Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Trace zero matrix that can't be written as $AB - BA$? According to this paper, every trace zero matrix over a field can be written in the form $AB - BA$. However, here's a basic counterexample:
Let $A = diag(a, -a)$ for some nonzero number a. Then $A = BC - CB$ if and only if a system of equations involving entries ... | $$B=\left(\begin{array}{rr}0&1\\0&0\end{array}\right).$$
$$C=a\left(\begin{array}{rr}0&0\\1&0\end{array}\right).$$
$$BC-CB=\left(\begin{array}{rr}a&0\\0&-a\end{array}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/466018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How can I evaluate $\lim_{x \rightarrow 0} \frac{\left(\sqrt{2+x}-\sqrt{2} \right)}{ x}$? I am having trouble with this practice problem on limits:
$$\lim_{x \rightarrow 0} \frac{\left(\sqrt{2+x}-\sqrt{2} \right)} {x}$$
The answer is $\sqrt{2} \over 4$, but I'm having trouble seeing how the answer was reached. Any he... | I assume that you made typing error (probably there should write $\sqrt{2+x}$, not $\sqrt{2x}$). Then, the result you have stated is a limit of the following expression:
\begin{eqnarray*}
\frac{\sqrt{2+x}-\sqrt{2}}{x}&=&\frac{\sqrt{2+x}-\sqrt{2}}{x}\frac{\sqrt{x+2}+\sqrt{2}}{\sqrt{x+2}+\sqrt{2}},\\
\frac{\sqrt{2+x}-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/466669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculate the determinant of the matrix $(a_{ij})$ where $a_{ij}=a+b$ when $i=j$, and $a_{ij}=a$ otherwise The matrix is $n\times n$ , defined as the following:
$$
a_{ij}=\begin{cases}
a+b & \text{ when } i=j,\\
a & \text{ when } i \ne j
\end{cases}.
$$
When I calculated it I got the answer of: $b^n +nab^{n-1... | Let the $n \times n$ matrix $A$ be given by $$A = \begin{pmatrix} a+b & a & a & \cdots & a & a \\ a & a+b & a & \cdots & a & a \\ a & a & a+b & \cdots & a & a \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ a & a & a & \cdots & a+b & a \\ a & a & a & \cdots & a & a+b \end{pmatrix} $$
Subtracting row $i+1$ fro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
what is the taylor expansion of $\ln(\sin x/x)$ what is the Taylor expansion of the function
$$f(x) = \ln \frac{\sin x}x$$
around the point $x=0$? Ignore powers of $x$ which are greater than $6$.
Here is my method:
$$\ln(1+x)=x-\frac{x^2}2 + \frac{x^3}3 -\frac{x^4}4,$$
so we should get the function
$$g(x)=\frac{\sin x... | As you suggest, it will be useful to know the expansion
$$\ln(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots\tag{1}$$
(at least for $|t|\lt 1$).
We also need the power series expansion for $\sin x$, which is
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots.$$
So the function which is $1$ at $x=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Limits of integration, surface joined by paraboloid and plane I need help with part of this old exam question:
Q:
C is the intersection line between the paraboloid $z=x^2+y^2$ and the plane $2x-2y+z=1$ travelled counter-clockwise from $(0,0,10)$. Calculate $\oint\limits_C F\cdot dr$. $F(x,y,z)=(z-3y)i+(3x+z)j+e^{x-y}k$... | Prof. solution is easier than the following ( see below ) !!!
You can find $x$ and $y$ in terms of $z$:
$$
z - \left(1 - z \over 2\right)^{2} = \left(x^{2} + y^{2}\right) - \left(x - y\right)^{2}
=
2xy
$$
$$
2xy = {-z^{2} + 6z -1 \over 4}
$$
$$
\left(x^{2} + y^{2}\right) + 2xy
=
z + {-z^{2} + 6z -1 \over 4}
=
{-z^{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/471370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Thinking through a Taylor error bound for arcsine In lecture, we went through solving a Taylor error bound for arcsine. I followed most of it except for where it talks about the odds divided by the evens divided by $2n+1$ gaining in accuracy by a factor of 1/10 for each successive term (see bolded sentence below), whic... | The assertion in the notes is, roughly speaking, false.
Let $\arcsin x=\sum a_{2n+1} x^{2n+1}$. Then
$$\lim_{n\to\infty}\frac{a_{2n+3}}{a_{2n+1}}=1.$$
More informally, in the long run we are picking up nothing "in the denominator." In the case $x=\frac{1}{10}$, for large $n$, almost all the gain in accuracy comes from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/473983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of the series $1^2-2^2+3^2-4^2+...-(2n)^2$ Find the sum of the series $$1^2-2^2+3^2-4^2+...-(2n)^2$$
I tried rewriting it as $$\sum_{r=1}^{2n}-1^{n+1}(r^2)$$ but it didn't help.
Also, looked at re-arranging as $$1^2+3^2+5^2+7^2+...+(2n-1)^2$$ and $$-2^2-4-6^2-8^2-...-(2n)^2$$
Still couldn't get to the give... | Hint 1: We have $$\sum_{k=1}^nk^2=\frac{1}{6}n(n+1)(2n+1).$$
Hint 2: Note that
$$1^2-2^2+3^2-4^2+\cdots-(2n)^2=1^2+2^2+\cdots+(2n)^2-2(2^2+4^2+\cdots+(2n)^2)$$
$$
=1^2+2^2+\cdots+n^2-2^3(1^2+2^2+\cdots+n^2)
=\sum_{k=1}^{2n}k^2-8\sum_{k=1}^{n}k^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Determinant of a $2 \times 2$ block matrix $\textbf{Problem}$: Let a $2n \times 2n$ matrix be given in the form $M=\left[ {\begin{array}{cc}
A & B \\
C & D \\
\end{array} } \right]$, where each block is an $n \times n$ matrix. Suppose that $A$ is invertible and that $AC=CA$. U... | Your proof seems fine to me.
As for a counterexample, consider
$$A = \begin{bmatrix}
2 & 0 & 1 & 0 \\
0 & 1 & 2 & 1 \\
2 & 1 & 1 & 1 \\
1 & 1 & 1 & 2
\end{bmatrix}.$$
In one hand, $\det A = -4$ (check here), and in the other hand, $\det (A_{11} A_{22} - A_{21}A_{12}) = 0$ (check here).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/476412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\ta... | Here is a geometric proof, for the case that all three angles are acute:
$QRUV$ are collinear because $B+90^\circ+(90^\circ-B)=180^\circ$.
$STV$ are collinear because $A+B+C=180^\circ$, so $\angle QSV=\angle UTV=C$.
Similar triangles $\triangle PQR\sim\triangle TRS$ and $\triangle RTU \sim \triangle SRQ$ give $\displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 6
} |
Solve $x+y+z = x^3 + y^3 + z^3 = 8$ in $\mathbb{Z}$ Solve $x+y+z = x^3 + y^3 + z^3 = 8$ in $\mathbb{Z}$
First I tried to transform this equation, substituting $x = 8-y-z$. So I end up with:
$$x^3 + y^3 + z^3 = 8$$
$$(8-y-z)^3 + y^3 + z^3 = 8$$
Using Wolfram Alpha I expanded this equation and tried to factorize it so f... | Hint: Calculate $(x-8)(y-8)(z-8)$.
There are several ways to do the calculation. Working from the last equation in the first method, observe that the second term can be factorized into $(y-8)(y+z)= -(y-8)(x-8)$, which gives us
$(x-8)(y-8)(z-8) = -168 $
Use this, along with the factors of 168 and that $x-8+y-8+z-8 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to apply polynomial long division to $x^3/(x^2+3x+2)$ So I have a basic question. I needed to simplify this expression
$$\frac{x^{3}}{x^{2} +3x +2}$$
I was to do a polynomial long division in order to simplify it, but I still do not understand how they reached this next expression (I know it can be more simplified... | $$\frac{x^3}{x^2+3x+2} = \frac{x(x^2+3x+2)-3x^2-2x}{x^2+3x+2} = x - \frac{3(x^2+3x+2)-7x-6}{x^2+3x+2}=x-3+\frac{7x+6}{x^2+3x+2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/481191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Creating a degree $n$ Taylor polynomial for $\sqrt{1+x}$ I have been asked to produce a general formula for the degree $n$ Taylor polynomial for $\sqrt{1+x}$ using a=0 as the point of approximation.
Given that $p_n(x)=f(a)+(x-a)f^\prime(a)+\frac{(x-a)^2}{2!}f^{(2)}(a)+...+\frac{(x-a)^n}{n!}f^{(n)}(a)=\sum_{j=0}^{n}\fra... | Note that $$1\cdot 3\cdot 5 \cdots (2j - 3) = \dfrac{(2j-3)!}{2\cdot 4 \cdot 6 \cdots (2j -2)} = \dfrac{(2j-3)!}{2^{j-1}\cdot 1\cdot 2 \cdot 3 \cdots (j -1)} = 2^{1-j}\dfrac{(2j-3)!}{(j-1)!}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/482273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How find this integral limit
Find this limit $$\lim_{n\to\infty}\dfrac{\displaystyle\int_{0}^{\frac{\pi}{2}}x\left(\dfrac{\sin{(nx)}}{\sin{x}}\right)^4dx}{n^2}.$$
This problem is post my teacher(tian275461, he is very popular in China), and he told me this problem answer is $\ln{2}$, but I can't prove it. Thank you.
| With the substitution $y = nx$, we have
\begin{align}
I_n &= \int_0^{n\pi/2} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y \\
&= \underbrace{\int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y}_{I_{n,1}} +
\underbrace{\int_{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/482583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Ancient Babylonian problem: solve the system $x + y = 50$, $x^2 + y^2 + (x - y)^2 = 1400$ $x + y = 50, x^2 + y^2 + (x - y)^2 = 1400$. [Hint: Subtract the square of the first equation from twice the second equation to get a quadratic in $x - y$.] I have gotten it reduced to $x^2 + y^2 -3xy = 550$ but am not sure how t... | As you have now figured out the answer on your own, I'll post the way I was hinting at in a comment.
I would say your book is pointing you towards the hard way. It's easier just to solve for $y$ in the first equation and substitute that into in the second, giving you a quadratic in $x$, trivially solved by inspection o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How prove this inequality $(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$ Now my question let $a_{1},a_{2},\cdots,a_{n}$ are positive numbers,and $a_{n+i}=a_{i},i=1,2,\cdots$,show that
$$(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$$
my teacher (tian275461) have prove this
$$(a^{1.5}+b^{1.5}+c^{1.5})^2\ge (a+b... | The conjecture is false.
(See my previous post for sufficient extra conditions such that the conjecture holds.)
Here is a counterexample. Let n even (n odd works as well).
Let $a_1=a_2 = \dots = a_{n/2} = A$ and $a_{n/2 +1 } = a_{n/2 +2 } = \dots = a_{n} = B$. Let, without loss of generality, $B<A$.
Then the conjectur... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Sums of rising factorial powers Doodling in wolfram, I found that
$$
\sum^{k}_{n=1}1=k
$$
The formula is pretty obvious, but then you get
$$
\sum^{k}_{n=1}n=\frac{k(k+1)}{2}
$$
That is a very well known formula, but then it gets interesting when you calculate
$$
\sum^{k}_{n=1}n(n+1)=\frac{k(k+1)(k+2)}{3}\\
\sum^{k}_{n=... | Partial sums of sequences $a_n$ that can be expressed as polynomials in $n$ are easily found using discrete calculus.
We start with the discrete version of the Fundamental Theorem of Integral Calculus:
\begin{align*}
\sum\limits_{n=1}^k a_n &= \sum\limits_{n=1}^k (\Delta b)_n \\
&= (b_2 - b_1)+(b_3 - b_2)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
} |
sum of legs of inscribed right triangle Consider a right and isoceles triangle ABC inscribed in a circle such as its hypotenuse forms the diameter AB of the circle (the right vertex is thus at the "apex" of the circle). If we procede to an infinitesimal displacement of the "apex" along the circumference $\ ds$ with sub... | Let's assume your circle to be the unit circle. Then you have
\begin{align*}
y &= \sqrt{1-x^2} \\
y^2 &= 1-x^2 \\
a(x) &= \sqrt{(1+x)^2+y^2} = \sqrt{1+2x+x^2+1-x^2} = \sqrt{2\left(1+x\right)} \\
b(x) &= \sqrt{(1-x)^2+y^2} = \sqrt{1-2x+x^2+1-x^2} = \sqrt{2\left(1-x\right)} \\
D(x) &= a(x)+b(x) = \sqrt{2\left(1+x\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Electric field of a spherical shell I hope this question is appropriate for this site, if not, just leave a comment and I will delete.
I am interested in knowing how to derive the electric field due to a spherical shell by Coulomb's law without using double integrals or Gauss Law.
Relevant equations are -- Coulomb's l... | We assume that the sphere of radius $R$ centered at 0. Let us assume an observation
point $o$ above the north pole of the sphere (by symmetry this should provide a
good answer). We consider rings from the bottom up to the north polo $(0,0,R)$.
The ring at a high $z$, $-R \le z \le R$ has a radius
$\rho=\sqrt{R^2 - z^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/486293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integrating $\int_0^{\frac{\pi}{2}}\cos(2x)\sin(3x)\,dx$ How would I integrate the following
$\int_0^{\frac{\pi}{2}}\cos(2x)\sin(3x)\,dx$
I did the following
$\frac{1}{2}\int \left(\sin(5x)-\sin(x)\right)\,dx$
$\frac{1}{10}-\cos(5x)+cos(x)$
$\frac{1}{10}(0)+0-0=0$
But I am not sure If I did it right.
| Not quite. Using the identity
$$2 \cos{a}\sin{b} = \sin{(a + b)} - \sin{(a - b)}$$
we find that
\begin{align}
\int_0^{\pi/2} \cos{(2x)} \sin{(3x)} &= \frac 1 2 \int_0^{\pi/2} \sin{5x} - \sin{(-x)} \\
&= \frac 1 2 \int_0^{\pi/2} \sin{5x} + \sin{x} \\
&= \frac 1 2 \left(\frac{-1}{5} \cos{5x} - \cos{x}\right)_0^{\pi/2} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/486816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solutions to $\sqrt{x}+y=6,x^2+y^2=90$ $$\begin{gather}
\sqrt{x}+y=6 \tag{1} \\
x^2 + y^2 = 90 \tag{2}
\end{gather}$$
WE have to solve for $x$ and $y$(Note that 9 is an obvious value of x) My friend asked me this question earlier today, which he said he had made up himself. However, he himself has no idea how to solve... | So we have $$\sqrt{x}=6-y$$ which we substitute into the second equation to give $$(6-y)^4+y^2=90.$$ This is equivalent to finding the roots of the polynomial $$f(y):=(6-y)^4+y^2-90.$$ The Fundamental Theorem of Algebra implies this has exactly $4$ (not necessarily distinct) roots.
You mentioned one root, $3$, in you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Help me solve this polynomial problem I have to find $a$,$b$ and $c$ so that this expression is a polynomial:
$$\frac{x^3+ax^2+bx+c}{x-1} + \frac{x^3+bx^2+cx+a}{x-2} + \frac{x^3+cx^2+ax+b}{x-3}$$
Please give me hints. Thanks!
| Hint: $x^3 + ax^2 + bx + c$ must have a root at $1$. This is because $(x - 1)$ must be a divisor of $x^3 + ax^2 + bx + c$, or else there would be a term involving the quantity $\frac{1}{x - 1}$. In a similar manner, $x^3 + bx^2 + cx + a$ has a root at $2$ and $x^3 + cx^2 + ax + b$ has a root at $3$. This gives us three... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Proof of Bézout's Theorem Gostaria de saber como provar usando divisibilidade o teorema de Bezout
$(a,b)=d\Longrightarrow \exists f,g\in\mathbb{Z^*}$ tal que $af+gb=d$
I'd like to know how to, using divisibility, prove Bézout's Theorem:
Given integers $a,\ b$, if $(a,b)=d$ then there exist $f,g\in\mathbb{Z^*}$ such... | First of all, it is only an implication in one sense, in English it is incorrectly given as an equivalence.
I very much like the constructive proof. Suppose without loss of generality $a \ge 0$ and $b > 0$, and write the obvious
\begin{equation*}
\begin{matrix}
a &=& a \cdot 1 &+& b \cdot 0\\
b &=& a \cdot 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/492864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find all real matrices $A$ such that $A^2 = \mathrm{diag}(1,1,2,3,5,8,13)$ Let $A \in \mathcal{M}_{7 \times 7} (\mathbb{R})$ such that
$$A^2= \begin{pmatrix} 1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&2&0&0&0&0 \\ 0&0&0&3&0&0&0 \\ 0&0&0&0&5&0&0 \\ 0&0&0&0&0&8&0 \\ 0&0&0&0&0&0&13\end{pmatrix} $$
How many matrices, which satisfy... | Hint: $$\begin{pmatrix} \cos \varphi & \sin \varphi & 0 & \ldots & 0 \\ \sin \varphi & -\cos \varphi & 0 & \ldots & 0 \\ 0 & 0& 0& \ldots & 0\\ \vdots & \vdots& \vdots& \ddots & \vdots\\ 0 & 0& 0& \ldots & 0 \end{pmatrix}^2 = ?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit of $s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$ \begin{align*}S_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)\end{align*}
how to calculate the limit $s_n$?
\begin{align*}\lim_{n\to \infty } \, S_n\end{align*}
| \begin{align}
{1 \over \sqrt{n}}\,\sum_{k = 1}^{n}{1 \over \sqrt{k\,}}
&=
{1 \over \sqrt{n}}\,\sum_{k = 1}^{n}{1 \over \sqrt{n\xi_{k}\,}}\,n\Delta\xi
=
\sum_{k = 1}^{n}{1 \over \sqrt{\xi_{k}\,}}\,\Delta\xi
\sim
\int_{1/n}^{1}{{\rm d}\xi \over \xi^{1/2}}
=
\left.\vphantom{\LARGE A}\;2\xi^{1/2}\right\vert_{1/n}^{1}
\\[3m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$
Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.
I have tried it by substituting $x = \exp(2i\pi/5)$
but it is getting complicated.
| Replacing the earlier sequence of hints with a solution now. The old aswer is scratched.
We know that $0=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$. As $x-1\neq0$ this implies that
$$1+x+x^2+x^3+x^4=0.\qquad(*)$$
From $x^5=1$ we also deduce that $x^4=x^{-1}$. This allows a rewrite:
$$
S=(x^2-x^3)(x^4-x)=(x^2-x^3)(x^{-1}-x)=(x-x^2)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
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} |
Find the constant term in the development of $(2x^2 + \frac{1}{4x})^{99}$ I am struggling with the following problem:
Find the constant term in the development of $(2x^2 + \frac{1}{4x})^{99}$
This is a combinatorics problem where I should find what the constant term is when $x^0$.
I have made the following development... | You found that the constant term was the $k=66$ term, which for your setup is correct; you get a coefficient of
$$\binom{99}{66}2^{-99}\;,$$
which is equal to $$\binom{99}{99-66}2^{-99}=\binom{99}{33}2^{-99}\;.$$
Don’t forget that $\dbinom{n}k=\dbinom{n}{n-k}$ always.
If you switch the rôles of the two terms before you... | {
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"source": "stackexchange",
"question_score": "5",
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How to find $\lim\limits_{n\to\infty}\sum\limits_{j=1}^{n^2}\frac{n}{n^2+j^2}$ find the limit value
$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}$$
this following is my methods:
let $$S_{n}=\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\sum_{j=1}^{n^2}\dfrac{1}{1+\left(\dfrac{j}{n}\right)^2}\dfrac{1}{n}$$
since
$$\... | Here is a method in the same spirit as what you have done. The main difference is that we explicitly estimate the difference between your sum and $\int_0^n\frac{{\rm d}x}{1+x^2}$.
Since $f$ is monotonely decreasing we have (see for example integral test on Wikipedia)
$$\left|\sum_{j=1}^n \frac{n}{n^2+(j+nm)^2} - \int_{... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Is this Cayley table correctly computed? If so, is it correct way of presentation of this dihedral group? Is the following table for $D_4$ correct?
$$\begin{array}{c|c|c|c|c|c|c|c|c|}
* & 1 & g & g^2& g^3& f& fg & fg^2 & fg^3 \\ \hline
1 & 1 & g & g^2 & g^3 & f & fg & fg^2 & fg^3 \\ \hline
g & g & g^2 & g^3 & 1 &... | A presentation of $D_4$ is $\langle g,f \mid g^4=1, f^2=1, gf=fg^{-1}\rangle$. So we have the $8$ elements $\lbrace 1,g,g^2,g^3,f,gf=fg^3,g^2f=fg^2,g^3f=fg\rbrace$, and the multiplication table follows from the relations - as far as I can see, your table is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of infinite series How to find sum of the following series
$$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\cdots={1\over 6} + \sum_{n=1}^\infty{\Pi_{i=1}^n{3i+2}\over (n+1)!6^{n+1}}$$
Please give me some hints. Thanks in advance.
| The general term of the series reads:
$$
\frac{1}{2}\frac{\prod_{k=1}^n (3k-1)}{\prod_{k=1}^n 6 k} = \frac{1}{2^{n+1}} \prod_{k=1}^n \left(1-\frac{1}{3k}\right) = \frac{1}{2^{n+1}} \frac{\left(2/3\right)_n}{n!} = \frac{1}{2^{n+1}} \frac{\Gamma\left(n+2/3\right)}{n! \cdot \Gamma\left(2/3\right)}
$$
The series thus r... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Simultaneous Equations That Should Be Inconsistent Has a Unique Solution Find the values of $k$ for which the simultaneous equations do not have a unique solution for $x, y$ and $z$.
Also show that when $k = -2$ the equations are inconsistent
$$kx + 2y +z =0$$
$$3x + 0y -2z =4$$
$$3x - 6ky -4z =14$$
Using a determinant... | Firstly, a non-zero determinant doesn't imply there are no solutions. The system of equations
$$
\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 1 \\
2 & 2 & 2 & 2 \\
3 & 3 & 3 & 3 \\
\end{array}
\right]
$$
has an infinite number of solutions, for example, and the relevant determinant will be zero. So, we can't be sure the s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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calculate a trigonometric expression related to $\sin(\pi/5)$ I tried to solve a question which led to
$$4 \sin\frac{\pi}{5} \sin\frac{3\pi}{5} = \sqrt{5}$$
But, I got stuck to prove that.
Is there any easy solution?
| Using this,
$\sin5x=16\sin^5x-20\sin^3x+5\sin x$
Let $\sin5x=0\implies 5x=n\pi$ where $n$ is any integer
$\displaystyle\implies x=\frac{n\pi}5$
Clearly, $\displaystyle\sin\frac{\pi}5, \sin\frac{3\pi}5$ satisfy $16\sin^5x-20\sin^3x+5\sin x=0$
Again, if $\sin x=0,x=m\pi$ where $m$ is any integer
$\implies \displaystyle\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Squares modulo a product of distinct odd primes Let $p,q$ be distinct odd primes. Is it true that a number $a$ is a square modulo $pq$ if and only if $a$ is a square modulo $p$ and modulo $q$?
| If $x^2\equiv a\pmod{pq}$, then it is immediate that $x^2\equiv a\pmod{p}$ and $x^2\equiv a\pmod{q}$. Thus a square modulo $pq$ is a square modulo each of $p$ and $q$.
Next we prove the converse, if $a$ is a square modulo $p$ and a square modulo $q$, then $a$ is a square modulo $pq$.
So we want to show that if there e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}$ How to prove that
$$\sum_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}=\frac1{\sin^2x}-\frac1{x^2}$$
| Often when $\sin(x)$ and $x$ appear in a formula such as this, it is based on the fact that
$$
\lim_{a\to0}\frac{\sin(ax)}{a}=x\tag{1}
$$
Since the summands are $\dfrac{4^{-k}}{\cos^2(2^{-k}x)}$ , it appears that an identity involving a double angle formula is being exploited. Therefore, with an eye toward a telescopin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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A book has a few pages on which page numbers are written. Someone has torn one page out of it and now average of all page numbers is $\frac{105}{4}$ I couldn't relate this question to any of the topics specifically , I found this in a miscellaneous math problems book(non-calculus) . Here's how it goes, A book has a fe... | More on how to deduce the value of $n$. The average page number before a page is torn is
$$ \frac{1+2+\cdots+n}{n}=\frac{n+1}{2}$$
Since two consecutive page numbers are removed, let the two page numbers be $2k-1$ and $2k$, where $2\le 2k\le n$, then
$$\begin{align}
n\cdot\frac{n+1}{2}-(n-2)\frac{105}{4} =& 2k-1+2k\\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Independence of $\frac{1-\cos(x)+k\sin(x)}{\sin(x)+k(1+\cos(x))}$ from $k$. How to one can show that the value of the following expression
$$\frac{1-\cos(x)+k\sin(x)}{\sin(x)+k(1+\cos(x))}$$
doesn't depend to values of $k$?
| Writing $x = 2y$, we obtain
$$\begin{align}
\frac{1-\cos x + k\sin x}{\sin x + k(1+\cos x)} &= \frac{(1-\cos (2y)) + k\sin (2y)}{\sin(2y) + k(1+\cos(2y))}\\
&= \frac{2\sin^2 y + 2k\sin y\cos y}{2\sin y\cos y + 2k\cos^2 y}\\
&= \frac{\sin y}{\cos y}\cdot\frac{\sin y+k\cos y}{\sin y + k\cos y}\\
&= \tan y
\end{align}$$
u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Rate of convergence for sequences I have numerically determined that the sequence $\{f_x\} = \frac{\sin(x^2)}{x^2}$ approaches $1$ (as $x$ approaches $0$) faster than the sequence $\{g_x\} = \frac{\sin^2(x)}{x^2}$. However, I am stuck for determing the rate of convergence of each sequence.
Should I use the small angle ... | Use Taylor expansion:
$$
f_x=\frac{1}{x^2}\Bigl(x^2-\frac{x^6}{6}+O(x^{10})\Bigr)=1-\frac{x^4}{6}+O(x^8).
$$
$$
g_x=\frac{1}{x^2}\Bigl(x-\frac{x^3}{6}+O(x^{5})\Bigr)^2=1-\frac{x^2}{3}+O(x^4).
$$
Added detail on the expansion of $g_x$:
$$
\Bigl(x-\frac{x^3}{6}+O(x^{5})\Bigr)^2=x^2-2\cdot x\cdot\frac{x^3}{6}-2\cdot x\cdo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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When computing $\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}$, I don't get -2. This limit:
$\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}$
is supposed to be equal to -2. My textbook and Wolfram Alpha both state that. However, I can't seem to get same exact result.
Here's what I tried to do:
$$\lim_{x\to -\infty... | Hint:
Think about the signs of the numerator and denominator in the last step. If $x\to -\infty$, then what happens to $2x + 7$? If you take the positive square root on the denominator, then what is the sign of the quotient?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Binomial Coefficient Identities I am trying to simplify the following fraction, which I think is equal to 1 but I am not sure.
$$\frac{\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}}{\frac{\left(\begin{ar... | The Fraction above does not simplify down to 1 as I had thought it would. It simplifies down to $\frac{k}{b}\cdot\frac{r+b}{n}$.
If we go back to Step 2 of the original question, we have
$$\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)}{\left(\begin{array}{c}
b\\
k
\end{array}\right)}\cdot\frac{\left(\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/505411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the inverse of $x^6+x^4+x+1$ in the finite field $\mathbb{Z}_2[x]/(x^8+x^4+x^3+x+1)$
How can I find the inverse of $x^6+x^4+x+1$ in the field $\mathbb{Z}_2[x]/(x^8+x^4+x^3+x+1)$?
can someone help me please to solve the problem.
actually I want to know the process of calculation.
I know that $1+(x^8+x^4+x^3+x+1)$... | You can use Extended Euclidean algorithm.
It only take three steps:
$x^8+x^4+x^3+x+1=(x^2+1)(x^6+x^4+x+1)+x^2$
$x^6+x^4+x+1=(x^4+x^2)x^2+x+1$
$x^2=(x+1)(x+1)+1$
You also needs to express the remainders using $f(x)=x^8+x^4+x^3+x+1$ and $g(x)=x^6+x^4+x+1$.
Let us make this into a table.
Each line contains coefficients ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to evaluate $\int_{0}^{\pi} \log(2+\cos x)dx$? I tried integrating it by part but it didn't work. I konw it can be solved by the Gauss Mean Value theorem .Is there some elementary method for evaluate it or just some others? It seems there are at least four differential ways(page 4 (17)) but I cannot think out any(t... | Another way: differentiate under integral sign!
Let $\displaystyle\quad J(a) = \int_0^\pi \log(a + \cos x) dx,\quad$ and $t = \tan\frac{x}{2}$, we have:
$$\begin{align}
J'(a)
= & \int_0^\pi \frac{dx}{a + \cos x}
= \int_0^{\infty} \frac{\frac{2 dt}{1+t^2}}{a + \frac{1-t^2}{1+t^2}}
= \frac{2}{a+1}\int_0^{\infty}\frac{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/505714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Is $\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}?$, for $a,b,c,d>0$? Is $$\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}$$
for $a,b,c,d>0$
If it is true, then can we generalize?
EDIT:typing mistake corrected.
EDIT, WILL JAGY. Apparently the real question is
Is $$\color{magenta}{\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}}$$
for $a,b,... | A slightly different approach:
Multiply both sides by (c+d), which we can do without altering the inequality because c and d are positive:
$$ a+b < \frac{a(c+d)}{c} +\frac{b(c+d)}{d}$$
$$ a+b < \frac{ac}{c} +\frac{ad}{c} +\frac{bc}{d} +\frac{bd}{d}$$
$$ a+b < a + \frac{ad}{c} +\frac{bc}{d} +b$$
$$ a+b < a+b +\frac{ad}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/506152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution of Equation $ 1 + x + \frac{1\cdot3}{2!}x^2 + \frac{1\cdot3\cdot5}{3!}x^3 + \ldots = \sqrt{2} $ I have an equation whose left side is a infinite series . I can solve the equation if I am able to find a close form of the series . The equation is as follows :
$$ 1 + x + \frac{1\cdot3}{2!}x^2 + \frac{1\cdot3\cd... | Compare the series with $1+na+\frac{n(n-1)a^2}{2!}\cdots$ we get
$na=x$ or $$n^2a^2=x^2\dots(1)$$
also $$\frac{n(n-1)a^2}{2!}=\frac{1.3x^2}{2!}\cdots(2)$$
Divide $(1)$ and $(2)$ and solve for $n$, substitute in first and solve for $a$
you will get the sum to be $(1-2x)^{\frac{-1}2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/509693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Simplify $(A+C)(AD+AD) + AC + C$ using Boolean algebra I have solved the equation like this:
(A + C)(AD + AD) + AC + C
=(A + C)(AD) + (A + C)(AD) + AC + C
=AAD + ACD + AAD + ACD + AC + C
=AAD + AAD + ACD + ACD + AC + C
=AAD + ACD + AC + C
=AAD + ACD + C(A+1)
=AAD + ACD + C
=AD + ACD + C
=AD(1+C) + C
=AD + C
Am I rig... | I would have taken the following steps:
$$
\begin{align}
(A + C)(AD + AD) + AC + C
&= (A + C)(AD) + (AC + C)\\
&= AD + ADC + C\\
&= AD + C\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/513518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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cubic equations which have exactly one real root Question is to check :
For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .
the way in which i have proceeded is :
let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$
i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$
But, $a^3+a+c... | You essentially split off the linear factor belnging to a real root and showed per determinant that the remaining quadratic has no real root.
This is fine but does not readily generalize to higher degrees.
Instead, it is probably easier to show that $f$ is injective: Assume $a,b$ are two real roots, i.e. $f(a)=f(b)=0$.... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Given $x,y,z >0$, $1/x+1/y+1/z = 4$, prove that $ 1/(2x+y+z)+1/(x+2y+z) +1/(x+y+2z) \le 1$ Given $x,y,z >0$, $1/x+1/y+1/z = 4$, prove that
$$ 1/(2x+y+z)+1/(x+2y+z) +1/(x+y+2z) \le 1 .$$
Any hints or direction will be appreciated.
| By C-S
$$\sum_{cyc}\frac{1}{2x+y+z}\leq\frac{1}{(2+1+1)^2}\sum_{cyc}\left(\frac{2^2}{2x}+\frac{1^2}{y}+\frac{1^2}{z}\right)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/517323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Writing u as a linear combination of the vectors in S. Write vector
u = $$\left[\begin{array}{ccc|c}2 \\10 \\1\end{array}\right]$$
as a linear combination of the vectors in S. Use elementary row operations on an augmented matrix to find the necessary coefficients.
S = {
$v1$$\left[\begin{matrix}1\\2\\2\end{matrix}\ri... | $a\left[\begin{matrix}1\\2\\2\end{matrix}\right] + b\left[\begin{matrix}4\\2\\1\end{matrix}\right]+
c\left[\begin{matrix}5\\4\\1\end{matrix}\right]=\left[\begin{matrix}2\\10\\1\end{matrix}\right]$.
Putting this into matrix form gives:
$\begin{bmatrix}1&4&5&2\\2&2&4&10\\2&1&1&1\end{bmatrix}\to\begin{bmatrix}1&4&5&2\\1... | {
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"timestamp": "2023-03-29T00:00:00",
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Proof of divergence of $1/2 + 1/3 + 1/5 + 1/7 + 1/11 +....$ What is an elementary proof to the fact that $\frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} + \dots$ diverges. ($p_i$ denotes the $i$th prime.)
| $$
\begin{align}
\log \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \log \left( \prod_p \frac{1}{1-p^{-1}}\right)
= \sum_p \log \left( \frac{1}{1-p^{-1}}\right) = \sum_p - \log(1-p^{-1}) \\
& {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\
& {} = \left( \sum_{p}\frac{1}{p}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How would I evaluate this limit? $\lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}}$ How would I evaluate the following limit by hand?
$\lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}}$
Thanks in advance.
| $$\begin{align} \lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}} & = \lim_{x \to 3+}\; \frac{\left(\sqrt{x - 3}\;\right)^2}{\sqrt{(x - 3)(x+3)}}\\ \\
& = \lim_{x \to 3+}\;\frac{\left(\sqrt{x-3}\;\right)^2}{(\sqrt{x-3}\;)(\sqrt{x+3}\;)}\\ \\
& = \lim_{x \to 3+}\;\frac{\sqrt{x - 3}}{\sqrt{x+3}}\\ \\
& = \frac 06\; = \;0\end{alig... | {
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"timestamp": "2023-03-29T00:00:00",
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} |
Evaluate the limit $\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$ I need to evaluate the following limit:
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
I have multiplied both sides by the conjugate $\sqrt{x^2+5}+3$ but am getting $x^2-4$ as the denominator. Is this the correct way to go about it?
| You are on the right track.
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}=\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3} \frac{\sqrt{x^2+5}+3}{\sqrt{x^2+5}+3}=\lim_{x\to 2} \frac{(x-2)(\sqrt{x^2+5}+3)}{(x-2)(x+2)}$ simplifying
$$\lim_{x\to 2} \frac{(\sqrt{x^2+5}+3)}{(x+2)}$$ and evaluating at $2$ we get
$$\lim_{x\to 2} \frac{(\s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Is it true that $\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2} \rfloor=\lfloor\sqrt{9n+7}\rfloor$? I know it's true that
$$\lfloor\sqrt{n}+\sqrt{n+1} \rfloor=\lfloor\sqrt{4n+1}\rfloor,\forall n\in \mathbb N. \tag 1$$
Is it true that
$$\left[\sqrt{n}+\sqrt{n+1}+\sqrt{n+2} \right ]=\left[\sqrt{9n+7}\right], \tag 2$$
$$\left[\s... | For $0<a<b$ we have $\sqrt b-\sqrt a=\frac{b-a}{\sqrt b+\sqrt a}$ and hence the estimate
$$\frac{b-a}{2\sqrt b}<\sqrt b-\sqrt a<\frac{b-a}{2\sqrt a}.$$
Therefore
$$x:=\sqrt {n}+\sqrt {n+1}+\sqrt{n+2} =3\sqrt{n+1}+(\sqrt{n+2}-\sqrt {n+1})-(\sqrt {n+1}-\sqrt n)$$
is estimated by
$$3\sqrt {n+1}+\frac1{2\sqrt{n+2}}-\frac... | {
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"source": "stackexchange",
"question_score": "15",
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Is my answer correct? (And what's the name of the rule?)$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!}$ Want to know if I solved this problem correctly:
$$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!} =\lim_{n \to \infty} \frac{1 \cdot 2 \ldots(n-1)n(n+1)(n+2)(n+3) - 1 \cdot ... | If $m=n$, and $a_n,b_m\neq 0$, then you can say the following:$$\lim_{x \to \infty}\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots+b_0}=\frac{a_n}{b_m}.$$
To prove this, you can factor out the $x$ terms from the numerator and denominator like so: $$\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots+b_0}=\frac{x^n}{x^m}\cdot\frac{a_n+\cdots... | {
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"source": "stackexchange",
"question_score": "3",
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Finding the limit of $\frac{\sqrt{x}}{\sqrt{x}+\sin\sqrt{x}}$ How would one find the limit of
$\displaystyle\lim_{x\to 0}\frac{\sqrt{x}}{\sqrt{x}+\sin\sqrt{x}}$
I know I have to use the L'Hospital rule.
$\displaystyle\lim_{x\to 0}\frac{\frac{1}{2}x^{-1/2}}{\frac{1}{2}\frac{1}{\sqrt{x}}+\frac{1}{2}\frac{1}{\sqrt{x}}\co... | Continuing from where you left off:
Simply cancel the common factor of $\frac {1}{2 \sqrt x}$ from numerator and denominator:
$$\frac{\frac{1}{2\sqrt x}}{\frac{1}{2}\frac{1}{\sqrt{x}}+\frac{1}{2}\frac{1}{\sqrt{x}}\cos\sqrt{x}} = \dfrac 1{1 + \cos \sqrt x}$$ Now evaluate the limit as $x \to 0$. You should arrive at a li... | {
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"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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How find this maximum $S_{\Delta ABC}$ in $\Delta ABC$,and $\angle ABC=60$,such that $PA=10,PB=6,PC=7$,
find the maximum $S_{\Delta ABC}$.
My try:let $AB=c,BC=a,AC=b$, then
$$b^2=a^2+c^2-2ac\cos{\angle ABC}=a^2+c^2-2ac$$
then
$$S_{ABC}=\dfrac{1}{2}ac\sin{60}=\dfrac{\sqrt{3}}{4}ac$$
Then I can't
| let $B$ is on the $O$ of polar system, $P(6,\alpha)$
so $C$ is on circle $p^2+6^2-2*6*cos(\beta- \alpha)=7^2$ and $\beta=0$
$A$ is on circle $p^2+6^2-2*6*cos(\beta- \alpha)=10^2$ and $\beta=\dfrac{\pi}{3}$
$a=BC=6cos\alpha+\sqrt{(6cos\alpha)^2+13}, c=BA=6cos(\dfrac{\pi}{3}-\alpha)+\sqrt{(6cos(\dfrac{\pi}{3}-\alpha))^2... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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How to evaluate the series: $ \frac {20} {(5-4)(5^2-4^2)} + \frac {20^2} {(5^2-4^2)(5^3-4^3)} + \frac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $ How to evaluate this infinte summation?
$$ \dfrac {20} {(5-4)(5^2-4^2)} + \dfrac {20^2} {(5^2-4^2)(5^3-4^3)} + \dfrac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $$
Telescopic cancellati... | Notice that:
$$\frac{20^k}{(5^k-4^k)(5^{k+1}-4^{k+1})}=\frac{4^k}{5^k-4^k}-\frac{4^{k+1}}{5^{k+1}-4^{k+1}},$$
in order to prove that:
$$\sum_{k=1}^{+\infty}\frac{20^k}{(5^k-4^k)(5^{k+1}-4^{k+1})} = 4.$$
| {
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"source": "stackexchange",
"question_score": "3",
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Find the orthogonal trajectories of the following family of curves: $x^3y - xy^3 = \alpha$ The answer is $-6x^2y^2 + x^4 + y^4 = \beta$, but I get $-3x^2y^2 + \frac{x^4}{4} + \frac{y^4}{4} = \beta$ and I can't seem to find my mistake. My work:
Differentiaing the given equation, we get
$$3x^2y + x^3\frac{dy}{dx} - y^3 ... | I t seems that we have $$y'=\frac{3xy^2-x^3}{y^3-3x^2y}$$ instead, which is an homogenous OE. We have $$\frac{du}{\frac{3u^2-1}{u^3-3u^2}-u}=\frac{dx}{x}~\equiv~\frac{(-u^3+3u^2)du}{u^4+3u^2-3u^2+1}=\frac{dx}{x},~~~~~~u=y/x,x\neq 0$$ in which the right ODE is a separable one.
| {
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"url": "https://math.stackexchange.com/questions/534331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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When does the locus $|z-z_1|^2 +|z-z_2|^2=k$ describe a circle?
Prove that the locus $|z-z_1|^2 +|z-z_2|^2=k$ of a moving point $z$ on the Argand plane is a circle when $|z_1-z_2|^2\leqslant 2k$.
What i have tried out-
$|z-1|^2+|z-2|^2=k$
let $z=x+iy$
$$\\ \implies2x^2-6x+5+2y^2-k=0$$
This is the equation of a circle... | Try with $z=x+iy, z_1=a+ib, z_2=c+id$
On rearrangement we have, $$x^2+y^2-x(a+c)-y(b+d)+\frac{a^2+b^2+c^2+d^2-k}2=0$$
$$\implies \left(x-\frac{a+c}2\right)^2+\left(y-\frac{b+d}2\right)^2=\frac{2k-2a^2-2b^2-2c^2-2d^2+(a+c)^2+(b+d)^2}4=\frac{2k-\{(a-c)^2+(b-d)^2\}}4$$
For real circle we need $2k-\{(a-c)^2+(b-d)^2\}\ge0$
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Is there any simple method to calculate $\sqrt x$ without using logarithm Suppose that we don't know logarithm, then how we would able to calculate $\sqrt x$, where $x$ is a real number? More generally, is there any algorithm to calculate $\sqrt [ n ]{ x } $ without using logarithm? More simple techniques would be nice... | Every positive number $x$ can be always written as the sum of two other numbers (or the difference).
Say that one of the two is a perfect square $n^2$, then we can always say that
$$x = n^2 + q$$
Where $q$ is the obvious remainder. Since we can also adopt the difference convention, it's better to write
$$x = n^2 \pm q... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 13,
"answer_id": 12
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Solving two simultaneous equations Suppose that $x$, $y$ and $z$ are three integers (positive,negative or zero) such that we get the following relationships simultaneously
*
*$x + y = 1 - z$ and
*$x^3 + y^3= 1 - z^2$
Find all such $x$, $y$ and $z$.
I have tried to determine the values of $x$ and $y$ by fixing... | If $z = 1$, then $y = -x$, and all triples $(x,-x,1)$ are solutions. So let's suppose $z \neq 1$. Then we can divide
$$x^2 - xy + y^2 = \frac{x^3+y^3}{x+y} = \frac{1-z^2}{1-z} = 1+z. \tag{3}$$
Adding $(3)$ to $(1)$ yields
$$x^2 - xy + y^2 + x + y = 2.$$
Multiplying by $2$ and adding $2$ to both sides yields
$$\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/538347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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system of equation (3 unknown, 3 equations) I want to solve a system of equations, but I seem to get it wrong.
Problem: see picture, and note that I should tell for which a I have
*
*no solution
*one solution
*$\infty$ solutions
Attempt:
$$\begin{matrix}i \\ ii \\ iii\end{matrix}\left\{\begin{matrix}x+y-az=3\\ a... | Using Gaussian Elimination (your way works too) and reducing the augmented system, we get:
$$\begin{bmatrix}1 & 0 & 0 & -\dfrac{a}{a-3}\\0 & 1 & 0 & \dfrac{-a^2+a-9}{a-3} \\ 0 & 0 & 1 & \dfrac{a+3}{-a + 3}\end{bmatrix}$$
We found that $a = 1$ and $a = 3$ appeared to be problematic.
If we dig deeper, we actually find:
... | {
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"url": "https://math.stackexchange.com/questions/539570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to prove $\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1$ Prove the following equation.
\begin{eqnarray}
\\\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1\\
\end{eqnarray}
I can't prove it by many methods I use.
Please give me some hints.
Thank you for your attention
| I'll start from your left hand side instead of R.H.S.
All terms contain integer powers of $\sin^2x$ and $\cos^2x$. Rewrite the l.h.s to
$$(\sin^2 x)^3+3(\sin^2x)(\cos^2 x)+(\cos^2x)^3$$
Cubic? It feels like $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ but notice the lack of $a^2b$ and $ab^2$. To make up for the missing powers, I'm te... | {
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"source": "stackexchange",
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Number Theory Contest Problem Given that $x, y$ are positive integers with $x(x + 1)\mid y(y + 1)$, but neither $x$ nor $x + 1$ divides either of $y$ or $y + 1$, and $x^2+ y^2$ as small as possible, find $x^2+ y^2$.
I have tried looking at the values, and it seems that neither $x$ or $x+1$ or the $y$'s are prime.
| Yes, we don't want $x$ or $x+1$ to be a prime power. The first candidate is $x=14$. Then $y=20$ works.
Added: Suppose we find consecutive composites integers $x$ and $x+1$ neither of which is a prime power. Let $x=ab$ and $y=cd$, where $a$ and $b$ are relaively prime, as are $c$ and $d$, and none is equal to $1$. Cons... | {
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"source": "stackexchange",
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Show that $(\sqrt{y^2-x}-x)(\sqrt{x^2+y}-y)=y \iff x+y=0$
Let $x,y$ be real numbers such that
$$\left(\sqrt{y^{2} - x\,\,}\, - x\right)\left(\sqrt{x^{2} + y\,\,}\, - y\right)=y$$
Show that $x+y=0$.
My try:
Let
$$\sqrt{y^2-x}-x=a,\sqrt{x^2+y}-y=b\Longrightarrow ab=y$$
and then
$$\begin{cases}
y^2=a^2+(2a+1)x+x^2\c... | We need to prove that $x=y$, where
$$\left(\sqrt{y^{2}+x}+x\right)\left(\sqrt{x^{2} + y}- y\right)=y$$ or
$$\left(\sqrt{y^{2}+x}+x\right)\left(\sqrt{x^{2} + y}- y\right)=\left(\sqrt{x^{2}+y}+x\right)\left(\sqrt{x^{2} + y}- x\right)$$ or
$$\left(\sqrt{y^{2}+x}+x\right)\left(\sqrt{x^{2} + y}- y\right)-\left(\sqrt{x^{2}+y... | {
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"url": "https://math.stackexchange.com/questions/543607",
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"source": "stackexchange",
"question_score": "25",
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$\int_0^{2\pi} \sqrt{1-\cos(x)}\,dx = 4\sqrt{2}$. Why? According to the textbook, and Wolfram Alpha the above is correct.
Here is the step by step procedure from Wolfram Alpha for evaluating the indefinite integral:
Take the integral: $$\int\sqrt{1-\cos(x)}\,dx$$ For the integrand $\sqrt{1-\cos(x)}$, substitute $u=1-\... | Your $u$-substitutions should be injective on their interval of evaluation. Otherwise, you risk running into exactly this sort of issue.
Note that $$\begin{align}|\sin x| &= \sqrt{\sin^2 x}\\ &= \sqrt{1-\cos^2 x}\\ &= \sqrt{1-\cos x}\sqrt{1+\cos x}\\ &=\sqrt{1-\cos x}\sqrt{2-(1-\cos x)},\end{align}$$ so if you want to ... | {
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"source": "stackexchange",
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What is the minimum value of $a+b+\frac{1}{ab}$ if $a^2 + b^2 = 1$? For the case when $a,b>0,$ I used AM-GM Inequality as follows that:
$\frac{(a+b+\frac{1}{ab})}{3} \geq (ab\frac{1}{ab})^\frac{1}{3}$
This implies that $(a+b+\frac{1}{ab})\geq 3$. Hence, the minimum value of $(a+b+\frac{1}{ab})$ is 3
But the answer is $... | If you want to use AM-GM for this, you need to ensure that the equality condition can be met along with the constraint, by "balancing coefficients". Illustrated below:
$$a+b+\frac1{ab} = a+b +\frac{1}{2\sqrt2 ab}+\left(1-\frac1{2\sqrt2}\right)\frac1{ab}$$
Now by AM-GM,
$$ a+b +\frac{1}{2\sqrt2 ab}\ge \frac{3}{\sqrt2... | {
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"source": "stackexchange",
"question_score": "5",
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How to prove this inequality?$\frac{1}{n+1} + \frac{1}{n+2} + \cdots+\frac{1}{n+n} + \frac{1}{4n} > \ln 2$ $$\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n} > \ln 2$$
$n$ is positive integer.
Thank you !
| Let
$$a_n=\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n}$$
Then
$$a_{n+1}-a_n=\frac{1}{2n+1}+\frac{1}{2n+2}+\frac{1}{4n+4}-\frac{1}{n+1}-\frac{1}{4n}\\
=\frac{1}{2n+1}+\frac{2}{4n+4}+\frac{1}{4n+4}-\frac{4}{4n+4}-\frac{1}{4n}\\
=\frac{2}{4n+2}-\frac{1}{4n+4}-\frac{1}{4n} <0 $$
Thus, $a_n$ is a... | {
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"url": "https://math.stackexchange.com/questions/549686",
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"source": "stackexchange",
"question_score": "4",
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Limit at Infinity $\lim\limits_{m\to\infty}\frac{\sum\limits_{k=1}^m\cot^{2n+1}\left(\frac{k\pi}{2m+1}\right)}{m^{2n+1}}$ How can I prove the following equality?
$$\lim_{m\to{\infty}}\frac{\displaystyle\sum_{k=1}^m\cot^{2n+1}\left(\frac{k\pi}{2m+1}\right)}{m^{2n+1}}=\frac{2^{2n+1}\zeta(2n+1)}{\pi^{2n+1}}$$
| We can use Paramanand Singh's idea of sandwiching $\cot x$ between $\frac1x$ and something close to that regardless of whether the exponents are odd or even.
The inequality we will use is
$$\frac1x - \frac{x}{2} < \cot x < \frac1x\tag{1}$$
for $0 < x \leqslant \frac{\pi}{2}$. To see $(1)$, note
$$\sin x - x\cos x = \in... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove $\frac{2}{3}<\frac{3x^6+15x^2+2}{2x^6+15x^4+3}\le\frac{3}{2}$
Let $x\in(0,1]$, show that
$$\dfrac{2}{3}<\dfrac{3x^6+15x^2+2}{2x^6+15x^4+3}\le\dfrac{3}{2}$$
My try: since
$$\begin{align}\dfrac{3x^6+15x^2+2}{2x^6+15x^4+3} &=\dfrac{\dfrac{3}{2}(2x^6+15x^4+3)+15x^2-\dfrac{45}{2}x^4-\dfrac{5}{2}}{2x^6+15x^4+3}... | It is enough to note that for all $x\in\mathbb{R}$
$$
\dfrac{3}{2}-\dfrac{3x^6+15x^2+2}{2x^6+15x^4+3}=\dfrac{5}{2}\cdot\dfrac{(3x^2-1)^2}{2x^6+15x^4+3}\geq 0
$$
and
$$
\dfrac{3x^6+15x^2+2}{2x^6+15x^4+3}-\dfrac{2}{3}=\dfrac{5x^2(x^2-3)^2}{3(2x^6+15x^4+3)}\geq 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/554230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper
Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$
and find the sum if it exists.
I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long wa... | You are on the right track. Fix $N \geq 1$. Then
$$\begin{align}
\sum_{n=1}^{N} \frac{1}{n \cdot (n+1) \cdot (n+2)}
&= \sum_{n=1}^N \left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4}\right)
= \frac{1}{2}\sum_{n=1}^N \frac{1}{n}-\sum_{n=1}^{N}\frac{1}{n+1}+\frac{1}{2}\sum_{n=1}^{N} \frac{1}{n+2}\\
&= \frac{1}{2}\sum_{n=1}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 10,
"answer_id": 4
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Does the integral $\int _0^\infty \frac 1{1+(x \cos x)^2}\,dx$ converge? Question
$$\int _0^\infty \frac 1{1+(x \cos x)^2}dx$$
Does this converge?
Thoughts
Tried bounding it with $\frac 1{\cos^2x}$ for $x>1$ but not much success.
| Note that
*
*$(x \cos x)^{2} \leq n^{2}\pi^{2} \cos^{2} x$ on each interval $[(n-1)\pi, n\pi]$, and
*$\sin^{2} x \leq x^{2} $ for any $x$.
So it follows that
\begin{align*}
\int_{0}^{\infty} \frac{dx}{1+(x \cos x)^{2}}
&\geq \sum_{n=1}^{\infty} \int_{0}^{\pi} \frac{dx}{1+n^{2}\pi^{2}\cos^{2} x}
= 2 \sum_{n=1}^{\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How find this $\sum_{i=1}^{5}\tan^4{\frac{i\pi}{11}}$ show that:$$\tan^4{\dfrac{\pi}{11}}+\tan^4{\dfrac{2\pi}{11}}+\tan^4{\dfrac{3\pi}{11}}+\tan^4{\dfrac{4\pi}{11}}+\tan^4{\dfrac{5\pi}{11}}=2365$$
my try:
I think first we can find this value：$$x=\tan{\dfrac{\pi}{11}}+\tan{\dfrac{2\pi}{11}}+\tan{\dfrac{3\pi}{11}}+\tan{... | Using this for odd $n=2m+1$ and setting $\tan(2m+1)x=0\implies x=\frac{r\pi}{2m+1}$ where $0\le r\le 2m$
So, the roots of $$\tan^{2m+1}x-\binom{2m+1}2\tan^{2m-1}x+\binom{2m+1}4\tan^{2m-3}x-\cdots=0$$ are $\tan\frac{r\pi}{2m+1}$ where $0\le r\le 2m$
Discarding $\tan 0=0,$ the roots of $$\tan^{2m}x-\binom{2m+1}2\tan^{2... | {
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"url": "https://math.stackexchange.com/questions/565061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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generating function and lucas numbers Defines the sequence $l$.
$l_n = l_{n−1} + l_{n−2}$, for all $n \ge 2$.
So the first few members of the sequence are: $2, 1, 3, 4, 7, 11, 18\;.$
Find the generating function for this sequence and deduce the following formula
$$l_n =\left(\frac{1 − \sqrt5}2\right)^n +\left(\frac{1 +... | Let $$g(x)=h_0+h_1x+h_2x^2+\cdots+h_nx^n+\cdots$$ $$-xg(x)=-h_0x-h_1x^2-h_2x^3-\cdots-h_nx^{n+1}-\cdots$$ $$-x^2g(x)=-h_0x^2-h_1x^3-h_2x^4-\cdots-h_nx^{n+2}-\cdots.$$ Adding we obtain $$(1-x-x^2)g(x)=h_0+(h_1-h_o)x+(h_2-h_1-h_0)x^2+\cdots.$$ Since $l_0=2$ and $l_1=1$ we now have $$g(x)={2\over 1-x-x^2}-{x\over 1-x-x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/565480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solving the improper integral $\int_0^\infty \frac{x^{1/3}}{1+x^2} \mathrm dx$ I'm trying to solve:
$\displaystyle \int \limits_0^\infty \dfrac{x^{1/3}}{1+x^2} \mathrm dx$
I have tried contour integration with $C_R^+$ and the real line like this:
$\displaystyle \int \limits_T \dfrac{z^{1/3}}{1+z^2} \mathrm dz = \int \l... | Hint: using the substitution $x=u^{\frac{3}{2}}$ your integral will be equal to $$\frac{3}{2}\int\frac{u}{u^3+1}du$$ which can be evaluated by decomposing partial fractions. Additional note: if you divide the integral as an integral from $0$ to $1$ and an integral from $1$ to $\infty$, it is clearly that the first one ... | {
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"source": "stackexchange",
"question_score": "2",
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Prove that $\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$ for $n\geq 2$ Prove, for all $n\geq 2$ that
$$\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$$
Let us prove the inductive base for $n = 2$
$$\rm{LHS} = 1^2 + 2^2= 1 + 4 = 5$$
$$\rm{RHS} = \binom{3}{2} + 2\binom{3}{3} = 3 + 2\cdot 1 = 5$$
$$\rm{LHS}... | There is a typo in one of your steps, it should be as follows (look at the red ${\color{red} + }$) :
$$\frac{3(k+2)! + 2k(k+2)! -3k(k{\color{red}+}1)! - 2k(k-1)(k+1)!}{6k!} \,\,(*)$$
And then, I do not see how you got from there to
$$\frac{(k+1)!(3k + 6 - 2k^2 - 2k) + (k-1)!(2k^3 + 2k^2 - 3k)}{6k!}$$
which in fact is ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$
Calculate all integer solutions $(x,y,z)\in\mathbb{Z}^3$ of the equation $x^2+y^2+z^2 = 2xyz$.
My Attempt:
We will calculate for $x,y,z>0$. Then, using the AM-GM Inequality, we have
$$
\begin{cases}
x^2+y^2\geq 2xy\\
y^2+z^2\geq 2yz\\
z^2+x^2\geq 2zx\\
\end{cases... | Suppose that $(x,y,z)$ is a solution. An even number of these must be odd. If two are odd, say $x$ and $y$, then $x^2+y^2$ has shape $4k+2$, and therefore so does $x^2+y^2+z^2$, since $z^2$ is divisible by $4$. But $2xyz$ has shape $4k$.
So $x,y,z$ are all even, say $2u,2v,2w$. Substituting we get $u^2+v^2+w^2=4uvw$.
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Proof: Divisible by 15 I have to proof that $16^m - 1$ is divisible by $15$. Is my following proof correct?
$$\begin{align}
16^m - 1=&\frac{16^{m+1}}{16}-1\\
=&\frac{16^{m+1}-16}{16} \\
=&(16^{m+1}-16)\cdot\frac{1}{16} \\
=&\underbrace{(16^{m+1}-16)}\cdot\frac{1}{15(1+1/15)} \\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,... | $$
16^{n} - 1 = 15\,{16^{n} - 1 \over 16 - 1} = 15\sum_{k = 0}^{n - 1}16^{k}
$$
$$
\color{#0000ff}{\large{16^{n} - 1 \over 15} = \sum_{k = 0}^{n - 1}16^{k}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/569563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Understanding the solution of a telescoping sum $\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$ I'm having trouble understanding infinite sequence and series as it relates to calculus, but I think I'm getting there.
For the below problem:
$$\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$$
The solution shows them breaking this up into a s... | $$\sum_{n=1}^{\infty}\frac{3}{n(n+3)}=\lim_{k\to\infty}\sum_{n=1}^{k}\frac{3}{n(n+3)}=$$
$$=\lim_{k\to\infty}\left(\frac{11}{6}-\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)\right)=\frac{11}{6}$$
because
$$\sum_{n=1}^{k}\frac{3}{n(n+3)}=\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+3}\right)=\sum_{n=1}^{k}\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/569663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve for: $2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4$ Solve for: $$2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4$$
My try:
$2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarro... | The domain of the first inequation is $x>3$ or $x\leq 3$, so the range of your $t$ is $t\geq 0$. Then you can use $t\geq 0$ as the domain to solve $t^2+3t−4\leq0$, which gets $0\leq t \leq 1$. At last, you just go back to solve $0 \leq t \leq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/572996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculation of $\int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$ Calculation of $\displaystyle \int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$
$\bf{My\; Try}::$ Using $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$
Let $\displaystyle I = \int_{0}^{\pi}\frac{1}{\left(5+\frac{4-4\tan^2 \frac{x}{2}}{1+\tan^2 \... | Using Weierstrass substitution, we get
$$\int_0^\pi \frac{dx}{(5+4\cos x)^2}=\int_0^\infty\frac{1}{(5+4\frac{1-t^2}{1+t^2})^2}\frac{2dt}{1+t^2}=2\int_0^\infty \frac{t^2+1}{(t^2+9)^2}dt=\int_{-\infty}^\infty \frac{t^2+1}{(t^2+9)^2}dt$$
Now let
$$f(z)=\frac{z^2+1}{(z^2+9)^2}$$
Then, $f$ has a poles of order 2 at $\pm 3i$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/574099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ Recently, I encountered a problem about infinite series.
So my question is how to know whether the infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ is convergent?
| $\sum_{n≥2} \frac{1}{n(\log n)}=\sum_{n≥2} \int_{1}^{\infty} \frac{1}{n^x} dx$
$=\int_{1}^{\infty} \zeta_2(x) dx$ where [$\zeta_2(x)=\zeta(x)-1$]
So, this is the area under the curve $\zeta(x), x≥1, y≥0$.
Now, for $x>1$,
$\zeta_2(x) > {\frac{1}{2^x} + 2\frac{1}{4^x}+ 4\frac{1}{8^x} +......}$
$=\frac{1}{2}(\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/574503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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"answer_id": 5
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How prove this $\tan{\frac{2\pi}{13}}+4\sin{\frac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$ Nice Question:
show that: The follow nice trigonometry
$$\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$$
This problem I have ugly solution, maybe someone have nice methods? Thank you
My ugly solution:
let $$A=... | The following argument is more or less a duplicate in this paper:
Let $x=e^{2\pi i/13}$. Then $$i\tan{2\pi/13}=\frac{x^2-1}{x^2+1}=\frac{x^2-x^{26}}{x^2+1}$$
(recall that $x^{13}=1$)
$$=x^2(1-x^2)(1+x^4+x^8+x^{12}+x^3+x^7)$$
$$=(x+x^2+x^5+x^6+x^9+x^{10}-x^3-x^4-x^7-x^8-x^{11}-x^{12})$$
$$4i\sin{6\pi/13}=2(x^3-x^{10})$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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All roots of the quartic equation $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real Problem
Prove that all roots of $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real. Here $a,b \in \mathbb R$, and $a \neq 0$.
Source
This is one of the previous year problem of Regional Math Olympiad (India). I had a hard time ... | Let $f(x) = x^4 + x^3 + x^2 + bx + a$. Then the quartic
$ax^4 + bx^3 + x^2 + x + 1$ is $x^4 f(1/x)$, and has four real roots
iff $f$ does. But $f''(x) = 12x^2 + 6x + 2 > 0$ for all $x$.
Therefore $f$ is convex upwards and hence can have at most two real roots, QED.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Solve equation $ 1+2^x=3^y$ Find integers $x$ and $y$ such that$$ 1+2^x=3^y.$$
It is obvious that $x = y = 1$ and $x = 3, y = 2$ are solutions.
I think others are not.
How to show that?
| Case 1: $y$ is odd. Then $3^y \equiv 3 \mod 8$. When is $2^x \equiv 2 \mod 8$?
Case 2: $y = 2 z$ is even. Then $2^x = 3^{2z} - 1 = (3^z - 1)(3^z + 1)$, so
$3^z - 1$ and $3^z + 1$ are powers of $2$. What powers of $2$ differ by $2$?
By the way, this proof of a special case of Catalan's conjecture dates back to Gers... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Evaluate $\int\frac{\cot{x}}{1+\sin{x}+\cos{x}} \mathrm dx$ Find this integral:
$$\int\dfrac{\cot{x}}{1+\sin{x}+\cos{x}}\mathrm dx$$
My try: since
$$1+\sin{x}+\cos{x}=2\cos^2{\dfrac{x}{2}}+2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}$$
$$\cot{x}=\dfrac{1-\tan^2{\dfrac{x}{2}}}{2\tan{\dfrac{x}{2}}}$$
so
$$\dfrac{\cot{x}}{1+\sin... | Now divide the numerator & the denominator by $\displaystyle \sec^2\frac x2=1+\tan^2\frac x2$
So, we have $I=\displaystyle\int\frac{1-\tan^2\frac x2}{2\tan\frac x2(2+2\tan\frac x2)}\sec^2\frac x2dx$
Putting $\displaystyle \tan\frac x2=u$
$$I=\int\frac{1-u^2}{2u(1+u)}du=\frac12\int\frac1u du-\frac12\int du$$ assuming $1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/580331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the $n$ times iterate of $f(x)=\frac{x}{\sqrt{1+x^2}}$? We were asked to determine the composition $f \circ f \circ f \circ \cdots \circ f $, $n$ times, where $f(x)=\dfrac{x}{\sqrt{1+x^2}}$.
Has anyone an idea?
| $$f(x)=\frac{x}{\sqrt{1+x^2}}$$
we see that
$$f^2(x)=f(f(x))=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+(\frac{x}{\sqrt{1+x^2}})^2}}=\frac{x}{\sqrt{1+2x^2}}$$
...... continuing this way
$$f^n(x)=\frac{x}{\sqrt{1+nx^2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$ Question is to Evaluate :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$$
what all i could do is :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\in... | Hint : $\displaystyle f(x)=\sum\frac{x^n}{(n+1)!}\iff f'(x)=\sum\frac{n\cdot x^{n-1}}{(n+1)!}\quad,\quad S=f'(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/581603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Please help me with this inequality $a,b,c > 0$ (no other conditions)
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\sqrt{3\left(a^2+b^2+c^2\right)}$
I tried this:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{\left (a+b+c\right)^2}{a+b+c}=a+b+c$
then
$a+b+c\geq\sqrt{3\left(a^2+b^2+c^2\right)}$ which is not cor... | As the inequality is homogeneous, WLOG we can set $a^2+b^2+c^2 = 3$ to get the equivalent inequality:
$$\frac{a^2}b + \frac{b^2}c + \frac{c^2}a \ge 3$$
$$\iff \left(\frac{a^2}b + \frac{b^2}c + \frac{c^2}a \right)^2 \ge 9 \iff \sum_{cyc} \frac{a^4}{b^2} \ge 3 \iff \sum_{cyc} a^2 \left(\frac{a^2}{b^2}+\frac{b^2}{a^2}-2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find an integral solution for a polynomial Given odd prime p that $\binom{-2}{p}=1$, show that there are integers x,y such that $x^2+2y^2=p$.
What i did: I move $2y^2$ to the right, then $x^2=p-2y^2$
If $a^2=-2$ (mod p) then this equation turn into $x=ay$ (mod p)
I don't know how to continue from here, should i just ... | There are two distinct ways I know of to prove such statements.
Proof 1 (elementary, using Pigeonhole Principle):
Since $(\frac{-2}{p})=1$, there exists $t$ s.t. $t^2 \equiv -2 \pmod{p}$.
Consider the integers $a+tb$, for $0 \leq a, b \leq \lfloor \sqrt{p} \rfloor$. There are $(1+\lfloor \sqrt{p} \rfloor)^2>p$ such i... | {
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Integration: Method of partial fractions - Any standard method of finding constants in hard to solve expressions? I've been computing many indefinite integrals using the method of partial decomposition.
The integrals are usual on the form $$\int \frac {x^2-29x+5} {(x-4)^2(x^2+3)} dx$$
which is equal to
$$\int \frac {A}... | One can find partial fractions this way:
\begin{equation*}
f(x)=\frac{x^{2}-29x+5}{(x-4)^{2}(x^{2}+3)}=\frac{A}{(x-4)}+\frac{B}{%
(x-4)^{2}}+\frac{Cx+D}{x^{2}+3}.
\end{equation*}
\begin{equation*}
B=\left. (x-4)^{2}f(x)\right] _{x=4}=\left. \frac{x^{2}-29x+5}{(x^{2}+3)}%
\right] _{x=4}=-5.
\end{equation*}
\begin{eqnarr... | {
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"url": "https://math.stackexchange.com/questions/587129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\left(\frac1\alpha-\frac1\beta\right)^2$ for $p(x)=x^2+x-2$ If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x)=x^2+x-2$,
then $\left(\frac1\alpha-\frac1\beta\right)^2 is:$
A) $\frac94$
B) $\frac{-9}4$
C) $\frac25$
D) $\frac{-2}5$
This is a homework question from our school's home assignment for c... | $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$
| {
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Can $x^3+3x^2+1=0$ be solved using high school methods? I encountered the following problem in a high-school math text, which I wasn't able to solve using factorization/factor theorem:
Solve $x^3+3x^2+1=0$
Am I missing something here, or is indeed a more advanced method necessary to solve this particular cubic? The ans... | You might be able to get away with using some "miracle" substitutions that solve the cubic (which can be generalised to solve a general cubic).
Begin with
$$x^3+3x^2+1=0 \tag{$\dagger$}$$
*
*Substitute $x = y - 1$ into $(\dagger)$ to obtain
$$y^3 - 3y + 3 = 0. \tag{$\star$}$$
*Let $y = z + \frac{1}{z}$ so that $(\s... | {
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How to integrate $\int\frac{\sqrt{1+x^2}}{x}\,\mathrm dx$ I want to know how to integrate
$$\int\frac{\sqrt{1+x^2}}{x}\,\mathrm dx$$
Could anyone solve it?
Thanks
| $$\int\frac{\sqrt{1+x^2}}{x}\,dx=\int\frac{x\sqrt{1+x^2}}{x^2}\,dx$$
Let $u=x^2+1, du =2x dx$. Then
$$\int\frac{\sqrt{1+x^2}}{x}\,dx= \frac{1}{2}\int\frac{\sqrt{u}}{u^2-1}\,du$$
If $v=\sqrt{u}$ the n $u=v^2, du=2vdv$. Thus
$$\int\frac{\sqrt{1+x^2}}{x}\,dx=\int\frac{v^2}{v^4-1}\,dv$$
By Partial Fraction Decomposition
$... | {
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Evaluating the integral $\int\frac{2x^3+x^2+1}{x^2+x-2}\space dx$ As suggested in the title, I am given the following integral:
$$\int\frac{2x^3+x^2+1}{x^2+x-2}\space dx$$
I have tried to solve it several times, but I wind up with the wrong answer, although its close. I assume there is a blunder somewhere.
Applying lon... | The remainder after long division is $5x - 1$; when you do long division, make sure you account for the 'missing' $x$ in the numerator: $2x^3 + x^2 + 0x + 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to find the minimum of $a+b+\sqrt{a^2+b^2}$ let $a,b>0$, and such
$$\dfrac{2}{a}+\dfrac{1}{b}=1$$
Find this minimum
$$a+b+\sqrt{a^2+b^2}$$
My try: since
$$2b+a=ab$$
so
$$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$
then I can't
maybe this problem can use AM-GM or Cauchy... | Now that someone has worked out the place where the minimum is achieved ($a:b = 4:3$), one can purposely use Cauchy-Schwarz/AM-GM.
$$(a^2+b^2)(16+9) \ge (4a+3b)^2 \Rightarrow \sqrt{a^2+b^2} \ge \frac{4a+3b}{5}$$
Then
$$a+b+\sqrt{a^2+b^2} \ge \frac{9a+8b}{5}$$
Again Cauchy-Schwarz,
$$(9a+8b)(\frac{4}{a} + \frac{2}{b}) \... | {
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A high school contest question Compute the largest root of $$x^4 −x^3 −5x^2 +2x+6$$
I cannot solved it by factor,what else I may do.
Such a question, how to think.
| Note that $x^4-x^3-5x^2+2x+6=(x^4-5x^2+6)-x(x^2-2)=(x^2-2)(x^2-3)-x(x^2-2)=(x^2-2)(x^2-x-3)$.
So we have two possible positive roots, $\sqrt{2}$ and $\frac{1+\sqrt{13}}{2}$. Notice that $\sqrt{13}>3$, so $\frac{1+\sqrt{13}}{2}>2>\sqrt{2}$. Thus, the largest root is $\frac{1+\sqrt{13}}{2}$.
| {
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"url": "https://math.stackexchange.com/questions/595057",
"timestamp": "2023-03-29T00:00:00",
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How find this integral $\int\frac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$ Question:
Find the integral
$$I=\int\dfrac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$$
my solution:
let $\sqrt{x}+\sqrt{x+1}=t\tag{1}$
then
$$t(\sqrt{x+1}-\sqrt{x})=1$$
$$\Longrightarrow \sqrt{x+1}-\sqrt{x}=\dfrac{1}{t}\tag{2}$$
$(1)-(2)$ we have
$$2\sqrt{x}=t-\dfrac{1}{... | since
\begin{align*}I&=\int\dfrac{\sqrt{x}+1-\sqrt{x+1}}{(\sqrt{x}+1)^2-(\sqrt{x+1})^2}dx=\int\dfrac{\sqrt{x}+1-\sqrt{x+1}}{2\sqrt{x}}dx=\dfrac{1}{2}\int\left(1+\dfrac{1}{\sqrt{x}}-\sqrt{1+\dfrac{1}{x}}\right)dx\\
&=\dfrac{1}{2}\left(x+2\sqrt{x}-\int\sqrt{1+\dfrac{1}{x}}dx\right)
\end{align*}
since
$$\int\sqrt{1+\dfra... | {
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"timestamp": "2023-03-29T00:00:00",
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Identity for the sum of three squares equal to sum of three squares I have been searching on Internet about a general solution to the Diophantine equation $A^2+B^2+C^2=D^2+E^2+F^2$. However, I haven't found anything. Can anyone help me? I'm looking for a identity or something like that. Very important is that must be a... | Diophantine equation:
$A^2+B^2+C^2=D^2+E^2+F^2$
Has a solution:
$A=-p^2+2(t+k-q-2a)ps+2(4t^2+4k^2-2a^2-5q^2+q(k+t)+7a(q-t-k)-tk)s^2$
$B=-p^2+2(t+a-q-2k)ps+2(4t^2+4a^2-2k^2-5q^2+q(t+a)+7k(q-a-t)-ta)s^2$
$C=-p^2+2(k+a-q-2t)ps+2(4k^2+4a^2-2t^2-5q^2+q(k+a)+$$7t(q-a-k)-ka)s^2$
$D=-p^2+2(t+k+a-q)ps+2(4t^2+4k^2+4a^2-5q^2+q(t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/600167",
"timestamp": "2023-03-29T00:00:00",
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Convert 2 + 5 +8+ . . . + (6n-1) = n(6n+1) to Sigma Notation 2 + 5 + 8 + . . . + (6n-1) = n(6n+1)
This is what I have so far.
The sum of (3j-1) from j=1 to something I`m not sure of.
| $$\sum\limits_{i=1}^{2n} (3i-1) = n(6n+1)$$
This will give you as terms of the series: \begin{align*}3(1)-1&=2\\3(2)-1&=5\\\ldots\\3(2n)-1&=6n-1\end{align*}
EDIT: the problem statement purported that $\sum_{i=1}^{2n} (3i-1) = n(6n-1)$. I will just prove below that this is wrong, and $\sum_{i=1}^{2n} (3i-1) = n(6n+1)$.
... | {
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"timestamp": "2023-03-29T00:00:00",
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showed that $x^2 + y^2$ = $1$ if ........... Please help me .how can I solve this problem:
If $x \sin^3 \alpha$ + $y \cos^3 \alpha$ = $\sin\alpha\cos\alpha$ and $x \sin\alpha-y\cos\alpha$ = $0$ then showed that $x^2 + y^2$ = $1$
| from the second equation, you get $y = x\tan \alpha$
Substitute this in first equation and you will get $x = \cos \alpha$
which means $y= \sin \alpha$
$$x^2+y^2 = \sin^2 \alpha + \cos^2 \alpha = 1$$
proved
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{x_1}{1+x_2+x_3+\ldots+x_n}+\frac{x_2}{1+x_1+x_3+\ldots+x_n}+\ldots+\frac{x_n}{1+x_1+x_2+\ldots+x_{n-1}}\ge\frac{n}{2n-1}$. If $x_1,x_2,\ldots,x_n>0$ and $x_1+x_2+\ldots+x_n=1$, prove that $$\frac{x_1}{1+x_2+x_3+\ldots+x_n} + \frac{x_2}{1+x_1+x_3+\ldots+x_n} +\ldots + \frac{x_n}{1+x_1+x_2+\ldots +x_{n-... | Hint: Use Jensen's inequality for f(x) = x/(2 -x)
| {
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"timestamp": "2023-03-29T00:00:00",
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integer $a$ for which $x^3-3x+a = 0$ has $3$ distinct real roots in $x\in (0,1)$ If the equation $x^3-3x+a=0$ has $3$ distinct roots between $0$ and $1$ then the integer value of $a$ is
Can we solve it without Calculus.
$\bf{My\; Try}::$ Let $x=2\cos \phi$, where $-1\leq \cos \phi\leq 1\Rightarrow 0\leq \phi \leq \pi$(... | If we factor the polynomial as $(x - b)(x - c)(x - d)$, we have
$$a = -bcd$$
But since $0 < b, c, d < 1$, we have that $|a| < 1$; the only integer satisfying this is zero, which certainly doesn't work.
| {
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"timestamp": "2023-03-29T00:00:00",
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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.