Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to construct an example where $\sum_{n=1}^{+\infty} a_n$ diverges and $\prod_{n=1}^{+\infty} (1+a_n)$ converges?
My Problem: Construct an example where
$$ \sum_{n=1}^{+\infty} a_n \quad \text{diverges}, \qquad \text{and}\qquad \prod_{n=1}^{+\infty} (1 + a_n) \quad \text{converges}$$
My thoughts:
By '$\prod _{n=1}... | Some author says that $\prod_{n=1}^{\infty}(1 + a_n)$ is convergent only if
$$ \lim_{N\to\infty} \prod_{n=1}^{N} (1+a_n) \in \mathbb{C}\setminus\{0\}. $$
This definition has some advantages over the more lenient version that allows $0$ as a possible value of infinite product, see the related Wikipedia article for insta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$. Suppose $a$ and $b$ are elements of the group $G$. If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$.
I'm trying to prove it as follow.
Here's my solution:
If $a^2=e$ so $a=a^{-1}$. So we have $b^3=ab^2a$.
$$b^2=ab^3a$$
So I thought if I can show that $b^3b^2=b^2 b^3=e$ I ca... | We are given that $b^3 = ab^2a$. If we cube this equation, we obtain that $b^9 = ab^6a = ab^3 \cdot b^3a$. Since from the given equation we deduce that $ab^3 = b^2a$ and $b^3a = ab^2$, we can make these substitutions to obtain $b^9 = b^2a \cdot ab^2 = b^4$. Hence, $b^5 = 1$, as desired.
Edit: The idea above shows that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the volume of the following region $E= (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 ≤1, \sqrt{2}(x^2 + y^2) ≤z≤ \sqrt{6}(x^2 + y^2) $
Find the volume of the following region $E= \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 ≤1, \sqrt{2}(x^2 + y^2) ≤z≤ \sqrt{6}(x^2 + y^2) \} $
I figured that the region E is formed... | Sticking to cylindrical coordinates, you solve a quadratic to deduce that $z=\sqrt6r^2$ and $r^2+z^2=1$ intersect when $r=\dfrac1{\sqrt3}$. Similarly, $z=\sqrt2r^2$ and $r^2+z^2=1$ intersect when $r=\dfrac1{\sqrt2}$.
When $0\le r\le \dfrac1{\sqrt3}$, we have $\sqrt2 r^2\le z\le \sqrt6 r^2$. Then when $\dfrac1{\sqrt3}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Evaluate $\int_{0}^{1}K\left ( \sqrt{1-x^2} \right ) K\left ( \sqrt{1-x^4} \right )\text{d}x$ How can we get a closed expression for this integral,
$$
\int_{0}^{1}K\left ( \sqrt{1-x^2} \right )
K\left ( \sqrt{1-x^4} \right )\text{d}x$$
where an complete elliptic integral $K(x)$ defined by $\int_{0}^{1} \frac{1}{\sqrt... | We have
$$
\int_{0}^{1}K\left ( \sqrt{1-x^2} \right )
K\left ( \sqrt{1-x^4} \right )\text{d}x
=\frac{\Gamma\left ( \frac14 \right )^4}{16}
{}_4F_3\left ( \begin{array}{c|}
\frac14,\frac14,\frac14,\frac14\\
\frac12,\frac12,1
\end{array}\text{ }1 \right )-\frac{\Gamma\left ( \frac34 \right )^4}{4}
{}_4F_3\left ( \beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4622179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How fast does the sequence $a_n := \int_{1/8}^{1/2} \Bigl(\frac{1}{2}+x^2\Bigr)^{1/2+n} dx$ decays as $n \to \infty$? The question is as in the title.
The sequence $\{a_n \}$ define by
\begin{equation}
a_n := \int_{1/8}^{1/2} \Bigl(\frac{1}{2}+x^2\Bigr)^{1/2+n} dx
\end{equation}
decays to $0$ as $n \to \infty$, which i... | If you perform the substitution $u = -\ln(1/2+x^2)$ you have that
\begin{equation}
a_n = \int_{1/8}^{1/2} \left( \frac{1}{2} + x^2\right)^{1/2+n} = \frac{1}{2} \int_{\ln(4/3)}^{\ln(64/33)} \frac{e^{-3u/2}}{\sqrt{e^{-u}-1/2}} e^{-nu} du
\end{equation}
This can be solved by Laplace's method. The minimum of $p(u)=u$ is at... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding Taylor Series of $ f(z) = \frac{z^2-2z+5}{(z-2)(z^2+1)}$ in $1<|z|<2$. I have to find Taylor series of
$f(z)=\frac{z^2-2z+5}{(z-2)(z^2+1)}$, for $1<|z|<2$.
I have started with
\begin{eqnarray*}
f(z)&=&\textstyle\frac{z^2 -2z +5}{(z-2)(z^2+1)}\\[4pt]
&=&\textstyle\frac{z^2+1-2(z-2)}{(z-2)(z^2+1)}\\[4pt]
&=&\text... | Since the poles of $f$ are $\pm i$ and $2$, the radius of convergence of the Taylor series of $f$ around $0$ is $1$ (the smallest distance from $0$ to the poles).
It is then divergent for $1<|z|<2$.
For $|z|<1$ we have $f(z)=\sum\limits_{k\geqslant0}a_kz^k$,
where $a_k=\begin{cases}
\displaystyle -\frac1{2^{k+1}}+2(-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4625086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A sphere of constant radius $r$ passes through the origin $O$ and cuts the axes in $A,B,C.$ Find the locus of the foot of the perpendicular from $O$ A sphere of constant radius $r$ passes through the origin $O$ and cuts the axes in $A,B,C.$ Find the locus of the foot of the perpendicular from $O$ to the plane $ABC$.
M... | I checked your computations. I agree with implicit equation that can be written as well under the form:
$$(x^2+y^2+z^2)^2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)=4r^2\tag{1}$$
I will take now on WLOG the value $r=\frac12$ in order that the RHS in (1) is $1$.
(1) can be given the equivalent form avoiding "... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove $\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b} \geq \frac{9}{2}$ for $a,b,c\geq1$ Prove $\dfrac{ab^2+2}{a+c} +\dfrac{bc^2+2}{b+a} +\dfrac{ca^2+2}{c+b} \geq \dfrac{9}{2}$ for $a,b,c\geq1$.
I tried using the Titu Andreescu form of the Cauchy Schwarz inequality and got to this point:
$\dfrac{(b\sqrt{a}+... | A solution proceeding from where the OP has stopped:
Let's assume $a=x^2, b=y^2, c=z^2$. We will show:
$$(y^2x+z^2y+x^2z)^2+18 \geq9(x^2+y^2+z^2).$$
To do so, consider the function $$f(x)=(y^2x+z^2y+x^2z)^2+18-9(x^2+y^2+z^2),$$
where $x,y,z \geq 1$. We have:
$$f^{\prime}(x)=2(y^2x+z^2y+x^2z)(y^2+2xz)-18x \\ \geq 2(x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
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Can we evaluate $\int \frac{1}{\sin ^ {2n+1} x+\cos ^ {2n+1} x} d x?$ After investigating the integral
$$\boxed{\quad \int \frac{1}{\sin ^5 x+\cos ^5x} d x \\=\frac{4}{5}\left[-\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{\sin x-\cos x}{\sqrt{2}}\right)+\frac{1}{\sqrt{\sqrt 5-2}} \tan ^{-1}\left(\frac{\sin x-\cos x}{\sqr... | Here is a generalized approach, which decomposes the integrand with $a_k=\frac{\pi k}{2n+1}$
\begin{align}
\frac{2n+1}{\sin^{2n+1}x+\cos^{2n+1}x}
= &\ \frac{2^n}{\sin x+\cos x}\\ &+2^{n+1}\sum_{k=1}^n(-1)^{k}\frac{\cos^{n}2a_k}{\sec a_k }\frac{\sin x+\cos x}{1+\cos 2a_k\sin2x} \\
\end{align}
Then, the decomposed terms ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Evaluate $\sum_{r=1}^{\infty} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$ I was only able to observe that:
$\dfrac{r^2 - 1}{r^4 + r^2 + 1} = \dfrac{r^2 - 1}{(r^2 + r + 1)(r^2 - r + 1)}$
This hints at telescoping, but I would need an $r$ term in the numerator.
The original question was
Evaluate $\sum_{r=1}^{\infty} \dfrac{r^3 + (... | Write \begin{align}&\frac{r^2-1}{r^4+r^2+1} = \frac{Ar+B}{r^2-r+1}+\frac{Cr+D}{r^2+r+1}\\ \iff &r^2-1 = (Ar+B)(r^2+r+1)+(Cr+D)(r^2-r+1)\tag{1}\end{align}
You can now expand the RHS as a polynomial in $r$ and by direct comparison of coefficients on the LHS and the RHS get a $4\times4$ linear system.
Alternatively, let $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solving $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$ using complex numbers My teacher gave this question to me while teaching complex numbers:
If $\;\sin A\!+\!\sin B\!+\!\sin C=\cos A \!+\!\cos B\!+\!\cos C=0$, then find:
*
*$\cos 2A + \cos 2B + \cos 2C$
*$\cos 3A + \cos 3C + \cos 3C$
*$\cos(... | As noted by copper.hat, the conditions on cosine and sine imply that $$e^{iA} + e^{iB} + e^{iC} = 0.$$ These vectors all lie on the unit circle $|z| = 1$ in the complex plane. Thus, the sum of the first two vectors must be of length one, so as to be cancellable by the third. This can only occur if the angle between the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Inverse related inverse trigonometric functions Prove that for $x<0$,
$\sin^{-1}\sqrt{1-x^2}=\cos^{-1}(-x)$
This question is related to Evaluate ${\cos ^{ - 1}}\sqrt {1 - {x^2}} = $ for $x<0$ but want to solve it differently.
I checked it on
If I use $x=\cos\theta$ then i get $\sin^{-1}|\sin\theta|$ but not able to us... | Let $\arccos(-x) = \theta$. Then $\theta$ is the unique angle in the interval $[0, \pi]$ such that $\cos\theta = -x$. Since $x < 0$, $-x > 0$, so $\arccos(-x) = \theta \in [0, \pi/2)$.
\begin{align*}
\sin^2\theta + \cos^2\theta & = 1\\
\sin^2\theta & = 1 - \cos^2\theta\\
& = 1 - (-x)^2\\
& =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4636380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Closed form of $\sum_{x=1}^{n} \dfrac{x-1}{e^x}$ Is there a closed form for this series?
$$\sum_{x=1}^{n} \dfrac{x-1}{e^x}$$
Ultimately, I want to find its convergence i.e.
$$\lim_{n \to \infty} \sum_{x=1}^{n} \dfrac{x-1}{e^x}$$
By the integral test, the series converges and for arbitrary large $n$ my code tells me it... | Define
$$f(y)=\sum_{x=1}^n y^x=\frac{y(1-y^n)}{1-y}$$
Differentiate $f$ with respect to $y$ and manipulate as follows:
$$\begin{aligned}
\frac{\text{d}f}{\text{d}y}=\sum_{x=1}^nxy^{x-1}&=\frac{1-(n+1)y^n+n y^{n+1}}{(1-y)^2}\\
\sum_{x=1}^nxy^x&=y\cdot\frac{1-(n+1)y^n+n y^{n+1}}{(1-y)^2}\\
\sum_{x=1}^n(x-1)y^x&=y\cdot\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4637461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Coefficient of a term in an expanded polynomial I came across this question while solving problems on functions. Find the coefficient of $x^{203}$ in the expansion of the following expression: $(x-2)((x)(x+1)(x+2)(x+3)...(x+202))$.
The solution given in the text is:
$(x-2)((x)(x+1)(x+2)(x+3)...(x+202)) = (x-2)(x^{203} ... | Observe that $$x(x+1)=x^2+x\\x(x+1)(x+2)=x^3+3x^2+2x\\x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2+6x\\x(x+1)(x+2)(x+3)(x+4)=x^5+10x^4+35x^3+50x^2+24x\\\dots$$
There seems to be a pattern forming regarding all coefficients, don't you agree?
But we are primarily interested in the coefficients of the second highest power of $x$.
We n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4638282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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2023 MIT Integration Bee Regular Season Problem 6 How do we solve $$\int\left(\frac{x^6+x^4-x^2-1}{x^4}\right)e^{x+\frac{1}{x}}dx$$? This is an integral in the 2023 MIT integration bee. I thought that maybe this is a consequence of the chain rule, but that is not the case. Then I thought there is a product rule hiding ... | First, factor the integrand:
$$\frac{x^6+x^4-x^2-1}{x^4}=\frac{(x^4-1)(x^2+1)}{x^4}=\left(\frac{x^2+1}{x}\right)^2\left(1-\frac1{x^2}\right).$$
Now, let $y=x+1/x$; we have $dy=(1-1/x^2)dx$, so our integral becomes
$$\int \left(\frac{x^2+1}{x}\right)^2e^ydy=\int y^2e^ydy.$$
A standard integration by parts gives
$$\int y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4639264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Generating Functions - Counting how many solutions for an exact $x^n$ Having trouble with this problem:
$$ x_1+x_2+x_3+x_4+x_5+x_6 = 13 $$
$$ x\ne 3 $$
Thanks in advance for any further help.
Being able to get to this point:
$$\begin{split}
f(x) &= (x^0+x^1+x^2+x^4+x^5+\cdots)^6\\
&= [(-x^3)+{x^0+x^1+x^2+(x^3)+x^4+x^5+... | It seems the coefficient needs to be manually extracted. Using the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ we obtain
\begin{align*}
\color{blue}{[x^{13}]}&\color{blue}{\left(\frac{1}{1-x}-x^3\right)^6}\\
&=[x^{13}]\sum_{k=0}^6\binom{6}{k}\frac{(-1)^kx^{3k}}{(1-x)^{6-k}}\\
&=\binom{6}{0}[x^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Putnam, A3 (2016) Let $f: \mathbb R\rightarrow \mathbb R$ satisfying \begin{equation*} f(x) + f\bigg(1-{1\over x}\bigg) = \arctan x\end{equation*}
for all $x\neq 0$. I want to evaluate \begin{equation*} \int_0^1 f(x) dx \end{equation*}
I know that $\tan\bigg(f(x) + f\bigg(1-{\cfrac 1x}\bigg)\bigg) = x$ and $-\pi/2 < f(... | Answer being revised
Noting that on $]0,\pi/2[$ :
$$\arctan(\frac{1}{x})+\arctan(x)=\dfrac{\pi}{2} $$
Then your relation can be writen
$$ f(x)+f(1-1/x)=\arctan(x) $$
$$ f(1-1/x)+f(x) = \dfrac{\pi}{2}-\arctan(\frac{1}{x})=\dfrac{\pi}{2}-(f(1/x)+f(1-x))$$
First
\begin{align*}
\int_0^1 f(1-1/x)+f(1/x)dx& = \dfrac{\pi}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Formula for $\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ b} \ln \left(a^2+b^2+a b \sqrt{2}\right) $ In my post, I had proved that $$
\int_0^{\infty} \frac{\ln \left(x^2+a^2\right)}{b^2+x^2} d x=\frac{ \pi}{b} \ln (a+b) \tag*{(*)}
$$
To go further, I guess that
$$\int_0^{\infty} \frac{\ln ... | To make it more general, consider the case of
$$I_n=\int_0^{\infty} \frac{\log \left(x^n+a^n\right)}{x^2+b^2} \,d x$$ where $a$ and $b$ are positive. Let $x=a t$ to get
$$I_n=\frac{\pi \log (a)}{2 b}n+\frac 1a\int_0^{\infty} \frac{\log \left(t^n+1\right)}{t^2+c^2} \,d x \qquad\text{with} \qquad c=\frac ba$$ Using t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Using Integration by Parts to find leading order approximation of exponential integral I'm trying to understand how finding leading order approximation of exponential integrals.
Here is my integral: $$\int^{+\infty}_0 e^{-xt}\ln(1+t^2)dt$$
I need to use Integration by parts to then find the leading order asymptotic app... | You correctly derived that
$$
f(x) = \int_0^\infty e^{-xt} \ln(1+t^2) \, dt
= \frac 2x \int_0^\infty e^{-xt} \frac{t}{1+t^2} \, dt
$$
for $x > 0$. This does not yet reveal the asymptotic behavior for $x \to \infty$, so let's integrate by parts again:
$$
f(x) = -\frac{2}{x^2} e^{-xt} \frac{t}{1+t^2} \big|_{t_0}^{t=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to evaluate $\int \frac1{(x^2+a^2)^{m}}dx$ The integral $$\int\frac{1}{(x^2+a^2)^m}dx$$ can be expressed by a recursive formula of $$\frac{1}{2a^2(m-1)}\frac{x}{(x^2+a^2)^{m-1}} + \frac{2m-3}{2a^2(m-1)}\int\frac{dx}{(x^2+a^2)^{m-1}}$$ I do not understand how integration by part leads to this result. Specifically, s... | Integrate by parts as follows
\begin{align}
&\int\frac{1}{(x^2+a^2)^m}dx\\
= &\int \frac1{2a^2(m-1)x^{2m-3}}\ d\left[\bigg(\frac{x^2}{x^2+a^2} \bigg)^{m-1}\right]\\
=&\ \frac{1}{2a^2(m-1)}\frac{x}{(x^2+a^2)^{m-1}} + \frac{2m-3}{2a^2(m-1)}\int\frac{1}{(x^2+a^2)^{m-1}}dx
\end{align}
Thus
$$u= \frac1{2a^2(m-1)x^{2m-3}},\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Taylor series estimation for the first four non-zero terms Question:
Find the Taylor series expansion (up to the first 4 non-zero terms) of the function $f(x) = \sqrt[3]{x+27}$ about the point $a=0$. Use the series to estimate $\sqrt[3]{27.1}$, up to 4 significant figures.
My Attempt:
Given the function $f(x) = \sqrt[3... | Yes you are correct. You can use a CAS software such as the open source and free software maxima to check your computation.
(%i1) taylor((x+27)^(1/3), x, 0, 5);
(%o1) $3+\frac{x}{27}-\frac{x^2}{2187}+\frac{5\,x^3}{531441}-\frac{10\,x^4}{43046721}+\frac{22\,x^5}{3486784401}+\cdots $
Note that you can start by factoring ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4650635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Non-differentiability at $x=0$ Question:
Let $f(x) =\left\{ \begin{array}{ll} x^2 + x \sin \left(\frac{1}{x}\right) & x\neq0, \\ 0& x=0. \end{array} \right. $
Discuss whether $f(x)$ is differentiable at $x=0$. If yes, find $f'(0)$.
Attempt:
To determine whether $f(x)$ is differentiable at $x=0$, we need to compute the ... | Since there exist two sequences
$a_n=\dfrac1{\frac{\pi}2+2n\pi}>0\;,\quad n\in\Bbb N\;,$
$b_n=\dfrac1{\frac{3\pi}2+2n\pi}>0\;,\quad n\in\Bbb N\;,$
such that $\;a_n\to0\;,\;b_n\to0\;$ and
$\lim\limits_{n\to\infty}\dfrac{f(a_n)-f(0)}{a_n-0}=\lim\limits_{n\to\infty}\left[a_n+\sin\left(\dfrac1{a_n}\right)\right]=1\;,$
$\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4651292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that the area of a circumscribed hexagon is twice the area of a triangle Let $\Delta ABC$ be an acute triangle and denote the circumscribed circle of $\Delta ABC$ $\Gamma$ with midpoint $O$. Let $A_1, B_1, C_1$ be the points on $\Gamma$ where the lines $AO, BO, CO$ intersect $\Gamma$. Show that the area of the he... | Let's assume $R$ is the circumradius of $\triangle ABC$. Then, we have:
$$S_{\triangle ABC}=\frac{1}{2} |AC||AB| \sin A=\frac{1}{2} \times 2R \sin B \times 2R\sin C \times \sin A \\ = 2R^2 \sin A\sin B \sin C.$$
Now, let's compute the area of $\triangle A_1BC$. We have:
$$S_{\triangle A_1BC}=\frac{1}{2}|A_1B||A_1C| \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4651888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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We can divide $7^{17} - 7^{15}$ by? We can divide $7^{17} - 7^{15}$ by?
The answer is $6$, but how?
Thanks in advance.
| HINT $\rm\quad X^{n+2} - X^n \;=\; (X^2 - 1)\: X^n \;=\; (X-1)\: (X+1)\: X^n$
Often number identities are more perceptively viewed as special cases of function or polynomial identities. For example, Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Ph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$? A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral:
$$\int\limits_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$$
Well, can anyone prove this without using Res... | This one I found in The American Mathematical Monthly from 1951 in the article 'A simple evaluation of an improper integral' written by Waclaw Kozakiewicz.
Theorem (Riemann). If $f(x)$ is Riemann integrable in the interval $a \leq x \leq b$, then:
$$\lim_{k \to +\infty} \int_a^b f(x) \sin kx \; dx = 0 \;.$$
Next, notic... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "265",
"answer_count": 32,
"answer_id": 25
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Funny identities Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?
| $$ \sin \theta \cdot \sin \bigl(60^\circ - \theta \bigr) \cdot \sin \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \sin 3\theta$$
$$ \cos \theta \cdot \cos \bigl(60^\circ - \theta \bigr) \cdot \cos \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \cos 3\theta$$
$$ \tan \theta \cdot \tan \bigl(60^\circ - \theta \bigr) \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "281",
"answer_count": 63,
"answer_id": 8
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Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However... | Let $\int \frac{1}{\sin x \cos x}dx=\int \frac{2}{2 \sin x \cos x}dx$
=$\int \frac{2}{\sin 2x}dx$
Let $\tan x=t$
$\sin 2x=\frac{2t}{1+t^2}$
$\tan^{-1} t=x$
$\frac{1}{1+t^2}dt=dx$
$2\int \frac{dx}{\sin 2x} =2\int \frac{1+t^2}{2t}\cdot\frac{dt}{1+t^2}$
=$\int\frac{dt}{t}$
=$\ln t+C$
=$\ln (\tan x)+C$
| {
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"url": "https://math.stackexchange.com/questions/9075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Floor of Square Root Summation problem I have problem calculating the following summation:
$$
S = \sum_{j=1}^{k^2-1} \lfloor \sqrt{j}\rfloor.
$$
As far as I understand the mean of that summation it will be something like $$1+1+1+2+2+2+2+2+3+3+3+3+3+3+3+\cdots$$
and I suspect that the last summation number will be $(k-1... | Let $\lfloor \sqrt{n} \rfloor = a$, then the following sum holds:
$$\sum_{0\le k < n} \lfloor \sqrt{k} \rfloor = (n+1)a - \frac{a^3}{3} -
\frac{a^2}{2} - \frac{a}{6}.$$
(Edited.)
You might also enjoy the slightly more tricky sum:
$$\sum_{0\le k < n} \lfloor k^{1/3} \rfloor = nb - \frac{b^2}{4} - \frac{b^3}{2} - \fr... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ?
| The cosine formula states
$$a^2 = b^2+ c^2 – 2bc \cos A$$
and since
$$\cos A = \frac{ \cot^2 (A/2) – 1}{ \cot^2(A/2)+1}$$
we have
$$a^2 = b^2 + c^2 -2bc \frac{ (b+c)^2 – a^2}{(b+c)^2 + a^2},$$
which reduces to $a^4 = (b^2 – c^2)^2.$
And so taking the square root $a^2 =b^2 – c ^2,$ where $b$ is the larger side.
Henc... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Solving $\log _2(x-4) + \log _2(x+2) = 4$ Here is how I have worked it out so far:
$\log _2(x-4)+\log(x+2)=4$
$\log _2((x-4)(x+2)) = 4$
$(x-4)(x+2)=2^4$
$(x-4)(x+2)=16$
How do I proceed from here?
$x^2+2x-8 = 16$
$x^2+2x = 24$
$x(x+2) = 24$ Which I know is not the right answer
$x^2+2x-24 = 0$ Can't factor this
| After $(x-4)(x+2)=16$, you get $x^2-2x-24=0$ (the coefficient of $x$ is $-2$ not $2$). So $x=\frac{2\pm \sqrt{100}}{2}$ by the quadratic formula. So $x=6$ or $x=-4$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Evaluation of the sum $\sum_{k = 0}^{\lfloor a/b \rfloor} \left \lfloor \frac{a - kb}{c} \right \rfloor$ Let $a, b$ and $c$ be positive integers. Recall that the greatest common divisor (gcd) function has the following representation:
\begin{eqnarray}
\textbf{gcd}(b,c) = 2 \sum_{k = 1}^{c- 1} \left \lfloor \frac{kb}{c}... | This isn't a complete solution but rather a reformulation for a large number of cases involving the problem above. Write $n = m \lfloor y \rfloor + r$, where $0 \leq r < \lfloor y \rfloor$ and $n, m , r \in \mathbb{Z}_{\geq 0}$. Then for real $x, y > 0$, we have
\begin{eqnarray}
\sum_{k = 0}^{n} \left \lfloor x + \frac... | {
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How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series:
$$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} ... | This is similar to the currently accepted answer, but more straightforward. You can use the trig identity
\begin{equation*}
\sin(\alpha + \beta) - \sin(\alpha - \beta) = 2\sin \beta \cos \alpha.
\end{equation*}
Let $a_n = a + 2dk$ be an arithmetic sequence of difference $2d$, and set $b_n
= a_n - d = a + d(2k - 1)$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "182",
"answer_count": 8,
"answer_id": 5
} |
If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$ then $\tan x + \cot x=?$ Hello :)
I hit a problem.
If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$, then how much is $\tan x + \cot x$?
| Hint: write $\tan(x) + \cot(x)$ in terms of $\sin(x)$ and $\cos(x)$, with a common denominator. Square $\sin(x) + \cos(x) = {\sqrt{3} + 1 \over 2}$ and compare.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Basic Recursion Im trying to write recursive formulas for sequences but it seems like there are different techniques depending on what type of sequence I'm dealing with. for example I want to the sequence:
$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +.... \frac{1}{n}$
What I see when I look at this is all the next te... | As Leonidas suggests, your formula doesn't work (let's try it for $n=3$):
\begin{align*}
f(2) & = f(1) + \frac{f(1)}{2} = 1 + \frac{1}{2} = \frac{3}{2} \\
f(3) & = f(2) + \frac{f(2)}{2} = \frac{3}{2} + \frac{\frac{3}{2}}{2} = \frac{9}{4} \neq 1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4}.
\end{align*}
Your last summand i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simplify this to the given answer? I'm trying to simplify:
$$\frac{x^2}{x^2-4}-\frac{x+1}{x+2}$$
but I can't get to the answer:
$$\frac{1}{x-2}$$
How to do it?
| First, do the operation by finding the least common denominator. Since $x^2-4 = (x-2)(x+2)$, the least common denominator is already $x^2-4$. Then do some simple algebra:
\begin{align*}
\frac{x^2}{x^2-4} - \frac{x+1}{x+2} &= \frac{x^2 - (x+1)(x-2)}{x^2-4} = \frac{x^2-(x^2-x-2)}{x^2-4}\\
&= \frac{x+2}{x^2-4} = \frac{x+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Striking applications of integration by parts What are your favorite applications of integration by parts?
(The answers can be as lowbrow or highbrow as you wish. I'd just like to get a bunch of these in one place!)
Thanks for your contributions, in advance!
| My favorite example of integration by parts (there are other nice tricks as well in this example but integration by parts starts it off) is this:
Let $I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^n(x) dx$.
$I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/20397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "167",
"answer_count": 20,
"answer_id": 13
} |
How can I find the roots of this complex equation? I have to find the domain of
$\frac{z^2-1}{z^2+z+1}$
So, there can't be any $z$ that makes null the denominator,
$z^2+z+1$ = 0
and after decomposing $z$,
$x^2-y^2+x+2xyi+yi=0$
How to solve that quadratic complex equation?
| Three points:
*
*You did the computation wrong. Writing $z=x+iy$, then
$$z^2+z+1 = (x+iy)^2 + (x+iy) + 1 = (x^2-y^2 + x + 1) + i(2xy+y).$$
*If $(x^2-y^2+x+1) + i(2xy+y) = 0$, with $x,y$ real numbers, then you need $x^2-y^2+x+1=0$ and $2xy+y=0$. You solve these the way you usually solve equations for real numbers.
S... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving the value of a limit using the $\epsilon$-$\delta$ definition I'm trying to solve the problem of showing that
$$\lim_{x\to6}\left(\frac{x}{4}+3\right) = \frac{9}{2}$$
using the $\epsilon$-$\delta$ definition of a limit.
| Given an $\epsilon\gt 0$, you want to show that provided $x$ is close enough to $6$, without being equal to $6$, then $\frac{x}{4}+3$ will be $\epsilon$-close to $\frac{9}{2}$. Well, first thing is to figure out how close $\frac{x}{4}+3$ is to $\frac{9}{2}$:
$$\left|\left(\frac{x}{4}+3\right) - \frac{9}{2}\right| = \le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Which is bigger: $9^{9^{9^{9^{9^{9^{9^{9^{9^{9}}}}}}}}}$ or $9!!!!!!!!!$? In my classes I sometimes have a contest concerning who can write the largest number in ten symbols. It almost never comes up, but I'm torn between two "best" answers: a stack of ten 9's (exponents) or a 9 followed by nine factorial symbols. Both... | I find it quite sad that none of the answers has actually formally shown that
$$9^{9^{9^{9^{9^{9^{9^{9^{9^9}}}}}}}} > 9!!!!!!!!!$$
The proces is a bit long, but it it worth it in my opinion. First, a simple calculation shows $9! < 9^8$. Then $9!! < (9^8)^{9^8} = 9^{8\cdot9^8}$. Then $$9!!!< \left(9^{8\cdot9^8}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/23228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
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Why does the sum of a division applied to individual items not equal the division applied to the sum of those items? When $a_2/a_1 = b_2/b_1$, $a_1 \neq b_1$, we have
$$\frac{a_{1}}{a_{2}/a_{1}}+\dfrac{b_{1}}{b_{2}/b_{1}}=
\frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}.$$
So why when $a_2/a_1 \... | Because for $a_{1},a_{2},b_{1},b_{2}\neq 0$ we have the following equivalent inequalities:
$$\frac{a_{1}}{\dfrac{a_{2}}{a_{1}}}+\dfrac{b_{1}}{\dfrac{b_{2}}{b_{1}}}\neq
\frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}\Leftrightarrow \frac{a_{2}}{a_{1}}\neq \frac{b_{2}}{b_{1}}.$$
This can be shown ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find limit of $\sqrt[n]{a^n-b^n}$ as $n\to\infty$, with the initial conditions: $a>b>0$ With the initial conditions: $a>b>0$;
I need to find $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}.$$
I tried to block the equation left and right in order to use the Squeeze (sandwich, two policemen and a drunk, choose your favourite) theor... | Another way: Note that $a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})$. Since $b < a$, each term in the sum is bounded by $a^{n-1}$ and we have
$$a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1} < na^{n-1}$$
Since the sum is at least the first term, we also have
$$a^{n-1} + a^{n-2}b + ... + ab^{n-2} +... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Simple formula for integer polynomial with $2\sin(2\pi/n)$ as a root? Is there a simple formula an integer polynomial that $2\sin(2\pi/n)$ satisfies?
For $2\cos(2\pi/n)$ the answer is relatively nice. For any given $n$, we have $2\cos(2\pi/n)= z + z^{-1}$ where $z = e^{2 \pi i/n}$ satisfies a cyclotomic polynomial of ... | I'll work with $\sin \frac{2 \pi}n$ instead of $2 \sin \frac{2 \pi}n$.
Since $\sin x = \cos (\frac \pi 2 - x)$, $\sin \frac{2 \pi}n = \cos (\frac {\pi}2 - \frac {2\pi}n) = \cos \frac{n-4}{2n}\pi$.
The Chebyshev polynomials of the first kind $T_k(x)$ have the property that $T_k(\cos x) = \cos kx$. They have integer co... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Multiplication in Permutation Groups Written in Cyclic Notation I didn't find any good explanation how to perform multiplication on permutation group written in cyclic notation. For example, if
$$
a=(1\,3\,5\,2),\quad b=(2\,5\,6),\quad c=(1\,6\,3\,4),
$$
then why does $ab=(1\,3\,5\,6)$ and $ac=(1\,6\,5\,2)(3\,4)$?
| I didn't find this method written as an answer here, so I am just adding it.
You might convert these cyclic notations to Cauchy's two-line notations. Carefully observing your $a$ and $b$, the permutation groups can be written in the following form:
$$a=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
3 & 1 & 5 & 4 & 2 & 6
\end... | {
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"url": "https://math.stackexchange.com/questions/31763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 4,
"answer_id": 3
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Infinitely many integer solutions for the equations $x^3+y^3+z^3=1$ and $x^3+y^3+z^3=2$ How do you show that the equation $x^3+y^3+z^3=1$ has infinitely many solutions in integers? How about $x^3+y^3+z^3=2$?
| You can reduce the first equation to $$x^3 = -y^3, z = 1$$ with obvious infinite solutions.
This paper details other families of solutions.
The second equation has solutions $(x,y,z)\equiv (6t^3+1, 1-6t^3, -6t^2)$ which (AFAIK) you find by construction (i.e you have to guess it).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $\alpha$ is an acute angle, show that $\int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$ If $\alpha$ is an acute angle, show that $\displaystyle \int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$
My attempt:
Write $x^2+2x\cos{\alpha}+1 = (x+\cos{\alpha})^2+1-\cos^2{\a... | Use $1 + \cos \alpha = 2\cos^2 (\frac{\alpha}{2})$
and $\sin \alpha = 2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/33059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 3,
"answer_id": 1
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Integral of $\frac{1}{(1+x^2)^2}$ I am in the middle of a problem and having trouble integrating the following integral:
$$\int_{-1}^1\frac1{(1+x^2)^2}\mathrm dx$$
I tried doing partial fractions and got:
$$1=A(1+x^2)+B(1+x^2)$$
I have no clue how to solve this since it is obvious there is no way to cancel out either $... | $$
\begin{aligned}
\int \frac{1}{\left(1+x^{2}\right)^{2}} d x\stackrel{IBP}{=} &-\frac{1}{2} \int \frac{1}{x} d\left(\frac{1}{1+x^{2}}\right) \\=&-\frac{1}{2 x\left(1+x^{2}\right)}+\frac{1}{2} \int\left(\frac{1}{1+x^{2}}-\frac{1}{x^{2}}\right) d x \\=&-\frac{1}{2 x\left(1+x^{2}\right)}+\frac{1}{2} \tan ^{-1} x+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/35924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 10,
"answer_id": 8
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Infinite limits on integration: acceptable notation So you have an integral like
$$
\int_{-\infty}^\infty{ \frac{dx}{1+4 x^2} }
$$
Schaum's Calculus 5e recommends you write this as
$$
\lim_{a \to -\infty} \int_a^b{ \frac{dx}{1+4 x^2} } + \lim_{c \to \infty} \int_b^c{ \frac{dx}{1+4 x^2} }
$$
Where b is chosen as... | To see why they have this approach, try
$$\int_{-\infty}^\infty{ \frac{8x \;dx}{1+4 x^2} } $$
using the recommended method [the indefinite integral is $\log(1+4 x^2) + \text{constant}$], and then using your suggestion, and alternatively what would happen if you had to work out
$$\lim_{c \to +\infty} \int_{-c}^c { \fr... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at nex... | You want to show that $$(n-1)^3+n^3+(n+1)^3 \equiv 0 \pmod 9.$$ There are $9$ cases to check,
*
*$(0-1)^3+0^3+(0+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$
*$(1-1)^3+1^3+(1+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$
*$(2-1)^3+2^3+(2+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$
*$(3-1)^3+3^3+(3+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$
*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/37805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
} |
What is the relation? The function $x^2 = y\quad$ limits two areas $A$ and $B$:
$A$ is further limited with the line $x= a$, $a\gt 0$.
$A$ rotates around the $x$-axis, which gives Volume $A = Va$.
$B$ is limited with the line $y=b$, $b\gt 0$.
$B$ rotates around the $y$-axis, which gives Volume $B = Vb$.
What are the r... | If we take the region bounded by the $y$-axis, the $x$-axis, the line $x=a$ (with $a\gt 0$), and the parabola $y=x^2$, and rotate it about the $x$-axis, the volume of the resulting solid of revolution is easily computed (using, for example, discs perpendicular to the $x$-axis) to be
$$\text{Volume A} = \int_0^a \pi(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/40533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem: Sum of absolute values of polynomial roots Can you please give me some hints as to how I might approach this problem? Thanks!
Given the polynomial $f = 2X^3 - aX^2 - aX + 2, a \mathbb \in R$ and roots $x_1, x_2$ and $x_3,$ find $a$ such that $|x_1| + |x_2| + |x_3| = 3.$
Edit: We know $-1$ is one of the roots... | We first have the following factorization,
$ 2x^3 - ax^2 - ax + 2 = (x+1)(2x^2 - (2+a)x + 2).$
Suppose $x_1 = -1$, then $|x_2| + |x_3| = 2$ and they are solutions to the quadratic equation $2x^2 - (2+a)x + 2 = 0$. Hence,
$ x_2 + x_3 = 1 + a/2 $ and
$ x_2 x_3 = 1$.
By observing the second equation, $x_2$ and $x_3$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/40701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
How many function are there? Let $f_k \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear map defined by
$$
f_k \begin{pmatrix} 1 \\ 2 \\ k \end{pmatrix}=
\begin{pmatrix} 2+k \\ 3 \\ 0 \end{pmatrix},
\quad
f_k \begin{pmatrix} 2 \\ k+1 \\ -1 \end{pmatrix}=
\begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix},
\quad
... | First, consider the three vectors,
$$\left(\begin{array}{c}1\\2\\k\end{array}\right),\qquad \left(\begin{array}{c}2\\k+1\\-1\end{array}\right),\qquad \left(\begin{array}{c}-3\\1\\5\end{array}\right).$$
If they are linearly independent, then we can define a linear transformation from $\mathbb{R}^3$ to any vector space s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/42023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Power series of $\ln(x+\sqrt{1+x^2})$ without Taylor The answer is $$x-\frac{ 1}{2}\frac{ x^3}{3}+\frac{ 1\cdot 3}{2\cdot 4}\frac{ x^5}{5}-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{ x^7}{7}+\cdots$$ But I can't see how. Unfortunately, "how" can't be using Taylor's formula, because that isn't introduced until the ne... | The first part of the derivate you calculated can be transformed to
$$\frac{ 1}{x+\sqrt{1+x^2}} =
\frac{ 1}{x+\sqrt{1+x^2}} \frac{ x-\sqrt{1+x^2}}{x-\sqrt{1+x^2}} =
\sqrt{1+x^2}-x$$
and therefore the derivate is
$$\sqrt{x^2+1} - \frac{x^2}{\sqrt{x^2+1}}$$
The term $\frac{x}{\sqrt{x^2+1}}$ is the derivate of $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/42295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Understanding algebraic manipulation If $a+b+c \neq 0 $ where $a,b$ and $c$ are three non-zero distinct integers, then find the value of:
$$\frac{ab+ca}{a^2+ab+ca} + \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc}$$
What confusing me here, is the not so obvious hint which is given with the problem,which says that tha... | All you need to do is "simplify" the fractions, for example by dividing "top" and "bottom" of the first one by $a$, of the second by $b$, of the third by $c$. We get
$$\frac{b+c}{a+b+c}+\frac{a+c}{a+b+c}+\frac{a+b}{a+b+c}$$
We have a common denominator $a+b+c$. The numerators add up to $2(a+b+c)$. Cancel. We get $2$.
C... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Evaluating $\lim\limits_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$
How would you evaluate the following limit:
$$\lim_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$$
I tried to use this formula: $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$, It didn't work.
Any hints?
| Hint: use $a^6 - b^6 = (a - b)(a^5 + a^4 b + a^3 b^2 + a^2 b^3 + a b^4 + b^5)$ with $a =(x^6 + x^5)^{1/6} = x (1 + 1/x)^{1/6}$ and $b = (x^6 - x^5)^{1/6} = x (1 - 1/x)^{1/6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/45970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
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Laplacian of a Smooth Function $f(\vec x) = 1/\|\vec x \|$ for $\| \vec x \| \geq 1$ This is a multivariable calculus problem from a past prelim exam. I have an answer for this written up (posted below), but it seemed rather time-intensive. If there is a slicker way to approach this problem, I'd appreciate seeing it. T... | *
*In rectangular coordinates, for $\|\vec{x}\| \geq 1$,
$
\begin{align*}
f(\vec{x}) &= \frac{1}{\sqrt{x^2+y^2+z^2}} \\
\nabla f &= \left\langle \frac{-x}{(x^2+y^2+z^2)^{3/2}}, \frac{-y}{(x^2+y^2+z^2)^{3/2}}, \frac{-z}{(x^2+y^2+z^2)^{3/2}} \right\rangle \\
&= \frac{-1}{(x^2+y^2+z^2)^{3/2}} \left\langle x,y,z \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/48961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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about $\operatorname{SO}(2)$ group When
$$
A =\begin{pmatrix} a & b \\ c & d \end{pmatrix},\quad a,b,c,d \in \mathbb{R}, \quad A^2 -2aA + I = 0
$$
Is $ A \in \operatorname{SO}(2)\; $ if $A^n \in SO(2)\;$ for some $n \in \Bbb N$?
$\operatorname{SO}(2)$ is special orthogonal group.
| I'm a bit surprised since nobody tried to find a counter-example! Let $p$ be a nonzero real number and
$$A = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -p \\ p^{-1} & 1 \end{pmatrix}.$$
Then we have $A^2 - \sqrt{2}A + I = O$ and $A^4 = -I \in SO(2)$, but still $A \notin SO(2)$ unless $p = \pm 1$.
But this is partially true... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/56411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$ Prove :
$$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$$
For $a,b,c,d>0$
| I know many time has passed, however i came up with a solution and I hope it is correct: here you are:
Let us denote
$$\begin{array}{c}\frac{1}{4}\frac{(b+c)^3}{2}(8(a^2+d^2)^3-(a+d)^3(a+b)^3)+\\\frac{1}{4}\frac{(a+d)^3}{2}(8(b^2+c^2)^3-(c+d)^3(c+b)^3)=(\clubsuit)\end{array}$$
And we want to prove $(\clubsuit)\geq0.$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
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When should the antiderivative of a rational function be defined as a piecewise function? I'm doing basic problems on antiderivatives and there seems to be an inconsistency in my book.
The instructions for these problems are:
Find the most general antiderivative of the function.
Number 11 is:
11. $f(x) = \dfrac{10}{... | The person who wrote out the solution may have thought that putting absolute value signs around $x$ takes care of the problem. It doesn't. Congratulations on reading with care.
The answer for Question $19$ ought to be something like
$$F(x) = \begin{cases} \frac{1}{2}x^2 - \ln x - \dfrac{1}{x^2} + C_1 & \text{if } x > 0... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Asymptotic behaviour of sums of consecutive powers Let $S_k(n)$, for $k = 0, 1, 2, \ldots$, be defined as follows
$$S_k(n) = \sum_{i=1}^n \ i^k$$
For fixed (small) $k$, you can determine a nice formula in terms of $n$ for this, which you can then prove using e.g. induction. For small $k$ we for example get
$$\begin{ali... | The standard way to sum $\sum\limits_{r=1}^n r^k$ gets this estimate.
We have:
$$
(r+1)^{k+1} - r^{k+1} = (k+1) r^{k} + \frac{k(k+1)}{2} r^{k-1} + O(r^{k-2}).
$$
Summing this equation for $r = 0, 1, 2, \ldots, n$, we get:
$$
(n+1)^{k+1} + O(1) = (k+1) \sum_{r=0}^nr^{k} + \frac{k(k+1)}{2} \sum_{r=0}^n r^{k-1} + \sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/63986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 2,
"answer_id": 1
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$a/b=c$, how to move the numerator to the right side of the equation Given an equation of the form $a/b=c$, how do you move $a$, the numerator, over to the other side so that you get $b =c?$ (where $?$ denotes my ignorance of what the right side should look like).
| $$
\begin{align}
\frac{a}{b} & = c \\ \\
b \cdot \frac{a}{b} & = c \cdot b \\
\require{cancel} \\
\cancel{b} \cdot \frac{a}{\cancel{b}} & = c \cdot b \\ \\
a & = cb \\ \\
\frac{a}{c} & = \frac{cb}{c} \\ \\
\frac{a}{c} & = \frac{\cancel{c}b}{\cancel{c}} \\ \\
\frac{a}{c} & = b
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/64903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Proving that for reals $a,b,c$, $(a + b + c)^2 \leq 3(a^2+b^2+c^2)$
Proving that for reals $a,b,c$, $(a+b+c)^2\leq 3(a^2+b^2+c^2)$.
This is a homework question and I have no clue where to even start on this. I don't know if I am just tired or what but I can't get anywhere. I've expanding both sides and seeing if tha... | We need to expand both sides of inequality:
$(a+b)^2+c^2+2(a+b)c\leq 3a^2+3b^2+3c^2$
$a^2+b^2+c^2+2ab+2ac+2bc\leq 3a^2+3b^2+3c^2$
$2ab+2ac+2bc\leq 2a^2+2b^2+2c^2$
$0\leq a^2+b^2-2ab+a^2+c^2-2ac+b^2+c^2-2bc$
$0\leq (a-b)^2 +(a-c)^2
+(b-c)^2$ , what is obvious true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/65695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How can I differentiate this? I was trying to differentiate this: $\frac{1}{2} \sin^{-1} \frac{2x}{1+x^2}$ but I am really stuck.I obtained an answer that does not match with the one given in the book.Your help is appreciated.[Edit: I used the formula $\frac{d}{dx} \sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$ to get $\frac{1+x^2... | The derivative of the arcsine function is
$$\frac{d}{du}\arcsin(u) = \frac{1}{\sqrt{1-u^2}}.$$
So, using the Chain Rule, you would have
$$\frac{d}{dx}\frac{1}{2}\arcsin\left(\frac{2x}{1+x^2}\right) = \frac{1}{2}\left(\frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}}\right)\left(\frac{2x}{1+x^2}\right)'.$$
The deriva... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Show that $x^{35}+\dfrac{20205}{2+x^{17}+\cos^2x}=100$ has no root $x\in \mathbb{R}$ Show that $x^{35}+\dfrac{20205}{2+x^{17}+\cos^2x}=100$ has no root $x\in \mathbb{R}$.
By plotting graph I have seen that there are no roots for $x$. Can somebody prove it theoretically?
| Here's an answer, but I am sure there should be better ones.
Assume $x \in \mathbb R$ to be a root of the given equation. Rearrange the equation as
$$
(x^{35}-100) (x^{17}+2+\cos^2 x) = - 20205.
$$
Let $A := x^{35}-100$ and $B := x^{17}+2+\cos^2 x$. Since $AB < 0$, exactly one of $A$ and $B$ is positive and the othe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/70838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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Digamma function integral Does anyone how to get a finite value to this integral ?
$ \int\nolimits_{0}^{\infty} dx \frac{ \Psi (1/4+ix/2) +\Psi (1/4-ix/2)}{x^{2}+1/4} $
i have tried residue theorem but i got nonsenses :( can anyone help
or give some advice
i believe that the integral should be equal to the $ \Psi (3/... | Use $\sum_{n=1}^\infty \left( \frac{1}{n+a-1} - \frac{1}{n}\right) = -\gamma - \Psi(a)$, where $\Psi(a)$ is di-gamma function. Using this representation for the di-gamma function and integrating term-wise:
$$ \begin{multline}
\mathcal{I} = \int_0^\infty \frac{ \Psi(\frac{1}{4} + i \frac{x}{2}) + \Psi(\frac{1}{4} - i ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/71230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Compute $1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2$ by mathematical induction I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side.
Please guide me how to do it further.
$1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2 = \frac{1}{3}n(2n-1)(2n+1)$.
Sol:
$P(n... | NOTE: I am not saying anything different, before someone start commenting that my answer is not any different.
$$
\begin{align*}
1^2 + 3^2 + 5^2 + \cdots + (2k+1)^2 &= 1^2+3^2+5^2+\cdots + (2k+1)^2\\
&= 1^2 + 3^2 + 5^2 + \cdots + (2k-1)^2 + (2k+1)^2\\
&= \frac{1}{3}k(2k-1)(2k+1) + (2k+1)^2\\
&=\frac{k(2k-1)(2k+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/72636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find critical points of a function: $y = x \sqrt{4 - x^2}$ I am having trouble finding the critical points of this function, I was wondering if someone could help me out.
So I have a function $y = x\sqrt{4 - x^2}$. I know that I have to take the derivative then set that equal to 0 to find the critical points.
$$\begi... | Put your last expression together as one fraction by getting a common denominator. The right hand expression will be
$$\frac{4-x^2}{\sqrt{4-x^2}}$$
so the overall fraction is
$$\frac{-x^2 + 4 - x^2}{\sqrt{4-x^2}} = \frac{4 - 2x^2}{\sqrt{4-x^2}} = \frac{-2(x^2 - 2)}{\sqrt{4-x^2}}$$
which is 0 when $x = \pm \sqrt 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/75500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Limit of $\left(\frac{2\sqrt{a(a+b/(\sqrt{n}+\epsilon))}}{2a+b/(\sqrt{n}+\epsilon)}\right)^{n/2}$ I'm having a hard time characterising the behavior of the following expression:
$$\lim_{n\rightarrow\infty}\left(\frac{2\sqrt{a(a+b/(\sqrt{n}+\epsilon))}}{2a+b/(\sqrt{n}+\epsilon)}\right)^{\frac{n}{2}}$$
with the following... | Let the desired limit be denoted by $L$. On taking logarithms we get
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{2\sqrt{a(a + b/(\sqrt{n} + \epsilon))}}{2a + b/(\sqrt{n} + \epsilon)}\right)^{n/2}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(\frac{2\sqrt{a(a + b/(\sqrt{n} + \epsilon))}}{2a + b/(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/77629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
quotient rule difficulties I'm trying to use the quotient rule to differentiate $\frac{r}{\sqrt{r^2+1}}$ but I'm getting the wrong answer. So far I have
$$\begin{align*}
\frac {d}{dr} \frac{r}{\sqrt{r^2+1}} &= \frac {\sqrt{r^2+1} \frac {d}{dr} r - r \frac {d}{dr} \sqrt{r^2+1}} {(\sqrt{r^2+1})^2} \\\\\\\\
&= \frac {\... | Your answer is right
$$ (r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2}= \frac{1}{\sqrt{r^2+1}} -\frac{r^2}{(r^2+1)^{3/2}}
= \frac{r^2+1}{(r^2+1)^{3/2}} -\frac{r^2}{(r^2+1)^{3/2}}$$
You can finish it from here....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Showing that there are infinitely many integer solutions to $x^2+y^2=z^2$ Let $x,y,z \in \mathbb{N}$ with $x,y$ relatively prime, $x$ even and $x^{2}+y^{2}=z^{2}$. Show that there are infinitely many $(x,y,z)$ triplets which satisfy these conditions.
| I we use Euclid's formula
$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$
we can see that
$$
A^2+B^2=\big((m^2-k^2)+(2mk)^2\\
=(m^4-2m^2k^2+k^4)+4m^2k^2\\
=(m^4+2m^2k^2+k^4)\\
=(m^2+k^2)^2=C^2
$$
One problem with Euclid's formula is that it generates trivial triples
like $f(1,0)=(1,0,1)\quad f(3,0)=(9,0,9)\quad \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/78162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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If $\gcd(a,b)=1$ , and $a$ is even and $b$ is odd then $\gcd(2^{a}+1,2^{b}+1)=1$? How to prove that:
$\gcd(a,b)=1 \Rightarrow \gcd(2^{a}+1,2^{b}+1)=1$ ,where $a$ is even and $b$ is odd natural number
For example:
$\gcd(2^8+1,2^{13}+1)=1 , \gcd(2^{64}+1,2^{73}+1)=1$
I know that Knuth showed that:
$\gcd(2^{a}-1,2^{b}-1... | We post a second answer that, instead of doing details, recycles old results. The OP mentioned the following standard result, but was not sure it was useful:
$$\gcd(2^{x}-1,2^{y}-1)=2^{\gcd(x,y)}-1.$$
It is useful! By the result, if $\gcd(a,b)=1$, then
$$\gcd(2^{2a}-1,2^{2b}-1)=2^{2}-1=3.$$
Thus
$$\gcd((2^{a}+1)(2^a-1... | {
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"timestamp": "2023-03-29T00:00:00",
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} |
integrate square of $\arctan x$. Tricky $$\int \left(\frac{\tan^{-1}x}{x-\tan^{-1}x}\right)^{2}dx$$
I ran across an integral I am having a time solving.
The solution merely works out to $\displaystyle\frac{1+x\tan^{-1}x}{\tan^{-1}x-x}$, but for the life of me I can not find a suitable method to tackle it.
Does anyone ... | The denominator suggests that maybe an integration by parts will help where $v=\frac{1}{x-\arctan x}$. This may make for a strange looking $dv$ that is not readily apparent from the given integral.
$$\frac{d}{dx}\left(\frac{1}{x-\arctan x}\right)=\frac{-\frac{x^2}{1+x^2}}{(x-\arctan x)^2}$$
So let's try $dv=\frac{-\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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Hard algebra problem Given
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
Now, it is necessary to find
$$\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}=?$$
Is this possible and how? a,b,c are given constants.
I think, ? is probably a complicated function $F=F(a,b,c)$
| The expression
$$\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}$$ isn't constant on the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1.$$ Notice that I get different values on the points $(x,y,z) = (a,0,0)$ and $(x,y,z)=(0,b,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/80438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determinant of a specific circulant matrix, $A_n$ Let
$$A_2 = \left[ \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]$$
$$A_3 = \left[ \begin{array}{ccc} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$$
$$A_4 = \left[ \begin{array}{cccc} 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0\end{array... | Here is an elementary way to compute the determinant of $A_n$:
Add row 2 to row 1, add row 3 to row 1, ..., and add row $n$ to row 1, we get
$$\det(A_n)=\begin{vmatrix}
n-1 & n-1 & n-1 & \cdots & n-1 \\
1 & 0 & 1 &\cdots & 1 \\
1 & 1 & 0 &\cdots & 1 \\
\vdots & \vdots & \vdots & \ddots & \vd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/81016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
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Which is the "fastest" way to compute $\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $?
I am looking for the "fastest" paper-pencil approach to compute $$\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $$
This is a quantitative aptitude problem and the correct/required answer is $3.75$
In addition, I am also interested... | $\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} =\frac{10}{4} \sum_{i=1}^{10} \frac{i}{2^i}-\frac{5}{4}\sum_{i=1}^{10} \frac{1}{2^i}$
Second $\sum$ is obviously $1-\frac{1}{1024}$.
First $\sum$:
$\frac{1}{2} +$
$\frac{1}{4} + \frac{1}{4} +$
$\frac{1}{8} + \frac{1}{8} + \frac{1}{8} +$
$\vdots$
$\frac{1}{1024} + \frac{1}{... | {
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"source": "stackexchange",
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How can I prove this equation: $(1+ \frac{1}{n})^n = 1+ \sum\limits_{k=1}^n{\frac{1}{k!}}1(1-\frac{1}{n})\cdot(1-\frac{2}{n})…(1-\frac{k-1}{n})$
Possible Duplicate:
Proving sequence equality using the binomial theorem
$(1+ \frac{1}{n})^n = 1+ \sum\limits_{k=1}^n{\frac{1}{k!}}\cdot1\cdot(1-\frac{1}{n})\cdot(1-\frac{2... | $$\frac{n!}{n^k(n-k)!}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{n\,n\,n\cdots n}=\left(1-\frac0n\right)\,\left(1-\frac1n\right)\,\left(1-\frac2n\right)\cdots\left(1-\frac{k-1}n\right).
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then $a + b = c + d$ I came across this problem today. I would be interested to see if anyone knows a proof for it:
If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then show that $a + b = c + d$.
| If negative values of the variables are allowed then there are counterexamples. The simplest is the following:
$$a=2,\quad b=2,\quad c=\sqrt{3}-1-\sqrt{2\sqrt{3}}\doteq -1.129,\quad d=\sqrt{3}-1+\sqrt{2\sqrt{3}}\doteq2.593$$
with $a+b=4$, $c+d\doteq1.464\ .$
In the following we assume $0\leq a\leq b$ and $0\leq c\leq ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\frac{1+\sqrt{m}}{2}; (19)^{1/3}$ are algebraic integral
This question is from a very old exercise sheet, I do not know what the notation $\mathbb{Z}_{\mathbb{C}}$ means:
*
*$1\equiv 1 \mod 4 $ in $\mathbb{Z}$. Show that $\frac{1+\sqrt{m}}{2} \in \mathbb{Z}_{\mathbb{C}}$ 2. Let $a= (19)^{1/3}$. Show that $... | Your approach seems too complicated in both cases. Try these simpler approaches:
*
*Let $\alpha =\frac{1+\sqrt{m}}{2}%$. Then $2 \alpha -1 = \sqrt{m}$. Can you now find a monic equation for $\alpha$? Make sure you use the hypothesis on $m \bmod 4$.
*Let $\beta=\frac{1+a+a^{2}}{3}$. Note that $\beta = \frac{a^3-1}{3... | {
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"timestamp": "2023-03-29T00:00:00",
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Find a couple of integers such that the third power of a given natural can be written as the difference of the squares of those integers Given a natural number $n$, find inegers $a, b$ such that $n^3=a^2-b^2$. I've tried, but I'm a bit rusty. Please Help
| Note that
$$n^3=\left(\frac{n^2+n}{2}\right)^2-\left(\frac{n^2-n}{2}\right)^2.$$
Comment: The magic identity that solved the problem in fact did not (for me) come by magic. Given a number $K$, we want to find numbers $a$ and $b$ such that $a^2-b^2=K$. So we want $(a+b)(a-b)=K$. This means that $a+b$ and $a-b$ are tw... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Deducing Markov inequality from reverse Markov inequality? Let $x$ be random variable, such that $E(x)=0,E(x^2)=1$ and $P(x^2\geq s^2)\geq\displaystyle\frac{C}{s^t}$, where $C>0,s\geq 1 , t>0$.
Let $m<n$ and $m,n$ are natural numbers very big. Let also $L\geq 1$ .
Consider
$1-(1-\frac{n}{2}P(x^2\geq Ln))^m$.
Assume (... | Probably it is better reformulate the question I've posted. Please tell me if my solution is correct and where I have mistakes. Thank you for advice.
Let x be random variable with mean zero and variance 1. And such that for $C>0$, $t>0$ and natural n, $P(x^2\ge n)\geq \frac{C}{n^t}$.
We want to show that if $t\ge 2$, t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use Lagrange Remainder Theorem to Prove Inequality I'm supposed to use Lagrange Remainder Theorem to prove that
$$1 + \frac{x}{2} - \frac{x^2}{8} < \sqrt{1+x} < 1 + \frac{x}{2} \text{ } \text{ if } x>0$$
Obviously, the left and right hand sides look like Taylor series, but how would you do this using Lagrange, it isn't... | For the RHS inequality you can even use Mean Value Theorem (which, of course, is a special case of Taylor theorem - where the reminder term contains first derivative).
Let $f(x) = \sqrt{1+x}$. Then the theorem gives that there exists some $\theta \in (0,x)$ such that
$$
\frac{f(x)-f(0)}{x-0} = \frac{\sqrt{1+x} - 1}{x}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Isotropy over $p$-adic numbers Over what $p$-adic fields $\mathbb{Q}_p$ is the form $\langle3, 7, -15\rangle$ isotropic?
| Everything is in Cassels, Rational Quadratic Forms, pages 41-44, then pages 55-59. In particular Lemma 2.5 on page 59. Your forms is isotropic at the primes $\{2,5,\infty\}.$ It is anisotropic at primes $\{3,7\}.$ And isotropic at everything else.
To give the simplest looking version, what happens if $$ 3 x^2 + 7 y^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $ n\geq1 $ such that 7 divides $n^n-3$ Find $ n\geq1 $ such that 7 divides $n^n-3$.
Here is what I found:
$ n\equiv 0 \mod7, n^n\equiv 0 \mod7,n^n-3\equiv -3 \mod7$ no solution.
$ n\equiv 1 \mod7, n^n\equiv 1 \mod7,n^n-3\equiv -2 \mod7 $ no solution.
$ n\equiv 2 \mod7, n^n\equiv 2^n \mod7, n^n-3\equiv 2^n-3 \mod7$... | Since the question only asks (at least the way I read it) to find some $n\geq 1$ such that $n^n-3\equiv 0 \bmod 7$, you can exploit the fact that a power $a^b \bmod 7$ only depends on the class of $a\bmod 7$ and the class of $b\bmod 6$.
Thus, if you pick $n\equiv 3 \bmod 7$ such that $n\equiv 1 \bmod 6$, then:
$$n^n\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/93165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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approximation formula for the integral Get an approximation formula for the following integral:
$$
\sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
$$
| $$\sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
$$
First observe the powers of $cos$ and $sin$
$$ cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) = cos^{2n-2k}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) =
(1-sin^{2})^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/94337",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Minimal polynomial of $1 + 2^{1/3} + 4^{1/3}$ I am attempting to compute the minimal polynomial of $1 + 2^{1/3} + 4^{1/3}$ over $\mathbb Q$. So far, my reasoning is as follows:
The Galois conjugates of $2^{1/3}$ are $2^{1/3} e^{2\pi i/3}$ and $2^{1/3} e^{4\pi i /3}$. We have $4^{1/3} = 2^{2/3}$, so the image of $4^{1/... | We can conclude that your polynomial has rational coefficients even without doing any expanding. The splitting field for the polynomial, which is $\mathbb{Q}(2^{1/3},\omega)$ (here $\omega = e^{2\pi i/3}$) has Galois group $S_3$ generated by the automorphisms $\sigma$ fixing roots of unity and sending $2^{1/3}$ to $2^{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Product in GF(16) i need some help with a product in GF(16), where it is seen as an extension of GF(4)={0, 1, x, x+1} (where $x^2 = x + 1$ ) with the irreducible polynom $f(y) = y^2 + y + x$
So the elements in the fields are: $\{0, 1, x, x+1, y, y + 1, y + x, y + x + 1, xy, ... \}$
I'm try to multiply $((x+1)y + x) * (... | Use the identities $x^2 = x+1$, $y^2 = y+x$, and $2=0$. After that you just have to compute. Here it goes explicitly for you to be confident :
\begin{align}
((x+1)y+x) \times (xy + x + 1) & = ((x+1)y+x)xy + ((x+1)y+x)x + (x+1)y+x \\
& = (x^2y^2+xy^2+x^2y)+(x^2y + xy + x^2) + xy + y + x \\
& = x^2 y^2 + xy^2 + x^2 + y ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Primes of the form : $x^2+4\cdot y^2$ How to prove following statement :
Conjecture:
Odd prime $p$ is expressible as : $p=x^2+4\cdot y^2$ , $x,y > 0$
if and only if : $p\equiv 1 \pmod {12}$ or $p\equiv 5 \pmod {12}$ .
Similar statements (without a proof) related to the Fermat's theorem on sums of two squares can be f... | HINT $\rm\ (\Rightarrow)\ \ odd\ p\ =\ x^2+4\ y^2\ \Rightarrow\ x\ odd,\ so\ \ mod\ 4\!:\ x\ \equiv\: \pm1\ \Rightarrow\ p\ \equiv\ x^2\:\equiv\ 1 $
Thus $\rm\ p\equiv 1\pmod 4\ \Rightarrow\ p\equiv 1,5,9\pmod{12}\:,\:$ but not $\:9\:$ since $\rm\:3\:|\:p\ \Rightarrow\ p = 3\ne x^2 + 4\ y^2$
$\rm(\Leftarrow)\ $ By F... | {
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What type of distribution would rand()/rand() produce? If rand() is a function which produces a linearly distributed random number over a range not containing zero, then what type of distribution would rand() / rand() produce?
I know it would center at 1, and there would be a few extreme values very close to zero or ve... | Let $U_1$ be a uniform random variable with domain $[a,b]$ for $b>a>0$, and let $U_2$ be another identically distributed independent uniform variable.
Let $X= \frac{U_1}{U_2}$. It is clear that $\frac{b}{a} \geqslant X \geqslant \frac{a}{b}$ with probability 1. Let $x$ be from such an interval. Then
$$ \begin{eqnarray}... | {
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"timestamp": "2023-03-29T00:00:00",
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If $f(x)$ and $f(x)-x$ have only one real root, then $f(f(x))-x$ has only one real root. First edition was: Let $f(x)$ be a polynomial such that $f(x)$ and $f(x)-x$ have only one real root. How to prove, without derivatives, that
$f(f(x))-x$ also has only one real root?
Second edition: Let $f(x)=a_3 x^3 + a_2 x^2 + a... | As Jonas Meyer has pointed out, the problem is actually not true.
As suggested, let $f(x)= -x$. Then $f(x) -x = -x -x = -2x$. Considering $f(f(x)) -x= f(-x) -x = -(-x) -x = 0$. That is to say, for $f(x)=-x$, we have shown $f(x) -x$ has only one real root, namely $x=0$, while $f(f(x))-x=0$ has infinitely many real roots... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is there a formula for $(1+i)^n+(1-i)^n$? I'm wondering if there is a formula for the value of $(1+i)^n+(1-i)^n$?
I calculated the first terms starting with $n=1$ to be, in order, $2$, $0$, $-4$, $-8$, $-8$, $0$, $16$, $\dots$
So it seems to be some sequence of positive and negative powers of $2$ with $0$s thrown in. I... | This is not a particularly elegant solution, but an alternative route is to simply note that $(1 \pm i)^2 = \pm 2i$. I will show a lot of steps, but this method involves only very easy calculations. If $n$ is even, then
$$
(1\pm i)^n = \left((1+i)^2\right)^{ \frac{n}{2}}= (\pm 2i)^{n/2} = \left\{ \begin{array}{ccl}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/99884",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "15",
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Solving $x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$ Could someone please explain how to solve this:
$$x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$$
not the answer only, but a step-by-step solution.
I tried to solve it, with the help of Khan Academy, but still I have no idea how to correctly solve it.
Thank you so much in advance!
| First note that $x= \pm 2$ are the roots of the equation. Hence using the remainder theorem, we can proceed as follows:
$$ x^3(x-2) - 8x^2(x-2) +5x(x-2) + 50(x-2)=0$$
$$\qquad \Rightarrow (x-2)(x^3 -8x^2 +5x +50) =0$$
$$ (x-2)( x^2(x+2) -10x(x+2) +25(x+2)) =0 $$
$$\qquad \Rightarrow (x^2-4)(x^2-10x+25)=0 $$
Hence we c... | {
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"timestamp": "2023-03-29T00:00:00",
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Integral( using Euler substitution) I try to solve this integral, but without success.
Can you help me please?
$$\int \frac{1}{2x+\sqrt{4x^{2}-x+1}}\,dx$$
Thanks a lot!
| Hyperbolic substitution
Note $4 x^2 - x +1 = 4\left(x - \frac{1}{8} \right)^2 + \frac{15}{16}$. Therefore, let's perform a $u$-substitution, $x = \frac{1}{8} + \frac{\sqrt{15}}{8} \sinh(t)$. Then
$$
4 x^2 - x +1 = \frac{15}{16} \cosh^2(t)
$$
and $\mathrm{d} x = \frac{\sqrt{15}}{8} \cosh(t) \mathrm{d} t$, therefore... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve the ordinary differential equation? $\displaystyle\frac{d²y}{dx^2}+ \frac{4}{y}\left(\frac{dy}{dx}\right)^2+2=0$
with $y(0) = 1$ and $\displaystyle\frac{dy}{dx} = 0$ for $x = 0$.
| your equation is $yy''+4(y')^{2}+2y=0$
$z=y^5$
$z'=5y^{4}y'$
$z''=20y^{3}(y')^{2}+5y^{4}y''=5y^{3}(4(y')^{2}+yy'')$
$(4(y')^{2}+yy'')=\frac{z''}{5y^{3}}$
If we put it to your equation
$\frac{z''}{5y^{3}}+2y=0$
$z''=-10y^{4}=-10z^{4/5}$
$z'z''=-10z^{4/5}z'$
$\int z'z'' dx=-10\int z^{4/5}z'dx$
$ (z')^{2}/2 =-(50/9) z^{9/... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Groups of units of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ On page 230 of Dummit and Foote's Abstract Algebra, they say: the units of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ are determined by the integers $a,b$ with $a^2+ab+b^2=\pm1$ i.e. with $(2a+b)^2+3b^2=4$, from which is is easy to see the group of u... | Let $\omega=\frac{1+\sqrt{-3}}{2}$. Notice that the norm of an element $a+b\omega$ in $\mathbb{Z}[\omega]$ is $$(a+b\omega)(a+b\overline{\omega})=a^2+ab+b^2.$$ Since the norm is multiplicative, all units must have norm $1$, otherwise they would not be invertible. This means we are looking for solutions to $a^2+ab+b^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
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value of $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\cdots$? Is there anything known about the value of the series $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\frac{1}{1+2+3+4+5+6}+\cdots$ ?
| The denominators $a_n$ are
$$
a_n = \sum_{k=1}^{n}k = \frac{n(n+1)}{2}
$$
so
$$
\sum_{n=1}^{\infty} \frac1{a_n} =
\sum_{n=1}^{\infty} \frac{2}{n(n+1)} =
2\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 2
$$
If you don't have the resources to check this on a computer (out to infinity),
you will ha... | {
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"timestamp": "2023-03-29T00:00:00",
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} |
Solving the exponential equation: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ I have this exponential equation that I don't know how to solve:
$3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ with $x \in \mathbb{R}$
I tried to factor out a term, but it does not help. Also, I noticed that:
$2 \cdot 9^... | Set $a = 2^x, b = 3^x$, then $6^x = (2 \cdot 3)^x = 2^x \cdot 3^x = ab$ and $2^{2x+2} = 2^2 \cdot 2^{2x} = 4 \cdot 2^x \cdot 2^x = 4a^2$, $9^{x+1} = 9 \cdot 9^x = 9 \cdot 3^x \cdot 3^x = 9b^2$. So we get $$12a^2 - 35ab + 18b^2 = 0.$$ I think.
What can you do with that?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/108447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
What is the name for the polynomials of the form : $ P_n(x)=2^{-n} \cdot ((x+\sqrt {x^2-4})^n+ (x-\sqrt {x^2-4})^n)$? Polynomials of the form :
$ T_n(x) =2^{-1} \cdot ((x+\sqrt {x^2-1})^n+ (x-\sqrt {x^2-1})^n)$
are known as Chebyshev polynomials of the first kind .
Consider the polynomials of the form :
$P_n(x)=2^{-n... | I don't think they do have names (although they bear a superficial resemblance to the Lucas polynomials), but note that your polynomials are mere rescalings of the Chebyshev polynomial of the first kind:
$$\begin{align*}
2^{-n}((x+\sqrt{x^2-4})^n+(x-\sqrt{x^2-4})^n)&=\left(\frac{x}{2}+\sqrt{\frac{x^2}{4}-1}\right)^n+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/110570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding domain of $\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$ How can I find the domain of:
$$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$
I think the hard part will be to find:
$$\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} \ge 0$$
So far I have: not sure how to preceed:
$... | $(x^2-1)$ is negative iff $x\in(-1,1)$. Similarly, $(x^2-2)$ is negative for $x\in(-\sqrt{2},\sqrt{2})$, $(x^2-3)$ for $x\in(-\sqrt{3},\sqrt{3}), \dots,(x^2-6)$ for $x\in(-\sqrt{6},\sqrt{6})$.
Consider the open intervals $(-\infty,-\sqrt{6}), (-\sqrt{6},-\sqrt{5}),(-\sqrt{5},-2),\dots$ separately. Don't forget to look... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding derivative of $\sqrt[3]{x}$ using only limits I need to finding derivative of $\sqrt[3]{x}$ using only limits
So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator
$$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3... | $$\lim_{h \to 0} \frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h} $$
$$=\lim_{h \to 0}
\frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h}
\cdot \frac{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}}{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}} $$
$$=\lim_{h \to 0} \frac{x+h-x}{h((x)^{2/3} + x^{1/3}(x+h)^{1/3} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Show $4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$ Show
$$4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$$
The hint says "note $\sin{\frac{3\pi}{5}} = \sin{\frac{2\pi}{5}}$" and "use double/triple angle or otherwise"
So I have
$$4 \cos^2{\frac{\pi}{5}} - 2 (2 \cos^2{\frac{\pi}{10}} - 1) - 1$$
$$4 \... | Use the identities $$\sin 2\theta=2\cos\theta\sin\theta$$
and $$\eqalign{\sin 3\theta &=\sin(2\theta+\theta)\cr
&= \sin2\theta \cos\theta+\sin\theta\cos2\theta\cr
&= 2\sin\theta\cos^2\theta +\sin\theta(2\cos^2\theta-1)\cr
}
$$
to write the identity
$$\textstyle\sin{2\pi\over5}=\sin(\pi-{2\pi\over5})=\sin{3\pi\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/113466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Writing $3\sin(x)+4\cos(x)$ in the form of $r \sin(x-a)$ I need to write $3\sin(x) + 4\cos(x)$ in the form of $r\sin(x-a)$.
Expanding $r\sin(x-a): r\sin(x)\cos(a)-r\sin(a)\cos(x)$
Comparing the two forms (The original equation and the expanded form): $3=r\cos(a)$ and $4=-r\sin(a)$.
Getting $r$:
$$\begin{align*}
3^2 + ... | The two values of $a$ differ by $180^{\circ}$. Since $\sin(y\pm 180^{\circ}) = -\sin(y)$, you want to pick $r$ and $a$ in such a way that $r\sin(x-a)$ has the same sign as $3\sin x+ 4\cos x$.
When $x=0$, $3\sin x + 4\cos x=4$ is positive; so you want to pick $r$ and $a$ so that $r\sin(-a)$ is positive. This happens wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/116209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
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