Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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What can be said about the series $\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]$ This is a sequel to this question.
I recently was browsing through Hansen's "A Table of Series and Products", and I miraculously found the sum that I was looking for:
$$
\sum_{n=1}^\infty K_{0}\left( n z \r... | Yes it is correct, note that
$$\frac{1}{\sqrt{ n^2 + x^2 }}=-\frac1n\left(1+\frac{x^2}{n^2}\right)^{-\frac12}=\frac1n\left(1-\frac{x^2}{2n^2}+o\left(\frac{1}{n^2}\right)\right)=\frac1n-\frac{x^2}{2n^3}+o\left(\frac{1}{n^3}\right)$$
thus
$$\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]=\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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An algebraic inequality involving $\sum_{cyc} \frac1{(a+2b+3c)^2}$ I was reading through the proof of an inequality posted on a different website and the following was mentioned as being easily proven by AM-GM:
Let $a,\ b,\ c>0$, then $$\frac{1}{(a+2b+3c)^2} + \frac{1}{(b+2c+3a)^2} + \frac{1}{(c+2a+3b)^2} \le \frac{1}{... | A full expanding gives
$$\sum_{cyc}(36a^6+80a^5b+104a^5c+21a^4b^2+189a^4c^2+58a^3b^3+82a^4bc-266a^3b^2c-122a^3c^2b-182a^2b^2c^2)\geq0,$$
which is true by AM-GM.
Also, SOS helps.
Let $2a+b=3x$, $2b+c=3y$ and $2c+a=3z$.
Thus, $a=\frac{x-2z+4y}{3}$, $b=\frac{y-2x+4z}{3}$, $c=\frac{z-2y+4x}{3}$,
$$ab+ac+bc=\frac{1}{9}\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $Q=X^2+5X+7$ divides $P=(X+2)^m+(X+3)^{2m+3}$ for any $m\in\Bbb N$ Let $$P=(X+2)^m+(X+3)^{2m+3}$$ and $$Q=X^2+5X+7.$$ I need to show that $Q$ divides $P$ for any $m$ natural.
I said like this: let $a$ be a root of $X^2+5X+7=0$. Then $a^2+5a+7=0$.
Now, I know I need to show that $P(a)=0$, but I do not know i... | Let us restate the value of $P$ first.
$P = (X+2)^m+(X+2)^{2m+3}+1^{2m+3} = (X+2)^m+(X+2)^{2m+3}+1 =$
$(X+2)^m+(X+2)^{2m+3}+(X+2)^0 = (X+2)^{3m+3} = (X+2)^{(m+1)^{3}}$
And, Let us restate the value of $Q$.
$Q = X^2+4X+4+X+3 = X^2+4X+4+X+2+1 = (X+2)^2+(X+2)^1+(X+2)^0$
Do you notice the relationship between $P$ and $Q$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the following limit algebraically? I've been trying to answer this for a while and I know it's a simple question relative to most questions that are posted here.
$$
\lim_{x\rightarrow -2}\: \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)}
$$
If we substitute -2 for $ x $ we get $ 0/0 $, an indeterminate form. I figure... | General tip (update) : When you can see that the denominator is equal to zero for a value $x=a$ which is the $x\to a$ of the limit, then you should try factoring on both the numerator and the denominator the factor $(x-a)$, such as you can get rid of the $\frac{0}{0}$ issue. This particular example though can also be d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
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How do I interpret this sum? So if the sum of $n$ integers $\ge 1$ equal $\frac{n(n+1)}2$. Then my book goes on and says $1 + 2 + 3 +\ldots + 2n = \frac{2n(2n + 1) }2$.
I'm confused about what $1 + 2 + 3 + \ldots +2n$ means. If the sequence is $1, 2, 3, 4$ then where does $2n$ have to do with the $n$th number?
| Compare:
$$\color{red}{1+2+\cdots +n}=\frac{n(n+1)}{2}$$
and
$$\color{red}{1+2+\cdots+n}+\color{blue}{(n+1)+(n+2)+\cdots+2n}=$$
$$\color{red}{1+2+\cdots+n}+\color{blue}{(1+2+\cdots+n)+n\cdot n}=$$
$$\color{red}{\frac{n(n+1)}{2}}+\color{blue}{\frac{n(n+1)}{2}+n^2}=\frac{2n(2n+1)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a_{n+1}=\frac{10}{a_n}-3$ and $a_1=10$, find $\lim_{n \to \infty} a_n$
Let $a_{n+1}=\dfrac{10}{a_n}-3$, $a_1=10$ then find the limit $\lim\limits_{n \to \infty} a_n$
My Try :
$$a_2=-2 \ \ ,a_3=-8 \ \, a_4=-4.25 \ \ a_n <0$$
thus visthe monotone convergence theorem $$\lim\limits_{n \to \infty}=l$$
so: $$l=\dfrac{1... |
When you have a sequence of the form $a_{n+1}=f(a_n)$ that apparently does not lead to a closed formula for $a_n$, then you have to study the function $f(x)$.
When you graph the curve $y=f(x)=\dfrac{10}x-3$ in blue and $y=x$ in red, you notice there are two intersection points.
These are called fixed points of $f$ sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\int(1+\frac{1}{x})^{2}\frac{dx}{x^2}$ This seems to be a straightforward integral:
$$\int(1+\frac{1}{x})^{2}\frac{dx}{x^2}$$
By substitution substitution, $u = 1+\frac{1}{x}, du = -\frac{1}{x^2}$
$$-\int u^2= -\frac{u^3}{3}+C = -\frac{(1+\frac{1}{x})^3}{3}$$
Then I took the derivative:
$$\frac{d}{dx}-\frac{(1+... | You assumed $u=1+\dfrac 1x$. So far so good.
But you didn't replace $dx$ by $du$ correctly.
You must not forget that $du= -\dfrac 1{x^2} dx$
So,
$$\frac{d(1+\dfrac 1x)^3}{dx} = \frac{d(u^3)}{du} \color{red}{\cdot \frac{du}{dx}}$$
Always remember that
$$\color{blue}{\frac{d}{dx} f(x)= \frac{d}{du} f(x) \cdot \frac {du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factor of a Mersenne number Why is it true that if 7 divides 91 then $(2^7-1) $ divides $(2^{91}-1)$?
1) $2^{91}-1$
$7|91 \implies (2^7-1)|(2^{91}-1)$
$\implies 2^7-1$ is factor
2) $2^{1001}-1$
$7|1001 \implies (2^7-1)|(2^{1001}-1)$
$\implies 2^7-1$ is factor
| It may be illustrative to write the numbers out in binary. I'll use $2^{21} - 1 = (2^7)^3 - 1$ instead of $2^{91} - 1$, since it's shorter:
$$\begin{aligned}
2^{21} - 1
&= \underbrace{111111111111111111111}_{21\text{ digits}}\,\vphantom1_2 \\
&= \underbrace{1111111}_{7\text{ digits}}\,\underbrace{1111111}_{7\text{ dig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
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Consider polynomial $X^3-3X+1$ If $\alpha$ is a root $\alpha^3-3 \alpha+1=0 $ Consider polynomial $$ X^3-3X+1$$
If $\alpha$ is a root
$$\alpha^3-3 \alpha+1=0 $$
showing $\alpha^2-2$ is also a root
set $X=\alpha^2-2$
$$ (\alpha^2-2)^3-(\alpha^2-2)+1=\alpha^6-9\alpha^4+26 \alpha^2-24$$
Let us look at $\alpha^6$
$$\begi... | A possible shortcut to the problem
Notice that $$\alpha^3-3 \alpha+1=0\implies \alpha^2=3-\frac1\alpha\implies\alpha^2-2=1-\frac1\alpha$$ Now to check whether this is a root, $$\begin{align}\left(1-\frac1\alpha\right)^3-3\left(1-\frac1\alpha\right)+1&=1-\frac3\alpha+\frac3{\alpha^2}-\frac1{\alpha^3}-3+\frac3\alpha+1\\&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Manipulating a functional equation
Consider a function $f$ such that $$f(x)f(y)=f(xy)+f\left (\frac {x}{y}\right )$$ then find $$f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )- f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )$$
And the options are as follows
A) $2f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )$
B) ... | Presumably, it's given that $f$ is non-constant, else we can have $f = 0$ or $f=2$.
So suppose $f$ is non-constant.
Choosing $x$ such that $f(x) \ne 0$, and letting $y=1$, we get $f(1) = 2$.
Letting $x=1$, we get $f(y) = f\left(\frac{1}{y}\right)$.
Noting that
$$
\left(\frac{3-2\sqrt 2}{\sqrt 2 + 1}\right )\left (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $\binom{2n}{n} > n2^n , \forall n \ge 4 $ I'm trying to prove the following statement: $\binom{2n}{n} > n2^n , \forall n \ge 4 $
This is my attempt at an inductive proof:
Let $P(n)$ be the following proposition: "$\binom{2n}{n} > n2^n , \forall n \ge 4 $"
Base case:
$\binom{2*4}{4} = 70 > 4*2^4 = 64$ so ... | $$\frac{(2n)!}{(n+1)!(n-1)!} =\frac n{n+1}\cdot{2n\choose n}\ge \frac{n^2}{n+1}2^n$$
A more direct proof for a better claim:
$2n\choose n$ is the largest of $2n+1$ summands in the expansion of $(1+1)^{2n}$, hence
$$ {2n\choose n}\ge\frac1{2n+1}\cdot 4^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of a standard dissection I am looking for the limit of the following expression $$\lim_{k\rightarrow\infty}\frac{(\frac{1}{k}+\frac{\epsilon}{j})^j(\frac{1}{k}-\frac{\epsilon}{k-j})^{k-j}}{(\frac{1}{k})^k}$$
with $j<k$ and some $\epsilon>0$.
What I have done so far:
We can rewrite it as $$\lim_{k\to \infty}{(1+\f... | Note that for $k\to +\infty$
*
*$(\frac{1}{k}+\frac{\epsilon}{j})^j =\left(\frac{1}{k}\right)^{j}(1+\frac{\epsilon k}{j})^j $
*$(\frac{1}{k}-\frac{\epsilon}{k-j})^{k-j}=\left(\frac{1}{k}\right)^{k-j}(1-\frac{\epsilon k}{k-j})^{k-j}$
thus
$$\frac{(\frac{1}{k}+\frac{\epsilon}{j})^j (\frac{1}{k}-\frac{\epsilon}{k-j}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Which of the two sums gives a better approximation to $\pi^2/6$? Which of the following two sums
$$\sum_{n=1}^{1000000}\frac1{n^2} \quad \text{or} \quad 1+\sum_{n=1}^{1000}\frac1{n^2(n+1)}$$
gives a better approximation to $\pi^2/6?$
I tested this on MATLAB and surprisingly obtained as a result the second sum. However... | $$\eqalign{\sum_{k+1}^\infty\frac{1}{n^2(n+1)}&<\sum_{k+1}^\infty\frac{1}{(n-1)n(n+1)}\cr
&=\frac12\sum_{k+1}^\infty\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)\cr
&=\frac{1}{2k(k+1)}}$$
$$\eqalign{\sum_{k^2+1}^\infty\frac{1}{n^2}&>\sum_{k^2+1}^\infty\frac{1}{n(n+1)}\cr
&=\sum_{k^2+1}^\infty\left(\frac{1}{n}-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Converting polar equations to cartesian equations. Where $$r=\sin(3\theta)$$ and $$y=r\sin(\theta),~x=r\cos(\theta),~r^2=x^2+y^2$$ I have started by saying that $$ r=\sin (2\theta) \cos (\theta) +\sin (\theta) \cos (2\theta) \\ r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta) (1-2\sin ^2 (\theta)) \\r=2\sin (\theta) \... | or you use $$\sin(3x)=3\sin(x)-4\sin(x)^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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continued fraction of $\sqrt{41}$
Show that $\sqrt{41} = [6;\overline {2,2,12}]$
here's my try:
$$\sqrt{36}<\sqrt{41}<\sqrt{49}\implies6<\sqrt{41}<7\implies\lfloor\sqrt{41}\rfloor=6$$
$$\sqrt{41}=6+\sqrt{41}-6=6+\frac{1}{\frac{1}{\sqrt{41}-6}}$$
$$\frac{1}{\sqrt{41}-6}=\frac{\sqrt{41}+6}{41-36}=\frac{\sqrt{41}+6}{5}=... | [This old question popped up in the feed today (years later), and I can see that there's an accepted answer. Case closed. Though, reading the question and its answers, I can't help but feeling that the problem could (should?) have been tackled differently. I'm leaving this answer in the hope it might inspire and help s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635420",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$ I am trying to evaluate the following integral
$$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$$
with $0< k < 1$.
My attempt
By performing the substitution
$$y=\... | I have completed the computation of ASM.
For $0<k<1$,
$\displaystyle I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)\,dx$
Let $\rho=\dfrac{1-k}{1+k}$
Observe that, since $0<k<1$ then $0<\rho<1$.
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$\begin{align}I(k)&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Constructing a Hypergeometric Function I am asked to find values for $a,b$ and $c$ such that $$ \frac{1}{2} ((1+x)^{2\alpha}-(1-x)^{2\alpha}) = 2\alpha x\ _2F_1(a,b;c;x^2)$$
I have attempted the following:
$$\frac{1}{2} ((1+x)^{2\alpha}-(1-x)^{2\alpha}) = \frac{1}{2}\sum^{\infty}_{k=0}\binom{2\alpha}{k}x^k-\sum^{\inft... | In the expansion
\begin{align}
f(x)&=\frac{1}{2}\sum_{k=0}^\infty \binom{2\alpha}{k}\left( x^k-(-x)^k \right)\\
&=x\sum_{n=0}^\infty \binom{2\alpha}{2n+1}x^{2n}\\
&=x\sum_{n=0}^\infty c_nX^{n}
\end{align}
with $X=x^2$, the ratio of two successive terms of the series is
\begin{align}
\frac{c_{n+1}}{c_n}\frac{X^{n+1}}{X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ as $x$ goes to $0$ As plugging $0$ in $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ makes the function becomes undetermined form of $\frac{0}{0}$. I tried applying L'Hospital's rule but it became messy and did not look helpful if I do further differentiation. So I tried finding the ... | $$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{\dfrac 1{x^2}-\dfrac{\log(1+x^2)}{x^4}}{\dfrac{\sin^2x}{x^2}}$$
Taking $z=x^2$, note that the numerator becomes $$\dfrac 1z-\dfrac{\log(1+z)}{z^2}=\dfrac{z-\log(1+z)}{z^2}\to\dfrac{1-\frac 1{1+z}}{2z}\to\dfrac{1}{2(z+1)^2}\to\frac 12$$
as $z=x^2\to 0$ as $x\to 0$ where the fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove the following algebraically $\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$ I came across the following proof in my textbook that was used as a end of chapter review. How can I prove the following algebraically?
$$\left( \begin{array}{c} 2n \\ ... | It is not correct, and it should go like this:
$$\binom{2n}{2}=\frac{2n(2n-1)}{2}=n(2n-1)=2n^2-n=n(n-1)+n^2=2\frac{n(n-1)}{2}+n^2=2\binom{n}{2}+n^2$$
It looks that your work starts well and ends well but there is a bit of a muddle in the middle.
| {
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"url": "https://math.stackexchange.com/questions/2639327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Positive Integer Divison Proof Prove that $6|n(n + 1)(n + 2)$ for any integer $n ≥ 1$
I have attempted to do this but never seem to prove it.
| If $n$ is even: $n = 2k \implies n(n+1)(n+2) = 2k(2k+1)(2k+2) = 2k(4k^2+6k+2)= 8k^3+12k^2+4k = 6k^3+12k^2 + 6k+2k^3-2k$. Observe that if $3 \nmid k\implies k^2-k = 0\pmod 3$ by Fermat Little's theorem. Thus $2k^3 - 2k = 0\pmod 6$. Thus $6 \mid n(n+1)(n+2)$. If $n$ is odd, then $n = 2k+1\implies n(n+1)(n+2) = (2k+1)(2k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 6
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Formula for a sequence defined on $K_1(x,y) := y+0$ if $x \geq y$ and $y-1$ otherwise Define $K_1:[0,1]^2\rightarrow\mathbb{R}$ as
$$K_1(x,y) := x - \frac{1}{2} - \begin{cases}
\ +(x - y - \frac{1}{2}) & \text{if $x \geq y$},\\
\ -(y - x - \frac{1}{2}) & \text{otherwise}
\end{cases}$$
then with
$$K_n(x,y) := \int_0^1K_... | Functions $K_n(x, y), n\in\mathbb N,$ are considered in the area
$$\mathbb S = \{(x, y) \in [0,1]^2\}.\tag1$$
Besides this, can be used step function
$$h(x) =
\begin{cases}
1, \text{ if }x \in (0, 1],\\
0, \text{ otherwize}
\end{cases}\tag2$$
for brief notation of 2D intervals method.
In particular, we can present t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
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Prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer.
Work so far:
(1) For n = 1:
$2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$
Check if div... | Because $$2^{5n+1}+5^{n+2}=2\cdot32^n-2\cdot5^n+27\cdot5^n=2(32-5)(32^{n-1}+...+5^{n-1})+27\cdot5^n,$$
which is divided by $27.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can ... | While others have pointed towards factorizing the expression, I would like to say that it isn't always easy for one to notice that the expression can be factorized. Also, the series has some property - the coefficients are gradually increasing or decreasing.
Let's try to find the sum of the series because that would s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 4
} |
How can I solve this system of equations?? I have this system of equations and I need values for $c_1, c_2, c_3, x_1, x_2, x_3$.
\begin{align*}
c_1 + c_2 + c_3 &= 2\\
c_1x_1 + c_2x_2+ c_3x_3 &= 0\\
c_1x_1^2 + c_2x_2^2+c_3x_3^2 &= \frac{2}{3}\\
c_1x_1^3 + c_2x_2^3+c_3x_3^3 &= 0\\
c_1x_1^4 + c_2x_2^4 + c_3x_3^4 &= \frac{... | Rewriting the system as
\begin{align}
\sum_{j=1}^3 c_j\,x_j^i&=v_i,\quad i=0,\dots,5
\tag{1}\label{1}
,
\end{align}
we can apply Prony's method
as follows.
Solve the linear system
\begin{align}
\left[\begin{matrix}
v_0 & v_1 & v_2 \\
v_1 & v_2 & v_3 \\
v_2 & v_3 & v_4
\end{matrix}\right]
\cdot
\left[\begin{matrix... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the value of $\sum\limits_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}$? How do I calculate the value of the series $$\sum_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}= \frac{1}{1\cdot2\cdot3}+\frac{1}{4\cdot5\cdot6}+\frac{1}{7\cdot8\cdot9}+\cdots?$$
| By making use of the integral
$$\int_{0}^{1} \frac{(1-x)^2}{1-x^3} \, dx = \frac{1}{2} \, \left(\frac{\pi}{\sqrt{3}} - \ln 3 \right)$$
one can take the following path.
\begin{align}
S &= \sum_{k=0}^{\infty} \frac{1}{(3k+1)(3k+2)(3k+3)} \\
&= \sum_{k=0}^{\infty} \frac{\Gamma(3k+1)}{\Gamma(3k+4)} = \frac{1}{2} \, \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2644864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Finding $\lim x_n$ when $\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$
Let $x_n$ be the unique solution of the equation $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$
Find $\lim_{n \to \infty} x_n$
I think that the limit must be $\frac{1}... | If $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$
then we have
$$\begin{align}x_n&=\frac{\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)}{\ln\left( 1+\frac{1}{n}\right)}-n\\&\sim \frac{\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)}{\left(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Is there any way to prove $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $ by induction since $ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $ we have that for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $
my problem i... | This is unlikely. The bare induction step would be
$$S_n\le\frac{\pi^2}6\implies S_n+\frac1{n^2}\le\frac{\pi^2}6,$$ which obviously doesn't hold. There is not enough information in the inductive hypothesis.
Any information on the asymptotics of the series will be of the form $\dfrac{\pi^2}6-\epsilon(n)$, which contains... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
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Determinate all the positive integers $x$ such that $x^4+6x^3+11x^2+3x+11$ is a perfect square. Determinate all the positive integers $x$ such that $x^4+6x^3+11x^2+3x+11$ is a perfect square.
My try
With pure intuition I can say that there is no $ x $ that meets this condition, i tried by brute force some numbers and ... | Suppose that exist $n\in\Bbb N$ such that $$x^4+6x^3+11x^2+3x+11=n^2$$
See $\text{mod } 3$ the equation and you can prove that $n^2\not\equiv 2 \text{ mod } 3$ for all $n\in\Bbb N$. In fact, if $n=3k+a$ with $a=0,1,2$ then $n^2=9k^2+6ka+a^2\equiv a^2 \text{ mod } 3$ and this values are $0,1,1$ respectively.
Now,
If $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Determining $A+B$, given $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A - \cos B = \sqrt{\frac12}$. Different approaches give different answers. The question:
Determine $A + B$ if $A$ and $B$ are acute angles such that:
$$\sin A + \sin B = \sqrt{\frac{3}{2}}$$
$$\cos A - \cos B = \sqrt{\frac{1}{2}} $$
Here are th... | $$\sqrt3\cos A-\sin A=\sqrt3\cos B+\sin B$$
$$\implies\cos(A+30^\circ)=\cos(B-30^\circ)$$
$$A+30^\circ=360^\circ n\pm(B-30^\circ) $$
$$-\implies A+B=360^\circ n$$ which is impossible as $0<A+B<180^\circ$
$+\implies A-B=360^\circ n-60^\circ$
As $-90<A-B<90,n=0$
Now use http://mathworld.wolfram.com/ProsthaphaeresisFormu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Radical with pattern Let $a =111 \ldots 1$, where the digit $1$ appears $2018$ consecutive times.
Let $b = 222 \ldots 2$, where the digit $2$ appears $1009$ consecutive times.
Without using a calculator, evaluate $\sqrt{a − b}$.
| $a= \overbrace{11\ldots11}^{2018} = \dfrac{10^{2018}-1}{9}$
$b= \overbrace{22\ldots22}^{1009} = 2\cdot\dfrac{10^{1009}-1}{9}$
$\Rightarrow a-b = \dfrac{10^{2018}-1}{9}-2\cdot\dfrac{10^{1009}-1}{9} = \dfrac{10^{2018}-2\cdot10^{1009}+1}{9} = \left(\dfrac{10^{1009}-1}{3} \right)^2$
$\Rightarrow \sqrt{a-b} = \dfrac{10^{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Probability of 3 of a kind - Probability of Pair Each of five, standard, six-sided dice is rolled once. What is the probability that there is at least one pair but not a three-of-a-kind (that is, there are two dice showing the same value, but no three dice show the same value)?
Here is my thinking. There are $6$ possib... | There are a total of $6^5=7776$ possible sets of dice rolls. To get a pair without a three-of-a-kind, we can either have one pair and the other three dice all showing different numbers, or we have two pairs and the die showing something different.
In the first case, there are $6$ ways to pick which number makes a pair ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
"Chain Rule" for Finite Difference Operator I am trying to prove using difference calculus the following formula;
$$\Delta(a+bx)^{(n)}=bn(a+bx)^{(n-1)}$$
which is akin to the chain rule for continuous Calculus. Here we have that
$$x^{(n)}=x(x-1)...(x-n+1)$$
and
$$\Delta f(x)=f(x+1)-f(x)$$
It seemed daunting at first ... | As you mentioned, the algebra works out when using the definition you gave in the edit.
Let $h(x) = f(x)^{(n)} = f(x)f(x-1)\cdots f(x-n+1)$. Then
\begin{align}
\Delta h(x) &= f(x+1) f(x) f(x-1) \cdots f(x-n+2) - f(x)f(x-1)\cdots f(x-n+1) \\
&= f(x) f(x-1) \cdots f(x-n+2)(f(x+1) - f(x-n+1)) \\
&= (f(x+1) - f(x-n+1)) f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2660442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Inequality with hypergeometric function I'm trying to prove the following inequality:
$$_1F_2\left(\frac{a}{2},\frac{3}{2},\frac{a}{2}+1;-\frac{\pi^2}{4}\right)\ge \frac{a+6}{(a+2)(a+3)}$$
where $a$ is a positive real number.
I wrote Euler's Integral Transform for Hypergeometric Functions, but it gaves me only more co... | Start with
$$f(a)=
{_1F_2}\left( \frac a 2; \frac 3 2, \frac a 2 +1; -\frac {\pi^2} 4 \right) =
\frac a \pi \int_0^1 t^{a-2} \sin \pi t \,dt,$$
which is a special case of
$${_1F_2}(a; b_1, b_2; z) =
\frac {\Gamma(b_2)} {\Gamma(a) \Gamma(b_2-a)}
\int_0^1 t^{a-1} (1-t)^{b_2-a-1} {_0F_1}(; b_1; z\,t) dt.$$
Expanding $\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $10^{n+1} - 9^{n+1} = 9^n + 9^{n-1}10 + 9^{n-2}10^2 + ... + 10^n$ In this video, James Grime shows that the number of $\mathbb{N}$ less than $10^{n+1}$ that have at least one $3$ among their digits is given by this recurrence relation:
$$T_{n+1} = 9T_{n} + 10^{n}\; where\; T_0=1$$
But later in the video, he says ... | With $a^0=1$, $b^0=1$ for $a, b\ne 0$:
$(a-b)(a^{n}b^{0}+a^{n-1}b^{1}+a^{n-2}b^2+...+a^{1}b^{n-1}+a^{0}b^{n})$
$=a^{n+1}b^{0}+a^{n}b^{1}+a^{n-1}b^2+...+a^{2}b^{n-1}+a^{1}b^{n}-(a^{n}b^{1}+a^{n-1}b^{2}+...+a^{1}b^{n}+a^{0}b^{n+1})$
$=a^{n+1}b^{0}+a^{n}b^{1}+a^{n-1}b^2+...+a^{2}b^{n-1}+a^{1}b^{n}-a^{n}b^{1}-a^{n-1}b^{2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integral calculating issue If
$$ \int \frac{1}{x+1}\,\mathrm{d}x = \ln|x+1|+C, $$
then why
$$ \int \frac{1}{x+\frac{1}{3}}\,\mathrm{d}x \not= \ln\left|x+\frac{1}{3}\right|+C? $$
| $$\int\frac{f'}{f}=\ln(f)$$
So both of your answers are correct. But you can do the second this way too:
$$\frac{1}{x+1/3}=\frac{1}{x+1/3}\frac{3}{3}=\frac{3}{3x+1}$$
And it's integral can be written as
$$\ln(3x+1)$$
But the $2$ results are the 'same' (Their difference is only a constant):
$$\ln(3x+1)-\ln(x+1/3)=\ln\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Eigenvalue and Eigenvector of $\small\pmatrix{0 & 0 \\ 0 & -7}$ I need help working out the eigenvectors for this matrix.
$ \begin {pmatrix} 0 & 0 \\ 0 & -7 \end{pmatrix} $
The original matrix is $ \begin {pmatrix} 5 & 0 \\ 0 & -2 \end{pmatrix} $ , eigenvalues are 5,-2,
but I am not sure how to about the eigenv... | No, from the first equation, $x$ and $y$ are free. From the second equation, $y=0$. So your eigenvector is
$$
\begin{bmatrix}1\\0 \end{bmatrix}
$$
as you can check, the equation is satisfied.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Geometric images of complex numbers $z$ such that triangle with vertices $z, z^2,z^3$ is a right angled triangle.
Find the geometric images of complex numbers $z$ such that triangle with vertices $z, z^2,z^3$ is a right angled triangle.
My try:
Let the complex number $z=x+yi$
Hence we need to find the locus of the po... | Let the squared-sides of the triangle be
$$\begin{align}
a^2 &\;:=\; |z^1-z^2|^2 \;=\; |z|^2 |1-z|^2 \\
b^2 &\;:=\; |z^2-z^3|^2 \;=\; |z|^4 |1-z|^2 \\
c^2 &\;:=\; |z^3-z^1|^2 \;=\; |z|^2 |1-z|^2 |1+z|^2
\end{align}$$
Trivially, we have a Pythagorean triple if $z=0$ or $z=1$. Otherwise, we can divide each of $a^2$, $b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\Delta$ I'll state the question from my book below:
If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of
$$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}... | You solved it all right , just took cosA.cos C in denominator instead of sinA.SinC
$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}$
$R_2 \to R_2 - R_1$
$R_3 \to R_3 -R_1$
$= \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B-\sin^2A & \cot B-\cot A & 0 \\ \sin^2C-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Intersection of two paraboloids Consider two paraboloids. The first one is given by $x^2 + y^2 = z+5$. So, it intersects the x-y plane in the circle $x^2+y^2=5$. The second paraboloid is exactly the same as the first one, only shifted in the x-y plane. It's equation becomes $(x-1)^2+(y-1)^2=z+5$. From the figure below,... | Alternatively, exploiting symmetry profitably, you can simply express the equations as
$$
\eqalign{
& x^{\,2} + y^{\,2} = \left( {x - 1} \right)^{\,2} + \left( {y - 1} \right)^{\,2} = z + 5 \cr
& \left( {x - 1/2 + 1/2} \right)^{\,2} + \left( {y - 1/2 + 1/2} \right)^{\,2} = \left( {x - 1/2 - 1/2} \right)^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the Cartesian equation of the locus described by $|z+2-7i| = 2|z-10+2i|$
Find the Cartesian equation of the locus described by $|z+2-7i| =
2|z-10+2i|$ Write your answer in the form $(x+a)^2+(y+b)^2=k$.
This was a question from my end of year exams just gone and I'm unsure as to where I have gone wrong :(. If an... | HINT
Note that in general the equation
$$|z+a+bi|=k|z+c+di|\iff\frac{|z+a+bi|}{|z+c+di|}=\frac{d_1}{d_2}=k$$
describes a circle known as Circle of Apollonius.
Thus as an alternative you could find C and D by the given condition and then the center $O$ and radius $R$ of the circle
$$O=\frac{C+D}{2} \quad R=OD=OC$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2667076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove that $24|n^2-1$, if $(n,6)=1$ $n^2-1 = (n-1)(n+1)$
Then $24|(n-1)(n+1)$
$(n,6)=1$: $\exists a,b\in\mathbb{Z}$ that $n = 6\cdot a+b$
Investigate the residues, which arise when dividing the number n by two and three:
$\frac{6\cdot a+b}{3} = \frac{6\cdot a}{3}+\frac{b}{3} = 2\cdot a+\frac{b}{3}$
$\frac{6\cdot a+b}{2... | If $\gcd(n,6)=1$, then $n$ is odd and $n^2 \equiv 1 \bmod 8$.
If $\gcd(n,6)=1$, then $\gcd(n,3)=1$ and $n^2 \equiv 1 \bmod 3$.
Thus $24 = lcm(8,3)$ divides $n^2-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2... | Hint: Alternately, by AM-GM, note $x^4+\frac14x^2\geqslant x^3$ and $\frac12x^2+\frac12\geqslant x$, so $x^4-x^3+x^2-x+1 \geqslant \frac14x^2+\frac12>\frac12$ as equality is not possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
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Show that $f(x)=x^{5/3}-kx^{4/3}+k^2x$ is increasing for $k\neq0$ So to show the function is increasing/decreasing we differentiate and show it is more than zero/less than zero:
We have
$$f(x)=x^{5/3}-kx^{4/3}+k^2x$$
Hence,
$$f'(x)=\frac{5}{3}x^{2/3}-\frac{4k}{3}x^{1/3}+k^2$$
But how do I show
$$\frac{5}{3}x^{\frac{2}{... | Notice that you now have a new second order polynomial equation with $k$ as a variable instead of x.
You can compute the discriminant $\Delta$:
$$\Delta = (\frac{4x^{\frac{1}{3}}}{3})^2 - 4\frac{5x^\frac{2}{3}}{3} $$
$$\Delta = \frac{16x^{\frac{2}{3}}}{9} - \frac{60x^\frac{2}{3}}{9} = -\frac{44x^\frac{2}{3}}{9} < 0$$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Verifying $\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ$ $$\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ.$$
Ηere is what I have so far
$$\sin 4θ = 2\sin 2θ \cos 2θ = 4\sin θ \cos θ \cos 2θ.$$
Not sure if this is the correct path I should take to solve this problem. I have been stuck hard for about an hour now.
| Note that:
${\sin n \theta = \dbinom{n}{1}\cos^{n-1}\theta\sin \theta- \dbinom{n}{3}\cos^{n-3}\theta \sin^3 \theta + \dbinom n 5\cos^{n-5}\theta\sin ^{5}\theta...}\\= \color{blue}{\displaystyle\sum_{r=0, 2r+1\le n}(-1)^r\dbinom{n}{2r+1}\cos^{n-2r-1}\theta \sin^{2r+1}\theta} $
For proof, see this.
Therefore,
$\sin 4\t... | {
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"question_score": "2",
"answer_count": 5,
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If $x=\frac{-1}{2} , y=\frac{3}{4} , z=\frac{-3}{2}$ . Find $x^3 \div y^2 z^2$ If I have Question like that :
If $ x=\frac{-1}{2} \ , \ y=\frac{3}{4} \ , \ z=\frac{-3}{2} \ $
Find a numerical Value for $$x^3 \div y^2z^2$$
First If we divide $x^3$ over $y^2$ and then multiplying the result we get $\frac{-1}{2}$
, but if... |
Find a numerical Value for $$x^3 \div y^2z^2$$
I find this notation a bit ambiguous, although I would assume the intended interpretation is:
$$\frac{x^3}{y^2z^2} \tag{$*$}$$
The division operator $\div$ isn't used much in mathematics (at least not 'later on') and if you would write something like $x^3/y^2z^2$, the st... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Why does Jacobi method only converge for one of these two equivalent ways of stating a problem? Consider the system \begin{align}a=b-2\\ b=2a+14\end{align} Implementing a simple Jacobi method and initializing at $(a,b)=(0,0)$, updating by $a^{k+1}=b^{k}-2$, $b^{k+1}=2a^k+14$, we get: $(-2,14),(12,10),(8,38),(36,30),(28... | The matrix that defines the "original" linear equations is $$A_1 = \begin{pmatrix}1 & -1 \\ -2 & 1 \end{pmatrix}$$ for unknowns $\begin{pmatrix} a \\ b \end{pmatrix} $ and right-hand-side $b_1 = \begin{pmatrix} -2 \\ 14 \end{pmatrix} $. The error propagation matrix $C$ such that
$$e^{k+1} = C e^k$$ for the Jacobi metho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $ \frac{(99)!!}{(100)!!} < \frac{1}{10}$ How to prove that $ \dfrac{(99)!!}{(100)!!}=\dfrac{1\cdot3\cdot5\cdot7\cdot9 \cdots99}{2\cdot4\cdot6\cdot8\cdot10\cdots100} < \dfrac{1}{10}$
Any hint to prove it?
| $$ \ln \left( \frac{99!!}{100!!} \right) = \ln \left(\frac{99}{100} \cdot \frac{97}{98} \cdot \ldots \cdot \frac{1}{2}\right) = \sum_{k=1}^{50} (\ln(2k-1)- \ln(2k)) $$
with $f(k) = \ln(2k-1)-\ln(2k)$ a concave function of $k$ for $k \ge 1$.
Thus
$$\eqalign{ \ln \left( \frac{99!!}{100!!} \right) &= \frac{f(1)}{2} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Binomial Questions
The Spring test will contain $20$ multiple choice questions, each with five responses, and only one response is correct for each question. No marks are deducted for incorrect answers.
*
*If a student randomly guesses the answer to every question, calculate the
probability that he fails the e... | $(i)$
If the student gets $8$ questions correct, then they score a $40$%, but the question asks for the probability that the student scores $below$ $40$%. Thus we have
$$\sum_{k=0}^7 {20 \choose k}\left(\frac{1}{5}\right)^k\left(\frac{4}{5}\right)^{20-k}\approx0.968$$
$(ii)$
You tried to calculate the probability that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding all positive integers $x,y,z$ that satisfy $2^x=3^y7^z+1$
Find all positive integers $x,y,z$ that satisfy $$2^x=3^y7^z+1$$.
I think that $(x,y,z)=(6,2,1)$ is the only solution, But how can I prove this?
| We have to show that for $x\ge 7$, the number $2^x-1$ has a prime divisor different from $3$ and $7$.
If $x$ has a prime factor $p\ge 5$, then $2^p-1$ divides $2^x-1$ and all prime factors of $2^p-1$ must be of the form $2kp+1$, hence there must be a prime factor greater than $7$.
Otherwise $x$ must be divisible by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Calculating $\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$. I have a difficulty in calculating this limit:
$$\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3},$$
I have tried $\tan x = \frac{\sin x}{\cos x}$, then I unified the denominator of the numerator of the given limit problem finally I got $$\lim_{x \rig... | If we let $A_n$ be the up/down numbers we have:
$$\tan x = \sum_{n=0}^\infty\frac{A_{2n+1}}{(2n+1)!}x^{2n+1}$$
$$\sin x = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$
Thus the first few coefficients of $\tan x - \sin x$ are:
$$0 + \frac{1}{2}x^3 + \frac{1}{8}x^5 + \cdots$$
Thus if we divide by $x^3$ we get a const... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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On finding $\sup\left\{3(-1)^n-\frac{1}{n^2+1}\right\}$ How does one find $\sup(A)$ where $A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$?
I've tried as follows, but I'm not so sure.
$\displaystyle 3(-1)^n-\frac{1}{n^2+1} \le 3-\frac{1}{n^2+1} \le 3. $ So $3$ is an upper bound for $A$.
Let $\epsilon >0$... | This sequence $$ A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$$
has a subsequence $$ \left\{3-\frac{1}{4n^2+1}: n \in \mathbb{N}\right\}$$ which is monotonically increases to $x=3$
Since x=3 is an upper bound for the set $$ A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$$
We ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Studying the extrema of $f(x,y) = x^4 + y^4 -2(x-y)^2$
Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ such that $f(x,y) = x^4 + y^4 - 2(x-y)^2$. Study its extrema.
So here was my approach.
We have $$\frac{\partial f}{\partial x}(x,y) = 4(x^3 -x + y),\frac{\partial f}{\partial y}(x,y)= 4(y^3 -y + x) $$
I have to find $(x... | To solve the system you can add:
$$\begin{cases} x^3-x+y=0 \\ y^3-y+x=0 \end{cases} \Rightarrow x^3+y^3=0 \Rightarrow x=-y \Rightarrow \\ x^3-x-x=0 \Rightarrow x(x^2-2)=0 \Rightarrow x=0;\pm \sqrt{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2697218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Fourier series applied I have a question concerning the Fourier series:
I started with the following:
$$\cos (\alpha x)= \frac{1}{2}a_0+ \sum_{k=1}^{\infty}a_k\cos(kx).$$
I proved that this series with Fourier coefficients is equal to:
$$\cos \alpha x= \frac {\sin \alpha \pi}{\alpha \pi}+ \frac{2 \alpha}{\pi}\sin \alph... | Because the series $\displaystyle \sum\limits_{k = 1}^\infty (-1)^{k - 1} \frac{\cos kπ}{k^2 - α^2} = -\sum\limits_{k = 1}^\infty \frac{1}{k^2 - α^2}$ converges, then\begin{align*}
\cos απ &= \frac{\sin απ}{απ} + \frac{2α}{π} \sin απ \sum_{k = 1}^\infty (-1)^{k - 1} \frac{\cos kπ}{k^2 - α^2}\\
&= \frac{\sin απ}{απ} - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2697828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculate $f(x)$ at a specific point Question:
Calculate $f(x) = \frac{49}{x^2} + x^2$ at points for which $\frac{7}{x}+x =3$
My attempt:-
I tried to find the value of $x$ and insert in $f(x)$
$$\frac{7}{x}+x =3$$
$$7+x^2 =3x$$
$$x^2 -3x + 7=0$$
$$x = \frac{3+\sqrt{9-7*4}}{2}$$
Now $x$ is coming out to be irrational an... | Make use of
$$
\left(\frac{7}{x}+x \right)^2=\frac{49}{x^2}+x^2+14
$$
i.e.
$$
3^2-14=\frac{49}{x^2}+x^2=-5
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding the the derivative of $y=\sqrt{1-\sin x}; 0A question I'm attempting is:
Find the derivative of $ y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$.
I did this:
$y = \sqrt {1 - \sin x} = \sqrt {\cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}} = \sqrt { (\sin \frac{x}{2}-\cos \frac{x}{2})^2} = \... | $$y=\sqrt {1-\sin x}$$
$$\ln y= \frac 12 \cdot \ln (1-\sin x)$$
$$\Rightarrow \frac 1y \frac {dy}{dx}=\frac 12\left (\frac {-\cos x}{1-\sin x}\right) $$
$$\Rightarrow \frac {dy}{dx}=\frac {-\cos x}{2\sqrt {1-\sin x}}$$
On rationalisation of denominator this turns out to be $$-\frac 12 (\sin x/2 +\cos x/2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Validity of proving identities by showing LHS-RHS =0 or using reversible steps? When proving $\mathrm{LHS}=\mathrm{RHS}$, the most common way of doing it is by manipulating it in such a way to show that $\mathrm{LHS}$ equals to some expression which equals to $\mathrm{RHS}$. But what about these methods:
Method 1:
Show... | One of the related questions here, seems to be more suitable for method $2$:
$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x} (*)$$
Let $a =\sin x$, $b=\cos x$, then:
$(*)\Leftrightarrow \frac{1+a}{b}=\frac{1+a+b}{1-a+b}$
$\Leftrightarrow (1+a)(1-a+b) = b(1+a+b)$
$\Leftrightarrow 1-a^2+b(a+1) = b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Differentiate $y=\sin^{-1}\left(2x\sqrt{1-x^2}\right),\quad\frac{-1}{\sqrt{2}}
Find $\dfrac{\mathrm dy}{\mathrm dx}$ if
$y=\sin^{-1}\left(2x\sqrt{1-x^2}\right),\quad\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$
I can solve it as follows:
$$
\begin{align}
y'&=\frac{1}{\sqrt{1-4x^2(1-x^2)}}\frac{d}{dx}\Big(2x\sqrt{1-x^2}\... | At one point you have $y=\sin^{-1}(\sin 2\alpha)$. It means that $-\pi/2\le y\le \pi/2$, so your only allowed solution is $n=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sum_{k=0}^\infty \frac{(-1)^k}{3k+2} = \frac{1}{9}\left(\sqrt{3}\pi-3\ln\,2\right)$ Just for fun:
How can we prove (calculate) that $\sum_{k=0}^\infty \frac{(-1)^k}{3k+2} = \frac{1}{9}\left(\sqrt{3}\pi-3\ln\,2\right)$ ? Can we use (9) from:
http://mathworld.wolfram.com/DigammaFunction.html
(9): $\sum_{k=0}^\inf... |
We can evaluate the series of interest without appealing to the Digamma Function.
Note that we can simply write
$$\begin{align}
\sum_{n=0}^{2N}\frac{(-1)^{n}}{3n+2}&=\sum_{n=0}^N\left(\frac{1}{6n+2}-\frac{1}{6n+5}\right)\\\\
&=\sum_{n=0}^N\int_0^1\left( x^{6n+1}-x^{6n+6}\right)\,dx\\\\
&=\int_0^1 x\left(\frac{1-x^{6N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $2$ points $A$ and $B$, and a point $Q$ on the circle. Find the minimized value of $\frac{2}{3}\overline{QA} + \overline{QB}$. Problem
Given $2$ points $A(0,3)$ and $B(4,0)$ on the plane, and a point $Q$ on the circle $x^2+y^2=4$.
Find the minimum value of $\frac{2}{3}\overline{QA} + \overline{QB}$.
My Idea
Let ... | You were on the right track. It is geometrically pretty evident that the minimizing $Q$ lies in the first quadrant, so you just have to solve
$$ \frac{\sin t}{\sqrt{5-4\cos t}}=\frac{\cos t}{\sqrt{13-12\sin t}}\tag{1}$$
for $t\in\left(0,\frac{\pi}{2}\right)$. With such constraint $(1)$ boils down to
$$ \sin^2(t)(13-12\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a$ be the real root of the equation $x^3+x+1=0$ Let $a$ be the real root of the equation $x^3+x+1=0$
Calculate $$\sqrt[\leftroot{-2}\uproot{2}3]{{(3a^{2}-2a+2)(3a^{2}+2a)}}+a^{2}$$
The correct answer should be $ 1 $. I've tried to write $a^3$ as $-a-1$ but that didn't too much, I guess there is some trick here :s
| The expansion of what is under square root gives
$$A=9a^4+6a^3-6a^3-4a^2+6a^2+4a=$$
$$a (9a^3+2a+4)=a (9 (-a-1)+2a+4)=$$
$$-a (7a+5)=$$
this must be equal to
$$B=(1-a^2)^3=1-3a^2+3a^4-a^6=$$
$$$$
the difference is
$$B-A=1+4a^2+5a+3a^4-(a+1)^2=$$
$$3a^2+3a+3a^4=3a (a^3+a+1)=0$$
It is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Can you explain the absolute value sign when integrating $\int_{-1}^{1}dx \int_{0}^{x^2} \sqrt{x^2-y} dy$? I'm doing homework on the following integral:
$$\int_{-1}^{1}dx \int_{0}^{x^2} \sqrt{x^2-y} dy$$
And here is the answer:
I try to think about it, but I couldn't answer why there is an absolute value sign in the t... | Let's compute $\int_0^{x^2} \sqrt{x^2 - y} \, dy$. Applying the substitution $u = x^2 - y$, we have $du = -dy$ and hence
$$ \int_0^{x^2} \sqrt{x^2 - y} \, dy = -\int_{x^2}^0 \sqrt{u} \, du = \int_0^{x^2} \sqrt{u} \, du = \left[ \frac{2}{3} u^{\frac{3}{2}} \right]^{u=x^2}_{u=0} = \frac{2}{3} \left( (x^2)^{\frac{1}{2}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $a=x \lfloor x \rfloor$ How can I solve for $x$ given $a=x \lfloor x \rfloor$
Where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$, and where $a$ is a rational number.
What I've done
\begin{align}
\frac{a}{x} &= \lfloor x \rfloor \\
\implies \frac{a}{x} & \le x < \frac{a}{x}+1 \... | If $\lfloor x \rfloor = n$, you want $x = a/n$ and $n \le a/n < n+1$. Thus (assuming $n > 0$)
$n^2 \le a < n^2 + n$. Now $(n+1)^2 = n^2 + 2 n + 1 > n^2 + n$. So:
Given $a \ge 1$,
take $n = \lfloor \sqrt{a} \rfloor$. If $a \ge n^2 + n$ there is no solution.
Otherwise, $x = a/n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the minimum of a three variate function We consider the function
$$f(x,y,z)=(7(x^2+y^2+z^2)+6(xy+yz+zx))(x^2y^2+y^2z^2+z^2x^2)$$
Find
$$m=\min\{f(x,y,z):xyz=1\}$$
Using the AM-GM inequality it is clear that
$$m_+=\min\{f(x,y,z):xyz=1,x,y,z>0\}=13\times 9=117.$$
But $f(-1,-1,1)=45$ so clearly $m<m_+$. Numerically, ... | Since $\sum\limits_{cyc}(7x^2+6xy)>0$ and $\sum\limits_{cyc}x^2y^2>0$, we see that the minimal value is non-negative.
Let $m$ be a minimal value.
Thus, $$\sum_{cyc}(7x^2+6xy)\sum_{cyc}x^2y^2\geq mx^2y^2z^2.$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative and $xyz=w^3$.
Hence, $$(7(9u^2-6v^2)+18v^2)(9v^4-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Differentiate $y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$, $-1\leq x\leq1$
Find $\frac{dy}{dx}$ if $y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$, $-1\leq x\leq1$
The solution is given as $y'=0$ in my reference. But that doesn't seem to be a complete solution as the graph of the function is:
My Attempt
Let $x=\sin\alpha\implies \alp... | $y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$, $-1\leq x\leq1$
Let, $x=\sin\alpha\implies \alpha=\sin^{-1}x$, We have $-\pi/2\leq\alpha\leq\pi/2\implies|\cos\alpha|=\cos\alpha$
$$
\begin{align}
y&=\sin^{-1}(\sin\alpha)+\sin^{-1}(|\cos\alpha|)=\sin^{-1}(\sin\alpha)+\sin^{-1}(\cos\alpha)\\&=\sin^{-1}(\sin\alpha)+\sin^{-1}(\sin(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Russian MO 2004 $\sqrt{a} + \sqrt{b} + \sqrt{c} \geq ab + bc + ca$ I have a doubt on a proof included in "Secrets in Inequalities" by Pham Kim Hung. The exercise is to prove $$\sqrt{a} + \sqrt{b} + \sqrt{c} \geq ab + bc + ca$$ for a, b, c whose sum is 3.
His approach is the following
He observes that:
$$2(ab + bc + ca)... | By AM-GM
$$\sum_{cyc}(a^2+2\sqrt{a})\geq3\sum_{cyc}\sqrt[3]{a^2\cdot(\sqrt{a})^2}=$$
$$=3\sum_{cyc}a=9=(a+b+c)^2=\sum_{cyc}(a^2+2ab),$$
which gives $$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+ac+bc.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\frac{dy}{dx}$ if $y=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0
Find derivative of $f(x)=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0<x<1$
Let $x=\sin a$ and $\sqrt{x}=\cos b$
Then I'll get:
$$
y=\sin^{-1}[\sin a\cos b-\cos a\sin b]=\sin^{-1}[\sin(a-b)]\\
\implies\sin y=\sin(a-b)\\
\implies y=n\pi+(-1)... | $$F(x)=\sin^{-1}x-\sin^{-1}\sqrt x=\sin^{-1}x+\sin^{-1}(-\sqrt x)$$
Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
$$\displaystyle F(x) =\begin{cases}
\arcsin( x\sqrt{1-x} -\sqrt x\sqrt{1-x^2}) \;\;;x^2+x \le 1 \;\text{ or }\; x^2+x > 1, -x\sqrt x< 0\iff x>0\\
\pi - \arcsin( x\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Integration by Substitution of Fraction involving e
Find $\int\frac{2}{e^{2x}+4}$ using $u=e^{2x}+4$
The answer is $\frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c$
I must have made a mistake somewhere as my answer is not the same. Apologies the question may be too specific, but I am teaching myself calculus.
$\int\frac{2}{e... | You have made a basic, but common algebraic error:
$$\int\frac{1}{u^2-4u}du \neq\int u^{-2}-\frac{1}{4}u^{-1}du$$
You cannot split denominator like that. To continue your method, use partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Getting the volume of a sphere from an infinite product I read an article about squeezing pi from a menger sponge. You can see the original here: Squeezing pi from a menger sponge
The author finds an infinite product which, he says, approximates the volume of a sphere with radius 1. The author mentions that the infinit... | You can derive this result as follows:
First, notice that:
$$(2n+1)^3-3(2n+1)+2 = 12n^2 + 8n^3 = 4n^2(2n+3)$$
Then, the product in your expression can be written as:
$$
\begin{align*}
\prod_{n=1}^\infty \frac{(2n+1)^3-3(2n+1)+2}{(2n+1)^3} &= \prod_{n=1}^\infty \frac{4n^2(2n+3)}{(2n+1)(2n+1)(2n+1)} \overset{1}{=} \prod_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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With an appropiate substitution of the Taylor polynomial around $0$ give an estimation of $sin 1$ with the accuracy of $10^{-4}$. I know that the reminder is $R_n=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ and that the Taylor polynom of the sinus is $x-\dfrac {x^{3}} {3!}+\dfrac {x^{5}} {5!}-\ldots +\left( -1\right) ^{n-1... | Let $A=1-1/3!+1/5!-1/7!.$ $$\text {Then }\quad A-\sin 1=1/9!-(1/10!-1/11!)-(1/12!-1/13!)-...<1/9!$$ $$ \text {and }\quad A-\sin 1=(1/9!-1/10!)+(1/11!-1/12!)+... >0.$$ So $0<A-\sin 1<1/9!=1/362,880.$
Or we could say that with $x=1,$ $a=0$ and $f(x)=\sin x,$ we have $x-a=1, $ so for some $c\in (a,x)=(0,1)$ we have $$A-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Calculating the largest possible area of a rectangle inscribed in an ellipse So i got the equation $4x^2 + 9y^2 = 3600$
What i've done so far is:
$A= (2x)(2y) = 4xy$
Then I find the expression of $y$
$9y^2= 3600 -4x^2$
$y = \pm \sqrt{3600 -4x^2 / 9} = 2/3(\sqrt {900 - x^2} \quad 2/3(900 -x^2)^{1/2}$
Then i set
$A = 4... | Alternative way by Lagrange's multipliers
*
*$4y=8\lambda x$
*$4x=18\lambda y$
since $\lambda=0$, $y=0$, $x=0$ don't lead to any solution we can divide and obtain
*
*$\frac y x = \frac 49\frac x y \implies 9y^2=4x^2 \\\implies 8x^2=3600 \implies x^2=450\implies x=\pm15\sqrt 2\implies y=\pm 10 \sqrt 2$
and th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Evaluate $\int_0^\infty \frac{x^2}{x^4 + 6x^2 + 13}dx$ In the context of the residue theorem, I have this integral to evaluate. The function is even, and $|\int_0^\pi\frac{R^2e^{2i\theta}iRe^{i\theta}}{R^4e^{4i\theta}+6R^2e^{2i\theta} + 13}d\theta| \leq \int_0^\pi2\frac{R^3}{R^4}d\theta \to 0$, so the problem is to fin... | Let us try to avoid useless computations: $x^4+6x^2+13=(x^2+\alpha)(x^2+\beta)$ for a couple of conjugated complex numbers $\alpha,\beta$ with positive real part and such that $\alpha\beta=13$ and $\alpha+\beta=6$. By partial fraction decomposition we have
$$ \int_{0}^{+\infty}\frac{x^2}{(x^2+\alpha)(x^2+\beta)}\,dx = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the $\lim\limits_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$ The task is to find $$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$
What I've tried is dividing both the numerator and the denominator by $x$, but I just can't calculate it completely.
I know it should be s... | From
$$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$
one can factor an $x$ from each term as follows:
\begin{align}
\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} &= \frac{x \left(1 + \sqrt{1 + \frac{a^{2}}{x^2}} \right)}{x \left( 1 + \sqrt{1 + \frac{b^2}{x^2}} \right)} =
\frac{1 + \sqrt{1 + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Closed form for fixed $m$ to $\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$ $I=\displaystyle\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$
Attempt:
$\dfrac{ A_0 }{ x }+\dfrac{ A_1 }{ x +1 }+\dfrac{ A_2 }{ x + 2 }...+\dfrac{ A_m }{ x +m } =\dfrac{1}{x(x+1)(x+2)(x+3)...(x+m)}$
But things got very messy.
I also thought that 1)a... | Maybe the following reduction will be helpful:
$$x(x+1)(x+2)...(x+(m-2))(x+(m-1))(x+m)=x(x+m)(x+1)(x+(m-1))(x+2)(x+m(m-1))...=(x^2+mx)(x^2+mx+1(m-1))(x^2+mx+2(m-2))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Interior Area of Circle of Circles
I am looking for the area of the white region interior to a set of circles with radius A, oriented on the edge of a larger circle with radius B, spaced apart from each other with distance C. You can assume that C is less than 2 times A, so that each smaller circle overlaps with its n... |
It's more convenient
to use the number of circles $n$
instead of the distance between their centers.
Then
\begin{align}
\phi&=\tfrac\pi{n}
,
\end{align}
the distance between the centers $|C_iC_{i+1}|=2\,R\sin\tfrac\phi2$
and
the total area $S$ of the interior region
consists of $n$ areas $S_p$ of petals $P_1OP_2$.
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How and the best way to resolve this matrix equation I have the information that: distribution of probabilities is
$$ P_{X} =\{\frac{11}{24},\frac{7}{24},\frac{1}{8},\frac{1}{8}\}$$
And the matrix has the form;
$$A= \left[
\begin{array}{cccc}
p_{1}&p_{2}&p_{2}&p_{4}\\
p_{2}&p_{1}&p_{2}&p_{5}\\
p_{3}&p_{3}&p_{2}&p_{5... | Hints: write the system as a matrix equation, say $Ax=b$, where $x=(p_1,\cdots,p_5)^t$ is a column vector of variables, $b$ is a constant column vector, consider the argumented matrix $B=[A \ b].$\begin{bmatrix}
1&0&1&1&1&\frac{3}{4}\\
\frac{11}{24}&0&0&\frac{1}{8}&0&\frac{17}{48}\\
\frac{7}{24}&0&0&0&\frac{1}{8}&\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the condition that one of the lines $ax^2+2hxy+by^2=0$ Find the condition that one of the lines $ax^2+2hxy+by^2=0$ may coincide with one of the lines $a_1x^2+2h_1xy+b_1y^2=0$.
My Attempt:
Here,
$$ax^2+2hxy+by^2=0$$
$$(\dfrac {y}{x})^{2}+\dfrac {2h}{b}.(\dfrac {y}{x})+\dfrac {a}{b}$$
Let $y=mx$ be a line represent... | Given lines are $ax^2+2hxy+by^2=0$ and $a_1x^2+2h_1xy+b_1y^2=0$
Let $y=mx$ and then we get,
First let us consider the equation $ax^2+2hxy+by^2=0$
$$ax^2+2hx(mx)+bm^2x^2=0$$
$$x^2(a+2hm+m^2b)=0$$
$$m^2b+2hm+a=0........(1)$$
Now consider the equation $a_1x^2+2h_1xy+b_1y^2=0$
$$a_1x^2+2h_1x(mx)+b_1m^2x^2=0$$
$$x^2(m^2b_1+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If sides $a$, $b$, $c$ of $\triangle ABC$ are in arithmetic progression, then $3\tan\frac{A}{2}\tan\frac {C}{2}=1$
If sides $a$, $b$, $c$ of $\triangle ABC$ (with $a$ opposite $A$, etc) are in arithmetic progression, then prove that
$$3\tan\frac{A}{2}\tan\frac{C}{2}=1$$
My attempt:
$a$, $b$, $c$ are in arithmetic ... | Hint:
$$\dfrac21=\dfrac{\cos\dfrac{A-C}2}{\cos\dfrac{A+C}2}$$
Apply Componendo and Dividendo
$$\dfrac{2+1}{2-1}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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How can I do the Euclidian's algorithm and the Extended Euclidean algorithm at the same time? This is what my lecture notes have but I cannot find anything like it online and there is no explanation in the notes. The example given is for 903 and 444.
Thank you.
| I like to write these as (simple) continued fractions.
$$ \gcd( 903, 444 ) = ??? $$
$$ \frac{ 903 }{ 444 } = 2 + \frac{ 15 }{ 444 } $$
$$ \frac{ 444 }{ 15 } = 29 + \frac{ 9 }{ 15 } $$
$$ \frac{ 15 }{ 9 } = 1 + \frac{ 6 }{ 9 } $$
$$ \frac{ 9 }{ 6 } = 1 + \frac{ 3 }{ 6 } $$
$$ \frac{ 6 }{ 3 } = 2 + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How can I evaluate $\lim_{n \rightarrow \infty} \int_n^\infty \frac{n^2 \arctan {\frac{1}{x}}}{x^2+n^2}\ dx$? I'm here wondering if this integral that our math teacher gave us (students) is even possible to evaluate? I just started to study real analysis so I find this very disturbing. Here you go, and if anyone has an... | Sorry for overlooking some constant, the following is for the integral $\displaystyle\int_{n}^{\infty}\dfrac{n\tan^{-1}(1/x)}{x^{2}+n^{2}}dx$, the integral in question will be addressed in the second part.
By change of variable $u=x/n$, then the integral is
\begin{align*}
\int_{1}^{\infty}\dfrac{\tan^{-1}(1/(nu))}{1+u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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Sum of the first n terms of series $\frac{x^{3n}}{3n(3n-1)(3n-2)}$ I need to find sum of first n terms of series $\sum_{1}^{\infty} {\frac{x^{3n}}{3n(3n-1)(3n-2)}}$. I tried but I just don't know how to transform it into any known form of power series.
EDIT: I tried partial fraction decomposition and telescoping, somet... | $$ \frac{1}{3n(3n-1)(3n-2)} = \frac{1}{2}\left[\frac{1}{3n-2}-\frac{2}{3n-1}+\frac{1}{3n}\right] = \frac{1}{2}\int_{0}^{1}\left(z^{3n-3}-2 z^{3n-2}+z^{3n-1}\right)\,dz $$
hence
$$ \sum_{n\geq 1}\frac{x^{3n}}{3n(3n-1)(3n-2)} = \frac{1}{2}\int_{0}^{1}\sum_{n\geq 1}x^{3n} z^{3n-3}(1-z)^2\,dz=\frac{1}{2}\int_{0}^{1}\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....\forall x \in \mathbb{R}$ is convergent? Is $\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....\forall x \in \mathbb{R}$ convergent? If it is convergent find the sum of the series.
Gives series $$\sin x-\frac{1}{2}\sin ^2x+\frac... | It converges to the limit $\frac{\sin x}{1+\frac{1}{2}\sin x}$ ;$a=\sin x$ and $r$(common ratio)$=-\frac{1}{2}\sin x.$ Since $|a|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find $y^{(y^2-6)}$? $$\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = y$$ $$x≠2$$ $$y^{(y^2-6)} = ?$$What is $y^{(y^2-6)}$? Could you please explain to me how to solve this question step by step?
| Calling $z = 3^{x-2}$ we have
$$
3\left(\frac{1}{1-z}+\frac{1}{1-z^{-1}}\right) = y \Rightarrow y = 3
$$
so finally
$$y^{y^2-6} = 3^{9-6} = 27$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Evaluate $\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdots(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}$ Evaluate $$\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}.$$
I can't figure out where to start. I tried using logarithms but I couldn't get a pattern going. Any advice will be helpful, thanks in ad... | Hint:
$$\frac{(5+6)+5^2}{3^1}=12$$
$$\frac{(5+6)(5^2+6^2)+5^4}{3^2}=144=12^2$$
$$\frac{(5+6)(5^2+6^2)(5^4+6^4)+5^8}{3^4}=20736=12^4$$
There is a pattern there. See if you can prove that the pattern continues.
One way is to generalize how something like $(5+6)(5^2+6^2)+5^4$ squares to become equal to $(5+6)(5^2+6^2)(5^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Why two different results to surface areal calculation using Guldin? Guldin's first rule, also known as Pappus first centroid theorem states that a linear figure creates an area when rotated that is the product of the distance the centroid is moved and the length of the figure.
I should therefore get the same resulting... | The two plane figures below are not the same:
In partricular, their centroids are not in the same spot.
The centroid of the filled-in triangle at right is at (3, 3) as you found, but the centroid of the empty triangle at left is not.
The centroid of the empty triangle is at the weighted mean of the midpoints of the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I factorise $r^4+r^2+1$? How do I factorise $r^4+r^2+1$ ?
$(r^2+r+1)(r^2-r+1)$ gives $r^4+r^2+1$
But how to split it into these factors?
I generally find roots and then write the factors, but $r^4+r^2+1$ seems to have no real root.
Thanks!
| $$x^4+x^2+1=\frac{x^6-1}{x^2-1}=\frac{(x^3-1)(x^3+1)}{(x-1)(x+1)}
=(x^2+x+1)(x^2-x+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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An inequality involving three consecutive primes Can you provide a proof or a counterexample to the following claim :
Let $p,q,r$ be three consecutive prime numbers such that $p\ge 11 $ and $p<q<r$ , then $\frac{1}{p^2}< \frac{1}{q^2} + \frac{1}{r^2}$ .
I have tested this claim up to $10^{10}$ .
For $p>5$ we get $\pi... | The inequality holds for all $p$ large enough. Let $a>1$ be such that $a^{-2}+a^{-4}=1$ and $p_n$ be the $n$-th prime. By the Prime Number Theorem there is an $N$ such that $p_{n+1}<a\,p_n$ for all $n\ge N$.If $p\ge p_N$, then $q<a\,p$ and $r<a\,q<a^2\,p$ and
$$
\frac{1}{q^2}+\frac{1}{r^2}>\frac{1}{a^2\,p^2}+\frac{1}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 1
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If $x$, $y$, and $z$ are real numbers such that $x+y+z=8$ and $x^2+y^2+z^2=32$, what is the largest possible value of $z$? I tried swapping $z$ from the first equation to the second, and got $$x^2 + x y - 8 x + y^2 - 8 y + 16=0$$ Not sure where to go from there, and if I'm on the right track at all.
| $$\max z$$
subject to $$x+y+z=8$$
$$x^2+y^2+z^2=32$$
The Lagrangian is $$z-\lambda(8-x-y-z)-\mu(32-x^2-y^2-z^2)$$
Differentiating wrt x:
$$8\lambda+2\mu x=0\tag{1}$$
Differentiating wrt $y$:
$$8\lambda+2\mu y=0\tag{2}$$
Differentiating wrt $z$:
$$1+8\lambda+2\mu z=0\tag{3}$$
If $\mu=0$, equation $(1)$ and $(2)$ gives u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2750329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
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Maximum number of circles tangent to two concentric circles In a recent math contest, the following question arose:
Two concentric circles of radii 1 and 9 make a ring. In the interior of this ring $n$ circles are drawn without overlapping, each being tangent to both of the circles of the ring. What is the largest pos... | Note that $\displaystyle \frac{\sqrt{2}}{2}<\frac{4}{5}<\frac{\sqrt{3}}{2}$. We have $\displaystyle \sin45^\circ<\sin\frac{\theta}{2}<\sin60^\circ$.
$90^\circ<\theta<120^\circ$.
Hence $\displaystyle 4>\frac{360^\circ}{\theta}>3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2750964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$. I know this question has been answered before, but I have a slightly different different question.
I saw the solution of this question in my book and the author has solved it by substituting $x-1=y$ and then equating the coefficients of $y^2$, $y^1$ and $y^0$ ... | Let the quotient be $Q(x)$ and the remainder be $ax^2+bx+c$. Then
$$(x+1)^n=(x-1)^3Q(x)+ax^2+bx+c$$
If we put $y=x-1$, then we have
$$(y+2)^n=y^nQ(y+1)+a(y+1)^2+b(y+1)+c$$
We have
$$(y+2)^n=2^n+\binom{n}{1}2^{n-1}y+\binom{n}{2}2^{n-2}y^2+\textrm{terms involving higher powers of }y$$
So, we have
$$a(y+1)^2+b(y+1)+c=2^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Show that the solutions from one quadratic equation are reciprocal to the solutions of another quadratic equation From Sullivan's Algebra & Trigonometry book: Chapter 1.2; Exercise 116:
Show that the real solutions of the equation $ax^2+bx+c=0$ are the
reciprocals of the real solutions of the equation $cx^2+bx+a=0$.... | To ensure that the equations are quadratic, we need the assumption that $a,c\ne0$.
If $\alpha$ is a root of $ax^2+bx+c=0$, then $a\alpha^2+b\alpha+c=0$.
Note that $\alpha\ne0$, otherwise, $c=0$.
So $\displaystyle c\left(\frac{1}{\alpha}\right)^2+b\left(\frac{1}{\alpha}\right)+a=\frac{a\alpha^2+b\alpha+c}{\alpha^2}=0$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find greatest value of $a^2+b^2$ If $f(x)=x^3+3x^2+4x+ a \sin x + b\cos x ~ \forall x \in \mathbb{R}$ is an injection then the greatest value of $a^2+b^2$ is _______?
To ensure injection, we must ensure that there is no maxima/minima in any interval which is equivalent to $f'(x)\neq 0$.
Note that $f'(x)=3x^2+6x+4+a \c... | Introducing $t=x+1$, rewrite the requirement on $f'(x)$ as:
$$
3t^2+1+A\cos(t+\phi) \ge 0.
$$
To maximize the value of $A=\sqrt{a^2+b^2}$ the expression should attend its minimal value as far from $t=0$ as possible, which corresponds to the choice $\phi=0$.
In this case for sufficiently large $A$ two symmetric minima o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$f(x) = \frac{x^3}{6}+\frac{1}{2x}$, $\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$ $$f(x) = \frac{x^3}{6}+\frac{1}{2x}$$
$$\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$$
Let's start by deriving the function, we have
$$f'(x) = \dfrac{x^4-1}{2x^2}$$
Hence we get
$$\int^{3}_{1} \sqrt{1 +\Big[\dfrac{x^4-1}{2x^2}\Big]^2}\, dx =... | Hint: Note that $$1+\left(\frac{x^4-1}{2x^2}\right)^2=\frac{(x^4+1)^2}{4x^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving inequalities (1)
Let $a, b, c > 0$ $;abc=8$ ,
$$\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}
+ \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}}
+ \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \ge \frac43
$$
(ref: original image)
I tried using AM-GM to eliminate the square root on the bottom but I am stuck, what is the general strategy here a... | By AM-GM $$1+a^3=(1+a)(1-a+a^2)\leq\left(\frac{1+a+1-a+a^2}{2}\right)^2=\frac{(2+a^2)^2}{4}.$$
Thus, it's enough to prove that
$$\sum_{cyc}\frac{a^2}{(2+a^2)(2+b^2)}\geq\frac{1}{3}$$ or
$$3\sum_{cyc}a^2(2+c^2)\geq\prod_{cyc}(2+a^2)$$ or
$$\sum_{cyc}(a^2b^2+2a^2)\geq72,$$ which is true by AM-GM:
$$\sum_{cyc}(a^2b^2+2a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof of Motzkin numbers recurrence I know the Motzkin numbers are given as $$M_n=\sum_{k=0}^{\lfloor n/2\rfloor}{n \choose 2k}C_k,$$ where $$C_k=\frac{1}{1+k}{2k\choose k}.$$
The recurrence relation is given as $M_n=M_{n-1}+\sum_{k=0}^{n-2}M_kM_{n-2-k}$. Is there a direct way to prove this recurrence using either Pasc... | One possible approach starts from
$$M_n = \sum_{k=0}^{\lfloor n/2\rfloor} {n\choose 2k} C_k.$$
This is
$$\sum_{k=0}^{\lfloor n/2\rfloor} {n\choose n-2k} C_k
\\ = \sum_{k=0}^{\lfloor n/2\rfloor} C_k [z^{n-2k}] (1+z)^n C_k
= [z^n] (1+z)^n \sum_{k=0}^{\lfloor n/2\rfloor} C_k z^{2k}.$$
Now when $2k\gt n$ there is no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Order of accuracy for trapezoidal integration I have a question regarding the order of accuracy for certain function using the trapezoidal formula.
I know that from theory the formula is second order accurate, but when working with matlab I get different answers.
I calculated the order of accuracy by plotting the logar... | You can compute the error in the trapeze method by using integration by parts, let $f:(0,1) \to \Bbb R$.
Then the error is the sum of local errors:
$$E=\sum_{k=0}^{n-1} \left(\int_\frac{k}{n}^\frac{k+1}{n}\left(f(t)-\frac{f\left(\frac{k}{n}\right)+f\left(\frac{k+1}{n}\right)}{2} \right) dt \right)$$
integrating by part... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Quick way to determine if a piecewise-defined function is injective/surjective. Let $f(x): \mathbb{R} \to \mathbb{R}$ be the map determined by:
$f(x)=\begin{cases} x &x \ge 2\\
\frac{x^3}{4} &-1\le x < 2\\
x &x < -1\end{cases}$
Is there an easy way to determine if this function is injective and surjective?
Injectivit... | injectivity: If $x \geq 2$, then $f(x) = x \geq 2$. Moreover, if $f(x_1) = f(x_2)$, then $f(x_1) = x_1 = x_2 = f(x_2)$.
If $x < - 1$, then $f(x) = x < -1$. Moreover, if $f(x_1) = f(x_2)$, then $f(x_1) = x_1 = x_2 = f(x_2)$.
It remains to establish that if $-1 \leq x < 2$, then $-1 \leq x < 2$ and that, if this is th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral: $\int\frac{x^2}{\sqrt{4-x^2}} dx$ So, I am really just having one issue with this integral, but I will go through the steps I have taken.
Consider $$\int\frac{x^2}{\sqrt{4-x^2}} dx$$
First, I set $x = 2\sin\theta$, found $dx = 2\cos\theta\space d\theta$ and plugged this back in, making $$4\int \frac{\sin^2\t... | Hint
Just use that $$\sin 2\theta=2\sin \theta\cos\theta=2\left(\frac x2\right)\left(\sqrt {1-\frac {x^2}{4}}\right) =\frac {x\sqrt {4-x^2}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Infinite sum converging to 2 How do I compute
$$\sum_{r=1}^{\infty} \frac{8r}{4r^4 +1}$$
Calculating first few terms tells me that the sum converges to 2.
I have also tried squeezing the term.
| Partial fraction expansion gives us$$\begin{align*} & \sum\limits_{r=1}^n\frac {8r}{4r^4+1}=\sum\limits_{r=1}^n\frac 2{2r^2-2r+1}-\sum\limits_{r=1}^n\frac 2{2r^2+2r+1}\\ & =\left[2+\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2-2n+1}\right]-\left[\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2+2n+1}\right]\end{align*}$$Notice h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Does $\lim_ {(x,y)\to (0,0 )} \frac{x^3+y^3}{x^2 + y^2}$ exist? My solution is the following:
approaching by the y-axis:
$\lim_ {(y)\to (0),(x=0)} =\lim_ {(y) \to (0)}=\frac{0+y^3}{0^2+y^2}=y=0$
approaching by $y=x$
$\lim_ {(y)\to (0),(y=x)} =\lim_ {y=x}=\frac{x^3+x^3}{x^2+x^2}=\frac{2x^3}{2x^2}=x=0$
So I think,that th... | HINT:
$$
\frac{x^3+y^3}{x^2+y^2}=x\frac{x^2}{x^2+y^2}+y\frac{y^2}{x^2+y^2}
$$
But your method doesn't answer the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Curve of degree four touching the line Let the curve $y=x^4+Ax^3+Bx^2+Cx+D$ touches the line $y=px+q$ at x=2 and x=3 where A, B, C, D, p, q $\in$R. If the area bounded by the curve and the line is $\frac{1}{\alpha}$ then the number of divisors of $\alpha$ are.
My approach was that I used $\frac{dy}{dx}=m=p$
Then I subs... |
\begin{align}
f_1(x)&=x^4+Ax^3+Bx^2+Cx+D
,\\
f_2(x)&=px+q
.
\end{align}
We have a system of four equations
\begin{align}
f_1(2)&=f_2(2)
,\\
f_1(3)&=f_2(3)
,\\
f'_1(2)&=f'_2(2)
,\\
f'_1(3)&=f'_2(3)
,
\end{align}
which allows to express $A,B,C,D$
in terms of given numbers $p,q$
as
\begin{align}
A &= -10,\quad
B = 37... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2767626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.