Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$R$ be the row reduced echelon form of a $4 \times 4$ real matrix $A$ Let $R$ be the row reduced echelon form of a $4 \times 4$ real matrix $A$ and
let the
third column of $R$ be $\left[\begin{array}{l}0 \\ 1 \\ 0 \\ 0\end{array}\right]$. Then which is true?
P): If $\left[\begin{array}{l}\alpha \\ \beta \\ \gamma \\ 0\... | For $P$ \begin{aligned}
&{\left[\begin{array}{llll}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
0 \\
-2 \\
2 \\
0
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0\\
0
\end{array}\right]} \\
\\
\\
&\Rightarrow \gamma=2
\end{aligned}
The statement $P$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\int^1_0 \frac{x\,\mathrm dx}{((2x-1)\sqrt{x^2+x+1} + (2x+1)\sqrt{x^2-x+1})\sqrt{x^4+x^2+1}}$ $$\int^1_0 \frac{x\,\mathrm dx}{((2x-1)\sqrt{x^2+x+1} + (2x+1)\sqrt{x^2-x+1})\sqrt{x^4+x^2+1}}$$
I've tried :
*
*Substitution:
*
*$t = x^2$
*$x = \cos{t}$
*Definite Integral Properties like $a+b-x$ (which mess... |
Let $\mathcal{I}$ denote the value of the following definite integral:
$$\mathcal{I}:=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left[\left(2x+1\right)\sqrt{x^{2}-x+1}+\left(2x-1\right)\sqrt{x^{2}+x+1}\right]\sqrt{x^{4}+x^{2}+1}}\approx0.2440169.$$
Observe that $\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)=x^{4}+x^{2}+1$ and
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Expected number of leaves of a planar tree with $n$ nodes I want to find out the expected number of leaves of a planar tree with $n$ nodes using the bivariate generating function for the number of leaves in a planar tree.
For the (univariate) generating function we have $$P(z)=\frac{z}{1-P(z)}$$Solving the quadratic eq... | The combinatorial class for plane trees with leaves marked is
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{P} = \mathcal{Z} \times \mathcal{U} +
\mathcal{Z}\times \textsc{SEQ}_{\ge 1}(\mathcal{P})$$
This gives the functional equation
$$P(z) = uz + z \frac{P(z)}{1-P(z)}$$
or
$$z = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\displaystyle I=\iiint_V(x+y+z)^2 \, dx \, dy \, dz$, where $V$ is the region bounded by $x^2+y^2+z^2=3a^2$ and $x^2+y^2=2az$ Compute $\displaystyle I=\iiint_V(x+y+z)^2 \,dx \,dy \,dz$, where $V$ is the region bounded by $x^2+y^2+z^2=3a^2$ and $x^2+y^2=2az$.
What I did was I found the intersection of the two s... | As Ninad Munshi put in his comments, the way question reads, it could mean two different regions. I am considering the region as,
$x^2+y^2+z^2 \leq 3a^2$ and $2az \geq x^2+y^2$.
First find the radius at the intersection of both surfaces.
$x^2+y^2+z^2=3a^2$ and $2az = x^2+y^2$.
$z^2+2az = 3a^2 \implies z = a$. So using... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How many non-congruent triangles can be formed by the vertices of a regular polygon of $n$ sides On a circle there are $n$ evenly spaced points (i.e., they form a regular polygon). How many triangles can be formed when congruent triangles are considered the same?
This question is closely related to
How many triangles c... | For each triangle $A_iA_jA_k$, the angles $\angle A_iOA_j$, $\angle A_jOA_k$ and $\angle A_kOA_i$ are respectively equal to $|j-i|\frac{2\pi}{n}$, $|j-k|\frac{2\pi}{n}$ and $|k-i|\frac{2\pi}{n}$ with $1\le i\ne j\ne k \le (n-1)$.
Let denote
$$
\begin{cases}
x=|j-i|\\
y=|k-j|\\
z=|k-i|
\end{cases}
$$
Then
$$
\begin{cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding the range of $a$ for which line $y=2x+a$ lies between circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting either
Find the range of parameter $a$ for which the variable line $y = 2x + a$ lies between the circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting or touching ei... | Your condition of tangency is assuming your circles are centered at the origin.
It derives for setting $y = mx + a$ and $x^2 +y^2 = r^2$ and substituting $x^2 + (mx+a)^2 = r^2$. Solving for $x$ and getting two values based an the discriminate $4a^2m^2 -4(a^2 - r^2)(1+m^2)$ and knowing if this is a tangent line the two... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Let $x^2 −(m−3)x+m=0,(m\in R)$ be a quadratic equation. Find the value of $m$ for which at least one root is greater than $2$
Let $x^2 −(m−3)x+m=0,(m\in R)$ be a quadratic equation. Find the value of $m$ for which at least one root is greater than $2$.
Discriminant$=m^2-10m+9\ge0\implies m\in(-\infty,1]\cup[9,\infty)... | The quadratic polynomial $ \ x^2 − (m−3)x + m \ \ $ represents an "upward-opening" parabola, for which the "vertex form" is $ \ \left(x \ − \ \frac{ m−3}{2} \right)^2 \ + \ \left( m \ - \ \frac{ [m−3]^2}{4} \right) \ \ . $ It may also be noted that $ \ f(1) \ = \ 4 \ \ , $ independent of the value of $ \ m \ \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding a general solution of pde $(x^2-y^2-u^2)\cdot u_x+(2xy)\cdot u_y=2xu$ how can I solve this partial dif. equation. I try to use Langarange methods
This is my solution;
$$\frac {dx}{x^2-y^2-u^2}=\frac {dy}{2xy}=\frac {du}{2xu}= λ$$
$$\frac {dy}{2xy}=\frac {du}{2xu}$$
$$\frac {du}{u}-\frac {dy}{y}=0$$
$$\ln u-\ln ... | There is a another way and this
$$\frac{du}{2xu}=\frac{xdx+ydy+udu}{x.(x^2-y^2-u^2)+y.2xy+u.2xu}$$
$$\frac{du}{2xu}=\frac{xdx+ydy+udu}{x^3-x.y^2-x.u^2+2xy^2+2xu^2}$$
$$\frac{du}{2xu}=\frac{xdx+ydy+udu}{x^3+x.y^2+x.u^2}$$ and if we simplify the equation
$$\frac{du}{2x.(u)}=\frac{xdx+ydy+udu}{x(x^2+y^2+u^2)}$$
$$\frac{du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Asymptotic expansion of $\sum_{k=1}^n {n \choose k} (-1)^k \frac 1 {1-x^k} $ I'm trying to find the limit (and asymptotic expansion as $n\to\infty $) of
$$\sum_{k=1}^n {n \choose k} (-1)^k \frac 1 {1-x^k} $$
for $0<x<1$.
So far, I have no idea...
I found this question when dealing with the expected value of $\max (X_1,... | Not a complete answer, but an idea of where to look and some lower bounds.
Let $$p_k(x)=\frac{1-x^k}{1-x}=1+x+\dots+x^{k-1}$$
Then:
$$\frac{1}{1-x^k}=\frac{1}{k(1-x)}+\frac{(k-1)+(k-2)x+\cdots+x^{k-2}}{kp(x)}$$
Also, $$\begin{align}\sum_{k=1}^{n}(-1)^k\binom{n}k\frac1k&=\int_0^1\frac{(1-t)^n-1}{t}\,dt\\
&=\int_0^1 \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Number of integer solutions of $a^2+b^2=10c^2$ Find the number of integer solutions of the equation $a^2+b^2=10c^2$.
I can only get by inspection that $a=3m, b=m,c=m$ satisfies for any $m \in Z$.
Is there a formal logic to find all possible solutions? Any hint?
Also i tried taking $a=p^2-q^2$, $b=2pq$ and $10c^2=(p^2+q... | We first transform the given equation $a^2+b^2=10c^2$ as follows:
Consider $x=\frac{a}{c}, y=\frac{b}{c}$ (assuming we are interested in $c \neq 0$ case). Then we have to find rational points on the circle
$$x^2+y^2=10.$$
Now $(x,y)=(3,1)$ is an obvious solution. To find other rational solutions, consider the line that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Comparing numbers of the form $c+\sqrt{b}$ (eg, $3+3\sqrt{3}$ and $4+2\sqrt{5}$) without a calculator It is easy to compare to numbers of the form $a\sqrt{b}$, simply by comparing their squares, for example $3\sqrt{3}$ and $2\sqrt{5}$.
But what if we have $a=3+3\sqrt{3}$ and $b=4+2\sqrt{5}$ for example?
How to compare ... | I think you can use Maclaurin series
For example, $3+3\sqrt{1+2}\approx3+3\cdot(1+2/2)=9$ and $4+2\sqrt{1+4}\approx4+2\cdot (1+4/2)=10$
Clearly, $10>9$ so $4+2\sqrt{5}>3+3\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Computing the character table of $SL_2(\mathbb{Z}/3)$ I'm self studying some group representation theory and I've hit a wall with Exercise 3.10 (Computing the character table of $G = SL_2(\mathbb{Z}/3)$) in Fulton and Harris' book. Any advice or hints would be really appreciated!
Here's what I know so far:
(1) The repr... | I'm rusty in representation theory, but I'll give it a try.
The characters $W,W',W''$ at $A_1,A_2,B_1,B_2$ can't all be zero, since this would break the orthogonality relation (not sure but this feels true).
Therefore one of them must be non-zero, without loss of generality a character $W$. This implies that $W'=W\otim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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How do I calculate the sum of sum of triangular numbers? As we know, triangular numbers are a sequence defined by $\frac{n(n+1)}{2}$. And it's first few terms are $1,3,6,10,15...$. Now I want to calculate the sum of the sum of triangular numbers. Let's define
$$a_n=\frac{n(n+1)}{2}$$
$$b_n=\sum_{x=1}^na_x$$
$$c_n=\sum_... | Notice that after $k$ summations, the formula is
$$\binom{n+k-1}{n-1}.$$
As we can check, by the Pascal identity
$$\binom{n+k-1}{n-1}-\binom{n-1+k-1}{n-2}=\binom{n-1+k-1}{n-1},$$
which shows that the last term of a sum (sum up to $n$ minus sum up to $n-1$) is the sum of the previous stage ($k-1$) up to $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
$\textbf{My attempts} :$
From the condition, we can write $$1+2n^2+3m^2\equiv 0(\mod 5)\tag 1$$
Now, since $5$ is pr... | You can make it a bit easier. The cool thing working $\mod 5$ is that squares are always $\equiv 0$ or $\equiv \pm 1$ (you see this by just entering all possibilities). If you know that
$$
1+2n^2+3m^2 \equiv 0 \mod 5
$$
test out the possibilities of $n,m \mod 5$ to see what works:
\begin{align}
n^2\equiv 0, m^2\equiv 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding the discriminant of a quadratic equation from the given information on the roots of a quadratic equation I recently came accross an old question that I solved during my school days. Which is
If $\alpha, \beta$ are two real roots of a quadratic equation $ ax^2+bx+c=0 $ and $\alpha+\beta, \alpha^2+\beta^2, \alph... | There is another approach that might be taken that largely avoids the issue of factorization. While it is tempting upon seeing the expressions $ \ \alpha + \beta \ , \ \alpha^2 + \beta^2 \ , \ \alpha^3 + \beta^3 \ $ to consider an argument based on Newton's identities, that leads to a great deal of writing (and a bit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$\sqrt{x+1}, \sqrt{x+2}, \sqrt{x+3}$ cannot be all three rational numbers. $x$ is a rational number. Then,
$\sqrt{x+1}, \sqrt{x+2}, \sqrt{x+3}$ cannot be all three rational numbers. Any idea? Thanks!
I tried to solve this problem and I arrived to prove that in natural set (positive integers) we cannot have:
$m^2 + n^2 ... | Note that your question is equivalent to showing that for a rational $x$, all of $\sqrt{x-1}$, $\sqrt{x}$ and $\sqrt{x+1}$ cannot be rational. If possible, on the contrary, let that be true.
Let $x=\frac{a^2}{b^2}$ with $(a,b)=1$ so that
\begin{align}
\sqrt{x-1}&=\frac{\sqrt{a^2-b^2}}{b}\\
\sqrt{x}&=\frac{a}{b}\\
\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Determine which of the following are metrics on $X$ Let $d$ and $d'$ be metrics on a non empty set $X$. Then which of the following are metrics on $X$.
*
*$\rho_1 (x,y)=d(x,y)d'(x,y)$ for all $x,y \in X$
*$\rho_2 (x,y)=max\{5,d(x,y)\}$ for all $x,y \in X$
*$\rho_3 (x,y)$=max$\{d(x,y),d'(x,y)\}$ for all $x,y \in X... | Option 1 is false. For a counterexample, take $X = \mathbb R$ and let $d = d'$ be the usual metric $d(x, y) = |x - y|$. Then observe that:
$$
\rho(4, 3) + \rho(3, 2)
= |4 - 3|^2 + |3 - 2|^2
= 2
< 4
= |4 - 2|^2
= \rho(4, 2)
$$
Option 2 is false, since $\rho(7, 7) = 5$, instead of $0$.
Option 4 is false. For a counterexa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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The product of two roots of $x^4-18x^3+kx^2+200x-1984$ is $-32$; what is $k$?
In the quartic equation, $x^4-18x^3+kx^2+200x-1984$, The product of two roots is $-32$. What is the value of $k$?
This is my effort: Without losing generality, let's say $ab=-32$. Then since $abcd=-1984$, $cd=62$. $bcd+acd+abd+abc=ab(c+d)+c... | Good attempt!
$$94X\color{red}-576=-200$$
$$X=\frac{376}{94}=4$$
That is now, we know $a+b=4$ and $c+d=18-4=14$.
Also $ab=-32$ and $cd=62$.
While we can solve for the roots, we can observe that we can make use of some tricks:
\begin{align}k&=ab+ac+ad+bc+bd+cd \\&=(ab+cd) + (ac+ad+bc+bd)
\\
&=(ab+cd) + (a+b)(c+d)\\&=(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that: $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+5\ge (a+b)(b+c)(c+a)$ Let $a,b,c>0$ satisfy $a+b+c=3$
Prove that: $$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+5\ge (a+b)(b+c)(c+a)$$I have my solution, we need to prove:
$$3\sqrt[3]{a}+3\sqrt[3]{b}+3\sqrt[3]{c}+15\ge 3(a+b)(b+c)(c+a)$$
We know that: $$3(a+b)(b+c)(c+a)=(a+b+c)^... | Another way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $v^2\leq u^2=1$ and by Schur $w^3\geq4uv^2-3u^3=4v^2-3$.
Id est, by AM-GM and C-S we obtain:
$$5-\prod_{cyc}(a+b)+\sum_{cyc}\sqrt[3]a=5-(9uv^2-w^3)+\sum_{cyc}\frac{3a}{3\sqrt[3]{a^2}}\geq$$
$$\geq5-9v^2+4v^2-3+3\sum_{cyc}\frac{a}{2a+1}=2-5v^2+3\sum_{cyc}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Eliminating $x,y,z$ from given set of equalities
Given that: $x-2y+z=a$, $x^2-2y^2+z^2=b$, $x^3-2y^3+z^3=c$ and $\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=0$. Eliminate $x,y,z$ from given set of equations. [Hint: Use $x^2+z^2=(x+z)^2-2xz$ and $x^3+z^3=(x+z)(x^2+z^2)-xz(x+z)$]
Now using hints as given, I can solve $y$ in te... | We leave it to the OP to show that if $a = 0$ then $b = 0$ and $c =0$ and $x=y=z$.
Else
$\tag 1 \Large{y = \frac{b-a^2}{3a}}$
and by the 4th equation $b \ne a^2$.
and since
$x^3+z^3=(x+z)(x^2+z^2)-xz(x+z)= (x+z)\big((x+z)^2-2xz\big)-xz(x+z)$
and
$xz = \frac{1}{2} (x+z)y$
we have
$x^3+z^3= (x+z)\big((x+z)^2-(x+z)y\big)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to make use of angle sum and difference identities to find the value of sine and cosine?
Calculate: $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$
What is the easiest way to find $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$ (without a calculator)? If I know that $\fra... | 1 equation and 2 unknows is, generally, an overdetermined system.
In this case, pick any solution for $\dfrac{5\pi}{12} = \dfrac{p_1\pi}{q_1} + \dfrac{p_2\pi}{q_2}$, and sum $\pm \dfrac{p_3\pi}{q_3}$ in a smart way
$$\dfrac{5\pi}{12} = \left( \dfrac{p_1\pi}{q_1} + \dfrac{p_3\pi}{q_3} \right) + \left( \dfrac{p_2\pi}{q_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find the value of $\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)$ The following question is taken from the practice set of JEE Main exam.
Find the value of $$\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+..... | $R=\lim_{ n \to \infty} \left(\frac{1^2+1}{-n^3}+\frac{2^2+2}{-n^3}+\frac{3^2+3}{-n^3}+...+\frac{n^2+n}{-n^3}\right)=\lim_{ n \to \infty} \left(\frac{n(n+1)(2n+1)+3n(n+1)}{-6n^3}\right)=\frac{-1}{3}$
$P=\lim_{ n \to \infty} \left(\frac{1^2+1}{n-n^3}+\frac{2^2+2}{n-n^3}+\frac{3^2+3}{n-n^3}+...+\frac{n^2+n}{n-n^3}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Probability two blocks have exactly 2 out of 4 properties the same?
I have 120 blocks. Each block is one of 2 different materials, 3 different colors, 4 different sizes, and 5 different shapes. No two blocks have exactly the same of all four properties. I take two blocks at random. What is the probability the two bloc... | Yes.
Since no two blocks have all four properties identical, and $2\cdot 3\cdot 4\cdot 5=120$, there are exactly enough blocks for every combination of properties; thus each arrangement has identical probability. Hence you can use this method.
The denominator counts the ways to select 2 among these 120 combinations, o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $f(x) = x^2 - 2x + 5$ what is $f^{-1}(x)$?
If $f(x) = x^2 - 2x + 5$ what is $f^{-1}(x)$?
with the condition : $x > 1$
I solved this problem in this way:
$f(x) = x^2 - 2x + 5 -1 +1 \longrightarrow (x-2)^2 + 1 = f(x) $
$f^{-1}(x) = \sqrt{x-1} + 2$
But I saw someone else solved it in this way:
$f(x) = x^2 - 2x +... | After the edit, in the given interval $f(x)$ is bijective and hence invertible. But your method is incorrect because:
$$(x-2)^2+1=x^2-4x+5\neq f(x)$$
Your friend is right, but I believe in the second-last step you meant $f(x)=(x-1)^2+4$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by limit definition $\lim _{x\to 3}\left(\frac{x^2-9}{x^3-3x^2-2x+6}\right)=\frac{6}{7}$. Prove, using the limit definition, that
$\lim _{x\to 3}\left(\frac{x^2-9}{x^3-3x^2-2x+6}\right)=\frac{6}{7}$.
I tried to do it this way, but I can’t move forward. Can anyone help me how to continue?
By definition we have
$\f... | Note that\begin{align}\frac{x^2-9}{x^3-3x^2-2x+6}-\frac67&=\frac{-6 x^2+7 x+33}{7 \left(x^2-2\right)}\\&=\frac{-6x-11}{7(x^2-2)}(x-3).\end{align}So, take $\delta_1>0$ such that $|x-3|<\delta_1\implies|x-3|<1\iff2<x<4$. Then$$7(x^2-2)>7(2^2-2)=14\quad\text{and}\quad|-6x-11|=|6x+11|<35.$$ Therefore$$|x-3|<\delta_1\implie... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $k^7/7 + k^5/5 + 2k^3/3 - k/105$ is an integer I tried to prove this using induction.
Let $k=1$; then the equation gives
$$1/7 + 1/5 +2/3 – 1/105 = 105/105 = 1,$$
which is an integer. So it is true for $k=1$. Now let it be true for $n>k$. This gives
$$105|(15n^7 + 21n^5 + 70n^3 – n).$$
For $(n+1)$ we have
$$105|(... | The generating function is $$\sum_{k\ge 0} f(k)x^k=\frac{320}{(x-1)^3}+\frac{3672}{(x-1)^5}+\frac{1540}{(x-1)^4}+\frac{4584}{(x-1)^6}+\frac{1}{(x-1)}+\frac{2880}{(x-1)^7}+\frac{720}{(x-1)^8}+\frac{29}{(x-1)^2}$$ with $$f(k)=\frac12 k^7+\frac15 k^5+\frac23 k^3-\frac{1}{105}k$$. Since all the factors of this polynomial o... | {
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"url": "https://math.stackexchange.com/questions/4203071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}.$
Evaluate: $$\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}.$$
I'm learning limits for the first time and this is an exercise problem from my book. Here is my solution to the problem:
Let $S=1\cdot3+2\cdot4+\dots+n(n+2)\\ =(... | Another way using summation notation.
$\begin{array}\\
s(n)
&=\dfrac{\sum_{k=1}^n k(k+2)}{n^3}\\
&=\dfrac{\sum_{k=1}^n (k^2+2k)}{n^3}\\
&=\dfrac{\sum_{k=1}^n (k^2+2k+1-1)}{n^3}\\
&=\dfrac{\sum_{k=1}^n ((k+1)^2-1)}{n^3}\\
&=\dfrac{\sum_{k=1}^n (k+1)^2}{n^3}-\dfrac{\sum_{k=1}^n 1}{n^3}\\
&=\dfrac{\sum_{k=2}^{n+1} k^2}{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
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Computing $\sum_{k = 1}^{n} \dfrac{k}{2^k} $ in different ways I am trying to compute the following sum \begin{align} S = \sum_{k = 1}^{n} \dfrac{k}{2^k} \end{align}
I have computed this sum and found that
$$ \bbox[5px,border:2px solid red]
{
S = 2 - \dfrac{n+2}{2^{n}}
}
$$
For prooving this let's rewrite the sum in t... | Here is a generating function approach. We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series.
We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^n}\color{blue}{\frac{k}{2^k}}
&=[z^n]\sum_{q=0}^{\infty}\left(\sum_{k=0}^q\frac{k}{2^k}\right)z^q\tag{1}\\
&=[z^n]\frac{1}{1-z}\,\sum_{q=0}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Using Rearrangement Inequality .
Let $a,b,c\in\mathbf R^+$, such that $a+b+c=3$. Prove that $$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$
$Hint$ : Use Rearrangement Inequality
My Work :-$\\$
Without Loss of Generality let's assume that $0\le a\le b\le c$ then we can infer that $c^2\ge b^... | Since triples $(a^2,b^2,c^2)$ and $\left(\frac{1}{b+c},\frac{1}{a+c},\frac{1}{a+b}\right)$ have the same ordering,
by Rearrangement and C-S we obtain: $$\sum_{cyc}\frac{a^2}{b+c}\geq\frac{1}{3}\sum_{cyc}a^2\sum_{cyc}\frac{1}{b+c}\geq\frac{1}{3}\sum_{cyc}a^2\cdot\frac{(1+1+1)^2}{\sum\limits_{cyc}(b+c)}=\frac{a^2+b^2+c^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$(1+A)^n \ge (1+a_1)\cdot(1+a_2)\cdots(1+a_n) \ge (1+G)^n$ $A$ and $G$ are arithmetic mean and the geometric mean respectively of $n$ positive real numbers $a_1$,$a_2$,$\ldots$,$a_n$ . Prove that
*
*$(1+A)^n$$\ge$$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$$\ge$$(1+G)^n$
*if $k$$\gt$$0$, $(k+A)^n$$\ge$$(k+a_1)\cdot(k+a_2)\cdo... | Adding on to @notallwrong's answer:
For the case $n = 3$, you want to know how many times say $a_1$ appears. We have to make triples. Choose $a_1$. The remaining two can be chosen in $\binom{n-1}{2}$ ways.
For the case $n = 4$, you want to know how many times say $a_1$ appears. We have to make quadruples. Choose $a_1$.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For which value of $t \in \mathbb R$ the equation has exactly one solution : $x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}$ For which value of $t \in R $ the equation has exactly one solution : $x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}$
Here $t \neq n\pi, t \neq (2n+1)\frac{\pi}{... | $$x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}\\
\implies x=\frac{-\frac{2}{\sqrt{\cos t}} \pm\sqrt{\frac{4}{\cos t}
-4(1)\big(\frac{1}{\sin t}-2\sqrt{2}\big)}}{2}\\
=\frac{-1}{\sqrt{\cos t}}\pm \sqrt{\sec {t}-\csc{t} +2\sqrt{2}}$$
Now, the problem is finding what values of $t$ yield
$\quad
\sec t - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this complicated trigonometric equation for the solution of this triangle.
In $\triangle ABC$, if $AB=AC$ and internal bisector of angle $B$ meets $AC$ at $D$ such that $BD+AD=BC=4$, then what is $R$ (circumradius)?
My Approach:- I first drew the diagram and considered $\angle ABD=\angle DBC=\theta$ and... | Indeed you are proceeding correctly. The equation can be solved as follows:
$$\sin 2\theta \sin \theta=\sin 4\theta(\sin 3\theta-\sin 2\theta) {\tag 1}$$
Now $\sin 3\theta-\sin 2\theta=2\cos \frac {5\theta}{2} \sin \frac {\theta}{2}$.
Also $\sin \theta=2\cos \frac {\theta}{2} \sin \frac {\theta}{2}$, and $\sin 4\theta=... | {
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"source": "stackexchange",
"question_score": "3",
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Question on partial answer of the generalized Catalan's conjecture (case $n =2$) Edited question.
Let $n,m\in\Bbb N$ be integers greater or equal to $2$. If $3\leq m<n$ then there is no $m,n$ such that $m^n -n^m =2$. If $n<m$ and $m$ is prime then there is no $m,n$ such that $m^n-n^m =2$. If $m = 2$ then there is no $... | This is a particular case of $m^n = n^m+2$ with $m,n\geq 2$. First note if both numbers are even then $m^n$ and $n^m$ are multiples of $4$, so we have $m,n \geq 3$. Let $n = m+k$ with $k\geq 1$. ( You can see that if $m$ is the larger one it wont work because the left side will be smaller than the right).
Taking logs g... | {
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"question_score": "3",
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"answer_id": 1
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Integration and summation: prove $399< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{40000}}<400$ is false. I need help proving that the following statement is false:
$$
399 < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}}
+ \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{40000}} < 400.
$$
I t... | Your sum is equal to
$$1 + \sum_{n=2}^{40000} \frac{1}{\sqrt{n}} < 1 + \int_1^{40000} \frac{1}{\sqrt{x}} \; dx = 1+ \left.2\sqrt{x}\right|_{x=1}^{40000} = 1+(2\sqrt{40000} - 2) = 399.$$
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve $\csc^2(x)=\csc^2(\frac x 2)+\csc^2 (\frac x 4)$
Solve $\displaystyle\csc^2(x)=\csc^2\left(\frac x 2\right)+\csc^2\left(\frac x 4\right)$ , $\forall
x\in[0,\pi/2] $.
My attempt:
$$\frac{1}{\sin^2\theta}=\frac{1}{\sin^2\frac \theta 2}+\frac{1}{\sin^2\frac \theta 4}$$
then $$1=\frac{\sin^2\theta}{\sin^2\frac \t... | Starting from your last equation
$$1=\left(4+16 \cos ^2\left(\frac{\theta }{4}\right)\right) \cos ^2\left(\frac{\theta
}{2}\right)$$
$$\cos ^2\left(\frac{\theta }{4}\right)=x \implies \theta=4 \cos ^{-1}\left(\sqrt{x}\right)\implies 1=4 (1-2 x)^2 (4 x+1)$$ So, we need to solve the cubic equation
$$64 x^3-48 x^2+3=0$... | {
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Is my logic valid in showing that any two consecutive terms of the Fibonacci sequence are coprime Assume terms $a_n$ and $a_{n+1}$ share some common factor $x$ so that $a_n = xm$ for some integer $m$ and $a_{n+1} = xk$ for some integer $k$.
Since $a_{n-1}$ = $a_{n+1} - a_n = xk - xm = x(k-m)$, $a_{n-1}$ is a multiple o... | Alternative:
One has $\begin{pmatrix} 1 & 1 \\ 1 & 0\end{pmatrix}^n = \begin{pmatrix} f_{n+1} & f_{n} \\ f_{n} & f_{n-1} \end{pmatrix}$ and so taking the determinant $(-1)^n = f_{n-1}f_{n+1} - f_n^2$. This shows $1$ is a linear combination of $f_{n+1}$ and $f_n$ and they are therefore coprime.
| {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Finding $a+b+c+d$, where $ab+c+d=15$, $bc+d+a=24$, $cd+a+b=42$, $da+b+c=13$ Let $a,b,c,d \in \mathbb{R}$. Consider the following constraints:
\begin{cases} ab+c+d=15 \\ bc+d+a=24 \\ cd+a+b=42 \\da+b+c=13 \end{cases}
Calculate the value of $a+b+c+d$.
It is easy to use the Gröbner basis to get the value:
\begin{cases}
10... | Playing around with the four LHSs, I tried to obtain polynomials with some symmetry. I first noted that
$$\left((ab+c+d)+(cd+a+b)\right)-\left((bc+d+a)+(da+b+c)\right)=(b-d)(a-c)$$
then after replacing the $+$ sign with $\cdot$, we get
$$(ab+c+d)\cdot(cd+a+b)-(bc+d+a)\cdot(da+b+c)=(b-d)(a-c)(a+b+d+c-1).$$
Putting all t... | {
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"source": "stackexchange",
"question_score": "7",
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Integrate via substitution: $x^2\sqrt{x^2+1}\;dx$ I need to integrate the following using substitution:
$$
\int x^2\sqrt{x^2+1}\;dx
$$
My textbook has a similar example:
$$
\int \sqrt{x^2+1}\;x^5\;dx
$$
They integrate by splitting $x^5$ into $x^4\cdot x$ and then substituting with $u=x^2+1$:
$$
\int \sqrt{x^2+1}\;x^4\c... | Proceeding with your method
$\begin{align} \int x^2\sqrt{x^2+1}\;dx &=\int \sqrt{x^2+1}\;x\cdot x\;dx\\
&=\frac1{2}\int \sqrt{u}\;\sqrt{u-1}\;du\;\left( \text{ let $\begin{align} u&=x^2+1 \\ du&=2xdx\end{align}$}\right)\\
&=\frac1{2}\int \sqrt{u^2-u} \;du\\
&=\frac1{2} \int \sqrt{\left(u-\frac1{2}\right)^2-\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
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Limit of a function in terms of exponents
Let $f(n)$ denote the $n$th term of the sequence $2, 5, 10, 17, 26,\cdots$ and $g(n)$ denote the $n$th term of the
sequence $2, 6, 12, 20,30,\cdots$.
Let $F(n)$ and $G(n)$ denote respectively the sum of n terms of the above sequences.
$$\lim_{n\to\infty}\left(\frac{F(n)}{G(n)... | The expressions you got for $F(n),G(n),f(n),g(n)$ are correct.
Now,
$\displaystyle \begin{align} \lim_{n \to\infty}\left(\frac{F(n)}{G(n)}\right)^n-\left(\frac{f(n)}{g(n)}\right)^n &= \lim_{n \to\infty}\left(\frac1{2} \frac{2n^2+3n+7}{n^2+3n+2}\right)^n-\left(\frac{n^2+1}{n^2+n}\right)^n\\
&= \lim_{n \to\infty}\left(\f... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Two ways of computing $\lim_{a \to 0^+} \left( \frac{1}{(x+ia)^2} - \frac{1}{(x-ia)^2} \right)$ gives contradictory result In this question, we assume that $\lim_{a\to 0^+}$ is implicit.
It is well-known that
$$\frac{1}{x-ia} - \frac{1}{x+ia} = 2\pi i \delta(x),$$
where $\delta(x)$ is the Dirac delta distribution. Diff... | Your observation is not wrong, but you have to keep in mind that you take the limit $a\rightarrow 0$ after all. I addition your product is in indeterminate form, since plugging in $x=0$ gives something of the form $\infty\cdot 0$.
If you really want to follow the route of factorization, rather than considering $$\frac{... | {
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"source": "stackexchange",
"question_score": "2",
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Given the function $f(x,y)=\frac{5xy^2}{x^2+y^2}$, argue analytically, if the discontinuity of $f$ at $(0,0)$ is avoidable or unavoidable. It is clear that $f(0,0)$ does not exist, and therefore the function is discontinuous. Now to know if the discontinuity is avoidable or inevitable, we must see that the limit at tha... | Your first sentence is wrong; since $f$ is undefined at $(0,0)$, it is neither continuous nor discontinuous at that point.
Otherwise, you proved correctly that $\lim_{(x,y)\to(0,0)}f(x,y)=0$ and that therefore you can extend $f$ in such a way that it becomes continuous at $(0,0)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Summation of square root fractional part Is there a way to calculate the following sum for large $c>K$, closed-form or approximation?
$$\sum_{n=1}^{c}\left\{\sqrt{K+n^2}\right\}$$Where K and n are positive integers and $\{\}$ indicates the fractional part.
A similar question: Fractional part summation
| The given sum can be written as
$$
\Sigma = \Sigma_1 - \Sigma_2
$$
where
$$
\Sigma_1 = \sum_{n=1}^{c} \sqrt{n^2 + K}
$$
and
$$
\Sigma_2 = \sum_{n=1}^{c} \lfloor\sqrt{n^2 + K}\rfloor.
$$
Now using $H_n = \log n + \gamma + \mathcal{O}(1/n)$ where $H_n$ is the $n$-th harmonic number,
\begin{align*}
\sum_{n=1}^{c} \sqrt{n^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Complex number related problem Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______
My solution is as follow
${z_1} = 2{e^{i{\theta _1}}};{z_2} = 2{e^{i{\theta _2}}};{z_3} = 2{e^{i{\theta _3}}}$ &... | Consider $$u=z_2/z_1,v=z_3/z_1$$ and then we have $$|u-1|=|v-1|\tag{1}$$ and $$u\neq v, |u|=|v|=1\tag{2}$$ and $$|1+u+v|=1\tag{3}$$ Then from first equation we get $$(u-1)(\bar{u} - 1)=(v-1)(\bar{v}-1)$$ or $$u+\bar{u}=v+\bar{v}\tag{4}$$ This means that $u, v$ have same real part and using $(2)$ we can see that they ar... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Pairs of integers $ (x,m)$ for which $\sqrt[3]{\sqrt[3]{x-2}+m}+\sqrt[3]{-\sqrt[3]{x-2}+m}=2$ hold?
Find all pairs of integers $(x,m)$ for which $$\sqrt[3]{\sqrt[3]{x-2}+m}+\sqrt[3]{-\sqrt[3]{x-2}+m}=2$$ hold.
I have used this property :
Property:
if $$a+b+c=0 \implies a^3+b^3+c^3=3abc, $$ I come up to the followin... | Generally solutions $(x,m)$ can be split into three cases:
*
*$(x,m)=\left(\frac{1}{243} \left(486-\sqrt{3} (m-1)^{3/2} (m+8)^3\right),m\right)$
*$(x,m)=\left(\frac{1}{243} \left(486+\sqrt{3} (m+8)^3 (m-1)^{3/2}\right)),m\right)$
*$(x,m)=(2,1)$
In the first two cases $m\ge2$ and we choose $m$ such that the resultin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Calculating $\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$
$$\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$$
I tried factoring as $${(a^2-x^2)(a^2+x^2)}$$ $$\text{and}$$ $$(x^2+a^2+ax)(x^2+a^2-ax)$$ but didn't get much after this. I sew that the expression is symmetric in a,x but I don't know how to use that.
| Hint:
First,
$$\dfrac{a^4 - x^4}{x^4 + a^2x^2 + a^4} = \dfrac{a^2x^2 + 2a^4}{x^4 + a^2x^2 + a^4} - 1 = a^2\dfrac{x^2 + 2a^2}{x^4 + a^2x^2 + a^4} - 1$$
Second,
$$\dfrac{x^2 + 2a^2}{x^4 + a^2x^2 + a^4} = \dfrac{x^2 + 2a^2}{\left(x^2 -ax + a^2\right)\left(x^2 + ax + a^2\right)}$$
Now, perform partial-fraction decompositio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$ Let $a,b,c>0$:
Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$
My solution:
We have: $\left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y} \\\... | Another way.
$$\frac{1}{3}-\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)}=\sum_{cyc}\left(\frac{1}{9}-\frac{a^2}{(2a+b)(2a+c)}\right)=$$
$$=\frac{1}{9}\sum_{cyc}\frac{2ab+2ac+bc-5a^2}{(2a+b)(2a+c)}=\frac{1}{18}\sum_{cyc}\frac{(c-a)(5a+b)-(a-b)(5a+c)}{(2a+b)(2a+c)}=$$
$$=\frac{1}{18}\sum_{cyc}(a-b)\left(\frac{5b+c}{(2b+c)(2b+a)}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Calculating value of $\delta$ for a function $y=\frac1{(1-x)^2}$ unbounded at $x\rightarrow 1$ The function $y=\frac 1{(1-x)^2}$ is unbounded at $x\rightarrow 1$.
How can I calculate the value of $\delta$ such that $y> 10^6$ if $\left | x-1 \right |< \delta$ ?
Taking $y=f(x)=10^6$, putting in ${y=\frac1{(1-x)^2}}$, we ... | We have that for $A,B>0$ (i.e. dividing both sides by $AB$)
$$A>B \iff \frac 1 A < \frac 1 B$$
therefore
$$\frac{1}{(1-x)^2}>10^6 \iff (1-x)^2<\frac1{10^6}=10^{-6}$$
form which, using that $\sqrt{A^2}=|A|$ and (taking the square roots both sides)
$$A^2<B \iff |A|<\sqrt B$$
we find
$$(1-x)^2<10^{-6} \iff |1-x|<10^{-3}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Let $x^3+\frac{1}{x^3}$ and $x^4+\frac{1}{x^4}$ are rational numbers. Show that $x+\frac{1}{x}$ is rational.
$x^3+\dfrac{1}{x^3}$ and $x^4+\dfrac{1}{x^4}$ are rational numbers. Prove that $x+\dfrac{1}{x}$ is rational number.
My solution:
$x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-1\right... | I wanted to write this answer with the thought that it might be cleaner than my previous answer.
We have,
$$\begin{align}&\begin{cases}x+\frac 1x=u,\thinspace u \in \mathbb R\\
x^2+\frac{1}{x^2}=v,\thinspace v\in\mathbb R\\
x^3+\frac{1}{x^3}=p,\thinspace p\in\mathbb Q\\
x^4+\frac{1}{x^4}=q,\thinspace q\in\mathbb Q\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 2
} |
Proving divisibility by $9$ for $10^n + 3 \times 4^{n+2} + 5$. I am trying to prove that for all $n$, $9$ divides $10^n + 3 \times 4^{n+2} + 5$. I first tried induction on $n$, but couldn't get anywhere with the induction step. I then tried to use modular arithmetic. The furthest I could get was:
As $10 \equiv 1 \mod ... | We can construct one of the possible proofs as follows:
$$\begin{align}&10^n + 3 \times 4^{n+2} + 5 \mod 9
\\
\iff &10^n + 3 \times 4^{n+2} -4\mod 9\\
\iff &10^n-1^n+3 \times 4^{n+2} -3\mod 9\\
\iff &3\left(4^{n+2}-1\right)\mod 9\\ \iff
&4^{n+2}-1^{n+2} \mod 3 =0.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Why is this approach wrong?
If $\alpha$ and $\beta$ are the roots of $x^{2} - 4 x - 3$, then find $$\frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}$$
Solution
This is the approach that gives wrong answer.
\begin{array}{l}
x^{2}-4 x-3=0 \\
(x-4)^{2}=19 \\
So, {(\alpha-4)^{2}} {\&(\beta-4)^{2}}=19 \\
\text { Hence, }... | You did not factorise $x^2-4x-3$ correctly. Actually, $x^2-4x-3=(x-2)^2-4-3=(x-2)^2-7$. We find that $(x-2)^2=7$, and so $x-2=\pm\sqrt{7}$. Therefore, $x-4=-2\pm\sqrt{7}$, and $(x-4)^2=11\pm4\sqrt{7}$. Hence,
$$
\frac{1}{(\alpha-4)^2}+\frac{1}{(\beta-4)^2}=\frac{1}{11+4\sqrt{7}}+\frac{1}{11-4\sqrt{7}}=\frac{11-4\sqrt{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4239918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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What happens to $w=\frac{x^2y^3}{z^4}$ , if $x,y,z$ increase by $\%1$ , $\%2$ , $\%3$ respectively? If $x,y,z$ increase by $\%1$ , $\%2$ , $\%3$ respectively , then $w=\dfrac{x^2y^3}{z^4}$ ............. approximately.
$1)\ \%\ 3\text{ decrease}$
$2)\ \%\ 4\text{ decrease}$
$3)\ \%\ 3\text{ increase}$
$4)\ \%\ 4\text{ ... | You can use partial elasticities, also know as condition numbers:
\begin{align*}
\varepsilon_w \approx & \frac{x w'_x}{w} \varepsilon_x+\frac{y w'_y}{w} \varepsilon_y + \frac{z w'_z}{w} \varepsilon_z\\
= & 2\varepsilon_x + 3 \varepsilon_y - 4 \varepsilon_z= 2 + 3\times 2-4 \times 3 = -4
\end{align*}
So, the answer is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Largest possible side of a triangle when $\cos(3P)+ \cos(3Q)+ \cos(3R) = 1$.
In triangle $PQR$, $\cos(3P)+ \cos(3Q)+ \cos(3R) = 1$. Two sides of the
triangle have lengths of $15 cm$ and $18 cm$. If the length of the
third side of the triangle PQR is $\sqrt{m}$ cm, then the largest
possible value that $'m'$ can take is... | Let $p$, $q$ and $r$ be sides-lengths of the triangle.
Thus, $$\sum_{cyc}\left(4\left(\frac{p^2+q^2-r^2}{2pq}\right)^3-3\cdot\frac{p^2+q^2-r^2}{2pq}\right)=1$$ or
$$\prod_{cyc}((p+q-r)(p^2+q^2+pq-r^2))=0,$$ which says that one of the measured angles of the triangle is equal to $120^{\circ}$.
Can you end it now?
I got $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding smooth behaviour of infinite sum Define
$$E(z) = \sum_{n,m=-\infty}^\infty \frac{z^2}{((n^2 + m^2)z^2 + 1)^{3/2}} = \sum_{k = 0}^\infty \frac{r_2(k) z^2}{(kz^2 + 1)^{3/2}} \text{ for } z \neq 0$$
$$E(0) = \lim_{z \to 0} E(z) = 2 \pi$$
where $r_2(k)$ is the number of ways of writing $k$ as a sum of two squares o... | §1. Basic results
We have to investigate the sum
$$E_s(z) = \sum_{n,m=-\infty}^\infty \frac{z^2}{((n^2 + m^2)z^2 + 1)^{3/2}}\tag{1.1}$$
Writing
$$\frac{z^2}{((n^2 + m^2)z^2 + 1)^{3/2}}=\int_{0}^{\infty} z^2 \sqrt{\frac{4 t}{\pi} }e^{-t(z^2 (m^2 + n^2) + 1)}\,dt\tag{1.2}$$
and doing the sums under the integral the doubl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Integration of $\sin^2x$ without using double angle identity of $\cos 2x$ I want to integrate $\sin^2x$ without using the double angle identity of $\cos2x$.
Here's what I tried:
$$
\int \sin^2x dx
= \int \sin x \tan x \cos x dx
$$
Let $u = \sin x$ => $du = \cos x dx$
And if $u = \sin x$, $\tan x = \frac{u}{\sqrt{1-u^2}... | If $t=\sqrt{1-u^2}$, then$$\mathrm dt=-\frac u{\sqrt{1-u^2}}\,\mathrm du\quad\text{and}\quad u^2=1-t^2.$$So,\begin{align}\int\frac{u^2}{\sqrt{1-u^2}}\,\mathrm du&=-\int u\frac{-u}{\sqrt{1-u^2}}\,\mathrm du\\&=-\int\sqrt{1-t^2}\,\mathrm dt\\&=-\frac12\left(t\sqrt{1-t^2}+\arcsin (t)\right).\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Solve for $\sqrt{x}+\sqrt[4]{y}=\sqrt{p}$ with $x$, $y$ are positive integers and $p$ is a prime number For that given problem, I have constructed a solution to it:
Squaring both sides, we get $x+\sqrt{y}+2\sqrt[4]{x^2y}=p$. Suppose $\sqrt{y}$ and $\sqrt[4]{x^2y}$ are integers, we can let $y=y'^2$, $y'$ being a positi... | See that $\sqrt[4]y = (\sqrt p - \sqrt x)$ so
$$ y = (\sqrt p -\sqrt x)^4 = p^{2} - 4 \, p \sqrt{px} + 6 \, p x - 4x \, \sqrt{px} + x^{2} =p^2 +6px - \sqrt{px}(4x+4p) +x^2 $$ So you can easily see that $\sqrt{xp}$ is an integer.
Now $\sqrt y = (\sqrt p -\sqrt x)^2 = p + x - 2\sqrt{px}$ so $\sqrt y$ is an integer.
Als... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Q: Epsilon-delta proof for $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$ In proving $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$, I have done this working:
Let $ϵ>0$.
$0<|x-1|< \Rightarrow|\frac{x}{(x^2+1)}-\frac{1}{2}|<$
$ |\frac{x}{(x^2+1)}-\frac{1}{2}|< = |\frac{2x-x^2+1}{2(x^2+1)}| = |\frac{-2(x-1)^2}{x(x-1)^2+4x}| = |\... | You have\begin{align}\frac x{x^2+1}-\frac12&=\frac{-x^2+2x-1}{2x^2+2}\\&=\frac{-(x-1)^2}{2x^2+2}.\end{align}Now, you alwas have $2x^2+2\geqslant2>1$. So$$\left|\frac{-(x-1)^2}{2x^2+2}\right|<(x-1)^2\tag1$$and therefore, given $\varepsilon>0$, if you take $|x-1|<\sqrt{\varepsilon}$, it will follow from $(1)$ that$$\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to show that $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$ I tried like this:
Let $y=a^{2x}-2\Rightarrow a^{2x}=y+2\Rightarrow 2x\ln a=\ln\left(y+2\right)\Rightarrow x=\dfrac{\ln\left(y+2\right)}{2\ln a}$
Also if $x\longrightarrow0,$ then $y\longrightarrow a^{2(0)}-2=-1.$
But we I put each and every t... | The secret here is to express $\dfrac{a^{2x}-2}{x^x}$ a function of the $\dfrac{e^{y}-1}{y}$, with $y=(2\cdot \ln a)\cdot x$, and function the $x\cdot \ln x$.
Note that
\begin{align}
\dfrac{a^{2x}-2}{x^x}
=&
\dfrac{a^{2x}-1}{x^x}-\dfrac{1}{x^x}
\\
=&
\dfrac{(e^{\ln a})^{2x}-1}{2x}\cdot\dfrac{2x}{x^x}-\dfrac{1}{x^x}
\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Does Diophantine equation $1+n+n^2+\dots+n^k=2m^2$ have a solution for $n,k \geq 2$? When studying properties of perfect numbers (specifically this post), I ran into the Diophantine equation
$$
1+n+n^2+\dots+n^k=2m^2, n\geq 2, k \geq 2.
$$
Searching in range $n \leq 10^6$, $k \leq 10^2$ yields no solution. So I wonder ... | Here is an elementary proof that $1+n+n^2+n^3=2m^2$ has only integer solution $(n,m)=(-1,0)$. This is the minimal case referred in the general proof and this post gives an alternative without using the elliptic curves.
Since $(n+1)(n^2+1)=2m^2$ and gcd of the expressions on the left is $2$ (by parity argument $n$ must ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Solving two equations in polar coordinates Problem:
Find the points of intersection of the following two curves.
\begin{align*}
r &= 1 - \cos \theta \\
r^2 &= \cos \theta
\end{align*}
Answer:
Note: While I do not show a plot, I did produce one. It indicated that there were only two solutions.
Now we solve for the poin... | There are indeed only two valid solutions (up to $2\pi$ periodicity). Note that $|\cos(\theta)|\leq 1$ so we discard the case $\cos(\theta)=\frac{3+\sqrt{5}}{2}$. Then $\cos(-x)=\cos(x)$ does not introduce new solutions, since we have
$$\cos(\theta)=\frac{3-\sqrt{5}}{2}\implies\theta=\pm\cos^{-1}\left(\frac{3-\sqrt{5}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Converting sets into intervals with variables I am trying to convert the set to interval form where center is $c$ and the radius is $r$. I have not had any problem doing the first example but I cannot find a way to get the $c$ value for the second question. How would I be able to solve the $c$ for the second question a... | It is ideal to solve these problems directly.
First problem
Note that for the first problem, you have to find all $x\in\mathbb{R}$ such that
$$ |x^2 -1| > 2 \tag{1}$$
Let $x\in\mathbb{R}$ be a real number that satisfies (1). By definition of the modulus sign, this condition is equivalent to the statement that
$$
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to color a cube with exactly 6 colors using Polya enumeration theory I have seen in other questions (Painting the faces of a cube with distinct colours) , and found for myself there are 30 colorings of the cube with exactly 6 colors. My issue is when I use Polya enumeration theory I take the number of colorings up ... | To find the number of ways to color the faces of a cube with exactly $6$ colors using Polya's Enumeration Theorem:
First, as you already know, the cycle index of the symmetries of the face permutations of a cube is
$$Z = \frac{1}{24}(x_1^6 + 6x_1^2x_4+3x_1^2x_2^2 + 8x_3^2 + 6x_2^3)$$
The figure inventory for $6$ colors... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find value of $x^{5} + y^{5} + z^{5}$ given the values of $x + y + z$, $x^{2} + y^{2} + z^{2}$ and $x^{3} + y^{3} + z^{3}$ If$$x+y+z=1$$
$$x^2+y^2+z^2=2$$
$$x^3+y^3+z^3=3$$
Then find the value of $$x^5+y^5+z^5$$
Is there any simple way to solve this problem ? I have tried all my tricks tried to multiply two equations ,... | This form of equations can be solved systematically using Newton's identities.
The problem at hand is a special case of $3$ variables. Let $e_1,e_2,e_3$ be the three elementary symmetric polynomial associated with $x,y,z$:
$$\begin{align}
e_1 &= x + y + z\\
e_2 &= xy+ yz + zx\\
e_3 &= xyz
\end{align}$$
This is equivale... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Probability of One Event Less Than Probability of Second Event I am having a bit of trouble proving some cases when one probability is smaller than the other probability for all positive integers $a, b$, so some suggestions would be appreciated.
Here is the problem:
For all $a, b \in \mathbb{Z}^+$, if $P(A) = \dfrac{a... | Incomplete solution:
$$
P(A)=P(B)-\left[1-P(B)\right]\cdot\frac{a+b}{a^{2}+2ab+b^{2}-a-b}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving or estimating $\int_{-\infty}^\infty \frac{\exp(-a/(x^2+b))}{x^2+c}dx$ Can the integral
$$\int_{-\infty}^\infty \frac{\exp(-a/(x^2+b))}{x^2+c}dx$$
be made explicit ($a,b,c>0$)? I'm also asking those of you who have access to CAS which can solve it.
In case it can't, is there a very good upper bound? I need a be... | Hint:
$$\begin{align}\int_{-\infty}^\infty\dfrac{e^{-\frac{a}{x^2+b}}}{x^2+c}~dx&=2\int_0^\infty\dfrac{e^{-\frac{a}{x^2+b}}}{x^2+c}~dx
\\&=2\int_0^\frac{\pi}{2}\dfrac{e^{-\frac{a}{b\tan^2x+b}}}{b\tan^2x+c}~d(\sqrt b\tan x)
\\&=2\sqrt b\int_0^\frac{\pi}{2}\dfrac{e^{-\frac{a}{b\sec^2x}}\sec^2x}{b\sec^2x+c-b}~dx
\\&=2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Find the sequence $a_k$ for generating function $\left(\frac{1-x^3}{1-x}\right)^n$ We know that $\frac{1}{(1-x)^n} = \sum_{k=0}^\infty \binom{k+n-1}{n-1} x^k$
I also worked out $(1-x^3)^n$ using the binomial theorem and got $(1-x^3)^n = \sum_{i=0}^\infty (-1)^i \binom{n}{n-i} x^{3i}$
I'm not sure what to do with these ... | Hint:
\begin{align*}
\left(\frac{1-x^3}{1-x}\right)^n=\left(\frac{(1-x)\left(1+x+x^2\right)}{1-x}\right)^n=\left(1+x+x^2\right)^n
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What did I do wrong in my proof of $4\nmid (n^2+3)\implies 2\nmid (n^4-3)$? A homework problem is asking me to prove the following statement:
For any $n\in\mathbb{Z}$, $2\mid (n^4-3)$ if and only if $4\mid (n^2+3)$
Quick note: we're not allowed to use the theory of congruences in our solution.
I figured the best way ... | You already noted that if $n^2+3= 4k+2$ then it's a problem. So we should show that $r=2$ is not possible.
Now that $2|n^2+3\implies n$ is odd. But odd squares are of the form $4l+1.$
Hence $n^2+3=4l+4.$
So if $2|n^2+3 $ then $4|n^2+3$ too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to calculate arccos of complex number I want to know what can I do to calculate the acos of a complex number. I looked at this answer on this site but that only work for imaginary numbers.
Given the complex number:
$$ \arccos(2 + 3i) $$
How can I calculate it?
Thanks in advance!
| $z=\arccos(2+3i)$
$2+3i=\cos z=(e^{iz}+e^{-iz})/2$; $e^{iz}+e^{-iz}=4+6i$; $u^2-(4+6i)u+1=0$ where $u=e^{iz}$; $$u={4+6i\pm\sqrt{(4+6i)^2-4}\over2}={4+6i\pm\sqrt{-24+48i}\over2}=2+3i\pm\sqrt{-6+12i}$$ so we need to find $\sqrt{-6+12i}$.
Let $\sqrt{-6+12i}=r+si$. Then $-6+12i=r^2-s^2+2rsi$, so $$r^2-s^2=-6,\qquad 2rs=12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4270736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the partial fraction: $\frac{x^2}{(x-1)^2(x-2)^2(x-3)}$ I'm trying to find the partial fraction for the following $\frac{x^2}{(x-1)^2(x-2)^2(x-3)}$ by the process of long-division, here's what I have tried:
Leaving $(x-3)$ as it is in the denominator and aiming for $(x-1)$ and $(x-2)$.
First step:
$\frac{1^2}{(x-1... | You've made some error in your simplification.
$$
\frac{x^2}{(x-1)^2(x-2)^2(x-3)} - \frac{1}{-2(x-1)^2} - \frac{4}{-(x-2)^2}$$
$$
\begin{align*}
&= \frac{x^2+\frac12(x-2)^2(x-3)+4(x-1)^2(x-3)}{(x-1)^2(x-2)^2(x-3)} \\
&= \frac{9x^3-45x^2+72x-36}{2(x-1)^2(x-2)^2(x-3)} \\[1ex]
&= \frac{9(x^3-5x^2+8x-4)}{2(x-1)^2(x-2)^2(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expanding $(1-i)^\frac{1}{3}$ using De Moivre's formula I want to rewrite $(1-i)^ \frac 1 3$ using de Moivre's formula.
I defined $z := 1 - i$, then $r_z = \sqrt{2}$ and
$1 = \sqrt2\cos\theta$ and $-1 = \sqrt2 \sin\theta \Rightarrow \theta = -\frac \pi 4$
So:
$$1 - i = \sqrt{2}(\cos(\frac \pi 4) + i\sin(\frac \pi 4))$$... | You have only take the principal root, indeed we have (there is also a typo for the angle in the $\sin$ term):
$$1 - i = \sqrt{2}\left(\cos\left(\frac \pi 4+2n\pi\right) + i\sin\left(-\frac \pi 4+2n\pi\right)\right)$$
and therefore the three possible roots are
$$2^ \frac 1 6 \left(\cos\left(\frac \pi {12}\right) + i\si... | {
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"source": "stackexchange",
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Show that $1998 < \sum_1^{10^6}\frac{1}{\sqrt{n}}<1999$ Show that $1998 < \sum_1^{10^6}\frac{1}{\sqrt{n}}<1999$ by first recalling the inequality:
$$2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1})$$ when $n=1, 2, 3...$
What I have tried to show using the inequality:
By the first step of induction to se... | You don't need induction here. We have
$$n(n+1)=n^2+n<n^2+n+\frac14=\left(n+\frac12\right)^2$$
Take square roots:
$$\sqrt n\sqrt{n+1}<n+\frac12$$
Multiply by $2/\sqrt n$ and rearrange:
$$2(\sqrt{n+1}-\sqrt n)<\frac{1}{\sqrt n}$$
Do the same with $n(n-1)$ to get the right-hand side of the inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $2\tan^{-1}\left(\sqrt{\frac{a}{b}}\tan\frac{x}{2}\right)=\sin^{-1}\frac{2\sqrt{ab}\sin x}{\left(b+a\right)+\left(b-a\right)\cos x}$ $$
2\tan ^{-1}\left(\sqrt{\frac{a}{b}}\tan \frac{x}{2}\right)=\sin ^{-1}\frac{2\sqrt{ab}\sin x}{\left(b+a\right)+\left(b-a\right)\cos x}
$$
I know within inverse of trigonometric ... | This should be the next step of your attempt: $$\sin ^{-1}\dfrac{\sqrt{ab}\sin x}{b\left(\frac{1+\cos x}{2}\right)+a\left(\frac{1-\cos x}2\right)}$$ using half-angle formula.
The result follows immediately.
| {
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"source": "stackexchange",
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How to solve $\int_{0}^{1} \frac{(x-1)\ln x}{(x+1)(x^2+2\cosh \alpha+1)} \,dx$ I have this unpleasant integral that appear in my last exam which i was not able solve
For $\alpha \in \mathbb{R}$. Evaluate $$I=\int_{0}^{1} \frac{(x-1)\ln x}{(x+1)(x^2+2\cosh \alpha+1)} \,dx$$
Anyway my attempt is that I try to do some p... | It is not so unpleasant. To make things easier (at least to me), consider
$$\frac{x-1}{(x+1)(x^2+k)}=\frac{x-1}{(x+1) (x-a) (x-b)}$$ and, as you did, using partial fraction decomposition, this becomes
$$\frac{a-1}{(a+1) (a-b) (x-a)}-\frac{2}{(a+1) (b+1) (x+1)}+\frac{b-1}{(b+1) (b-a) (x-b)}$$ One integration by parts ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to get the maximum value of the sum in two sides in two right triangles? The problem is as follows:
The figure from below shows two triangles intersected on point $E$. Assume $AE=3\,m$ and $ED=1\,m$. Find the maximum value of $AB+EC$
The given choices in my book are as follows:
$\begin{array}{ll}
1.&\sqrt{10}\,m... | Let $AM=CD$ and $AN=DE$
Since $\angle AMN=\angle CDE$,(cyclic quadrilateral), $\triangle AMN \cong \triangle DCE$, then, $AN=ED=1$
Let $AB=x$, and, $EC=MN=y$
Since $\triangle ABE \sim \triangle AMN$, (AAA Similarity), $BE=3y$
FIRST ALTERNATIVE SOLUTION:
$x^2+9y^2=9$, in $\triangle ABE$
$x=\sqrt{9-9y^2}$
To find max$(x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Children's Fruit Division How many ways can $11$ apples and $9$ pears be divided between 4 children so that each child receives five fruits? (Apples are identical. just like pears).
Solution: $f\left(x,y\right)=\left(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5\right)^4$
$f\left(x,y\right)=\left(\frac{x^6-y^6}{x-y}\right)^4$
$f\lef... | Yes that is correct. But a simpler solution exists given each kid must get equal number of fruits. We can first distribute $9$ pears (or $11$ apples) to four children such that none of them get more than $5$ pears. We then note that as each kid must have five fruits, once we distribute pears, the distributions of appl... | {
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Is there a natural number $n$ for which $\sqrt[n]{22-10\sqrt7}=1-\sqrt7$ Is there a natural number $n$ for which $$\sqrt[n]{22-10\sqrt7}=1-\sqrt7$$
My idea was to try to express $22-10\sqrt7$ as something to the power of $2$, but it didn't work $$22-10\sqrt7=22-2\times5\times\sqrt7$$ Since $5^2=25, \sqrt7^2=7$ and $25+... | To paraphrase the excellent answer given in the comments:
Because $\sqrt{7}$ is a pure quadratic surd, if there is such a natural number $n$, then also
$$\sqrt[n]{22+10\sqrt{7}}=1+\sqrt{7}.$$
As $(1+\sqrt{7})(1-\sqrt{7})=1^2-\sqrt{7}^2=-6$ it follows that
\begin{eqnarray*}
-6&=&(1+\sqrt{7})(1-\sqrt{7})\\
&=&\sqrt[n]{22... | {
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"timestamp": "2023-03-29T00:00:00",
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Possible division by zero in integration by u-substitution When I was solving the indefinite integral $\int x^2(x^3-7)^3\ dx$, my working was as follows:
Let $u=x^3-7$, then $dx=\frac{1}{3x^2}du$.
Hence, $\int x^2u^3\frac{1}{3x^2}\ du=\frac{1}{3}\int u^3\ du=\frac{u^4}{12}+C=\frac{(x^3-7)^4}{12}+C$.
However, during the... | You may avoid dividing by zero if instead of $$dx=\frac{1}{3x^2}du$$ consider $$ du = 3x^2 dx $$ and substitute $$ x^2dx = \frac {1}{3} du$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When does $\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=2mx+4$ have $2$ real solutions? $$\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=2mx+4$$
My solution:
I consider only real numbers, so $|x|\geq1$. After adding the fractions I get the following quadratic ... |
I consider only real numbers, so $|x|\geqslant1.$ After adding the fractions I get the following quadratic equation: $$4x^2-mx+1=0$$
Can you show us the steps of how you got to that equation? Because that is not the equation I got. In general, when calculating $$\frac{a+b}{a-b}+\frac{a-b}{a+b},$$
you have $$\begin{al... | {
"language": "en",
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"source": "stackexchange",
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Given triangle ABC with $\alpha=2\beta$ and $b, c, a$ is forming an arithmetic sequence. Find $\alpha, \beta, \gamma$. Given $\triangle ABC$ with $\alpha=2\beta$ and $b, c, a$ is forming an arithmetic sequence ($b$ is the $1$'st term, and $a$ is the last term), find $\alpha, \beta, \gamma$.
My attempts so far:
Let $c=... | From $b,c,a$ forming an arithmetic sequence,
$$c-b=a-c$$ $$2c=a+b$$ Applying sine rule, $$2\sin C=\sin A+\sin B$$
Using the given data $A=2B$, we can rewrite the above as, $$2\sin 3B=\sin2B+\sin B$$ $$\require{cancel}2\cdot \cancel{2\sin\frac{3B}{2}}\cos\frac{3B}2=\cancel{2\sin\frac{3B}{2}}\cos\frac B2\qquad\small{(\be... | {
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In equation involving absolute value one solution might get rejected, why is that? For example the equation $|2x-1|+3=4x$,
We can get two equations out of that, $2x-1=4x-3$ and $2x-1=-(4x-3)$. Solve the first one we get $x=1$ and for the second one $x=2/3$. But when you substitute 2/3 into the original equation, you ge... | The absolute value of a number cannot be negative.
Observe that $2x - 1 \geq 0 \implies x \geq \dfrac{1}{2}$.
Hence, $|2x - 1| = 2x - 1$ if $x \geq 1/2$. On the other hand, if $x < 1/2$, then $2x - 1 < 0$, so $|2x - 1| = -(2x - 1) = -2x + 1 > 0$. Therefore, we obtain the piecewise definition
$$
|2x - 1| = \begin{case... | {
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Compute the series $\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}$ Compute the series $$\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}.$$
How do I go about with the index notation, for example to arrange the series instead as $\sum_{n=1}^{\infty}a_n $?
I have tried to simplify the expression as:
$$\begin{align}&\sum_{n=3}^{\infty} \fr... | $$
\begin{align}
\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}&=\sum_{n=3}^{\infty} \frac{n^4}{n^4}\frac{1-\frac{1}{n^4}}{n^2}\\
&=\sum_{n=3}^{\infty} \frac{1-\frac{1}{n^4}}{n^2}\\
&=\sum_{n=3}^{\infty} \frac{1}{n^2}-\frac{1}{n^6}\\
\end{align}$$
Now note, since each limit exists (see below, for the value)
$$\sum_{n=3}^{\infty... | {
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Prove uniqueness of solution in ODE I need to find function $f(x)$, such that $x > 0$ and $f(x) > 0$, so that the area limited by the function, the $y$ and $x$ axes and $x = a$ ($a > 0$) is $f^3(x)$ for every $a$, and prove that there is only one such function.
Let $y' = f(x)$ and $y = \int^{x}{f(t)dt}$ I get the ODE $... | From the question we get that $y = \sqrt{\frac{8}{27}}(x+C)^{3/2}$ so $f(x) = y'= \sqrt[3]{y}=\sqrt{\frac{2}{3}(x+C)}$ we know that $y(a) = \int_{0}^{a} f(x)dx = (y'(a))^3$ hence
$$
\int_{0}^{a} f(x)dx = \int_{0}^{a} \sqrt{\frac{2}{3}(x+C)}dx = \sqrt{\frac{8}{27}}(x + C)^{3/2}|_0^a = \sqrt{\frac{8}{27}}(a + C)^{3/2} -... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there any formula for $\operatorname{arctan}{(a+b)}$ Is there any formula for $$\operatorname{arctan}{(a+b)}$$
I know couple of formulas for trigonometric functions. But I don't know if such formulas exists for inverse trigonometric functions. I don't have a clue where to start with. Any hint will be appreciated.
| The following formula is well known and you can find it here.
$$\arctan(a) + \arctan(b) = \arctan\left(\frac{a+b}{1-ab}\right)$$
(of course up to a multiple of $\pi$ with $ab\neq 1$). Then
$$\arctan(a+b) + \arctan(a-b) = \arctan\left(\frac{2 a}{1-a^2+b^2}\right)$$
$$\arctan(a+b) - \arctan(a-b) = \arctan\left(\frac{2 b}... | {
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How many negative real roots does the equation $x^3-x^2-3x-9=0$ have? How many negative real roots does the equation $x^3-x^2-3x-9=0$ have ?
My approach :-
f(x)= $x^3-x^2-3x-9$
Using rules of signs, there is 1 sign change , so there can be at most 1 positive real root
f(-x)= $-x^3-x^2+3x-9$
2 sign changes here, indicat... | Actually, in working through my Calculus oriented answer, I accidentally thought of a somewhat convoluted precalculus approach to solving the problem.
$f(x)= x^3-x^2-3x-9.$
How many negative real roots does the equation $f(x) = 0$ have ?
$$\text{Let} ~~g(x) = x^3 - x^2 - 3x + 3 = (x - 1) \times (x^2 - 3).\tag1$$
Pa... | {
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If $X_{1},\ldots,X_{n}$ are independent Bernoulli random variables, calculate $E[S_{\tau_{2}}]$ Let $X_{1},\ldots,X_{n}$ be independent Bernoulli random variables. $P(X_{i}=1)=p$, $P(X_{i}=0)=q=1-p$. We denote $S_{n}=X_{1}+\cdots+X_{n}$. There are $\tau_{1}=\min \{n: X_{n}=1\}$ and $\tau_{2}=\min \{n: X_{n}=X_{n-1}=1\}... | Sketch (almost solution):
Put $$A_n = \{ \text{ in a set $\{ X_1, X_2, \ldots, X_n\}$ there's no two consecutive units \}} \}$$ $$= \{ \not \exists 1 \le i \le n-1: X_i = X_{i+1} \} = \{ \tau_2 > n\}.$$
Put $a_n = P(A_n, X_n = 0 )$ and $b_n = P(B_n, X_n=1)$. We have
$$b_{n+1} = P(A_{n+1}, X_{n+1} = 1) = P(A_{n+1}, X_... | {
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A non standard way to a probability problem We have teams $A$ and $B$.Every team has equal chance to win. A team wins after they have won $4$ matches. Team $A$ won the first match, what is the probability, that team $B$ won the whole game.
My try:
So team $A$ needs $3$ more wins to win the game. Team $B$ needs $4$ more... | To more simply calculate the odds of this problem, we can relate it to flipping a coin. A heads gives team $A$ the victory, while a tails gives team $B$ the victory. To solve the problem, we are looking for the probability that at least $4$ of $6$ flips are tails. This is not difficult to calculate.
Total possible outc... | {
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In Desmos, function fills entire screen unless $y$ is used in some denominator In Desmos, I am graphing the equation
$$(g^2 - 4x)y^2 + (2g - 4x - 4)y + 1 = 0$$
which uses the helper function
$$g = x + 1 - \frac{x^2 - 2x + 1}{x + 1 + 2\sqrt x}$$
which fills up the entire screen, even though it should be a smooth curve. ... | There are quite some redundancy in the way your implicit function was formulated. Note that the rational function in $g$ can be simplified.
$$
\frac{x^2 - 2x + 1}{x + 1 + 2\sqrt x}= \left(\frac{x-1}{\sqrt{x} + 1} \right)^2 = \left(\frac{(\sqrt{x} + 1)(\sqrt{x} - 1)}{\sqrt{x} + 1} \right)^2 = (\sqrt{x} - 1)^2
$$
Put thi... | {
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Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square. Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square.
I have solved this problem and will post the solution as soon as possible. Hope everyone can check my solution! Thanks ver... | Disclaimer: This is only a partial answer. I may complete this later.
Let $x$, $y$ and $z$ be three positive integers such that
$$(xy+x+2)(xy+1)=z^2,\tag{1}$$
The greatest common divisor $d$ of the two factors divides $1+x$ and $1-y$, so we have
$$x=ud-1\qquad\text{ and }\qquad y=vd+1,$$
for coprime integers $u$ and $v... | {
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Let the roots of the polynomial $P(x) = x^3 − x + 1$ be a, b, c. Construct the polynomial with roots $a^4, b^4, c^4$ and degree 3 I know that $P(x) = x^3 − x + 1$ has a single real root. But I guess it's irrelevant to find it.
What I thought was to change $x$ into ${x}^{\frac{1}{4}}$ so it would satisfy the condition. ... | Maybe I was wrong in calculations, one needs to check.
Vieta: $a+b+c=0$, $ab+ac+bc=-1$, $abc=-1$.
Let's do some algebraic manipulations to obtain the same for $a^4$, $b^4$, $c^4$.
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac) \Rightarrow a^2+b^2+c^2=2$$
$$(ab+ac+bc)^2=a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c)\Rightarrow a^2b^2+a^2c^2+b^2... | {
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Showing that $\sqrt{2+\sqrt3}=\frac{\sqrt2+\sqrt6}{2}$ So I was playing around on the calculator, and it turns out that $$\sqrt{2+\sqrt3}=\frac{\sqrt2+\sqrt6}{2}$$
Does anyone know a process to derive this?
| Both sides are clearly positive, and squaring the left hand side shows that
$$\begin{align} \left(\frac{\sqrt{2}+\sqrt{6}}{2}\right)^2&=\frac{\sqrt{2}^2+2\sqrt{2}\sqrt{6}+\sqrt{6}^2}{2^2}\\&=\frac{2+4\sqrt{3}+6}{4}\\&=2+\sqrt{3}.
\end{align}$$
| {
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Proof verification: $\lim_{x\to 2} \frac{\sqrt{x^2 + 5} - 3}{x - 2} = \frac23$ The question is as follows:
Prove that $\displaystyle\lim_{x\to 2} \dfrac{\sqrt{x^2 + 5} - 3}{x - 2} = \dfrac23$.
My proof is:
Fix $\varepsilon > 0$.
Note that $$\begin{array}{rcl}\left|\dfrac{\sqrt{x^2 + 5} - 3}{x - 2} - \dfrac23\right... | Yes, you have the main idea correct.
Though, there seems to be too much formalism without explaining in words, which may obscure your proof.
One key step you make is to rewrite the function as
$$
\frac{\sqrt{x^2+5}-3}{x-2}=\frac{x^2+5-9}{(x-2)(\sqrt{x^2+5}+3)}
=\frac{x+2}{\sqrt{x^2+5}+3}\;.
$$
At this point, you can s... | {
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Extracting coefficients I stuck at the following problem:
Let
\begin{equation}
f(z) = \frac{1 - \sqrt{1 - 4z}}{2}.
\end{equation}
Find $[z^n]$ of $f(z)^r$ for $r \in \mathbb{N}$. Where $[z^n]f(z)$ is the $n$-th coefficient of the power series $f(z) = \sum_{n \geq 0}{a_n z^n}$ therefore $[z^n]f(z) = a_n$.
So far I got
\... | Too long for a comment, but not a full flesh answer.
$f(z)$ can be considered as the root of this quadratic equation:
$$f(z)^2-f(z)+z=0,\tag{1}$$
Let us multiply (1) by $f(z)$, giving:
$$f(z)^3-f(z)^2+zf(z)=0\tag{2}$$
Adding (1) and (2) gives an expression for $f(z)^3$.
More generally, iterating $r-2$ times the process... | {
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What is wrong with the application of L'Hopital's rule to $\lim_{x\rightarrow 2}\frac{x^2 - 7x + 10}{x^2 + x - 6}$? The question is
$$\lim_{x\rightarrow 2}\frac{x^2 - 7x + 10}{x^2 + x - 6}$$ If we plug in $x = 2$ we see that both quadratrics have a zero there. "Logically" this is a L'Hopital question, so if I take th... | It holds $x^2-7x+10=(x-2)(x-5)$ and $x^2+x-6=(x-2)(x+3)$. Hence, we have $$\frac{x^2-7x+10}{x^2+x-6}=\frac{(x-5)(x-2)}{(x+3)(x-2)}=\frac{x-5}{x+3}$$
for all $x\notin\{-3,2\}$. Hence, the pole $x=2$ is removable, because we can extend the domain of the lefthandside by $x=2$, such that the function is still continuous. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the number of different colorings of the faces of the cube with 2 white, 1 black, 3 red faces? I tried using Polya theorem same in this https://nosarthur.github.io/math%20and%20physics/2017/08/10/polya-enumeration.html guide, using S4 group for cube faces. But i Have a $\frac{1}{24}12w^2*b*r^3$.
Please help to me ... | The cycle index polynomial for rotational symmetries of the faces of a cube is
$$
\tfrac1{24}(z_1^6+12z_2^3+8z_3^2+3z_1^2z_2^2)
$$
Therefore, in order to count the number of colorings with one black, three red, and two white faces, we need to find the coefficient $b^1r^3w^2$ in
$$
\tfrac1{24}[(b+w+r)^6+12(b^2+r^2+w^2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Signature of $q(x,y)=5x^2-6xy+y^2$ a quadratic form Let $q:R^2\rightarrow R$ a quadratic form such that $q(x,y)=5x^2-6xy+y^2$
find the signature of $q$
I know that $q_A=$ $\begin{pmatrix}
5 & -3\\
-3 & 1
\end{pmatrix}$
and how $q(1,0)=5\neq0$ then $(1,0)$ is an element of a orthogonal basis $\beta=\{(1,0),v\}$
and the... | $$ Q^T D Q = H $$
$$\left(
\begin{array}{rr}
1 & 0 \\
- \frac{ 3 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
5 & 0 \\
0 & - \frac{ 4 }{ 5 } \\
\end{array}
\right)
\left(
\begin{array}{rr}
1 & - \frac{ 3 }{ 5 } \\
0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rr}
5 & - 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to find first integral in a given quasilinear partial differential equation? I have to solve PDE
$$y(x+y)\dfrac{\partial z}{\partial x}+x(x+y)\dfrac{\partial z}{\partial y}=2(x^2-y^2)+xz-yz.$$
From this equation I get the system of differential equations:
$$\dfrac{dx}{y(x+y)}=\dfrac{dy}{x(x+y)}=\dfrac{dz}{2(x^2-y^2... | $$y(x+y)\dfrac{\partial z}{\partial x}+x(x+y)\dfrac{\partial z}{\partial y}=2(x^2-y^2)+xz-yz.$$
You correctly wrote :
$$\dfrac{dx}{y(x+y)}=\dfrac{dy}{x(x+y)}=\dfrac{dz}{2(x^2-y^2)+xz-yz}.$$
And you correctly found a first characteristic equation
$$x^2-y^2=C_1$$
For a second characteristic equation :
$$\dfrac{dx}{y(x+y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Legendre relations for elliptic integrals with null imaginary part I am trying to compute the transformation of $K(k)$ as $k \to 1/k$, where $K(k)$ is just the Legendre integral
$$K(k) := F\left(\frac{\pi}{2}, k\right).$$
The Digital Library of Mathematical Functions [1] gives a relation for this transformation that re... | Let $0<k<1$, then
$$ K\left(\frac{1}{k}\right) = \int_{0}^{1}\frac{dt}{\sqrt{\left(1-t^2\right)\left(1-\frac{1}{k^2}t^2\right)}}dt $$
Make the following change of variable: $\displaystyle w^2 \mapsto \frac{1}{k^2}t^2$
$$K\left(\frac{1}{k}\right) = k\int_{0}^{\frac{1}{k}}\frac{dt}{\sqrt{\left(1-k^2w^2\right)\left(1-w^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$100$ numbers are written around the circle
$100$ positive numbers are written in a circle. The sum of any two
neighbors is equal to the square of the number following them
clockwise. Find all such sets of numbers.
Let us introduce the notation $n=100$, not forgetting that $n$ is even. Let the numbers start in clockw... | Here is my explanation of the solution.
To me, it's a case where "minor details are skipped", as opposed to Paul's "big jump / stopped trying to explain / logic leap".
However, the exposition could be significantly improved on by stating where the condition $a_i+a_{i+1} = a_{i+2}^2$ was repeatedly used.
Let us introd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $2Q(x)$ can be written as the sum of three perfect squares (a) Show that $x^3-3xbc+b^3+c^3$ can be written in the form $(x+b+c)Q(x)$, where $Q(x)$ is a qudaratic expression.
(b) Show that $2Q(x)$ can be written as the sum of three expressions, each of which is a perfect square.
My attempt at solution: (a) Thr... | I will put the Hessian matrix of $x^2 + y^2 + z^2 - yz - zx -xy$ and display , with square matrices $PQ = QP = I,$ $P^T HP = D$ is diagonal, so $Q^T D Q = H$
I will type in the final outcome first. The $Q^T DQ$ version reads
$$ \left( x - \frac{y}{2} - \frac{z}{2} \right)^2 + \frac{3}{4} (y-z)^2 = x^2 + y^2 + z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Formula for distance between Incenter and Orthocenter. On pg. 205 of Johnson's Advanced Euclidean Geometry, he gives three formulas, preparatory to a proof of Feuerbach's theorem.
$$OI^2=R^2-2R\rho$$
$$IH^2=2\rho^2-2Rr$$
$$OH^2=R^2-4Rr$$
where $O$ is the circumcenter, $I$ the incenter, $H$ the orthocenter; $R$ the radi... | Here is a try to get the formula quickly.
Disclaimer: I will use more or less standard notations in a triangle $\Delta ABC$. The side lengths are $a,b,c$, the incenter is $I$, the inradius is $r$ (not $\rho$), the circumradius is $R$, the area is $S$. Let $AA'$, $B'$, $CC'$ be the heights intersecting in $H$. The inr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $\frac{1+2x}{1+\sqrt{1+2x}}+\frac{1-2x}{1-\sqrt{1-2x}}$ for $x=\frac{\sqrt3}{4}$. Find the value of $$\dfrac{1+2x}{1+\sqrt{1+2x}}+\dfrac{1-2x}{1-\sqrt{1-2x}}$$ for $x=\dfrac{\sqrt3}{4}$.
I have no idea why I can't solve this problem. I tried to simplify the given expression, but I was able to reach on... | You can dodge a certain amount of work by re-writing the given expression using $ \ u \ = \ \sqrt{1 + 2x} \ $ and $ \ v \ = \ \sqrt{1 - 2x} \ \ $ to produce
$$ \frac{1 \ + \ 2x}{1 \ + \ \sqrt{1 + 2x}} \ + \ \frac{1 \ - \ 2x}{1-\sqrt{1-2x}} \ \ \rightarrow \ \ \frac{u^2}{1 \ + \ u} \ + \ \frac{v^2}{1 \ - \ v} \ \ = \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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.