Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Height of a cube edge from the floor
A cube $ABCD.EFGH$ has side length $2a$ cm. Point $A$ is lifted $a$ cm from the floor, point $C$ is still on the floor, point $B$ and point $D$ are on the same height from the floor. What is the height of point $E$ from the floor?
I'm sorry for my bad English, but I try to illustr... | Let $p = \angle ACA'$, $q = \angle ECA$ (in the original figure), where $A,C,A'$ are the same as in Math Lover's answer.
Then $\tan p = \frac{1}{\sqrt 7}, p = \tan^{-1} \frac{1}{\sqrt 7}$ and $\tan q = \frac{2}{2 \sqrt 2} = \frac{1}{\sqrt 2}, q = \tan^{-1} \frac{1}{\sqrt2}$. Now using the arctangent addition formula:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove combinatorically - $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$ $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$
I have no idea which story to form, I thought about $n$ students assuming n=4
picking 4 students for a committee with 3 roles. since each students has 3 options.
then on the other side.. $2^{0}\binom{4}{0} + 2... | Define $A_n$ to be the set of strings of length $n$ made of $\{0, 1, 2\}$. It is clear that $|A_n| = 3^n$. We will prove that $|A_n| = \sum_{k=0}^n 2^{k}\binom{n}{k}$.
For $k=0, 1, \dots, n$ partition $A_n$ into $k+1$ disjoint sets $B_{n,k}$ each consisting of strings that contain exactly $k$ $2$s. It is clear that
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3951876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$ Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
This is how i started solving this limit:
*
*$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
*$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$
*$\left(\frac {4x+3}{... | Being
$$\lim_{x \to \infty} \left(\frac{3 x + 2}{4 x + 3}\right)^{x} = \lim_{x \to \infty} e^{\ln{\left(\left(\frac{3 x + 2}{4 x + 3}\right)^{x} \right)}}= e^{\lim_{x \to \infty} x \ln{\left(\frac{3 x + 2}{4 x + 3} \right)}}=e^{\lim_{x \to \infty} x \color{blue}{\ln{\left(\lim_{x \to \infty} \frac{3 x + 2}{4 x + 3} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
How can I prove this reduction formula for $\int^1_0{x(1-2x^4)^n}dx$ The exercise in my textbook states
You are given that $$I_n=\int^1_0{x(1-2x^4)^n}dx$$
Show that $$I_n=\frac{(-1)^n}{4n +2} + \frac{2n}{2n+1}I_{n-1}$$
I have started out by splitting the power
$$\int^1_0{x(1-2x^4)^n}dx=\int^1_0{x(1-2x^4)(1-2x^4)^{n... | Let $J = \int_{0}^{1}x^{5} (1 - 2x^{4})^{n-1} dx$.
Integration by parts gives: $$I_n=\int^1_0{x(1-2x^4)^n}dx = \frac{(-1)^{n}}{2} + 4nJ$$.
Separating the integral as you have done gives:
$$I_n=\int^1_0{x(1-2x^4)^n}dx =\int^1_0{x(1-2x^4)(1-2x^4)^{n-1}}dx=I_{n-1} -2 J$$
Solving for $J$ in the first and substitute it into... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find minimal polynomial of given matrix I want to find the minimal polynomial of this matrix
$$\left[\begin{array}{ccccc}
2 & 0 & 0 & 0 & 0\\
4 & 2 & 0 & 0 & 0\\
1 & 4 & 2 & 0 & 0\\
0 & 0 & 0 & 3 & 0\\
0 & 0 & 0 & 0 & 3
\end{array}\right]$$
I found it charastic polynomial $\chi_{A}=\left(2-x\right)^{3}\left(3-x\right)^... | The minimal polynomial is $(X-2)^3(X-3)$. This a triangular matrix with a $2\times 2$ diagonal block whose eigenvalue is $3$ thus whose minimal polynomial is $X-3$. There exists a block which is a $3\times 3$ matrix which has $2$ on the diagonal. The minimal polynomial of this block is $(X-2)^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculating the residues What are the residues at $s=0,\pm 1$ for the following function
$$I=\frac{1}{e^{2\pi is}-1}\frac{e^{isu}}{s^2(s^2-1)}?$$
Now it appears to me that, two pieces, $e^{2\pi is}$ and $s^2(s^2-1)$, diverge in the denominator. What are the residues then?
| Since you asked only for the residues and not how to get them:
$$\text{res}\left(I,0\right)=-\frac{i \left(3 u^2-6 \pi u+2 \pi ^2-6\right)}{12 \pi }$$
$$\text{res}\left(I,1\right)=\frac{e^{i u} (2 u+5 i-2 \pi )}{8 \pi }$$
$$\text{res}\left(I,-1\right)=\frac{e^{-i u} (-2 u+5 i+2 \pi )}{8 \pi }$$
To get the first resid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3961310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving the system $3a=(b+c+d)^3$, $3b=(c+d+e)^3$, ..., $3e=(a+b+c)^3$ for real $a$, $b$, $c$, $d$, $e$ $$\begin{align}
3a&=(b+c+d)^3 \\
3b&=(c+d+e)^3 \\
3c&=(d+e+a)^3 \\
3d&=(e+a+b)^3 \\
3e&=(a+b+c)^3
\end{align}$$
I tried to use inequality for
$((\sum \alpha)/n)^r\leq \sum \alpha^r/n$ by taking $\alpha=a_1+a_2+a_3$ w... | Notice that for all $x,y$ we have $$x^2+xy+y^2 = {1\over 2}(x^2+(x+y)^2+y^2)\geq 0$$
Now we have:
$$3(a-b) = (b-e)\color{red}{\Big(} \underbrace{b+c+d}_x )^2+\underbrace{(b+c+d)(c+d+e)}_{xy}+(
\underbrace{ c+d+e}_y)^2\color{red}{\Big)} $$
so $${\rm sign} (a-b) = {\rm sign} (b-e) $$ and similary for:
\begin{align} {\rm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.
Is there a shorter/simpler solution... | This answer provides some motivation for a high school student learning the binomial theorem and problem solving techniques; we also get a 'feeling' for homogeneous polynomials.
Employing the binomial theorem we naturally write out (including the quadratic),
$\tag 1 x + y \color\red{ = a} $
$\tag 2 x^2 + y^2 = (x+y)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Given triangle ABC and its inscribed circle O, AO = 3, BO = 4, CO = 5, find the perimeter of ABC As title,
I could only figure out that
$$a^2 + r^2 = 9$$
$$b^2 + r^2 = 16$$
$$c^2 + r^2 = 25$$
$$r^2 = \frac{abc}{a+b+c}$$
But couldn't get how to derive $2(a+b+c)$.
| Not sure if there is a more straightforward solution but in meantime,
$3 \sin\frac{A}{2} = 4 \sin\frac{B}{2} = 5 \sin\frac{C}{2} = r$
$\frac{C}{2} = 90^0 - \frac{A+B}{2}$
So, $\cos \frac{A+B}{2} = \frac{r}{5}$
Expanding and writing values in terms of $r$ and squaring both sides, we simplify it to a cubic
$\frac{\sqrt {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Summation of integration terms If $I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $ then
(A) $I>\ln99$
(B) $I<\ln99$
(C) $I<\frac{49}{50}$
(D) $I>\frac{49}{50}$
The official answer is B and D
My approach is as follow
$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k +... | Hint: $$\sum_1^{98} (k+1)\ln\frac{(k+1)^2}{k(k+2)} \\ = \sum_1^{98} (k+1) \left( \ln\frac{k+1}{k} - \ln \frac{k+2}{k+1} \right) \\ = 2(\ln 2-\ln\frac 32) + 3(\ln \frac 32 -\ln \frac 43) +4(\ln \frac 43 -\ln \frac 54) +\dots +99(\ln \frac{99}{98} -\ln \frac{100}{99}) \\ =2\ln 2 + \big( \ln\frac 32 +\ln \frac 43 +\dots +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution $\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.
Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substitute... | With a u-substitution:
$\int x^3\sqrt{x^2-9} \ dx\\
\int \frac 12 x^2\sqrt{x^2-9} (2x\ dx)\\
u = x^2 - 9\\
x^2 = u+9\\
du = 2x\ dx\\
\frac 12\int (u+9)\sqrt{u}\ du\\
\frac 12(\frac 23 (9u^\frac 32) + \frac 25 u^\frac 52)+ C\\
3(x^2-9)^\frac 32 + \frac 15(x^2-9)^\frac 52 + C$
With a trig substitution:
$\int x^3\sqrt{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
show this $\lceil \frac{n}{1-a_{n}}\rceil =n+1$ let $a_{n}$ be squence such $a_{1}=2-\dfrac{\pi}{2}$, and
$$a_{n+1}=2a_{n}+\dfrac{a_{n}-1}{n},n\in N^{+}$$
show that
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
I try:since
$$f(1)=\lceil \dfrac{1}{\dfrac{\pi}{2}-1}\rceil=2$$
and $$a_{2}=2a_{1}+a_{1}-1=3a_{1}-1=5-\dfrac{... | The series $\frac{\pi}{2}=\frac{1}{2}\sum\limits_{n=0}^\infty \frac{(n!)^22^{n+1}}{(2n+1)!}=1+\frac{1}{3}+\frac{1\cdot 2}{3\cdot 5}+\frac{1\cdot2\cdot3}{3\cdot5\cdot7}+\cdot\cdot\cdot$
causes $2-\frac{\pi}{2}$ to have nice properties with the given sequence $a_{n+1}= \frac{2n+1}{n}a_n-\frac{1}{n}.$
$a_1 = 2-\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Find $a$ such that $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions
Find $a$ so that equation $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions.
My approach: Since that $$\log_{2}(x)=\frac{\ln(x)}{\ln(2)} \quad \text{and} \quad \log_{\... | Somewhat oddly, this becomes easier by first eliminating $a$. Using your substitution for $t$, let $b=\sqrt{a+t}$. Then $a=b^2-t$ and the equation becomes
$$t^2 - 2t = b^2 - t - b$$
Or $$
0 = t^2 - b^2 - (t - b)
= (t-b)(t+b-1)
$$
At this point it might be tempting to get back to $t$ by replacing $b$ with $\sqrt{a+t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3966258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Difficult Cauchy Problem Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the following inequality:
$$
\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c}
$$
I tried using the fact that $a^2 + b^2 + c^2 = 3abc$ but I could only think of one case where $a=b=c$. I also tried... | By Cauchy-Schwarz,
\begin{align}
\left(\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \right)(a+b+c) & \geq \left(\dfrac{a}{bc} + \dfrac{b}{ca} + \dfrac{c}{ab}\right) ^2 \\
& = \left (\dfrac{a^2+b^2+c^2}{abc} \right)^2 \\
& = \left (\dfrac{3abc}{abc} \right)^2 \\
& = 9. \\
\end{align}
Thus, dividin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
What I Tried: We have :-
$$ (a^3 - 503a - 500)^5 = [a(a^2 - 3) - 500(a - 1)]^5$$
$$= \Bigg(\Bigg[\frac{1 + \sqrt{2009}}{2}\Bigg]\Bigg[\frac{1009 + \sqrt{200... |
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
I see no reason for elegance. Since one of the factors in the numerator is $1$, computing $a^3$ is is straightforward.
$$a^3 = \left(\frac{1}{8}\right) \times
\left[
1 + 3\sqrt{2009} + 3(2009) + 2009\sqrt{2009}
\right]$$
$$=~
\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Prove specially rearranged alternating harmonic series converges to $\frac 12 \ln{\frac{4p}{q}}$ By Leibnitz's test the alternating series is convergent.
$\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac 1n
=\frac 12 \ln{\frac{4×1}{1}}$
\begin{align}
& \left(1-\frac 12- \frac 14\right)+\left(\frac 13-\frac 16-\frac 18\r... | $t_{(p+q)n}$
\begin{align}
& =\left(1+\frac 13+...+\frac{1}{2p-1}-\frac 12-\frac 14-...-\frac{1}{2q}\right)+\left(\frac{1}{2p+1}+...+\frac{1}{4p-1}-\frac{1}{2q+2}-...-\frac{1}{4q}\right)+...+\left(\frac{1}{2p(n-1)+1}+...+\frac{1}{2pn-1}-\frac{1}{2q(n-1)+2}-...-\frac{1}{2qn}\right)\\
&=\left(1+\frac 13+...+\frac{1}{2pn-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\int x^{2}\sqrt{a^{2}+x^{2}}\,dx$. Is there another way to solve it faster? I have to calculate this integral:
\begin{align} \int x^{2}\sqrt{a^{2}+x^{2}}\,dx \qquad\text{with}
\quad a \in \mathbb{R} \end{align}
My attempt:
Using, trigonometric substitution
\begin{align}
\tan \theta &= \frac{x}{a}\\ \Longrightarrow ... | Integrate by parts
\begin{align}
I & =\int x^{2}\sqrt{a^{2}+x^{2}}\,dx
=\int \frac x3d(a^2+x^2)^{3/2} \\
&=\frac13x( a^2+x^2)^{3/2}-\frac13I-\frac{a^2}3\underset{=J}{\int \sqrt{a^2+x^2}dx} \\
&=\frac14x( a^2+x^2)^{3/2}-\frac{a^2}4J\tag 1
\end{align}
Integrate $J$ by parts again
\begin{align}
J & =\int \sqrt{a^{2}+x^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Eliminate $\theta$ and prove $x^2+y^2=1$ We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$
And want to prove $x^2+y^2=1$
My works:
I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\co... | From your solution,
\begin{gather*}
x\cos \theta =y\sin \theta \\
y=x\cot \theta \\
\end{gather*}
Substitute the corresponding value of y into any of the given two equations, and you should be able to get the values of x any y both.
Let's see what we get by substituting it into the first equation:
\begin{gather*}
2x=y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Express roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta, $
If roots of the equation $ax^2+bx+c=0$ are $\alpha, \beta, $ find roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta$
Here's what I have tried so far,
I know that $\alpha+ \beta=\frac{-b}{a} $ and $\alpha \beta=\fra... | This is somewhat along the lines of the approach Quanto takes, but causes the roots to emerge in a different fashion.
Dividing the given quadratic equation $ \ ac·x^2 \ - \ b(c+a)·x \ + \ (c+a)^2 \ = \ 0 \ \ $ through by $ \ ac \ $ produces
$$ x^2 \ - \ \frac{b(c+a)}{ac}·x \ + \ \frac{(c+a)^2}{ac} \ = \ 0 \ \ \Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
if $a_{n+1}=xa_{n}+ya_{n-1}$ then have this results :$a_{m+n}=Xa_{m}a_{n+1}+Ya_{m-1}a_{n}$？ it is well known:
if $F_{1}=1,F_{2}=1$,and $F_{n+1}=F_{n}+F_{n-1}$.then we have
$$F_{m+n}=F_{m}F_{n+1}+F_{m-1}F_{n}$$
prove 1
and I have know this :if $a_{0}=0,a_{1}=1$ and $x,y$ be give postive integers,and $$a_{n+1}=xa_{n}+ya... | So, the answer is indeed, yes, such a relation exists (with some caveats), and we can prove it via induction, just as in the 1st answer you have linked in the question.
MAJOR EDITS: My original answer was incorrect, as pointed out by @mathlove. Here is a revised answer.
First, let us prove the induction step. For it, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3979611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Difficulties solving this integral: $ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $ by differentiation under the integral sign So in the book Advanced Calculus Explored, by Hamza E. Asamraee. The next integral appears as an exercise to solve by differentiating under the integral sign:
$$ \int_0^1 \frac{\ln(x+1)} ... | $$f(x)=\frac{1}{(x+a)(x^2+1)}=$$
$$\frac{A}{x+a}+\frac{Bx+C}{x^2+1}$$
we find that $$A=\frac{1}{a^2+1}$$
and
$$Bi+C=\frac{1}{a+i}=\frac{a-i}{a^2+1}$$
So,
$$B=-\frac{1}{a^2+1}\;,\;C=\frac{a}{a^2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$. If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.
I found this in a Facebook group.
I start by doing the math in the LHS:
$8(1-x-y+xy-z+xz+yz-xyz) = 8(1-2+xy+xz+yz-xyz) = 8(xy+xz+yz-1-xyz)$.
Then we set $z=2-... | As $0 \leq x, y, z \leq 1, (1-x), (1-y), (1-z) \geq 0$
Using A.M - G.M,
$(1-x) + (1-y) = z \geq 2 \sqrt{(1-x)(1-y)} \ $ (as $ \ x + y + z = 2$)
Similarly for $x, y$.
Multiplying them, we have
$xyz \geq 8 (1-x)(1-y)(1-z)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3982410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$ It is also given that $abc = 1$.
I used AM-GM inequality to reach till
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab... | An alternative way: By the rearrangement inequality we have $$\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{b}}+\frac{a}{\sqrt{c}}+\frac{b}{\sqrt{c}}\geq \frac{a}{\sqrt{a}}+\frac{a}{\sqrt{a}}+\frac{b}{\sqrt{b}}+\frac{b}{\sqrt{b}}+\frac{c}{\sqrt{c}}+\frac{c}{\sqrt{c}}=2(\sqrt{a}+\sqrt{b}+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Equivalence class for the following relation A relation of $\mathbb{R}$ is defined as $a\sim b : a^4-b^2=b^4-a^2$
Show that $\sim$ is equivalence relation (I have done this part)
Determine the equivalence class $[-1]_\sim$
Prove or disprove: Every equivalence class in $\mathbb{R}/\sim$ contains exactly 2 real numbers.
... | Rearrange: $a^4 - b^4 = b^2 - a^2$
But $a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)$ (DOTS)
So $(a^2 - b^2)(a^2 + b^2) = b^2 - a^2$
(1) Assume $a^2 - b^2 \neq 0$
Then divide both sides by $a^2 - b^2$
$a^2 + b^2 = -1$
Since $a^2$ and $b^2$ are both positive in $\Bbb R$, this is a contradiction.
(2) Therefore $a^2 - b^2 = 0$, or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Given a directional derivative to find a point Let $ f(x,y,z)$ be differentiable, and assume that
$$ f(x, y, x^2 + y) = 3x - y $$
(for all values of $x,y$).
Also, given the direct-derivative of the point $A=(0, 12, 12)$ and the direction vector is $(1, 0, 1)$ is equal to $3$, I need to find the gradient of $f$ at $A$.
... | Using the chain rule:
$$\frac{ \partial}{ \partial x} f(x,y,x^2 + y) = f_x(x,y,x^2 + y) \cdot \frac{ \partial}{ \partial x} (x) + f_y (x,y, x^2 + y) \cdot \frac{ \partial }{ \partial x}(y) + f_z(x,y,x^2 + y) \cdot \frac{\partial }{\partial x} (x^2 + y) = f_x(x,y,x^2 + y) + 2x \cdot f_z(x,y,x^2 + y) = \frac{\partial}{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Expanding Dawson's integral in a series of Hermite functions We want to show that
$$F(x)=\sqrt{\pi } \sum
_{n=0}^{\infty }
\frac{(-1)^n H_{2
n+1}(x)}{2^{3 n+3} \Gamma
\left(n+\frac{3}{2}\right)}$$
where $F(x)$ is the Dawson Integral ($F(x)=\exp \left(-x^2\right)
\int_0^x \exp
\left(u^2\right) \, du$) ... | We will proceed to evaluate the integral appearing above in $c_{2 n+1}$.
The Dawson Integral can be expressed in the form
$$F(x)=\int_0^{\infty } \exp
\left(-y^2\right) \sin (2 x
y) \, dy$$
Hence,
$$\int_{-\infty }^{\infty } \exp
\left(-x^2\right) F(x) H_{2
n+1}(x) \,
dx=\int_0^{\infty } \exp
\left(-y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
how to compute a series whose terms are a rational function times an exponential function? How can I compute the following series?
\begin{equation}
\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n
\end{equation}
I manipulated the term and got
\begin{equation}
\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\r... | $$\frac{n+10}{2n^2+5n-3}=\frac{3}{2 n-1}-\frac{1}{n+3}$$
Consider the first sum
$$S_1=\sum_{n=1}^\infty\frac{x^n}{2 n-1}=y\sum_{n=1}^\infty\frac{y^{2n-1}}{2 n-1}=y\tanh ^{-1}(y)=\sqrt{x} \tanh ^{-1}\left(\sqrt{x}\right)$$ With $x=-\frac 13$ we then have
$$S_1=-\frac{\pi }{6 \sqrt{3}}$$
For the second sum
$$S_2=\sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3987217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain. $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$
I know I have to switch the $f(x)$ and the $y$:
$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\fra... | As IV_ said, your inverse function should be $y=\pm\frac{3}{2}\arcsin(\sqrt{x})-\frac{1}{2}$.
Your original function $ f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ is equal to a sine function squared. If you graph f(x), you can see that its biggest value is $1$, and it achieves its smallest value at $x=(-3\pi+1)/2$ because ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why is this definite integral divergent? The integral given is
$$\int^{\frac{\pi}{3}}_0 \frac{5}{1+ \cos(4x)}dx$$
The steps I have taken to solve it are first applying the double angle formula for the cosine function
$$\int^{\frac{\pi}{3}}_0 \frac{5}{1+ (2\cos^2(2x)-1)}dx $$
Then simplifying
$$\int^{\frac{\pi}{3}}_0 \f... | You properly found that
$$\int\frac{5}{1+ \cos(4x)}dx=\frac{5}{4} \tan (2 x)$$ Now, write
$$I=\int^{\frac{\pi}{3}}_0 \frac{5}{1+ \cos(4x)}dx=\int^{\frac{\pi}{4}-\epsilon}_0 \frac{5}{1+ \cos(4x)}dx+\int_{\frac{\pi}{4}+\epsilon}^{\frac \pi3} \frac{5}{1+ \cos(4x)}dx$$ $$I=\frac{5}{4} \cot (2 \epsilon )+\frac{5}{4} \cot (2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
coordinate form of an affine transformation I am having trouble with this seemingly simple question: Write the standard coordinate form of an affine transformation in $\mathbb{A}^{2}(\mathbb{R})$ that maps
the point (1, −2) to the point (0, 10), and the lines $10x_{1} − 4x_{2} = 1$ and $3x_{1} − 3x_{2} = −7$ to the lin... | Here is a different method. Let $A$ be the looked for matrix.
First of all, let us determine the intersection of the "initial" axes and of the "final" axes. It is rather easy to find that the first two axes intersect at $(14/9,35/9)$ and the second pair of axes at $(15,9)$.
Therefore, $(15,9,1)$ is the image of $(14/9,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3992573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
On the integral $\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$ I came across this integral
$$\mathcal{J} = \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$$
According to W|A it equals $\frac{1}{2}$. However, I cannot find a way to crack it. It smells like a Beta integral , but I do not see any obvious subs. ... | This is an elliptic integral of the first kind, so there is no analytical way to evaluate it.
More on Elliptical Integrals: https://en.wikipedia.org/wiki/Elliptic_integral
In any case, we might try to give your integral a numerical result because it's bounded between $0$ and $1$.
We can expand the whole integrand in Ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3994657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How to find all polynomials that follows the equation $P(P'(x)) = P'(P(x))$ I want to find all polynomials that has the following property
$$P(P'(x)) = P'(P(x))$$
where P(x) is a polynomial.
Can you please tell me a way to solve this problem?
| Let $ax^{n}$ be the leading coefficient, then the leading coefficient of both sides of the equation are equal giving:
$a(anx^{n-1})^{n} = an(ax^{n})^{n-1}$
So $a^{n+1}n^{n} = a^{n}n$
So $a= \frac{1}{n^{n-1}}$
So one such family is $\frac{x^{n}}{n^{n-1}}$
Now consider the constant coefficient K and the linear coefficien... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is the answer to this summation correct? Consider the summation below:
$$\sum_{k=1}^\infty(2k+1)x^{2k}$$
The problem asks to find what the mentioned summation is equal to. The solution provided in the book starts like the following:
$$x^3+x^5+x^7+...=\frac{x^3}{1-x^2}$$
and then takes the derivative of both sides and u... | Let's make some order.
As @Crostul and @cosmo5 pointed out in the comments we clearly have to suppose $|x| < 1$ otherways the series diverges.
Then using only some known properties of geometric series:
$$ x^3 + x^5 + \dots + x^{3+2n} = \sum_{n=0}^N x^{3 + 2j} = x^3 \cdot \sum_{n=0}^N (x^2)^j = x^3 \cdot \frac{1-(x^2)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3999479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrals of the type $\int\frac{dx}{\left(a^2+x^2\right) \sqrt{a^2+b^2+x^2}}$ I'm interested in understanding the sequence of substitutions needed to obtain
\begin{align}
\int\frac{dx}{\left(a^2+x^2\right) \sqrt{a^2+b^2+x^2}}
=\frac{\tan ^{-1}\left(\frac{b x}{a \sqrt{a^2+b^2+x^2}}\right)}{a b}\, ,
\end{align}
with $a... | Your first choice is a better start than it seems, because Bioche's rules advise continuing with$$u=\sin\xi=\frac{\sqrt{a^2+b^2}\tan\xi}{\sqrt{a^2+b^2}\sec\xi}=\frac{x}{\sqrt{a^2+b^2+x^2}}.$$Then$$\frac{d\xi}{(a^2+ (a^2+b^2)\tan^2\xi)\cos\xi}=\frac{\cos\xi d\xi}{a^2\cos^2\xi+(a^2+b^2)\sin^2\xi}=\frac{du}{a^2+b^2u^2}.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show $\cos\frac\gamma2=\sqrt{\frac{p(p-c)}{ab}}$
Show that the equality $$\cos\dfrac{\gamma}{2}=\sqrt{\dfrac{p(p-c)}{ab}}$$ holds for a $\triangle ABC$ with sides $AB=c,BC=a, AC=b$, semi-perimeter $p$ and $\measuredangle ACB=\gamma$.
I have just proved that $$l_c=\dfrac{2ab\cos\dfrac{\gamma}{2}}{a+b}$$
Is this a wel... | Form half-angle trig-identity
$$\cos\gamma=2\cos^2\frac{\gamma}{2}-1\iff \cos \frac{\gamma}{2}=\sqrt{\frac{1+\cos\gamma}{2}}$$
Substituting the value from cosine formula: $\cos \gamma=\frac{a^2+b^2-c^2}{2ab}$
$$\cos \frac{\gamma}{2}=\sqrt{\frac{1+\frac{a^2+b^2-c^2}{2ab}}{2}}$$
$$=\sqrt{\frac{(a+b)^2-c^2}{4ab}}$$
$$=\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4002119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Largest consecutive integers with no prime factors except $2$, $3$ or $5$? The number $180$ has a special property. Its prime factors are only $2$, $3$, and $5$. However the number $220$ does not have this special property because one of its prime factors is $11$.
In the first one-hundred natural numbers, the numbers w... | Only one of the two consecutive numbers can be divisible by $2$. Similarly, only one is divisible by $3$ and only one is divisible by $5$. Depending on how we pair the prime powers we therefore have six possibilities.
The cases $p^Aq^B=r^C-1$ can all be solved in the same way.
$2^A3^B=5^C-1 $
If $A\ge4$ then modulo $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
a random variable has the density $f\left(x\right)=a+bx^{2}$ . Determine and b so that its mean will be 2/3 A random variable has the density $f\left(x\right)=a+bx^{2}$ with $0<x<1$. Determine and b so that its mean will be 2/3.
I'm a little confued trying to get a and b. This is what I already tried:
$$P\left(0<X<1\... | $$E[X]=\int_0^1 xf(x)dx=\int_0^1(ax+bx^3)dx=a\frac{x^2}{2}+b\frac{x^4}{4}\bigg|_0^1=\frac{2a+b}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Min $P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1}$
Given $x,y,z$ are positive numbers such that $$x^2y+y^2z+z^2x=3$$
Find the minium of value: $$P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1} $$
My Attempt:
$$P= x^3y+y^3z+z^3x - \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{... | Too long for a comment.
Considering the minimization problem
$$
\max_{x,y,z} -f(x,y,z)\ \ \text{s. t.}\ \ g(x,y,z) = 0
$$
with
$$
\cases{
f(x,y,z) = \frac{x^5 y}{x^2+1}+\frac{x z^5}{z^2+1}+\frac{y^5 z}{y^2+1}\\
g(x,y,z) = x^2 y + y^2 z + z^2 x - 3
}
$$
and according to the notes, the point $x=y=z=1,\lambda=\frac 56$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4009304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Famous or common mathematical identities that yield $1$ To me the most common identity that comes to mind that results in $1$ is the trigonometric sum of squared cosine and sine of an angle:
$$
\cos^2{\theta} + \sin^2{}\theta = 1 \tag{1}
$$
and maybe
$$
-e^{i\pi} = 1 \tag{2}
$$
Are there other famous (as in commonly us... | Limits
$$\lim_{x\to0}\frac{\sin x}{x}=1$$
This is actually a special case of the more general identity: If $f(0)=0$ and $f'(0)=1$, then
$$\lim_{x\to0}\frac{f(x)}{x}=1$$
which includes things like
$$\lim_{x\to0}\frac{\arctan x}{x}=1$$
Series
Any geometric series (which converges) will do as long as the first term, $a$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$ $x,y \in \mathbb{R}\setminus\{0\}$
Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.
My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{... | $$x^3 + y^3 + 3 x^2 y^2 = x^3 y^3\tag{1}$$
$$\frac{x+y}{xy}\tag{2}$$
$$x^3 + y^3 + 3 x^2 y^2 - x^3 y^3=0$$
can be factorized$^1$ as
$$(x+y-xy) \left(x^2 y^2+x^2 y+x^2+x y^2-x y+y^2\right)=0$$
$x^2 y^2+x^2 y+x^2+x y^2-x y+y^2=0$ has solution$^2$ given by $x=y=-1$ we have $$\frac{x+y}{xy}=-2$$, the solutions of $(1)$ are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding the value of $ax^4+by^4$
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.
Here is my attempt:
$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unkno... | At the core of contest problems like
Given $x^n+1/x^n=p$, evaluate $x^m+1/x^m$
Given $x^n+y^n=p$, evaluate $x^m+y^m$
for most likely, integers $m,n$, lies the recurrence relation
$$x^{n+1}+y^{n+1}=(x+y)(x^n+y^n)-xy(x^{n-1}+y^{n-1})$$
Our question seems a bit generalized but in the same spirit, the following recurre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Establishing $\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha$ without the use of Werner or prostaferesis formulas I have this identity:
$$\bbox[5px,border:2px solid #138D75]{\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha} \tag 1$$
If I write this like as:
$$(\cos 4\alpha+\cos 2\alpha)\cdot (\cos 4\alp... | I think that we can establish the identity without too much work; it certainly isn't short though. Your statement is equivalent to $$\cos^2 4x-\cos^2 2x=-\sin 6x\sin2x$$We'll start with left hand side; my strategy will be to make everything in terms of $\cos 2x$. I will make use of the following identities:
$$\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $x+y+z \geq xyz$ Prove that $x^2 + y^2 + z^2 \geq xyz$. $x,y,z \in \mathbb R$
My attempt:
Notice that if $x^2 + y^2 + z^2 \geq xyz$ and $x+y+z \geq xyz$. So :
$$x^2 + y^2 + z^2 \geq x+y+z $$
But this is actually not always true, take the case when $ 0<x,y,z <1$ And you will see that this is not working.
Maybe I’m wr... | First, we prove for all positive $x,y,z$.
If all of $x,y,z\geq1$ we have
$$x^2\geq x ,\\ y^2\geq y \\ z^2\geq z$$ $$⇒x^2+y^2+z^2\geq x+y+z \geq xyz\\$$
If at least one of $x,y,z$ is less than $1$ then take (WLOG) $x\geq y\geq z>0$ & $1>z>0$
$$\frac{x}{y}\geq1 $$
$$⇒\frac{x}{yz}\geq1 $$
$$⇒\frac{x}{yz}+\frac{y}{xz}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proof of $\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}$ I found some combinatorial identities in my old notebooks, but I cannot recall how I derived them. Can anyone help provide the most elementary/elegant proofs for the following identity? Specifically, the connec... | Proving the equality between the two closed form is easier:
$$\begin{align*}
(2n+1)\binom{2n}n &=(2n+1)\frac{(2n)!}{n!n!}\\
&= \frac{(2n+1)!}{n!n!}\\
&= \frac{(n+1)^2}{(2n+2)}\cdot\frac{(2n+2)!}{(n+1)!(n+1)!}\\
&= \frac{n+1}{2}\binom{2(n+1)}{n+1}
\end{align*}$$
By induction, consider the $n+1$ case of the summation for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$ Problem. (Nguyễn Quốc Hưng) Let $0\le a,b,c\le 3;ab+bc+ca=3.$ Prove that $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$$
I have one solution but ugly, so I 'd like to find another. I will post my solution in the answer.
| $\textbf{Solution.}$ (Khang Nguyen) We have $$(3-a)(3-b)(3-c)\ge 0 \rightarrow abc+9(a+b+c)\le 36$$
Assume that $$(b-1)(c-1)\ge 0 \rightarrow 36\ge 8a+(9+a)(b+c)\Rightarrow b+c\le \dfrac{(36-8a)}{9+a}$$
Therefore
\begin{align*}
\text{LHS}&=\sqrt{2a+b+c+2\sqrt{(a+b)(a+c)}}+\sqrt{b+c}\\&
\le \sqrt{2a+\dfrac{36-8a}{9+a}+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4028695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\int_0^1 \frac{x\ln(x+1)}{x^2+1} dx$ without using complex numbers One can find the antiderivate with help of the partial fraction method introducing complex numbers:
$$\frac{x\ln(x+1)}{x^2+1}=\frac{1}{2}\left(\frac{\ln(x+1)}{x-i}+\frac{\ln(x+1)}{x+i}\right).$$
The result is a complicated function containi... | Rewrite the integral as
$$I=\int_0^1 \frac{x\log (x+1)}{x^2+1}dx =\int_0^1\int_0^1\frac{x^2\:dydx}{(yx+1)(x^2+1)}$$
A partial fraction and integration order reverse later we get
$$\int_0^1\int_0^1 \frac{xy}{(y^2+1)(x^2+1)}-\frac{1}{(y^2+1)(x^2+1)} +\frac{1}{xy+1}-\frac{y^2}{(y^2+1)(xy+1)}\:dxdy = \int_0^1\frac{y}{2(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Kernel, Basis and Column Space If a matrix $A$ has row reduced form
$A=$$\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}$
then the kernel of the linear map $T : \mathbb{R}^4 → \mathbb{R}^3$ defined by $T(x) = Ax$ has basis $$(−3,1,0,0), (−3, 0, 1, 1).$$
How do I prove or disprove the claim regarding the k... | The reduced matrix is $A=\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}$ and if we want to find the generator of the Kernel we have to solve the system $A\underline x=\underline0$:
$$A=\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}\underline x=\underline 0\implies \begin{pmatrix} 1&3&0&3 \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}$ for $x\geq 1$? Intuitively when $x$ gets bigger, $\sqrt{x+1}$ will get closer to $\sqrt{x}$, so their difference will get smaller.
However, I just cannot get a proper proof.
| \begin{align}
\sqrt{x} - \sqrt{x-1}
&= \dfrac{(\sqrt{x} - \sqrt{x-1})(\sqrt{x} + \sqrt{x-1})}{\sqrt{x} + \sqrt{x-1})} \\
&= \dfrac{x-(x-1)}{\sqrt{x} + \sqrt{x-1}} \\
&= \dfrac{1}{\sqrt{x} + \sqrt{x-1}}
\end{align}
\begin{align}
\sqrt{x+1} - \sqrt{x}
&= \dfrac{(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Prove $f(x) = x^2 + 3$ is continuous at $x=3$ Prove that $f(x) = x^2 + 3$ is continuous at $x=3$.
I have tried using $\delta = \sqrt{\epsilon + 9} - 3$.
I tried to split $|x^2-9| = |(x-3)(x+3)|$ and tried to make $x+3$ in terms of $\delta$.
But I get $\delta^2 + 6\delta$.
I don't really get what I'm doing wrong or righ... | You need to solve the equation
$$0<|x-3|<\delta\implies|(x^2+3)-(3^2+3)|<\epsilon$$ for $\delta$.
As the function $x^2+3$ is monotonic around $x=3$, you can solve the equation
$$|x^2-9|=\epsilon$$ and use any $\delta$ such that $(3-\delta,3+\delta)$ is wholly contained between the two solutions, $x=\sqrt{9\pm\epsilon}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there a known closed form solution to $\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx$? $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
Related question
Is there a known closed form solution to
\begin{ali... | Not a complete answer, but an elaboration of what I said in the comments.
To evaluate $$\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx$$
Write it as $$\int_1^\infty\frac{\ln\left(x^{-2n}+1\right)+\ln\left(x^{2n}\right)}{1+x^2} dx = \int_1^\infty\frac{\ln\left(x^{-2n}+1\right)}{1+x^2} dx+2n\int_1^\infty\frac{\ln\left(x\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Question on Rudin Theorem 3.31 Proof On page 65, we have:
$e-s_n = \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+\dots < \frac{1}{(n+1)!}(1 + \frac{1}{n+1}+\frac{1}{(n+1)^2}+\dots)=\frac{1}{n!n}$
Given $s_n = \sum_{k=0}^{n}{\frac{1}{k!}}$.
What I don't understand is, where does $\frac{1}{n!n}$ come from?
| Note that since for $ |a| < 1$ (Geometric series formula)
$$
\sum_{i=0}^{\infty}a^i = \frac{1}{1-a},
$$
choosing $a := \frac{1}{n+1}$, on the right hand side it follows that
$$
\frac{1}{(n+1)!}\left(\sum_{i=0}^{\infty}\left(\frac{1}{n+1}\right)^i\right) = \frac{1}{(n+1)!}\left(\frac{1}{1 - \frac{1}{n+1}}\right) = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4039516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve for integer values of $x,y,z$: $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$ Solve for integer values of $x,y,z$;
$$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3.$$
My attempt:
*
*Note that all $x, y, z$ are non-zero, otherwise a denominator would be zero.
*Mutliplying by $xyz$ gives:
$$
x^2y^2+x^2z^2+y^2... | Hint: If $(x,y,z)$ is a solution, then show that so is $(x,-y,-z)$ and other cyclical permutations. This then severely restricts the values of $|x|, |y|, |z|$ can take after you have derived the inequality:
$$x+y+z\leq 3. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
limit $\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}}$ How can I find the limit to infinity of this function? As this is a $0/0$ equation, I tried using the L'Hôpital's rule in this but ended up making it more complex. I've also tried rationalising the denominator but it didn't lead to anywhe... | Assume WLOG $x\geq0$. Thus, $$\frac{\sqrt{x+2}-\sqrt{x+1}}{\sqrt{x+1}-\sqrt x}\cdot\frac{\sqrt{x+1}+\sqrt x}{\sqrt{x+1}+\sqrt x}$$$$=(\sqrt{x+1}+\sqrt x)(\sqrt{x+2}-\sqrt{x+1})\cdot\frac{\sqrt{x+2}+\sqrt{x+1}}{\sqrt{x+2}+\sqrt{x+1}}$$$$=\frac{\sqrt{x+1}+\sqrt x}{\sqrt{x+2}+\sqrt{x+1}}=1-\frac{\sqrt{x+2}-\sqrt x}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The value of the second-order Eulerian polynomials at x = -1/2. Recently, the second-order Eulerian polynomials
$ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n $
have been discussed on MSE [ a ,
b ].
$$ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n =
\sum_{k=0}^n \left\langle\!\!\left\lang... | Here are some comments. Seeking to invert
$$-\frac{1}{9} + \frac{2}{3} x + \frac{1}{9} \exp(3x) = z$$
we consult Wikipedia on
LambertW
to find that the closed form solution to
$$x = a + b \exp(cx)$$
is given by
$$x = a - \frac{1}{c} W(-bc \exp(ac)).$$
We thus write
$$x = \frac{3}{2} z + \frac{1}{6} - \frac{1}{6} \exp(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the probability that two randomly selected subsets of $\{1,2,3,4,5\}$ have exactly 2 elements common in their intersection Clearly, the total number of subsets possible is $2^5$
For two elements to be common, both subsets need to have at least two elements, so we can form quite a lot of cases which satisfy both co... | Let's say instead of choosing two subsets from a single set, we choose subsets from the two sets $A=\{1,2,3,4,5\}$ and $B=\{1,2,3,4,5\}$, so number of subsets that can be chosen from $A$ are $2^{5}$ and similarly total number of subsets chosen from $B$ are $2^5$.
So, in a total of $2^5 \cdot 2^5=2^{10}$ ways, we can ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve $\sin(x) = 4 \sin(60 ^{\circ} -x)\sin(6 ^{\circ} )\sin(18 ^{\circ} ) $ Solve
$$\sin(x) = 4 \sin(60 ^{\circ} -x)\sin(6 ^{\circ} )\sin(18 ^{\circ} ) $$
under degrees (not radians).
There is a well known identity
$$4 \sin x \sin(60 ^{\circ} −x) \sin(60 ^{\circ} +x) = \sin3x$$
but it doesn't quite help here. I can gu... | Apply $\sin(60-x)=\frac{\sqrt3}2\cos x -\frac12\sin x$ and
rearrange the equation to isolate $x$
\begin{align}
\cot x &=\frac{1+2 \sin6\sin18} {2\sqrt3\sin 6\sin 18}
\end{align}
Then, substitute $ \sin 18 =\frac{\sin36}{2\cos18}
= \frac{\sin72}{4\cos18\cos36}
= \frac1{4\cos36}$
to obtain
\begin{align}
\cot x &=\frac{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4047686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Question about the exponential generating function for $T_n=T_{n-1}+(n-1)\cdot T_{n-2}$ So I came across this step when finding an expontential generating function using this recurrence relation,
$$
T_{n}=T_{n-1}+(n-1) \cdot T_{n-2}
$$
Since $T_{0}=T_{1}=1$
For $n=0$
$$
\sum_{n \geq 0} \frac{T_{0} x^{0}}{0 !}=1
$$
For ... | Just compare the two lines, focusing on
$$\sum_{n\geq 1} \frac{T_{n-1}}{n!}x^n = \color{red}{\underbrace{\frac{T_{1-1}}{1!}x^1}_{n=1}} + \color{blue}{\underbrace{\frac{T_{2-1}}{2!}x^2}_{n=2} + \underbrace{\frac{T_{3-1}}{3!}x^3}_{n=3} + \cdots}$$
$$x + \sum_{n\geq 2} \frac{T_{n-1}}{n!}x^n = \color{red}{x} + \color{blue}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4048133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Leap year probability A year in the 2020s (i.e., from 2020 to 2029 inclusive) is selected uniformly at random, a month
is selected uniformly at random from that year, and a day of the month is selected uniformly at
random from that month.
(a) What is the (exact) probability that the day is the 29th of February?
->I use... | Your answer to a is correct. You must choose a leap year, and then you must choose February, and then you must choose the 29th.
For part (b), $$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$ Setting $A = $ "February" and $B = $ "29th", we find $P(A) = 1/12$, regardless of the year chosen.
Finding $P(B)$ is the sum of all the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $x\geq 0,$ what is the smallest value of the function $f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$ If $x\geq 0,$ what is the smallest value of the function
$$f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$$
I tried doing it by completing the square in numerator and making it of the form
$$\frac{4(x+ 1)^2+ 9}{6(1+ x)}$$
and then, I pu... | After completing the squares you can use the inequality between the arithmetic and geometric mean (AM-GM) in the form $a^2+b^2 \geq 2ab$:
Hence,
$$\frac{(2(x+1))^2+3^2}{6(x+1)}\stackrel{AM-GM}{\geq}\frac{2\cdot 2(x+1)\cdot 3}{6(x+1)}=2$$
Equality is reached for $2(x+1)=3 \Leftrightarrow x = \frac 12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the Closed Form of a Recursion I wanted to find the closed form of the recursion: $$a_n = \frac{1}{n+1} + \frac{n-1}{n+1} a_{n-1}$$ where $a_0 = 0$.
So far, my progress has been to multiply both sides by $\frac{n(n+1)}{2}$ which gives $$\frac{n(n+1)}{2} a_n = \frac{n}{2} + \frac{n(n-1)}{2} a_{n-1}.$$
Now, if we... | I think all of $a_n = \frac{1}{2}$
You can easily prove it using induction also.
$a_1 = \frac{1}{2}$
Assume $a_m = \frac{1}{2}$
To prove $a_{m+1} = \frac{1}{2}$
$a_{m+1} = \frac{1}{m+2} + \frac{m+1-1}{m+2} a_m$
$a_{m+1} = \frac{1}{m+2} + \frac{m}{m+2} \frac{1}{2}$
$a_{m+1} = \frac{1}{2}$
Since the relation is true for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4061569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to go about finding whether this limit exists: $\lim_{x \to 0} x [[\frac{1}{x}]]$? I am trying to solve few challenge questions on Real Analysis from Kaczor and Nowak's Problems in Mathematical Analysis, to become more proficient and stimulate thinking. I'd like someone to (a) verify if my proof is correct (b) is t... | Hint: $\frac 1x-1\lt [[\frac 1x]]\le 1/x $ for all $x\ne 0$
By Squeeze theorem, $\lim_{x\to 0^+} x[[\frac 1x]]=1=\lim_{x\to 0^-} x[[\frac 1x]]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4062394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\sum_{k=1}^{a}\frac{-1-H_k}{k(1-e)^k}$ Question :
My attempt:
Let $a=17399172$
$$\begin{align}
&\sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(\sum_{j=0}^{k}\left(\ln\left(e^{C_j^k}\right)\right)\right)(1-e)^k} \\
&= \sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(\sum_{j=0}^{k}\left(C_j^k\right)\right)(1-e)^k} \\
&= \su... | This is not a full answer but I will elaborate on @Dr.WolfgangHintze comments. We have
$$
\begin{align}
S
&= \sum_{k=1}^{a}\frac{-1-H_k}{k(1-e)^k}\\
&= -\sum_{k=0}^{a-1}(1+H_{k+1})\frac{(\frac{1}{1-e})^{k+1}}{k+1} \\
&= -\int_0^{\frac{1}{1-e}}\sum_{k=0}^{a-1}(1+H_{k+1})t^k\,\mathrm dt\\
&=: -\int_0^{\frac{1}{1-e}}S^\as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4062541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Big O for error terms The following link from wikipedia explains the Big O notation really good. I have only one problem, which is to formalize the usage of Big O notation for error terms in polynomials. In the example give here we have
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=1+x+\frac{x^2}{2!}+\m... | Always remember that $\mathcal{O}(g)$ is a set. So it is more appropriate to say that when $x\to a, f\in \mathcal{O}(g)$ which means that there exists an $\epsilon\gt 0 $ for which you can find a $\delta\gt 0$ such that $0\le |f(x)|\le \epsilon |g(x)|$ for all $0\lt|x-a|\lt \delta$.
Your question 1): Why $\frac{x^3}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4065119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_... | One more option that apparently hasn't been mentioned yet: in your original question you mentioned being stuck on the term
$$\sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
This can be rewritten as
$$\frac{1}{2}\sum_{n=1}^{ \infty } \left(-\frac{1}{3}\right)^n$$
which can also be evaluated as a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln... | Consider the function $\displaystyle f(x) = \frac{1-x^2}{(1+x^2)^2} $. The Maclaurin series of $f$ is given by:
$$\displaystyle \sum_{k \ge 0} [k(-1)^k x^{2k} - k(-1)^k x^{2k-2}]$$
The integral is equal to $\displaystyle \int_0^1 f(x)\log{x}\,\mathrm{d}x $. Using the series expansion you should get
$$\displaystyle I =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
} |
Show that a number belongs or does not belong to a Cantor set I have trouble showing that $\frac{10}{31}$ and $\frac{45}{69}$ belong/do not belong to the Cantor set. I tried to do the following
$$\sum_{n = x}^\infty \frac{2}{3^n} = \frac{2}{3^x} + \frac{2}{3^{(x + 1)}} + \frac{2}{3^{(x + 2)}} + \dots = \frac{2}{3^x} \l... | Neither one is in the Cantor set. $\dfrac{10}{31} = 0.02220101110012..._{3}$, and $\dfrac{45}{69} = \dfrac{15}{23} = 0.12212110220122..._{3}$. Both has $1$ in their ternary presentation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My sol... | You can crudely bound $1+\frac{1}{2} + \cdots + \frac{1}{n}$ by
$$\sum_{k=1}^n \frac{1}{k} \le \sum_{k=1}^{\lceil\sqrt{n}\rceil} \frac{1}{k} + \sum_{k=\lceil \sqrt{n} \rceil}^n \frac{1}{k} \le \lceil \sqrt{n}\rceil \cdot 1 + n \cdot \frac{1}{\sqrt{n}} \le 3\sqrt{n},$$
so that when you divide this by $n$, the expression... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
Equation of 2nd order kernel of arc-cosine kernel In the paper "Kernel method for deep learning" (Cho and Saul, NIPS 2009), a part $J_n(\theta)$ of definition of the arc-cosine kernel is given as
$$J_n(\theta) = (-1)^n (\sin \theta)^{2n+1} \left( \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \right)^n \left( \... | Most likely you made a computational mistake somewhere.
\begin{align}
J_2(\theta) &= (-1)^2 (\sin\theta)^5 \left( \frac{1}{\sin \theta}\frac{\partial}{\partial \theta}\right)^2\left( \frac{\pi - \theta}{\sin \theta}\right)\\
&= (\sin^5 \theta) \left( \frac{1}{\sin \theta}\frac{\partial}{\partial \theta}\right)\frac1{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How would I integrate $\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx$? $$\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx$$
I have known this equals to $\frac{\pi^2}{16}$, but I cannot prove it.
| Note
\begin{align}
I=&\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx\\
= &\int_0^1 \frac{\frac12\ln(1-x^2)+\ln(1+ \frac x{\sqrt{1-x^2}})}{x}\,dx=\frac12I_1+I_2\tag1
\end{align}
where
\begin{align}
I_1&=\int_0^1 \frac{\ln(1-x^2)}{x}\,dx\overset{x^2\to x}=\frac12 \int_0^1 \frac{\ln(1-x)}{x}\,dx\\
&= \int_0^1 \frac{\ln(1-x)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Abundant products of iterations of Euler's totient function Let $a_0(n) = n$ and $a_{i+1}(n) = \varphi(a_i(n))$ for $i\geq 0$, where $\varphi(n)$ is Euler's totient function (the number of positive integers less than or equal to $n$ and coprime with $n$). Denote $f(n) = \prod_{k=0}^{\infty}a_k(n)$.
Determine all positi... | Let's observe that:
\begin{align*}
a_0(n) &= n \,, \\
a_1(n) &= \varphi(n) \,, \\
a_2(n) &= \varphi(a_1(n)) = \varphi(\varphi(n)) = \varphi^{(2)}(n)\,, \\
... \\
a_k(n) &= \varphi^{(k)}(n)\,,
\end{align*}
where $\varphi^{(k)}(n)$ is $k$ times composition of the totient function. This gives us the function $f(n)$ as
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all natural solutions that satisfy $2^ + 3^ = ^2$ It looks like an easy question but I couldn't find a way to solve it. I found (0,1,2),(3,0,3),(4,2,5) by trial and error and I'm kinda sure they are the only answers but I'm not sure how to prove it.
| If $y=0$, we have $2^x=z^2-1=(z+1)(z-1)$, for which the only solution is $x=z=3$. That is, we have $z+1=2^a$ and $z-1=2^b$ with $a+b=x$. But then $2=(z+1)-(z-1)=2^a-2^b$, for which the only solution is $a=2$ and $b=1$, from which $x=z=3$ follows.
If $y\ge1$, then $2^x$ is a square mod $3$, which implies $x$ is even. Wr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Use binomial series for $\sqrt{1+x}$ to find the binomial series for $\frac{1}{\sqrt{1+x}}$ The question specifically asks me to use the binomial series for $\sqrt{1+x}$ to find the series for $\frac{1}{\sqrt{1+x}}$.
First, I need to find the derivative of the binomial series $\sqrt {1+x}$
I first constructed the binom... | The expanded form of $\frac{d}{dx}\sqrt{1+x}$ is correct, the compact one is not. You could do $$\frac{d}{dx}\left(1+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n!}\frac{1\cdot3\cdots(2n-3)}{2^n}x^n\right)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(n-1)!}\frac{1\cdot3\cdots(2n-3)}{2^n}x^{n-1}$$ without expanding (at $n=1$ we have an e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4085145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A Difficult Area Problem involving a Circle and a Square A few days ago, I encountered the following problem:
After a little bit of thinking, I managed to come up with the following solution:
*
*Rotate the square $90^\circ$ clockwise and let the new bottom left corner of the square be $(0,0)$.
*The circle inscribed... | Here is an alternate solution,
We assume right bottom vertex to be the origin then, equations of circles are
Circle S: $x^2+y^2 = 100$
Circle T: $(x+5)^2+(y-5)^2 = 25$
Solving both equations, intersection points are $A \left(\frac{5}{4}\left(\sqrt7-5\right), \frac{5}{4}\left(\sqrt7+5\right)\right)$ and $B\left(-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Find the number of three-digit numbers in which exactly one digit $3$ is used? Find the number of three-digit numbers in which exactly one digit $3$ is used?
The number is of one of the forms $$\_\text{ }\_\text{ }3 \\ \_\text{ } 3 \text{ } \_ \\ 3\text{ }\_ \text{ }\_$$ There are $V_9^2-V_8^1=9\times8-8=64$ possibilit... | We have $9$ digits other than $0$, but $0$ can't be the first digit. So for both the first and second, we have $8 \cdot 9=72$ possibilities each. For the last one we have $9 \cdot9=81$ possibilities.
Alternatively, there are $9 \times 10 \times 10 = 900$ three-digit numbers. $8 \times 9 \times 9 = 648$ of them have no ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $\{a_n\}$ converges.
Suppose $a_1,a_2>0$ and
$a_{n+2}=2+\dfrac{1}{a_{n+1}^2}+\dfrac{1}{a_n^2}(n\ge 1)$. Prove
$\{a_n\}$ converges.
First, we may show $\{a_n\}$ is bounded for $n\ge 3$, since $$2 \le a_{n+2}\le 2+\frac{1}{2^2}+\frac{1}{2^2}=\frac{5}{2},~~~~~~ \forall n \ge 1.$$
But how to go on?
| Another approach: Let $L \approx 2.3593$ be the unique solution of $L = 2 + 2/L^2$ in the interval $[2, 2.5]$, and $b_n = a_n - L$. We want to show that $b_n \to 0$.
The recursion formula becomes
$$
b_{n+2} = \frac{1}{(L+b_{n+1})^2} + \frac{1}{(L+b_{n})^2} - \frac{2}{L^2} \, .
$$
We estimate
$$
\left| \frac{1}{(L+b_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4094333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
irrational integral $ \int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}\, dx$ I have to solve this irrational integral $$ \int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}\, dx$$
It seems that the most convenient way to operate is doing the substitution
$$ x= \frac{t^2}{3-2t}$$
according to the rule,
obtaining the integral:
$$ \... | Note that the substitution $x=\frac{t^2}{3-2t}$ leads to
$ \sqrt{x^2+3x}=\frac{3t-t^2}{3-2t}, \>\>\>\>\> dx = -\frac{2(t^2-3t)}{(3-2t)^2}dt
$
and
$\int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}dx
= -\int \frac{1+\frac{3t-t^2}{3-2t}}{2-\frac{3t-t^2}{3-2t}}\, \frac{2(t^2-3t)}{(3-2t)^2}dt
= -\int \frac{3+t-t^2}{t^2-7t+6}\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4094912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int\frac{x}{(x^4-1)\cdot\sqrt{x^2+1}} \, dx$ Integrate $\int\frac{x}{(x^4-1)\cdot\sqrt{x^2+1}} \, dx$
I tried substituting $x=\tan(t)$ in order to get away with square root. ($\:dx=\frac{1}{\cos^2(t)}dt\:$)
$\sqrt{x^2+1}=\frac{1}{\cos(t)}\:\:$ and $\:\:x^4-1=\frac{\sin^4(t)-\cos^4(t)}{\cos^4(t)}$
Now after ... | Let $\sqrt{x^2+1}=y, dy=\dfrac x{\sqrt{x^2+1}}$
$$\implies x^2=y^2-1$$
$$\int\dfrac{dy}{(y^2-1)^2-1}=\int\dfrac{dy}{y^2(y^2-2)}$$
Now use,
$$\dfrac2{y^2(y^2-2)}=\dfrac{y^2-(y^2-2)}{y^2(y^2-2)}=\dfrac1{y^2-2}-\dfrac1{y^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4098156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove the series generated by the sequence diverges Consider the positive sequence $\{x_n\}_{n\ge 1}$ given by
$$\begin{cases}
x_1&=1,\\
x_{n+1}^2+x_{n+1}x_n^2-x_n^2&=0,\ \forall n\ge 1.
\end{cases}$$
More specifically, the second equation together with condition that the sequence is positive-valued mean that $$ x_{n+1... | This is the answer from one of my friends.
We will prove that $x_n\ge 1/n$ by induction. Firstly, notice that
\begin{align*}
x_{n+1}=\dfrac{-x_n^2+\sqrt{x_n^4+4x_n^2}}{2}=\dfrac{4x_n^2}{2(x_n^2+\sqrt{x_n^4+4x_n^2})}=\dfrac{2}{1+\sqrt{1+\dfrac{4}{x_n^2}}}.
\end{align*}
Consider the function $f(x)=\dfrac{2}{1+\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} = \sum_{n\geq 0} \frac{(xy)^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}.$ Prove that
$$\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} = \sum_{n\geq 0} \frac{(xy)^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}.$$
Here I am trying the following
\begin{align*}
\prod_{i\geq 1}\frac{1}{... |
We obtain
\begin{align*}
\color{blue}{\prod_{k=1}^\infty\frac{1}{1-xy^{2k-1}}}&=\prod_{k=0}^\infty\frac{1}{1-xy^{2k+1}}\\
&=\prod_{k=0}^{\infty}\frac{1}{1-\left(xy\right)\left(y^{2}\right)^k}\\
&=\frac{1}{\left(xy;y^2\right)_{\infty}}\tag{1}\\
&=\sum_{n=0}^{\infty}\frac{1}{\left(y^2;y^2\right)_n}(xy)^n\tag{2}\\
&=\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Finding the intersection of $x= y^2 + y - 2$ and $y = -x^2 - \frac32x + 1 $ Goodmorning, I am struggling in finding the points of intersection of the following parabolas:
$$\begin{align}
x &= y^2 + y - 2 \\[4pt]
y &= -x^2 - \frac32x + 1
\end{align}$$
I know that these two can be solved either algebraically or with matr... | $\begin{align}
x &= y^2 + y - 2 \\[4pt]
y &= -x^2 - \frac32x + 1
\end{align}$
The first one can be written as,
$x = (y-1) (y+2)$ ...(i)
The second one is $y = -x^2 - \frac32x + 1 \implies y-1 = - x^2 - \frac{3}{2}x$ ...(ii)
Plugging in $y-1$ from (ii) into (i),
$x = (- x^2 - \frac{3}{2}x) (- x^2 - \frac{3}{2}x + 3)$
$x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculating a limit. Is WolframAlpha wrong or am I wrong? What I'm trying to solve:
$$\lim _{x\to -\infty \:}\frac{\left(\sqrt{\left(x^2+14\right)}+x\right)}{\left(\sqrt{\left(x^2-2\right)}+x\right)}$$
What I put into WolframAlpha:
(sqrt(x^2+14)+x)/(sqrt(x^2-2)+x)
My result: $1$, which I get by simply dividing bot the ... | The limit is as $x$ goes to $-\infty,$ so you can't just divide by $x.$ Both the numerator and the denominator have the form $\infty -\infty.$ WolframAlpha is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Solve for integers $x, y$ and $z$: $x^2 + y^2 = z^3.$
Solve for integers $x, y$ and $z$:
$x^2 + y^2 = z^3.$
I tried manipulating by adding and subtracting $2xy$ , but it didn't give me any other information, except the fact that $z^3 - 2xy$ and $z^3+2xy$ are perfect squares.
This doesn't give us much information to w... | Let $C$, $D$, $S$ and $T$ be integers, and define
\begin{eqnarray*}
x&=&ab^3X=(C^2+D^2)(CS^3-3DS^2T-3CST^2+DT^3),\\
y&=&ab^3Y=(C^2+D^2)(DS^3+3CS^2T-3DST^2-CT^3),\\
z&=&ab^2Z=(C^2+D^2)(S^2+T^2).
\end{eqnarray*}
Then a routine verification shows that $x^2+y^2=z^3$. I will show that every solution is of this form. Moreove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$ \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x $ I am trying to evaluate this antiderivative $$
\int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x
$$
What i have done:
$$
\begin{split}
I
&= \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x \\
&= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{4+x+1+4 \sqrt{x+1}-3+x} \cdot d x\\... | Too large for a comment in the other answer.
Let us first try to identify the correct substitution
For real calculus, we need $x+1\ge0\text{ and }3-x\ge0\iff-1\le x\le3$
$$\iff-\dfrac12-\dfrac12\le\dfrac{x-1}2\le\dfrac32-\dfrac12$$
WLOG $\dfrac{x-1}2=\cos2t, 0\le2t\le\pi\implies x=2\cos2t+1$
$\sqrt{x+1}=+(2\cos t),\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4111706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the radius of convergence of $\sum_{n=0}^{\infty}{\ln\left(\cos{\frac{1}{3^n}}\right)x^n}$ I need to find the radius of convergence of
$$\sum\limits_{n=0}^{\infty}{\ln\left(\cos{\frac{1}{3^n}}\right)x^n}$$.
I came up with the following solution:
Since $\cos{\frac{1}{3^n}\sim1-\frac{1}{2(3^n)^2}}$, then
$$\ln\left(... | As a few commentors have pointed out: Yes, this looks good. Ratio test is always going to be your friend on questions such as these. Let's go ahead and work it out so you're sure you got the right final answer:
\begin{align}
\frac{a_{n+1}}{a_n} &= \frac{{\frac{x^{n+1}}{2(3^{n+1})^2}}}{{\frac{x^n}{2(3^n)^2}}}\\
&= {\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4112047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $\frac{x^2-10x+15}{x^2-6x+15}=\frac{4x}{x^2-12x+15}$ Solve the equation $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}.$$
First we have $$x^2-6x+15\ne0$$ which is true for every $x$ ($D_1=k^2-ac=9-15<0$) and $$x^2-12x+15\ne0\Rightarrow x\ne6\pm\sqrt{21}.$$ Now $$(x^2-10x+15)(x^2-12x+15)=4x(x^... | If you take out $x$ (clearly $x\ne 0$) we get: $$\dfrac{x(x-10+{15\over x})}{x(x-6+{15\over x})}=\dfrac{4x}{x(x-12+{15\over x})}.$$
Cancel $x$ and let $t=x+ {15\over x}$, then we have $${t-10\over t-6} = {4\over t-12}$$ or $$t^2-22t+120 = 4t-24$$
and so on...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Using polar coordintaes to evaluate $\int \int {\sqrt {\frac {1-x^2-y^2} {1+x^2+y^2}} }\ dx \ dy$ I want evaluate the following integral using polar coordinates.
$$ \int \int {\sqrt {\frac {1-x^2-y^2} {1+x^2+y^2}} }\ dx \ dy $$ over the positive quadrant of the circle $$x^2+y^2=1$$
I used the substitution $x=r\cos\the... | Integrate by parts as follows
$$\int {\sqrt{\frac{1-u}{1+u}}}du
= \int {\sqrt{\frac{1-u}{1+u}}}d(1+u)
=\sqrt{1-u^2}+\int \frac{du}{\sqrt{1-u^2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Study the series of $\int_0^{\frac{1}{n^a}}\sin{(\sqrt[3]{x})}\,dx$ with respect to $a>0$ I have to study the series $\sum a_n$ with $a_n=\int_0^{\frac{1}{n^a}}\sin{(\sqrt[3]{x})}\,dx$, with respect to $a>0$.
I have thought to use the asymptotic criterion for series.
In particular I can observe that:
$\sin{x}\sim x-\fr... | Here is a simpler proof of the equivalent: let $f(t) = \frac{\sin(t)}{t}$. It is well known that $|f(t)|\le 1$ for all $t\in{\mathbb R}$ and that $f(0)=1$. The substitution $x = \frac{y}{n^a}$ in the integral gives
\begin{equation}
a_n = \left(\frac{1}{n^a}\right)^{4/3}\int_0^1 f\left(\left(\frac{y}{n^a}\right)^{1/3}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4122996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$ Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$
And Prove it by induction.
Attempt:
$$A(x)= \sum_{k=0}^n (k+1)(n-k+1)$$
$$\sum_{n=0}^\infty(\sum_{k=0}^n (k+1)(n-k+1))x^n$$
The series of differences is :$\{(n+1),2(n),3(n... | Unclear if this is off-topic. The OP specifically requires that generating functions be used.
Other responses have covered this constraint, so I'll show an alternative approach.
Known (or routinely proven by induction) that
*
*$\displaystyle \sum_{i = 1}^n i = \left(\frac{1}{2}\right)(n+1)(n).$
*$\displaystyle \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4124684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
An immediate alternative to a trigonometry problem (high school) I have a right triangle, $AH\perp BC$, where $\cos \beta=\sqrt 5/5$ and $\overline{AH}+\overline{CH}+\overline{HB}=7$.
I have to found the area.
My steps (or solution):
$$\mathcal A(\triangle ABC)=\frac 12\overline{AB}\cdot \overline{BC} \sin \beta$$
wit... | $\cos\beta=\dfrac{\sqrt5}{5}$ and $\sin\beta=\dfrac{2\sqrt5}{5}.$ Let $\overline{AH}=x,$ then you can obtain values $$\overline{AB}= \dfrac{\sqrt{5}x}{2},\qquad \overline{BC}= \dfrac{5x}{2},\qquad \overline{AC}=\sqrt{5}x$$ and $$\overline{AH}=x,\qquad \overline{BH}=\dfrac{x}{2},\qquad \overline{CH}=2x$$ using simple tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Given 3 points. What's the normal to the plane that contains these 3 points?
Given are these points $$v_1=\begin{pmatrix} 1\\ 2\\ 3
\end{pmatrix}, v_2=\begin{pmatrix} 2\\ 3\\ 3 \end{pmatrix},
v_3=\begin{pmatrix} 2\\ 4\\ 4 \end{pmatrix}$$
Determine the normal to the plane which contains these 3 points.
My proble... | “The” normal to a plane typically really does refer to any vector that is normal to the plane. In other words, your suggested answer (or any non-zero scalar multiple of it) is perfectly fine.
No need to express the normal generally as $$k\begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix}\quad(k\in\mathbb R\setminus\{0\}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating limits using Taylor expansions The limits are
$$\lim_{x\to 0}(\frac{\cos{x}-e^{-x^2/2}}{x^4})$$
$$\lim_{x\to 0}\frac{e^x\cdot\!\sin{x}-x(1+x)}{x^3}$$
Probably wrong things that I've tried
$\lim_{x\to 0}\frac{\cos{x}-e^{-x^2/2}}{x^4}=\lim_{x\to 0}(\frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^5)-1+\frac{x^2}{2}+\... | $$\forall\;x\in\Bbb R\;,\;\;\;e^x=\sum_{n=0}^\infty\frac{ x^n}{n!}=\implies e^{-x^2/2}=1-\frac{x^2}2+\frac{x^4}{4\cdot2!}-\ldots$$
and since
$$\cos x=1-\frac{x^2}2+\frac{x^4}{4!}-\ldots$$
we get
$$\cos x-e^{-x^2/2}=1-\frac{x^2}2+\frac{x^4}{24}-\ldots-\left(1-\frac{x^2}2+\frac{x^4}8-\ldots\right)=-\frac{x^4}{12}+\ldots$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Showing $\tan70° = \tan20° + 2\tan50°$ Q. Prove that $\tan 70° = \tan 20° + 2\tan 50°$.
My approach:
$ LHS = \tan 70° = \dfrac1{\cot 70°} = \dfrac1{\tan 20°}$
$ \begin{align}
RHS
&= \tan 20° + 2\tan 50° \\
&= \tan 20° + 2\tan (20+30)° \\
&= \tan 20° + \dfrac{2(\tan 20° + 1/√3)}{1 - \tan 20°/√3} \\
&=\dfrac{2 + 3√3 \t... | You can use the identity $2 \cot 2 A = \cot A - \tan A$ which makes the job easy.
$\tan 20° + 2\tan 50° = \tan 20° + 2\cot 40°$
$= \tan 20° + \cot 20° - \tan 20^0 = \cot 20^0 = \tan 70^0$
Otherwise proceed from $\tan 20° + 2\tan 50° = \tan 20° + 2\cot 40°$ as below -
$\displaystyle \tan 20^0 + 2 \cot 40^0 = \tan 20° + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding distance from Local maximum of $f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant
What is the distance of the local maximum of the function
$f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant?
$1)1\qquad\qquad2)\sqrt2\qquad\qquad3)2\qquad\qquad3)2\sqrt2$
It is a question from timed exam so the fastest answ... | $x^2 - 4x + 4 = (x-2)^2$ is a perfect square, so rewrite $f(x)$ as $x + \sqrt{4 - (x-2)^2}$. We want to get $\sqrt{4 - 4 \cos^2 u} = 2 \sin u$ in the given domain, which leads to the parametrisation:
$$(x,y) = (2 + 2 \cos t, 2 + 2 \cos t + 2 \sin t).$$
Now $2 + 2 \cos t + 2 \sin t = 2(1 + \sqrt{2} \cos(t - \pi/4))$, wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Given $x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=3,$ can we conclude that $x^n+y^n+z^n\in\mathbb{Q}$ for all $n\in\mathbb{N}$. Is there an explicit form? I had seen a fun problem that is exactly the problem in the question, except it was a specific case of this. Turns out, if $x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=3,$ then $x^5+y^... | None of the calculations we need are mysterious, but one might need some experience dealing with symmetric functions and their relations to know what to look for.
We denote $x^n+y^n+z^n$ by $p_n$. You can verify the following inductive formula:
$p_n = (x+y+z)p_{n-1}-(xy+yz+zx)p_{n-2}+(xyz)p_{n-3}$
We can calculate the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Algebra trick clarification question I have the equation $x = \sqrt{3} + \sqrt{2} \,\,$ and the trick is that $\frac{1}{x} = \sqrt{3} - \sqrt{2}$. I don't see how the minus sign appears just by taking 1 over x?
| \begin{align}
\frac1x &= \frac1{\sqrt{3}+\sqrt{2}}\\
&= \left(\frac1{\sqrt3 + \sqrt2} \right) \left( \frac{\sqrt{3}-\sqrt{2}}{\sqrt3-\sqrt2}\right)\\
&= \frac{\sqrt3-\sqrt2}{3-2}
\end{align}
In general if $x = \sqrt{a}+\sqrt{b} \ne 0, a \ne b$, then $\frac1x = \frac{\sqrt{a}-\sqrt{b}}{a-b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What kind of transformation is this matrix refering to?
Given is the 3D transformation $$M_1=\begin{pmatrix} 0 & 1 & 0\\ 3 &
0 & 0\\ 0 & 0 & 1 \end{pmatrix}$$
What kind of transformation is $M_1$?
I've used several different points \begin{pmatrix}
x\\
y\\
z
\end{pmatrix}
and did operations with it $$\begin{pmatri... | $$M_1=\begin{pmatrix} 0 & 1 & 0\\ 3 &
0 & 0\\ 0 & 0 & 1 \end{pmatrix}$$
$$=\begin{pmatrix} 1 & 0 & 0\\ 0 &
3 & 0\\ 0 & 0 & 1 \end{pmatrix}
\begin{pmatrix} 0 & 1 & 0\\ 1 &
0 & 0\\ 0 & 0 & 1 \end{pmatrix}$$
Looks like a reflection in $y=x$ together with a scaling.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Changing the order of integration to find the volume My goal is to set up the triple integral that will solve the volume $\iiint \ xyz \ dV $ if S if the region bounded by the cylinders $x^2 + y^2 = 25$ and $ x^2+ z^2 = 25$ and the 1st octant with dV = dxdydz.
With order dV = dzdydx, the value of the volume is $\frac{1... | The second $S$ is incorrect. On one surface we have $0\le x\le\sqrt{25-y^2}$ and on the other $0\le x\le\sqrt{25-z^2}$, which one do you choose? Since we have to take the intersection of the regions enclosed by these surfaces, we take the intersection of these inequalities as well, i.e.$$\begin{align*}0\le x\le\sqrt{25... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Centroid of the region bounded by y = f(x) and x = f(y) curve How do i find the centroid of the region if $x = f(y)$ is involved ?
For example : find the centroid of the region bounded by $y = x^2$ and $x = y^2$
And how do i find it if there is curve intersection that is below x axis ?
For example : find the centroid o... | In this case you can first try to sketch the region, as shown below.
From the sketch we learn that it suffice to consider the first quadrant where $ x, y \geq 0 $.
Thus the equation $ x = y^2 $ can be converted to $ y = \sqrt{x} $.
Put $ f(x) = \sqrt{x} $, $ g(x) = x^2 $, $ a = 0 $, $ b = 1 $, and using the formulae y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$ for positive reals $a$, $b$, $c$ Question $5$ of BMO1 $2008$:
For positive real numbers $\;a,\;b,\;c,\;$ prove that $$\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
I noticed that the right side can be grouped, but did not get further.... | And now, a geometric argument.
If any of the terms on the right hand side are nonpositive, the inequality follows trivially. If they are all positive, $a,b,c$ form the legs of a triangle:
By Heron's formula, the area of this triangle is
$$
\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}
$$
The area of the triangle is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Limit of the series $\sum_{k=1}^\infty \frac{n}{n^2+k^2}.$ I am trying to evaluate $$\lim_{n\to \infty} \sum_{k=1}^\infty \frac{n}{n^2+k^2}.$$ Now I am aware that clearly $$\lim_{n\to \infty} \sum_{k=1}^n \frac{n}{n^2+k^2} = \int_0^1 \frac{1}{1+x^2}dx = \tan^{-1}(1) = \frac{\pi}{4},$$ but I do not know what to do if ea... | Denote $\displaystyle \mathcal{S}(n) = \sum_{k \ge 1} \frac{n}{n^2+k^2}$. We have
\begin{aligned} \mathcal{S}(n) & = \sum_{k\geq 1}\frac{n}{ n^2+k^2} \\& = \frac{1}{2}\sum_{k\geq 1}\frac{2n }{ n^2+k^2} \\& = \frac{1}{2}\sum_{k\geq 1} \frac{d}{dn}\log\left(1+\frac{n^2}{k^2}\right) \\& = \frac{1}{2}\frac{d}{dn} \log \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"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.