Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Why does $\cos(\pi/2-(\pi/2-x)) = \cos(x)$? In proving $\sin(\pi/2 - x) = \cos(x)$, in my book its given that
$$\sin \left(\frac{\pi}{2} - x\right) = \cos \left( \frac{\pi}{2} - \left(\frac{\pi}{2} - x \right)\right) = \cos x$$
So I understand this but im confused about how $\cos x$ is being obtained from $\cos(\pi/2-(... | \begin{align*}
\sin\left(\frac{\pi}{2}-x\right) &= \left(\sin \frac{\pi}{2}\right)\cos x - \cos \frac{\pi}{2} \sin x \\
&= \cos x &\left(\cos \frac{\pi}{2} = 0 \text{ and } \sin \frac{\pi}{2} =1 \right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation... | Is this method allowed ?
$$\left[\begin{array}{rr|rr}
3 & 2 & 36 \\
5 & 4 & 64
\end{array}\right]
\sim
\left[\begin{array}{rr|rr}
1 & \frac{2}{3} & 12 \\
5 & 4 & 64
\end{array}\right]
\sim \left[\begin{array}{rr|rr}
1 & \frac{2}{3} & 12 \\
0 & \frac{2}{3} & 4
\end{array}\right] \sim \left[\begin{array}{rr|r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
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"answer_id": 6
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Using the binomial expansion to approximate a square root.
Find the first four terms of the expansion $\sqrt{1-4x}$ in ascending powers of x.
Hence, approximate $\sqrt{6}$ to four decimal places.
I've expanded it to form this: $1-2x-2x^2-4x^3$
When approximating the value of $\sqrt{6}$, bearing in mind that for the... | Here we look at some arguments why taking out a factor $2.5$ is a convenient choice.
Problem: We want to (manually) approximate $\sqrt{6}$ by using the first few terms of the binomial series expansion of
\begin{align*}
\sqrt{1-4x}&= \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}(-4x)^n\qquad\qquad\qquad\qquad |x|<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Calculate $\lim{a_{n}}$ for $a_{n}=((1+\frac{1}{n^{2}})^{n^{2}}(1-\frac{1}{n})^{n}(1+\frac{1}{n}))^{n}$ $$a_{n}=((1+\frac{1}{n^{2}})^{n^{2}}(1-\frac{1}{n})^{n}(1+\frac{1}{n}))^{n}=(e^{\ln((1+\frac{1}{n^{2}})^{n^{2}})+\ln((1-\frac{1}{n})^{n})+\ln((1+\frac{1}{n})^{n})})^{n}$$Then I can use Taylor's theorem - because func... | You know $$\left(1 + \frac{1}{n}\right)^n \to e$$ as $n \to \infty$. As for
$$\left(1 + \frac{1}{n^2}\right)^{n^3} \left(1 - \frac{1}{n}\right)^{n^2}$$
take logarithms and use $\log(1 + x) = x - x^2/2 + \dots$ as $x \to 0$ to get
$$ n^3 \log\left(1+\frac{1}{n^2}\right) + n^2 \log\left(1 - \frac{1}{n}\right)= n - \dots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $a,b$ are even integers I'm trying to prove the following theorem:
Let $a,b\in\mathbb{Z}$ . Then $$a^{2}+b^{2}\equiv0\pmod 4 \iff a \;\text{and}\; b\;\text{are even}$$
I always struggle to prove some number is odd or even. How to prove it?
I thought of using the $(a+b)^2=a^2+2ab+b^2$ formula but not sur... | $$a^{2}+b^{2}\equiv0\pmod 4 \iff(a+b)^2-2ab\equiv 0\pmod{4}.$$
If both $a$ and $b$ are odd, then:
$$\begin{align}a+b\equiv 0\pmod{2} &\Rightarrow (a+b)^2\equiv 0\pmod{4}, \text{but}\\
2ab\equiv 0 \pmod{2} &\Rightarrow \qquad 2ab\not\equiv 0\pmod{4}\end{align}$$
If $a$ is odd and $b$ is even, then:
$$\begin{align}2ab&\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show With Induction that $2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$ Show with induction that
$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$
My solution:
Base case 1: n = 1
LHS = $(1+1)\cdot2^{1-1} = 2$
RHS = $1\cdot2^{1}= 2$
Case 2: n = p
When... | $LHS_{p+1} = RHS_{p+1}$
$p\cdot2^{p} + (p+2)\cdot2^{p}=(p+1)\cdot2^{p+1}$
$2^{p}(p+(p+2) = (p+1)2^{p}\cdot2$
$2p+2=2p+2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$
Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$
My try:
$$\sin x=x+r_{1}(x), r_{1}(x)=o(x)$$ $$\tan x=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x), r_{2}(x)=o(x^5)$$ So: $$\sin \tan x=(x+\frac{x^3}{3}+\frac{2}{15}x^5)+r_{3}(x+\f... | The organized and detailed solution must be:
$$\sin y=y-\frac{y^3}{6}+\frac{y^5}{120}+O(y^7);\\
\sin \overbrace{\tan x}^{y}=\overbrace{\tan x}^{y}-\frac{(\overbrace{\tan x}^{y})^3}{6}+\frac{(\overbrace{\tan x}^{y})^5}{120}+O((\overbrace{\tan x}^{y})^7)=\\
\tan x=\color{blue}{x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)};\\
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Representing a function as a power series and finding the coefficients of the terms? My prblem asks me to find the coefficients of the first five terms of the power series representation of the function:
$$f(x)=\frac{10}{17+x}$$
$$f(x)=10*\frac{\frac{1}{17}}{1-\frac{-x}{17}}$$
$$\sum_{n=0}^{\infty}10*\frac{1}{17}*(\fra... | The coefficients of the terms of a Maclaurin power series for a function $f(x)$ are numbers $a_0,a_1,a_2,\dots$ such that $$f(x)=\sum_{n=0}^\infty a_nx^n$$ for all $x$ within the interval of convergence of the series. You got very close to finding them, but unfortunately, didn't quite get all the way there, because you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit for $e$ and $\frac{1}{e}$ My question concerns the derivation of this: $$e^r = \lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n \ \ ...(1).$$
One of the definitions of $e$ is as follows:
$$e = \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n.$$
Then, textbooks usually derive equation (1) in the ... | Let $P= \left (1 - \frac 1 n \right )^n.$ Then $\ln P = \frac {\ln \left (1 - \frac 1 n \right )} {\frac 1 n}.\ (*)$
Now as $n \rightarrow \infty$ then $(*)$ is a $\frac 0 0$ form. Applying L'Hospital we have $$\begin{align*} \lim\limits_{n \rightarrow \infty} \ln P & = \lim\limits_{n \rightarrow \infty} \frac {\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
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Does digit $6$ always lead to $\ 25921=161^2\ $? Consider prime numbers with the property that the product of the factorials of the digits plus $1$ is a perfect square, for example the prime $$30241$$ leads to the square $$3!\cdot 0!\cdot 2!\cdot 4!\cdot 1!+1=17^2$$
If the prime number starts with digit $6$, the only s... | A comment on strategy too long for the comment section, but not a complete answer:
You want to find $6!\prod a_i! +1=x^2$ where $a_i$ are single digits. Note that the single digits are $0,1,2,3,2^2,5,2\cdot 3,7,2^3,3^2$, so the factorials will all either be $1$ or will have only the prime factors $2,3,5,7$.
Rearranging... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Evaluation of a sum by means of Poisson sum formula and digamma function I have the following series:
$$\sum_{n=-\infty}^{\infty}\frac{1}{(2n+1)^2\pi^2+a^2}=\frac{1}{2a}\tanh\left(\frac{a}{2}\right)$$
and on the text it is written that it can be proven by means of either Poisson sum formula or digamma function. However... | To make it short, start with $$(2n+1)^2\pi^2+a^2=( (2 n+1)\pi -i a)\, ( (2 n+1)\pi +i a)$$ Using partial fraction decomposition
$$\frac 1 {(2n+1)^2\pi^2+a^2}=\frac i {2a}\left(\frac 1{(2 n+1)\pi +i a } -\frac 1{(2 n+1)\pi -i a } \right)$$ make
$$S_m=\sum_{n=-m}^{m}\frac{1}{(2n+1)^2\pi^2+a^2}$$ $$S_m=\frac{2 \pi a+\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Solve for x: $x^3-\lfloor x\rfloor=5$
Solve for x:$$x^3-\lfloor x\rfloor=5$$
My Attempt:
$x^3-5=\lfloor x\rfloor$
Now, $x-1<\lfloor x\rfloor\leq x$
$x-1<x^3-5\leq x$
Not able to proceed from here
| Consider the three functions $ y = x-1 $, $ y = x ^ 3-5 $ and $ y = x $. Compare the portion of the graph of $ y = x ^ 3-5 $ that is between the graphs of $ y = x-1 $ and $ y = x $. What can you say about that?
The possible solutions of the equation $x^3-\lfloor x\rfloor=5 $ live in the set of solutions of the inequati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Primality test for numbers of the form $N=k \cdot 3^n-1$ Can you provide proof or counterexample for the claim given below?
Inspired by Lucas-Lehmer-Riesel primality test I have formulated the following claim:
Let $P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$ . Let $N= k \cdot 3^{n}-1 $ where $k$ is a po... | This is a partial answer.
This answer proves that if $N$ is prime, then $S_{n-2}\equiv -2\pmod N$.
Proof :
Let us define $s,t$ as $$s:=\frac{a-\sqrt{a^2-4}}{2},\qquad t:=\frac{a+\sqrt{a^2-4}}{2}$$ where $st=1$.
Here, let us prove by induction on $i$ that
$$S_i\equiv s^{3^{i+2}k/2}+t^{3^{i+2}k/2}\pmod N\tag1$$
We see t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3196971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
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5 digits numbers such that when the sum of digits divided by 4 leaves remainder 2. How many 5 digits numbers such that when the sum of digit divided by 4 leaves remainder 2.
Example:-
Consider a 5 digit number-
$(x1,x2,x3,x4,x5)$
Then $(x1+x2+x3+x4+x5)$ must be of form $(4n+2)$
I tried this
(x+x²+x³...+x^9)(1+x+x²+x³..... | Call digits $0$ and $1$ small, and digits $2-9$ large.
Given $n\ge 1$, we divide the $n$-digit numbers into two sets: $S$, which contains all numbers consisting entirely of the small digits $0$ and $1$; and $L$, which contains all other numbers. Note that:
*
*$|S|=2^{n-1}$, because each digit apart from the initial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $5^n\mid2^{5^{n-1}}+3^{5^{n-1}}$ for $n \gt 0$. Show that $5^n\mid 2^{5^{n-1}}+3^{5^{n-1}}$ for $n \gt 0$.
I tried using Euler's Theorem: $2^{4\cdot5^{n-1}} \equiv 1 \mod{5^n}$ and $3^{4\cdot5^{n-1}} \equiv 1 \mod{5^n}$ but I can't find a way to use it.
I also tried showing that the sum of the remainders of ... | Hint:
If $a+b=c5^r$ where $a,b,c$ are integers
$$(a+b)^5=a^5+b^5+\binom51ab(a^3+b^3)+\binom52a^2b^2(a+b)$$
$$\implies a^5+b^5=(a+b)^5-5ab(a^3+b^3)-10a^2b^2(a+b)$$ is clearly divisible by $5^{r+1}$
Set $a=2,b=5d-2,$ where integer $d$ is not divisible by $5$ then $r=1$
By induction, we can establish the proposition
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Real roots of the equation $\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$
Find the set of real roots of the equation$$\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$$
My Attempt
$$
2x+3>0, 5x+4>0, 2x+3,5x+4\neq1\implies x>-4/5\;\&\;x\neq -1\;\&\;x\neq -3/5
$$
$$
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)(2x... | One can show why this case yields exactly one solution, as follows.
First, the expression on $\text{LHS}$ of the equation $$\log_{(5x+4)}{(2x+3)}=-\frac 13$$ uniquely defines a real number provided that $1\ne 5x+4>0$ and $2x+3>0.$ Thus, any solution of the equation must also satisfy the conditions $$x>-0.8, x\ne -0.6.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Calculate $\sum_{n=1}^{\infty}\left(\frac{1}{3}\frac{2}{5}\cdots\frac{n}{2n+1}\frac{1}{n+1}\right)$
Denote $a_n=\frac{1}{3}\cdot\frac{2}{5}\cdots\frac{n}{2n+1}\cdot\frac{1}{n+1}$.
Please prove
$$\sum_{n=1}^{\infty}a_n=\frac{\pi^2}{8}-1$$
This is the answer given by my friend. He also used the integral $\int_{-\pi}^{\... | Notice we can decompose $a_n$ as a product of two factors
$$a_n = \frac{1}{n+1}\prod_{k=1}^n \frac{k}{2k+1} = \frac{1}{n+1}\frac{n!}{(2n+1)!!}$$
and both factors can be represented as integrals.
$$\begin{align}
\frac{1}{n+1} &= \int_0^1 x^{n} dx\\
\frac{n!}{(2n+1)!!} &= \frac{2^n n!^2}{(2n+1)!} = \frac{2^n\Gamma(n+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a linear system of reciprocals. Solve for $\begin{cases}\frac{1}{x} +\frac{1}{y}+\frac{1}{z}=0\\\frac{4}{x} +\frac{3}{y}+\frac{2}{z}=5\\\frac{3}{x} +\frac{2}{y}+\frac{4}{z}=-4\end{cases}$
I turn the equations into $\begin{cases}yz+xz+xy=0\\4yz+3xz+2xy=5xyz\\3yz+2xz+4xy=-4xyz\end{cases}$
Not sure if I am doing f... | Observe that the system's matrix of the reciprocal of the unknowns is:
$$A=\begin{pmatrix}1&1&1&0\\
4&3&2&5\\
3&2&4&-4\end{pmatrix}\stackrel{R_2-4R_1,\,R_3-3R_1}\longrightarrow\begin{pmatrix}1&1&1&0\\
0&-1&-2&5\\
0&-1&1&-4\end{pmatrix}\stackrel{R_3-R_2}\longrightarrow$$
$$\begin{pmatrix}1&1&1&0\\
0&-1&-2&5\\
0&0&3&-9\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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$-7\cos(x) + 24\sin(x) = -25\sin(x - 16.26..) Why?$ Im confused as to why the $x$ angle is $-16.26..$ and not $360-16.26$ as well as why its $-25$ and not $25$?
I dont understand why its $-25sin(x-16.26..)$ and not $25sin(x+343.74..)$
Can someone please help me! thank you.
| Here. I found this paper that explains why this method works, but I will try and summarize.
This formula is derived from the trig identity $$\cos(x-\alpha) = \cos(x)\cos(\alpha) + \sin(x)\sin(\alpha)$$
Basically,
$$R\cos(x-\alpha) = R\cos(x)\cos(\alpha) + R\sin(x)\sin(\alpha)$$
This means for an equation $b\sin(x)+a\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Writing $3.8473221018630726$ in the form $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$. I attempted the following question on Brilliant which has to do with finding roots of a cubic polynomial. I was successful in finding what the only real root is but I am facing a problem rewriting the root in the sought expression.
The equ... | From the numerical solution (which is pretty "small"), you can try to check a few small values for $a,b,c$... which I did:
$$3.8473221018630726=\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}$$
EDIT: Actually this had to be solved in the following way (for one real solution):
$$x^3 - 3x^2 - 3x - 1 = 0$$
$$2x^3 - (x+1)^3 = 0$$
$$2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $\sum_\text{cyc}\frac{a}{b^2}\ge 3\sum_{cyc}\frac{1}{a^2}$ for $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
$a$, $b$ and $c$ are three positives such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1$. Prove that $$\dfrac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2} \ge 3 \cdot \left(\frac{1}{a... | Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, $u\geq0$, $v\geq0$ and we need to prove that
$$\sum_{cyc}\frac{1}{a}\sum_{cyc}\frac{a}{b^2}\geq3\sum_{cyc}\frac{1}{a^2}$$ or
$$\sum_{cyc}(a^4c^3+a^4c^2b-2a^2b^3c)\geq0,$$ which is true because
$$\sum_{cyc}(a^4c^3+a^4c^2b-2a^2b^3c)=4(u^2-uv+v^2)a^5+(5u^3-2u^2v+7uv^2+5v^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Determine the nature of the singularity of $\frac{z^{3}}{\cos{(z^{5})} - 1}$ at $0$. How can I find if the singularity of $f(z) = \frac{z^{3}}{\cos{z^{5}} - 1}$ at $0$ is essential, removable, or a pole?
I have considered the the Laurent series, and it appears that there are an infinte number of negative powers, so I w... | $$cos (z^5) = 1 - \frac{(z^5)^2}{2!} + \frac{(z^5)^4}{4!} + ....$$
So, $$cos(z^5)-1 = (1 -\frac{(z^5)^2}{2!} + \frac{(z^5)^4}{4!} + .... - 1) = -\frac{(z^5)^2}{2!} +\frac{(z^5)^4}{4!} - ...$$
Now, $$\frac{z^3}{cos(z^5)-1} = \frac{z^3}{-\frac{(z^5)^2}{2!} +\frac{(z^5)^4}{4!} - ...} = \frac{1}{-\frac{z^7}{2!} +\frac{(z^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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$\frac{23x - 11x^2}{(2x-1)(9-x^2)}$ when resolved into partial fractions is equal to? $\frac{23x - 11x^2}{(2x-1)(9-x^2)}$ when resolved into partial fractions is equal to?
I solved it using $\frac{23x - 11x^2}{(2x-1)(9-x^2)} = \frac{A}{2x-1} + \frac{B}{3-x} + \frac{C}{3+x}$, but this is very long method to solve, as af... | $$\frac{23x-11x^2}{(2x-1)(9-x^2)}=\frac{23x-11x^2}{(2x-1)(3-x)(3+x)}$$
The denominator of $A$ vanishes when $x=\frac12$, omit the vanishing term, we have
$$A=\frac{23\left( \frac12\right)-11\left( \frac12\right)^2}{(3-\frac12)(3+\frac12)}$$
The denominator of $B$ vanishes when $x=3$, again, we have
$$B = \frac{23(3)-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do i integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$? Integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$
I put $x^2+3x+2=t,$ and notice that $2x+3 dx=dt$, but the $dx$ is above! Please help me!
| Hint
For $(x+2)(x+1)>0$
We need $x>-1$ or $x<-2$
$2x<-4$ or $>-2$
$2x+3<-1$ or $>1$
WLOG $2x+3=\sec t,-\dfrac\pi2<t<\dfrac\pi2$
$2dx=\sec t\tan t\ dt$
$x^2+3x+2=\dfrac{(2x+3)^2-1}4=\dfrac{\tan^2t}4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Prove $x^4+y^4+x^2+y^2+x^3y+y^3x\geq 0$ I tried 2 ways, first, take $x$ and $y$ $\geq0$ then obviously true. Take $x$ and $y$ both $\leq0$ same thing.
Now $y<0<x$ and wlog $|y|\leq x,$ then for big $x$ small $|y|$ we have $-x^4\leq x^3y$ and $-x^2\leq y^3x$ but if both $|y|$ and $x$ are big I got stuck.
So I decided to... | Multiply by $4$ and use $(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$:
$$4(x^4+y^4+x^2+y^2+x^3y+y^3x)\geq 0 \iff \\
(x+y)^4+4x^2+4y^2+\underbrace{3(x^4+y^4)-6x^2y^2}_{\ge 0 \ by \ AM-GM}\ge 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving roots will be real If the roots of the equation $x^2+bx+c=0$ are real.
Show that the roots of the equation $x^2+bx+c(x+a)(2x+b)=0$
are again, real for every real number $a.$
| Not true.
Take $c=-1$. Then the discriminant of $x^2+bx+c$ is $b^2 + 4 > 0$ and so the roots are real.
The discriminant of $x^2+bx+c(x+a)(2x+b)$ is $4 a (a - b)$, which is negative if $0 < a < b$.
Therefore, the roots of $x^2+bx+c(x+a)(2x+b)$ are not real when $a=b/2$ and $b>0$, for instance. Indeed, in this case, the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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When recreating the quadratic formula by completing the square of $ax^2+bx+c=0$ I cannot shorten the right hand side I am attempting to derive the quadratic formula by completing the square on the generic generic expression:
$$ax^2+bc+c=0$$
I'm struggling with the right hand side of the equation which, for the step I... | "The common denominator will be the product of the denominators so $4a^3$" -- this is wrong.
The common denominator is $4a^2$, not $4a^3$. You multiply denominators only when they don't have a common factor; here they do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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What is the area of a triangle with sides $\sqrt{5}$, $\sqrt{10}$, $\sqrt{13}$? I found a "fun algebra problem" that asks you to find the area of a triangle whose sides are $\sqrt{5}$, $\sqrt{10}$, $\sqrt{13}$. After some algebra hell trying to work with Heron's formula, I plugged the question into Wolfram and it spit... | For another interesting approach, consider the Law of Cosines, $a^2 = b^2 + c^2 - 2bc\cos(\alpha)$. If we let $a = \sqrt{13}$, $b = \sqrt{5}$, and $c = \sqrt{10}$, then we find that $13 = 15 - 10\sqrt{2}\cos(\alpha)$, and thus that $\cos(\alpha) = \sqrt{2}/10$. Using $\cos(\alpha)$, we can calculate $\sin(\alpha)$ thro... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Show that $\int_0^{\pi/3} \frac{1}{(\cos\theta + \sqrt{3}\sin\theta)^2}\,d\theta = \frac{\sqrt3}{4}$
Show that $$\int_0^{\pi/3} \frac{1}{(\cos\theta + \sqrt{3}\sin\theta)^2}\,d\theta = \frac{\sqrt3}{4}$$
So I have attempted this, but have got a different answer so I have obviously done something wrong but i can't spo... | You just missed a factor of two in the beginning and have a sign error in the last step.
\begin{align*}
\int_0^{\pi/3}\frac{1}{(\cos t +\sqrt{3}\sin t)^2} dt=&\frac 14 \int_0^{\pi/3} \frac{1}{\left(\frac 12 \cos t + \frac{\sqrt{3}}{2} \sin t\right)^2} dt = \frac 14 \int_0^{\pi/3}\frac{1}{\cos^2(t -\frac{\pi}{3})}dt\\
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$ Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$
After few calculations I get that if I take $\frac{3}{2}x$ then after integral I get $\frac{3}{4}x^2+ C$ and
$$ \int_{\pi/6}^{\pi/2} \frac{3}{4}x^2+ C = \pi^2 / 6$$ so I shou... | The maximum value of the integrand in the given range is when $x=\pi/2$ and the integrand is also $\pi/2$ so we just get that
$$\int_{\pi/6}^{\pi/2}\frac{x}{\sin{(x)}}\mathrm{d}x\le\frac\pi2\left(\frac\pi2-\frac\pi6\right)=\frac{\pi^2}6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find maximize of the function $\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}$ Let $a,b\in R^+$ such that $ab+bc+ca=1$. Find the maximize of $$P=\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}$$
By Wolframalpha i can see that if $a=b=2-\sqrt 3;c=\sqrt 3$ we will have $P=\dfrac 1 4$
WLOG $a\le b$. I proved that $$P=f(... | Find $c$:
$$ab+bc+ca=1 \Rightarrow c=\frac{1-ab}{a+b}$$
Sub to $P$:
$$\begin{align}P&=\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}=\\
&=\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{(a+b)^2}{(a^2+1)(b^2+1)}=\\
&=\frac{(a+b)(a-1)(b-1)}{(a^2+1)(b^2+1)}\end{align}$$
Now it is symmetric and the maximum is achieved at $a=b$:
$$P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that the number $3^{3^n} + 1$ has at least $2n + 1$ prime factors. For any natural $n,$ prove that $3^{3^n} + 1$ has at least $2n + 1$ prime factors.
My idea was to use induction:
*
*for $n = 1$:
$$f(1) = 3^3 + 1 = 28 = 7*2^2$$
*let it be true for $n = k$, then for $n = k + 1$:
$$f(k + 1) = 3^{3^{k + 1}} + 1 ... | $$\large(3^{2\times3^k} - 3^{3^k} + 1) = (3^{3^k}-3^{(3^k+1)/2}+1)(3^{3^k}+3^ {(3^k+1)/2}+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Proving $\frac{1+\csc^2A\tan^2C}{1+\csc^2B\tan^2C}=\frac{1+\cot^2A\sin^2C}{1+\cot^2B\sin^2C}$
Prove
$$\frac{1+\csc^2A\tan^2C}{1+\csc^2B\tan^2C}=\frac{1+\cot^2A\sin^2C}{1+\cot^2B\sin^2C}$$
I chose to manipulate the left hand side of the equation, by firstly replacing $\cot^2A$ with $\csc^2A-1$ according to the ident... | Convert $tan^2(C)$ to $\frac{sin^2(C)}{cos^2(C)}$
$$\frac{1+\frac{\csc^2A\sin^2C}{cos^2C}}{1+\frac{\csc^2B\sin^2C}{cos^2C}}$$
Multiply all by $cos^2(C)$
$$\frac{cos^2C+\csc^2A\sin^2C}{cos^2C+\csc^2B\sin^2C}$$
Convert $cos^2C$ to $1-sin^2C$
$$\frac{1-sin^2C+\csc^2A\sin^2C}{1-sin^2C+\csc^2B\sin^2C}$$
Rearrange
$$\frac{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3227974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can't solve $\int_{0}^{\pi} \frac{x}{1 + \cos^2x} dx$ I tried this :-
Let $$I =\int_{0}^{\pi}\frac{x}{1 + \cos^2x}dx\tag{1}$$
then $$I = \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2(\pi-x)}dx= \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2x}dx\tag{2}$$
Adding (1) and (2), we get
$$
2I = \int_{0}^\pi\frac{\pi}{1 + \cos^2x}dx\\
= \pi\in... | As you have found ever the dicontinuing of $\tan$ is the problem, so you can solve it with your method after this substitution
$$\int_{0}^{\pi}\dfrac{1}{1+\cos^2x}\ dx=2\int_{0}^{\pi/2}\dfrac{1}{1+\sin^2t}\ dt= \color{blue}{\dfrac{\pi}{\sqrt{2}}}$$
where we use the substitution $x=t+\dfrac{\pi}{2}$. Other way is comple... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to determine the base of $\ker\phi$ for polynomial function? Given is a base defined as $$B:=(x\mapsto1,x\mapsto x,x\mapsto x^2,x\mapsto x^3 ,x\mapsto x^4)$$ A set V defined as
$$V:= \{ f: \mathbb{R} \mapsto \mathbb{R}\ |\ \exists\ {a_0},...{a_4} \in \mathbb{R}\ : f(x)=\sum_{i=0}^{4}{a_ix^i} \ \forall \ x \in \math... | You can ease the computation of the associated matrix by building the matrices of $f(x)\mapsto f''(x)$, $f\mapsto xf'(x)$ and $f(x)\mapsto f(x-1)$ so
$$
\begin{pmatrix}
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 6 & 0 \\
0 & 0 & 0 & 0 & 12 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
-
\begin{pmatrix}
0 & 0 & 0 & 0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$ .
Prove that $$4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$$
I think midway calculations are not easy formulas, I can't find what kind of formula transformation to solve! I think the best solution here is using right triangle . . . I have one solution... | We also know a well-known indentify as follow for $\theta= \dfrac{\pi}{7}$
$$1+ 2\cos 2\theta+ 2\cos 4\theta+ 2\cos 6\theta= 0$$
On the other hand, we have $7- (\!4\sin 2\theta- \tan \theta\!)^{2}= (\!\sec \theta\!)^{2}(\!1+ 2\cos 2\theta+ 2\cos 4\theta+ 2\cos 6\theta\!)= 0$
Furthermore $4\sin 2\theta- \tan \theta= \ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear dependence of 3 vectors in $\mathbb{R}^4$ Let $a,b,c \in \mathbb{R},$ $\vec{v_1}=\begin{pmatrix}1\\4\\1\\-2 \end{pmatrix},$ $\vec{v_2}=\begin{pmatrix}-1\\a\\b\\2 \end{pmatrix},$ and $\vec{v_1}=\begin{pmatrix}1\\1\\1\\c \end{pmatrix}.$ What are the conditions on the numbers $a,b,c$ so that the three vectors are l... | Just build a matrix with these vectors as rows and perform row reduce. The vectors will be linearly dependent if at least one row is made of zeros. The idea is that the rank of a matrix is the maximum number of linearly independent rows (or columns), hence, the rows will be linearly dependent if and only if $r(A) < 3$.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show power series converges for every $x$.
Let $$f(x) = 1 + a_{1}x^{1}+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+...$$ be a solution of the differential equation $f'(x)=xf(x).$
Now I need to explain that the power series that define $f(x)$ converges for every $x$.
This problem is introduced by two smaller proofs which I will ... | With $a_{2n}= \frac{1}{2}a_{2n-2}$ you can write your series as the sum of two geometric series. First take n even. Define $b_n= a_{2n}$. Then $b_{n}= \frac{1}{2}b_{n-1}$ so this is a geometric series: $b_0= a_0$, $b_1= a_2= \frac{1}{2}a_0$, $b_2= a_4= \frac{1}{2}a_2= \left(\frac{1}{2}\right)^2a_0$, etc.
And similar... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factoring $(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$ and use the result to solve an equation I managed to prove that $(x+a+b+c)$ is a factor of
$$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$$
Then I was asked to use the result to solve
$$(x+2)(x-3)(x-1)+4=0$$
I know by comparison, $a=2, b=-3, c=-1$, and thus $(x-2)$ is a factor, but I can'... | Probably not more efficient than straightforwardly factoring out, but it works as well:
$$\begin{eqnarray}\frac{(x+2)(x−3)(x−1)+4}{x-2} & = & \frac{(x-2+4)(x−3)(x−1)}{x-2}+\frac{4}{x-2} \\
& = & (1+\frac{4}{x-2})(x−3)(x−1)+\frac{4}{x-2} \\
& = & (x−3)(x−1) + \frac{4(x−3)(x−2+1)}{x-2}+\frac{4}{x-2}\\
& = & (x−3)(x−1) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Rolling two dice: different probabilities Today my professor said that if I roll two identical dice at the same time then there will be $21$ outcomes and the probability of getting sum of $7$ is $1/7$:
\begin{array}{c c c c c c}
\{1,1\}, & \{1,2\}, & \{1,3\}, & \{1,4\}, & \{1,5\}, & \{1,6\},\\
& \{2,2\}, & \{2,3\}, & \... | The second situation is easiest, so let's start there. In the second situation, you throw two six-sided dice. There are 6*6=36 different outcomes (for example [1,6] and [6,1] are two different outcomes), and six of them have a sum of 7. So 6/36 = 1/6 outcomes have a sum of 7, as you said.
In the first situation, the s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x^2+kx+1$ is a factor of $px^5+qx^2+r$ prove that $(p^2-r^2)(p^2-r^2+qr)=q^2p^2$ if $x^2+kx+1$ is a factor of $px^5+qx^2+r$ prove that $(p^2-r^2)(p^2-r^2+qr)=q^2p^2$
My try
Since this is a factor I tired finding A, B, C & D such that,
$(x^2+kx+1)(Ax^3+Bx^2+cx+D)$ =$px^5+qx^2+r$
if I can find k in terms of p, q & r ... | $\mathrm{px^5+qx^2+r\\if~x^2+kx+1~is ~a ~factor\\ [x^2+kx+1][Ax^3+Bx^2+Cx+D]\equiv px^5+qx^2+r\\
by~expansion:\\x^5:Ax^5=px^5 ~\underbrace{A=p} \cdots(i) \\x^4: Bx^4+kAx^4=0,~\underbrace{B=-kA} \cdots(ii) \\x^3: Cx^3+kBx^3+Ax^3=0; \underbrace{C+kB+A=0} \cdots(iii) \\x^2:~Dx^2+kCx^2+Bx^2=qx^2;~\underbrace{D+kC+B=q} \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to "see" this algebraic transformation used to derive the exponential form of the hyperbolic cosine? In this page, the exponential form of cosh is derived from the geometric definition (which is based on the unit hyperbola). Here is one step of the derivation, fully expanded out for novices like me:
$$a = ln(b + \s... | Let $$\cosh(a)=b$$ and solve for $a$
$$e^a+e^{-a}=2b\implies e^{2a}-2be^a+1=0$$
Solve for $e^a$ with quadratic formula $$ e^a= b\pm \sqrt {b^2-1}$$
Choose the plus sign and take logarithm and you have the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation: $x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$ Here is the equation:
$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$$
The equation looks very simple. But, it's a little misleading (for me).
$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}\Longrightarrow x(x^2-b^2)-a^2x(x^2-b^2)+b^2+a(x^... | Note that
$$\frac{b^2}{b^2-x^2}=-\frac{b^2}{x^2-b^2},$$
so, the equation becomes
$$x-a^2x+\frac{b^2}{x^2-b^2}+a=\frac{x^2}{x^2-b^2}$$
Passing $\frac{b^2}{x^2-b^2}$ to the RHS you obtain
$$x-a^2x+a=1.$$
This equation becomes $x(1-a^2)+a-1=0$.
The LHS becomes, $x(1-a)(1+a)-(1-a)=(1-a)\Big(x(1+a)-1\Big).$
So, if $a=1$, yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$ $f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$, where $f^{(9)}$ is the $9$-th derivative of $f$.
Domain of $f$ is $\mathbb{R} - \{-1, 2\}$. I've got that $f(x) = \frac{1}{1 - (-x)} + \frac{1}{1 - \frac{1}{2}x} = \sum_{n=0}^\infty ((-1)^n + 2^{-n})x^... | How about:
$$f(x) = (1+x)^{-1}+2(2-x)^{-1}$$
$$f'(x) = -(1+x)^{-2}+2(2-x)^{-2}$$
$$f^{(2)}(x) = 2(1+x)^{-3}+4(2-x)^{-3}$$
.
.
.
$$f^{(9)}(x) = -9!(1+x)^{-10}+2\cdot 9!(2-x)^{-10}$$
Now, plug in 1:
$$f^{(9)}(1) = 2\cdot 9!-\dfrac{9!}{2^{10}}$$
You may want to check I did not miss a negative sign, but this should be corr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Combinatorial proof of $\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$ and then use result to find a formula for $1^2 + 2^2 + \ldots + n^2$ a) Give a combinatorial proof that for every $n \geq r \geq 1$ that:
$$\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$$
And use (a) to concoct a formula for $1^2 + 2^2... | The starting-point is $\binom{i}{j}+\binom{i}{j+1}=\binom{i+1}{j+1}$, of which the combinatorial proof is famous. We can then telescope:$$\sum_{i=0}^n\binom{i}{r-1}=\sum_{i=0}^n\left(\binom{i+1}{r}-\binom{i}{r}\right)=\binom{n+1}{r}.$$The case $r=3$ gives $$\frac{n(n+1)(n-1)}{6}=\binom{n+1}{3}=\sum_{i=0}^n\binom{i}{2}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $0.9999^{\!101}<0.99<0.9999^{\!100}$ Prove
$$0.9999^{\!101}<0.99<0.9999^{\!100}$$
I think its original idea is
$$(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})} \tag{I can't prove!}$$
For $x=100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,0.99^{\!-99}<1.01^{\!100}$.
Furthermore $... | Here is an outline of a proof that $(1-x)^{\left(1-\frac{1}{x}\right)}<(1+x)^{\left(\frac{1}{x}\right)}$ when $0<|x|<1,$
using some of your tags such as derivatives and logarithms.
Let $f(x)=\dfrac{(1+x)^{\frac1x}}{(1-x)^{(1-\frac1x)}}=\dfrac{(1-x^2)^{\frac1x}}{1-x}.$
Note that $f(0)$ is not defined, but $\lim_{x\to0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3242298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Angle formed by orthocenter, incenter and circumcenter of a triangle $>135^\circ$? If $H$ is the orthocenter, $I$ the incenter and $O$ the circumcenter of a triangle , the I need to show that the angle $HIO>135^\circ$
With the assumptions of $OI^2=R^2-2Rr$, $OH^2=9R^2-(a^2+b^2+c^2)$, $HI^2=2r^2-4R^2\cos A\cos B \cos C... | It is not always true that $\angle{HIO}\gt 135^\circ$.
A counterexample is $(a,b,c)=(85,13,88)$.
$\qquad\qquad$
For $(a,b,c)=(85,13,88)$, we have
$$\begin{align}R&=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=\frac{2431}{4\sqrt{186}}
\\\\r&=\frac{1}{2}\sqrt{\frac{(-a+b+c)(a-b+c)(a+b-c)}{a+b+c}}=\frac{40\sqrt{186}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Prove a summation equals to one How to prove that
$$
S = \sum_{k=0}^{K-1} \binom{k+K-1}{K-1} \frac{a^k b^K + a^K b^k}{(a+b)^{k+K}} \mathop{=}\limits^{?} 1,
$$
For $K=2$, I have S = $\frac{a^2+b^2}{(a+b)^2} + \frac{2ab(a+b)}{(a+b)^3}$, which can be boiled down as $S = \frac{(a+b)(a^2+b^2+2ab)}{(a+b)^3} = 1$.
Also, Wolf... | Starting from
$$S = \sum_{q=0}^{K-1} {K-1+q\choose K-1}
\frac{a^q b^K + a^K b^q}{(a+b)^{q+K}}$$
we get two pieces
$$\frac{b^K}{(a+b)^K}
\sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{a^q}{(a+b)^q}
\\ + \frac{a^K}{(a+b)^K}
\sum_{q=0}^{K-1} {K-1+q\choose K-1} \frac{b^q}{(a+b)^q}.$$
This is
$$\frac{b^K}{(a+b)^K}
[z^{K-1}] \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Domain of $f(x,y) = {\sqrt{x+y-1 \over x-y+1}}$ How I can get the domain of the function
$$ f(x,y) = \sqrt{x+y-1 \over x-y+1}?$$
I know that is: $x+1 \neq y$ and $x^2 \ge (y-1)^2$
But I don't know how to get the second condition.
| Hint: Where defined, $f(x,y)$ equals
$$f(x,y)=\sqrt{\frac{x+y-1}{x-y+1}\cdot\frac{x-y+1}{x-y+1}}=\frac{\sqrt{(x+(y-1))\cdot(x-(y-1))}}{|x-y+1|}=\frac{\sqrt{x^2-(y-1)^2}}{|x-y+1|}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove that the only solution to $X^4+4=pY^4$. is $X=\pm 1,Y=\pm 1,p=5$ for $p \ge 3$ prime.
Let $p$, be a prime, $p\ge 3$. Consider the equation $X^4+4=pY^4$. Prove that the only solution is $X=\pm 1,Y=\pm 1,p=5$
Hint: $x^4+4=(x^2-2x+2)(x^2+2x+2)$. I've tried Fermat's Little Theorem and the linear Diophantine Equatio... | Notice that $\text{gcd}(X^2+2X+2,X^2-2X+2)=\text{gcd}(X^2+2X+2,4X)=2$ or $1$. If $2|X$ then 2 must also divide $Y$ and thus: $4X^4+1=4Y^4$ a contradiction.
Then $\text{gcd}(X^2+2X+2,X^2-2X+2)=1$ and since $(X^2+2X+2)(X^2-2X+2)=pY^4$ this implies:
1) $X^2+2X+2={Y_1}^4$ and $X^2-2X+2=p{Y_2}^4$ or
2) $X^2+2X+2=p{Y_1}^4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine the convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$ by comparison test It's simple to evaluate the sum as follows
\begin{align*}
\sum_{n=1}^{\infty}\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)&=\lim_{n \to \infty}\sum_{k=1}^{n}\left[\left(\sqrt{k+2}-\sqrt{k+1... | MVT:
$A:=\sqrt{n+2}-\sqrt{n+1}=$
$ (1/2)\dfrac{1}{√s},$ $n+1 <s <n+2$;
$B:= \sqrt{n+1}-√n= (1/2)\dfrac{1}{√t}$, $n <t <n+1$;
MVT:
$A-B=(1/2)(-1/2)\dfrac{1}{r^{3/2}}(s-t)$, $t <r <s$;
Finally :
$|A-B| < (1/4)\dfrac{1}{n^{3/2}} \cdot 2$, since
$r >n$, and $s-t <2$.
Comparison test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Checking a proof that the equality $ax^2+bx+c=0$ is never true for integer $a , b$ and $c$ if $ x = 2^{\frac{1}{3}} $ The solution of $ax^2+bx+c=0$ is $x=-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$ . Since $ x^3=2$ then after raising both sides to the third power we get $$ 2=-\frac{b^3}{8a^3}+\frac{\left(b^2-4ac\right)^{\... | The solution seems alright, but it is too complicated. An easier solution would be to notice that $x^3 - 2$ is satisfied by $\sqrt[3]{2}$. Then considering $x^3 - 2$ and $ax^2 + bx + c = 0$, as elements of $\mathbb{Q}[x]$ we have that they have a common divisor. However, $x^3 - 2$ is irreducible (by Eisenstein criteri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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} |
Calculate the maximum value of $\frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy}$ where $x, y, z > 0$ and $x^2 + y^2 + z^2 = 3xyz$.
$x$, $y$ and $z$ are positives such that $x^2 + y^2 + z^2 = 3xyz$. Calculate the maximum value of $$\large \frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 +... | We have that $$\frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy} \le \frac{x^2}{2x^2\sqrt{yz}} + \frac{y^2}{2y^2\sqrt{zx}} + \frac{z^2}{2z^2\sqrt{xy}}$$
$$ = \frac{1}{2}\left(\frac{1}{\sqrt{xy}} + \frac{1}{\sqrt{yz}} + \frac{1}{\sqrt{zx}}\right) = \frac{z\sqrt{xy} + x\sqrt{yz} + y\sqrt{zx}}{2xyz} \le \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$
Question-Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$
I was playing around with the formulae
$$(a+b+c)^3=a^3 +b^3+c^3+3(a+b)(b+c)(c+a)$$ and,
$$a^3 +b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
and noted that if $(a+b+c)\equiv0$(mod 6)$\implies a^3 +b^3+c... | a^3+b^3+c^3-a-b-c = (a-1)a(a+1)+(b-1)b(b+1)+(c-1)c(c+1)
(a-1)a is divisible by 2.
(a-1)a(a+1) is divisible by 3.
Hence, (a-1)a(a+1) is divisible by 6.
Similarly, (b-1)b(b+1) and (c-1)c(c+1) are also divisible by 6.
Since a^3+b^3+c^3 is divisible by 6 and (a-1)a(a+1)+(b-1)b(b+1)+(c-1)c(c+1) is divisible by 6,
a+b+c is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Prove that $5^n + 6^n -1$ is divisble by 10 for all positive integers n. I know the base case is:
$$ n=1,
5^1+6^1-1=10,10|10
$$
But I am struggling about how to go about the inductive step.
| $5^n-1$ is even implies that $5^n+6^n-1$ is divisible by $2$,
$6^n-1=(6-1)(6^{n-1}-1+..)$ ($6=1$ mod $5$ implies that $6^n=1$ mod $5$ and $6^n-1=0$ mod $5$) implies that $5^n+6^n-1$ is divisible by $5$.
Recursive proof
$5^{n+1}+6^{n+1}-1=5.5^n+(5+1)6^n-1=5(5^n+6^n-1)+6^n+4$
Suppose that $6^n=6$ mod $10$, $6^{n+1}=36=6 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Factorize $x^4 + y^4 -x^2 y^2$ over $\mathbb C$ As in the title, I'd like to factorize $x^4 + y^4 - x^2y^2$ into irreducible factors over $\mathbb C$ (i.e. linear factors).
Attempts:
First I tried doing
$$x^4 + y^4 - x^2y^2 = (x^2 - y^2)^2 + x^2y^2 \\ = (x^2 - y^2) - (ixy)^2 \\ = (x^2 - y^2 - ixy)(x^2 - y^2 + ixy) $$
A... | A smart way to approach this is to solve the equation
$$x^2-y^2+ixy=0,$$ find the solutions $x_\pm(y)$ and then write
$$x^2-y^2+ixy=(x-x_+(y))(x_-(y)).$$
In this case we have that the solutions are
$$x_i(y)=\frac{-iy\pm\sqrt{(iy)^2+4y^2}}{2}=\frac{-iy\pm\sqrt{3}y}{2}=\frac{-1\pm\sqrt{3}}{2}\cdot y$$
thus the factorisat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Solve the inequality $ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $
Solve the inequality
$$ 2x^{2} + x - 4 \sqrt{2x^{2} + x + 4} < 1 $$
Attempt:
I get that $2x^{2} + x + 4 > 0$ for all $x$. Let $y = 2x^{2} + x$, then the inequality becomes
$$ y - 4 \sqrt{y + 4} < 1 $$
$$ 0 < \frac{1 + 4 \sqrt{y+4} - y}{y}$$
so from ... | It's $$2x^2+x+4-4\sqrt{2x^2+x+4}+4<9$$ or
$$\left(\sqrt{2x^2+x+4}-2\right)^2<9$$ or since $\sqrt{2x^2+x+4}>0$, it's
$$\sqrt{2x^2+x+4}<5$$ or
$$2x^2+x-21<0$$ or
$$(x-3)(2x+7)<0$$ or
$$-3.5<x<3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
value of $2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$
If $ x^3+bx^2+cx+1=0$ has only real root $\alpha $.
Where $(b<c)$. Then $\displaystyle 2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$ is
Plan
$$\tan^{-1}\bigg(\frac{2\csc \alpha}{1-\csc^2\alpha}\bigg)+\tan^{-1}\bigg(2\sin \alpha\sec^... | Note that
$$
2\sin\alpha\sec^2\alpha=\frac{2\sin\alpha}{\cos^2\alpha}=\frac{2\sin\alpha}{1-\sin^2\alpha}
$$
If
$$
f(x)=\arctan\frac{2x}{1-x^2}
$$
then
$$
f'(x)=\frac{2}{1+x^2}
$$
so $f(x)$ differs from $2\arctan x$ by a constant over $(-1,1)$. Since $f(0)=0=2\arctan0$, we can say that
$$
\arctan\frac{2\sin\alpha}{1-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Unable to solve $ \int \frac{x + \sqrt{2}}{x^2 + \sqrt{2} x + 1} dx $? This comes from a bigger problem :-
$$ \text{Evaluate } \int\frac{dx}{1+x^4} $$
After making $ \int \frac {dx}{1+x^4} = \frac{dx}{(1+x^2)^2 - (\sqrt{2}x)^2} $ and then applying partial fraction method, I got :-
$$ \int \frac{dx}{1 + x^4} =\frac{1}{... | $\displaystyle \frac{x+\sqrt{2}}{x^2+\sqrt2 x+1}=\frac{x+\sqrt{2}}{\left(x+\frac{\sqrt{2}}{2}\right)^2+\frac12}$.
Let $\displaystyle x+\frac{\sqrt{2}}2=\frac{\sqrt2}2\tan\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Geometric consequence of absolute vectors If $| \mathbf{a} + \mathbf{b} | = | \mathbf{a} | + | \mathbf{b} |$, where $\mathbf{a}$ and $\mathbf{b}$ are vectors, what is the geometrical significance of this?
My first thought was that the vectors $\mathbf{a}$ and $\mathbf{b}$ must be in the first quadrant, only having pos... | Let $\theta$ be the angle between $\mathbf a$ and $\mathbf b$. Using properties of the dot product,
$$|\mathbf a+\mathbf b|=|\mathbf a|+|\mathbf b|$$
$$\implies|\mathbf a+\mathbf b|^2=|\mathbf a|^2+|\mathbf b|^2+2|\mathbf a||\mathbf b|$$
$$\implies(\mathbf a+\mathbf b)\cdot(\mathbf a+\mathbf b)=\mathbf a\cdot \mathbf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Common volume of three cylinders with unequal radii I would like to solve it please for the case where the radii can be a similar size - so the case where this statement is NOT true: $$ \mathbf r_1^2 \mathbf \geq \mathbf r_2^2 \mathbf + \mathbf r_3^2 $$
How do you solve for the common volume of 3 cylinders with unequ... | Let $a<b<c$ with $c<\sqrt{a^2+b^2}$ be the radii of the three cylinders
$$x^2+y^2\leq a^2,\qquad x^2+z^2\leq b^2,\qquad y^2+z^2\leq c^2\ ,$$
and denote by $S$ the part of the intersection lying in the positive octant. The set $S$ is standing on the quarter disc
$$S':\qquad x\geq0,\quad y\geq 0, \quad x^2+y^2\leq a^2$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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establish that every prime number p of the form $ 8k + 1$ or $8k +3$ can be written as $ p = a^2 + 2b^2$ for some choice of integers a and b. The question is:
Establish that every prime number p of the form $ 8k + 1$ or $8k +3$ can be written as $ p = a^2 + 2b^2$ for some choice of integers a and b.
And the Hint says:
... | Comment:An experimental approach; following conditions cover a set of these types of primes:
For $8k+1$:
Let $k=k_1+2$ ⇒ $p=8(k_1+2)+1=8k_1+8+8+1=2[4(k_1+1)]+3^2$
If $k_1+1=c^2$ then we have:
$p=3^2+2\times (2c)^2$
Examples; $k_1=3$ ⇒ $p=41=3^2+2 (2\times 2)^2=3^2+2\times 4^2$
$k_1=15$ ⇒ $c^2=15+1=4^2$ ⇒ $p=137=3^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Extreme values of function with one condition What is the simple way to find extreme values of a function $f(x,y) = 5*x^2 + 3*x*y + y^2$ with conditions $x^2 + y^2 = 1$.
I tried to make partial derivatives and initialize them with $0$, but the result is not good.
| As araomis's comment suggested, you can try using polar co-ordinates with $r = 1$. This is because the unit circle $x^2 + y^2 = 1$ can be parametrized by $x = \cos(\theta)$ and $y = \sin(\theta)$ for $\theta \in [0,2\pi)$. Using this, along with the unit circle, then
$$f(x,y) = 5x^2 + 3xy + y^2 = 4x^2 + 3xy + 1 \tag{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding angle in the isosceles triangle There is the following triangle. I infer that the triangle is isosceles. But I cannot go further.
| Without trigonometry
Let $D$ be the symmetric of $C$ with respect to $A$, $H$ and $K$ the orthogonal projections of $A$ and $C$ on $BC$ and $AB$.
$BCD$ is a right triangle. By the midpoint theorem, $CK=2MN=6$ and $MN=MA=4$ so $BK=2$. By Pythagoras theorem, $BC=\sqrt{BK^2+CK^2}=2\sqrt{10}$ and $AH=\sqrt{AB^2-BH^2}=3\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Integrate squared gaussians I need to compute the integral
$$
\int_{-\infty}^\infty e^{-a^4 - (1-a-b)^4} da
$$
for which it seems there is no analytic solution. Trying with Mathematica also gives no solution.
Do you have any suggestion, or possible method to solve this? (e.g. smart change of variables, ...)
| First let's rewrite the integral a bit. Introducing $1-b=2c$, we have:
\begin{align}
&\int_{-\infty}^\infty e^{-a^4-(1-a-b)^4}\mathrm{d}a=\int_{-\infty}^\infty e^{-a^4-(2c-a)^4}\mathrm{d}a=\int_{-\infty}^\infty e^{-(c+a-c)^4-(c+c-
a)^4}\mathrm{d}a=\\
&=\int_{-\infty}^\infty e^{-(c+x)^4-(c-x)^4}\mathrm{d}x=
=\int_{-\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Solution Of Diophantine Equations
Find all integer solutions for the equation:
$$(x+y)(y+z)(z+x)=txyz$$
such that gcd$(x, y)=1$, gcd$(y, z)=1$ and gcd$(z, x)=1$
Now we can write our equation as
$$(\frac{x+y}{y})(\frac{y+z}{z})(\frac{x+z}{x})=t$$
This gives $$(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=t$$
Now a... | Hint:
Say $x,y,z$ are positive. We have $x|y+z$ and $y|x+z$ and $z|x+y$. We can assume $x\geq y\geq z$. Then $y+z= kx$ where $k=1$ or $k=2$.
If k=1we have $(2y+z)(2z+y)=tyz$ and so $y|2z^2$ so $y=1$ or $y=2$.
If k=2 we have $(3y+z)(3z+y) =2tyz$ so $y| 3z^2$ so $y=1$ or $y=3$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find the number of elements of the cyclic subgroup $\left\langle -\frac{1}{ 2}+\frac{\sqrt 3}{2} i \right\rangle$ of $\mathbb{C}^*$. Find the number of elements of the cyclic subgroup $\left\langle -\frac{1}{ 2}+\frac{\sqrt 3}{2} i \right\rangle$ of $\mathbb{C}^*$.
Answer:
Since $\left\langle -\frac{1}{ 2}+\frac{\sqrt... | Note that
$\cos \left ( \dfrac{2\pi}{3} \right ) = -\dfrac{1}{2}, \tag 1$
and
$\sin \left ( \dfrac{2\pi}{3} \right ) = \dfrac{\sqrt 3}{2}; \tag 2$
thus,
$-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} = \cos \left ( \dfrac{2\pi}{3} \right ) + i \sin \left ( \dfrac{2\pi}{3} \right ) = e^{2 \pi i /3}; \tag 3$
it follows that, for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $A=\begin{bmatrix} 1 & 2\\ 3& 4 \end{bmatrix}$ then det$(A^3-6A^2+5A+3I)=3$
Let $A=\begin{bmatrix}
1 & 2\\
3& 4
\end{bmatrix}$ then det$(A^3-6A^2+5A+3I)=3$
det$(A^3-6A^2+5A+3I)=$det$((A^2-5A-2I)(A-I)+2A+I)= $det$(2A+I)=3$, Since a matrix satisfies its characteristic polynomial. Is this right?
| Here is another method. Assume that $\lambda_i\in\mathbb{C}$, $i=1,2$ are the two igenvalues of $A$. Let $f(x)\in \mathbb{C}[x]$ be the characteristic polynomial of $A$. Then $$f(x)=\det(A-xI)=x^2-5x-2.$$
The matrix $A$ can be diagonalized as
$$A=P\begin{bmatrix} \lambda_1 & 0\\ 0& \lambda_2 \end{bmatrix}P^{-1},$$
whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate using Stokes' Theorem To evaluate $\oint_{C} -y^3dx+x^3dy+z^3dz,$ where $C$ is the intersection of cylinder $x^2 + y^2 =1$ and plane $x+y+z=1$. The orientation of $C$ is counter-clockwise motion in the $xy$ plane.
Now I have computed $\nabla \times\mathbf{F} = \left(0,0,3\left(x^2+y^2\right)\right).$ I am havi... | Let $\mathbf{F}=\left(-y^3,x^3,z^3\right).$
So we have
\begin{align*}
\oint_{C} -y^3dx+x^3dy+z^3dz&=\oint_C\mathbf{F}\cdot d\mathbf{r} \\
&=\iint_S(\nabla\times\mathbf{F})\cdot dS \\
&=\iint_S(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}\,dA \\
&=\sqrt{3}\iint_S\left(x^2+y^2\right)\,dA.
\end{align*}
My hunch is that we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to integrate $\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4 x} \,dx$ So I followed the explanations made in this post and I got that:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \frac{\sqrt{2}}{2}\arctan\left(\frac{\sqrt{2}}{2}\tan\left(2x\right)\right) + C$$
But when I try to use the Leibniz-Newton formula and evaluate... | Given integral $$ I = \int^{2 \pi}_0 \cfrac{1}{\sin^4 x + \cos^4 x} dx = 4 \int^{\pi/2}_0 \cfrac{dx}{\sin^4 x + \cos^4 x} $$
By repeatedly using $ \int^{2a}_0 f(x) dx = \int^a_0 f(x) dx + \int^a_0 f(2a-x) dx$
On multiplying numerator and denominator by $ \sec^4(x)$, followed by substitution of $\tan(x) = u$ $$ \begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Prove that $\prod\limits_{n= 2}^{\infty} \frac{n^{3}+ 1}{n^{3}- 1}= \frac{3}{2}$, which is a beautiful value! Prove that $\prod\limits_{n= 2}^{\infty} \dfrac{n^{3}+ 1}{n^{3}- 1}= \dfrac{3}{2}$, which is a beautiful value ! A solution due to a p u as follow
$$\begin{align}
\prod\limits_{n= 2}^{\infty} \frac{n^{3}+ 1}{n^... | Here is a much simpler way. Observe that the partial product
$$
\prod_{n=2}^{m} \frac{n^3+1}{n^3-1} = \frac{3 (m^2+m)}{2(m^2+m+1)}.
$$
Then when $m\to\infty$, the product converges to $\displaystyle \frac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given a diagonalizable matrix $A$ and polynomial $f$, prove $f(A)$ is diagonalizable I have been given a diagonalizable matrix $A \in K^{n
\times n}$ and a polynomial $f \in K[X]$ for a field $K$. I need to prove that $f(A)$ is diagonalizable. Because the matrix $A$ is a arbitrary matrix, it follows:
$$A = \left(\begi... | I think you can do it as this. Let
$$f(X) = X^n + a_{n-1}X^{n-1} + \cdots +a_1 X + a_0, \quad \in K[X]$$
be your polynomial and let $A$ diagonalize as
$$A = TBT^{-1}.$$
Then you have
\begin{align*}
f(A) &=A^n + a_{n-1}A^{n-1} + \cdots + a_1A + a_0I \\
&=TB^nT^{-1} + a_{n-1}TB^{n-1}T^{-1} + \cdots + a_1 TBT^{-1} + a_0 I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve$(10+6\sqrt3)^{\frac{1}{3}}-(-10+6\sqrt3)^{\frac{1}{3}}$ We need to solve the following equation $y=(10+6\sqrt3)^{\frac{1}{3}}-(-10+6\sqrt3)^{\frac{1}{3}}$ and it is equal to 2 while I am getting the value in excel I am not able to solve it manually eventhough the values are conjugate
I tried $y=a-b$
$a=(10+6\sqrt... | Substituting the values of $a$ and $b$, and the relation $y=a-b$, into $y^3=a^3-b^3-3ab(a-b)$ yields
$$y^3=10+6\sqrt3+10-6\sqrt3-3((6\sqrt3+10)(6\sqrt3-10))^{1/3}y$$
$$y^3=20-3(3\cdot36-100)^{1/3}y$$
$$y^3=20-6y$$
$$y^3+6y-20=0$$
$$(y-2)(y^2+2y+10)=0$$
Since the quadratic factor has no real roots, $y=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\,n\bmod 42\,$ given $\,7\mid 3^n+4n+1$ I have been working on a problem relating to remainders: "The number $3^n+4n+1$ is divisible by $7$. Find the remainder of n when it is divided by $42$." I have been unable to find a strategic way to solve this problem and so far, I have only found the remainders, $4$, $8$,... | Strategy:
$4n + 1 \equiv k \pmod 7$ will have a solution $n_k$ determined by $n \mod 7$. And for $k = 0,... 7$ there will be $7$ solutions $n_0, ... n_6$.
By Fermats little Theorem, $3^6\equiv 1 \pmod 7$ for $3^n \equiv j\pmod 7$ will have a solution $m_j$ determined by $n \pmod 6$. If we assume $3$ is a primitive r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve the following equation: $\sqrt {\sin x - \sqrt {\cos x + \sin x} } = \cos x$ Solve the following equation:
\begin{array}{l}{\sqrt{\sin x-\sqrt{\cos x+\sin x}}=\cos x} \\ \text{my try as follows:}\\{\sin x-\sqrt{\cos x+\sin x}=\cos ^{2} x} \\ {\sin x-\cos ^{2} x=\sqrt{\cos x+\sin x}} \\ {\sin ^{2} x+\cos ^{4} x-2... | We do this in one period $[-\pi, \pi)$.
According to the equation, $\cos x = \sqrt \cdots \geqslant 0$, then $x \in [-\pi/2, \pi/2]$. Also inside the square root, $\sin x \geqslant \sqrt \cdots \geqslant 0$, then $x\in [0, \pi) \cup \{-\pi\}$. Thus $x \in [0, \pi/2]$ at least.
To be more exact, we consider
$$
\sin x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to choose your $u$ value properly to solve $\int\frac{1}{5+3\cos (x)} \cdot dx$. We are given the following facts:
*
*$\cos(x) = \frac{1-u^2}{1+u^2}$,
*$dx = \frac{2du}{1+u^2}$
Using 1. and 2. along with u substitution solve the following integral.
$$\int\frac{1}{5+3\cos (x)} \cdot dx$$
I re-wrote the integral... | Write $$4+u^2=4\left(1+\left(\frac{u}{2}\right)^2\right)$$ and substitute $$t=\frac{u}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find every $a>0$ such that $\int_2^{\infty} \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} dx$ converges We know that $\forall x \in [2,+\infty)$,
$$\frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} \leq \frac{2x^6}{(x-1)^2(x+1)^2x^a}$$
And it's easy to prove that $\int_2^{\infty} \frac{2x^6}{(x-1)^2(x+1)^2x^a} dx$ converges if an... | Let $f(x)$ be the given integrand and $g(x)=x^{2-a}(1+\cos^{2}x)$. You can easily verify that $\frac {f(x)} {g(x)} \to 1$ as $ x \to \infty$. Hence $\int_2^{\infty} f(x)dx$ is finite iff $\int_2^{\infty} g(x)dx$ is finite. Now using the fact that $x^{2-a} \leq g(x) \leq 2x^{2-a}$ we can conclude that $\int_2^{\infty} g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\frac{3(a + b + c)^2}{25(ab + bc + ca)(a^2 + b^2 + c^2)} \le \sum_{cyc}\frac{1}{(3a + b + c)(c + 4a)} \le \frac{\sqrt{3(ab + bc + ca)}}{25abc}$
Given postives $a, b, c$. Prove that $$\frac{3(a + b + c)^2}{25(ab + bc + ca)(a^2 + b^2 + c^2)} \le \sum_{cyc}\frac{1}{(3a + b + c)(c + 4a)} \le \frac{\sqrt{3(ab + bc + ca)}}... | For the proof of the right inequality we can use BW.
Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, we need to prove that:
$$3(ab+ac+bc)\geq625a^2b^2c^2\left(\sum_{cyc}\frac{1}{(3a+b+c)(c+4a)}\right)^2,$$ which is something obvious:
https://www.wolframalpha.com/input/?i=3%28xy%2Bxz%2Byz%29-625x%5E2y%5E2z%5E2%281%2F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the minimum $K$ such that $\sum_{cyc} \frac{a}{\sqrt{a+b}} \leq K\sqrt{a+b+c}$ The question is to find the minimum $K$ such that $$\sum_{cyc} \frac{a}{\sqrt{a+b}} \leq K\sqrt{a+b+c}$$ holds true for all non-negative $a,b$ and $c$.
My attempt:
I used Cauchy-Schwarz to get
$$\left(\sum_{cyc} \frac{a}{\sqrt{a+b}}\ri... | For $a=3$, $b=1$ and $c=0$ we obtain $k\geq\frac{5}{4}.$
We'll prove that $\frac{5}{4}$ is a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{a}{\sqrt{a+b}}\leq\frac{5}{4}\sqrt{a+b+c}.$$
Now, by C-S
$$\sum_{cyc}\frac{a}{\sqrt{a+b}}=\sqrt{\left(\sum_{cyc}\frac{a}{\sqrt{a+b}}\right)^2}\leq\sqrt{\sum_{cyc}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Sum of all the numbers in the grid.
A square containing numbers
$$
\begin{array}{|c|c|c|}
\hline 1 & 2 & 3 \\ \hline
1 & 2 & 2 \\ \hline
1 & 1 & 1 \\\hline
\end{array}
\qquad \qquad\qquad
\begin{array}{|c|c|c|c|}
\hline 1 & 2 & 3 & 4 \\ \hline
1 & 2 & 3 & 3 \\ \hline
1 & 2 & 2 & 2 \\ \hline
1 & 1 & 1 & 1 \\\hline
\e... | Consider the L-shaped figure $L_i$. $L_1$ adds $1.19$, $L_2$ adds $2.17$, $L_3$ adds $3.15$ and so on till $L_{10}$ adds $10.1$ for a total of $385$. $1.19+2.17+3.15+...+10.1=385$
Edit: Another way is to look at it as a 3d pyramid where each cube adds a $1$ to the sum. The ground floor has $10^2$ cubes, first floor has... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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A circle of radius $r$ is inscribed into a triangle. A circle of radius $r$ is inscribed into a triangle $ABC$. Tangent lines to this circle parallel to the sides of the triangle cut out three smaller triangles, $\triangle A_cB_cC$, $\triangle A_bBC_b$, $\triangle AB_aC_a$. The radii of the circles inscribed in these s... |
Let $|CE|=H_c$ and $|CD|=h_c$ be the altitudes
of similar triangles $\triangle ABC$ and $\triangle A_cB_cC$.
Then
\begin{align}
\frac{|CD|}{r_c}
&=
\frac{|CE|}{r}
\tag{1}\label{1}
,\\
\frac{H_c-2r}{r_c}
&=
\frac{H_c}{r}
\tag{2}\label{2}
,\\
H_c &= \frac{2r^2}{r-r_c}
\tag{3}\label{3}
.
\end{align}
Similarly, two other ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding $k$ such that $(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$
$(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$ . Find $k$
....couldnt figure out how to find $k$
I tried assuming $(x^2 + kx + 1)= (x - 1)^2 $ where $k = (-2) $ comsidering that $(x-1) $ is a factor of the above polynomial...... | Observe that there is no cubic term, and the independent term is $3$. Hence the factorization must be
$$(x^2+kx+1)(x^2-kx+3)=x^4+(4-k^2)x^2+2kx+3.$$
By identification $k=4$.
The "standard" solution is by equating the remainder of the long division,
$$(8+14k-k^3)x+16-k^2$$ to zero. The only solution is $k=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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If $4 = \frac{3}{a_1} = \frac{3}{a_2}+a_1= \frac{3}{a_3} + a_2 = ... = \frac{3}{a_{n+1}} + a_n$ Prove that $a_n= \frac{3^{n+1} - 3}{3^{n+1}-1}$ If $$4 = \frac{3}{a_1} = \frac{3}{a_2}+a_1= \frac{3}{a_3} + a_2 = \space ... \space = \frac{3}{a_{n+1}} + a_n$$
Prove that $$a_n= \frac{3^{n+1} - 3}{3^{n+1}-1}$$
I tried to pro... | You want to prove
$$a_n= \frac{3^{n+1} - 3}{3^{n+1}-1} \tag{1}\label{eq1}$$
From $4 = \frac{3}{a_1}$, you have $a_1 = \frac{3}{4}$, which matches \eqref{eq1} when $n = 1$ as it gives $a_1 = \frac{9 - 3}{9 - 1} = \frac{6}{8} = \frac{3}{4}$. Assume \eqref{eq1} holds for all $n \le k$ for some integer $k \ge 1$. Now, from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find Sum of Sum of Partial Harmonic Series: $\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} $? How do I find the Sum of the Sum of Finitely Many Harmonic Series: $\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} $?
According to maple it is:
$\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} = \left( \left( m+1 \right) ^{2}-m-4 \r... | The iterated sum actually evaluates to
$$\sum_{b = 1}^m \sum_{k = 1}^b \frac{1}k = \left(m + 1\right)H_m - m.$$
You can show by regrouping terms that for a general sum
$$\sum_{a = 1}^x \sum_{b = 1}^a f(b) = \sum_{k = 1}^x \left(x - k + 1\right) f(k).$$
So plugging in $f(b) = \frac{1}b$, we obtain
\begin{align}
\sum_{a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Simplify $\left[ 2 \cos \left( \frac{\pi}{2n} \right) \right]^n$ In relation to this question, consider the quantity
$$\left[ 2 \cos \left( \frac{\pi}{2n} \right) \right]^n$$
when $n \in \mathbb{N}$ and $n \geq 2$.
Under these conditions, it should be a real number.
Is it possible to exploit the presence of $n$ both in... | Don't see a reason to write in terms of imaginary numbers.
Distribute the power of $n$:
$$\left[ 2 \cos \left( \frac{\pi}{2n} \right) \right]^n$$
$$= 2^n \left[\cos \left( \frac{\pi}{2n} \right) \right]^n$$
Using the half-angle identity Gerry mentioned:
$$\cos(\frac{x}{2})=\sqrt{\frac{1+\cos x}{2}}$$
Where $x=\pi/n$:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Expressing $\sum_{n = 1}^\infty \sum_{k = 1}^n \frac{1}{n^4 k\,2^k}$ as a finite sum involving $\zeta(\cdot)$, $Li_k(\cdot)$, $\pi$, and $\ln 2$ While working on the integral posted here, through a large amount of skulduggery, I managed to arrive at the following intriguing sum
$$\begin{align}\sum_{n = 1}^\infty \sum_... | We have $$S=\sum_{n\geq1}\frac{1}{n^{4}}\sum_{k=1}^{n}\frac{1}{k2^{k}}=\sum_{n\geq1}\frac{1}{n^{4}}\int_{0}^{1}\frac{1-x^{n}2^{-n}}{2-x}dx$$ $$=\int_{0}^{1}\frac{1}{2-x}\left(\zeta\left(4\right)-\mathrm{Li}_{4}\left(\frac{x}{2}\right)\right)dx.$$
Now, since we can split the integral, we obtain $$
S=\zeta\left(4\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
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Prove that if $x > 3$ and $y < 2$, then $x^2 − 2y > 5$
Suppose $x > 3$ and $y < 2$. Then $x^2 − 2y > 5$.
My attempt:
*
*$x>3 \implies x^2 > 9$
*$ y < 2 \implies 5 + 2y < 9$
It follows that $5+2y < 9 < x^2$. From this we can see that $5+2y < x^2 \implies x^2 -2y > 5$
Therefore, if $x>3$ and $y<2$, then $x^2 − 2... | Yes, your proof is correct. Here is an alternative:
$x^2> 9$ and since $-y>-2$ we have $-2y>-4$, so $$x^2-2y>9-4 =5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum value of $(\sin x+ \csc x)^2 + (\sec x + \cos x)^2$ I made it till $5+\tan^2x + \cot^2x$, but I don’t know how to proceed further.
| Now, by AM-GM
$$5+\tan^2x+\cot^2x\geq5+2=7.$$
The equality occurs for $\tan^2=\cot^2x,$ which says that we got a minimal value.
But for your problem we can use C-S:
$$\left(\sin{x}+\frac{1}{\sin{x}}\right)^2+\left(\cos{x}+\frac{1}{\cos{x}}\right)^2=5+\frac{1}{\sin^2x}+\frac{1}{\cos^2x}=$$
$$=5+(\sin^2x+\cos^2x)\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Evaluate $\int_{-3}^3 \sqrt{9-x^2} dx$ using the limit definition of a definite integral I can't evaluate $\int_{-3}^3 \sqrt{9-x^2} dx$ using the limit definition I need some help here's what I did:
Formula to use:
$$\int_a^bf(x)dx= \lim_{n\rightarrow \infty} \sum_{k=1}^n f(c_k)\Delta x_k$$
Get the variables:
$$\Delta ... | Let's simplify the integral
\begin{align*}
\int_{-3}^3 \sqrt{9-x^2} dx &= \int_{-3}^3 \sqrt{9(1-\frac{x^2}{9})} dx = 3 \int_{-3}^3 \sqrt{1-\left(\frac{x}{3}\right)^2} dx = 9\int_{-1}^1 \sqrt{1-x^2} dx
\end{align*}
We have
\begin{align*}
\frac{2}{n} \sum_{k=0}^{n-1} \sqrt{1-\left(-1 + k \times \frac{2}{n}\right)^2} &= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Differentiation of $\sin^{-1}(1/\sqrt{1+x})$ $$\arcsin \frac{1}{\sqrt{1+x}}.$$ I tried differentiating it but every time my answer came wrong. The answer in my book is $$- \frac{1}{2(x+1)\sqrt x}.$$
| The function
$$
f(x) = \arcsin{\frac{1}{\sqrt{1+x}}}
$$
can be seen as a composite function $f(x) = u(s(x))$, where $u(x) = \arcsin{x}$ and $s(x)= \frac{1}{\sqrt{1+x}}$. Their derivatives are (I will assume these as given - I won't prove these)
$$
u'(x) = \frac{1}{\sqrt{1-x^2}}
$$
$$
s'(x) = -\frac{1}{2(x+1)^{2/3}}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
im trying to use AM-GM
$\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}}
\le ... | For $x_i>0$ with $i=1,\dots,n$, the given inequality is equivalent to
$$\frac{1}{\frac{x_1+x_2+\dots+x_n}{n}} \le \frac{\frac{1}{x_1}+ \frac{1}{x_2} + \dots+ \frac{1}{x_n}}{n}.$$
Now note that $x\to 1/x$ is convex in $(0,+\infty)$ and use Jensen's inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$.
I have tried:
$$\frac{10!}{(3!\times7!)} \times (-x + 2)^7 \times (x^2)^3 $$
But got an incorrect answer $-15360$.
| You just missed one term. Note that by using the binomial theorem we get
$$(x^2-x+2)^{10}=(x(x-1)+2)^{10}=\sum_{k=0}^{10}\binom{10}{k}2^{10-k}x^k(x-1)^k$$
Hence, since $[x^3](x^k(x-1)^k)=0$ for $k\not\in \{2,3\}$, we have that
$$\begin{align}[x^3](x^2-x+2)^{10}&=[x^3]\sum_{k=2}^{3}\binom{10}{k}2^{10-k}x^k(x-1)^k\\&=\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
On the Clausen triple $8\rm{Cl}_2\left(\frac{\pi}2\right)+3\rm{Cl}_2\left(\frac{\pi}3\right)=12\,\rm{Cl}_2\left(\frac{\pi}6\right)$ While doing research on the Clausen function, I came across this nice identity,
$$8\operatorname{Cl}_2\left(\frac{\pi}2\right)+3\operatorname{Cl}_2\left(\frac{\pi}3\right)=12\operatorname{... |
In general, assuming convergence, we have
$$ \sum_{n=1}^\infty \sin\left(\frac{\pi}{6}\,n\right)\, a_{n} = \sum_{n=1}^\infty \sin\left(\frac{\pi}{3}\,n\right)\, a_{2n}
+ \sum_{n=1}^\infty \sin\left(\frac{\pi}{2}\,n\right)\, \left(\frac12 a_{n}+\frac32 a_{3n} \right)$$
To see this, we partition the naturals into e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Find $n$ if $\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$ Find $n$ if $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$$
In this video they show a shortcut and say $n=-1/2$ without any explanation.
Key observation here is that the geometric mean of $9$ and $4$ is $6$.
It seems numerator and denominator are partial sums of geometric ser... | $ \frac{9^{n+1} + 4^{n+1}}{9^n + 4^n} = 6 $
$ 9^{n+1} + 4^{n+1} = 6\cdot 9^n + 6 \cdot 4^n $
$3 \cdot 9^n = 2 \cdot 4^n $
$ 3^{2n+1} = 2^{2n+1}$
$(\frac{3}{2})^{2n+1} = 1$
So, $2n+1 = 0 $
$ n = -\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
Why does $\int_{-\infty}^{+\infty }\arctan\left(\frac{1}{1+x^2}\right)dx$ have a real value when the indefinite integral uses $i$? WolframAlpha gives a real closed form for this definite integral:
$$\int_{-\infty}^{+\infty } \arctan\left(\dfrac{1}{1+x^2}\right)dx = \sqrt{2\left(\sqrt{2}-1\right)}\;\pi$$
Yet, the formul... | A pretty integral, and I'm not sure I would be able to guess the primitive right away.
Still, substitution is always a good way to simplify things:
$$I=\int_a^b \arctan\left(\dfrac{1}{1+x^2}\right)dx, \\ 0 \leq a<b$$
$$x= \tan t$$
$$I=\int_{\arctan a}^{\arctan b} \arctan\left(\cos^2 t\right)\frac{dt}{\cos^2 t}$$
$$u =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Are There Infinitely Many Square Triangular Numbers? Toying around with triangular numbers, it seems that there are
already 4 squares below 1000. At least I get this one:
?- between(1,1000,N), T is N*(N+1)//2, sqrtrem(T,_,0).
N = 1,
T = 1 ;
N = 8,
T = 36 ;
N = 49,
T = 1225 ;
N = 288,
T = 41616 ;
No
Are there infinitel... | To reproduce work
done many times before.
If
$T_n = m^2$,
since
$T_n = \dfrac{n(n+1)}{2}$,
this becomes
$n(n+1) = 2m^2$.
Completing the square,
$n^2+n+\frac14
=2m^2+\frac14
$
or
$(n+\frac12)^2
=2m^2+\frac14
$.
Clearing fractions,
this is
$(2n+1)^2
=8m^2+1
$
or
$(2n+1)^2-8m^2
=1
$.
This is a case
of a Pell equation
$x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $xyz$ if $x-\frac{1}{x}=y$, and $y-\frac{1}{y}=z$, and $z-\frac{1}{z}=x$ If
$$x-\frac{1}{x}=y, \qquad y-\frac{1}{y}=z, \qquad z-\frac{1}{z}=x$$
find the value of $xyz$.
This is how far I proceeded:
$x+y+z=z-1/z+x-1/x+y-1/y=>1/x+1/y+1/z=0
=>xy+yz+zx=0$
Also from question,
$x^2-1=xy,y^2-1=yz,z^2-1=zx$.
Adding $x^2... | Eliminating $x$ and $z$ between the equations we obtain:
$$3y^6-9y^4+6y^2-1=0.$$
By symmetry we know that $x$ and $z$ must also satisfy this equation. Hence $x^2$, $y^2$ and $z^2$ are roots of $$3y^3-9y^2+6y-1=0.$$
Therefore $(xyz)^2=\frac{1}{3}$ and hence $xyz=\pm\frac{1}{\sqrt 3}$.
In addition I think this may be one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solve the equation $4 \sin x \cos (2x) \sin (3x) = 1$ for $0^{\circ} < x < 180^{\circ}$ I have tried using identities but none of them worked for me.
Eg: $4 \sin x(1 - 2 \sin^2 x)(3 \sin x - 4 \sin^3 x) = 1$
| $$4 \sin(x)\cos(2x)\sin(3x) - 1 = 32 \sin(x)^6 - 40 \sin(x)^4 + 12 \sin(x)^2 - 1$$
Letting $\sin(x) = s$, this is
$$ 32 s^6 - 40 s^4 + 12 s^2 - 1 = (2s+1)(2s-1)(8s^4 - 8s^2+1)$$
which has roots
$$ s = \pm \frac{1}{2},\; \pm \frac{\sqrt{2\pm\sqrt{2}}}{2} $$
The values of $x$ in $(0,\pi)$ are $$\frac{\pi}{8},\; \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Is there any set of numbers that, when multiplied with anything from $\mathbb{N}$, results in an uneven number? I noticed the (very easy to see) fact that, whenever you multiply any natural number $n$ with anything having a prime factor of $2$, you get an even number.
For example:
$ 3 \times 2 = 6$, even
$5 \times 6 = ... | Any real number $r$ is equal to the limit of a series of this form:
$$
n + \frac{a_1}{2} + \frac{a_2}{2^2} + \frac{a_3}{2^3} +
\cdots + \frac{a_i}{2^i} + \cdots, \tag1
$$
where $n$ is an integer and each $a_i$ is either $0$ or $1.$
Moreover, if $r$ is any number in your desired set $\mathbb G,$ we can rule out the case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the asymptote of the function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ We have a function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ and when $x$ goes towards $-\infty$, we have an asymptote
$y = -2x - 3/2$.
How we get this asymptote?
| You can apply simple algebraic transformations, being careful to always have positive expressions under the square root. For $x<0$, and thus $|x|=-x$, the expression is equal to
\begin{align}
=\frac{-x\sqrt{-x}}{\sqrt{3-x}}-x
&=\frac{x}{\sqrt{3-x}}(-\sqrt{-x}-\sqrt{3-x})\\
\\
&=-2x+\frac{x}{\sqrt{3-x}}(-\sqrt{-x}+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.