Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Compute integral $\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$ While solving a more general problem, I came across the following integral
$$I= \int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$$
To start off, I defined $t:=x^2/2$ to get
\begin{align*}
I= \int_{-\infty}^\infty x^{4n} e^{-x^2/2} \, dx &= 2\int_0^\infty... | You're really done after the first integration by parts: as $\displaystyle \Gamma(z) = \int_0^\infty x^{z-1} e^{-x}\,dx,$
$\displaystyle 2 (\sqrt{2})^{4n - 1} \int_{0}^{\infty} t^{2n -\frac{1}{2}} e^{-t} dt =2 (\sqrt{2})^{4n - 1} \int_{0}^{\infty} t^{2n + \frac{1}{2}-1} e^{-t} dt =2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Question about the proof for "The rational line Q does not have the completeness property" Referring to the attached png file, I have questions about the inequality:
$$(n^2)/(2n+1) > (p^2)/(2q^2 - p^2)$$
How is this inequality derived? How did the n get into the inequality? Where did $n^2/(2n+1)$, $p^2/(2q^2 - p^2)$,... | Yeah....
So we have $(\frac pq)^2 < 2$ and we want to show $\frac pq$ is not an upper bound so we need to show there is $S > \frac pq$ so that $(\frac pq)^2 < S^2 < 2$.
$S$ is rational, so there is a rational $r$ so that $S =r\frac pq$ and $r > 1$.
We need to show that $r^2\frac {p^2}{q^2} < 2$ or that $r^2 p^2 < 2q^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Compute close-form of $\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}$ I came across the improper trigonometric integral recently shared in a Chinese web forum
\begin{align}
\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}\\
\end{align}
which amuses me because of its appear... | This feels more like a comment to Travis Willse's excellent answer, which essentially solves the problem, but here's the rest of the calculation:
So far we have ($u = \frac{\tan \left(\frac{\pi}{8} - \frac{t}{2}\right)}{\sqrt{2}-1}$)
\begin{align}
I &\equiv \int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sin(t) + \cos(t) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 0
} |
Solving a Polynomial Equation by Factoring. $n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.
| Factor out a $64$ to get:
$$64 \left(\frac{n^3}{64} + \frac{12n^2}{64} + \frac{48n}{64} +1\right) = 64 \left(\frac{n^3}{64} + 3 \cdot\frac{n^2}{16} + 3 \cdot \frac{n}{4} +1\right)$$
$$= 64 \left( \left(\frac{n}{4}\right)^3 + 3\left(\frac{n}{4}\right)^2 + 3 \left(\frac{n}{4}\right) + 1 \right)$$
$$= 64(n/4 + 1)^3 = (4 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 5
} |
Locus of a point of intersecting lines The perpendicular from the centre of the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ to the tangent at any point $P$ meets the line joining point $P$ to the focus $(ae,0)$ at $Q$. Prove that the locus of $Q$ is $$x^2+y^2-2aex=b^2$$
My attempt
Tangent to an ellipse at $P(\theta)$
$$bx\cos\th... | You are almost there. Now equating both,
$\displaystyle \frac{a}{b\cos\theta} x = \frac{b}{\left(a\cos\theta-ae\right)} (x-ae)$
$(x-ae+ae) (a^2\cos\theta-a^2e) = b^2\cos\theta(x-ae)$
$(x-ae) (a^2 \cos \theta - b^2 \cos\theta - a^2e)= a^3e^2 - a^3e \cos\theta$
We will use the fact that $a^2 - b^2 = a^2e^2$ now and again... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4061022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Decomposition of a sum of cubes I have to decompose the following polynomial:
$$(x+2y)^3+(x-2y)^3$$
Since I have the sum of two cubes, I have thought:
$$(x+2y)^3+(x-2y)^3=(x+2y+x-2y)[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]=2x\color{red}{[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]}$$
Now in my book the result is $2x(x^2+12y^2)$, but I can... | Your function is odd in $x$ and even in $y$, to wit
$[(-x)+2y]^3+[(-x)-2y]^3=-[(x-2y)^3+(x+2y)^3]$
$[x+2(-y)]^3+[x-2(-y)]^3=[(x-2y)^3+(x+2y)^3]$
So after factoring out $x$ your remaining quadratic factor must be even in both $x$ and $y$. Therefore your red expression can only contain three nonzero terms $ax^2+by^2+c$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Sequence queue - converges or diverges? I have the next sequence queue: $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$. Does the queue converges or diverge?
My attempt:
I have tried to show that $\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$ for any $n>0$, and then by the comparation test we get that since $\sum_{n=1}^... | We are trying to prove that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$
converges.
Let us use the Limit Comparison Test. That is, if $a_n>0$ and $b_n>0$, then if $$\lim_{n\to \infty}\frac{a_n}{b_n}=L$$ for some real number $L$. Then, if $b_n$ converges, by the limit comparison test, $a_n$ must also c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Expected Value of Binomial Distribution What is the expected value of the absolute value of the difference between the number of incoming tails and the number of incoming heads when a coin is tossed 5 times?
Here is what I think : Let $X$ is random variable of number of incoming heads then
$p=\frac{1}{2}$ and $n=5$... | Since $X\sim Binom(5,\frac{1}{2})$, the range of $X$ is $\{0,1,2,3,4,5\}$, which implies that $2X - 5$ is not necessarily always positive. Therefore in your proof, it is not true that $E(\left|2X - 5\right|) = E(2X - 5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number of possible solutions to $2 \le |x-1||y+3| \le 5$, where $x$ and $y$ are negative integers
If $$2 \le |x-1||y+3| \le 5$$ and both $x$ and $y$ are negative integers, find the number of possible combinations of $x$ and $y$ .
Below is my solution approach :-
As $x$ is a negative integer, hence $|x-1|$ in the $2 \... | Both $|x-1|$ and $|y+3|$ are non-negative integers. As the product must lie between $2$ and $5$, the only candidates are:
$$(1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (3,1), (4,1), (5,1)$$
However, we can't force $|x-1|=1$ with a negative $x$, so this leaves just five possible solutions:
$$(2,1), (2,2), (3,1), (4,1), (5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent. Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.
Attempt:
Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have
\begin{align*}
... | Just for the fun.
In the same spirit as @Alexander Conrad, we have the infinite series representation
$$\log \left(\cos \left(\frac{1}{n}\right)\right)=\sum_{k=1}^\infty (-1)^k\frac{ 2^{2 k-3} (E_{2 k-1}(1)-E_{2 k-1}(0)) }{k (2 k-1)!}\,n^{-2 k}$$ where appear Euler polynomials.
Truncated to any order, we can transform ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $a$ and $b$ have no greatest common divisor in $\mathbb{Z}[\sqrt-5]$
Show that in the ring $R=\mathbb{Z}[\sqrt{-5}]$, $a=3\cdot 7 \cdot(1+2\sqrt{-5})$ and $b=(1+2\sqrt{-5})\cdot 7\cdot(1+2\sqrt{-5})$ have no greatest common divisor.
Since $N(a)=3^3\cdot 7^3$ and $N(b)=3^2\cdot 7^4$ so it is clear that any c... | Note that
$$
\gcd\left(3\cdot7\left(1+2\sqrt{-5}\right),7\left(1+2\sqrt{-5}\right)^2\right)=7\left(1+2\sqrt{-5}\right)\gcd\left(3,\left(1+2\sqrt{-5}\right)\right)
$$
Since $N(3)=9$ and $N\!\left(1+2\sqrt{-5}\right)=21$, we must have
$$
\left.N\!\left(\gcd\left(3,\left(1+2\sqrt{-5}\right)\right)\right)\,\middle|\ 3\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find limits for uniformly distributed random variable given equations of expectations My question is this:
Let $\eta$ be uniformly distributed random variable over the interval $[a, b]$.
Let E$\eta^2$ = 1 and E$\eta$ = -E$\eta^3$. Find a and b.
This is my approach:
$
\eta\sim Unif[a,b]\Rightarrow
f_\eta(x)=
\... | We have $$4(b+a)=-2(b^2+a^2)(a+b)$$ If $b+a\neq0$ then $$b^2+a^2=-2$$ which is absurd. Therefore, $$a=-b$$ Substitute this into $$\frac{b^2+ab+a^2}3=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Representing all rational numbers between $\dfrac{1}{2}$ and $1$ How do I show that
$$\dfrac{2^{\left\lfloor\frac12 a_1\right\rfloor} + 2^{\left\lfloor\frac12 a_2\right\rfloor} + \ldots + 2^{\left\lfloor\frac12 a_n \right\rfloor}}{2^{\left\lceil\frac12 a_1\right\rceil} + 2^{\left\lceil\frac12 a_2\right\rceil} + \ldots... | We can observe that $2^{\lfloor a/2 \rfloor}$ and $2^{\lceil a/2 \rceil}$ are simply powers of $2$ which are either equal or consecutive. Hence:
$$2^{\lfloor a/2 \rfloor} \leqslant 2^{\lceil a/2 \rceil} \leqslant 2 \cdot 2^{\lfloor a/2 \rfloor}$$
Taking these bounds in our expression shows that any rational expressed i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 2
} |
Show $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6+\sqrt{14})$ Sifting through my old Galois Theory notes, I found a proof that $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6 + \sqrt{14})$, but my proof was chicken-scratch so I can no longer decipher it.
Can somebody come up with a way of showing this?
| You're probably missing a $+$
The key is to convert it to the form $a^2+2ab+b^2$. Here you have $\sqrt{21}$ inside the outer square root, so first multiply and divide by $\sqrt 2$. Now $a$ and $b$ can be found easily as $7+3 = 10$ and $7\cdot 3 = 21$
$\begin{align}\sqrt{5+\sqrt{21}}& = \frac1{\sqrt2}\sqrt{10+2\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find $(a, b, c)$ that satisfies the system of equations $2a + 3b -4c = 7, a-b+2c=6$ $\begin{align}
2a + 3b -4c = 7, \\
a - b + 2c = 6
\end{align}$
I haven't seen a three-variable system of equations with only two equations before. I try to make the problem into a two-variable system.
Solving for $a$ in the first equat... | Once you get the equations for b and c in terms of a, then you are done. Just say that all ordered pairs (a,b,c) that work are in the form ($k$, $-4k+19$, $\frac{-5k+25}{2}$) for any real number k. If you plug back into the two formulas, you will see that they will both simplify to c=c for some constant c, meaning any ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Has the equation $x^2-21 = 17y$ integer solutions? Has the equation $x^2-21 = 17y$ integer solutions?
Attempt:
I saw this: The equation $x ^ 2 + py + a = 0$ can be solved as an integer precisely, if $-a$ is a quadratic remainder modulo p.
I get: $x^2-17y-21=0$
Now i have to show $21$ is quadratic remainder modulo $-17$... | $x^2-21=17y$ has integer solutions precisely when $x^2-4=17(y+1)$ has integer solutions, which occurs precisely when it's possible to solve $x^2 \equiv 4 \pmod{17}$. This has solutions, of course, when $x \equiv \pm 2 \pmod{17}$.
Checking: $x =2, x^2-21=-17=17(-1); x=15, x^2-21=204=17(12)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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An upper bound for $\{r_n\}$ which satisfies $r_n = r_{n-1} + \frac{K}{r_{n-1}}$ I have a sequence $\{r_n\}$ of positive numbers defined recursively by
$$\tag{1} r_n = r_{n-1} + \frac{K}{r_{n-1}}.$$
for some positive $K$. The sequence is clearly strictly increasing and tends to $+\infty$ as $n\to \infty$. Also, by indu... | tl;dr: You cannot prove the upper bound you want, as the true growth of the sequence is "slightly more" than $2Kn$:
$$
r_n^2 = 2Kn + \frac{K}{2}\log n + O(1)
$$
First, for the sake of "elegance", set $a_n := \frac{r_n}{\sqrt{K}}$ for all $n\geq 0$, so that the recurrence relation becomes
$$
a_n = a_{n-1} + \frac{1}{a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4085112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Advanced complex numbers/roots of unity Let $a, b, c, d$ be real numbers, none of which are equal to $-1$, and let $\omega$ be a complex number such that $\omega^3 = 1$ and $\omega \neq 1.$ Given that,
$$
\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega} + \frac{1}{d + \omega} = \frac{2}{\omega},
$$
ho... | I will write $u$ instead of $\omega$, so that $u$ is the primitive third root of unity, and the other one is $u^2=\bar u$.
Below, $v$ will be an element in the list $L=\{1,u,u^2\}$ of all third roots of unity.
The given relation can be rewritten:
$$
2=u\sum\frac 1{a+u}=\sum\frac u{a+u}=\sum\frac 1{1+au^2}
\ ,
$$
where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
The equation ${4 \choose k}=6$ Find the solution of the equation $${4 \choose k}=6.$$
So $k$ must be a natural number $(n\in \mathbb{N})$ and we can find that when $k=1 \rightarrow {4\choose 1}=4$ and when $k=2\rightarrow {4\choose2}=6$, so the solution of the equation is actually $k=2$. Can we solve the problem withou... | For this case, you can use Pascal's triangle and binomial expansion.
$$
Pascal's\ \ triangle:
\\
\newcommand\pad[1]{\rlap{#1}\phantom{50}}
\begin{matrix}
n=0:&&&&&&&1\\
n=1:&&&&&&1&&1\\
n=2:&&&&&1&&2&&1\\
n=3:&&&&1&&3&&3&&1\\
n=4:&&&\pad{1}&&\pad{4}&&\pad{6}&&\pad{4}&&\pad{1}\\
n=5:&&\pad{1}&&\pad{5}&&\pad{10}&&\pad{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Solving $\tan^{-1}\left(\frac{2x+1}{x+1}\right)+\tan^{-1}\left(\frac{2x-1}{x-1}\right)=2\tan^{-1}\left(1+x\right)$
Find all $x$ that satisfy the equation $$\tan^{-1}\left(\frac{2x+1}{x+1}\right)+\tan^{-1}\left(\frac{2x-1}{x-1}\right)=2\tan^{-1}\left(1+x\right)$$
Attempt:
Taking tangent on both sides, and using the id... | Too large for comment.
Like Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$\dfrac{(2x+1)(2x-1)}{(x+1)(x-1)}$ will be $\le1$
$\iff0\ge\dfrac{(2x+1)(2x-1)}{(x+1)(x-1)}-1=\dfrac{3x^2}{x^2-1}$
$\iff x^2-1\le0\iff-1\le x\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
$a_1Give that $a_1<a_2<a_3<a_4$ are positive integers such that $$\sum_{i=1}^4\frac{1}{a_i}=\frac{11}{6}$$, find the value of $a_4-a_2$.
My try:
Since $11$ is prime, atleast one of $a_1,a_2,a_3,a_4$ should be divisible by $6$.
Now we should express the fraction $\frac{11}{6}$ as $\frac{A+B+C+D}{E}$ such that $E$ is div... | Use the identity $\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$ with $n=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\angle DAB = 10^\circ$, $\angle CAD = 20^\circ$, $\angle CDB = 140^\circ$, $CD = BD$. Find $\angle ABD$ $\angle DAB = 10^\circ$
$\angle CAD = 20^\circ$
$\angle CDB = 140^\circ$
The triangle $\triangle BCD$ is isosceles. $CD = BD$
Find the angle $x$
The answer is $30º$.
Playing with GeoGebra, I found that $AC = AD$. B... | Set $\angle ACD= y, x+y=110^\circ$
For $\triangle ADC$ and $\triangle ADB$:
\begin{align}\frac{DB}{AD}=\frac{\sin 10^\circ}{\sin x}=\frac{CD}{AD}=\frac{\sin 20^\circ}{\sin (110^\circ -x)}\end{align}
\begin{align}\frac{\sin 10^\circ}{\sin 20^\circ}=\frac{sin x}{\cos(x-20^\circ)}\end{align}
\begin{align}\tan x=\frac{\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4090521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Given that $3\sin x + 2\sin y = 4$. Find the maximum value of $3\cos x + 2\cos y$.
Given that $3\sin x + 2\sin y = 4$. Find the maximum value of $3\cos x + 2\cos y$.
I wonder if calculus can be used to solve this problem? If not how to solve this? I am thinking about $(3\cos x + 2\cos y)^2=9\cos^2x+12\cos x\cos y + 4... | Although Koro's solution is my favourite one, you can also solve it just using AM-GM as follows:
*
*Set $s=\sin x, t = \sin y$
*$\Rightarrow$ to maximize is $3\sqrt{1-s^2}+2\sqrt{1-t^2}$ subject to $3s+2t=4$ for $(s,t)\in [0,1]^2$
Since the objective function is non-negative you can maximize the square of it:
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Any pattern in the following sequence I am practicing finding patterns and sequences with numbers and found one that I cannot seem to find the logic.
$$
\frac{2}{3} ,\frac{4}{3} ,\frac{4}{5} ,\frac{6}{5} ,\frac{6}{7},\frac{6}{8}...
$$
The last term 6/8 breaks my understanding of a possible pattern.
If the last term did... | For
$$\{2,4,4,6,6,6\}$$ there are $27$ sequences in $OEIS$ and for
$$\{3,3,5,5,7,8\}$$ there are $30$ other.
Now, for
$$\frac{2}{3} ,\frac{4}{3} ,\frac{4}{5} ,\frac{6}{5} ,\frac{6}{7},\frac{6}{8}...$$ it is hard to say
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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recurrence relation - how to solve using generating functions How to solve this recurrence relation using generating functions (or in another way)?
$$\begin{cases} A(0)=0\\A(b)=1\\A(n) = pA(n+1)+qA(n-1)
\end{cases}\\ n=1,2,3,4,...,b-1.
$$
| Your recurrence relation can be written as
$$A(n+2) = \frac{A(n+1) - qA(n)}{p}, \qquad n=0,1,2,...,b-2.$$
which can be regarded as a generalization of the Fibonacci numbers. Let us set $A(1)=A$: notice that as far as the recurrence relation is concerned, the value of $A$ is arbitrary and the sequence proceeds indefinit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4099010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a$,$b$ and $y$ are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$ If a b and y are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$
This is what I have done so far, but apparently it is incorrect. I want to know why.
$(b+1/y)(y+1/a)(a+1/b)$
$(by+1/y)(ay+1/a)(ab+1/b)$
$(aby^2+1/ay)(ab+1/b)$
which... | You can also calculate the expression directly using the fact that
$p(x) = 3(x-a)(x-b)(x-y)\Rightarrow 3aby = 1$.
To do so, note that
\begin{eqnarray*} P
& = & (b+1/y)(y+1/a)(a+1/b) \\
& = & \frac{(by+1)(ay+1)(ab+1)}{aby} \\
& \stackrel{aby=\frac 13}{=} & 3\left(\frac 1{3a}+1\right)\left(\frac 1{3b}+1\right)\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Determine all positive integers $n$ such that $n^2+1$ divides $2^{n!}-1$ For this question, I have tried to start from reading similar question (How many rationals of the form $\frac{2^n+1}{n^2}$ are integers? from this post How many rationals of the form $\large \frac{2^n+1}{n^2}$ are integers?)
I notice that $2^{2!}-... | In the link in comment it is proved that $m^2-1\big|2^{m!}-1$ . Let $m=n^2$, then $m^2-1=n^4-1\big|2^{m!=n^2!}-1$ and $n^2+1\big|(n^4-1)$. For example :
$n=4 \rightarrow 4^2+1=17\big|4^4-1$ and we have:
$2^{4!}-1=2^{24}-1=[(2^4)^3-1][(2^4)^3+1]$
$(2^4)^3+1=(2^4+1)[(2^4)^2-(2^4)+1]=17k$
therefore:
$4^2+1\big|2^{4!}-1$
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Help with $ \lim \frac{n\pi}{4} - \left( \frac{n^2}{n^2+1^2} +\frac{n^2}{n^2+2^2} + \cdots + \frac{n^2}{n^2+n^2} \right) $ I have proved that $$\lim_{n\to \infty} \left( \frac{n}{n^2+1^2} +\frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) = \frac{\pi}{4},$$
by using the Riemann's sum of $\arctan$ on $[0,1]$. Now... | Let $f(x):=\frac{1}{1+x^2}$ so$$L=\lim_{n\to\infty}L_n,\,L_n:=n\int_0^1f(x)dx-\sum_{i=1}^nf(i/n).$$Taylor-expanding $f$,$$L_n=\sum_{k\ge1}\frac{f^{(k)}(0)}{k!}c_{nk},\,c_{nk}:=n\int_0^1x^kdx-\frac{1}{n^k}\sum_{i=1}^ni^k\sim-\tfrac12$$by Falhaber's formula. So$$L=-\tfrac12\sum_{k\ge1}\frac{f^{(k)}(0)}{k!}=-\tfrac12(f(1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4111276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Trouble with this one step involving arcsin So I'm solving a very basic trigonometric equation and after substituting the function for u and solving the quadratic equation, I ended up with $\sin\left(2x+\frac{\pi}{3}\right)=\frac{1}{4}$.
I need to take the arcsin of both sides if I'm not mistaken, but after inputting t... |
For $\alpha \in (-1,1)$ the equation $\sin(x) = \alpha$ has solutions
$$ x = \arcsin(\alpha) + 2 k \pi \qquad \text{and}\qquad x = \pi - \arcsin(\alpha) + 2 k \pi $$
Hence, in your case, the solutions are
$$
2x + \frac{\pi}{3}= \arcsin\left(\frac{1}{4}\right)+ 2 k \pi \implies x = -\frac{\pi}{6} +\frac{1}{2}\arcsin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that $a^ab^bc^cd^d\geq\frac{1}{16}(ab+c+d+1)^2$
Given $a, b, c, d\in\mathbb{R^+}$ such that $abcd = 1$. Prove that
$a^ab^bc^cd^d\geq\frac{1}{16}(ab+c+d+1)^2$
I've tried,
$a^ab^bc^cd^d=(abcd)^ab^{b-a}c^{c-a}d^{d-a}=b^{b-a}c^{c-a}d^{d-a}\geq\frac{1}{16}(ab+c+d+1)^2$
i'm stucked
| Since $x \mapsto x\ln x$ is convex on $x > 0$, and $a + b + c + d \ge 4\sqrt[4]{abcd} = 4$, we have
\begin{align*}
a\ln a + b\ln b + c\ln c + d\ln d
&\ge 4 \cdot \frac{a + b + c + d}{4} \ln \frac{a + b + c + d}{4}\\
&\ge 4 \ln \frac{a + b + c + d}{4}.
\end{align*}
Also, we have
$$ab + c + d + 1 \le \frac{1}{4}(a + b)^2... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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When is $E(X-\mu)^2 \ne E(X^2)-\mu^2$? Consider the probability density function
$$f(x)=\frac{x}{18}, \quad x \in [0,18]$$
Find $E(x)$ and $Var(x)$.
Solution:
a.)
\begin{align*}
E(x)
&=\int_{-\infty}^{\infty} x f(x) dx \\
&=\int_{0}^{18} x \left(\frac{x}{18}\right) dx \\
&=\frac{1}{18} \int_0^{18} x^2 dx \\
&=108
\end... | The probability density function integrates to 9 in the interval given, that is why it does not work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Calculating $\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}$ Please help me calculate the:
$$\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}$$
My attempt is: First method: for $t>0$ we have $e^t\geq \frac{t^3}{3}$ now i have:
$$(x^2+y^2)e^{-(x+y)}\leq\frac{3(x^2+y^2)}{(x+y)}\rightarrow 0, (x,y)\r... | Aside from what I take to be a typo (the denominator under the $3(x^2+y^2)$ should be $(x+y)^3$, not just $(x+y)$), your first method looks fine. It could be elaborated to
$$(x^2+y^2)e^{-(x+y)}\le{3(x^2+y^2)\over(x+y)^3}={3x^2\over(x+y)^3}+{3y^2\over(x+y)^3}\le{3x^2\over x^3}+{3y^2\over y^3}={3\over x}+{3\over y}\to0+0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Does $\int_1^{\infty}\frac{x^2\tan^{-1}(ax)}{x^4+x^2+1}dx$ have closed form? I have been trying to find the closed form for integral below $$\int_1^{\infty}\frac{x^2\tan^{-1}(ax)}{x^4+x^2+1}dx ,\; \; a>0 $$
My progress to this integral $$\cong\frac{\pi^2}{8\sqrt 3}+\frac{\pi}{8}\log(3)-\frac{\pi}{6a\sqrt 2}+\frac{1}{2... | We have \begin{align}\int_1^{\infty}\frac{x^2\arctan ax}{x^4+x^2+1}\,dx&=\int_0^1\frac{\arctan a/u}{u^2(1/u^4+1/u^2+1)}\frac{du}{u^2}\\&=\int_0^1\frac{\pi/2-\arctan u/a}{u^4+u^2+1}\,dx\\&=\frac\pi2\left(\frac14\log3+\frac{\pi\sqrt3}{12}\right)-\int_0^1\frac{\arctan bu}{u^4+u^2+1}\,du\end{align} using $u^4+u^2+1=(u^2+u+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maximize $x^2y$ where $x^2+y^2=100$. (Looking for other approaches)
For positive numbers $x,y$ we have $x^2+y^2=100$. For which ratio of $x$ to $y$, the value of $x^2y$ will be maximum?
$1)\ 2\qquad\qquad2)\ \sqrt3\qquad\qquad3)\ \frac32\qquad\qquad4)\ \sqrt2$
It is a problem from a timed exam so I'm looking for alte... | Alternatively, by AM-GM inequality, $x^2y = 2\sqrt{\dfrac{x^2}{2}\cdot \dfrac{x^2}{2}\cdot y^2}\le 2\sqrt{\dfrac{\left(\dfrac{x^2}{2}+\dfrac{x^2}{2}+ y^2\right)^3}{27}}= 2\cdot \sqrt{\dfrac{100^3}{27}}= \dfrac{2,000}{3\sqrt{3}}=\dfrac{2,000\sqrt{3}}{9}\implies (x^2y)_{\text{max}} = \dfrac{2,000\sqrt{3}}{9}$ at $\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Set of primes in the spectrum of a rational number. Let $x \in \mathbb{Q}^{+}$ (numerator and denominator are in the reduced form). Can we comment anything about the set $\left\{ \left\lfloor nx \right\rfloor \; : \; n \in \mathbb{N} \right\}$? Is there any structure in this set of integers especially the possible se... | Let $m = \lfloor x \rfloor $. Let $r = x-m$ so $0\le r < 1$. $q$ is rational so let $j,k \in \mathbb N$ so that $\gcd(j,k) = 1$ and $r = \frac jk$. In other words..... Let $x = m + \frac jk$ were $m$ is integers and $\frac jk$ is a non-negative rational in reduced terms less than $1$.
$ nx = nm + nj\cdot \frac 1k ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Angle between tangents to the curve $x^2+3y^2=9$
Tangents drawn from the point $(\alpha,\alpha^2)$ to the curve $x^2+3y^2=9$ include an acute angle between them, then find $\alpha$.
My attempt is by using the equation for pair of tangents from an external point to the curve. If the curve is $S=0$ then the equation fo... | $$y=mx+c$$ will be tangent of $$\dfrac{x^2}{3^2}+\dfrac{y^2}3=1$$ if $c^2=9m^2+3$
$$(y-mx)^2=c^2=9m^2+3$$
$$ m^2(9-x^2)+2mxy+3-y^2=0$$
As these pass through $(\alpha,\alpha^2),$
$$ m^2(9-\alpha^2)+2m\alpha^3+3-\alpha^4=0$$
which is a quadratic equation in $m$
$$m_1+m_2=-\dfrac{2\alpha^3}{9-\alpha^2}, m_1m_2=\dfrac{3-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Possible pattern involving $x$ in the continued fraction expansion of $\frac{1}{\sqrt[3]{x^3+1}-x}$ Consider the expression $$\frac{1}{\sqrt[3]{x^3+1}-x}$$
Plugging in $10$ for $x$ and using $W|A$, we find that the continued fraction expansion is $[300; \mathbf{10}, 450, 8, ...]$.
For $x=11$, it's $[363; \mathbf{11}, 5... | Let $$\sqrt[3]{x^3+1} - x = a$$ Then $$a^3 + 3a^2x + 3ax^2 = 1$$ This shows that $a < \frac{1}{3x^2}$. Now we have everything we need to work out the continued fraction:
$$\begin{align*} \frac{1}{a} &= 3x^2 + a(3x + a)\\
&=3x^2 + \frac{1}{\frac{1}{a}\frac{1}{3x+a}}\\
&=3x^2 + \frac{1}{\frac{3x^2 + 3ax + a^2}{3x+a}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Summation over $a+b+c=5$ Let $a,b,c$ be positive integers. Compute
$$\sum_{a+b+c=5} (6-a)(6-b)(6-c).$$
The first thing I notice is symmetry, so that I can permute $3!=6$ ways, but i'm not really sure how that works with the condition $a+b+c=10.$ The other method is to fix $a$, but that is reall time-consuming and unfe... | This problem is somewhat specific and nice, since when using symmetry it reduces to the summation of two terms.
We obtain
\begin{align*}
&\color{blue}{\sum_{\substack{a+b+c=5\\a,b,c\geq 1}}(6-a)(6-b)(6-c)}\\
&\qquad=6\sum_{\substack{a+b+c=5\\1\leq a<b<c\leq 3}}(6-a)(6-b)(6-b)\tag{1}\\
&\qquad\quad +3\sum_{\substack{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Problem with integration by substitution.
Using the substitution $u=1-x$, compute the integral of
$\int{x(1-x)^2}dx.$
My Work: Let $u=1-x.$ Then $\mathrm dx=-\mathrm du$ and $x=1-u,$ so
$$\int{x(1-x)^2}\,\mathrm dx\\
=-\int{(1-u)u^2}\,\mathrm du\\
=\frac{u^4}{4}-\frac{u^3}{3}+C\\
=\frac{(1-x)^4}{4}-\frac{(1-x)^3}{3}+... | Well
\begin{align}
\int{x(1-x)^2}dx &= \int x^3-2x^2+x \mathrm{d}x \\
&=\frac{1}{4}x^4-\frac{2}{3}x^3+\frac{x^2}{2} +C_1
\end{align}
Whereas with $u=1-x \, \implies \mathrm{d}u=-\mathrm{d}x$ and
\begin{align}
\int (1+u)^2u (- \mathrm{d}u) &= -\int u^3+2u^2+u \mathrm{d}u \\
&=-\frac{u^4}{4}-\frac{2}{3}u^3-\frac{u^2}{2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given 2d vectors $A$, $B$, $C$, $D$, $P$, find $x$ such that the line through $Ax+B(1-x)$ and $Cx+D(1-x)$ passes through $P$ I'm trying to parametrize the quad with an x and y coordinate (from 0 to 1) so that I can properly apply a texture to it for a computer graphics application. I came up with the idea for this myse... | Give all vectors a third coordinate of $0$ so that, as members of $\Bbb{R}^3$, they're compatible with cross products. Also, translate all points by $-P$ to move $P$ to the origin. Then, for $P$, $Q$, $R$ to be collinear, "all we have to do" is force
$$R\times Q = O \tag{0}$$
where $O:=(0,0,0)$. To simplify some calcul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\left\lfloor \frac{1}{1^{\frac{2}{3}}}+\frac{1}{2^{\frac{2}{3}}}+\frac{1}{3^{\frac{2}{3}}}+...+\frac{1}{1000^{\frac{2}{3}}}\right\rfloor$ Let $$M=\left\lfloor \frac{1}{1^{\frac{2}{3}}}+\frac{1}{2^{\frac{2}{3}}}+\frac{1}{3^{\frac{2}{3}}}+...+\frac{1}{1000^{\frac{2}{3}}}\right\rfloor$$.
Find $M$
My Attempt
Let $f(x... | Let $$S = \sum_{r=1}^{1000} r^{-2/3}.$$ Then because for each positive integer $r$, $$r^{-2/3} = \int_{x=r}^{r+1} \lfloor x \rfloor^{-2/3} \, dx > \int_{x=r}^{r+1} x^{-2/3} \, dx,$$ it follows that
$$S > \sum_{r=1}^{1000} \int_{x=r}^{r+1} x^{-2/3} \, dx = \int_{x=1}^{1001} x^{-2/3} \, dx = 3(1001^{1/3} - 1) > 3(1000^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Convolution of two independent uniform variables Question: Let's say $X$ and $Y$ are two independent random variables with $Uniform (0,3)$ distribution. What is the probability density function of $X+Y$?
In solution, it is told to use convolution method by many sources since the pdf of the sum of two independent random... | Let $Z = X + Y$, then probability distributionn function of $Z$ is defined as
$$
\begin{aligned}
F_Z(z) &= \text{Pr}(Z \leq z) = \text{Pr}(X + Y \leq z) = \text{Pr}(X \leq z - Y) = \int\limits_{-\infty}^{+\infty}F_X(z-y)f_Y(y)dy = \\
&= \frac{1}{3}\int\limits_{0}^{3}F_X(z-y)dy =
\left|
\begin{array}{l}
u = z-y \Leftri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4146514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What method is required to find out "for what values of k is $4x^2+kx+\frac14$ a perfect square?"
For what values of $k$ is this expression a perfect square?
$$4x^2+kx+\frac14$$
I'm thinking maybe the perfect square trinomial is supposed to be used, as I had to use that formula on the problem prior to this one. Howev... | We consider a polynomial in $x$
\begin{align*}
p(x)=4x^2+kx+\frac{1}{4}
\end{align*}
which has degree $2$, constant term $\frac{1}{4}$, linear term $kx$ and square term $4x^2$.
We are looking for a representation as perfect square
\begin{align*}
q(x)=(ax+b)^2
\end{align*}
We recall two polynomials are equal, i.e. $p=q$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solve the inequality $\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$ $$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$$
The answer for this inequality is given as $x ∈ (- \infty , -1) \cup(0, 1) \cup (2, \infty)$ but when I solve it, I am only able to get $x ∈ ( - \infty, -1)$.
How should I solve... | Solve the Inequality :$$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2} \\$$
$$\frac{-1}{x(x+1)} \le \frac{-1}{(x-2)(x-1)} \\$$
Consider two scenarios :
scenario 1 : Less than
$$\frac{-1}{x(x+1)}~< \frac{-1}{(x-2)(x-1) }$$.
$$\frac{-1~(x^2-x-2x+2)}{x(x+1)}+1~<0$$
$$\frac{(4~x~-2)}{x~(~x+1)}~<0$$
$$x \in (-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given the following set of equations, find: $xy + 2yz + 3zx$.
The positive reals x, y, z satisfy the equations
$$x^2 + xy + \frac{y^2}{3}=25$$
$$\frac{y^2}{3}+z^2 = 9$$
$$z^2+ zx + x^2 = 16$$
Find $$xy + 2yz + 3zx$$
My understanding:
*
*What struck me first were the squares $9, 16, 25$. This is the “Egyptian tria... | Adding the second and third equation and subtracting first equation , we get
$2z^2+xz=yx$
Substitute the value for $y$ in the first equation, we get
$$3x^4+7x^2z^2+3x^3z+4z^4+4z^3x=75x^2$$
which easily gets factored into
$$(x^2+z^2+xz)(3x^2+4z^2)=75x^2$$
$$(16)(3x^2+4z^2)=75x^2$$
One possible way to avoid all the algeb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Reverse the order of integration of $\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \,dx\,dy$. Q: Reverse the order of integration of $\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \, dx\,dy$.
I imagined that $x$ should varie between $\dfrac{1}{9}$ and $\sqrt{\dfrac{2}{3}}$ and $y$ varie betwee... | Three piece decomposition $\int\limits_\frac{1}{9}^\frac{4}{9}\int\limits_\frac{1}{3}^\sqrt{x}+\int\limits_\frac{4}{9}^\sqrt{\frac{1}{3}}\int\limits_\frac{1}{3}^\frac{2}{3}+\int\limits_\sqrt{\frac{1}{3}}^\sqrt{\frac{2}{3}}\int\limits_{x^2}^\frac{2}{3}f(x,y)dydx$
Draw a picture. You will see it immediately.
| {
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"url": "https://math.stackexchange.com/questions/4152434",
"timestamp": "2023-03-29T00:00:00",
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Given a unitary vector $(x_1,\ldots,x_n)^T$ and a vector $y=(x_2,\ldots, x_n)^T$, prove the columns of a created matrix form an orthonormal basis
Let $X=\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}\in M_{n1}(\Bbb F),x_1\ne-1$ be a unit vector wrt the standard scalar product and let $Y=\begin{bmatrix}x_2\\x_3\\\vd... | As given, $\ M\ $ is not always unitary. Take
$$
X=\begin{bmatrix}\frac{i}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}
\end{bmatrix}\ .
$$
then
$$
M=\begin{bmatrix}\frac{i}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&\frac{1}{2+i\sqrt{2}}-1
\end{bmatrix}=\begin{bmatrix}\frac{i}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Q is to prove that integer just above ($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers. Q is to prove that integer just above($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers.
In Q , by integer just above means that:
For an example , which is the in... | Note that $0\lt \sqrt 3 -1 \lt 1$
Now look at the numbers $a=1+\sqrt 3, b=1-\sqrt 3$ with $a+b=2, ab=-2$ which are roots of the quadratic $x^2-2x-2=0$
Then with $v_n=a^n+b^n$ we have $v_{n+2}-2v_{n+1}-2v_n=0$ (spot the coefficients) or $v_{n+2}=2(v_{n+1}+v_n)$ now $|b^n|\lt 1$ and indeed $0\lt b^{2n}\lt 1$ so the diffe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Rate of change of radius of spherical balloon
A spherical balloon is filled with gas at a rate of $4 \text{ cm}^3/\text{s}$. What rate is the radius $r$ changing with respect to the time when the vol $V=36π \text{ cm}^3$?
[ans:$\frac 19, \frac{\pi}{9} $or 9$\pi$]
I tried this one with implicit differentiation:
$\frac... | Just as another way to approach the calculation, since the volume of the balloon is given, rather than its radius:
You had the result from implicit differentiation,
$$ \frac{dV}{dt} \ \ = \ \ \frac43 \pi \cdot \frac{d}{dt} \ [r^3] \ \ = \ \ \frac43 \pi \cdot 3r^2 · \frac{dr}{dt} \ \ = \ \ 4 \pi r^2 · \frac{dr}{dt}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4163944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
"Find the volume of the ellipsoid $x^2+\frac{y^2}{100}+\frac{z^2}{4} = 1$" was the question and here's my attempt. So, I asked this question and got downvotes and someone saying I should present my attempt. Here lies my thought process:
I know the volume is $\iiint dxdydz$ on any W$\subset R^3$ and I "just" had to find... | What you did is correct. Translating it into an integral in Cartesian coordinates, it becomes$$\int_{-1}^1\int_{-10\sqrt{1-x^2}}^{10\sqrt{1-x^2}}\int_{-2\sqrt{1-x^2-\frac{y^2}{100}}}^{2\sqrt{1-x^2-\frac{y^2}{100}}}1\,\mathrm dz\,\mathrm dy\,\mathrm dx.$$You can compute it as follows:\begin{align}\int_{-1}^1\int_{-10\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How can you quickly use the graph of $3x-6y=4$ to produce the graph of $3x-6y=10$? To start with, I am asked to graph the lines $3x-2y=k$ for $k=1,2,3$ and $4$.
I find that the equations produce different parallel lines.
For each line where $k=2,3$ and $4$, the coordinates points are $0.5k$ below the coordinates points... | You can consider the $x$-intercepts instead: for the first line, it is the point $\bigl(\frac23, 0\bigr)$, and for the second line the point$\bigl(\frac53,0\bigr)$. The difference is equal to $1$, so you obtain the second line through a horizontal translation with vector $(1,0)$ of the first.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Calculate $\iint \sqrt{2-x^2-y^2} \, dx \, dy$ over the circle $x^2+y^2=x\sqrt{2}$ so as my question states I have to calculate
$$\iint_\limits A \sqrt{2-x^2-y^2}\,dx\,dy$$ where $A$ is the circle defined by
$$x^2+y^2=x\sqrt{2}$$
After introducing polar coordinates $(x = r\cos\theta, y= r\sin\theta)$, I get that:
The l... | Your are almost there. You need to integrate over the circle $x^2 + y^2 \leq \sqrt2 x$.
So as you said, $0 \leq r \leq \sqrt2 \cos\theta, - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$
The integrand is $\sqrt{2-x^2-y^2} = \sqrt{2-r^2}$
and as $r \leq \sqrt2 \cos\theta, \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Local maxima, Local minima and saddle points I am trying to find the local maxima, minima, saddle points and choose the global maxima and minima of this function;
$f(x,y) = e^{-x^2-3y^2}.(x+y^3)$
I tried to find the partial derivatives for x and y. I find the equations and then I use $f_x = 0$ and $f_y = 0$ to find t... | We have
$$\begin{align} f_x &= e^{-x^2-3 y^2}-2 x e^{-x^2-3 y^2} \left(x+y^3\right) \\ f_y &= 3 y^2 e^{-x^2-3 y^2}-6 y e^{-x^2-3 y^2} \left(x+y^3\right) \end{align}$$
If we simultaneously set $f_x = f_y = 0$, we can divide out $e^{-x^2-3 y^2}$ and we are left with
$$1 - 2x(x+y^3) = 0 \\ 3 y^2 - y(x + y^3) = 0$$
Simulta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is a symmetric matrix a subspace of nxn matrices? Let $M$ be a vector space for all 2x2 matrices.
Show that the set of all the symmetric matrices $M2$={$B ∈ M2 : B=B^t$} is a subspace of M.
My solution:
*
*The null matrix is symmetric
$$B=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$
*Let $B$, $C$ $∈$ $M2$, where $B = ... | A simpler way to prove this is to note that if $A,B$ are symmetric matrices then since $(\lambda A)^T = \lambda A^T = \lambda A$ and $(A+B)^T = A^T+B^T = A +B$, so are $\lambda A$ and $A+B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\log_2(1+b(ax^2+c)^{-2})$ I am wondering how to calculate $\int \log_2(1+b(ax^2+c)^{-2})$, where $a,b,c>0$.
I have used wolframalpha to calculate it, and the answer is like
While it gives me a nice result, I am wondering how to get it. (Wolfram|Alpha does not provide Step-by-Step Solution for this problem)
... | $$I=\int\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\,dx$$ Use integration by parts to get rid of the logarithm
$$u=\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\implies du=-\frac{4 a b x}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}$$
$$I=x\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve for $x: \dfrac{x+6}{x-6}\left(\dfrac{x-4}{x+4}\right)^2+\dfrac{x-6}{x+6}\left(\dfrac{x+9}{x-9}\right)^2\lt\dfrac{2x^2+72}{x^2-36}$
Solve for $x: \dfrac{x+6}{x-6}\left(\dfrac{x-4}{x+4}\right)^2+\dfrac{x-6}{x+6}\left(\dfrac{x+9}{x-9}\right)^2\lt\dfrac{2x^2+72}{x^2-36}$
I tried taking LCM on LHS, but it's getting ... | The inequality is satisfied if
$$\begin{align} x & \;<\; -6\tag{1a}\\
\text{or}\quad\frac 65\big(1-\sqrt{26}\,\big)\,=\, -4.92 \;<\; x & \;<\; 0\;\;\text{but }\; x\neq -4\tag{1b}\\
\text{or}\quad 6 \;<\; x & \;<\; 7.32\,=\,\frac 65\big(1+\sqrt{26}\,\big)\,.\tag{1c}
\end{align}$$
To arrive at this result, exploit Paress... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Draw the Ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ a tangent plane which cuts off equal segments on the coordinate axis. Draw the Ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ a tangent plane which cuts off equal segments on the coordinate axis.
First, I want to find the ... | Ellipsoid is given by, $ \displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
If a plane is tangent to ellipsoid at point $(x_1, y_1, z_1)$, its equation is given by,
$\displaystyle \frac{x x_1}{a^2} + \frac{y y_1}{b^2} + \frac{z z_1}{c^2} = 1$
If $d \gt 0$, we also know that the plane passes through ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What is $\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y}$? I want to understand this multivariate limit.
WolframAlpha says $\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} = 0$
But what if I take a curve $y = -x$, then the limit doesn't exist, right? Wouldn't this make the limit inexistent?
I tried to prove using sandwic... | Method $-1.$
$$\begin{align}\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} &=2\lim_{(x,y) \to (0,0)} \dfrac{\sin^2\left(\frac x2\right)}{x+y}\end{align}$$
Let $y(x):=r\sin^2\left(\frac x2\right)-x$, as $x\to 0$ and $r\in\mathbb R\setminus \left\{0\right\}$ then we get
$$\begin{align}2\lim_{(x,y(x)) \to (0,0)} \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Integrate using $t$ substitution integrate this $\int \frac{1}{\sqrt{x+1}+\sqrt{x}}$
I took $\sqrt{x} = t$
Differentiated it
So $\frac{1}{2\sqrt{x}}=\frac{dt}{dx}$
Then I don't know how to proceed
I did some steps and it got too complicated
I also took $\frac{1}{\sqrt{x+1}+\sqrt{x}} = t$
But that also got complicated
... | A COMPLICATED SOLUTION:
Let $$\sqrt{1+x}+\sqrt{x}=u, \quad \sqrt{1+x}-\sqrt{x}=1/u \implies 2\sqrt{x}=u-1/u~~~~(*)$$
So we get $dx=\frac{(u^4-1)}{4u^2} du$
So $$I=\int \frac{dx}{\sqrt{1+x}+\sqrt{x}}=\frac{1}{2}\int (1-\frac{1}{u^4}) dx$$
$$\implies I= \frac{1}{2} \left(u+\frac{1}{3u^3}\right) =\frac{1}{2}\left(\sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to un-simplify a general term I have a general term of $\frac{3n-1}{5n+7}$ and I want to find out which term $\frac{7}{12}$ is. I have a number in mind that I got from just trial and error but how would I apply a formula to find this. I am treating it as an arithmetic sequence with a common difference of 3 on top a... | You can't treat $\frac{3n - 1}{5n + 7}$ as an arithmetic progression. For reference, let $a_{i} = \frac{3i - 1}{5i + 7}$. Then,
\begin{align*}
a_{n} - a_{n - 1} &= \frac{3n - 1}{5n + 7} - \frac{3(n - 1) - 1}{5(n - 1) + 7} \\
a_{n} - a_{n - 1} &= \frac{3n - 1}{5n + 7} - \frac{3n - 3 - 1}{5n - 5 + 7} \\
a_{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $\int_a^b \sin\left(x+\frac{1}{x}\right) dx <3.$ Show that $$\int_a^b \sin\left(x+\frac{1}{x}\right) dx <3$$ for all $a,b \in \mathbb{R}.$
Here is my solution:
\begin{align*}
\int_a^b \sin\left(x+\frac{1}{x}\right) dx&=\int_a^b \left(\sin x\cos \frac{1}{x}+\cos x\sin \frac{1}{x}\right) dx\\
&\le \int_a^b( \si... | Partial answer :
We use the triangle inequality so we want to show :
$$f(a,b)=\left|\int_{b}^{a}\cos\left(x\right)\sin\left(\frac{1}{\left(x\right)}\right)dx\right|+\left|\int_{b}^{a}\sin\left(x\right)\cos\left(\frac{1}{\left(x\right)}\right)dx\right|<3$$
With the inequality due to Ostrowski (see my last comment) we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
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Asymptotics for $\sum_{k=0}^{n-1}(-1)^k \cot(\frac{2k+1}{4n}\pi)\log{(2\sin(\frac{2k+1}{4n}\pi))} $ I have been able to establish the identity
$$
I(n):=\int_0^1 \frac{y^{n-1}(1+y-y^{n+1}) - 1}{(1-y)(1+y^{2n})}\ dy = \frac{1}{n} \sum_{k=0}^{n-1}(-1)^k\cot(\frac{2k+1}{4n}\pi)\log{(2\sin(\frac{2k+1}{4n}\pi))}
$$
However, ... | Note that
$$
I(n) = \int_0^1 {\frac{{(t^n - 1)^2 }}{{(t - 1)(t^{2n} + 1)}}dt} + \int_0^1 {\frac{{t^{n - 1} }}{{t^{2n} + 1}}dt} .
$$
Here
$$
\int_0^1 {\frac{{t^{n - 1} }}{{t^{2n} + 1}}dt} \le \int_0^1 {t^{n - 1} dt} = \frac{1}{n}.
$$
Now
\begin{align*}
\int_0^1 {\frac{{(t^n - 1)^2 }}{{(t - 1)(t^{2n} + 1)}}dt} &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove if number is rational or irrational I've been asked to prove if $\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}}$ is a rational number. I've tried a proof as follows:
Suppose the number is rational, so it can be written as the quotient of 2 numbers $a$ and $b$
\begin{align*}
\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + ... | Alternative approach:
In a problem like this, my first step is to clear the denominator.
$\displaystyle \frac{\sqrt{3 + \sqrt{5}}}{\sqrt{10} + \sqrt{2}} \times \frac{\sqrt{10} - \sqrt{2}}{\sqrt{10} - \sqrt{2}} = \frac{1}{8} \times \left[\sqrt{3 + \sqrt{5}}\right] \times \left[\sqrt{10} - \sqrt{2}~\right].$
Therefore, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Line $y = mx$ through the origin that divides the area between the parabola $y = x-x^2$ and the x axis into two equal regions. There is a line $y = mx$ through the origin that divides the area between the parabola $y = x-x^2$ and the x axis into two equal regions. Find m.
My solution:
When I compute my answer, I get ... | This is my solution.
First, we recognize that the quadratic has roots at $(0,0)$ and $(1,0)$. So, to first solve for the area under the quadratic:
$\int_0^1 x-x^2.dx=\left[\frac{x}{2}-\frac{x^3}{3}\right]_0^1$
$=\frac{1}{6}$
So, half of this area is $\frac{1}{12}$
Let $x=a$ be the abscissa of the intersection point of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find $\sin^5 x + \cos^5 x$
Given $$\cos \left[\sqrt{\left(\sin x + \cos x\right)\left(1 - \sin x \cos x \right)}\right] \\{}= \sqrt{\cos \left(\sin x + \cos x \right) \cos \left(1 - \sin x \cos x\right)}.$$
Find $\sin^5 x + \cos^5 x.$
My try:
\begin{align*}
\sin^5x+\cos^5x
&=(\sin x+\cos x)(\sin^4x-\sin^3x\cos x+\si... | I wanted to develop the idea expressed in comments that LHS is bounded below by $\cos(1)$ while RHS is bounded above by $\cos(1)$ and that we get equality for $x=0\text{ or }\frac{\pi}2$ only.
*
*The LHS part and equality case is not too complicated to prove:
$\require{cancel}\begin{align}f(x)
&=\Big(\cos(x)+\sin(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $n$ can be expressed as $2x^2+3y^2$ with integers $x$ and $y$, then so can $7n$. The problem is:
Let us say that a positive integer $n$ is obtainable if there exist integers $x$ and $y$ such that $n=2x^2+3y^2$. If $n$ is obtainable, prove that $7n$ is also obtainable.
I tried doing that
$$2(x+a)^2+3(y+b)^2=14x^2+2... | I would modify your substitution to avoid linear term.
$$2(x+ay)^2 + 3(y+bx)^2 = 14x^2+21y^2$$
$$2(x^2+2axy+a^2y^2)+3(y^2+2bxy+b^2x^2)=14x^2+21y^2$$
$$(2+3b^2)x^2 +(4a+6b)xy + (2a^2+3)y^2=14x^2+21y^2$$
Can we find $(a,b)$ such that the equality holds for all $x,y$? I will leave the investigation to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Finding $x_1+x_2$, where the $x_i$, and $y$ and $z$, are positive integers satisfying $x^2+y^2+z^2=2(xy+1)$ and $x+y+z=2022$
Let $x$, $y$, and $z$ be positive integers satisfying
$$(x)^2+(y)^2+(z)^2=2(xy+1) \quad\text{and}\quad x+y+z=2022$$
If $x_1$ and $x_2$ are 2 distinct solutions for $x$, what is the value of $x_1... | Considering achile hui comment we have:
$z=1$ and $(x-y)^2=1$ and:
case 1):$\begin {cases}x-y=1\\x+y+1=2022\end{cases}$
which gives $x=1011$ and $y=1010$
case 2):$\begin {cases}x-y=-1\\x+y+1=2022\end{cases}$
which gives $x=1010$ and $y=1011$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving Diamond & Shurman Exercise 3.7.1, about conjugacy class of $\Gamma_0^{\pm}(N)$. I am reading chapter 3 of A First Course in Modular Forms but have troubles in Exercise 3.7.1 (c) and (d).
(c) Show that the $\Gamma_0^{\pm}(N)$-conjugacy class of $\gamma \in \Gamma_0(N)$ is the union of the $\Gamma_0(N)$ conjugac... | Let
$$\Gamma_0^-(N)=\left\{ \begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\in\Gamma_0^\pm(N):ad-bc=-1 \right\},$$
so the $\Gamma_0^\pm(N)$-conjugacy classes of $\gamma$ is the union of $\Gamma_0(N)$-conjugacy classes of $\gamma$ and $\Gamma_0^-(N)$-conjugacy classes of $\gamma$. Let $\alpha=\begin{bmatrix}
a & b\\
c &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4199234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Proving on the equation $\frac1{x+1}+\frac1{2(x+2)}+\frac1{3(x+3)}+\cdots+\frac1{n(x+n)}=1$ Consider the equation:
\begin{gather*}
\frac{1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + \cdots + \frac{1} {n(x+n)} = 1.\tag1
\end{gather*}
Prove that:
*
*For every $n \geq 2$, equation $(1)$ always has a unique positive s... | Since you already find a solution for the first question. I write an answer for the second one. Observe that
$$\sum_{k=1}^n\frac{1}{k\cdot (k+1)}=\sum_{k=1}^n\frac{1}{k}-\frac{1}{k+1}=1-\frac{1}{n+1}.$$ Hence
$$\sum_{k=1}^{\infty}\frac{1}{k\cdot (k+1)}=1.$$
This means for $x=1$ the left handside of your equation tends ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Ambiguity in Integration done by substitution. I was solving this integral $$I= \int x\sqrt{1-x^2} \mathrm dx$$ My solution is:
Let $x=\sin y$, then $|\cos y|=\sqrt{1-x^2}$ I now use method of substitution for the cases $\cos y\geq 0$ and $\cos y<0$. The correct answer is $-\frac{(1-x^2)^{3/2}}{3} +C$. I am able to obt... | I suspect you forgot to include the sign when you substituted $\cos y$ back to get a function in terms of $x$ at the end. Here is the complete derivation.
I'm assuming that you did the substitution and integration correctly: if $x=\sin y$, then $dx = \cos y\,dy$, so we get
$$\int x\sqrt{1-x^2}\,dx = \int\sin y|\cos y|\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the point on the curve $x^{2} - yz=1$ that is closest to the origin The problem statement is as follows:
Find the point on the curve $x^{2} - yz=1$ that is closest to the origin.
To do this we minimize $D^{2} = x^{2} + y^{2} + z^{2}$ with the above constraint.
Attempted solution:
Isolate $z$ in the constraint:
$$
... | In the case of a quadratic form compared with the standard (squared) distance $x^2 + y^2 + z^2,$ from Lagrange multipliers: the extremal directions are the eigenvectors of the Hessian matrix of the other form, here $x^2 - yz \; . \; \;$ In this case the eigenvalues are three distinct integers, one may take eigenvector... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given $a,b,c$ are sides of a triangle, Prove that :- $\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$
Given $a,b,c$ are sides of a triangle, Prove that :-
$$\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$$
What I Tried:- I was able to solve the left hand side inequality. From RMS-AM Inequa... | We need to prove that:$$2(a^2+b^2+c^2)<(a+b+c)^2,$$ which is true because
$$(a+b+c)^2-2(a^2+b^2+c^2)=\sum_{cyc}(2ab-a^2)=$$
$$=\sum_{cyc}(ab+ac-a^2)=\sum_{cyc}a(b+c-a)>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Alternative series expansion for the elliptic integral of the second kind I found the following alternative series expansion for the complete elliptic integral of the second kind (E):
$ E(k) = \frac{(1+k')\pi}{4} \left\{ 1+\frac{1}{2^2}\left(\frac{1-k'}{1+k'}\right)^2+ \frac{1^2}{2^2\cdot 4^2} \left(\frac{1-k'}{1+k'}\r... | Your path is not a dead end! We have $\frac{2k'}{(1+k')^2} = \frac{1}{2}\left[1 - \left(\frac{1-k'}{1+k'}\right)^2\right]$, so your last line implies
\begin{align}
\frac{4 \operatorname{E}(k)}{\pi (1+k')} &= \sum \limits_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\left[\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4209421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\int^{\infty}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ diverges First of all, I noticed that $\sin^2(\pi(x+\frac{1}{x}))$ "approaches" $\sin^2(\pi x)$ as $x \to \infty$. In other words, it seems that the area under one period of $\sin^2(\pi(x+\frac{1}{x}))$ should converge to the value of the area under one period of $... | Here's another approach that uses your intuition that $x+1/x\sim x$ for large $x$ by doing a change of variable. Denoting the integrand by $f(x)$, we have
$$
\int_0^{\infty} f(x)\,dx = \int_0^1 f(x)\,dx +\int_1^{\infty}f(x)\,dx
$$Set $y=x+1/x$, or $x = \frac{y\mp \sqrt{y^2-4}}{2}$, with $-$ in the first region and $+$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4211726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Comparing $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the values
I want to compare $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the actual value of square roots.
I tried to apply square on both side but it still carries the root terms.
Any trick I was missing here?
| Use the Taylor expansions:
$$\sqrt{1+x}=\sqrt{x} + \frac1{2\sqrt x} - \frac1{8x\sqrt x}+\frac1{16x^2\sqrt x}-\frac5{128x^3\sqrt x}+O(x^{-4})\\
\sqrt{2+x}=\sqrt{x} + \frac1{\sqrt x} - \frac1{2x\sqrt x}+\frac1{2x^2\sqrt x}-\frac5{8x^3\sqrt x}+O(x^{-4})$$
Now we will plug in and calculate (manually possible, though tediou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Evaluating integral $\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx$ I come across a problem, which is to evaluate: $$\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx$$
My attempt so far is:
$$\begin{align}\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx
&= \int_0^{2\pi}\frac{1}{2-\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}}\,dx\\
&= \int_0^{2\p... | This is the step-by-step process I mentioned in the link to deal with the discontinuity at $x=\pi$.
Using Weierstrass substitution $t = \tan\left(\frac{x}{2}\right)$, where $-\pi < x < \pi$ or $\pi < x < 3\pi$, then $\sin(x) = \frac{2t}{1+t^2}$, $dx = \frac{2}{1+t^2}dt$, there is a discontinuity at $x=\pi$ that needs t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Hard and non-trivial inequality with three variable reals For all reals $a$, $b$, $c$, show that $$a^2+b^2+c^2 \geq a\sqrt[\leftroot{-1}\uproot{1}4]{\frac{b^4+c^4}{2}} + b\sqrt[\leftroot{-1}\uproot{1}4]{\frac{c^4+a^4}{2}} + c\sqrt[\leftroot{-1}\uproot{1}4]{\frac{a^4+b^4}{2}}.$$
I tried to use Holder inequality:
$$\left... | It's enough to prove our inequality for non-negatives $a$, $b$ and $c$.
Now, by C-S $$\sum_\text{cyc}a\sqrt[4]{\frac{b^4+c^4}{2}}\leq\sum_\text{cyc}a\sqrt{b^2-bc+c^2}\leq\sqrt{\sum_\text{cyc}a\sum_\text{cyc}a(b^2-bc+c^2)}\leq\sum_\text{cyc}a^2$$
because $$\sqrt[4]{\frac{b^4+c^4}{2}}\leq\sqrt{b^2-bc+c^2}$$ it's just $$(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$x_1$ , $x_2$ are roots of $x=5-x^2$. Find the equation with roots $\frac1{(x_1+1)^3}$ and $\frac1{(x_2+1)^3}$.
Suppose $x_1$ , $x_2$ are the roots of the equation $x=5-x^2$.Then
$\dfrac1{(x_1+1)^3}$ and $\dfrac1{(x_2+1)^3}$ are roots of which
equation?
$1)125x^2+16x=1$
$2)125x^2=16x+1$
$3)125x^2=12x+1$
$4)125x^2+12x=... | As you can see the roots given for the new equation are of the format $\frac{1}{(x+1)^3}$ hence you can use root transformation, either you directly transform $$x \rightarrow \frac{1}{(x+1)^3}$$ or you can do $$x\rightarrow \frac{1}{1+x}$$ and then calculate a bit, I'm going to do using the second method as it reduces ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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What is the possible number of unique google meet codes? This is a sample google meet code: kgk-imsy-era
Any idea how many such unique codes can be there?
Given that a particular letter in the code may or may not repeat. If it repeats, we do not know how may times it'll repeat.
Is it something like
$$^{10}C_{x} \cdot 2... | I do not know about the procedure of google meeting codes , but i will write my answer according to given information by you.
We have two option such that a particular letter repeat or not repeat.
If it the particular letter does not repeat :
Firstly ,select this letter by $C(26,1)=26$ ways and select the rest by $C(25... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find all primes $r$ and $s$ such that $ r^7+s=s^3+r $
Find all primes $r$ and $s$ such that $\;r^7+s=s^3+r\;.$
I have tried to factorise the expression since we are dealing with primes so:
$$ s(s^2-1)= r(r^6-1)=r(r^2-1)(r^4+r^2+1).$$
Since $ s^3>r^7$, then $\;s^2-1>r^4-1$.
That is all what I have found, thank you in ... | This isn't a full-fledged answer yet but it is too long for a comment: On the one hand,
$$r|(s^2-1) \Rightarrow s=kr \pm 1$$ $$\Rightarrow s \equiv_r \pm 1.$$
On the other hand, $s \in \theta(r^{\frac{7}{3}})$ and $s$ divides one of $r^3-1$, $r^3+1$ gives $as = r^3\pm 1$ for some $a \in O(r^{3-\frac{7}{3}}) =O(r^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$3\times 2^m + 4 = n^2$ I have been trying to solve this problem but I couldn't find any way to find the answers. Find all natural numbers m and n such that:
$3\times\ 2^m + 4 = n^2$
This is a question from the Moroccan Maths Olympiad (2019.)
I tried bringing $4$ to the other side and factorising to get: $3\times 2^m=(... | Assume that $m, n \in \mathbb{N}$. From the equation, you can deduce that $$2 \mid n^2 \implies 2 \mid n\implies n = 2k \implies 3\cdot 2^m + 4 = 4k^2\quad(1)$$ Observe that $m = 1$ yields no integer solution, and $m = 2 \implies k = 2 \implies n = 4$.Thus consider $m \ge 3$, divide both sides of $(1)$ by $4$, you hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Finding $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$
What is the limit of $\,\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$ ?
My thought was $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\left(\frac{1}{\sin^2 x} - \frac{1}{x^2}\right) = \lim\limits_{x... | Without L'Hopital, we can rewrite the limmand as
$$\frac{x-\sin x}{x^3}\cdot\frac{x+\sin x}{x}$$
The limit of the product is the product of the limits when both exist. The term on the right goes to $2$, so all there is left is to evaluate the term on the left. By using the identity
$$\sin 3x = 3\sin x - 4\sin^3x$$
noti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the sum of the series $1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$ The following is a question which has been bugging me for quite a while,
Find the sum of the series
$$1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$$
Where $n$ is odd
I started by ... | Let f(n) be the sum of squares from 1 up to $n^2$. Then the answer to your question is f(n) + 4 f((n-1)/2). Find a formula for f(n) and substitute it. It should be close to 1.5 f(n) since you are adding about half of the terms twice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$u_1 + \sqrt{u_1u_2} + \sqrt[3]{u_1u_2u_3} + \cdots \leq e(u_1+u_2+u_3 +\cdots)$. $\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\lrp}[1]{\left(#1\right)}$
IMO Longlisted Problem 28.
Let $u_1, \ldots, u_n$ be a positive real numbers.
For each $k$ in $\set{1, \ldots, n}$ define $v_k=\sqrt[k]{u_1 \cdots u_k}$.
Show that
$... | The Polya's proof in my writing.
Let $c_k=\frac{(k+1)^k}{k^{k-1}}.$
Thus, $c_1c_2...c_k=(k+1)^k$ and by AM-GM we obtain:
$$\sum_{k=1}^n\sqrt[k]{u_1 u_2 ... u_k}=\sum_{k=1}^n\frac{\sqrt[k]{u_1c_1 u_2c_2 ... u_kc_k}}{k+1}\leq\sum_{k=1}^n\frac{u_1c_1+u_2c_2+...+u_kc_k}{k(k+1)}=$$
$$=\sum_{k=1}^n u_kc_k\sum_{i=k}^n\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Deducing an inequality from a Taylor series expansion Consider the inequality
$$
1-\frac{x}{2}-\frac{x^2}{2} \le \sqrt{1-x} < 1-\frac{x}{2}
$$
for $0 < x < 1$. The upper bound can be read off the Taylor expansion for $\sqrt{1-x}$ around $0$,
$$
\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \dots
$$
by... | We need to prove that
$$(2-x-x^2)^2\leq4(1-x),$$ which is
$$x^2(3+x)(1-x)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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If $a^{2}+b^{2} \leq 4$, prove that $a+b \leq 4$ \begin{equation}
\text { If } a^{2}+b^{2} \leq 4 \text { prove that } a+b \leq 4 \text { }
\end{equation} What I have tried:
\begin{equation}
a^{2} \leq a^{2}+b^{2} \leq 4 \text { and } b^{2} \leq a^{2}+b^{2} \leq 4
\end{equation}
\begin{equation}
|a| \leq 2 \text { and... | By rearrangement inequality we have
$$2ab\le a^2+b^2\le 4$$
and
$$(a+b)^2=a^2+b^2+2ab\le 8\implies |a+b|\le 2\sqrt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Why does a boundedness proof of $\exp(x)$ depend on the sign of $x$? To prove $\lim\limits_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}$ exists, we prove that the sequence
$$f_n=\left(1+\frac{x}{n}\right)^n$$
is bounded and monotonically increasing toward that bound.
Proof Attempt:
We begin by showing $f_n=\left(1+\fra... | *
*You application of Bernoulli's inequality is correct as long as $y_n=-\frac{x}{(n+x)(n+1)}\geq-1$. In reference (2) which you mention in your posting, Mark Viola assumes $x> -1$, in which case $y_n>-1$ for all $n\geq1$ and so, the sequence $f_n$ (in your notation) is monotone increasing for all $n\in\mathbb{N}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $f(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+ax+1$ and $f(x) = 0$ has two distinct negative roots and equal positive roots, find least integral value of a If $f(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+ax+1$ be a polynomial where a and b are real numbers and $f(x) = 0$ has two distinct negative roots and equal positive roots, fi... | You have
$$\begin{aligned}
f(x)&=x^4+ax^3+bx^2+ax+1\\
&= x^2\left(x^2+ax+b+\frac{a}{x}+\frac{1}{x^2}\right)\\
&=x^2\left(\left(x + \frac{1}{x}\right)^2+ a\left(x + \frac{1}{x}\right)+b-2\right)
\end{aligned}$$
As $x=0$ is not a root of $f(x)$, $x$ is a root of $f(x)$ if and only if $X =x + \frac{1}{x}$ is a root of $g(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$ Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms
My Attempt:
The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac... | Try this.
$$
F(x) := \sum_{i=0}^\infty\frac{x^{4i+3}}{(4i+1)(4i+2)(4i+3)},\quad |x| \le 1
$$
Problem: compute $F(1)$.
Differentiate, to get geometric series
$$
F'''(x) = \sum_{i=0}^\infty x^{4i} = \frac{1}{1-x^4},\quad |x|<1
$$
Now integrate this three times, using $F(0) = F'(0) = F''(0)=0$.
$$
F(x) = \frac{x^2-1}{4}\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
Finding the range of $f(x)=\log_2\left(4^{x^2}+4^{(x-1)^2}\right)$
If the range of the function $f(x)=\log_2\left(4^{x^2}+4^{(x-1)^2}\right)$ is $\left(\frac pq,\infty\right)$, where $p, q$ are in their lowest form then find $(p+q), (p-q)$
Looks like the domain is $\mathbb R$.
If $x^2=0$, $(x-1)^2=1\implies f(0)=\log... | You have
$$\begin{align}4^{x^2}+4^{{(x-1)}^2}
&≥2\sqrt{4^{x^2+(x-1)^2}}\\
&=2^{{x^2+(x-1)^2}+1}\\
&≥2^{2\left(x-\frac 12\right)^2+\frac 32}\\
&≥2^{\frac 32}\end{align}$$
$$\log_2\left(4^{x^2}+4^{(x-1)^2}\right)≥\log_2{2^{\frac 32}}=\frac 32.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Solve $x+3<\sqrt{x+33}$ : Solution Verification First of all the part inside the square root should be greater than or equal to $0$. So,
$x+33 \ge0$
$\Rightarrow x \ge -33$
Now I solved the inequality,
$x+3<\sqrt{x+33}$
$\Rightarrow(x+3)^2<{x+33}$
$\Rightarrow x^2+9+6x<{x+33}$
$\Rightarrow x^2+5x-24 < 0$
$\Rightarrow ... | HINT
You can only square both sides iff both sides and the argument of the square root are nonnegative.
Under such conditions, the proposed inequation is equivalent to
\begin{align*}
x + 3 < \sqrt{x + 33} & \Longleftrightarrow
\begin{cases}
(x+3)^{2} < x + 33\\\\
x+3 \geq 0\\\\
x + 33 \geq 0
\end{cases}\\\\
& \Longleft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find all odd functions $f:\mathbb R\to\mathbb R$ that satisfy $f(x+1)=f(x)+1$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$
Find all functions: $f:\mathbb R\to\mathbb R$ that satisfy all the following three conditions:
*
*$f(-x)=-f(x)$
*$f(x+1)=f(x)+1$
*$f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$
I assume $f(... | Well, I just saw Dan Vellerman's proof (hint actually), which is very elegant, and which I did not think of.
I will nevertheless add the proof that $f(x)=x$ on $\mathbb{Q}$ because the induction techniques used might be useful for some.
Let's prove that $f(x)=x$ on $\mathbb{Q}$.
*
*Notice that $f(0) = -f(-0) = -f(0)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Derivative of $\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ I'm a calculus beginner. I was asked to find the derivative of the function: $$\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}.$$
I'm able to solve it in the following way:
I first calculate the derivative of $\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ and get $\frac{1}{\sqrt 2}(1-\... | Let $f(x) = \sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt{2}} $. We have, for $\theta = \sin^{-1} x \in[-{\pi\over 2},{\pi\over 2}]$:
$$ f(\sin \theta) = \sin^{-1}\frac{\sin\theta+\sqrt{1-\sin^2\theta}}{\sqrt{2}} = \sin^{-1}\frac{\sin\theta+\cos\theta}{\sqrt{2}} = \sin^{-1}\big(\sin(\theta+{\pi\over 4})\big)$$
That is
$$ f(\sin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Solve the equation $2\sin{ix} -3i\cos{ix}=3i$ I have got an answer for this, but I know that it is wrong because it contains $\cos^{-1}\left(-\frac{3}{\sqrt{5}}\right)$, which is undefined. So I was wondering if the issue is with the method I am using or if I have made a mistake with the numbers.
I rewrote the equation... | $$2\sin{ix} -3i\cos{ix}=3i<=>$$
$$ <=>\sqrt{4+9i^2} \sin \left(ix -\arcsin {\frac{3i}{\sqrt{4+9i^2}}}\right)=3i<=>$$
$$ <=>i \sqrt{5} \sin \left(ix -\arcsin {\frac{3}{\sqrt{5}}}\right)=3i<=>$$
$$ <=>\sin \left(ix -\arcsin {\frac{3}{\sqrt{5}}}\right)=\frac{3} {\sqrt{5}}<=>$$
$$ <=>\left[\begin{array} {1} ix -\arcsin {\f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $ab+bc+ca=1$ what is the minimum value of $10a^2+10b^2+c^2$?
$a,b,c$ are positive real numbers such that $ab+bc+ca=1$. What is the minimum value of $10a^2+10b^2+c^2$?
I want to solve this problem without using Lagrange multipliers or calculus. I tried the following with some basic inequalities:
From AM-GM inequali... | Noting that
$$20 a^{2}+20 b^{2}+2 c^{2}-8(a b+b c+c a)=(2 a-2 b)^{2}+(4 b-c)^{2}+(4 a-c)^{2} \geqslant 0,
$$
we have
$$
\begin{aligned}
\quad 2\left(10 a^{2}+10 b^{2}+c^{2}\right)\geqslant 8 \quad \Rightarrow \quad 10 a^{2}+10 b^{2}+c^{2} \geqslant 4
\end{aligned}
$$
The minimum value of $10 a^{2}+10 b^{2}+c^{2}$ is $4... | {
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"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Finding equation of tangent line to $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ at the point $(0,\frac{1}{2})$
Find the equation of the tangent line to $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ at the point $(0,\frac{1}{2})$
This is in the context of learning implicit differentiation.
First, I apply $\frac{dy}{... | Notice that $\frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}}$. This can be seen because, by definition of an inverse function, $\sin(\arcsin(x)) =x$. So differentiating both sides gives
$$
1 = \frac{d}{dx}x =\frac{d}{dx}\sin(\arcsin(x)) \overset{\color{purple}{\text{Chain rule}}}{=} \cos(\arcsin(x))\frac{d}{dx}\arcsin(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that for every positive integer $n$, $1/3 + 1/9 + \cdots + 1/{3^n} < 1/2$ Base case is $n=1$: $\frac {1}{3} < \frac{1}{2}$. So the base case holds.
Inductive hypothesis for $n = k$: $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} < \frac{1}{2}$
Inductive Step for $n = k + 1$:
$$ \left( \frac{1}{3} + \frac{1}... | Also, we can use a formula for the sum of a geometric progression.
It's $$\frac{1}{3}\cdot\frac{1-\left(\frac{1}{3}\right)^{n}}{1-\frac{1}{3}}<\frac{1}{2}.$$
Can you end it now?
| {
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"url": "https://math.stackexchange.com/questions/4245646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
What is maximum value of $f\left(x\right)=x\sqrt{x}-\frac{1}{\sqrt{1-x^2}}$?
Find maximum value of $$f\left(x\right)=x\sqrt{x}-\frac{1}{\sqrt{1-x^2}}$$
Domain of the function is $[0,1)$. Equating the derivative by zero we get,
$$\frac32\sqrt x-\frac{x}{(1-x^2)^{\frac32}}=0$$
$$\frac94x=\frac{x^2}{(1-x^2)^3}$$
$$9(x^2... | You end with the problem of solving for $x$ the sextic equation
$$9 x^6-27 x^4+27 x^2+4 x-9=0$$ and, by inspection (as @NickD already commented) the solution is close to $0.6$.
Make $x=y+\frac 35$ to make
$$9 y^6+\frac{162 y^5}{5}+\frac{108 y^4}{5}-\frac{648 y^3}{25}-\frac{1728
y^2}{125}+\frac{53972 y}{3125}+\frac{6... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Let $x\geq y\geq z>0$. Prove that $ \frac {x^{2}y}{z} + \frac {y^{2}z}{x} + \frac {z^{2}x}{y}\geq x^{2} + y^{2} + z^{2}.$
Let $x\geq y\geq z>0$. Prove that $$ \frac {x^{2}y}{z} + \frac {y^{2}z}{x} + \frac {z^{2}x}{y}\geq x^{2} + y^{2} + z^{2}.$$
The problem is from Vietnamese MO 1991 and has been posted here before. ... | (Note:
This was done without looking at any of the links.
It probably duplicates some of them.)
Let
$y = (1+b)z,
x = (1+a)y,
a, b \ge 0$.
$x = (1+a)(1+b)z$.
$\begin{array}\\
RHS
&=z^2((1+a)^2(1+b)^2+(1+b)^2+1)\\
LHS
&=\dfrac {x^{2}y}{z} + \dfrac {y^{2}z}{x} + \dfrac {z^{2}x}{y}\\
&=\dfrac {((1+a)(1+b)z)^{2}(1+b)z}{z} +... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Need help solving a exponent+multiplication problem The question goes as follows:
$$\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}$$
Of course the question tests for knowledge in logic and simplification as who would keep calculating this sum. I observed th... | You found:
$$(x^4+4y^4)=(x^2+2y^2+2xy)(x^2+y^2-2xy)$$
Just plug in:
$(x, y)=(10, 3), (4, 3)$
for fraction:
$\frac{10^4+324}{4^4+324}=\frac{178\times 58}{58\times 10}$
$(x, y)=(22, 3), (16, 3)\rightarrow \frac{22^4+324}{16^4+324}=\frac{634\times 370}{370\times178} $
continues this procedure you finally get:
$\frac{178}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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