Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding an explicit formula for $a_n$ defined recursively by The sequence $a_n$ is defined recursively by $a_0=0$, $a_n=4a_{n-1}+1$. I must use generating functions to solve this. $n\geq1$.
I have found a pattern:
$$\sum_{n=1}^\infty(4a_{n-1}+1)x^n = x+5x^2+21x^3+85x^4+341x^5+\ldots$$
If we subtract 1 from each term, r... | A way to go about this is to define
$$f(x) = \sum_{n=0}^{\infty} a_n x^n$$
Then
$$x f(x) = \sum_{n=0}^{\infty} a_n x^{n+1} = \sum_{n=1}^{\infty} a_{n-1} x^n$$
Using your recurrence, we find that
$$(1-4 x) f(x) = a_0 + \sum_{n=1}^{\infty} (a_n-4 a_{n-1}) x^n = a_0 + \sum_{n=1}^{\infty} x^n = a_0 + \frac{x}{1-x}$$
Note t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/777808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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The proof of $e^x \leq x + e^{x^2}$ How can we prove the inequality $e^x \le x + e^{x^2}$ for $x\in\mathbb{R}$?
| Note the inequality $e^t \ge t + 1$ for all $t \in \mathbb{R}$.
In particular $e^{x^2} \ge x^2 + 1$.
If $x \le -1$, then
$$
e^{x^2} - e^x + x \ge x^2 + 1 - e^0 + x = x(x+1) \ge 0.
$$
If $-1 < x < 1$, then
\begin{align*}
e^{x^2} - e^x + x
&\ge x^2 + x + 1 - \sum_{k \ge 0} \frac{x^k}{k!} \\
&= \frac{x^2}{2!} - \sum_{k \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
How find the equation $\cot x=\frac{\sin 20^\circ - \sin 80^\circ \cos 20^\circ}{\sin 80^\circ \sin 20^\circ}$ let $x\in R$, and such
$$\cot x =\frac{\sin 20^\circ -\sin 80^\circ \cos 20^\circ}{\sin 80^\circ \sin 20^\circ}$$
Find $x$
my idea:
$$\cot x=\csc 80^\circ - \cot 20^\circ$$
then I can't
| $$\sin80^\circ=2\sin40^\circ\cos40^\circ=2(2\sin20^\circ\cos20^\circ)\cos40^\circ$$
$$\implies\csc80^\circ-\cot20^\circ=\frac{1-4\cos^220^\circ\cos40^\circ}{\sin80^\circ}$$
Now $\displaystyle N= 1-4\cos^220^\circ\cos40^\circ=1-2\cos40^\circ(2\cos^220^\circ)$
Using Double angle formula $\cos2A=2\cos^2A-1,$
$\displaystyl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/782921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Do the differences of perfect squares apply to perfect cubes and more? I'm curious about a special property of squares. The difference between perfect squares starting from 0 are 1,3,5,7,9..., where each difference goes up by 2. I want to know if there are any patterns for perfect cubes or quartics. Are the differences... | For squares:
$$1$$
$$\begin{array}{cc}
\color{red}{2} & \color{red}{2} \\
1 & \color{red}{2}
\end{array}$$
$$\begin{array}{ccc}
\color{blue}{3} & \color{blue}{3} & \color{blue}{3} \\
\color{red}{2} & \color{red}{2} & \color{blue}{3} \\
1 & \color{red}{2} & \color{blue}{3}
\end{array}$$
$$\begin{array}{cccc}
\color{gr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/786977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
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Algebraic maximum and minimum based on a constraint Suppose $a,b,c$ are real numbers such that $a^2b^2 + b^2c^2 + c^2a^2 = k$, where $k$ is a constant.
Then the set of all possible values of $abc(a+b+c)$ is?
I attempted writing the constraint in the form of $(abc)^2(1/a + 1/b + 1/c)$ and then tried using $(a+b+c)(1/a+... | *
*It is known that $x^2+y^2+z^2\geq xy+yz+zx$. Note $x=ab,y=bc,z=ca, A=abc(a+b+c)$ it follows that $(ab)^2+(bc)^2+(ca)^2 \geq abc(a+b+c)<=>k\geq A.$
*$\dfrac{k}{2}+A = \dfrac{1}{2}[(ab)^2+(bc)^2+(ca)^2]+ abc(a+b+c)= \dfrac{1}{2}[(ab+bc)^2+(bc+ca)^2+(ca+ab)^2] \geq 0=>A\geq -\dfrac{k}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$(x^n)^2=x^n$ for all $n\geq 1$. $x$ is idemotent ring where $x^2= x$ I am asked to prove the above
I have been trying a lot but my teacher did not like they way I solved it,he did not even tell me what is wrong with it.i was thinking if $n=3$ then $x^{2n}$ will be equal to $x^6$ is equal to $$x\cdot x\cdot x\cdot x\c... | Hint: $(x^n)^2=x^{2n}=(x^2)^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $g(x) > 0$ If $f(x)$ is a quadratic expression such that $f(x) > 0,\ x\in\mathbb{R}$ and if $g(x)= f(x) + f'(x) + f''(x)$, then prove that $g(x) > 0, \ x\in\mathbb{R}$.
| Let $f(x) = ax^2 + bx + c$, then $f'(x) = 2ax + b$, and $f''(x) = 2a$. So:
$g(x) = ax^2 + (2a + b)x + 2a + b + c$.
Let $x = -\dfrac{c}{b}$, then $f\left(-\dfrac{c}{b}\right) = a\cdot \dfrac{c^2}{b^2} > 0$ implies that $a > 0$.
So: $\triangle_g = (2a + b)^2 - 4a(2a + b + c) = -4a^2 + (b^2 - 4ac) < 0$, since
$b^2 - 4ac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/789848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Eliminating a parameter when intersecting a manifold with a hyperplane In the Euclidean space $\mathbb R^4$ we look at the intersection of the equations$$x^2 + y^2 = 1 \\ z^2 + w^2 = 1$$
sometimes known as the Clifford torus. This is known to be a 2-dimensional manifold, with global parameterization in 2 parameters, gi... | I think that algebraic methods give better result.
Rewrite this system of equations a little differently.
$\left\{\begin{aligned}&G^2+E^2=K^2\\&D^2+F^2=K^2\end{aligned}\right.$
Solutions of this system are of the form:
$G=2XY$
$E=X^2-Y^2$
$D=2ZR$
$F=Z^2-R^2$
$K=X^2+Y^2=Z^2+R^2$
We need solutions will be taken from the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/790211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is there an elegant way to simplify $\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}$ I wonder how to solve this equation: $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)$$ in an elegant/short... | $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=\\\frac{\frac{\sin(x+20^{\circ })}{\cos(x+20^{\circ })}-\sin(x+20^{\circ })}{\frac{\sin(x+20^{\circ })}{\cos(x+20^{\circ })}+\sin(x+20^{\circ })}=\\\frac{\sin(x+20^{\circ })}{\sin(x+20^{\circ })}\frac{\frac{1}{\cos(x+20^{\circ })}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/790632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Complex Numbers Exercise If $a,b,c$ are complex numbers with $a+b+c=0$ and $\|a\|=\|b\|=\|c\| = r>0$ then prove that $$a^{2^n} + b^{2^n} + c^{2^n} = 0$$
Any ideas?
Thanks!
| We have
$$
a + b + c = 0
$$
Taking the conjugate and multiplying by $abc$,
$$
abc \overline{a} + abc \overline{b} + abc \overline{c} = 0
$$
i.e.
$$
\newcommand{\abs}[1]{\left| #1 \right|}
bc \abs{a}^2 + ac \abs{b}^2 + ab \abs{c}^2
$$
But then $\abs{a} = \abs{b} = \abs{c}$, so
$$
ab + bc + ca = 0
$$
It follows that
$$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/793702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find bounded solutions of this ODE The function $y(x)$ satisfies the following ode:
$L[y]=\frac{d}{dx}(x^2\frac{dy}{dx})-x^2y= f(x)$
By substituting $y(x)=\frac{z(x)}{x}$ we can solve the homogeneous problem $(L[y]=0)$ and obtain the answer
$y(x)=\frac{Ae^x}{x}+\frac{Be^{-x}}{x}$.
How would you show that the solution ... | Hint on the first question:
$$f(x) = \frac{\sinh(x)}{x} = \frac{1}{2x}(e^x - e^{-x})$$
We can show that this function has limit at $x=0$ as follows:
$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$
$e^{-x} = \sum_{n=0}^\infty \frac{(-1)^n x^n}{n!}$
\begin{align*}\frac{\sinh(x)}{x} &= \frac{1}{2x} \left(\sum_{n=0}^\infty \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/797852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Algebraic solution for the intersection point(s) of two parabolas I recently ran through an algebraic solution for the intersection point(s) of two parabolas $ax^2 + bx + c$ and $dx^2 + ex + f$ so that I could write a program that solved for them. The math goes like this:
$$
ax^2 - dx^2 + bx - ex + c - f = 0 \\
x^2(a -... | Well, you somehow lost the factor of 4, otherwise seems valid. As you are going to write a program for this, you should also write a solution gfor the case $a=d$ and deal with the case the radical is zero or does not exist.
Actually, you did not have to write all this, as people usually know the solution of quadratic e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/799519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to prove $\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$ Question:
let $a,b,c>0$,show that:
$$\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$$
maybe th... | We can use also, AM-GM, C-S and uvw.
Indeed, $$\sum_{cyc}\sqrt{\frac{ab}{2a^2+bc+ca}}=\sum_{cyc}\frac{4ab}{2\sqrt{4ab(2a^2+bc+ca)}}\geq$$
$$\geq\sum_{cyc}\frac{4ab}{2a^2+bc+ca+4ab}=\sum_{cyc}\frac{4b^2}{\frac{b}{a}(2a^2+bc+ca+4ab)}\geq$$
$$\geq\frac{4(a+b+c)^2}{\sum\limits_{cyc}\frac{b}{a}(2a^2+bc+ca+4ab)}.$$
Thus, it'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/800082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$ Hi how can we prove this integral below?
$$
I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}
$$
I tried to use
$$
I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx
$$
and now tried changing variables to $y=x(1-x)$ ... | Consider the integral
\begin{align}
I = \int_{0}^{1} \frac{\ln^{2}(x)}{1 - x + x^{2}} \ dx
\end{align}
Now consider the factorization of $x^{2} - x + 1$ which is $(x - a)(x-b)$ where $a$ and
$b$ are $e^{\pi i/3}$ and $e^{-\pi i/3}$, respectively. With this in mind it is seen that
\begin{align}
\frac{1}{x^{2} - x + 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 8,
"answer_id": 5
} |
Inequality $\sum\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum \frac{1}{x + n^2} $
$x\geq0$, then, we have
$$\sum_{n=1}^{\infty}\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{x + n^2} $$
The problem is not easy, even $x=1$. Any help will be appreciated
| The problem is easy for $ 0 \leq x \leq 1$. We have $x \leq n^2$, and hence $ \frac{ x}{x+n^2} \leq \frac{1}{2}$. Thus,
$$ \sum_{n=1}^{\infty} \frac{ x}{(x+n^2)^2} \leq \sum_{n=1}^{\infty} \frac{1}{2} \times \frac{1}{x+n^2}.$$
It remains to verify that we have strict inequality in at least one case.
It is interesting ... | {
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"url": "https://math.stackexchange.com/questions/803497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Do these integrals have a closed form? $I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$ The following integrals look like they might have a closed form, but Mathematica could not find one. Can they be calculated, perhaps by differentiating under the integral sign?
$$I_1 = \int_{-\infty }^{\infty } \... | For the first one we need:
$$\int _{-1/2}^{1/2}\!{{\rm e}^{2\,iax}}{da}={\frac {\sin \left( x
\right) }{x}}\tag{1}$$
$$ \frac{1}{\cosh \left( x \right)}=-2\,\sum _{n=1}^{\infty
} \left( -1 \right) ^{n}{{\rm e}^{- \left| x \right| \left( 2\,n-1
\right) }}\tag{2}$$
$$\int _{-\infty }^{\infty }\!{{\rm e}^{2\,iax}}{{\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/804483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 3
} |
Summation with Ceilinged Logarithmic Function According to Johann Blieberger's paper - "Discrete Loops and Worst Case Performance" (1994):
$$
\sum_{i = 1}^{n}\left \lceil \log_2{(i)} \right \rceil = n\left \lceil \log_2{(n)} \right \rceil - 2^{\left \lceil \log_2{(n)} \right \rceil} + 1
$$
Now, I was wondering if someo... | I could come with a solution with an incisive precision for any base a by doing this:
$$
\\
T(n) = \sum_{i= 1}^{n}i\left \lceil \log_a(i) \right \rceil = \left[\sum_{i= 1}^{\left \lfloor \log_a(n) \right \rfloor}i \left(\sum_{j = a^{i - 1} + 1}^{a^i} j\right)\right] + \left \lceil \log_a(n) \right \rceil\sum_{i = a^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$ \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ$ How can I find the following product using elementary trigonometry?
$$ \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ.$$
I have tried using a substitution, but nothing has worked.
| First, let's re-arrange these terms so that we can make use of the hints in other answers.
$$\tan(1^\circ) \cdot \tan(89^\circ) \cdot \tan(2^\circ) \cdot \tan(88^\circ) \cdot\cdot\cdot \tan(44^\circ) \cdot \tan(46^\circ) \cdot \tan(45^\circ)$$
Here, we can see a clear pattern of $$\tan(x) \cdot\tan(90^\circ-x)$$ repeat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
solving a system of equations dealing with Lorentz transformations Can anyone help me to find the solutions of this system of equations:
$$c^2x^2-v^2y^2=c^2$$
$$y^2-c^2z^2=1$$
$$vy^2+c^2zx=0$$
I know the answer:
$$x= \frac{1}{ \sqrt{1- \frac{ v^{2} }{ c^{2} } } } $$
$$y= \frac{1}{ \sqrt{1- \frac{ v^{2} }{ c^{2} } } } $... | $$c^2x^2-v^2y^2=c^2$$
$$y^2-c^2z^2=1$$
$$vy^2+c^2zx=0$$
mutliplying the second by $x^{2}$ we find
$$
y^2x^2 - c^2z^2x^2 = x^2
$$
using the third equation we find
$$
y^2x^2 - c^2z^2x^2 = y^2x^2 - c^2\left(\frac{vy^2}{c^2}\right)^2 = x^2
$$
or
$$
y^2x^2 - \frac{v^2}{c^2}y^4 = x^2
$$
now we can use the first equation to y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to solve the following? $ x^3+1=2{(2x-1)}^{1/3} $. Find all the real solutions of
$$x^3+1=2{(2x-1)}^{1/3} $$
I tried to cube both sides but got messed up with a nine degree equation! Please help.
Thanks in advance!
| We are given
$$x^3+1=2{(2x-1)}^\frac{1}{3}$$
Substitute $(2x-1)=y$
$\implies x^3=2y-1$
Also,
$y^3=2x-1$ [Using the initial substitution]
Subtracting the above two equations, we get,
$x^3-y^3=-2(x-y)$
$\implies x=y$
or,
$x^2+xy+y^2=-2$
But since $x^2+y^2 \ge xy$ and L.H.S. is thus positive,
$\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/813541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Functions and inequalities I have no idea about this question. Please give a hand whoever can. I am a beginner.
Let $f(x)= x^2 - 10$.
Given $\epsilon > 0$, find the value "$a$" so that $$|x -3 |< \min\{2,a\} \implies |f(x)-10|<\epsilon.$$
| Assume that $x \neq 3$ to begin with:
$|f(x) - 10| = |x^2 - 10 - 10| = |x^2 - 20| < \epsilon \iff 20 - \epsilon < x^2 < 20 + \epsilon$.
Now let $t = x - 3$, then $x = t + 3$. So:
$20 - \epsilon < (t+3)^2 < 20 + \epsilon$. Thus:
$\sqrt{20 - \epsilon} < |t+3| < \sqrt{20 + \epsilon}$. Thus:
$-\sqrt{20 + \epsilon} < t + 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/814588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Set of linear equations Find eigenvalues and eigenvectors of the matrix:
$\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix}$
$\begin{pmatrix} 1-\lambda & 0 & -2 \\ 1 & 3-\lambda & -1 \\ -1 & 0 & 2-\lambda \end{pmatrix}$
I have found the eigenvalues, which are: 0 and 3.
For the 0-eig I have found thei... | I would gladly write a comment but I have no enough rep. yet.
It seems that your matrix (3x3) has only 2 eigenvalues which means that you can not find the third eigenvector. You have to find a generalized eigenvector, which seems to be your solution.
Regarding your second matrix, I believe that eig = 0 will give you (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/814777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove $\int\frac{12x\sin^{-1}x}{9x^4+6x^2+1}dx=-\frac{2\sin^{-1}x}{3x^2+1}+\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$ How to prove
$$\int\frac{12x\sin^{-1}x}{9x^4+6x^2+1}dx=-\frac{2\sin^{-1}x}{3x^2+1}+\tan^{-1}\left(\frac{2x}{\sqrt{1-x^2}}\right)+C$$
where $\sin^{-1}x$ and $\tan^{-1}x$ are inverse of trig functio... | $$\int \frac{12\sin^{-1} x}{(3x^2+1)^2}\,dx=\sin^{-1}x\int \frac{12x}{(3x^2+1)^2}\,dx -\int \left(\frac{1}{\sqrt{1-x^2}}\int \frac{12x}{(3x^2+1)^2}\,dx\right)\,dx$$
To evaluate $\displaystyle \int \frac{12x}{(3x^2+1)^2}$, use the substitution $3x^2+1=u \Rightarrow 6x\,dx=du$ to get: $\frac{-2}{3x^2+1}$, i.e
$$\int \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/817934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Reciprocal Identities Help. $\displaystyle\frac{\sin^2A + 2\cos A - 1}{2 + \cos A - \cos^2A}$ = $\displaystyle\frac{1}{\sec(A)+1}$
Can someone help me with this one? I can't seem to get it right, I get lost, right side has me stumped. Sorry; there's the right side.
| Use the identity $\sin^2 x = 1 - \cos^2 x$. Then, factor the left hand side:
$$\frac{\sin^2A + 2\cos A - 1}{2 + \cos A - \cos^2A} = \frac{1 - \cos^2 A+ 2 \cos A - 1}{(2 - \cos A)(1 + \cos A)} = \frac{\cos A(2 - \cos A)}{(2-\cos A)(1 + \cos A)}$$
Now cancel the common factor, to get $$\frac{\cos A}{1 + \cos A}$$
Now, if... | {
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If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4,$ then $(a-b)^2=\;$? If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4$, what is the value of $(a-b)^2$? I think $a^2+b^2=36$, please confirm and is it possible to to figure out one of the variables?
| Hints:
$\frac{1}{a^2}+\frac{1}{b^2} = 4 \rightarrow b^2 + a^2 = 4(ab)^2$
$(a-b)^2 = (a^2+b^2) -2(ab)$
edit:
To solve for a particular variable, you can use $ab=3 \rightarrow a = \frac{3}{b}$ to eliminate a variable. For example $a^2+b^2 = \frac{9}{b^2}+b^2$
| {
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prove that $ a^2 b^2 \left( {a^2 + b^2 - 2} \right) \ge \left( {a + b} \right)\left( {ab - 1} \right) $ good evening I want to show that
$(1)a;b\in\mathbb {R^*_+}:a^2 b^2 \left( {a^2 + b^2 - 2} \right) \ge \left( {a + b} \right)\left( {ab - 1} \right)
$
$
\begin{array}{l}
\frac{a}{b} + \frac{b}{a} \ge 2 \\
\frac{... | By way of simplifying the situation, let x = ab and y = a+b. Note that as a and b range over all positive reals, x and y range over all pairs for which x>0 and y > 2 √x because if x lies on the hyperbola ab = x in the first quadrant of the a-b plane, then the least value for a+b will be when a=b and then it is 2 √x. S... | {
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Gram-Schmidt process when metric is not an Euclidean Can we use a Gram-Schmidt process for finding an orthogonal vector if a metric on a space is not an Euclidean one?
These are the vectors and the metric.
If we have the next two vectors:
\begin{align}
\notag
A_{1}=c_{1}y_{1}^{2}y_{2}^{2}u(u+2t)(u+y_{2}^{2})\frac{\part... | In the exact same way, but instead of using the dot product, you use whatever inner product you're given.
Example: Consider $\mathbb{R}^2$ with the inner product
$$\langle (x_1, x_2), (y_1, y_2)\rangle = 2x_1y_1 + x_1y_2 + x_2y_1 + 3x_2y_2.$$
Given the basis $\{(1, 0), (0, 1)\}$ of $\mathbb{R}^2$, what do we get when... | {
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Improper integral comparison test Having this integral
$$\int_1^{\infty}\frac{3x^2+2x +1}{x^3+6x^2+x+4}$$
In order to do the comparison test at some point it gets like
$$\frac{3x^2+2x +1}{x^3+6x^2+x+4}\geq \frac{1}{4x}$$
How is $\frac{1}{4x}$ found ? It doesnt seem obvious to me.
EDIT
Also how is this found
$$0\leq\fra... | For positive $x$, the top is $\ge 3x^2$. For $x\ge 1$, the bottom is $\le x^3+6x^3+x^3+4x^3$.
Edit: For your added question, the range of values of $x$ is not specified. However, if $x\ge 1$, then the top is $\le 4x$. The bottom is $\ge \sqrt{x^5}$. So for $x\ge 1$, the whole thing is $\le \frac{4x}{x^{5/2}}$, which ... | {
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How to integrate the inverse of sin How does one integrate $\int\ {\sin^{-1}(x)}$, using integration by parts, where:
$$ u={\sin^{-1}}, du=\frac{1}{\sqrt{1-x^2}},dv=dx, v=x ? $$
This is a partial solution, and I do not quite understand how to move forward from here$$\int\ {\sin^{-1}(x)}= x\sin^{-1} - \int\left( \frac{... | Let $$ J= \int\left( \frac{x}{\sqrt{1-x^2}} \right) dx.$$
Then $$ J=\int\left( \color{green}{x} \cdot \color{red}{\frac{1}{\sqrt{1-x^2}}} \right) \color{green}{dx}.$$
Let $t=1-x^2 \implies \frac{dt}{dx}=-2x \iff \color{green}{xdx}= -\frac{1}{2}dt$.
Then $J=-\frac{1}{2}\int \color{red}{\frac{1}{\sqrt{t}}}dt=-\frac{1}{2}... | {
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Find positive integers $(x,n)$ such that $x^{n} + 2^{n} + 1$ is a divisor of $x^{n+1} +2^{n+1} +1$ Find all positive integers $(x,n)$ such that $x^{n} + 2^{n} + 1$ is a divisor of $x^{n+1} +2^{n+1} +1$
I encountered this question in one of my monthly assignments. Unfortunately, I don't know how to proceed about thi... | Check that $n=1$ gives two solutions $x=4$ and $x=11$. From now on $n>1$.
For each single case $x=1$, $x=2$ check that there is no solution.
Now we will consider
$$
x(x^n+2^n+1)-(x^{n+1}+2^{n+1}+1)=2^n(x-2)+x-1
$$
instead of $x^{n+1}+2^{n+1}+1$.
Check that $x=3$ gives no solution. From now on $x>3$ and $n>1$, hence
$$... | {
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Two ways to show that $\sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ Show that: $\large \sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ on: $0<x<\frac {\pi}2$
I tried to solve it in two ways and got a little stuck:
One way is to use Cauchy's MVT, define $f,g$ such that $f(x)=\sin x -x +\frac {x^3}{3!}$ and $g(x)=\frac ... | Use increasing functions.
For example
$$ 1- \frac{x^2}{2} <\cos x$$
take
$$f(x)=\cos x +\frac{x^2}{2}-1$$ then
$$f^{\prime}=-\sin x + x >0$$ so $f$ is increasing, and $f(0) <f(x)$ gives the desired inequality. Next take $x-\frac{x^3}{3} < \sin x$ then go back to the corresponding $cos$ expansion and finally your re... | {
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Finding an equation of circle which passes through three points How to find the equation of a circle which passes through these points $(5,10), (-5,0),(9,-6)$
using the formula
$(x-q)^2 + (y-p)^2 = r^2$.
I know i need to use that formula but have no idea how to start, I have tried to start but don't think my answer is... | $\begin{vmatrix}
x^2+y^2&x&y&1\\
5^2+10^2&5&10&1\\
(-5)^2+0^2&-5&0&1\\
9^2+(-6)^2&9&-6&1\\
\end{vmatrix}=
\begin{vmatrix}
x^2+y^2&x&y&1\\
125&5&10&1\\
25&-5&0&1\\
117&9&-6&1\\
\end{vmatrix}
=
0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving a simple matrix polynomial Does there exist a $2\times 2$ Matrix $A$ such that
$A-A^2=\begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$ ?
| Assume such an $A$ exists. Let $\lambda_1$ and $\lambda_2$ be the eigenvalues of $A$. Then, the eigenvalues of $A-A^2$ are $\lambda_1-\lambda_1^2$ and $\lambda_2-\lambda_2^2$.
The eigenvalues of $\begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix}$ are $\dfrac{7 \pm \sqrt{5}}{2}$. Hence $\lambda_1-\lambda_1^2 = \dfrac{7 - \s... | {
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Prove this identity... $$\frac{\sin 2x}{1+\cos 2x} \times \frac{\cos x}{1+\cos x}=\tan\frac{x}{2}$$
This is what I've done:
$$\frac{2\sin x \cos x}{1+\cos^2 x-\sin^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{2\sin x \cos x}{2\cos^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{\sin x}{1+\cos x}$$
I have no idea what ... | $$\frac{\sin x}{1+\cos x}=\frac{\sin 2\frac{x}{2}}{1+\cos 2\frac{x}{2}}=\frac{2 \sin (\frac{x}{2})\cos(\frac{x}{2})}{1+\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find length of $CD$ where $\measuredangle BCA=120^\circ$ and $CD$ is the bisector of $\measuredangle BCA$ meeting $AB$ at $D$ $ABC$ is a triangle with $BC=a,CA=b$ and $\measuredangle BCA=120^\circ$. $CD$ is the bisector of $\measuredangle BCA$ meeting $AB$ at $D$. Then the length of $CD$ is ____ ?
A)$\frac{a+b}{4}$
B)$... | Solution.
$\triangle ABC=\triangle ACD+\triangle BCD$
$\Longrightarrow$ $~\dfrac{ab\sin 120^{\circ}}{2}=\dfrac{b\cdot CD\sin 60^{\circ}}{2}+\dfrac{a\cdot CD\sin 60^{\circ}}{2}$
$\therefore$ $CD=\dfrac{ab}{a+b}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Product in terms of $n$ of $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \cdots \cdot \frac{2n-1}{2n}$ What is the following product in terms of $n$?
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \cdots \cdot \frac{2n-1}{2n}$$
Thank you.
| Given
$$
(1/2)(3/4)(5/6)(7/8)\cdots([2n-1]/2n)
$$
Thus
$$
\frac{1}{2} \frac{3}{4} \frac{5}{6} \cdots \frac{2n-1}{2n} = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots [2n-1] \cdot 2n}{2^{2n} \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdots 2n)^2 } = \frac{[2n]!}{2^n (n!)^2}
$$
| {
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Problem of quadratic equation If $\alpha$ be a root of $ 4x^2 +2x -1 = 0 $ , prove that the other root is $4\alpha^3 - 3\alpha$ . I have tried to do it but of no success.[$4\alpha^3 -2\alpha$ = $\dfrac {-1}{2}$ and $4\alpha^4 - 3\alpha^2$ = $\dfrac {-1}{4}$ ] .How to prove it?
| Write $4x^3-3x=(4x^2+2x-1)(x-1/2)-(x+1/2)$.
Plug $x=\alpha$ and get $4\alpha^3 - 3\alpha=-\alpha-1/2$, which is the other root because the sum of the roots is $-1/2$.
| {
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When is the sum of two squares the sum of two cubes When does $a^2+b^2 = c^3 +d^3$ for all integer values $(a, b, c, d) \ge 0$. I believe this only happens when: $a^2 = c^3 = e^6$ and $b^2 = d^3 = f^6$. With the following exception:
*
*$1^3+2^3 = 3^2 + 0^2$
Would that statement be correct?
Is there a general formu... | *
*$a=x^3-3x^2y-3xy^2+y^3,b=x^3+3x^2y-3xy^2-y^3,c=d=x^2+y^2.$
*$a=3(x^3-3xy^2),b=3(3x^2y-y^3),c=x^2+y^2,d=2(x^2+y^2)$
$\cdots$
See this post.
| {
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Simple Differentiation Problem Involving Area Radius and Circumference A stone is dropped into a pool of water, and the area covered by the spreading ripple increases at a rate of $4 m^2 s^{-1} $.
Calculate the rate at which the circumference of the circle formed is increasing 3 seconds after the stone is dropped.
My m... | Start from the equation $$\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt}$$ using $A(r) = \pi r^2$. We get $dA/dr = 2 \pi r$ and $$4 = 2 \pi r \frac{dr}{dt} \text{ or } r \frac{dr}{dt} = \frac{2}{\pi}.$$ Write $$r \, dr = \frac{2}{\pi} \, dt$$ and integrate, obtaining $$\int_0^r \rho \, d \rho = \frac{r^2}{2} = \int_0^t \f... | {
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"url": "https://math.stackexchange.com/questions/849920",
"timestamp": "2023-03-29T00:00:00",
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How to prove $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty$ How to prove $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty.$$
I try to do like $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{n+m=N}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{m=1}^{N-1... | $$
\begin{align}
\sum_{m=1}^\infty\sum_{n=1}^m\frac1{m^2+n^2}
&\ge\sum_{m=1}^\infty\sum_{n=1}^m\frac1{2m^2}\\
&=\frac12\sum_{m=1}^\infty\frac1m\\
&=+\infty.
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Determinant involving recurrence Evaluate
$$\left| A \right| = \left| {\matrix{
{x + y} & {xy} & 0 & \cdots & \cdots & 0 \cr
1 & {x + y} & {xy} & \cdots & \cdots & 0 \cr
0 & 1 & {x + y} & \cdots & \cdots & 0 \cr
\cdots & \cdots & \cdots & \cdots & \cdots & \vdots \cr
0 & ... | Hint: compute $D_1 = x+y, D_2 = x^2 + xy + y^2$.
Then, prove that $D_n = x^n + x^{n-1}y + \dots + xy^{n-1} +y^n$ via induction.
Eventually, as $(x^n + x^{n-1}y + \dots + xy^{n-1} +y^n)(x-y) = x^{n+1} - y^{n+1}$, you are done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you simplify an algebraic expression? Please explain how to simplify an expression that is similar to this one
$\displaystyle\frac{a+3}{6}+\frac{a-4}{4}+\frac{a+2}{-3}$
| Depending on what "simplify" means (teachers and test writers should avoid this word, because it is often not specific enough to avoid ambiguity), my last expression below might be what is intended, in contrast to what Varum Iyer and mfl ended with. I've also carried out the steps a little differently (and in great det... | {
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Prove $\prod_{i=1}^{n-1} \sin(i\pi/n) = 2^{1-n} n$ without complex functions. Note that $i$ here refers to indexing variable, not $\sqrt{-1}$.
$$\prod_{i=1}^{n-1} \sin\left(\frac{i \pi}{n}\right) = 2^{1-n} n$$
This formula was used here to give an 'elementary' proof of product of diagonals = N. Mathworld is the only pl... | Thanks to Lahtonen's suggestion to use Chebyshev's polynomials, I was able to answer the question. I used this paper as a reference on the properties of the polynomials.
I'll prove $\frac{\sin((n+1)\theta)}{\sin(\theta)} = U_n(\cos(\theta))$ using induction (the paper used complex numbers to prove it).
$$
\frac{\sin(\t... | {
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Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ et... | If we denote $s:=a+b+c$, we have to minimize
$$\frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}$$
for $a,b,c>0$.
Using new variables $x:=a/s$, $y:=b/s$, $z:=c/s$ we can see, that this is equivalent to minimizing
$$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}$$
for $x+y+z=1$.
If we notice that the function
$$f(x)=\frac{x}{1-x}=-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/855283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Evaluation of a trigonometric series $$
\mbox{Question: Evaluate}\quad
\tan^{2}\left(\pi \over 16\right) + \tan^{2}\left(2\pi \over 16\right) + \tan^{2}\left(3\pi \over 16\right) + \cdots + \tan^{2}\left(7\pi \over 16\right)
$$
What I did:
Well I know that $\tan^{2}\left(7\pi/16\right)$ is the same as
$\cot^{2}\left(\p... | Like Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$,
$$\tan16\theta=\frac{\binom{16}1t-\binom{16}3t^3+\cdots+\binom{16}{13}t^{13}-\binom{16}{15}t^{15}}{\cdots}\text{ where }t=\tan\theta$$
If $\displaystyle\tan16\theta=0,16\theta=r\pi$ where $r$ is any integer
$\displaystyle\implies\theta=\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find maximum of $P$ Let $$P = \frac{{{x^2}}}{{{x^2} + yz + x + 1}} + \frac{{y + z}}{{x + y + z + 1}} - \frac{{1 + yz}}{9}.$$
Find maximum of $P$ where $x, y,z$ are nonnegative real numbers such that ${x^2} + {y^2} + {z^2} = 2$.
I guess $\max P=\dfrac{5}{9}$ when $x=y=1, z=0$. But I can not prove.
| Step 1. Prove that
$$
x^2+yz+x+1\geq x(x+y+z+1)
$$
by using the expansion of $(x-y-z)^2\geq 0$ and $x^2+y^2+z^2=2$.
Step 2. Observe that
$$
x+y+z\leq\sqrt{2(x^2+(y+z)^2)}=2\sqrt{1+yz}
$$
Step 3. By Step 1, we have
$$
P\leq\frac{x^2}{x(x+y+z+1)}+\frac{y+z}{x+y+z+1}-\frac{1+yz}{9}=1-Q
$$
with
$$
Q=\frac{1}{x+y+z+1}+\fra... | {
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"timestamp": "2023-03-29T00:00:00",
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find an elementary matrix $E$ such that $EA=B$ Let matrix
$$A = \begin{pmatrix} 1 & 2 & 0 \\ -3 & 1 & 1 \\ 0 & 4 & 2 \end{pmatrix} $$
and
$$B = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 7 & 1 \\ 0 & 4 & 2 \end{pmatrix} $$
Find an elementary matrix $E$ such that $EA= B$.
I try with many
$$E = \begin{pmatrix} -1 & -2 & 0 \\... | An elementary matrix is one which differs from the identity matrix by one elementary row operation.
Note that $B$ is the matrix $A$ with three times the first row added to the second. So if we take the matrix
$$E=\begin{pmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$
and now consider
$$EA=\begin{pmatrix}1 & ... | {
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Going from closed form to recurrence relation If I had a closed form for a sequence that I suspect to represents a recurrence relation how would I determine the recurrence relation?
In particular, I have the sequence
$$a_n = \frac{1}{4}\bigg(\frac{3}{2}+\sqrt{2}\bigg)^n + \frac{1}{4}\bigg(\frac{3}{2}-\sqrt{2}\bigg)^n$$... | Suppose we want to find a recurrence relation such that
$a_n = \left(\frac{3}{2}+\sqrt{2}\right)^n$ and
$a_n = \left(\frac{3}{2}-\sqrt{2}\right)^n$ are both solutions, which has the form
$$
a_n -b a_{n-1} -c a_{n-2} = 0
$$
If we assume that $a_n = r^n$ for some $r$ is a solution, we'd have
$$
r^2(a_{n-2}) - br(a_{n-2})... | {
"language": "en",
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"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding domain of $\sqrt{4-2\sqrt{x^2 - 1}}$ Find the (maximum) domain of: $\sqrt{4-2\sqrt{x^2 - 1}}$
Well, I guess it is $2 - \sqrt2 \sqrt{x^2 - 1}$ = $2- \sqrt{2(x^2 - 1)}$
from here, $x^2 - 1 \ge 0$ and then $ x \ge 1$ and $ x \ge -1$
What do you guys think?
| $$x^2-1 \geq 0 \Rightarrow x^2 \geq 1 \Rightarrow x \geq 1 \text{ or } x \leq -1$$
Also:
$$4-2 \sqrt{x^2-1} \geq 0 \Rightarrow 4 \geq 2 \sqrt{x^2-1} \Rightarrow 2 \geq \sqrt{x^2-1} \Rightarrow 4 \geq x^2-1 \Rightarrow 5 \geq x^2 \\ \Rightarrow \sqrt{-5} \leq x \leq \sqrt{5}$$
So:
$$1 \leq x \leq \sqrt{5} \ \text{ or... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How I could evaluate this :$ \sum_{n=1}^{\infty}({-1})^{n+1}n(tan^{-1}s-s+\frac{s^3}{3}+....({-1})^{n+1} \frac{s^{2n+1}}{2n+1}) $? let $s$ be a complex variable which $Re(s)>0$. Evaluate : $ \sum_{n=1}^{\infty}({-1})^{n+1}n(tan^{-1}s-s+\frac{s^3}{3}+....({-1})^{n+1}
\frac{s^{2n+1}}{2n+1}) $
I would be interest for an... | For $n \in \mathbb{Z}_{+}$, let $a_n$ be the expression
$\displaystyle\;\tan^{-1}s-s+\frac{s^3}{3} + \cdots + (-1)^{n+1}$. It has an integral representation:
$$\begin{align}
a_n &= \int_0^s \left[\frac{1}{1+t^2} - \left( 1 + ( -t^2) + \cdots + (-t^2)^n\right)\right]dt\\
&= \int_0^s \left[\frac{1}{1+t^2} - \frac{1-(-t^2... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I solve $y' = \sin(x - y)$? How can I solve the differential equation:
$$y'=\sin(x-y)$$
Could I do this?
$$\frac{dy}{dx}= \sin x \cos y - \sin y \cos x$$
But, how would I continue?
| Let $z = x - y$. Thus, $\dfrac{dy}{dx} = 1 - \dfrac{dz}{dx}$ and
$$
\dfrac{dy}{dx} = \sin(x - y) \quad \Rightarrow \quad \dfrac{dz}{dx} = 1 - \sin z \quad \Rightarrow \quad \int dx = \int \dfrac{dz}{1 -\sin z}
$$
The next step is to change $u = \tan(z/2)$ so that $dz = \dfrac{2du}{1 + u^2}$. Note that
$$
\sin z = \dfra... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$. Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$.
My attempt:
$p\mid a^2+ab+b^2 \implies p\mid (a-b)(a... | By lil' Fermat, $a^{3k+1}\cong b^{3k+1}\bmod p\implies a\cdot
b^{3k}\cong b\cdot b^{3k}\bmod p$. Thus $\begin{align} a\cong b\bmod p\implies a=b+pk \implies a^2+ab+b^2=(b+pk)^2+(b+pk)b+b^2=3b^2+3bpk+p^2k^2\end{align}$. But this expression is divisible by $p$. So $3b^2=-3bpk-p^2k^2+pl\implies p|3b^2\implies p|b\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Equation $\frac {x^2-x-6}{x^2+x -2} - \frac {2x-4}{x-1}$ I am trying to answer the problem as to how to show $\dfrac {x^2-x-6}{x^2+x -2} - \dfrac {2x-4}{x-1} = -1$
I have the key with the answer; the left-hand side is supposed to reduce to $ -1$ and have confirmed this is correct but which steps do I take to reach that... | Note that $$\dfrac{x^2 - x - 6}{x^2 + x-2} = \frac{(x-3)(x+2)}{(x+2)(x - 1)} = \frac {x-3}{x-1}$$
Now, we have a common denominator: $$\frac{(x-3)}{(x-1)} - \frac{2x-4}{(x-1)}=$$ $$= \frac{(x-3) - (2x-4)}{(x-1)} = $$ $$ = \dfrac{-x+1}{x-1} = \dfrac{-(x-1)}{x-1} = -1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/871674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric Identities help How do you solve this? I can't figure out what I should do.
$$\sin ^4\left(A\right)+\cos ^2\left(A\right)=\cos ^4\left(A\right)+\sin ^2\left(A\right)$$
Also, why is this equal zero? Can someone explain how that simplifies to be zero?
$$\frac{\left(\frac{1}{\cos \left(x\right)}\right)-1}{... | For the 1st problem there's a little trick: add zero.
\begin{array}{lll}\sin^4A+\cos^2A&=&\sin^4A+(-\cos^4A+\cos^4A)+\cos^2A\\
&=&(\sin^4A-\cos^4A)+\cos^4A+\cos^2A\\
&=&(\sin^2A-\cos^2A)(\sin^2A+\cos^2A)+\cos^4A+\cos^2A\\
&=&(\sin^2A-\cos^2A)+\cos^4A+\cos^2A\\
&=&\cos^4A+\sin^2A
\end{array}
For the other problem, the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/874346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How prove this $3^{\frac{5^{2^n}-1}{2^{n+2}}}\equiv (-5)^{\frac{3^{2^n}-1}{2^{n+2}}}\pmod {2^{n+4}}$ Question:
show that:
$$3^{\frac{5^{2^n}-1}{2^{n+2}}}\equiv (-5)^{\frac{3^{2^n}-1}{2^{n+2}}}\pmod {2^{n+4}},n\geq 1$$
My idea: since
I have prove
$$5^{2^n}-1\equiv 0\pmod {2^{n+2}}$$
$$3^{2^n}-1\equiv 0\pmod {2^{n+2}}$... | We can check with an easy induction that the orders of $3$ and $-5$ modulo $2^{n+4}$ are $2^{n+2}$, so in order to check the equality it is enough to know $\frac {a^{2^n}-1}{2^{n+2}}$ modulo $2^{n+2}$, hence $a^{2^n}-1$ modulo $2^{2n+4}$ (where $a=3$ and $5$)
Suppose $a$ is odd and define for $n\ge 1$ the sequence $(u_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving the following trigonometric proportion $$\frac{a\sin(B-C)}{b^2-c^2}=\frac{b\sin(C-A)}{c^2-a^2}=\frac{c\sin(A-B)}{a^2-b^2}$$
A, B, C are angles of a triangle and a, b, c are the sides of a triangle.
I tried using various things such as sine rule and then replacing the various rations in terms of sides of triangl... | By the Law of Sines, we have
$$a = d \sin A \qquad b = d \sin B \qquad c = d \sin C$$
where $d$ is the diameter of the triangle's circumcircle. As a result, the relation to prove becomes (after eliminating some $d$s)
$$\frac{\sin A \;\sin(B-C)}{\sin^2B-\sin^2C} = \frac{\sin B \;\sin(C-A)}{\sin^2C-\sin^2A} = \frac{\sin ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\tan x $ if $\sin x+\cos x=\frac12$ It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $.
I tried all identities I know but I have no idea how to proceed. Any help would be appreciated.
| Hint :
\begin{align}
\sin x+\cos x&=\frac12\\
\sin x&=\frac12-\cos x\\
\sin^2 x&=\left(\frac12-\cos x\right)^2\\
1-\cos^2 x&=\frac14-\cos x+\cos^2 x\\
2\cos^2 x-\cos x-\frac34&=0\\
8\cos^2 x-4\cos x-3&=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 0
} |
Polynomial $f(x)$ degree problem. Suppose the polynomial $f(x)$ is of degree $3$ and satisfies $f(3)=2$, $f(4)=4$, $f(5)=-3$, and $f(6)=8$. Determine the value of $f(0)$.
How would I solve this problem? It seems quite complicated...
| Let $g(x)=f(x+3)$. Then the difference table for $g$ is given by
$2\;\;\; 4\;\; -3\;\; 8$
$\;\; 2 \;\; -7\;\; 11$
$\;\;\;-9\;\; 18$
$\;\;\;\;\;\;27,$
so $f(0)=g(-3)=2+2\binom{-3}{1}-9\binom{-3}{2}+27\binom{-3}{3}=2+2(-3)-9(6)+27(-10)=-328.$
Alternatively, $f(x)=g(x-3)=2+2\binom{x-3}{1}-9\binom{x-3}{2}+27\binom{x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
taylor series of $\ln(1+x)$? Compute the taylor series of $\ln(1+x)$
I've first computed derivatives (up to the 4th) of ln(1+x)
$f^{'}(x)$ = $\frac{1}{1+x}$
$f^{''}(x) = \frac{-1}{(1+x)^2}$
$f^{'''}(x) = \frac{2}{(1+x)^3}$
$f^{''''}(x) = \frac{-6}{(1+x)^4}$
Therefore the series:
$\ln(1+x) = f(a) + \frac{1}{1+a}\frac{... | Note that
$$\frac{1}{1+x}=\sum_{n \ge 0} (-1)^nx^n$$
Integrating both sides gives you
\begin{align}
\ln(1+x)
&=\sum_{n \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\\
&=x-\frac{x^2}{2}+\frac{x^3}{3}-...
\end{align}
Alternatively,
\begin{align}
&f^{(1)}(x)=(1+x)^{-1} &\implies \ f^{(1)}(0)=1\\
&f^{(2)}(x)=-(1+x)^{-2} &\implies f^{(2)... | {
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 2
} |
How prove $x^3+y^3+z^3-3xyz\ge C|(x-y)(y-z)(z-x)|$
let $x,y,z\ge 0$,and such
$$x^3+y^3+z^3-3xyz\ge C|(x-y)(y-z)(z-x)|$$
Find the maximum of the $C$
witout loss of we assume that $$x+y+z=1$$
I think
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=(x+y+z)^3-3(xy+yz+xz)(x+y+z)=1-3(yz+xz+xy)$$
then $$(x-y)(y-z)(x-z)... | This is the standard technique for this. This one is a bit tricky, but anyways. Let WLOG $x\le y\le z$ and let $y=a+x,z=b+x$. Now plug this into
$$(x+y+z)\frac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2)\ge C|(x-y)(y-z)(z-x)|$$
$$\iff(3x+a+b)(a^2-ab+b^2)\ge Cab(a-b)\quad \text{since } a\ge b$$
Now this is linear in $x$. Hence we m... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
The limit of $((1+x)^{1/x} - e+ ex/2)/x^2$ as $x\to 0$ $$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
*
*by directly substituting $x=0$ i got $\infty$
*by using L-H's rule, i got $-1/8$
the given options are
$a)\frac{24e}{11}$
$b)\frac{11e}{24}$
$c)\frac{e}{11}$
$d)\frac{e}{24}$
may b... | We have $$\begin{aligned}F(x)\,&= (1 + x)^{1/x} - e + \frac{ex}{2}\\
&= \exp\left(\frac{\log(1 + x)}{x}\right) - \exp\left(1 + \log\left(1 - \frac{x}{2}\right)\right)\\
&= \exp\left\{1 + \log\left(1 - \frac{x}{2}\right)\right\}\left\{\exp\left(\frac{\log(1 + x)}{x} - 1 - \log\left(1 - \frac{x}{2}\right)\right) - 1\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/879430",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$
If $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$$
Additional info:We can use AM-GM and Cauchy inequaliti... | By AM-GM, we have:
$$\frac{a^3}{b^2} + a\ge \frac{2a^2}{b}\Rightarrow \frac{a^3}{b^2}\geq \frac{2a^2}{b} -a$$
Similar: $$\frac{b^3}{c^2}\geq \frac{2b^2}{c} -b$$
$$\frac{c^3}{a^2}\geq \frac{2c^2}{a} -c$$
Now adding them together and using C-S, we have:
$$\frac{a^3}{b^2}+\frac{b^3}{c^2} + \frac{c^3}{a^2}$$
$$\geq (\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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What are the roots of $\sin(ax) + \sin((a + 2)x)$? I was playing around with $\sin(5x) + \sin(7x)$, wondering where the roots of the function are. I graphed it on wolframalpha and from the list of solutions I guessed that the solutions to $\sin(5x) + \sin(7x) = 0$ are exactly the roots of $\sin(6x)$.
After playing arou... | We transform that sum into a product. An easy way to deduce it is: We begin with $\sin X + \sin Y$. Let $X = c + d$ and $Y = c - d$. This gives us $c = \frac{X + Y}{2}$ and $d = \frac{X - Y}{2}$. This way: $$\begin{align} \sin X + \sin Y &= \sin(c+d)+ \sin(c-d) \\ &= \sin c \cos d + \sin c \cos d + \sin c \cos d - \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/884295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluating limit $\lim_{x\to0_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}$ Find the limit $$\lim_{x\to0_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}.$$
First I've rewritten it as $$\sqrt{ \frac{2x(1-x)}{\sin x}} \frac{\log(1-x^2)}{1-\cos x}$$
Now we can use the approximation $\sin x =... | You've found the right approach. I would use your knowlege about the Taylor approximations to intelligently split the limit up into 2 simpler limits then be more formal about solving it.
$$\begin{align}
\lim_{x\to0_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}} &= \lim_{x\to0_+} \left(\sqrt{ \frac{2x(1-... | {
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"url": "https://math.stackexchange.com/questions/884731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\sum_{k=0}^{511}\frac{\sin\frac\pi{2^{11}}}{\sin\frac{(4k+1)\pi}{2^{12}}\sin\frac{(4k+3)\pi}{2^{12}}}$ I need to evaluate
$$\sum_{n=0}^{511}\frac{\sin\frac\pi{2^{11}}}{\sin\frac{(4n+1)\pi}{2^{12}}\sin\frac{(4n+3)\pi}{2^{12}}}$$
Please give me some hint!
The final answer is $2^{10}$.
By CuriousGuest's answer, ... | Using the partial fractional decomposition, the sum can be restated as
$$\huge {\sum_{k=0}^{511} \cot{\frac{4k+1}{2^{12}}\pi} - \cot{\frac{4k+3}{2^{12}}\pi}}$$
Summing in reverse order gives
$$\huge {\sum_{k=0}^{511} \tan{\frac{4k+3}{2^{12}}\pi} - \tan{\frac{4k+1}{2^{12}}\pi}}$$
Using the identity $\cot{\theta} - \tan{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do I go from this $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$? So I am doing $\int\frac{x^2-3}{x^2+1}dx$ and on wolfram alpha it says the first step is to do "long division" and goes from $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$. That made the integral much easier, so how would I go about doing that in a clear... | Notice that:
$$
\frac{x^2 - 3}{x^2 + 1} = \frac{(x^2 + 1) - 4}{x^2 + 1}
= \frac{x^2 + 1}{x^2 + 1} - \frac{4}{x^2 + 1}
= 1 - \frac{4}{x^2 + 1}
$$
The hard part is the first step. Basically, we want the denominator to show up in the numerator somehow so that they cancel out nicely. So we apply wishful thinking and try to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/885999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
How to prove this determinant is $\pi$?
prove or disprove
$$\pi=\begin{vmatrix}
3&1&0&0&0&\cdots\\
-1&6&1&0&0&\cdots\\
0&-1&\dfrac{6}{3^2}&1&0&\cdots\\
0&0&-1&\dfrac{3^2\cdot 6}{5^2}&1&\cdots\\
0&0&0&-1&\dfrac{5^2\cdot 6}{3^2\cdot 7^2}&\cdots\\
0&0&0&0&-1&\dfrac{3^2\cdot 7^2\cdot 6}{5^2\cdot 9^2}&\cdots\\
0&0&0&0&0&... | A proof can be found on pages 11-13 of An Elegant Continued Fraction for $\pi$. Try deriving from the Leibnitz formula $1 - \frac{1}{3} + \frac{1}{5} - ... = \frac{\pi}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
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Prove that $ \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $ Can someone please help me with this question?
$ \large \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $
My steps so far:
$ \large \frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{x-2}.4^{x}}{2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{2... | $\frac{4^{2x-2}}{2^{x-2}} = \frac{(2^2)^{2x-2}}{2^{x-2}} = 2^{(4x-4)-(x-2)} = 2^{3x-2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Trigonometric Proof: Question:
If $m\cos\alpha-n\sin\alpha=p$ then prove that $m\sin\alpha+n\cos\alpha=\pm \sqrt{m^2+n^2-p^2}$
My Efforts:
$(m\cos\alpha-n\sin\alpha)^{2}=p^2$
$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$
Now i think we need to add something on both side but i can't figure out what to ... | $$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$$
multiply by $-1$ and add $m^2+n^2$ to both sides :
$$m^2+n^2-m^2\cos^2\alpha-n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=m^2+n^2-p^2$$
$$m^2(1-\cos^2\alpha)+n^2(1-\sin^2\alpha)-2mn\cos\alpha\ \sin\alpha=m^2+n^2-p^2$$
$$m^2(\sin^2\alpha)+n^2(\cos^2\alpha)-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluation of limit How to evaluate the value of this limit?
$$\lim_{x\to 2} \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{x^2 - 4}}$$
Actually I'm struck at algebraic part. Please guide..
| $$\eqalign{ \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{x^2 - 4}} &= \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{(x+2)(x-2)}} \\&= \frac{1}{\sqrt{x+2}} + \frac{\sqrt x - \sqrt2}{\sqrt{(x+2)(x-2)}} \\&= \frac{1}{\sqrt{x+2}} + \frac{(\sqrt x - \sqrt2)(\sqrt x + \sqrt 2)}{\sqrt{(x+2)(x-2)}(\sqrt x + \sqrt 2)} \\&= \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Is $\tan(x)$ differentiable for $x\in ( -\pi/2 , \pi/2 )$ This is an assignment question and in class we taught the definition that:
A function $f(x)$ is differentiable if we can find $f(x+h) - f(x) = Kh +h E(x,h)$, where $K=f'(x)$ and $E(x,h) \rightarrow 0$ as $h \rightarrow 0$
E.g. $f(x) = x^3 \Rightarrow f(x+h) -... | We can get our hands dirty with more trig identities:
\begin{align*}
\frac{(\tan h)(1+\tan^2 x)}{1-\tan x\tan h}
&= \frac{\frac{\sin h}{\cos h}(\sec^2 x)}{1 - \frac{\sin x\sin h}{\cos x \cos h}} \cdot \frac{\cos x \cos h}{\cos x \cos h} \\
&= \frac{(\sin h)(\sec^2 x)(\frac{1}{\sec x})}{\cos x \cos h - \sin x\sin h} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The inflection points of $f(x)=(x^2-4x+1)e^{-x}$ I got the function $f(x)=(x^2-4x+1)e^{-x}$. The task is to find the inflection points. The correct answer is $x=4-\sqrt{5}$ and $ x=4+\sqrt{5} $. I got the second derivative to $f(x)$.
But when I equal it to zero I got $x=2+\sqrt{5}$ and $ x=2 - \sqrt{5} $.
What am I do... | We have:
$$f(x) = P(x) e^x,$$
so $f'' = P'' e^x + 2 P' e^x + e^x P = e^x (2 + 4 x - 8 + x^2- 4x +1 ) = (x^2- 5) e^x$. Since $e^x > 0 $ for $x \in (-\infty,\infty)$, the solution of $f''(x) = 0$ is given by $x^2 -5 =0 $, so the inflection points are $x_i = \pm \sqrt{5}$.
Hope this helps.
Cheers!
Edit: have a look of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/889473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int_0^\infty \frac{x^n - 2x + 1}{x^{2n} - 1} \mathrm{d}x=0$ Inspired by some of the greats on this site, I've been trying to improve my residue skills. I've come across the integral
$$\int_0^\infty \frac{x^n - 2x + 1}{x^{2n} - 1} \mathrm{d}x=0$$
where $n$ is a positive integer that is at least $2$.
With non-... | Resolving the integrand into two partial fractions, we have
$$\displaystyle I=\int_0^{\infty} \frac{x^n-2 x+1}{\left(1+x^n\right)\left(1-x^n\right)} d x=\int_0^{\infty}\left(\frac{x}{1+x^n}-\frac{1-x}{1-x^n}\right) d x=J-K\tag*{} $$
For the integral $J$, we are going to transforms it into a Beta function by letting $y=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
} |
The other ways to calculate $\int_0^1\frac{\ln(1-x^2)}{x}dx$
Prove that
$$\int_0^1\frac{\ln(1-x^2)}{x}dx=-\frac{\pi^2}{12}$$
without using series expansion.
An easy way to calculate the above integral is using series expansion. Here is an example
\begin{align}
\int_0^1\frac{\ln(1-x^2)}{x}dx&=-\int_0^1\frac{1}{x}... | After substituting $y=x^2$, we obtain
$$
\int_0^1\frac{\ln(1-x^2)}{x}\ dx=\frac12\int_0^1\frac{\ln(1-y)}{y}\ dy
$$
Using the fact that
$$
\frac{\ln(1-x)}{x}=-\int_0^1\frac{1}{1-xy}\ dy
$$
then
$$
\frac12\int_0^1\frac{\ln(1-x)}{x}\ dx=-\frac12\int_{x=0}^1\int_{y=0}^1\frac{1}{1-xy}\ dy\ dx.
$$
Using transformation variab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 4
} |
Dealing with absolute values after trigonometric substitution in $\int \frac{\sqrt{1+x^2}}{x} \text{ d}x$. I was doing this integral and wondered if the signum function would be a viable method for approaching such an integral. I can't seem to find any other way to help integrate the $|\sec \theta|$ term in the numerat... | $ x = \sinh u$
$$S = \int \frac{\sqrt{1+x^2}}{x}\mathbb{d}x = \int \frac{\cosh^2 u}{\sinh u }\mathbb{d}u = \int \Big\{\mathbb{csch}\ u +\sinh u \Big\}\mathbb{d}u \\
=-\ln\Big(\mathbb{csch}\ u +\mathbb{coth}\ u\Big)+\cosh u + C\\
$$
$$\cosh \sinh^{-1} x = \frac{1}{2}\bigg\{x+\sqrt{x^2+1} + \frac{1}{x+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How prove $(x+\sqrt{x^{2}-1})^{n}+(x-\sqrt{x^{2}-1})^{n}\leq 2(1+n(x-1))^{n}$ for $n\in\mathbb{N}$? Let $x\ge 1$. How prove that $(x+\sqrt{x^{2}-1})^{n}+(x-\sqrt{x^{2}-1})^{n}\leq 2(1+n(x-1))^{n}$ for $n\in\mathbb{N}$?
| The easy way:
Let $f(x)=RHS-LHS$.
$f'(x)=2n^2(1+n(x-1))^{n-1}-n(1+\frac{x}{\sqrt{x^2-1}})(x+\sqrt{x^2-1})^{n-1}-n(1-\frac{x}{\sqrt{x^2-1}})(x-\sqrt{x^2-1})^{n-1}$
$f'(x)=2n^2(1+n(x-1))^{n-1}-\frac{n}{\sqrt{x^2-1}}(x+\sqrt{x^2-1})^{n}+\frac{n}{\sqrt{x^2-1}}(x-\sqrt{x^2-1})^{n}$
$f'(x) > 2n^2(1+n(x-1))^{n-1}-\frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/893802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Simplify the following compound fraction: $$\frac{2x+1}{\frac{3}{x^2}+\frac{2x+1}{x}}$$
My calculator says the final answer is $$\frac{x^2(2x+1)}{2x^2+x+3}$$
Please show the work. Thanks.
| The answer could be considered complete, but, as I said in my comments above, you could simplify a bit more since the numerator has a higher degree than the denominator, i.e.:$$
\frac{x^2(2x+1)}{2x^2+x+3}=\frac{2x^3+x^2}{2x^2+x+3}=\frac{2x^3+x^2+3x-3x}{2x^2+x+3}
$$$$=\frac{x(2x^2+x+3)-3x}{2x^2+x+3}=x-\frac{3x}{2x^2+x+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Counting words with specific requirements How many words of length exactly 6 can be created from letters A,B,C,D,E,F, so that each word has not more than one A, not more than one B, not less than one C and not less than one D?
I've tried using simple binomial coefficient, but it is clear that using this method one woul... | We can break this up into 4 cases:
1) If the word has 0 A's and 0 B's, then there are $4^6$ possible words, and subtracting the words with no C's and no D's gives $4^6-2(3^6)+2^6=2702$ possibilities.
2) If the word has 1 A and 1 B, then there are 6 choices for the A, 5 choices for the B, and then $[4^4-2(3^4)+2^4]$ way... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for positive reals Solve for positive reals $x,y$
$(x+y)(1+\frac{1}{xy})+4=2(\sqrt{2x+1}+\sqrt{2y+1})$
I started by accumulating the terms of $x$ and then used AM-GM inequality but unsuccessfully....
| First:
$$\begin{align}
(x+y)\left(1+\frac{1}{xy}\right)+4 &= x+y+\frac{1}y+\frac{1}x +4\\
&= \left(x+2+\frac{1}{x}\right)+\left(y+2+\frac{1}{y}\right) \\
&=\frac{1}{x} \left(x+1\right)^2 + \frac{1}{y}\left(y+1\right)^2
\end{align}$$
So, define $$f(x)=\frac{1}{x}\left(x+1\right)^2-2\sqrt{2x+1}$$
Then you want to find t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to prove that $\tan^2(\frac\theta2)= \tan^2(\frac\alpha2)\tan^2(\frac\beta2)$? I'm unable to solve this question:
$\cos(\theta)=\dfrac{\cos(\alpha)+\cos(\beta)}{1+\cos(\alpha) \cos(\beta)}$
Prove: $\tan^2\left(\frac\theta2\right)= \tan^2\left(\frac\alpha2\right)\tan^2\left(\frac\beta2\right)$
I have tried the foll... | $$ \tan ^2 {\frac {\theta}{2}} = \frac{\sin^2 {\frac {\theta}{2}} }{\cos^2 {\frac {\theta}{2}}} = \frac{ \left({\dfrac{1 - \cos \theta}{2}}\right)}{\left({\dfrac{1 + \cos \theta}{2}}\right)} \;\; \text{(since $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta= 1 - 2 \sin^2 \theta = 2 \cos^2\theta - 1$ )}$$
Now substituting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/901991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx$ Integrate: $$\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx$$
My attempt :
I substituted $\sec{x} + \tan{x} $ as $t^2$
Then, $$ (\sec{x} \cdot \tan{x} + \sec^2x) dx =2tdt$$
$$\sec{x}( \tan{x} + \sec{x}) dx =2tdt$$
$$\sec{x}\cdot t^2 dx =2tdt$$
But th... | If we use Weierstrass substitution, the problem becomes$$I=\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx=2 \int \frac{ \sqrt{\frac{1+t}{1-t}} \left(1+t^2\right)}{\left(1-t^2\right)^2}\,dt$$ Now, setting $$\sqrt{\frac{1+t}{1-t}}=z$$ the integral becomes $$I=\int (z^2+\frac{1}{z^2})\,dz=\frac{z^3}{3}-\frac{1}{z}$$ No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/902581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int\frac{\sqrt{9-x^2}}{x^2}\mathrm dx$ I am trying to solve $$\int\frac{\sqrt{9-x^2}}{x^2}\mathrm dx$$
My answer is slightly different to the memo:
$x=3\sin\theta\quad\iff\quad\theta=\arcsin\left(\frac x 3\right)\\
\text dx=3\cos\theta\ \text d\theta\\$
$\begin{align}I&=\int\frac{3\sqrt{1-\sin^2\theta}}{3\si... | $$\int\frac{\sqrt{9-x^2}}{x^2}dx=\vert x=3\sin t\Rightarrow dx=-3\cos t dt\vert=\int\frac{\sqrt{9-9\sin^2t}}{9\sin^2 t}\cdot (-3\cos t) dt=$$$$\int\frac{\sqrt{9(1-\sin^2 t)}}{9\sin^2 t}\cdot (-3\cos t) dt=\int\frac{-3\sqrt{1-\sin^2 t}}{9\sin^2 t}\cdot 3\cos t dt=$$
$$-\int\frac{\cos^2 t}{\sin^2 t }dt=-\int\cot^2 t dt=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/902969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why is $ A_1 x + ... + A_n x^n $ a solution of $ \sum_0^{n} (-1)^n \frac{x^n}{n!} \frac{d^n y}{d x^n} = 0 $? I was playing(/fiddling) around with some maths and I saw this pattern(
where $ A_n $ is a constant.):
$ A_1 x $ is a soultion of:
$$ \frac{y}{x} - \frac{dy}{dx} = 0 $$
$ A_1 x + A_2 x^2 $ is a solution of:
$$ \... | Use induction. Suppose the property holds for
$$ A_1 x + A_2 x^2 + A_3 x^3 + ... + A_n x^n $$
Prove that it works for $ n+1$ terms by writing the expression, use the fact that $\dfrac{\mathrm d^{n+1}y}{\mathrm dx^{n+1}}f = \dfrac{\mathrm d^{n}y}{\mathrm dx^{n}}\left(\dfrac{\mathrm dy}{\mathrm dx}f\right)$ and then sho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find $n$ between $100$ and $1000$ so that $2^n+2$ is divisible by $n$ Find $n$ such that $n$ divides $2^n + 2$. Also, $n$ should be between $100$ and $1000$.
It can be easily seen that $n$ is not a multiple of $4$. By brute force I have figured out that answer is $946$, but I don't know how to proceed further.
| See OEIS sequence A006517 and references there.
EDIT: $n$ must be even, but not divisible by $4$, and it's not hard to show
that $n=2p^j$ for prime $p$ doesn't work, so $n$ is divisible by at least two odd primes.
Let's say we guess that $n = 2 p q$ for odd primes $p < q$. Since $n/2 < 500 < 23^2$, we must have $p < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do I simplify $\frac{\sqrt{21}-5}{2} + \frac{2}{\sqrt{21} - 5}$?
How do I simplify the following equation?
$$\frac{\sqrt{21}-5}{2} + \frac{2}{\sqrt{21} - 5}$$
I have no idea where to start. If I multiply either fraction by its denominator I will still end up with a square root. I know the end result should be $-5... | $$\frac{\sqrt{21}-5}{2}+\frac{2}{\sqrt{21}-5}=\frac{\sqrt{21}-5}{2}+\frac{2(\sqrt{21}+5)}{(\sqrt{21}-5)(\sqrt{21}+5)}=\frac{\sqrt{21}-5}{2}+\frac{2\sqrt{21}+10}{-4}=\frac{2\sqrt{21}-10}{4}-\frac{2\sqrt{21}+10}{4}=\frac{2\sqrt{21}-10-2\sqrt{21}-10}{4}=-5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Dice, balls and boxes probability problem (conditional probability) Problem
Suppose there are two boxes $A$ and $B$ such that $A=\{\text{5 red balls and 3 white balls}\}$, $B=\{\text{1 red ball and 2 white balls}\}.$
A dice is thrown, if the result is $3$ or $6$, a ball from the box $A$ is extracted and then placed it ... | You can use
$P(R)=P(R|D_{3,6})P(D_{3,6})+P(R|D_{3,6}^{c})P(D_{3,6}^{c})=\left(\frac{5}{8}\cdot\frac{1}{2}\right)\cdot\frac{1}{3}+\left(\frac{1}{3}\cdot\frac{2}{3}\right)\cdot\frac{2}{3}=\frac{109}{432}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve the quadratic equation $x^2-1=2$?
Solve $x^2-1=2$
I have no idea how to do this can somebody please help me? I have tried working it out and I could never get the answer.
| I feel from the comments that you lack some understanding of equations in general, not just quadratic equation since this problem should be quite simple. Let's try and fix that!
First off, what does it mean to solve the equation $x^2-1=2$? It means to find the value(s) that we can plug into the $x$ such that the equali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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$(a_n)$ is bounded $\implies \sum a_n \frac{1}{n^2}$ converges Show that if $(a_n)$ is a bounded sequence, then
$$\sum \frac{1}{n^2} a_n $$ converges
I only could prove that this sequence is bounded, but not its convergence.
Thanks in advance.
| $$V \le a_k \le U\tag{k > 0}$$
$$\frac{1}{k^2}V \le \frac{1}{k^2}a_k \le \frac{1}{k^2}U$$
$$\sum_{k=0}^\infty \frac{1}{k^2}V \le \sum_{k=0}^\infty \frac{1}{k^2}a_k \le \sum_{k=0}^\infty \frac{1}{k^2}U$$
$$\frac{\pi^2}{6}V \le \sum_{k=0}^\infty \frac{1}{k^2}a_k \le \frac{\pi^2}{6}U$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/909053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
What is the limit regarding $a$ What is the limit of :
$$ \lim_{x\to 0} \frac{\sin(ax) - \ln(1-2x)}{e^{ax}-1-2x-2x^{2}}$$
I did this with Maclaurin, because my exam is about solving these with MacLaurin.
Gave $$\lim_{x\to 0} \frac{ax-\frac {(ax)^{2}}{3!} +2x +4x^{2}}{1+ax+(ax)^{2} -1-2x-2x^2} = \frac {a+2}{a-2}$$
The ... | $\displaystyle\lim_{x\to 0}\frac{\sin(ax)-\ln(1-2x)}{e^{ax}-1-2x-2x^2}=\lim_{x\to 0}\frac{ax-\frac{(ax)^3}{3!}+\cdots-[-2x-\frac{(2x)^2}{2}-\frac{(2x)^3}{3}+\cdots]}{1+ax+\frac{(ax)^2}{2!}+\frac{(ax)^3}{3!}+\cdots-1-2x-2x^2}$
$=\displaystyle\lim_{x\to 0}\frac{(a+2)x+2x^2+(\frac{8}{3}-\frac{a^3}{6})x^3+\cdots}{(a-2)x+(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/910059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\ge xyz(x+y+z)^3$
if $x,y,z$ are positive real numbers,Prove:$$3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\ge xyz(x+y+z)^3$$
Additional info:$\sum_{cyc}$ denotes sums over cyclic permutations of the symbols $x,y,z$. I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM be... | The inequality is equivalent to
$$3\sum x^3y^3 +2\sum xy^4z+3x^2y^2z^2\ge 3\sum(xy^3z^2+x^2y^3z) =3\sum xy^3z(x+z)$$
AM-GM gives
$$x^3y^3+xy^4z+x^2y^2z^2\ge 3 x^2y^3z$$
$$y^3z^3+xy^4z+x^2y^2z^2\ge 3xy^3z^2$$
Summing up we have
$$2\sum x^3y^3+2\sum xy^4z+6x^2y^2z^2\ge 3\sum (x^2y^3z+xy^3z^2),$$
or
$$\sum x^3y^3+\sum x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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For What $a$ The Linear Equations Have Sloutions $2x+ay-z=-2$
$x-3z=-3$
$x+2y+az=-1$
I have thought about reducing a matrix so in the end I will have an equation with $a$ then I can determine for which $a$ the are one solution/infinite solutions/no solution, but it does not seems to work
I have row reduced the matrix... | I got the Row Reduced Form as:
$ \left( \begin{array}{ccc|c}
1 & 2 & a & -1 \\
0 & 1 & \frac{a+3}{2} & 1 \\
0 & a-5 & \frac{-5(a+1)}{2} & -1 \end{array} \right)$
So the determinant comes as $(-3a+10-a^2)$
For no solution, determinant is zero, so, $(-3a+10-a^2)=0$
$\Rightarrow a=-5, a=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/911876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How do I solve $x^5 +x^3+x = y$ for $x$? I understand how to solve quadratics, but I do not know how to approach this question. Could anyone show me a step by step solution expression $x$ in terms of $y$?
The explicit question out of the book is to find $f^{-1}(3)$ for $f(x) = x^5 +x^3+x$
So far I have reduced $x^5 +x^... | Solving it for a general $y$ will be difficult, but solving for $y=0$ is more doable.
If $x^5+x^3+x=0$, then because the left hand side is divisible by $x$, either $x=0$ or we can divide by $x$ to get $(x^2)^2+x^2+1=0$. This is a quadratic in $x^2$, so by the quadratic formula, $x^2=\frac{-1\pm \sqrt{-3}}{2}.$
Thus, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 5
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How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$
let $x,y,z>0$, find the minimum of the value
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
I think we can use AM-GM inequality to find it.
$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4... | Here is another way, using Holder's inequality:
$$\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz} \geqslant \frac{(\sqrt[4]{5y \cdot 5\cdot x\cdot z}+\sqrt[4]{2\cdot2z\cdot3y\cdot3x})^4}{xyz} \\ = (\sqrt5+\sqrt6)^4=241+44\sqrt{30}$$
with equality iff $\dfrac{5y}2=\dfrac5{2z}=\dfrac{x}{3y}=\dfrac{z}{3x} \iff (x, y, z) = \left(1, \... | {
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Evaluating $\lim\limits_{(x,y)\rightarrow(1,1)} \frac {\sin(x) - \sin (y)} {x-y}$ I am taking a calculus exam in less than one week, and I've stumbled upon this expression.
$$\lim\limits_{(x,y)\rightarrow(1,1)} \frac {\sin(x) - \sin (y)} {x-y}$$
Which I know to be cos(1), but I cannot seem to find the inequalities to... | You can use this identity:
$$\sin x - \sin y = 2\sin \frac{x - y}{2}\cos \frac{x + y}{2}.$$
Then your limit becomes:
$$\lim_{(x, y) \to (1,1)} \frac{2\sin \frac{x - y}{2}\cos \frac{x + y}{2}}{x - y} =$$
$$2 \lim_{(x, y) \to (1,1)} \frac{\sin \frac{x - y}{2}\cos \frac{x + y}{2}}{x - y} =$$
$$\frac{1}{2} \cdot2 \lim_{(x,... | {
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The Shortest Distance Between 2 Points On The Earth
Assuming that the earth is a perfect sphere with radius 6378 kilometers, what is the expected straight line distance through the earth (in km) between 2 points that are chosen uniformly on the surface of the earth?
| View the sphere as the surface of revolution of the circle $$x^2+y^2=r^2 \tag 1$$ about the $x$-axis. Differentiating both sides of $(1)$ we get
$$
2x\,dx+2y\,dy=0\quad\text{or}\quad x\,dx+y\,dy=0\quad\text{or}\quad\frac{dy}{dx}=\frac{-x}{y}.
$$
The distance from $(r,0)$ to $(x,y)$ is $\sqrt{(x-r)^2+y^2}$, by the Pyth... | {
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When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:
Here's how this question works. To motivate what I'll be doing,
consider \begin{equat... | Colloquially (as I find sometimes works with my daughters at high school)
When we knock out $14$s the remainder is $10$. We knock out $7$s - well the $14$s all go because $14 = 2\times 7$ and we're left with $10=7+3$.
| {
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How can I accurately compute $\sqrt{x + 2} −\sqrt{x}$ when $x$ is large? How can the values of the function $f(x) = \sqrt{x + 2} −\sqrt{x}$ be computed accurately when $x$ is large?
I have tried using Matllab. I am not able to understand when $x$ will be large.
| If directly computing $\sqrt{x+2}-\sqrt{x}$ is giving you problems, you can instead try using
$\sqrt{x+2}-\sqrt{x} = \dfrac{(\sqrt{x+2}-\sqrt{x})(\sqrt{x+2}+\sqrt{x})}{\sqrt{x+2}+\sqrt{x}} = \dfrac{(x+2)-x}{\sqrt{x+2}+\sqrt{x}} = \dfrac{2}{\sqrt{x+2}+\sqrt{x}}$.
For large $x$, this is approximately $\dfrac{2}{\sqrt{x... | {
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Finding $\large\zeta_7\left(\zeta_3\right)^5$ where $\large\zeta_n=\cos{\frac{2\pi}{n}}+i\sin{\frac{2\pi}{n}}$ $\large\zeta_7\left(\zeta_3\right)^5$ where $\large\zeta_n=\cos{\frac{2\pi}{n}}+i\sin{\frac{2\pi}{n}}$
I am having trouble getting a final answer that makes sense to me. Here is what I tried:
$\large\zeta_7\le... | For this you need to know that $$(A \text{ cis } B) \cdot (C \text{ cis } D) = (AC) \text{ cis } (B + D)$$
And that
$$(A \text{ cis } B)^n = \left(A^n \text{ cis } Bn\right)$$
Your problem is
$$\large\zeta_7\left(\zeta_3\right)^5$$
$$\left(1 \text{ cis } \frac{2\pi}{7}\right)\left(1 \text{ cis } \frac{2\pi}{3}\right)^... | {
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Integrate $\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+C$ Wolfram gives this nice result:
$$\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+\text{constant}$$
I have tried writing $\cos 2x = \cos^2x - \sin^2x $ and doing Weierstrass substitution $\tan (x/2) = t$ but its getting ver... | Alternative method (akin to finding an integrating factor by inspection):
$\cos x = \cos (2x - x) = \cos 2x \cos x + \sin 2x \sin x$.
Then the integral becomes:
$$\begin{align}
\int \dfrac{\cos 2x \cos x + \sin 2x \sin x}{\cos^{\frac 3 2} 2x} dx
& = \int \dfrac{\cos x}{\sqrt {\cos 2x}} + \dfrac{\sin x \sin 2x}{\cos^\fr... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Show that two expressions are equivalent I am trying to prove a hyperbolic trigonometric identity and I ran into the following expression:
$$\frac{\left (\sqrt{x^2+1}+x \right )^2+1}{2\left ( \sqrt{x^2+1} + x \right )} \quad.$$
This expression is supposed to be equivalent to
$$\sqrt{x^2+1} \quad.$$
I tried to algebrai... | What you got is correct . Then, multiply $$\frac{1}{\sqrt{x^2+1}+x}$$ by $$\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x}(=1).$$
| {
"language": "en",
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Fractions in Questions and Answers
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