Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $f(1)=1\;,f(2)=3\;,f(3)=5\;,f(4)=7\;,f(5)=9$ and $f'(2)=2,$ Then sum of all digits of $f(6)$
$(1):$ If $P(x)$ is a polynomial of Degree $4$ such that $P(-1) = P(1) = 5$ and
$P(-2)=P(0)=P(2)=2\;,$Then Max. value of $P(x).$
$(2):$ If $f(x)$ is a polynomial of degree $6$ with leading Coefficient $2009.$ Suppose
furthe... | Question 1 directly states that $P(-2)=P(0)=P(2)=2$. Why on earth are you then saying that they're roots of the polynomial?
What you can say is that $x=-2,0,2$ are roots of $P(x)=2$, or equivalently $P(x)-2=0$. So we can say that, if $P(x)$ is 4th degree, then $P(x)-2$ is as well, and hence $P(x)-2=A(x-\alpha)(x-2)x(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Domain, range and zeros of $f(x,y)=\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}$ Given the following function with two variables:
\begin{equation}
\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}
\end{equation}
I need to find a) the domain for the above function.
Can anyone give me a hint on how to find the domain in f?
... | You factored your first equation correctly. Now determine which values of $y$, $x$, yield a zero denominator: $$y^2(x-3)(x-1) = 0 \implies y = 0, \text{ or } x = 3, \text{ or } x = 1$$
So all ordered pairs of the form $(x, 0), (3, y), (1, y), \; x, y \in \mathbb R$ must be omitted from the domain. One way to see this... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{n\rightarrow\infty}(1+\frac{1}{n})(1+\frac{2}{n})^{\frac{1}{2}}\cdots(1+\frac{n}{n})^{\frac{1}{n}}$ Evaluate
$$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{\frac{1}{2}}\cdots \left(1+\frac{n}{n}\right)^{\frac{1}{n}}$$
solve:
$$ \exp\left\{\lim_{n\rightarrow\infty} \frac... | $\dfrac{\ln (1+x)}{x} = 1 -\dfrac{x}{2} + \dfrac{x^2}{3} - \dfrac{x^3}{4} + ....$, thus:
$\displaystyle \int_{0}^1 \dfrac{\ln (1+x)}{x}dx = 1 - \dfrac{1}{2^2} + \dfrac{1}{3^2} - \dfrac{1}{4^2} - ... = \displaystyle \sum_{k=1}^\infty \dfrac{1}{k^2} - 2\cdot \dfrac{1}{2^2}\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^2} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\lim_{(x,y)\rightarrow (0,0)} \frac{(xy)^3}{x^2+y^6}$ $$\lim_{(x,y)\rightarrow (0,0)} \frac{(xy)^3}{x^2+y^6}$$
I don't really know how to do, but I was trying to do like that:
$a=x$,
$b=y^2$
then I was trying to do this
$$\lim_{(x,y)\rightarrow (0,0)} \frac{ab}{a^2+b^2}$$
then I don't know no more how to d... | If the correct wording is
$$\lim_{(x,y)\rightarrow (0,0)} \frac{x(y^3)}{x^2+y^6}$$
it is a good idea to change into :
$a=x$,
$b=y^3$
then is there a "limit" to :
$$\lim_{(a,b)\rightarrow (0,0)} \frac{ab}{a^2+b^2}$$
$$\frac{ab}{a^2+b^2}=\frac{1}{\frac{a}{b}+\frac{b}{a}}$$
if $a$ and $b$ both tend to $0$ then $\frac{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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if $6^m+2^{m+n}\cdot 3^w+2^n = 332\;,$ Then $m^2+mn+n^2 = \;,$ where $m,n,w \in \mathbb{Z^{+}}$ Suppose $m.n$ are positive integers such that $6^m+2^{m+n}\cdot 3^w+2^n = 332\;,$ Then $m^2+mn+n^2 = $
$\bf{My\; Try::}$ Given $6^m+2^{m+n}\cdot 3^w+2^n=332\Rightarrow 2^m\cdot 3^n+2^{m}\cdot 3^n\cdot 3^w+2^n=332=2^2\cdot 13... | You made some arithmetic errors in what you did (the initial factorization is incorrect, plus $332 = 4\cdot 83$, not $4\cdot 133$) , but you are on the right track.
$$
6^m + 2^{m+n}\cdot 3^w + 2^n = 332 \Rightarrow
2^m\cdot 3^m + 2^m\cdot 2^n \cdot 3^w + 2^n = 332.
$$
Now assuming $m\le n$ (if it doesn't work, you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Evaluating $I_{\alpha}=\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^\alpha}dx$ using complex analysis Again, improper integrals involving $\ln(1+x^2)$
I am trying to get a result for the integral $I_{\alpha}=\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^\alpha}dx$ - asked above link- using some complex analysis, however, I couldn't f... | Let $\alpha = 1 + p$. Then we have
$$
\int_0^{\infty}\frac{\ln(1+x^2)}{x^{1+p}}dx
$$
where $0<p<2$. Let $u = \ln(1+x^2)$ and $dv = x^{-1-p}dx$. Then $du = \frac{2x}{1+x^2}$ and $v = \frac{-1}{px^p}$.
$$
uv\big|_0^{\infty}\to 0
$$
for $p\in(0,2)$.
$$
\int_0^{\infty}\frac{\ln(1+x^2)}{x^{1+p}}dx=\frac{2}{p}\int_0^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Trigonometric substitution in the integral $\int x^2 (1-x^2)^{\frac{9}{2}} \ \mathrm dx$ I'm trying to solve
$$\int_{-1}^{1} x^2(1-x^2)^{\frac{9}{2}} \, dx$$
The hint said to use the substitution $x=\sin y$
I got $$\int_{-\pi/2}^{\pi/2} \sin^2y \cos^{\frac{11}{2}} y \,dy$$
|
$$x=\sin y$$
$$\int x^2 ( 1-x^2)^{\frac{9}{2}}\,dx = \int x^2 (\sqrt{1-x^2})^9 \,dx= \int \sin^2 y (\sqrt{1-sin^2 y})^9 \cos y\, dy $$
$$= \int \sin^2 y (\sqrt\cos^2 y)^9 \cos y\, dy= \int \sin^2 y (\cos y)^9 \cos y\, dy$$
$$\int \sin^2 y \cos^{10} y \,dy$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$ using complex analysis How do I compute
$$\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$$
What I am doing is take
$$f(z)=\frac{(\log z)^2}{1+z^2}$$
and calculating
$\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{1+z^2}$
which came out to be $\dfrac{\pi}{2}-\df... | We have
$$\int_1^{\infty} \dfrac{\ln^2(x)}{1+x^2}dx = \int_1^0 \dfrac{\ln^2(1/x)}{1+1/x^2} \dfrac{-dx}{x^2} = \int_0^1 \dfrac{\ln^2(x)}{1+x^2}dx$$
Hence, our integral is
$$I = 2\int_0^1 \dfrac{\ln^2(x)}{1+x^2}dx$$
From complex analysis, since $\dfrac1{1+z^2}$ is analytic in the open unit disk, we have $$\dfrac1{1+z^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
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Prove that $ \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+\cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$ for $n\in \mathbb N$ I want to prove that if $n \in \mathbb N$ then $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}.$$
I think I am stuck on two fronts. First, I don't... | We'll use induction to prove this
given statement is true for $n=1$
then we assume that it's true for $n=k$
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{k}{(k+1)!} = 1 - \frac{1}{(k+1)!}$$
and then by adding $\dfrac{k+1}{(k+2)!}$ on both sides we get
$$\begin{align}
\frac{1}{2!} + \frac{2}{3!} + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Proving and Finding a limit I need to find the following limit and prove using the definition of limits.
$$\lim_{x\to1} {x \over x+1} = \frac 1 2$$.
Following the definition:
$$\forall \epsilon \exists \delta : \lvert x - c \rvert < \delta \Rightarrow \lvert F(x) - L \rvert < \epsilon$$
$$\left\lvert \frac{x}{x+1} - ... | Showing the versatility of this problem.
$$\begin{align}|x - 1| < \delta \leq \frac{1}{2} &\Leftrightarrow -\frac{1}{2} < x - 1 < \frac{1}{2} \Leftrightarrow \frac{1}{2} < x < \frac{3}{2} \\ &\Leftrightarrow \frac{3}{2} < x + 1 < \frac{5}{2} \Rightarrow \frac{2}{5} < \frac{1}{x+1} < \frac{2}{3} \end{align}$$
In partic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum GCD of two polynomials Consider $f(n) = \gcd(1 + 3 n + 3 n^2, 1 + n^3)$
I don't know why but $f(n)$ appears to be periodic. Also $f(n)$ appears to attain a maximum value of $7$ when $n = 5 + 7*k $ for any $k \in \Bbb{Z}$.
Why?
And how would one find this out?
Given two polynomials $p_1(n)$ and $p_2(n)$, how d... | If integer $d$ divides both $1+n^3,1+3n+3n^2$
$d$ must divide $n(1+3n+3n^2)-3(1+n^3)=3n^2+n-3$
$d$ must divide $1+3n+3n^2-(3n^2+n-3)=2n+4$
$d$ must divide $3n(2n+4)-2(1+3n+3n^2)=6n-2$
$d$ must divide $3(2n+4)-(6n-2)=14$
So, $(1+n^3,1+3n+3n^2)$ must divide $14$
Now, $3n^2+3n+1=6\dfrac{n(n+1)}2+1\equiv1\pmod2\implies 3n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$ How to solve this integral?
$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$
my attempt
$$
\int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}=
\int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\
\r... | The natural course of action whenever you see $x^2 + y^2$ in a multiple integration problem is to convert to polar coordinates.
Set $x = r\cos\theta$, $y = r\sin\theta$, $0 \leq \theta <2\pi$. Then $dx~dy = r~dr~d\theta$, and the integrand becomes
$$
\frac{r}{(r^2 + a^2)^{3/2}}~dr~d\theta.
$$
This integral doesn't depe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Sinha's Theorem for Equal Sums of Like Powers $x_1^7+x_2^7+x_3^7+\dots$ Sinha’s theorem can be stated as, excluding the trivial case $c = 0$, if,
$$(a+3c)^k + (b+3c)^k + (a+b-2c)^k = (c+d)^k + (c+e)^k + (-2c+d+e)^k\tag{1} $$
for $\color{blue}{\text{both}}$ $k = 2,4$ then,
$$a^k + b^k + (a+2c)^k + (b+2c)^k + (-c+d+e)^k ... | (Too long for a comment.)
I simplified your expression and found they are ternary quadratic forms. (Why didn't you just simplify them? Maple and Mathematica can do it easily.) So,
$$R^n+Q^n+T^n = X^n+Y^n+Z^n,\quad for\; n =2,4\tag{1}$$
$$\begin{align}R =& -2 k^2 - 2 k s + s^2 + 3 k t - t^2\\
Q =&\; k^2 - 2 k s - 2 s^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Irrational number inequality : $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$ it is easy and simple I know but still do not know how to show it (obviously without simply calculating the sum but manipulation on numbers is key here.
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$$
| In general, for $n \ge 3$,
$$\sum_{k=1}^{n} \frac{1}{\sqrt{k}} \gt 2(\sqrt{n+1}-1) \gt \sqrt{n}$$
by using
$$\frac{1}{\sqrt{n}} = \frac{2}{2\sqrt{n}} \gt \frac{2}{\sqrt{n+1} + \sqrt{n}} = 2(\sqrt{n+1} - \sqrt{n})$$
and having a telescopic sum.
For $n \ge 3$ we also have that
$$ 2\sqrt{n+1} - 2 = \sqrt{n+1} + (\sqrt{n+1... | {
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"url": "https://math.stackexchange.com/questions/1037112",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find equation of Tangent line at $(4, 1)$ on $5y^3 + x^2 = y + 5x$ Can someone help me find equation of tangent line at $(4, 1)$ on
$5y^3 + x^2 = y + 5x$
$Y=f(x)$
I dont know how to isolate the $Y$
| $$\dfrac{d}{dx}5y^3 + x^2 = y + 5x$$
$$=y'15y^2 + 2x = y' + 5$$
$$=y'15y^2-y' = -2x + 5$$
$$=y'(15y^2-1) = -2x + 5$$
$$=y' = \dfrac{-2x + 5}{(15y^2-1)}$$
Plugging in the point $(4,1)$ gives
$$y'=-\dfrac{3}{14}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving of $\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{144n^2-144n+35}$ This is a homework for my son, he needs the proving.I tried to solve it by residue theory but I couldn't.
$$\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{14... | Another way to evaluate the sum is to use this. Note that
$$ 576n^2 - 576n + 95=24^2(n^2-n+\frac{95}{24^2})=24^2[(n-\frac{1}{2})^2+(\frac{7i}{24})^2] $$
and
$$ 144n^2 - 144n + 35=12^2(n^2-n+\frac{35}{12^2})=12^2[(n-\frac{1}{2})^2+(\frac{i}{12})^2] $$
and hence we have
\begin{eqnarray*}
&&\sum_{n=1}^\infty \left(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Calculus - Functions How do I go about this question? Also how exactly will its graph be?
$$ f(x) = 1 + 4x -x^2 $$
$$g(x) = \begin{cases}
\max f(t) & x \le t \le (x+1) ;\quad 0 \le x < 3 \\
\min (x+3) & 3 \le x \le 5
\end{cases}
$$
Verify continuity of $g(x)$ for all $x$ in $[0,5]$
| here is a way to think of solving this problem geometrically. note that $1 + 4x - x^2$ is a parabola symmetric about $x = 2,$ opens downward, a has a local max at $x = 2, y = 5.$ think of the function $g$ as the global max on the moving window $x, x+1]$ of length $1.$ for $0 \le x \le 1,$ this max occurs at the right e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$ This is the assignment I have:
Find a closed form for the equations
$1^3 = 1$
$2^3 = 3+5$
$3^3 = 7+9+11$
$4^3 = 13+15+17+19$
$5^3 = 21+23+25+27+29$
$...$
Hints. The equations are of the form $n^3 = a1 +a2 +···+an$, where
$a_{i+1} = a_... | Let us derive the $r$ term of $1,3,7,13,21$
Let $S_r=1+3+7+13+21+\cdots+T_r$
$S_r-S_r=1+(3-1)+(7-3)+(13-7)+(21-13)+\cdots+T_r-T_{r-1}-T_r$
$\implies T_r=1+(2+4+6+8+$ up to $r-1$th term $)$
$=1+\dfrac{(r-1)}2\{2\cdot2+(r-2)2\}=1+r^2-r$
Now the $m$th row will be, $$\sum_{r=0}^{m-1}\{(m^2-m+1)+2r\}=(m^2-m+1)\sum_{r=0}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int \frac{x^4}{(x-1)(x^2+1)}dx$ Evaluation of $\displaystyle \int \frac{x^4}{(x-1)(x^2+1)}dx$
$\bf{My\; Try::}$ Let $$\displaystyle I = \int\frac{x^4}{(x-1)(x^2+1)}dx = \int \frac{(x^4-1)+1}{(x-1)(x^2+1)}dx = \int\frac{(x-1)\cdot (x+1)\cdot (x^2+1)}{(x-1)(x^2+1)}+\int\frac{1}{(x-1)(x^2+1)}dx$$
So $\disp... | Hint
Using partial fraction decomposition from the very beginning, we have $$\frac{x^4}{(x-1)(x^2+1)}=\frac{-x-1}{2 \left(x^2+1\right)}+x+\frac{1}{2 (x-1)}+1$$ $$\frac{x^4}{(x-1)(x^2+1)}=\frac{-x}{2 \left(x^2+1\right)}+\frac{-1}{2 \left(x^2+1\right)}+x+\frac{1}{2 (x-1)}+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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If $\tan\theta$, $2\tan\theta+2$, $3\tan\theta +3$ are in geometric progression then find … Problem:
If $\tan\theta$, $2\tan\theta+2$, $3\tan\theta +3$ are in geometric progression then find the value of $$\frac{7-5\cot\theta}{9+4\sqrt{\sec^2\theta -1}}$$
Solution :
Since $\tan\theta$, $2\tan\theta+2$, $3\tan\thet... | Simply because if $\tan \theta = -1$, then the "progression" becomes $-1, 0, 0$. So we have to rule it out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can this series be put in a generalized form? I asked a similar question here But this one seems to not work out so nicely...
I started looking at the series,
$$
S = \frac{1}{2}+\frac{2}{3}-\frac{3}{4}-\frac{4}{5}+\frac{5}{6}+\frac{6}{7}-\frac{7}{8}-\frac{8}{9}+\cdots
$$
Which is equivilent to
$$
S = \sum_{k=0}^{\infty... | This is nowhere near a complete answer, so I've marked it community wiki.
$$\frac{n-1}{n} + \frac{n}{n+1} = \frac{n^2 -1}{n(n+1)} + \frac{n^2}{n(n+1)} = \frac{2n^2-1}{n^2+n}$$
This means every pair of consecutive terms in your original series that have the same sign sum to $$\pm \frac{2n^2 -1}{n^2+n}$$ for some $n \i... | {
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"url": "https://math.stackexchange.com/questions/1044041",
"timestamp": "2023-03-29T00:00:00",
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Integral of $\frac{1}{\sqrt{x^2-x}}dx$ For a differential equation I have to solve the integral $\frac{dx}{\sqrt{x^2-x}}$. I eventually have to write the solution in the form $ x = ...$ It doesn't matter if I solve the integral myself or if I use a table to find the integral. However, the only helpful integral in an in... | $\sqrt{x}=z \implies\dfrac{dx}{2\sqrt{x}}=dz$
$\displaystyle\int\dfrac{dx}{\sqrt{x^2-x}}=\displaystyle\int\dfrac{dx}{\sqrt{x(x-1)}}=2\displaystyle\int\dfrac{ dz}{\sqrt{z^2-1}}$
$z=\sec \theta \implies dz=\sec \theta \tan \theta\ d\theta$
$\therefore \displaystyle\int\dfrac{ dz}{\sqrt{z^2-1}}=\displaystyle\int{\sec \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $g(a) = b, g(b) = c, g(c) = a$
Let
$$f(x) = x^3 - 3x^2 + 1$$
$$g(x) = 1 - \frac{1}{x}$$
Suppose $a>b>c$ are the roots of $f(x) = 0$. Show that $g(a) = b, g(b) = c, g(c) = a$.
(Singapore-Cambridge GCSE A Level 2014, 9824/01/Q2)
I was able to prove that
$$fg(x) = -f\left(\frac{1}{x}\right)$$
after which I have co... | Rearranging $g(x) = 1 - \frac{1}{x},$ we obtain $x = \frac{1}{1-g(x)}$. Substitute into $f$ and we obtain $$\left(\frac{1}{1-g(x)}\right)^3 - 3\cdot\left(\frac{1}{1-g(x)}\right)^2+1.$$ The numerator of this turns out to be: $$1 - 3(1-g(x)) + (1-g(x))^3 = -(g^3(x) - 3g^2(x) +1) \equiv -f.$$ for $x\equiv g(x).$ Hence we... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Show that $x^a+x-b=0$ must have only one positive real root and not exceed the $\sqrt[a]{b-1}$ If we take the equation $$x^3+x-3=0$$ and solve it to find the real roots, we will get only one positive real roots which is $(x=1.213411662)$. If we comparison this with $\sqrt[3]{3-1}=1.259921$, we will find that $x$ is les... | If $a > 0$, you can use the Intermediate Value Theorem.
If $a < 0$, let $v$ be the minimum value of $x^a + x$ on $(0,\infty)$. Then $x^a + x - b$ has no positive real roots if $b < v$, one if $b = v$ and two if $b > v$.
If $a$ is complex, $x^a + x - b$ is real only for a discrete set of positive real $x$'s, so there ... | {
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"url": "https://math.stackexchange.com/questions/1047032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Compute $\bar z - iz^2 = 0$ $\bar z - iz^2 = 0, i = $ complex unit.
I've found 2 solutions to this, like this:
$x - iy - i(x+iy)^2 = 0$
$i(-x^2+y^2-y)+2xy+x=0$
$2xy + x = 0$ --> $y=-\frac{1}{2}$
$x^2 - y^2 + y = 0$ --> $x_1=\sqrt\frac34$ $x_2=-\sqrt\frac{3}{4}$
Solution 1: $z = \sqrt\frac34 - \frac12i $
Solution 2: $z ... | Did's suggestion is excellent: setting $z = re^{i\theta}$ we have $\bar z = re^{-i\theta}$ and $z^2 = r^2e^{2i\theta}$ whence
$re^{-i\theta} = i r^2 e^{2i\theta}; \tag{1}$
using
$i = e^{i \pi/2} \tag{2}$
yields
$r e^{-i\theta} = r^2 e^{i(2\theta + \pi/2)}; \tag{3}$
taking moduli,
$r = r^2 \Rightarrow r = 0, 1; \tag{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it w... | Non-inductive derivation:
\begin{align}
\sum_{k=1}^n(3k-2) &= \sum_{k=1}^n3k -\sum_{k=1}^n2\\
&= 3\left(\sum_{k=1}^n k\right) -2n\\
&= \frac{3(n)(n+1)}{2} - \frac{4n}{2}\\
&=\frac{3n^2-n}{2}\\
&= \frac{n(3n-1)}{2}\\
\end{align}
This, of course, relies on one knowing the sum of the first $n$ natural numbers, but that's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 2
} |
Evaluation of $ \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ Evaluation of $\displaystyle \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ where $\lfloor x \rfloor $ represent floor function of $x$.
My Try:: Here $\displaystyle f(x) = \frac{x^2}{\sin x\cdot \tan x}$ is an... | Note that the function under consideration is even and hence it is sufficient to consider $x \to 0^{+}$. Now we need to compare $x^{2}$ and $\sin x \tan x$ for $x > 0$. Let $$f(x) = x^{2} - \sin x\tan x$$ Then $f(0) = 0$ and $$f'(x) = 2x - \sin x\sec^{2}x - \sin x$$ Then we have $f'(0) = 0$. Further $$f''(x) = 2 - \sec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find limit of $\frac {1}{x^2}- \frac {1}{\sin^2(x)}$ as x goes to 0 I need to use a taylor expansion to find the limit.
I combine the two terms into one, but I get limit of $\dfrac{\sin^2(x)-x^2}{x^2\sin^2(x)}$ as $x$ goes to $0$. I know what the taylor polynomial of $\sin(x)$ centered around $0$ is… but now what do I... | Outline: If we can write down the Taylor series for $\sin^2 x$, or just the first couple of terms, we will be finished.
This can be done from $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ by ordinary squaring, treating the series for $\sin x$ casually as a "long" polynomial. Maybe it will feel better, however, to use the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Four cards are drawn without replacement. What is the probability of drawing at least two kings? Four cards are drawn without replacement. What is the probability of drawing at least two kings?
is the below my answer correct or not?!
since cards are drawn without replacement, (4/52)(3/51)(48/50)*(47/49)=1/240=0.0042
1-... | There are $\binom{4 \cdot 13}{4} = \binom{52}{4} = 270725$ cases of "drawing four cards", the ways to get $2$ out of $4$ kings is $\binom{4}{2}$, to get $3$ is $\binom{4}{3}$, for $4$ it is $\binom{4}{4}$. In all:
$$\binom{4}{2} + \binom{4}{3} + \binom{4}{4}
= 11$$
The probability is thus $\frac{11}{270725} = 4.063... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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If $g(x) = \max(y^2-xy)(0 \leq y\leq 1)\;,$ Then minimum value of $g(x)$ If $g(x) = \max\limits_{0 \leq y\leq 1}(y^2-xy)$, then minimum value of $g(x)$
$\bf{My\; try::}$ We can write $\displaystyle f(y) = y^2-xy = y^2-xy+\frac{x^2}{4}-\frac{x^2}{4} = \left(y-\frac{x}{2}\right)^2-\frac{x^2}{4}$
Now when $y=0\;,$ Then e... | It seems like you picked $y=x/2$ as a possible maximum value of $\left(y-\frac{x}{2}\right)^2-\frac{x^2}{4}$ instead of a possible minimum. You're almost there. Here's a full argument:
To evaluate $\max\limits_{0 \leq y\leq 1}(y^2-xy)$, you need to first find its stationary point by taking $f(y) = y^2-xy$ and solving $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the value of the sum $$\arctan\left(\dfrac{1}{2}\right)+\arctan\left(\dfrac{1}{3}\right)$$
We were also given a hint of using the trigonometric identity of $\tan(x + y)$
Hint
$$\tan\left(x+y\right)\:=\:\dfrac{\tan x\:+\tan y}{1-\left(\tan x\right)\left(\tan y\right)}$$
| Let $x = \arctan \dfrac{1}{2}$ and $y = \arctan \dfrac{1}{3}$. Then, $\tan x = \dfrac{1}{2}$ and $\tan y = \dfrac{1}{3}$.
Using that formula, you can easily compute $\tan(x+y)$. Do you see how to get $x+y$ from that?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Differential equation, substitution By means of the substitute $y = v(x)Y (x)$, where $Y (x)$ is to be specified, solve the differential equation:
$$\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{y^2}{x}$$
with $y=2$ at $x=1$
Anyone can solve it for me with explaining the steps please, I have no idea how to do it. Thank you very ... | $$
y' = v'Y + vY' = -\frac{vY}{x} + \frac{v^2Y^2}{x}
$$
if we re-write
$$
v'Y = -vY' -\frac{vY}{x} + \frac{v^2Y^2}{x} = -\left(Y' + \frac{Y}{x}\right)v + \frac{v^2Y^2}{x}
$$
set the brackets term to zero i.e.
$$
Y' = -\frac{Y}{x}\implies Y(x) = \frac{C}{x}
$$
then we obtain
$$
v' = \frac{1}{Y}\frac{v^2Y^2}{x} = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Using the half/double angle formulas to solve an equation I am completely stumped by this problem: $$\cos\theta - \sin\theta =\sqrt{2} \sin\frac{\theta}{2} $$
I know that I should start by isolating $\sin\dfrac{\theta}{2}$
and end up with $$\frac {\cos\theta - \sin\theta}{\sqrt{2}} = \sin\frac{\theta}{ 2}$$
From here o... | Given : $\dfrac{\cos\theta - \sin\theta}{\sqrt{2}} = \sin\dfrac{\theta}{ 2}$
using $\sin(\frac{\theta}{2})=\sqrt{\frac{1-\cos(\theta)}{2}}$ and squaring both sides,
\begin{align*}
\frac{\sin^2(\theta)+\cos^2(\theta)-2\sin(\theta)\cos(\theta)}{2} & = \frac{1-\cos(\theta)}{2}\\
1-2\sin(\theta)\cos(\theta) & = 1-\cos(\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integrate a periodic absolute value function \begin{equation}
\int_{0}^t \left|\cos(t)\right|dt = \sin\left(t-\pi\left\lfloor{\frac{t}{\pi}+\frac{1}{2}}\right\rfloor\right)+2\left\lfloor{\frac{t}{\pi}+\frac{1}{2}}\right\rfloor
\end{equation}
I got the above integral from https://www.physicsforums.com/threads/closed-for... | We can use the Fourier series of the square wave. We have:
$$\operatorname{sign}\left(\sin\frac{s}{2}\right)=\frac{4}{\pi}\sum_{k=0}^{+\infty}\frac{1}{2k+1}\sin\frac{(2k+1)s}{2},$$
and since:
$$\frac{1}{2k+1}\int_{0}^{t}\sin\frac{t-s}{2}\sin\frac{s}{2}\sin\frac{(2k+1)s}{2}\,ds=\\ = \frac{4\cos(t/2)}{(2k-1)(2k+1)^2(2k+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that : $f(\sin x)+f(\cos x) \ge 196, \forall x\in\left(0;\frac{\pi}{2}\right)$ Given: $$f(\tan2x)=\tan^{4}x+\frac{1}{\tan^{4}x}, \forall x\in\left(0;\frac{\pi}{4}\right)$$
Prove that :$f(\sin x)+f(\cos x) \ge 196, \forall x\in\left(0;\frac{\pi}{2}\right)$
Could someone help me ?
| since
$$t=\tan{2x}=\dfrac{2\tan{x}}{1-\tan^2{x}}=-\dfrac{2}{\tan{x}-\dfrac{1}{\tan{x}}},t>0$$
so
$$\tan{x}-\dfrac{1}{\tan{x}}=-\dfrac{2}{t}$$
and
$$\tan^4{x}+\dfrac{1}{\tan^4{x}}=\left[\left(\tan{x}-\dfrac{1}{\tan{x}}\right)^2+2\right]^2-2$$
so
$$f(t)=\left[\dfrac{4}{t^2}+2\right]^2-2=\dfrac{16}{t^4}+\dfrac{16}{t^2}+2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How find $a,b$ if $\int_{0}^{1}\frac{x^{n-1}}{1+x}dx=\frac{a}{n}+\frac{b}{n^2}+o(\frac{1}{n^2}),n\to \infty$ let
$$\int_{0}^{1}\dfrac{x^{n-1}}{1+x}dx=\dfrac{a}{n}+\dfrac{b}{n^2}+o(\dfrac{1}{n^2}),n\to \infty$$
Find the $a,b$
$$\dfrac{x^{n-1}}{1+x}=x^{n-1}(1-x+x^2-x^3+\cdots)=x^{n-1}-x^n+\cdots$$
so
$$\int_{0}^{1}\dfrac... | Here's a possible to find the coefficients:
From
$$
n\int_0^1 \frac{x^{n-1}}{1+x} d\;x =\left[\frac{x^n}{1+x} \right]^1_0 + \int_0^1 \frac{x^n}{(1+x)^2} d\;x =1/2 + \int_0^1 \frac{x^n}{(1+x)^2} d\;x
$$
Since the last term tends to zero as $n$ increases, we get $a =1/2$.
The same process allows to determine $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such th... | hint: You may want to reconsider the way you split the terms at the end.
Note that $64(2^{3k - 1}) + 45(3^k) = 9(2^{3k - 1} + 5(3^k)) + 55(2^{3k - 1})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 0
} |
Generic rotation to remove Quadratic Cross-product Show that if $b\neq 0$, then the cross-product term can be eliminated from the quadratic $ax^2 + 2bxy + cy^2$ by rotating the coordinate axes through an angle $\theta$ that satisfies the equation $$
\cot{2\theta}=\frac{a-c}{2b}.
$$
I know that $ax^2 + 2bxy + ... | Let $u$ and $v$ be the rotated coordinates. That is,
$$
\begin{pmatrix} u \\ v \end{pmatrix} =
\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}.
$$
You want to be able to substitute a quadratic with no $uv$ term for the quadratic
$ax^2 + bxy + cy^2.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Question about sines of angles in an acute triangle
Let $\triangle ABC$ be a triangle such that each angle is less than $ 90^\circ $.
I want to prove that $\sin A + \sin B + \sin C > 2$.
Here is what I have done:
Since $A+B+C=180^{\circ}$ and $0 < A,B,C < 90^\circ$, at least two of $A,B,C$ are in the range 45 < x <... | I have found a simpler solution.
Observing the graph of $y = \sin x$ and $y = \frac{2}{\pi}x$, $x\in[0,\frac{\pi}{2}]$.
We can see that $\sin x\ge\frac{2}{\pi}x$.
Let $\bigtriangleup ABC$ be a triangle such that each angle is less than $90^{\circ}$. So we have:
$A\in(0,\frac{\pi}{2}),B\in(0,\frac{\pi}{2}),C\in(0,\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Avoiding extraneous solutions When solving quadratic equations like $\sqrt{x+1} + \sqrt{x-1} = \sqrt{2x + 1}$ we are told to solve naively, for example we would get $x \in \{\frac{-\sqrt{5}}{2},\frac{\sqrt{5}}{2}\}$, even though the first solution doesn't work, and then try all the solutions and eliminate the extraneo... | Try to square both sides.
$$ \sqrt{x+1} + \sqrt{x-1} = \sqrt{2x+1} \Leftrightarrow $$
$$ \Leftrightarrow \left ( \sqrt{x+1} + \sqrt{x-1} \right )^{2} = \left ( \sqrt{2x+1} \right )^{2} \Leftrightarrow $$
$$ \Leftrightarrow \left ( \sqrt{x+1} \right )^2 + 2\sqrt{x+1}\sqrt{x-1} + \left ( \sqrt{x-1} \right )^2 = 2x+1 $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rules for whether an $n$ degree polynomial is an $n$ degree power Given an $n$ degree equation in 2 variables ($n$ is a natural number)
$$a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n=y^n$$
If all values of $a$ are given rational numbers, are there any known minimum or sufficient conditions for $x$ and $y$ to have:
... | (The OP suggested a connection to this post.)
Because this question is too broad, it spreads thin and may be vague. I suggest it be limited so coefficients $a_i$ are rational, and $x,y$ are also rational.
Having said that, two nice results are discussed in Kevin Brown's website.
I. Deg 2: The sum of $24$ consecutive s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\int_{0}^{\pi/2}\frac{x\sin x\cos x\;dx}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}$ How to evaluate the following integral
$$\int_{0}^{\pi/2}\frac{x\sin x\cos x}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}dx$$
For integrating I took $\cos^{2}x$ outside and applied integration by parts.
Given answer is $\dfrac{\pi}{4ab^... | The case $a^2=b^2$ being simple, let's just consider, by symmetry, the case $a>b>0$.
Observe that
$$
\partial _x \left(\frac{1}{a^2 \cos^2x+b^2 \sin^2 x}\right)=2(a^2-b^2)\frac{\cos x\sin x}{(a^2 \cos^2x+b^2 \sin^2 x)^2}
$$
then, integrating by parts, you may write
$$
\begin{align}
I(a,b)&=\int_0^{\pi/2}\frac{x\cos x\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
Limits using Maclaurins expansion for $\lim_{x\rightarrow 0}\frac{e^{x^2}-\ln(1+x^2)-1}{\cos2x+2x\sin x-1}$ $$\lim_{x\rightarrow 0}\frac{e^{x^2}-\ln(1+x^2)-1}{\cos2x+2x\sin x-1}$$
Using Maclaurin's expansion for the numerator gives:
$$\left(1+x^2\cdots\right)-\left(x^2-\frac{x^4}{2}\cdots\right)-1$$
And the denominator... | Let's begin by writing out some of the expansions.
$$ \begin{align}
e^{x^2} &= 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6!} + \cdots \\
\ln(1 + x^2) &= x^2 - \frac{x^4}{2} + \frac{x^6}{3} + \cdots \\
\cos(2x) &= 1 - \frac{4x^2}{2} + \frac{16x^4}{4!} - \frac{64x^6}{6!} + \cdots \\
2x\sin x &= 2x^2 - \frac{2x^4}{3!} + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Sum of $k$-th powers Given:
$$
P_k(n)=\sum_{i=1}^n i^k
$$
and $P_k(0)=0$, $P_k(x)-P_k(x-1) = x^k$ show that:
$$
P_{k+1}(x)=(k+1) \int^x_0P_k(t) \, dt + C_{k+1} \cdot x
$$
For $C_{k+1}$ constant.
I believe a proof by induction is the way to go here, and have shown the case for $k=0$. This is where I'm stuck. I have look... | From the definition of $P_k(n)$ we get
$$
P_k(n) = \sum_{i=1}^n i^k \Rightarrow
P_0(n) = \sum_{i=1}^n 1 = n.
$$
We now use complete induction over $k$ to proof the statement $S(k)$
$$
P_k(x) = k \int\limits_0^x P_{k-1}(t) \, dt + C_k \, x \quad (*)
$$
Base case
For $k=1$ we have $S(1)$:
$$
1 \int\limits_0^x P_0(t) \, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
In triangle ABC, Find $\tan(A)$.
In triangle ABC, if $(b+c)^2=a^2+16\triangle$, then find $\tan(A)$ . Where $\triangle$ is the area and a, b , c are the sides of the triangle.
$\implies b^2+c^2-a^2=16\triangle-2bc$
In triangle ABC, $\sin(A)=\frac{2\triangle}{bc}$, and $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$,
$\implies \ta... | $$b^2+c^2-a^2=16\Delta - 2bc \Rightarrow \frac{b^2+c^2-a^2}{2bc}=\frac{8\Delta}{bc}-1 \Rightarrow \cos A=4\sin A-1$$
$$\Rightarrow 2\cos^2\frac{A}{2}-1=8\sin\frac{A}{2}\cos\frac{A}{2}-1 \Rightarrow \tan\frac{A}{2}=\frac{1}{4}$$
Hence,
$$\tan A=\frac{2\cdot \frac{1}{4}}{1-\frac{1}{4^2}}=\boxed{\dfrac{8}{15}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Advanced Integration techniques: Quadratic Expressions and U-Substitution Find $$\int \frac{2x-1}{x^2-6x+13}dx $$
In the final steps after a u-substitution, one arrives at $$\int \frac{2u}{u^2+4}du + \int\frac{5}{ u^2+4}du$$
The next step is arriving at $$\ln(u^2+4) + 5\arctan(\frac{u}{2}) + C$$
How does $\int$ $dx(2u... | Letting $u = x - 3$ we have that $du = dx$ and $2u + 5 = 2x -1$.
$$\begin{align}\int \frac{2x- 1}{x^2-6x + 13}dx &= \int \frac{2x- 1}{(x-3)^2 + 4}dx\\&=\int \frac{2u + 5}{u^2 + 4}du\\&=\int \frac{2u}{u^2 + 4}du + \int \frac{5}{u^2 + 4}du\\&=\ln |u^2 + 4| + \frac{5}{2}\arctan\Big(\frac{u}{2}\Big) \end{align}$$
Because ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
No. of different real values of $x$ which satisfy $17^x+9^{x^2} = 23^x+3^{x^2}.$ Number of different real values of $x$ which satisfy $17^x+9^{x^2} = 23^x+3^{x^2}.$
$\bf{My\; Try::}$Using Hit and trial $x=0$ and $x=1$ are solution of above exponential equation.
Now we will calculate any other solution exists or not.
I... |
$$17^x+9^{x^2} = 23^x+3^{x^2}$$
Clearly $0,1$ are the two roots.I would prefer rough sketching the graph:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Power series for the rational function $(1+x)^3/(1-x)^3$
Show that $$\dfrac{(1+x)^3}{(1-x)^3} =1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$$
I tried with the partial frationaising the expression that gives me
$\dfrac{-6}{(x-1)} - \dfrac{12}{(x-1)^2} - \dfrac{8}{(x-1)^3} -1$
how to proceed further on this havi... | $\dfrac{(1+x)}{(1-x)^3}^3 =\dfrac{6}{(1-x)} - \dfrac{12}{(1-x)^2} + \dfrac{8}{(1-x)^3} -1=6\sum_{n=0}^\infty x^n-12\sum_{n=0}^\infty (n+1)x^n+4\sum_{n=0}^\infty (n+1)(n+2)x^n=1+\sum_{n=1}^\infty (6-12n-12+4n^2+12n+8)x^n=1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1078092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
AlgebraII factoring polynomials equation: $2x^2 - 11x - 6$
Using the quadratic formula, I have found the zeros: $x_1 = 6, x_2 = -\frac{1}{2}$
Plug the zeros in: $2x^2 + \frac{1}{2}x - 6x - 6$
This is where I get lost. I factor $-6x - 6$ to: $-6(x + 1)$, but the answer says otherwise. I am also having trouble factoring ... | $$
2x^2-11x - 6 = 2(x -\bullet)(x-\bullet)
$$
The two "$\bullet$"s are the two roots, $6$ and $\dfrac{-1}2$. So you have
$$
2(x-6)\left(x-\frac {-1} 2 \right) = (x-6)\ \underbrace{2\left(x+\frac 1 2 \right)} = (x-6)(2x+1).
$$
So $2\left(x + \dfrac 1 2\right)$ becomes $2x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Number of $ 6 $ Digit Numbers with Alphabet $ \left\{ 1, 2, 3, 4 \right\} $ with Each Digit of the Alphabet Appearing at Least Once Find the number of 6 digit numbers that can be made with the digits 1,2,3,4 if all the digits are to appear in the number at least once.
This is what I did -
I fixed four of the digits to ... | Instead of dividing into cases, we use the Principle of Inclusion/Exclusion.
There are $4^6$ strings of length $6$ over our $4$-letter alphabet. Now we count the bad strings, in which one or more digits are missing.
There are $3^6$ strings with the digit $1$ mising, and also $3^6$ with $2$ missing, with $3$ missing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding singularities in circle of convergence of $f(z)$ and showing taylor series diverges there $$f(x)=\arctan(x)$$
I know that $$\dfrac{1}{1+x^2}=1-x+x^2-x^3+\cdots=\sum_{i=0}^\infty (-x)^i$$
Also: $$\arctan(x)=\int \dfrac{1}{1+x^2}dx = \int \sum_{i=0}^\infty (-x)^i dx = x - 0.5x^2+1/3x^3+\cdots=\sum_{i=0}^\infty \... | Observe: $$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +\cdots + (-1)^kx^{2k} + \cdots\to \int \dfrac{dx}{1+x^2} = \sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{2k+1}.$$ When $ x = \pm i$, the series equals $\pm i\cdot \displaystyle \sum_{k=0}^\infty \dfrac{1}{2k+1} = \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show without differentiation that $\frac {\ln{n}}{\sqrt{n+1}}$ is decreasing
Show that the function $\displaystyle \frac {\ln{n}}{\sqrt{n+1}}$ is decreasing from some $n_0$
My try: $\displaystyle a_{n+1}=\frac{\ln{(n+1)}}{\sqrt{n+2}}\le \frac{\ln{(n)+\frac{1}{n}}}{\sqrt{n+2}}$
so we want to show that $\ln{n}\cdot(\sq... | First, you can show very easily without calculus that $k!>4^k$ whenever $k\ge18$: we then have
$$\eqalign{k!
&=(1\times2\times3)\times(4\times5\times\cdots\times15)\times(16\times17\times18)\times19\times\cdots\times k\cr
&>(1\times1\times1)\times(4\times4\times\cdots\times4)\times(4^2\times4^2\times4^2)\times4\tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Simplification a trigonometric equation $$16 \cos \frac{2 \pi}{15} \cos\frac{4 \pi}{15} \cos\frac{8 \pi}{15} \cos\frac{14 \pi}{15}$$
$$=4\times 2 \cos \frac{2 \pi}{15} \cos\frac{4 \pi}{15} \times2 \cos\frac{8 \pi}{15} \cos\frac{14 \pi}{15}$$
I am intending in this way and then tried to apply the formula, $2\cos A \cos ... | use this well know identity
$$2\sin{x}\cos{x}=\sin{2x}$$
so
\begin{align*}\cos{x}\cos{2x}\cos{4x}\cos{8x}&=\dfrac{2\sin{x}\cos{x}\cos{2x}\cos{4x}\cos{8x}}{2\sin{x}}\\
&=\dfrac{\sin{2x}\cos{2x}\cos{4x}\cos{8x}}{2\sin{x}}\\
&=\dfrac{2\sin{2x}\cos{2x}\cos{4x}\cos{8x}}{4\sin{x}}\\
&=\dfrac{\sin{4x}\cos{4x}\cos{8x}}{4\sin{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The locus of points $z$ which satisfy $|z - k^2c| = k|z - c|$, for $k \neq 1$, is a circle Use algebra to prove that the locus of points z which satisfy $|z - k^2c| = k|z - c|$, for $k \neq 1$ and $c = a + bi$ any fixed complex number, is a circle centre $O$.
Give the radius of the circle in terms of $k$ and $|c|$.
I s... | Here are the steps
$$ \left|z - k^2c\right| = k\left|z - c\right| $$
$$ \left|z - k^2c\right|^2 = k^2\left|z - c\right|^2 $$
$$ \left(z - k^2c\right)\left(\overline{z - k^2c}\right) = k^2(z - c)\left(\overline{z-c}\right) $$
$$ \left(z - k^2c\right)\left(\overline{z} - k^2\overline{c}\right) = k^2(z - c)\left(\overline... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Is $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$? Plotting both functions $\ln(1+\frac{1}{x-1})$ and $\frac{1}{x}$ in $[2,\infty)$ gives the impression that $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$.
Is it possible to prove it?
| Expand $\ln(1+\frac{1}{x-1})=\ln(\frac{x}{x-1})$ using Taylor series. You will get:
$$\ln(\frac{x}{x-1})=\frac{1}{x-1}-\frac12\left(\frac{1}{x-1}\right)^2+O(x^3)$$
Now you have to show that:
$$\frac{1}{x-1}-\frac12\left(\frac{1}{x-1}\right)^2\geq\frac1x$$
$$\frac{2x-3}{2\cdot(x-1)^2}\geq\frac1x$$
Which is equivalent to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Compute the Jacobi bracket $R(x,y,z)=(x,y,z)$ and $\Theta(x,y,z)=(xz,yz,-(x^2+y^2))$
Show $[R,\Theta]=\Theta$
$[R,\Theta]=(R\cdot\nabla)\Theta-(\Theta\cdot\nabla)R$
$[R,\Theta]=(\frac{d}{dx}x+\frac{d}{dy}y+\frac{d}{dz}z)(xz,yz,-(x^2+y^2))-(\frac{d}{dx}xz+\frac{d}{dy}yz-\frac{d}{dz}(x^2+y^2)(x,y,z)$
$=((\frac{d}{dx}x+\f... | Thanks to Ted Shifrin's post, I have found the mistakes I was making. Here is the correct working:
$R(x,y,z)=(x,y,z)$, $\Theta(x,y,z)=(xz,yz,-(x^2+y^2))$
$[R,\Theta]=(R\cdot\nabla)\Theta-(\Theta\cdot\nabla)R$
$[R,\Theta]=(x\frac{d}{dx}+y\frac{d}{dy}y+z\frac{d}{dz})(xz,yz,-(x^2+y^2))-((xz\frac{d}{dx}+yz\frac{d}{dy}-(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int \frac{ax^2-b}{x\sqrt{c^2x^2-(ax^2+b)^2}} dx$ Integrate:
$$\int \frac{ax^2-b}{x\sqrt{c^2x^2-(ax^2+b)^2}} dx$$
What should be the substitution here so that it becomes simpler??? Please help
| This question follows a series of substitution, I'd say it's very tedious. However, yet simple.
Here's a hint:
First substitute. $x^2=t$, giving $dx=\dfrac{dt}{2x}$
Giving us, the integral as :
$$\large\int\dfrac{(at-b)}{2t\sqrt{c^2t-(at+b)^2}}.dt$$
The same can be re written as,
$$\large\int\dfrac{a}{2\sqrt{c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How find the maximum possible length of OC, where ABCD is a square, and AD is the chord of the circle? Given a circle $o(O(0,0), r=1)$. How to find the maximum possible length of $OC$, where $ABCD$ is a square, and $AD$ is the chord of the circle?
I have no idea how to do this, can this be proved with simple geometry?
| This might not exclusively be using simple geometry, but here is a solution.
Without loss of generality, we can assume that the point $D$ is located at $(1,0)$, and that the point $A$ is located with a nonnegative $y$ coordinate.
We can construct a function which gives the $y$ coordinate of the $A$ point given the $x$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Closed form for integral of inverse hyperbolic function in terms of ${_4F_3}$ While attempting to evaluate the integral $\int_{0}^{\frac{\pi}{2}}\sinh^{-1}{\left(\sqrt{\sin{x}}\right)}\,\mathrm{d}x$, I stumbled upon the following representation for a related integral in terms of hypergeometric functions:
$$\small{\int... | $$
\newcommand{\as}{\sinh^{-1}}
\newcommand{\at}{\tan^{-1}}
\begin{align}
I &:= 2 \int_0^{\pi/2} \frac{x \as x}{\sqrt{1-x^4}}
\\&= \int_0^{\pi/2} \as\sqrt{ \sin x }
\\&= \frac 1 2\int_0^{\pi} \as\sqrt{ \sin x }
\\&= \frac 1 2\int_0^{\infty} \frac{2}{1+t^2}\as\sqrt{ \frac{2t}{1+t^2}}
\\&= \left.\at x \as\sqrt{ \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How do I integrate: $\int\sqrt{\frac{x-3}{2-x}} dx$? I need to solve:
$$\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$
What I did is:
Substitute: $x=2\cos^2 \theta + 3\sin^2 \theta$. Now:
$$\begin{align}
x &= 2 - 2\sin^2 \theta + 3 \sin^2 \theta \\
x &= 2+ \sin^2 \theta \\
\sin \theta &= \sqrt{x-2} \\
\theta &=\sin^{-1}\sqrt{... | Hint:
For solving the integral, notice that
$$t^2 = \frac{x-3}{2-x} \Rightarrow dx = -\frac{2t}{(t^2+1)^2}dt.$$ Hence,
$$\int \sqrt{\frac{x-3}{2-x}}dx = -2\int\frac{t^2}{(t^2+1)^2}dt. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
A simple Partial Differential Equation? I thought... I was looking at the Partial Differential Equation involving function:
$$ z(x,y)$$
$$ \frac{\partial z}{\partial x} + c \frac{\partial z}{\partial y} = 0 $$
Which fairly intuitively has a solution:
$$ z = f\left( x - \frac{1}{c} y \right)$$
Now I was considering the... | After playing around with wolfram alpha I found that
$$ \sum_{i=1}^{\infty} \left[ \frac{1}{i} \begin{pmatrix} 2i-2 \\ i-1 \end{pmatrix} u^i \right] = \frac{1}{2} (1 - \sqrt{1 - 4u} )$$
Therefore
$$g = a_0 + a_1 x - 2a_1 x \frac{1}{2} \left(1 - \sqrt{1 - 4\frac{y}{2x^2}} \right) = $$
$$ a_0 + a_1x - a_1x \left(1 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculate the limit of $(1+x2^x)/(1+x3^x)$ to the power $1/x^2$ when $x\to 0$ I have a problem with this: $\displaystyle \lim_{x \rightarrow 0}{\left(\frac{1+x2^x}{1+x3^x}\right)^\frac{1}{x^2}}$.
I have tried to modify it like this: $\displaystyle\lim_{x\rightarrow 0}{e^{\frac{1}{x^2}\ln{\frac{1+x2^x}{1+x3^x}}}}$ and ... | Since
$$
2^x=1+\log(2)x+O\!\left(x^2\right)\\
\Downarrow\\
\log\left(1+x2^x\right)=x+\left(\log(2)-\tfrac12\right)x^2+O\!\left(x^3\right)
$$
and
$$
3^x=1+\log(3)x+O\!\left(x^2\right)\\
\Downarrow\\
\log\left(1+x3^x\right)=x+\left(\log(3)-\tfrac12\right)x^2+O\!\left(x^3\right)
$$
Therefore,
$$
\frac1{x^2}\log\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Find $\sin^3 a + \cos^3 a$, if $\sin a + \cos a$ is known
Given that $\sin \phi +\cos \phi =1.2$, find $\sin^3\phi + \cos^3\phi$.
My work so far:
(I am replacing $\phi$ with the variable a for this)
$\sin^3 a + 3\sin^2 a *\cos a + 3\sin a *\cos^2 a + \cos^3 a = 1.728$. (This comes from cubing the already given state... | Note
$$
x^3+y^3 = (x+y)\left(x^2-xy +y^2\right)
$$
Now we can see that if we set
$$
x =\sin a\\
y = \cos a.
$$
Then we get
$$
\sin^3a+\cos^3a = (\sin a +\cos a)\left(1-\sin a\cos a\right)
$$
We can utilise
$$
(\sin a+ \cos a)^2 = 1 +2\sin a\cos a
$$
Combining all the above will yield
$$
\sin^3a+\cos^3a = (\sin a +\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 0
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Finding value of 1 variable in a 3-variable $2^{nd}$ degree equation The question is: If $a,b,\space (a^2+b^2)/(ab-1)=q$ are positive integers, then prove that $q=5$. Also prove that for $q=5$ there are infinitely many solutions in $\mathbf N$ for $a$ and $b$. I simplified the equation as follows:-$$\frac {a^2+b^2}{ab-... | I take it that you have looked into the details of Vieta Jumping wiki link, so if we know the curve $x^2+y^2 - qxy + q = 0$ has a single integer solution then there are infinitely many, generated by the pair of iterations:
$$(x,y) \mapsto (qx-y,x)$$ $$(x,y) \mapsto (y,x)$$
on the graph of the curve.
Further observe tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Help on proving a trigonometric identity involving cot and half angles Prove: $\cot\frac{x+y}{2}=-\left(\frac{\sin x-\sin y}{\cos x-\cos y}\right)$.
My original idea was to do this:
$\cot\frac{x+y}{2}$ = $\frac{\cos\frac{x+y}{2}}{\sin\frac{x+y}{2}}$, then substitute in the formulas for $\cos\frac{x+y}{2}$ and $\sin\fra... | You should know the following factorisation formulae:
*
*$\sin p +\sin q= 2\sin\dfrac{p+q}2 \cos\dfrac{p-q}2$
*$\sin p-\sin q= 2\sin\dfrac{p-q}2 \cos\dfrac{p+q}2$
*$\cos p +\cos q= 2\cos\dfrac{p+q}2 \cos\dfrac{p-q}2$
*$\cos p-\cos q= -2\sin\dfrac{p+q}2 \sin\dfrac{p-q}2$
(They're derived from : \begin{align*}
&\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that: $\frac{1}{1+\log_ab+\log_bc}+\frac{1}{1+\log_bc+\log_ca}+\frac{1}{1+\log_ca+\log_ab}\leq1$
Let $a, b, c>1$. Show that:
$$\frac{1}{1+\log_ab+log_bc}+\frac{1}{1+\log_bc+\log_ca}+\frac{1}{1+\log_ca+\log_ab}\leq1.$$
My attempt:
We noted $\log_bc=x, \log_ca=y, \log_ab=z$ with $xyz=1$ and we have reduced ineq... | Following OP's simplification of the inequality:
$\displaystyle \begin{align} xy^2+x^2y+yz^2+y^2z+xz^2+x^2z &= \sum\limits_{cyc} (x^2y + xy^2) \\ &=\sum\limits_{cyc} (x^{6/3}y^{3/3} + x^{3/3}y^{6/3}) \\ & \ge \sum\limits_{cyc} (x^{5/3}y^{4/3} + x^{4/3}y^{5/3}) \tag{1}\\ &=\sum\limits_{cyc} x^{5/3}(y^{4/3} + z^{4/3})\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$ If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$
I tried Schturm's method for quite some time, and Cauchy Schwarz for numerators because of the given product condition.
| $f(x,y,z) = 2xy+2yz+2zx -x-y-z-3$ subject to $xyz=1$.
$f_x = \lambda zy \to 2xy+2xz - x = \lambda$.
$f_y = \lambda xz \to 2xy + 2yz - y = \lambda$.
$\Rightarrow (2z-1)(x-y) = 0$.
If $z=\dfrac{1}{2} \to f(x,y,z) = f(x,y,\frac{1}{2}) = 2xy + y+x-x-y-\dfrac{7}{2}=2(2)-\dfrac{7}{2} = \dfrac{1}{2}>0$
Thus assume $x=y$, and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
The roots of a certain recursively-defined family of polynomials are all real Let $P_0=1 \,\text{and}\,P_1=x+1$ and we have $$P_{n+2}=P_{n+1}+xP_n\,\,n=0,1,2,...$$
Show that for all $n\in \mathbb{N}$, $P_n(x)$ has no complex root?
| For $x = u+iv$ define the pair of sequences $u_n,v_n$ such that $P_n(x) = u_n+iv_n$ are the real and imaginary parts of $P_n(x)$.
Then, $u_{n+1}+iv_{n+1} = u_n+iv_n + (u+iv)(u_{n-1}+iv_{n-1})$
$\implies u_{n+1} = u_n + uu_{n-1} - vv_{n-1}$ and $v_{n+1} = v_n + uv_{n-1} + vu_{n-1}$
Assume $a+ib$ (where, $a,b \in \mathbb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Mean value theorem with trigonometric functions Let $f(x) = 2\arctan(x) + \arcsin\left(\frac{2x}{1+x^2}\right)$
*
*Show that $f(x)$ is defined for every $ x\ge 1$
*Calculate $f'(x)$ within this range
*Conclude that $f(x) = \pi$ for every $ x\ge 1$
Can I get some hints how to start? I don't know how to start provin... | The expression for $f(x)$ is defined whenever
$$
-1\le\frac{2x}{1+x^2}\le1
$$
that is
\begin{cases}
2x\le 1+x^2\\
2x\ge -(1+x^2)
\end{cases}
or
\begin{cases}
x^2-2x+1\ge0\\
x^2+2x+1\ge0
\end{cases}
which is satisfied for every $x$.
Computing the derivative of
$$
g(x)=\arcsin\frac{2x}{1+x^2}
$$
just requires some patien... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{n \rightarrow \infty} \frac{1-(1-1/n)^4}{1-(1-1/n)^3}$ Find
$$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}$$
I can't figure out why the limit is equal to $\dfrac{4}{3}$ because I take the limit of a quotient to be the quotient of their limits.
I'm taking th... | If you expand the binomials, you get $\dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}=\dfrac{1-\left(1-\dfrac{4}{n}+\dfrac{6}{n^2}-\dfrac{4}{n^3}+\dfrac{1}{n^4}\right)}{1-\left(1-\dfrac{3}{n}+\dfrac{3}{n^2}-\dfrac{1}{n^3}\right)}=\dfrac{\dfrac{4}{n}-\dfrac{6}{n^2}+\dfrac{4}{n^3}-\dfrac{1}{n^4}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How do I solve this, first I have to factor $ 2x\over x-1$ + $ 3x +1\over x-1$ - $ 1 + 9x + 2x^2\over x^2-1$? I am doing calculus exercises but I'm in trouble with this
$$\frac{ 2x}{x-1} + \frac{3x +1}{ x-1} - \frac{1 + 9x + 2x^2}{x^2-1}$$
the solution is
$$ 3x\over x+1$$
The only advance that I have done is fact... | Notice that $$\frac{A}{B} + \frac{A'}{BC} = \frac{C}{C}\frac{A}{B} + \frac{A'}{BC} = \frac{AC + A'}{BC}$$
Then
$$\begin{align}\frac{(x+1)}{(x+1)}\frac{2x}{(x-1)}+&\frac{(x+1)}{(x+1)}\frac{3x+1}{(x-1)}+\frac{1 + 9x + 2x^2}{(x-1)(x+1)}\\&=\frac{2x(x+1) + (3x+1)(x+1) - (1+9x + 2x^2)}{(x-1)(x+1)} \\&= \frac{2x^2 + 2x + 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why is $\sum_{m=0}^k \frac{k^2+k}{2} = \frac{(k^2+k)(k+1)}{2}$? I recieved an answer like this :
$S= \displaystyle \sum_{m=0}^k \dfrac{k^2+k}{2} - k\displaystyle \sum_{m=0}^k m + \displaystyle \sum_{m=0}^k m^2=\dfrac{(k^2+k)(k+1)}{2}-\dfrac{k\cdot k(k+1)}{2}+\dfrac{k(k+1)(2k+1)}{6}=k(k+1)\left(\dfrac{k+1}{2} - \dfrac{k... | Since $\displaystyle \sum_{x=0}^n C =(n+1) C$. you are summing constant $n+1$ times.
Note that you are summing over $m$ and $\dfrac {k^2+k}{2}$ is independent of $m$.
Note that $\displaystyle \sum_{x=0}^n C = \underbrace{C+C+ \dots +C}_{k+1 \text{times}} = (n+1) C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1104109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding determinant of $n \times n$ matrix I need to find a determinant of the matrix:
$$
A = \begin{pmatrix}
1 & 2 & 3 & \cdot & \cdot & \cdot & n \\
x & 1 & 2 & 3 & \cdot & \cdot & n-1 \\
x & x & 1 & 2 & 3 & \cdot & n-2 \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot... | If $c_i$ is $i$th column of your second determinant, do $c_n= c_n-c_{n-1}$, $c_{n-1}=c_{n-1}-c_{n-2}$, ..., $c_2=c_2-c_1$ to get: $$\left|\begin{array}{ccccccc}
1-x & x & 0 & 0 & \cdots & 0 & 0\\
0 & 1-x & x & 0 & \cdots & 0 & 0\\
0 & 0 & 1-x & x & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1104569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
complex algebra Calculate: \begin{align*}
&\left( 1+\frac{1+i}{2} \right) \left( 1+ \left( \frac{1+i}{2} \right)^2 \right) \left( 1+{\left( \frac{1+i}{2} \right)^2}^2 \right) \ldots \\
&\left( 1+{\left( \frac{1+i}{2} \right)^2}^k \right) \ldots \left( 1+{\left( \frac{1+i}{2} \right)^2}^{2001} \right) \text{.}
\end{alig... | Multiply by $$\left(1-\frac{1+i}{2}\right)$$ compute, and divide by that number at the end.
The thing is that $$\left(1-\frac{1+i}{2}\right)\left(1+\frac{1+i}{2}\right)=\left(1-\left(\frac{1+i}{2}\right)^2\right)$$ then $$\left(1-\left(\frac{1+i}{2}\right)^2\right)\left(1+\left(\frac{1+i}{2}\right)^2\right)=\left(1-\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Number theory: prove that if $a,b,c$ odd then $2\gcd(a,b,c) = \gcd(a+b,b+c, c+a)$ Please help! Am lost with the following:
Prove that if $a,b,c$ are odd integers, then $2 \gcd(a,b,c) = \gcd( a+b, b+c, c+a)$
Thanks a lot!!
| Let $d=\gcd(a+b,b+c,c+a)$.
$$d\mid a+b,b+c\implies d\mid (a+b)-(b+c)\implies d\mid a-c$$
$$\begin{cases}d\mid a-c\\d\mid a+c\end{cases}\implies d\mid (a+c)-(a-c)\implies d\mid 2c$$
Similarly, we have $d\mid 2a$, $d\mid 2b$.
$$\begin{cases}d\mid 2a\\d\mid 2b\\d\mid 2c\end{cases}\implies d\mid \gcd(2a,2b,2c)\implies d\mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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why is $\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}}$=-3? Exercise taken from here: https://mooculus.osu.edu/textbook/mooculus.pdf (page 42, "Exercises for Section 2.2", exercise 4).
Why is $\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}}$=-3*? I always find 3 as the solution. I tried two approaches:
Approach 1:
$$
\lim_{x\... | $\sqrt{x^2}=|x|$ for real $x$
For $x\le0, |x|=-x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Summation of an infinite series The sum is as follows:
$$
\sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1}\\
$$
This is how I started:
$$
= \frac{1}{6}\sum_{n=1}^{\infty} n \left ( \frac{5}{6} \right )^{n-1} \\
= \frac{1}{5}\sum_{n=1}^{\infty} n \left ( \frac{5}{6} \right )^{n}\\\\... | Letting $a = d = 1/6$ and $r = 5/6$, our sum is:
$$
S = a + (a + d)r + (a + 2d)r^2 + (a + 3d)r^3 + \cdots
$$
Scaling by $r$, we find that:
$$
rS = ar + (a + d)r^2 + (a + 2d)r^3 + \cdots
$$
Subtracting the two equations (by collecting like powers of $r$), we obtain:
$$
(1 - r)S = a + dr + dr^2 + dr^3 + \cdots = a + dr(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integral $\int \frac{x+2}{x^3-x} dx$ I need to solve this integral but I get stuck, let me show what I did:
$$\int \frac{x+2}{x^3-x} dx$$
then:
$$\int \frac{x}{x^3-x} + \int \frac{2}{x^3-x}$$
$$\int \frac{x}{x(x^2-1)} + 2\int \frac{1}{x^3-x}$$
$$\int \frac{1}{x^2-1} + 2\int \frac{1}{x^3-x}$$
now I need to resolve one... | HINT:
$$\int\frac{x+2}{x^3-x}=\int\frac{-2}{x}+\int\frac{\frac{3}{2}}{x-1}+\int\frac{\frac{1}{2}}{x+1}$$
Can u do it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the real and imaginary part of the following I'm having trouble finding the real and imaginary part of $z/(z+1)$ given that z=x+iy. I tried substituting that in but its seems to get really complicated and I'm not so sure how to reduce it down. Can anyone give me some advice?
| Just multiply the fraction by the complex conjugate of $z+1$, that is,
\begin{equation*}
\frac{z}{z+1} = \frac{x+iy}{1+x+iy} = \frac{1+x-iy}{1+x-iy} \frac{x+iy}{1+x+iy} = \frac{(1+x)x+y^2+iy }{(1+x)^2+y^2} =
\end{equation*}
\begin{equation*}
= \frac{(1+x)x+y^2}{(1+x)^2+y^2} + i \frac{y}{(1+x)^2+y^2}
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The maximum of $\binom{n}{x+1}-\binom{n}{x}$ The following question comes from an American Olympiad problem. The reason why I am posting it here is that, although it seems really easy, it allows for some different and really interesting solutions. Do you want to give a try?
Let $n$ be one million. Find the maximum valu... | Note that
$$
\begin{align}
\binom{n}{x+1}-\binom{n}{x}
&=\left[\frac{n-x}{x+1}-1\right]\binom{n}{x}\\
&=\frac{n-2x-1}{x+1}\binom{n}{x}\tag{1}
\end{align}
$$
$(1)$ is positive for $x\lt\frac{n-1}2$ and negative for $x\gt\frac{n-1}2$.
Furthermore,
$$
\begin{align}
&\hphantom{}\left[\binom{n}{x+1}-\binom{n}{x}\right]-\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Find a parametrization of the intersection curve between two surfaces in $\mathbb{R^3}$ $x^2+y^2+z^2=1$ and $x^2+y^2=x$. Find a parametrization of the intersection curve between two surfaces in $\mathbb{R}^3$
$$x^2+y^2+z^2=1$$ and $$x^2+y^2=x.$$
I know that $x^2+y^2+z^2=1$ is a sphere and that $x^2+y^2=x$ is a circular... | +1) The polar equation of $x^2+y^2=x$ is $r=cos\theta$ and so we have $x=rcos\theta$ and $y=rsin\theta$ as parametrics (with $r$ known) Substituting into the first equation, you can solve for $z$. Note that $1-cos^2\theta=sin^2\theta$ Hope this helps, if not or incorrect, I will take it off.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $\frac{a+1}{b}+\frac{b}{a}$ is an integer then it is $3$. If $\frac{a+1}{b}+\frac{b}{a}$ is an integer for positive integers $a,b$ then prove that this integer is $3$.
I reduced the to prove that if $\frac{c^2+d^2+1}{cd}$ is an integer then it is $3$ where $c,d\in{\mathbb{N}}$. And this is equivalent to prove that t... | Let
$$\dfrac{a^2+a+b^2}{ab}=k,k\in N^{+}$$
so
$$b^2-ka\cdot b+a^2+a=0$$
case 1
if $a\le b$ and $b$ is minimum
Now Assmue that $b>a$,
then $b_{1}=ka-b\in Z$,and
$$b\cdot b_{1}=a^2+a$$
so $b_{1}$ is postive integer number.
then
$$b_{1}=\dfrac{a^2+a}{b}<\dfrac{a^2+a}{a}=a+1$$
since $b_{1}>b\ge a+1$ impossible.
so $a=b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Use integration by substitution I'm trying to evaluate integrals using substitution. I had
$$\int (x+1)(3x+1)^9 dx$$
My solution: Let $u=3x+1$ then $du/dx=3$
$$u=3x+1 \implies 3x=u-1 \implies x=\frac{1}{3}(u-1) \implies x+1=\frac{1}{3}(u+2) $$
Now I get $$\frac{1}{3} \int (x+1)(3x+1)^9 (3 \,dx) = \frac{1}{3} \int \fr... | I know the question was about integration by substitution, but if you want another way, you can also do it using tabular integration by parts
$$x^2 + 2 \quad \quad \quad (x-1)^7$$
$$2x \quad \quad \quad \frac 1 8 (x-1)^8$$
$$2 \quad \quad \quad \frac 1 {72} (x-1)^9$$
$$0 \quad \quad \quad \frac 1 {720} (x-1)^{10}$$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Number of real roots of $2 \cos\left(\frac{x^2+x}{6}\right)=2^x+2^{-x}$ Find the number of real roots of
$ \cos \,\left(\dfrac{x^2+x}{6}\right)= \dfrac{2^x+2^{-x}}{2}$
1) 0
2) 1
3) 2
4) None of these
My guess is to approach it in graphical way. But equation seems little difficult.
| Rearranging we get $$(2^x)^2-2\cos\dfrac{x^2+x}6(2^x)+1=0\ \ \ \ (0)$$ which is a Quadratic equation in $2^x$
For real $2^x,$ the discriminant must be $\ge0$
i.e., $$\left(2\cos\dfrac{x^2+x}6\right)^2-4=-4\sin^2\dfrac{x^2+x}6\ge0\ \ \ \ (1)$$
As $\sin\dfrac{x^2+x}6$ is real, $$\sin^2\dfrac{x^2+x}6\ge0\iff-4\sin^2\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $ Let $a,b,c\in\mathbb{R^+}$ such that $ a^2 + b^2 + c^2 = 3 $. Prove that
$$
(a+b+c)(a/b + b/c + c/a) \geq 9.
$$
My Attempt
I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.
| We'll prove that $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\geq\frac{3}{2}\sum\limits_{cyc}\left(\frac{a}{b}+\frac{a}{c}\right)$.
Let $\frac{a}{b}=x$, $\frac{b}{c}=y$ and $\frac{c}{a}=z$.
Thus, we need to prove that $2(x+y+z)^2\geq3(x+y+z+xy+xz+yz)$,
where $x>0$, $y>0$ and $z>0$ such that $xyz=1$
or $\sum\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Independence of path in a closed curve line integral
Let $f(t)$ be a continuous function. Let $C$ be a smooth closed curve. Show that $$\oint\limits_C xf(x^2 + y^2)\,dx + y f(x^2 + y^2)\,dy = 0$$ Hint: Remember that $f(t)$ has a primitive function $F(t)$. Use this fact to construct a potential function for the vector ... | Let $g(u) := \frac{1}{2}\int_0^u f(t)\, dt$. Then $$\frac{\partial}{\partial x} g(x^2 + y^2) = \frac{1}{2}f(x^2 + y^2) \frac{\partial}{\partial x}(x^2 + y^2) = xf(x^2 + y^2)$$
and similarly $$\frac{\partial}{\partial y}g(x^2 + y^2) = yf(x^2 + y^2).$$ Therefore, the vector field $xf(x^2 + y^2)\vec{i} + yf(x^2 + y^2)\vec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solution of 2nd order linear ODE with regular singular points, and complex exponents at singularity The steady state temperature distribution of a rod given by:
\begin{equation}
\frac{\textrm{d}p(x)y'}{\textrm{d}x} - y = 0,\; 0 \leq x \leq 1,\; \text{and} \;y(0) = 0,
\end{equation}
where $y(x)$ is the s... | $$y''+\frac{s}{x}y'-x^{-s}y=0$$
This is an ODE of the Bessel kind, but not on standard form. In order to make it standard, let $y(x)=x^aF(b\:x^c)=x^aF(X)$. The goal is to transform it to :
$$F''(X)+\frac{1}{X}F'(X)-\left(1+\frac{\nu^2}{X^2}\right)F(X)=0$$
The calculus is arduous. The result is :
$$a=\frac{1-s}{2} \: ; ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 1,
"answer_id": 0
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Chinese Remainder Theorem for non prime-numbers. Let's say I want to find x such that x leaves remainder 2 when divided by 3 and x leaves remainder 3 when divided by 5.
x % 3 = 2
x % 5 = 3
We break down the problem to:
x % 3 = 1
x % 5 = 0
Therefore,
5k % 3 = 1
2k % 3 = 1
k = 2
10, when remainder = 1
20, when remainder ... | It has nothing to see with primeness of the moduli, but to their coprimeness.
The systematic method consists in finding Bézout's coefficients for $7$ and $3$: $\enspace-1\cdot 7+2\cdot 4=1$
(For bigger numbers you can use the Extended Euclidean Algorithm).
The solution to the system of congruences \begin{equation*}\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove this: $a^4+b^4+2 \ge 4ab$? How to prove this: $a^4+b^4+2 \ge 4ab$?
$a$ and $b$ are reals.
| Notice that:
$$(a^2+b^2)^2 = a^4 + b^4 + 2a^2b^2$$
So, then we have that: $(a^2 + b^2)^2 - 2a^2b^2 + 2 \ge 4ab$
It follows that:
$$(a^2 + b^2)^2 \ge 2a^2b^2 + 4ab +2 -4,\ (a^2 + b^2)^2 \ge 2(ab+1)^2 -2^2 = \left(\sqrt{2}(ab+1) + 2\right)\left(\sqrt{2}(ab+1) - 2\right)$$
Our inequality now becomes:
$$(a^2 + b^2)^2 \ge\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Show that the arithmetic mean is less or equal than the quadratic mean I tried to solve this for hours but no success.
Prove that the arithmetic mean is less or equal than the quadratic mean.
I am in front of this form:
$$
\left(\frac{a_1 + ... + a_n} { n}\right)^2 \le \frac{a_1^2 + ... + a_n^2}{n}
$$
With rewriting ... | for sake of contradiction suppose:
$\sqrt{\frac{a_1^2+ \cdots +a_n^2}{n}}<\frac{a_1+ \cdots +a_n}{n}\\
\frac{a_1^2+ \cdots +a_n^2}{n}<\frac{(a_1+ \cdots +a_n)^2}{n^2}\\
\frac{a_1^2+ \cdots +a_n^2}{n}<\frac{a_1^2+ \cdots +a_n^2}{n^2}+2\frac{\sum_{sym}a_1a_2}{n^2}\\
\\
(n-1)(a_1^2+\cdots+a_n^2)<2\sum_{sym}a_1a_2 \quad \q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
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Does $\sum_{n\ge0} \cos (\pi \sqrt{n^2+n+1}) $ converge/diverge? How would you prove convergence/divergence of the following series?
$$\sum_{n\ge0} \cos (\pi \sqrt{n^2+n+1}) $$
I'm interested in more ways of proving convergence/divergence for this series.
My thoughts
Let
$$u_{n}= \cos (\pi \sqrt{n^2+n+1})$$
Let's firs... | From the Taylor expansion
$$
(1+x)^{1/2}=1+\frac{x}{2}-\frac{x^2}{8}+O(x^3)
$$
we get
$$\begin{align}
\sqrt{n^2+n+1}&=n\Bigl(1+\frac{1}{n}+\frac{1}{n^2}\Bigr)^{1/2}\\
&=n\Bigl(1+\frac{1}{2}\Bigl(\frac{1}{n}+\frac{1}{n^2}\Bigr)-\frac{1}{8}\Bigl(\frac{1}{n}+\frac{1}{n^2}\Bigr)^2+O\Bigl(\frac{1}{n}+\frac{1}{n^2}\Bigr)^3\B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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need help on this fraction equation $2/5 = 2/3 - r/5$ $$\frac{2}{5} = \frac{2}{3} - \frac{r}{5}$$
I'm trying to find $r$. Can anyone give me a step by step?
| First note that
$$ \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm b}{c}$$
And
$$ \frac{a}{b}\pm c=\frac{a\pm bc}{c} $$
So now we have
$$\frac{2}{5} = \frac{2}{3} - \frac{r}{5} $$
$$\frac{2}{5} + \frac{r}{5} = \frac{2}{3} $$
$$\frac{2+r}{5} = \frac{2}{3} $$
$$ 2+r = \frac{2\times 5}{3} $$
$$ r = \frac{10}{3}-2 =\frac{10-(2\times... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Representation of Quaternion group in $GL(2,3)$ I am working with the representation of the quaternion group in $GL(2,3)$ generated by $A=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}, B=\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}, C=\begin{pmatrix} -1 & 1\\ 1 & 1 \end{pmatrix}$.
Now $C=PBP^{-1}$ where $P= \begin{pma... | Let's suppose they are conjugate by
$$M = \left(\begin{array}{rr}
x & y\\
z & w \end{array} \right).$$
Then, we have
$$MB = AM \Rightarrow \left(\begin{array}{rr}
x + y & x - y\\
z+w & z-w \end{array} \right) = \left(\begin{array}{rr}
-z & -w\\
x & y \end{array} \right).$$
So, $z = -x- y, w = y - x$ and putting these ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find a solution: $3(x^2+y^2+z^2)=10(xy+yz+zx)$ I'm something like 90% sure that this diophantine equation has nontrivial solutions:
$3(x^2+y^2+z^2)=10(xy+yz+zx)$
However, I have not been able to find a solution using my calculator. I would greatly appreciate if someone could try to find one using a program. Or maybe yo... | Because the equation is homogenous, the integer solutions can be derived from the rational solutions, in other words swapping between projective and affine form.
I prove below that the set of non-zero rational solutions are common rational multiples of the following (which, conversely, satisfies the equation identicall... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $x^3 - 9x^2 + 18x - 12$
If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $$x^3 - 9x^2 + 18x - 12.$$
This is not a homework problem. I'm not even a student. I'm going through an old textbook. I know this is a simple problem. Can't seem to crack it though.
| Let $\,a = \sqrt[3]3.\, $ $\, (\!\overbrace{x\!-\!3}^{\Large a^2+a}\!)^2 = \overbrace{a^4}^{\Large 3a}\!+\overbrace{2a^3}^{\Large 6}+a^2\,$
Therefore $\ \color{#0a0}{x^2\!-6x\!+\!3\, =\, a^2\!+\!3a}.\ $ By very simple arithmetic we have:
$\begin{align} f(x)\, & =\ (x-3)\,(\color{#0a0}{x^2\!-6x+3})\, -\, 3(x+1)\ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to integrate $\frac{1}{x\sqrt{1+x^2}}$ using substitution? How you integrate
$$\frac{1}{x\sqrt{1+x^2}}$$
using following substitution? $u=\sqrt{1+x^2} \implies du=\dfrac{x}{\sqrt{1+x^2}}\, dx$
And now I don't know how to proceed using substitution rule.
| If $u=\sqrt{1+x^2}$ then $u^2 = 1+x^2$, so $x^2= u^2-1$. Then you have
\begin{align}
& \int \frac 1 {x\sqrt{1+x^2}} \,dx = \int \frac {x} {x^2\sqrt{1+x^2}} \,dx \\[8pt]
= {} & \int\frac{du}{u^2-1} = \int\frac{du}{(u-1)(u+1)}.
\end{align}
Then use partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 0
} |
Sum $\sum_{n=2}^{\infty} \frac{n^4+3n^2+10n+10}{2^n(n^4+4)}$ I want to evaluate the sum $$\large\sum_{n=2}^{\infty} \frac{n^4+3n^2+10n+10}{2^n(n^4+4)}.$$ I did partial fraction decomposition to get $$\frac{1}{2^n}\left(\frac{-1}{n^2+2n+2}+\frac{4}{n^2-2n+2}+1\right)$$ I am absolutely stuck after this.
| You are almost there.
So once you have obtained $$\frac{1}{2^n}\left(\frac{-1}{n^2+2n+2}+\frac{4}{n^2-2n+2}+1\right)$$ then observe the following:
$n^2+2n+2=(n+1)^2+1$ and $n^2-2n+2=(n-1)^2+1$. Which means the given sum becomes
$$\sum\limits_{n=2}^\infty \left[\frac{-1}{2^n\{(n+1)^2+1\}}+\frac{1}{2^{n-2}\{(n-1)^2+1\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Find all real solutions for $x$ in $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 .$ Find all real solutions for $x$ in $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 .$
I have found out that the answers were 0,1, and -1. But I used sort of a guess-and check way.
$2(2^x-1)x^2+(2^{x^2}-2)x=2^{x+1}-2$
I expanded it into:
$(x^2-1... | Mistake when you divide by 2. You get:
$$2^{x^2-1} = 2^{(x+1)(x-1)} = 2^{{x+1}^{x-1}}$$
Now maybe it gets easier?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(z^3-z)(z+2)$ is divisible by $12$ for all integers $z$ I am a student and this question is part of my homework.
May you tell me if my proof is correct?
Thanks for your help!
Prove that $(z^3-z)(z+2)$ is divisible by $12$ for all integers $z$.
$(z^3-z)(z+2)=z(z^2-1)(z+2)=z(z-1)(z+1)(z+2)=(z-1)(z)(z+1)(z+2)... | That's correct. Alternatively it is divisible by $\,24\,$ by integrality of binomial coefficients
$$\,(z+2)(z+1)z(z-1)\, =\, 4!\ \dfrac{(z+2)(z+1)z(z-1)}{4!}\, =\, 24{ {z+2\choose 4}}\qquad\qquad$$
Similarly $\,n!\,$ divides the product of $\,n\,$ consecutive naturals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int\frac{dx}{x^5+1}$ Calculate $\displaystyle\int\dfrac{dx}{x^5+1}$
| Hint:
$$
\begin{split}
x^5+1&=(1 + x)(x^4 - x^3 + x^2 - x + 1) \\
&=(1 + x)(x^2 + ((-1 + \sqrt 5) / 2)x + 1)(x^2 + ((-1 - \sqrt 5) / 2)x + 1)
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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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.