Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Can I simplify $n\cdot r\cdot \sin(90^\circ-\frac{180^\circ}{n})\sqrt{r^2-r^2\sin^2(90^\circ-\frac{180^\circ}{n})}$ further? I need to write a simplified formula for this:
$$A_i = n\cdot r\cdot \sin\left(90^\circ-\frac{180^\circ}{n}\right)\sqrt{r^2-r^2\sin^2\left(90^\circ-\frac{180^\circ}{n}\right)}$$
I am not very co... | $$n\cdot r\cdot \sin\left(90^\circ-\frac{180^\circ}{n}\right)\sqrt{r^2-r^2\sin^2\left(90^\circ-\frac{180^\circ}{n}\right)}$$
$$=n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)\sqrt{r^2-r^2\cos^2\left(\frac{180^\circ}{n}\right)}$$
$$ =n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)\sqrt{r^2\left(1-\cos^2\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $n$ such that there are $11$ non-negative integral solutions to $12x+13y =n$
What should be the value of $n$, so that $12x+13y = n$ has 11 non-negative integer solutions?
As it is a Diophantine equation, so we check whether the solution exists? it exists if $gcd(12,13)|n$ that is $1|n$ and hence integer solution... | $12x + 13y = n$
$12(-n)+13(n)=n$ implies $x(t)=-n+13t$ and $y(t)=n-12t$
$x(t) \ge 0 \implies t \ge \dfrac{n}{13}$
$y(t) \ge 0 \implies t \le \dfrac{n}{12}$
So $\dfrac{n}{13} \le t \le \dfrac{n}{12}$
The existence of $11$ solutions implies
$$\text{$\dfrac{n}{13} \le t_0$ and $t_0 + 10 \le \dfrac{n}{12}$} \tag{1}$$
That... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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If $\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1$, then what's the value of $4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin x}\right)$? I was trying to solve this problem:
If
$$\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1$$
then what is the value of
$$S=4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin... | http://comentario.fariasbrito.com.br/vest/index.php?vid=68 There you go (it's the seventh question).
It's a pretty interesting problem, but I think when you're searching problems from IME or ITA, you should first check the sites of courses that are about them if the problems aren't from before 2000 (like the one I sent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Integrate $\int \sqrt{\frac{1+x}{1-x}}dx$ using $x = \cos(u) $ I have to integrate $\int \sqrt{\frac{1+x}{1-x}}$ using $x = \cos(u)\Rightarrow dx = -\sin(u) du$
My attempts:
$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)(1-x)}{(1-x)(1-x)}}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx$$
$$\i... | Substitute
$$t=\sqrt{\frac{1+x}{1-x}}$$ then we get
$$x=\frac{t^2-1}{t^2+1}$$
$$dx=\frac{4t}{(t^2+1)^2}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the epsilon- N definition of the limit verify that: $ \lim_{n\to\infty} \frac{n}{\sqrt{9n^2+4n+3}} = \frac{1}{3}$
Here is what I have so far:
$\mathrm{x}_\mathrm{n}=\frac{n}{\sqrt{9n^2+4n+3}},$ $a=\frac{1}{3}$
$ \forall{\epsilon}$ $\exists{\mathrm{N}_\mathrm{\epsilon}}$ $|$ $\forall{n}\in{\mathbb{N}}$ $n>\mathrm{... | I can only guess that the task is about showing that $$ \lim_{n \to \infty} \frac{n}{\sqrt{9n^2+4n+3}} = \frac{1}{3} $$
Here's one way:
It holds that $\frac{n}{\sqrt{9n^2+4n+3}} = \frac{n}{\sqrt{n^2\left(9+\frac{4}{n}+\frac{3}{n^2}\right)}} = \frac{n}{\sqrt{n^2}{\sqrt{9+\frac{4}{n}+\frac{3}{n^2}}}} $
Since $n \ge 1$ it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Derivation of the pdf of RV $X (\left|X \right| + 1)$ How to derive the probability density function $p_Y(y)$ in case, when $Y = X (\left|X \right| + 1)$, where $X \sim \mathrm{Uniform}(-1, 1)$.
I know that if $X < 0$ and $(\left|X \right| + 1) X < y$ then $X < \dfrac{1}{2} \left(1- \sqrt{1 - 4y} \right)$ and also if ... | Your results so far are correct and you are almost there.
Note that the density of $X$ is symmetric with respect to zero as in $\Pr\{ X < 0 \} = \Pr\{ X > 0 \} = \frac12$. Meanwhile, $X < 0$ actually means $-1 < X < 0$ by definition, and similarly $X > 0$ implicitly means $0 < X < 1$.
Also note that $Y$ has the same ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Finding value of $\int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx$ without contour Integration
Finding value of $\displaystyle \int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx\;\;, a>b>0$ without contour Integration
Try: Let $\displaystyle I =\int^{2\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx=2\int^{\pi}_{0}\frac{\sin^2 x}{a-b\cos... | $b<a$ or the integral will not converge.
If you don't want to use complex analysis, Try the substitution:
$t = \tan \frac {x}{2}\\
x = 2\arctan t\\
dx = \frac {2}{1+t^2}\ dt\\
\sin x = \frac {2t}{1+t^2}\\
\cos x = \frac {1-t^2}{1+t^2}$
Due to the discontinutity of the $\tan$ function, this will work better if we integr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Factorization of $(1+x+x^2+x^3)^2 - x^3$
Factorize : $(1+x+x^2+x^3)^2 - x^3$
I've tried to expand it but the equation will be even more complicated, anyone can give me some hints to solve it without expanding it (or it is necessary to expand it)?
| A strange but efficient way that uses the geometric series formula 3 times:
$$(1+x+x^2+x^3)=\frac{1-x^4}{1-x}$$
Your polynomial is then:
$$p(x)=(1+x+x^2+x^3)^2-x^3=\frac{(1-x^4)^2-x^3(1-x)^2}{(1-x)^2}$$
$$=\frac{1-2x^4+x^8-x^3+2x^4-x^5}{(1-x)^2}=\frac{1-x^3-x^5+x^8}{(1-x)^2}=$$
$$=\frac{(1-x^3)(1-x^5)}{(1-x)(1-x)}=(1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of the form $\int_a^b \frac{\ln(c+dx)}{P(x)}dx$ I found here a "great theorem" which states that: $$\int_a^b \frac{\ln(c+dx)}{P(x)}dx =\frac{\ln((ad+c)(bd+c))}{2}\int_a^b\frac{dx}{P(x)}$$
I don't know how to prove this, but I am pretty sure that we should work by symmetry with a substitution of the form $\frac... | \begin{align}J=\int_a^b \frac{\ln(c+dx)}{P(x)}dx\end{align}
Formally,
1)"Clean up" the logarithm. Perform the change of variable $u=c+dx$,
\begin{align}J=\frac{1}{d}\int_{c+da}^{c+db} \frac{\ln u}{P\left(\frac{u-c}{d}\right)}du\end{align}
2) Change of the bounds of the integral to new ones, m,M such that $m\times M=1$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Prove $\frac{2n+\sin(n)}{n+2}$ converges to 2. Here's my attempt.
Let $\epsilon >0.$ Then $N \geq \frac{5}{\epsilon}$, with $n \geq N \implies$
$| \frac{2n+\sin(n)}{n+2} -2|=| \frac{2n+\sin(n)-2n-4}{n+2}|=| \frac{\sin(n)-4}{n+2}|= \frac{|\sin(n)-4|}{n+2} \leq \frac{5}{n+2} \leq \frac{5}{n} \leq \epsilon$
| Alternatively,
$$
\frac{2n-1}{n+2}
\le
\frac{2n+\sin(n)}{n+2}
\le
\frac{2n+1}{n+2}
$$
and both bounds go to $2$ as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Showing that the inequality $\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2} \leq u_n \leq \frac{\sqrt{2}}{n+1}$ stands.
If $\displaystyle u_{n}=\int_{0}^{1} t^n \sqrt{t+1} dt$, where $n\geq 1$, prove that
$$\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2} \leq u_n \leq \frac{\sqrt{2}}{n+1}.$$
I have already proven the second inequa... | Since $t\to\sqrt{t+1}$ is concave, it follows that, for $t\in [0,1]$,
$$1+(\sqrt{2}-1)t\leq \sqrt{t+1}\leq \sqrt{2}.$$
Hence
$$\frac{1}{n+1}+\frac{\sqrt{2}-1}{n+2}\leq \int_{0}^{1} t^n \sqrt{t+1} dt\leq \frac{\sqrt{2}}{n+1}.$$
It remains to show that
$$
\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2}\leq
\frac{1}{n+1}+\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Show that $\frac {1}{3+x^2+y^2} + \frac {1}{3+y^2+z^2} +\frac{1} {3+x^2+z^2}\leq \frac {3}{5} . $ Let $x, y, z>0$ s.t. $x+y+z=3$.
Show that $$\frac {1}{3+x^2+y^2} + \frac {1}{3+y^2+z^2} +\frac{1 } {3+x^2+z^2}\leq \frac {3}{5}\ . $$
My idea: $$3 + x^2 + y^2 \geq 1 + 2x+ 2y=7-2z $$
I notice that $f (t)=\frac {1}{7-2t} $... | The first inequality.
We need to prove that
$$\sum_{cyc}\frac{1}{3+x^2+y^2}\leq\frac{3}{5}$$ or
$$\sum_{cyc}\left(\frac{1}{3+x^2+y^2}-\frac{1}{3}\right)\leq\frac{3}{5}-1$$ or
$$\sum_{cyc}\frac{x^2+y^2}{x^2+y^2+3}\geq\frac{6}{5}.$$
Now, by C-S twice we obtain:
$$\sum_{cyc}\frac{x^2+y^2}{x^2+y^2+3}\geq\frac{\left(\sum\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Expanding $4\sin^3(x)$ using the Complex Exponential How would I go about proving:
$$
4 \sin^3(x)=3\sin(x)-\sin(3x)
$$
Using the complex exponential, i.e.
$$
e^{ix}=\cos(x)+i\sin(x)
$$
| $$(\cos x+i\ \sin x)^n=\cos nx+i\ \sin nx....(1)$$
Since we want $\sin^3x$ we will expand $(\cos x+i\ \sin x)^3$
$(\cos x+i\ \sin x)^3=\cos^3x-3\cos x\sin^2x+i\ (3\cos^2x\sin c-\sin^3x)$
From $(1)$ we have
$$\cos3x+i\ \sin3x=\cos^3x-3\cos x\sin^2x+i(3\cos^2x\sin x-\sin^3 x)$$
Now equate the imaginary parts and we ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed-form expression for infinite series related to a Gaussian Consider the following infinite series, where $x$ is indeterminate and $r$ is held constant:
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^2} + \frac{x^3}{r^3} + ...$
It is relatively easy to see that the above, for $\frac{x}{r} < 1$, converges to
$\displ... | Interestingly, if we change the definition slightly, we get something related to the Jacobi theta function.
If we start with this series:
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^4} + \frac{x^3}{r^9} + ...$
We can make the following substitutions:
$q = \frac{1}{r}$
$x = \exp(2\pi i z)$
to obtain
$\displaystyle 1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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minimum value of $\bigg||z_{1}|-|z_{2}|\bigg|$
If $z_{1}\;,z_{2}$ are two complex number $(|z_{1}|\neq |z_{2}|)$ satisfying
$\bigg||z_{1}|-4\bigg|+\bigg||z_{2}|-4\bigg|=|z_{1}|+|z_{2}|$
$=\bigg||z_{1}|-3\bigg|+\bigg||z_{2}|-3\bigg|.$Then minimum of $\bigg||z_{1}|-|z_{2}|\bigg|$
Try: Let $|z_{1}|=a$ and $|z_{2}|=b$ an... | The minimum for $|a-b|$ is at $a=b$
$$
|a-c|-a = |b-c|-b \Rightarrow a+b=c
$$
so the intersections for
$$
a+b = 4 \cap a = b \Rightarrow a = 2\\
a+b = 3 \cap a = b \Rightarrow a = \frac 32
$$
so the minimum is at $a = b = \frac 32$ or $a = b = 2$ which is $0$
Attached in red $a+b=4$ in blue $a+b=3$ and in lightblue $a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Bound on $\sum\limits_{n=0}^{x}{\sin{\sqrt{n}}}$ Using Desmos and Mathematica, I was able to find a function $g(x)$ that seemingly estimated the function
$$f(x)=\sum_{n=0}^{x}{\sin{\sqrt{n}}}$$
I found that
$${g(x)=2\sqrt{x}*\sin{\left({{\sqrt{4x+{\pi}^2}-\pi}\over{2}}\right)}}\approx f(x)$$
Furthermore, I found tha... | (Too long for a comment)
We have
\begin{align*}
f(x)
= \sum_{k=1}^{n} \int_{0}^{\sqrt{k}} \cos x \, dx
= \int_{0}^{\infty} \left( \sum_{k=1}^{n} \mathbf{1}_{\{ x < \sqrt{k}\}} \right) \cos x \, dx.
\end{align*}
Notice that $\sum_{k=1}^{n} \mathbf{1}_{\{ x < \sqrt{k}\}} = (n - \lfloor x^2 \rfloor)_+$, where $a_+ = \max... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 1
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Induction proof: $\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}$ $<2$ Prove by induction the following.
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}<2.$$
Caveat: The $<$ will be hard to work with directly. Instead, the equation above can be written in the form,
$$\frac{1}{2^1}... | This question is circling in rounds.
Because, as everybody knows, the sum tends to $2$ so that in
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}+ r_n=2,$$ the expression of $r_n$ must be exact!
If you try with an inequality such as
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given that f(x) is a quadratic function and that f(x) is only positive when x lies between -1 and 3, find f(x) if f(-2) = -10. What i did:
$f(x)=ax^2+bx+c$
$f(-2)=4a-2b+c=-10$
$f(0) =c > 0$
$f(1) =a+b+c > 0$
$f(2) =4a+2b+c > 0$
I thought using $b^2-4ac = 0$ for $f(-2)$ but its wrong since I am getting c = -6.
ANS: ... | So $-1$ and $3$ are zeroes of this function so we can write it in factor form:
$$ f(x)=a(x+1)(x-3)$$ Now use $f(-2)=10$ to get $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2962306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving that $\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}=\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$ Trying to show using a different approach that $\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx =\frac{\pi^2\sqrt 3}{9}-\frac{8}{3}G\, $ I have stumbled upon this series: $$\sum_{n=1}^\infty \frac{\sin\left... | Integrate by parts
\begin{align}
\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx
=& \int_0^1 \ln x \>d\left( \tan^{-1} \frac{\sqrt x} {1-x}
-\frac1{\sqrt3 } \tanh^{-1} \frac{\sqrt {3x}} {1+x} \right)\\
=&\frac1{\sqrt3 }I_1 - I_2\tag1
\end{align}
where
\begin{align}
I_1&=\int_0^1 \frac{\tanh^{-1} \frac{\sqrt {3x}} {1+x}}x dx\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
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For which real $a$ does$ \sum_{n=1}^{\infty} \left\{ e-\left(1+\frac{1}{n}\right)^{n+a} \right\}$ converge or diverge? For which real $a$ does
$ \sum_{n=1}^{\infty} \left\{ e-\left(1+\frac{1}{n}\right)^{n+a} \right\}$
converge or diverge?
This is a generalization of
$\sum_{n=1}^{\infty} \left\{ e-(1+\frac{1}{n})^n \ri... | I think you are correct.
Note that
$$\begin{align}\log \left(1 + \frac{1}{n} \right)^{n +a} &= \frac{n+a}{n+1/2}(n + 1/2)\log\left(1 + \frac{1}{n} \right)\\&=\frac{n+a}{n+1/2}(n + 1/2)\left(\frac{1}{n} - \frac{1}{2n^2} +\mathcal{O}\left(\frac{1}{n^3} \right) \right) \\ &= \frac{n+a}{n+1/2}\left(1 + \mathcal{O}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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The rth term in $(1+x)^{1/x} = e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ Let
$$(1+x)^{\frac {1}{x}} = e.G(x)$$
Taking logarithm on both sides,
$$\frac {1}{x} \log {(1+x)} = 1 + \log {G(x)}$$
Putting in the Taylor expansion for $\log {(1+x)}$ we have,
$$\frac {1}{x}(x - \frac {1}{2}x^2 + \frac... | This is a slight rewriting of a problem solved in Math Overflow 77389. That answer can be put in the form
$$ (1+x)^{1/x} = e\,\sum_{n=0}^\infty a_n x^n \text{ with } a_n=\sum_{k=0}^n \frac{S_1(n+k, k)}{(n+k)!} \sum_{m=0}^{n-k}\frac{(-1)^m}{m!} $$
$S_1(n,k)$ is the Stirling number of the first kind, StirlingS1[n,k], in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Sum of polynomials over a ring Please refer to question 1. in the below link (pg 469)
Ques link
I think the question is incorrect as $f(x)$ contains the term $-5$ which doesn't belong to the ring $Z8[x]$ or I should go ahead by changing $f(x)$ to $f(x)= 4x+3$ and similarl changes for $g(x)$ and them add both using modu... | $-5$ is an element of $\mathbb{Z}_8$.
As you mentioned we have $-5=3=11=19=27=-13=-21=...$ in $\mathbb{Z}_8$.
$f(x)=4x-5=4x+3$
$g(x)=2x^2-4x+2=2x^2+4x+2=2(x^2+2x+1)=2(x+1)^2$
We get:
$(f+g)(x)=4x+3+2x^2+4x+2=2x^2+8x+5=2x^2+5$
and
$(f\cdot g)(x)=(4x+3)(2(x+1)^2)=8x(x+1)^2+6(x+1)^2\stackrel{8=0}{=}6(x+1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$?
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$ then $\frac{d}{dx}=?$
Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivati... | Notice that with
$$f(x):=\frac{1+x}{1-x}$$ we have
$$f(-x)=\frac1{f(x)}$$ so that
$$\arctan\sqrt{f(-x)}=\text{arccot}\sqrt{f(x)}=\frac\pi2-\arctan\sqrt{f(x)}.$$
This shows that to an additive constant, the initial function is odd and its derivative must be even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$, with $p$ being a prime number greater than $3$.
For $p \equiv 1 \mod 4$, there exists an integer $j$ such that $p\mid j^2+1$ (since $-1$ is a quadratic residue of $p$), therefore $R... | We can also use the symmetric polynomials.
Let $A=(1^2+1)(2^2+1) \cdots ((p-1)^2+1)(p^2+1)$. Then, $A \equiv (1^2+1)(2^2+1) \cdots ((p-1)^2+1) \mod p$, since $p^2+1\equiv 1 \mod p$.
Consider the symmetric polynomial $\displaystyle\prod_{i=1}^{\frac{p-1}2}(1+x_i)=\sum_{j=0}^{\frac{p-1}2}e_j(x)$, where $e_j$ is the $j$-t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Sum to $n$ terms the series $\frac{1}{3\cdot9\cdot11}+\frac{1}{5\cdot11\cdot13}+\frac{1}{7\cdot13\cdot15}+\cdots$.
Q:Sum to n terms the series :
$$\frac{1}{3\cdot9\cdot11}+\frac{1}{5\cdot11\cdot13}+\frac{1}{7\cdot13\cdot15}+\cdots$$
This was asked under the heading using method of difference and ans given was
$$S_... | Let
$$S_n=\sum_{k=1}^n\frac{1}{(2k+1)(2k+7)(2k+9)}.$$
By the partial fraction decomposition:
$$\frac{1}{(2k+1)(2k+7)(2k+9)}=\frac{1/48}{2k+1}
-\frac{1/12}{2k+7}+\frac{1/16}{2k+9}$$
Then, after letting $O_n=\sum_{k=1}^n\frac{1}{2k+1}$, we have that
\begin{align}
S_n&=\frac{O_n}{48}-\frac{O_{n+3}-O_3}{12}+\frac{O_{n+4}-O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find $a$, $b$, $c$, $d$ such that $(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$ How would I find $a$, $b$, $c$, and $d$ in :
$$(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$$
I have already worked out that $a=2$, but I may be wrong. I am not sure how to find the values of the other letters.
Thanks in advance!
| Just do it:
$(ax+b)^2(x+c)$ expands to $(a^2x^2 + 2abx + b^2)(x+c) = a^2x^3 + 2abx^2 + b^2 x + a^2cx^2 + 2abcx + b^2c= a^2x^3 + (2ab + a^2c)x^2 + (b^2 +2abc)x + b^2 c$
So
$a^2x^3 + (2ab + a^2c)x^2 + (b^2 +2abc)x + b^2 c= 4x^3 + dx^2 + 55x - 100$
So you get three sets of equations:
$a^2 = 4$
$2ab + a^2c = d$
$b^2 +2abc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Primes of the form $p=2a^2-1$ satisfying $p^2=2b^2-1$. Are there any prime numbers $p$ for which there exist integers $a$ and $b$ such that
$$p=2a^2-1\qquad\text{ and }\qquad p^2=2b^2-1,$$
other than $p=7$?
The fact that $p^2=2b^2-1$ implies that
$$(p+b\sqrt{2})(p-\sqrt{2}b)=-1,$$
and hence that $p+b\sqrt{2}$ is a unit... | Clearly $p\ne 2$ and $p>b>a$. Write $$ p(p-1)= 2(b-a)(a+b)$$
Case 1: $$p\mid a+b\implies a+b = kp \implies k<2$$
So $a+b=p$ and $p-1 = 2(b-a)$. From here we get $p=4a-1$ so $$4a-1 = 2a^2-1\implies a=2$$
so $p= 7$.
Case 2: $$p\mid b-a\implies p \leq b-a <p $$
Which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$ Q:Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$.My book solve it leting the roots of the equation be $\alpha,\alpha+3,\beta$ then find the equation whose roots are $\alpha-3,\alpha,\beta-3$.And i know... | Just to give another approach, if the roots are $\alpha$, $\alpha+3$, and $\beta$, then, by Vieta, we have
$$\alpha+(\alpha+3)+\beta=-1/2$$
and
$$\alpha(\alpha+3)+\alpha\beta+(\alpha+3)\beta=-7/2$$
The first of these can be rewritten as $2\alpha+\beta=-1/2-3=-7/2$, which means the second implies
$$\alpha^2+3\alpha+2\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find bases B3 and B2 for $\Bbb R^3$ and $\Bbb R^2$ given a linear transformation and its matrix A linear transformation T is defined by
T: $\Bbb R^3$ $\rightarrow$ $\Bbb R^2$ $\Rightarrow$ T$\begin{pmatrix}x\\y\\z\end{pmatrix}$ = $\begin{pmatrix}2x+y\\y+2z\end{pmatrix}$
Find bases $\mathscr B_3'$ and $\mathscr B_2'$ fo... | I do not think your approach is correct.
We can simplify the answer by taking the ordered bases
$$\begin{align}
\mathscr B_3' & = \left\{
\begin{pmatrix}1\\0\\0\end{pmatrix},
\begin{pmatrix}0\\1\\0\end{pmatrix},
\begin{pmatrix}0\\0\\1\end{pmatrix}\right\},\\\\
\mathscr B_2' & = \left\{
\begin{pmatrix}\beta_{11}\\ \beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Convergence to $\sqrt{2}$ It is a very good way to approximate $\sqrt{2}$ using the following;
Let $D_{k}$ and $N_{k}$ be the denominator and the numerator of the $k$th term, respectively.
Let $D_1=2$ and $N_1=3$, and for $k\geq2$, $D_k=D_{k-1}+N_{k-1}$, and $N_{k}=D_{k-1}+D_{k}$.
To clarify, here is the sequence;
$\fr... | You have a pair of linear recurrence relations. You can write it as $$\begin {pmatrix} N_k\\D_k \end {pmatrix}=\begin {pmatrix} 1&2\\1&1 \end {pmatrix}\begin {pmatrix} N_{k-1}\\D_{k-1} \end {pmatrix}$$ You find the eigenvalues and eigenvectors of the matrix. The eigenvector $\begin {pmatrix} \sqrt 2\\1 \end {pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How many Eisenstein integers modulo 3 are there? I'm trying to find a system of representatives for the Eisenstein integers modulo 3. What would the set S be and how can I determine the number of solutions in S of the equation x^2=0 mod 3?
| The ring of Eisenstein integers is isomorphic to $\Bbb{Z}[X]/(X^2+X+1)$, so the ring of Eisenstein integers mod 3 is isomorphic to
\begin{eqnarray*}
(\Bbb{Z}[X]/(X^2+X+1))/(3)&\cong&(\Bbb{Z}[X]/(3))/(X^2+X+1)\\
&\cong&\Bbb{F}_3[X]/(X^2+X+1)\\
&\cong&\Bbb{F}_3[X]/((X-1)^2).
\end{eqnarray*}
This is a ring of $9$ elements... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
I am using $\tan(A-B)=1$, so $A-B=n\pi+\pi/4$ and $A+B = 2n\pi\pm\pi/6$. Solving these I am getting $B =7\pi/24$ and $A =37\pi/24$.
The book I am refe... | First of all, your values for $A$ and $B$ satisfy the given formulae and $B$ is positive, so their proposed minimum for $B$ is definitely wrong.
One thing you did miss in your reasoning is that there are two families of solutions to $\sec\left(\theta\right) = \frac{2}{\sqrt{3}}$; there should be a $\pm$ with $\frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Transition Matrix from B to C
If $B=\{ 2+x,1+2x\}$ and $C=\{ 1+x, 1-x\}$ are 2 basis for $P_1$, and $v=-3x+4$ find $[v]_B$, $_BP_C$ and $[v]_C$.
my attempt:
Since $B$ is a basis for $V$, then any $v\in B$ can be written "uniquely" as a linear combination of the vectors of $B$.
then $v=-3x+4$ can be written uniquely ... | $b_1 = 2+x = a_{11} c_1 + a_{21} c_2\\
2+x = a_{11}(1+x) + a_{21}(1-x)\\
a_{11}+a_{21} = 2\\
a_{11}-a_{21} = 1\\
a_{11} = \frac 32, a_{21} = \frac 12$
and
$b_2 = 1+2x = a_{12} c_1 + a_{22} c_2\\
a_{12} + a_{22} = 1\\
a_{12} - a_{22} = 2\\
a_{12} = \frac 32, a_{22} = - \frac 12$
$_CP_B = \begin{bmatrix} \frac 32 &\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all $a$ and $b$ that make this value rational: $ \left(\sqrt 2 + \sqrt a\right)/\left(\sqrt 3 + \sqrt b\right)$
$$ \frac{\sqrt 2 + \sqrt a}{\sqrt 3 + \sqrt b} $$ Find all integers $a$ and $b$ that make this value rational.
Looking at it, you will easily see that $a=3$, $b=2$ will make this value $1$.
Are there... | Note that $b=3$ does not give any solution since $$\frac{\sqrt 2 + \sqrt a}{\sqrt 3 + \sqrt 3}=\frac{\sqrt{6}+\sqrt{3a}}{6}$$and $\sqrt{6}+\sqrt{3a}$ cannot be rational number (See this question for why). Therefore, we can rationalize it.
Rationalize the value and expanding gives$$\frac{\sqrt 2 + \sqrt a}{\sqrt 3 + \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do I calculate the value of $\tan(1/2)$ for the De Moivre formula I have to solve $(2+i)^3$ using the trigonometric representation. I calculated the modulus but I don't know how to calculate $\varphi$ when it is equal to $\tan(1/2)$.
How do I calculate $\varphi$
Also, is there a fast way to solve these in analogous... | There is no real advantage in computing powers this way over the algebraic method. There would be an advantage if the argument is a “known angle”.
You can surely write $2+i=\sqrt{5}(\cos\varphi+i\sin\varphi)$, where
$$
\cos\varphi=\frac{2}{\sqrt{5}}\qquad\sin\varphi=\frac{1}{\sqrt{5}}
$$
Then, yes,
$$
(2+i)^3=5\sqrt{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Is $x(1 - 2x) \le \frac{1}{8}$ and further, is $x(1 - ax) \le \frac{1}{4a}$ It is clear that $x(1-x) \le \frac{1}{4}$
Does it likewise follow that $x(1-2x) \le \frac{1}{8}$?
Here's my reasoning:
(1) For $x < \frac{1}{4}$, $x(1-2x) < \frac{1}{8}$
(2) For $\frac{1}{4} < x < \frac{1}{2}$, $x(1-2x) < \frac{1}{8}$
(3) Fo... | If you already know that $x(1-x) \le 1/4$ for all $x$, then it also holds for $ax$. Divide the inequation you get by $a$ and you obtain the result you wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Sine Fourier Series for $\min(x, 1 - x)$. I am currently taking a class on differential equations and only Analysis and Linear Algebra I and II were expected. In those subjects we never covered Fourier series.
In an exercise I am now expected to solve
$$
\sum_{k = 1}^{\infty} a_k \sin(\pi k x)
= \min(x, 1 - x)
$$
for t... | The sine functions $\sin(k \pi x)$ are orthogonal. Thus forming the same scalar product on both sides of the sine series expansion $\sum_n a_n\sin(nπx)=\min(x,1-x)$, that is, multiplying with $\sin(k \pi x)$ and integrating, filters out the term with the coefficient $a_k$.
$$
a_k\int_0^1\sin^2(k \pi x)\,dx=\int_0^1\min... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2984136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the inverse Laplace of $ \ \frac{2}{(s-1)^3(s-2)^2}$. Find the inverse Laplace of $ \ \frac{2}{(s-1)^3(s-2)^2}$.
Answer:
To do this we have to make partial fractions as follows:
$ \frac{2}{(s-1)^3(s-2)^2}=\frac{A}{S-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)}+\frac{J}{(s-1)^3}+\frac{K}{(s-2)^2}+\frac{L}{s-2}$
Am I righ... | Another way is using convolution
\begin{align}
\frac{2}{(s-1)^3(s-2)^2}
&= \frac{2}{(s-1)^3}\cdot\frac{1}{(s-2)^2} \\
&= {\cal L}\left(t^2e^t\right){\cal L}\left(te^{2t}\right) \\
&= \int_0^xt^2e^t(x-t)e^{2x-2t}\ dt \\
&= e^{2x}\int_0^xe^{-t}(xt^2-t^3)\ dt \\
&= \color{blue}{2e^{2x}(x-3)+e^x(x^2+4x+6)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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What am I doing wrong solving this system of equations? $$\begin{cases}
2x_1+5x_2-8x_3=8\\
4x_1+3x_2-9x_3=9\\
2x_1+3x_2-5x_3=7\\
x_1+8x_2-7x_3=12
\end{cases}$$
From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)
$$\left[\begin{array}{ccc|c}
2 & 5 & -8 & ... | You do (in the third matrix): $$L3-L4=(0, -3, 1 \mid -5)-(0, -13, 9 \mid -19)=(0, 10, -8 \mid 12)$$ but you have $(0, 10, 8 \mid -12)$ instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
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Need help showing that $\zeta (x)$= ${1\over 1-2^{1-x}}$ $\eta (x)$ This question is I am working on is the extension of the domain of zeta function.$\zeta(x)$=$1+\frac{1}{2^x}+\frac{1}{3^x}+\frac{1}{4^x}+\frac{1}{5^x}+\frac{1}{6^x}+ \cdots$$\eta(x)$=$1-\frac{1}{2^x}+\frac{1}{3^x}-\frac{1}{4^x}+\frac{1}{5^x}-\frac{1}{6... | As i said in comment:
$\zeta (x) - \eta (x) = \sum\limits_{n=1}^{\infty}\frac{1}{n^x} - \sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^x}=\sum\limits_{n=1}^{\infty}(\frac{1}{n^x} - \frac{(-1)^{n-1}}{n^x})=2\sum\limits_{n=1}^{\infty}\frac{1}{(2n)^x}=2^{1-x}\sum\limits_{n=1}^{\infty}\frac{1}{n^x}=2^{1-x}\zeta (x) $
Hence:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the length of the closed curve $x^{2/3} + y^{2/3} = 4$ I realise that this function forms a closed curve, and the range of both $x$ and $y$ are: $-8 \leq x, y \leq 8$.
I began by differentiating the function implicitly, arriving at a expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$:
\begin{align}
\frac{\math... | With parametric equations you have $$x=8cos^3 \theta, y=8\sin ^3 \theta $$
Then you integrate $$L=4 \int _{0}^{\pi/2} \sqrt {(\frac {dx}{d\theta})^2 +(\frac {dx}{d\theta})^2} d\theta $$
The derivatives are straight forward and you can finish it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Does the integral $\int_{0}^{\infty} \frac{x^3\,\cos{(x^2-x)}}{1+x^2}$ diverge Does the integral
$$J:=\int_{0}^{\infty} \frac{x^3\,\cos{(x^2-x)}}{1+x^2}dx $$
diverge ?
If we integrate by parts we find
$$J=\lim_{a\rightarrow +\infty} \frac{a^3}{(1+a^2)(2a-1)}\cos{(a^2-a)} -\\
\int_{0}^{\infty}\sin{(x^2-x)}
\left( \frac... | Also you can do that: an improper integral $\lim_{r\to\infty}\int_0^r f(x)\, dx$ converges if and only if for each $\epsilon>0$ there is some $M>0$ such that $\left|\int_a^b f(x)\, dx\right|<\epsilon$ for all pairs $a,b\ge M$.
Now note that $\cos(x)\ge\sqrt 2/2$ when $x\in(-\pi/4+2k\pi,\pi/4+2k\pi)$ for any $k\in\Bbb Z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate squared trigonometric function I'm trying to integrate $\int_a^b \left( \frac{1}{1+x^2} \right)^2 dx$
I know that $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$, but how can I integrate with the squared part?
I've tried substitution with no success.
| Hint. Note that
$$\left( \frac{1}{1+x^2} \right)^2 = \frac{1-x^2+x^2}{(1+x^2)^2}= \frac{1}{1+x^2} -\frac{x^2}{(1+x^2)^2}=\frac{1}{1+x^2} +\frac{x}{2}\left(\frac{1}{1+x^2}\right)'.$$
then integrate by parts the last term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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Additional methods for integral reduction formula So I have successfully found a reduction formula for
$$I_{m,n}=\int\frac{dx}{\sin^m(ax)\cos^n(ax)}$$
Went as follows:
$$\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\int\csc^m(ax)\sec^n(ax)dx=\int\csc^m(ax)\sec^{n-2}(ax)\sec^2(ax)dx\\
\begin{vmatrix}u=\csc^m(ax)\sec^{n-2}(ax)\\... | The idea of this trick is similar to IBP but this ruduces a lot in calculations (because differentiation is "easier" than integration).
Take $f_{m,n}(x)= \csc^m(ax)\sec^n(ax)$. Take the derivative with respect to $x$:
\begin{align*}
f'_{m-1,n-1}&=a(n-1)\underbrace{(1-\cos^2(ax))f_{m,n}}_{f_{m-2,n}}-a(m-1)f_{m,n-2}\\
f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the general solution to the ODE $x\frac{dy}{dx}=y-\frac{1}{y}$ I have been working through an ODE finding the general solution and following the modulus through the equation has left me with four general solutions, as shown below. Online ODE solvers, however, have only calculated the two (positive and negative), o... | Your solutions are $\pm\sqrt{1\pm Kx^2}$, where $K$ is positive. But this does not differ from $\pm\sqrt{1+Kx^2}$ where $K$ is unconstrained.
From
$$\frac12\log|y^2-1|=\log|x|+C$$ you draw
$$\log|y^2-1|=\log(e^Cx^2)$$ and
$$|y^2-1|=e^Cx.$$
Then
$$y^2=1\pm e^Cx^2$$ which you can very well rewrite as
$$y^2=1+Dx^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Finding a cubic polynomial with Cayley-Hamilton Theorem I have two matrices:
$A = \begin{bmatrix}
1 &2 \\
-1 &4
\end{bmatrix}
$
and
$B = \begin{bmatrix}
0 &-1 \\
2 &3
\end{bmatrix}
$
I need to find a monic cubic polynomial $g$ such that $g(A)=g(B)=0$, the zero matrix.
I understand through the Cayley-Hamilton Theorem t... | the least common multiple of $(t-3)(t-2)$ and $(t-1)(t-2)$ is
$$ (t-1)(t-2)(t-3) $$
As with natural numbers, the LCM of two polynomials is their product divided by their gcd. If you did not notice the common root, the euclidean algorithm can find the gcd
$$ \left( x^{2} - 5 x + 6 \right) $$
$$ \left( x^{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_0^1\frac{\ln(1+x-x^2)}xdx$ without using polylogarithms.
Evaluate $$I=\int_0^1\frac{\ln(1+x-x^2)}xdx$$ without using polylogarithm functions.
This integral can be easily solved by factorizing $1+x-x^2$ and using the values of dilogarithm at some special points.
The motivation of writing this post i... | No need multiple-integrals, no need Feynman's trick
\begin{align}J&=\int_0^1 \frac{\ln(1+x-x^2)}{x}dx\overset{\text{IBP}}=-\int_0^1 \frac{(1-2x)\ln x}{1+x(1-x)}dx\\
&=-\int_0^{\frac{1}{2}} \frac{(1-2x)\ln x}{1+x(1-x)}dx-\int_{\frac{1}{2}}^1 \frac{(1-2x)\ln x}{1+x(1-x)}dx\\
&\overset{u=x(1-x)}=-\int_0^{\frac{1}{4}}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How to solve the recursive equation $y^{(n+2)} +(n+1)y^{(n+1)} +\tfrac{n(n+1)}{2} y^{(n)}=0$ I encounter the problem when I try to get the Taylor series of $\arctan x$ at $x=1$. It seems that the method for expanding it at $x=0$ does not work anymore (which is obtained by observing that $\arctan' x=\frac{1}{1+x^2}=\sum... | If you simply want to find the Taylor series about $x=1$ of $\text{arctan}(x)$, then you can use $$\frac{\text{d}}{\text{d}x}\,\text{arctan}(x)=\frac{1}{1+x^2}=\frac{1}{1+(z+1)^2}=\frac{1}{2\text{i}}\left(\frac{1}{1-\text{i}+z}-\frac{1}{1+\text{i}+z}\right)\,,$$
where $z:=x-1$. This gives
$$\frac{\text{d}}{\text{d}x}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Binomial Theorem with Three Terms $(x^2 + 2 + \frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
| To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In oth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Laurent series of $ \frac{z-12}{z^2 + z - 6}$ for $|z-1|>4$
How do you find the Laurent series for $f(z) = \dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?
I know that $f(z) = \dfrac{z-12}{z^2 + z - 6} = \dfrac{-2}{z-2} + \dfrac{3}{z+3}$
It is easy for me to extract a series for $\dfrac{3}{z+3}$, but have no idea ho... | Use the fact that\begin{align}\frac{-2}{z-2}&=\frac{-2}{-1+(z-1)}\\&=\frac2{1-(z-1)}\\&=-2\sum_{n=-\infty}^{-1}(z-1)^n\end{align}and that\begin{align}\frac3{z+3}&=\frac3{4+(z-1)}\\&=\frac34\times\frac1{1+\frac{z-1}4}\\&=-\frac34\sum_{n=-\infty}^{-1}\frac{(-1)^n}{4^n}(z-1)^n.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Solve for $z$, which satisfy $\arg(z-3-2i) = \frac{\pi}{6}$ and $\arg(z-3-4i) = \frac{2\pi}{3}$. solve for $z$, which satisfy $\displaystyle \arg(z-3-2i) = \frac{\pi
}{6}$ and $\displaystyle\arg(z-3-4i) = \frac{2\pi}{3}$.
So I'm first assuming $z=x+iy$, then putting it in the place of $z$ and putting real parts togeth... | $$\arg(z-3-2i)=\frac{\pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^\circ$ with positive $x$ axis.
And $$\arg(z-3-4i)=\frac{2\pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^\circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$ Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
My try:
By Lagrange Multiplier method we have
$$L(x,y,z,\lambda, \mu)=(x^2+6y^2+4z^2)+\lambda(x+2y+z-4)+\mu(2x^2+y^2-16)$$
For $$L_x=0$$ we get
$$2x+\lambda+4\mu x=0 \tag{1}$$
For $$L_y... | Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $\frac{11}2y^2+4z^2$. But then your first equation becomes $\lambda+4\mu x=0$, which is much simpler.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate and Simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$ I am trying to evaluate and simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$.
I am getting $\frac{11}{10}$ but the answer is $\frac{3-4\sqrt{3}}{10}$
My Process:
$\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$
$\cos[\cos^{-1}(\frac{3}{5})] + \co... | You cannot separate out the $\cos$ function as you have done in step two.
You can remember this identity.
$$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$
Here $\arccos(\frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)
Using thus result you should get the desired answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
An inequality to finalize a proof Related to this Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$ and my second answer I have to prove this :
Let $a,b,c$ be real positive numbers then we have : $$\sum_{cyc}\left(\frac{a^3}{13a^2+5b^2}\right)\left(\frac{b^3}{13b^2+5c^2}\right)\geq \frac{ab+bc+c... | It's wrong. Try $a=b=1$ and $c=2$.
We have:
$$LS=\frac{1}{18\cdot33}+\frac{8}{33\cdot57}+\frac{8}{57\cdot18}=\frac{155}{11286}$$ and
$$RS=\frac{5}{324}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the value of $a$
Find $a$ for which $f(x) = \left(\frac{\sqrt{a+4}}{1-a} -1\right)x^5-3x+\ln5 \;$ decreases for all $x$ with $a\neq 1$ and $a\geq -4$.
My try:
For $f(x)$ to decrease
$$5x^4\left(\frac{\sqrt{a+4}}{1-a} -1\right) -3 <0\implies x^4\left(\frac{\sqrt{a+4}}{1-a} -1\right) < 3/5 $$
How can I proceed... | This holds true for every $x$ only if $$\frac{\sqrt{a+4}}{1-a} -1\le 0$$or $${\sqrt{a+4}\over 1-a}\le 1$$Surely $a>1$ is one region of answer. For $a<1$ we have $$\sqrt {a+4}\le 1-a\to a+4\le a^2-2a+1\to a^2-3a-3\ge 0\to\\a\ge {3+\sqrt {21}\over 2}\\a\le {3-\sqrt {21}\over 2}$$therefore the region of the answer is $$[-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solution to infinite product $\prod_{p-primes}^{\infty} \frac{p}{p-1}$ I want to find the $\prod_{p-primes}^{\infty} \frac{p}{p-1}$. This question stems from a question from amc 12a 2018. It goes as follows: Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum... | An alternative approach: $\frac{p}{p-1}\geq 1+\frac{1}{p}$ gives
$$ \prod_{p}\frac{p}{p-1}\geq \prod_{p}\left(1+\frac{1}{p}\right) =\!\!\!\!\!\!\!\! \sum_{\substack{n\geq 1\\n\text{ squarefree}}}\!\!\!\!\!\!\!\frac{1}{n} $$
but since the set of squarefree numbers has a positive density in $\mathbb{N}$ ($\frac{6}{\pi^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why is my solution incorrect for solving these quadratic equations? $$\frac2x -\frac5{\sqrt{x}}=1 \qquad \qquad 10)\ \frac3n -\frac7{\sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable,
$y = \sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get:
$\displaystyle \fr... | You have to pay attention to your domain. In the first equation you get a positive and a negative value for $y$, while $\sqrt{x}$, which is your substitution, can only be positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
For pair of st. lines , length of line joining feet of perpendiculars from $(f,g)$ to them is$\sqrt {4.\frac {(h^2-ab)(f^2+g^2)}{(a-b)^2+4h^2}}$ Consider a pair of straight lines through the origin,
$$ax^2+2hxy+by^2=0$$
This can be written as,
$$y=m_{1,2}x$$
where $m_{1,2}=-\frac {a}{h±\sqrt {h^2-ab}}$. Now, suppose a ... | Write the lines like $(Ax+By)(A'x+B'y)=AA'x^2+2\frac{AB'+A'B}{2}xy+BB'y^2=0$
Then the system of one line and its perpendicular through $(f,g)$ is $\langle Ax+By,Ay-Bx-Ag+Bf\rangle$ and has solution $(x,y)=(\frac{B}{A^2+B^2}(Bf-Ag),\frac{A}{A^2+B^2}(Ag-Bf))$ and similarly for the primed, which gives the distance $$\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve differential equation $\frac{dx}{dt}=a-b x-cx(1-x)=cx^2-x(b+c)+a$ I want to solve the following first- order nonlinear ordinary differential equation:
$\frac{dx}{dt}=a-b x-cx(1-x)=cx^2-x(b+c)+a$
where a,b and c are constants. I rewrote the equation:
$\leftrightarrow 1=\frac{1}{cx^2-x(b+c)+a}\frac{dx}{dt}\\
\leftr... | After completing the square the integral has the form:
$\int \frac{1}{c(x-\frac{b+c}{2c})^2+a-\frac{(b+c)^2}{4c}} dx=\frac{1}{c}\int \frac{1}{(x-\frac{b+c}{2c})^2+\frac{a}{c}-\frac{(b+c)^2}{4}} $
By defining $y:=x-\frac{b+c}{2c}$ and $p^2:=\frac{a}{c}-\frac{(b+c)^2}{4}$ we get:
$\frac{1}{c}\int \frac{1}{(x-\frac{b+c}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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limit of $\sqrt{x^6}$ as $x$ approaches $-\infty$ I need to solve this limit:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}$$
The answer is $-3$, but I got 3 instead. This is my process:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6(9-\frac {5}{x^2})}... |
I've been told that in the third step the $\sqrt{x^6}$ should be equal $\textbf{-}\sqrt{x^3}$, but I didn't understand why.
Because $\sqrt{a^2}=a$ is only true if $a \ge 0$; for $a \le 0$, you have $\sqrt{a^2}=-a$.
You can summarize this as follows (and remember by heart!), for all $a$ you have:
$$\boxed{\sqrt{a^2}=|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Primes dividing $x^2+xy+y^2+1$
Is there infinitely prime numbers $p$ such that $p$ divides $x^2+xy+y^2+1$ for some integers $x,y \in \Bbb Z$ ?
I can show that every prime $p$ divides $x^2+y^2+1$ for some integers $x,y \in \Bbb Z$, see Show that for any prime $ p $, there are integers $ x $ and $ y $ such that $ p|... | If $p\ne2$ then $p\mid(x^2+xy+y^2+1)$ iff $p\mid(4x^2+4xy+4y^2+4)$
that is iff $p\mid((2x+y)^2+3y^2+4)$. The same trick as before proves that there
are $y$ and $z$ with $p\mid(z^2+3y^2+4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{n\rightarrow\infty} \left(\frac 1e\left( 1+\frac1n+\frac c{n^2}\right)^n\right)^n$
Find
$$\lim_{n\rightarrow\infty} \left(\frac 1e\left( 1+\frac1n+\frac c{n^2}\right)^n\right)^n.$$
According to Wolfram the limit is $e^{c-1/2}$. I have simplified the expression to
$$ \exp \lim \left[ n^2 \ln \left(1 + \f... | Put $x=\dfrac 1n+\dfrac c{n^2}$, then
\begin{align*}
n^2f(x)-n&=\color{blue}{n^2}\left(\dfrac 1n+\color{blue}{\dfrac c{n^2}}-\frac12\left(\dfrac 1n+\dfrac c{n^2}\right)^2+o(\frac1{n^2})\right)-n\\
&=\color{blue}c-\frac12+o(1).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\frac{\pi^2}9$ I am in the middle of proving that
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac{\pi^2}{18}$$
And I have reduced the series to
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac12\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt$$
But this integral is giving ... | Firstly observe that, for $x$ real,
\begin{align}(1-x)^2-(1-x)+1&=(1-2x+x^2)-1+x+1\\
&=x^2-x+1
\end{align}
\begin{align}J&=\int_0^1 \frac{\ln(x^2-x+1)}{x(x-1)}\,dx\\
&=-\int_0^1 \frac{\ln(x^2-x+1)}{1-x}\,dx-\int_0^1 \frac{\ln(x^2-x+1)}{x}\,dx
\end{align}
In the first integral perform the change of variable $y=1-x$,
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Proving a three variables inequality Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)\ge 64$$
My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c \ge 63$$
I applied Cauchy-Schwarz on $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ getting that $a+b+c\ge9$.
Then I al... | I think you mean $$a,b,c>0$$ in this case we have
$$\frac{a+b+c}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ so $$a+b+c\geq 9$$ and we get also $$\frac{bc+ac+ab}{3}\geq \sqrt[3]{(abc)^2}$$ so $$abc\geq 27$$ and $$ab+ac+bc\geq 27$$ putting things together we have
$$(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1\geq 27... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3026163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Find limit $\lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$ $\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}\displaystyle -\sqrt{2x^{4}}\right)$
$\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^... | $$
\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}-\sqrt{2x^4}=
\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4}}}=
\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}\cdot\frac{x^{2}\sqrt{x^{4} +1} +x^{4}}{x^{2}\sqrt{x^{4} +1} +x^{4}}\\=\frac{x^4}{\big(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Minimizing distance from a point to a parabola Problem: The point on the curve $x^2 + 2y = 0$ that is nearest the point $\left(0, -\frac{1}{2}\right)$ occurs at what value of y?
Using the distance formula, I get my primary equation: $L^2 = (x-0)^2 + \left(y-\left(-\frac{1}{2}\right)\right)^2$
However, when using the se... | Note that $x\in (-\infty,+\infty), \ y\in (-\infty, 0]$. Your approach $2$ has implicit constraint on $y\le 0$ and you must check the border too. Hence $y=0$ is an optimal point.
Also note for $y=\frac 12$, $x^2=-1$, which gives complex roots as in the approach $1$: $x^2+1=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving this equation: $3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}$
Solve this equation:
$$3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}\qquad (1)$$
I tried to make both sides of the equation have a same base and I started:
$$(1)\Leftrightarrow 3^{\log_{4}x}.\sqrt{3}+ \frac{3^{\lo... | You may also continue as follows:
$$\begin{eqnarray*}
4 \cdot 3^{2 \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
4 \cdot 9^{\log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
4 \cdot 4^{\log_4{9} \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
(\sqrt{x})^{\log_4{9} -1} &= & \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Derivative of piecewise function with $\sin\frac{1}{x}$ term I was going through my calculus book, and I am not sure I understand this part
$f(x) = \begin{cases} \frac{x^2}{4}+x^4\sin(\frac{1}{x}) &\text{if $x\neq0$ } \\ 0 &\text{if $x=0$ } \end{cases}$
$ f'(x) = \begin{cases} \frac{x}{2}-x^2\cos(\frac{1}{x})+4x^3\sin(... | hint
$$f''(0)=\lim_{x\to 0,x\ne 0}\frac{f'(x)-f'(0)}{x-0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)$ As part of a recent question I posted, I decided to try and generalise for a power of $2$ to any $r \in \mathbb{R}$. As part of the method I took, I had to solve the following integral:
\begin{equat... | NOT A FULL SOLUTION:
I've been working with special cases of the integral.
Here we will consider $r = 2m$ where $m \in \mathbb{N}$. In doing so, we observe that the roots of the denominator are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:
\begin{align}
x^{2m} + 1 = 0 \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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Proof by mathematical induction that, for all non-negative integers $n$, $7^{2n+1} + 5^{n+3}$ is divisible by $44$. I have been trying to solve this problem, but I have not been able to figure it out using any simple techniques. Would someone give me the ropes please?
Prove for $n=1$
$7^3 + 5^4 = 968 = 44(22)$
Assume ... | Base case : n=1✓
Hypothesis : $44$ divides $7^{2n+1} +5^{n+3}$.
Step $n+1$:
$7^{2(n+1)+1}+ 5^{(n+1)+1}=$
$7^{2n+1}7^2+ 5^{n+1}5=$
$7^{2n+1}(44+5)+5^{n+1}5=$
$(44)7^{2n+1} +5(7^{2n+1}+5^{n+1}).$
First term divisible by $44$, so is the second term by hypothesis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the geometric locus $z \in \mathbb C$ so that $\frac{z+2}{z(z+1)}\in \mathbb R$
Find the geometric locus of the set of $z \in \mathbb C$ so that
$$\frac{z+2}{z(z+1)}\in \mathbb R$$
Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)
My attempt: With the notation $z=a+bi$, the solution pr... | Using
$$
z = \sqrt{x^2+y^2}e^{\arctan\frac yx}
$$
we have
$$
\frac{z+2}{z(z+1)} = \frac{\rho_1 e^{i\phi_1}}{\rho_2 e^{i\phi_2}\rho_3 e^{i\phi_3}}
$$
and we seek for
$$
\arctan\frac{y}{x+2}-\arctan\frac yx -\arctan\frac{y}{x+1} = 0
$$
then
$$
\tan\left(\arctan\frac{y}{x+2} -\arctan\frac{y}{x+1}\right) = \frac yx
$$
or
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Global extrema of $f(x,y)= e^{-4x^2-9y^2}(2x+3y)$ on the ellipse $4x^2+9y^2 \leq 72$ I'm asked to find the global extremas of the following function :$$f(x,y)= e^{-4x^2-9y^2}(2x+3y)$$ on the ellipse $4x^2+9y^2 \leq 72$
So for the inside, I have $\nabla f=(2e^{-4x^2-9y^2}+(-16x^2-24xy)e^{-4x^2-9y^2}, 3e^{-4x^2-9y^2}+(-3... | From the lagrangian
$$
L(x,y,\mu,\epsilon) = f(x,y) + \mu(g(x,y)-72+\epsilon^2)
$$
here $\epsilon$ is a slack variable to transform the inequality constraint into an equality.
The stationary conditions are
$$
\left\{
\begin{array}{rcl}
8 \lambda x-8 (2 x+3 y) x+2& = & 0 \\
18 \lambda y-18 (2 x+3 y) y+3& =& 0 \\
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a diagonal form of the quadratic form $f(x) = \sum_{i = 1}^nx_i^2 + \sum_{i < j}x_ix_j$ Find a diagonal form of the quadratic form
$$f(x) = \sum_{i = 1}^nx_i^2 + \sum_{i < j}x_ix_j.$$
It turned out to be such a problem:
How to change quadratic form $f(x) = \sum_{i = 1}^nx_i^2 + \sum_{i < j}x_ix_j$ into diagonal f... | This is too long for the comment box.
The essence of your exercise is given in the following example.
Suppose we have $f(x, y, z) = x^2 + y^2 + z^2 + xy + xz + yz.$ We complete the square by getting rid of one of the variables, say we pick the first one ($x$ in my notation). So you get,
$$\begin{align*}
f(x, y, z) &= (... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$
Find the value of this :$$1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$$
Try: We can write the above series as
$${S} = \int^{1}_{0}\bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+\cdots\big... | The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+\sqrt 2 x+1)(x^2-\sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $\frac{1}{4}\left(2\tan^{-1}x+\sqrt{2}\left(\tan^{-1}\left(\sqrt{2}x+1\right)-\tan^{-1}\left(1-\sqrt{2}x\right)\right)\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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An inequality for convex function from $\mathbb R$ to $\mathbb R$ If $f(x): \mathbb R \to \mathbb R$ is a convex function, prove that
$$f(x_1) + f(x_2) + f(x_3) + 3 f(\frac{x_1 + x_2 + x_3}{3}) \geq 2 f(\frac{x_1 + x_2}{2}) + 2 f(\frac{x_1 + x_3}{2}) + 2 f(\frac{x_2 + x_3}{2})$$
What I have tried:
If two of $x_1$, $x_2... | A proof of Popoviciu's inequality follows from Jensen's Inequality for convex functions. Wlog, assume $x \le y \le z$ there are two possibilities. Either $\displaystyle y < \frac{x+y+z}{3}$ or the reverse.
In the first case, $\displaystyle \frac{x+y+z}{3} < \frac{x+z}{2}< z$ and $\displaystyle \frac{x+y+z}{3} < \frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Simplify the determinant of a $4 \times 4$ matrix. I have to find the determinant of the following 4x4 matrix:
$\quad A=\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bmatrix}$
So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$det\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bma... | It should be: $$\begin{vmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{vmatrix}\rightarrow \begin{vmatrix}3&0&1&0\\0&2&0&0\\0&0&\color{red}{3-{1\over 3}}&0\\0&0&0&-4\end{vmatrix}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can $3p^4-3p^2+1$ be square number? I know $4p^4+1$ can't be square number, cause $4p^4<4p^4+1<4p^4+4p^2+1$ for all natural number p.
But I don't know the way to prove $3p^4-3p^2+1$ can(not) be square number. Is there a well known way to prove it?
| Partial solution if $p$ is prime.
Write $$3p^4-3p^2+1=n^2\implies 3p^2(p^2-1) = (n-1)(n+1)$$
If $p\ne 2$ (which is not a solution) then $p^2\mid n-1$ or $p^2\mid n+1 $
First case: If $p^2\mid n-1$ then $n+1\mid 3p^2-3$ so $ n-1= p^2k$ and $n+1\leq 3p^2-3$.
If $k\geq 3$ then $$3p^2-3\geq n+1 >n-1 \geq 3p^2$$ which is i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047428",
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"source": "stackexchange",
"question_score": "1",
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How to evaluate $\int \frac{x^3}{\sqrt {x^2+1}}dx$ Evaluate $$\int \frac{x^3}{\sqrt {x^2+1}}dx$$
Let $u={x^2}+1$. Then $x=\sqrt {u-1}$ and $dx=\frac{1}{2\sqrt{u-1}}du$.
Therefore, $$\int \frac{(u-1)\sqrt{u-1}}{\sqrt u}\frac{1}{2\sqrt {u-1}}du$$
$$=\frac{1}{2}\int {\sqrt u}-{\frac {1}{\sqrt u}du}$$
$$={\frac{2}{3}}{{(x^... | You made a small mistake on the last line; integrating $\frac{1}{2}u^{1/2}-\frac{1}{2}u^{-1/2}$ should give $\frac{1}{3}u^{3/2}-u^{1/2}+C$, not $\frac{2}{3}u^{3/2}-\frac{1}{2}u^{1/2}+C$. When you want to double-check an antiderivative, it's worth differentiating it to see if you get the expected integrand.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $\log_5 (2x+1)=\log_3 (3x-3)$. I am trying to resolve the equation $$\log_5 (2x+1) = \log_3 (3x-3)$$ and then of sketch the functions $y=\log_5 (2x+1)$ and $y=\log_3 (3x-3)$ get the solution $x=2$. There is an method that do not use the graphic method? Thanks for your suggestions. Any ideas will be appreciated.... | If you want a purely algebraic solution, although rather tedious, you can solve the equation as follows:
$$\log_5(2x+1) = \frac{\log_3(2x+1)}{\log_3 5} \tag{1}$$
$$\frac{\log_3(2x+1)}{\log_3 5} = \log_3(3x-3) \tag{2}$$
$$\left(3^{\log_3(2x+1)}\right)^{\frac{1}{\log_3 5}} = 3x-3 \tag{3}$$
$$(2x+1)^{\frac{1}{\log_3 5}} =... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given c... | $|z| =|z+1/z-1/z| \le$
$ |z+1/z| +|1/z| =1+|1/z|;$
$r:=|z|$;
$r-1/r \le 1;$
$r^2-r -1\le 0;$
$(r-1/2)^2 -1/4-1\le 0;$
$r-1/2 \le (1/2)√5;$
$r \le (1/2)(1+√5)$.
Is the maximum attained?
| {
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"url": "https://math.stackexchange.com/questions/3048864",
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"source": "stackexchange",
"question_score": "1",
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How to get the value of $a + b + c$? $(0 \leq a < b < c) \in Z$,
$a + b + c + ab + ac + bc + abc = 1622$
$a + b + c = ?$
I've assumed that $a = 0$, by that we can get rid of this part $ab + ac + abc$
Now $bc + b + c = 1622$.
But I found that was useless and got stuck.
| Note that $$1+a+b+c+ab+bc+ca+abc=(1+a)(1+b)(1+c)$$
$$(1+a)(1+b)(1+c)=1623=1\cdot3\cdot541$$
$$\implies a=0,\ b=2,\ c=540.$$
Note that $0\leq a<b<c$. Those three numbers, 1, 3 and 541 are the only possible factorisation that could satisfy this criteria
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I solve $\lim \limits_{x \to \frac{π}{3}} \frac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$ Alright, I know, there are easier ways to solve this, like L'hopitals Rule etc.
But I'm not solving it for the answer, just doing it for the fun so I tried using substitution method.
Put $t= x- \dfrac{π}{3}$
$\lim \limits_{x... | $\lim \limits_{x \to \frac{π}{3}} \dfrac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}=2\lim \limits_{x \to \frac{π}{3}} \dfrac{\sin x - \sin \cfrac \pi 3}{\cos \frac{3x}{2}}$
$=4\lim \limits_{x \to \frac{π}{3}} \dfrac{\cos (\cfrac x2+\cfrac \pi 6) \sin (\cfrac x2 - \cfrac \pi 6)}{\sin (\frac \pi 2-\frac {3x}{2})}$
$=4\lim \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Prove that $\sin a + \sin b + \sin(a+b) = 4 \sin\frac12(a+b) \cos \frac12a \cos\frac12b$ Helping my son with his trigonometry review. We know that
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$
We also know that
$$\sin a \cos b = \frac12 \left(\sin(a-b) + \sin(a+b)\right)$$
And we have
$$\sin a + \sin b = 2 \sin\fra... | \begin{align}
\sin a + \sin b + \sin (a+b)&=2 \sin\left(\frac{a+b}2\right) \cos\left(\frac{a-b}2\right)+2 \sin\left(\frac{a+b}2\right) \cos\left(\frac{a+b}2\right)\\
&=2 \sin\left(\frac{a+b}2\right)\left(\cos\left(\frac{a-b}2\right)+\cos\left(\frac{a+b}2\right)\right)
\end{align}
Now use the identity $$\cos A +\cos B=... | {
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"source": "stackexchange",
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Finding sum multinomial
I did put x=$w, w^2 ,i ,-i$ but nothing of type is fetting
formed. How come 1/2 is remaining constant.
That means because of some substitution,
$2a_o= a_1+ a_2$ is happening.
Also tried putting x=ix.
| We consider
\begin{align*}
f(x)&:=(x^{2016}+x^{2008}+2)^{2010}=\sum_{j=0}^na_jx^j\\
g(x)&:=\frac{1}{3}\left(f(x)+f(xe^{2\pi i/3})+f(xe^{4\pi i/3}\right)=\sum_{{j=0\ }\atop{\ \ j\equiv 0(3)}}^n a_jx^j
\end{align*}
We obtain
\begin{align*}
&\color{blue}{a_0-\frac{1}{2}a_1-\frac{1}{2}a_2+a_3-\frac{1}{2}a_4-\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054900",
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"source": "stackexchange",
"question_score": "2",
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Remainder of the polynomial: What is wrong with this approach?
Find the remainder when $P(x)=5x^6 + x^5 - 2x^3 - x^2 + 1$ is divided
by $x^2+1$.
$$
P(x)=(x^2+1)\cdot Q(x)+R(x)
$$
When $x^2=-1$,
$$
P(x)=-5+x+2x+1+1 = 3x-3
$$
Exactly, why the method above, does not work for the following question?
Find the remainder... | The first method does work:
When $x^3=x$, we have $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x= x+x+x+x+x = 5x$.
When $x^3=x$, we have $P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)= x(x^2+x^2+x^2+x^2+1) = x(4x^2+1)=4x^3+x=5x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving used Real Based Methods: $\int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt$ In working on integrals for the past couple of months, I've come across different cases of the following integral:
\begin{equation}
I\left(x,a,k,n,m\right) = \int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt
\end{equation}
Where $x,a\in \m... | NOT A SOLUTION:
I've found some special cases on this site that I will list (this will evolve as I find more special (but generalised) cases:
*
*Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$
*Evaluate the integral $ \int _0^{+\infty} \frac{x^m}{(a+bx^n)^p}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Showing $\sum_{k=1}^{nm} \frac{1}{k} \approx \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{m} \frac{1}{k}$ Since $\log(nm) = \log(n) + \log(m)$, and $\sum_{k=1}^n \frac{1}{k} \approx \log n$ for large $n$, we would expect that $$\sum_{k=1}^{nm} \frac{1}{k} \approx \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{m} \frac{1}{k}$$
wh... | Let $$S = \sum_{k=1}^n \frac{1}k + \sum_{k=1}^m \frac{1}k.$$
Note that we can write $\sum_{k=1}^{nm} \frac{1}k$ as
$$\begin{align*}
1 + \cdots + \frac{1}n \\
\frac1{n+1} + \cdots + \frac{1}{2n} \\
\cdots \\
\frac{1}{n(m-1)} + \cdots + \frac{1}{nm}
\end{align*} $$
Label the rows of the above table from $1$ upto $m$. Not... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Several ways to prove that $\sum\limits^\infty_{n=1}\left(1-\frac1{\sqrt{n}}\right)^n$ converges I believe there are several ways to prove that $\sum\limits^{\infty}_{n=1}\left(1-\frac{1}{\sqrt{n}}\right)^n$ converges. Can you, please, post yours so that we can learn from you?
HERE IS ONE
Let $n\in\Bbb{N}$ be fixed su... | Noting that $\ln(\tfrac{1}{1-x}) = x + x^2/2 + O(x^3)$,
\begin{align*}
\frac{\left(\,1-\frac{1}{\sqrt{n}}\,\right)^n}{e^{-\sqrt{n}}}
%
&= \frac{\exp \left(\, -n\ln \frac{1}{1-\frac{1}{\sqrt{n}}}\,\right)}{e^{-\sqrt{n}}}\\
%
&= \frac{\exp \left(\, -n\left(\tfrac{1}{\sqrt{n}} + \tfrac{1}{2n} + O\left(\tfrac{1}{n^{3/2}}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060671",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Why is $\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ $15\sqrt{5}$ and not $15\sqrt[4]{25}$? I have an expression I am to simplify:
$$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$
I arrived at $15\sqrt[4]{25}$. My textbook tells me that the answer is in fact $15\sqrt{5}$. Here is my thought process:
$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}... | It turns out that $\sqrt[4]{25}=\sqrt{5}$. This is because $25=5^2$, so that $\sqrt[4]{25}=\sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=\sqrt{5}.$ So, you are correct, as is the book.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Evaluating an improper integral $\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$ I tried to solve the integral: $$\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$$
using $ x = \sqrt{\tan(u)}$ and $dx = \frac{ \sec^2(u)}{2\sqrt{\tan(u)}} du,$
but I ended up with an even worse looking integral $$ \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\... | We could do it with contour integration.
take the contour from 0 to R along the real axis.
$\int_0^R \frac {x^2}{(x^4+1)^2} \ dx$
The quater circle.
$\int_0^{\frac \pi 2} \frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) \ dt$
$\lim_\limits{R\to \infty} \frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) = 0$
And down the imaginar... | {
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"url": "https://math.stackexchange.com/questions/3061224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Changing variable - Integration goes wrong. I was trying to do the integration
$$I=\frac{\pi}{2}\int_0^\pi \frac{dx}{a^2\cos^2x + b^2\sin^2x}$$
If I divide throughout by $\cos^2x$ and use substitution ($t=\tan x$),
I obtain$$I=\frac{\pi}{2}\int_0^0\frac{dt}{a^2+(bt)^2} $$
This would be evaluated to be $0$.
But however ... | The reason why your substitution doesn't work is due to the singularity at $\frac {\pi}2$. Indeed, $\tan\left(\frac {\pi}2\right)$ isn't defined. However, you can remedy this by splitting up the integral along the singularity$$\begin{align*}\int\limits_0^{\pi}\frac {\mathrm dx}{a^2\cos^2x+b^2\sin^2x} & =\int\limits_0^{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove $a,b,c$ in A.P if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$
In $\Delta ABC$, if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
\sin A=\frac{2.5}{6}.\frac{36}{61}=\frac{60}{61}\\
\sin C=\frac{2.2}{5}.\frac... | You can first deduce
$$
\tan\frac{B}{2}=\tan\left(\frac{\pi}{2}-\frac{A+C}{2}\right)=\cot\frac{A+C}{2}=
\frac{1-\tan\frac{A}{2}\tan\frac{C}{2}}{\tan\frac{A}{2}+\tan\frac{C}{2}}=\frac{2/3}{37/30}
=\frac{20}{37}
$$
Therefore
$$
\sin A=\frac{2(5/6)}{1+25/36}=\frac{60}{61}
$$
Similarly,
$$
\sin B=\frac{1480}{1769}\qquad \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Logarithmic integral $ \int_0^1 \frac{x}{x^2+1} \, \log(x)\log(x+1) \, {\rm d}x $ At various places e.g.
Calculate $\int_0^1\frac{\log^2(1+x)\log(x)\log(1-x)}{1-x}dx$
and
How to prove $\int_0^1x\ln^2(1+x)\ln(\frac{x^2}{1+x})\frac{dx}{1+x^2}$
logarithmic integrals are connected to Euler-sums. In view of the last link I'... | Integrate as follows
\begin{align}
& \hspace{5mm}\int_0^1 \frac{x \ln x\ln(1+x) }{1+x^2}dx\\
&= \frac12 \int_0^1 \frac{x \ln x\ln(1-x^2) }{1+x^2}\overset{x^2\to x}{dx}
-\frac12\int_0^1 \frac{x \ln x\ln\frac{1-x}{1+x}}{1+x^2}\overset{\to x}{dx}\\
&= \frac18 \int_0^1 \frac{\ln x\ln(1-x) }{1+x}dx
-\frac14\int_0^1 \frac{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Super hard system of equations
Solve the system of equation for real numbers
\begin{split}
(a+b) &(c+d) &= 1 & \qquad (1)\\
(a^2+b^2)&(c^2+d^2) &= 9 & \qquad (2)\\
(a^3+b^3)&(c^3+d^3) &= 7 & \qquad (3)\\
(a^4+b^4)&(c^4+d^4) &=25 & \qquad (4)\\
\end{split}
First I used the identity
$$(a^2+b^2)(c^2+d^2)=(ac-bd)... | Hint: $ac=x, bc=y, ad=u, bd=v$, then the equations are
$x+y+u+v=1$
$x^2+y^2+u^2+v^2=9$
$x^3+y^3+u^3+v^3=7$
$x^4+y^4+u^4+v^4=25$
Use Newton-Girard to compute the elementary polynomials.
Then you have the polynomial $P(z)= (z-x)(z-y)(z-u)(z-v)$ with variable $z$.
Solve the quartic equation $P(z)=0$, and there you have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
"answer_id": 3
} |
Non zero solution of $3x\cos(x) + (-3 + x^2)\sin(x)=0$ How can I find exact non-zero solution of $3x\cos(x) + (-3 + x^2)\sin(x)=0$. Simple analysis and the below plot show that the equation has an infinite number of non-zero solutions.
| There's probably no closed-form solution for the real roots you're looking for, although you have a chance with the complex solutions.
Let's notice that $3x \cos x$ and $\left( x^2 - 3 \right) \sin x$ have the same trigonometric argument. That means that the Harmonic Addition Theorem can be used: we can rewrite this as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find maximum value of $\frac xy$
If $x^2-30x+y^2-40y+576=0$, find the maximum value of $\dfrac xy$.
First I completed the squares and got $(x-15)^2+(y-20)^2=7^2$, which is the equation of a circle.
I think I need to use some properties but I don't know what to do next.
| Let $\frac{x}{y}=k$.
Thus, $x=ky$ and the equation
$$k^2y^2-30ky+y^2-40y+576=0$$ or
$$(k^2+1)y^2-2(15k+20)y+576=0$$ has real roots, which says that the discriminant must be non-negative.
Id est, $$(15k+20)^2-576(k^2+1)\geq0$$ or
$$351k^2-600k+176\leq0$$ or
$$\frac{44}{117}\leq k\leq\frac{4}{3}.$$
The value $\frac{4}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Prove the identity for $\tan3\theta$
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$
Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\c... | Use that $$\tan(3x)=\frac{\tan(x)+\tan(2x)}{1-\tan(x)\tan(2x)}$$ and then
$$\tan(x+x)=\frac{2\tan(x)}{1-\tan^2(x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove all elements $x_n \geq \sqrt{2}.$ Given $x_1 = 2,$ and $$x_{n+1} = \frac{1}{2}\bigg(x_n + \frac{2}{x_n} \bigg),$$ show that for all $n \in \mathbb{N},$ $x_n \geq \sqrt{2}.$
I tried the following. Suppose, for contradiction, that $x_{n+1} < \sqrt{2}.$ Then, $$\frac{1}{2}\bigg(x_n + \frac{2}{x_n} \bigg) < \sqrt{2},... | This is immediate from AM-GM: $\frac{1}{2}\bigg(x + \frac{2}{x} \bigg)$ is the arithmetic mean of $x$ and $2/x$, which is always greater than or equal to the geometric mean $\sqrt{x\cdot\frac{2}{x}}=\sqrt{2}$ (for $x>0$).
Alternatively, if you don't know AM-GM, you can reach the same conclusion easily with a bit of cal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
We throw $5$ dice: What is the probability to have $4$ different numbers? We throw $5$ dice: What is the probability to have $4$ different numbers?
I know it is $$\frac{6\cdot 5\cdot 4\cdot 3}{6^5}.$$
I wanted to use an other argument, but it look to not work : I take $\binom{6}{4}$ numbers. Then I have $$\frac{6\cdot... | Notice that for a favorable outcome, exactly two of the five dice must display the same number, while each of the other three dice must display different numbers.
Suppose the dice are five different colors so that we can distinguish between them. There are $\binom{5}{2}$ ways for two of the five dice to display the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Convergence of $\int_0^{\frac{\pi}{2}} \frac{\exp({-1/x)}}{\sqrt{ \sin x}}\, \mathrm{d}x$ Let $$f(x)=\frac{\exp(-1/x)}{\sqrt{ \sin x}}\, \mathrm{d}x.$$
$f(x)$ seems to not have any problems at $x=\frac {\pi}{2}$, but at $x=0$.
So I should understand behavior of $f(x)$ at $x=0$ by evaluating $\lim_{x\to0+}f(x)$.But I a... | Using the continuity of the square root, we can conclude
\begin{equation*}
I
:=\lim_{x \searrow 0} \frac{e^{-\frac{1}{x}}}{\sqrt{\sin(x)}}
= \sqrt{ \lim_{x \searrow 0} \frac{e^{-\frac{2}{x}}}{\sin(x)}}
= \sqrt{ \lim_{x \searrow 0} \frac{1}{e^{\frac{2}{x}} \sin(x)}}
\end{equation*}
Now, we can apply L'Hôpital to the fol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.