Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find all values of $z$ such that $z^3=-8i$ If I am asked to find all values of $z$ such that $z^3=-8i$, what is the best method to go about that?
I have the following formula:
$$z^{\frac{1}{n}}=r^\frac{1}{n}\left[\cos\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)+i\sin\left(\frac{\theta}{n}+\frac{2\pi k}{n}\righ... | Short way write it as $$z^{ 3 }=-8i\\ z^{ 3 }=8{ e }^{ i\frac { 3\pi }{ 2 } }\\ z=\sqrt [ 3 ]{ 8 } { e }^{ i\left( \frac { \frac { 3\pi }{ 2 } +2\pi k }{ 3 } \right) }=2{ e }^{ i\left( \frac { 3\pi +4\pi k }{ 6 } \right) },k=0,1,2\quad $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Functions satsify $f'=f^{-1}$ with $f^{-1}$ is compostional inverse of$ f$ let $f$ be a function such that :$f:\mathbb{C}\to \mathbb{C}$ and $f^{-1}$ is the compositional inverse of $f$, I seek for the analyticity of $f$ at $0$, then my question here is :
Question:
Are there functions satisfy:$f'=f^{-1}$ with $f... | If
$f'=f^{-1}$
and $f(x) = ax^b$,
then
$f'(x) = ab x^{b-1}$
and
$f^{(-1)}(x)
=(x/a)^{1/b}
$
so
$(x/a)^{1/b}
=ab x^{b-1}
$.
If $b \ne 0$,
then,
raising to the $b$ power,
$x/a
=(ab)^b x^{b(b-1)}
$
or
$a^{-b-1}b^{-b}
= x^{b(b-1)-1}
$.
For the right side
to be constant,
we must have
$0
=b(b-1)-1
=b^2-b-1
$
so
$b
=\dfrac{1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
prove that the following function is convex at least I want a hint Let we have the following function
$$ψ:R^{*+} \to R$$
$$x\to ψ(x)=x^3$$
How can I prove that $ψ$ is a convex function by using the definition ?
I meant that I have to prove that
$$ψ(λx+(1-λ)y) \le λψ(x)+ (1-λ)ψ(y)$$ where $λ \in [0,1] $ $x$ ... | Take $a,b$ positive reals. Then for $a \ge b$, we have $a^2+ab \ge 2b^2$, that is, $a^2 + ab+b^2 \ge 3b^2$. Multiplying by $(a-b)$, we get $a^3 - b^3 \ge 3b^2(a-b)$.
For $a\le b$, we have $a^2+ab \le 2b^2$, that is, $a^2+ab+b^2 \le 3b^2$. Here $(a-b) \le 0$, so multiplying by $(a-b)$, we have $a^3-b^3 \ge 3b^2(a-b)$.
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{n\to \infty} \sum_{k=1}^n \frac{k}{n^2+k^2}$
Find $\lim_{n\to \infty} \sum_{k=1}^n \frac{k}{n^2+k^2}$
Since $\frac{k}{n^2+k^2}\leq \frac{k}{k^2+k^2}=\frac{1}{2k}$, then $\sum_{k=1}^n \frac{k}{n^2+k^2}\leq \sum_{k=1}^n \frac{1}{2k}=\frac{1}{2} \sum_{k=1}^n\frac{1}{k}$.
Now we send $n$ to infinity, then sin... | Just for the fun.
Consider $$S_n= \sum_{k=1}^n \frac{k}{n^2+k^2}=\sum_{k=1}^n \frac{k}{(n^2+ik)(n-ik)}=\frac i2 \sum_{k=1}^n\left(\frac{1}{n+i k}-\frac{1}{n-i k} \right)$$ Using generalized harmonic numbers $$S_n=\frac{1}{2} \left(-H_{-i n}-H_{i n}+H_{(1-i) n}+H_{(1+i) n}\right)$$ Using asymptotics $$H_p=\gamma +\log ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help with a Binomial Coefficient Identity The following (conjectured) identity has come up in a research problem from representation theory that I am working on:
$$\binom{i}{n-1}
=\frac{(n+k+i)!}{(n-1)!(k+i)!} \sum_{j=0}^{n-1} \frac{(-1)^j }{(n+k+i-j)}\binom{n-1}{j} \binom{2n+k-j-2}{n-1}$$
where $k, i, n$ are positive ... | Let us re-write this as follows:
$${q\choose n} =
(n+1) {n+1+k+q\choose n+1}
\sum_{j=0}^n \frac{(-1)^j}{n+1+k+q-j}
{n\choose j} {2n+k-j\choose n}
\\ = (n+1+k+q) {n+k+q\choose n}
\sum_{j=0}^n \frac{(-1)^j}{n+1+k+q-j}
{n\choose j} {2n+k-j\choose n}.$$
where clearly $q\ge n.$ We have
$${n+k+q\choose n} {n\choose j}
= \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2207043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve $2\cos(x)^2 - 5\sin(x) - 4 = 0$ on the interval $0 \leq x \leq 2\pi$ This is homework.
I got it down to either $\sin(x) = -2$ or $\sin(x) = -1/2$ but I have absolutely no idea where to go from here.
| $2\cos(x)^2 - 5\sin(x) - 4 = 0$ on the interval $0 \leq x \leq 2\pi$
$2(1-\sin^2x)-5\sin x-4=0$
$2\sin^2x +5\sin x+2=0$
$(2\sin x+1)(\sin x+2)=0$
Then either $\sin x = -2$ which has no solutions since $|\sin x |\leq1$
or $\sin x = -\frac{1}{2}$
Now, let's consider the graph of $\sin x$:
So we have $ x = -\frac{\pi}{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
My approach is: Let the roots of $ax^2+bx+c=0$ are $\alpha$ and $\beta$. So, $$\alpha+\beta=-\frac{b}{a}\\ \alpha\... | You can check the other answers for (possible faster) alternatives, but you were doing fine.
So, my final equations are: $\dfrac{b}{a}=\dfrac{34}{9}$ and $\dfrac{4a-2b+c}{a}=\dfrac{7}{9}$. My mind says I'm pretty close to the solution but I can't find it out.
Notice that the system
$$\left\{ \begin{array}{l}\dfrac{b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Right Triangle Problem with Angle Bisector Theorem Right triangle $ABC$ has $AC = 8$ and $CB = 6$. $M$ is the midpoint of $AB$. Pick point $N$ on line $CM$ with $M$ between $C$ and $N$ such that $∠CAB = ∠BAN$. Compute $MN$. Express your answer as a common fraction.
I figured out that $CM$, $BM$, and $AM$ were $5$, but... | I think it is better to leave the trigonometry out of the solution.
Since $\angle BCA = 90^{\circ}$ one can conclude that $AM = BM = CM$. By Pythagoras one gets $AB = 10$ so $AM = BM = CM = 5$. Triangle $ACM$ is isosceles and $$\angle \, ACN = \angle \, ACM = \angle\, CAM = \angle \, CAB = \angle \, BAN = \angle \, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Coordinate system $(a,\tau)$ for butterfly circles? Consider this picture
which shows a coordinate system $(a,\tau)$ on the Cartesian coordinate frame $(x,y)$, which is very similar to the bipolar coordinate system, with isosurfaces (i.e. circles) $\tau$ and their respective foci $(-a,0)$ and $(a,0)$.
Can anyone help ... | We have one family of curves $(x-a)^2+y^2=a^2$. Calculating the partial derivatives
\begin{eqnarray*}
\left( \frac{\partial a}{\partial x},\frac{\partial a}{\partial y} \right) = \left( \frac{x^2-y^2 }{2 x^2},\frac{y}{x} \right)
\end{eqnarray*}
We need a vector that is orthogonal to this
\begin{eqnarray*}
\left( \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that if $a\neq b$, then we have for the $n\times n$-matrix $\textrm{det}=\frac{a^{n+1}-b^{n+1}}{a-b}$. The Problem
Show that if $a\neq b$, then we have for the $n\times n$-matrix
$$\textrm{det}\begin{pmatrix} a+b & ab & 0 & \ldots & 0 & 0 \\
1 & a+b & ab & \ldots & 0 & 0 \\
0 & 1 & a+b & \ldots & 0 & 0 \\
\ldots &... | Let's call $A_n$ the matrix for which you want to compute the determinant and let's see how it can be constructed iteratively:
$$ A_{n+1} = \begin{pmatrix} a+b & ab & 0 & \ldots & 0 & 0 \\
1 & a+b & ab & \ldots & 0 & 0 \\
0 & 1 & a+b & \ldots & 0 & 0 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
0 & 0 & 0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Equation $2n^2=m^2+1$ Given the equation
$$2n^2=m^2+1$$
Is there a general solution?
since $m$ is odd for example $$2\cdot5^2=7^2+1$$ $$2\cdot29^2=41^2+1$$
Any hint would be appreciation.
| If
$2n^2=m^2+1$
then
$2(3n+2m)^2 = 18n^2 + 14nm +8m^2$
$=9(m^2 + 1) + 14nm + 8m^2$
$=9(m^2+1) + 14nm + 8(2n^2-1)$
$=16n^2 + 14nm + 9m^2 + 1$
$=(4n+3m)^2 + 1$
so we have (1,1) -> (5,7) -> (29,41) etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2217729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sign of determinant when using $det A^\top A$ We have been given matrix:
$$A =
\begin{pmatrix}
a& b& c &d \\
b &−a& d& −c\\
c& −d &−a& b \\
d &c& −b& −a\\
\end{pmatrix}
$$
...and have been asked to calculate $\det(A)$ using $AA^T$.
We see that:
$$AA^T=\begin{pmatrix}... | This might be an overkill but it is a useful observation nonetheless. You have a matrix $A = A(a,b,c,d)$ which depends on four real parameters and you computed $\det(A)^2$. From your computation, it is clear that $\det(A) = 0$ iff $a = b = c = d = 0$. Set $U = \mathbb{R}^4 \setminus \{ (0,0,0,0) \}$. If $(a,b,c,d) \in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2217821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
if $\lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x} = 1$ then $6(\alpha+\beta)=?$ Let $\alpha, \beta \in \mathbb{R}$ be such that $$\lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x} = 1$$ then what is the value of $6(\alpha+\beta)$
| $$
\begin{aligned}
\lim _{x\to 0}\left(\frac{x^2\sin \left(\beta \:x\right)}{\alpha \:x-\sin \:x}\right)
& = \lim _{x\to 0}\left(\frac{x^2\left(\beta x+o\left(x\right)\right)}{\alpha \:x-\left(x-\frac{x^3}{3!}+o(x^3)\right)}\right)
\\& = \beta \:\cdot \lim _{x\to 0}\left(\frac{6x^2}{6α+x^2-6}\right) (\text{so } \color{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2218158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
$c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$ so prove $e - c_n \le \frac{1}{n! * n}$ $c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$, so $e - c_n \le \frac{1}{n! * n}$
I absolutely have no idea how to solve it, could anyone tell me the approach?
| We know
$$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \cdots$$
Fixing a positive integer $n$, let
\begin{align*}
x &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!}\\[4pt]
r &= \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \frac{1}{(n+3)!} + \cdots\\[4pt]
\end{align*}
Thus, we have
$$e = x + r$$
and w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integral $\int {t+ 1\over t^2 + t - 1}dt$
Find : $$\int {t+ 1\over t^2 + t - 1}dt$$
Let $-w, -w_2$ be the roots of $t^2 + t - 1$.
$${A \over t + w} + {B \over t+ w_2} = {t+ 1\over t^2 + t - 1}$$
I got $$A = {w - 1\over w - w_2} \qquad B = {1- w_2\over w - w_2}$$
$$\int {t+ 1\over t^2 + t - 1}dt = A\int {1\over t+ w} ... | Another approach would be to rewrite the integrand as
$$\int\frac{2 t+1}{2 \left(t^2+t-1\right)}\,dt+\int\frac{1}{2 \left(t^2+t-1\right)}\,dt$$
Then let $u=t^2+t-1$ such that $du=2t+1\, dt$ so the integral becomes
$$\frac 12 \int \frac 1u\,du+\frac 12\int\frac{1}{ \left(t^2+t-1\right)}\,dt \;=\; \frac {\ln(|t^2+t-1|)}2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Show the value of determinant is 0 I dont have any idea how to show the value of the following determinant is 0 without expanding the determinants
$$
\begin{vmatrix}
1 & a & a^2-bc \\
1 & b & b^2-ca \\
1 & c & c^2-ab \\
\end{vmatrix}
$$
Any ideas?
| Subtract the first row from each of the other rows:
$$
\begin{vmatrix}
1 & a & a^2-bc \\
1 & b & b^2-ca \\
1 & c & c^2-ab \\
\end{vmatrix}
=\begin{vmatrix}
1 & a & a^2-bc \\
0 & b-a & (b-a)(b+a+c) \\
0 & c-a & (c-a)(c+a+b) \\
\end{vmatrix}\\
$$
Note that the second and third rows are both multiples of $\begin{bmatrix}0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Knowing that $\cos(\frac{3\pi}{2}+\alpha)=-\frac{1}{3}$ and $\tan \alpha > 0$, find the exact value of $\tan \alpha$ First I tried to solve for $\alpha$:
$$\cos(\frac{3\pi}{2}+\alpha)=-\frac{1}{3} \Leftrightarrow \alpha = \arccos(-\frac{1}{3})+2k\pi-\frac{3\pi}{2} \lor \alpha = -\arccos(-\frac{1}{3})+2k\pi-\frac{3\pi}{... | $\cos\left(\frac{3\pi}{2}+a\right)=\cos\left(\frac{3\pi}{2}\right)\cos(a)-\sin\left(\frac{3\pi}{2}\right)\sin(a)=-\sin(a)=-\frac{1}{3}=>\sin(a)=\frac{1}{3}$
$\cos(a)=\sqrt{1-\sin^2{a}}=\sqrt{1-\frac19}=\sqrt{\frac89}$
$\tan(a)=\frac{\frac{1}{3}}{\sqrt{\frac89}}=\frac{1}{\sqrt{8}}=\boxed{\frac{\sqrt{2}}{4}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing that $f(x)=\frac x{x+1}$ is the unique function satisfying $f(x)+f\left(\frac1x\right)=1$ and $f(2x)=2f\big(f(x)\big)$ We are given a function $ f : \mathbb Q ^ + \to \mathbb Q ^ + $ such that $$ f ( x ) + f \left( \frac 1 x \right) = 1 $$
and
$$ f ( 2 x ) = 2 f \big( f ( x ) \big) \text . $$
Find, with proof, ... | For each coprime $n, m \in \mathbb N_+$, $f\left( \dfrac nm\right)=\dfrac{n}{n+m}$ is a solution.
To prove this, we induct on $n+m$. The base case is clearly true. Suppose it's true for $n+m\leq k$ for some natural number $k$ and consider $n+m=k+1$.
We consider two cases:
Case 1, $n>m$:
Using the first condition and Ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
"answer_id": 2
} |
Area enclosed by the curve $\lfloor x + y\rfloor + \lfloor x - y\rfloor = 5$
What is the area enclosed by the curve $$\lfloor x + y\rfloor + \lfloor x - y\rfloor = 5$$ $$x\ge y, \forall x, y \ge 0$$
$\lfloor x\rfloor$ stands for the Greatest Integer Function.
I think that the curve will start from $\left(\cfrac{5}{2}... | Blue tiles:
$$
\lfloor{x + y}\rfloor + \lfloor{x - y}\rfloor = 5
$$
Gray triangle
$$
y > 0, \quad x \le y
$$
Find the area of the blue tiles within the gray triangle.
The black dots are the vertices of the first square:
$$
\left( \begin{array}{c}
3 \\ 0
\end{array} \right), \quad
%
\left( \begin{array}{c}
3.5 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Showing that roots of a quadratic polynomial are of opposite signs.
Prove that, for all values of $k$, the roots of the quadratic polynomial $x^2 - (2 + k) x - 3$ are real. Show further that the roots are of opposite signs.
For the first part I was able to demonstrate such by using the discriminant of the quadratic, ... | Well, the solutions to the equation are
$$x=\frac{2+k}{2}\pm\frac12\sqrt{(-2-k)^2-4\cdot1\cdot-3}$$
or, simplified,
$$x=\frac{2+k}2\pm\frac12\sqrt{(2+k)^2+12}$$
and this is equal to
$$x=\frac{2+k}2\pm\sqrt{\left(\frac{2+k}2\right)^2+3}$$
The important step here is
$$\sqrt{\left(\frac{2+k}2\right)^2+3}>\sqrt{\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2228975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Problem on number of ways of writing $n$ as a sum of distinct integers vs a sum of odd integers I found this problem while doing a few problems on partitions of an integer:
Prove that the number of ways of writing $n$ as a sum of distinct positive integers is equal to the number of ways of writing $n$ as a sum of odd p... | I think this was proved by Euler.
The number of partitions of $n$ as a sum of odd positive integers is the coefficient of the $x^n$ in the expansion of :
$$\frac{1}{(1-x)(1-x^3)(1-x^5)(1-x^7)\dots} $$
and the number of ways of writing $n$ as a sum of distinct positive integers is the coefficient of $x^n$ in
$$(1+x)(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Algebraic simplification problem
I must simplify this expression:
$$\frac{a^2+ac}{a^2c-c^3}-\frac{a^2-c^2}{a^2c+2ac^2+c}+\frac{2c}{c^2-a^2}-\frac{3}{a+c}$$
I managed to simplify it to:
$$\frac{( a+1)^2-(a+c)^2}{c(a+c)(a+1)^2}$$
However, I am stuck now. Therefore, I would like some help to simplify it further.
| Hint
$$\frac{a(a+c)}{c(a-c)(a+c)}-\frac{(a-c)(a+c)}{ac(a+c)+c(a+c)}+\frac{2c}{(c-a)(a+c)}-\frac{3}{a+c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Help to use change of variables to solve the double integral I'm trying to evaluate:
$$\iint_D \left(\sqrt{a^2-x^2-y^2}-\sqrt{x^2+y^2}~\right)dxdy$$
where $D_{xy}$ is the disk $x^2+y^2\le a^2$.
The exercise is to use change of variables to solve this integral.
My solution
I chose $\varphi (r,\theta)=(ra\cos\theta,r... | Just for fun, I figured I'd show that the answer is indeed zero by a direct substitution without any evaluation of an integral.
Let
$$
(x,y)=\frac{\sqrt{a^2-\hat{x}^2-\hat{y}^2}}{\sqrt{\hat{x}^2+\hat{y}^2}} (\hat x,\hat y)
$$
Now, a quick calculation will confirm that the determinant of the Jacobian is $-1$ (and so the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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mathematical induction with exponent $$\frac 13 + \frac 1{3^2} + \frac 1{3^3} + \dots + \frac 1{3^n} + = \frac 12 \times \left( 1 - \frac{1}{3^n} \right)$$
Step 1 - $n=1$
$$\begin{align}
\frac 1 {3^1} & = \frac 1 2 \times \left( 1 - \frac 1 {3^1} \right) \\
\frac 1 3 & = \frac 1 2 \times \left( 1 - \frac 1 3 ... | You must prove that $$\sum_1^k\frac{1}{3^i} + \frac{1}{3^{k+1}} = \frac{1}{2}\left(1-\frac{1}{3^{k+1}}\right).$$
Applying the assumption that $$\sum_1^k\frac{1}{3^i} = \frac{1}{2}\left(1-\frac{1}{3^k}\right)$$, you can get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\int {e^{3x} - e^x \over e^{4x} + e^{2x} + 1} dx$
$$I = \int {e^{3x} - e^x \over e^{4x} + e^{2x} + 1} dx$$
Substituting for $e^x$,
$$I = \int {u^2 - 1 \over u^4 + u^2 + 1} du = \int { u^4 + u^2 + 1 + - 2 - u^4 \over u^4 + u^2 + 1} du = u - \int {u^4 + 2 \over u^4 + u^2 + 1} du $$
Now I don't know anything I can do ... | Observe that $$u^4 + u^2 + 1 = (u^4 + 2u^2 + 1) - u^2 = (u^2 + 1)^2 - u^2 = (u^2 + u + 1)(u^2 - u + 1).$$ From this, we try a partial fraction decomposition of the form $$\frac{Au + B}{u^2 - u + 1} + \frac{Cu + D}{u^2 + u + 1} = \frac{u^2 - 1}{u^4 + u^2 + 1},$$ and after multiplying out and comparing like coefficients... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2234797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a+1)$ if $\log_{b}a+\log_{c}b+\log_{a}c=13$ and $\log_{a}b+\log_{b}c+\log_{c}a=8$ Let $a,b$, and $c$ be positive real numbers such that
$$\log_{a}b + \log_{b}c + \log_{c}a = 8$$
and
$$\log_{b}a + \log_{c}b + \log_{a}c = 13.$$
What is the value of
$$(\log_{a}b... | We have the following:
$$(log_ab+1)(log_bc+1)(log_ca+1)=(\frac {logb}{loga}+1)(\frac {logc}{logb}+1)(\frac {loga}{logc}+1)$$
Expanding yields:
$$(\frac {logc}{loga}+\frac {logb}{loga}+\frac {logc}{logb}+1)(\frac {loga}{logc}+1)$$
$$=(\frac {loga}{loga}+\frac {logc}{loga}+\frac {logb}{logc}+\frac {logb}{loga} +\frac {lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 0
} |
Prove that $\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$ is a closed and bounded set.
Prove that the following set $$V=\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$$ is closed and bounded.
The set $V$ is closed. In fact, for any continuous function $f:\mathbb R^n\rightarrow \mathbb R$, the set $\{\... | The equation $x^2 - x y + y^2 = 1$ can be rewritten as follows
$$\begin{bmatrix} x\\ y\end{bmatrix}^{\top} \begin{bmatrix} 1 & -\frac 12\\ -\frac 12 & 1\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix} = 1$$
Computing the spectral decomposition,
$$\begin{bmatrix} 1 & -\frac 12\\ -\frac 12 & 1\end{bmatrix} = \frac 12 \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , ... | Hint:
For $x\ne2$,
$$|3x+1|<|x-2|.$$
Then we need to distiguish three cases:
*
*$x\le-\frac13\to-3x-1<-x+2$, or $x>-\frac32$,
*$-\frac13\le x\le 2\to3x+1<-x+2$, or $x<\frac14$,
*$2\le x\to 3x+1<x-2$ or $x<-\frac32$, which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
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On Chinese remainder theorem Suppose we know $x\equiv 3 \bmod 11$ and $x\equiv 7\bmod 13$ and $0<x<143$ holds then CRT gives that $x=3\times 13[13^{-1}\bmod 11] + 7\times 11[11^{-1}\bmod 13]=39\times 6 + 77\times 6=696$ gives a solution but it is not within $0$ and $143$. So we take $696\bmod 143$ and choose $124$ as ... | Some people do CRT theorem problems like this. If $x\equiv 3\pmod{11},$ then $x = 3+11k$. Plug that into $x \equiv 7 \pmod{13}$ to get $3+11k \equiv 7 \pmod{13}$ which reduces to
$$11k \equiv 4 \pmod{13}$$
$$-2k \equiv 4 \pmod{13}$$
$$k\equiv -2 \equiv 11 \pmod{13}$$
So $x = 3+11\cdot 11 =124.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2244885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Multiples of six in an interval How many multiples of $6$ are there between the following pairs of numbers?
$0$ and $100$ and $-6$ and $34$.
*
*$16$ and $6$
*$17$ and $6$
*$17$ and $7$
*$16$ and $7$
My attempt:
Total number of integer between $0$ to $100 = 101$
So, total number of multiplier between $0$ to $100 =... | There are no multiples of $m$ bewteen $km+1$ and $km + (m-1)$ and the only multiple between $km$ and $km + (m-1)$ is $km$.
So....
The multiples of $m$ between $km + 1$ and $jm$ will be the $lm$ for $k < l \le j$ or $0 < l-j \le j-k$ and there will be $j-k$ of them.
The multiples of $m$ between $km$ and $jm$ will be th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Number Theory: Find possible $n$ values that $n(n+1)(5n+2)$ is divisible by $2017^{2018}$ Assume $m= 2017^{2018}.$ How can I find possible value(s) of $n$ that $n < m$ and $n(n+1)(5n+2)$ is divisible by $m$.
Any hints?
| $2017$ is prime. $\gcd(n,n+1) = 1$, $\gcd(5n+2,n) = \gcd(2,n) = \{1,2\}$, $\gcd(5n+2, n+1) = \gcd(-3, n+1) = \{3,1\}$. So if $2017^{2018}|n(n+1)(5n+2)$ then either
$2017^{2018}|n$ and $n = k*m$.
or $2017^{2018}|n+1$ and $n= k*m - 1$.
or $2017^{2018}|5n + 2$ and $n = \frac{km- 2}5$. This requires $km \equiv 2 \mod 5$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some ... | Cauchy- Schwarz works:
$$x^2+y^2+z^2=\frac{1}{3}(1^2+1^2+1^2)(x^2+y^2+z^2)\geq\frac{1}{3}(x+y+z)^2=\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 15,
"answer_id": 3
} |
How do I prove using the definition of limit that $\lim_{(x,y)\rightarrow (0,0)}\frac {x^2+y^3}{y^2 - x + xy}$? I want to use the definition of limit in order to prove $\lim_{(x,y)\rightarrow (0,0)}\frac {x^2+y^3}{y^2 - x + xy}$, but I almost crack my head open trying to do this, it seems to be easy way but y cant find... | The limit does not exist. You can write $$
y^{2}+xy-x=\left( y+\frac{1}{2}x+\frac{1}{2}\sqrt{x\left( x+4\right)
}\right) \left( y-\frac{1}{2}\sqrt{x\left( x+4\right) }+\frac{1}
{2}x\right).
$$
Take $y=x^{2}+\frac{1}{2}\sqrt{x\left( x+4\right) }-\frac{1}{2}x$ with $x>0$, to get
\begin{align*}
& \frac{x^{2}+\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $A= 2 \cdot \pi/7$ then show that $ \sec A+ \sec 2A+ \sec 4A=-4$ If $A= 2 \times \pi/7$ then how to show,
$$\sec A+ \sec 2A+ \sec 4A=-4$$ I have tried using formula for $\cos 2A$ but I failed.
| Let us consider $\Phi_7(x)=\frac{x^7-1}{x-1}$. It is a palyndromic polynomial, hence $\frac{\Phi_7(x)}{x^3}$ can be written as a polynomial of $x+\frac{1}{x}$. Given that $\Phi_7(x)$ vanishes at the primitive seventh roots of unity, we get a polynomial that vanishes at $\left\{\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Given: $\int f(x)\sin x \cos x dx = \log(f(x)){1\over 2( b^2 - a^2)}+C$, find $f(x)$.
$$\int f(x)\sin x \cos x dx = \log(f(x)){1\over 2( b^2 - a^2)}+C$$
On differentiating, I get,
$$f(x)\sin x\cos x = {f^\prime(x)\over f(x)}{1\over 2( b^2 - a^2)}$$
$$\sin 2x (b^2 - a^2) = {f^\prime( x)\over (f(x))^2} $$
On integr... | $f(x) = {2\over b^2 \cos^2x -b^2\sin^2 x- a^2\cos^2x+ a^2\sin^2 x}$
$= {2\over b^2 \cos^2x -b^2(1-\cos^2 x)- a^2(1-\sin^2x)+ a^2\sin^2 x}$
$= {2\over b^2 \cos^2x -b^2 + b^2\cos^2 x- a^2 +a^2\sin^2x+ a^2\sin^2 x}$
$= {2\over -(a^2 + b^2 ) + 2b^2\cos^2x +2a^2\sin^2x}$
As you can see one extra term $-(a^2+b^2)$ in denomin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the Maximum value of $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$ if $x$, $y$ and $z$ are positive real numbers such that $x+y+z=4$ Find the maximum value of $$S=\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$$
I tried as follows.
The given expression can be rewritten as
$$S=\s... | $$\mathbf{\color{green}{New\ version\ of\ 10.02.2018}}$$
Let us find the greatest value of $\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$ under the condition $x+y+z=4.$
Substitutions
$$a=\sqrt{x+y},\quad b=\sqrt{y+z},\quad c=\sqrt{z+x}$$
lead to the task on the greatest value of $a+b+c$ under the condition $a^2+b^2+c^2=8.$
Station... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Showing that $\displaystyle\prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$ for any fixed $x \in \mathbb{R}_{>0}$ How can I show that
$$\displaystyle\prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$$ when $x$ is a positive real number ?
| $$ \prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$$
Proof:
$$\frac{1+kx}{2+kx}=1+\frac{-1}{2+kx} \leqslant\exp \left( \frac{-1}{2+kx} \right) \leqslant \exp \left( \frac{-x^{-1}}{k} \right)$$
So,
$$ \prod_{k = 1}^{N} \frac{1+kx}{2+kx} \leqslant \prod_{k = 1}^{N} \exp \left( \frac{-x^{-1}}{k} \right) = \exp(-x^{-1}H_N)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Number Theory: Prove that $7^{7^n}+1$ can be shown as product of at least $2n+3$ prime numbers Prove that for each natural number $n$, $7^{7^n}+1$ can be represented as product of at least $2n+3$ prime numbers. (Prime numbers are not necessarily different)
Any hints how to start the proof?
| Aurifeuillean factorization gives
$$7^{14k+7} + 1 = (7^{2k+1}+1)\cdot(C-D)(C+D)$$
with $$C,D=(7^{6k+3} + 3\cdot 7^{4k+2} + 3\cdot 7^{2k+1} + 1),(7^{5k + 3} + 7^{3k+2} + 7^{k+1})$$
so for $14k+7=7^{n+1}$:
$$\begin{align*}7^{7^{n+1}} + 1 \;= &\;\;(7^{7^n}+1)\\
&\cdot\left[(7^{3\cdot7^n} + 3\cdot 7^{2\cdot7^n} + 3\cdot 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $m\in\mathbb N$, $n\in\mathbb N$, and $f(0)$ where $f(x)=ax^3+bx^2+cx+d$ $(a,b,c,d\in\mathbb Z)$, $f(mn)=1$, $f(m)=n^2$, $f(n)=m^2$, ...
Question: Find $m\in\mathbb N$, $n\in\mathbb N$, and $f(0)$ where ($m,n\gt 1$)
*
*$f(x)=ax^3+bx^2+cx+d$ $(a,b,c,d\in\mathbb Z)$
*$f(mn)=1$
*$f(m)=n^2$
*$f(n)=m... | By 2., we can use the Factor Theorem to write
$$
f(x)-1=q(x)(x-mn)
$$
for polynomial $q$ with integer coefficients (and degree at most $2$, though that doesn't directly figure into this proof).
From 3. it follows that
$$
n^2-1=q(m)(m-mn)
$$
and so (since $n>1$)
$$
q(m)=\frac{n^2-1}{m-mn}=-\frac{n+1}{m}
$$
Similarly,
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Find the radius of convergence of $\sum_{n=1}^{\infty} \left(\frac{5^{n} + (-1)^{n}}{n^{3}}\right)(x-2)^{n}$ Find the radius of convergence of the following:
$$\sum_{n=1}^{\infty} \left(\frac{5^{n} + (-1)^{n}}{n^{3}}\right)(x-2)^{n}$$
My attempt:
I used Ratio Test and managed to get until
$$\lim_{n\to\infty} \bigg|\fr... | $$\sum_{n=1}^\infty\dfrac{5^n+(-1)^n}{n^3}(x-2)^n=\sum_{n=1}^\infty\dfrac{\{5(x-2)\}^n}{n^3}+\sum_{n=1}^\infty\dfrac{\{(-1)(x-2)\}^n}{n^3}$$
Can you apply Ratio Test separately?
OR
$$\lim_{n\to\infty} \bigg|\frac{5^{n+1} + (-1)^{n+1}}{5^{n} + (-1)^{n}} \left(\frac{n}{n+1}\right)^{3} (x-2)\bigg|$$
$$=\lim_{n\to\infty}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Two-row notation and disjoint cycles $$\pi = (12)(13)(826)(58)$$
I am asked to express $\pi$ as a product of disjoint cycles.
I am given the answer $(513268)$, why is it not $(513)(268)$
| $\begin{array}{|c|c|c|c|c|c|c|c|}
\hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline (58)x & 1 & 2 & 3 & 4 & \color{blue}8 & 6 & 7 & \color{blue}5 \\
\hline (826)(58)x & 1 & \color{blue}6 & 3 & 4 & \color{blue}2 & \color{blue}8 & 7 & 5 \\
\hline (13)(826)(58)x & \color{blue}3 & 6 & \color{blue}1 & 4 & 2 & 8 & 7 & 5 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Finding a sum of $1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$ I need someone to find a mistake in my soliution or maybe to solf it much more easily... I have got a sum $$1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$$ and need to evaluate it. So here's my... | Similar to the answer by Jack D'Aurizio and yours, but let $$f(x) = \sum_{n=0}^\infty \frac{1}{3n+1}\cdot x^{3n+1}\implies f'(x) = \sum_{n=0}^\infty x^{3n} =\frac{1}{1-x^3}$$
Hence,
$$f(x) = \int\frac{1}{1-x^3}dx = \frac{1}{6}\left(\ln(x^2+x+1)-2\ln(1-x)+2\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right)+C$$ B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
If $a,b,c$ are positive, prove that $\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a} \geq \frac{9}{a+b+c}$ If $a,b,c$ are positive real numbers, prove that
$$\frac{2}{a+b}+\frac{2}{b+c}+ \frac{2}{c+a}≥ \frac{9}{a+b+c}$$
| By Cauchy-Schwartz. $$\sum_{cyc}\frac{1}{a+b}\sum_{cyc}(a+b)\geq (1+1+1)^2=9\implies 2\cdot\sum_{cyc}\frac{1}{a+b}\geq2\cdot \frac{9}{\sum_{cyc}(a+b)}=\frac{9}{a+b+c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving a Line to bisect a line in a Triangle From point $A$ tangents $AB$ and $AC$ to a circle are drawn ($B$ and $C$ tangent points); $PQ$ is a diameter of the circle; line $L$ is tangent to the circle at point $Q$. Lines $PA$, $PB$, and $PC$ intersect line $L$ at points $A_1, B_1, C_1$. Prove that $A_1B_1 = A_1C_1$.... | Not a complete solution yet, but this is the beginning of pure "bashing" with coordinates.
Choose a coordinate system so that $B_1 = (1,0)$ and $C_1 = (0, -1).$ Then, it suffices to prove that $A_1 = (0,0).$
Let $P = (a,2b)$ and $Q = (a,0)$, so the equation of the circle is $(x-a)^2+(y-b)^2 = b^2$. This is simplified ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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2D Geometry. Find equation of a circle
The line $y=x$ touches a circle at $P$ so that $OP=4\sqrt{2}$ where $O$ is the origin. The point $(-10,2)$ lies inside the circle and the length of the chord $x+y=0$ is $6\sqrt{2}$.
Find the equation of the circle.
How to approach this sum?
My approach to the sum: I found that t... | The only points on $y=x$ which are $4 \sqrt{2} $ away from the origin are $(4,4)$ and $(-4,-4)$.
The perpendicular distance from the centre of the circle to the chord $x+y=0$ is $4 \sqrt{2} $.
So the radius of the circle is $\sqrt{(4\sqrt{2}) ^2 +(6\sqrt{2} \div 2)^2 } =5\sqrt{2} $.
The centre of the circle is $(4,4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a, b, c$ are positive and $a+b+c=1$, prove that $8abc\le\ (1-a)(1-b)(1-c)\le\frac{8}{27}$ If $a, b, c$ are positive and $a+b+c=1$, prove that $$8abc\le\ (1-a)(1-b)(1-c)\le\dfrac{8}{27}$$
I have solved $8abc\le\ (1-a)(1-b)(1-c)$ (by expanding $(1-a)(1-b)(1-c)$)
but do not get how to show that $(1-a)(1-b)(1-c)\le\fra... | For the 1st one: $(1-a)(1-b)(1-c) = (b+c)(a+c)(b+a) \ge 2\sqrt{bc}\cdot 2\sqrt{ac}\cdot 2\sqrt{ba} = 8abc$, and the 2nd one is by AM-GM as shown by the other answer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Continued Cosine Product. Is there a way to evaluate,
$$
\large \cos x \cdot \cos \frac{x}{2} \cdot \cos \frac{x}{4} ... \cdot \cos \frac{x}{2^{n-1}} \tag*{(1)}
$$
I asked this to one of my teachers and what he told is something like this,
Multiply and divide the last term of $(1)$ with $\boxed{\sin \frac{x}{2^{n-1... | How does step 2 happen:
multiplying the last factor by $\frac {\sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}}$
gives us
$\cos x \cdot \cos \frac x2\cdots \cos \frac x{2^{n-2}}\cdot \frac{\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}}$
Double angle formula.
$\cos \frac{x}{2^{n-1}} \cdot ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Equality involving greatest integer I am trying to find the integral solutions to the equation
$$[x][y]=x+y$$
and show that all non-integral solutions lie on exactly two lines. Here $[x]$ denotes greatest integer function.
For integer solutions I solved the equation $xy=x+y.$
Upon solving it I got $y=\frac{x}{x-1}$ an... | First suppose $x,y$ are integers.
\begin{align*}
\text{Then}\;\;&\lfloor x \rfloor \lfloor y \rfloor = x + y\\[4pt]
\iff\; &xy = x + y\\[4pt]
\iff\; &xy - x - y = 0\\[4pt]
\iff\; &xy - x - y + 1 = 1\\[4pt]
\iff\; &(x-1)(y-1) = 1\\[4pt]
\iff\; &x-1 = 1\;\;\text{and}\;\;y-1=1\\[0pt]
&\;\;\;\;\text{or}\\[0pt]
&x-1 = -1\;... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$a+b+c+d+e=79$ with constraints How many non-negative integer solutions are there to $a+b+c+d+e=79$ with the constraints $a\ge7$, $b\le34$ and $3\le c\le41$?
I get that for $a\ge7$ you do $79-7=72$, $\binom{72+5-1}{5-1}=\binom{76}4$. For $b\ge35$ I think it's $\binom{47}4$ and I'm not too sure what it is for $3\le c\le... | So we are looking for non-negative solutions for to $a+b+c+d+e=79$ with the constraints $a\ge7$, $b\le34$ and $3\le c\le41$
I always like, in this kind of problem, to work with generating functions. Everything variable gets a polynomial such that its powers correspond to the restraints, and such that he requested solut... | {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Differentiation - Power rule Find derivative of this function
$$g(t)=-3t(6t^4-1)^{4/3}$$
I have tried it till the answer:
$ -3t.\frac{4}{3} (6t^4 - 1)^{\frac{4}{3}-1} .\frac{d}{dt}(6t^4 -1) $
$-3t x \frac{4}{3} (6t^4-1)^\frac{1}{3} ( 6X4 t^3 -0 ) $
$ -4t (6t^4 -1)^\frac{4}{3} (24t^3) $
$-96t(6t^4-1)^\frac{1}{3} $
Ho... | As there are two terms. You need to use product rule.
Product rule -
$(UV)' = UV' + U'V$
So it should be -
$g(t) = -3t(6t^4-1)^\frac{4}{3}$
$-3t.\frac{4}{3} (6t^4 - 1)^{\frac{4}{3}-1} .\frac{d}{dt}(6t^4 -1) - 3.\frac{d}{dt} t.(6t^4-1)^\frac{4}{3}$
Take common terms and simplify.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3$ I have to prove that $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3 $$
is always true for real numbers $a, b, c>0$ with $abc=1$.
Using the AM-GM inequality I got as far as $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq \frac{b}... | Now I see. Using Especially Lime's comment, we can see that $$\left(\frac{1+ab}{1+a}\right)\left(\frac{1+bc}{1+b}\right)\left(\frac{1+ca}{1+c}\right)=1.$$
Hence, we can set $$x:=\frac{1+ab}{1+a}, y:=\frac{1+bc}{1+b}$$ and our inequality becomes $$x+y+\frac{1}{xy}\geq 3$$ which is equivalent to $$\frac{x+y+\frac{1}{xy}}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$
If $\cos^2\theta+\cos\theta = 1$, express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$.
We have \begin{align*}\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\the... | $S=\sin^8x+\sin^6x+\sin^4x+\sin^2x-2=\cos^4x+\cos^3x+\cos^2x+\cos x-2$
Now $S=(\cos^2x+\cos x-1)\cos^2x+2(\cos^2x+\cos x-1)-\cos x=?$ as $\cos^2x+\cos x-1=0$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Simpson's Rule in Numerical Integration I have a problem in proving that the simple Simpson’s rule $$\int_a^b f(x) dx \approx \frac{(b − a)}6[ f (a) + 4 f \left(\frac{a + b}{2}\right) + f (b)]$$
is exact for all cubic polynomials.
I am a little bit confused about using the change of variable $$x = a + t (b − a)$$ for... | If the change of variables bothers you, you don't really need it: it just makes some of the formulas somewhat simpler, because it means you can take $a=0$.
If $f(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3$,
$$ \eqalign{\int_a^b f(x)\; dx &= \left. c_0 x + c_1 \frac{x^2}{2} + c_2 \frac{x^3}{3} + c_3 \frac{x^4}{4} \right|_{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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If $a\cos(θ)=b$ $\cos( θ+2π/3)=c\cos(θ+4π/3)$, prove that $ab+bc+ca=0.$ THE PROBLEM: If $a\cos(θ)=b$ $\cos( θ+2π/3)=c\cos(θ+4π/3)$, prove that $ab+bc+ca=0.$
MY THOUGHT PROCESS: We have to prove that $ab+bc+ca=0$.
One method using which we can do this is, if we can somehow obtain the equation $k(ab+bc+ca)=0$ we can ded... | I don't know whether I am just repeating your first solution.
Let $a\cos\theta=k$. Then
\begin{align*}
k(ab+bc+ca)&=abc\cos\left(\theta+\frac{4\pi}{3}\right)+abc\cos\theta+abc\cos\left(\theta+\frac{2\pi}{3}\right)\\
&=abc\left[\cos\left(\theta+\frac{2\pi}{3}\right)+2\cos\left(\theta+\frac{2\pi}{3}\right)\cos\frac{2\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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showing $ 1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0 $ How to show that the following infinite series
$$
1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0?
$$
The above series is of the form $\sum_{n \ge 1} \frac{f(n)}{n}$,... | sum of all odd terms: $X = (1 + 1/3 + 1/5 + ... )$
sum of all terms (4n+2): $Y = -3/2(1 + 1/3 + 1/5 + ...) = -3X/2$
Sum of remaining terms = $Z = 1/4 + 1/8 + 1/12 + ... = 1/4(1 + 1/2 + 1/3 + 1/4 + 1/5 + ...)$
splitting into odd and even terms: $Z = 1/4(1+ 1/3 + 1/5 +...) + 1/8(1 + 1/2 + 1/3 + ...)$
So $Z = X/4 + Z/2 \i... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ . Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ .
...
Let $\cot^{-1} x=z$
$$x=\cot z$$
Then,
$$\sin \cot^{-1} (x)$$
$$=\sin z$$
$$=\dfrac {1}{\csc z}$$
$$=\dfrac {1}{\sqrt {1+\cot^2 z}}$$
$$=\dfrac {1}{\sqrt {1+x^2}}$$
| So your final expression is: $\cos\tan^{-1}\left(\dfrac{1}{\sqrt{1+x^2}}\right)$
Now we know that:
$\tan^{-1}p=\cos^{-1}\left(\dfrac{1}{\sqrt{1+p^2}}\right)\\
\implies\tan^{-1}\left(\dfrac{1}{\sqrt{1+x^2}}\right)=\cos^{-1}\left(\dfrac{1}{\sqrt{1+{\left(\dfrac{1}{\sqrt{1+x^2}}\right)}^2}}\right)=\cos^{-1}\left(\dfrac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluate the following triple summation
Evaluate $$\sum_{k=0}^{\infty} \sum_{j=0}^{\infty} \sum_{i=0}^{\infty} \frac{1}{3^i3^j3^k}$$ for $i\neq j\neq k$
My Attempt
Evaluate the summation with no restriction on $i,j,k$. Let this be $a_0$
When $i= j=k$. Let this be $a_1$.
When $i\neq j=k$, let this be $a_2$.
The... | I have the same answer.
\begin{align*}
\sum_{k\in\mathbb{N}} \sum_{j\in\mathbb{N}\setminus\{k\}} \sum_{i\in\mathbb{N}\setminus\{j,k\}} \frac{1}{3^i3^j3^k}&=\sum_{k\in\mathbb{N}} \sum_{j\in\mathbb{N}\setminus\{k\}} \left(\frac{1}{3^j3^k}\sum_{i\in\mathbb{N}\setminus\{j,k\}} \frac{1}{3^i}\right)\\
&=\sum_{k\in\mathbb{N}}... | {
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"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find all solutions to $|x^2-2|x||=1$ Firstly, we have that $$\left\{
\begin{array}{rcr}
|x| & = & x, \ \text{if} \ x\geq 0 \\
|x| & = & -x, \ \text{if} \ x<0 \\
\end{array}
\right.$$
So, this means that
$$\left\{
\begin{array}{rcr}
|x^2-2x| & = & 1, \ \text{if} \ x\geq 0 \\
|x^2+2x| & = & 1, \ \tex... | \begin{align*}
|x^2-2|x||&=1\\
x^2-2|x|&=1\quad\text{or}\quad -1\\
|x|^2-2|x|-1&=0\quad\text{or}\quad |x|^2-2|x|+1=0\\
|x|&=1+\sqrt{2} \quad\text{or}\quad 1\qquad(|x|=1-\sqrt{2}<0\text{ is rejected})\\
x&=\pm(1+\sqrt{2}) \quad\text{or}\quad \pm1
\end{align*}
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Let $A$ be a $n\times n$ matrix with entries $a_{ij}=i+j $ . Calculate rank of $A$ Let $A$ be a $n\times n$ matrix with entries $a_{ij}=i+j $ . Calculate rank of $A$.
My work : I noticed that A is symmetric . Hence all of its eigen vectors are real .. That is all i have got .
Your help will be highly appreciated .Than... | If $n=1$, $r(A) = 1$.
Otherwise if $n>1$, then $r(A) = 2$. Notice that
$$A =
\begin{pmatrix}
2 & 3 & \cdots & n+1 \\
3 & 4 & \cdots & n+2 \\
&&\cdots&\\
n+1&n+2&\cdots &2n
\end{pmatrix}
$$
Use elementary row operations to subtract the 1st row from the $i$th row for $2 \leq i \leq n$. Then we get
$$ A' = \begin{pmat... | {
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"source": "stackexchange",
"question_score": "2",
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How to find the inverse Laplace transform of the following fraction? Please help me to find the inverse Laplace transform of the following:
$$\frac{s^2+6s+9}{(s-1)(s-2)(s-3)}$$
Do I have to divide the fractions and use partial fractions decomposition?
Like this:
$$\frac{s^2}{(s-1)(s-2)(s-3)}=\frac{A}{s-1}+\frac{B}{s-2}... | The partial expansion is:
$$\frac{s^2 + 6 s + 9}{(s - 3) (s - 2) (s - 1)}=\frac{18}{s-3}-\frac{25}{s-2}+\frac{8}{s-1}$$
Inverting term-by-term we get:
$$\mathcal{L}_s^{-1}\left(\frac{18}{s-3}\right)=18 e^{3 n}$$
$$-\mathcal{L}_s^{-1}\left(\frac{25}{s-2}\right)=-25 e^{2 n}$$
$$\mathcal{L}_s^{-1}\left(\frac{8}{s-1}\right... | {
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"source": "stackexchange",
"question_score": "1",
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Evaluate $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right)$ How do you evaluate $$\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right)$$ using the identity $$\sin(\pi z)=\pi z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)?$$
I assume I'll have to express $1+\frac{1}{n^2}+\frac{1}{n^4}$ as $\... | We have $1+z+z^2=\Phi_3(z)=(z-\omega)(z-\bar{\omega})=(1-\omega z)(1-\bar{\omega}z)$, with $\omega=\exp{\frac{2\pi i}{3}}$.
By considering $z=\frac{1}{n^2}$ and $\eta=\exp\frac{\pi i}{3}$ we have that
$$ \frac{\sin(\pi w)}{\pi w}=\prod_{n\geq 1}\left(1-\frac{w^2}{n^2}\right)\tag{1} $$
implies:
$$ \prod_{n\geq 1}\left(1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Tail Probability for Student's t-Distribution Let $n$ be a positive integer, $t < 0$, and let $F_n$ the cumulative distribution function of Student's t-distribution with $n$ degrees of freedom. Then it is a well known fact (which I always found stated without proof) that the sequence $(F_n(t))_{n=1}^{\infty}$ is strict... | I finally found the proof, by following Henry's suggestion.
For every positive integer $m$, let $f_m(t)$ be the density function of Student's t-distribution with $m$ degrees of freedom, that is
\begin{equation}
f_m(t)=\frac{\Gamma \left(\frac{m+1}{2} \right)}{\sqrt{m \pi} \Gamma \left( \frac{m}{2} \right)} \left( 1 + \... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Minimum value of $\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$ Given $a,b,c \in \mathbb{R^+}$ such that $a+b+c=12$
Find Minimum value of $$S=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$$
My Try: By Cauchy Schwarz Inequality we have
$... | Just another way: By QM-AM inequality,
$$S=\sum \left(a+\frac1b\right)^2 \geqslant \frac13 \left(\sum a + \sum \frac1b\right)^2 \geqslant \frac13\left(12 + \frac34\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the locus of $\frac{\pi}{3}\le \arg(-z+3+i)\le \frac{5\pi}{3}$ I tried simplifying this first:
$$\frac{\pi}{3}\le \arg(-z+3+i)\le \frac{5\pi}{3} \Leftrightarrow \\
\frac{\pi}{3}\le \arg(-(z+(3-i)))\le \frac{5\pi}{3} \Leftrightarrow \\
\frac{\pi}{3}\le \arg(z+(3-i))+\pi \le \frac{5\pi}{3} \Leftrightarrow \\
???$$
M... | Here is how I was taught how to do this (I think it might be a slightly longer method):
$$ \frac{\pi}{3}\le \arg(-z+3+i)\le \frac{5\pi}{3} $$
Split this inequality into two inequalities:
$$ \frac{\pi}{3} \le \arg(-z+3+i) ~~\text{and}~~ \arg(-z+3+i) \le \frac{5\pi}{3}$$
Let's solve the first inequality first: Let $z=x+i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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equation involving absolute value function has $4$ solutions
If the equation $|x^2-5|x|+k|-\lambda x+7 \lambda = 0$ has exactly $4$ solution, Then $(\lambda,k)$ is
Attempt:$x^2-5|x|+k = \lambda x-7 \lambda$
$\star$ For $x>0,$ then $x^2-5x+k=\lambda x- 7 \lambda.$
$\star$ For $x\leq 0,$ then $x^2+5x+k=\lambda x- 7 \l... | No harm in doing it in a case by case basis.
You have
$x^2\pm 5x +k = \pm(\lambda x-7 \lambda)$
That's four equations to solve:
1) $x^2+ 5x +k = \lambda x-7 \lambda\implies x^2 + x(5 - \lambda) + (k+7\lambda) = 0$
2) $x^2- 5x +k = \lambda x-7 \lambda \implies x^2 + x(-5 - \lambda) + (k+7\lambda) = 0$
3) $x^2+ 5x +... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find $xyz$, given that the value of $x^2+y^2+z^2$, $x+y+z=x^3+y^3+z^3=7$
Given that $$x^2+y^2+z^2=49$$ $$x+y+z=x^3+y^3+z^3=7$$
Find $xyz$.
My attempt,
I've used a old school way to try to solve it, but I guess it doesn't work.
I expanded $(x+y+z)^3=x^3+y^3+z^3+3(x^2y+xy^2+x^2z+xz^2+yz^2)+6xyz$
Since I know substit... | If we identify $x,y,z$ as the roots of a cubic polynomial $av^3+bv^2+cv+d$ with $a\neq0$ then there are recursions in terms of $a,b,c,d$ for the power sums
$$P_i:=x^i+y^i+z^i.$$
These are the newton identities. As seen in this link
\begin{align}
P_0=&+3\\
P_1=&-\dfrac{b}{a}\\
P_2=&-\dfrac{b}{a}P_1-\dfrac{c}{a}2\\
P_3... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Use integration to find which function has Fourier series $\frac{4}{\pi} \sum\limits^{\infty}_{n = 1, n\text{ odd}} \frac{\sin(nt)}{n^3}$ Problem
It can be shown that the function
$$
f(t) = \begin{cases}
\dfrac{\pi}{2} + t & , -\pi < t < 0,\\
\dfrac{\pi}{2} - t & , 0 < t < \pi,
\end{cases}
$$
has Fourier series $FS_f(... | Using the property that the function evaluated at $t=0$ is the average of the left and right piece-wise function evaluated at zero and
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 \, n + 1)^3} = \frac{\pi^3}{32}$$
then the following is obtained:
Integrating
$$
f_{c}(t) = \begin{cases}
\frac{\pi}{2} + t & , -\pi < t < 0,\\
\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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how many $3\times 3$ matrices with entries from $\{0,1,2\}$.
How many $3 × 3$ matrices $M$ with entries from $\left\{0, 1, 2\right\}$ are there for which taken from
the sum of the main diagonal of $M^TM$ is $5$.
Attempt: Let $M = \begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$. where $a,b,c,d,e,f... | Positioning the $0$'s, we get $\dbinom94=126$ for the $'000011111'$ case.
For the $'000000012'$ case, we have $\dbinom97=36$ ways to position the $0$'s, we need to multiply this by $2!$ to account for the two cases $12, 21$.
Total: $126+72=198$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Given that $a\cos\theta+b\sin\theta+A\cos 2\theta+B\sin 2\theta \leq 1$ for all $\theta$, prove that $a^2+b^2\leq 2$ and $A^2+B^2 \leq 1$ I tried to solve this question but things got too much complicated and hence my efforts were completely futile.
Let $a,A,b,B \in \mathbb{R}$ and $$F(\theta)= 1- a\cos \theta - b\sin... | Here is another (?) approach: define $c=a-bi$, $d=A-Bi$. Then we are given that
$$
\operatorname{Re}(cz+dz^2)\le 1,\quad\forall z\in\Bbb C\colon |z|=1.\tag{1}
$$
Replacing $z$ with $-z$ and with $iz$ in (1) we get two more inequalities
\begin{eqnarray}
z\mapsto -z:\quad\operatorname{Re}(-cz+dz^2)\le 1,\quad\forall z\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
| If you suspect there exists a factorization over $\mathbb{Q}[x]$ into polynomials of degree 2 and 3, but you just don't know their coefficients, one way is to write it down as
$x^5 + x + 1 = (x^2 + ax + b)(x^3 + cx^2 + dx + e)$ where the coefficients are integers (by Gauss' lemma). And then expand and solve the system.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
} |
Manipulating a Summation Series
Summation: How does$$\begin{align*} & \frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k+\frac n{2n+1}\tag1\\ & =\sum\limits_{k=1}^n\frac 1{2k-1}-\frac 12\sum\limits_{k=1}^n\frac 1k\tag2\\ & =\sum\limits_{k=1}^{2n}\frac ... | \begin{align}
&\sum\limits_{k=1}^n\frac 1{2k+1}\\
= & \ \sum\limits_{k=1}^{n-1}\frac 1{2k+1} + \frac{1}{2n+1} \\
= & \ 1 + \sum\limits_{k=1}^{n-1}\frac 1{2k+1} + \frac{1}{2n+1} -1\\
= & \ \sum\limits_{k=0}^{n-1}\frac 1{2k+1} + \frac{1 -2n-1}{2n+1}\\
= & \ \sum\limits_{l=1}^{n}\frac 1{2l-1} - \frac{2n}{n+1} \tag{where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solutions of $\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$
:Solutions of:
$$\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$$
My try :
$$\lfloor 2x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac12 \rfloor $$
$$\lfloor 3x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac13 \rfloor... | HINT:
If $0<b < 1$ then $\lfloor x + b \rfloor = \lfloor x \rfloor$ if $x-\lfloor
x\rfloor + b < 1$ or $\lfloor x \rfloor + 1$ otherwise
you might have to look at the different cases that arise
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is the $n^{th}$ derivative of $\sin(x)$ just a translation of $\sin(x)$? I noticed that
$$\frac d{dx}\sin x=\cos x=\sin\left(x+\frac\pi2\right)$$
$$\frac{d^n}{dx^n}\sin x=\sin\left(x+\frac{\pi n}2\right)$$
Does this hold for any positive real value of $n$?
If so, does anybody have any reasoning behind why it's just a ... | Indeed, we know that
$$
\frac{d}{dx} \sin{x} = \cos{x} = \sin{\left( x + \frac{\pi}{2} \right) }
$$
and then
$$
\frac{d^2}{dx^2} \sin{x} = \cos{\left(x + \frac{\pi}{2} \right)}= \sin{\left(x + \frac{2\pi}{2} \right)}
$$
Proof by induction:
Let's assume that
$$
\frac{d^k}{dx^k} \sin{x} = \sin{\left( x+ \frac{k\pi}{2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Probability for 5 dice (high d6 result +1 per additional die 4 or higher) I think I am calculating these odds properly. Hopefully someone can confirm that I am, or tell me where I am going wrong if I'm not.
When rolling 5 regular six-sided dice, the total result I am looking for is the value of the highest die +1 for e... | You get $6 + 1$ if you throw one six, one $4$ or $5$ and three dice lower than $4$, or if you throw two sixes and three dice lower than $4$. The probability of the former equals:
$$\frac{{5 \choose 1}1^4{4 \choose 1}2^1{3 \choose 3}3^3}{6^5} = \frac{1080}{7776}$$
The probability of the latter equals:
$$\frac{{5 \choose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Circle equation - diametric form - polar coordinates. A line segment joining $(a,\alpha)$, $(b,\beta)$ in polar coordinates is the diameter of a circle. I want to find the equation of this circle. It can be done by converting into the Cartesian system but I want to find the equation without moving out of polar.
| General equation of circle:
$$r^2-2rr_0\cos (\theta-\phi)+r_0^2=R^2$$
Now
\begin{align}
(r_0 \cos \phi,r_0 \sin \phi) &=
\left(
\frac{a\cos \alpha+b\cos \beta}{2},
\frac{a\sin \alpha+b\sin \beta}{2}
\right) \\
r_0^2 &=
\left( \frac{a\cos \alpha+b\cos \beta}{2} \right)^2+
\left( \frac{a\sin \alpha+b\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Inclusion exclusion distribution problem Prompt: How many ways are there to give 20 different presents to 4 different children, so that no child gets exactly 6 presents. All presents are different.
Here's what I tried doing
Since there are 20 presents to be distributed among 4 children,
$$C_1 + C_2 + C_3 + C_4 = 20$$... | The presents are different so the stars and bars formula doesn't work here. For example, the number of ways to distribute $20$ different presents to $4$ children is $4^{20}$ since each present can go to $4$ possible children (think of stamping the name of the child on each present).
Let $U$ be the set of all distribut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Mass Point Geometry Question In the triangle $ABC$, let $E$ be a point on $BC$ such that $BE : EC =
3: 2$. Pick points $D$ and $F$ on the sides $AB$ and $AC$ , correspondingly, so that
$3AD = 2AF$ . Let $G$ be the point of intersection of $AE$ and $DF$ . Given that $AB = 7$
and $AC = 9$, find the ratio $DG: GF$.
I have... |
Let $D'\in AB$ be such that $CD'\parallel FD$ and let $G'=AE\cap CD'$.
By similarity, $\frac{DG}{GF}=\frac{DG'}{G'C}$. Let $H=BG'\cap AC$. By Ceva's theorem
$$ 1=\frac{BE}{EC}\cdot \frac{CH}{HA}\cdot \frac{AD'}{D'B} = \frac{3}{2}\cdot\frac{CH}{HA}\cdot\frac{\frac{2}{3}b}{c-\frac{2}{3}b} $$
hence $\frac{CH}{HA}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximization of multivariable function $ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$ subject to constraint This is purely out of curiosity. I have this set of a high school entrance Math exam in Vietnam (a special high school for gifted kids, the exam was held 3 days ago, maybe 4, t... | The functional form of $M$ suggests us to change variable to $u = a+1, v = b+1, w = c + 1$.
In terms of them, the expression of $M$ simplifies to
$$M = \frac{u}{1+u^2} + \frac{v}{1+v^2} + \frac{w}{1+w^2}$$
The constraint $abc + ab + bc + ca = 2$ can be rewritten as
$$(a+1)(b+1)(c+1) - (a+b+c) - 1 = 2\quad\iff\quad u + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2315548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Infinite zeros in infinite series The problem:
Given that
$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \ldots $$
Prove
$$\frac{\pi}{3} = 1 + \frac{1}{5} - \frac{1}{7} - \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \ldots$$
My solution:
We know
$$
\begin{align}
\frac{\pi}{4} & = 1 - \f... | Both the given identities follow from the fact that $\frac{\pi}{\text{something}}$ is related with the integral over $(0,1)$ of a rational function. The first identity is a consequence of
$$\frac{\pi}{4} = \arctan(1)=\int_{0}^{1}\frac{dx}{1+x^2} = \int_{0}^{1}\left(1-x^2+x^4-x^6+\ldots\right)\,dx $$
and for the second... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2320456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 0
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Polynomial remainder problem. A polynomial $f(x)$ is such that upon division by $(x-3)$ it leaves a remainder of $15$ and upon division by $(x-1)^2$, it leaves a remainder of $2x+1$.
What is the remainder when $f(x)$ is divided by $(x-3)(x-1)^2$?
| Hint. We have that
$$f(x)=A(x)(x-3)+15=B(x)(x-1)^2+2x+1=C(x)(x-3)(x-1)^2+r(x)$$
where $r(x)=ax^2+bx+c$ (because the degree of $(x-3)(x-1)^2$ is $3$) and $A,B,C$ are polynomials.
In order to find $a$, $b$, and $c$, let $x=1$ and $x=3$.
Consider also the derivative of $f$ at $x=1$.
Therefore we obtain the equations:
$$r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2320784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How do you integrate $\frac{\cos^5(x)}{\sin^5(x) + \cos^5(x)}$?
$$\int_{0}^{\frac{\pi}{2}} \frac{\cos^5(x)}{\sin^5(x) + \cos^5(x)} \,dx$$
I tried by dividing the terms in both the numerator and denominator by $\cos^5x$ but still cant find my way.
| The problem was changed and my following work is not necessary.
Let $\tan{x}=t$.
Hence, $dt=\frac{1}{\cos^2x}dx=(1+t^2)dx$.
Thus,
$$\int\frac{\cos^5x}{\sin^5x+\cos^5x}dx=\int\frac{1}{1+\tan^5x}dx=\int\frac{1}{(1+t^2)(1+t^5)}dt.$$
Let $t+\frac{1}{t}=u$.
Hence, $$1+t^5=(1+t)(1-t+t^2-t^3+t^4)=t^2(1+t)\left(t^2+\frac{1}{t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} = -7$ for $x$ I have been scratching my head for solving this equation but I am unable to do this. Even I am unable to get how to use the hint. A way to solve this would be of great help
Solve for $x$ :-
$$\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+... | \begin{align}
\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} &= -7\\
\frac{a+b-x}c+\frac{c}{c} + \frac{a+c-x}b+\frac{b}{b} + \frac{b+c-x}a+\frac{a}{a} -\frac{4abc}{a+b+c} &= -7+3\\
\frac{a+b+c-x}c + \frac{a+b+c-x}b + \frac{a+b+c-x}a -\frac{4abc}{a+b+c} &= -4\\
(a+b+c-x)\left(\frac{1}{a}+\frac{1}{b}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\cos6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1 $ So far I've done this:
LHS $ =\cos^{2}3x-\sin^{2}3x$
$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$
$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$
I can tell I'm going in the right direction but how should I proceed further?
EDIT ... | Simpler way!
Use the fact $\displaystyle (\cos(x)+i\sin(x))^6 = \cos(6x)+i\sin(6x)$.
Expand, use $\sin^2(x)=1-\cos^2(x)$ to simplify higher powers of sine in the real terms, and then organizing by real terms, we get
$\cos(6x)+i\sin(6x)=$
$=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)+32i\sin(x)\cos(x)^5-32i\sin(x)\cos(x)^3+6i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
$u$-substitution always evaluates to $0$ Consider the integral
$$ \int_a^b f(x) \,dx $$
now make the $u$-substitution $u \mapsto c + (x-a)(x-b)$. The resulting integral is
$$ \int_c^c h(u) \,du $$
where $h(u)$ is the integrand $f$ after the substitution, however, regardless of $f$ the integral $\int_c^c du = 0$. Looki... | Assume $a<b$.
If $u=c+(x-a)(x-b)$. Then
\begin{align}
u&=x^2-(a+b)x+ab+c\\
&=\left(x-\frac{a+b}{2}\right)^2+ab-\left(\frac{a+b}{2}\right)^2+c\\
&=\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2+c\\
x&=\frac{a+b}{2}\pm\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}
\end{align}
When $\displaystyle x<\frac{a+b}{2}$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 2
} |
Why multiply a matrix with its transpose? This might be a very stupid question, but I do not seem to understand why I would multiple a matrix with its transpose. I am not a mathematician, but I am very interested in understanding the practical usage of equations:
Imagine I have three products sales Apple, Orange and P... | It can be a part of a bigger question (context needed, for example, linear regression as mentioned by lhf in the linked post).
Assume that the vector (Q) of quantity apple sales and the vector (R) of revenues on four days are:
$$Q=\begin{pmatrix}10\\ 5\\ 4 \\ 8\end{pmatrix}; \ \ R=\begin{pmatrix}20\\ 10\\ 8\\ 16\end{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
} |
Proof/Derivation of Closed form of Binomial Expression $\sum\limits_{k=0}^{2n}(-1)^k\binom{2n}{k}^2$ The binomial expression given as follows:
$$\sum_{k=0}^{2n}\left(-1\right)^{k}\binom{2n}{k}^{2}$$
results nicely into the following closed form:
$$(-1)^{n}\binom{2n}{n}$$
I wish to know how exactly is it being done? I h... | I will use the notation $[x^k]\,f(x)$ for denoting the coefficient of $x^k$ in the Taylor/Laurent expansion of $f(x)$ around the origin. We have:
$$ S(n)=\sum_{k=0}^{2n}(-1)^k \binom{2n}{k}^2 = \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}\binom{2n}{2n-k}=\sum_{\substack{a,b\geq 0 \\ a+b=2n}}(-1)^a\binom{2n}{a}\binom{2n}{b} $$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find a degree 2 homogeneous linear recurrence relation given a formula Given the formula $a_n = (-1)^n + 2^{n+1}$, how do you find a degree 2 homogeneous linear recurrence relation?
| Let $a_{n+2} = a_{n+1}x + a_ny$. This gives
$$
(-1)^{n+2} + 2^{n+3} = ((-1)^{n+1} + 2^{n+2})x + ((-1)^n + 2^{n+1})y
$$
Separating this into two equations, one for the $-1$ term and one for the $2$ term, we get
$$
\cases{(-1)^{n+2} = (-1)^{n+1}x + (-1)^{n}y\\2^{n+3} = 2^{n+2}x + 2^{n+1}y}
$$
Dividing the first equation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find the number of integers 1≤x≤210 which are divided by two of the numbers 3,5,7 I went with;
Numbers that can be divided by $3,5$ can also be divided with $15$. These are $15,30,45,60,75,90,105,120,135,150,165,180,195,210$
Numbers that can be divided by $3,7$ can also be divided with $21$. These are $21,42,63,84,105,... | $$210= 3 \times 5 \times 7 \times 2$$
$$\frac{210}{3 \times 5} = 2 \times 7$$
$$\frac{210}{3 \times 7} = 2 \times 5$$
$$\frac{210}{5 \times 7} = 3 \times 2$$
If a number is divisible by all $3$ numbers, it will appear in each list, hence we need to substract $2 \times$ number of multiples of $3 \times 5 \times 7$.
Hen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Inequality in three variables $a,b,c$ given $abc\geq 1$
Let $a,b,c$ be three real positive numbers such that $abc\geq 1$. Prove that $$\frac{a}{b} + \frac{b}{c} +\frac{c}{a} ≥ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$
Here's what I have done. The expression is equivalent to
$$a^2c+b^2a+c^2b \geq ab+bc+ac .$$
Deno... | By AM-GM we obtain:
$$\sum_{cyc}\frac{a}{b}=\frac{1}{3}\sum_{cyc}\left(\frac{a}{b}+\frac{2b}{c}\right)\geq\sum_{cyc}\sqrt[3]{\frac{ab}{c^2}}=\sum_{cyc}\sqrt[3]{\frac{abc}{c^3}}\geq\sum_{cyc}\frac{1}{a}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving $\int \frac{1}{6+(x+4)^2} dx$. $\int \frac{1}{6+(x+4)^2} dx = 6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx$
Now $u=\frac{(x+4)}{\sqrt{6}}$ and $du=\frac{1}{\sqrt{6}}dx$.
$6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx=\frac{6}{\sqrt{6}}\int\ \frac{1}{1+u^2}=\frac{6}{\sqrt{6}}$ arctan$(u)$=$\frac{6}{\sqrt{6}}$ arctan$(\frac{... | Anyway everyone should know that
$$\int\frac{\mathrm dx}{x^2+a^2}=\frac1a\arctan\frac xa.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Finding $\int_0^7 g(r)~dr$ from $ g(r)=\begin{cases} \sqrt{9-(r-3)^2}, & 0\le r<3 \\ r+2, & 3\le r<5 \\ \sqrt{4-(r-5)^2}, & 5 \le r \le 7\end{cases} $ I have some piece-wise integration that I for some reason am not getting.
$$
g(r) =
\begin{cases}
\sqrt{9-(r-3)^2} && 0 \le r \lt 3 \\
r+2 && 3 \le r \lt 5 \\
\sqrt{4-(... | I have no idea why you separated it like how you did it, this is incorrect:
$$\int_0^7 g(r)~dr= \int_0^2 \sqrt{9-(r-3)^2}~dr+\int_3^4 (r+2)~dr+\int_5^7 \sqrt{4-(r-5)^2}~dr$$
Doing your method will give this area instead:
Splitting definite integrals in two does not work like when you split summations!
$$\sum_{k=0}^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
convergence of $\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^\alpha dx $ For which value $\alpha$ does the following integral converges:
$\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^\alpha dx $
I tried to use: $ \sqrt{x^2 + 2x + 2} - 1 = \frac{x^2 + 2x + 1}{\sqrt{x^2 + 2x + 2} + 1} $
I wanted to simplify i... | $I_a =\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^a dx
$
Use the standard method of
$a-b
=\dfrac{a^2-b^2}{a+b}
$.
$\begin{array}\\
\sqrt{x^2 + 2x + 2} -( x + 1)
&=\dfrac{(x^2 + 2x + 2)-(x+1)^2}{\sqrt{x^2 + 2x + 2} + x + 1}\\
&=\dfrac{(x^2 + 2x + 2)-(x^2+2x+1)}{\sqrt{x^2 + 2x + 2} + x + 1}\\
&=\dfrac{1}{\sqrt{x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help to understand polynomial factoring I'm following some proof, but got stuck at how the factoring works. I can follow this part:
$$\begin{align*}
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\
&= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\
\end{align*}$$
The next two steps are not so clear to me ... | Just looking at the numerators,
$$k^2(k+1)^2+4(k+1)^3 =k^2(k+1)^2+4(k+1)(k+1)^2$$
$$ = (k+1)^2(k^2+4(k+1)) = (k+1)^2(k^2+4k+4)$$
$$=(k+1)^2(k+2)^2$$
For the second equality, I factored out the $(k+1)^2$ that appears in both terms; something of the form $ba+ca$ can be rewritten as $a(b+c)$. For the third equality, I ju... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve the following quadratic equation: $4\tan^2(\theta)x^4-4x^2+1=0$ $$4\tan^2(\theta)x^4-4x^2+1=0$$
It can be seen that as $\theta$ goes to $0$, the leftmost term disappears and $x=\pm \frac{1}{2}$.
But when I try to solve this using the quadratic formula I get:
$$x=\pm \sqrt{\frac{1\pm \sqrt{1-\tan^2(\theta)}}{2\tan... | $$\lim_{\theta\to0}\frac{1-\sqrt{1-\tan^2\theta}}{2\tan^2\theta}
= \lim_{\theta\to0}\frac{(1-\sqrt{1-\tan^2\theta})(1+\sqrt{1-\tan^2\theta})}{(2\tan^2\theta)(1+\sqrt{1-\tan^2\theta})} = \lim_{\theta\to0}\frac{\tan^2\theta}{(2\tan^2\theta)(1+\sqrt{1-\tan^2\theta})}=\lim_{\theta\to0}\frac{1}{2(1+\sqrt{1-\tan^2\theta})}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding minima and maxima to $f(x,y) = x^2 + x(y^2 - 1)$ in the area $x^2 + y^2 \leq 1$ I'm asked to find minima and maxima on the function
$$f(x,y) = x^2 + x(y^2 - 1)$$
in the area
$$x^2 + y^2 \leq 1.$$
My solution:
$$\nabla (f) = (2x + y^2 - 1, 2xy)$$
Finding stationary points
$$2xy = 0$$
$$2x + y^2 - 1 = 0$$
gets... | Just as you check the boundary of the 2_D region, you also have to check the endpoints of your interval. $f(x)=x^2-x^3$ on the interval $[-1,1]$. You find that $x=-1, y=0$ is your max.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove the sequence of general term $\sum\limits_{k=1}^{2n+1} \frac{(-1)^{k+1}}{k}$ converges
Given the sequence $$a_n :=\sum_{k=1}^n \frac{(-1)^{k+1}}{k}$$
Prove the sequence $(a_{2n+1})_{n \geq 0}$ converges.
My thoughts
I've proven that the sequence $(a_{2n+1})_{n \geq 0}$ is monotone decreasing. So now I want t... | To show that $a_{2n+1} > \frac12$, note that
$$
a_{2n+1} = \sum_{k=1}^{2n+1} \frac{(-1)^{k+1}}{k} = \underbrace{1 - \frac12}_{=\frac12} + \underbrace{\frac13 - \frac14}_{>0} + \underbrace{\frac15 - \frac16}_{>0} + \ldots + \underbrace{\frac{1}{2n-1} - \frac{1}{2n}}_{>0} + \underbrace{\frac{1}{2n+1}}_{>0} > \frac12.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x... | You have proved that the LHS is
$$
LHS=\frac{1-\cos x}{1+\cos x}
$$
And if you know the formula for $\tan \frac{x}{2}$ you have done (bisection).
If not use:
$$
\tan^2 \frac{x}{2}=\frac{\sin ^2 \frac{x}{2}}{\cos ^2 \frac{x}{2}}=\frac{\frac{1-\cos x}{2}}{\frac{1+\cos x}{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
| Work out $a=z+1/z$. Note that $z^3+1/z^3=a^3-3a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Given triangle $ABC$ with $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians. Find the area of $ ABC $
Given triangle $ABC$ with $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians.
Find the area of $ ABC $
This problem is too difficult for me , the teacher said it is a challenge pr... | Let $AB=c$, $AC=b$ and $BC=a$.
Thus, $a^2=b^2+c^2-bc$, $\frac{1}{2}\sqrt{2a^2+2c^2-b^2}=9$ and $\frac{1}{2}\sqrt{2a^2+2b^2-c^2}=6$, which gives
$$\frac{2(b^2+c^2-bc)+2c^2-b^2}{2(b^2+c^2-bc)+2b^2-c^2}=\frac{9}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$a-b \nmid a^2+b^2$ if $a>b+2$ and $\gcd(a,b)=1$ Any two co-prime number $a,b$ with $a>b+2$ we have $a^2+b^2$ is not divisible by $a-b$, $a,b \in \mathbb{N}$.
But how to prove this?
| Since $a-b$ divides $a^2-b^2$, if it also divides $a^2+b^2$ then $a-b$ divides $2b^2$.
Now use that $a$ and $b$ are coprime to see that $a-b$ and $b^2$ are coprime. So by Euclides' lemma $a-b$ divides $2$. This contradicts the condition $a-b>2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.