Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Stuck with solving inequality to find the product of highest and lowest integer solutions The inequality in question is
$$\sqrt{(5+2\sqrt{6})^{2x}}+\sqrt{(5-2\sqrt{6})^{2x}}\leq98$$
This time our job is to find the product of highest and lowest integer solutions.
My attempt
$$\sqrt{(5+2\sqrt{6})^{2x}}+\sqrt{(5-2\sqrt{... | As $(5+2\sqrt6)(5-2\sqrt6)=1$
if $(5+2\sqrt6)^x=a>0$
$$a+\dfrac1a\le98\iff a^2-98a+1\le0$$
$a^2-98a+1=0\implies a=\dfrac{98\pm40\sqrt6}2=49\pm20\sqrt6=(5\pm2\sqrt6)^2$
$\implies(5+2\sqrt6)^{-2}=(5-2\sqrt6)^2\le a\le(5+2\sqrt6)^2 $
$\implies-2\le x\le2$
| {
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"url": "https://math.stackexchange.com/questions/2795546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$a_0=0$, $a_{n+1}=a_n-\frac{1}{2}(a_n ^2-a)\rightarrow \sqrt{a}$ using banach's theorem Let's have a sequence $a_0=0$, $a_{n+1}=a_n-\frac{a_n ^2-a}{2}$. This sequence converges for $0<a<1$ to $\sqrt{a}$.
One way of proving this is to prove that $T(x)=x-\frac{1}{2}(x ^2-a)$ is a contraction mapping.
$$x-\frac{x ^2-a}{2}... | $T$ is not a contraction on $\left[0,\frac{a+1}2\right]$.
We have
\begin{align}
|Tx - Ty| &= \left|x - \frac12(x^2 - a) - y + \frac12(y^2 -a )\right| \\
&= \left|x-y - \frac12(x^2 - y^2)\right| \\
&= \left|x-y - \frac12(x-y)(x+y)\right| \\
&= \left|(x-y)\left(1 - \frac{x+y}2\right)\right| \\
&= \left|1-\frac{x+y}2\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding the Jordan canonical form of A and Choose the correct option Let $$ A = \begin{pmatrix}
0&0&0&-4 \\ 1&0&0&0 \\ 0&1&0&5 \\ 0&0&1&0
\end{pmatrix}$$
Then a Jordan canonical form of A is
Choose the correct option
$a) \begin{pmatrix}
-1&0&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$b) \begin{pmatri... | HINT
Since all options are compatible with the check on det(A)=4, we need to determine the eigenvalues by $det(A-\lambda I)=0$ and the evaluate again the given options.
Note also that b is not a Jordan normal form.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of coefficients of even powers of x in $(1+x)^5 (1+x^2)^5$? the exact question is asking the sum of coefficients of even powers of x in the expansion of $(1+x+x^2+x^3)^5$ and in the solution this expression is simplified to $(1+x)^5(1+x^2)^5$. The solution to this question says this expression is equal to $(1+5x+10... | The sum of coefficients of even power in $P(x)$ equals $$\frac{P(1)+P(-1)}{2}$$ in your case it's $$\frac{0+2^5\cdot 2^5}{2}=2^9=512$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rearrangement inequality and minimal value of $\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}$ For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to ... | Correct me if wrong:
Another option.
Let $x \in (0,π/2)$.
$f(x):=$
$2 \dfrac{\sin^4 x}{2\sin x \cos x} +2 \dfrac{\cos^4 x}{2 \sin x \cos} =$
$\dfrac{2}{\sin 2x}×$
$((\sin^2x+\cos^2x)^2 -(1/2)\sin^2 2x)$
$=\dfrac{2}{\sin 2x}(1-(1/2)\sin^2 2x)$;
Let $y:= \sin 2x$, then $0 <y \le 1,$
and consider $g(y): = 2(1/y - (1/2)y)$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Solve the equation $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$
Solve $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$.
I am able to reduce the LHS to $\sqrt{x}=3^{\log_4(x)} \cdot \dfrac{4}{3}$. Squaring both sides do not seem to lead to a result. Do you know how to proceed?
| Easy step by step
$$\begin{align}
3^{\log_4x+\frac{1}{2}}+3^{\log_4x-\frac{1}{2}}&=\sqrt{x} \\
\sqrt{3}\cdot 3^{\log_4x} + (\sqrt 3)^{-1} \cdot 3^{\log_4x}&=\sqrt x \\
3 \cdot 3^{\log_4x}+3^{\log_4x}&=\sqrt{3x} \\
4\cdot3^{\log_4x}&=\sqrt{3x} \\
3^{\log_4x}&=\frac{\sqrt{3x}}{4} \\
4^{(\log_4 3) \cdot (\log_4 x)} &=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Prediction of index in Lexicographical ordering of binary sequences? Suppose all $n$ sequences of 0's and 1's were displayed in Lexicographical order. Can the index of sequences that sum to $k$ be predicted?
| $\textbf{Better stated question}$: given all possible sequences of $n$ coin flips (a total of $2^n$ many), arrange them lexicographically. Can we find the index of those sequences that have $k$ heads?
$\textbf{Solution}:$ First let us consider a small case example where $n = 10$ and $k = 4$:
\begin{cases}
0000001111 & ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show differential equation solution is arc of great circle I have to show that the solution of a differential equation is an arc of a great circle. The differential equation is as follows $($in spherical coordinates$)$:
$$\frac{\sin ^2\theta\phi '}{(1+\sin ^2\theta (\phi ')^2)^{\frac{1}{2}}}=C$$ where $C$ is an arbit... | Let $u=\cot \theta$, then $1+u^2=\csc^2 \theta$ and $du=-\csc^2 \theta \, d\theta$.
\begin{align}
\frac{d\phi}{d\theta} &= \frac{d\theta}{\sin \theta \sqrt{\sin^2 \theta-C^2}} \\
&= \frac{C\csc^2 \theta}{\sqrt{1-C^2\csc^2 \theta}} \\
d\phi &= -\frac{C\,du}{\sqrt{1-C^2(1+u^2)}} \\
&= -\frac{C\,du}{\sqrt{(1-C^2)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $f(x)=(2+x+ax^2)\ln (1+x)-2x$ with $f(x)$ achieving its maximum value at $x=0$. Find $a$. Problem
Let $f(x)=(2+x+ax^2)\ln (1+x)-2x$ with $f(x)$ achieving its maximum value at $x=0$. Find $a$.
Solution
At first, let's review a theorem, which states as follows:
Let $f^{(n)}(x)$ denote the $n-$order derivative of $f(... | Here's a nicer solution: using the Maclaurin series of $\log(1+x)$,
$$
-2x+\left(a x^2+x+2\right) \log (x+1) = -2x+\left(a x^2+x+2\right) \left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O\left(x^5\right)\right)
$$
$$
=\left(a+\frac{1}{6}\right) x^3+\left(-\frac{a}{2}-\frac{1}{6}\right)
x^4+O\left(x^5\right)
$$By y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5$
Let $P(x)$ be a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5.$ Determine the value of $P(6)$.
Let $P(x) = ax^4 + bx^3 + cx^2 + dx + e$. For $n=1,2,3,4,5$ I have plugged it into this polynomial and got the follo... | Hint: We can write $P(x)$ in the form of
$$P(x) =c_1(x-2)(x-3)(x-4)(x-5)\ +\ c_2(x-1)(x-3)(x-4)(x-5)\ +\dots+\ c_5(x-1)(x-2)(x-3)(x-4)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Finding out the last digit of number $3a/4$ knowing the last two digits of $a$ Natural number $a$ ends with digits $16$. If $a$ is not divisible by $8$, what is the last digit of $3a/4$?
Solution:
$$7$$
How do we arrive at that solution?
| Note that $a = 100k + 16$ where $k$ is a positive integer.
Recall the divisibility criteria for $ 8$ : a number is divisible by $8$ if and only if its last three digits are.
Let $b$ be the last digit of $k$. Then, the last three digits of $a$ are $\overline{b16}$.
When is $\overline{b16}$ divisible by $8$? It is not d... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $\lim_{x\to1}=\frac{x^2+x-2}{1-\sqrt{x}}$? let $f(x)=\dfrac{x^2+x-2}{1-\sqrt{x}}$
How do I solve this limit?
$$\lim_{x\to1}f(x)$$
I can replace the function with its content
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}$$
Then rationalizing the denominator
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}\cdot\dfrac{1... | Observe that
$$ x^2 + x - 2 = (x-1)(x+2). $$
Therefore
\begin{align} \lim_{x\to 1}\frac{(x^2+x-2)(1+\sqrt{x})}{1-x}
&= \lim_{x\to 1} \frac{(x-1)(x+2)(1+\sqrt{x})}{1-x} \\
&= \lim_{x\to 1} \frac{-(1-x)(x+2)(1+\sqrt{x})}{1-x} \\
&= -\lim_{x\to 1} (x+2)(1+\sqrt{x}) \\
&= -(1+2)(1+\sqrt{1}) && \text{($\ast$)}\\
&= -6. \end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}} $ $$\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}} $$
I tried writing it as $\lim\limits_{x \to 0} \dfrac{\dfrac{\tan^{12}x}{x^{12}}-1}{x^{2}} $ thinking that if I applied L'hospital two times I would reach my result but with no luck. Any suggestions or tricks... | $\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}}
$
$\begin{array}\\
\lim_{x \to 0} \dfrac{\tan^{12}x-x^{12}}{x^{14}}
&=\lim_{x \to 0} \dfrac{(\tan(x)/x)^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\sin(x)/(x\cos(x))^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\frac{\sin(x)}{x\cos(x)})^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819047",
"timestamp": "2023-03-29T00:00:00",
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Without relating the problem to polynomials, how can we solve this inequality? $a,b,c$ are real;
$a<b<c$ ;
$a+b+c=6$ ;
$ab+bc+ca = 9$.
Prove that $0<a<1<b<3<c<4$.
I solved this by treating $a,b,c$ as roots of $x^3 - 6x^2 + 9x + d$ for some $d$, but I have not been able to solve it in a more direct way.
| I want to note that solving this problem using polynomials may be the most natural and, in fact, direct way to do. You can for example prove each inequality individually. To show $a>0$, you have $$0 < (b-c)^2=(b+c)^2-4bc=(6-a)^2-4\big(9-a(b+c)\big)=(6-a)^2-4\big(9-a(6-a)\big)\,,$$
which gives $3a(4-a)>0$, whence $a>0... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\cosh(x)\cosh(y) \geq \sqrt{x^2 + y^2}$ I'm trying to prove the inequality :
$\cosh(x)\cosh(y) \geq \sqrt{x^2 + y^2}$
I played a bit with Geogebra and it looks true. I've tried proving it via convexity but so far I'm unsuccesful.
Thank you for your help.
| By Taylor's theorem with the Lagrange form of the remainder,
$$ \cosh{x} = 1 + \frac{x^2}{2}\cosh{\xi} $$
for some $\xi$ with $0 \leq \lvert\xi\rvert \leq \lvert x\rvert$. Hence $\cosh{x} \geq 1+\frac{1}{2}x^2$. Thus
$$ \cosh{x}\cosh{y} \geq (1+\tfrac{1}{2}x^2)(1+\tfrac{1}{2}y^2) \geq 1 + \frac{1}{2}(x^2+y^2), $$
since... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Connection Formula for Hypergeometric Function 2F1 Suppose I have the function $_2F_1\left(a,b;c;x^2\right)$ with $a=\frac{3}{4}+\frac{k}{4}$, $b=\frac{3}{4}-\frac{k}{4}$ and $c=\frac{1}{2}$.
I want to know the behaviour about $x=1.\,$
I go to DLMF equation 15.10.21 and choose
$$
w_1\left(x^2\right) = {\frac {\Gamma \l... | Let
$$F(x) = {_2F_1}\left( \frac {3+k} 4, \frac {3-k} 4; \frac 1 2; x \right),
\quad k \geq 0.$$
If $(3-k)/4$ is an integer, $F(x)$ becomes a polynomial and we have
$$F(x) = F(1) + O(|1-x|) =
\frac {(-1)^{(k-3)/4} \sqrt \pi \,\Gamma \left( \frac {k+5} 4 \right)}
{\Gamma \left( \frac {k-1} 4 \right)} + O(|1-x|).$$
Other... | {
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"timestamp": "2023-03-29T00:00:00",
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Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$? Please provide proof
The answer in the back of the book is different to both my calculations and also the online calculator I crosschecked my answer with...
The Question:
"Differentiate with respect to $x$:"
$
(x^3+2x^2+x)^4
$
My Answer: $4(x^3+2x^2+x)^3(3x^2+4x+1... | Note that $(x^3+2x^2+x)=x(x^2+2x+1)=x(x+1)^2$.
Also, $3x^2+4x+1=(x+1)(3x+1)$
Apply those factorizations to your answer, and collect like factors together; you should get the book's answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the inverse of $x^3+x^2+1$ in $\Bbb F_2[x]/(x^4+x^2)$ with the Euclidean algorithm My first thought was successful: $x^4+x^2=x^2(x^2+1)$ and $x^3+x^2+1=x^2(x+1)+1$ so it is its own inverse because $(x^2(x+1)+1)^2\equiv x^4(x+1)^2+1\equiv x^4(x^2+1)+1\equiv1.$
The given solution claims to use the Euclidean algor... | A possible way to "visualize" what happens, in analogy with the same algorithm for (the ring of) integers, is to (formally) use the formalism of continued fractions,
so let us write:
$$
\begin{aligned}
\frac{x^4+x^2}{x^3+x^2+1}
&=
(x+1)+\frac{x+1}{x^3+x^2+1}
\\
&=
(x+1)+\frac1{\frac{x^3+x^2+1}{x+1}}
\\
&=
(x+1)+\frac1{... | {
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"url": "https://math.stackexchange.com/questions/2824411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\sin (x/2)+2\sin (x/4)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$
Given that $$f (x)=\sin (x/2)+2\sin (x/4)$$
Show that $f (x)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$.
My try
$f (x)=\sin (x/2)+2\sin (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}... | Put $\theta = \dfrac{x}{4} \implies \text{ We prove :} f(\theta) = \sin (2\theta)+2\sin(\theta) \le \dfrac{3\sqrt{3}}{2}, \theta \in [0,\pi/2]$ . Put $t = \sin(\theta)\implies t \in [0,1], f(t) = 2t\sqrt{1-t^2}+ 2t\implies f'(t)=2\sqrt{1-t^2}-\dfrac{2t^2}{\sqrt{1-t^2}}+2= 0 \iff t = 0, \dfrac{\sqrt{3}}{2}$ . We have $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Approximate integral: $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx.$
Approximate integral: $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx.$
My attempt:
Let I = $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx$
$u=7-x\implies I=\frac 72 \int_3^4 \frac 1{\sqrt{(7x-10-x^2)^3}}dx$ Now I have to approximate this...
I got to this poi... | I will try to solve the problem.
The integrand has the form $P_1 \cdot Q_2 ^ {- 3/2} $.
It is logical to assume that the antiderivative has the form $ R_1 \cdot Q_2 ^ {- 1/2} $.
Differentiate and equate.
$\bigg((ax+b)(7x-10-x^2)^{-\frac{1}{2}}\bigg)'=x\cdot (7x-10-x^2)^{-\frac{3}{2}}$
So $\displaystyle a(7x-10-x^2)^{-\... | {
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"timestamp": "2023-03-29T00:00:00",
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Bounds for an integral I am trying to show that $$\frac{1}{5} < \int_5^8 \frac{2x-7}{2x+5} dx <1$$ since for the integral $$5\le x \le 8 \rightarrow 15\le 2x+5 \le 21$$$$-\frac{12}{15}\le-\frac{12}{2x+5}\le -\frac{12}{21}\rightarrow \frac{3}{15}\le 1- \frac{12}{2x+5}\le \frac{3}{7}$$ By taking integral $$\frac{3}{5}\le... | We wish to show that $$\int_5^8{\frac{2x-7}{2x+5}dx}<1.$$ Let $g:(0,\infty)\to\mathbf R$ be given by $$g(x)=1-\int_5^x{1-\frac{12}{2w+5}dw}.$$ Then $$g'(x)={12\over{2x+5}}-1$$ and $$g''(x)={-24\over{(2x+5)^2}}<0\,\, \forall x.$$ Thus, $g$ is concave. Also, $g'=0$ at $x=3.5$, so that $g$ has its turning point $\left( g'... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Showing that $4b^2+4b = a^2+a$ has no non-zero integer solutions? The problem is
Show that $4b^2+4b = a^2+a$ has no integer solutions where none of $a, b$ are zero.
I have a solution but I think there must be some better ways:
My Solution:
$$4b^2+4b = a^2+a$$
$$(2b+a)(2b-a)+4b-a= 0$$
Now letting $x = 2b + a$ and $y =... | Multiply by $4$ and complete the square on both sides. This gives
$(4b+2)^2-4=(2a+1)^2-1$
$(4b+2)^2-(2a+1)^2=3$
What are the only two squares differing by exactly $3$?
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$ If the circle $(x-a)^2+(y-b)^2=r^2$ and the line $y=mx+c$ do not meet:
Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$
These are the steps I have thus taken (although they may be wrong/useless):
*
*Rearranged the circle equation for y:
$y=\sqrt{r^2-x^2+2ax-a^... | The line is outide the circle hance putting the line equation inside the formula for the circle is larger then $r^2$ hence
$$
(x - a)^2 + (mx + c - b)^2 > r^2
$$ l.h.s is minimized for $2(x-a) + 2m(mx + c - b) = 0$ or for $x = (m(b - c) + a)/(m^2+1))$. let $f=m^2+1$, then the expression is minimized by
$$
(m(b - c) + a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Q: Proving Existence I'm currently stuck on a problem right now for my Intro to Proofs Class. The problem says:
Let $a,b ∈ ℕ$. Prove that if $a+b$ is even, then there exists nonnegative integers $x$ and $y$ such that $x^2-y^2= ab$.
So far I've tried it directly, and by contrapositive and came to a similar road block.
D... | Well, brainstorm first.
$x^2 - y^2 = ab$
$(x - y)(x + y) = ab$. Can I say $x - y = a$ and $x+y = b$? Why or why not? Well, first off $x - y$ would have to equal the smaller of $a$ or $b$ and $x + y$ would have to be the larger. But we can assume without loss of generality that $a \le b$. So can we say that $x, y$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Finding critical points of $f(x,y)= \sin x+\sin y + \cos(x+y)$
Find the critical points of function$$
f(x,y)=\sin x + \sin y + \cos(x+y),$$
where $0<x<\dfrac{\pi}{2}$, $0<y<\dfrac{\pi}{2}$.
What I have done:
$$f_{x}=\cos(x)-\sin(x+y),\\
f_{y}=\cos(y)-\sin(x+y).$$
From $f_{x}=0$, $\cos(x)=\sin(x+y)$. From $f_{x}=0$,... | Hint:
$$\frac{df}{dx}=\cos x - \sin(x+y) = 0$$
$$\frac{df}{dy}=\cos y - \sin(x+y) = 0$$
Exploding the $\sin$s
$$\cos x - \sin x \cos y - \cos x \sin y = 0$$
$$\cos y - \sin x \cos y - \cos x \sin y = 0$$
Dividing the first with $\cos x$, the second with $\cos y$:
$$1 - \tan x \cos y - \sin y = 0$$
$$1 - \sin x - \tan x... | {
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"url": "https://math.stackexchange.com/questions/2832009",
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"source": "stackexchange",
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"answer_count": 2,
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Then general values of $\theta$ in inverse Trigo sum
If $\displaystyle \theta = \tan^{-1}\bigg(2\tan^2 \theta\bigg)-\frac{1}{2}\sin^{-1}\bigg(\frac{3\sin 2 \theta}{5+4\cos 2 \theta}\bigg).$ Then general values of $\theta$ is
Try: Let $\alpha =\tan^{-1}\bigg(2\tan^2 \theta\bigg)\Rightarrow \tan \alpha =2 \tan^2 \thet... | Hint:
$$\dfrac{3\sin2t}{5+4\cos2t}=\dfrac{6\tan t}{9+\tan^2t}=\dfrac{2\cdot\dfrac{\tan t}3}{1+\left(\dfrac{\tan t}3\right)^2}=\sin2y$$ where $3\tan y=\tan t$
$$\implies\sin^{-1}\dfrac{3\sin2t}{5+4\cos2t}=\begin{cases}2\tan^{-1}\dfrac{\tan t}3 &\mbox{if }-1\le\dfrac{\tan t}3\le1\\
\pi-2\tan^{-1}\dfrac{\tan t}3 & \mbox{i... | {
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"source": "stackexchange",
"question_score": "2",
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Prove that $\cos{x} \cdot \cos{\sqrt{3}x} = 1$ has exactly one solution at $x=0$
Prove that $\cos{x} \cdot \cos{\sqrt{3}x} = 1$ has exactly on solution at $x=0$
I've recently asked a question on periodicity of the $f(x) = \cos{x}\cdot\cos{\sqrt{3}x}$
One of my statements was:
But this equation has only one solution... | $$\cos \sqrt 3x \cdot \cos x
= \frac 12 \cos(\sqrt 3 + 1)x + \frac 12 \cos(\sqrt 3-1)x$$
So $\cos \sqrt 3x \cdot \cos x = 1 \iff
\cos(\sqrt 3 + 1)x + \cos(\sqrt 3-1)x = 2$
This can only happen when $\cos(\sqrt 3 + 1)x = \cos(\sqrt 3-1)x = 1$
So, for some integers $m$ and $n$,
\begin{align}
(\sqrt 3 + 1)x... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Confusion about roots Solving $$z^4 + 1 = 0$$ where $z = x+iy$, I tried by factorizing the expressiong into $$z^4 + 1 = (z^2 - 2z + 2)(z^2 + 2z + 2).$$ But then solving $$z^2 - 2z + 2 = 0$$ for $z$ gives the solutions $z = 1 \pm i$, and solving $$z^2 + 2z + 2 = 0$$ for $z$ gives the solutions $z = -1 \pm i$. The actual... | Your starting factorization is not correct. It should be:
$$z^4+1 = z^4+2z^2+1-2z^2 = (z^2+1)^2-(\sqrt{2}z)^2=...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is
If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$
Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is
Try:
From $$(a_{1}-a_{2})^2+(a_{2}-a... | Since $$a_{1}+a_{2}+a_{3}+a_{4}=0,$$ we can denote $$a_1+a_3=-(a_2+a_4)=t.$$Thus,
\begin{align}
&(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2\\=&2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})-2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})\\
=&2-2(a_{1}+a_{3})(a_{2}+a_{4})\\
=&2+2t^2.
\end{align}
Notice that
$$a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842697",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Evaluate the Determinants A? Evaluate the Determinants
$$A=\left(\begin{matrix}
1 & 1 & 0 & 0 & 0\\
-1 & 1 & 1 & 0 & 0\\
0 & -1& 1 & 1 & 0\\
0& 0 & -1 & 1 & 1\\
0 & 0 & 0 & -1 & 1\\
\end{matrix}\right)$$
My attempst : I was thinking abouts the
Schur complement https://en.wikipedia.org/wiki/Schur_co... | Given $A=\left(\begin{matrix}
1 & 1 & 0 & 0 & 0\\
-1 & 1 & 1 & 0 & 0\\
0 & -1& 1 & 1 & 0\\
0& 0 & -1 & 1 & 1\\
0 & 0 & 0 & -1 & 1\\
\end{matrix}\right)$
To find the determinant you need to find the upper triangular matrix and then multiply the diagonal elements of the matrix.
$$=\begin{vmatrix}
1 & 1 & 0... | {
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"answer_count": 4,
"answer_id": 3
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$a+b+c=1.$ Show that $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Assume $a,b,c>0$ and $a+b+c=1.$ Show that
$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq
\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Here's what I tried:
$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+b}+\frac{a+b+c}... | From Jensen's Inequality with the convex function $f(t):= \frac{1}{1-t}$ for $t\in (0,1)$, we have
$$\frac{a}{1-c}f(a)+\frac{b}{1-c}f(b)\geq f\left(\frac{a^2}{1-c}+\frac{b^2}{1-c}\right)\,.$$
Since $f$ is increasing, the Power-Mean Inequality $\frac{a^2+b^2}{2}\geq \left(\frac{a+b}{2}\right)^2$ implies that
$$f\left(\f... | {
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"source": "stackexchange",
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Calculation of the modulus of elasticity of a stretched string I posted concerning this question a little while ago, not asking for the answer but for an understanding of the setup. Well, I thought I would solve it but I seem to be unable to obtain the required answer. If anyone could help I'd be very grateful.
Here's ... | Assuming the Hooke's law
$$
F = \lambda\left(\frac{l-l_0}{l_0}\right)
$$
Calling
$$
|AC|_0 = a\frac 43\\
|CB|_0 = a\frac 47\\
|AC| = 2a\sin(\frac{\pi}{3})\\
|CB| = 2a\sin(\frac{\pi}{6})\\
F_{AC} = \lambda\left(\frac{|AC|-|AC|_0}{|AC|_0}\right)(-\cos(\frac{\pi}{6}),\sin(\frac{\pi}{6}))\\
F_{CB} = \lambda \left(\frac{|CB... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ a. at $\frac{1}{4}<|z|<1$ b. $|z|>1$
I know that $\mathbb{C}$ is a closed algebraic field so we can write the polynomial has a product of first degree polynomials, so we will have to guess one root and divide and find the others. but ... | Using Rouche's theorem, in $|z|\leq\dfrac14$ we consider $f(z)=z^3-2z^2$ and $g(z)=\dfrac14$ then
$$|f(z)|=|z^3-2z^2|\leq|z|^3+2|z|^2=\dfrac{9}{64}<|g(z)|$$
then $z^3-2z^2+\frac14=0$ hasn't zero in $|z|\leq\dfrac14$.
In $|z|\leq1$ with $f(z)=z^3+\dfrac14$ and $g(z)=-2z^2$ then
$$|f(z)|=|z^3+\dfrac14|\leq\dfrac54<2=|g... | {
"language": "en",
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"source": "stackexchange",
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"answer_count": 3,
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Trying to evaluate $\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{dx}{1+x^3}$ I would like to work this out:
$$I=\large\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{\mathrm dx}{1+x^3}$$
Making a sub: $u=x^3$, $dx=\frac{du}{3x^2}$
$$I=\frac{1}{3}\int_{0}^{\infty}\frac{\ln(1+u)}{u^{2/3}(1+u)^2}\mathrm du$$
Making a s... | The substitution $t = (1 + x^3)^{-1}$ yields
$$ I = \frac{1}{3} \int \limits_0^1 - \ln(t) \left(\frac{t}{1-t}\right)^{2/3} \, \mathrm{d} t = f'\left(\frac{2}{3}\right) \, , $$
where
$$ f(\alpha) \equiv - \frac{1}{3} \int \limits_0^1 \frac{t^\alpha}{(1-t)^{2/3}} \, \mathrm{d} t = - \frac{1}{3} \operatorname{B} \left(\fr... | {
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"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
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Show that the system has a limit cycle. The full question reads:
Show that the system with $\dot{x} = x-y-(x^2 + \frac{3}{2} y^2)x$ and $\dot{y} = x+y-(x^2 + \frac{1}{2} y^2)y$ has a limit cycle.
I wanted to check the correctness of my method. My thought was to find the equilibrium points (If I am correct, the only one... | Yes, your approach looks good to me. If $r^2 = x^2 + y^2$ then we have
$r\dot{r} = x\dot{x} + y\dot{y} = x^2(1-x^2-\frac{3}{2}y^2) + y^2(1-x^2-\frac{1}{2}y^2)$
$\Rightarrow r(1-x^2-\frac{3}{2}y^2) \le \dot{r} \le r(1-x^2-\frac{1}{2}y^2)$
and since $1-\frac{3}{2}r^2 \le 1-x^2 - \frac{3}{2}y^2 \le 1-x^2-\frac{1}{2}y^2 \l... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find number of integers satisfying $x^{y^z} \times y^{z^x} \times z^{x^y}=5xyz$ Find number of integers satisfying $$x^{y^z} \times y^{z^x} \times z^{x^y}=5xyz$$
My try:
we can rewrite the equation as
$$x^{y^z-1} \times y^{z^x-1} \times z^{x^y-1}=5$$
Then since all are integers we get
$x=5$, $y=1$ ,$z=1$ and like wise ... | The factor of $5$ must come from somewhere, so $x, y$ or $z$ must have a factor of $5$ (let's just assume it's $x$, as the expression is symmetric). The power of $x$, i.e. $y^z - 1$, must be strictly positive. If there were any other prime factor of $x$, because $x$ is raised to a positive integer power, it would have ... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got a... | Just for a variation, using Lagrange’s method:
$$
f(a,b,t)=\frac{b+1}{a+b-2}-t(a^2+b^2-1)
$$
Then
\begin{align}
\frac{\partial f}{\partial a}&=-\frac{b+1}{(a+b-2)^2}-2at \\[6px]
\frac{\partial f}{\partial b}&=\frac{a-3}{(a+b-2)^2}-2bt
\end{align}
If these equal $0$, then
$$
-\frac{b+1}{a(a+b-2)^2}=\frac{a-3}{b(a+b-2)^2... | {
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"source": "stackexchange",
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"answer_id": 3
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Can we conclude that two of the variables must be $0$?
Assuming $$a^2+b^2+c^2=1$$ and $$a^3+b^3+c^3=1$$ for real numbers $a,b,c$, can we conclude that two of the numbers $a,b,c$ must be $0$ ?
I wonder whether mathworld's result that only the triples $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ satisfy the given equation-system ... | Since $|a|,|b|,|c|\in[0,1]$ (because $a^2+b^2+c^2=1$), we have
$$
a^2(1-a)+b^2(1-b)+c^2(1-c)=0
$$
The LHS is a sum of non-negative terms; to be zero, all the terms must be zero. This means that
$$
a^2(1-a)=b^2(1-b)=c^2(1-c)=0
$$
and therefore $|a|,|b|,|c|\in\{0,1\}$. Along with $a^2+b^2+c^2=1$, we get that exactly one... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Minimizing in 3 variables
Find the least possible value of the fraction $\dfrac{a^2+b^2+c^2}{ab+bc}$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3\sqrt[3]{-2ac/b(a+c)}$
And I cant somehow move on.
| Yet another approach: The expression does not change if $(a, b, c)$
are multiplied by a common factor, so we can assume that $a+c=2$. Then
$$
\frac{a^2+b^2+c^2}{ab+bc} = \frac 12 \left (\frac{a^2+(2-a)^2}{b} + b
\right) \ge \sqrt{a^2 + (2-a)^2} = \sqrt{2(a-1)^2 + 2} \, ,
$$
using $AM \ge GM$. It follows that
$$
\fra... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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prove this inequality $(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$ Let $x,y,z>0$,show that
$$(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$$
I have prove this inequality
$$(x+y-z)(y+z-x)(x+y-z)\le xyz$$
because it is three schur inequality
$$\Longleftrightarrow x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x)$$
how to solve ... | It's enough to prove our inequality for $\prod\limits_{cyc}(x+y-z)\geq0.$
Now, if $x+y-z\leq0$ and $x+z-y\leq0$ then $x+y-z+x+z-y\leq0$ or $2x\leq0$,
which is a contradiction.
Thus, we can assume that $x+y-z>0$, $x+z-y>0$ and $y+z-x>0.$
Now, let $x+y-z=c$, $x+z-y=b$ and $y+z-x=a$.
Thus, we need to prove prove that
$$27... | {
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"url": "https://math.stackexchange.com/questions/2853415",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Solve Recurrence Relation $a_k=(a_{k-1})^2-2$ $$a_k=\left(a_{k-1}\right)^2-2$$
$a_0=\frac{5}{2}$
Then find $$P=\prod_{k=0}^{\infty} \left(1-\frac{1}{a_k}\right)$$
My try:
I rewrote the Recurrence equation as
$$a_k+1=(a_{k-1}-1)(a_{k-1}+1)$$ $\implies$
$$\frac{1}{a_{k-1}-1}=\frac{a_{k-1}+1}{a_k+1}$$ $\implies$
$$\frac{a... | Hint: There are three steps. First, for each $k=0,1,2,\ldots$, show that $$a_k=2^{2^k}+\frac{1}{2^{2^k}}\,.$$
Second, write $$1-\frac{1}{a_k}=\frac{a_k-1}{a_k}=\left(\frac{a_{k+1}+1}{a_k+1}\right)\frac{1}{a_k}\,,$$
for all $k=0,1,2,\ldots$.
Finally, show that
$$\prod_{k=0}^n\,a_k=\frac{2}{3}\left(2^{2^{n+1}}-\frac{1}{... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Determining $\sin(2x)$
Given that
$$\sin (y-x)\cos(x+y)=\dfrac 1 2$$
$$\sin (x+y)\cos (x-y) = \dfrac 1 3 $$
Determine $\sin (2x)$.
As stated in my perspective, the question does not make any sense. We know that the double angle identity for $\sin(2x)$ is given by
$$\sin(2x) = 2\sin\cos$$
Let us try simpiflying the s... | You were on the right track but the equations contain the product, not the difference.
\begin{align}
\frac12 &= \sin (y-x)\cos(x+y)\\
&=(\sin y\cos x - \sin x \cos y)(\cos x\cos y - \sin x\sin y) \\
&= \sin y\cos y - \cos x \sin x
\end{align}
\begin{align}
\frac13 &= \sin (x+y)\cos(x-y)\\
&=(\sin x\cos y + \sin y \cos ... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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knowing: $\cos x+\sin x=\frac{5}{4}$, obtain: $\cos(4x)$
knowing: $\cos x+\sin x=\frac{5}{4}$,
obtain: $\cos(4x)$
$$\cos x+\sin x=\frac{5}{4}$$
$$\sin^2x+\cos^2x+2\sin x\cos x=\frac{25}{16}$$
$$\sin2x=\frac{25}{16}-\frac{16}{16}=\frac{9}{16}$$
$$\cos4x=1-2\sin^22x=1-2\Bigl(\frac{9}{16}\Bigr)^2=\frac{47}{128}$$
Taken ... | Another way to do this is as follows-
$\cos x+\sin x=\frac{5}{4}$, obtain: $\cos(4x)$
$$\Rightarrow 4\cos x + 4\sin x = 5$$
$$\Rightarrow 16 \cos^2x +25-40 \cos x = 16 \sin x$$
$$\Rightarrow 15 \cos^2x -56 \sin x+25=0$$
$$\Rightarrow \cos x = \frac{\sqrt{409}}{15}$$
Now, by knowing the value of x, the cosine value of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
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Give an asymptotic developement of $I_n=\int_0^1 (x^{n}-x^{n-2})\ln(1+x^n)dx.$ Let $$I_n=\int_0^1 (x^n+x^{n-2})\ln(1+x^n)dx.$$
Give an asymptotic developement of $I_n$ at order $O(\frac{1}{n^3})$ when $n\to \infty $.
I wanted to use the fact that $$\sum_{k=1}^\infty \frac{(-1)^{k+1} x^{kn}}{k},$$
and thus $$I_n=\int_0^... | Assuming that
you can reverse the order,
$\begin{array}\\
I_n
&=\int_0^1(x^n+x^{n-2})\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^{kn}dx\\
&=\sum_{k=1}^\infty \int_0^1(x^n+x^{n-2})\frac{(-1)^{k+1}}{k}x^{kn}dx\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k} \int_0^1(x^n+x^{n-2})x^{kn}dx\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k} \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Calculate boundary of two domains $D_1 =\{ -1 \le x \le 1-\sqrt{y^2+z^2}\}$
$D_2 =\{ x^2+y^2+z^2\le 8,z\ge 2\}$
Is it correct to say that the boundary of $D_1$ is :
$B(D_1) =\{ (x-1)^2=y^2+z^2,(x+1)^2=y^2+z^2\}$
and the boundary of $D_2$ :
$B(D_2) =\{z\ge 2 ,x^2+y^2+z^2=8 \} \cup \{x^2+y^2=4\}$
?
Especially the last ... | $C= \{x \in \mathbb R^3 ; x^2+y^2=4\}$ is not a subset of $B(D_2)$. $C$ is the right circular cylinder of axis $Oz$ and with radius $2$. $C$ is unbounded while $D_2$ is bounded. What is true is that
$$B(D_2) = \{z\ge 2 ,x^2+y^2+z^2=8 \} \cup \{z=2, x^2+y^2 \le 4\}.$$
Your $B(D_1)$ isn't correct either. To "compute" th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2860296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the positive value of $x$ satisfying the given equation $${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = \sec(y)$ but couldn't even come close to the solution.
| Let's see how insurmountable this sextic equation is.
Start by deriving the equation. First multiply by $\sqrt{x}$, square both sides, and isolate the remaining radical:
$(x^2-1)+2\sqrt{(x^2-1)(x-1)}+(x-1)=x^3$
$2\sqrt{(x^2-1)(x-1)}=x^2-x^2-x+2$
Square again, expand and collect to get our monster:
$x^6-2x^5-x^4+2x^3+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Reduction of Order via Substitution
Suppose $u_1=\sin{x^2}$ is a solution of
$$xu''-u'+4x^3u=0\Rightarrow u''-x^{-1}u'+4x^2u=0 \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
I am trying to find a second, linearly independent, solution (say $u_2$) to the above equation using reduction of order.
Now, using the formula
$$u_2=u_... | The ODE for $w$ is actually separable:
\begin{align*}
\sin(x^2)\frac{dw}{dx} + \left(4x\cos(x^2) - \frac{\sin(x^2)}{x}\right)w & = 0 \\
\int \frac{1}{w}\, dw & = \int \frac{1}{\sin(x^2)}\left(\frac{\sin(x^2)}{x} - 4x\cos(x^2)\right)\, dx \\
\int\frac{1}{w}\, dw & = \int\frac{1}{x} - 4x\cot(x^2)\, dx \\
\end{align*}
Ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int_{-1}^{1} \cot^{-1} \left(\frac{1}{\sqrt{1-x^2}}\right) \cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-(x^2)^{|x|}}}\right)$
$$\int_{-1}^{1} \left(\cot^{-1} \dfrac{1}{\sqrt{1-x^2}}\right) \left(\cot^{-1}\dfrac{x}{\sqrt{1-(x^2)^{|x|}}}\right)= \dfrac{\pi^2(\sqrt a-\sqrt b )}{\sqrt c}$$
, where a,b, c are na... | I usually see $\cot^{-1}(-x)=-\cot^{-1}(x)$, but working with what is stated in the question,
$$
\begin{align}
&\int_{-1}^1\cot^{-1}\left(\frac1{\sqrt{1-x^2}}\right)\cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-\left(x^2\right)^{|x|}}}\right)\,\mathrm{d}x\tag1\\
&=\int_0^1\cot^{-1}\left(\frac1{\sqrt{1-x^2}}\right)\cot^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $... | $$
\cos x-\cos(3x) = \cos x\left(\sin^2x+3\sin^2 x\right)
$$
then
$$
\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} = \frac{x^2}{x^2}\cos x\left(\frac{\sin^2x+3\sin^2 x}{\sin (3x^2)-\sin (x^2)}\right)=\cos x\left(\frac{\left(\frac{\sin x}{x}\right)^2+3\left(\frac{\sin x}{x}\right)^2}{\frac{3\sin(3x^2)}{3x^2}-\frac{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 7
} |
Evaluate the intgral $\int{\frac{dx}{x^2(1-x^2)}}$ (solution verification) I have to solve the following integral
$$\int{\frac{dx}{x^2(1-x^2)}}$$
What I've got:
\begin{split}
\int{\frac{dx}{x^2(1-x^2)}} &=\int{\frac{(1-x^2+x^2)dx}{x^2(1-x^2)}}\\
&=\int{\frac{dx}{x^2}}+\int{\frac{dx}{1-x^2}}\\
&=\int{\frac{dx}{x^2}}+\in... | Don't forget $\dfrac1i$ before $\arctan$
$$\int{\frac{dx}{x^2(1-x^2)}}=-x^{-1}+\dfrac{1}{i}\arctan{xi}+C$$
Also better to write
$$\int\dfrac{1}{1-x^2}dx=\dfrac12\ln\dfrac{1+x}{1-x}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Find the values of $x$ and $y$ that satisfies $\sin(x+y)=\sin x+\sin y$. I know that in general the following equality does not hold: $\sin(x+y)=\sin x + \sin y$. However, I have been looking for specific values of $x$ and $y$ that satisfies the given equation. This is what I have done so far:
$\sin(x+y)=\sin x\cos y +... | We know that
$$\sin(x+y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right)$$
and by sum to product formula we have
$$\sin(x)+\sin(y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
therefore
$$\sin(x+y)=\sin(x)+\sin(y) $$
$$\iff 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
compute $\frac{1}{2}(\frac{1}{3.5........(2n-1)} -\frac{1}{3.5........(2n+1)})$ compute the summation
$\sum_ {n=1}^{\infty} \frac{n}{3.5........(2n+1)}= ?$
My attempts : i take $a_n =\frac{n}{3.5........(2n+1)}$
Now =$\frac{1}{2}$ .$\frac{(2n+ 1) - 1}{3.5........(2n+1)}= \frac{1}{2}(\frac{1}{3.5........(2n-1)} -\fra... | You don't need to compute $a_n = \frac{1}{2}(\frac{1}{3.5…(2n-1)} -\frac{1}{3.5…(2n+1)})$
You need to compute $\sum_{n=1}^K a_n$.
Which is $\frac 12(\frac 11 - \frac 13) + \frac 12(\frac 13-\frac 1{3\cdot 5})+.....+ \frac{1}{2}(\frac{1}{3.5…(2K-3)} -\frac{1}{3.5…(2K-1)}) + \frac{1}{2}(\frac{1}{3.5…(2K-1)} -\frac{1}{3.5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Coin change with fixed denominations with duplicate values Suppose I have $\{1, 1\} \{2, 2, 2\} \{5\} \{10\}$ coins.
How many ways I can get 15 by adding coins?
If I use DP then it counts 7, because $\{1+2+2+10\} \{1+2+2+10\} \{1+2+2+10\} \{1+2+2+10\} \{1+2+2+10\} \{1+2+2+10\}$ and $\{10 + 5\}$. But actual result is $\... | This is a supplement to the already given answer showing that calculating the coefficient is not that cumbersome. We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a polynomial.
We obtain
\begin{align*}
\color{blue}{[x^{15}]}&\color{blue}{(1+x+x^2)(1+x^2+x^4+x^6)(1+x^5)(1+x^{10})}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $4\cos^4x-2\cos2x-1/2\cos4x$ is independent of x $4\cos^4x-2\cos2x-1/2\cos4x$
I don't know how to proceed. $\cos4x=2\cos^2(2x)-1$ Is this useful at all?
| Hint: using $\cos2x = 1-2\sin^2(x)$, $\sin(2x) = 2\sin(x)\cos(x)$ and $\sin^2(x) + \cos^2(x) = 1$ to get the following and simplify it:
$$S = 4\cos^4(x)-2(1-2sin^2(x))+1/2(1-2sin^2(2x))$$
$$S = 4\cos^4(x)-2+4\sin^2(x)+1/2 - 4sin^2(x)cos^2(x)$$
$$S = 4\cos^4(x)-2+1/2 + 4\sin^2(x)(1-\cos^2(x)) = 4\cos^4(x)-2+1/2 + 4\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to solve $\frac{|x+2|}{x-1}>\frac{x+1}{2x+1}$? I am working on the following problem:
$$\frac{|x+2|}{x-1}>\frac{x+1}{2x+1}$$
Here's what I have done so far:
$$|x+2|>\frac{x+1}{2x+1}\times(x-1)$$
$$-\left(\frac{(x+1)(x-1)}{(2x+1)}\right)<x+2<\frac{(x+1)(x-1)}{(2x+1)}$$
This is where I stopped. I am not entirely sure... | We need to consider 2 cases
*
*for $x+2\ge 0 \implies x\ge -2$ we need to solve
$$\frac{x+2}{x-1}>\frac{x+1}{2x+1}$$
*for $x+2< 0 \implies x< -2$ we need to solve
$$\frac{-x-2}{x-1}>\frac{x+1}{2x+1}$$
then the final solution is given by the union of the solution obtained for each case.
For case 1 we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the quotient and the remainder of $(n^6-7)/(n^2+1)$
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.
So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?
| Yet another way:
$$
\begin{align}
n^6-7 &= \big(\left(n^2\color{red}{+1}\right)\color{red}{-1}\big)^3-7 \\
&= \left(n^2+1\right)^3-3\left(n^2+1\right)^2+3\left(n^2+1\right)-1 -7 \\
&= \left(n^2+1\right)\big(\left(n^2+1\right)^2-3\left(n^2+1\right)+3\big) - 8 \\
&= \left(n^2+1\right)\left(n^4-n^2+1\right) - 8
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Evaluate: $u =\int_0^\infty\frac{dx}{x^4 +7x^2+1}$ Evaluate:
$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$
Attempt:
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2... | I'm not sure you have done anything wrong. You have $a+b$ in your answer, but $a$ is specifically $\sqrt{\frac{7-\sqrt{45}}{2}}=\frac{3-\sqrt{5}}{2}$. And $b=\frac{3+\sqrt{5}}{2}$. So $a+b=3$, making your answer agree with what you are expecting.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$
$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating... | Here is another straightforward way to obtain the limit.
Using the integral relation
$$\arctan u=\int_0^u{dt\over1+t^2}$$
we have
$$\begin{align}
(x+2)\arctan(x+2)-x\arctan x
&=2\arctan(x+2)+x\int_0^{x+2}{dt\over1+t^2}-x\int_0^x{dt\over1+t^2}\\
&=2\arctan(x+2)+x\int_x^{x+2}{dt\over1+t^2}
\end{align}$$
Now $2\arctan(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Find the roots of $3x^3-4x-8$ It is given that $\alpha$, $\beta$ and $\gamma$ are the roots of the polynomial $3x^3-4x-8$.
I have been asked to calculate the value of $\alpha^2 + \beta^2 + \gamma^2$.
However I am unsure how to find these roots, seeing as though I haven't been given a root to start with.
I began by ide... | Yet another approach: $\,\dfrac{1}{\alpha}, \dfrac{1}{\beta}, \dfrac{1}{\gamma}\,$ are the roots of $\,8x^3+4x^2-3\,$, so $\,\dfrac{1}{\alpha}+ \dfrac{1}{\beta}+ \dfrac{1}{\gamma}=-\dfrac{1}{2}\,$.
But $\,3\alpha^3=4\alpha+8 \iff \alpha^2=\dfrac{4}{3} + \dfrac{8}{3\alpha}\,$, and therefore: $$\alpha^2+\beta^2+\gamma^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Intersection of 3 planes along a line I have three planes:
\begin{align*}
\pi_1: x+y+z&=2\\
\pi_2: x+ay+2z&=3\\
\pi_3: x+a^2y+4z&=3+a
\end{align*}
I want to determine a such that the three planes intersect along a line. I do this by setting up the system of equations:
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
x+ay+2z&... | Hint
See, that from your system of equations you can obtain two lines:
$$l_1:\begin{cases}x=(a-2)t+1\\y=t\\z=(1-a)t+1\end{cases}$$
$$l_2:\begin{cases}x=\frac{5-a}{3}+\frac{a^2-4}{3}t\\y=t\\z=\frac{a+1}{3}+\frac{1-a^2}{3}t\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Express $z = \dfrac{3i}{\sqrt{2-i}} +1$ in the form $a + bi$, where $a, b \in\Bbb R$.
Express $$z = \frac{3i}{\sqrt{2-i}} +1$$ in the form $a + bi$, where $a, b \in\Bbb R$.
I figure for this one I multiply by the conjugate of $\sqrt{2 +1}$? But I’m still struggle to achieve the form $a+bi$.
| Given $$\dfrac{3i}{\sqrt{2-i}}+1=\dfrac{3i+\sqrt{2-i}}{\sqrt{2-i}}$$
Now $$=\dfrac{3i+\sqrt{2-i}}{\sqrt{2-i}}\times \dfrac{\sqrt{2+i}}{\sqrt{2+i}}$$$$=\dfrac{3i\sqrt{2+i}+\sqrt{5}}{\sqrt{5}}$$
$$=\dfrac{3i\sqrt{2+i}}{\sqrt{5}}+\dfrac{\sqrt{5}}{\sqrt{5}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Probability of at least two being grey Say we have $11$ grey and $15$ white mice, so $26$ in total in a container we can't see. We want to take $5$ of them home. What is the probability of at least two of them being grey?
*
*2 grey: The probability of the first being grey is $\frac{11}{26}$ the second grey is $\frac... | The probability that at least two are grey is one minus the probability that at most one is grey.
The probability that none are grey is : $\dfrac{15}{26}\dfrac{14}{25}\dfrac{13}{24}\dfrac{12}{23}\dfrac{11}{22}$ or in short $\left.\dbinom {15}5\middle/\dbinom{26}5\right.$
The probability that one is grey is : $\dfrac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Find the roots of $x^3 -6x^2 +13x -12$ I am trying to find the roots of $$\tag{1} x^3 -6x^2 +13x -12$$ by applying the method outlined here (I think this Cardan’s method).
Letting $y= x-2$, we can transform $(1)$ into $$\tag{2} y^3 + y -2=0.$$
It is clear that $1$ is a root of $(2)$, and, from the formula $$u - \frac{h... | $$x^3+x-2= 0$$
We can eliminate the $x^1$ by using a quadratic tschirnhausen transformation $y= x^2+mx+n$, to get the cubic in a binomial form
$$(y-(x_{1}^{2}+mx_{1}+n))(y-(x_{2}^{2}+mx_{2}+n))(y-(x_{3}^{2}+mx_{3}+n)) = 0$$
Collect the new variable $y$
$$y^3+(\dots \dots)y^2+(\dots \dots)y+(\dots \dots) = 0$$
Solving f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$a+\sqrt{b} = b + \sqrt{a}$$
Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x... | Calling
$$
a = x^2\\
b = y^2
$$
the problem reads now
$$
\max (x^2+y^2)\;\;\mbox{s.t.}\;\; x^2-y^2=x-y\Rightarrow x+y=1
$$
now the problem is reduced to:
Find $r$ in $C\to x^2+y^2-r^2=0$ such that $C$ is tangent to $x+y = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
Sum of a repeating pattern of numbers Consider a pattern, in which 5 repeats 3 times & 4 repeats 1 time.
$$5,5,5,4,5,5,5,4,5,5,5,4,...$$
Help me find terms count, when sum of the pattern is at least S.
For example,
if S=29, pattern would be $5+5+5+4+5+5=29$, terms count is 6.
if S=20, pattern would be $5+5+5+4+5=24$, t... | Note that you have cycles of $$5+5+5+4$$ and may be a few extra terms.
We have $$5+5+5+4=19$$ so the first step is to divide your S by 19 and then fit the remainder into your cycle.
For example if you have $S=123$, then $123= 6(19)+9$ so you have 6 cycles and the remainder of 9 indicates that you can fit two $5$ into... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How can I factorize this: "$X^3 + X^2 + X - 3$" I am going to elementary school & I am living in one of those deprived areas of Africa.
I can solve mathematical questions like this:
$$X^3 + X^2 + X +1 = X^2(X+1)+(X+1) = (X+1)(X^2+1)$$
Or even
\begin{align}X^2 − 2X + X^2 - X + 1 &= (X^2 - 2X + 1) + (X^2 - X) \\
&= (X - ... | For this kind of problem it is worth knowing about the factor theorem
This says if $a$ is a root of your polynomial $f(x)$ i.e. $f(a)=0$ then $x-a$ is a factor of, i.e. divides, the polynomial.
In these kinds of problems it is worth trying a few values such as $\pm1, \pm2$.
In your example $f(1)=0$ so $x-1$ divides you... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find two $2\times2$ real matrix $A$ and $B$ such that $A$, $B$ , $A+B$ are all invertible with $(A+B)^{-1}=A^{-1}+B^{-1}$ Find two $2\times2$ real matrices $A$ and $B$ such that $A$, $B$ , $A+B$ are all invertible with $(A+B)^{-1}=A^{-1}+B^{-1}$
Tried to write the matrices as
$$A=\pmatrix {a&b\\c&d},B=\pmatrix {e&f\\g... | You get $I = (A^{-1}+B^{-1})(A+B) = I + A^{-1}B + B^{-1}A + I$. Thus we get:
$$(A^{-1}B) + (A^{-1}B)^{-1} = -I$$
$$(A^{-1}B)^2 + I = -(A^{-1}B)$$
So $A^{-1}B$ satisfies the polynomial $x^2 + x + 1 = 0$. Take any matrix satisfying this polynomial; for example you can take
$$A^{-1}B = \begin{bmatrix} -1 &1 \\ -1&0 \end{... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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matrix identity extends to real entries? Let a matrix be $(m+n)$ by $(m+n)$ be all reals with determinant $1.$ Call the matrix $S.$ Name the top $m$ rows $A$ and the next $n$ rows $V.$ Now, $S$ has an inverse $T,$ $ST = I.$ Name the left $m$ columns $W^T$ and the right $n$ columns $B^T.$
We get two Gram matrices, firs... | Example with integer coefficients I had partly typed up. Detail that could be important, could be irrelevant: here, i did not work out a complete matrix $S$ that contained $A$ as the top two rows. i went directly to putting $AR$ in Hermite form by column operations. Hmmm: with the square matrix $R$ far below, the top t... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Summation of Binomial Distribution 2 So we have
$\sum _{k=0}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k} = (0.3+0.7)^n=1^n=1$
Now I want to solve
$\sum _{k=\frac{n}{2}+2}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}$
This is what I tried doing
$\sum _{k=0}^n\binom{n\:}{k}\:\left(0.3\right)^... | Hint: $$\sum _{k=0}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}=1$$
if $n$ is even
$$S=\sum _{k=0}^{\frac{n}{2}-1}\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}=\sum _{k=\frac{n}{2}+1}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}$$
leaving one term in between $$\binom{n\:}{\frac{n}{... | {
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"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove $\ \frac{z-1}{z+1} $ is imaginary no' iff $\ |z| = 1 $ Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $
Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so
$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \c... | Another approach
$${\bf Re}\dfrac{z-1}{z+1}={\bf Re}\dfrac{(z-1)(\bar{z}+1)}{|z+1|^2}=\dfrac{|z|^2-1}{|z+1|^2}=0 \iff |z|=1$$
| {
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"answer_count": 5,
"answer_id": 4
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Converting complex numbers into Cartesian Form 3 When calculating the real and imaginary parts of the complex number, do we take the angle as shown or the magnitude of it? I thought that we would just take the angle as shown, but apparently not, according to my textbook (unless its a typo):
$$z=2\sqrt{3}\operatorname{... | You are correct and the text book is wrong.
$$4\:\cos\left(-\frac{\pi }{6}\right)\:+\:4i\:\sin\left(-\frac{\pi }{6}\right)\:=\:2\sqrt{3}-2i\\
2\sqrt{3}\cos\left(\frac{\pi }{3}\right)+2\sqrt{3}i\:\sin\left(\frac{\pi }{3}\right)=\:\sqrt{3}\:+3i\\
\sqrt{3}\:+3i\:+\:2\sqrt{3}-2i\:=\:3\sqrt{3}+i$$
is correct.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $10^n \gt 6n^2+n$ Show for all $n \in \mathbb{N}$ $(n\geq1):$ $10^n \gt 6n^2+n$
My solution:
Base case: For $n=1$
$10^1 \gt 6 \cdot 1^2+1$
Inductive hypothesis:
$10^n \gt 6n^2+n \Rightarrow 10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
Inductive step:
$10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
$\Rightarrow$ $10^{n+1} \gt 6(n^2+... | For the inductive step we can proceed as follow
$$10^{n+1}=10\cdot 10^n \stackrel{Ind.Hyp.}>10\cdot(6n^2+n) \stackrel{?}>6\cdot(n+1)^2+(n+1)$$
and the latter requires
$$10\cdot(6n^2+n) \stackrel{?}>6\cdot(n+1)^2+(n+1)$$
$$60n^2+10n \stackrel{?}>6n^2+12n+6+n+1$$
$$54n^2-3n-7 \stackrel{?}>0$$
which can be shown to be ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Unable to prove that $\sqrt{i} + \sqrt{-i}$ is a real number. I did this :-
$$
\sqrt{i} = \sqrt{\frac{1}{2}.2i} = \sqrt{\frac{1}{2}.(1 + 2i - 1)} = \sqrt{\frac{1}{2}.(1 + 2i + i^2)} = \sqrt{\frac{1}{2}.(1+i)^2} = \frac{1}{\sqrt{2}}(1+i) \\= \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \mbox{ ---------
1} \\ \mbox{Also, }... | Note that $i=e^{(\pi/2) i}$ and $-i=e^{(-\pi/2) i}$
Thus $\sqrt i = e^{(\pi/4) i}$ and $\sqrt -i = e^{(-\pi/4) i}$
Adding the two we get $$\sqrt i+\sqrt -i =e^{(\pi/4) i}+e^{(-\pi/4) i}=2 \cos (\pi/4)=\sqrt 2$$ Which is real.
| {
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"url": "https://math.stackexchange.com/questions/2899061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Extremum Problem
How big is the circle?
My first steps:
*
*$ x^2 + y^2 =r^2$
*$ f(x)=y=e^{-x^2}$
Substitute $y^2$ in $x^2 + y^2 =r^2.$
So, $x^2 + e^{-2x^2} =r^2$
Is this way correct? Because after calculating the first derivative $2x -4xe^{-2x^2}$ and so on my solution at the end is $x^2 = \dfrac... | Your solution is right. Clearly the circle and the curve meet at $(x,y)$ in first quadrant where $$x^2+y^2=r^2\\y=e^{-x^2}\\-2xe^{-x^2}=\dfrac{-x}{\sqrt{r^2-x^2}}$$which means that $$2y=\dfrac{1}{\sqrt {r^2-x^2}}=\dfrac{1}{y}$$therefore $$y=\dfrac{\sqrt 2}{2}$$and $$x=\sqrt{\dfrac{\ln 2}{2}}$$which means that $$r=\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$
If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $$\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$$
Here's how I tried it ... | Squaring & adding we get
$$\cos^2(a+b+c)+\sin^2(a+b+c)=(\cos a+\cos b+\cos c)^2+(\sin a+\sin b+\sin c)^2$$
$$\implies1=1+1+1+2\sum_{\text{cyc}(a,b,c)}\cos(a-b)$$
Observe that we actually don't need $x+y+z=xyz$
The sufficient condition is $$ x+y+z=\cos p+i\sin p\iff|x+y+z|=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$
Calculate
$$\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$$
My Attempt:
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\l... | You can rewrite the function as
$$
\frac{\log\cos(x)}{\log\cos(x/2)}\frac{\log\sin(x/2)}{\log\sin(x)}
$$
The first fraction is easy to deal with:
$$
\log\cos x=\log(1-x^2/2+o(x^2))=-\frac{x^2}{2}+o(x^2)
$$
and similarly
$$
\log\cos\frac{x}{2}=\log(1-x^2/8+o(x^2))=-\frac{x^2}{8}+o(x^2)
$$
Therefore
$$
\lim_{x\to0}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Induction. Am I missing something or is there a mistake in the question? $$\sum_{k=1}^n k*3^k=\frac {3(3^n(2n-1)+1)} 4 $$
So let f(n)= $\sum_{k=1}^n k*3^k $
and g(n)=$\frac {3(3^n(2n-1)+1)} 4$
By induction hypothesis, $f(n+1) = f(n) + (n+1)3^{n+1} \overset{\text{i.h.}}{=} g(n) + (n+1) 3^{n+1} = g(n+1).$
$$\frac{3(3^{n... | You can calculate $$\frac{3^{n+2}(2n+1)-3-3^{n+1}(2n-1)-3}{4}=\frac{3^{n+1}(3(2n+1)-2n+1)}{4}=\frac{3^{n+1}(6n+3-2n+1)}{4}=\frac{3^{n+1}(6n+3-2n+1)}{4}=…$$
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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prove that $\lim _{x\to 1}f(x)=3$ if $f(x)=\frac{x^3-1}{x-1}$ Prove that $\lim _{x\to 1}f(x)=3$ where $f:(0,\infty)\to \mathbb{R}$ is given by $f(x)=\frac{x^3-1}{x-1}$.
I proved by definition of the limit
$$|f(x)-3|=\left|\frac{x^3-1}{x-1}-3\right|=|x^2+x-2|=|x-1||x+2|\leq |x-1|(|x|+2)$$
how to processed from this
| Let $\varepsilon > 0$. We have
$$\left|\frac{x^3-1}{x-1} - 3\right| = \left|\frac{x^3-3x+2}{x-1}\right| = \left|x^2+x-2\right| = |x-1||x+2| \le |x-1|(|x-1| + 3)$$
Solving the quadratic inequation $y^2+3y - \varepsilon < 0$ gives $y \in \left\langle \frac{-3-\sqrt{9+4\varepsilon}}2, \frac{-3+\sqrt{9+4\varepsilon}}2\righ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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How we derive $y \geq (b+1)$ from $(b+1)b^n < (b+1)^n b$? In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page 5 that-
$y \geq (b+1)$ since, $(b+1)b^n < (b+1)^n b$ and it was given that $(b+1)x^n - by^n =1$.
but $(b+1)x^n - by^n =1 \implies (b+1)x^n > by^n $, and toge... | The given equation is-
$(b+1)x^n - by^n =1 \cdots (1)$.
Case $1$: If $x=y$,
then equation (1) becomes $(b+1)x^n - bx^n =1 \implies bx^n+x^n - bx^n =1 \implies x^n =1 \implies x=1 $, but it is given that $(x, y) \neq (1, 1)$. So, $x \neq y$.
Case $2$: If $x>y$,
then, let $x=y+c$ (where $c>0$, $c$ is an integer since ... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x ... | Just using that $\frac{1}{1+\epsilon}=1-\epsilon+O(\epsilon^2)$ and $\sum_{i=1}^n i=\frac{n(n+1)}{2}$
So $$\frac{x}{x^2+c}=\frac{1}{x}\frac{x^2}{x^2+c}=\frac{1}{x}\frac{1}{1+\frac{c}{x^2}}=\frac{1}{x}\left(1-\frac{c}{x^2}+O(\frac{c}{x^4})\right)$$
$$\sum_{c=1}^x\frac{x}{x^2+c}=\frac{1}{x}\sum_{c=1}^x1-\frac{c}{x^2}+O(\... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Find the value of p when the equation has 3 real roots( Derived from a symmetric line question)
Question statement: Two points M and N are on a parabola $y^2=2px$ ($p\neq0$) such that they are symmetric points about a straight line $x+y=1$. Determine the range of $p$.
What I did was I reflected that parabola over $x+... | Let $M(a,b),N(c,d)$ where $a\not=c$ or $b\not=d$.
These points are on the parabola $y^2=2px$, so
$$b^2=2pa\iff a=\frac{b^2}{2p}\tag1$$
$$d^2=2pc\iff c=\frac{d^2}{2p}\tag2$$
Also, the two points are symmetric points about a straight line $x+y=1$, so the midpoint of the line segemnt $MN$ is on the line $x+y=1$, and the l... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integral $\int_0^\infty dp \, \frac{p^5 \sin(p x) e^{-b p^2}}{p^4 + a^2}$: any clever ideas? I am trying to solve the following integral, with $a>0,$ $b>0$:
$I \equiv \int_0^\infty dp \, \frac{p^5 \sin(p x) e^{-b p^2}}{p^4 + a^2} $
By expanding the $\sin$, I get
$I = \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!} \int_0^\i... | We have
$$\frac {p^5 e^{-b p^2} \sin x p} {p^4 + a^2} =
p e^{-b p^2} \sin x p \left( 1 +
\frac {i a} {2 (p^2 - i a)} - \frac {i a } {2 (p^2 + i a)} \right), \\
\frac d {db} \left( e^{i a b} \int_0^\infty
\frac {p e^{-b p^2} \sin x p} {p^2 - i a} dp \right) =
-e^{i a b} \int_0^\infty p e^{-b p^2} \sin x p \,dp,$$
and,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Volume of a tetrahedron whose 4 faces are congruent. Suppose that I have a tetrahedron such that all four faces consist of congruent triangles, says with the lengths $a,b$ and $c$ for each side. Is there a beautiful method to compute its volume?
PS. The reason for me tagging calculus and linear algebra is that I figur... | If the given triangle is acute-angled, define positive numbers $x, y, z$ by:
\begin{align*}
x^2 & = \tfrac{1}{2}(b^2 + c^2 - a^2), \\
y^2 & = \tfrac{1}{2}(c^2 + a^2 - b^2), \\
z^2 & = \tfrac{1}{2}(a^2 + b^2 - c^2).
\end{align*}
Let $\mathbf{u}, \mathbf{v}, \mathbf{w}$ be mutually orthogonal vectors of unit length, and ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\sum_{k=2}^\infty \frac{1}{k^2 - 1}$ I'm wondering if someone could check on this working:
$$\sum_{k=2}^\infty \frac{1}{k^2 - 1} = \sum_{k=2}^\infty \frac{1}{(k - 1)(k + 1)} = \frac{1}{2}\sum_{k=2}^\infty \frac{1}{k-1} - \frac{1}{k+1}$$
This is a nice telescoping thing that has a $k$-th partial sum that loo... | You have
$$
...= 1 + \frac{1}{2} - \frac{1}{k+1} - \frac{1}{k+1}
$$
and there should be
$$
= 1 + \frac{1}{2} - \frac{1}{\color{cyan}{k}} - \frac{1}{k+1}
$$
but it is easy to correct and even with this mistake does not change the result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is this proof correct (Rationality of a number)? Is $\sqrt[3] {3}+\sqrt[3]{9} $ a rational number? My answer is no, and there is my proof. I would like to know if this is correct:
Suppose this is rational. So there are positive integers $m,n$ such that $$\sqrt[3]{3}+\sqrt[3]{9}=\sqrt[3]{3}(1+\sqrt[3]{3})=\frac{m}{n}$$... | Alternatively, denote $x=\sqrt[3] {3}+\sqrt[3]{9}$ and cube it to get a cubic equation with integer coefficients:
$$x^3=3+3^2(\sqrt[3]3+\sqrt[3]{9})+9 \Rightarrow \\
x^3-9x-12=0.$$
According to the rational root theorem, the possible rational roots are: $\pm (1,2,3,4,6,12)$. However, none of them satisfies the equation... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Calculate $A^5 - 27A^3 + 65A^2$, where $A$ is the matrix defined below. If $A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$, find $A^5 - 27A^3 + 65A^2$
$$A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$$
Let $\lambda$ be its eigenvalue, then
$$(A-\lambda I) = \begin{bmatrix} 0-\lambd... | Let
$p(x) =x^5 -27x^3 + 65x^2; \tag 1$
one can always compute $p(A)$ directly, but perhaps the following requires less arithmetical work:
Let
$\chi(x) = x^3 - 5x^2 +6x - 5 \tag 2$
be the characteristic polynomial of $A$.
By the division algorithm for polynomials, there exist unique
$q(x), r(x) \in \Bbb Q[x] \tag 3$
suc... | {
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"url": "https://math.stackexchange.com/questions/2909758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find the sum of the power series $\sum_{n=1}^\infty n*(n+1)*x^n$ $\sum_{n=1}^\infty n*(n+1)*x^n$
Hello everyone, I need help in solving the question above.
I started with $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$, if differentiated it once so it became $\frac{1}{(1-x)^2} =\sum_{n=1}^\infty nx^{n-1} $ but from here I don... | Using
$$\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$$
then
$$\sum_{n} n \, x^{n+1} = x^2 \, \frac{d}{dx} \, \frac{1}{1-x}$$
and
$$\sum_{n} n \, (n+1) \, x^n = \frac{d}{dx} \left( x^{2} \, \frac{d}{dx} \frac{1}{1-x} \right).$$
After some work it is found that:
\begin{align}
\sum_{n=0}^{\infty} x^{n} &= \frac{1}{1-x} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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positive integer solutions to $x^3+y^3=3^z$ I am seeking all positive integer solutions to the equation $x^3+y^3=3^z$.
After doing number crunching, I think there are no solutions. But I am unable to prove it.
Attempt
If $x$ and $y$ have common divisor $d$, we have $d^3(m^3+n^3)=3^z$. So $d$ must be a power of $3$, and... | You already proved that we can assume that $x,y$ are relatively prime.
We can find easly solution for $n\leq 2$. Let $n\geq 3$.
So $x+y = 3^a$ and $x^2-xy+y^2=3^b$ for some nonegative integers $a+b=n$.
If $b\geq 2$ then $9\mid x^2-xy+y^2$ and $3\mid x+y$, so $$9\mid (x+y)^2-(x^2-xy+y^2)=3xy\implies 3\mid xy$$
A contr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim\limits_{x \to\infty} (\frac {2+3x}{2x+1})^{x+1}$ without L'Hopital $$
\lim_{x\to\infty}\left(\frac {2+3x}{2x+1}\right)^{x+1}
$$
Not sure how to deal with this, I've tried doing the following
$$
\lim_{x\to\infty}\left(\frac {2+3x}{2x+1}\right)^{x}\cdot \lim_{x\to\infty}\left(\frac{2+3x}{2x+1}\right)
$$
Then I... | We have
\begin{align}
\lim_{x\to\infty}\left(\frac {2+3x}{1+2x}\right)^{1+x}
&=
\lim_{x\to\infty}\left(\frac {(1+2x)+(1+x)}{1+2x}\right)^{1+x}
\\
&=
\lim_{x\to\infty}\left(1+\frac {1+x}{1+2x}\right)^{1+x}
\\
&=
\lim_{x\to\infty}\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{1+x}
\\
&=
\lim_{x\to\infty}\left[\left(1+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 5
} |
The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\... | The second way is preferable and since the product "$\infty \cdot \frac12$" is not an indeterminate form from here
$$\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
we can conclude that the sequence diverges.
Note that also with the first method from here
$$\lim_{n\to\infty}\frac{1-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 1
} |
Proof by induction, $1/2 + ... + n/2^n < 2$ So I'm having trouble with proving this homework question by induction.
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{n-1}{2^{n-1}} + \frac{n}{2^n} <2
$$
I know how to prove that the series converges to 2 (using things like the ratio method), but actually using induction is ... | Hint:
$$
\begin{align}
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} + \frac{k+1}{2^{k+1}}
&=\qquad\;\;\frac12\Big(\frac{1}{2^1} + \frac{2}{2^2} + \dots +\frac{k-1}{2^{k-1}} + \frac{k}{2^k}\Big) \\
&\quad+\Big(\frac{1}{2^1} + \frac{1}{2^2}+\frac1{2^3} + \dots + \;\;\frac{1}{2^k}\;\;+\frac1{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Fourier series of $f(x)=\sin^2x\cos^2x$ at $(-\pi,\pi)$ Find Fourier series of $f(x)=\sin^2 x \cos^2 x$ at $(-\pi,\pi)$
$f(x)$ is even so we only have to evaluate $a_0,a_n$
$$
a_0 = \frac{1}{\pi}
\int_{-\pi}^{\pi} \sin^2 x \cos^2 x dx
= \frac{1}{4\pi}
\int_{-\pi}^{\pi}\sin^2(2x)
= \frac{1}{4\pi}
... | As a more effective way we have that
$$
\begin{split}
f(x) &= \sin^2 x \cos^2 x=\frac12(1-\cos 2x)\cdot \frac12(1+\cos 2x) \\
&=\frac14(1-\cos^2 2x)=\frac14\left[1-\frac12\left(1+\cos 4x\right)\right]=\frac18-\frac18\cos 4x
\end{split}
$$
Using the same expression in the integral obviously we obtain
$$a_n= \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}$ How can I prove this equality:
$$
\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}= \frac{-1}{n(n+1)(n+1)!}
$$
| Given $$\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}$$
$$=\frac{1}{(n+1)n!} + \frac{1}{(n+1)(n+1)n!} - \frac{1}{n(n!)}$$
$$=\dfrac{1}{n!}\left(\frac{1}{(n+1)} + \frac{1}{(n+1)(n+1)} - \frac{1}{n}\right)$$
$$=\dfrac{1}{n!}\left(-\dfrac{1}{n(n+1)}+ \frac{1}{(n+1)(n+1)}\right)$$
$$=\dfrac{-1}{n(n+1)(n+1)!}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{\left(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\right)^2}dx$ The goal is to show (preferably without contour integration, as my knowledge is pretty limited there, but if you can do it that way there is no problem to share it) that: $$I= \int_0^1 \frac{x\ln\left(\... | We will use $3$ instances of Schroder's formula which evaluate Gregory coefficients https://en.wikipedia.org/wiki/Gregory_coefficients#Computation_and_representations
\begin{eqnarray*}
-G_2=\int_0^{\infty} \frac{dx}{(1+x)^2(\pi^2+(\ln x)^2)} =\frac{1}{12} \\
G_3=\int_0^{\infty} \frac{dx}{(1+x)^3(\pi^2+(\ln x)^2)} =\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solving the equation $\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$ for $x$ or $y$ One day, while making and doing math problems, I came across this equation:
$$\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$$
After some simple steps, I found $g$, but I couldn't find $x$ or $y$.
He... | Multiply $$\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$$ by $x y^2$ to obtain
$$g y^3 + 2 y^4 = g x y + 2 x^2.$$
By grouping the terms together this becomes $g y (y^2 - x) + 2 (y^4 - x^2) = 0$ and factoring the last term leads to
$$(y^2 - x)(2 y^2 + g y + 2 x) = 0.$$
Solving to $x$ yields two possi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$ The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$.
I tried the following concept, how to sum 9 using 1-9 only without repetition.
1:9, 2:1+8,3:7+2, 4: 6+3, 5: 5+4,
6:1+2+6 ,7:1+3+5, 8:2+3+4
The answer is... | As Theo Bendit has already commented this is not solved by Newton's binomial theorem. The solution is the same way that is used to prove Newton's bionomian theorem that is the proper use of combinatorial principles. Let
$$
p(x)=(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)(1+x^6)(1+x^7)(1+x^8)(1+x^9)
$$
We have
\begin{align}
p(x)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is a fast way to evaluate $\cos(\pi/4 + 2\pi/3)$ and $\cos(\pi/4 + 4\pi/3)$ (maybe using inspection) Given
$\cos(\pi/4 + 2\pi/3)$
and
$\cos(\pi/4 + 4\pi/3)$
I wish to show that these values are given by $(\pm(\sqrt{6}) - \sqrt{2})/4$.
What is a quick way to arrive at these exact solutions (perhaps even by insp... | Is it really that slow?
$\cos (\frac{\pi}{4} + \frac{2\pi}{3}) =\cos \frac{\pi}{4}\cos \frac{2\pi}{3} - \sin \frac{\pi}{4}\sin\frac{2\pi}{3} = (\frac {\sqrt{2}}{2})(\frac {-1}2) - (\frac {\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) = \frac {-\sqrt 2 - \sqrt 6}{4}$
That seems pretty straight forward to me.
$\cos (\frac{\pi}{4} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding $\lim_{x\to-\infty}\sqrt{x^2-5x+1}-x$ results in loss of information Let $f(x) = \sqrt{x^2-5x+1}-x$
Find $\lim_{x\to\infty}f(x)$
$$\lim_{x\to\infty} \sqrt{x^2-5x+1}-x$$
$$\lim_{x\to\infty} \dfrac{x^2-5x+1-x^2}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac{-5x+1}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac... | Note that for $x\to -\infty$
*
*$\sqrt{x^2-5x+1} \to +\infty$
*$-x \to +\infty$
and therefore
$$\sqrt{x^2-5x+1}-x \to +\infty$$
As an alternative by $x=-y\to -\infty$ with $y\to +\infty$ we have
$$\lim_{x\to-\infty}\sqrt{x^2-5x+1}-x=\lim_{y\to +\infty}\sqrt{y^2+5y+1}+y \to +\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Application of Fermat's little theorem to check divisibility Using Fermat's little theorem to prove: $(i)19\mid 2^{2^{6k+2}}+3$, where $k=0,1,2.....$$(ii)13\mid 2^{70}+3^{70}$My Approach: I couldn't think of how to go with $(i)$ but i tried $(ii)$ to show $2^{70} \equiv 0\pmod {13}$ and $3^{70} \equiv 0\pmod {13}$.Sinc... | $n|a + b$ does not mean that $n|a$ and $n|b$. It means that if $a \equiv k \mod n$ then $b \equiv -k \mod n$.
So if $2^{70}\equiv 10 \mod 13$ (which it is) then $13|2^{70} + 3^{70}$ if $3^{70} \equiv -10 \mod 13$.
Does $3^{70}\equiv -10 \mod 13$?
Well If $0 < a < 13$ then $a^{12} \equiv 1 \mod 13$ and $a^{60} = (a^{12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve a system of equation with absolute value and rotation Solve
$\begin{cases} \left|x_1-x_2\right|=\left|x_2-x_3\right|=...=\left|x_{2018}-x_1\right|,\\ x_1+x_2+...+x_{2018}=2018. \end{cases}$
I think there must be such a way to solve systems of equation with the form of rotation and absolute value like this.
I have... | Here is the answer to verify my comment.
Let $|x_{1} - x_{2}| = C \geq 0$ and note that
\begin{align}
|x_{1} - x_{2}| = C \implies x_{1} = \pm C + x_{2}
\end{align}
Similarly, we have
\begin{align}
x_{1} &= \pm C + x_{2} \\
x_{2} &= \pm C + x_{3} \\
& \; \; \vdots \\
x_{2017} &= \pm C + x_{2018} \\
x_{2018} &= \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.