Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Is there a reason the prime factors of $|M_{24}|$ are all one less than the factors of $24$? Wikipedia says of the Mathieu group $M_{24}$, a $5$-transitive permutation group acting on $24$ points,
$$ |M_{24}|= 2^{10}\cdot3^3\cdot5\cdot7\cdot11\cdot23. $$
The prime factors $2,3,5,7,11,23$ are all one less than one of th... | The fact that $M_{24}$ is 5-transitive forces its order to be
$$
|M_{24}| = 24 \cdot (24-1) \cdot (24-2) \cdot (24-3) \cdot (24-4) \cdot |H|,
$$
where $H$ is the subgroup fixing any ordered list of five distinct points. (In fact, we have $|H| = 48$.) It follows that $|M_{24}|$ is divisible by
\begin{align*}
24 \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
modulus, floor, quotient $a = qM + r,\; q = \big\lfloor \frac{a}{M} \big\rfloor,\; 0 \leq r < M,\; r = a \bmod M$ are not self-consistent when $M < 0$ This question is different than other $\bmod M$ where $M < 0$ questions I've found because those don't specifically ask about the 5 relations I ask about below.
When de... | Based on @BillDubuque's comment/Wikipedia link, I made a Desmos that explains 5 types of division and their corresponding quotient, remainder, and Modulo operation.
Summary: Modulo operation, mod(x, M), is defined in terms of remainder of integer division, which in turn, is defined in terms of quotient of integer divis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find exact value of integral $\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}}dx$? When I first encountered the integral $\displaystyle \int_{0}^{\infty} \frac{1}{x^{4}-x^{2}+1} d x$, I am very reluctant to solve it by partial fractions and search for any easier methods. Then I learnt a very useful tr... | Too long for a comment
You could be interested by
$$\displaystyle I_{n}=\int_{0}^{t} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{n}}=t \, F_1\left(\frac{1}{2};n,n;\frac{3}{2};e^{+\frac{i \pi }{3}} t^2,e^{-\frac{i \pi
}{3}} t^2\right)$$ where appears the Appell hypergeometric function of two variables and the roots of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 0
} |
Proving that $\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}$ when $n$ is very large This is an example from Mathematical Methods in the Physical Sciences, 3e, by Mary L. Boas.
My question is,
\begin{equation}
\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}
\end{equation}
can also be written as,
\begin... | $x:=1/n$ and consider small $x$.
The expression reads
$x^2-\dfrac{x^2}{(1+x)^2}=$
$x^2(1-(1+x)^{-2})=$
$x^2(1-[1-2x+O(x^2)])=$
$2x^3+O(x^4)$, and we are done.
Used: Binomial expansion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
A typical inequality: $\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}$
For $x, y, z\in (0, \infty)$ prove that: $$\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}.$$
My attempts to apply media inequality or other inequalities have been unsuccessful. In desperation I did the calculations an... | Alternative proof:
(pqr method)
Let $p = x + y + z, q = xy + yz + zx, r = xyz$.
The desired inequality is written as
$$\frac{8}{27} \le \frac{p^3 r - 6pqr + q^3 + 8r^2}{q^3} = \frac{p^3 r - 6pqr + 8r^2}{q^3} + 1. \tag{1}$$
We split into two cases:
Case 1 : $p^3 r - 6pqr + 8r^2 \ge 0$
Clearly, (1) is true.
Case 2: $p^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4381204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluate $\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x$. Latest Edit
By the contributions of the writers, we finally get the closed form for the integral as:
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x =-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$
I first evaluat... | Alternative solution:
Using the identity ($q > 0$)
$$
\frac{1}{q^n} = \frac{1}{\Gamma(n)}\int_0^\infty \mathrm{e}^{-qy}y^{n - 1}\mathrm{d} y,
$$
we have
\begin{align*}
I_n &= \int_0^\infty \frac{\ln x}{(x^2 + 1)^n}\mathrm{d} x\\
&= \int_0^\infty \left( \frac{1}{\Gamma(n)}\int_0^\infty e^{-y(1 + x^2)}y^{n - 1}\mathrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 5
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n-rooks n-colors problem The problem goes as follows: $n$ colors are used to color the squares in a $n\times m$ chess board so that each color is used exactly $m$ times. Can you always place $n$ rooks on the board, so that the rooks don't attack each other and there is exactly one rook for every color?
Visual example f... | Proof that this is impossible for $n=m$
The proof is disappointingly simple. You can just have every row be numbers $[1,n]$ in order, except the last one which is shifted one down.
Example for $n=4$:
\begin{matrix}
1 & 1 & 1 & 4 \\
2 & 2 & 2 & 1 \\
3 & 3 & 3 & 2 \\
4 & 4 & 4 & 3 \\
\end{matrix}
The proof that this is i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4386111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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How do I evaluate $ \int_{0}^{\frac{\pi}{2}} \sin ^{a} x d x, \textrm{ where }0 \leq a\leq 1$? We tackle the integral by converting it into a Beta function by letting $y=\sin^2x$, then
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \sin ^{a} x d x &=\int_{0}^{1} \frac{y^{\frac{a}{2}} d y}{2 y^{\frac{1}{2}}(1-y)^{\frac{1... | First, there is an antiderivative : for $0\leq x \leq \frac \pi 2$ $$I=\int \sin ^{a}( x)\, dx=-\cos (x) \,\,
_2F_1\left(\frac{1}{2},\frac{1-a}{2};\frac{3}{2};\cos ^2(x)\right)$$ In terms of summation
$$I=-\frac{1}{\Gamma \left(\frac{1-a}{2}\right)}\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1-a}{2}\right)}{(2 n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\alpha$ such that $b$ is in the column space of $A$ Let $A=\begin{bmatrix}
1 & 0 & -2 & 1\\
0 & 4 & -1 & 0\\
1 & 0 & -1 & 0\\
2 & 4 & -4 & 1
\end{bmatrix} $ and let $b= \begin{bmatrix}
1\\
-3\\
4\\
\alpha
\end{bmatrix}.$ Find $\alpha$ such that $b$ is in the column space of $A$.
I’ve found $rref(A)= \begin{bmatri... | You have to find the RREF of the augemnted matrix
$A=\begin{bmatrix}
1 & 0 & -2 & 1&1\\
0 & 4 & -1 & 0&-3\\
1 & 0 & -1 & 0&4\\
2 & 4 & -4 & 1&\alpha
\end{bmatrix} $
And then see that for what value of $\alpha$ do you get rank of the augmented matrix to be $3$. (That is the last row is all zeroes). That would imply that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given diagonals in a parallelogram. Find sides. Given a parallelogram with $d_1 = AC = 26$ cm, $d_2 = BD = 18$ cm and $\sin \displaystyle \angle AOD = \frac{12}{13}$.
Find $AB = a$ and $AD = b$.
What I did in order to solve it
*
*Using the formula for the area
$S = \frac{d_1d_2\sin \displaystyle \Phi}{2} = \frac{26... | You already noted that the diagonals are divided by half in the center. However, you're making it far too complicated by using the area.
Instead, take a look at the triangle $\triangle AOD$. You already know the length of two sides, $a=9$ and $c=13$ and their inclosed angle $\beta=\operatorname{arcsin} \frac{12}{13}\ap... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4395766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Elementary inequalities exercise - how to 'spot' the right sum of squares? I have been working through CJ Bradley's Introduction to Inequalities with a high school student and have been at loss to see how one could stumble upon the solution given for Q3 in Exercise 2c.
Question:
If $ad-bc=1$ prove that $a^2+b^2+c^2+d^2... | You could use the identity:
$$(ap+br+cs)^2+(aq+bs-cr)^2$$
$$=(a^2p^2+b^2r^2+c^2s^2+2abpr+2acps+2bcrs)+(a^2q^2+b^2s^2+c^2r^2+2abqs-2acqr-2bcrs)$$
$$=(a^2p^2+b^2r^2+c^2s^2+2abpr+2acps)+(a^2q^2+b^2s^2+c^2r^2+2abqs-2acqr)$$
$$=a^2(p^2+q^2)+(b^2+c^2)(r^2+s^2)+2ab(pr+qs)+2ac(ps-qr)$$
If $a^2=b^2+c^2$, dividing by $a^2$ leave... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4396645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Summation of cosines
Find the sum $\displaystyle\sum_{n=1}^\infty\left(\frac{2\cos n}{\sqrt{n}}-\frac{1}{\sqrt{n}}\right)\left(\frac{2\cos n}{n}\right)^3\left(\frac{2\cos n}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)$.
I tried in many ways, like applying Fourier series, GP formula, $\cos^3x$, conversions but nothing worked ... | Doing some algebraic manipulation we get that your sum $S$ is equal to
\begin{align*}
S &= 8\sum_{n \ge 1}\frac{\cos^3(n)}{n^4}\left(2\cos(n) -1 \right)\left(2\cos(n) +1 \right)\\
& =8\sum_{n \ge 1}\frac{\cos^3(n)}{n^4}\left(4\cos^2(n) -1 \right)\\
& = 32\sum_{n \ge 1}\frac{\cos^5(n)}{n^4} -8\sum_{n \ge 1}\frac{\cos^3(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that $\frac{3}{5} + \frac{4}{5}i$ number in multiplicative complex numbers field(apart from $0)$ has infinite order Show that $\frac{3}{5} + \frac{4}{5}i$ number in multiplicative complex numbers field(apart from $0)$ has infinite order and prove that $\frac{1}{\pi}\arctan(\frac{4}{3})$ is irrational.
by contradic... | As you observed, $(3+4i)^2 = -7 + 24i \equiv 3 + 4i \pmod 5$. Therefore $$(3+4i)^n =(3+4i)^{n-2} (3+4i)^2 \equiv (3+4i)^{n-2} (3+4i) = (3+4i)^{n-1} \pmod 5$$
for all $n > 1$, so $(3+4i)^n \equiv 3+4i \pmod 5$ by induction. Hence $(3+4i)^n \neq 5^n$ for $n > 0$.
If you know some algebraic number theory, you can say the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Differentiating $y = x - \frac2x + \frac3{x^2}$ Another easy question for you guys.
I'm differentiating the below to find the equation of the tangent at $(-3,-2)$
$$y = x - \dfrac{2}{x} + \dfrac{3}{x^2}$$
I simplified to:
$$ y = x - 2x^{-1} + 3x^{-2}$$
Then differentiated to get:
$$ \frac{dy}{dx} = 1 + 2x^{-2} - 6x^{... | Your calculation up to
$$\frac{dy}{dx} = 1 + \frac{2}{x^2} - \frac{6}{x^3}$$
is correct. However, placing $x=-3$ we get
$$m=1+\frac{2}{(-3)^2}-\frac{6}{(-3)^3}=\frac{13}{9}$$
and so the tangent's equation is given by
$$y - (-2) = \frac{13}{9}(x-(-3))$$
i.e.
$$y = \frac{13}{9}x + \frac{7}{3}$$
or equivalently,
$$9y - 13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Lyapunov function for a second order system involving trigonometric functions I am studying the stability of the following system:
\begin{aligned}
\dot{x}_{1} &= -x_{1}^{2} - \sin x_{2}\\
\dot{x}_{2} &= x_{1} - \frac{\cos x_{2}}{x_{1}}\\
\end{aligned}
The system itself has 2 equilibrium points:
\begin{equation}
\be... | From the stable equilibrium point $\left(\frac{1}{\sqrt[4]{2}}, - \frac{\pi}{4}\right)$, the state variable transformation can be constructed as
$$
\begin{aligned}
y_{1} &= x_{1} - \frac{1}{\sqrt[4]{2}} \\
y_{2} &= x_{2} + \frac{\pi}{4} \\
\end{aligned}
$$
for which there is an inverse
$$
\begin{aligned}
x_{1} &= y_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Derivative of $f(x) = \csc(x)\cot(x)$ using first principles. How to find derivative of $f(x) = \csc(x)\cot(x)$ using First principle of derivative?
I tried the following method.
$f(x) = \csc(x)\cot(x) = \dfrac{\cos(x)}{\sin^2(x)}$
Now using the limit,
$f'(x) = \lim\limits_{h\to0}\dfrac{\dfrac{\cos(x+h)}{\sin^2(x+h)} ... | Your $\sin x$ in the last denominator you found should be $\sin^2x$.
Act on the numerator:
$$
\cos(x+h)\sin^2x=\cos x\sin^2x\cos h-\sin^3x\sin h
$$
and
\begin{align}
\cos x\sin^2(x+h)
&=\cos x(\sin x\cos h+\cos x\sin h)^2 \\
&=\cos x\sin^2x\cos^2h+2\sin x\cos^2x\sin h\cos h+\cos^3x\sin^2h
\end{align}
Now subtract and c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Infinite sums of squares $$\sum_{n=0}^{\infty} \frac {k^2(1-k)^2}{(n+k)^2(n+1-k)^2}$$
Here can anyone help me to solve this question,I can't think of any logic like telescopic, coefficient compare etc .
It would be helpful if anyone could provide the solution
| The series diverges for integer $k$, Assuming $k \in \mathbb{R} \setminus \mathbb{Z}$, the series sums to $$\frac{k^2(1-k)^2}{(1-2k)^2}\left[\frac{\pi^2}{\sin^2(\pi k)} - \frac{2\pi\cot(\pi k)}{1-2k}\right]\tag{*1}$$
Let $\alpha = k$ and $\beta = 1 - k$, the sum at hand equals to
$$\begin{align}\verb/sum/
&= \alpha^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Generating function for the number of strings following a certain pattern. Let $f_n$ denote the number of strings of length $n$ using letters $A,B,C$ and $D$ such that there are even number of $A's$; $1,2,3$ or $4 \ B's$; either $2$ or $5$ $C's$ and at least $1$ $D$.
So $f_0 = f_1 = f_2 = f3 = 0$, $f_4 = 12$ and $f_5 ... | $\color{blue}{\text{Generating Function:}}$
I think that exponential generating functions is the best choice for your question.
*
*If there are even number of $A's$ , then E.G.F of $A$ is : $$1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...+\frac{x^{2k}}{(2k)!}+..=\frac{e^x + e^{-x}}{2}$$
*E.G.F of $B's$ : $$\big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $y’’ – 4y’ + 5y = 4e^{2x}\sin(x)$ using $\mathcal D$ operator Hello – I am working through the following question and get stuck at step 6. Could someone please advise in simple terms which I can hopefully understand. Thanks
$$y'' – 4y' + 5y = 4e^{2x}\sin(x)$$
Step one – Order equation so that differential operato... | For simplicity, we denote the particular solution as $y_p(t)$.
Then
$$
y_p(t) = {1 \over D^2 - 4 D + 5} \left( 4 e^{2 x} \sin x \right) =
4 {1 \over D^2 - 4 D + 5} \left( e^{2 x} \sin x \right)
$$
Thus, we have
$$
y_p(t) = 4 e^{2 x} {1 \over (D + 2)^2 - 4 (D + 2) + 5} \ \sin x
= 4 e^{2 x} {1 \over (D^2 + 4 D + 4) - (4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Is this nice-looking inequality actually trivial?
Let, $x,y,z>0$ such that $ xyz=1$, then prove that
$$(xy+yz+xz)(x^2+y^2+z^2+xy+yz+xz)≥2(x+y+z)^2 $$
I tried to use the inequality
$$x^2+y^2+z^2≥xy+yz+xz$$
Then I got,
$$(xy+yz+xz)(x^2+y^2+z^2+xy+yz+xz) ≥2(xy+yz+xz)^2≥2(x+y+z)^2\implies xy+yz+xz≥x+y+z$$
Which is corre... | Let $S=x+y+z, T=xy+xz+yz$ and note that $S,T \ge 3$ by AMGM and $xyz=1$ and also $S^2 \ge 3T$ by C-S
The inequality is: $T(S^2-T) \ge 2S^2$
Case 1: $S \ge T$ then $S^2(T-2) \ge S^2 \ge T^2$ which rewrites to the required inequality
Case2: $S \le T$ then using $S^2-T \ge 2T$ we get $T(S^2-T) \ge 2T^2 \ge 2S^2$ so we are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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MCQ about the product of $4$ consecutive odd numbers The product of four consecutive odd numbers must be …
(A) A multiple of 3, but not necessarily of 9.
(B) A multiple of 5 .
(C) A multiple of 7.
(D) A multiple of 9.
(E) A multiple of 3×5×7×9= 945.
Let $n\in \mathbb{Z}$, then
$(2n-1)(2n+1)(2n+3)(2n-3)$
$= (... | We use the fact that every number $n\neq 3$. can be written as $3k+1$ or $3k+2$. We chek this in:
$A=(2n-1)(2n+1)(2n+3)(2n+5)$
*
*$n=3k+1\Rightarrow 2n+1=6k+3\Rightarrow 3|A$
$2n+3=6k+5 \Rightarrow 5 \nmid A$
$2n+5=6k+7 \Rightarrow 7 \nmid A$
2): $n=3k+2$
$2n-1=6k \Rightarrow 3\mid A$
$2n+3=6k+7 \Rightarrow 7\nmid A$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4424201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to find the inverse function of the function $f$, defined as $f(x)=\frac{x}{1-x^2}$ for $-1 \lt x \lt 1$ I want to find the inverse function of the function $f$, defined as $f(x)=\frac{x}{1-x^2}$ for $-1 \lt x \lt 1$.
Letting $y=f(x)$, we first note that $x=0 \iff y=0$. Next, we consider the case when $y \neq 0$:
... | I will retain the notation found in the original question...starting from:
$$\left|x+\frac{1}{2y} \right|=\sqrt{1+\frac{1}{4y^2}}$$
This expression means that for any $y \in \mathbb R \setminus \{0\}$:
$x=\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y} \ \lor\ x=(-1)\cdot\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Making a matrix with an unknown diagonalizable I'm stuck on a question about diagonalizability of matrices and need some help moving forward.I think I made some progress but I'm also not sure if its any good. Let $a \in \mathbb{R}$, $A$ is a matrix. For which values of $a$ is the matrix $A$ diagonalizable.
$$A:= \begi... | Since the matrix $A$ is triangular, the eigenvalues coincide with the diagonal elements of $A$. Therefore, it is given by $p(x)=(x-1)(x-2)^2(x-7)$. Since the only repeated eigenvalue is 2, we need to make sure that the geometric multiplicity of this eigenvalue is equal to 2 to make the matrix diagonalizable.
So, we hav... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
How to deal with odd $m$ in integral $\int_{0}^{\frac{\pi}{4}}(\sin^{6}m x+\cos^{6}m x) \ln (1+\tan x) d x $ Latest edit
*
*Thanks to @Quanto for settling down the question by proving the odd one as:$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)$$
By our resu... | Similar to the even case, recognize
$$\sin ^{6}mx+\cos ^{6}mx = \frac58+\frac38 \cos4mx $$ and rewrite the integral as follows
\begin{align}
I_{2n+1}=&\int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(2 n +1)x+\cos ^{6}(2 n +1)x\right] \ln (1+\tan x) d x \\
=&\> \frac58\int_{0}^{\frac{\pi}{4}}\ln \overset{\frac\pi4-x \to x}{(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Find minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$
Find the minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$.
I can prove that $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1,$ because:
$$ab+bc+ca+abc=4 \\ \implies (a+2)(b+2)(c+2)=(a+2)(b+2)+... | Complement to @richrow's nice answer
Using $\sqrt{u} \ge \frac{2u}{1 + u}$ for all $u\ge 0$ (easy to prove using AM-GM),
we have
$$\sqrt a = \sqrt 2 \sqrt{a/2}
\ge \sqrt 2 \frac{2 \cdot a/2}{1 + a/2}
= \sqrt 2 \frac{2a}{2 + a} = 2\sqrt 2
- 4\sqrt 2 \frac{1}{2 + a}.$$
Thus, we have
$$\sqrt{a} + \sqrt{b} + \sqrt{c}
\ge 6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Show that, if $t_{3n}=x$, $t_{3n+1} = y$ and $t_{3n+2} = z$ for all values of $n$, then $p^3+q^3+3pq-1=0$
A sequence of numbers $t_0,t_1,t_2,...$ satisfies $$t_{n+2} = pt_{n+1} + qt_n, ~~~n\geq0$$
where $p$ and $q$ are real. Throughout this question, $x,y,z$ are non-zero real numbers.
Show that, if $t_{3n}=x$, $t_{3n... | The approach looks correct to me, since all the steps either follow from previous step or from given information in the question. Here I provide another solution which might be of interest for you based on linear algebra.
(FYI I am currently revising for my linear algebra exam in a few weeks, so it is a cool revision e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Algebraic Proof help with real numbers Let $a$ and $b$ be distinct non-zero real numbers
Show that $-\frac{4ab}{(a-b)^2} = 1 - (\frac{a+b}{a-b})^2$
It's been 15 years since I have been in high school but I came across this problem and wanted to solve it.
$-\frac{4ab}{(a-b)^2} = -\frac{4ab}{a^2+b^2-2ab}$
And now I am st... | Recall that
$(a+b)^2=a^2+2ab+b^2\tag{1}$$(a-b)^2=a^2-2ab+b^2\tag{2}$
Subtracting $(1)$ from $(2)$
$$4ab=(a+b)^2-(a-b)^2$$
$$\begin{align}-\dfrac{4ab}{(a-b)^2}&=-\dfrac{(a+b)^2-(a-b)^2}{(a-b)^2}\\&=-\left(\dfrac{a+b}{a-b}\right)+\dfrac{(a-b)^2}{(a-b)^2}\\&=1-\left(\dfrac{a+b}{a-b}\right)^2\qquad\blacksquare\quad\text{Q.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
My Attempt:$$y^2=\frac{x^2+2x-3}{x^2+4x+4}=z\\\implies x^2(z-1)+x(4z-2)+4z+3=0$$
Discriminant greater than equal to zero, so, $$(4z-2)^2-4(z-1)(4z+3)\ge0\\\implies z\le\frac43\\\implies -\frac2{\sqrt3}\le y\le\frac2{\... | HINT
A little bit different approach from the other users.
To begin with, notice that
\begin{align*}
\frac{\sqrt{(x - 1)(x + 3)}}{x + 2} & = \frac{\sqrt{x^{2} + 2x - 3}}{x + 2}\\\\
& = \frac{\sqrt{(x^{2} + 4x + 4) - (2x + 4) - 3}}{x + 2}\\\\
& = \frac{\sqrt{(x + 2)^{2} - 2(x + 2) - 3}}{x + 2}\\\\
& = \frac{\sqrt{[(x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Maclaurin series of $\frac{x^2}{1- x \cot x}$ I wonder if there is an explicit formula for the Maclaurin expansion of $\frac{x^2}{1 - x \cot x}$. We know an explicit formula for $1- x \cot x$.
Due to the continued fraction formula for $\tan x$, we know that all of the coefficients after the first are negative.
$\bf{Add... | Using Bessel functions, we find
\begin{align*}
\frac{{x^3 }}{{1 - x\cot x}} & = 3 - x\frac{{J_{5/2} (x)}}{{J_{3/2} (x)}} = \frac{3}{2} + x\frac{{J'_{3/2} (x)}}{{J_{3/2} (x)}} = 3 - 2x^2 \sum\limits_{k = 1}^\infty {\frac{1}{{1 - (x/j_{3/2,k}^2 )^2 }}}
\\ & = 3 - 2x^2 \sum\limits_{k = 1}^\infty {\sum\limits_{n = 0}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}} = 0$ without substituting I found this possible solution:
Let $r^2=x^2 + y^2, x = r \cos(\theta)$ and $y=r\sin(\theta)$. Then:
$$
\begin{split}
\lim_{(x,y) \to (0,0)} \frac{x^2}{\sqrt{x^2 +y^2}}
&= \lim_{r \to 0} \frac{(r \cos(\theta))^2}{\sqrt{r^2}... | You can consider that the following generalized limit:
$$\lim_{(x,y) \to (0,r)} \frac{x^2}{\sqrt{x^2 +y^2}}=0.$$
Observe that,
$$\begin{align}\frac{x^2}{\sqrt{x^2}}&=\frac{x^2}{|x|}=|x|\\
&≥\frac{x^2}{\sqrt{x^2 +y^2}}\\
&≥0\end{align}$$
This implies
$$\begin{align}0&=\lim_{(x,y) \to (0,0)}|x|\\
&=\lim_{(x,y) \to (0,0)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to prove the product of first n consecutive odd numbers is a square less than another square? I have observed for first few values of consecutive odd numbers, the result is always of the form:
$m^2 - n^2$, where $m$ and $n$ are two distinct positive integers. That is:
$$1\cdot 3\cdot 5 \cdot 7\cdots (2k-1) = m^2 - ... | Note that any product of two distinct odd numbers or two distinct even numbers can be written as a difference of squares. Let $p<q$ be both odd or both even. Then $(q-p)/2$ and $(q+p)/2$ are distinct integers, and
$$
((q+p)/2)^2 - ((q-p)/2)^2=(q^2/4+pq/2+p^2/4)-(q^2/4-pq/2+p^2/4)=pq.
$$
So this is a fairly weak condit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find extreme values of ellipse I have the curve equation
$$
Ax^2 + By^2 + Cxy = 1
$$
which represents ellipse with center in (0, 0) and rotated some angle.
How can I find max X and Y values (not semi-axes) on this ellipse? Points are marked on the picture.
There x-es and y-es can not be moved left or right side, so I ... | Another way, use polar co-ordinates $x=r \cos t, y= r \sin t$ to get
$r^2(t)=\frac{1}{A \cos^2 t+ B\sin ^2 t+C \sin t \cos t}$
Let $u(t)=1/r^2(t)$ $\implies u(t)-\frac{A+B}{2}= \frac{A-B}{2} \cos 2t + \frac{c}{2}\sin 2t= \sqrt{\frac{(A-B)^2}{4}+\frac{c^2}{4}}~ \sin(2t+k)$
$\implies \frac{A+B}{2}-\sqrt{\frac{(A-B)^2}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Other ways to factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ To factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ I used the fact that $x=-y$ and $y=z$ and $x=-z$ make the expression zero. Hence it factorize to $\lambda (x+y)(y-z)(x+z)$ and we can check that the number $\lambda$... | $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)=xy\left(x+y\right)+yz(y+\color{red}{x}-\color{red}x-z)-xz\left(x+z\right)=xy(x+y)+yz(y+\color{red}x)-yz(\color{red}x+z)-xz(x+z)=y(x+y)(x+z)-z(x+z)(x+y)=(x+y)(x+z)(y-z)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Let p be a prime. Prove that $x^4 + 4 y^4 = p$ has integer solutions if and only if $p = 5$. In this case, find the solutions for x and y. I have tried to factor
$$x^4 + 4 y^4 = p$$
using the Sophie Germain's Identity, as someone suggested in the comments, which yields:
$$((x+y)^2 + y^2)((x-y)^2 + y^2) = p$$
Since p is... | The only prime ramified in the ring of Gauss integers $\mathbb Z[i]$ it is known to be $2=i^3((1+i)^2$ so we have since $i^4=1$ $$x^4+4y^4=(x^2+i2y^2)(x^2-i2y^2)=(x^2+(1+i)^2y^2)(x^2-(1+i)^2y^2)$$
then the factorization in $\mathbb Z[i]$ of $x^4+4y^4$ becomes
$$(x+i(1+i)y)(x-i(1+i)y)(x+(1+i)y)(x-(1+i)y)$$ which gives
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate $C=\sin3\alpha\cos\alpha$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$. Calculate $$C=\sin3\alpha\cos\alpha$$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$.
My idea was to find $\sin\alpha$ and $\cos\alpha$. Then we have $\sin3\alpha=3\sin\alpha-4\sin^3\alpha$. So $$\tan2\alpha=2=\dfrac{2\tan\a... | Consider a right triangle with acute angle $2\alpha$ and side lengths
\begin{align}
\text{opposite} &= 2\\
\text{adjacent} &= 1\\
\text{hypotenuse} &= \sqrt{2^2 + 1^2} = \sqrt{5}
\end{align}
Hence,
\begin{align}
\tan(2\alpha) &= \frac{2}{1} = 2\\
\cos(2\alpha) &= \frac{1}{\sqrt{5}}\\
\sin(2\alpha) &= \frac{2}{\sqrt{5}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show that $\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$ How to show that
$$\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$$
without evaluating each integral individually?
I came up with this problem while working on some problem where ... | Performing integration by parts,
\begin{align*}
\int_{0}^{1} \frac{\log(1-x)}{1+x^2} \, \mathrm{d}x
&= \underbrace{\left[ \log(1-x) \left(\arctan x - \frac{\pi}{4}\right) \right]_{0}^{1}}_{=0} - \int_{0}^{1} \frac{\frac{\pi}{4} - \arctan x}{1 - x} \, \mathrm{d}x.
\end{align*}
Now substituting $x = \frac{1-t}{1+t}$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculate the integral $\int\limits_{0}^{8}\frac{dx}{x^2+\sqrt[3]{x}}$ My attempt:
$$\phi =\frac{1+\sqrt{5}}{2}, \; \bar{\phi }=\frac{1-\sqrt{5}}{2}\Rightarrow y^4-y^3+y^2-y+1=\left ( y^2-\phi y+1 \right )\left ( y^2-\bar{\phi }y+1 \right )$$
\begin{multline*}
\int\limits_{0}^{8}\frac{dx}{x^2+\sqrt[3]{x}}\overset{y=\sq... | There is an antiderivative (have a look here) which can be simplified and write
$$I=\int\frac{dx}{x^2+\sqrt[3]{x}}$$
$$I=\frac{3 \log \left(x^{2/3}+\frac{1}{2} \left(\sqrt{5}-1\right)
\sqrt[3]{x}+1\right)}{4 \sqrt{5}}+\frac{3}{20} \log \left(x^{2/3}+\frac{1}{2}
\left(\sqrt{5}-1\right) \sqrt[3]{x}+1\right)-\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4463639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Integrate $\frac{x}{\sin 2x}$ \begin{align}\int_{{\pi}\over5}^{{3\pi}\over10}\frac{x}{\sin2x}\,dx\end{align}
This integral came up while learning integration using Leibnitz rule. What has been tried is taking the integral as:
\begin{align}I(a)=\frac{1}{2}\int_{\pi\over5}^{3\pi\over10}\frac{\arctan (a \tan x)}{\sin x\co... | A simpler approach using the following property of definite integral:
$$\int_{a}^ {b} f(x) dx=\int_{a} ^{b} f(a+b-x)dx$$
\begin{align}
I&=\int_{\pi \over 5}^{3 \pi \over 10}\frac{x}{\sin{2x}}\:dx\\
&=\int_{\pi \over 5}^{3 \pi \over 10}\frac{{\pi \over 5}+{3 \pi \over 10}-x}{\sin\left({2\left({\pi \over 5}+{3 \pi \ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Spivak, Ch. 13 "Integrals", Prob. 2: Show $\int_0^b x^4 dx=\frac{b^5}{5}$ by finding unique number such that $L(f,P)\leq \int_0^b x^4 \leq U(f,P)$? Consider the problem (Spivak, Chapter 13 "Integrals", problem 2) of showing that $\int_0^b x^4 dx=\frac{b^5}{5}$ by finding the unique number $\int_0^b x^4$ such that
$$L(f... | Since those 2 inequalities derive from equally spaced partition of $[0,b], P_n = \{t_0, t_1, \ldots, t_n \},$ where $t_i-t_{i-1}=\frac{n}{b}$, it must be true that $n \geq 1$.
Let's show that
\begin{equation}
\frac{1}{n^5}(n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6}) > 1.
\end{equation}
By doing some algebra we get:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve long integration by partial fraction decomposition problems faster? Some problems are just too time consuming for short exam times what is the fastest way to solve problems like this one for example $$\int \frac{5x^4-21x^3+40x^2-37x+14}{(x-2)(x^2-2x+2)^2}dx$$
| If you're going for a purely partial fraction decomposition approach, then I'm afraid there isn't much you can do to expedite the process. Partial fraction decomposition is just a naturally tedious and long-winded thing to do.
$$\frac {5x^4-21x^3+40x^2-37x+14}{(x-2)(x^2-2x+2)^2}=\frac A{x-2}+\frac {Bx+C}{x^2-2x+2}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability regarding random marbles chosen from a bag I am unsure if my thought process for these questions is correct.
Susie has a bag of marbles containing 3 Red, 7 Green, and 10 Blue marbles.
*
*What is the probability of picking 5 marbles and getting at least one red marble? Calculate the probability (a) with... | You solved both parts of the first problem correctly.
For the second part of the first problem, where you are selecting without replacement, you could also express the answer in the form
$$\Pr(\text{at least one red}) = 1 - \Pr(\text{no reds}) = 1 - \frac{\dbinom{17}{5}}{\dbinom{20}{5}}$$
where the denominator counts t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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${f}\begin{pmatrix}x \\y \\\end{pmatrix}$=$\begin{pmatrix}x^2-y^2 \\2xy \end{pmatrix}$ is differentiable at each point I need help with understanding some steps of a task. I tried to solve the uncertainty by myself by using "Approach zero" but the problem is, that I am not able to write these kind of functions in "Appr... | Here is a sketch of a solution: Let $\mathbf{x}=[x, y]^\intercal$ and $\mathbf{h}=[h, k]^\intercal$.
Notice that
$$(x+h)^2-(y+k)^2=x^2-y^2+2xh-2yk+h^2-k^2$$
and
$$2(x+h)(y+k)=2xy+2xk+2yk+2hk$$
Thus
\begin{align}
\mathbf{f}(\mathbf{x}+\mathbf{h})&=\begin{pmatrix}x^2-y^2\\ 2xy\end{pmatrix} + \begin{pmatrix} 2xh-2yk\\ 2xk... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Conjecture: $\frac1\pi=\sum_{n=0}^\infty\left((n+1)\frac{C_n^3}{2^{6n}}\sum_{k=0}^n(-1)^k{n\choose k}{\frac{(n-k)(k-1)}{(2k-1)(2k+1)}}\right)$ Let $C_n$ denote the $n$-th Catalan number defined by
$${\displaystyle C_{n}={\frac {1}{n+1}}{2n \choose n}=\prod \limits _{k=2}^{n}{\frac {n+k}{k}}\quad \left(n\geqslant 0\rig... | $$\sum _{k=0}^{n}(-1)^{k}{n \choose k}{\frac {(n-k)(k-1)}{(2k-1)(2k+1)}}=\frac{\sqrt{\pi } \,\,\Gamma (n+2)}{4 \,\Gamma \left(n+\frac{1}{2}\right)}$$
$$ A_{n}={\frac {C_{n}^{3}}{2^{6n}}}\frac{\sqrt{\pi } \,\,\Gamma (n+2)}{4 \,\Gamma \left(n+\frac{1}{2}\right)}=\frac{\Gamma \left(n+\frac{1}{2}\right)^2}{4 \pi \Gamma (n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Integral with Weierstrass substituion, limits gone wrong I'm trying to evaluate the following integral with a Weierstrass substitution:
$$
\int_{\pi/3}^{4\pi/3} \frac{3}{13 + 6\sin x - 5\cos x} \text{d}x
$$
This comes out to about 0.55 when evaluated numerically. I thought it would be possible to rewrite the integral u... | The Weierstrass substitution is valid for $x\in(-\pi,\pi)$ or $x\in(\pi,3\pi)$, but not for an interval containing a neighborhood of $\pi$. As $x$ tends to $\pi$ from below, $z$ tends to $+\infty$. As $x$ tends to $\pi$ from above, $z$ tends to $-\infty$.
$$
\begin{align}
&\int^{4\pi/3}_{\pi/3}\frac3{13+6\sin(x)-5\cos(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there any method other than Feynman’s Integration Technique to find $ \int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x?$ We are going to find the formula, by Feynman’s Integration Technique, for
$$\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x,$$
where $a+c$ $\... | This isn't really an answer as such, but as a general comment, the next trick you should know is how to "peel off" the logarithm. Let's guess that for some functions $g$ and $h$:
$$\int \log f(x)\,dx = h(x) \log f(x) + g(x)$$
Taking the derivative of both sides:
$$\log f(x) = h'(x) \log f(x) + h(x) \frac{f'(x)}{f(x)} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Log rules for calculating joint entropy This question is probably not so hard for you.
Why is the entropy equal to:
$$
H(x,y)=2\log_2(5)-\frac{8}{25}\log_2(2)-\frac{6}{25}\log_2(3),
$$
for the following joint distribution?
$$
p(x,y) =
\frac{1}{25}
\begin{bmatrix}
1&1&1&1&1\\
2&1&2&0&0\\
2&0&1&1&1\\
0&3&0&2&0\\
0&0&1&1... | As far as I understand you need help with logarithms simplification? Because I think your calculations are almost (possible typo) ok:
My answer is: $H(x,y)=\frac{11}{25}\log_2(\frac{1}{25})-\frac{8}{25}\log_2(\frac{2}{25})-\frac{6}{25}\log_2(\frac{3}{25}).$
I thing you miss minus sine $H(x,y)=-\frac{11}{25}\log_2\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$. $$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$$
Edit : $D> 0$.
My work:
Let $x = D\tan \theta$
$$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}dx$$
$$=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2... | Keep it easy: for any $A>0$
$$ f(A) = \int_{0}^{+\infty}\frac{dx}{x^2+A}\stackrel{x\mapsto z\sqrt{A}}{=} \frac{1}{\sqrt{A}}\int_{0}^{+\infty}\frac{dz}{z^2+1}=\frac{\pi}{2\sqrt{A}}.\tag{1}$$
By differentiating with respect to $A$ we get:
$$ -\frac{\pi}{4 A\sqrt{A}} = -\int_{0}^{+\infty}\frac{dx}{(x^2+A)^2} \tag{2}$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4477389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Distribution of amount of green balls drawn Urn of type A contains 4 green and 3 blue balls. Urn of type B contains 4 blue and 5 green balls.
There are three urns of type A and two urns of type B. We pick one urn at random (out of 5) and we draw 3
balls (with a single draw) out of the chosen urn. Let X denote the numbe... | I think you may have some unneeded terms. You want $\frac35$ times some hypergeometric probability and $\frac25$ times some hypergeometric probability as you seem to have found. But then you included $\binom{3}{N}$ which I do not understand. You also had a $7$ instead of a $9$ in the bottom right denominator.
Perhap... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$ with various solutions.
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
Solutions in the answers.
$\ \\ \ \\ \ \\ \ \\$
Edit) Since this question is closed, I'll add more contexts for this question.
This identity is called "Brahmagupta-Fibonacci identity", which the comment... | The identity is trivially satisfied if $\,a=b=0\,$ or $\,c=d=0\,$. Otherwise $\,a^2+b^2 \ne 0\,$ and $\,\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2 = 1\,$, so there exists an angle $\,\alpha\,$ such that $\,\frac{a}{\sqrt{a^2+b^2}} = \sin \alpha\,$, $\,\frac{b}{\sqrt{a^2+b^2}} = \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving inequality using root of unity Let $\omega$ be a complex cube root of unity. It can be shown that if $a,b,c \in \mathbb{R}$, then $$(a+b+c)(a+b\omega+c\omega^2)(a+b \omega^2 + c \omega)= a^3+b^3+c^3-3abc$$
I was wondering if this could be extended to prove the AM/GM inequality for $n=3$. All that needs to be do... | Proof of AM-GM:
Using $\omega^3=1, \omega^2+\omega+1=0$, We know that $(a+b\omega+c\omega^2)(a+b \omega^2 + c \omega)=a^2+b^2+c^2-ab-bc-ca$, then
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\ge 0$
If $a,b,c>0$, then $a+b+c>0$ and $\frac{a^2+b^2}{2}\ge ab$ etc give $a^2+b^2+c^2\ge ab+bc+ca$
Hence $\frac{a^3+b^3+c^3}... | {
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"url": "https://math.stackexchange.com/questions/4481472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compute $\int_0^{\frac\pi2} \frac1{\sin(x+\frac\pi3)\sin(x+\frac\pi6)} dx$ Compute the following definite integral:
$$\int_0^{\frac\pi2} \frac1{\sin(x+\frac\pi3)\sin(x+\frac\pi6)} dx$$
I know that I can rewrite the denominator as shown below:
$$\int_0^{\frac\pi2} \frac1{\frac12\left(\cos(\frac\pi6) - \cos(2x+\frac\pi2)... | Here is some trick:
$\int_0^{\frac\pi2} \frac4{\sqrt{3} + 2\sin(2x)} dx$=$\int_0^{\frac\pi2} \frac4{\sqrt{3} + 4\sin(x)\cos(x)} dx$=$\int_0^{\frac\pi2} \frac{4\sqrt{3}\sec^2{x}}{\sec^2{x}(3 + 4\sqrt{3}\sin(x)\cos(x))} dx$=$\int_0^{\frac\pi2} \frac{4\sqrt{3}}{(3(1+\tan^2{x}) + 4\sqrt{3}\tan{x})} d(\tan{x})$
=$-\int_0^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\lim_{x \to 2^{+}} \frac{x - 2}{\sqrt{x^{2} - 4}} + 1 = 1$ using $\epsilon$-$\delta$ definition I want to prove that $$\lim_{x \to 2^{+}} \frac{x - 2}{\sqrt{x^{2} - 4}} + 1 = 1$$ but I am not sure if my proof is valid because I did some algebraic manipulation. Can you please verify my proof?
Given $ \epsilon ... | You should have $\delta<4\varepsilon^2$, so that $\frac{\sqrt\delta}2\leqslant\varepsilon$. So, take $\delta=\min\left\{3,4\varepsilon^2\right\}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all the real values of $a,b,c,e,d,f$ for which $A$ is diagonalizable $\newcommand{\span}{\text{span}}$
The matrix in question is:
$$A=\begin{pmatrix}
2&0&0&0\\
a&-1&0&0\\
b&c&-1&0\\
d&e&f&2
\end{pmatrix}$$
This was on a test that I had and I got it wrong because I needed to add more conditions, what I... | Ummm. A matrix is diagonalizable if and only if the minimal polynomial is squarefree.
Your characteristic polynomial is $ (\lambda +1)^2 (\lambda - 2)^2.$ Both factors occur in the minimal polynomial with exponent either $1$ or $2.$ We need both exponents $1,$ so that the minimal poly becomes $ (\lambda +1) (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving the equation $x^2+(\frac{x}{x+1})^2=\frac54$
Solve the equation$$x^2+\left(\frac{x}{x+1}\right)^2=\frac54$$
I noticed that for $0< x$, both $x^2$ and $\left(\dfrac x{x+1}\right)^2$ are increasing functions so their sum is also increasing and it has only one root which is $x=1$ (by inspection). But I'm not sur... | $x^2 + \left(\frac{x}{x+1}\right)^2 = \frac{5}{4}$
Let $y=-x\qquad ⇒ y^2 + \left(\frac{y}{1-y}\right)^2 = \frac{5}{4}$
Let $z=\frac{x}{x+1}\qquad ⇒ z^2 + \left(\frac{z}{1-z}\right)^2 = \frac{5}{4}$
By inspection, $(x=1)$ is a solution.
$(x=1) → (y=-1) → (z=-1)$
Solve $z=-1$, back for x, we have $\;\left(x=-\frac{1}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Conditions for $A\cos^2(x)+B\sin^2(x)+C\sin(x)\cos(x)+D\cos(x)+E\sin(x)+F = 0$ to have (a) real root(s) I am doing research on Linear Algebra and encounter a function in the form like below:
$$A\cos^2(x)+B\sin^2(x)+C\sin(x)\cos(x)+D\cos(x)+E\sin(x)+F = 0$$
What are the conditions that the coefficients of the equation n... | The conic section given in Zarrax's answer,
$$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$
can be reordered into a simple quadratic in terms of either $x$ or $y$:
$$Ax^2 + (Cy + D)x + (By^2 + Ey + F) = 0$$
$$By^2 + (Cx + E)y + (Ax^2 + Dx + F) = 0$$
Plugging the coefficients into the quadratic formula, we get:
$$x = \frac{-(Cy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I solve a recurrence relation of the type $a_{n+1} = K(n)a_{n} + P(n)a_{n-1}$ where $K(n)$ and $P(n)$ are rational functions? I've been stuck on a problem for days now and was able to boil it down to this recurrence relation. I researched a bit about ways to solve it -- generating functions, characteristic polyn... |
$a_{n+1} = \dfrac{2n+5}{n+2} a_{n} + \dfrac{(n+1)(n+3)^2}{n+2} a_{n-1}$
Rewriting $\,2n+5 = (n+3)^2 - (n+2)^2\,$:
$$
\quad\quad a_{n+1} = \frac{(n+3)^2 - (n+2)^2}{n+2} a_{n} + \frac{(n+1)(n+3)^2}{n+2} a_{n-1}
\\ \iff\quad\quad a_{n+1} + (n+2) a_n = \frac{(n+3)^2}{n+2} \big(a_n + (n+1) a_{n-1}\big)
$$
With $\,b_n = a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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if $~f(x)=x^2+10x+20~$, find the number of solutions of $f(f(f(f(x))))=0$ or $f^4(x)=0$ where $f^n$ means f is composed to itself n times. The answer given is 2. I tried to bash the question by $~f^3(x)=\alpha~$ where $~\alpha~$ is the root. Then find the solution, and keep on continuing. It is a long process, so I di... | To do it the very hard way
$a^2 + 10a + 20 = k$ was solutions $a =\frac{-10\pm\sqrt{100-4(20-k)}}2=-5\pm \frac {\sqrt{20 + 4k}}2=-5\pm \sqrt{5+k}$.
So if $f^4(f^3(x)) = 0$ we have solutions $f^3(x)=-5\pm \sqrt{5}$.
To solve $f^3(x) =f(f^2(x))=-5\pm \sqrt{5}$ we have solutions $f^2(x) = -5\pm\sqrt{5 + (-5\pm\sqrt 5)}=-5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim\limits_{n\to\infty}\big(\frac{n}{2}+\min\limits_{x\in\mathbb{R}}\sum_{k=0}^n{\cos(2^k x)}\big)$ What is the exact value of the following limit?
$$L=\lim_{n\to\infty}\left(\frac{n}{2}+\min_{x\in\mathbb{R}}\sum_{k=0}^n{\cos(2^k x)}\right)$$
Experimenting on desmos suggests the following claims:
$\sum_{k=0... | Long comment. I believe this type of question is extremely hard, if not impossible by the current technology, to answer. However, let me share some observation.
Define $\varphi_n$ by the $n$th sum:
$$ \varphi_n(x) = \frac{n}{2} + \sum_{k=0}^{n} \cos(2^k x) $$
We will consider the behavior of $\varphi_n (x)$ near $x = \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is... | Here $x$ is common, hence we can "cancel" it to get $x^9+x^{10}+x^{11}+x^{12}+ \cdots +x^{19}$ & $x^2+1$
By your method, let this be $f(x) = (x^2+1)g(x)+ax+b$
At $x=i$, we get $x^9+x^{10}+x^{11}+x^{12}+ \cdots +x^{19} = 0 + ai+b$
At $x=-i$, we get $x^9+x^{10}+x^{11}+x^{12}+ \cdots +x^{19} = 0 - ai+b$
LHS in both can be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to find the maximum area of a quadrilateral, when three of the sides add up to 24? If there's a quadrilateral ABCD, and the sides AB + BC + CD add up to 24, how can I find the maximum area of the quadrilateral formed, what length should the four sides of the quadrilateral be, and what should the interior angles be?... | Denote the four sides of the quadrilateral by $w = DA$, $x = AB$, $y = BC$, $z = CD$.
Let $d = AC$ be the length of one diagonal. This splits the quadrilateral into two triangles, $\triangle ABC$ and $\triangle ACD$.
Use Heron's formula to find the area of the two triangles.
$$A = \frac{1}{4}\sqrt{(x + y + d)(-x + y +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of all real roots of $\frac{{3{x^2} - 9x + 17}}{{{x^2} + 3x + 10}} = \frac{{5{x^2} - 7x + 19}}{{3{x^2} + 5x + 12}}$?
The sum of real roots of $\dfrac{{3{x^2} - 9x + 17}}{{{x^2} + 3x + 10}} = \dfrac{{5{x^2} - 7x + 19}}{{3{x^2} + 5x + 12}}$ is____
How do I proceed with this type of problem.
Cross multiplying and s... | Cross-multiplying gives:
$$(3x^2 - 9x + 17)(3x^2 + 5x + 12) = (x^2 + 3x + 10)(5x^2 - 7x + 19)$$
$$9x^4 + 15x^3 + 36x^2 - 27x^3 - 45x^2 - 108x + 51x^2 + 85x + 204 = 5x^4 - 7x^3 + 19x^2 + 15x^3 - 21x^2 + 57x + 50x^2 - 70x + 190$$
$$9x^4 - 12x^3 + 42x^2 - 23x + 204 = 5x^4 + 8x^3 + 48x^2 - 13x + 190$$
$$4x^4 - 20x^3 - 6x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the sequence of three real numbers
If $a,b,c$ are non zero real numbers satisfying $$(ab+bc+ca)^3=abc(a+b+c)^3$$ then prove that $a,b,c$ are terms in $G.P$
My work:
I assumed that they are in $G.P$ and so assumed $b=ak$ and $c=ak^2$ for some arbitrary $k$. After that I expanded both sides of the equality and go... | We can think of the problem in terms of polynomials too. Assume that $a, b, c$ are the roots of the cubic polynomial '$ P(x) = x^3 + ux^2 + vx + w$. We get
$$-u = a+b+c$$
$$v = ab+bc+ca$$
$$-w = abc$$
Substituting these in the original condition, we have
$$v^3 = (-w)(-u)^3 = wu^3$$ or
$$ w = \frac{v^3}{u^3}$$
$$P(x) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How many methods are there to find $\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}, \textrm{ where }a>b>0$ When I deal with this simple integral, I found there are several methods. Now I share one of them.
Letting $x\mapsto \pi-x $ yields $$I=\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}=\int_{0}^{\pi} \frac{d \thet... | One more
$$I=\int_{0}^{\pi} \frac{dx}{a+b \cos x}=\int_{0}^{\pi}\frac{dx}{a+b(2\cos^2(x/2)-1)}, a>b.$$
Use $t=x/2$, then
$$I=\int_{0}^{\pi/2} \frac{2dt}{a-b+2b \cos^2 t}=\int_{0}^{\pi/2} \frac{2\sec^2t ~dt}{(a-b)\sec^2 t+2b}=2\int_{0}^{\pi/2} \frac{\sec^2 t~dt}{(a+b)+(a-b)\tan^2 t}$$
Let $\tan t=u$, then
$$I=\frac{2}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the limit of $\frac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$ Find the limit $$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$$
For the numerator we have $1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}=\dfrac{1-\... |
How could we actually see (and show) that the terms in the sum(s) are not $n$, but $+1$
The sum is $$1+\frac13+…\frac{1}{3^n}=\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+…+\frac{1}{3^n}$$$$=\sum_{i=0}^n 3^{-n}.$$ The index $i$ starts from the value $\textbf 0$ (and not $1$) and ends at $n$. So the number of terms is $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x^2+2(\alpha-1)x-\alpha+7=0$ has distinct negative solutions... Let $\displaystyle{ \alpha }$ be real such that the equation $\displaystyle{ x^2+2(\alpha-1)x-\alpha+7=0 }$ has two different real negative solutions. Then
*
*$ \ \displaystyle{ \alpha<-2 }$ ;
*$ \ \displaystyle{ 3<\alpha<7 }$ ;
*it is impossible... | You made a slight algebra error.
$$a = 1; b = 2(\alpha - 1); c=-\alpha + 7$$
$$x = \frac{-2(\alpha - 1) \pm \sqrt{(2(\alpha - 1))^2 - 4(-\alpha + 7)}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4(\alpha^2 - 2\alpha + 1) + 4(\alpha - 7)}}{2}$$
$$x = \frac{-2\alpha + 2 \pm \sqrt{4\alpha^2 - 8\alpha + 4 + 4\alpha - 28}}{2}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Question about continuity and the Epsilon-Delta definition So, I' currently working through an example:
Show that $\lim \limits_{x \to 0}\frac{x^3}{x-4}\cos{(\frac{1}{x})}=0$ using the $\epsilon-\delta$ definition.
So far I have the following working:
For any $\epsilon>0$, we must find a $\delta>0$ such that if $|x-0|<... | The first answer is correct. But this one goes (i believe), more in your original direction, but without restricting $\delta$.
In this first part $x<4$ :
For $f = |\frac{x^3}{x-4}|$, We call $f_+ = |\frac{x^3}{x-4}|$ for $x>0$ and $f_- = |\frac{x^3}{x-4}|$ for $x<0$ (and both are $0$ on the rest). We have $f = f_+ + f_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Show that $4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$. The problem asks us to show that the following equation holds true. $$4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$$
This equation has been verified on my calculator.
Perhaps some basic trigonometric formulas will be enough to solve th... | I prove: $4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) - 1=0$ without using special values at $18^\circ$ or $36^\circ$
Start:
$$\begin{align} \text{LHS} &=4\sin^2(24^{\circ})+\frac{2\sin(24^{\circ})\sin(24^{\circ})}{\cos(12^\circ)} - 1\\
\\
&=4\sin^2(24^\circ)\cdot\frac{\cos(12^\circ)+\cos(60^\circ)}{\cos(12^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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The sum of $n$ terms in $1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots$ I am just confused, considering we can take $1 \cdot 2$ as the first term then we get the $n$th term as $n(n+1)$ so the sum of $n$ terms would be $\frac{n(n+1)(2n+1)}{6}$ + $\frac{n(n+1)}{2}$ but let's assume $0 \cdot 1$ as the first term then the $n$th ter... | In first case. ${\color{green}S_n}=\frac{n(n+1)(n+2)}{6}+\frac{n(n+1)}{2}={\color{red}{\frac{n(n+1)(n+2)}{3}}}$
In second case, $S'_{n+1}=\frac{(n+1)(n+2)(2n+3)}{6}-\frac{(n+1)(n+2)}{2}={\color{red}{\frac{n(n+1)(n+2)}{3}}}={\color{green}S_n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Prove $0\le ab+bc+ca-abc\le2$ A question asks
Let $a,b,c$ be non-negative reals such that $$a^2+b^2+c^2+abc=4$$
Prove that$$0\le ab+bc+ca-abc\le2$$
The lower bound can be proven as follows:
If $a,b,c>1$, then $$a^2+b^2+c^2+abc>4$$
This implies that at least one of $a,b,c$ is less than or equal to $1$, and we can assu... | If one of $a,b,c$ is zero or two of them are equal, the inequality holds (almots) trivially.
Consider the function $ f(a,b,c) = ab+bc+ca-2abc $, with $a,b,c\ge 0, g(a,b,c) = 1$, where $g(a,b,c) = a^2 + b^2 + c^2 + abc$. The domain is a compact subset of $\mathbb{R}^3$ so $f$ must admit maximum. Assume the maximum is st... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the limit of the sequence $a_{n+1}=a_n + \frac{1}{2^n a_n}$ with $a_1=1$ Find the limit of the sequence $a_{n+1}=a_n + \frac{1}{2^n a_n}$ with $a_1=1$.
I could only estimate the upper and lower bound for $a_n$. My result is
$$\sqrt{3} \leq \lim_{n\to \infty} a_n \leq \sqrt{\frac{79}{24}}$$.
Is there any way to com... | Write $L = \lim_{n \to \infty} a_n$. Squaring gives
$$a_{n+1}^2 = a_n^2 + \frac{1}{2^{n-1}} + \frac{1}{2^{2n} a_n^2}$$
which gives, ignoring the last term
$$L^2 \ge 1 + \sum_{k=0}^{\infty} \frac{1}{2^k} = 3$$
which I presume is where your lower bound of $\sqrt{3}$ comes from. Since $a_n$ is monotonically increasing, we... | {
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"source": "stackexchange",
"question_score": "4",
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Two different answers for diameter $x$ I am solving this question shown below:
Two circles embedded between two parallel lines whose distance is $x$ and three other circles with indicated diameters touch both the bigger ones. Find $x$
My try:
I used the following facts:
Fact $1:$ If two circles touch externally, dista... | As mentioned in a comment, OP's error was in assuming that point $M$ was on segment $OQ$, an impossibility due to the vertical asymmetry of the figure.
There's probably a slick solution here involving inversive geometry, but here's one that follows OP's trigonometric lead:
We take the large circles $\bigcirc O$ and $\... | {
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Contradictions when do integral $\unicode{x222F} ~f(x,y,z)~dS $, given $f(tx,ty,tz)=t^4 f(x,y,z)$ and $f_{xx}+f_{yy}+f_{zz}=x^2+y^2+z^2$ If $f(tx,ty,tz)=t^4 f(x,y,z)$ and $f_{xx}+f_{yy}+f_{zz}=x^2+y^2+z^2$, then compute the surface integral on the unit sphere $x^2+y^2+z^2=1$:
$$\unicode{x222F} ~f(x,y,z)~dS $$
One parti... | Both integrals should have a value of $\dfrac\pi5$.
Using the first choice of $f_p$:
$$\iint\limits_{x^2+y^2+z^2=1} \frac{x^4+y^4+z^4}{12} \, dS \\ = \int_0^{2\pi} \int_0^\pi \frac{\cos^4(\theta)\sin^4(\phi) + \sin^4(\theta)\sin^4(\phi) + \cos^4(\phi)}{12} \sin(\phi) \, d\phi \, d\theta$$
Now,
$$\begin{align*}
I &= \i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a^2 + b^2 = c^2$, $ab = 6d^2$, $(a,b) = 1$ has no solutions $(a,b,c,d) \in \mathbb N^4$
Prove that $$a^2 + b^2 = c^2$$
$$ab = 6d^2$$
$$\gcd(a,b) = 1$$
has no solutions $(a,b,c,d) \in \mathbb N^4$.
Put $a = p^2 - q^2, b = 2pq, c= p^2 + q^2$ where $(p,q) = 1$, and $p - q \equiv 1 \bmod 2$ (i.e., one is odd, the other ... | For case 2, we could use the trick of Fermat's Infinite Descent.
Observe that $$\left(\dfrac{x_3+x_4}{2}\right)^2+\left(\dfrac{x_3-x_4}{2}\right)^2=x_1^2$$, and $3|x_3-x_4$. So we let $\dfrac{x_3+x_4}{2}=m^2-n^2, x_3-x_4=2mn, x_1=m^2+n^2$, where $m,n$ are positive integers and $(m,n)=1$. Then we have $x_3=m^2+2mn-n^2$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given $\left|e^{i\frac{\phi}{2}}\left(e^{i\frac{\phi}{2}}-e^{-i\frac{\phi}{2}}\right)\right|$ show that it's equal to $2|\sin{\frac{\phi}{2}}|$ Given $\left|e^{i\frac{\phi}{2}}\left(e^{i\frac{\phi}{2}}-e^{-i\frac{\phi}{2}}\right)\right|$ I want to show that it's equal to $2\left|\sin{\frac{\phi}{2}}\right|.$
My work.
$... | Continuing from $|i\sin\phi - 2\sin^2{\frac{\phi}{2}}|$ we have $$\sqrt{\sin^2\phi + 4\sin^4\frac{\phi}{2}} = \sqrt{4\sin^2\frac{\phi}{2}\cos^2\frac{\phi}{2} + 4\sin^4\frac{\phi}{2}} = \sqrt{4\sin^2\frac{\phi}{2}(\cos^2\frac{\phi}{2} + \sin^2\frac{\phi}{2}}) = \sqrt{4\sin^2\frac{\phi}{2}} = 2|\sin \frac{\phi}{2}|$$
| {
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"source": "stackexchange",
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Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1... | $$\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)=\frac{\sin^2\left(\dfrac{3\pi}{4}+\alpha\right)}{\cos^2\left(\dfrac{3\pi}{4}+\alpha\right)}=\frac{\left(\frac{-\sqrt{2}}{2}\cos\alpha+\frac{\sqrt{2}}{2}\sin\alpha\right)^2}{\left(\frac{-\sqrt{2}}{2}\cos\alpha-\frac{\sqrt{2}}{2}\sin \alpha\right)^2}=\frac{\frac{1}{2}\cos^2\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\fr... | Let $g(x)=Ax^2+Bx+C$,if $A>0$, then $g(x)\ge 0$
Case 1: $\forall x \in \Re$, if $B^2\le 4AC$.
Case 2: For $x \in [p,q]$, if the position of the min (bottom) of $g(x)$, $x_0=-\frac{B}{2A}$ should be s.t $x_0 \le p$ or $x_0\ge q$.
Here $$f(x)=4x^2-2ax+a^2-5a+4>0$$ in $x\in (0,2)$ can be met in both the cases.
Case 1: $4a... | {
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"source": "stackexchange",
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What is the inverse for: $x^2+4x+5,x\ge-2$ What is the inverse for: $x^2+4x+5,x\ge-2$
I have tried to solve it using the quadratic formula:
$x^2+4x+5=y$
$x^2+4x+5-y=0$
$x=-2\pm\sqrt{y-1}$
$y\ge1$
The inverse should be: $x=-2+\sqrt{y-1}$
I don't understand why it sould be $+\sqrt{y-1}$ and not $-\sqrt{y-1}$
Can someone... | Try plugging in a number. Since we assume $x\geq -2$, plugging in $x=3$ gives us $f(3)=26$. We want $f^{-1}(26)=3$. This only happens when $f^{-1}(y)=-2+\sqrt{y-1}$. In case we were to take $x<-2$, the inverse would have been $-2-\sqrt{y-1}$. This phenomena occours since $x^2+4x+5$ is not globally invertible, and hence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4533890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$using implicit differentiation
Proof that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$ using implicit differentiation
My workings:
$y=\tan^{-1} x$,
$x= \tan y$
$\frac{d}{dx} (x) = \frac{d}{dx} \tan y$
$1 = \sec^2 y \frac{dy}{dx}$
$\frac{dy}{dx} = \frac{1}{\sec^2 ... | Use the basic trigonometric identity
$$
y'=\cos^2y=\frac{\cos^2y}1=\frac{\cos^2y}{\cos^2y+\sin^2y}=\frac1{1+\tan^2y}=\frac1{1+x^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Probability of getting white balls from urns QUESTION: An urn $A$ contains $3$ white balls and $4$ black balls. $2$ balls are drawn from urn $A$ and without seeing the colour and we introduce them in urn $B$ (which was empty). Later, a ball is drawn randomly from each urn.
a) Find out the probability that the ball draw... | Given the solutions, we are hinted that the procedure of choosing $1$ or $2$ white balls does not affect the answer.
$$ \frac {3 \choose 1} {7 \choose 1} = {3 \over 7} $$
$$ \frac {3 \choose 2} {7 \choose 2} = {1 \over 7} $$
This makes sense.
Let's think to a deck with red an black cards. It does not matter how much we... | {
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"source": "stackexchange",
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find two primes less than $500$ that divide $20^{22}+1$
Find two primes less than $500$ that divide $20^{22}+1$.
Note that $20^2+1=401$ divides the required number (since for any integer a, if k is odd, then $a+1$ divides $a^k+1$). Also, any prime dividing the given number must be congruent to $1$ modulo $4$ since $-... | I claim that the two primes are $89$ and $401$. The $401$ is kind of obvious because $20^{22}=(20^2)^{11}\equiv (-1)^{11}\equiv -1\pmod{401}$. The $89$ is a bit more subtle. Note that $2^{44}\pmod{89}$ is just $\left(\frac{2}{89}\right)$, the Legendre symbol. It's easy to calculate by quadratic reciprocity that this is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4537818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Converse to Jensen's inequality for $1/x$ on a positive bounded interval? Consider the function $f(x) = 1/x$ on the interval $I = [a, b]$, where $0 < a \leq b$.
By Jensen's inequality, we have for any $\{x_j \}_{j=1}^n \subset I$,
$$
f(\overline{x}) \leq \frac{1}{n} \sum_{i=1}^n f(x_i).
$$
Above, $\overline{x} = (1/n)... | We want to determine an upper bound for the function
$$
h(x_1, \ldots, x_n) = \frac{1}{n^2} \sum_{k=1}^n x_k\sum_{k=1}^n \frac{1}{x_k}
= \sum_{k, l=1}^n \frac{x_k}{x_l}
$$
on the hypercube $[a, b]^n$. $h$ is convex in each variable, so that the maximum of $h$ is attained at a corner point of the hypercube. Therefore i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Angle Between Two Vectors Discrepancy Here is something that is bugging me: Consider the vectors
$$ \mathbf{v} = 2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k} \qquad \text{ and } \qquad \mathbf{w} = 2\mathbf{j} - \mathbf{k}. $$
We have two different formulas for the angle $\theta$ between these vectors:
$$ \theta = \cos^{-1}... | Both functions only give one specific inverse, but others exist. If you remove the inverse functions and convert it to a simple system of equations, you actually have a valid solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A question on the conditions under which $\left|\begin{smallmatrix}x+a&b&c\\a&x+b&c\\a&b&x+c\end{smallmatrix}\right|=A$ is non singular The question is if
$$
\begin{vmatrix}
x+a & b & c \\
a & x+b & c \\
a & b & x+c \\
\end{vmatrix}=A
$$
find the conditions under which it's nonsingular
on performing some ERTs
I got
$$
... | You made a sign error in the computation of the determinant; it should be
$-x^3 - x^2(a+b+c)$. The solutions of $-x^3 - x^2(a+b+c) = 0$ are then $x = 0$ and $x = -(a+b+c)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution the ode $x^2y''+4xy'+2y=f(x)$ I'm trying to find a solution for the second order ode $x^2y''+4xy'+2y=f(x)$, where $f\in\mathcal{C}^1(\mathbb{R})$.
I already found that the solution for the homogeneous part is equal to
$y_h(x)=\frac{C_1}{x^2}+\frac{C_2}{x}$. But now I'm stuck trying to get the nonhomogeneous so... | If you say that $y = f(x)$ you can solve it like this (homogeneous):
step 2: Assume a solution to this Euler-Cauchy equation will be proportional to for some constant $λ$. Substitute $y(x) := x^{λ}$ into the differentialequation.
$$
\begin{align*}
f(x) &= x^{2} \cdot y'' + 4 \cdot x \cdot y' + 2 \cdot y\\
y &= x^{2} \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Tricky limitting sum Find the sum of the following infinite series, given $|x|<1$
$$2+4x+\frac{9}{2}x^2+\frac{16}{3}x^3+\frac{25}{4}x^4+\frac{36}{5}x^5+\frac{49}{6}x^6+\frac{64}{7}x^7+\frac{81}{8}x^8+ \ldots $$
I have tried turning this into a geometric series, but I just didn't even know where to begin. I also tried r... | The series appears to take the form
$$ f(x) = 2 + \sum_{n=1}^{\infty} \frac{(n+1)^2}{n} \, x^n $$
which can be seen in the form:
\begin{align}
f(x) &= 2 + \sum_{n=1}^{\infty} \frac{(n+1)(n+2)}{n} \, x^n - \sum_{n=1}^{\infty} \frac{(n+1) \, x^n}{n} \\
&= 2 + \frac{d^2}{d x^2} \, \sum_{n=1}^{\infty} \frac{x^{n+2}}{n} - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Write $ z = \frac{(1-i)^3(√3+i)}{4i}$ to polar form Write the complex number in polar form:
$$ z = \frac{(1-i)^3(\sqrt 3+i)}{4i}$$
So my try goes as follows:
\begin{align}
\frac{(1−i)^3(\sqrt 3+i)}{4i} &=
\frac{(1−i)^3(\sqrt 3+i) \times -4i}{16}\\& =
\frac{(1-3i-3+i)(\sqrt 3+i)\times-4i}{16}\\& =
\frac{(-2-2i)(\sqrt 3... | The beat way to change to polar form is to change individual terms to polar form first
So
$$(1-i)^3=\left(\sqrt{2}e^{-\frac{\pi i}{4}}\right)^3=2\sqrt{2}e^{-\frac{3\pi i}{4}}$$
$$\sqrt{3}+i=2e^{\frac{\pi i}{6}}$$
and
$$4i=4e^{\frac{\pi i}{2}}$$
Which means that
$$z=\sqrt{2}e^{-\frac{13\pi i}{12}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the system $ \left\lbrace \begin{array}{ccc} \sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\ x+y &=& 2xy \end{array}\right. $ I have to solve the following system:
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\
x+y &=& 2xy
\end{array}\right.
$$
I've showed that it is equivalent to
$... | To solve the second equation system.
From $x+y=2xy$. We have $y=\frac{x}{2x-1}$.
Then use $\sqrt{1-x} = 2 + \sqrt{1-y}$ (I use the second equation, x=y=1 is excluded.)
$y-x-4 = 4 \sqrt{1-y}$
$\frac{x}{2x-1}-x-4 = 4 \sqrt{1-\frac{x}{2x-1}}$
$\frac{-2x^2-6x+4}{2x-1} = 4 \sqrt{\frac{x-1}{2x-1}}$
$(x^2+3x-2)^2 = 4(x-1)(2x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Solving $\frac{dy}{dx} = \frac{xy+3x-2y+6}{xy-3x-2y+6}$ I'm stuck with this problem...
$$\frac{\operatorname{d}y}{\operatorname{d}x} = \frac{xy+3x-2y+6}{xy-3x-2y+6}$$
I have tried separating variables in the following way:
$$\frac{\operatorname{d}y}{\operatorname{d}x} = \frac{x(y+3)2(-y+3)}{y(x-2)3(-x+2)}$$
$$\left(\fr... | $\begin{align}\frac{xy+3x-2y+6}{xy-3x-2y+6}&=\frac{6x+xy-3x-2y+6}{xy-3x-2y+6}\\
&=\frac{6x+x(y-3)-2(y-3)}{x(y-3)-2(y-3)}\\
&=\frac{6x+(x-2)(y-3)}{(x-2)(y-3)}\\
&=\frac{6x}{(x-2)(y-3)}+1\\
\end{align}$
So, your non-linear ODE is
$$y'=\frac{6x}{(x-2)(y-3)}+1$$
or
$$(y-3)(y'-1)=\frac{6x}{x-2}.$$
If we make a change of var... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Finding $\lim_{x\to 0}\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}$ How to evaluate the following limit without L'Hôpital's rule
?$$\lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)$$
My attempt:
$$\begin{align}L &= \lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)\tag1\\& = \lim_{x\to 0}\... | $$\lim_{x \rightarrow 0} \left( \frac{1}{\arcsin(x)^2} - \frac{1}{x^2} \right) = \lim_{x \rightarrow 0} \left( \frac{x^2 - \arcsin(x)^2}{\arcsin(x)^2 x^2} \right)$$
Taylor series of $ \arcsin(x) = x + x^3/6 + O(x^5) $ at $x \rightarrow 0$, in this way
$$\lim_{x \rightarrow 0} \left( \frac{x^2 - x^2 - x^4/3 - x^6/36}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How can we generalize factorisation of $(a+b)^n-(a^n+b^n)$
How can we generalize factorisation of
$$(a+b)^n-(a^n+b^n)\,?$$
where $n$ is an odd positive integer.
I found the following cases:
$$(a+b)^3-a^3-b^3=3ab(a+b)$$
$$(a+b)^5-a^5-b^5=5 a b (a + b) (a^2 + a b + b^2)$$
$$(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$$
Exap... |
A generalisation for non-negative integer $n$ is provided in this answer. The expansion there can be written as
\begin{align*}
&\,\,\color{blue}{(a+b)^n-\left(a^n+b^n\right)}\\
&\quad\,\,\color{blue}{=ab(a+b)\sum_{k=1}^{\lfloor n/2\rfloor}\left(\binom{n-k}{k}+\binom{n-k-1}{k-1}\right)(-ab)^{k-1}(a+b)^{n-2k-1}}\tag{1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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The number of distinct powers of x that appear in the expansion of $(x^{7}+x^{11}+x^{14})^{10}$ is
The number of distinct powers of x that appear in the expansion of
$(x^{7}+x^{11}+x^{14})^{10}$ is
Let the power of $x^7$ is a, $x^{11}$ is b and $x^{14}$ is c
then by multinomial a+b+c=10
so we have to find the number ... | First of all, notice that the answer to the question is equal to that of $(1+x^4+x^7)^{10}$, and then the answer is equal to the number of distinct sums of the form $4a+7b$, where $0\le a+b\le10$. If $a+b=k$, it is easy to notice that $4a+7b$ produces $k+1$ numbers, hence, we already have $1+2+...+11=66$ numbers. Howev... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4564425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Stationary points of a cubic function If t is a positive constant, find the local maximum and minimum values of the function
$f(x) = (3x^2 - 4)\left(x - t + \frac{1}{t}\right)$
and show that the difference between them is $\frac{4}{9}(t + 1/t)^3$.
Find the least value of this difference as $t$ is varied.
My attempt:
\b... | Note that you can factor the derivative as
$$ \frac{\rm d}{{\rm d}x} f = \tfrac{1}{t} ( 3x -2 t) ( 3 t x + 2) = 0 $$
which gives you the $x$ values for the extrema
$$ \begin{gathered} x = \frac{2 t}{3} & x = -\frac{2}{3 t} \end{gathered} $$
For the next part, you can do a variable substitution with $ u = t + \frac{1}{t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $x+y$, given $xy= 1$, $x^2+y^2=5$, $x^3+y^3=8$ This problem is from a math competition, but I think is wrong:
Find the value of $x+y$ if:
$$\begin{align}
xy &= 1 \\
x^2 + y^2 &= 5 \\
x^3+y^3 &= 8
\end{align}$$
Solution (I think is wrong):
$x^3 + y^3 = (x + y)(x^2-xy+y^2) = (x+y)(5-1) = 4(x+y)$
So we have:
$... | Let us call $a := x + y$ and $b := xy$. Then the proposed system of equations is equivalent to
\begin{align*}
\begin{cases}
b = 1\\\\
a^{2} - 2b = 5\\\\
a^{3} - 3ab = 8
\end{cases} \Longleftrightarrow
\begin{cases}
b = 1\\\\
a^{2} = 7\\\\
a^{3} - 3a = 8
\end{cases} & \Longleftrightarrow
\begin{cases}
b = 1\\\\
a = \pm\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Find all the solutions for $f\left(x\right) =2f\left(\frac{1}{x}\right)-\frac{2x^{2}-1}{x^{2}+1}$ The function is $f:\left(0,\infty\right)\rightarrow\mathbb{R}$, I tried to do it like that, first I saw that: $$f\left(x\right) =2f\left(\frac{1}{x}\right)-\frac{2x^{2}-1}{x^{2}+1}$$ and then decided to try to put $\frac{1... | $$f(x)=2f\left(\frac 1x\right)-\frac{2x^2-1}{x^2+1}\\f\left(\frac 1x\right)=2f(x)-\frac{2-x^2}{x^2+1}$$ Adding and simplifying we have
$$f(x)+f\left(\frac 1x\right)=1$$ It follows
$$\boxed{f(x)= \frac{1}{x^2+1}}$$ and it is easy to verify that in fact
$$f(x)=2f\left(\frac 1x\right)-\frac{2x^2-1}{x^2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequalities showing monotonicity imply another inequality I came across a proof that shows that $\lim\limits_{n \rightarrow \infty} (1+ \frac{1}{n})^{n\cdot (\frac{1}{n+1} +\dots + \frac{1}{2n})} =2$
First it is shown that $((1+\frac{1}{n})^{n+1})_{n=1}^{\infty}$ is monotonically decreasing and $((1+\frac{1}{n})^n)_{n... | By starting from
$$
\left(1 + \frac1n\right)^{n} \le \left(1 + \frac1{n+k}\right)^{n+k}
$$
and raising both sides to the power $\frac1{n+k}$, we get
$$
\left(1 + \frac1n\right)^{\frac{n}{n+k}} \le \left(1 + \frac1{n+k}\right). \tag{1}
$$
Also, by starting from
$$
\left(1 + \frac1{n+k}\right)^{n+k+1} \le \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $? I need to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $. I've checked numerically that this is true, but I haven't been able to prove it.
I've tried trigonometric substitutions. Let $\tan u= t:$
$$\int_0^1 (1+t^2)^{\frac 7 2} dt = \int_0^{\frac... | Alternative proof:
Clearly, we have
$\sqrt{1 + t^2} \le 1 + t^2/2$. Thus, we have
$$(1 + t^2)^{7/2} \le (1 + t^2)^3 (1 + t^2/2) = \frac12t^8 + \frac52 t^6 + \frac92 t^4 + \frac72 t^2 + 1.$$
Thus, we have
\begin{align*}
\int_0^1 (1 + t^2)^{7/2} \, \mathrm{d} t
&\le \int_0^1 \left( \frac12t^8 + \frac52 t^6 + \frac92 t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Extracting sequence from generating function I was provided the following generating function, and was unsure how to use it. I have never seen an example where the function “involved” itself.
The generating function is
$F(z)^8$
Where
$$F(z)=z+z^6 F(z)^5+z^{11} F(z)^{10}+z^{16} F(z)^{15}+z^{21} F(z)^{20}$$
Any help appr... | A method that will take a while but gives the first few terms for the equation
$$ F(x) = x + x^6 F(x)^5 + x^{11} F(x)^{10} + x^{16} F(x)^{15} + x^{21} F(x)^{20} $$
is to let $F(x) = a_{0} + a_{1} \, x + a_{2} \, x^2 + \cdots$.
First notice that the right-hand side does not have a constant term. This leads to $F(x)$ bei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve the differential equation $y'=\frac{y-xy^2}{x+x^2y}$ $$y'=\frac{y-xy^2}{x+x^2y}$$
This is the equation I want to solve.
My idea is to substitute $xy$ with $u,$ $u=xy$ and $\frac{du}{dx}=xy'+y.$
So, the equation becomes
$y'=\frac{y(1-u)}{x(1+u)}.$
What I am struggling with is how to deal with the $x$ and $y... | Solve
\begin{gather*}
\boxed{y^{\prime}-\frac{y-x y^{2}}{x +x^{2} y}=0}
\end{gather*}
Using $u=yx$ the above becomes
\begin{align*}
u' &= \frac{2 u}{x \left(u +1\right)}
\end{align*}
This is separable. Integrating gives
\begin{align*}
\left(\frac{u +1}{u}\right)\mathop{\mathrm{d}u}&= \left(\frac{2}{x}\right)\mathop{\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How resolve $y''+y=\sin(x)$ by power series around the point $x_0=0$ The ODE I tried to solve is:
$$y''+y=\sin(x)$$
using the power series method around the point $x_0=0$, with the conditions:
$$y(0)=0,\qquad y'(0)=0$$
Let
\begin{align}
y(x) &= \sum_{n=0}^{\infty} a_{n} \, x^n \\
y^{'}(x) &= \sum_{n=0}^{\infty} (n+1) \... | The difficulty as stated in the work of the proposed problem the left-hand side should be broken into even and odd index to match the right-hand side. Doing so gives the desired recurrence equations and thus solution.
This solution takes a similar but different path. Let
$$ y(x) = \sum_{n=0}^{\infty} a_{n} \, x^n $$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Better algorithm to find numerical value of one over squareroot Description: To find the value of $1/\sqrt{a}$, basically we find the root $x^{\star}$ of the function $f(x)$
$$
f(x) = x^2 - \dfrac{1}{a}
$$
One of the ways to find it is by using Newton's method and iterating from an initial value $x_0$.
$$
x_{k+1} = x_{... | Householder's method of order $d$ to solve $f(x) = 0$ gives
$$x \to H_d(x) = x + d \frac {\left(1/f\right)^{(d-1)}(x)}{\left(1/f\right)^{(d)}(x)}$$
with rate of convergence $d + 1$.
Using the (wx)Maxima computer algebra system, for
$$f(x) = x^{-2} - a$$
I get
$$\begin{aligned}
H_1(x) &= \frac{(3 - a x^2)x}{2} \\
H_2(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4581057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"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.