Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Definite integral of the product of modified Bessel function of the first kind, exponential, and a power term Is there a solution to the following integral?
$$
\int_0^{\infty} t^{-0.5}e^{-at}I_{l}\left(kt\right)dt,\;\;\;a,k>0
$$
Here, $I_{l}$ is the modified Bessel function of the first kind, and $a,k$ are constants. I... | $$
\int_0^{\infty}{t^{-\frac{1}{2}}I_l\left( kt \right) e^{-at}dt}
\\
=\sum_{n=0}^{\infty}{\frac{1}{n!\Gamma \left( n+l+1 \right)}\left( \frac{k}{2} \right) ^{2n+l}}\int_0^{\infty}{t^{-\frac{1}{2}+2n+l}e^{-at}dt}
\\
=\sum_{n=0}^{\infty}{\frac{1}{n!\Gamma \left( n+l+1 \right)}\left( \frac{k}{2} \right) ^{2n+l}\frac{\Gam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Euler's partition method. How does someone use it? I came across partitions recently and am not very much informed about it but I have a question regarding Euler's method for this. I came to know about this formula from a YouTube video so, it may not be the full equation.
Symbols:
P(n) -> Partition of n
$\pi$ -> Produc... | This example supplements MAAvL's answer and is an expansion of dust05's comment. The following is a bit pedantic, with including exponents 1 and writing $(1+x^2+x^{2+2}+\cdots)$ rather than $(1+x^2+x^4+\cdots)$, for example, but the idea is to make the connection to partitions very concrete.
\begin{align*}
&\sum_{n \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\int_A e^{x^2+y^2-z^2-w^2} \, dx\,dy\,dz\,dw$ Calculate $\int_A e^{(x^2+y^2-z^2-w^2)}\,dx\,dy\,dz\,dw$ where $A=\{x, y, z, w \in \mathbb{R} \mid x^2+y^2+z^2+w^2\leq1\}$
attempt:
$$\int_A e^{x^2+y^2-z^2-w^2} \,dx\,dy\,dz\,dw
= \int_0^1 \int_{\mathbb{S}^3_r} e^{x^2+y^2-z^2-w^2} \;\mathrm{d}S \;\mathrm{d}r
... | Note that $$A=\{x^2+y^2\leq1, z^2+w^2\leq1-(x^2+y^2)\}$$
So the integral becomes
$$I=\int_{\{x^2+y^2\leq1\}}e^{x^2+y^2}\int_{\{z^2+w^2\leq1-(x^2+y^2)\}}e^{-(z^2+w^2)}$$
Where the inner integral is by Fubini again:
$$(*) = \int_{-\pi}^{\pi}\int_0^cre^{-r^2}drd\theta$$
where $c=\sqrt{1-(x^2+y^2)}$.
Now, this equals $\pi(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum and asymptotic expansion of $\sum_{u=1}^{[N/2]} \sum_{v=1}^{[\sqrt{N}]} \left[{|u^2-2v| \le N}\right] \left[{|-2uv| \le N}\right]$ Where $[x]$ in the summation limits is the floor function. In the summation body this is the Iverson bracket
Find the sum and asymptotic expansion of $\sum_{u=1}^{\lfloor{N/2}... | Lets look at a table between perfect squares. Say from N = 16 to 24. Then
N
Sum
Difference
16
16
0
17
17
0
18
17
1
19
18
1
20
18
2
21
19
2
22
19
3
23
20
3
24
20
4
This pattern repeats for each perfect square sequence.
So with ${S}_{1} \left({N}\right) = \sum_{u=1}^{\lfloor{N/2}\rfloor} \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Integration using mean value theorem of $\int_0^{2\pi} \frac{2+\cos\phi}{5+4\cos\phi}\,\mathrm{d}\phi=\pi$ I am going through some worksheets to study for an exam and one question is to use the mean value theorem:
$$u(a)=\frac{1}{V_r}\int_{B_r(a)}u$$
with a suitable harmonic function to show that:
$$
\int_0^{2\pi} \fra... | We can try Poisson kernel.
$$
P_r(\theta)=\frac{1-r^2}{1-2r\cos\theta+r^2} = \Re \left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\right).
$$
To have $\cos \theta$ on the numerator, we use $1+P_r(\theta)$. Then
$$
1+P_r(\theta) = \frac{2-2r\cos\theta}{1-2r\cos\theta+r^2} = 1+\Re \left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric equality proof $ \cos^2(\omega t) + \cos^2(\omega t + \delta) = \sin^2\delta + 2\cos(\omega t + \delta)\cos(\omega t)\cos(\delta)$ Looking to prove
$x = A\cos(\omega t)\\
y = A\cos(\omega t + \delta)
\\
\\$
YIELDS
$x^2-2xy\cos(\delta)+y^2=A^2\sin^2(\delta)$
Specifically we're trying to express the equatio... | By setting $\omega t = x, \omega t + \delta = y$, it is equivalent to showing:
$\cos^2 x + \cos^2 y = \sin^2(y-x) + 2\cos y\cos x \cos (y-x) $
$\cos^2 x + \cos^2 y - \sin^2(y-x) \\= \cos^2 x + \cos^2 y - (\sin y\cos x - \cos y\sin x)^2 \\= \cos^2 x + \cos^2 y - \sin^2 y\cos^2 x - \cos^2 y\sin^2 x + 2\sin y\cos x \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Inequality with induction. I was wondering if my solution to this problem is correct:
Show that for every integer $n\ge1$ the following is true:
$\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\le\frac{1}{\sqrt{3n+1}}$
Base case:
$\frac{2\cdot1-1}{2}\le\frac{1}{\sqrt{3*1+1}}$
$\frac{1}{2}\le\frac{1}{2}$
$\Rightarrow \text{which ... | To conclude for the induction step we should use the induction hypothesis that is
$$\prod\limits_{k=1}^{n+1}\frac{2k-1}{2k}=\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\cdot\left(\frac{2n+1}{2n+2}\right)\stackrel{Ind.Hyp.}\le\frac{1}{\sqrt{3n+1}}\cdot\left(\frac{2n+1}{2n+2}\right)\stackrel{?}\le \frac{1}{\sqrt{3(n+1)+1}}$$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4229967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the limit $\lim_{(x,y)\to(0,0)} \frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x})$ exists using $\epsilon \text{ and } \delta$ definition Prove that the limit $\lim_{(x,y)\to(0,0)} \frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x})$ exists using $\epsilon \text{ and } \delta$ definition.
I have so far reduced it algebr... | Hint:$\left|\frac{y^4}{\sqrt{3x^2+y^2}}\right|\leqslant |y^3|$
Addition:
Let's take $\forall \varepsilon >0$. As we have limit $(x,y)\to (0,0)$, then we should find $\delta>0$, such, that $\sqrt{x^2+y^2}< \delta$ implies $|f|<\varepsilon$. As we have $|y|\leqslant \sqrt{x^2+y^2}< \delta$, then it gives, that $\delta^3<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim_{n \to \infty}\left(\frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}\right)$
Compute the limit:
$$\lim_{n \to \infty}\left(\frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}\right)$$
I tried applying the sandwich rule, constructing an upper and lower bound for the sequence, but I ... | We have $$\frac{n^2(n+1)}{2(n^3+n)}=\frac{n}{n^3 +n } +...+\frac{n^2}{n^3 +n }\leq \frac{n}{n^3 +1 } +...+\frac{n^2}{n^3 +n }\leq \frac{n}{n^3 +1 } +...+\frac{n^2}{n^3 +1 }=\frac{n^2(n+1)}{2(n^3+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to show $\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \Sigma_{k=1}^n \sqrt{k-1} = \frac{2}{3}$? How to show $$\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \sum_{k=1}^n \sqrt{k-1} = \frac{2}{3}$$
In fact, this is the lower sum of the integral of $\sqrt{x}$ from 0 to 1. So the value of the above must be $... | A solution based on Stolz-Cesàro theorem. Let $f(x)=x^{\frac{3}{2}}$
\begin{align*}
\lim_{n \to \infty} \frac{\sum_{i=1}^{n-1}\sqrt{i}}{n\sqrt{n}} = \lim_{n \to \infty} \frac{\sqrt{n}}{(n+1)\sqrt{n+1} - n\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{n}}{(1 + \frac{1}{n})^{\frac{3}{2}} - 1} = \frac{1}{f'(1)} = \frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Determinant of an interesting Toeplitz matrix Let $ab=1$. Find
$$\begin{vmatrix} c & a & a^2 & ... & a^{n-1} \\ b & c & a & \dots & a^{n-2} \\ b^2 & b & c& \dots &a^{n-3} \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ b^{n-1} & b^{n-2} & b^{n-3} & \dots & c \end{vmatrix}$$
I tried to decompose by line however it doe... | Let's add a third answer for fun. Since $ab=1$, we can write
$$ \det
\begin{pmatrix}
c&a&a^2&\cdots&a^{n-1} \\
b& c & a &\dots& a^{n-2}\\
b^2 & b & c &\cdots & a^{n-3}\\
\vdots &\vdots & \vdots & &\vdots\\
b^{n-1} & b^{n-2} &b^{n-3}& \cdots &c
\end{pmatrix}
=\frac{1}{b}\cdot\frac{1}{b^2}\cdots\frac{1}{b^{n-1}}\det
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solving cubic equations with sine and cosine sums. I was playing with math, and then I tried to rewrite some cubic equation with sine power reduction formula
$$y^3 + my^2 + ny + d = 0.$$
Let
$$y = \sin(x).$$
Then
$$y^2 = \frac{1 - \cos(2x)}{2},$$
$$y^3 = \frac{3\sin(x) - \sin(3x)}{4}.$$
So the equation will be
$$\frac{... | This approach has been used to resolve the diminished cubic $x^3 = 3px - 2q$
$x = 2\sqrt p\cos\theta\\
8p\sqrt p\cos^3 \theta = 6p\sqrt p \cos\theta + 2q\\
2p\sqrt p(4\cos^3 \theta - 3\cos\theta) = 2q\\
\cos 3\theta = \frac {q}{p\sqrt p}$
We detour into the complex plane if the original cubic does not have 3 real root... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Infinite product involving triangular numbers $\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}$ The question is about the value for the following infinite product involving triangular numbers $T_n=\frac{n(n+1)}2$:
$$\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}=\prod_{n=1}^{\infty} \frac{n(n+1)}{n(n+1)+2}=\prod_{n=1}^{\infty} \fr... | For complex $a, b, c, d$ with non-negative real part we have
$$
\prod_{n=1}^{N-1}\frac{(n+a)(n+b)}{(n+c)(n+d)} = \frac{\Gamma(c+1)\Gamma(d+1)}{\Gamma(a+1)\Gamma(b+1)} \cdot \frac{\Gamma(a+N)\Gamma(b+N)}{\Gamma(c+N)\Gamma(b+N)} \, .
$$
If additionally $a+b=c+d$ then Stirling's formula shows that the second factor conv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Finding the range of $f(x)=\log_2\left(4^{x^2}+4^{(x-1)^2}\right)$
If the range of the function $f(x)=\log_2\left(4^{x^2}+4^{(x-1)^2}\right)$ is $\left(\frac pq,\infty\right)$, where $p, q$ are in their lowest form then find $(p+q), (p-q)$
Looks like the domain is $\mathbb R$.
If $x^2=0$, $(x-1)^2=1\implies f(0)=\log... | By Jensen for the convex function $f(x)=4^{x^2}$ we obtain:
$$4^{x^2}+4^{(x-1)^2}\geq2\cdot4^{\left(\frac{x+1-x}{2}\right)^2}=2\sqrt2.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Algebraic operations between two expressions: how to include a constant outside a fraction into the numerator of that fraction I'm looking at a problem where the derivative of a function is defined as:
$$f'(x) = a-\frac{3x^2(1+x^2)-2x(x^3)}{(1+x^2)^2}$$
Then at the next line, the expression is expressed with the numera... | You have\begin{align}f'(x)&= a-\frac{3x^2(1+x^2)-2x(x^3)}{(1+x^2)^2}\\&=\frac{a(1+2x^2+x^4)-3x^2-3x^4+2x^4}{(1+x^2)^2}\\&=\frac{a+(2a-3)x^2+(a-1)x^2}{(1+x^2)^2}.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I handle the integation which contains $~ \left( 1+ t ^{2} \right) ^{2} ~$ and $~ \left( 1- t ^{2} \right) ^{2} ~~$? $$ \left( a,d \in \mathbb R_{> 0} \right) ~~\wedge~~ \left( a < d \right) $$
I have to find out the value of the below integral .
$$ \alpha := \int_{0 }^{1 } \frac{ 2 a \left( 1- t ^{2} \r... | Substitute $t=\tan \frac x2$
\begin{align}
&\int_{0 }^{1 } \frac{ 2 a \left( 1- t ^{2} \right) ^{2} }{ \left( 1+ t ^{2} \right) \left\{ d ^{2} \left( 1+ t ^{2} \right) ^{2} - a ^{2} \left( 1- t ^{2} \right) ^{2} \right\} } \,dt \\
=& \int_0^{\frac\pi2}\frac{a \cos^2x}{d^2 - a^2\cos^2x}dx
= \frac1a\left(-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $a,b,c$ be non-negative real numbers .Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $ Let $a,b,c$ be non-negative real numbers
Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
My idea is to use the $(p,q,r)$ method:
$p=a+b+c$
$q=ab+bc+ca$
$r =... | $uvw$ helps. See here: https://math.stackexchange.com/edit-tag-wiki/5758
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality it's $$9u^2-6v^2+\sqrt2w^3+3+2\sqrt2\geq3(2+\sqrt2)u,$$ which is a linear inequality of $v^2$.
Thus, by $uvw$ it's enough to prove our inequality for equality case of two variable... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\alpha, \beta$ are roots of $x^2+px+q=0, f(x)=(\alpha+\beta)x-\left(\frac{\alpha^2+\beta^2}2\right)x^2+\left(\frac{\alpha^3+\beta^3}3\right)x^3-...$
If $\ln(1+x)=x-\dfrac {x^2}2+\dfrac {x^3}3-\dfrac{x^4}4+...$ and $\alpha$ & $\beta$ are roots of the equation $x^2+px+q=0$. If $f(x)=(\alpha+\beta)x-\left(\dfrac{\alpha^... | $f(x)=(\alpha+\beta)x-\left(\dfrac{\alpha^2+\beta^2}2\right)x^2+\left(\dfrac{\alpha^3+\beta^3}3\right)x^3-\left(\dfrac{\alpha^4+\beta^4}4\right)x^4+...$
$f(x)=\ln(1+\alpha x)+\ln(1+\beta x)$
$f(x)=\ln(1+(\alpha+\beta)x+\alpha\beta x^2)$
$f(x)=\ln(1-px+qx^2)$
Can you continue??
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluating $ \sum_{r=0}^\infty { 2r \choose r} x^r $ I recently came across this question during a test
Evaluate the following $$ \sum_{r=0}^ \infty \frac {1\cdot 3\cdot 5\cdots(2r-1)}{r!}x^r $$
I converted it to the follwing
into $\sum_{r=0}^\infty { 2r \choose r} x^r $
Then I tried using Beta function to further ... | Use the Duplication formula for Gamma function
$$
\Gamma \left( {2\,z + 1} \right) = \Gamma \left( {2\,\left( {z + 1/2} \right)} \right)
= 2^{\,2\,z} \frac{{\Gamma \left( {z + 1} \right)\Gamma \left( {z + 1/2} \right)}}{{\Gamma \left( {1/2} \right)}}
$$
So
$$
\begin{array}{l}
\left( \begin{array}{c}
2r \\ r \\
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I thi... | The solution $-\frac CB$ is the limit at $0$ of one of those two solutions. Suppose that $B\geqslant0$. Then $\sqrt{B^2}=B$ and\begin{align}\lim_{A\to0}\frac{-B+\sqrt{B^2-4AC}}{2A}&=\frac12\lim_{A\to0}\frac{\sqrt{B^2-4AC}-B}A\\&=\frac12\left.\frac{\mathrm d}{\mathrm dA}\sqrt{B^2-4AC}\right|_{A=0}\\&=-\frac C{\sqrt{B^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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Proving a curve has no curve point I am trying to show that the curve $x^2 + y^2 - 3 = 0$ has no rational point, where a rational point is defined as a solution $(x,y) \in \mathbb{Q}^2$ to this equation.
My attempt was to proceed by contradiction. Assume there is such a point $(x,y) \in \mathbb{Q}^2$ and let $x = \frac... | You can rearrange the equality to get $a^2 d^2 = 3 b^2 d^2 - b^2 c^2 = b^2(3 d^2 - c^2)$ which means that $b^2 | a^2 d^2 \implies b | ad$, but $\gcd(a, b) = \gcd(c, d) = 1$ means that this implies $b | d$, and by a similar argument $d | b$, but that only happens if $b = d$.
From there, you can reduce your equation to $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for every positive integer $n$, $1/3 + 1/9 + \cdots + 1/{3^n} < 1/2$ Base case is $n=1$: $\frac {1}{3} < \frac{1}{2}$. So the base case holds.
Inductive hypothesis for $n = k$: $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} < \frac{1}{2}$
Inductive Step for $n = k + 1$:
$$ \left( \frac{1}{3} + \frac{1}... | As noticed you went in the wrong direction, for the induction step it is better to start from the induction hypothesis, for example as follows
$$\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} < \frac{1}{2}$$
$$\implies 1+\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} <1+ \frac{1}{2}$$
$$\implies \frac{1}{3} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\int \frac{{\rm d}x}{1+e\cos x}$
Find $$\int \frac{{\rm d}x}{1+e\cos x}$$
*
*When $e$ lies between $0$ and $1$
*When $e$ is greater than $1$
I got $\int \frac{dx}{1+e\cos x}$ when $e$ lies between $0$ and $1$, by substituting $\tan\frac{x}{2} =t$.
$$\int \frac{dx}{1+e\cos x} = \frac{2}{\sqrt{1-e^2}} \tan^{-... | \begin{align}
& \int \frac{dx}{1+e\cos x} = \int \frac{\frac{2\,dt}{1+t^2}}{1+e\cdot\frac{1-t^2}{1+t^2}} \\[8pt]
= {} & \int \frac{2\,dt}{(1+t^2) + e(1-t^2)} \\[8pt]
= {} & \int \frac{2\,dt}{(1+e) + (1-e)t^2} \\[8pt]
= {} & \frac 1 {1+e} \int \frac{dt}{1 + \frac{1-e}{1+e}\cdot t^2} \\[8pt]
& \text{Then if } e<1 \text{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
PLANE TYPE COLUMN BASED QUESTION
My approach is as follow
$D = \left| {\begin{array}{*{20}{c}}
{2\left( {a + b} \right)}&{3\left( {b + c} \right)}&{4\left( {a + c} \right)}\\
{2\left( {b + c} \right)}&{3\left( {a + c} \right)}&{4\left( {a + b} \right)}\\
{2\left( {a + c} \right)}&{3\left( {a + b} \right)}&{4\left( {b ... | $\bullet$ If $a+b+c=0$ and $ a^2 + b^2 + c^2 \neq ac + ba + cb $ $\implies $ $T=0$ so the equations have infinitely many solutions
$\bullet$ If $a+b+c\neq0$ and $ a^2 + b^2 + c^2 = ac + ba + cb $ $\implies $ $T=0$ and $a=b=c\neq0$ so the equations represent identical planes
$\bullet$ If $a+b+c\neq0$ and $ a^2 + b^2 + c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Sum of digits of $2^n - k$ in binary Denote the sum of digits of $n$ in binary as $s_2(n)$.
Trying to figure the relation $f(n,k) = s_2(2^n - k)$ for positive integers $n,k$ such that $2^n > k$. I see no pattern here and was thinking there should be at least a good lower bound as a function of $n,k,s_2(k)$.
It is clear... | Note that
$$
s_2((2^n-1)-k)=n - s_2(k),
$$
since subtracting from $11\cdots 1$ has the effect of complementing every bit of $k$. Therefore,
$$
s_2(2^n-k)=s_2((2^n-1)-(k-1))=\boxed{n-s_2(k-1).}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find volume of a solid figure that lies between: $x^2+y^2+z^2\leq 4$, $2x^2+2y^2-2z^2\geq 1$, $2z^2\geq x^2+y^2$ Find volume of a solid figure that lies between:
$x^2+y^2+z^2\leq 4$,
$2x^2+2y^2-2z^2\geq 1$,
$2z^2\geq x^2+y^2$
I just really can't figure out the limits of integration... any hint would be great
-----edit-... | Let's consider the region above $z = 0$. Due to symmetry the volume bound is same above and below $z = 0$.
The region is defined by,
a) $x^2 + y^2 + z^2 \leq 4$
In spherical coordinates, $\rho \leq 2$ (the sphere)
b) $2x^2+2y^2-2z^2\geq 1$
In spherical coordinates, $ - 2 \rho^2 \cos 2\phi\geq 1$
c) $2 z^2 \geq x^2 + y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4252443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\... | If $k>0$, then $a<b\iff ak<bk$. However, if $k<0$, then $a<b\iff ak\color{red}{>}bk$.
So when multiplying an inequality by a constant, we should be certain whether that constant is positive or negative. In the case of $-1<\frac{7}{x+4}$, multiplying by $x+4$ is not a good idea because we don't know if $x+4<0$ or $x+4>0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
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Question about calculating a series involving zeta functions On this page it had shown that the sum of $\frac{1}{n^3(n+1)^3}=10-\pi^2$. I'm curious about, what is the value of $$\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$
For some positive integer $k$.
According to partial fraction expansion, we can show that $$\frac1{n^3(n+... | Here's a smooth and Elementary way of solving the problem .
I'll use the fact that $$\frac{1}{a.b}=\frac{1}{(b-a)}{\left(\frac{1}{a}-\frac{1}{b}\right)}$$
So , $$\sum_{n=1}^{\infty}\left(\frac{1}{n(n+k)}\right)^3=\sum_{n=1}^{\infty}\frac{1}{k^3}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)^3$$ $$\Rightarrow \frac{1}{k^3}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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Volume of regular tetrahedron in a cube
Given a cube with side length $a$, a regular tetrahedron is constructed such that two vertices of the tetrahedron lie on the cube’s body diagonal and the other two vertices lie on the diagonal of one of the faces of the cube. Determine the volume of the tetrahedron.
All I can d... | The key insight is to first determine the minimum distance between the body diagonal and the face diagonal. Without loss of generality let the cube have unit side length and take the body diagonal to be the segment joining $(0,0,0)$ to $(1,1,1)$; then there are six face diagonals that do not intersect the body diagona... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find number of arrangements of a cube if sum of numbers on each face must be same
Each vertex of a cube is to be labelled with an integer 1 through 8, without repetition, such that sum of numbers of the four vertices of a face is the same for each face. Arrangements that can be obtained through rotations of the cube ... | Note that the sum of the numbers on each face must be $18$ because $\frac{1+2+\cdots+8}{2}=18$.
So now consider the opposite edges (two edges that are parallel but not on the same face of the cube); they must have the same sum value too. Now think about the points $1$ and $8$. If they are not on the same edge, they mus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
| Having gotten $ab=cd$, take a linear combination with $a^2+b^2=c^2+d^2$:
$\color{blue}{ab=cd}$
$\color{brown}{a^2+b^2=c^2+d^2}$
$\color{brown}{a^2}\color{blue}{-2ab}\color{brown}{+b^2}=\color{brown}{c^2}\color{blue}{-2cd}\color{brown}{+d^2}$
Those polynomials are squares:
$(a-b)^2=(c-d)^2$
$a-b=\pm(c-d)$
Then if $a-b=+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
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Solve the following equation: $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$ I have tried many ways but not success
$\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$
$\Leftrightarrow \sqrt{x}(\sqrt{x+1}-\sqrt{x-1})=x+1$
or $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1 \Leftrightarrow \sqrt{\dfrac{x}{x+1}}-\dfrac{\sqrt{x(x-1)}}{x+1}=1 \Leftrightarrow \dfrac{x}{x+1}=1... | Using your first approach, square both sides:
$$x \left((x+1) - 2\sqrt{x^2-1} + (x-1)\right) = (x+1)^2.$$ Now rearrange and collect like terms to isolate the square root:
$$2x \sqrt{x^2 - 1} = 2x^2 - (x+1)^2 = x^2 - 2x - 1.$$
Next, square again:
$$4x^2 (x^2 - 1) = (x^2 - 2x - 1)^2.$$
The resulting quartic has a rather... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4265678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve this probability Anna writes down a random sequence created by the following process. She repeatedly rolls a fair
6-sided die. If the number she rolls is larger than all of the numbers she has previously rolled (if
any), then she writes the new number down and then continues rolling. Otherwise, she does ... |
What is $Pr(X=2|Y=3)$
Use Bayes Theorem.
In order for $Y$ to equal $3$, several things have to have happened:
*
*You rolled zero, one, or two rolls below a $3$.
*Then, you rolled a $3$.
*Then, you rolled a number below a $4$.
$$P(X = 2| Y = 3) = \frac{P(X = 2 \wedge Y = 3)}{P(Y = 3)}.$$
I will repeat some of yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4267324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum of the series $3x+8x^2+15x^3 + ....$ I'm trying to express the following series $3x+8x^2+15x^3 + ....$ as a sum and hope to find its sum for $|x| < 1$
Here is what I have so far:
To me the series looks to be the derivative of the following form:
$1 + \frac{3}{2}x^2+\frac{8}{3}x^3+\frac{15}{4}x^4 ...$
Given... | If you check your $\frac{x(1+x)}{(1-x)^3}-\frac{x}{1-x}$ you will find it is $3x^2+8x^3+15x^4 + \cdots$ so you need to remove a factor of $x$ to get $\frac{(1+x)}{(1-x)^3}-\frac{1}{1-x}$ though I would not write it that way
If you know the coefficient of $x^n$ is $n(n+2)$ then one approach could be to manipulate $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Issue solving an integral I'm having issues solving the following integral:
$$ \int{\frac{e^x}{e^{2x}-e^x-2}}dx $$
From what I can tell, I should substitute $u = e^x$ and $du = e^xdx$ and end up with
$$ \int{\frac{1}{u^2 - u - 2}}du $$
where I can then use partial fraction decomposition to simplify the integrand as fol... | It turns out that$$\operatorname{arctanh}'(x)=\frac1{1-x^2}.$$But\begin{align}\frac1{u^2-u-2}&=\frac1{\left(u-\frac12\right)^2-\frac94}\\&=-\frac49\frac1{1-\left(\frac13-\frac{2u}3\right)^2},\end{align}and therefore\begin{align}\left(\frac23\operatorname{arctanh}\left(\frac13-\frac{2u}3\right)\right)'&=-\frac49\frac1{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Cycle notation and transpositions For example, consider the permutation
$$ \pi=\left(\begin{matrix}
1&2&3&4&5\\
4&3&2&5&1
\end{matrix}\right).$$
You can write it with two cycles as
$$ \pi =(145)(23).$$
Now I want to write $\pi$ as a product of transpositions. I know one way to do that is
$$ \pi =(14)(45)(23)$$
because... | Multiplication of cycles can be written as composition of permutations (order right to left). This way we can see the commonality when representing $\pi$ as multiplication of transpositions as well as multiplication of other cycles.
We obtain
\begin{align*}
\color{blue}{\pi}&\color{blue}{=(1\,4\,5)(2\,3)}\\
&=(1\,4\,5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all complex numbers which make the following equations true: $ |z+1| =1 $ and $ |z^2+1| =1 $ Find all complex numbers which make the following equations true:
$$ |z+1| =1 $$
$$ |z^2+1| =1 $$
Solution:
If $ |z+1| =1 $ holds true, then
$$z+1 = 1.e^{i2n\pi}$$
$$z = 1.e^{i2n\pi}-1$$
$$z = 1.e^{i2n\pi}-1e^{i2m\pi}$$
If... | If we interpret the absolute-value expressions as descriptions of loci of points at specified distances from particular locations, then
$ |z + 1| \ = \ 1 \ \ $ is the set of points at unit distance from $ \ z \ = \ -1 \ \ $ [the unit circle centered at that point] and
$ |z^2 + 1| \ = \ |(z + i)·(z - i)| \ = \ 1 \ \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that: $\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\sqrt{3(a^2+b^2+c^2-ab-bc-ca)}\ge \frac{3}{2}$ Problem:
Given non- negative real numbers such that: $ab+bc+ca>0: a+b+c=3.$ Prove that: $$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\sqrt{3(a^2+b^2+c^2-ab-bc-ca)}\ge \frac{3}{2}$$
My approach:
... | The following stronger inequality is also true.
Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$ and $ab+ac+bc\neq0$. Prove that:
$$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\frac{7}{27}\sqrt{a^2+b^2+c^2-ab-bc-ca}\ge \frac{3}{2}.$$
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying a derivative of a function involving roots. I am working on a question where I need to differentiate $$y=\{x+\sqrt{1+x²}\}^\frac{3}{2}\tag1$$
Using the chain rule I have found the first derivative $$\frac{dy}{dx} =\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{1}{2}\{1+\frac{x}{\sqrt{1+x²}}\}\tag2$$
Which can be writte... | Since it is desired to have an expression for the derivative that includes $ \ y \ \ , $ we will want to keep in mind that
$$ y \ \ = \ \ [ \ x \ + \ \sqrt{1+x^2} \ ]^{3/2} \ \ \Rightarrow \ \ x \ + \ \sqrt{1+x^2} \ \ = \ \ y^{2/3} $$
and thus $ \ [ \ x \ + \ \sqrt{1+x^2} \ ]^{1/2} \ = \ y^{1/3} \ \ . $
The first d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the complex integral as follows Let $C$ be a contour that is formed by an arc $y = x^2$ from $(0,0)$ to $(1,1)$, and a line segment from $(1,1)$ to $(0,1)$. Find the integral $\int_C f(z) dz$ where $f(z) = 2xy+ i(-x^2+y^2)$.
Attempt:
Let $C_1$ be the arc $y=x^2$ and $C_2$ be the line segment such that $C= C_1 ... | Since $f(z)=-iz^2$, $f$ is indeed analytic. If we "add" another path $C_3$ to the contour, namely the one from $(0,1)$ to $(0,0)$. Then $C$ is closed. Now we get $$\int_C f(z)dz=0$$ by Cauchy. Furthermore $$0=\int_C f(z)dz=\int_{C_1}f(z)dz+\int_{C_2}f(z)dz+\int_{C_3}f(z)dz$$ and $$\int_{C_1}f(z)dz+\int_{C_2}f(z)dz=-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4278035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The smallest value of the expression $4x^2y^2+x^2+y^2-2xy+x+y+1$ What is the smallest value that $4x^2y^2+x^2+y^2-2xy+x+y+1$ can take with real numbers $x$ and $y$?
I suspect the following transformation can be done: $(2xy-1/2)^2 + (x+1/2)^2 + (y+1/2)^2 + 1/4$.
| After $(2xy-1/2)^2 + (x+1/2)^2 + (y+1/2)^2 + 1/4$, use a change of variables $x - 1/2 = u, y - 1/2 = v$, and $(2xy-1/2) = 2(u + 1/2)(v + 1/2) - 1/2$ $ = 2(uv + u/2 + v/2 + 1/4) - 1/2$ to get:
$$(2uv+u+v)^2 + u^2 + v^2 + 1/4$$
$$=(2uv+u+v)^2 + (u+v)^2 - 2uv + 1/4$$
$$= (p+q)^2 + q^2 - p + 1/4$$
where $p=2uv, q = u+v$.
F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4278610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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An uncommon continued fraction of $\frac{\pi}{2}$ I'm currently stuck with the following infinite continued fraction:
$$\frac{\pi}{2}=1+\dfrac{1}{1+\dfrac{1\cdot2}{1+\dfrac{2\cdot3}{1+\dfrac{3\cdot 4}{1+\cdots}}}}$$
There is an obscure clue on this: as one can derive the familiar Lord Brouncker’s fraction below
$$
\fra... | We know that $$\sin^{-1}x=\int\underbrace{\color{red}{\frac{1}{\sqrt{1-x^2}}}}_{\text{apply binomial theorem}}dx=\int1+\sum_{n=0}^{\infty}\frac{2n-1}{2^{n+1}}x^{2n}dx$$and if you solve this further you'll get $$\sin^{-1}x=x+
\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+......$$ $$= x+\frac{1}{2}\cdot\frac{x^3}{3}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4279046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
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How would you prove this by induction? I've been trying to solve this one using induction for quite a while but I don't get to the solution. Any tips would be apreciated.
Let $a_1 = 47$, $a_2=80$ and for $n \geq 3$, $a_n =4a_{n-1} - 4a_{n-2}+3(n-2)^2$. Prove that for any positive integer $$ a_n=2^n(3+n)+3n^2+12n+24$$
| By Induction
Prove basis (trivial).
Induction step:
$\begin{align}
a_{n+1}=&4a_{n}-4a_{n-1}+3(n-1)^2 \\
=&4.2^n(3+n)+12n^2+48n+96 \\
-&4.2^{n-1}(2+n)-12(n-1)^2-48(n-1)-96 \\
+&3n^2-6n+3
\end{align}$,
by induction hypothesis.
$\implies a_{n+1}=2^{n+1}(3+n+1)+3(n+1)^2+12(n+1)+24$, by simplifying
$\therefore a_n=2^n(3+n)+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Max of $ax+by$ given that $x^2+y^2 \le 1$ If a and b a positive real numbers find the maximum value of ax+by in terms of a and b given that $x^2+y^2 \le 1$.
I started by letting $k = ax+by$ giving the line $y=\frac{-ax}{b} + \frac{k}{b}$. Now we want to find the largest value of k such that the line still intersects th... | Applying Cauchy-Schwarz Inequality
$$(ax+by)^2\leq (a^2+b^2)(x^2+y^2)\leq (a^2+b^2) \times 1 =(a^2+b^2)$$
$$ax+by \leq \sqrt{a^2+b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4284805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Write with one radical $\sqrt[4]{2^6}\cdot\sqrt{3^3}$
Write with one root (radical) $$\sqrt[4]{2^6}\cdot\sqrt{3^3}$$
In this lesson we have learnt that when the roots exist then $$\sqrt[n]{a}=\sqrt[nk]{a^k}$$
Using that here, we have $\gcd(4,2)=2,$ so
$$\sqrt[4]{2^6}\cdot\sqrt[2\cdot2]{\left(3^{3}\right)^2}=\sqrt[4]{... | Both answers are mathematically equal. What is different is just observation.
We see that,
$$\begin{align}\sqrt[4]{2^6}\cdot\sqrt{3^3}&=2^{\frac 64}\times 3^{\frac 32}\\
&=2^{\frac 32}\times 3^{\frac 32}\\
&=(2\times 3)^{\frac 32}\\
&=6^{\frac 32}\\
&=\sqrt{6^3}\\
&=\sqrt{6^2\times 6}\\
&=6\sqrt{6}\end{align}$$
Also,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Cone Section: Intersection between a plane ax + by + cz = d and two cones of height $h$, raidus $r$ whose apexes lie at the centre, along the $z$ axis A double cone can be described by circles that increase with $|z|$ and have radius $R$ at $h = |z_R|$. It holds:
\begin{align}
\frac{R^2}{h^2}\,z^2 = x^2 + y^2
\end{... | First, you need to express the unit normal to the plane in spherical coordinates:
$n = \dfrac{1}{\sqrt{a^2 + b^2+c^2}} (a, b, c) = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) $
Comparing the expressions, we have
$ \theta = \cos^{-1} \dfrac{c}{\sqrt{a^2 + b^2+c^2} }$
$ \phi = \text{ATAN2}(a,b) $
It follo... | {
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Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now... | Since I saw it working, I wanted to write this method for readers as well.
We see that, the polynomial $4x^2-2xy-4x+3y-3$ contains the term of $y$, but not the term of $y^2$.
Thus, we can come to the following conclusion:
If factorization is possible, then only one multiplier polynomial contains the term of $«y»$, bu... | {
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When is the sum of the products of any three distinct numbers less than $n$, divisible by $n$? Define $A$ as the set of all integer triples $(a,b,c)$, such that $0<a<b<c<n$.
$$A=\left\{(a,b,c)\in \mathbb{Z^3}:1\le a<b<c\le n-1 \right\}$$
Define $S$ as the sum of the product $abc$ for each triplet in $A$.
$$S=\displayst... | Set
$$A_{3,n} := \{(a,b,c) \in \mathbb{Z}^3: 1 \leq a < b<c \leq n-1 \}$$
$$S_{3,n} := \sum_{(a,b,c)\in A_{3,n}}abc$$
Note that $$6S_{3,n} = \sum_{1 \leq a,b,c \leq n-1, a\neq b , b \neq c, a \neq c}abc$$
$$= (\sum_{j=1}^{n-1}j)^3 - 3(\sum_{j=1}^{n-1}j^2)(\sum_{j=1}^{n-1}j)+2\sum_{j=1}^{n-1}j^3$$
$$=\frac{(n-1)^3n^3}{8... | {
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How to find the exact value of the integral $ \int_{0}^{\infty} \frac{d x}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}}$? $\textrm{I first reduce the power two to one by Integration by Parts.}$
$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6... | It is a nice solution for sure.
As you showed the problem is to compute
$$I=\int \frac {dx}{x^6+1}$$
My working
Writing
$$x^6+1=(x^2+1)(x^4-x^2+1)=(x^2+1)(x^2-a)(x^2-b)$$ Using partial fraction decomposition
$$\frac 1 {(x^2+1)(x^2-a)(x^2-b)}=$$ $$\frac{1}{(a+1) (a-b) \left(x^2-a\right)}-\frac{1}{(b+1) (a-b)
\left(x... | {
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Finding a pair of commutative stochastic $3 \times 3$ matrices Could you please give me an example of a pair $\left(A,B\right)$ of commutative stochastic $3 \times 3$ matrices with real entries given $A \neq B$ and excluding the identity matrix, symmetric matrices and doubly stochastic matrices?
| What it we take $B$ a convex combination of powers of $A$? In generic cases, if two matrices commute $A$, $B$ commute one $B$ will be a polynomial in the other $A$ ( if the eigenvalues of $A$ are distinct).
$\bf{Added:}$ Here is a stochastic matrix $A$ with eigenvalues $1$, $\frac{1}{6}$, $\frac{1}{6}$
$$A=\left( \beg... | {
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How to prove $n{n+9 \choose 8} \bmod 10 = 5 \Rightarrow n \bmod 16 \in \{1,3,5,15\}$ In reading this question I noticed that the values of $n$ there for which $s(n)$ there has the last digit equal to $5$ are a subset of OEIS A103127. Combining that observation with another comment in the above question, plus a little n... | Consider mod 2: $ n { n+ 9 \choose 8} \equiv 1 \pmod{2}$.
We must have both
*
*$ n \equiv 1 \pmod{2} $, and
*$ n + 9 \equiv 8, 9, 10, 11, 12, 13, 14, 15 \pmod{16} $ by Lucas theorem.
This simplifies to $ n \equiv 15, 1, 3, 5 \pmod{16}$.
This answers the first part of your question.
Consider mod 5: $ n { n + 9 \ch... | {
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cubic polynomial for $L^2$ approximation.
Consider $f(x) = \sin \pi x$. Find the cubic polynomial giving $L^2$ approximation to $f$ on $[0,1]$.
I suspect I need to find a polynomial of the form $p(x) = \sum\limits_{i=0}^3 c_ix^i$ and calculate this (EDIT: below):
$$\int_0^1\left|f(x) - p(x)\right|^2 dx$$
(EDIT)Follow... | The question is quite badly worded, in my opinion. I suppose it is asking you to find the cubic polynomial that is the best approximation of $f(x)$ in the $L2$ norm on the interval $[0,1]$.
In other words, you need to find the cubic polynomial $p$ that makes the error
$$
\int_0^1(f(x)-p(x))^2\,dx
$$
as small as possibl... | {
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Proof of identity related to sum of binomial coefficients and powers of $2$ Given two positive integers $a,b$, prove that $\left(\sum_{n=a}^{a+b-1}\binom{n-1}{a-1}2^{-n}\right)+\left(\sum_{n=b}^{a+b-1}\binom{n-1}{b-1}2^{-n}\right)=1$
The context comes from this MSE problem, where I indirectly showed that this is true w... | The first term is
$$\sum_{n=a}^{a+b-1} {n-1\choose a-1} 2^{-n}
= 2^{-a} \sum_{n=0}^{b-1} {n+a-1\choose a-1} 2^{-n}
\\ = 2^{-a} [z^{b-1}] \frac{1}{1-z} \frac{1}{(1-z/2)^a}
= (-1)^a \mathrm{Res}_{z=0}
\frac{1}{z^b} \frac{1}{1-z} \frac{1}{(z-2)^a}.$$
Now residues sum to zero and the residue at infinity is zero by
inspecti... | {
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What's the measure of the segment $BC$ in the rhombus below? For reference: Given a rhombus $ABCD$, on $BC$
mark the point $P$ such that : $BP= 3PC$ and $AP^2+ 3DP^2 = 38$.
Calculate $BC$.(answer: $2\sqrt2$)
My progress:
$BP = 3CP\\
AP^2+3DP^2 = 38\\
AB=BC=CD=AD$
Th. Stewart:
$\triangle ABC:\\
AC^2.BP+AB^2.CP=AP^2BC+B... | I would have simply placed the figure on a coordinate plane such that $$\begin{align}
C &= (x,0), \\
B &= (0,y), \\
A &= (-x,0), \\
D &= (0,-y), \\
\end{align}
$$
hence $$P = (\tfrac{3}{4}x, \tfrac{1}{4}y),$$
and
$$AP^2 = \left(\tfrac{7}{4} x\right)^2 + \left(\tfrac{1}{4}y\right)^2 = \frac{49x^2 + y^2}{16}, \\
DP^2 = ... | {
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Determinant of a $4 \times 4$ matrix
If $$ A = \begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix} $$ calculate $\det(A)$.
If you calculate
$$AA^t=\begin{pmatrix}a^2+b^2+c^2+d^2&0&0&0\\0&a^2+b^2+c^2+d^2&0&0\\ 0&0&a^2+b^2+c^2+d^2&0\\0&0&0&a^2+b^2+c^2+d^2\end{pmatrix}$$
$$\det(AA^t)=(a^2+b^2+c^2+d^2)^... | Because the coefficient of $a^4$ (which is the product of the main diagonal) is $+1$ not $-1$.
| {
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Factoring by grouping: how to deal when the factors extracted are of the form $(x+k)$ and $ (x-k) $. EDITED : the original post contained a mistake regarding the factorization of the numerator.
The source of the exercice I'm trying to do is : Barton's College Practice Placement Test, Q. 17 https://www.barton.edu/pdf/m... | One way is to try to get the factors of the numerator into the denominator:
$$4x^2 -1 = (2x-1)(2x+1)$$
and
$$2x^2 +5x -3 = (2x^2 -x) + (6x-3) = x(2x-1) + 3(2x-1) = (2x-1)(x+3).$$
Another possibility (if you do not wish to study the discriminant formula) is to complete the square:
\begin{align}
2x^2 + 5x -3 &= 2\left(x^... | {
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Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=xyz$. Prove the following $xyz\geq27,xy+yz+zx\geq27,x+y+z\geq9$ Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=xyz$.
Prove $$xyz\geq27,\\xy+yz+zx\geq27,\\x+y+z\geq9.$$
Here's what I tried:
$$\begin{align} \frac{x^3+y^3+z^3}{3}&\ge\sqrt[3]{x^3y^3z... | (1)
$xyz=x^2+y^2+z^2 \ge 3\sqrt[3]{x^2y^2z^2}$
So $\sqrt[3]{xyz}\ge3$
$xyz\ge27$
(2)
$xy+yz+zx\ge3\sqrt[3]{x^2y^2z^2}\ge3\sqrt[3]{27^2}=27$
(3)
$x^2+y^2\ge2xy$
$x^2+z^2\ge2xz$
$y^2+z^2\ge2yz$
add and divided by 2
$\Rightarrow x^2+y^2+z^2\ge xy+yz+zx$
$\Rightarrow (x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz\ge3(xy+yz+zx)\ge3*27$
... | {
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Proving that $x^2-2xy+6y^2-12x+2y+41\ge 0$ where $x,y \in\Bbb{R}$ We use the result that $AX^2+BX+C \ge 0, \forall X ~ \in \Bbb{R} $ if $A>0$ and $B^2-4AC \le 0$
for proving that $f(x,y)=x^2-2xy+6y^2-12x+2y+41\ge 0$, where $x,y \in\Bbb{ R}.$
Let us re-write the quadratic of $x$ and $y$ as a quadratic of $x$ as
$$f(x,y)... | In the original variables, your polynomial is $(x-y-6)^2 + 5(y-1)^2 $ and so is non-negative, equal to zero only when $y=1$ and then $x=7,$ so $(7,1)$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- 1 & 1 & 0 \\
- 6 & - 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 0 ... | {
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"answer_id": 2
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System of non-linear equations (ISNMO, 1991) Let $x,y,z$ be real numbers. Solve the following system of equations: $$\begin{cases}
\frac{3(x^2+1)}x=\frac{4(y^2+1)}y=\frac{5(z^2+1)}z; \\
xy+xz+yz=1.
\end{cases}$$
I tried to solve this system using the following method:
Since $$\frac{3(x^2+1)}x=\frac{4(y^2+1)}y; \Right... | Expanding on OP's work and my comments:
From the first double equality it is obvious that $x,y,z\ne 0$, and $x, y$ and $z$ have the same sign (since the numerators are positive). So one also has $x+y,y+z$, and $z+x\ne 0$
Then as in OP's idea, using the given:
$$xy+yz+xz=1 \tag{1}$$
One gets:
$$x^2+1=x^2+xy+yz+xz=x(x+y)... | {
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Solve the equation $12x^5+16x^4-17x^3-19x^2+5x+3=0$ Solve the equation $$12x^5+16x^4-17x^3-19x^2+5x+3=0$$ The divisors of $3$ are $\pm1;\pm3$ and the divisors of $12$ are $\pm1;\pm2;\pm3;\pm4;\pm6;\pm12$, so the possible rational roots are $$\pm1;\pm\dfrac12;\pm\dfrac13;\pm\dfrac14;\pm\dfrac16;\pm\dfrac{1}{12};\pm3;\pm... | We first try the simplest $x= \pm 1$, then $12+16-17-19+5+3=0 \Rightarrow x-1 $is a factor and $-12+16+17-19-5+3=0 \Rightarrow x+1 $ is also a factor.
By synthetic division, we have
$$
\begin{aligned}12 x^{5}+16 x^{4}-17 x^{3}-19 x^{2}+5 x+3 =(x+1)(x-1)\left(12 x^{3}+16 x^{2}-5 x-3\right)
\end{aligned}
$$
Now we can t... | {
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How can i evaluate this integral with or without using CAS $\int_0^{\infty}\frac{\mathrm dx}{(x^2+1)\cosh(\pi x)}$ $$\int_0^{\infty}\frac{\mathrm dx}{(x^2+1)\cosh(\pi x)}$$
Firstly i used substitution $\pi x=t; \mathrm dx=\frac{\mathrm dt}{\pi}$
$$\pi \int_0^{\infty}\frac{\mathrm dt}{(t^2+\pi^2)\cosh t}$$
writing $\cos... | Similar to the question already mentioned by @Laxmi Narayan Bhandari, considering the infinite series
$$\text{sech}(z)=\pi\sum_{k=0}^\infty (-1)^k\frac{ (2 k+1)}{\pi ^2 \left(k+\frac{1}{2}\right)^2+z^2}$$
Making the problem more general
$$I=\int_0^\infty \frac{\text{sech}(a x)}{b x^2+1}\,dx \quad \text{with} \quad a>... | {
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Finding a root of a degree 5 polynomial
There are positive integers $m$ and $n$ such that $m^2 −n = 32$ and $$\sqrt[5]{m+\sqrt{n}} + \sqrt[5]{m-\sqrt{n}}$$ is a real root of the
polynomial $$x^5 −10x^3 + 20x −40$$ Find $m+n$.
Okay I write the other $4$ roots as $x_{1}, x_{2}, x_{3}, x_{4}$ and get $\sqrt[5]{m+\sqrt{n... | Your second idea is better $$(a+b)^5 = a^5+b^5 + 5ab(a^3+b^3)+10a^2b^2(a+b)$$
So $$x^5 = 2m + 10(a+b)\Big((a+b)^2-6\Big)+40(a+b)$$ or $$x^5+2m+10x^3-60x+40x$$
so $$x^5-10x^3+20x -2m =0$$ and thus $m=20$ so ...
| {
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What are the minimum and maximum of $f(x) = x^6 - 5x^4 + 5x^2 - 1$? I am stuck in obtaining the critical points:
$f'(x) = 6x^5 - 20x^3 + 10x$
$f''(x) = 30x^4 - 60x^2 + 10$
For critical points, we have: $f'(x) = 0$
$2x(3x^4 - 10x^2 + 5) = 0$
Now, how should i solve them to get the critical points. And what should the mi... | $2x(3x^4 - 10x^2 + 5) = 0$
$x_1=0$
$3x^4 - 10x^2 + 5 = 0$
Let $y=x^2$ then $3y^2 - 10y + 5 = 0$ and $y_{1,2}=\frac{5\pm\sqrt{10}}{3}$
and
$x_2=-\sqrt{\frac{5-\sqrt{10}}{3}}$, $x_3=\sqrt{\frac{5-\sqrt{10}}{3}}$, $x_4=-\sqrt{\frac{5+\sqrt{10}}{3}}$, $x_5=\sqrt{\frac{5+\sqrt{10}}{3}}$
So the maximum is $\infty$
Function i... | {
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"answer_id": 1
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What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left... | When you got$$\require{cancel}\frac{\cancel{-x^{2}}+\cancel{x^{2}}-\frac{x^{4}}{3}+o\left(x^{3}\right)}{-\frac{x^{4}}{2}+o\left(x^{5}\right)},$$you divided both the numerator and the denominator by $x^4$, getting$$\frac{-\frac13+o(x^3)}{-\frac12+o(x^5)}.$$This is not correct, because $\frac{o(x^3)}{x^4}$ is not $o(x^3)... | {
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"source": "stackexchange",
"question_score": "3",
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What's the approach for this inequality question? The question:
$a,b,c > 0; ab+bc+ca =3$, Prove that
$\sum_{cyc}\frac{a}{\sqrt{a^{3}+5}}\leq\frac{\sqrt{6}}{2}$
The sum is cyclic over $a,b,c$
I've looked at the problem for a long time but still can't think of an approach for this, so how can I solve this?
| We have
\begin{align*}
\sum_{\mathrm{cyc}} \frac{a}{\sqrt{a^3 + 5}}
&= \sum_{\mathrm{cyc}} \frac{a}{\sqrt{\frac{a^3}{2} + \frac{a^3}{2} + \frac12 + \frac92}}\\
&\le \sum_{\mathrm{cyc}} \frac{a}{\sqrt{3\sqrt[3]{\frac{a^3}{2} \cdot \frac{a^3}{2} \cdot \frac12} + \frac92}}\\
&= \sum_{\mathrm{cyc}} \frac{2a}{\sqrt{6a^2... | {
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How many ways can $720$ be decomposed into a product of two positive integers?
How many ways can $720$ be decomposed into a product of two positive integers?
My solution: There are $5 \cdot 3 \cdot 2 = 30$ ways to choose the exponents a, b, c, such that $2^a \cdot 3^b \cdot 5^c= 720$. Soon there are $30$ dividers. As... | I believe this may be another approach:
We want $ab=720$. Since we have $2^5 \cdot 3^3 \cdot 5^2$ and we want this to form the product of 2 primes such that it results in $720$, then:
$$\binom{5}{2} \cdot \binom{3}{2} \cdot \binom{2}{2}= 30.$$ But since we ended up counting the possibilities twice (for example: if $a=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve $(y+1)dx+(x+1)dy=0$ Solve $(y+1)dx+(x+1)dy=0$
$$\frac{dy}{y+1}=-\frac{dx}{x+1}$$
then we get $\ln|y+1|=-\ln|x+1|+c$
$$\ln(|(y+1)(x+1)|)=c$$
$$|(y+1)(x+1)|=e^c=c_1$$
but answer is $y+1=\frac{c}{x+1}$ can you help to find where is my mistake?
| $(y+1)dx+(x+1)dy=0$
$M(x, y) =y+1$
$N(x, y) =x+1$
Then, $\frac {\partial M}{\partial y}=1=\frac {\partial N}{\partial x}$
Hence, the differential equation is exact.
Choose, $u(x, y) $ be such that
$u_x =M $ and $u_y =N$
Then, $u(x, y) =\int {M {dx}}=x(y+1)+h(y) $
$u_y (x, y)=x+h'(y) $
$x+h'(y)=x+1 $
$h(y) =y$
Hence, $u... | {
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} |
Ways to find $\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\cdots$
$$\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}+\cdots$$
is equal to?
My approach:
We can see that the $n^{... | If you look at the Binomial expansion of
$$(1-x)^{-\frac{1}{2}}$$ you get :-
$$\sum_{r=0}^{\infty}\frac{\binom{2r}{r}x^{r}}{4^{r}}$$
So $$\int_{0}^{1}(1-x)^{-\frac{1}{2}}dx=\sum_{r=0}^{\infty}\frac{\binom{2r}{r}}{4^{r}(r+1)}$$
So you get $$\frac{1}{2}\sum_{r=0}^{\infty}\frac{\binom{2r}{r}}{4^{r}(r+1)}=\frac{1}{2}\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Why am I getting a different value for $\sin\left(2\tan^{-1}\frac{4}{3}\right)$ than my calculator? The expression:
$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$
Way 1:
If I punch the above expression in my calculator, I get $\frac{24}{25}$.
Way 2:
$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$
$$\si... | One has to keep in mind, which quadrant the angle lies in from the beginning, and not derive it from calculations. Your calculations will be prone to mistakes because trigonometric and inverse trigonometric functions are not one-to-one functions.
Notice that $\tan^{-1} 4/3 > \tan ^{-1} 1 = \pi/4$. So, $\theta = 2\tan^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4336034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determinant formula for coordinates of circumcenter and orthocenter of a triangle I've come across the following formulas for coordinates of circumcenter $O=(x_O,y_O)$ and orthocenter $H=(x_H,y_H)$ of a triangle, in a formula book, stated without derivation. For a triangle with vertices $(x_i,y_i),$ $i \in \{1,2,3\}$,
... | Each point verifies
\begin{eqnarray*}
(x_k-x_O)^2 + (y_k-y_O)^2 &=& r^2 \\
x_k^2+y_k^2+
x_O^2+y_O^2-
2(x_kx_O+y_ky_O) &=&
r^2 \\
2(x_kx_O+y_ky_O)
+ (r^2-x_O^2-y_O^2) &=& x_k^2+y_k^2
\end{eqnarray*}
This yields to the linear system
$$
\begin{pmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{pmatrix}
\begin{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4336754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
how to calculate the cube root of this complex number Can anyone help me solve this exercise?
calculate the cube roots of $\frac{1}{(2-2i)}$ I started by rationalising by doing $\frac{1}{(2-2i)}$= $\frac{1}{(2-2i)}$ * $\frac{(2+2i)}{(2+2i)}$ --->
$\frac{(2(1+i)}{8}$ ----> $\frac{(1+i)}{4}$
then how can I continue?
than... | suppose $z=\dfrac{1+i}{4}=\dfrac{1}{2\sqrt{2}}\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$. Let $z_1=r(\cos\alpha +\sin\alpha)$ be a root of $z_1^3=z$. Then,
$$r^3(\cos3\alpha+i\sin3\alpha)=\dfrac{1}{2\sqrt{2}}\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$$
Comparing moduli we get $r^3=\dfrac{1}{2\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$ Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$.
So far,
\begin{align*}
x^2 + y^2 + 121 + 2xy + 22x + 22y &= x^2 + y^2 + 121\\
2xy + 22x + 22y &= 0\\
(2x+22)y &= -22x\\
(x+11)y &= -11x
\end{align*}
At least 1 of $x,y$ must be a mu... | Hint for another way:Very useful the "Simon's trick" cited above by @John Omielan to solve this kind of diophantine equations.We want to give another kind of solution.
From $11=\dfrac{-xy}{x+y}$ we have $-xy=11K$ and $x+y=K$ so we can put $y=11M$ from which $$-xM=K\text { and }x+11M=K\Rightarrow \begin{cases}x=\dfrac{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find 2nd order recurrence relation with given solution Let $(a)_{n\geq0}$ with $a_n=a_{n-1}+6a_{n-2}$ and $a_0=1, a_1=4$.
The sequence $(b)_{n\geq0}$ shall satisfy $$b_n=\sum_{k=0}^{n}\binom{n}{k}a_k$$
I have to find a 2nd order recurrence relation which has $(b_n)_{n\geq0}$ as solution. How does one approach this task... | For every $n \geq 0$, one has
\begin{align*} b_n&=\sum_{k=0}^{n}\binom{n}{k}a_k\\
&=\sum_{k=0}^{n}\binom{n}{k}\frac{6\cdot3^k+(-1)^{k+1}2^k}{5}\\
&=\frac{6}{5}\sum_{k=0}^{n}\binom{n}{k}3^k - \frac{1}{5}\sum_{k=0}^{n}\binom{n}{k}(-2)^k\\
&=\frac{6}{5}(3+1)^n - \frac{1}{5}(1-2)^n\\
&=\frac{6 \cdot 4^n - (-1)^n}{5}\\
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4345392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Conditions for positivity of this symmetric real matrix I have the following real symmetric matrix $M$ of size 3:
\begin{align}
M =
\begin{pmatrix}
a & b & c \\
b & d & e \\
c & e & f
\end{pmatrix},
\end{align}
for real parameters $a$, $b$, $c$, $d$, $e$, $f$. What are the necessary and sufficient conditions on these... | $D_k : k$-th order principal minor of $A$
$A_{n×n}$ is positive matrix if $D_k\ge 0\space \forall k=1, 2,...,n$
$D_1 : det[a]_{1×1}= a\ge 0$
$D_2 : det \begin{align}
\begin{pmatrix}
a & b \\
b & d \\
\end{pmatrix}
\end{align}=ad-b^2\ge 0$
$\begin{align} D_3&= det
\begin{pmatrix}
a & b & c \\
b & d & e \\
c & e & f\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4354379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $a,b\in \mathbb{R}$ such that $ab=2$. Find the max value of $\frac{3}{2\left(a+b\right)^2}$ and $a, b$ where max is attained, without calculus. My thinking:
By Arithmetic and Geometric mean inequality (AGM):
$ab\le \left(\frac{a+b}{2}\right)^2$
We know $ab=2$
$\rightarrow$ $2\le \left(\frac{a+b}{2}\right)^2$
$\righ... | Why you found the maximal value of $a, b$? The question is asking $\frac3{2(a+b)^2}$!
Using AM-GM is, yes, the best approach. Here is my answer.
Since using AM-GM, $8=4ab\le(a+b)^2$, so $\frac1{(a+b)^2}\le\frac18$.
So the maxima of $\frac3{2(a+b)^2}$ is $\frac32\cdot\frac18=\frac3{16}$ when $a=b=\pm\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Compute the sum $ \sum_1^{\infty} \ln{ \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} }$ I need to prove that the sum $ \sum_{1}^{\infty} \ln \left( \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} \right) $ converges and equals $ \ln(\frac{4}{3})$. I tried ex... | Let's do it carefully:
$\ln(\sum_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$
$\ln(\prod_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$
$\ln(\prod_{n=1}^k \frac {n+1}{n}\prod_{n=1}^k\frac {3n+1}{3n+4})$.
Now $\prod_{n=1}^k\frac {n+1}n$ clearly equals $k+1$ as it's just the telescoping series $\require{cancel} \frac {\cancel 2}1\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4357875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Trouble integrating square of a sine I have the following expression which I am trying to evaluate:
$$ \frac{2}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2 kx$$ where $k = \frac{\pi}{L}$.
According to my calculator, the answer should be $\frac{1}{6}$ but I can't seem to get this result by manual integration.
Here's what... | There are two errors: First,
$$
\frac{1}{L} \bigl[ x \bigr]^{\frac{L}{3}}_{\frac{L}{6}} = \frac 16
$$
and not $1/3$. Second,
$$
\left[\frac{1}{2 \pi} \sin \frac{2 \pi}{x} \right]^{3}_{6}
= \frac{1}{2 \pi} \left( \sin \frac{2\pi}{3}- \sin \frac{ \pi}{3}\right) = 0 \, .
$$
It seems that you added the terms instead of su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Diagonalisation of stochastic matrices Suppose that $(X_n)_{n≥0}$ is a Markov chain on a state space $I = {1, 2}$ and stochastic matrix
$$P = \begin{bmatrix} \frac{1}{4} & \frac{3}{4} \\ \frac{1}{3} & \frac{2}{3} \end{bmatrix}$$
(a) Find the eigenvalues
$$|P-\lambda I| = \begin{vmatrix} \frac{1}{4}-\lambda & \frac{3}{4... | You're almost there. We have
\begin{align}
P^n &=
\begin{bmatrix}
\frac{1}{4} & \frac{3}{4} \\ \frac{1}{3} & \frac{2}{3}
\end{bmatrix}^n =
\begin{bmatrix}
1 & -9 \\ 1 & 4
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\ 0 & -\frac{1}{12}
\end{bmatrix}^n
\begin{bmatrix}
\frac{4}{13} & \frac{9}{13} \\ -\frac{1}{13} & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Two partial fraction approaches, one is wrong, the other is right, why? I want to do a partial fraction on
\begin{equation}
\frac{z}{(z-4)(z+\frac{1}{2})}
\end{equation}
Method one, which apparently is wrong:
\begin{equation}
\frac{z}{(z-4)(z+\frac{1}{2})}=\frac{A}{z-4}+\frac{B}{z+\frac{1}{2}}
\end{equation}
\begin{arr... | Both methods are algebraically correct. That is, both methods yield new functions that are the same as the original function.
Your method is the correct way to carry out the partial fraction decomposition algorithm. Moreover, your method results in the outcome desired from the partial fraction algorithm—a sum of ration... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Mixed Poisson distribution with Lindley mixing I want to calculate the unconditional distribution of $x$
$$p(x|p)
= \int_0^\infty p(x,\theta|p) d\theta = \int_0^\infty p(x|\theta)p(\theta|p) d\theta$$
where $x|\theta \sim Poisson(\lambda)$ and $\lambda$ is a random variable that follows the Lindley distribution $$p(\t... | Assuming your work up to the last line (before edit) is correct we have with a little algebraic manipulation
\begin{align}
f_X(x|p)
&=\frac{p^2}{(p+1)x!}\int_0^\infty e^{-\theta(1+p)}\theta^{x}(\theta+1)\,\mathrm d\theta\\
&=\frac{p^2}{(p+1)^{x+2}}\int_0^\infty (1+\theta)\frac{(p+1)^{x+1}}{\Gamma(x+1)}\theta^{(x+1)-1}e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find exact value of integral $\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}}dx$? When I first encountered the integral $\displaystyle \int_{0}^{\infty} \frac{1}{x^{4}-x^{2}+1} d x$, I am very reluctant to solve it by partial fractions and search for any easier methods. Then I learnt a very useful tr... | With CAS and help of MellinTransfrom:
$$\int_0^{\infty } \frac{1}{\left(x^4-x^2+1\right)^n} \, dx=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \mathcal{M}_a\left[\frac{1}{\left(x^4-a x^2+1\right)^n}\right](s) \,
dx\right](1)=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \frac{(-1)^{-s} x^{-2 s} \left(1+x^4\right)^{-n+s} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
} |
Find all positive integers s.t. $\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$ Find all positive integers $a, b, c, d$ such that :
$$\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$$
The original problem came from atomic electron transitions :
I would like to find out non-trivial positiv... | (Remark: The following is inspired by Equation of 1/x^2 on AoPS. However, a bit more work is needed to characterize all solutions of the given equation. It does not suffice to start with Pythagorean triples.)
Let $a, b, c, d$ be positive integers satisfying
$$ \tag{$1$}
\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
$\sum \frac{a}{b+c+d}\le \frac{2\sum a^2}{\sum ab}$ if $\sum a =4$ Let $a, b, c, d$ positive real numbers such that $a+b+c+d=4$. Prove that
$$\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\le
\frac{2(a^2+b^2+c^2+d^2)}{ab+ac+ad+bc+bd+cd}.$$
My idea is to cancel the denominators, expand the inequality an... | Fill in any missing gaps.
*
*Multiplying by $(ab+ac+ad+bc+bd+cd)$, show that the inequality is equivalent to $$\sum \frac{ abc+abd+acd}{b+c+d} \leq \sum a^2. $$
*Show that $$ \frac{ abc+abd+acd}{b+c+d} \leq \frac{ab+ac+ad}{3} \leq \frac{3a^2+b^2+c^2+d^2}{6}$$
*Hence, summing up the cyclic inequalities, the conclus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $x^2\cdot f(x)+f(1-x)=2x-x^4$ $\forall\; x\in \mathbb R$
Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that: $\;x^2\cdot f(x)+f(1-x)=2x-x^4,\;\forall\; x\in \mathbb R.$
My solution:
Replace $x$ by $(1-x)$ and by eliminating $f(1-x)$.
I ob... | Your solution is incomplete, because the solution is not required to be continuous. Your system of equations
$$
\begin{pmatrix}
x^2 & 1 \\
1 & (1-x)^2
\end{pmatrix}
\begin{pmatrix}
f(x) \\
f(1-x)
\end{pmatrix}
=
\begin{pmatrix}
2x-x^4 \\
2(1-x) - (1-x)^4
\end{pmatrix}
$$
has infinitely many solutions if $x=\varphi =\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4371595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find prime number $p$ such that $19p+1$ is a square number.
Find prime number $p$ such that $19p+1$ is a square number.
Now, I have found out, what I think is the correct answer using this method.
Square numbers can end with - $1, 4, 9, 6, 5, 0$.
So, $19p+1$ also ends with these digits.
Thus, $19p$ ends with - $0, 3,... | It's generally not a good idea to try to solve problems like this by looking at decimal digits. Pay more attention to properties of divisibility.
You want $19p+1 = x^2$ for a positive integer $x$, so
$$
19p = x^2 - 1 = (x+1)(x-1).
$$
Comparing the two factorizations $19p$ and $(x+1)(x-1)$ is the key point.
Since $x^2 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How we may express four squares whose difference is each a square in terms of (preferably solid) geometry? The problem of finding four squares whose difference is each a square is much more exhaustive as I thought. A quest up to $2^{34}$ yields nothing. The largest almost solution found in the range up to $2^{34}$ is $... | Suppose you have positive integers satisfying
\begin{array}{lllllll}
z^2-y^2&=&a^2&\qquad\qquad&y^2-x^2&=&d^2\\
z^2-x^2&=&b^2&\qquad\qquad&y^2-w^2&=&e^2\\
z^2-w^2&=&c^2&\qquad\qquad&x^2-w^2&=&f^2
\end{array}
Then
\begin{eqnarray*}
b^2&=&z^2-x^2=(z^2-y^2)+(y^2-x^2)=a^2+d^2,\\
c^2&=&z^2-w^2=(z^2-y^2)+(y^2-w^2)=a^2+e^2,\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What are the possible values of the angles of triangle $ABC$?
In a triangle $ABC$, let $AP$ be the bisector of $\angle BAC$ with $P$ on the side $BC$, and let $BQ$ be the bisector of $\angle ABC$ with $Q$ on the side $CA$.
We know that $\angle BAC=60^\circ$ and that $AB + BP = AQ + QB$.
What are the possible values ... |
Hints: As can be seen in figure there are two key points you have to show:
1- Q is on perpendicular bisector of BC.
2- Triangle BPI is isosceles.
This is only possible construction.Using bisectors theorem may help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculating arc length of $y=\frac{x^5}{10}+\frac{1}{6x^2}$, $x \in [1, 2]$ I want to calculate arc length of a function: $y=\frac{x^5}{10}+\frac{1}{6x^2}$, $x \in [1, 2]$. I calculate the derivative of $y$ and multiply it by itself to prepare for the formula:
$((\frac{x^5}{10}+\frac{1}{6x^2})')^2 = \frac{9x^{14}-12x^7... | This exercise resembles a set of contrived examples commonly used in textbooks when students are studying arc length. For example:
Let $y=\dfrac{x^5}{10}+\dfrac{1}{6x^3}$ with $x^3$ rather than $x^2$ in the second term.
[In these contrived examples, the exponent in the denominator of second term must be two less than ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4381453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the maclaurin series of 4x/(8+x^3) So I've done multiple attempts at this practice problem, but my result seems wrong. I'm new to this chapter, and I think I miss something fundamental, but I can't tell what. Can anybody see where I'm going wrong?
The practice problem
Find the maclaurin series of f(x) by using kno... | I shall use the fact that if, on some interval $(-a,a)$ (with $a>0$) you have $f(x)=\sum_{n=0}^\infty a_nx^n$, then the Maclaurin series of $f$ is $\sum_{n=0}^\infty a_nx^n$. In this case, you have\begin{align}\frac1{8+x^3}&=\frac18\cdot\frac1{1+\left(\frac x2\right)^3}\\&=\frac18\left(1-\left(\frac x2\right)^3+\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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About lineal dependence of sets of vectors in $\mathbb{R}^{4}$ Which of these sets of vectors in $\mathbb{R}^{4}$ are linearly dependent?
\begin{eqnarray}
&\text{(a)}& (1,2,1,-2), (0,-2,-2,0), (0,2,3,1),(3,0,-3,6)\\
&\text{(b)}& (4,-4,8,0),(2,2,4,0),(6,0,0,2),(6,3,-3,0)\\
&\text{(c)}& (4,4,0,0),(0,0,6,6),(-5,0,5,5)... | Let $S:=\{v_{1},v_{2},\ldots,v_{n}\}\subseteq V$ with $V$ a vector space $n-$dimensional over a field $\mathbb{F}$ so $S$ is linearly independent iff $\det(v_{1},v_{2},\ldots,v_{n})\not=0$. For a) since $\det\begin{bmatrix} 1 & 0 & 0 &3 \\ 2 & -2 & 2 & 0\\ 1 & -2 & 3 & -3\\ -2 & 0 & 1 & 6\end{bmatrix}=-24\not=0$ so lin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to prove $2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$
Prove $$2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$$
I got this question when I was going through some basic trigonometric identities as follows
$2(\cos\frac{\pi}{5}-\cos\frac{2\pi}{5}) =1\tag1$
very much straightforward to recognise
$2(\cos\... | Let's use the identity $(-1)^a\cos x=\cos(a\pi-x)$. Therefore,
$$\begin{align}&2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} \\= &2\sum_{k=1}^n\cos\left((k-1)\pi-\frac{k\pi}{2n+1}\right)\\=&-2\sum_{k=1}^n\cos\frac{2nk\pi}{2n+1}\end{align}$$
Take a factor $\csc\left(\frac{n\pi}{2n+1}\right)$ and then use product to su... | {
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"url": "https://math.stackexchange.com/questions/4389329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the remainder when $f(x) = \sum_{i=0}^5 x^i$ is divided into $f(x^{12})$ I am looking for confirmation that this proof works. Feel free to be very pedantic.
We wish to find the remainder when $f(x) = \sum_{i=0}^5 x^i$ is divided into $f(x^{12})$. Notice that $f(x)(x - 1) = x^6 - 1$, so let $r$ be a root of $f$. Th... | You are almost there. Let us write $Q(x) := f(x^{12})$. Then indeed: $$Q(x) = f(x^{12})=f(x)q(x)+r(x),$$ where $r$ is a polynomial of degree at most $4$.
You noted that $Q(a)=6$ for all $a$ satisfying $a^6=1$, and that the roots of $f$ are $\{a; a^6=1$; $a \not = 1\}$, so that $Q(a)=6$ for each $a$ such that $f(a)=0$, ... | {
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"url": "https://math.stackexchange.com/questions/4389478",
"timestamp": "2023-03-29T00:00:00",
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How do you prove: $\dfrac{1}{\sin ^2 (\frac{2\pi}{9})} - \dfrac{1}{\sin ^2(\frac{4 \pi}{9})} = 8\sin(\frac{\pi}{18})$ I have to prove:
$\displaystyle \tag*{} \alpha={\dfrac{1}{\sin ^2 (\frac{2\pi}{9})} - \dfrac{1}{\sin ^2(\frac{4 \pi}{9})}}= 8\sin(\frac{\pi}{18})$
I tried to make a common denominator of $\alpha$ and us... | Let $c_n:=\cos\frac{n\pi}{9},\,s_n:=\sin\frac{n\pi}{9}$ so we want to prove$$\alpha:=\frac{s_6}{s_2s_4^2}=8c_4.$$Since $s_{9-n}=s_n$ and $s_3=\frac{\sqrt{3}}{2}$,$$\frac{\alpha}{8c_4}=\frac{\sqrt{3}}{8s_1s_2s_4}.$$Now we just need to prove$$s_1s_2s_4=\frac{\sqrt{3}}{8},$$which is a duplicate.
| {
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"url": "https://math.stackexchange.com/questions/4390254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $\prod_{n=0}^\infty (1 + x^{2^{n}}) = \frac{1}{1-x}$ I would like to verify the truth of my proof that $$\large\prod_{n=0}^\infty (1 + x^{2^{n}}) = \frac{1}{1-x}.$$
Proof. The proof is rather straightforward, just using some observations that two consecutive factors form a sum of a geometric sequence.
\beg... | Hint:
You can make the step $$\prod_n\frac{a_n}{b_n}=\frac{\prod_n a_n}{\prod_n b_n}$$
if and only if the sums
$$A=\sum_n\log a_n,\qquad B=\sum_n\log b_n$$
converge absolutely. Set $a_n=x^{4\cdot2^{2n}}-1$, $b_n=x^{2^{2n}}-1$, and see for which values of $x$ we have absolute convergence of $A, B$.
Edit:
As was pointed... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding the maximum of $ \sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x}$ for $0\le x \le 13$ using the Cauchy-Schwarz inequality gives two different answers I am trying to find the maximum of the expression $$\sqrt{x+27} + \sqrt{13-x} + \sqrt{x} \qquad \text{for } 0 \le x \le 13.$$
Clearly using Cauchy is the way to go here,... | How to "guess":
The form is making something nice, so I try to guess something to make $x+27,13-x,x$ are all squares, and indeed $x=9$, son the previous three numbers are $6^2,2^2,3^2$. So, this can have
$$(\sqrt{x+27} + \sqrt{13-x} + \sqrt{x})^2\le(6+2+3)(\frac{x+27}6+\frac{13-x}2+\frac{x}3)=121$$
So we can get $\sqrt... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $x^3+y^3+z^3+(x+y+z-1)^2\ge1+3xyz$
Let $x,y,z\ge0$ satisfy $\max\left \{ x,y,z \right \}\ge 1$. Prove that $$x^3+y^3+z^3+(x+y+z-1)^2\ge1+3xyz$$
My attempts:
From the condition we can deduce $x+y+z\ge 1$
The inequality can be written as $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+(x+y+z)^2-2(x+y+z)\ge0$$ or $$x^2+y^2+z... | Remark: I rewrote the proof using substitution $p = y + z, q = yz$.
WLOG, assume that $x \ge 1$. Let $p = y + z, q = yz$. We have $p^2 \ge 4q$.
We have
\begin{align*}
&x^3 + y^3 + z^3 + (x + y + z - 1)^2 - 1 - 3xyz\\[5pt]
=\, & x^3 + p^3 - 3pq + (x + p - 1)^2 - 1 - 3xq\\[5pt]
=\,& x^3 + p^3 + (x + p - 1)^2 - 1 - 3(p... | {
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"url": "https://math.stackexchange.com/questions/4405576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What´s the length of the segment AC in the triangle below? For reference: In the triangle $\angle A$ is right and $D$ is a point on the side $AC$ such that the segments $BD$ and $DC$ have length equal to $1 m$. Let $F$ be the point on the side $BC$ so that $AF$ is perpendicular to $BC$. If the segment $FC$ measures $1m... | $\triangle ABF\sim \triangle AFC$
We have:
$$\frac {c+m}{c+AD}=\frac {c+AD}{c}$$
So:
$$(c+AD)^2=c^2+cm\Rightarrow AD=\sqrt{c^2+mc}-c$$
Therefore:
$$AC=c+AD=\sqrt{c^2+mc}$$
| {
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"url": "https://math.stackexchange.com/questions/4407067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Idea for recursively generating formulas for sums of powers of integers. Is it correct? I am not even a beginner mathematician, being unable to understand simple proofs others easily grasp. Still I noodle around and I think I stumbled on a fun way to determine the formula for sums of powers of integers, i.e. $f(n) = 1^... | I've been informed by someone that this is a well-known result and not value added. I apologize to the community.
| {
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"url": "https://math.stackexchange.com/questions/4408797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
\frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
... | one example would be, if $\omega$ is a complex cube root of unity, then
$$(1+\omega+\omega^2)^2=1+\omega^2+\omega^4$$
| {
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"url": "https://math.stackexchange.com/questions/4411628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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Prove that $\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$
Prove that
$$J_n=\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$$
For $n=2$, it is OK. For general $n$, it seems impossible by integration by parts. A... | You can probably use the following formula :
$\cos((n-2)x)=\sum_{j=0}^{[(n-2)/2]}(-1)^j\binom{n-2}{2j}\cos^{(n-2)-2j}(x)\sin^{2j}(x)$, which can be obtained by using Newton binom with $e^{i(n-2)x}=(\cos x+i\sin x)^{n-2}$.
You take the sum out of the integral and use adapted changes of variables.
| {
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"url": "https://math.stackexchange.com/questions/4413657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Fractions in Questions and Answers
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