Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find all the solutions in the complex field of the following equation: Firstly my apologies for anything that should be in LaTex format correctly, I gave it a valiant effort. I have been asked to solve the equation: $(1-z)^6 = (1+z)^6$ A hint given states: do not multiply out!. If the equation was in a more suitable fo... | Ignoring the given hint, you can solve it by expanding out terms.
Take the square root of both sides (we get two options):
$$(1-z)^6=(1+z)^6\Longleftrightarrow (1-z)^3=\pm(1+z)^3$$
*
*For $(1-z)^3=(1+z)^3$:
$$(1-z)^3=(1+z)^3\Longleftrightarrow$$
$$(1-z)^3=z^3+3z^2+3z+1\Longleftrightarrow$$
$$(1-z)^3-(z^3+3z^2+3z+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can you solve $y=\frac{a}{2}x^2\left(y'-\frac{1}{y'}\right)^2+x\left(y'-\frac{1}{y'}\right)+ax^2+c$? I've recently come across this differential equation, but I am having trouble proceeding toward a solution.
$y=\frac{a}{2}x^2\left(y'-\frac{1}{y'}\right)^2+x\left(y'-\frac{1}{y'}\right)+ax^2+c$
where $y'=\frac{dy}{dx}$ ... | I don't know if an implicit solution fits your needs, anyway with this solution you could at least find $x$ as a function of $y$.
Put $y'=\tan z$, $y=-\log k|\cos z|$, $z= \pm \arccos {e^{-y} \over k}$. So: $$
-\log k|\cos z|{\sin^2 z \over \cos^2 z}={a \over 2}{x^2 \over \cos^2 z}+x {\tan z \over \cos^2 z}+ax^2{\sin^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+... | To break a rational polymomial expression into parts, the degree of the numerator must be less than the degree of the denominator.
This is not the case with $\dfrac{x^4+1}{x^3+x^2}$. Using long division, we find
\begin{array}{rcccccccc}
& & x & - & 1 &\\
& & --- & --- & --- & --- & ---... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Irreducible representation of $S_3$ How can I show that this representation of $S_3$ is irreducible?
$$\rho\left(e\right)=\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right),\,\,\rho\left(a_{1}\right)=\frac{1}{2}\left(\begin{array}{cc}
-1 & -\sqrt{3}\\
\sqrt{3} & -1
\end{array}\right),\,\,\rho\left(a_{2}\right)=\f... | Well, if it were reducible, $\rho(a_3)$ and $\rho(a_4)$ would share an eigenvector. Which is clearly not the case, because the standard basis is the only (up to scalar multiplication and permutation) basis of eigenvectors of $\rho(a_3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integral with a parameter Hello I try solve a problem on integration
$\int_0^\frac{\pi}{2}\frac{\arctan(y\tan(x))}{\tan(x)}dx$
My stuck is that I use differentiation under integral sign and producing $ I'(y) = \int_0^ \frac{\pi}{2}\frac{dx}{1+(y\tan(x))^2}$ what can I do?
Thanks
The solution is $\frac{\pi}{2}\text{sg... | First, note that the integral given by
$$I(y)=\int_0^{\pi/2}\frac{\arctan(y\tan(x))}{\tan(x)}\,dx$$
is an odd function with $I(y)=-I(-y)$. Therefore, we examine the case for $y\ge 0$.
Next, the integral given by
$$I'(y)=\int_0^{\pi/2} \frac{1}{1+y^2\tan^2(x)}\,dx \tag 1$$
can be evaluated a host of ways.
In doing so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to calculate $\lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)^{-1}$? $$f (x) = \frac{1}{x^2} - \frac{1}{\sin^2 x}$$
Find limit of $\dfrac1{f(x)}$ as $x\to0$.
| You don't really need to work with $1/f(x)$; consider
$$
f(x)=
\frac{\sin^2x-x^2}{x^4}\frac{x^2}{\sin^2x}=
\frac{\sin x-x}{x^3}\frac{\sin x+x}{x}\frac{x^2}{\sin^2x}=
\frac{\sin x-x}{x^3}\left(\frac{\sin x}{x}+1\right)
\left(\frac{x}{\sin x}\right)^2
$$
Since the limits of the second and third factors are well known, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Which of the numbers is larger: $7^{94}$ or $9^{91} $? In this problem, I guess b is larger, but not know how to prove it without going to lengthy calculations. It is highly appreciated if anyone can give me a help.
Which number is larger
$$\begin{align}
&\textrm{(a)}\quad 7^{94}
&\quad\textrm{(b)}\quad 9^{91} ... | I voted for André's answer, but here's another approach, using a different bit of maths.
Note that $7^{94} = 7^3 \times 7^{91}$. $9^{91} = (7 \times \frac{9}{7})^{91}$, where $\alpha = \frac{9}{7} = 1 + \frac{2}{7} > 1$. So
$$
\frac{7^{94}}{9^{91}} = \frac{7^3}{\alpha^{91}}.
$$
What do we make of $\frac{7^3}{\alpha^{91... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 6
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Someone please explain cosine similarity equation to me? I'm trying to understand the cosine similarity in a simple and graphical way, very much similar to this question here but I do not understand how the person got to their answer.
| The vectors are $(x_1 \ x_2 \ x_3 \ x_4 \ x_5 \ x_6 \ x_7 \ x_8)^T=(2 \ 0 \ 1 \ 1 \ 0 \ \ 2 \ 1 \ 1)^T$ and $(y_1 \ y_2 \ y_3 \ y_4 \ y_5 \ y_6 \ y_7 \ y_8)^T=( \ 2 \ 1 \ 1 \ 0 \ 1 \ 1 \ 1 \ 1)^T$
The cosine of angle ($\alpha$) between the two vectors is $cos(\alpha)=\large{\frac{\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sin^2(10 ^\circ)-\sin^2(20^\circ)-\sin^2(40^\circ)=-\frac{1}{2}$ identity 10 degrees
$$\sin^2(10^\circ)-\sin^2(20^\circ)-\sin^2(40^\circ)=-\frac{1}{2}$$
$$\cos^2(10^\circ)-\cos^2(20^\circ)-\cos^2(40^\circ)=-\frac{1}{2}$$
Why are they both have same answer?
The only time they have same answer is at 45 degrees r... | Let $C=\cos^2A-\cos^2B-\cos^2C$
and $S=\sin^2A-\sin^2B-\sin^2C$
$\implies C+S=-1$
and $C-S=\cos2A-\cos2B-\cos2C=\cos2A-2\cos(B+C)\cos(B-C)$ using Prosthaphaeresis Formula
If $C-S=0, C=S=-\dfrac12\ \ \ \ (0)$
Now if $B+C=60^\circ$ or more generally, $360^\circ n\pm60^\circ, \ \ \ \ (1)$
$C-S=\cos2A-\cos(B-C)$ will be $0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1791258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of $\lim\limits_{x\rightarrow 0}\frac1x\left((1+2x+3x^2)^{1/x}-(1+2x-3x^2)^{1/x}\right) $
Evaluation of $$\lim_{x\rightarrow 0}\frac{(1+2x+3x^2)^{\frac{1}{x}}-(1+2x-3x^2)^{\frac{1}{x}}}{x} $$
$\bf{My\; Try::}$ Let $$l=\lim_{x\rightarrow 0}\frac{e^{\frac{\ln(1+2x+3x^2)}{x}}-e^{\frac{\ln(1+2x-3x^2)}{x}}}{x}$... | As usual this limit can also be evaluated without the use of L'Hospital's Rule and Taylor's series just by applying standard limits combined with the use of algebra of limits. We have
\begin{align}
L &= \lim_{x \to 0}\frac{(1 + 2x + 3x^{2})^{1/x} - (1 + 2x - 3x^{2})^{1/x}}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Show $\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$? Prove that:
$$\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$$
The LHS is irrational number and RHS is rational number. May ... | Let me try.
Note that $$(1+\sqrt[3]{2})^3 = 3(1+\sqrt[3]{2}+\sqrt[3]{4}).$$
Now we have:
$$LHS = \sqrt[3]{4} - \frac{1}{2}(\sqrt[3]{4} + \sqrt[3]{2} + 1) + \left(\frac{9}{4(\sqrt[3]{4}+\sqrt[3]{2}+1) }\right)^{\frac{1}{3}} = \sqrt[3]{4} - \frac{1}{2}(\sqrt[3]{4} + \sqrt[3]{2} + 1) + \left(\frac{27}{4(\sqrt[3]{2}+1)^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove the fractions aren't integers
Prove that if $p$ and $q$ are distinct primes then $\dfrac{pq-1}{(p-1)(q-1)}$ is never an integer. Is it similarly true that if $p,q,r$ are distinct primes then $\dfrac{pqr-1}{(p-1)(q-1)(r-1)}$ is also never an integer?
I think using a modular arithmetic argument here would help. I... | I will assume that
$p \le q$,
not just
$p < q$.
Also,
you don't need the assumption
about primality.
I will show that
the only solutions are
$p=q=2$
and
$p=q=3$.
If
$p=2$
then
$\dfrac{pq-1}{(p-1)(q-1)}
=\dfrac{2q-1}{q-1}
=\dfrac{2q-2+1}{q-1}
=2+\dfrac{1}{q-1}
$
which is not an integer
unless $q = 2$.
If
$p=3$
then
$\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
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Prove $A=B=\pi$ $\gamma=0.57721566...$
$\phi=\frac{1+\sqrt5}{2}$
Let,
$$A=\sum_{n=1}^{\infty}\arctan\left(\frac{(e+e^{-1})(\phi^{\frac{2n-1}{2}}+\phi^{\frac{1-2n}{2}})}{e^2+e^{-2}-1+\phi^{2n-1}+\phi^{1-2n}}\right)$$
$$B=\sum_{n=1}^{\infty}\arctan\left(\frac{(\gamma+\gamma^{-1})(\phi^{\frac{2n-1}{2}}+\phi^{\frac{1-2n}{2... | For any $a \ge 1$, let $f_n(a)$ be the expression
$$f_n(a) \stackrel{def}{=}\frac{(a+a^{-1})(\phi^{n-1/2} + \phi^{1/2-n})}{a^2 + a^{-2} - 1 + \phi^{2n-1} + \phi^{1-2n}}$$
The sums at hand equal to $\sum\limits_{n=1}^\infty \tan^{-1}f_n(e)$ and $\sum\limits_{n=1}^\infty \tan^{-1}f_n(\gamma^{-1})$ respectively.
For any $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
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Solution of $4 \cos x(\cos 2x+\cos 3x)+1=0$ Find the solution of the equation:
$$4 \cos x(\cos 2x+\cos 3x)+1=0$$
Applying trigonometric identity leads to
$$\cos (x) \cos \bigg(\frac{x}{2} \bigg) \cos \bigg(\frac{5x}{2} \bigg)=-\frac{1}{8}$$
But I can't understand what to do from here. Could some suggest how to proceed... | Hint:
$1)$ Note that:
*
*$\cos{2x}=2\cos^2{x}-1$, and
*$\cos{3x}=4\cos^3{x}-3\cos{x}$
$2)$ Using this, setup a cubic equation in $\cos{x}$.
$3)$ Now, solve the quartic equation by making the substitution $t=\cos{x}$.
$4)$ Finally, back-substitute and use $\cos{x}=\cos{\alpha} \Rightarrow x=2n\pi \ \pm \alpha$, where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
Question: If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
My solution: Consider
$$x^2 + y^2 + z^2 < 2(xy + yz + xz)$$
Notice that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$
Hence... | $x,y$ and $z$ are sides of a triangle. So, $y+z\gt x,\; x+y\gt z,\; z+x\gt y$.
Now you have to prove that $$x^2+y^2+z^2\lt2\left(\sum xy\right).$$
Just a simple look through the right hand term.
$$2\left(\sum xy\right)=2xy+2yz+2zx\\=x\color{blue}{(y+z)}+y\color{blue}{(z+x)}+z\color{blue}{(x+y)}\gt x\times \color{blu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Parametric equations for intersection between plane and circle So I was looking at this question Determine Circle of Intersection of Plane and Sphere
but I need to know how to find a parametric equation for intersections such as these. My particular question is to find a parametric equation for the intersection betwee... | The plane $ \zeta = \{(x,y,z) : x + y + z = 1\},$ is perpendicular to the vector
$\langle 1,1,1 \rangle$.
Two points on $\zeta$ are
$\mathbf O = \left( \dfrac 13,\ \dfrac 13,\ \dfrac 13 \right)
\text{ and }
\mathbf A = (0,0,1)$.
Since $\overrightarrow{\mathbf{OA}} = \dfrac 13 \langle -1, -1, 2 \rangle$,
the unit vect... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Find the value of $a^2-b^2+c^2$ Let $a$, $b$, and $c$ be real numbers such that $a − 7b + 8c = 4$ and $8a + 4b − c = 7$. What is the
value of $a^2-b^2+c^2$ ?
| You can also do it solving the two equations for $b$ and $c$ to be expressed as functions of $a$. This would give $$b=\frac{12}{5}-\frac{13 a}{5}\qquad c=\frac{13}{5}-\frac{12 a}{5}$$ Now, replace in the expression, expand and simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Elasticity of the function Please find the elasticity of the function f(x) in x0:
$$f(x) = \frac{ax^2} {1+ x^2} $$ $$ x0=10$$
Could you please check is done correct ?
$$E_xf(x_0) = \frac{f'(x_0)·x_0} {f(x_0)} $$
$$f'(x)=\frac{2ax·(1+x^2)-2ax^2·2x} {(1+x^2)^2} = \frac{2ax} {(1+x^2)^2} $$
$$E_xf(x) = \frac{\frac{2ax} {(... | As you have mentioned before, the function for elasticity of a function with respect to $x$ is
$$E_x(f(x)) = \frac{f'(x) \cdot x}{f(x)}.$$
We see by the quotient rule that
$$f'(x) = \frac{(2ax)(1+x^2) - (ax^2)(2x)}{(1+x^2)^2}.$$
Thus,
$$E_x(f(x)) = \frac{\frac{(2ax)(1+x^2) - (ax^2)(2x)}{(1+x^2)^2} \cdot x}{\frac{ax^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How should I try to evaluate the integral $\int_a^b \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ I've tried to evaluate $\displaystyle\int_{-r}^r \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ on my own, but I have encountered a problem I cannot get around.
The indefinite integral $\sqrt{\frac{r^2}{r^2-x^2}} \sqrt{r^2-x^2} \tan ^{... | I think you can just find a common denominator and get to $\int \frac{r}{\sqrt{r^2-x^2}}dx=\int \frac{1}{\sqrt{1-(\frac{x}{r})^2}}$. Then you are a u-sub away from finding it as $r\arcsin(\frac{x}{r}) + C $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Joint distribution weird result
We have 3x3 board Each board cell has $0.5$ chance to be white
(there is no dependency between different cell colors)
Let X = number of white rows (a row with only white cells on it)
Let
N = total number of white cells in the whole board
Find $P\{X=0,N=3\}$
Find $P\{X=0 |N=... | First note that
$$P(X=0,N=3)=P(X=0\,|\,N=3) P(N=3)$$
We will find $P(X=0 \,|\,N=3)$ then $P(N=3)$ separately, then multiply them together.
Part 1: $P(X=0\,|\,N=3)$
With three white squares, there can be at most one white row. This means that if $N=3$, then $X=0$ or $X=1$ are the only possibilities. Therefore
$$P(X=0\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Prove that $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \frac{\pi}{2}\frac{1}{4^{n}}\binom{2n}{n}$ and also find value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
$\bf{My\; Try::}$ Let $$\displaystyle I_{n} = \int_{0}^{\fr... | Maybe it is interesting to see that $$S=\sum_{n\geq0}\dbinom{2n}{n}x^{n}=\sum_{n\geq0}\frac{\left(2n\right)!}{n!^{2}}x^{n}$$ and now note that $$\frac{\left(2n\right)!}{n!}=2^{n}\left(2n-1\right)!!=4^{n}\left(\frac{1}{2}\right)_{n}=\left(-1\right)^{n}4^{n}\left(-\frac{1}{2}\right)_{n}
$$ where $(x)_{n}$ is the Pochham... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
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Extending the ordered sequence of 'three-number means' beyond AM, GM and HM I want to create an ordered sequence of various 'three-number means' with as many different elements in it as possible.
So far I've got $12$ ($8$ unusual ones are highlighted):
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{ \frac{\sqrt{x^2+... | Let $\mathbf{x}$ denote a vector of positive real numbers (you can take it for your purposes to be 3 such numbers). Let $\mathbf{w}$ denote a vector of positive real numbers of the same length (i.e. 3-dimensional in this case), with $\sum_i w_i=1$; the simplest option is all $w_i$ being equal. Define $M_p:=\left(\sum_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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$\int\frac{\sin x}{\sqrt{1-\sin x}}dx=?$ Calculate this integral $\displaystyle\int\dfrac{\sin x}{\sqrt{1-\sin x}}dx=?$
Effort;
$1-\sin x=t^2\Rightarrow \sin x=1-t^2\Rightarrow \cos x=\sqrt{2t^2-t^4}$
$1-\sin x=t^2\Rightarrow-\cos x dx=2tdt\Rightarrow dx=\frac{2t}{\sqrt{t^4-2t^2}}dt$
$\displaystyle\int\frac{1-t^2}{t}... | I might be missing something, but I think this is possible to do with a prolonged substitution:
$$
\int \frac{\sin(x)}{\sqrt{1-\sin(x)}} \ dx.
$$
Let $u = 1-\sin(x)$, then $$ du = -\cos(x) dx \implies -\sec(x) du = dx.$$
Then we get $$ \int (1-u) \frac{1}{\sqrt{u}} (-\sec(x) du).$$
We need to find $\sec(x)$ in terms o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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How is $\lim_{x \to a}\left(\frac{x^n - a^n}{x - a}\right) = n\times a^{n-1}$? In my book this is termed as a theorem and the proof given is as follows :-
$$\begin{align}
\lim_{x \to a}\left(\frac{x^n - a^n}{x - a}\right)
&=\lim_{x \to a}\left(\frac{(x - a)*(x^{x-1} + x^{n-2}*a + x^{n-3}*a^2 + x^{n-4}*a^3 + \cdots + x^... | No, this isn't a binomial expansion. This is just long division. To check that it makes sense, we can just expand it out:
\begin{align*}
&(x - a)(x^{n - 1} + ax^{n - 2} + a^2x^{n - 3} + \cdots + a^{n - 2}x + a^{n - 1}) \\
&= x(x^{n - 1} + ax^{n - 2} + a^2x^{n - 3} + \cdots + a^{n - 1}) \\
&~~~~~~~~~~~~~~~~~- a(x^{n - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving by induction that $\sum\limits_{i=1}^n\frac{1}{n+i}=\sum\limits_{i=1}^n\left(\frac1{2i-1}-\frac1{2i}\right)$ I have a homework problem to prove the following via induction:
$$\sum_{i=1}^n \frac{1}{n+i} = \sum_{i=1}^n \left(\frac{1}{2i-1} - \frac{1}{2i}\right) $$
The base case is true.
So far I've done the follo... | As Thomas Andrews notes, the key part lies in reindexing the sum (so you can apply the inductive hypothesis, IH). More explicitly, the following is the core part of the inductive argument:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{k+1+i}&=\sum_{i=2}^{k+2}\frac{1}{k+i}\tag{reindex}\\[1em]
&= \sum_{i=1}^k\frac{1}{k+i}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Show that: $97|2^{48}-1$ Show that: $97|2^{48}-1$
My work:
$$\begin{align}
2^{96}&\equiv{1}\pmod{97}\\
\implies (2^{48}-1)(2^{48}+1)&=97k\\
\implies (2^{24}-1)(2^{24}+1)(2^{48}+1) &=97k\\
\implies (2^{12}-1)(2^{12}+1)(2^{24}+1)(2^{48}+1)&=97k\\
\implies (2^6-1)(2^6+1)(2^{12}+1)(2^{24}+1)(2^{48}+1) &=97k
\end{align}$... | $$97\equiv 1\pmod8$$
Thus, $2$ is a quadratic residue $\bmod 97$. So, there exists $a$ such that $a^2\equiv 2 \pmod {97}$. Thus, $2^{48}\equiv a^{96}\equiv 1\pmod{97}$, as desired.
This solution will work for any prime number $p$ that is $\pm 1\pmod{8}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$ Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n(n$ arbitrary positive integer)
First of all I checked a few cases for small $n$'s and in all cases the rightmost digit was $5$, so maybe this is the case for all values of $n$.
Then I tho... | Given that the order of any of the components here will divide $4$, since that is the Carmichael function value for $10$, it is only necessary to check the values for $n=\{1,2,3, 4\}$. The results follow a pattern across the range of values (all $\bmod 10$):
$$\begin{array}{c|c}
n & 1^n & 2^n & 3^n & 4^n & 5^n & 6^n & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Prove that $-2, -3, -5, -7$ are quadratic non-residues modulo p Prove that if $p$ is prime and $p\equiv 173 \pmod{1680}$, then $-2, -3, -5, -7$ are quadratic non-residues modulo p.
| Since $p\equiv1\pmod4,$ it follows that $\left(\dfrac{-1}{p}\right)=1.$ Then $p\equiv5\pmod8,$ so $\left(\dfrac{2}{p}\right)=-1.$
Also, $\left(\dfrac{-3}{p}\right)=(-1)^{(p-1)/2}(-1)^{(p-1)/2}\left(\dfrac{p}{3}\right)\equiv p\pmod3.$ But $p\equiv-1\pmod3,$ and hence $\left(\dfrac{-3}{p}\right)=-1.$
Now $\left(\dfrac{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Contour integral for finding $\int_0^\infty \frac{\ln x}{(x+a)^2+b^2} \, dx$ I can't prove the following result:
$$\displaystyle\int_0^\infty \frac{\ln x}{(x+a)^2+b^2} \, dx=\frac{\ln \sqrt{a^2+b^2}}{b}\arctan\frac{b}{a} \text{ for all } a,b \in \mathbb{R}.$$
Well, I consider $\displaystyle\int_C \frac{\operatorname{Ln... | An alternative, real-analytic solution by symmetry only.
$$ \int_{0}^{+\infty}\frac{\log x}{x^2+2ax+(a^2+b^2)}\stackrel{x\mapsto u\sqrt{a^2+b^2}}{=}\frac{1}{\sqrt{a^2+b^2}}\int_{0}^{+\infty}\frac{\log u+\log\sqrt{a^2+b^2}}{u^2+\frac{2a}{\sqrt{a^2+b^2}}u+1}\,du $$
and since the polynomial $p(u)=u^2+\frac{2a}{\sqrt{a^2+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Given that $\tan 2x+\tan x=0$, show that $\tan x=0$ Given that $\tan 2x+\tan x=0$, show that $\tan x=0$
Using the Trigonometric Addition Formulae,
\begin{align}
\tan 2x & = \frac{2\tan x}{1-\tan ^2 x} \\
\Rightarrow \frac{2\tan x}{1-\tan ^2 x}+\tan x & = 0 \\
\ 2\tan x+\tan x(1-\tan ^2 x) & = 0 \\
2+1-\tan ^2 x & = 0 ... | Notice that the following conditions are equivalent:
\begin{align*}
\tan x + \tan y &= 0\\
\tan x &= -\tan y\\
\tan x &= \tan(-y)\\
x &= -y + k\pi\\
x+y&= k\pi
\end{align*}
If we use $y=2x$ we get
\begin{align*}
3x &= k\pi\\
x &= k\frac\pi3
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Find the equations of the tangent planes to the sphere $x^2+y^2+z^2+2x-4y+6z-7=0,$ which intersect in the line $6x-3y-23=0=3z+2.$ Find the equations of the tangent planes to the sphere $x^2+y^2+z^2+2x-4y+6z-7=0,$ which intersect in the line $6x-3y-23=0=3z+2.$
Let the tangent planes be $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+... | Two easy points on the required line are $(0,-\frac{23}{3},-\frac{2}{3}),(\frac{23}{6},0,-\frac{2}{3})$. For these to lie in the plane $Ax+By+Cz+D=0$ we require $23A-4C+6D=0,-46B-4C+6D=0$. Subtracting gives $A=-2B$, and we also have $3D=23B+2C$.
The sphere has equation $(x+1)^2+(y-2)^2+(z+3)^2=21$, so it has centre $(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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proof of $\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}$ for large $n$ I'm trying to conclude this question in which I just need to prove that:
$$0<\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}$$
and then I'll have that, for large $n$:
$$\begin{align}
\left\lvert\cos\left(\sqrt... | Cross-multiply, square, subtract ${1\over n}$ from both sides to get
$$
0\le 2x^2+2x\sqrt{x^2+{1\over n}}
$$
Assuming $x\in[0,1]$ the above identity is trivial as each term on the RHS is non-negative. Now backtrack and get your desired identity.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does this formula ${n^2+3n\over 2}+{2(n+1)(n+2)-1\over 2(n+1)(n+2)}$ generates Pythagorean triples for all n? The idea came from this site
another formula for generating Pythagoras Triples
Let $n\ge1$
$2{11\over 12}, 5{23\over24}, 9{39\over 40},\cdots$ is generated from ${n^2+3n\over 2}+{2(n+1)(n+2)-1\over 2(n+1)(n+2)}... | Let $A = n(n+3)$ and $B = (n+1)(n+2)$. The $n$-th element of generated sequense is
$$\frac{A}{2} + \frac{2B - 1}{2B} = \frac{AB + 2B - 1}{2B}$$.
We want to prove that $(AB + 2B - 1)^2 + 4B^2 = (AB + 2B + 1)^2$. Indeed, after moving summand $(AB+2B-1)^2$ to the right side we have
$$4B^2 = 4(AB + 2B)$$
which leads to $B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what values of $n$ is $n(n+2)$ a triangle number? A triangle number is a number which can be written in the form $\frac{m(m+1)}{2}$ for some natural number $m$. For what values of $n$ is $n(n+2)$ a triangle number ?
Using a a brute force method I could generate first $24$ such values of $n$ , but beyond that its t... | You are trying to solve $n(n+2)=\frac{m(m+1)}{2}$. This can be written as $(n+1)^2-1 =\frac{(2m+1)^2-1}{8}$, or:
$$(2m+1)^2-8(n+1)^2 = -7$$
The equation $x^2-8y^2=-7$ is called a Pell-like equation, and, since it has the solution $(x,y)=(1,1)$, it has infinitely many solutions. Specifically, if $(x,y)$ is a solution, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Functional Equation of iterations
Problem: Let $f : \mathbb{Q} \to \mathbb{Q}$ satisfy $$f(f(f(x)))+2f(f(x))+f(x)=4x$$ and $$f^{2009}(x)=x$$ ($f$ iterated
$2009$ times).
Prove that $f(x)=x$.
This is a contest type problem so it is supposed to have an elegant solution.
My approach: I considered $f^{(k)}(x)=a_k$ ... | I'm not sure that my solution is elegant.
So let $g(x) = f(x) - x$. We may rewrite the first equation from task as
$$
f(f^2(x)) - f^2(x) + 3f^2(x)-3f(x) + 4f(x) - 4x = 0,
$$
or using $g(x)$
$$
g(f^2(x)) + 3g(f(x)) + 4g(x) = 0.
$$
Let $a_n = g(f^n(x))$. Then $a_n + 3a_{n-1} + 4a_{n-2} = 0$. We may make this recurrent re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Use integration by parts $\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$ $$I=\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$$
Clearly $$-2I=\int^{\infty}_{0} \ln x \cdot \frac{-2x }{(1+x^2)^2} dx$$
My attempt :
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \left(\f... | $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
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Angle between 3 points I have three points $(x_1, y_1), (x_c, y_c), (x_3, y_3)$, where I know $(x_1, y_1), (x_c, y_c)$, the angle $\theta$, and $c$ on the dash line in the following figure. How to calculate the point $(x_3, y_3)$?
I think of this form:
$$
\theta = arccos\left(\frac{a\cdot b}{||a||\cdot ||b||}\right)
$$... | Let's approach the problem via a simplified case, where $(x_c,y_c)$ is at origin, and $(x_1, y_1)$ is on the $x$ axis at $(a, 0)$:
Obviously, we can calculate $a$ from the original coordinates,
$$a = \sqrt{\left(x_1 - x_c\right)^2 + \left(y_1 - y_c\right)^2}$$
We have three unknowns, $x$, $y$, and $b$ (the distance fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int x^2 \sqrt{1-x^2}\ dx$ I hope I can find a way to integrate this formula without resorting to numerical techniques.
\begin{equation}
\int x^2 \sqrt{1-x^2}\ dx
\end{equation}
I am not sure if there's actually a closed form for this or not?
I tried integration by parts, but it seems not working! Here's my... | Here is a solution with integration by parts and rearranging terms:
First, integrating by parts and adding and subtracting a $1$ in the numerator,
$$
\begin{aligned}
I=\int x^2\sqrt{1-x^2}\,dx & = \frac{x^3}{3}\sqrt{1-x^2}+\int\frac{x^3}{3}\frac{x}{\sqrt{1-x^2}}\,dx\\
&=\frac{x^3}{3}\sqrt{1-x^2}-\int\frac{x^3}{3}\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $x$ and $y$ are non-negative integers for which $(xy-7)^2=x^2+y^2$. Find the sum of all possible values of $x$. I am not able to reach to the answer. I have used discriminant as $x$ and $y$ are both integers but it didn't give any hint to reach to answer. I am not able to understand how should I deal with these type... | $$(xy)^2−12xy+49=(x+y)^2\longleftrightarrow(xy−6)^2+13=(x+y)^2.$$ Hence $$13=(x+y+xy−6)(x+y−xy+6).$$ It follows the only possibility (after discarding $−13$ and $−1$) is that $$x+y+xy−6=13,$$ $$x+y−xy+6=1.$$ This implies clearly $x+y=7$ which gives the candidates $(0,7),(1,6),(2,5),(3,4)$ and symmetrics, and it is veri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Trigonometric Roots of a Polynomial After wondering on this question, I wondered how would you be able to find the roots of a polynomial, in the form $y=x^3+ax^2+bx+c$ if they are the sums of cosines?
I'm wondering if it can, too, be expressed in the form $b\cos\left(A\right)$ with $A$ larger than $2\pi$.
EDIT: I belie... | After research, I found a relatively easy method. Given the cubic $y=x^3+ax^2+bx+c$, one can find the roots by finding $p$ and $q$ where $$p=\frac {3b-a^2}{9}\tag{1}$$$$q=\frac {9ab-27c-2a^3}{54}\tag{2}$$
And plugging them into $$\cos(\theta)=\frac {q}{\sqrt{-p^3}}\tag{3}$$So the solutions are given $$\begin{cases}x_1=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$ Can someone point me in the right direction how to solve this?
$3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$
I guess I have to get to logarithms of the same base. But how? What principle should I use here?
Thx
| Using $a^{-b}=\frac{1}{a^b}$, The equation can be rewritten as
$$\frac{3^{x+2}}{4^{x+3}}+\frac{3^{x+4}}{4^{x+3}}=\frac{40}{9}$$
$$\implies \frac{9}{64}\left(\frac{3}{4}\right)^x+\frac{81}{64}\left(\frac{3}{4}\right)^x=\frac{40}{9}$$
$$\implies \left(\frac{3}{4}\right)^x=\frac{256}{81}$$
$$\implies x=-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\... | Here is an alternative approch using complex analysis. Since both integrands are even one can start with integrating around the singularities in the positive half-plane using the residue theorem which yields
$$
\begin{align*}
\frac{1}{2}\int_{-\infty}^\infty \frac{1}{x^4+x^2+1}\,\mathrm dx
&= \pi\mathrm i\left(\opera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
"answer_id": 4
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Prove that $\frac{2^a+1}{2^b-1}$ is not an integer
Let $a$ and $b$ be positive integers with $a>b>2$. Prove that $\frac{2^a+1}{2^b-1}$ is not an integer.
This is equivalent to showing there always exists some power of a prime $p$ such that $2^a+1 \not \equiv 0 \pmod{p^a}$ but $2^b-1 \equiv 0 \pmod{p^a}$. How do we pr... | if $b$ divides $a$ then $2^b - 1$ divides $2^a-1$.
We can use the Euclidean algroithm to find $q,r$ such that $a = qb + r$ with $r < b$.
$2^a+1 = (2^{qb})(2^r) + 2^r - 2^r + 1 = (2^{qb} - 1)(2^r) + 2^r + 1$
$2^b-1$ divides $(2^{qb} - 1)2^r$ leaving a remainder $2^r+1$
If $r<b$ and $b>1, 2^r + 1 < 2^b - 1$ and $2^b -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Surface area of the part of a sphere above a hexagon I want to calculate the surface area of the part of a half-sphere, which lies above a regular 6-gon. (Radius $r=1$)
More formally,
Let $G$ be the region on the $XY$-Plane, bounded by the points $\{P_k=(\cos(\frac{2\pi k}{6}), \sin(\frac{2\pi k}{6}))\}$ for $k=1, ...... | Hint:
By simmetry, the area of the surface of the half sphere which lies above the equilateral triangle $OP_6P_1$ is $\frac16S_G$.
Now, $\triangle OP_6P_1$ is the set bounded by the $x$ axis, and the lines $y=\sqrt{3}x$ and $y=\sqrt{3}(1-x)$. In polar coordinates this region is
$$\left\{(\theta,r)\in\mathbb{R}^2:0\leq\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Distribution of throws of die rigged to never produce twice in a row the same result A die is “fixed” so that each time it is rolled the score cannot be the same as the preceding score, all other scores having probablity 1/5. If the first score is 6, what is the probability that the nth score is 6 and what is the proba... | Let $p_n$ be the probability the $n$-th toss is a $6$. Note that $p_1=1$. We obtain a recurrence for $p_n$.
The probability the $n$-th toss is not a $6$ is $1-p_n$. Given that, the probability the $n$-th toss is a $6$ is $\frac{1}{5}$.
Thus
$$p_{n+1}=\frac{1}{5}(1-p_n)=\frac{1}{5}-\frac{1}{5}p_n.$$
The homogeneous r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solve definite integral: $\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$ I need to solve:
$$\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$$
Here is my steps, first of all consider just the indefinite integral:
$$\int \arctan(\sqrt{x+2})dx = \int \arctan(\sqrt{x+2}) \cdot 1\ dx$$
$$f(x) = \arctan(\sqrt{x+2})$$
$$f'(x) = \frac{1}{1+x+2} \... | You can set $t=\sqrt{x+2}$. For $x=-1$ we have $t=1$, for $x=1$ we have $t=\sqrt{3}$. Moreover $x=t^2-2$, so $dx=2t\,dt$. Therefore the integral is
$$
\int_{1}^{\sqrt{3}}2t\arctan t\,dt=
\underbrace{\Bigl[t^2\arctan t\Bigr]_1^{\sqrt{3}}}_A-
\underbrace{\int_{1}^{\sqrt{3}}\frac{t^2}{1+t^2}\,dt}_B
$$
(by parts). Then
$$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 5
} |
I want to show that, ${\Phi\tan{9^\circ}-\phi\tan{27^\circ}\over \sin^2{9^\circ}-\sin^2{27^\circ}}=4$ $\phi$: golden ratio, $\Phi={1\over \phi}$
I want to show that,
$${\Phi\tan{9^\circ}-\phi\tan{27^\circ}\over \sin^2{9^\circ}-\sin^2{27^\circ}}=4$$
Using $\sin^2{x}={1\over 2}(1-\cos{2x})$
$${\Phi\tan{9^o}-\phi\tan{27^... | Noting that
$$ \phi=2\cos(\frac{\pi}{5})=2\sin(\frac{3\pi}{10}), \Phi=2\sin(\frac{\pi}{10})$$
from https://en.wikipedia.org/wiki/Golden_ratio, so one has
\begin{eqnarray}
{\Phi\tan{9^\circ}-\phi\tan{27^\circ}\over \sin^2{9^\circ}-\sin^2{27^\circ}}&=&{2\sin(\frac{\pi}{10})\tan(\frac{\pi}{20})-2\sin(\frac{3\pi}{10})\tan(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
I want to show that $\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$ I want to show that
$$\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$$
Expand $(x^4-x+\pi)^2=x^4-2x^3+2x^2-2x\pi+\pi{x^2}+\pi^2$
Let see (substitution of $y=x^2$)
$$\int_{-\infty}... | First note that because of
$$
\int_{\mathbb{R}}dx\frac{x^4}{(x^4-x^2+1)^2}=\int_{\mathbb{R}}dx\frac{x^4\overbrace{-x^2+1+(x^2-1)}^{=0}}{(x^4-x^2+1)^2}=\\
\int_{\mathbb{R}}dx\frac{1}{(x^4-x^2+1)}+\int_{\mathbb{R}}dx\frac{x^2-1}{(x^4-x^2+1)^2}=\\
\int_{\mathbb{R}}dx\frac{1}{(x^4-x^2+1)}+\int_{\mathbb{R}}dx\frac{-1}{(x^4-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
} |
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + 1/β$ and $β + 1/α$. Quadratic equation question, as specified in the title.
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + \frac{1}{β}$ and $β + \frac{1}{α}$.
I gather that ... | $(x-\alpha)(x-\beta) = x^2 - (\alpha + \beta)x + \alpha\beta=0\\
(\alpha + \beta) = 6\\
\alpha\beta = 7$
$(x-\alpha - \frac 1\beta)(x-\beta - \frac 1\alpha) = x^2 - (\alpha + \beta + \frac 1\alpha + \frac 1\beta)x + (\alpha\beta + 2+ \frac {1}{\alpha\beta})=0\\
x^2 - (\alpha + \beta + \frac{(\alpha + \beta)}{\alpha\bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$
Prove that $\displaystyle \sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$.
I tried using the partial fraction decomposition $a_j = \frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}$, but I don't see... | Note that we can write
$$\begin{align}
\frac{1}{m(m+1)(m+2)}&=\frac12\left(\frac{(m+2)-m}{m(m+1)(m+2)}\right)\\\\
&=\frac12\left(\frac{1}{m(m+1)}-\frac{1}{(m+1)(m+2)}\right)
\end{align}$$
So, letting $a_m=\frac{1}{m(m+1)}$, we see that
$$\begin{align}
\frac{1}{m(m+1)(m+2)}&=\frac12\left(a_m-a_{m+1}\right)
\end{align}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is:
Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer.
I'm stuck at the basis step.
If I started with $1$. I get the right hand side is $1... | As you want to prove for $n$ nonnegative and integer, your basis is $n=0$
You just have to prove that $\frac{3(5^{0+1} -1)}{4} = 3$ and it is true, because
$\frac{3(5^{0+1} -1)}{4} = \frac{3(5 -1)}{4} = \frac{3.4}{4} = 3$
If you want to dofor $n = 1$:
For $n=1$, you have that:
$3 + 3.5 = 18$ and $\frac{3(5^{1+1} -1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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How to solve the limit of this sequence $\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$ $$\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$$
I have tried to split the subset into telescopic series but got no result.
I also have tried to use the sque... | $$\frac{1}{6(2n-1)(3n+1)}=\frac{1}{(6n-3)(6n+2)}=\frac{1}{5}\left(\frac{1}{6n-3}-\frac{1}{6n+2}\right)=\frac{1}{5}\int_0^1\left(x^{6n-4}-x^{6n+1}\right)$$
by using partial fractions and noting that $\int_0^1x^a\,dx=\frac{1}{a+1}$. Then:
$$\sum_{n=1}^N\frac{1}{6(2n-1)(3n+1)}=\frac{1}{5}\int_0^1\sum_{n=1}^N\left(x^{6n-4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1839102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. So I have to prove
$$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$
I rearranged it
$$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$
My idea from there is somehow using the AM-GM inequality. Not sure how t... | Note that: $$a^2c^2 + a^2b^2 \ge 2a^2bc \quad\text{ by AM-GM}$$ Now add all the cyclic inequalities and you'll get the wanted inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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What is the area of the triangle having $z_1$, $z_2$ and $z_3$ as vertices in Argand plane? What is the area of the triangle having $z_1$, $z_2$ and $z_3$ as vertices in Argand plane?
Is it
$$\frac{-1}{4i}[z_1(z_2^* - z_3^*)-z_1^*(z_2-z_3)+{z_2(z_3^*)-z_3(z_2^*)}]$$
where $w^*$ denotes the complex conjugate?
| Let $z_j = x_j + iy_j$, $j = 1, 2, 3$. The area of the triangle is given by
\begin{align*}
\frac{1}{2} \begin{vmatrix}
1 & x_1 & y_1 \\
1 & x_2 & y_2 \\
1 & x_3 & y_3
\end{vmatrix}&= \frac{1}{2} \begin{vmatrix}
1 & x_1+iy_1 & y_1 \\
1 & x_2+iy_2 & y_2 \\
1 & x_3+iy_3 & y_3
\end{vmatrix}\\
&= \frac{1}{4i} \begin{vmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
System of Equation
Solve the system of equations
$$-x_1+x_2+x_3=a$$ $$x_1-x_2+x_3=b$$ $$x_1+x_2-x_3=c$$
I have tried writing the augmented matrix of the system of equations above and reducing it into echelon form but that didn't work out. Please help.
| Let's do the Gaussian elimination:
\begin{align}
\left[\begin{array}{ccc|c}
-1 & 1 & 1 & a \\
1 & -1 & 1 & b \\
1 & 1 & -1 & c
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & -1 & -1 & -a \\
1 & -1 & 1 & b \\
1 & 1 & -1 & c
\end{array}\right]
&& R_1\gets -R_1
\\
&\to
\left[\begin{array}{ccc|c}
1 & -1 & -1 & -a \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate $\sum_{n=1}^{\infty}a_n$? If
$$a_{n}=1-\frac{1}{2}+\frac{1}{3}-\cdots +\frac{\left ( -1 \right )^{n-1}}{n}-\ln 2$$
then how to eveluate
$$\sum_{n=1}^{\infty}a_n$$
does it converge?
| By using Taylor expansion, we have
\begin{align}
a_{n}=(-1)^{n+1}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)
\end{align}
which alternates signs and converges to zero (note that $|a_{n}|<\frac{1}{n+1}$.) Also,
\begin{align}
|a_{n}|&=\left(\frac{1}{n+1}-\frac{1}{n+2}\right)+\left(\frac{1}{n+3}-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Trinomial Pascal's Triangle I know that there's a trinomial theorem (and a multinomial theorem), but I was wondering if there was a similar structure for trinomials as there is for binomials, like Pascal's triangle.
Thanks in advance!
| Binomial Coefficients
Pascal's Triangle gives the coefficients of the binomial
\begin{align*}
(1+x)^n\qquad\qquad n\geq 0
\end{align*}
we obtain for $n=0,\ldots,4$
\begin{array}{ccccccccccl}
&&&&&1&&&&\qquad&\qquad(1+x)^0=1\\
&&&&1&&1&&&\qquad&\qquad(1+x)^1=1+1x\\
&&&1&&\color{blue}{2}&&\color{blue}{1}&&\qquad&\qquad(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisble by $11$. How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisible by $11$.
My attempt-
A number is divisible by 11 if the alternating sum of its digit... | Since the raw digit sum of $1..9$ will be $45$, an odd number, we clearly can't have two equal sums for the even-position and odd-position digits. So we need to have one set of digits sum to $17$ and the others to $28$, so that the difference is divisible by $11$.
This can be done either way around; $28$ as the sum of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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To prove the given inequality Question:-
If $a,b,c$ are positive real numbers which are in H.P. show that $$\dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b} \ge 4$$
Attempt at a solution:-
I tried it by AM-GM inequality, but got stuck at a step. My attempt was as follows:-
$$\dfrac{ \dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b}}{2} \ge ... | The theorem you are asked to prove is not true in general. The additional condition needed is that $a \geq b$. And then you can do the proof without AM-GM.
Assuming $(a,b,c)$ are in H.P. in the order $$a = \frac{1}{x}, b = \frac{1}{x+d}, c = \frac{1}{x+2d}$$
then the LHS is
$$
\frac{4x^2+8dx+6d^2}{x^2+2dx}
$$
Consider... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Constant such that $\max\left(\frac{5}{5-3c},\frac{5b}{5-3d}\right)\geq k\cdot\frac{2+3b}{5-c-2d}$ What is the greatest constant $k>0$ such that
$$\max\left(\frac{5}{5-3c},\frac{5b}{5-3d}\right)\geq k\cdot\frac{2+3b}{5-c-2d}$$
for all $0\leq b\leq 1$ and $0\leq c\leq d\leq 1$?
The right-hand side looks like a weighted... | Note: This is the pretty much the same solution as that of NP-hard.
Let $x=5-3c$, $y=5-3d$, $t=\dfrac{5}{3k}$.
Then we are looking for the minimum $t$ for which:
$min(x,\dfrac{y}{b}) \leq t\dfrac{(x+2y)}{(3b+2)}$.
with $2 \leq y \leq x \leq 5$ and $0 \leq b \leq 1$.
Consider the case $x \leq \dfrac{y}{b}$.
Then suffici... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Show that $x f \cdot f'' + f \cdot f' - x (f')^2 > 0$ for certain $f(x) = 1 + x - c x^2 + x^3 + x^4$. Consider the polynomial $f(x) = 1 + x - c x^2 + x^3 + x^4$ where $c \ge 0$. Suppose that $|f(z)| < f(|z|)$ for every complex number $z \notin [0, \infty)$. How can we show that $$F(x):= x f(x)\cdot f^{\prime\prime}(x) ... | The expression that should be positive for $0< x\leq1$ computes to
$$g(x,c):=1+9x^2 +20x^3+9x^4+x^6-cx(4+x+x^3+4x^4)\ .$$
We then can reformulate the question as follows: For which $c\geq0$ do we have
$$p(x):={1+9x^2 +20x^3+9x^4+x^6\over x(4+x+x^3+4x^4)}>c\qquad(0<x\leq1)\ ?$$
This amounts to finding the minimum of $p(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove by induction that $4^{2n}-3^{2n}-7$ is divisible by 84 for any n starting from 1? How to prove by induction that $4^{2n}-3^{2n}-7$ is divisible by 84 for any n starting from 1 ?
Take n=1 and prove for the base case, assume its true for some n, then the third step went like this: $$4^{2n+2}-3^{2n+2}-7=84k=4... | for $n=1$ we have $4^{2}-3^{2}-7=0$ and $0$ is divisible by $84$.
Let $4^{2k}-3^{2k}-7$ is divisible by $84$. We show $4^{2k+2}-3^{2k+2}-7$ is divisible by $84$.
$$4^{2k}-3^{2k}-7=84q$$
$$4^{2k+2}-3^{2k}(9+7)-7\times 16=84(16q)$$
$$4^{2k+2}-3^{2k+2}-7=84(16q+1)+7(3+3^{2k})$$
$$4^{2k+2}-3^{2k+2}-7=84(16q+1)+21(1+3^{2k-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric identities: $ \frac{1+\cos(a)}{1-\cos(a)} + \frac{1-\cos(a)}{1+\cos(a)} = 2+4\cot^2(a)$ I don't really know how to begin, so if I'm missing some information please let me know what it is and I'll fill you guys in :).
This is the question I can't solve:
$$
\frac{1+\cos(a)}{1-\cos(a)} + \frac{1-\cos(a)}{1+\... | Simplify:
$$\frac{1 + \cos a}{1 - \cos a} + \frac{1-\cos a}{1 + \cos a} \equiv \frac{(1+ \cos a)^2 + (1-\cos a)^2}{(1-\cos a)(1+\cos a)}$$
The denominator is the difference of two squares, expand the numerator:
$$\begin{align} \frac{2 + 2\cos^2 a}{1 - \cos^2 a} &\equiv \frac{2(1 + \cos^2 a)}{\sin^2 a} \\ & \equiv \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Find $\sin \theta $ in the equation $8\sin\theta = 4 + \cos\theta$ Find $\sin\theta$ in the following trigonometric equation
$8\sin\theta = 4 + \cos\theta$
My try ->
$8\sin\theta = 4 + \cos\theta$
[Squaring Both the Sides]
=> $64\sin^{2}\theta = 16 + 8\cos\theta + \cos^{2}\theta$
=> $64\sin^{2}\theta - \cos^{2}\theta= ... | Replacing $cos\theta=\pm\sqrt{1-sin^2\theta}$ in your equation, $8\sin\theta = 4 + \cos\theta$, and considering $u=sin\theta$, we have
$8u=4\pm\sqrt{1-u^2}$ which leads to $64u^2 +16 - 64u=1-u^2\to 65u^2-64u+15=0$.
Now, you only need to solve this equation to find $u$ that is $sin\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integrate $\int \frac{x^2-2}{(x^2+2)^3}dx$ $$\int \frac{x^2-2}{(x^2+2)^3}dx$$
What I did :
Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.
Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $
Is there any other easy methods ?
Some substitution ?
| A standard method for dealing with a numerator that is a power of positive definite quadratic polynomial is the following.
Observe that for any positive integer $n$ we have
$$
D\frac{x}{(x^2+2)^n}=-\frac{(2n-1)x^2-2}{(x^2+2)^{n+1}}.\qquad(*)
$$
You can write the right hand side as a linear combination of $1/(x^2+2)^n$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Commutativity and $(a + b)^2$ I read "note that if a and b are commutative, $(a + b)^2 = a^2 + 2ab + b^2$".
Could someone explain how we need commutativity for this to happen?
| Expand it termwise:
$$(a+b)^2 = (a+b) \cdot (a+b) = a \cdot a + \color{blue}{a \cdot b} + \color{red}{b \cdot a} + b\cdot b$$
Commutativity assures us that $a\cdot b = b\cdot a$ so that the expression becomes
$$(a+b)^2 = a^2 + 2ab + b^2$$
Without commutativity, we have to leave our answer as it is in the first expres... | {
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"url": "https://math.stackexchange.com/questions/1855169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Another formula for the angle bisector in a triangle I have seen in an old geometry textbook that the formula for the length of the angle bisector at $A$ in $\triangle\mathit{ABC}$ is
\begin{equation*}
m_{a} = \sqrt{bc \left[1 - \left(\frac{a}{b + c}\right)^{2}\right]} ,
\end{equation*}
and I have seen in a much older ... | Here is a proof of the formula:
Let $AD$ be the angle bisector at $A$ (where $D \in BC$).
The area of $\triangle ABC$ is equal to the area of $\triangle ABD$ $+$ the area of $\triangle ACD$:
$$\frac{1}{2} bc \sin\alpha=\frac{1}{2} b m_a \sin(\alpha/2) + \frac{1}{2} c m_a \sin(\alpha/2)$$
$$m_a = \frac{2bc}{b+c}\cos(\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solve in integers $x,y$ the equation $\left\lfloor\frac{xy-xy^2}{x+y^2} \right\rfloor=a$
Solve in integers the equation
$$\left\lfloor\frac{xy-xy^2}{x+y^2} \right\rfloor=a$$
My work so far:
1) If $a=1$, then $x\in\{-1;-2;-3\}$.
i) $x=-1 \Rightarrow y\ge-3$
ii) $x=-2 \Rightarrow y\ge3$
iii) $x=-3$. No solution.
2) $... | HINT.-$x+y^2\ne 0$. Consider two cases $x+y^2\gt 0$ and $x+y^2\lt 0$.
$$\boxed{x+y^2\gt 0}$$
You have $$a(x+y^2)\le xy-xy^2\lt a(x+y^2)+(x+y^2)$$ This gives two quadratic inequalities in $y$
$$\begin{cases}(a+x)y^2-xy+ax\le 0\\(a+x+1)y^2-xy+ax+x\gt 0\end{cases}$$
Now for $x$ fixed, $(\alpha, \beta)$ and $(\gamma,\delt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Integrate $\int\sin^2x\cos4x\,dx$ I'm having a difficult time solving this integral.
I tried integrating by parts:
$\int\sin^2x\cos4x\,dx$
$u=\sin^2x$, $dv=\cos4x\,dx$
I used the power reducing formula to evaluate $\sin^2x$
$du = 2\sin x\cos x\,dx$, $v=1/4\sin4x$
$uv - \int\ v\,du$
$\dfrac{1}{4}\sin^2x\sin4x - \dfrac{... | Use following trigonometric identities to solve this question,
$\sin^2x=(\frac{1-\cos2x}{2})$
& $\cos A\cos B=\frac{\cos(A+B)+\cos(A-B)}{2}$.
Given integration
$$I=\int \sin^2x\cos 4x\ dx$$
$$put\ \sin^2x=\frac{1-\cos 2x}{2}$$
$$I=\int \Big(\frac{1-\cos 2x}{2}\Big)\cos 4x\ dx$$
$$\Rightarrow I= \frac{1}{2}\int \Big(\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Solving a system of linear congruences
Find all positive integer solutions to \begin{align*}x &\equiv -1 \pmod{n} \\ x&\equiv 1 \pmod{n-1}. \end{align*}
I rewrote the system as $x = nk_1-1$ and $x = (n-1)k_2+1$. Thus, we have $nk_1-1 = (n-1)k_2+1$ and so $n(k_1-k_2) = 2-k_2 \implies n = \frac{2-k_2}{k_1-k_2}$. How do... | First note that $x_0=2n-1$ is a solution.
Now, if $x$ is a solution, then $x\equiv x_0\pmod n$ and $x\equiv x_0\pmod {n-1}$. Thus, $n|x-x_0$ and $(n-1)|x-x_0$. Also, since $\gcd(n,n-1)=1$, we have $n(n-1)|x-x_0$.
Moreover, for all non-negative integer $k$, we can check that $x=2n-1+kn(n-1)$ is a solution.
Thus, all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find the probability of getting two sixes in $5$ throws of a die.
In an experiment, a fair die is rolled until two sixes are obtained in succession. What is the probability that the experiment will end in the fifth trial?
My work:
The probability of not getting a $6$ in the first roll is $\frac{5}{6}$
Similarly for ... | Here is simplified solution.
In order experiment to end at 5th trial, the last two rolls must be 6. So we have,
NNN66 $Pr=5^3/6^5$
6NN66 $Pr=5^2/6^5$
N6N66 $Pr=5^2/6^5$
$(N <6)$
Add all probabilities.
PS. The last two roll must be (6,6), so we get $\frac {1}{6^2}$
The third roll can't be 6, so we get $\frac {5}{6}$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Solve an overdetermined system of linear equations I have doubt to solve this system of equations
\begin{cases}
x+y=r_1\\
x+z=c_1\\
x+w=d_1\\
y+z=d_2\\
y+w=c_2\\
z+w=r_2
\end{cases}
Is it an overdetermined system because I see there are more equations than unknowns.
Can we just solve this system in a simple way?
| Your system is described by the augmented matrix
$$
A=
\left[\begin{array}{rrrr|r}
0 & 1 & 1 & 0 & r_{1} \\
0 & 1 & 0 & 1 & c_{1} \\
1 & 1 & 0 & 0 & d_{1} \\
0 & 0 & 1 & 1 & d_{2} \\
1 & 0 & 1 & 0 & c_{2} \\
1 & 0 & 0 & 1 & r_{2}
\end{array}\right]
$$
Row-reducing the system gives
$$
\DeclareMathOperator{rref}{rref}\rr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Problem calculating the sine of a matrix Given the matrix $A=\begin{pmatrix}-\frac{3\pi}{4} & \frac{\pi}{2}\\\frac{\pi}{2}&0\end{pmatrix}$, I want to calculate the sine $\sin(A)$.
I do so by diagonalizing A and plugging it in the power series of the sine:
\begin{align}
\sin (A) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!... | $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Difficult Functions Evaluation Problem I have a question about finding the value of a certain function that I cannot wrap my head around.
The question is: Given a function $f(x)$ satisfying $$f(x) + 2f\left(\frac{1}{1-x}\right) = x,$$ Then find $f(2).$
So far, I have tried plugging 2 into the original equation to yi... | Given $$f(x)+2f\left(\frac{1}{1-x}\right) = x...................(1)$$
Replace $\displaystyle x\rightarrow \frac{1}{1-x}\;,$ We get
$$f\left(\frac{1}{1-x}\right)+2f\left(\frac{x-1}{x}\right) = \frac{1}{1-x}.......(2)$$
Again replace $\displaystyle x\rightarrow \frac{1}{1-x}\;,$ We get
$$f\left(\frac{1}{1-x}\right)+2f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve recurrence relation $a(n) = -a(n - 2) + \cos({n} \cdot {\frac{\pi}{2}})$ Given recurrence equation
$$a(n) = -a(n - 2) + \cos({n} \cdot {\frac{\pi}{2}})$$
find the closed form solution.
Here is my attempt.
First solve the homogeneous equation:
$$a^{(0)}(n) = -a^{(0)}(n - 2)$$
My solution is:
$$a^{(0)}(n) = k_1 \co... | The $\cos$ seems to add a lot of complexity. Why not consider instead separately the even and odd subsequences, which do not "interact" and — rewriting the recurrence relation — satisfy:
$$\begin{align}
a_{2n+2} &= -a_{2n} + \cos (n+1)\pi = -a_{2n} - (-1)^n\\
a_{2n+1} &= -a_{2n-1}
\end{align}$$
for all $n$. Solving thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x+y=10^{200}$ then prove that 50 divides $x$ Let $x$ be a positive integer and $y$ is another integer obtained after rearranging the digits of $x$. If $x+y=10^{200}$ then prove that $x$ is divisible by 50.
My attempt
Since $y$ is the digit rearrangement of $x$ so $x$ $\cong$ $y$ $\bmod{9}$ from there we get $x$ $\c... | We can assume $x$ and $y$ are non-zero. So, with suitable initial $0$-padding, each has $200$ digits.
If $x$ ends in $00$ we are finished. Suppose now that $x$ ends in $0$ but not $00$. Then the next to last digits of $x$ and $y$ are $10$'s complements of each other, and non-zero. Each of $x$ and $y$ has $198$ digits t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Prove the following trigonometric result If $\theta_1,\theta_2(0\leq\theta_1,\theta_2<2\pi)$ are two solutions of $\sin(\theta+\phi)=\frac{1}{2}\sin(2\phi)$, prove that
$$\frac{\sin(\theta_1)+ \sin(\theta_2) }{ \cos(\theta_1)+ \cos(\theta_2)} =\cot\phi$$
I have tried with the following process:
Since
$\theta_1,\thet... | $$\sin(\theta+\phi)=\frac{1}{2}\sin(2\phi)$$
$$\sin\theta \cos \phi + \sin\phi \cos \theta = \sin \phi \cos\phi$$
$$\sin\theta \cot \phi + \cos \theta = \cos\phi$$
$$\sin\theta \cot \phi - \cos \phi = -\cos\theta$$
Squaring both sides, we get
$$\sin^2\theta \cot^2 \phi + \cos^2 \phi - 2\sin\theta \cot \phi \cos \phi= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
For what value of $(a+b)$ will all roots of $f(x)=x^4-8x^3+ax^2+bx+16$ be positive? I was thinking of using Descartes' rule of signs, from which I find there are at most 2 positive roots and 2 negative roots of the given equation.
Also, $f(\infty)>0$ and $f(0)>0$ imply that either there are no real roots or 2 real root... | Let us consider first the case when $b=-32$.
$$f_1(x)=x^4-8x^3+ax^2-32x+16=0\iff a=\frac{-x^4+8x^3+32x-16}{x^2}$$
Now, let
$$g(x):=\frac{-x^4+8x^3+32x-16}{x^2}$$
Then,
$$g'(x)=\frac{-2 (x-2)^3 (x+2)}{x^3}$$
So, $g(x)$ is increasing for $x\lt -2$ or $0\lt x\lt 2$, and is decreasing for $-2\lt x\lt 0$ or $x\gt 2$ with $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
On real part of the complex number $(1+i)z^2$
Find the set of points belonging to the coordinate plane $xy$, for which the real part of the complex number $(1+i)z^2$ is positive.
My solution:-
Lets start with letting $z=r\cdot e^{i\theta}$. Then the expression $(1+i)z^2$ becomes $$\large\sqrt2\cdot|z|^2\cdot e^{{i}\... |
what am I missing in my solution
After having $\cos\left(2\theta+\frac{\pi}{4}\right)\gt 0$, you have
$$-\dfrac{\pi}{2}\lt\left(2\theta+\dfrac{\pi}{4}\right)\lt\dfrac{\pi}{2}$$
which is incorrect.
To make it easy to understand why this is incorrect, let $\alpha:=2\theta+\frac{\pi}{4}$.
Then, we want to solve
$$\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find Area Enclosed by Curve I want to find the area enclosed by the plane curve $x^{2/3}+y^{2/3}=1$. My attempt was to set $x=\cos^3t, \ y=\sin^3t$ so:$$x^{2/3}+y^{2/3}=\cos^2t+\sin^2t=1$$
Then the area is $$2A=\oint_Cxdy-ydx=3\oint_C\cos^3ty'dy+\sin^3tx'dx=3\int_0^{2\pi}\cos^2t\cdot \sin^2tdt=\frac{3\pi}{4}\implies A=... | Notice that we have symmetry about the $y$ and $x$ axes. Hence we can use the formula
$$y=(1-x^{2/3})^{3/2}$$
Which gives the top half, integrate from $x=0$ to $1$, and then multiply by $4$, exploiting the symmetry. Hence we have
$$A=4\int_0^1(1-x^{2/3})^{3/2}dx$$
Letting $x=\cos(u)^{6/2}=\cos^3(u)$, $dx=-3\cos^2(u)\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that $\log(1+y) \approx y- \frac {y^2}2 + \cdots$ without Taylor Series
For small $y$, prove that $\log(1+y)\approx y -\frac {y^2}2 + \cdots $
I have no idea to solve it.
| One approach directly based on a definition of $\log x$ is as follows. We use the following definition of $\log x$ $$\log x = \lim_{n \to \infty}n(x^{1/n} - 1)$$ Replacing $x$ by $(1 + x)$ and for $|x| < 1$ using binomial theorem we get
\begin{align}
\log (1 + x) &= \lim_{n \to \infty}n((1 + x)^{1/n} - 1)\notag\\
&= \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the value of $\sum_{r=0}^{\left\lfloor\frac{n-1}3\right\rfloor}\binom{n}{3r+1}$
Show that $$\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\ldots=\dfrac{1}{3}\left[ 2^{n-2} + 2\cos{\dfrac{(n-2)\pi}{3}}\right]$$
My solution:-
$$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\ldots=\sum_{r=0}^{n}{\bi... | Let $\alpha=e^{2\pi i/3}=\frac{-1+i\sqrt3}2$, that is $\alpha^3=1$, then if $k\not\equiv0\pmod3$, $\alpha^k+1+\alpha^{-k}=\frac{\alpha^{3k}-1}{\alpha^k\left(\alpha^k-1\right)}=0$. If $k\equiv0\pmod3$, then $\alpha^k+1+\alpha^{-k}=1+1+1=3$. That is,
$$
\frac{\alpha^{k-1}+1+\alpha^{1-k}}3=\left\{\begin{array}{}
1&\text{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Length of side of biggest square inscribed in a triangle I have seen that the length of each side of the biggest square that can be inscribed in a right triangle is half the harmonic mean of the legs of the triangle. I have not seen a rigorous explanation for it, though. I would appreciate such an explanation. (This... | Let we assume that the legs $CB$ and $CA$ of our right triangle $ACB$ have lengths $a$ and $b$.
First case: we consider the largest inscribed with a vertex at $C$. Assuming that its side length is $l$,
by triangle similarities we have
$$ c = \sqrt{a^2+b^2} = l\cdot\left(\frac{c}{b}+\frac{c}{a}\right), $$
from which $l=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Sum to infinity of trignometry inverse: $\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)$ If we have to find the value of the following (1)
$$
\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)
$$
I know that
$$
\arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r+1}2 \right)-\arctan \left(\frac{r-1}2... | $$\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r-1}2\right)=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=-1}^{N-2}\arctan\left(\frac{r+1}2\right)$$
$$=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\arctan\left(0\right)-\arctan\left(\frac{1}2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Show that $3^n+4^n+\dots+(n+2)^n=(n+3)^n$ has no answers for $n\ge 6$. Considering
$$3^n+4^n+\dots+(n+2)^n=(n+3)^n$$
Clearly $n=2$ and $n=3$ are solutions of this equation and this equality does not hold for $n=4$ and $n=5$.
How can I show this equation has no solutions for $n>5$.
Thanks.
| A direct proof for $n\ge7$:
One can very easily show by induction $\sum_{k=1}^{n-1} k^n<n^n$. This implies \begin{align} \sum_{k=3}^{n+2}\left(\frac{k}{n}\right)^n &=1+\left(1+\frac1n\right)^n+\left(1+\frac2n\right)^n+\sum_{k=3}^{n-1} \left(\frac{k}{n}\right)^n \\ &< 2+e+e^2<\left(\frac{10}{7}\right)^7\le\left(1+\frac3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Solve $\cos^n x + \sin^n x =1 $ the solutions of this equation as a function of the value of $n$??
\begin{align}
\cos^n x + \sin^n x =1
\end{align}
I already found the solution if n is odd,
| If $n$ is odd, we can write it as:$n = 2k+1, k \ge 1$, $1 = |\sin^{2k+1}x + \cos^{2k+1}x| \le |\sin x|\cdot \sin^{2k}x + |\cos x|\cdot \cos^{2k}x \le \sin^{2k}x+\cos^{2k}x \le \sin^2 x + \cos^2x = 1$. thus $|\sin x| = 1, 0$ , and you can find $x$ from this. If $n = 1$, then $\sin x + \cos x = 1\implies (\sin x+\cos x)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integration by means of partial fraction decomposition I'm trying to solve this indefinite integral by means of partial fraction decomposition:
$\int\dfrac{x+1}{\left(x^2+4x+5\right)^2}\ dx$.
The denominator has complex (but not real) roots because $\Delta<0$; so, according with my calculus book, i try to decompose the... | Note that we can write the partial fraction expansion of $\frac{1}{(x+4x+5)^2}$ as
$$\frac{1}{(x+4x+5)^2}=\frac{1}{(x-r)^2(x-r^*)^2}=\frac{A}{x-r}+\frac{B}{(x-r)^2}+\frac{C}{x-r^*}+\frac{D}{(x-r^*)^2}$$
where $r=-2+i$ and $r^*=-2- i$.
Multiplying by $(x-r)^2$ and letting $x\to r$ reveals that $B=\frac{1}{(r-r^*)^2}=-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If the roots of the equation are integers then find the value of $k$. The question says roots of $x^2-kx+36=0$ are integers then the number of values for $k$ are?
I know roots are integral if discriminant is a perfect square of an integer, but using this I get infinite values for $k$. Which values should I reject?
| HINT:
Let $k^2-4\cdot1\cdot36=y^2\iff(k+y)(k-y)=144$
As $k+y\pm(k-y)$ are even, $k-y,k+y$ have the same parity, hence both must be even
$$\dfrac{k+y}2\cdot\dfrac{k-y}2=\dfrac{144}4=?$$
What are the positive divisors of $36?$
Also as $k+y\ge k-y,36=\dfrac{k+y}2\cdot\dfrac{k-y}2\le\dfrac{(k+y)^2}4$
$\implies k+y\ge\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac... | One should be careful when using identities related to the arctangent, because $\arctan\tan t=t$ only holds for $-\pi/2<t<\pi/2$. On the other hand, $\tan\arctan s=s$ holds without any condition.
Set
$$
\alpha=\arctan\frac{1}{\sqrt{2}},
\qquad
\beta=\arctan\frac{\sqrt{5-2\sqrt{6}}}{1+ \sqrt{6}}=
\arctan\frac{\sqrt{3}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Find the least squares solution for rank deficient system
Find the least squares solution to the system
$$x - y = 4$$
$$x - y = 6$$
Normally if I knew what the matrix $A$ was and what $b$ was I could just do $(A^TA)^{-1} A^Tb$, but in this case I'm not sure how to set up my matrices. How can I find the least square s... | Problem statement: underdetermined system
Start with the linear system
$$
\begin{align}
\mathbf{A} x &= b \\
%
\left[
\begin{array}{cc}
1 & -1 \\
1 & -1 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
x \\
y
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
4 \\
6
\end{array}
\right]
%
\end{align}
$$
The system... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find $\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $ Find $$\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $$
Any hints please?
Could'nt think of any approach till now...
| $$ \begin{aligned}
\int\frac{x^{2}}{(x\cos x - \sin x)(x\sin x + \cos x)}\,\mathrm{d}x&=\int\left(\frac{x\sin x + \cos x}{x\cos x - \sin x}\right)^{-1}\frac{x^{2}}{(x\cos x - \sin x)^{2}}\,\mathrm{d}x\\&=\int\left(\frac{x\sin x + \cos x}{x\cos x - \sin x}\right)^{-1}\,\mathrm{d}\!\left(\frac{x\sin x + \cos x}{x\cos x -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
What is the $\lim_\limits{x \to 1+}\left(\frac{3x}{x-1}-\frac{1}{2 \ln(x)}\right)$?
What is the limit of $$\lim_\limits{x \to 1+} \left(\frac{3x}{x-1}-\frac{1}{2 \ln(x)} \right)$$
I attemped the problem using L^Hopital's Rule.
My Work
$$\lim_\limits{x \to 1+} \left(\frac{3x}{x-1}-\frac{1}{2 \ln(x)} \right)$$
$$\lim_\... | It is easier to put $x = 1 + h$ and then $h \to 0^{+}$. We have then
\begin{align}
L &= \lim_{x \to 1^{+}}\left(\frac{3x}{x - 1} - \frac{1}{2\log x}\right)\notag\\
&= \lim_{h \to 0^{+}}\left(\frac{3 + 3h}{h} - \frac{1}{2\log (1 + h)}\right)\notag\\
&= \lim_{h \to 0^{+}}\frac{(3 + 3h)2\log(1 + h) - h}{2h\log(1 + h)}\not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Minimum of $ f(\alpha) = \left(1+\frac{1}{\sin^{n}\alpha}\right)\cdot \left(1+\frac{1}{\cos^{n}\alpha}\right)$
Minimum value of $\displaystyle f(\alpha) = \left(1+\frac{1}{\sin^{n}\alpha}\right)\cdot \left(1+\frac{1}{\cos^{n}\alpha}\right)\;,$ Where $n\in \mathbb{N}$ and $\displaystyle \alpha \in \left(0,\frac{\pi}{2}... | Thanks friends got it
Let $\displaystyle f(x) = \ln\left(1+\frac{1}{t}\right)\;, t>0$ Using Jesan Inequality function $f(t)$ is convex function.
So $$\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\frac{1}{y}\right)\geq 2\ln\left(1+\frac{2}{x+y}\right)$$
Where $x,y\in (0,1)$
So $$\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve $\int_{0}^{1}\frac{1}{1+x^6} dx$ Let $$x^3 = \tan y\ \ \text{ so that }\ x^2 = \tan^{2/3}y$$
$$3x^2dx = \sec^2(y)dy$$
$$\int_{0}^{1}\frac{1}{1+x^6}dx = \int_{1}^{\pi/4}\frac{1}{1+\tan^2y}\cdot \frac{\sec^2y}{3\tan^{2/3}y}dy = \frac{1}{3}\int_{1}^{\pi/4} \cot^{2/3}y\ dy$$
How should I proceed after this?
EDITED: C... | It is easier to solve this integral using partiel fraction decomposition
$\frac{1}{1+x^6} =\frac{1}{3(1+x^2)}+\frac{2-x^2}{3(x^4-x^2+1)}$
and solving the 3 integrals seperatly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
How to find the singular value decomposition of $A^TA$ & $(A^TA)^{-1}$ I want to find the singular value decomposition of $A^TA$ & $(A^TA)^{-1}$.
The singular value decomposition of $A$ is $$A=U \Sigma V^T$$
Basically, I want to find the singular values of $A^TA$ & $(A^TA)^{-1}$
$$A^TA=V{\Sigma}^2 V^T$$.
Does this mean... | Finding the singular value decomposition
Start with a matrix with $m$ rows, $n$ columns, and rank $\rho$:
$$
\mathbf{A} = \mathbb{C}^{m \times n}_{\rho}.
$$
Since the question concerns the normal equations, let's fix $\rho = m$ and $m\ge n$. The matrix $\mathbf{A}$ is tall and has full column rank.
The singular value... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Semisimple, connected Lie groups generated by unipotent elements. Let $G$ be a linear, semisimple Lie group with no compact factors.
The unipotent elements of $G$ are those that have only eigenvalue 1.
I've seen it asserted that $G$ is generated by its unipotent elements: see Exercise #2 $\S 4.5$ in Dave Witte Morris' ... | I think I did the case $SL_2(\mathbb{R}):$
Let
$$A=\begin{pmatrix}
a&b\\c&d
\end{pmatrix} \in SL_2(\mathbb{R}).$$
Notice that
$$A\cdot\begin{pmatrix}
0&1\\-1&0
\end{pmatrix} = \begin{pmatrix}
-b&a\\-d&c
\end{pmatrix}, $$
and
$$\begin{pmatrix}
0&1\\-1&0
\end{pmatrix}\cdot A = \begin{pmatrix}
c&d\\-a&-b
\end{pmatrix},$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
A horrid-looking integral $\int_{0}^{5} \frac{\pi(1+\frac{1}{2+\sqrt{x}} )}{\sqrt{10}\sqrt{\sqrt{x}+x}} $
$$
\mathbf{\mbox{Evaluate:}}\qquad
\int_{0}^{5} \frac{\pi(1+\frac{1}{2\sqrt{x}} )}{\sqrt{10}\sqrt{\sqrt{x}+x}}
\,\,\mathrm{d}x
$$
This is a very ugly integral, but appears to have a very simple closed form of: ... | Let $u = \sqrt{x} + x$. Then we have
$$ \frac\pi{\sqrt{10}} \int_0^5 \frac{1 + \frac{1}{2\sqrt x}}{\sqrt{\sqrt x + x}} \, dx = \frac\pi{\sqrt{10}} \int_0^{\sqrt{5}+5} \frac{1}{\sqrt{u}} \, du$$
If you want to be pedantic (and who doesn't?!) then we need to note that this is an improper integral because the integrand i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Does $2764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$? In the process of my numerical computations I have found a very special identity:
*
*$\;\;1264483 + 1707789 \,\sqrt[3]{7} - 1238313\,\sqrt[3]{7^2} = 9.313225746154785 \times 10^{-10}$
*$
-1500493 - 2026256\,\sqrt[3]{7} + 1469290\,\sqrt[3]{7^2}
= 9.... | Assume
$$ \tag 12764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$$
Multiply by $\sqrt[3]7$ to obtain
$$ \tag 22764976\,\sqrt[3]7 + 3734045\,\sqrt[3]{7^2} -18953221 = 0.$$
The system of linear equations
$$\begin{align}
3734045 x-2707603 y&=-12764976\\
22764976 x+3734045 y&=18953221
\end{align} $$
has rational... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
How to prove that $2 \min(|a−b|,|b−c|,|c−a|)$ ≤ $R$ How to prove that $2 \min(|a−b|,|b−c|,|c−a|)$ ≤ $R$
I know that min is for $\sqrt{7}-1,\sqrt{7},\sqrt{7}+1$, but how to prove this?
| We may assume without loss of generality that the side lengths are given by $2x-y,2x,2x+y$ with $y\leq x$ in order to fulfill the triangle inequality. With such assumptions, the area of $ABC$, by Heron's formula, is given by:
$$ \Delta = \frac{\sqrt{3}}{2}\sqrt{x^2(2x-2y)(2x+2y)} $$
and the circumradius is given by:
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the value of the constants If $\displaystyle
\frac{\left( \frac{2x^2}{3a} \right)^{n-1}}
{\left( \frac{3x}{a} \right)^{n+1}} =
\left( \frac{x}{4} \right)^3$, determine the values of the constants $a$ and $n$
I could find the value of $a$, i.e, $\displaystyle \frac{\sqrt{x^6 \times 3^{2n}}}{2^n x^n 2^5 }$ ... | \begin{align*}
\frac{\left( \frac{2x^2}{3a} \right)^{n-1}}
{\left( \frac{3x}{a} \right)^{n+1}} &=
\left( \frac{x}{4} \right)^3 \\
\frac{2^{n-1}}{3^{2n}a^{2n}} x^{n-3} &=
\frac{x^3}{2^6}
\end{align*}
Comparing $x$ terms,
$$n-3=3 \implies n=6$$
Comparing coefficient of $x^3$,
$$\frac{2^5 a^2}{3^{12}}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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.