is wrong with this integration of $ \int_0^{2\pi}\sin x /(1 + A \sin x)$ A worked solution of the integral has been provided as an answer to a previous question.
But I am still unclear why I get the wrong answer from the following method which uses a formula for the definite integral provided by Wolfram Alpha.
Wol... | Using the guidance provided by Chappers and Hans I have come up with this approach.
The Wolfram Alpha solution is:-
$$
\int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx
=
(1/A)\left( x - \frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} + C\right)^{2\pi}_0
$$
The antiderivative (on th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove (or provide a counterexample): no pair of primitive Pythagorean triples (a,b,c) and (2a,k,c) exists. A primitive Pythagorean triple is an ordered set of coprime integers (a,b,c) such that $a^2+b^2=c^2$. Show that the system of Diophantine equations
$$a^2+b^2=c^2$$
$$4a^2+k^2=c^2$$
have no solutions.
| $$c^2=a^2+b^2=4a^2+k^2$$
$$k^2+3a^2=b^2$$
All solutions of this Diophantine equation are defined parameterization.
$$a=2ps$$
$$k=p^2-3s^2$$
$$b=p^2+3s^2$$
Then the amount of;
$$a^2+b^2=4p^2s^2+p^4+6p^2s^2+9s^4=p^4+10p^2s^2+9s^4$$
A square can not be.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Integral using contour integration Here is the integral I want to evaluate:
$$\int_{0}^{2\pi} \frac{dx}{a+b \cos x }, \quad a>b >0$$
Apparently there are limitations as to what values $a, b$ are supposed to take but let us not concern about this. Since using the sub $u =\tan \frac{x}{2}$ (the Weiersstrass sub) results ... | $$\frac 1i \frac{2}{2 a z+b z^2+b}$$
This has a residue at
$$\frac{\sqrt{a^2-b^2}-a}{b}$$
Which is evaluated as
$$\frac{1}{i\sqrt{a^2-b^2}}$$
And the final answer is
$$\frac{2 \pi}{\sqrt{a^2-b^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
inequalities with fraction problem x $\frac{1}{x} > \frac{2x} {x^2 +2}$ solving this inequalities:
My long solution (wrong) :
multiplying $(x^2 + 2)^2 (x)^2 \dots$ (multiplying square of each denominator, getting rid of the > or < 0)
$x (x^2 +2)^2 > 2x(x^2) ( x^2 +2)$
$x(x^4+4x^2 +4)>2x^2(x^2+2)$
$x^5+4x^3+4x>2x^4+4x^2... | As both sides of the inequation have the same sign, it is equivalent to:
$$\frac{x^2+2}{x}=x + \frac2x> 2x\iff x<\frac2 x\iff\begin{cases}x^2<2&\text{if}\enspace x>0,\\x^2>2&\text{if}\enspace x<0\end{cases}$$
Thus the solutions are $\,(0< x<\sqrt 2)\,$ or $\,(x<-\sqrt 2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Asymptotic expansion computation I found a paper in which the following expression
$$\log\left(1 - \frac{1}{x}\right) + \frac{1}{x} + \frac{A}{x^2}+\epsilon x = 0$$
Where $\epsilon$ is a constant of order $10^{-2}$ to $10^{-5}$, $A$ is a constant of order unity, is approximated for $\frac{1}{x}<<1$ as
$$\epsilon x^4 +... | $$\begin{align}0 &=\ln\left(1 - \frac{1}{x}\right) + \frac{1}{x} + \frac{A}{x^2}+\epsilon x \\
&= -\frac1x - \frac1{2x^2}-\frac1{3x^3} +\cdots+\frac1x+\frac a{x^2} + \epsilon x \\
&=\epsilon x^4+(a-1/2)x -1/3+\cdots \end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim_{x\to 0}\frac{\sqrt[m]{\cos\alpha x}-\sqrt[m]{\cos\beta x}}{\sin^2x},m\in \mathbb{N}$
Evaluate $$\lim_{x\to 0}\frac{\sqrt[m]{\cos\alpha x}-\sqrt[m]{\cos\beta x}}{\sin^2x},m\in \mathbb{N}$$
I used L'Hospital's rule, but that didn't work. Could Taylor series be used? I don't know how to use it with irrat... | This can be done via the use of following standard limits $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1},\,\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2},\,\lim_{x \to 0}\frac{\sin x}{x} = 1$$ Let us proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt[m]{\cos \alpha x} - \sqrt[m]{\cos \beta ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Solution of a quartic equation. Suppose that the equation $x^4-2x^3+4x^2+6x-21=0$ is known to have two roots that are equal in magnitude but opposite in sign. Solve the equation.
This is what I have been thinking. Suppose $\zeta_1$ $\zeta_2$ are roots. Such that $|\zeta_1|=|\zeta_2|$. Then $(x-\zeta_1)(x-\zeta_2)$ div... | Let $P(x)= x^4−2x^3+4x^2+6x−21=0$ Let $a,-a $ are the two roots given. So, $P(a)=P(-a)=0$
\begin{align}
a^4−2a^3+4a^2+6a−21 &= &a^4+2a^3+4a^2-6a−21\\
\implies 4a^3-12a =0\\
\implies a(a^2-3)=0\\
a=0,a=\pm\sqrt3
\end{align}
But $a=0$ is not a root of the polynomial. Hence $\pm \sqrt3$ are the two roots of $P(x)$ becaus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Solve $x^8 \equiv 3 \pmod {13}$ I need to find all solutions to $x^8 \equiv 3 \pmod {13}$.
What I've tried: I know $2$ is a primitive root modulo $13$.
So it is equivalent to solve
$2^{8t} \equiv 2^4 \pmod {13}$
Then I get $t = 2 + 3k$.
I think I'm wrong... and if not, what are the final solutions??
| In $\mathbb{F}_{13}$,
$$ x^8-3 = x^8-16 = (x^4-4)(x^4+4)=(x^2-2)(x^2+2)(x^2+2x+2)(x^2-2x+2) $$
hence we just have to check which of the numbers $-2,2,-1$ are quadratic residues. Since $13$ is a prime of the form $8k+5$, the only quadratic residue among the previous ones is $-1$, so $x^8-3$ splits as:
$$ x^8-3 = (x-4)(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Could translate/explain this for me? I have this problem:
$$
10x^2 - 7x - 12 = 0
$$
And apparently the method to factoring it is to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the constant term, and whose sum is the same as the coefficient of $x$.
However, I can barely u... |
to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the constant term, and whose sum is the same as the coefficient of $x$.
Take the smallest, innermost noun phrases first:
to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the consta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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How to derive FWHM of sinc function So this is probably a simple question, but I am unable to get my head around it. If we have $\operatorname{sinc}(2 \pi v L)$, what is the width of that $\operatorname{sinc}$ in terms of $v$ at half the maximum point.
Normally I'd just say that the maximum is $1$, thus half point is..... | I cannot resist the pleasure of providing an approximate solution for $Lv \leq 1$. This approximation was made by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician. $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ So $$\frac{\sin(x)}x\simeq \frac{16 (\pi -x) }{5 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplify a Combinatorial Sum $\sum_{k=0}^\infty {a\choose k}{b\choose c-k}{d-k\choose e}$ Is there a way to simplify
$$\sum_{k=0}^\infty {a\choose k}{b\choose c-k}{d-k\choose e}$$
where $a,b,c,d,e$ are natural numbers?
In particular, I would like to see the case for $a=45, b=3,c=4,d=48,e=4$.
| The series can be cast into hypergeometric form and is given by
\begin{align}
\sum_{k=0}^\infty \binom{a}{k} \binom{b}{c-k} \binom{d-k}{e} = \binom{b}{c} \binom{d}{e} \, {}_{3}F_{2}(-a, -c, e-d; -d, b-c+1; 1)
\end{align}
Further reduction may be possible depending on the values of $\{a,b,c,d,e\}$.
Proof of result
\beg... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the volume formed by rotating the region bounded by $y = e^{-x} \sin x$, $x\ge 0$ about $y =0$.
Find the volume formed by rotating the region bounded by $y = e^{-x} \sin x$, $x\ge 0$ about $y =0$.
I tried to graph this using Wolfram Alpha, but it didn't help. I don't know how to start or graph this.
| $$f(x) = e^{-x} \sin x$$
Your volume can be found using cylindrical coordinates :
\begin{align}
\mathcal{V}\equiv\int\limits_0^\infty \int\limits_0^{2\pi}\int\limits_0^{f(x)} r dr d\theta dx &= 2\pi \int\limits_0^\infty\int\limits_0^{f(x)} r dr dx\\
&=\pi \int\limits_0^\infty f^2(x) dx =\pi \int\limits_0^\infty e^{-2x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Checking if a matrix is in the span of other matrices Problem: Expand the following set matrices \begin{align*} \left\{ \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix}, \begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix}, \begin{pmatrix} 0 & -3 \\ 5 & 4 \end{pmatrix}\right\} \end{align*} to a basis of $\mathbb{R}^{2 \times 2}$... | Instead of a blind guess you could check which standard basis element is not in the span of your set. (You should also check that your set is linearly independent if it is not already assumed so).
i.e. check that these systems have solutions
\begin{align*} \begin{cases} a + 2b &= 1 \\ -a + b -c &= 0 \\ 2a - b + 5c &= 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Derivative of the function $y = 2^{\sqrt{\tan x}}$ How to find derivative of the following function: $y = 2^{\sqrt{\tan x}}$ , $y' = ?$ I did the following $$\frac{d}{dx}2^{\sqrt{\tan x}} = 2^{\sqrt{\tan x}}\ln{2}(\sqrt{\tan}x)'$$ and stopped here. Can you guide me?
The solution is as follows in the book:
$$2^{\sqrt... | If $y = 2^{\sqrt{\tan x}}$ then $\displaystyle y = e^{\sqrt{\tan x} \ln 2}$. So, using the chain rule yields $$y' = \frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt{\tan x} \cdot \ln 2\right) \cdot e^{\sqrt{\tan x} \ln 2}$$
But, using the chain rule again yields $$\frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt{\tan x} \cdot \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How do i find $\tan(\theta)$ such that : $\frac{16}{\sin^6(\theta)} + \frac{81}{\cos^6(\theta)}=625$?? How do i find $\tan(\theta)$ such that :$$\frac{16}{\sin^6(\theta)} + \frac{81}{\cos^6(\theta)}=625$$?
Note : i used some trigono-form but sorry i didn't succed .
Thank you for any help.
| I cannot say that this is an elegant solution but oh well,
i will use $s$ for $sin$, $c$ for $cos$ and $t$ for tan
we know that $\frac{s}{c} = t$
so $\frac{16}{c^{6}*t^{6}} + \frac{81}{c^{6}} = 625$
$16 + 81t^{6} = 625c^{6}*t^{6}$
we know that $ 1 + t^{2} = \frac{1}{c^{2}}$
or $16 + 81t^{6} = \frac{625*t^{6}}{(t^{2} +1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How do I solve this trigonometric equation?
Solve the equation $$\sin\left(\frac{3\pi}{10}-\frac{x}{2}\right)=\frac{1}{2}\sin\left(\frac{\pi}{10}+\frac{3x}{2}\right).$$
I tried applying some Ratio properties, but they just made the equation nasty. Some hints please. Thanks.
| HINT:
Method $\#1:$
Let $\dfrac{3\pi}{10}-\dfrac x2=y\iff\dfrac x2=\dfrac{3\pi}{10}-y$
$\implies\dfrac\pi{10}+\dfrac{3x}2=\dfrac\pi{10}+3\left(\dfrac{3\pi}{10}-y\right)=\pi-3y$
So, we have $\sin y=\dfrac12\sin(\pi-3y)\iff2\sin y=\sin3y$
Now use $\sin3y=3\sin y-4\sin^3y$ and $\cos2B=1-2\sin^2B$
Method $\#2:$
As $\sin(18... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculate Infinite Limit I'm trying to calculate the limit and when I get to the last step I plug in infinity for $\frac 8x$ and that divided by -4 I get - infinity for my answer but the book says 0. Where did I go wrong?
$$
\frac {8x^3-x^2}{7+11x-4x^4}
$$
Divide everything by $x^4$
$$
\frac {\frac{8x^3}{x^4}-\frac{x^2... | Hint
When you have expressions which are ratios of polynomials and that $x\to \infty$, the best is to factor the highest powers in numerator and denominator. So, in your case $$A=\frac {8x^3-x^2}{7+11x-4x^4}=\frac{x^3\big(8-\frac 1x\big)}{x^4\big(\frac 7{x^4}+\frac {11}{x^3}-4\big)}$$ So, since $x$ is large, we can "ig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to prove that $7^{31} > 8^{29}$ How can I prove that $7^{31}$ is bigger than $8^{29}$?
I tried to write exponents as multiplication, $2\cdot 15 + 1$, and $2\cdot 14+1$, then to write this inequality as $7^{2\cdot 15}\cdot 7 > 8^{2\cdot 14}\cdot 8$. I also tried to write the right hand side as $\frac{8^{31}}{8^2}$.
| First, $7^5=16807$ and $2^{14}=16384$. And we have $\dfrac{7^{31}}{8^{29}}=\left(\dfrac{7^5}{2^{14}}\right)^6\times\dfrac78$.
Now, for $x\ge0$ and integer $n\ge0$, we have $(1+x)^n\ge1+nx$, hence
$$\left(\frac{7^5}{2^{14}}\right)^6\ge1+6\left(\frac{7^5}{2^{14}}-1\right)=1+\frac{6\times423}{2^{14}}=1+\frac{3\times423}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
Solving Riccati equation $v'=av^2+bv+c$ I am trying to solve Riccati equation
$$
v(t)'=av(t)^2+bv(t)+c
$$
where $a$,$b$,$c$ are real-valued constants. Wolfram|Alpha gives the solution (here), but I am not able to reproduce the result.
Thank you in advance.
| $$v(t)'=av(t)^2+bv(t)+c\Longleftrightarrow$$
$$\frac{dv(t)}{dt}=av(t)^2+bv(t)+c\Longleftrightarrow$$
$$\frac{\frac{dv(t)}{dt}}{av(t)^2+bv(t)+c}=1\Longleftrightarrow$$
$$\int\left(\frac{\frac{dv(t)}{dt}}{av(t)^2+bv(t)+c}\right)dt=\int 1dt\Longleftrightarrow$$
$$\frac{2\tan^{-1}\left(\frac{b+2av(t)}{\sqrt{-b^2+4ac}}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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A golden trigonometric diophantine equation After answering this question I reflected on the identity $$\cos\frac{\pi}{5}=\phi\cos\frac{\pi}{3}$$ and thought of looking for all the quadruplets of positive integers $(a,b,c,d)$ satisfying $$\cos \frac{a}{b}\pi=\phi\cos\frac{c}{d}\pi,\tag{1}$$ where both fractions are in ... | Suppose you put $a/b$ and $c/d$ in terms of a common denominator as $m/n$ and $k/n$, and let $\omega = \exp(\pi i/n)$, so
$$\eqalign{\cos(a\pi/b) &= \cos(m \pi/n) = (\omega^m + \omega^{-m})/2\cr
\cos(c\pi/d) &= \cos(k \pi/n) = (\omega^k + \omega^{-k})/2\cr}$$
The equation $\phi^2 = \phi + 1$ then becomes $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $n = 4k + 1$, does $4$ divide $n^2 -1$? How would I show that $4$ divides $n^2 -1$ if $n = 4k+1$?
Is there more than one way to solve this?
| For $k = 0$, $n = 1$, $n^2-1 = 0 = 4\cdot 0$.
Assume statement is true for some $k = h$. Let $(4h+1)^2-1 = 4a$.
For $k = h+1$,
$$\begin{align*}
(4h+5)^2 - 1 &= (4 + 4h+1)^2 - 1\\
&= 4^2 + 2\cdot4(4h+1) +(4h+1)^2-1\\
&= 4^2 + 2\cdot4(4h+1) + 4a
\end{align*}$$
which is divisible by $4$.
Similarly for $k=h-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
weighing avg by columns vs rows outputs different results First of all, I just know the basics of math, so please be patient.
I have an overall score for a company with the product1 and product2. However when I do the overall for each criteria like A, B, C, D, E, the sum of these criteria is not equal to the overall s... | Your $2.8$ is not correct. If you do the weighted average of $1,5\ \ 2 \ \ 3 \ \ 3,5\ \ 4$ with the weights you are using you get $3.0333333$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proof of $\sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}$ The title pretty much summarizes my question. I am trying to prove the following:
$$\displaystyle \forall N \in \mathbb{N}: \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}.$$
I tried proving this using induction. Startin... | you may find it easier to note that
$$
S=\sum_{n=1}^{2N} (-1)^{n-1}\frac1{n} = \sum_{n=1}^{N} \left(\frac1{2n-1} - \frac1{2n} \right)
$$
for then:
$$
S+\sum_{n=1}^{N} \frac1{n} = \sum_{n=1}^{2N} \frac1{n} = \sum_{n=1}^{N} \left( \frac1{n}+\frac1{N+n} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Is this matrix diagonalizable over $\mathbb{R}$ or $\mathbb{C}$? Problem: Let $A = \begin{pmatrix} 6 & 0 \\ -2 & 2 \end{pmatrix}$. Is this matrix diagonalizable over $\mathbb{R}$? If not, is it diagonalizable over $\mathbb{C}$? Compute the eigenvalues $\lambda$ and the corresponding eigenspaces $E_{\lambda}$.
Attempt a... | You miscalculated the determinant. You should have ended up with the characteristic polynomial
$$
p(x) = (x-6)(x-2)
$$
check your work, note that $0 \times (-2) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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3 Questions on number theory. We have to everything using number congruences, and I am just a beginner, I know a few theorems, and we have to solve these using basics.
1) If $n=a^4$ where $a \in \mathbb Z$ then prove that $n \equiv 0,1,5$ or $6(mod \ 10)$.
My work: $a^4 \equiv x(mod \ 10)$
We have to find $x$.
But ho... | Question 1: Because we can write any number $a \in \mathbb{Z}$ as $a=10\cdot b + c$ with $b,c \in \mathbb{N}$ we see that $a^4 = (10 \cdot b + c)^4 \mod 10$. It is easy to see that this forms a polynomial with all terms except for the $c^4$ part is a multiple of $10$. Since we are working in $\mod 10$ we see that $a^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find numbers $\overline{abcd}$ so that $\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}$ Find the numbers $\overline{abcd}$, with digits not null that satisfy the equality
\begin{equation}\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}\end{equation}
where
\begin{equation}\overline{abc... | As you observed, $$(d,a) \in \{(5,1), (8,3), (9,7), (6, 5)\}$$ but notice also that $d = a+k_3$, where $k_3$ is the carry resulting from $bcd+cd+d+1$. This quantity is at most $999+99+9+1 = 1108$ so the carry is either $0$ or $1$ and thus $d = a$ or $d=a+1$. Combined with the foregoing, this means that $(d,a) = (6, 5)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Using equation to find value of $1/x - 1/y$ $$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
What is the value of $\frac{1}{x}-\frac{1}{y}$?
I have already used that when $48$ divided by $10$ then it becomes $4.8$ and when $8$ divided by $10$ then it becomes $0.8$ by getting $10^x$ and $10^y$ to make i... | $$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
$$48^x = 10^{x+3} \; \; \; \& \; \; \; 8^y = 10^{y+3}$$
$$x+3 = x \log {48} \Rightarrow \; \; \; 1 + \frac{3}{x} = \log {48}$$
$$y+3 = y \log {8} \Rightarrow \; \; \; 1 + \frac{3}{y} = \log {8}$$
$$3 \times (\frac{1}{x} - \frac{1}{y}) = \lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Limits. Finding positive value Find a real number k such that the limit $$\lim_{n\to\infty}\ \left(\frac{1^4 + 2^4 + 3^4 +....+ n^4}{n^k}\right)$$ has as positive value.
If I am not mistaken every even $k$ can be the answer. But the answer is 5.
| Do you know Riemann Sum?
$\begin{eqnarray}
&& \lim_{n\to\infty} \left( \frac{1^4+2^4+3^4+\ldots+n^4 }{n^k } \right) \\
&=& \lim_{n\to\infty}n^{4-k} \left( \left(\frac1n\right)^4 +\left(\frac2n\right)^4 + \left(\frac3n\right)^4 + \ldots + \left(\frac nn\right)^4 \right) \\
\end{eqnarray} $
Should it converge to a finite... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the matrix $\mathbf{A}$ if $A\binom{7}{-1} = \binom{6}{2}.$ Find the $2\times2$ matrix $A$ where $A^2=A$ and
$$A\begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$$
I tried plugging in: $A= \begin{pmatrix}a&b\\c&d\end{pmatrix}$ but that became messy very quickly. I got the equations:
$7a... | As $A^2=A$, you also know that $A\begin{pmatrix}6\\2\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$. Since $\begin{pmatrix}7\\-1\end{pmatrix},\begin{pmatrix}6\\2\end{pmatrix}$ are linearly independant, this determines $A$ completely. We conclude that both columns of $A$ are multiples of $\begin{pmatrix}6\\2\end{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Prove by induction that $\frac{n^3}{3}+\frac{2n}{3}$ is an integer. The question that I am working on is:
Prove that $\dfrac{n^3}{3}+\dfrac{2n}{3} \in \mathbb Z \ \forall \ n \in \mathbb N$
The method that I think would be will work for this question is that I say that $3|(n^3+2n)$ and prove that.Would this be a go... | Look at this:
assuming
$\dfrac{k^3 + 2k}{3} \in \Bbb Z, \tag{1}$
we have
$\dfrac{(k + 1)^3 + 2(k + 1)}{3} = \dfrac{k^3 + 3k^2 + 3k + 1 + 2k + 2}{3} = \dfrac{(k^3 + 2k) + 3(k^2 + k +1)}{3}$
$= \dfrac{k^3 + 2k}{3} + k^2 + k + 1 \in \Bbb Z, \tag{2}$
by (1).
QED!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
} |
Solving $6 \cos x - 5 \sin x = 8$ My attempt:
Using the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi)$$
$$
\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8
$$
$$
\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}
$$
And... | Edit: In general, the maximum value of $a\cos A+b\sin A$ is $\sqrt{a^2+b^2}$
Hence the maximum value of $6\cos x-5\sin x$ is $\sqrt{6^2+(-5)^2}=\sqrt{61}$
But $RHS=8$ $\implies \sqrt{61}<8$ i.e. equality does not hold true for any value of $x$
Hence there is no solution of the given equation: $6\cos x-5\sin x=8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Baby Rudin claim: $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...$ converges This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence ... | Showing Convergence
Breaking the series into chunks of $3$ terms, which is okay since the terms tend to $0$, we get
$$
\begin{align}
\sum_{k=0}^\infty\left[\frac1{4k+1}+\frac1{4k+3}-\frac2{4k+4}\right]
&=\sum_{k=0}^\infty\left[\left(\frac1{4k+1}-\frac1{4k+4}\right)+\left(\frac1{4k+3}-\frac1{4k+4}\right)\right]\\
&=\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Differentiate the Function: $y=2x \log_{10}\sqrt{x}$ $y=2x\log_{10}\sqrt{x}$
Solve using: Product Rule $\left(f(x)\cdot g(x)\right)'= f(x)\cdot\frac{d}{dx}g(x)+g(x)\cdot \frac{d}{dx}f(x)$
and $\frac{d}{dx}(\log_ax)= \frac{1}{x\ \ln\ a}$
$(2x)\cdot [\log_{10}\sqrt{x}]'+(\log_{10}\sqrt{x})\cdot [2x]'$
$y'=2x\frac{1}{\sqr... | $$\log\sqrt{x} = \frac{1}{2}\log_{10} x= \frac{\ln x}{2\ln 10}$$
$$\frac{d}{dx}\ln x= \frac{1}{x}$$
using chain rule,
$$\frac{d}{dx}\left(2x\log\sqrt{x}\right)= 2x \frac{dx}{dx}\frac{\ln x}{2\ln 10} + \frac{\ln x}{2\ln 10}\frac{d}{dx} 2x $$
$$=\frac{2x}{2x\ln 10} + \frac{2\ln x}{2\ln 10} $$
$$ =\frac{1}{\ln 10}+ \log_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
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Wanted : for more formulas to find the area of a triangle? I know some formulas to find a triangle's area, like the ones below.
*
*Is there any reference containing most triangle area formulas?
*If you know more, please add them as an answer
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=... | $\require{cancel}$
Some errors
spotted in
Baker's papers, referenced in
the accepted answer.
Expression N20
is definitely wrong (even the dimension):
\begin{align}
20.\ &
\cancel{\color{blue}{\frac{R\,r}{\beta_a\beta_b\beta_c}\,
\left(\frac1a+\frac1b\right)
\left(\frac1b+\frac1c\right)
\left(\frac1c+\frac1a\right)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 6
} |
Show that $ \tan (A + \theta) $ can be simplified to $- \cot \theta$ as A tends to $\frac{\pi}{2}$ So far I have used the identity,
$$\tan\left(\frac{\pi}{2} + \theta\right) = \frac{\tan A + \tan \theta} {1 - \tan A \tan \theta}$$
As $A \to \frac{\pi}{2}$, $\tan A \to \infty$, so my reasoning is, $\infty + \tan \the... | Using:
$$\lim_{A \to \pi/2} \frac{- \tan A \tan \theta}{1 - \tan A \tan \theta} = 1:$$
$$\lim_{A \to \pi /2} \frac{\tan A + \tan \theta}{1 - \tan A \tan \theta} = \lim_{A \to \pi/2} \frac{\tan A}{1 - \tan A \tan \theta} + \frac{\tan \theta}{1 - \tan A \tan \theta} = \lim_{A \to \pi /2} \frac{\tan A}{-\tan A \tan \thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Limit of the sequence $\sin \left( {2\pi \sqrt {{n^2} + n} } \right)$ I would like to calculate the following limit: ${\lim _{n \to \infty }}\sin \left( {2\pi \sqrt {{n^2} + n} } \right)$
I am not sure if this limit exists...
| Hint: Note that
$$\sqrt{n^2+n}-\left(n+\frac{1}{2}\right)=\frac{\left(\sqrt{n^2+n}-\left(n+\frac{1}{2}\right)\right)\left(\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)\right)}{\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)}=\frac{-1/4}{\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)}.$$
Thus
$$\sin\left(2\pi\sqrt{n^2+n}\right)=\sin\left(2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Confusion about the integral $\int dT/(1-T^2)$ From some reference on the internet we have the following real valued function and its derivative:
$$
M(T) = \frac{\sqrt{1-T^2}}{1+T}
\quad \Longrightarrow \quad \frac{dM}{dT} = - M/(1-T^2)
$$
The reverse of differentiation is integration:
$$
\frac{dM}{dT} = - M/(1-T^2) \q... | $$
\int \frac{dT}{1-T} = -\ln(1-T)+\text{constant}.
$$
The initial minus sign comes from the chain rule. You omitted it.
(No absolute value sign is needed since $\sqrt{1-T^2}$ exists only when $1-T$ and $1+T$ are both $\ge0$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s... | Problem: For a given integer $N$, how many integers $n_1$ and $n_2$ larger than or equal to zero are there that satisfy the equation:
$$n_1 + n_2 = N$$
We note that it is the coefficient of $x^N$ in the expansion of
$$\left(\sum_{k=0}^{\infty}x^k\right)^2 = \frac{1}{(1-x)^2}$$
We can also solve the problem by noting th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 1
} |
partial fraction derivative question So I have this partial fraction derivative question. I know how to solve it, but for some reason I keep swapping two numbers. Here is the problem:
$$\int\frac{3-4x}{x^2+x}= \frac{A}{x}+\frac{B}{x+1}$$
$$(3-4x)=A(x)+B(x+1)$$
Let $x=-1$
$$(3-4(-1))=A(-1)+B(0)$$
$$-7=A$$
Let $x=0$
$$(3... | $$\frac{3-4x}{x^2+x}= \frac{A}{x}+\frac{B}{x+1}$$
$$\frac{A}{x}+\frac{B}{x+1}=\frac {A(x+1)}{x(x+1)}+\frac{Bx}{x(x+1)}$$
then we have to set:$$3-4x=(A+B)x+A$$
Then $A=3$ and $B=-7$ so our integral will be:
$$\int\frac{3-4x}{x^2+x}dx=\int\frac{3}{x}dx+\int \frac{-7}{x+1}dx $$
$=3\ln\mid x\mid-7\ln\mid(x+1)\mid+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is this problem wrongly built? Or is there a solution which I don't know how to arrive at? I was solving a Cauchy-Schwarz's inequality based problem. Given that $x^2+y^2+z^2=1$ I am supposed to show that $x+y+z \le 6$. After struggling for a while I realised that I could solve this inequality had the condition been $x^... | In the first part of your problem, assuming $x^2+y^2+z^2=1$ is actually no problem, because $x^2+y^2+z^2=1$ implies $x,y,z\le 1$, so $x+y+z\le 3<6$.
Also, using Cauchy-Schwarz you can slightly improve this bound to $x+y+z\le \sqrt{3}$.
Looking however also at your next problem, showing that $x^3+y^3+z^3\ge 24$, it see... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Using Lagrange's Method in Finding Extreme Values of $x^2 + y^2 + z^2$ for $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ (New to This Method) Did I do this hw question correctly (at least in theory, I do not expect anyone to check my algebra work)? In particular, did I solve for lambda and plug lambda back into m... | Setting $f_x=\lambda g_x, \;f_y=\lambda g_y, \;f_z=\lambda g_z$ gives
$\;\;\displaystyle2x=\lambda\cdot\frac{2x}{a^2}, \;\;2y=\lambda\cdot\frac{2}{b^2},\;\;2 z=\lambda\cdot\frac{2z}{c^2}$.
Therefore $\textbf{1)}$ $x=0$ or $\lambda=a^2\;\;\;$$\textbf{2)}$ $y=0$ or $\lambda=b^2\;\;\;$ $\textbf{3)}$ $z=0$ or $\lambda=c^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$ The problem is the following (Velleman's exercise 3.2.10):
Suppose that $x$ and $y$ are real numbers. Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$.
My approach so far: Suppose $x \neq 0$. Suppose $ y = \frac{3x^2+2... | $y = \frac{3x^2+2y}{x^2+2}$, multiplying both sides of the equation by $x^2+2$ results in an equivalent equation because that term is never $0$ (in the reals at least).
You end up with $yx^2 + 2y = 3x^2+2y$
subtract $2y$ from both sides (always legitimate).
$yx^2=3x^2$
Since $x\neq 0$ we can divide both sides by $x^2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $(m^2-m-2)x=m^2+4m+3$ Here's how I solve it
I think that m is the variable (am I right?).
Then
$$m^2x-mx-2x-m^2-4m-3=0$$
$$m^2(x-1)-m(x+4)-(2x+3)=0$$
$$D=x^2+8x+16+4(x-1)(2x+3)$$
$$=x^2+8x+16+4(2x^2-2x+3x-3)$$
$$=9x^2+12x+4$$
$$=(3x+2)^2$$
$$m=\frac{x+4\pm (3x-2))}{2(x-1)}$$
$$m_1=\frac{4x+2}{2x-2... | $m^2-m-2=(m-2)(m+1)$ and $m^2+4m+3=(m+1)(m+3)$
Clearly $m+1=0$ is a solution
Else $(m-2)x=m+3\implies x=\dfrac{m+3}{m-2}=1+\dfrac5{m-2}$
If $x$ is an integer, $(m-2)$ must divide $5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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coefficient of $x^{17}$ in the expansion of $(1+x^5+x^7)^{20}$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
find the coefficient of $x^{17}$ in the expansion of $(1+x^5+x^7)^{20}$
(A)3400 (B)3410 (C)3420 (D)3430 (E)3440
so ... | So if you think about
$$ (1 + x^5 + x^7)^{20} $$
That intuitively is just
$$ ((1 + x^5) + x^7) \times ((1 + x^5) + x^7) \times ((1 + x^5) + x^7) ... $$
Which can be expanded out term by term. By the Binomial Theorem as
$$ (1 + x^5)^{20} (x^7)^0 + \begin{pmatrix} 20 \\ 1\end{pmatrix}(1 + x^5)^{19}x^7 + \begin{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
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Digital Roots of Square Numbers Can anyone offer a proof of the following:
The digital root of a square number is always $1$, $4$, $7$ or $9$. (It is never $2$, $3$, $5$, $6$ or $8$.)
Digital root : Add the digits of a number until you get a single digit.
examples: The digital root of $144$ is $1+4+4 = 9$.
The digi... | Mod $9$: $0^2 \equiv 3^2 \equiv 6^2 \equiv 0$, $1^2 \equiv 8^2 \equiv 1$, $2^2 \equiv 7^2 \equiv 4$, $4^2 \equiv 5^2 \equiv 7$. That's all!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex integration by Cauchy's residue theorem
Evaluate the following integral by Cauchy's Residue Theorem
$$\int_C\frac{2z^2-z+1}{(2z-1)(z+1)^2}\,dz$$where , $C:r=2\cos \theta$ , $0\le \theta \le \pi.$
I have problem about the contour $C$.
Here, $r^2=4\cos^2 \theta=\frac{4x^2}{x^2+y^2}$ , as $\tan \theta =y/x$.
Th... | $$\int_{C}\frac{2z^2-z+1}{(2z-1)(z+1)^2}dz$$$$=\frac{1}{2}\int_{C}\frac{2z^2-z+1}{\left(z-\frac{1}{2}\right)(z+1)^2}dz$$
We see that the singularities $z=\frac{1}{2}$ & $z=-1$ are on the real axis inside the curve, $C: r=2\cos \theta$ hence there are the poles of first & second order respectively. Let's find out the re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $x,y\in \mathbb{Z}$ Solve for $x,y\in \mathbb{Z}$
$$x^{6}=y^{2}+53$$
I tried but I couldn't complete
| $$x^6 = y^2 + 53$$
$$x^6-y^2 = 53$$
From $x^6-y^2 = (x^3+y)(x^3-y)$
$$(x^3+y)(x^3-y) = 53$$
The only factors of $53$ are $1$ and $53$ so let: $$x^3-y = 1$$$$x^3+y = 53$$
Adding the $2$ equations we get: $$2x^3 = 54$$ $$x = 3$$$$y = 3^3-1 = 26$$
Our solutions are $x = 3$, $y = 26$.
But notice that in the original equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the least $N$ so there is no square
Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000 \cdot N$ contains no square of an integer.
Let $x^2$ appear before $1000N$ so:
$(x+1)^2 - x^2 > 1000 \implies x \ge 500$
Let $A = [1000N, 1000(N+1)]$
So I let $x=500$ then... | HINT :
As you wrote, we have $x\ge 500$.
Observe that
$$501^2=251001,\quad 502^2=252004,\quad 503^2=253009,\cdots$$
$$251\times 1000=251000,\quad 252\times 1000=252000,\quad 253\times 1000=253000,\cdots$$
Now we need to have
$$(500+a)^2-(250+a)\times 1000\gt 1000\Rightarrow a\gt 31$$
where $N=250+a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
Show that for any two positive real numbers $a$ and $b$, $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
My attempt: $(\sqrt a-\sqrt b)^2\geq0\\\frac{a+b}{2}\geq \sqrt{ab}$
$(a-b)^2\geq0\\a^2+b^... | Run this backwards:
$$\frac{a+b}2-\sqrt{ab}\geq\sqrt{\frac{a^2+b^2}2}-\frac{a+b}2\\
a+b\geq\sqrt{ab}+\sqrt{\frac{a^2+b^2}2}\\
a^2+2ab+b^2\geq ab+\frac{a^2+b^2}2+2\sqrt{ab\frac{a^2+b^2}2}\\
a^2+2ab+b^2\geq4\sqrt{ab\frac{a^2+b^2}2}\\
(a+b)^4\geq 8(a^3b+ab^3)\\
(a-b)^4\geq0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means t... | By $MA\geq MG$ we have $2xy\leq x^2+y^2=1$. Note that $(x+y)^2-(x^2+y^2)=2xy$ and this implies that $(x+y)^2\leq 2$ . So, $-\sqrt{2}\leq x+y\leq \sqrt{2}$. Then $f(x,y)\leq 1+\sqrt{2}$ with equality for $x=y=\frac{\sqrt{2}}{2}$ and $f(x,y)\geq 1-\sqrt{2}$ with equality for $x=y=-\frac{\sqrt{2}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $ \ \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} \ = \ 0 \ $ has a solution in $ \ (-1,1) $ If $a$ and $b$ are positive numbers, prove that the equation
$$\frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} = 0$$
has at least one solution in the interval $ \ (-1,1) \ $ .
The question is from the exercises sectio... | In the interval $(-1,1)$ your equation is equivalent to
$f(x):= a(x-1)(x^2+x+2)+b(x+1)(x^2+x-1)=0$
Now you can see that $f(x)$ is continuous and
$\lim\limits_{x\to -1}f(x)<0, \quad
\lim\limits_{x\to 1}f(x)>0$
Thus $f(x)=0$ has at least one solution (in the interval $(-1,1)$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Compare $A=\frac{1.0\,000\,004}{(1.0\,000\,006)^2}$ and $ B=\frac{(0.9\,999\,995)^2}{0.9\,999\,998}$ My work:
*
*$1.0\,000\,004 = 1+\frac{4}{10^7}=1+\frac{1}{125\cdot 10^6}$
*$ (1.0\,000\,006)^2=(1+\frac{6}{10^7})^2=(1+\frac{3}{5\cdot 10^6})^2$
*$ (0.9\,999\,995)^2=(\frac{9\,999\,995}{10^7})^2=(\frac{1\,999\,999}{... | Let $a=\frac{1}{10^7}$.
Then, $$A=\frac{1+4a}{(1+6a)^2},\quad B=\frac{(1-5a)^2}{1-2a}$$
Now,
$$\begin{align}A-B&=\frac{1+4a}{(1+6a)^2}-\frac{(1-5a)^2}{1-2a}\\&=\frac{(1+4a)(1-2a)-(1-5a)^2(1+6a)^2}{(1+6a)^2(1-2a)}\\&=\frac{3 a^2 (20a (1-15 a)+17)}{(1+6a)^2(1-2a)}\end{align}$$
This is positive, so $A\gt B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} ={a}(x-y)$ Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y)$. $a$ is a constant.
I have the final answer, which is $$\frac{ \mathrm{d}y}{ \mathrm{d}x} = \sqrt{\frac{1-y^2}{1-x^2}}.$$... | start with the defining equation and use difference of squares ...
$$ a(x-y) = \sqrt{1-y^2}+\sqrt{1-x^2} $$
$$ a \frac{x^2-y^2}{x+y} = \frac{x^2-y^2}{\sqrt{1-y^2}-\sqrt{1-x^2}} $$
$$ a (\sqrt{1-y^2}-\sqrt{1-x^2}) = x+y $$
$$ a \sqrt{1-y^2}-y =a\sqrt{1-x^2}+x $$
so
$$ \frac{a\sqrt{1-x^2}+x}{ a \sqrt{1-y^2}-y }=1... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ I was trying to integrate $\int \frac{a}{x(x^2-a^2)^{1/2}} dx$ and by applying on what i saw on the formula of inverse trigonometric functions, there is formula like $\frac{1}{a}Arcsec\frac{u}{a}$ = $\int \frac{du}{u\sqrt(u^2-a^) }$ so my answer is $\frac{1}{a}Arcsec\frac{x... | To obtain the solution that the OP quotes
$$
\int \frac{a}{x\sqrt{x^2-a^2}}dx
$$
we then have
$$
\int \frac{a}{x^2\sqrt{1-\left(\frac{a}{x}\right)^2}}dx
$$
Here I used the fact
$$
\sqrt{x^2-a^2} = \sqrt{x^2\left(1-\frac{a^2}{x^2}\right)} = \sqrt{x^2}\sqrt{\left(1-\frac{a^2}{x^2}\right)} = x\sqrt{1-\left(\frac{a}{x}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Quadratic formula does not work If I put the equation: $5x^2-x-4 =0$ in the quadratic formula, than I get
$x = 1$ or $x = \frac{-4}{5}$
but the real zeros are: $x = -1$ or $x = \frac{4}{5}$
Can somebody explain me if the quadratic formula fails or me?
| $5x^2-x-4$ does indeed have roots $x = 1, -\frac{4}{5}$, so you did it correctly.
To illustrate:
$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$=\frac{1 \pm \sqrt{1^2-4*5*(-4)}}{2(5)}$$
$$=\frac{1 \pm \sqrt{81}}{10}$$
$$= \frac{1\pm 9}{10}$$
$$x = 1, -\frac{4}{5}$$
As Peter said if you had meant $5x^2+x-4$ then your roots... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$ $\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$
$(A)0\hspace{1cm}(B)\frac{-\pi}{2}\hspace{1cm}(C)\frac{\pi}{2}\hspace{1cm}(D)\frac{7\pi}{2}$
I tried and got the answer but my answer is not matching the options given.Is my method n... | We use the facts that
$$\arccos(-x)=\pi-\arccos(x)$$
and
$$\arccos(x)+\arcsin(x)=\pi/2.$$
We have
\begin{align}
& \int_{-1/2}^{1/2}\left(\arcsin\left(3x-4x^3\right)-\arccos\left(4x^3-3x\right)\right)\mathrm{d}x \\
& \quad = \int_{-1/2}^{1/2}\left(\arcsin\left(3x-4x^3\right)-\pi+\arccos\left(3x-4x^3\right)\right)\mathrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understa... | When dealing with a sum like you are here (summing $n$ terms for some general expression), I would almost always recommend that you use $\Sigma$-notation, for it tidies up a lot of the algebraic mess you have to deal with in your induction proof. With that in mind, you may write your claim as follows.
Claim: For any $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Computing the limit of a series involving trigonometric identities Let $x_n\in(n\pi,(n+1)\pi)$ with $\tan(x_n)=x_n$. Can
$$s:=\sum_{n=1}^\infty \frac{1}{x_n^2}$$
be determined explicitly?
My ideas so far: First, I wrote $x_n$ as
$$x_n= \pi (n+z_n)$$
with $z_n\in(0,\frac{1}{2})$.
Using $\zeta(2)=\frac{\pi^2}{6}$ gives w... | At each $x_n$, the function $f(z) = \tan z - z$ has a simple zero. Therefore
$$g(z) = \frac{f'(z)}{z^2\cdot f(z)}$$
has a simple pole with residue $\frac{1}{x_n^2}$ there. Also, the negative solutions to $\tan z = z$ are the points $-x_n, \, n \in \mathbb{N}\setminus \{0\}$. The equation $\tan z = z$ has no non-real so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve trigonometric inequality $ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $ Solve this trigonometric inequality:
$$ \sin x \sin 2x - \cos x \cos 2x > \sin 6x $$
My steps:
$$ \cos x \cos 2x - \sin x \sin 2x < - \sin 6x $$
$$ \cos 3x < \sin (-6x)$$
$$ \cos 3x < \cos (\frac{\pi}{2}+6x) $$
From this we get:
$$ 3x > \d... | Your error is that in general $\cos A < \cos B$ does not imply that $A > B$. (If you don't see this right away, try $A = \pi$ and $B = \dfrac{3\pi}{2}$).
Instead, try the following approach:
$ \cos x \cos 2x - \sin x \sin 2x < -\sin 6x $
$\cos 3x < -\sin 6x$
$\cos 3x + \sin 6x < 0$
$\cos 3x + 2\sin 3x\cos 3x < 0$
$\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Limit of $\frac{\sin^2(x^p)}{x^q+x^r}$ as $x \to 0^+$
Suppose $p,q,r \in \mathbb{R}$ with $p > 0$. Write (without proof) the limit of the following expression: $$\lim_{x \to 0^+} \frac{\sin^2(x^p)}{x^q+x^r}$$ In particular, find and prove the limit of the expression if $p=3$, $q=6$, $r=7$.
I got (through countless gr... | It is well known that $$\frac{\sin\left(x\right)}{x}\longrightarrow1
$$ at $x\rightarrow0
$, then $$\lim_{x\rightarrow0^{+}}\frac{\sin^{2}\left(x^{3}\right)}{x^{6}+x^{7}}=\lim_{x\rightarrow0^{+}}\frac{x^{6}}{x^{6}+x^{7}}=1
$$ no need to split in two inequalities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Completing the square of $(x+a)(x+b)$ The problem is simple, to complete the square of $(x+a)(x+b)$. My calculations yield
$$\left(x+\frac{a+b}{2}\right)^2-\frac{(a+b)^2}{4}+ab,$$
But the textbook's answer is different ("problem 361", at the bottom of the page):
$$\left(x+\frac{a+b}{2}\right)^2-\frac{(a-b)^2}{4}$$
Did ... | $\displaystyle \bf{(x+a)\cdot (x+b)} = x^2+(a+b)x+ab$
$\displaystyle = \underbrace{x^2+(a+b)x+\left(\frac{a+b}{2}\right)^2}+\underbrace{ab-\left(\frac{a+b}{2}\right)^2} = \left[x+\frac{a+b}{2}\right]^2-\left(\frac{a-b}{2}\right)^2$
So your answer is Right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Being careful with terms of infinite sums $ cos(x): = \sum_{k=0}^\infty \frac{(-1)^nx^{2n}}{2n!}$
$=1- \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}...$
I would like to show that for $x \in [0,2]$
$cos(x) \leq 1- \frac{x^2}{2!}+\frac{x^4}{4!}$
This means the stuff from $-\frac{x^6}{6!}$ onwards must sum t... | The thing you are talking about is called rearrangement of terms of a series. For example
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5} + \dots = \sum_{k=1}^\infty (-1)^{k+1} \frac{1}{k} = \log2 \text{.}
$$
On the other hand
$$
1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $\int \frac{5x^4+4x^5}{(x^5+x+1)^2}$ $\displaystyle \int \frac{5x^4+4x^5}{(x^5+x+1)^2}$
Since in the denominator of the integrand,$(x^5+x+1)^2$ is there.So the answer must be in the form $\displaystyle \frac{f(x)}{(x^5+x+1)}$, but i could not figure out what $f(x)$ i should suppose.I could not imagine any other wa... | Let $\displaystyle \frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \frac{d}{dx}\left[\frac{ax+b}{(x^5+x+1)}\right] = \frac{(x^5+x+1)\cdot a-(ax+b)\cdot (x^5+x+1)}{(x^5+x+1)^2}$
Now Equate above equation, and get the value of $a$ and $b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\lim_{(x,y,z)\to(0,0,0)}\frac{xy+yz+zx}{x^2+y^2+z^4}$ $$\lim_{(x,y,z)\to(0,0,0)}\frac{xy+yz+zx}{x^2+y^2+z^4}$$
In order to calculate this limit, I did:
$$\lim_{(x,y,z)\to(0,0,0)}\left(y\frac{x}{x^2+y^2+z^4}+z\frac{y}{x^2+y^2+z^4}+x\frac{z}{x^2+y^2+z^4}\right)$$
then, in each of these terms, one part is limit... | Let $y=nx$, $z=mx$. So,
$$
\frac{xy+yz+xz}{x^2 + y^2 + z^4} = \frac{n+nm+m}{1 + n^2 + m^4x^2}.
$$
Limit depends on $n$ and $m$; so, it doesn't exists.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
To compute improper integral $\int_3^{5}\frac{x^{2}\, dx}{\sqrt{x-3}{\sqrt{5-x}}}$ I am given improper integral as
$$\int_3^{5}\frac{x^{2}}{\sqrt{x-3}{\sqrt{5-x}}}dx$$
DOUBT
I see that problem is at both the end points, so i need to split up the integral. But problem seems to me that when i split up integral one of te... | Notice, the following
$$x^2=A(x-3)(5-x)+B(8-2x)+C$$
On solving we get
$$x^2=-(x-3)(5-x)-4(8-2x)+17$$
Now, we have
$$\int_{3}^{5}\frac{x^2dx}{\sqrt{(x-3)}\sqrt{(5-x)}}$$
$$=\int_{3}^{5}\frac{(-(x-3)(5-x)-4(8-2x)+17)dx}{\sqrt{(x-3)}\sqrt{(5-x)}}$$
$$=-\int_{3}^{5}\frac{(x-3)(5-x)dx}{\sqrt{(x-3)}\sqrt{(5-x)}}-4\int_{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Modular maths: How do I find the remainder? How do I find the remainder of $5^{22} \pmod{25}$?
And also how do I find the remainder of $3^{16} + 7 \pmod{5}$?
| Well $5^3 = 125 \equiv -1 \pmod {7}$ because $7 \mid 125 - (-1)$
And since ${5}^{3} \equiv -1 \pmod {7}$ then ${(5^3)}^{7} \equiv (-1)^7 \pmod{7}$ and so $5^{21} \equiv -1 \pmod{7}$ and so $5^{22} \equiv -5 \pmod 7$. But since $-5 \equiv 2 \pmod 7$ then $5^{22} \equiv 2 \pmod 7$ as well
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How is $\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \leq \frac{\sqrt{|xy|}}{\sqrt{2}}$ $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \leq \frac{\sqrt{|xy|}}{\sqrt{2}}$$
Does this apply in general, because here in this example i have $x \to 0, y \to 0$. Some inequality is used I believe to prove this one, but I do not see which.
| $$\begin{align*}
(|x|-|y|)^2 &\ge 0\\
x^2 - 2|xy| + y^2 &\ge 0\\
x^2 + y^2 &\ge 2|xy|\\
\sqrt{x^2+y^2} &\ge \sqrt{2|xy|}\\
\frac{1}{\sqrt2} &\ge \frac{\sqrt{|xy|}}{\sqrt{x^2+y^2}}\\
\frac{\sqrt{|xy|}}{\sqrt2} &\ge \frac{|xy|}{\sqrt{x^2+y^2}}\\
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
Factoring Quadratic equation I am trying to factor $9x^2-6x+1$ after finding the roots, I am using the following formula $a(x-x_1)(x-x_2)$ in this case there is just one root ($\frac{1}{3}$)
How do I know that the answer is $(3x-1)^2$ and not just $(3x-1)$?
| Using the quadratic formula for $ax^2+bx+c=0$ we get: $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
In your case we have $x_1=\frac{1}{3}$ and $x_2=\frac{1}{3}$, hence we can write $$9x^2-6x+1=9(x-x_1)(x-x_2)=9\left(x-\frac{1}{3}\right)\cdot\left(x-\frac{1}{3}\right)=9\left(x-\frac{1}{3}\right)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding Volume of Rugby Ball I am asked to find the volume of rugby ball whose surface is given by the ellipsoid:
$$\frac{x^2}{4} + \frac{y^2}{4} + \frac{z^2}{9} = 1$$
I am having trouble figuring out which coordinate system I should use. Is it possible to solve the triple integral of the volume by just using cartesian... | Notice,
In general, the volume of ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ is $$=\frac{4\pi}{3}(abc)$$
Hence, the volume of ellipsoid $\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{9}=1\iff \frac{x^2}{2^2}+\frac{y^2}{2^2}+\frac{1^2}{3^2}=1$ is $$=\frac{4\pi}{3}(2\cdot2\cdot 3)$$ $$=\frac{48\pi}{3}$$ $$=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluation of $\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$
Evaluation of $$\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$$
$\bf{My\; Try::}$ Let $$\displaystyle \mathop{I = \int\frac{x^7+2}{(x^2+x+1)^2}}dx = \int\frac{(x^7-1)+3}{(x^2+x+1)^2}dx$$
$$\mathop{\displaystyle = \int\frac{x^7-1}{(x^2+x+1)^2}}+\displaystyle \in... | HINT:
Notice, we have $$\int\frac{x^7+2}{(x^2+x+1)^2}dx$$
$$=\int\frac{x^7+2}{\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)^2}dx$$
Let, $$x+\frac{1}{2}=t\implies dx=dt$$
$$=\int\frac{\left(t-\frac{1}{2}\right)^7+2}{\left(t^2+\frac{3}{4}\right)^2}dt$$
Use reduction to solve further
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$
How can I show that $\arctan (x) + \arctan(1/x) =\frac{\pi}{2}$?
I tried to let $x = \tan(u)$. Then
$$ \arctan(\tan(u)) + \arctan(\tan(\frac{\pi}{2} - x)) = \frac{\pi}{2}$$
but it does not seem useful.
I'd appreciate most a proof that gives intuition and / or uses geome... | Let $f(x)=\arctan(x)+\arctan(\frac1x)$. Then $f'(x)=\frac{1}{1+x^2}+\frac{1}{1+1/x^2}\frac{-1}{x^2}=\frac{1}{1+x^2}-\frac{1}{1+x^2}=0$ so $f(x)$ is constant for $x>0$. Then note that $f(1)=\frac{\pi}{2}$ and thus $f(x)=\frac{\pi}{2}$.
There is a discontinuity at $x=0$, so the derivative only makes sense for $x\neq 0$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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How to find $\#\{1\le x\le 5^k:5^k|(x^4-1)\}$? Find $\#\{1\le x\le 5^k:5^k|(x^4-1)\}$. I am not so sure how it is done, nor am I completely sure if it is about any specific $k$ or all of them together. What I did arrive at, not being really sure, is:
If $k=0$, the answer is $1$, I guess, so suppose $k>1$.
I need to ha... | First note that if $x$ works for $k+1$ then $x$ (reduced modulo $5^k$) also works for $k$, so every solution for $k+1$ comes from a solution for $k$.
Now suppose we have a solution for $k$, that is, $a^4-1\equiv0\bmod{5^k}$. Let $b=a+5^kt$; then there is exactly one value of $t$ in $\{\,0,1,2,3,4\,\}$ such that $b$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is $\sum_{n=1}^\infty \frac{m}{(n+m)^2}$ bounded for all $m\in\mathbb{N}$? I'm trying to figure out if there is a finite constant $C$ such that $\sum_{n=1}^\infty \frac{m}{(n+m)^2}\leq C$ for all $m\in\mathbb{N}$.
I can see that $\sum_{n=1}^\infty \frac{m}{(n+m)^2}\leq\sum_{n=1}^\infty \frac{m}{n^2}=mc$ for some finite... | By the integral comparison test,
$\sum_{n=1}^\infty \frac{1}{(n+m)^2}
=\sum_{n=m+1}^\infty \frac{1}{n^2}
\approx \int_m^{\infty} \frac{dx}{x^2}
=\frac{-1}{x}\big|_m^{\infty}
=\frac{1}{m}
$,
so
$\sum_{n=1}^\infty \frac{m}{(n+m)^2}
\approx 1
$.
To be more precise,
$\int_m^{\infty} \frac{dx}{x^2}
> \sum_{n=m+1}^\infty \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Solve in $\mathbb{Q}$ the equation $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$. Solve in $\mathbb{Q}$ the equation $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$
Somebody can help me? I dont remember how to do.
| Using Quadratic Formula
If $ax^2+bx+c=0\;,$ Then $\displaystyle x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Here $a = 1$ and $b=-\left(\sqrt{2}+1\right)$ and $c=\sqrt{2}$
So $$\displaystyle x = \frac{(\sqrt{2}+1)\pm \sqrt{2+1+2\sqrt{2}-4\sqrt{2}}}{2}$$
So $$\displaystyle x = \frac{(\sqrt{2}+1)\pm (\sqrt{2}-1)}{2} = \sqrt{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solving a rational equation with multiple and nested fractions
This is the equation to solve: $\dfrac{\dfrac{x+\dfrac{1}{2}} {\dfrac{1}{2}+\dfrac{x}{3}}}{\dfrac{1}{4}+\dfrac{x}{5}}=3$
What I did:
$x+\dfrac{1}{2}=\dfrac{2x+1}{2}$
$\dfrac{x}{3}+\dfrac{1}{2}=\dfrac{2x+3}{6}$
$\dfrac{x}{5}+\dfrac{1}{4}=\dfrac{4x+5}{20}$
... | $$\frac { \frac { 2x+1 }{ \frac { 2 }{ \frac { 3+2x }{ 6 } } } }{ \frac { 5+4x }{ 20 } } =3\quad \Rightarrow \frac { \frac { 2x+1 }{ 2 } \cdot \frac { 6 }{ 2x+3 } }{ \frac { 5+4x }{ 20 } } =3\quad \Rightarrow \frac { \frac { 2x+1 }{ 2x+3 } }{ \frac { 5+4x }{ 20 } } =1\Rightarrow \frac { 2x+1 }{ 2x+3 } =\frac { ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$ $(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$ Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$
$(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$
I tried to prove it but my answer is not correct.
For first part,As $0\leq x\leq2\... | $(i)$ Let $$\displaystyle I = \int_{0}^{2}\frac{1}{2+x^2}dx = \int_{0}^{1}\frac{1}{2+x^2}dx+\int_{1}^{2}\frac{1}{2+x^2}dx\geq \int_{1}^{2}\frac{1}{3}dx+\int_{0}^{1}\frac{1}{6}dx \geq \frac{1}{2}$$
And $$\displaystyle I = \int_{0}^{2}\frac{1}{2+x^2}dx = \int_{0}^{1}\frac{1}{2+x^2}dx+\int_{1}^{2}\frac{1}{2+x^2}dx\leq \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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If $x^4+7x^2y^2+9y^4=24xy^3$,show that $\frac{dy}{dx}=\frac{y}{x}$ If $x^4+7x^2y^2+9y^4=24xy^3$,show that $\frac{dy}{dx}=\frac{y}{x}$
I tried to solve it.But i got stuck after some steps.
$x^4+7x^2y^2+9y^4=24xy^3$
$4x^3+7x^2.2y\frac{dy}{dx}+7y^2.2x+36y^3.\frac{dy}{dx}=24x.3y^2\frac{dy}{dx}+24y^3$
$\dfrac{dy}{dx}=\dfrac... | Another way to view this is that the differential equation $ \ \frac{dy}{dx} \ = \ \frac{y}{x} \ $ is separable and has the general solution given by
$$ \ \int \ \frac{dy}{y} \ = \ \int \ \frac{dx}{x} \ \ \Rightarrow \ \ \ln |y| \ = \ \ln |x| \ + \ C \ \ \Rightarrow \ \ y \ = \ A \ x \ \ , $$
with $ \ A \ $ being any r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find the value of function at $\,x=5$ If $\,f(x)\,$ is a non-constant polynomial of $\,x\,$ such that $\,f\left(x^3\right)-f\left(x^3-2\right)=f^2\left(x\right)+12\,$ is true for all $\,x\,$ then find the value of $\,f\left(5\right).\,$
| Given
$$
f(x^3) - f(x^3-2) = f(x)^2 + 12. \tag 1
$$
Let $f(x) = P_n(x)$, where $P_n(x)$ is a polynomial of degree $n$.
LHS is of degree $3(n-1)$ and RHS is of degree $2n$.
Whence
$$
\bbox[16px,border:2px solid #800000] { f(x) = P_3(x) }. \tag 2
$$
We can differentiate $(1)$ and we obtain
$$
\begin{eqnarray}
3 x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proving uniqueness of solutions to $\sin^2A + \sin^2B = \sin (A+B)$ without using multivariable calculus In the course of solving a trigonometric problem (see $a^2+b^2=2Rc$,where $R$ is the circumradius of the triangle.Then prove that $ABC$ is a right triangle), in one approach the following equation needed to be solve... | Suppose $a,b\in(0,\frac\pi2)$ and
\begin{align*}
\sin^2 a + \sin^2 b &= \sin(a+b) \\
&= \sin a\cos b + \cos a \sin b \tag{1}
\end{align*}
Squaring both sides, we get
\begin{align*}
\sin^4 a + 2\sin^2 a\sin^2 b + \sin^4 b
&= \sin^2 a \cos^2 b + 2\sin a\sin b\cos a\cos b + \cos^2 a\sin^2 b \\
&= \sin^2 a (1-\sin^2 b) + 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How Euler get $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$. I saw on wikipedia that he consider $$\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}+...$$
The roots are given by $x=\pm n\pi$ and thus (to me)
$$\frac{\sin x}{x}=(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)...$$
How can he get
$$\frac{\sin x}{x}=\le... | Euler's proof is a bit adhoc. He uses the fact that for finite polynomials if:
$$(x-r_1)\dots(x-r_n)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$$
then $$\dfrac{1}{r_1}+\dots+\dfrac{1}{r_n}=\dfrac{-a_1}{a_0}$$
Then he assumes the result is true for power series, and uses:
$$\dfrac{\sin(\sqrt x)}{\sqrt x}=1-\dfrac{x}{3!}+\dfrac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Trying to solve the trig equation $\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$ The equation is
$$\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$$
My solution goes like this
$$
\begin{cases}
3+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x) \\
\frac{\sin(x)}{\sqrt 3}+3\cos(x) \g... | The roots of $4 t^2-3 \sqrt{3} t-3$ are $-\frac{\sqrt{3}}{4}$ and $\sqrt{3}$.
A slightly different approach to this problem could be the following:
\begin{align}
3+4\cos^2(x)-\frac{\sin^2(x)}{3}-\frac{6}{\sqrt 3}\sin(x)\cos(x)-9\cos^2(x)=\frac{1}{3} \left(-3 \sqrt{3} \sin (2 x)-7 \cos (2 x)+1\right)
\end{align}
With th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integration of the square root of a quadratic I am in the tricky situation of trying to integrate the following.
$$\sqrt{4 a^2 (y-b)^2+c^4}$$
$a, b$ and $c$ are all known constants.
Can anybody provide insight as to how to do this?
I have tried to rearrange to fit the form:
$$\int (ax+b)^{\alpha}dx = \dfrac1a \cdot... | Let $$\displaystyle I = \int \sqrt{4a^2(y-b)^2+c^4}dy\;,$$ Let $(y-b) = t\;,$ Then $dy = dt$
So Integral $$\displaystyle I = \int \sqrt{4a^2t^2+c^4}dt = 2a\underbrace{\int \sqrt{t^2+k^2}dt}_{J}\;,$$ Where $\displaystyle k= \frac{c^2}{2a}$
For calculation of Integral $J\;,$ We Use Integration by parts.
Now Let $$\disp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Finding the maximum of a function on $ \Bbb{S}^{7} $. I'm trying to find the maximum of the function
$$2 a^2 h+\sqrt{3} a d f+\sqrt{3} a e g+2 b^2 h-\sqrt{3} b d g+\sqrt{3} b e f\\+2 c^2
h+\sqrt{3} c d^2+\sqrt{3} c e^2-\sqrt{3} c f^2-\sqrt{3} c g^2\\-d^2 h-e^2 h-f^2
h-g^2 h-2 h^3$$
on the sphere $$a^2+b^2+c^2+d^2... | Let $$a=p\cos\chi,\, b=p\sin\chi\\
d=m\cos\phi,\,e=m\sin\phi\\
f=k\cos\theta,\,g=k\sin\theta\\
q^2=c^2+q^2$$
Then replace $h^3$ by $h(1-p^2-c^2-m^2-k^2)$
$$\text{Maximise }h\left[4p^2+4c^2+m^2+k^2-2\right]+\sqrt{3}pmk\cos(\chi-\phi+\theta)+\sqrt{3}c(m^2-k^2)$$
Let the angle be zero, $m=n\cos\alpha,k=n\sin\alpha$
$$\tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Intermediate digits of 34! Problem: Given that $34!=295232799cd96041408476186096435ab000000$. Find $a, b, c, d$.
$a, b, c, d$ are single digits.
I am able to find $a$ and $b$ but cant find $c, d$.
I did the prime factorisation of $34!$ using De-Polignac's formula.
I got $\frac{34!}{10^7}=2^{25}\times3^{15}\times7^4\tim... | Here $34!$ contain $2^{32}$ and $5^7$. So $2^{25}$ remaining and $2^{7}\cdot 5^7 = (10)^7$ form $7$ zeros at the end.
So $34! = 295232799cd9604140809643ab \times 10^7$ now after deleting $7$ zero,s ,
So we get $34! = 295232799cd9604140809643ab$
Now we use divisibility test for last $7$ digits using $2^7$.
Here $(a,b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
How I can evaluate $\lim_{(x,y) \rightarrow (0,0)} xy(\frac{1+xy}{x^3+y^3})^{1/3}$ I don't have idea how I can evaluate this double limit
$$\lim_{(x,y) \rightarrow (0,0)} xy \left(\frac{1+xy}{x^3+y^3} \right) ^{1/3}$$ could you help me please!
I try prove that $f$ is continuous: $f(x,y)=xy \left(\frac{1+xy}{x^3+y^3} ... | This is prossible?
$ 0\leq|xy\left(\frac{1+xy}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3y^3+x^4y^4}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3y^3}{x^3+y^3}+\frac{x^3y^3(xy)}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3}{x^3+y^3}y^3+\frac{x^3}{x^3+y^3}y^3(xy)\right)^{1/3}|\leq|(y^3+y^3(xy))^{1/3}|\leq (y^2|y|+y^4|x|)^{1/3}\leq(|y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How do you factor $3x^{3/2} -9x^{1/2}+6x^{-1/2}$? How do you factor $3x^{3/2} -9x^{1/2}+6x^{-1/2}$ ?
I factored out a 3 to get:
$3(x^{3/2} -3x^{1/2}+2x^{-1/2})$, but it seems this can be factored further.
| $$ 3 x^{\frac{3}{2}} -9 x^{\frac{1}{2}} + 6x^{-\frac{1}{2}}$$
Factor out $\sqrt{x}$ to get
$$\sqrt{x} (3 x - 9 + 6x^{-1})$$
Furthermore you get
$$\sqrt{x} 3(x - 3 + 2x^{-1})$$
and then
$$\sqrt{x} 3 \frac{(x-2)(x-1)}{x}$$
Simplify:
$$3 \frac{(x-2)(x-1)}{\sqrt{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimum value of $\cos x+\cos y+\cos(x-y)$ What is the minimum value of $$ \cos x+\cos y+\cos(x-y). $$ Here $x,y$ are arbitrary real numbers. Mathematica gives (with NMinimize) $-3/2$. But I don't know if this is correct and if so, how to prove it.
| Let $$\displaystyle z=\cos x+\cos y+\cos(x-y) = 2\cos\left(\frac{x+y}{2}\right)\cdot \cos\left(\frac{x-y}{2}\right)+2\cos^2 \left(\frac{x-y}{2}\right)-1$$
so we get $$\displaystyle 2\cos^2 \left(\frac{x-y}{2}\right)-2\cos\left(\frac{x+y}{2}\right)\cdot \cos \left(\frac{x-y}{2}\right)-(1+z) =0$$
Now Let $$\displaystyle ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Is the matrix $A$ positive (negative) (semi-) definite? Given, $$A = \begin{bmatrix}
2 &-1 & -1\\
-1&2 & -1\\
-1& -1& 2
\end{bmatrix}.$$
I want to see if the matrix $A$ positive (negative) (semi-) definite.
Define the quadratic form as $Q(x)=x'Ax$.
Let $x \in \mathbb{R}^{3}$, with $x \neq 0$.
So, $Q(x)=x'Ax = ... | If you want to proceed with this solution, you should complete the square. It is important that you "complete one variable completely every time". We write
$$\begin{aligned}
x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3&=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}x_2^2+\frac{3}{4}x_3^2-\frac{3}{2}x_2x_3\\
&=\B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How can this expression be simplified? How do I factorize $$a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)?$$
I've tried it in different ways but failed. Wish some one could help solving it out.
| Another proof. By Laplace expansion (along the last row):
$$ -\sum_{cyc}a^2(b-c) = \det\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix},\tag{1}$$
But the RHS is a Vandermonde matrix, whose determinant is well-known.
Gaussian elimination gives:
$$ D=\det\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
A trick for calculating $n^6$ that I don't understand I was doing a math exercise and it asked to find what are the possible units digits of $n^6$ knowing that $n\in\mathbb Z$. The solution said that because we are concerned only with finding what the units digits of $n^6$ could be, it is sufficient to take the sixth p... | What is the units digit of $340274513\times 384759374\,{}$?
$$
\begin{array}{ccccccccccccc}
& & & & 3 & 4 & 0 & 2 & 7 & 4 & 5 & 1 & 3 \\
& & & \times & 3 & 8 & 4 & 7 & 5 & 9 & 3 & 7 & 4 \\
\hline
& & & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & 2 \\
& & \cdot & \cdot & \cdot & \cdot & \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
if $\sin(x)=\frac{1}{3}$ and $\sec(y)=\frac{5}{4} $, what is $\sin(x+y)?$ if $\sin(x)=\frac{1}{3}$ and $\sec(y)=\frac{5}{4} $, what is $\sin(x+y)?$
Here is my thought process:
the identity says:
$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$
We know $\sin(x)=1/3$ and $\sec(y)=\frac{1}{\cos(y)} = \frac{4}{5}$, so
$\sin(x+y... | You have started off well. To get the values of $\cos x$ and $\sin y$, you should use the identity
$$\sin^2\theta + \cos^2\theta \equiv 1$$
I will show you how to find $\cos x$.
You can use the same method to find $\sin y$ yourself.
For example, if $\sin x = \frac{1}{3}$ then we have
\begin{eqnarray*}
\sin^2 x + \cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \frac{\sin ^3(\theta/2)}{\cos(\theta/2)\sqrt{\cos^3\theta+\cos^2\theta+\cos \theta}}d\theta$ I came about the following practice problem in a book of integration:-
$Q.$ Evaluate $$I=\int \frac{\sin ^3(\theta/2)}{\cos(\theta/2)\sqrt{\cos^3\theta+\cos^2\theta+\cos \theta}}d\theta$$ To do this, first I subs... | Let $$\displaystyle I = \int \frac{\sin ^3(\theta/2)}{\cos(\theta/2)\sqrt{\cos^3\theta+\cos^2\theta+\cos \theta}}d\theta = \frac{1}{2}\int\frac{2\sin^2 \frac{\theta}{2}\cdot 2\sin \frac{\theta}{2}\cdot \cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}\sqrt{\cos^3 \theta+\cos^2 \theta+\cos \theta}}d\theta$$
So we get $$\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
There are 30 tokens numbers from 0 to 30. Find the number of ways of choosing 3 tickets such that the sum of the numbers on the tokens is 30 Then the number of solutions is divisible by
(A) 2 (B) 3 (C) 5 (D) 7
One way to solve its by finding the number of unequal integral solutions of $x+y+z=30$.
The possible cases a... | start|range|combos of 2 to get 30 (Gauss method)
0 .....1-29 ...$\lfloor\frac{29}{2}\rfloor = 14$
1 .....2-27 ...$\lfloor\frac{26}{2}\rfloor = 13$
2 .....3-25 ..$\lfloor\frac{23}{2}\rfloor = 11$
3 .....4-23 ...$\lfloor\frac{20}{2}\rfloor = 10$
4 .....5-21 ...$\lfloor\frac{17}{2}\rfloor = 8$
5 .....6-19 ...$\lfloor\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
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.