Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the range of $A$ if $A=\sin^{20}x+\cos^{48}x$
Find the range of $A$ if $$A=\sin^{20}x+\cos^{48}x$$
$$
A'=20\sin^{19}x\cos x-48\cos^{47}x\sin x=0\implies5\sin^{19}x\cos x=12\cos^{47}x\sin x\\
\implies5\sin^{18}x=12\cos^{46}x
$$
How do I proceed further and prove that $A\in(0,1]$ ?
Is it possible to find the rang... | Apart from the trivial upper bound $A\le 2$, we have the stronger (and sharp - try $x=0$) bound
$$ \tag1A\le 1.$$
Consider $f(x):=(1-x)^{10}+x^{24}$ for $0\le x\le 1$. Then $f'(x)=24x^{23}-10(1-x)^9$ is strictly increasing (as each summand is) on $[0,1]$, hence has at most one root there. As $f'(0)=-10$ and $f'(1)=24$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
find the answer in terms of $a$ and $b$ only ($a, b$ are roots of $\ x^4 + x^3 - 1 = 0$ If $a$ and $b$ are the two solutions of $\ x^4 + x^3 - 1 = 0$ , what is the solution of $\ x^6 + x^4 + x^3 - x^2 - 1 = 0$ ?
Well I am not able to eliminate or convert $\ x^6$. Please help.
| Let $f(x)=x^4+x^3−1$ and $F(x)=x^6+x^4+x^3−x^2−1$. One has
$$F(x)=(x^2-x+2)f(x)-x^3-x+1\\F(x)=0\iff f(x)=\frac{-x^3-x+1}{x^2-x+2}$$
We look at the values for which $$\frac{-x^3-x+1}{x^2-x+2}=x^4+x^3-1$$
The problem suggests that these values are a simple function of $a$ and $b$. Proving with $a + b$ and with $ab$, thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Calculate side lenght of triangle from two rectangle on top of each other I want to calculate the side lenght of $b$. I have two rectangles with one at 0° (screen) and I have one rectangle at 20° (turned image). With respect to the middle point. Both rectangles have a height of 6 and a width of 8.
Beceause the image is... | Using the center where you've drawn the dot as the origin, and letting $c \approx .940$ and $s\approx .342$ denote the cosine and sine of $20$ degrees, respectively, the equation of the top horizontal line is
$$
y = 3,
$$
and the equation of the tilted top line is
$$
\pmatrix{-s \\ c} \cdot \pmatrix{x\\y} = 3,
$$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
My try
It can be verified that $\lim_{k \to \infty} S_{3k} < + \infty$ and $\lim_{k \to \infty} S_{3k} = \lim_{k \to \infty} S_{3k+1} = \lim_{k \to \infty} S_{3k+2}$.
So letting $a_n := S_{3n}$, $a_{n... | It seems to me the pattern is you are adding the following triplets: $\frac 1 {4k+1} +\frac 1 {4k+3}-\frac 1 {2^{k+1}} $.
If this converged we could rearrange the terms. The infinite sum of $\sum\frac {-1} {2^{k+1}}$ is $-1$ which is finite so that would mean the sum $\sum (\frac 1 {4k+1}+\frac 1 {4k+3})= \sum \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2967089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding dependencies such that $0 > \frac{2b^2r^2}{z}-\left(2r ^2-2br\sqrt{1-\frac{b^2}{z^2}}\right)z$ I'm trying to solve an inequality with 3 variables.
$$0 > \frac{2 b^2 r^2}{z} - \left(2 r ^2 - 2 b r \sqrt{1 - \frac{b^2}{z^2}}\right) z$$
Basically, I want to know under which dependencies the formula is less than ze... | Note that $b=z$ violates the strict inequality (the right-hand side becomes zero). Consequently, we have $0<b < z$, which allows us to write
$$b=z \sin\theta \tag{1}$$
for some $0^\circ < \theta < 90^\circ$. Then the square root reduces immediately to $\cos\theta$, and your inequality simplifies to
$$0 > 2 r^2 z \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2967619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find the number of real roots for a polynomial? How can you find of real roots for $P(x) = x^4 - 4x^3 + 4x^2 - 10$?
Using the Descartes' rule of signs:
The polynomial $P(x) = x^4 - 4x^3 + 4x^2 - 10$ has three sign changes between the first, second, third and fourth terms (the sequence of pairs of successive sign... | Let $\>x=t+1\>$ then
$$
\\P(x)=P(t+1)=(t+1)^4-4(t+1)^3+4(t+1)^2-10=
\\=(t^4+4t^3+6t^2+4t+1)-(4t^3+12t^2+12t+4)+(4t^2+8t+4)-10=
\\=t^4-2t^2-9
\\P(t+1)=0=>t^4-2t^2-9=0=>
$$
if $\>x\>$ is real then $\>t\>$ is real =>
$$
\\t^2=\dfrac{2+\sqrt{40}}{2}=1+\sqrt{10}
$$
Answer $\>x=1+\sqrt{1+\sqrt{10}}\>$ or $\>x=1-\sqrt{1+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove Fibonacci recurrence holds mod p? Let $$J_n \equiv c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^n - \Big( \dfrac{1-c}{2} \Big)^n \Big) \ \text{(mod p)},$$ with $c$ and $c^{-1}$ integers such that $c^2 \equiv 5 \ \text{(mod p)}$ and $cc^{-1} \equiv 1 \ \text{(mod p)}$. And $c$ is an odd integer.
It is easy t... | $$J_n \equiv c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^n - \Big( \dfrac{1-c}{2} \Big)^n \Big) \ \text{(mod p)}$$
In order to show that $$J_n = J_{n-1}+J_{n-2}$$ it suffices to show that both $ (\frac {1+c}{2})^n$ and $ (\frac{1-c}{2})^n$ satisfy the same relation.
Note that for $ (\frac {1+c}{2})^n =(\frac{1+c}{2})^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974392",
"timestamp": "2023-03-29T00:00:00",
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Finding $B^{225}$ without many computations I have that $B$ is a 4x4 matrix. $B-5I=\begin{pmatrix} -2 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 3 & 0 & -3 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. The question asks to find $B^{225}$ without performing many computations, leaving the answer as a product of 3 matrices (univerted matrice... | Let $C = B - 5I$. Notice it can be written as a outer product of two column vectors (or matrix product between a column and a row vector):
$$C = B - 5 I = u \otimes v = u v^T\quad\text{ where }\quad u = \begin{bmatrix}-2\\ 0 \\ 3 \\ 0\end{bmatrix}\quad\text{ and }\quad v = \begin{bmatrix}1 \\ 0 \\ -1 \\0\end{bmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How can one calculate $342342^{1001}$ mod $5$? How can one calculate $342343^2$ mod $3$? I know that the answer is $1$.
And $342342^{1001}$ mod $5$.
I know that
$
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
$
So 1001 = 250 + 250 + 250 ... | This is an observation but you can take it as an answer.
Let $n\in\mathbb{N}$ then $3|n\iff$ $3$ divides the sum of all the digits of $n$. Now consider $n=342343$ and sum of all of its digits $=19$. So clearly $n\equiv-2 (\mod 3)\implies n^2\equiv1(\mod 3)$ since $4\equiv 1(\mod 3)$.
For the second part observe this, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$ is a root for $x^3+\sqrt[3]{6}x^2-1$
Prove that
$$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$
is a root for
$$P(x)=x^3+\sqrt[3]{6}x^2-1$$
Source: list of problems for math contest preparation.
I have n... | Let $\alpha = \sqrt[3]{\frac 19}$ and $\beta = 1-\sqrt[3]2+\sqrt[3]{2^2}$, so that $c = \alpha\beta$. Define $$Q(x) = P(\alpha x) = \frac{x^3}9+\frac{\sqrt[3]2}3 x^2-1$$ and note that $P(c) = 0$ iff $Q(\beta) = 0$.
Now, notice that $(1+\sqrt[3]2)(1-\sqrt[3]2+\sqrt[3]{2^2}) = 1^3+\sqrt[3]2^3 = 3$, so $\beta = \frac{3}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987981",
"timestamp": "2023-03-29T00:00:00",
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Algebraic manipulation with indices The question is:
For a>0 and $\sqrt{a}+\frac{1}{\sqrt{a}}=3$, find the value of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$
So I first squared the given equation and got:
$$a+\frac{1}{a}+2=9$$
$$a+\frac{1}{a}=7$$
Then to get the form of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$:
$$(a+\frac{1}{a})(\sqrt{a}... | $\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}=\sqrt{a}+\frac{1}{\sqrt{a}}=3$ !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim\limits_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right].$ Problem
Evaluate
$$\lim_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right].$$
Attempt
First, we may obtain
$$\lim_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right]=\lim_{x ... | The key point is that $x\log \sin x \to 0$ and $\sin x \log x \to 0$ then by Taylor's series we have
*
*$x^{\sin x}=e^{\sin x \log x}=1+x\log x+\frac12x^2\log^2 x+\frac16x^3\log x(\log^2 x -1)+O(x^4\log^2 x)$
*$(\sin x)^{x}=e^{x \log (\sin x)}=1+x\log x+\frac12x^2\log^2 x+\frac16x^3(\log^3 x -1)+O(x^4\log x)$
then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)$
Evaluate $$\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)$$
My Solution
Denote
$$f(t):=\arctan t.$$
By Lagrange's Mean Value Theorem，we have
$$f\left(\frac{2x^2+5}{x^2+1}\r... | When $x\to \infty$,
$$\frac{2x^2+5}{x^2+1}=2+\underbrace{\frac{3}{x^2+1}}_{\to 0},\qquad \frac{2x^2+7}{x^2+2}=2+\underbrace{\frac{3}{x^2+2}}_{\to0},$$
and the Taylor approximation of $\arctan(2+t)$ around $0$ is given by
\begin{align*}
\arctan(2+t)=\arctan 2+(\arctan(t+2))'|_{t=0}t+o(t)=\arctan 2+\frac t5+o(t).
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $C_{3 \over 2}^n$ is bounded given $C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}$
Let:
$$
\begin{cases}
C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}\\
C_{a}^0 = 1
\end{cases}
$$
Prove $C_{3 \over 2}^n$ is bounded.
I've started with finding a reduced formula:
$$
C_{3\over 2}^n = \frac{{3\over 2}\left({... | After some pondering i'm going to try to answer this myself.
I've been thinking about considering two cases for $n$ is odd and $n$ is even. Start with even $n$. We know that for any $n \ge 2 \implies C_{3/2}^n > 0$ in case $n$ is even. Lets inspect a subsequence for even indices:
$$
C_{3/2}^n = \frac{1}{2^nn!}\prod_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Irrational equation $\sqrt{9-4x}=p-2x$
The equation
$$\sqrt{9-4x}=p-2x$$
has exactly 2 real and different solutions only if parameter $p$ belongs to which set?
So what I see here, to have the solutions be real in the first place,
$9-4x\ge0$
So $-4x\ge-9 \Rightarrow x\le{9\over4}$
$x\in\left(-\infty,{9\over4}\righ... | As already pointed out in a comment, $(p-2x)^2 = p^2-4px+4x^2$, which results in
$$4x^2 - 4(p-1)x+(p^2-9) = 0,$$
or
$$x = \frac{(p-1)\pm\sqrt{10-2p}}{2}.$$
For two real distinct roots, $10-2p > 0 \implies \color{red}{p < 5}$.
Also, $p \ge 2x$ implies $$(p-1)\pm\sqrt{10-2p} \le p \implies \sqrt{10-2p}\le 1 \implies \col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probably of winning with 2 dice (maximum of them) against another one could some of you help me to find out what is the probability of A) obtain with two dice a greather number than another die? B) and if the dice are 3 how can I do? Not the sum of the 2 dice, but the greatest value of those 2 against another die
| When rolling $1$ die, the probability of throwing "N or less" is $P(X \le N) =\frac{N}{6}$. When rolling $n$ dice, the probability of throwing "N or less" is $P_n(X\le N) = (\frac{N}{6})^n$.
So with $n$ dice, the probability of throwing "at least one N and nothing higher than N" is $P_n(N) = (\frac{N}{6})^n-(\frac{N-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\binom{n}{1}^2+2\binom{n}{2}^2+\cdots +n\binom{n}{n}^2=n\binom{2n-1}{n-1}$
Prove that
$$
\binom{n}{1}^2+2\binom{n}{2}^2+\cdots + n\binom{n}{n}^2
= n \binom{2n-1}{n-1}.
$$
So
$$
\sum_{k=1}^n k \binom{n}{k}^2
= \sum_{k=1}^n k \binom{n}{k}\binom{n}{k}
= \sum_{k=1}^n n \binom{n-1}{k-1} \binom{n}{k}
= n \sum... | We present a slight variation using formal power series and the
coefficient-of operator. Starting from
$$\sum_{k=1}^n k {n\choose k}^2
= \sum_{k=1}^n k {n\choose k} [z^{n-k}] (1+z)^n
\\ = [z^n] (1+z)^n \sum_{k=1}^n k {n\choose k} z^k
= n [z^n] (1+z)^n \sum_{k=1}^n {n-1\choose k-1} z^k
\\ = n [z^n] z (1+z)^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Inequality equation of a Triangle I have divided the Problem into two parts, a) and b):
a) Let $a$, $b$ and $c$ be the side lengths of a triangle with a perimeter of $4$.
I need to prove that
$a^2 + b^2 + c^2 + abc < 8$.
b) And is there a real number $d<8$ such that for each triangle with the perimeter $4$, the inequa... | a) Because of $a+b+c=4$, you have
$${a^2}+{b^2}+{c^2}={(a+b+c)^2}-2(ab+bc+ac)=4(a+b+c)-2(ab+bc+ac)$$
Therefore
$${a^2}+{b^2}+{c^2}+abc=4(a+b+c)-2(ab+bc+ac)+abc=8-(2-a)(2-b)(2-c)$$
By triangle-inequality $b+c>a$ $$\Rightarrow4=a+b+c>a+a=2a\Rightarrow2>a$$
Analugously you can prove that $2>b$ and $2>c$.
Thus
$${a^2}+{b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. Show that
$$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})<\frac{1}{10}$$
My attempt: This problem is from a text book where is introduced as: https://en.wikip... | Let $a=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}. . . .\frac{99}{100}$
and:
$b=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}. . . \frac{100}{101}$
It is clear that $a<b$ because each factor of $a$ is less than its corresponding factor in $b$ :
$\frac{1}{2}<\frac{2}{3}, \frac{3}{4}<\frac{4}{5}. . . \frac{99}{100}<\frac{100}{101}$
⇒ $a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$ In a proof, the author states that it is clear that:
Given $x\geq 1$ and $ n-x \geq 1$ and finally also $n\geq 2$
$${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$$
This is not immediately clear to me. Of course, If I have $n$ objects and I split them up in $x$ and $n-... | \begin{align}
{x\choose2} +{n-x\choose2}
&=\frac{(x-1)x}2 +\frac{(n-x)(n-x-1)}2 \\
&=\tfrac12 (x^2-x +n^2-nx-n -nx+x^2+x) \\
&=\tfrac12 (2x^2 -2nx+n^2-n) \\
&=\tfrac12 (2x(x-n) +n(n-1))
\end{align}
Given $1 \leq x \leq n-1$ (and $x-n \leq -1$), then
\begin{align}
&\leq \tfrac12 (2(n-1)(-1) +n(n-1)) \\
&=\frac{(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Prove that number $4p^2+1$ can show as sum of squares of three different numbers Let $p>3$ is prime number. Prove that number $4p^2+1$ can show as sum of squares of three different numbers.
Only what I know that every prime number $p>3$ can show as $p=6k+1$ or $p=6k-1$, such that $k \in \mathbb Z$.
If I put $p=6k+1$, t... | Notice that $5=1^2+2^2,$ so:
$4(6k+1)^2+1=144k^2+48k+5=(ak)^2+(bk+1)^2+(ck+2)^2=(a^2+b^2+c^2)k^2+(2b+4c)k+5$
$\therefore 144=a^2+b^2+c^2 \qquad$ and $\qquad 2b+4c=48$
By trial and error, I found that $a=4, b=8, c=8$.
So $4(6k+1)^2+1=(4k)^2+(8k+1)^2+(8k+2)^2$
Now for $p=6k-1,$
$4(6k-1)^2+1=144k^2-48k+5=(ak)^2+(bk-1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$? $$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
Solution
\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\
&= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\s... | Your problem arises from the fact that you used $\color{red}{\lim_\limits{x \to 0} \frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $\color{red}{\infty-\infty}$...
Only split an initial limit into a product if the individual limits are defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
What am I doing wrong finding the derivative of $\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$? $$y=\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$$
For convenience, let
$$A=\frac{3-x}{2}\sqrt{1-2x-x^2},$$
$$B=2\arcsin{\frac{1+x}{\sqrt{2}}}.$$
$$y'=A'+B'$$
$$A'=(-\frac{1}{2})(\sqrt{1-2x-x^2}... | You got A' wrong.
$$ A = \frac{3-x}{2}\cdot(\sqrt{1-2x-x^2}) = f\cdot g$$
where $ f =\frac{3-x}{2} $ and $ g = \sqrt{1-2x-x^2} $.
Looking at $$A'= f'g + g'f$$the first part
$$ f'g = (-\frac{1}{2})(\sqrt{1-2x-x^2})$$ is correct, but the second part is $$ g'f = (-2x -2)(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}})(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the sum: $\sum_{n=2}^\infty \frac{1}{n^2-1}$
Evaluate : $$\sum_{n=2}^\infty \frac{1}{n^2-1}$$
I've tried to rewrite the questions as $$\sum _{n=2}^{\infty \:\:}\left(-\frac{1}{2\left(n+1\right)}+\frac{1}{2\left(n-1\right)}\right)$$ but still couldnt't get any answer as when I substitute numbers $(n)$ into th... | $$\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac{1}{2(n-1)}-\frac{1}{2(n+1)}$$
Taking $\frac{1}{2}$ common ,
$$\sum_{n=2}^{\infty}\frac{1}{n^2-1}=\frac12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$
Write the first few terms of the series as :
$\frac{1}{2}( 1 - \frac{1}{3} +\frac{1}{2} - \frac{1}{4} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to find a matrix with given eigen values Find a $3\times 3$ matrix $B$ which contains $A$ as a sub-matrix and has eigen values $0$, $1$ only
where $$A=
\begin{bmatrix} -2& -12 \\1&5 \end{bmatrix}$$
I cant find a way how to construct $B$
I took the eigen values to be $0$, $0$, $1$ and then took $$B =
\begin{bmatrix... | Let's start with your $$B = \begin{bmatrix} -2& -12& a\\1&5 &b\\c&d&e\end{bmatrix}$$
Let's take the eigenvalues $0,1,1$. Then $\operatorname{Tr}(B)=3+e=2 \Rightarrow e=-1$. And the characteristic polynomial must be $\lambda(\lambda -1)^2=\lambda^3-2\lambda^2+\lambda$.
$$B = \begin{bmatrix} -2& -12& a\\1&5 &b\\c&d&-1\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to prove that $(-18+\sqrt{325})^{\frac{1}{3}}+(-18-\sqrt{325})^{\frac{1}{3}} = 3$ How to prove that $\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}} = 3$ in a direct way ?
I have found one indirect way to do so: Define $t=\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{... | Let first root be $A$ and second be $B$, then we need to find the value of $A + B = x$
*
*$A^3 + B^3 = -36$
*$AB = -1$
Expand the cube: $A^3 + B^3 = (A+B)*(A^2 - AB + B^2) = (A+B)*((A+B)^2-3AB) = x*(x^2+3) = x^3 + 3x = -36$.
Easy guess $x = -3$
Another approach: assume that you have a cube under the root, then, sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove the sequence $a_n =\frac{1\cdot 3\cdot…\cdot(2n-1)}{2\cdot4\cdot…\cdot2n}$ has a limit I have several questions to ask:
1) Show increasing, find the upper bound if you can of
$\sqrt{(n^2-1)}/n$.
$\sqrt{(n^2-1)}/n= |n|\sqrt{1-1/n^2}/n$ if $n$ is positive than $\sqrt1$ else $-\sqrt1$;
bound: $\sqrt{(n^2-1)}/n \le... | Cross-multiplication yields, for $k\ge1$,
$$
\left(\frac{2k-1}{2k}\right)^2\le\frac{2k-1}{2k+1}
$$
Therefore,
$$
\begin{align}
\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2
&\le\prod_{k=1}^n\frac{2k-1}{2k+1}\\
&=\frac1{2n+1}
\end{align}
$$
Thus,
$$
\prod_{k=1}^\infty\frac{2k-1}{2k}=0
$$
For $n\ge1$,
$$
\begin{align}
\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Show $11^{11}+12^{12}+13^{13} =10k$ without direct calculation Prove that $11^{11}+12^{12}+13^{13}$ is divisible by $10$.
Obviously you could just put that in to a calculator and see the results, but I was wondering about some of the other approaches to this? I have not studied modulus', so if you could explain it with... | You can easily prove it without any modular arithmetic. Just look at the last digits.
$$3^1 = \color{blue}{3} \quad 3^2 = \color{blue}{9} \quad 3^3 = 2\color{blue}{7} \quad 3^4 = 8\color{blue}{1} \implies 3, 9, 7, 1, 3, 9, 7, 1, ...$$
$$2^1 = \color{green}{2} \quad 2^2 = \color{green}{4} \quad 2^3 = \color{green}{8} \q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 11,
"answer_id": 5
} |
Compute $\int_{-1}^{1} 5x^{6}(1 - |x|^{5}) \mathop{dx}$
Compute $$\int_{-1}^{1} 5x^{6}(1 - |x|^{5}) \mathop{dx}$$
$$\int_{-1}^{1} 5x^{6}(1 - |x|^{5}) \mathop{dx}$$
$$=\int_{-1}^{1} 5x^{6}- 5x^{6}|x|^{5} \mathop{dx} $$
$$= \int_{-1}^{1} 5x^{6} - 5\int_{-1}^{1}x^{6}|x|^{5}\mathop{dx}$$
$$= \frac{10}{7} - 5\left(\int_{... | Another way:
As $x^6(1-|x|^5)$ is an even function,
$$I=\int_{-1}^15x^6(1-|x|^5)\ dx=2\int_0^15x^6(1-|x|^5)\ dx$$
$$\dfrac I{2\cdot5}=\int_0^1(x^6-x^{11})\ dx=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3018272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Co-ordinate Geometry : Circle Problem: Circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 8x + 11 = 0$ cut off equal intercepts on a line through the point $(-2, \frac{1}{2})$. Find the slope of the line.
Comments: First I write the family of lines passing through the given point i.e., $y - \frac{1}{2} = m (x + 2)$ then by using... | *
*Line $$2mx-2y+4m+1=0 \tag{1}$$
*First circle $$x^2+y^2=1$$
centre $O=(0,0)$, radius $r_1=1$
distance of $O$ from $(1)$: $$d_1=\left| \frac{4m+1}{2\sqrt{m^2+1}} \right|$$
*Second circle $$x^2+y^2-8x+11=0$$
centre $P=(4,0)$, radius $r_2=\sqrt{4^2-11}=\sqrt{5}$
distance of $P$ from $(1)$: $$d_2=\left| \frac{12m+1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3019642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof using AM-GM inequality The questions has two parts:
Prove
(i) $ xy^{3} \leq \frac{1}{4}x^{4} + \frac{3}{4}y^{4} $
and
(ii) $ xy^{3} + x^{3}y \leq x^{4} + y^{4}$.
Now then, I went about putting both sides of $\sqrt{xy} \leq \frac{1}{2}(x+y)$
to the power of 4 and it left me with
$$-x^{3}y \leq \frac{1}{4}x^{4}... | Because by AM-GM $$3y^4+x^4\geq4\sqrt[4]{\left(y^4\right)^3x^4}=4|xy^3|\geq4xy^3$$ and
$$x^4+y^4-x^3y-xy^3=(x-y)(x^3-y^3)=(x-y)^2(x^2+xy+y^2)\geq$$
$$\geq(x-y)^2\left(2\sqrt{x^2y^2}+xy\right)=(x-y)^2\left(2|xy|+xy\right)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3020293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $\frac{x}{y^\frac{n-1}{n}}$ is constant, how do I prove $\frac{dx}{x}=\frac{n-1}{n}\frac{dy}{y}$? From:
$\frac{x}{y^\frac{n-1}{n}}=constant$
To:
$\frac{dx}{x}=\frac{n-1}{n}\frac{dy}{y}$
It's from a turbomachinery lecture. The text I'm studying says that binomial expansion is used, but I don't get it.
| Let $x=c\cdot y^\frac{n-1}{n}$. Then
\begin{align}
\frac{dx}{dy}&=\frac{n-1}{n}c\cdot y^\frac{-1}{n}\\
&=\frac{n-1}{n}c\cdot y^\frac{-1}{n} \cdot \frac{y}{y}\\
&=\frac{n-1}{n}c\cdot y^\frac{n-1}{n} \cdot \frac{1}{y}\\
&=\frac{n-1}{n} \frac{x}{y}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\int_0^{\frac{n\pi}{4}} \frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{n\pi}{4} $
Prove that $$\int_0^\frac{n\pi}{4} \left(\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}\right) dx = \frac{n \pi}{4} $$
is true for all integers $n$.
Here I've looked at the case where $n=1$ as $t=\tan(x)$ is injective in that domai... | Note that the integrand function $f(x)=\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}$ is even with period $\pi/2$, so
$$\int_0^\frac{n\pi}{4} f(x) dx=
\frac{1}{2}\int_{-\frac{n\pi}{4}}^\frac{n\pi}{4} f(x) dx=
\frac{1}{2}\int_{0}^\frac{n\pi}{2} f(x) dx
=\frac{n}{2}\int_{0}^\frac{\pi}{2} f(x) dx.$$
Now, following your approach, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove that $({a\over a+b})^3+({b\over b+c})^3+ ({c\over c+a})^3\geq {3\over 8}$
Let $a,b,c$ be positive real numbers. Prove that $$\Big({a\over a+b}\Big)^3+\Big({b\over b+c}\Big)^3+ \Big({c\over c+a}\Big)^3\geq {3\over 8}$$
If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ and
$$\Big({1\over 1+x}\Big)^3+\Big({1\... | This is more of a comment, but I don't have the reputation. Use Lagrange multipliers. Solving, you find that the critical points occur when $xyz=1$ and $yz (1+x)^4 = xz (1+y)^4 = xy (1+z)^4$. I think the only solution is $x=y=z=1$. Clearly, it's a minimum and plugging in shows the bound.
Added:
We can rewrite th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate $\int\frac{1}{3+4x+4x^2}$?
How to evaluate
$$\int\frac{1}{3+4x+4x^2}\quad ?
$$
This is what I've done so far:
$$
\frac{1}{4} \int\frac{1}{x^2+x+\frac{3}{4}}
=\frac{1}{4} \int\frac{1}{(x+\frac{1}{2})^2 + \frac{1}{2}}
$$
$$
y = \frac{1}{a}\arctan\frac{u}{a},\quad \frac{dy}{du} = \frac{1}{a^2 + u^2}
$... | What can be done is a double U-Substitution
We have $$\int\frac{1}{3+4x+4x^2}dx$$
By completing the square we see $$\int\frac{1}{3+4x+4x^2}dx = \int\frac{1}{(2x+1)^2+2}dx$$
Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$\frac{1}{2}\int\frac{1}{u^2+2}du = \frac{1}{2}\int\frac{1}{2(\frac{u^2}{2}+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $A$ given $A^2$ and $B$ and $(A^2)^{-1} = A^{-1}B$ This excersice took place in class I had today. The exercise was the following:
Let the regular matrix $A$,$B$:
$$A^2 = \left[ \begin{matrix} 2 &-2 & 2\\ -2 & 2 & 2 \\ 4&4&-4 \end{matrix} \right] \hspace{2cm} B = \left[ \begin{matrix} 0 &-1 & 2\\ 0 & 2 & 0 \\ 1... | The problem is broken, here is a precise statement showing how it is broken.
Let $B=\left[ \begin{matrix} 0 &-1 & 2\\ 0 & 2 & 0 \\ 1&3&-1 \end{matrix} \right].$
For every invertible matrix $A$, at least one of the following are false:
*
*$A^2 = \left[ \begin{matrix} 2 &-2 & 2\\ -2 & 2 & 2 \\ 4&4&-4 \end{matrix} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Simplified form of $\cos^{-1}\big[\frac{3}{5}\cdot\cos x+\frac{4}{5}\cdot\sin x\big]$, where $x\in\big[\frac{-3\pi}{4},\frac{3\pi}{4}\big]$
Find the simplified form of $\cos^{-1}\bigg[\dfrac{3}{5}\cdot\cos x+\dfrac{4}{5}\cdot\sin x\bigg]$, where $x\in\Big[\dfrac{-3\pi}{4},\dfrac{3\pi}{4}\Big]$
My reference gives the ... | $$-\dfrac{3\pi}4\le x\le\dfrac{3\pi}4$$
$$\iff-\dfrac{3\pi}4-\cos^{-1}\dfrac35\le x-\cos^{-1}\dfrac35\le\dfrac{3\pi}4-\cos^{-1}\dfrac35$$
Now $\dfrac{3\pi}4-\cos^{-1}\dfrac35\le\pi$ as $\cos^{-1}\dfrac35>0>\dfrac{3\pi}4-\pi$
So, $\cos^{-1}\bigg[\cos\Big(x-\cos^{-1}\dfrac35\Big)\bigg]=x-\cos^{-1}\dfrac35$ if $x-\cos^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Quadratic form as a homogenous polynomial Let $Q(v)=v'Av$, where$v=\begin{pmatrix}x&y&z&w\end{pmatrix}\ \ A=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}$, then does there exist an invertible matrix $P$ such that $Q(Pv)=x^2+y^2-zw$? Note that all terms are in $\mathbb{R}$. I think we should use orthogo... | Note that $Q(Pv)=v^{\top}P^{\top}APv$, so one approach is to find $P$ such that
$$P^{\top}AP=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&-\frac{1}{2}\\0&0&-\frac{1}{2}&0\end{pmatrix}.$$
As the first two rows and columns of $P^{\top}AP$ are the same as those of $A$, it makes sense to have $P$ leave those alone, i.e. to take ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $x_1,x_2,\ldots,x_n$ are the roots for $1+x+x^2+\ldots+x^n=0$, find the value of $\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$
Let $x_1,x_2,\ldots,x_n$ be the roots for $1+x+x^2+\ldots+x^n=0$. Find the value of
$$P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$$
Source: IME entrance exam (... | We notice that the equation:
$$x^{n} + x^{n-1} + ... +x +1 = 0 $$
is a geometric sequence and can be rewritten as:
$$x^{n} + x^{n-1} + ... +x +1 = \sum_{i=0}^{n}x^n = \frac{1-x^{n+1}}{1-x} = 0 $$
From the denominator $x-1 $ we conclude that $x\ne1$, but we can rewrite $ 1 = e^{2\pi k} $. So from:
$$ 1-x^{n+1} = 0 \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3038472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 5
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Finding a distribution with a given correlation Below is a problem that I made up and my attempt at a solution to it. I am hoping that somebody here can help me finish it. I believe there is a unique
answer to the problem.
Thanks,
Bob
Problem:
Let $X$ and $Y$ be uniformly distributed independent variables on the inte... | As you show, Pearson's $r$ should be a function of $k$. We will find that function and solve for
$r(k) = \frac{1}{2}$.
$r(A,B)$ may be calculated from the covariance $cov(A,B)$ and the standard deviations $\sigma_A$ and $\sigma_B$ as $\frac{cov(A,B)} { \sigma_A \sigma_B}$.
The covariance itself may be calculated as $E(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3038725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all Pythagorean triples $x^2+y^2=z^2$ where $x=21$ Consider the following theorem:
If $(x,y,z)$ are the lengths of a Primitive Pythagorean triangle, then $$x = r^2-s^2$$ $$y = 2rs$$ $$z = r^2+z^2$$ where $\gcd(r,s) = 1$ and $r,s$ are of opposite
parity.
According to the previous theorem,My try is the followin... | We have $21=x=k(m^2-n^2),\, y=2kmn,\, z=k(m^2+n^2)$ where $m,n, k \in \Bbb N$ with $\gcd (m,n)=1$ and $m,n$ not both odd.
So $(m^2-n^2,k)\in \{(1,21),(3,7),(7,3),(21,1)\}.$ Now $m^2-n^2=1$ is impossible, so $(m,n,k)\in \{(2,1,7), (4,3,3),(11,10,1),(5,2,1)\},$ giving $$(x,y,z)\in \{ (21,28,35), (21,72, 75),(21,220, 22... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3040030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the Arc length of the parametric curve $$x=6t-6sint$$
$$y=6-6cost$$
Find the arc length of the parametric curve
$$Arc length = \int_{0}^{2\pi} \sqrt{(6-6cost)^2+(6sint)^2}dt\\
=\int_{0}^{2\pi} \sqrt{36-72cost+36cos^2t+36sin^2t}dt \\
=\int_{0}^{2\pi} 6 \sqrt{1-2cost+cos^2t+sin^2t}dt\\
=\int_{0}^{2\pi} 6\sqrt{2-2co... | Leaving the coefficient $6$ on the side, we have
$$s=\int_0^{2\pi}\sqrt{(1-\cos t)^2+(\sin t)^2}\,dt=\int_0^{2\pi}\sqrt{2-2\cos t}\,dt=2\int_0^{2\pi}\left|\sin\frac t2\right|\,dt.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3041693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How can we solve the diophantine equation? How can we find all the primitive solutions of the diophantine equation $x^2+3y^2=z^2$ ?
Some solutions are $(\pm n , 0 , \pm n)$ for $n\in \mathbb{N}$. How can we find also the other ones?
| We can follow the stereographic projection method used to find Pythagorean triples. We start by dividing through by $z^2$, giving us the equation
$$\left(\frac{x}{z}\right)^2 + 3\left(\frac{y}{z}\right)^2 = 1.$$
Or, in other words, we search for rational points on the ellipse
$$x^2 + 3y^2 = 1.$$
We'll take the point $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to prove that $\sum_{iIn a example about U-statistics, $h(x_1,x_2)=\frac 12(x_1-x_2)^2$, then
$$U_n=\frac{2}{n(n-1)}\sum_{i<j}\frac{(X_i-X_j)^2}{2}=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2$$
I don't know how to prove it completely.
| We know that (I found it here)
\begin{equation}
\left( \sum_{n=1}^N a_n \right)^2 = \sum_{n=1}^N a_n^2 + 2 \sum_{j=1}^{N}\sum_{i=1}^{j-1} a_i a_j
\end{equation}
So using the above identity
\begin{align}
\sum_{i=1}^{n}(X_i-\bar{X})^2
&=
\sum_{i=1}^{n}(X_i-\frac{1}{n}\sum_{j=1}^nX_j)^2\\
&=
\sum_{i=1}^{n}(X_i^2-\frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the value of $\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)$
Find the value of $\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)$
My Attempt
\begin{align}
\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21}\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{... | You started by noting $\cot^{-1}21=\tan^{-1}\frac{1}{21}$ etc. The problem came when you encountered a negative argument. Since all the functions of interest are odd, $\cot^{-1}(-8)=-\cot^{-1}8=-\tan^{-1}\frac{1}{8}$, no $\pi$ involved. (Another way to prove $\tan^{-1}\frac{1}{21}+\tan^{-1}\frac{1}{13}=\tan^{-1}\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Can't solve a quartic equation I'm trying to solve an algebraic question.The question wants me to solve $n^4+2n^3+6n^2+12n+25=m^2$.The question also states that n is a positive integer and the answer for $n^4+2n^3+6n^2+12n+25$ is a square number. Here's how I tried to solve it:
$$n^4+2n^3+6n^2+12n+25=\\n^4+6n^2+2n^3+12... | I believe (correct me if I'm wrong) that $a^2-b^2$ only has factors $1, a^2-b^2, (a-b), (a+b)$ if $a$ and $b$ are coprime. Using this, we have that:
$$n^4+2n^3+6n^2+12n+25=m^2 \to n(n^3+2n^2+6n+12)=m^2-25$$
If we assume that $m\neq 5k,k\in\Bbb Z$, by what I stated earlier we have that either $n=1$ and $(n^3+...)=m^2-25... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
AMC 1834 - If $a, b $ and $c$ are nonzero numbers such that (a+b-c)/c=(a-b+c)/b=(-a+b+c)/a and x=((a+b)(b+c)(c+a))/abc and x<0, then x=? I am getting stuck on this question:
If a, b, and c, are nonzero numbers such that $\frac{a+b-c}{c}$=$\frac{a-b+c}{b}$=$\frac{-a+b+c}{a}$ and x=$\frac{(a+b)(b+c)(c+a)}{abc}$ and x<0, ... | Hint: $$a=-\frac{b}{2},b=b,c=-\frac{b}{2}$$ solves your problem and $$x=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find minimal polynomial in $GF(2^3)$ I have $GF(2^3)$ field defined by $\Pi(x)=x^3+x+1$. From literature they say those are the minimal polynomial, but I can't understand the operative method to find them. Any explanation for a general method?
$$\begin{array}{lll}
\textbf{Elem.} & \textbf{Polyn.} & \color{red}{\... | The elements of $GF(8)$ are exactly the roots of the polynomial $X^8-X\in GF(2)[X]$. This polynomial decomposes into irreducible polynomials as follows,
$$X^8-X = X(X+1)(X^3+X+1)(X^3+X^2+1).$$
If the field $GF(8)$ is given as $GF(2)[X]/\langle X^3+X+1\rangle$ and $\alpha$ is a zero of $X^3+X+1$, then (as stated above),... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $\sum_{k=1}^{\infty}\frac{k^{n+1}}{k!}=eB_{n+1}=1+\cfrac{2^n+\cfrac{3^n+\cfrac{4^n+\cfrac{\vdots}{4}}{3}}{2}}{1}$ Through some calculation, it can be shown that
$$e = 1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{\vdots}{4}}{3}}{2}}{1}\tag{1}$$
$$2e = 1+\cfrac{2+\cfrac{3+\cfrac{4+\cfrac{\vdots}{4}}{3}}{2}}{1}\t... | After some work, I recalled the method from Brilliant.org. This is not a rigorous proof, but it offers some intuitive sense.
We will first prove the first equation. Note that
$$\begin{align}
1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1+\vdots}{5}}{4}}{3}}{2}}{1}&=1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1+\cfrac{1}{5}+\cf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Find the limit of $\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$ without using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ Find the limit of the sequence $$\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal ... | $$
\begin{align}
\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
&=\lim_{n\to\infty}n\left(\left(1+\frac1n\right)^{1/3}-\left(1+\frac1{n^3}\right)^{1/3}\right)\\
&=\lim_{n\to\infty}n\left(\left[1+\frac1{3n}+O\!\left(\frac1{n^2}\right)\right]-\left[1+O\!\left(\frac1{n^3}\right)\right]\right)\\
&=\lim_{n\to\infty}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to factor this quadratic expression? A bit confused on how to factor $2x^2 + 5x − 3 = 0$. Firstly, I multiplied $a \cdot c$, so $2(-3)=-6$, however couldn't find two numbers that will add up to $5$. Then, I thought of the factors could be $6(-1)=-6$, and $6+-1=5$. However, shouldn't the $a$ and $c$ values always be... | $$2x^2 + 5x − 3 = 0$$
$$ac = 2(-3) = -6$$
$$\text{$^-1\times 6 =\phantom .^-6 \ $ and $ \ ^-1+6 = 5$}$$
$-1$ and $6$ are correct. What you did after that is wrong.
Here are two methods that I know of for proceeding from $-1$ and $6$.
Method 1. Replace $5x$ with $-1x+6x$ and factor-by-pairing-off.
\begin{array}{c}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Evaluate $\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)$
$\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)$
What are the good/ clever methods to evaluate this limit?
I tried taking $\tan^{-1} (x+5) = \theta$ to avoid inverse functions but its not helpful and makes it even more complicated.
... | Use Taylor expansion:
$$\tan^{-1}(x+1)=\frac{\pi}{2} - \frac 1x + \frac 1{x^2} - \frac{2}{3 x^3} + O\left(\frac{1}{x^5}\right);\\
\tan^{-1}(x+5)=\frac{\pi}2 - \frac 1x + \frac5{x^2} - \frac{74}{3 x^3} + \frac{120}{x^4} + O\left(\frac 1{x^5}\right);\\
\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)=\\
\lim_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Inequality with $(x+y)(y+z)(z+w)(w+x)=1$ Let $x,y,z,w>0$ and such that
$$(x+y)(y+z)(z+w)(w+x)=1.$$
Show that
$$\sqrt[3]{xyz}+\sqrt[3]{yzw}+\sqrt[3]{zwx}+\sqrt[3]{wxy}\le 2.$$
I'm trying to use Holder's inequality
$$(\sqrt[3]{xyz}+\sqrt[3]{yzw})^3
\le (x+y)(y+z)(z+w)$$
$$(\sqrt[3]{zwx}+\sqrt[3]{wxy})^3\le (z+w)(w+x)(x+y... | This does not seem the best, but using only Holder's inequality, we can get
$$\begin{eqnarray}
S^{12}&\le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}\left(\prod_{\text{cyc}}(x+y)\right)^3 =2^{12},
\end{eqnarray}$$for $S= \sqrt[3]{xyz}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
if $x-y =\sqrt{x}-\sqrt{y}$ with $x\neq y$ then $(1+\frac{1}{x})(1+\frac{1}{y})\geq 25$? let $x\neq y$ be positive real numbers such that :$x-y= \sqrt{x}-\sqrt{y}$ , I have tried to prove this inequality $(1+\frac{1}{x})(1+\frac{1}{y})\geq 25$ that i have created but i didn't got it.
Attempt I have showed that:$(\frac{... | By AM-GM
$$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\left(1+\frac{4}{4x}\right)\left(1+\frac{4}{4y}\right)\geq$$
$$\geq\frac{5}{\sqrt[5]{(4x)^4}}\cdot\frac{5}{\sqrt[5]{(4y)^4}}=\frac{25}{\sqrt[5]{4^{8}\left(\sqrt{xy}\right)^8}}\geq\frac{5}{\sqrt[5]{4^8\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^{16}}}=25.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Factor polynomials I am having trouble with these expressions:
$$x^4 - 23x^2 + 1$$
$$2y^2 - 5xy + 2x^2 - ay - ax - a^2$$
$$2x^3 - 4x^2y - x^2z + 2xy^2 - y^2z +2xyz$$
I tried to consult a chapter on factoring in my textbook. It seems to suggest the following:
Check for expressions like $a^n - b^n$ or $(a + b)^n$.
Rememb... | $$x^4-23x^2+1=x^4+2x^2+1-25x^2=(x^2-5x+1)(x^2+5x+1).$$
$$2x^2-5xy+2y^2-ax-ay-a^2=$$
$$=2x^2-5xy+2y^2+\frac{1}{4}(x+y)^2-\frac{1}{4}(x+y)^2-a(x+y)-a^2=$$
$$=\frac{9}{4}(x-y)^2-\left(\frac{1}{2}(x+y)+a\right)^2=$$
$$=\left(\frac{3}{2}(x-y)-\frac{1}{2}(x+y)-a\right)\left(\frac{3}{2}(x-y)+\frac{1}{2}(x+y)+a\right)=$$
$$=(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}$ I have stumbled upon the following integral:$$I=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}=-\frac{\pi}{24}$$
Although I could solve it, I am not quite comfortable with the way I did it.
But first I will show the ... | From the identity
$$\Im\int_0^\infty e^{-(\pi-it)x}\,dx=\frac t{\pi^2+t^2}$$
we see that it suffices to compute the imaginary part of the integral
$$\int_0^\infty dx\int_{-\infty}^\infty dt\;
\frac{e^{\alpha t}}{(1+e^t)^2}e^{-\pi x}$$
where $\alpha=1/2+ix$. Now, the integral with respect to $t$ is easy by means of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 4,
"answer_id": 3
} |
Twin Prime Formula I have a function involving polynomials and the centre of the Binomial Triangle and I'd like to prove that the function produces a positive integer infinitely many times. I don't have any interest in what values the integers take so much, merely that they exist.
The function I have is:
$$f(n) = \dfra... | Here is a proof that if both $n$ and $n+2$ are prime ($n>3$), then the output is integer:
clearly it suffices to show that if $n+2$ is prime then $$ \left(4n^3+28n^2+84n+76\right)\binom{2n}{n} \equiv 2(n+2) \bmod{(n+2)^4}.$$
Let $p=n+2$ be prime, then this is equivalent to showing that:
$$\tag1 2\left(p^3+p^2+5p-3\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Trigonometric inequality $ 3\cos ^2x \sin x -\sin^2x <{1\over 2}$ I' m trying to solve this one. Find all $x$ for which following is valid:
$$ 3\cos ^2x \sin x -\sin^2x <{1\over 2}$$
And with no succes. Of course if we write $s=\sin x$ then $\cos^2 x = 1-s^2$ and we get $$6s^3+2s^2-6s+1>0$$
But this one has no ration... | We consider the inequality you found:
$6s^3+2s^2-6s+1>0$, for $s=\sin x$
We compare left side with following equation:
$8s^3-4s^2-4s+1=0$
Which have solutions: $s=\cos \frac{\pi}{7}, \cos \frac{3\pi}{7}, \cos \frac{5\pi}{7}$
We have:
$$8s^3-4s^2-4s+1>2s^3-6s^2-2s$$
That means we can write:
$2s^3-6s^2-2s<0$ for $s=\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
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On calculating the limit of the infinite product $\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})$
Let $S_n=\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})$. What is the value of $\lim_{n \to \infty} S_n$ ?
What I attempted:-
$\log S_n=\sum_{k=3}^n \log (1-\tan^4\frac{\pi}{2^k})$.
Since $\lim_{x \to 0} \frac{\tan x}{x}=1$, $\tan^4... | You obtained an approximation of the exact value. In order to find such exact vale, note that
$$(1-\tan^4(\alpha/2))=(1+\tan^2(\alpha/2))(1-\tan^2(\alpha/2))=
\frac{4}{\cos(\alpha)}\left(\frac{\tan(\alpha/2)}{\tan(\alpha)}\right)^2.$$
Hence, as $n$ goes to infinity,
$$\prod_{k=3}^n \left(1-\tan^4(\pi/2^k)\right)=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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How do people come up with solutions like this? I was going through some problems in high school textbook and stumbled on this problem.
I could be trying to solve it whole day and I wouldn't solve it. The solution seems to be too complicated for high school student (I marked the part red I can't understand). Can you ex... | Through practice, you become more familiar with various techniques which you can use to make simplification easier and faster.
The previous answer provides a clever substitution which simplifies the problem greatly, but you can solve the question rather easily even without it. In the example, you have
$$\frac{-2ab^{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Function Optimization with Non-linear Constraint, Lagrange Multipliers Fails I am trying to maximize the function $A(x,y)=\frac{1}{2}(x(12-x)+y(13-y))$ subject to the constraint $x^2+(12-x)^2-y^2-(13-y)^2=0$.
My attempt:
$\begin{align*} \nabla A=\frac{1}{2}\langle 12-2x,\,13-2y\rangle &= \lambda\langle4x-24,\, -4y+26\r... | Let $x(12-x)=a$ and $y(13-y)=b$.
Thus, the condition gives $b=a+12.5$.
Also, we have
$$a=x(12-x)\leq\left(\frac{x+12-x}{2}\right)=36$$ and
$$b=y(13-y)\leq\left(\frac{y+13-y}{2}\right)=42.25,$$
which gives
$$a=b-12.5\leq42.25-12.5=29.75.$$
Id est, $$A(x,y)=\frac{1}{2}(a+b)=a+6.25\leq29.75+6.25=36.$$
The equality occurs ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Not getting the right answer with alternate completing the square method on $\int\frac{x^2}{\sqrt{3+4x-4x^2}^3}dx$ So I've looked up how to do this problem and when they complete the square it's using the $(\frac{-b}{a})^2$ method where, in the denominator, they factor out the $4$, and then make:
$$x^2-x=x^2-x+{1\over4... | Find $I=\int\frac{x^2}{(3+4x-4x^2)^{3/2}}dx$.
First 4 steps:
*
*Change $x=t+\frac12$
$$I=\int\frac{t^2+t+\frac14}{8(1-t^2)^{3/2}}dt\\=\frac18\int\frac{t^2}{(1-t^2)^{3/2}}dt+
\frac18\int\frac{t}{(1-t^2)^{3/2}}dt+\frac1{32}\int\frac{1}{(1-t^2)^{3/2}}dt$$
*$$\int\frac{t}{(1-t^2)^{3/2}}dt=
\frac{1}{\sqrt{1-{{t}^{2}}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Minimum value of $\frac{4}{4-x^2}+\frac{9}{9-y^2}$ Given $x,y \in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$
$$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$
Using Differentiation we... | Using $xy = -1$, we can put the fractions over a common denominator as
$${4\over{4 - x^2}} + {9\over{9 - y^2}} = {{72 - 9x^2 - 4y^2}\over{37 - 9x^2 - 4y^2}} = 1 + {35\over{37 - (9x^2 + 4y^2)}}$$
So we need to minimize $9x^2 + 4y^2$. Since $9x^2 \times 4y^2$ is constant, the minimum occurs when $9x^2 = 4y^2$, i.e. $9x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integrating positive function on interval from -1 to 1 but result is negative I have the following function:
$$\int_{-1}^1 \frac1 {x^4} dx$$
and my result seems to be:
$$\frac {-2}3$$
Why is my result negative? My function is always positive, does it have to do with the fact that since the interval extends from -1 to... | Since
$\dfrac1{x^4}$
is not defined at
$x=0$,
one definition of
$\int_{-1}^1 \frac1 {x^4} dx
$
is
$\lim_{c \to 0^+} (\int_{-1}^{-c} \frac1 {x^4} dx+\int_{c}^1 \frac1 {x^4} dx)
$.
We have
$\int \dfrac{dx}{x^4}
=\int x^{-4} dx
=\dfrac{x^{-3}}{-3}
=-\dfrac{1}{3x^3}
$
so
$\begin{array}\\
\int_{-1}^{-c} \frac1 {x^4} dx
&=-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3077833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve recurrence relation: $T(n) = \frac{n}{n+1}T(n-1) + 1$ I am not able to solve this recurrence relation by substitution and variable change method.
$$T(n) = \frac{n}{n+1}T(n-1) + 1;\ \ T(1) = 1 $$
| (n+1)T(n) = nT(n-1) + n+1
Let
S(n) = (n+1)T(n) S(0) = T(0) = c
S(n) = S(n-1) + n+1
Since, S(n-1) = S(n-2) + n
therefore, S(n) = S(n-2) + n+1 + n
for k terms...
S(n) = S(n-k) + (n+1) +n +.....+ (n-k+2)
k = n
S(n) = S(0) + (n+1) + n + ...+ (n-k+2)
= c + (n+1) + n + (n-1) + ........ | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3078589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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What's the answer to $\int \frac{\cos^2x \sin x}{\sin x - \cos x} dx$? I tried solving the integral $$\int \frac{\cos^2x \sin x}{\sin x - \cos x}\, dx$$ the following ways:
*
*Expressing each function in the form of $\tan \left(\frac{x}{2}\right)$, $\cos \left(\frac{x}{2}\right)\,$ and $\,\sin \left(\frac{x}{2}\righ... | $$\int\frac{\cos^2x\sin{x}}{\sin{x}-\cos{x}}dx=$$
$$=\int\left(\frac{\cos^2x\sin{x}}{\sin{x}-\cos{x}}+\frac{1}{2}\sin{x}(\sin{x}+\cos{x})\right)dx-\frac{1}{2}\int\sin{x}(\sin{x}+\cos{x})dx=$$
$$=\frac{1}{2}\int\frac{\sin{x}}{\sin{x}-\cos{x}}-\frac{1}{2}\int\sin{x}(\sin{x}+\cos{x})dx=$$
$$=\frac{1}{2}\int\left(\frac{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
How to find the equation of the conic before applying the rotation? Given the rotation matrix:
$$Q=\begin{bmatrix}\frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\\-\frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}\end{bmatrix},$$ I want to find the equation of the conic $C$ given by $$x^2+4xy+4y^2-10\sqrt{5}x=0,$$ which I know is a parabo... | Your approach is entirely correct; plugging in your substitutions into the first equation for $C$ yields the result you claim it should be:
\begin{eqnarray*}
x^2+4xy+4y^2-10\sqrt{5}x&=&
\left(\frac{2}{\sqrt{5}}\tilde{x}+\frac{1}{\sqrt{5}}\tilde{y}\right)^2+4\left(\frac{2}{\sqrt{5}}\tilde{x}+\frac{1}{\sqrt{5}}\tilde{y}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Help with how to show aritmetic progression question. How can I show that if $(\chi_{n})$ is a aritmetic progression, then:
$$\frac{1}{ \sqrt{\chi_{1}} + \sqrt{\chi_{2}} } +
\frac{1}{ \sqrt{\chi_{2}} + \sqrt{\chi_{3}} } + \cdots +
\frac{1}{ \sqrt{\chi_{n-1}} + \sqrt{\chi_{n}} } =
\frac{n-1}{ \sqrt{\chi_{1}} + \sqrt{\... | $ \frac{1}{ \sqrt{\chi_{1}} + \sqrt{\chi_{2}} } +
\frac{1}{ \sqrt{\chi_{2}} + \sqrt{\chi_{3}} } + \cdots +
\frac{1}{ \sqrt{\chi_{n-1}} + \sqrt{\chi_{n}} } = \frac{\sqrt{\chi_{2}} - \sqrt{\chi_{1}} }{ \chi_2-\chi_1 } + \frac{\sqrt{\chi_{3}} - \sqrt{\chi_{2}} }{ \chi_3-\chi_2 } + \cdots +
\frac{\sqrt{\chi_n}-\sqrt{\chi_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3083085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Write a Limit to calculate $f'(0)$ Let $f(x) = \frac {2}{1+x^2} $
I need to write a limit to calculate $f'(0)$.
I think I have the basic understanding. Any help would be greatly appreciated.
d=delta and so far what I have is
$f'(x)$= lim (f(x+dx)-f(x))/dx
(dx)->0
((2/1+(x+dx)^2)-(2/1+x^2))/dx
((2/1+x^2+... | it is $$\frac{f(x+h)-f(x)}{h}=\frac{\frac{2}{1+(x+h)^2}-\frac{2}{1+x^2}}{h}$$
Can you finish?
Ok, another hint:
The numerator is given by $$-2\,{\frac {h \left( h+2\,x \right) }{ \left( {h}^{2}+2\,xh+{x}^{2}+1
\right) \left( {x}^{2}+1 \right) }}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3083591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding real $a$, $b$, $c$ such that $x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$ has $1+i$ as a zero, and one negative integer as a zero with multiplicity $2$
Find $a, b, c \in \mathbb{R}$, if one zero of $p$ is $1+i$ and $p$ has one negative integer zero with multiplicity $2$, where:
$$p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x... | Since all coefficients are reals, $p$ is divisible by $$(x-1+i)(x-1-i)=(x-1)^2+1=x^2-2x+2.$$
We have $$p(x)=x^5-2x^4+ax^3+bx^2-2x+c=$$
$$=x^5-2x^4+2x^3+(a-2)x^3-2(a-2)x^2+2(a-2)x+(b+2a-4)x^2-(2a-2)x+c=$$
$$=(x^2-2x+2)(x^3+(a-2)x)+(b+2a-4)x^2-(2a-2)x+c.$$
We need $$b+2a-4=k,$$ $$2a-2=2k$$ and $$c=2k,$$ which gives
$$b=3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$ when $a,b$ are integers? Let $a$ and $b$ be positive integers.
If $b$ is even, then we have $$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$$
I think the equality also hold when $b$ is odd. Wh... | The equality only depends on the parity of $a+b$, as it is the same as that of $a-b$. Then
$$\left\lfloor\frac02\right\rfloor+\left\lceil\frac02\right\rceil=0$$
and
$$\left\lfloor\frac12\right\rfloor+\left\lceil\frac12\right\rceil=1$$
are enough as a proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$ If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$
My attempt:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$
| Hint — using this changing variable $y = cos^2(x)$ —
$$y^3 + y^2-2y = 0 \Rightarrow y(y^2+y-2) = 0 \Rightarrow y(y+2)(y-1) = 0 $$
As $y = cos^2(x)$ and $0 \leq y \leq 1$:
$$y = 0, 1 \Rightarrow cos^2(x) = 0, 1 \Rightarrow x = 0, \frac{\pi}{2},$$
for $x \in [0, \frac{\pi}{2}]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Idea for $\lim\limits_{x \to \frac{\pi}{2}} \left( \tan \left( \tfrac{\pi}{4} \sin x\right)\right)^{1/ ( \tan (\pi \sin x))}$ $$\lim_{x \to \frac{\pi}{2}} \tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right).$$
Need help for this. I tried using trigonometric identities but nothing seems to fi... | Using trigonometric identities, by $y=\frac \pi 2 \sin x \to \frac \pi 2$ and $t=\frac \pi 2 -y \to 0$, we obtain:
*
*$\tan \frac{\theta}{2} = \frac{\sin \theta}{1 +
\cos \theta} \implies \tan \left(\frac{\pi}{4}\sin x\right)=\frac{\sin y}{1 +
\cos y}=\frac{\cos t}{1 +
\sin t}$
*$\tan (2\theta) = \frac{2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\int_{0}^{1}\frac{1+x+x^2}{1+x+x^2+x^3+x^4}dx$ Evaluate $$I=\int_{0}^{1}\frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=\frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$
So we get:
$$I=\int_{0}^{1}\frac{1-x^3}{1-x^5}dx$$
$$I=1+\int_{0}^{1}\frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from her... | Hint:
As the roots of the denominator are the complex fifth roots of unity, you find the factorization to be
$$1+x+x^2+x^3+x^4=\left(x^2+\frac{1+\sqrt5}2x+1\right)\left(x^2+\frac{1-\sqrt5}2x+1\right).$$
Then we can find a linear combination to obtain a quadratic polynomial with equal coefficients,
$$\left(\sqrt5+1\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
System of linear recurrences; finding an explicit description for both sequences involved. $$\left\{\begin{aligned} a_n &&= &&2a_{n-1} + b_{n-1} + a_{n-2} - b_{n-2} && n \ge 2 && (1)\\ b_n &&=&& b_{n-1} + b_{n-2} - a_{n-2} && n \ge 2&& (2)\end{aligned}\right.$$
with $a_0 = 5, a_1 = 3, b_0 = 0, b_1 = 3$.
From (2) I get:... | Hint:
Adding the two equations gives
$$
a_n + b_n = 2(a_{n-1} + b_{n-1}) \quad {\rm{for}} \quad n \ge 2
$$
so
$$
a_n + b_n = 2^{n-1}(a_{1} + b_{1}) = 6 \cdot 2^{n-1}
$$
Plugging this into the second recursion gives
$$
b_n = b_{n-1} + 2 b_{n-2} - (a_{n-2} + b_{n-2}) = b_{n-1} + 2 b_{n-2} -6 \cdot 2^{n-3}
$$
Likewis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Improving the bound for $\sigma(q^k)/q^k$ where $q^k n^2$ is an odd perfect number given in Eulerian form Let $x$ be a positive integer. (That is, let $x \in \mathbb{N}$.)
We denote the sum of divisors of $x$ as
$$\sigma(x) = \sum_{d \mid x}{d}.$$
We also denote the abundancy index of $x$ as $I(x)=\sigma(x)/x$.
If $N$... | It turns out that
$$I(q^k) < \sqrt[3]{\frac{2}{I(q^k)}}$$
implies
$$1 + \frac{1}{q} = I(q) \leq I(q^k) < \sqrt[4]{2}$$
from which we obtain
$$q > \bigg(\sqrt[4]{2} - 1\bigg)^{-1} \approx 5.2852135$$
thereby giving
$$q \geq 13$$
since $q$ is a prime satisfying $q \equiv 1 \pmod 4$. Thus, the implication
$$I(q^k) < \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$ 0=1 $ ? Where is the mistake? I just found this formula although it can be easily derived.
Let $ n $ be any integer then,
$$n=\sqrt{n^2-n+\sqrt{n^2-n+.....}}$$
So if I plug in $ 0 $ in this equation I get,
$$0=\sqrt{0-0+\sqrt{0-0+....}}$$
$$0=\sqrt{0+\sqrt{0+\sqrt{0+....}}}$$—————->1
But if I plug in $1$ in the equ... | Assuming convergence:
$$a=\sqrt{n^2-n+\sqrt{n^2-n+.....}}$$
$$a=\sqrt{n^2-n+a}$$
$$a^2=n^2-n+a$$
$$a^2-a=n^2-n$$
$$a(a-1)=n(n-1)$$
For the case $n=1$:
$$a(a-1)=0$$
Notice that $a$ has roots $1$ and $0$. Why did you assume $a$ to be one of them, $1$? All $a$ that satisfies $a=\sqrt{n^2-n+a}$ and is positive is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluate ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$
Evaluate $${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$$
It should be $$\frac{1}{12}n(n+1)(2n+1)$$ but I don't know how to prove that. I am also aware that $\lim\limits_{\theta \to 0} \dfrac{1-\cos \theta}{\theta ^2}=... | the series expansion at $x=0$ of $ \ \cos(ax)=1-\frac{a^2 x^2}{2}+o(x^4) \ \ $ and
$\cos(a x) \cos(bx)=\frac{1}{2}\cos(x(a-b))+\frac{1}{2}\cos(x(a+b))$
the series of $ \ \frac{1}{2}\cos(x(a-b))+\frac{1}{2}\cos(x(a+b))=\frac{1}{2}(1-\frac{(a-b)^2 x^2}{2}+o(x^4) )+\frac{1}{2}(1-\frac{(a+b)^2 x^2}{2}+o(x^4) )=$
$\cos(a x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Quick way of solving the contour integral $\oint \frac{1}{1+z^5} dz$ Consider the contour integral in the complex plane:
$$\oint \frac{1}{1+z^5} dz$$
Here the contour is a circle with radius $3$ with centre in the origin. If we look at the poles, they need to satisfy $z^5 = -1$. So the solutions of the poles are given ... | Partial fractions can work. I think this approach is pretty much equivalent to Mark Viola's residue one, but it's nice to see the agreement.
Let $\zeta$ be a primitive fifth root of unity. If you want to be explicit, let $\zeta = e^{2\pi/5}$. But the important thing is that
*
*the roots of $z^5+1$ are $-1$, $-\ze... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find all possible values of $x$ if $\frac{\sin\frac{x}{2}+\cos\frac{x}{2}-i\tan x}{1+2i\sin\frac{x}{2}}$ is real
If the expression $\dfrac{\sin\frac{x}{2}+\cos\frac{x}{2}-i\tan x}{1+2i\sin\frac{x}{2}}$ is real, find the set of all possible values of $x$.
My Attempt
$$
-\tan x-2\sin^2\frac{x}{2}-2\sin\frac{x}{2}\cos\f... | The second expression $\cos \frac{x}{2} +\sin \frac{x}{2}\cos x+\cos \frac{x}{2}\cos x=0$ reduces to :
$$\cos \frac{x}{2} + \cos x(\sin \frac{x}{2}+ \cos \frac{x}{2})=0$$
$$\cos \frac{x}{2} = -\cos x(\sin \frac{x}{2}+ \cos \frac{x}{2})$$
$$\cos x = -\frac{1}{1+\tan \frac{x}{2}}$$
Apply the half angle formula for $$\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why does the CRT formula yield a solution of a congruence system? I understand there is a method for solving simultaneous modular equations. For example;
$$x = 2 \mod{3}$$
$$x = 3 \mod{5}$$
$$x = 2 \mod{7}$$
We find numbers equal to the product of every given modulo except one of them - giving $5 \cdot 7$, $3 \cdot 7$ ... | Taking Bill Dubuque's graphic answer and graphically expanding on it:
$x = 2 \cdot\overbrace{ (5 \cdot 7) \cdot 2}^{\equiv 1 \pmod 3\\ \equiv 0 \pmod 5\\ \equiv 0 \pmod 7} + 3 \cdot \overbrace{(3 \cdot 7) \cdot 1}^{\equiv 0 \pmod 3\\ \equiv 1 \pmod 5\\ \equiv 0 \pmod 7} + 2 \cdot \overbrace{(3 \cdot 5) \cdot 1}^{\equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Convergence and divergence of a Complex Series I've been given the following series:
$$\frac{z}{1-z^2} + \frac{z^2}{1-z^4} + \frac{z^4}{1-z^8} + ...$$
and been told to investigate the convergence. Clearly this diverges if $z=1$ (possibly if $\vert{z}\vert = 1$?), but other than that I am at a loss as to how to proceed.... | We can show inductively that
$$\begin{align*}
\sum_{j=0}^n \frac{z^{2^j}}{1-z^{2^{j+1}}}&=\frac{\sum\limits_{k=1}^{2^{n+1}-1}z^k}{1-z^{2^{n+1}}}\\&=\frac{z-z^{2^{n+1}}}{(1-z)(1-z^{2^{n+1}})}\\&=\frac{z-1+1-z^{2^{n+1}}}{(1-z)(1-z^{2^{n+1}})}\\&=-\frac{1}{1-z^{2^{n+1}}}+\frac1{1-z}.
\end{align*}$$ The base case $n=0$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Asking about $\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{n^2+n+4372}{(2n+1)^7(n+1)}\right]$ $$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{n^2+n+4372}{(2n+1)^7(n+1)}\right]=\frac{61}{184320}\pi^7\tag1$$
Step 1:
$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{... | It comes from telescoping series:
$$
\frac{(-1)^n}n\left(\frac{1}{(2 n+1)^7 (n+1)}+\frac{1}{(2 n-1)^7 (n-1)}\right) = (-1)^n\left(\frac{1}{n-1} + \frac{1}{n}\right) +
(-1)^n\left(\frac{1}{n}+\frac{1}{n+1}\right)
- 4(-1)^n\left(\frac{1}{(2 n-1)^7}+\frac{1}{(2 n+1)^7}\right)
- 4(-1)^n\left(\frac{1}{(2 n-1)^5}+\frac{1}{(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Proving Table of Integral Integral (Trigonometric Substitution) I need help with the proof of the table of integral that:
$$\int \frac{\sqrt{a^2 + u^2}}{u} du = \sqrt{a^2 + u^2} + a \ln \left|\frac{\sqrt{a^2 + u^2} - a}{u}\right| + c$$
Must solve using trigonometric substitution.
I understand that you have to use $a^2\... | Let $u=a \tan x \tag{1}$ Then:- $u^2=a^2\tan^2 x \implies 2u \cdot du= 2a^2\tan x \cdot \sec^2x \cdot dx $
The integral reduces to :- $$ a\int \sec^2x \csc x \cdot dx $$
$$ = a\int (\tan^2x +1) \csc x \cdot dx = a\left(\int \sec x \tan x \cdot dx + \int \csc x \cdot dx \right)\\
= a\left(\sec x - \ln(\cot x + \csc ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why $\frac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \frac{\pi}{4}$? Since
$$\dfrac{d}{dx} \left( \dfrac{1}{2} \arctan(x) \right) = \dfrac{d}{dx} \left( \arctan(x-\sqrt{x^2+1}) \right) $$
then the format of their graphs are the same, but separated by a constant, which is $\dfrac{\pi}{4}$. That is,
$$\dfrac{1}{2} \ar... | I'm going to use $y$ instead of $x$.
Let $P(\theta) = (1, y)$. Then $$\theta = \arctan(y).$$
Then
$$\sin \theta = \dfrac{y}{\sqrt{1+y^2}}
\qquad \text{and} \qquad
\cos \theta = \dfrac{1}{\sqrt{1+y^2}}$$
Then \begin{align}
\tan \bigg( \frac 12 \theta - \frac{\pi}{4} \bigg)
&= \dfrac{\tan \frac 12 \theta ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Find a generating function for which $A(n)={n \choose 2}$ In the book I'm using, $A(x)$ denotes the formal power series (generating function), $A(x) = \sum a_ix^i$.
I'm really stuck on this problem. Thanks for any help.
My attempt after the given hint:
$$\begin{align}
A(x)&=\sum_{n\geq 0} \binom{n}{2}x^n\\
&=\sum_{n\ge... | Since $\binom{n}{2}x^n=\frac12 x^2\frac{d^2}{dx^2}x^n$, $$\sum_{n\ge 0}\binom{n}{2}x^n=\frac12 x^2\frac{d^2}{dx^2}\frac{1}{1-x}=\frac{x^2}{(1-x)^3}.$$We can double-check by the binomial theorem: the $x^n$ coefficient is $$\frac{(-1)^n}{(n-2)!}\prod_{j=1}^{n-2}(-2-j)=\frac{n!}{(n-2)!2!}=\binom{n}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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In a triangle $ABC$, $\cos 3A + \cos 3B + \cos 3C = 1$, then find any one angle.
In a triangle $ABC$, $\cos 3A + \cos 3B + \cos 3C = 1$, then find any one angle.
HINT: The answer is $\frac{2\pi}{8}$
I have tried these steps and got stuck in the middle.
$$ A +B + C = \pi$$
$$ A + B = \pi - C$$
$$\cos 3A + \cos 3B + \c... | Given $$\cos 3A+\cos3B+\cos3C=1$$
$$\implies\sum\cos3A=1$$
$$\implies\sin\dfrac{3A}{2}\cdot\sin\dfrac{3B}{2}\cdot\sin\dfrac{3C}{2}=0$$
Therefore,$$A=\dfrac{2\pi}{3}\mbox{ or }B=\dfrac{2\pi}{3}\mbox{ or }C=\dfrac{2\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove these two functions are always equal? If $a$ and $b$ are positive integers and $/$ stands for integer division, we have these two functions:
$$f(a,b) = (a + b - 1) / b$$
and
$$g(a,b) =
\begin{cases}
a/b, & \text{if $a \mod b = 0$} \\[2ex]
a / b + 1, & \text{if $a \mod b \neq 0$}
\end{cases}
$$
We can see... | Notice that we can express integer division in terms of vanilla-flavoured division using the floor function. For example, your function $f$ could be expressed
$$f(a,b)=\left\lfloor\frac{a+b-1}{b}\right\rfloor=\left\lfloor\frac{a-1}{b}+1\right\rfloor=\left\lfloor\frac{a-1}{b}\right\rfloor+1$$
and $g$ could be expressed
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Reduction Formulae Integral $x(1-x^3)$ The question asks us:
"If
$$u_n=\int_0^1x(1-x^3)^ndx$$
show that
$$u_n= \frac {3n}{3n+2}u_{n-1}$$
I've tried integration by parts using a coefficient of $1, x$ and even tried reducing the $1-x^3$ term into its factors but with no progress. Any help would be greatly apprecia... | \begin{align}
u_n &= \int_0^1 \frac{1}{2}x^2 n(1-x^3)^{n-1} 3x^2 dx \\
&= \frac{3}{2}n\int_0^1 x x^3 (1-x^3)^{n-1} dx \\
&= \frac{3}{2} n \int_0^1 x(x^3-1+1)(1-x^3)^{n-1}dx \\
&= -\frac{3}{2}n u_n + \frac{3}{2}n u_{n-1}
\end{align}
Therefore,
$$
u_n = \frac{3n}{2+3n} u_{n-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Integral of $\frac{1}{\sqrt{(z-z')^2 + s^2}}$ I have a question about the signs of the antiderivative when one integrate $\frac{1}{\sqrt{(z-z')^2 + s^2}}$.
According to Wolfram Alpha here and here:
If one evaluates $\int \frac{1}{\sqrt{(z-z')^2 + s^2}} dz$, they get $\log (z-z' + \sqrt{(z-z')^2 + s^2}) + C$.
Evalutatin... | Notice that
\begin{align}
-\log{(z-z'+\sqrt{(z-z')^2+s^2})} &= \log{\left(\frac{1}{\sqrt{(z-z')^2+s^2} -(z'-z)}\right)} \\
&= \log{\left(\frac{\sqrt{(z-z')^2+s^2} +(z'-z)}{(\sqrt{(z-z')^2+s^2} -(z'-z))(\sqrt{(z-z')^2+s^2} +(z'-z)}\right)} \\
&= \log{\left(\frac{\sqrt{(z-z')^2+s^2} +(z'-z)}{s^2}\right)} \\
&= \log{(\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
a,b,c are three real numbers where $a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$. Now $abc$ = ? Here will the answer be a number? I want to know whether it is possible to get a real number (not an algebraic expression) as the product of $a$, $b$ and $c$.
I tried for a long time and this is what I got.
$$3abc =... | We start with $$a + \frac 1b = b + \frac 1c = c + \frac 1a$$
This makes three equalities:
\begin{align}a + \frac 1b &= b + \frac 1c\tag{1}\\
b + \frac 1c &= c + \frac 1a\tag{2}\\
c + \frac 1a &= a + \frac 1b\tag{3}\end{align}
First consider equation $(1)$ $$a + \frac 1b = b + \frac 1c$$ which can be rearranged to give ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Stuck at proving whether the sequence is convergent or not I have been trying to determine whether the following sequence is convergent or not. This is what I got:
Exercise 1: Find the $\min,\max,\sup,\inf, \liminf,\limsup$ and determine whether the sequence is convergent or not:
$X_n=\sin\frac{n\pi}{3}-4\cos\frac{n\... | Computing the first $6$ terms alone does not prove that $\min X_n=-4$ nor $\max X_n=4$ but rather that $\min X_n\le-4$ and $\max X_n \ge 4$ as it does not say anything about the rest of the sequence.
Hint : $X_{n+6}=X_n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Radical equation - can I square both sides with more than 1 radical on one side? I'm familiar with equations like:
$\sqrt{x+1} - \sqrt{x+2} = 0 $
Has no solutions, it's just an example off the top of my head
Just move the negative square root to the other side, square both sides and solve.
$\sqrt{x+1} = \sqrt{x+2}$
$x... |
Hint: $(\sqrt{x+1} - \sqrt{x+2})^2 = (x+1) - 2\sqrt{x+1}\sqrt{x+2} + (x+2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Prove that $(a+b) (a^2 + b^2) (a^4 + b^4)...(a^{32} + b^{32}) = a^{64} - b^{64}$ if $b = a-1$ Prove that if $b = a-1$, then
$(a + b) (a^2 + b^2 ) (a^4 + b^4 ) ... (a^{32} + b^{32} ) = a^{64} - b^{64}$ .
I saw this in a website and it wrote this:
hint:
Write down the equality $1 = a+b$ and use the formula $k^2 -n^2 = (... | hint: $a+b = \dfrac{a^2-b^2}{a-b}, a^2+b^2= \dfrac{a^4-b^4}{a^2-b^2}, ...$, can you see the factors that cancel each other ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate : $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
Evaluate: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
My solution:
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}} & = \lim_{x\to -\infty} \frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\\
& = \fra... | Dividing both the numerator and the denominator by $x$ is always allowed, but the square root creates a trap:
$$\frac{\sqrt{a}}x=\text{sgn }x\sqrt{\frac a{x^2}}$$ because the square root is always a positive number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Limit involving inverse functions When I am given the limit
$$\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2+1}}$$
would it be possible to evaluate it giving some substitution?
L'Hospital's rule seemed an option but I ended up going in circles.
| You may proceed as follows:
*
*Set $\tan y = \sqrt{1+x^2}$ and consider $y \to \frac{\pi}{2}^-$
\begin{eqnarray*}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2+1}}
& = & \sqrt{\tan^2y -1}\frac{y}{\tan y} \\
& = & \frac{\sqrt{\sin^2 y - \cos^2 y}}{\sin y}\cdot y\\
&\stackrel{y \to \frac{\pi}{2}^-}{\longrightarrow} & \frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Question from the 2011 IMC (International Mathematics Competition) Key Stage III paper, about the evaluation of a quadratic equation
When $a=1, 2, 3, ..., 2010, 2011$, the roots of the equation $x^2-2x-a^2-a=0$ are $(a_1, b_1), (a_2, b_2), (a_3, b_3),\cdots, (a_{2010}, b_{2010}), (a_{2011}, b_{2011})$ respectively. Ev... | My first thought is to "complete the square". The given equation, $x^2- 2x- a^2- a$ is the same as $x^2- 2x= a^2+ a$. We can "complete the square" on the left by adding 1 to both sides: $x^2- 2x+ 1= (x- 1)^2= a^2+ a+ 1$. So $x- 1= \pm\sqrt{a^2+ a}$ and $x= 1\pm\sqrt{a^2+ a}$. Now pair the reciprocals: $\frac{1}{a_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Understanding conditional probability. Consider the problem $(b)$ (from Hwi Hsu):
This problem can be visually represented in either of the three very closely related ways:
Why is $\frac{1}{3}$ only the correct answer?
| 1) Probability of rolling doubles given no other information. $P(A)$
$\begin{array}( \color{blue}{(1,1)} & (1,2)& (1,3)&(1,4)&(1,5)&(1,6)\\
(2,1) & \color{blue}{(2,2)}& (2,3)&(2,4)&(2,5)&(2,6)\\
(3,1) & (3,2)& \color{blue}{(3,3)}&(3,4)&(3,5)&(3,6)\\
(4,1) & (4,2)& (4,3)&\color{blue}{(4,4)}&(4,5)&(4,6)\\
(5,1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2}dx$
How can we prove $$\int_0^1\frac{x^{2/3}(1-x)^{-1/3}}{1-x+x^2}\mathrm{d} x=\frac{2\pi}{3\sqrt 3}?$$
Thought 1
It cannot be solved by using contour integration directly. If we replace $-1/3$ with $-2/3$ or $1/3$ or something else, we can use contour integratio... |
Here we piggy back off the solution posted by @pisco, organize the analysis with detail on the definitions of $\arg(z)$ and $\arg(1-z)$, and finish by evaluating the resiudes enclosed by the closed "keyhole contour."
Let $f(z)$ be the function given by
$$f(z)=\frac{z^{2/3}(1-z)^{-1/3}}{z^2-z+1}$$
where choose the b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int_0^1 x^n\left\{\frac{k}{x}\right\}dx$ I am trying to solve the following integral containing fractional part function (denoted by $\{.\}$)
$$\int_0^1 x^n\left\{\frac{k}{x}\right\}dx,\ 0<k\le 1,\ n\in \mathrm N^*$$
For $n=0$, it is already known that $\int_0^1 \left\{\frac{k}{x}\right\}dx=k(1-\gamma-\ln ... | I will try to redo the work from beggining and try to color in red what it's not right, let me know if I am missing something too please.$${I(n,k)=\int_0^1 x^{n}\left\{\frac{k}{x}\right\}dx }\overset{\large x=\frac{k}{t}}=k^{n+1}\int_k^\infty \frac{\left\{t\right\}}{t^{n+2}}dt=$$
$$=k^{n+1} \int_k^1 \frac{t-\lfloor t \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.