Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Interesting four-sum inequality $n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right) \ge...$
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \lef... | (update 2019/08/09)
Let $H_n = \sum_{k=1}^n \frac{1}{k}$. Since $\frac{1}{k} \le -\ln (1 - \frac{1}{k}) $ for $k\ge 2$,
we have $H_n - 1 \le -\sum_{k=2}^n \ln (1 - \frac{1}{k}) = \ln n$ or $H_n \le \ln n + 1$.
Rewrite the inequality as
$$(n^2+2n-H_n)\left(n^2 - 2n - 3 + \frac{3}{n+1} + 3H_n\right) \le n^2(n+ H_n)\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Solve the initial value problem below using the method of Laplace transform I am able to get up to a certian point this problem but get stuck when doing the Laplace transforms. The problem is
Solve the initial value problem below using the method of Laplace transform :
$$y''-4y=8t-20e^{-2t}$$
with $y(0)=0$ and $y'... | Since you already have the expression for $y(s)$, you can further simplify each term using partial fractions.
First, we have $\frac{8}{s^2 (s^2 - 4)} = \frac{A}{s^2} + \frac{B}{s^2 - 4}$, which gives us $8 = A (s^2 - 4) + Bs^2$. Thus, $A = -2$ and $B = 2$.
Then we have $\frac{20}{(s + 2)(s^2 - 4)} = \frac{A}{s + 2} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing $\sqrt[3]{\cos\frac{2\pi}{9}}+\sqrt[3]{\cos\frac{4\pi}{9}}+\sqrt[3]{\cos\frac{8\pi}{9}} = \sqrt[3]{\frac{3\sqrt[3]9-6}{2}} $
Show that
$$\sqrt[3]{\cos\frac{2\pi}{9}}+\sqrt[3]{\cos\frac{4\pi}{9}}+\sqrt[3]{\cos\frac{8\pi}{9}} = \sqrt[3]{\frac{3\sqrt[3]9-6}{2}} $$
I tried to find a polynomial that had its roots,... | We make use of the general equality that $$\sf a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\tag1$$ where $\sf a,b,c=\sqrt[3]{\cos\frac{2\pi}9},\sqrt[3]{\cos\frac{4\pi}9},\sqrt[3]{\cos\frac{8\pi}9}$ such that $\sf a^3+b^3+c^3=0$ and $\sf abc=-\frac12$.
We can also substitute in that $\sf a^2+b^2+c^2=(a+b+c)^2-2(ab+bc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
What are the solutions for $ n(n+1)=p^2$ for n belongs to $N$ The following is my approach:
$ n^2+n =p^2$
$ n^2+n+\frac{1}{4} = p^2 + \frac{1}{4}$
$ (n+\frac{1}{2})^2 = p^2+\frac{1}{4}$
$ (n+\frac{1}{2}-p) (n+\frac{1}{2}+p) = \frac{1}{4}$
I am not able to proceed further from here. Any suggestion on what to do... | You are actually very close to the answer. All you really need to do is multiply both sides of $(n+{1\over2}-p)(n+{1\over2}-p)={1\over4}$ by $4$ and then note that the only integer solutions of $ab=1$ is $a=b=\pm1$. Here is a complete presentation:
Assuming only $n,p\in\mathbb{Z}$, we have
$$\begin{align}
n(n+1)=p^2&\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Complex root used to find $a$. Find $a$ if $z=2+i$ is a root of $2z^{2}+3z+2a-14+3i=0$
Workings:
$2(2+i)^{2}+3(2+i)+2a-14+3i=0$
$2(4+4i-1)+6+3i+2a-14+3i=0$
$8-8i-2+6+3i+2a-14+3i=0$
$-2+14i=-2a$
$1-7i=a$
Is this correct?
| Your answer is correct.
You could say $2a=14-3i-3z-2z^2=14-3i-3(2+i)-2(2+i)^2$
$=14-3i-3(2+i)-2(3+4i)=2-14i,$ so $a=1-7i$.
By Vieta's formulas, the sum of the roots is $-\dfrac32$, so the other root besides $2+i$ is $-\dfrac72-i$.
As noted in comments, these roots are not complex conjugates;
their product is $(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of the limit $b_n = n - \sqrt{n^2 + 2n}$ if $(1/n)\rightarrow 0$. Consider the sequence given by $b_n = n - \sqrt{n^2 + 2n}$. Taking $(1/n)\rightarrow 0$ as given, and using both the Algebraic Limit Theorem, and the fact that if $(x_n)\rightarrow x$ then $(\sqrt{x_n})\rightarrow \sqrt{x}$, show $\lim b_n... | Observe that $n - (n^2 + 2n)^{\frac{1}{2}}=n - n\left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}$.
By the binomial theorem, it follows that
$$
\left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}=1+\frac{1}{n}-\frac{1}{2n^2}+o(n^{-2})
$$
So
$$
n - n\left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}=n - \left(n + 1 - \frac{1}{2n} + o(n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve for $x, y \in \mathbb R$: $(x^2 - 1)^2 + 3 = \dfrac{6x^5y}{x^2 + 2}, 3y - x = \sqrt{\dfrac{4x - 3x^2y - 9xy^2}{x + 3y}}$
Solve the following system of equations $$\large \left\{ \begin{align} (x^2 &- 1)^2 + 3 = \frac{6x^5y}{x^2 + 2}\\ 3y - x &= \sqrt{\frac{4x - 3x^2y - 9xy^2}{x + 3y}}\end{align} \right.$$ such t... | $(x \le 3y \ne -x)$
Let $a = \dfrac{2}{x^2} > 0$ and $b = \dfrac{3y}{x}$. The system of equation becomes
$$\left\{ \begin{align} \left(\frac{2}{a} - 1\right)^2 + 3 = \frac{2b \cdot \left(\dfrac{2}{a}\right)^3}{\dfrac{2}{a} + 2}\\ b - 1 = \sqrt{\frac{2a - b - b^2}{1 + b}} \end{align} \right.$$
$$\iff \left\{ \begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $(1-\frac{1}{x-c})^{\epsilon x-c}\geq(1-\frac{1}{x})^{\epsilon x}$ How do I formally prove that $(1-\frac{1}{x-c})^{\epsilon x-c}\geq(1-\frac{1}{x})^{\epsilon x}$ which can be seen graphically. Here $\epsilon$ is a small positive constant $\ll1$, $x\gg0$ and $c>0$.
| Suppose that $x$ is very large and use Taylor expansion
$$\log \left(1-\frac{1}{x-c}\right)=-\frac{1}{x}-\frac{c+\frac{1}{2}}{x^2}+O\left(\frac{1}{x^3}\right)$$
$$(\epsilon x-c)\log \left(1-\frac{1}{x-c}\right)=-\epsilon +\frac{-c \epsilon +c-\frac{\epsilon }{2}}{x}+O\left(\frac{1}{x^2}\right)$$
$$\text{lhs}=\left(1-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Geometry : what is the $\phi$ angle, if area of yellow rectangle is equal with area of red triangle? I have a right triangle and in it area of yellow rectangle is equal with area of red triangle. How could prove that $\phi=45^{\circ}$?
$$\text{Area of Yellow Rectangle}=\text{Area of Red Triangle}$$
This problem ha... | Thanks for all.
first equality: area of red and yellow are equal.
$$ 2xy=vw \label{eq:1} \tag{1}$$
Now by using Thales' theorem in this triangle:
$$ \frac{v+y}{c}=\frac{b-x}{b} \Rightarrow v=c(\frac{b-x}{b})-y \label{eq:2} \tag{2}$$
$$ \frac{x+w}{b}=\frac{c-y}{c} \Rightarrow w=b(\frac{c-y}{c})-x \label{eq:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integration with $\ln(x)$ in the denominator
Find$$\displaystyle\int_1^\infty\frac{(x^2-1)(x^4-1)(x^6-1)}{\ln(x)(x^{14}-1)} dx$$
I tried simplifying the terms without logarithm
$x^2-1=(x-1)(x+1)\\x^{14}-1=(x^7-1)(x^7+1)$
to see if any substitution is possible but since the logarithm is in the denominator it is getti... | The following answer is similar to Song's answer except that it involves using a common integral representation of the digamma function:
$$\psi(z) = -\gamma + \int_{0}^{1} \frac{1-t^{z}}{1-t} \, \mathrm dt\, , \quad \mathcal \Re (z) >0$$
The integral representation of the tetragamma function used in Song's answer can b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Interesting relation derived from the identity $\sin^2 x + \cos^2 x \equiv 1$ Every one knows that
$$\sin^2 x + \cos^2 x \equiv 1.$$
It is also well known that
$$\sin x \equiv \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n + 1}}{(2n+1)!},\quad\cos x \equiv \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}.$$
One can rewrite t... | Here I provide explicit reasoning to make the identity in question obvious (note that I use the notation ${N \choose k} \equiv C(N,k)$).
One can easily note that
$$(2c)! \sigma(c) = \sum_{n = 0}^{c-1} \frac{(2c)!}{(2n+1)! (2c-(2n+1))!}
= \sum_{n = 0}^{c-1} \frac{r!} {b_{odd}! (r -b_{odd})!}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
extracting a coefficient from formal power series multiplication The Questions is Compute the value of $[x^n] \frac{(1+x)^n}{1-x} $
The solution is as follows:
$[x^n] \frac{(1+x)^n}{1-x} $ = $[x^n](1+x)^n(1 + x + x^2 + x^3 + . . . ) $
= $\sum_{k=0}^n [x^k](1+x)^n * [x^{n-k}](1 + x + x^2 + x^3 + . . . ) $
= $\sum_{k=... | $\begin{array}\\
S(n)
&=[x^n] \frac{(1+x)^n}{1-x}
\qquad\text{want coefficient of }x^n\\
&=[x^n](1+x)^n\sum_{m=0}^{\infty} x^m
\qquad\text{expand } \dfrac1{1-x}\\
&=\sum_{k=0}^n [x^k](1+x)^n [x^{n-k}]\sum_{m=0}^{\infty} x^m
\qquad\text{for each }0\le k \le n
\text{ look at separate terms in the product}\\
&=\sum_{k=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
This question in my opinion is difficult and have tried many things. I tried to expand both sides but that will not help since I will not be a... | With variable subsitutions $a = x + y$, $b = x - y$, the inequality becomes $$(x^2 + y^2)^3 - 8 (x^3 + xy^2)(x^2 - y^2 - 2x) \geq 0$$ in the domain $x \geq 0$. Denote the LHS $f(x, y)$. As $f(x, y) = f(x, -y)$, it suffices to prove $f \geq 0$ in the closed first quadrant.
First note that $$f(x, 0) = x^6 - 8x^5 + 16 x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
The equation of the line cutting the circle at $A$ and $B$ given that $AB=\sqrt{2}$ Given the circle $C: 2x^2 + 2y^2 + 2x + 2y - 13 = 0$. Line $L$, with slope $m$ and passing through the point $P(0,2)$, cuts the circle at points $A$ and $B$ such that $AB=\sqrt{2}$.
*
*Find the equation of $L$.
*Find the equation o... | Let $x$ the distance from the centre of the circle $(x+1/2)^2+(y+1/2)^2=7$, I have $x=\sqrt{7-\frac{1}{2}}=\frac{\sqrt{13}}{\sqrt{2}}$. The distance of the line from the centre $P(-1/2,-1/2)$ has to be equal to $x$, so I obtain: $$\frac{|\frac{m}{2}-\frac{1}{2}-2|}{\sqrt{m^2-4}}=\frac{\sqrt{13}}{\sqrt{2}}$$ Solving for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there a simple pattern to memorize the sine of $0^\circ$, $15^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $75^\circ$, $90^\circ$? We know there is a nice pattern to memorize the sine of $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$ as follows.
\begin{align}
\sin 0^\circ &= \tfrac12\sqrt0\\
\sin 30^\circ... | Hmm, this pattern works:
$$
\sin 0^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{4}}, \\
\sin 15^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{3}}, \\
\color{gray}{\sin 22.5^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{2}},} \\
\sin 30^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{1}}, \\
\sin 45^{\circ} = \frac{1}{2}\sqrt{2\pm\sqrt{0}}, \\
\sin 60^{\circ} = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\int_0^1\frac{1-x}{(\ln x)(1+x)}\ dx=\ln\left(\frac2{\pi}\right)$? A friend asked me to prove
$$\int_0^1\frac{1-x}{(\ln x)(1+x)}\ dx=\ln\left(\frac2{\pi}\right)$$
and here is my approach:
\begin{align}
I&=\int_0^1\frac{1-x}{\ln x}\frac1{1+x}\ dx\\
&=\int_0^1\left(-\int_0^1x^y\ dy\right)\frac1{1+x}\ dx\\... | Note that $$\int_0^1 x^y dy=\frac{x-1}{\ln x}$$using this result our integral becomes $$\begin{aligned} I &=\int_0^1\frac{x-1}{(x+1) \ln x}\\&=\int_0^1\left(\frac{1}{x+1}\int_0^1x^ydy\right)dx \\&=\int_0^1\int_0^1\frac{x^y}{1+x}dy dx\\&=\int_0^1\underbrace{\int_0^1\frac{x^y}{1+x}dx }_{I_1}dy \end{aligned}$$ Since the i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Find limit $\lim_{x \to \infty} (\ln (x+1))^{2} - (\ln x)^2$ Find limit $\lim_{x \to \infty} (\ln (x+1))^{2} - (\ln x)^2$.
My attempt
$\lim_{x \to \infty} (\ln (x+1))^{2} - (\ln x)^2$.
$= \lim_{x \to \infty} (\ln(x+1) - \ln (x))(\ln (x+1) + ln(x))$
$=\lim_{x \to \infty} (\ln (\frac{x+1}{x})(\ln x(x+1))$
It's $0 \times... | Write
$$\log(x+1)=\log \left(x\left(1+\frac{1}{x}\right) \right)=\log(x)+\log \left(1+\frac{1}{x}\right) $$ and, by Taylor,
$$\log \left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right)$$
So,
$$A=\log ^2(x+1)-\log ^2(x)=\left(\log(x)+\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right) \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 4
} |
What is the solution $f:\mathbb{R^3}\to \mathbb{C}$ to the functional equation $f(A)f(B)=f(A+B+A\times B)$? Given $3$-vectors $A$ and $B$ and an unknown function $f:\mathbb{R^3}\to \mathbb{C}$, are there any solutions to this functional equation?
$$f(A)f(B)=f(A+B+A\times B)$$
apart from the trivial solutions $f(A)=1$ o... | The functional equation has no non-trivial solution.
For any $a, b \in \mathbb{R}^3$ such that $|a| = |b|$ and $a \ne -b$. Let
$$A = \frac{a+b}{4} + \frac{2}{|a+b|^2}(a \times b)
\quad\text{ and }\quad
B = \frac{a+b}{4} - \frac{2}{|a+b|^2}(a \times b)
$$
Notice
$$\begin{align}A + B + A \times B
&= \frac{a+b}{2} - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $ If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $
Attempt:
Notice that $$ \tan(a+b+c) = \frac{\tan(a+b) + \tan(c)}{1 - \tan(a+b)\tan(c)} $$... | Writing $$\tan c=\tan (\pi/2 -(a+b))=\frac 1{tan(a+b)}=\frac {1-\tan a\tan b}{\tan a +\tan b}$$
So we have $$\tan a\tan b+\tan a\tan c+\tan b\tan c=\tan a\tan b+ \frac {\tan a-\tan^2 a\tan b}{\tan a +\tan b}+ \frac {\tan b-\tan a\tan^2 b}{\tan a +\tan b}= \frac {\tan^2 a\tan b+\tan a\tan^2 b+\tan a-\tan^2 a\tan b+\tan ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find $k^{th}$ power of a square matrix I am trying to find the $A^{k}$, for all $k \geq 2$ of a matrix, \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}
My approach:
$A^{2}=\begin{pmatrix} a^2 & ab+b \\ 0 & 1 \end{pmatrix}$
$A^{3}=\begin{pmatrix} a^3 & a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{4}=\begin{pmatrix} a^4 ... | I'm partial to the SVD for this. The eigenvalues are $a$ and $1$ associated with eigenvectors $(1,0)$ and $(b,1-a)$ respectively.
So, we get
$$\begin{bmatrix}
a & b \\
0 & 1
\end{bmatrix}^n= \begin{bmatrix}
1 & b \\
0 & 1-a
\end{bmatrix} \begin{bmatrix}
a^n & 0 \\
0 & 1
\end{bmatrix} \begin{bmatrix}
1 & \frac{-b}{1-a} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
How to solve given differential equation $(x^3y^3+x^2y^2+xy+1)ydx+(x^3y^3-x^2y^2-xy+1)xdy=0$? The equation to solve is:
$$(x^3y^3+x^2y^2+xy+1)ydx+(x^3y^3-x^2y^2-xy+1)xdy=0$$
I tried putting $xy=t$ but that just gave me this:
$$\frac{t^3-t^2-t+1}{t^3+t^2+t+1}dt=\frac{dx}{x}$$
I suppose there must be some clever factorin... | Grouping terms, we get
$$(x^3y^3+1)(ydx+xdy) - (x^2y^2+xy)(xdy-ydx) = 0$$
Now substitute $t=xy$ and $s=\frac{y}{x}$
$$\implies dt = ydx+xdy \hspace{20 pt} x^2 ds = \frac{t}{s}ds = xdy - ydx$$
turning the differential equation into
$$s(t^3+1)dt - t^2(t+1)ds = 0$$
which is now separable and yields
$$\int \frac{ds}{s} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
About $e^{i z} = \cos z + i \sin z$ in Michael Spivak "Calculus 3rd Edition". I am reading "Calculus 3rd Edition" by Michael Spivak.
The author wrote as follows (p. 555):
Moreover, if we replace $z$ by $i z$ in the series for $e^z$, and make a rearrangement of the terms (justified by absolute convergence), something p... | Technically a rearrangement of a series $\sum_{n=0}^\infty a_n$ is understood to be any series $\sum_{n=0}^\infty a_{f(n)}$ where $f : \mathbb N_0 \to \mathbb N_0$ is a bijection. Here $\mathbb N_0$ is the set of nonnegative integers.
In that sense no rearrangement of $e^{iz} = \sum_{n=0}^\infty \frac{(iz)^n}{n!}$ can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How am I supposed to find the area of the shaded quadrilateral? In the diagram (which is not drawn to scale) the small triangles each have the area shown. Find the area of the shaded quadrilateral.
|
It can be deduced that the area $[FED] = 7\cdot 4/14=2$.
Furthermore, examine the ratios below,
$$\frac{[FDC]}{[BDC]}=\frac{9}{18}=\frac{\frac{a}{a+b} \frac{d}{c+d}[ABC] }{ \frac{b}{a+b}[ABC]} =\frac ab \frac{1}{\frac cd +1} \tag{1}$$
$$\frac{[FDB]}{[FCB]}=\frac{6}{21}=\frac{\frac{c}{c+d} \frac{b}{a+b} [ABC] }{ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
If $a+b+c=2$ where $0 \leq {a,b,c} \leq 1$ find the range of $\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}$
If $a+b+c=2$ where $0 < {a,b,c} <1$ find the range of $$\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}.$$
I have tried using AM-GM but I was not able to solve it . I also assumed the expression equal to... | By Show that $\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z} \ge 8$.
$$\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}\geq 8$$
where the equality holds for $a=b=c=2/3$.
Moreover, for $t\geq 3$, by letting $b(t)=c(t)=1-1/t\in (0,1)$ and $a(t)=2/t\in (0,1)$, then $a(t)+b(t)+c(t)=2$ and
$$g(t):=\frac{a(t)}{1-a(t)}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3359503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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What is $\alpha^{4} + \beta^{4} + \gamma^{4}$? If
$$ \alpha + \beta + \gamma = 14 $$
$$ \alpha^{2} + \beta^{2} + \gamma^{2} = 84 $$
$$ \alpha^{3} + \beta^{3} + \gamma^{3} = 584 $$
What is $\alpha^{4} + \beta^{4} + \gamma^{4}$?
Attempt:
Notice that
$$ \alpha + \beta + \gamma = 14 \implies \alpha^{2} + \beta^{2} + \ga... | Define $A,B,C$ such that $(x-\alpha)(x-\beta)(x-\gamma) = x^3 - Ax^2 + Bx - C$.
For any $n \ge 1$, let $p_n = \alpha^n + \beta^n + \gamma^n$. In terms of $p_n$, the question becomes:
Given $p_1 = 14, p_2 = 84, p_3 = 584$, what is $p_4$?
There is a bunch of Newton's identities connecting $p_n$ and the coefficients $A,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\sin 3\theta=1/2$ on the interval $[0,2\pi]$. I don't understand where some solutions came from. I am learning precalculus, and I understand how to obtain first two solutions, but I don't understand where did last four solutions came from:
All values of $\theta$ in the interval $[0,2\pi]$ that satisfy $\sin 3... | Consider the diagram below.
When does $\sin\theta = \sin\varphi$?
The sine of a directed angle is the $y$-coordinate of the point where its terminal side intersects the unit circle. Clearly, one solution is when $\theta = \varphi$. By symmetry, $\sin\theta = \sin(\pi - \theta)$, so $\varphi = \pi - \theta$ is anothe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Confused about dealing with Big $\mathcal{O}$ in functions For the following defined functions
$$f(x) = \frac{1}{x}+\frac{n-1}{x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)$$
and
$$y(x) = c\left(\frac{1}{x}+\frac{m-1}{x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\right)$$
where $m > n > 0$ and $c$ is large.
I want to bou... | Regarding your first question
I thought that $ f(x) − y(x) \leq \frac{1}{x} + \frac{n−1}{x^2} −c (\frac{1}{x} + \frac{m−1}{x^2}) $
, since we are talking about an approximation!
Is this true?
No. Consider for example $ f(x) = \frac{1}{x} + \frac{n-1}{x^2} + \frac{c+1}{x^3}, g(x) = c(\frac{1}{x} + \frac{m-1}{x^2} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to solve $\ln(x^2)=x$? I am struggling to solve the equation $\ln(x^2)=x.$ I can't figure out how to isolate the variable. Exponentiating both sides leaves me with $x=e^{\frac{x}{2}}.$ I would really appreciate any guidance, or first steps.
Also, isn't one of the properties of logs that $\ln(x^2)=2\ln(x)?$ How can ... | Consider that you look for the zero's of function
$$f(x)=\log(x^2)-x$$ for which
$$f'(x)=\frac 2 x-1 \qquad \text{and} \qquad f''(x)=-\frac 2 {x^2} < 0 \qquad \forall x$$
The first derivative cancels at $x=2$ and $f(2)=\log(4)-2 <0$. By the second derivative test $x=2$ corresponds to a maximum; then, no solution for $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Solving $ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$ Evaluate:
$$ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$$
My approach :
$$\lim_{x\to0}{\frac{1}{x^2}-\frac{\cos^2(x)}{\sin^2(x)}}$$
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} $$
Using $$\lim_{x\to0}\frac{\sin^2(x)}{x^2}=1 $$
$$ \lim_{x\to0}\frac{\sin^2(x)-x^... | $$L=\lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\cot^2[x] \right)= \frac{1}{x^2}-\frac{1}{(x+x^3/3+O(x^5))^2}$$ Let us use $(1+z)^{\nu} \approx 1+\nu z, ~if ~$z$ <<1.$
$$\Rightarrow L=\frac{1}{x^2}-\frac{1}{x^2} (1+x^2/3+O(x^4)^{-2}=\frac{1}{x^2}-\frac{1}{x^2} (1-2x^2/3+O(x^4))= \lim_{x \rightarrow 0} \frac{2}{3}+O(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the coefficient of $x^{19}$ in the generating function of $f(x) = \frac{(1-x^{5})^5}{(1-x)^{10}}$
Find the coefficient of $x^{19}$ in the generating function development of $f(x) = \dfrac{(1-x^{5})^5}{(1-x)^{10}}$
My Attempt:
Let $\mathbf{(I)} \ (1-x^5){^5}$, then $(1-x^5){^5} = \sum_{k=0}^{5} \binom{5}{k} \lef... | You are almost correct! Since $\binom{5}{3}=10$ (not $15$), it should be
$$[x^{19}]f(x) = \binom{28}{9} -5\binom{23}{9} +10\binom{18}{9} -\color{red}{10}\binom{13}{9}=3300000.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Roots of x inside square root In finding the roots of
$ \sqrt{6-4x-x^2}=x+4$
I get that the roots of $x$ are $-5$ and $-1$.
However first I need to take into account that:
*
*$x+4 \ge 0$
and
*$6-4x-x^2 \ge 0$.
Considering the second equation gives me:
$x \ge -2+ \sqrt{10}$,
would the roots of $x$ I found in the f... | Let $x+4 \ge 0~~~(1)$ and for tje reality of the roots $$6-4x-x^2 \ge 0~~~~~~(2) \Rightarrow x^2+4x-6 \le 0 \Rightarrow -2-\sqrt{10} \le x \le -2+\sqrt{10}.$$ So that we can square the given equation, then we have
$$x^2+6x+5 \Rightarrow x=-5, -1$$. As per the conditiona (1,2), $x=-1$ is the correct root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to identify that a polynomial has factor $(x+y)^2$ About integer polynomial $f(x,y)$, conclusions such as "If $f(x,x)=0$, then $y-x$ is a factor" "If $f(x,wx)=0$, then $x^2+xy+y^2$" is a factor" are easy.
But what about $(x-y)^2$? How to verify, or identify that $f(x,y)$ has a factor $(x+y)^2$? Does differentiation... | Let $t:=x+y$.
Then we must have
$$P(x,t-x)=t^2Q(x,t-x).$$
E.g.
$$P(x,y)=x^3+x^2-xy^2+2xy-y^3+y^2+yx^2\to P(x,t-x)=2xt^2-t^3+t^2.$$
You can simply check that the polynomial has a double root at $t=0$,
$$\left.P(x,t-x)\right|_{t=0}=P(x,-x)=0,$$
$$\left.\frac{\partial P}{\partial y}(x,t-x)\right|_{t=0}=\frac{\partial P}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I have troubles with proving the induction step: $\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$ Let's assume that the unequality is right if $n = k$, that is:
$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k-1)}{2k} ≤ \frac{1}{\sqrt{2k+1}}$
Let's prove that t... | Using the induction hypothesis:
$$\frac{2k+1}{2k+2} ≤ \frac{\sqrt{2k+1}}{\sqrt{2k+3}}$$
Now let $u = 2k+2$:
$$\frac{u-1}{u} ≤ \frac{\sqrt{u-1}}{\sqrt{u+1}}$$
For $u ≥ 1$ or $2k + 2 ≥ 1$ which implies $k ≥ -\frac{1}{2}$, both sides are positive. So we can square both sides:
$$\frac{u^2-2u+1}{u^2} ≤ \frac{u-1}{u+1}$$
$$u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find the sum to n terms $$S=1^2+3^2+6^2+10^2+15^2+.......$$
My attempt is as follows:
$$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$
$$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$
$$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(n^4+n^2+2\cdot n^3\right)$$
Now to solve this one has to calculate $\sum_{n=1}^{n}n^4$ which will b... | You may use the hockey-stick identity, like in the computation of $\sum_{n=1}^{N}n^m$. We have
$$ \binom{n}{2}^2 = 6\binom{n}{4}+12\binom{n}{3}+7\binom{n}{2}+\binom{n}{1} $$
hence
$$\begin{eqnarray*} \sum_{n=1}^{N}\binom{n}{2}^2 &=& 6\binom{N+1}{5}+12\binom{N+1}{4}+7\binom{N+1}{3}+\binom{N+1}{2}\\&=&\color{red}{\frac{N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Convergence of $\sum_{n=1}^{\infty} \log (n\sin\frac{1}{n})$ Convergence of $\sum_{n=1}^{\infty} \log (n\sin\frac{1}{n})$
First, we have
$$ n \sin \frac{1}{n} = \frac{\sin \frac{1}{n}}{\frac{1}{n}} < 1 $$
for all positive integers $n$. (Since $\frac{\sin x}{x} < 1 \forall x > 0$.) Therefore,
$$ \sum_{n=1}^{\infty}\le... | We have $\sin \dfrac{1}{n} = \dfrac{1}{n}-\dfrac{1}{3! n^3} + o\left(\dfrac{1}{n^3}\right)$, then :
$$\log\left( n \sin \dfrac{1}{n} \right) = \log\left( 1-\dfrac{1}{3! n^2} + o\left(\dfrac{1}{n^2}\right) \right) = -\dfrac{1}{3! n^2} + o\left(\dfrac{1}{n^2}\right) + \mathcal{O}\left( \dfrac{1}{n^4} \right)$$
Then :
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3377929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the inequality is true Here is a question that I need to prove
Prove that for $a, b \geq 0$
$$a^8+b^8\geq a^3b^5+a^5b^3$$
So far I have managed to simplify to
$$(a^3-b^3)(a^5-b^5)\geq 0$$
| If $a \ge b$, then $a^3 \ge b^3$ and $a^5 \ge b^5$, which implies $a^3 - b^3 \ge 0$ and $a^5 - b^5 \ge 0$. Since a positive times a positive is positive (or $0$ times anything is $0$), $$(a^3 - b^3)(a^5 -b^5) \ge 0.$$
If instead $a < b$, then
$a^3 < b^3$ and $a^5 < b^5$, which implies $a^3 - b^3 < 0$ and $a^5 - b^5 < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
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Calculating $\iiint_E\sqrt{3x^2+3z^2}\,dV$
Evaluate the following integral :$$\iiint_E\sqrt{3x^2+3z^2}\,dV$$where $E$ is the solid bounded by $y=2x^2+2z^2$ and the plane $y=8$.
I put $x=r\cos\theta$ and $z=r\sin\theta$ so, $y=2r^2\implies8\leq y\leq2r^2$. Finding limits of $r$ I got $0\leq r\leq2\csc\theta$ and for $... | We do not have to convert to other coordinate system. Since for a fixed value of y, the cross section is a circle with radius of $\sqrt{\frac{y}{2}}$. Then the integral becomes
$$\int_{y=0}^{8} \int_{x=-\sqrt{\frac{y}{2}}}^{\sqrt{\frac{y}{2}}} \int_{z=-\sqrt{\frac{y}{2}-x^2}}^{\sqrt{\frac{y}{2}-x^2}} dz dx dy$$
$$= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$? Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$
I am not sure how to do it. I thought of squaring both sides and moving the $2$ to the other side, then squaring again etc. But ... | Let $x$ be the given iterated radical. Render
$x^2=2+\sqrt{3+\sqrt{4+\sqrt{5+\sqrt{6+...}}}}$
Now multiply $x$ by $\sqrt{3/2}$:
$(\sqrt{3/2})x=\sqrt{3+(3/2)\sqrt{4+\sqrt{5+\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+(9/4)\sqrt{5+\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+\sqrt{(405/16)+(81/16)\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+\sqrt{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Find the coefficient of $ x^6 $in the given expansion. Find the coefficient of $x^6$ in $\left[(1+x)(1+x^2)^2(1+x^3)^3 \cdots (1+x^n)^n\right]$.
Expansion
$$\left[\left(1+\binom11x \right)\left(1+\binom21x + \binom22x^2 \right)\left(1 +\binom31x+\binom32x^2+\binom33x^3 \right)\cdots \left(1 + \binom n1 x + \binom n2x^... | Here we give a supplement regarding the cases $1\leq n\leq 5$. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.
We obtain
Cases $n=1,2$:
\begin{align*}
[x^6](1+x)=[x^6](1+x)(1+x^2)^2\color{blue}{=0}
\end{align*}
Case $n=3$:
\begin{align*}
\color{blue}{[x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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$\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational
Prove that $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational
My attempt:-
Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational, then for some $x\in\mathbb{Q}$ we have $$\sqrt{6}-\sqrt{2}-\sqrt{3}=x$$
Rewriting this equation as $$\sqrt{6}-x=\sqrt{2}+\sqrt{3}$$
and now squaring this ... | In your proof, after $6-2x\sqrt{6}+x^2=5+2\sqrt{6}$ we have that
$$x^2+1=2(x+1)\sqrt{6}$$
If $x=-1$ then, from the above equation, it follows that $2=0$. Therefore $x$ is a rational number different from $-1$. After dividing by $2(x+1)\not=0$ we get
$$\sqrt{6}=\frac{x^2+1}{2(x+1)}\in \mathbb{Q}.$$
Contradiction! Hence ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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Trigonometric systems of equations I seek to express $DE$ and $UI$ and $IE$ in terms of $\theta$:
$$\sec { \theta } =DE+IE+UI \tag{1.1}\label{myeqone}$$
$$\frac { 1 }{ 4 } =UI(UI+IE) \tag{2.1}\label{myeqtwo}$$
$$(\frac { 1 }{ 2 } -\tan { \theta } )^{ 2 }=DE(DE+IE) \tag{3.1}\label{myeqthree}$$
I rewrote $(2)$ and $(3)... | For notational brevity, let $x=DE$, $y=IE$ and $z=UI$ and work with the set below,
$$x+y+z = \sec \theta \tag{1}$$
$$z(z+y) =\frac 14\tag{2}$$
$$x(x+y)=\left(\frac 12 -\tan\theta \right)^2\tag{3}$$
Take (3)-(2) and use (1),
$$x-z=\sin\theta(\tan\theta-1)\tag{4}$$
From (3)+(2),
$$x^2+z^2+y(x+z)=\left(\frac 12 -\tan\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3387557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality involving Harmonic mean Let $a, b \ge 0 $ and $x,y > 1$ Show $\frac{1}{1/x+1/y}(a+b) \le \max(ax,by)$
If $a = b = 0$ then this is clear so assume not both are zero.
This seems to be related to the harmonic mean but I am not quite getting it.
$\frac{1}{1/x+1/y}(a+b) = \frac{xy}{x+y}(a+b) = \frac{1}{2}H(x,y)(... | Cauchy Schwarz says for $a,b\in\mathbb{R}$ and $x,y\gt0$,
$$
(a+b)^2\le\left(a^2x+b^2y\right)\left(\frac1x+\frac1y\right)
$$
which gives, for $a+b\ge0$,
$$
\begin{align}
\frac1{\frac1x+\frac1y}(a+b)
&\le\frac{a(ax)+b(by)}{a+b}\\
&\le\max(ax,by)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3387697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finite summation $\sum _{k=0}^{n-1} \csc ^{2 m}\left(\frac{\pi (2 k+1)}{2 n}+y\right)$ How to prove
$$\sum _{k=0}^{n-1} \csc ^4\left(\frac{\pi (2 k+1)}{2 n}+y\right)=(n \sec (n y))^4-\frac{2}{3} \left(n^2-1\right) (n \sec (n y))^2$$
Moreover, is there a closed-form for higher order summations? Any help will be appreci... | Like How to prove: $11=10^{12}+10^{7}-45\sum_{n=1}^{999}\csc^4\frac{n\pi}{1000}$
Question regarding $f(n)=\cot^2\left(\frac\pi n\right)+\cot^2\left(\frac{2\pi}n\right)+\cdots+\cot^2\left(\frac{(n-1)\pi}n\right)$
Observe that $n\left(\frac{\pi (2 k+1)}{2 n}+y\right)=ny+k\pi+\dfrac\pi2$
As $\tan nx=\dfrac{\binom n1\cot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3387820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Reducible polynomials similar to Sophie Germain's identity Are there known any other simple nontrivial polynomials in two or more variables that can be factored like the one in Sophie Germain's identity?
$$x^4+4y^4=(x^2-2 x y+2 y^2) (x^2+2 x y+2 y^2)$$
Besides the well known (for $n \in \mathbb {Z} ^{+}$):
$$x^n-y^n=(x... | More formulas similar to Sophie Germain's identity can be derived from Aurifeuillean factorizations such as
$$x^6+27y^6=(x^2+3y^2)(x^2-3xy+3y^2)(x^2+3xy+3y^2),$$
$$x^{10}-3125y^{10}=(x^2-5y^2)(x^4-5x^3 y+15x^2 y^2-25x y^3+25y^4)(x^4+5x^3 y+15x^2 y^2+25x y^3+25y^4).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proof that inequality $|a| < |b-c+d|, |b| < |a-c+d|, |c| < |a-b+d|, |d|< |a-b+c| $ does not have solution in $\mathbb{R}$
Proof that inequality $$|a| < |b-c+d|, \\|b| < |a-c+d|, \\|c| < |a-b+d|,\\ |d|< |a-b+c| $$ does not have solution in $\mathbb{R}$.
Solution:
$$|a| < |b-c+d| \leq |b|+|c|+|d|$$
$$|b| < |a-c+d| \le... | There is no mistake, but the inequality you obtained is not useful!
Note that
$$|a| < |b-c+d|\Leftrightarrow (b−c+d)^2−a^2>0$$
and similarly we have that $(a-c+d)^2-b^2>0$, $(a-b+d)^2-c^2>0$, and $(a-b+c)^2-d^2>0$. Then by taking the their product, and by recalling that
$x^2-y^2=(x+y)(x-y)$, we find
$$\begin{align}((... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Solve $x\equiv 1\pmod2$, $x\equiv 2\pmod3$, $x\equiv 3\pmod4$, $x\equiv 4\pmod5$, $x\equiv 5\pmod6$ and $x\equiv 0\pmod7$ $$\begin{align*}
x&\equiv 1\pmod2\\
x&\equiv 2\pmod3\\
x&\equiv 3\pmod4\\
x& \equiv 4\pmod5\\
x&\equiv 5\pmod6\\
x&\equiv 0\pmod7\\
\end{align*}$$
So the solution says we can eliminate $x\equiv 5(\b... | Because of the special case
$$x\equiv -1\pmod {2,3,4,5,6}$$
we have a shortcut for handling these.
$$x+1\equiv 0\pmod {2,3,4,5,6}$$
It is way easier to work with $x+1$ here.
This gives us
$$x+1 \equiv 0 \pmod {LCM(2,3,4,5,6)=60}$$
Now we need to incorporate
$$x+1 \equiv 1 \pmod 7 $$
Which is easy, as $7$ is prime. List... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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If $\omega =\cos 40 + i\sin 40$ then $|\omega + 2\omega ^2 + 3\omega ^3....9\omega ^9|^{-1}$ If $\omega =\cos 40 + i\sin 40$ then $|\omega + 2\omega ^2 + 3\omega ^3....9\omega ^9|^{-1}$
All angles are in degrees
I will be writing $\omega$ as w to make it easier to type
From the value of $w$ it is clear that $w^9=1$
So... | If $S=\omega + 2\omega ^2 + 3\omega ^3+\cdots+9\omega ^9$
$(1-\omega)S=\omega + \omega ^2 + \omega ^3+\cdots+\omega ^9-9\omega ^{10}$
$=1+\omega + \omega ^2 + \omega ^3+\cdots+\omega^8-9w$ as $\omega ^9=1$
$=\dfrac{1-\omega^9}{1-w}-9w=-9w$
$$S=\dfrac{-9\omega}{1-\omega}=\dfrac9{1-\omega^{-1}}$$
$1-\omega^{-1}=1-(\cos2x... | {
"language": "en",
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"source": "stackexchange",
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Triple integral for the volume between $z=2$ and the top of the sphere $x^2 + y^2 + z^2 = 9$ Apparently we do something like this:
$z=2$, so $x^2 + y^2 + z^2 = 9$ tells us that $x^2+y^2=5$, and so our integral is:
$$\int_{-\sqrt{5}}^{5}\int_{-\sqrt{5-x^2}}^{\sqrt{5-x^2}}\int_2^{\sqrt{9-x^2-y^2}} 1\,dz\,dy\,dx$$
That ... | You want to project your solid onto the $xy$-plane first (Hint: look at the last two variables where you integrate). What does it look like?
Essentially, when $$z=2,$$ the solid has a region $R$ bounded by $$x^2+y^2=5,$$
a circle, which you pointed out already. From this region, we start to look at $y$ first and wh... | {
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Compute the value of the $30$th derivative of the function $g(x) = \sinh{\left(x^4\right)}$ at the origin, i.e. $g^{(30)}(0)$. Compute the value of the $30$th derivative of the function $g(x) = \sinh{\left(x^4\right)}$ at the origin, i.e. $g^{(30)}(0)$.
So we have the (Macluarin) series
\begin{equation*}
\sinh{(x)} = \... | Your second formula
$$g(x)=\sum_{n=0}^\infty{x^{8n+4}\over(2n+1)!}$$
is still correct. Now the $m^{\rm th}$ derivative of any power series $\sum_{k=0}^\infty a_kx^k$ at $0$ is equal to the coefficient $a_m$ belonging to $x^m$, times a combinatorial factor. In your case $m=30$. Since $8n+4\ne30$ for all $n\geq0$ it foll... | {
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"source": "stackexchange",
"question_score": "1",
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Under what conditions is the following matrix diagonalisable Let
$$A=\begin{bmatrix} 0&1/2&1/2 \\ 0&0&1/2 \\ 1&1/2&0 \end{bmatrix}.$$
Could anyone advise me on how to show $M=(1-m)A+mS,$ where $S_{ij}=\frac{1}{3},$ is not diagonalisable for $0\leq m < 1 \ ?$ Do we compute the characteristic polynomial of $M$, show it ... | The matrix $M$ is
$$M=\frac{1}{6}\left(
\begin{array}{ccc}
2 m & 3-m & 3-m \\
2 m & 2 m & 3-m \\
6-4 m & 3-m & 2 m \\
\end{array}
\right).$$
Now compute the characteristic polynomial
$$f(x)=\frac14(m-1-2x)^2 (x-1).$$
This shows that $M$ has eigenvalues $1$ and $\frac{m-1}{2}$. Now check that $\dim\ker\left(M-\left(\... | {
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"source": "stackexchange",
"question_score": "1",
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Showing continuity of $g(x, y)$ Let $g : \mathbb{R}^2 \to \mathbb{R}$. Let the domain be $y^2 \leq \frac{x}{3}$.
$g(x, y) = \begin{cases}
\frac{x^3}{y^2} \left (\sqrt{x} - \sqrt{y^2 + x} \right ) & y \neq 0\\
0 & y = 0
\end{cases} $
I want to show $g$ is continuous at $(0, 0)$.
We have $|g(x, y) - g(0,0)| = \frac{x^3}... | Write your term as $$\frac{x^3}{y^2}\frac{x+y^2-x}{\sqrt{x}+\sqrt{y^2+x}}$$
| {
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"source": "stackexchange",
"question_score": "1",
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Prove inequalities $\frac {36}{25} \le A(a) < 2$ Given the expression,
$$ A(a) = \frac{\left( 1 +a +\frac 1a\right)^2 }{\left(\frac12+a+a^2\right)\left(\frac12+\frac 1a + \frac{1}{a^2}\right)} $$
with $a > 0$.
Prove the following inequalities:
$$\frac {36}{25} \le A(a) < 2$$
Note that the bounds are rather tight. I en... | Here's a simple approach: let $b = 1 + a + 1/a$, so $b \geq 3$, and
$$\frac{1}{A(a)} = \frac{(ab - 1/2)(b/a - 1/2)}{b^2} = 1 - \frac{a + 1/a}{2b} + \frac{1}{4b^2} = 1 - \frac{b-1}{2b} + \frac{1}{4b^2} = \frac{1}{2} + \frac{1}{2b} + \frac{1}{4b^2}.$$
As $a$ varies over $(0, \infty)$, $b$ varies over $[3, \infty)$, and i... | {
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"source": "stackexchange",
"question_score": "1",
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Can anyone prove this function is concave? Can anyone help me prove the following function $$f(x) = \frac{ax}{1-x^a}-\frac{x}{1-x}$$ is concave for any parameter $a \geq 1$ when $0<x<1$ ?
I have tried to prove its second derivative is negative, but it becomes another difficult problem.
| AmerYR calculated $$f''(x)=\frac{a^2x^{a-1}\left(1+a+x^a(a-1)\right)}{(1-x^a)^3} - \frac{2}{(1-x)^3} \, .$$
As you noticed $f''(x)=0$ at $a=1$. So it suffices to show that $\partial_a f''(x) \leq 0$. In the comments I gave the derivative with respect to $a$ by $${\frac { \left( \left( \left( {a}^{2}-a \right) \ln \l... | {
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"source": "stackexchange",
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Maximum value of $a+b+c$ in an inequality Given that $a$, $b$ and $c$ are real positive numbers, find the maximum possible value of $a+b+c$, if
$$a^2+b^2+c^2+ab+ac+bc\le1.$$
From the AM-GM theorem, I have
$$a^2+b^2+c^2+ab+ac+bc\geq 6\sqrt[6]{a^4b^4c^4} = 6\sqrt[3]{a^2b^2c^2} \\
6\sqrt[3]{a^2b^2c^2} \le1 \\
a^2b^2c^2 \l... | $$ \begin{align}& a^2+b^2+c^2+ab+ac+bc\le1 \\
& (a+b+c)^2 -(ab+bc+ac) \le 1 \\
& (a+b+ c) \le \sqrt{1+(ab+bc+ac)} \quad \quad \color{red}{\text{ (1.)}} \end{align}$$
Also note that :
$$\frac {(ab+bc+ac)}{3} \ge \sqrt[3]{a^2b^2c^2}$$
Assuming you are correct :
$${(ab+bc+ac)} \ge 3 \cdot\frac16 \implies (ab+bc+ac) \ge... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int \frac{dx}{1-\sin x+\cos x}$ Solve :$$ \int \frac{dx}{1-\sin x+\cos x} $$
I tried:
$$\int \frac{\text{dx}}{(1+\cos x)-\sin x} \times \frac{(1+\cos x)+\sin x}{(1+\cos x)+ \sin x} dx=\int \frac{1+\cos x+\sin x}{(1+\cos x)^2-(\sin x)^2}$$ $$=\int \frac{1+\cos x+\sin x}{1-\sin x^2+2 \cos x+\cos x^2}=\int \frac{1+\cos ... | Start by multiplying the top and bottom by $\sec(x)$, which will result in the following:
$$\int{\frac{\sec x}{\sec x-\tan x+1}}$$
Now we can perform a $u-$substitution for the following:
$$u=(\sec x-\tan x)\\du=(\sec x)(-(\sec x-\tan x))dx$$
The resulting integral is then:
$$\int{\frac{-1}{(u+1)(u)}}$$
We can now make... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
The sum of the series
$$\frac{1}{\log_24}+\frac{1}{\log_44} + \frac{1}{\log_84}.....\frac{1}{\log_{2^n}4}$$
MY SOLUTION
We can write it as
$$\frac{\log2}{\log4} + \frac{\log4}{\log4} + ....\frac{\log 2^n}{\log4}$$
$$=\frac{1}{\log4}\left[\log 2 + \log 4....\log 2^n\right]$$
$$=\frac{1}{\log 4}\left [\log(2.4.8....2^... | You are almost right, we have that
$$\dots=\frac{1}{\log 4}\left [\log(2\cdot 4\cdot 8\cdot \ldots \cdot 2^n)\right]=\frac{1}{\log 4}\left [\log(2\cdot 2^2\cdot 2^3\cdot \ldots \cdot 2^n)\right]=$$
$$=\frac{1}{\log 4}\left [\log(2^{1+2+3+\dots+n}
)\right]=\frac{1}{\log 4}\left [\log(2^{\frac{n(n+1)}2})\right]=\frac{n(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402896",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Do the coefficients of these polynomials alternate in sign? Define polynomials $p_i(x)$ by the recurrence
\begin{align}
p_0(x)&=0 \\
p_1(x)&=1 \\
p_{2i}(x)&=p_{2i-1}(x)-p_{2i-2}(x) \\
p_{2i+1}(x)&=xp_{2i}(x)-p_{2i-1}(x) \\
\end{align}
The first few are given by
\begin{align}
p_0(x)&=0\\
p_1(x)&=1\\
p_2(x)&=1\\
p_3(x)&=... | Lord Shark's answer is good (but might contain some subtle mistakes) ... I reckon ...
\begin{eqnarray*}
p_{2i}(x) = \sum_{j=0}^{i-1} \binom{2i-j-1}{j} (-1)^j x^{i-j-1} \\
p_{2i-1}(x) = \sum_{j=0}^{i-1} \binom{2i-j-2}{j} (-1)^j x^{i-j-1}. \\
\end{eqnarray*}
Proof of the even formula ...
\begin{eqnarray*}
&p_{2i-1}(x) -... | {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Evaluate $\int _0^1\int _0^1\frac{\log \left(x^2+y^2\right)}{\sqrt{x+y}}dydx$ How to show
$$\int _0^1\int _0^1\frac{\log \left(x^2+y^2\right)}{\sqrt{x+y}}d\sigma=-\frac{128 a}{9}+\frac{16}{3} \sqrt{2 a} \tan ^{-1}\left(\sqrt{\frac{a}{2}}\right)+\frac{8}{3} \sqrt{2} \log (2)$$
Where $a=\sqrt{2}-1$? Any kind of help will... | One can exploit a symmetry and split the integral in half along the line $y=x$ then use the following modified polar coordinates:
$$x = s^{\frac{2}{3}}\cos\theta \hspace{10 pt} y = s^{\frac{2}{3}}\sin\theta$$
$$\implies 2\cdot \frac{2}{3}\int_0^{\frac{\pi}{4}} \int_0^{\sec^{\frac{3}{2}}\theta} \frac{\log\left(s^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I solve a problem with term $a^{n} + b^{n}$?
Given two non-zero numbers $x$ and $y$ such that $x^{2} + xy + y^{2} = 0$.
Find the value of
$$\left(\frac{x}{x + y}\right)^{2013} + \left(\frac{y}{x + y}\right)^{2013}$$.
I found out that $(x + y)^2 = xy$ and I'm stuck at $\frac{x^{2013} + y^{2013}}{(x + y)^{2013}}... | Let $a_n = \left(\dfrac{x}{x + y}\right)^{n} + \left(\dfrac{y}{x + y}\right)^{n}$.
Since
$$
\dfrac{x}{x + y}+\dfrac{y}{x + y}=1, \quad
\dfrac{x}{x + y}\cdot\dfrac{y}{x + y}=1
$$
they are the roots of $t^2=t-1$ and so
we get
$$
a_{n+2} = a_{n+1}-a_n,
\quad
a_0=2,
\quad a_1=1
$$
This sequence is periodic of period $6$:
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to simplify $\frac{2+\sqrt{3}}{6+4\sqrt{3}}$ among solving a problem I need to simplify: $$\frac{2+\sqrt{3}}{6+4\sqrt{3}}$$
I know it is $\frac{1}{2\sqrt{3}}$ but I dont know how to show that in mathematic way for example I can write $$\frac{2+\sqrt{3}}{2\sqrt{3}(\sqrt{3}+2)}$$ but still not obvious.Is there better... | We have that
$$\frac{2+\sqrt{3}}{6+4\sqrt{3}}=\frac{2+\sqrt{3}}{6+4\sqrt{3}}\cdot \frac{6-4\sqrt{3}}{6-4\sqrt{3}}=\frac{-2\sqrt{3}}{-12}=\frac{\sqrt 3}6$$
or by your idea
$$\frac{2+\sqrt{3}}{6+4\sqrt{3}}=\frac{2+\sqrt{3}}{2\sqrt{3}(\sqrt{3}+2)}=\frac1{2\sqrt{3}}=\frac1{2\sqrt{3}}\cdot\frac{2\sqrt{3}}{2\sqrt{3}}=\frac{\... | {
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"source": "stackexchange",
"question_score": "2",
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$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}$ Is this solution correct? Just want to double check whether all my operations are legal. The result seems to be correct but I want to make sure I didn't make any mistakes.
$\require{cancel}$
$$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=$$
$$=\lim_{x \... | L'Hopital's rule also works as the limit is in the form $\frac{0}{0}$:
$$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}
=\lim_{x \to 0} \frac{11x^{10}-6x+\cos x}{e^x + \sin x}
= \frac{0-0+1}{1 + 0}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409502",
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"source": "stackexchange",
"question_score": "3",
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Proving limits involving floor function.
Let $\theta $ be real, and $0<\theta<1$, define $g_n = 0$ if $\lfloor n\theta\rfloor=\lfloor(n-1)\theta \rfloor$ and $g_n=1$ otherwise. For example, take $\theta = \frac{3}{4}$, then $g_2 =1$, since $\left\lfloor \frac{3}{2}\right\rfloor = 1 \ne 0 =\left\lfloor \frac{3}{4} \rig... | Hint: Show that $g_n=\lfloor n\theta \rfloor-\lfloor (n-1)\theta \rfloor$, so that $\sum_{i=1}^n g_n=\lfloor n\theta\rfloor$ by a telescoping sum argument. So what you need to do is show that
$$\lim_{n\to\infty}\frac{\lfloor n\theta\rfloor}{n}=\theta$$
which shouldn't be too difficult for you since that is a series o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int \frac{1}{(x^2-4)^2}dx$ Calculate:
$$\int \frac{1}{(x^2-4)^2}dx.$$
I tried Partial Fractions method first I write:
$$\frac{1}{(x^2-4)^2}=\frac{A}{X-2}+\frac{Bx+C}{(x-2)^2}+\frac{D}{x+2}+\frac{Ex+F}{(x+2)^2}.$$
We have:
$$A(x-2)(x+2)^2+(Bx+C)(x+2)^2+D(x+2)(x-2)^2+(Ex+F)(x-2)^2=1.$$
$$(A+B+D+E)x^3+(4A-2A+4B+C-4D+2D... | HINT:
$x=2\text{sec}(\theta)$, $dx=2\text{sec}(\theta)\text{tan}(\theta)d\theta$ then:
$\int \frac{1}{(x^2-4)^2}dx=\int \frac{2\text{sec}(\theta)\text{tan}(\theta)}{(4\text{sec}^2(\theta)-4)^2}d\theta=\frac{1}{8}\int \frac{\text{sec}(\theta)\text{tan}(\theta)}{\text{tan}^4(\theta)}d\theta$
$$=\frac{1}{8}\int \text{ctg}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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what is the limit as $x$ approaches infinity of $x-\sqrt{x^2-x+2}$ $$
\lim_{x \to \infty} \left( x - \sqrt{x^2 - x +2 } \right)
$$
I've tried rationalizing the expression but after repeated applications of L'Hospital's rule, it doesn't feel like I'm getting anywhere.
| Yes rationalize is a good idea, indeed we obtain
$$x - \sqrt{x^2 - x +2 }=x - \sqrt{x^2 - x +2 }\cdot\frac{x + \sqrt{x^2 - x +2 }}{x + \sqrt{x^2 + x +2 }}=\frac{x^2-x^2+x-2}{x+ \sqrt{x^2 + x +2 }}$$
$$=\frac{x-2}{x+ \sqrt{x^2 + x +2 }}=\frac x x\frac{1-\frac 2x}{1+ \sqrt{1 + \frac1x +\frac2{x^2} }}\to \frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
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Complex numbers such that $(z-4)^4=-1$
Find all complex numbers $z$ in Cartesian form such that $(z-4)^4=-1$.
This is what I did: I squared root both side making $-1$ into $i$ and $(z-4)^2$ and I took square of both side again and had $i^2$ which is equal to $-1$ and the other side $(z-1)$. At the end I had $z=0$.
|
This is what I did: I squared root both side making $-1$ into $i$ and $(z-4)^2$
This is on the right track. Except, when you square root both sides, you need to do $\pm$. So instead of getting $i$ you get $\pm i$:
$$
(z - 4)^2 = \pm i
$$
and I took square of both side again and had $i^2$ which is equal to $-1$ and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413816",
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"source": "stackexchange",
"question_score": "2",
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Is $(n + 1)^n = n^n + (n + 1)$? I'm trying to solve my homework, but I don't know if I'm on the right way.
I have
$$\frac{(n+1)^n(n+1)}{2(2n+1)n^n}$$
and I want to cancel $(n+1)^n$ with $n^n$.
So I have a sequence of series and try to check the convergence with Ratio Test.
$$\lim_{n\to \infty}\frac{(n+1)^{n+2}}{n^n(2... | Always try with trivial cases at first, notably for $n=1$ we have
$$(n+1)^n=n^n+(n+1) \implies 2=1+2$$
Edit
Since the original problem seems related to the ratio test for convergence, we have that
$$\frac{(n+1)^n(n+1)}{2(2n+1)n^n}=\frac{(n+1)^n}{n^n}\frac{n+1}{4n+2}=\left(1+\frac1n\right)^n\frac{n+1}{4n+2}$$
then take... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Probability Counting Problem Confusion Im having difficulty understanding how to do the following problem using Combinations.
Q: An urn contains 2 red balls, 4 blue balls and 6 green balls. If three balls are drawn at random one at a time, without replacement, what is the probability that the first and last are the sam... | The book's answer can ignore order. Your answer cannot. You need specifically for the first and last ball to be the same. For the book's answer, they are only looking at the first and last ball, and they do not care which order they pull them so long as they are the same color. Once you add in the third ball, order mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Alternative proof of reduction formula for $\int {dx \over (a\sin x + b\cos x)^n }$
The problem i was working on is to prove:
$$
\begin{align}
J_n &= \int {dx \over (a\sin x + b\cos x)^n } \\
&={1\over (n-1)(a^2 + b^2)}\left({b\sin x - a\cos x\over (a\sin x + b\cos x)^{n-1}} + (n-2)J_{n-2}\right)
\end{align}
$$
Wh... | Note that,
$$\left({a\cos x - b\sin x\over a\sin x + b\cos x} \right)'
=-{a^2+b^2\over (a\sin x + b\cos x)^2}
$$
and apply the integration-by-parts,
$$J_n = \int {1 \over (a\sin x + b\cos x)^n } $$
$$ = -\frac{1}{a^2+b^2}\int {1 \over (a\sin x + b\cos x)^{n-2} } d\left({a\cos x - b\sin x\over a\sin x + b\cos x} \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Centroid of area enclosed by $x^n$ and $x^{1/n}$. I am interested in finding the centroid of the surface enclosed by the lines $y=x^n$ and $y=x^{1/n}$ where $n \in \mathbb N, n >1$. The exercise, gives the following sketch.
I know that in order to find the centroid, we need to calculate its $x$ and $y$ coordinates, na... | $$\iint_A x\mathrm{d}A = \int_0^1 \left( \int_{y^n}^{y^{1/n}} x \mathrm{d}x \right) \mathrm{d}y$$
$$= \int_0^1 \left[\frac {x^2}2 \right]_{y^n}^{y^{1/n}} \mathrm{d}y$$
$$= \frac 12 \int_0^1 \left[y^{\frac 2n}-y^{2n} \right] \mathrm{d}y$$
$$= \frac 12 \left[\frac{y^{\frac 2n+1}}{\frac 2n+1}-\frac{y^{2n+1}}{2n+1} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the set of values of $a$ ($a \in \mathbb{R}$) for which $\log_a(x^2 + 4) \ge 2$, $\forall x \in \mathbb{R}$? How can I find the set of values of $a$, with $a \in \mathbb{R}$ such that:
$$\log_a(x^2+4) \ge 2, \forall x \in \mathbb{R}$$
How should I approach this?
| Note if $0 < a < 1$ then $\log_a M$ is decreasing; If $M > 1$ then $\log_a M < 0$, $\log_a 1 = 0$ and if $0< M < 1$ then $\log_a M > 0$. And $M < N\iff M> N$.
So $\log_a (x^2 +4) \ge 2 \iff a^{\log_a x^2 +4 } = x^2 + 4 \le a^2 \iff x^2 \le a^2 -4$. But because $a^2 < 1$ then $a^2 -4 < -3$ and $x^3 < -3$ is impossibl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Combinatorics: Selecting pebbles making up a 9-sided shape Find the number of ways to simultaneously select $3$ pebbles from $9$ pebbles that are arranged in a circle such that no two of the three selected pebbles are adjacent to each other, and one selection is the same as another rotation if the former can be rotated... | We may also use the Polya Enumeration Theorem. Selecting three
non-adjacent pebbles leaves a total of six non-selected ones that must
be placed in three available slots with the cyclic group $C_3$ acting
on them and at least one pebble in every slot. The cycle index is
$$Z(C_3) = \frac{1}{3} a_1^3 + \frac{2}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Real part of a complex number
Compute $\operatorname{Re}\left(\frac{1}{z+1}\right)$ when when $|z| = 1$.
The only way I could think to go about this is to simply go by definitions. If $z\in \Bbb C$, then $z\bar z$ = $|z|^2$. Now $$z=\frac{|z|^2}{\bar z}$$and$$z=\frac{1^2}{\bar z}.$$ So $z$ must be the inverse of the ... | Let $z = a+bi$
$$\begin{align}Re\left(\frac{1}{z+1}\right) &= Re\left(\frac{1}{a+1+bi}\right) = Re\left(\frac{1}{a+1+bi}\cdot\frac{a+1-bi}{a+1-bi}\right) = \\ &= Re\left(\frac{a+1-bi}{(a+1)^2-(bi)^2}\right) = Re\left(\frac{a+1-bi}{(a+1)^2+b^2}\right) = \\ &= Re\left(\frac{a+1}{(a+1)^2+b^2} - \frac{b}{(a+1)^2+b^2}i\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Proof $a_{n}$ by induction This topic is a continuation of Prove that $a_{n} = a_{n-1} + a_{n-2}$.
I need to prove that $a_{n} = (\frac{5+3\sqrt{5}}{10})(\frac{1+\sqrt{5}}{2})^{n} + (\frac{5-3\sqrt{5}}{10})(\frac{1-\sqrt{5}}{2})^{n}$ for all $n \geq 1$.
This is how I tried to prove this.
Basic step: $a_{1} = (\frac{5+3... | We have $a_{n} = a_{n-1} + a_{n-2}$ with $a_{0} = 1$ and $a_{1}=2$.
Let $b_{n} = (\frac{5+3\sqrt{5}}{10})(\frac{1+\sqrt{5}}{2})^{n} + (\frac{5-3\sqrt{5}}{10})(\frac{1-\sqrt{5}}{2})^{n}$.
To prove that $a_n=b_n$, it is enough to prove that $b_{n} = b_{n-1} + b_{n-2}$ with $b_{0} = 1$ and $b_{1}=2$.
*
*$b_{0} = 1$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Cauchy-Schwarz inequality in $\mathbb{R}^{3}$ How to use Cauchy-Schwarz inequality in $\mathbb{R}^{3}$ to show that if $a+b+c=1$ then
$\displaystyle \left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right) \geq 8$?
Any idea how to take the vectors? Thanks a lot.
| With a little algebra, you can write $$\bigg{(}\frac{1}{a}-1\bigg{)}\bigg{(}\frac{1}{b}-1\bigg{)}\bigg{(}\frac{1}{c}-1\bigg{)} = \frac{1}{abc}(1-c-b+bc-a+ac+ab-abc) = \frac{1}{abc}(bc+ac+ab-abc) = \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1 $$
Now, take $x = \bigg{(}\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}}\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Systems of equations over reals
Solve over reals
$$2x^2+xy+x-3y+3=0$$
$$3xy+y^2-2x+6y-6=0$$
Adding or subtracting these equations doesn't help, and completing the square also doesn't work. I'm considering substituting $x+y$ and $xy$, but it also doesn't show much promise. Thanks!
| Another way to do it.
From the first equation, extract $y$
$$y=-\frac{2 x^2+x+3}{x-3}$$
Plug $y$ in the second equation to get
$$\frac{-2 x^4+5 x^3+49 x^2+51 x+9}{(x-3)^2}=0$$ By inspection $x=-1$ and $x=-3$ are roots of the numerator. So, we are left with
$$\frac{(x+1) (x+3) \left(2 x^2-13 x-3\right)}{(x-3)^2}=0$$ So,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find coefficient for $x^{10}$ in $x^3(x^2-3x^3-1)^6$ I'm trying to solve the following problem:
Find the coefficient for $x^{10}$ in $x^3(x^2-3x^3-1)^6$.
Can I use the multinomial theorem to solve it? I'm unsure how to start..
Thanks!
| Well it's simple
we are multiplying every term by $x^3$ so coefficient of $x^{10}$ should be coeffiecent of $x^7$ in the expansion of $(x^2-3x^3-1)^6$
Now, divide three terms into two groups $((x^2-3x^3)+(-1))^6$
To get $x^7$, the only combination possible is when $x^2$ is raised to the power two and $-3x^3$ is rased ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
cauchy's first theorem on limits of sequences Cauchy's first theorem on limits goes like this
If$\ <f_n> $ be a sequence of positive terms and $$ \lim_{\ n\to\infty}\ f_n=l$$ Then
$$ \lim_{ n\to\infty}\ [\ \frac{f_1+f_2+\dots+f_n}{n}]=l$$
Now this is an example of its application.
$Q)$ Find the value of $$ \lim_{\ n \... | Answer to the second question.
Observe that for all $n \ge k$,
$$ \frac{1}{2n} \le \frac{1}{n+k} \le \frac{1}{n}$$
Thus,
$$\left( {\frac{1}{{{{(2n)}^2}}} + \frac{1}{{{{(2n)}^2}}} + \ldots + n{\mkern 1mu} {\mkern 1mu} terms} \right) \le \left( {\frac{1}{{{{(n + 1)}^2}}} + \frac{1}{{{{(n + 2)}^2}}} + \ldots + \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3439806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Prime-free sequence
Prove that there exist infinitely many positive integers $k$ such that
the sequence $\{x_n\}$ satisfying
$$ x_1=1, x_2=k+2, x_{n+2}-(k+1)x_{n+1}+x_n=0(n \ge 0)$$ does not
contain any prime number.
I found a similar question: The sequence ${x_n}$ satisfies $x_{n+2}=(k+1)x_{n+1}-x_n, x_0=1, x_1=... | I tried using the Z transform. From the question, we conclude that $x_0 = -1$. And taking the Z transform we get:
$$ x_{n+2} - (k+1)x_{n+1}+x_n = 0 $$
$$ z^2X(z)-x_0z^2-x_1z-(k+1)(zX(z)-x_0z) + X(z) = 0 $$
$$ X(z) = \frac{(k+1)z+z-z^2}{z^2-(k+1)z+1} $$
if we let $a=k+1$ then after taking the inverse Z transform we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving the inequality $\sum_{\text{cyc}} \frac{a}{a^2+b^3+c^3}\le\frac1{5abc}$ Let $a,b,c> 0$ be three real numbers such that $a+b+c=1$. I want to prove that
$$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\le\frac{1}{5abc}.$$
My attempt: Using AM-GM on each denominator gives (here, LHS denotes the... | Multiply both sides with $abc$ to get that your inequality is equivalent to
\begin{equation}\tag 1\label 1\sum_{\text{cyc}} \frac{a^2bc}{a^2+b^3+c^3}\le\frac15.\end{equation}
Now note that, since $a+b+c=1$, we have $$a^2+b^3+c^3=a^2\cdot(a+b+c)+b^3+c^3=a^3+b^3+c^3+a^2b+a^2c.$$
Hence, by AM-GM inequality for 5 variables... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all positive integral solutions of $\tan^{-1}x+\cos^{-1}\frac{y}{\sqrt{y^2+1}}=\sin^{-1}\frac{3}{\sqrt{10}}$ Find all the positive integral solutions of, $\tan^{-1}x+\cos^{-1}\dfrac{y}{\sqrt{y^2+1}}=\sin^{-1}\dfrac{3}{\sqrt{10}}$
Assuming $x\ge1,y\ge1$ as we have to find positive integral solutions of $(x,y)$
$$\t... | Another way
As we need $x,y>0$
If $x/y<1$
$$\tan^{-1}x+\tan^{-1}(1/y)=\tan^{-1}\dfrac{xy+1}{y-x}$$
$$xy+1=3(y-x)$$
$$\iff x=\dfrac{3y-1}{y+3}=3-\dfrac{10}{y+3}$$
So, $y+3(>3), $ must divide $10$ and must honor $x<y$
If $x/y >1,x>y$ as $x,y>0$
$$\tan^{-1}x+\tan^{-1}\dfrac1y>\tan^{-1}y+\cot^{-1}y=\dfrac\pi2>\tan^{-1}3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all values of $n$ I have tried to prove this through mathematical induction but I can't seem to prove that the proposition works for $k+1$.
| You can extract the factor $7$ as follows using
*
*$(\star)$: $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$
Hence,
$$3^{2n-1} + 2^{n+1}= 3\cdot 9^{n-1} + 4\cdot 2^{n-1}$$
$$= 3\cdot 9^{n-1} + (7-3)\cdot 2^{n-1} = 3(\underbrace{9^{n-1} - 2^{n-1}}_{\stackrel{ (\star)}{=}7\cdot m}) + 7\cdot 2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Evaluating $\lim \limits_{x\to -\infty} x + \sqrt{x^2+2x}$ which step is wrong? So I'm trying to evaluate $\lim \limits_{x\to -\infty} x + \sqrt{x^2+2x}$
These are my steps:
I first rationalize the expression (square root trick) -
$$\lim \limits_{x\to -\infty} \frac{-2x}{x - \sqrt{x^2+2x}}$$
Then I simply divide by $x$... | What you got wrong is in the step "I simply divide by $x$", where you should get
$$
\frac{-2x}{x - \sqrt{x^2+2x}}=
\frac{-2}{1-\frac{1}{x}\sqrt{x^2+2x}}
=\frac{-2}{1{\color{red}+}\sqrt{\frac{x^2+2x}{x^2}}}
=\frac{-2}{1{\color{red}+}\sqrt{1+\frac{2}{x}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the closed form for the following power series I wish to find the closed form for this power series:
$1+\dfrac{1}{6}x+\dfrac{3}{40}x^2+\dfrac{5}{112}x^3+\dfrac{35}{1152}x^4...$
I have been able to spot the that the second term is the first term multiply with $\dfrac{1 \cdot 1}{2 \cdot 3}$, the third term is the se... | As noted in the comments we have for $|x|\lt1$
$$\arcsin{(x)}=x+\frac16x^3+\frac3{40}x^5+\frac5{112}x^7+\dots$$
Hence the provided power series is just
$$1+\frac16x+\frac3{40}x^2+\frac5{112}x^3+\dots=\begin{cases}1&x=0\\\frac{\arcsin{(\sqrt{x})}}{\sqrt{x}}&x\ne0\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\int_0^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $
Find out if the following integral diverges or converges:
$$
\int_0^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx
$$
First I split the integral as $\displaystyle \int_0^1 \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx + \int_1^\infty... | Your approach on $[0,1]$ seems fine. For the other part, you can verify that:
$$\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}} \le \frac{\ln (1 + x^2)}{\sqrt {x^6}}=\frac{\ln (1 + x^2)}{x^3} \quad \mbox{and} \quad \int_1^{\infty}\frac{\ln (1 + x^2)}{x^3}\,\mbox{d}x = \ln 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
What's the length of EC? Consider the following construction (the lenght of segment EC in the image is just an approximation):
I wanted to calculate the exact value of the length of EC.
Using the law of cosines, law of sines, the angle bisector theorem and some manipulation I arrived at:
$$\sqrt{16+\left(5-\frac{35 \... | Calling $A = \{x_a,y_a\}, B = \{0,0\}, C = \{x_c,0\}$ and equating
$$
\cases{
|A-B|^2 = 5^2\\
|B-C|^2 = 7^2\\
|C-A|^2 = 4^2
}
$$
we obtain
$$
\cases{
A = \{\frac{29}{7},-\frac{8\sqrt 6}{7}\}\\
C = \{7,0\}
}
$$
Calling now
$$
\vec v_{AC} = \frac{C-A}{|C-A|} =\left \{\frac{5}{7},\frac{2\sqrt 6}{7}\right\}\\
\vec v_{AB} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we h... | Use lagrange multipliers. $\nabla f=\langle 2x,2y,2z\rangle$ and $\nabla g=\langle x^{-2},y^{-2},z^{-2}\rangle$ where $f(x,y,z)=x^2+y^2+z^2$ and $g(x,y,z)= x^{-1}+y^{-1}+z^{-1}$. Solving $\nabla f=\lambda\nabla g$ yields $\lambda=2x^3=2y^3=2z^3$ and therefore $x=y=z$. Plugging this into $g(x,y,z)=1$ we have $x=y=z=3$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 6
} |
Show that following determinant is divisible by $\lambda^2$ and find the other factor. Show that $\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$ is divisible by $\lambda^2$ and find the other factor.
My attempt is as follows:-
$$R_1\rightarrow R_1+R_2+R_3$$
$$\be... | I got a new way to solve this question and its just beautiful:
Multiply $R_1$ by $a$, $R_2$ by $b$, $R_3$ by $c$
$$\dfrac{1}{abc}\begin{vmatrix}
a^3+a\lambda&a^2b&a^2c\\
ab^2&b^3+b\lambda&b^2c\\
ac^2&bc^2&c^3+c\lambda
\end{vmatrix}$$
Taking a common from first column, b from second column, c from third column
$$\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Sum of products of binomial coefficients: $\sum\limits_{k=0}^{19}\binom{18}k\binom{20}k$ Please help me find the sum given below
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}$
First I used formula $\binom{m}{k}=\binom{m-1}{k}+\binom{m-1}{k-1}$ twice and got
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}=\sum_{k=0}^{19}\binom... | Your sum is equal to $c_n = \sum_{k=0}^n \binom{18}{k} \binom{20}{n-k}$ for $n=20.$ Note that $c_n$ is the coefficient of $x^n$ in $(1+x)^{18} (1+x)^{20}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $\cos 2x-\cos 3x+\sin 4x = 0$ Solve Equation$$\cos 2x-\cos 3x+\sin 4x=0 $$
Attemp:Developping up to $\sin x,\cos x$, equation is :
$(\cos x-1)(4\cos^2x+2\cos x-1)=4\sin x(2\cos^3x-\cos x)$
Note that squaring will give solutions $x_i$ such that either $x_i$, either $-x_i$ is solution of the current equation.
So $(... | Factorirzing your given equation we get
$$2 \sin \left(\frac{x}{2}\right) \left(\sqrt{2} \sin
\left(\frac{5 x}{2}+\frac{\pi }{4}\right)+\cos
\left(\frac{x}{2}\right)+\cos \left(\frac{3
x}{2}\right)+\cos \left(\frac{7 x}{2}\right)\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Minimize $x + y + z$ subject to $x^2 + y^2 + z^2 \le 1$ and $x \ge 0$ I'm trying to solve this problem by KKT's condition:
$$\begin{align*}
\text{min} & \quad x + y + z \\
\text{s.t} & \quad x^2 + y^2 + z^2 &&\le 1 \\
& \quad x &&\ge 0
\end{align*}$$
The linear independence constraint constraint qualificat... | Alternate Solution :
$x^2 + y^2 + z^2 \le 1 $ represents a solid sphere of radius 1
Now, we need to find a point on sphere such that the value of $x+y+z$ is minimum
It is also given that, $x \ge 0 $ , So we need to only consider the 4 octants out of 8
Now, out of these 4 octants, a minimum occurs when both $y$ and $z$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find perfect number which in this form: $p^2q$ I want to find a perfect number in the form $p^2q$ where $p$ and $q$ are both prime numbers.
I get the diophantine equation :$p^2q=p^2+pq+p+q+1$ from the definition of the perfect number.
So I assume that there is one positive integer solution which is $(p,q)$ and $c=p+q$ ... | Solving $p^2q=p^2+pq+p+q+1$ for $q$ we get
$$
q = \frac{p^2 + p + 1}{p^2 - p - 1} = 1 + \frac{2p+2}{p^2 - p - 1}
$$
The last fraction needs to be an integer and so we need $2p+2\ge p^2 - p - 1$. This gives $p \in \{0,1,2,3\}$. The only solution is $p=2$ and $q=7$, which gives $p^2q=28$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Let $A$ be a real $2×2$ matrix such that $A^6=I$. The total number of possibilities for the characteristic polynomial of $A$ is: Let $A$ be a real $2×2$ matrix such that $A^6 = I$ (where $I$ denote the identity $2×2$ matrix). The total number
of possibilities for the characteristic polynomial of $A$ is:
Annihilating po... | This statement is false:
Characteristic polynomial divides any annihilating polynomial of degree greater than or equal to the degree of the characteristic polynomial.
$$f(t) = (t-1)(t^4+7t^3+105t-999)$$
Annihilates $A=I$. The characteristic polynomial of $A$ does not divide $f(t)$.
It seems like you may be confusing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$ The following problem is from CHKMO 2018 Problem 1:
If $ab+bc+ca\ge1$, prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$$
I tried to use Cauchy–Schwarz inequality, by try multiplying different... | Let $x=\dfrac{1}{a}, y=\dfrac{1}{b}, z=\dfrac{1}{c}$. It is easy to get
$$x^2+y^2+z^2\ge xy+yz+zx$$ which is $$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{a+b+c}{abc}$$
We try to prove $a+b+c\ge\sqrt{3}$. As $a^2+b^2+c^2\ge ab+bc+ca\ge1$,
$$(a+b+c)^2=a^2+b^2+c^2+2ab+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve $2xyy^\prime=x^2+2y^2$
Solve $2xyy^\prime=x^2+2y^2$
I tried to solve with substitution $v=y^2$
then $\frac{dy}{dx}=\frac{1}{2}v^{-1/2}\frac{dv}{dx}$
and substituting the given equation I get $$\frac{x^2+v}{2x\sqrt{v}}=\frac{1}{2\sqrt{v}}\frac{dv}{dx}$$
$$\frac{x^2+v}{x}=\frac{dv}{dx}$$
$$\frac{dv}{dx}=x+\frac{v... | Note that you have a linear equation
\begin{eqnarray*}
\frac{dv}{dx} +P v =Q
\end{eqnarray*}
where $P$ and $Q$ are functions of $x$ only. We multiply through by a function $R$ in order to make the LHS an exact diffeential of a product
\begin{eqnarray*}
R\frac{dv}{dx} +RP v =RQ \\
\frac{d}{dx} (Rv) = R \frac{dv}{dx} +R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $\binom{N+K-1}{K}=\sum_{i=1}^{N-1} (-1)^{i+1}\binom{N}{i}\binom{N-i+K-1}{K}$ using polynomial I find the identity, $$\binom{N+K-1}{K}=\sum_{i=1}^{N-1} (-1)^{i+1}\binom{N}{i}\binom{N-i+K-1}{K}$$
by counting the number of ways to distribute $K$ candies to $N$ people ($N>K)$.
For right-hand eq., I count the number o... | We can re-write the identity as
$$\sum_{q=0}^{N-1} (-1)^{q+1} {N\choose q} {N-q+K-1\choose K} = 0.$$
Equivalently (divide by $-1$),
$$\sum_{q=0}^{N-1} (-1)^q {N\choose q}
{N-q+K-1\choose N-1-q}
\\ = \sum_{q=0}^{N-1} (-1)^q {N\choose q}
[z^{N-1-q}] (1+z)^{N-q+K-1}
\\ = [z^{N-1}] (1+z)^{N+K-1}
\sum_{q=0}^{N-1} (-1)^q {N\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show, that $ I_{m,n}:=\int\limits_{0}^{1}x^m(1-x)^n dx \text{ holds }I_{m,n}=\frac{m!\,n!}{(m+n+1)!}$
Show, that for $m,n\in\mathbb{N}$ holds: $$\displaystyle I_{m,n}:=\int\limits_{0}^{1}x^m(1-x)^n dx \text{ holds }I_{m,n}=\frac{m!\,n!}{(m+n+1)!}$$
I tried to do integration by parts and got $$\frac { ( 1 - x ) ^ { n ... | $I_{m+1,n} = \frac{m+1}{n+1}\int_0^1 x^m(1-x)^{n+1}dx$ by IBP.
$= \frac{m+1}{n+1}\int_0^1 x^m(1-x)^{n}(1-x)dx = \frac{m+1}{n+1}(I_{m,n}-I_{m+1,n})$.
Thus $I_{m+1,n} = \frac{m+1}{m+n+2}I_{m,n}$. We may proceed by induction on $m$.
By symmetry we apply the same argument in $n$ and thus get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is $\underbrace{444\dots44}_{2n} + \underbrace{888\dots88}_{n} + 4$ never a perfect square? In this question, the questioner asked to prove that $$f(n)=\underbrace{444\dots44}_{2n} + \underbrace{888\dots88}_{n} + 4$$ is a perfect square for all $n\in\mathbb N$. However, I was not able to find any $n$ for which this... | Whenever I see something of the form $A + 2A\times\text{something of magnitude} + A\times \times\text{something of geometrically more magnitude}$, I figure it fits roughly into the form of $A(\text{something of mangitude} + 1)^2$.
So so $4_{2n}= \frac 49\times (10^{2n} -1)$ is roughly of magnitude $(10^n)^2$ and $8_{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx<0.5$ Prove:
$ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx<0.5$
My solution:
$( x+1>x)$
$$ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx< \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{1-x}}=0.5
$$
Is there a cl... | Multiply numerator and denominator by $$\sqrt[3]{(1+x)}^2
-\sqrt[3]{1+x}\sqrt[3]{1-x}+\sqrt[3]{(1-x)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim\limits_{x \to \infty} \left( \frac{\sqrt{x^2+2x-3}}{x+2} \right)^{3-2x}$ How can I find this limit?
$$\lim\limits_{x \to \infty} \bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$$
Firstly I thought I can use the limit:
$$\lim\limits_{x \to \infty} \bigg ( 1 + \dfrac{1}{x} \bigg )^x=e$$
by adding $1$ and s... | \begin{align*}
&\left(\dfrac{\sqrt{x^{2}+2x-3}}{x+2}\right)^{3-2x}\\
&=\left(\dfrac{\sqrt{x^{2}+2x-3}}{\sqrt{x^{2}+2x}}\right)^{3-2x}\left(\dfrac{\sqrt{x^{2}+2x}}{x+2}\right)^{3-2x}\\
&=\left(1-\dfrac{3}{x^{2}+2x}\right)^{3/2-x}\left(\dfrac{x}{x+2}\right)^{3/2-x}\\
&=\left(1-\dfrac{3}{x^{2}+2x}\right)^{-(x^{2}+2x)(x-3/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Determine $\lim_{x\to\infty} f(x)$, where $f(x)=\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}$
Determine $\lim_{x\to\infty} f(x)$ where
$$f(x)=\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}.$$
I used the binomial expansion method and my answer is $e^2$ but when I plot this function on Desmos an... | Take $$\ln f(x)=({x^3+3x})\ln\left(\frac{x^4+\sqrt{x^2 +1}}{x^4-x+2}\right)$$
and note
$$\frac{x^4+\sqrt{x^2 +1}}{x^4-x+2}
=\frac{1+\frac1{x^3}\sqrt{1+\frac1{x^2}}}{1-\frac1{x^3}+\frac2{x^4}}
=\frac{1+\frac1{x^3}+O(\frac1{x^5})}{1-\frac1{x^3}+\frac2{x^4}}
=1+ \frac2{x^3}+O(\frac1{x^4})$$
Then,
$$\lim_{x\to\infty} \ln f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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