Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Use the de Moivre theorem to evaluate the following $$\frac{(1+i)(\sqrt{3} + i)^3}{(1-\sqrt{3}i)^{3}} = 1-i$$
What confuses me is how would I do the numerator because I have two expressions.
| $1+i=\sqrt 2 (\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))$
$ \sqrt3+i=2(\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6})) $
$ 1-i\sqrt3=2(\cos(\frac{-\pi}{3})+i\sin(\frac{-\pi}{3})$
Then you can simply apply De Moivre's theorem:
The numerator becomes $8\sqrt2(\cos(\frac{9\pi}{12})+i\sin(\frac{9\pi}{12}))=8\sqrt2(\cos(\frac{3\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int_0^{\pi/2}\frac{\sin 2013x }{\sin x} \ dx\space$ How would you approach
$$\int_0^{\pi/2}\frac{\sin 2013x }{\sin x} \ dx\space?$$
The way I see here involves Dirichlet kernel. I wonder what else can we do, maybe some easy/elementary approaching ways. Thanks !
| Let $I=\displaystyle\int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx$
As $\displaystyle\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$
$\displaystyle I=\int_0^{\frac\pi2} \frac{\sin (2n+1)(\frac\pi2-x)}{\sin (\frac\pi2-x)} dx$
$\displaystyle =\int_0^{\frac\pi2} \frac{\sin \{n\pi+\frac\pi2-(2n+1)x\}}{\cos x} dx$
$\displaystyle =\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Quick way to iterate multiples of a prime N that are not multiples of primes X, Y, Z, ...? Is there a way to quickly iterate multiples of some prime $N$ while avoiding multiples of blacklisted primes $X$, $Y$, $Z$, ...? By quickly I mean is there a faster way than:
*
*Increment current number by N.
*Check if curren... | Up to $kN$ (inclusive) there are $k$ positive integer multiples of $N$, of which $\lfloor k/X \rfloor$ are multiples of $X$, etc. By inclusion-exclusion, the number that escape a "blacklist" of $3$ primes is
$$ k - \left\lfloor \frac{k}{X} \right\rfloor - \left\lfloor \frac{k}{Y} \right\rfloor - \left\lfloor \frac{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/272452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x)\,\mathrm dx$ What would you recommend me for the integral below?
$$
\int_{0}^{1}\ln(1 - x)\ln(x)
\ln(1 + x)\,\mathrm dx
$$
For instance, for the version without the last logarithm would work to use Taylor series, but in this case things are a bit more complicated and it doesn't s... | $$I=\int _0^1\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)\:dx$$
By using the the algebraic identity $ab=\frac{1}{4}\left(a+b\right)^2-\frac{1}{4}\left(a-b\right)^2$ we get:
$$I=\frac{1}{4}\underbrace{\int _0^1\ln \left(x\right)\ln ^2\left(1-x^2\right)\:dx}_{t=x^2}-\frac{1}{4}\int _0^1\ln \left(x\right)\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/274742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 3,
"answer_id": 2
} |
Getting the equation of an ellipse using the constant and the foci Find the equation of the ellipse with the foci at (0,3) and (0, -3) for which the constant referred to in the definition is $6\sqrt{3}$
So I'm quite confused with this one, I know the answer is $3x^2+2y^2=54$ through trial and errror, but is there any w... | Let $(x,y)$ be a point of ellipse then from definition
$$\sqrt{(x-0)^2+(y-3)^2}+\sqrt{(x-0)^2+(y+3)^2}=6\sqrt{3}$$
$$\sqrt{x^2+(y-3)^2}+\sqrt{x^2+(y+3)^2}=6\sqrt{3}$$
$$2x^2+2y^2+18+2\sqrt{(x^2+(y+3)^2)(x^2+(y-3)^2)}=108$$
$$x^2+y^2+9+\sqrt{(x^2+(y+3)^2)(x^2+(y-3)^2)}=54$$
$$(x^2+y^2-45)^2=(x^2+(y+3)^2)(x^2+(y-3)^2)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/279335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
There is a $3\times 3 $ orthogonal matrix with all non zero entries.? Is the statement true?
There is a $3\times 3 $ real orthogonal matrix with all non zero entries.
for orthogonality, $AA^T=A^TA=I_3$, please give me hint
| Gordon Pall found all rational orthogonal matrices, 3 by 3, in 1940, see PALL_PDF
Given an odd number
$$ n = a^2 + b^2 + c^2 + d^2,$$ the matrix
$$ \frac{1}{n} \begin{pmatrix} a^2 + b^2 - c^2 - d^2 & 2(-ad+bc) & 2(ac+bd) \\ 2(ad+bc) & a^2 - b^2 + c^2 - d^2 & 2(-ab+cd) \\ 2(-ac+bd) & 2(ab+cd) & a^2 - b^2 - c^2 + d^2 \en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/279397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
If $\prod x_k=1$, then $\prod\frac{1+x_k}{2}\leq ( \frac{x_1+\cdots +x_n}{n})^{n-1}$ Despite many attempts, no one at StackOverflow has succeeded in solving that old question about proving a deceptively simple-looking inequality.
I propose now a weaker and slightly simpler inequality (no more squares) which may perhaps... | As mentioned in comment, one can use equal variable method for $f(x) = \log \left(\frac{1+x}{2}\right)$. Note that
$$g(x) = f'(1/x) = \frac{x}{x+1}$$
is strictly concave for $x > 0$. By corollary 1.6, the maximum of LHS is reached when $0 < x_1 \leq 1 \leq x_2 = \cdots = x_n = t$. Substitute $x_1 = \frac{1}{t^{n-1}}$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/280914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
show $\ln \frac{1-x}{1+x}$ is $L^2$ but not $L^1$ using its taylor expansion I am trying to show that
$$
\ln{\Big|\frac{1-x}{1+x}\Big|}
$$
belongs to $L^2({\Bbb{R}})$ but not to $L^1({\Bbb{R}})$ by using it's taylor expansion (this is the entire statement of the problem). More important to me than the solution to thi... | Hint 1:
For $|x|\lt1$, we have the standard
$$
\log\,\left|\frac{1-x}{1+x}\right|=-2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\dots\right)
$$
For $|x|>1$, we have
$$
\begin{align}
\log\,\left|\frac{1-x}{1+x}\right|
&=\log\,\left|\frac{1/x-1}{1/x+1}\right|\\
&=\log\,\left|\frac{1-1/x}{1+1/x}\right|\\
&=-2\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$ Evaluate
$$\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$$
| As @rlgordonma has let us make use of the substitution $x = e^{-y}$ to get the integral as
$$\int_0^{\infty} dy e^{-y} \left ( \frac{(e^{-y} - (1-y))^2}{y^2 (1-e^{-y})^2} \right ) $$
which can be rewritten as $$\sum_{k=1}^{\infty} k \int_0^{\infty} dy \left ( \frac{(e^{-y} - (1-y))^2}{y^2} \right ) e^{- k y} $$
If we c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
"answer_count": 7,
"answer_id": 3
} |
Prove $\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab} \ge 1$ Let $a,b,c$ be positive real numbers such that $\dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \ge 1$. Prove that $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.
| Denote $X=\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}$ and $Y=\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}$. In fact, if $X\ge 1$, then $Y\ge \sqrt{3}$. ($\sqrt{3}$ is optimal, because when $a=b=c=\frac{1}{\sqrt{3}}$, $X=1$ and $Y=\sqrt{3}$.)
By Cauchy-Schwarz inequality,
$$3(a^2+b^2+c^2)\ge (a+b+c)^2.$$
It follows that
$$Y=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Evaluate the limit I am hung up on this limit: $\displaystyle\lim_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x}$
I must be missing something related to dealing with square roots but I can not for the life of me figure out what.
Here is my work so far:
$\displaystyle\lim_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x} = \lim_{x\... |
$$\lim\limits_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x}$$
Note that evaluating $\quad \displaystyle \lim \frac{\sqrt{1+x} + \sqrt{1-x}}{x}\;$ as $\;x\to 0\;$ gives you $\;2\;$ in the numerator, and $\;0\;$ in the denominator. So the task that ultimately remains is to evaluate the limits as $\;x \to 0^+\,$ and as $\;x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to expand $\tan x$ in Taylor order to $o(x^6)$ I try to expand $\tan x$ in Taylor order to $o(x^6)$, but searching of all 6 derivative in zero (ex. $\tan'(0), \tan''(0)$ and e.t.c.) is very difficult and slow method.
Is there another way to solve the problem?
Any help would be greatly appreciated :)
| Here's the long division method suggested by coffemath.
\begin{align}
\sin(x)
= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots
&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{(2n+1)}
&=& \sum_{n=0}^\infty a_{(2n+1)} x^{(2n+1)} \\
\cos(x)
= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots
&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/286529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
What will be the slope of $BC$? The vertex $A$ of triangle $\triangle ABC$ is $(3,-1)$. The equation of median $BE$ is $$6x+10y-59=0$$ and angle bisector $CF$ is $$x-4y+10=0.$$ Then what is the slope of $BC$?
Let slopes of $AC, CF, BC$ be $m1, m2, m3$ respectively, then
$$(m1-m2)/(1+m1*m2) = (m2-m3)/(1+m2*m3)$$
How to ... | Let's begin at point $C$.
As $C \in CF$ it follows that
$$C=(4y_C-10, y_C).\quad (1)$$
As $E$ is midpoint of $AC$ then the coordinates of E are the arithmetic mean of coordinates of $A$ and $C$, as it is shown in equation below:
$$E=(\frac{4y_C-7}{2}, \frac{y_C-1}{2}).\quad (2)$$
We know that $E \in BE$, therefore
$$6x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/287486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Asymptotic expansion of $ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $ I'm trying to compute the asymptotic expansion of
$$ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $$
Here is what I've done:
Change of variable $$ t= \tan x $$
$$ I_n = \int_0^1 \frac{t^n \mathrm dt}{1+t^2} = \int_0^1 \frac{(1-t)^n \mathrm dt}{t^2-2t+2}... | Not an independent answer but completing the circle start by others.
Several people here have derived the expression:
$$I_n = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{n+1+2k}$$
Using the digamma function $\psi(x) = \frac{\Gamma'(x)}{\Gamma(z)}$
which has the expression and asymptotic expansion:
$$\begin{align}
\psi(x) &= - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/290772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
What's the integration of $\int \sin^4 (x)dx$? What's the integration of $\int \sin^4 x \,dx$? I don't see the approach to this question.
I have a issue with this question as well:
$$\int \sin x \cos x (\sin x+\cos x) \,dx.$$
I simplify this to $\sin^2 x \cos x + \sin x \cos^2 x$
and set $u= \sin x$, $du = \cos x \,dx$... | Recall the identities
$$\sin^2(\theta) = \dfrac{1-\cos(2\theta)}2$$
and
$$\cos^2(\theta) = \dfrac{1+\cos(2\theta)}2$$
Hence,
$$\sin^4(x) = \left(\dfrac{1-\cos(2x)}2 \right)^2 = \dfrac{1 - 2 \cos(2x) + \cos^2(2x)}4 = \dfrac{1 - 2 \cos(2x) + \dfrac{1+\cos(4x)}2}4$$
Hence,
$$\sin^4(x) = \dfrac{3-4\cos(2x) + \cos(4x)}8$$
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/294013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to integrate $\frac{1}{1 + a^2 \tan^2x}$? Can you please help me out with evaluating this integral?
$$\int_0^{2\pi} \frac{1}{1 + a^2 \tan^2x}dx$$
| $$
\begin{align}
\int_o^{2\pi}\frac1{1+a^2\tan^2(x)}\mathrm{d}x
&=4\int_0^{\pi/2}\frac1{1+a^2\tan^2(x)}\mathrm{d}x\\
&=4\int_0^\infty\frac1{1+a^2u^2}\mathrm{d}\arctan(u)\\
&=4\int_0^\infty\frac1{1+a^2u^2}\frac1{1+u^2}\mathrm{d}u\\
&=\frac4{a^2-1}\int_0^\infty\left(\frac{a^2}{1+a^2u^2}-\frac1{1+u^2}\right)\mathrm{d}u\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/295524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving a second-order nonlinear ordinary differential equation Let's start with equation with two parameters $$y=a x^b.$$ Then we calculate $y'=a b x^{b-1}$ and solve $a=y/x^b$ from original. Substitute that to the derivative and $$y'=b \frac{y}{x}.$$ Then differentiate again and substitute $b=x y'/y$ and we get $$y''... | Hint: $(d/dx)(y'/y) = (y y'' - (y')^2)/y^2$
Specifically:
$$y y'' - (y')^2 = -\frac{y y'}{x}$$
$$\frac{y y'' - (y')^2}{y^2} = -\frac{y'}{x y}$$
$$\frac{d}{dx} \left ( \frac{y'}{y} \right ) = -\frac{1}{x} \frac{y'}{y}$$
$$\frac{d}{dx} \log{\left ( \frac{y'}{y} \right )} = -\frac{1}{x}$$
$$\log{\left ( \frac{y'}{y} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/296534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding $\frac{FC}{EG}$ from $FG+EG=DG,EG+DG=DA=2EC=AF-FG$ If $FG+EG=DG,EG+DG=DA=2EC=AF-FG$ .How to find $\frac{FC}{EG}$
| This is really more an algebraical than a geometrical exercise. Essentially we are given 4 equations in 6 variables. Then we add one more variable $FC$ and exclude $EG$ by introducing the ratio. So the total number of variables remains 6.
Therefore, we must infer the two missing relations from the given geometrical co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/298894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is the equation $x^2=x \sin x+ \cos x$ true? I came across a problem that says: The equation $x^2=x \sin x+ \cos x$ is true for:
$1.$No real value of $x$
$2.$Exactly one real value of $x$
$3.$Exactly two real values of $x$
$4.$Infinitely many real values of $x$ .
Thanks in advance for your time.
| Consider
$f(x) = x^2-x \sin x- \cos x$.
$f(0) = -1$,
$f(\pi/2) = \pi^2/4 - \pi/2 > 0$,
and
$f(-\pi/2) = f(\pi/2) > 0$,
so $f$ has at least two real roots.
$f(-x) = f(x)$
(since $f(-x) = (-x)^2 - (-x)\sin(-x) - \cos(-x)
= x^2 - x \sin(x) - \cos(x) = f(x)$,
so we only need look at $x > 0$.
Since $\sin$ and $\cos$ are bo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/299239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A log improper integral Evaluate :
$$\int_0^{\frac{\pi}{2}}\ln ^2\left(\cos ^2x\right)\text{d}x$$
I found it can be simplified to
$$\int_0^{\frac{\pi}{2}}4\ln ^2\left(\cos x\right)\text{d}x$$
I found the exact value in the table of integrals:
$$2\pi\left(\ln ^22+\frac{\pi ^2}{12}\right)$$
Anyone knows how to evaluate t... | I find a way to get the number using gamma functions, nothing is rigorous.
Consider the integral $I(\beta) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\cos x)^\beta dx$.
We know:
$$2 \frac{d^2}{d\beta^2} I(\beta) \bigg|_{\beta=0} = 4 \int_{0}^{\frac{\pi}{2}} \ln^2(\cos x) dx$$
is the integral we want. Introduce $u = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/300061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 2
} |
Prove the identity $1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$ Here is the problem:
$$1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$$
Can you help me prove that this is an trigonometric identity?
| $$\begin{align}
2\cdot \color{blue}{\cos^2\left(45^{\circ}-\frac{x}{2}\right)}&=\\
2\cdot \color{blue}{\frac{1+cos(double angle)}{2}}&=\\
2\cdot \color{blue}{\frac{1+cos(90^{\circ} - x)}{2}}&=\\
1+\cos\left(90^{\circ}-x\right) &= \\
1+\sin(x)&\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
How to find the minimal polynomial when given only the characteristic polynomial? I am trying to find all possible Jordan forms of a transformation with Characteristic Polynomial $(x-1)^2(x+2)^2$. How can I find its minimal polynomial? Or do I just assume the $2$ (minimal and characteristic pol.) coincide?
| Actually, in the general case to obtain the Jordan canonical form you need to find monic, non-constant polynomial $d_{i}$ such that $$d_{1} \mid d_{2} \mid \dots \mid d_{n},$$ with $$d_{1} d_{2} \cdots d_{n} = \text{the characteristic polynomial},$$ and then $d_{n}$ is the minimal polynomial. (See Smith normal form.) I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/301183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Proof about an inequality We have $a,b,c>0$, prove that:
$$\dfrac {1+\sqrt {3}} {3\sqrt {3}}\left( a^{2}+b^{2}+c^{2}\right) \left( \dfrac {1} {a}+\dfrac {1} {b}+\dfrac {1} {c}\right)\geq a+b+c+\sqrt {a^{2}+b^{2}+c^{2}}$$
| Because the inequality is homogeneous, we can assume that $a^2 + b^2 + c^2 = 1$.
Then by the power mean inequality
$$\frac{a+b+c}{3} \le \sqrt{\frac{a^2 + b^2 + c^2}{3}} = \frac{1}{\sqrt{3}}$$
and
$$\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \le \sqrt{\frac{a^2 + b^2 + c^2}{3}} = \frac{1}{\sqrt{3}}.$$
Thus
$$a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/305405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Given $y_n=(1+\frac{1}{n})^{n+1}$ show that $\lbrace y_n \rbrace$ is a decreasing sequence Given
$$
y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1.
$$
Show that $\lbrace y_n \rbrace$ is a decreasing sequence. Anyone can help ? I consider the ratio $\frac{y_{n+1}}{y_n}$ but I... | Note that
$$
\begin{align}
\frac{\left(1+\frac1n\right)^{n+1}}{\left(1+\frac1{n+1}\right)^{n+2}}
&=\left(\frac{n+1}{n}\right)^{n+1}\left(\frac{n+1}{n+2}\right)^{n+2}\\
&=\frac{n}{n+1}\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+2}\\
&=\frac{n}{n+1}\left(1+\frac1{n(n+2)}\right)^{n+2}\\
&\ge\frac{n}{n+1}\left(1+\frac{n+2}{n(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/306178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 7,
"answer_id": 1
} |
Integer solution $(x,y,z)$ to $xyz=24$
Find the total number of integer solutions $(x,y,z)\in\mathbb{Z}$ of the equation $xyz=24$.
I have tried $xyz = 2^3 \cdot 3$
My Process:
Factor $x$, $y$, and $z$ as
$$
\begin{cases}
x = 2^{x_1} \cdot 3^{y_1}\\
y = 2^{x_2} \cdot 3^{y_2}\\
z = 2^{x_3} \cdot 3^{y_3}
\end{cases}
$$
... | Your logic is correct for positive solutions. Now you need the sign pattern to be $+++,+--,-+-,--+$, four possibilities, so multiply by four.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
A hard log definite integral: $\int_0^{\pi/4}\ln^3\sin x\,\mathrm dx$ Show that:
$$\int_0^{\pi/4}\ln ^3\sin x\text{d}x=\frac34\text{Im}\left(\text{Polylog}\left(4,i\right)\right)-3\text{Im}\left(\text{Polylog}\left(4,\frac12+\frac{i}{2}\right)\right)-\frac{23}{128}\pi^3\ln 2+\frac32\text{Im}\left(\text{Polylog}\left(3,... | A related problem. Here is a different form for the solution in terms of the hypergeometric function
$$\int_0^{\frac{\pi}{4}}\ln ^3\sin x\text{d}x = -\frac{\pi\,\ln^3(2)}{32}-\frac{3}{8}\sqrt {2} \left(\ln^2( 2 ) + 4\,\ln
\left( 2 \right) +8 \right)\times$$
$${_7F_6\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/307593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that the difference of two consecutive cubes is never divisible by $3$. Here is my proof:
Let $n \in \Bbb Z$. Then, $n$ is of the form $2k$(even) or $2k + 1$(odd), for some $k \in \Bbb Z$.
Without loss of generality (not sure if I can use this), let $n = 2k$.
Then, $n + 1 = 2k + 1$.
$$\begin{align}
(n + 1)^3 - n^... | Let, number 1 = x then
number 2 = x+1
(If Difference of cubes is divisible by 3 then if we mod that value by 3 it should give reminder as 0.)
Now, [Difference of cubes] mod 3 ,
= [(x+1)^3 - x^3] mod 3
= [x^3 + 3(x)(1)(x+1) + 1 - x^3] mod 3
= [3(x)(x+1) + 1] mod 3
= 1 (because 3(x)(x+1) is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/308449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Using Taylor's Theorem to show that $\ln(1 + x^2) \leq x^2$ Can we show that if $\operatorname{abs}(x) \lt 1$, then $$\ln(1+x^2) \leq x^2\;,$$
using Taylor's Theorem?
I am thinking of expanding it about $x=0$ but I got something like
$$f(x) = -x^2 + \frac{x^4}{2} - \dots$$
Is my approach correct? Could you give me som... | Sure.
$$\ln(1 + x^2) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \ldots$$
In particular, the terms alternate sign. If we subtract $x^2$, we are left with
$$\ln(1 + x^2) - x^2 = -\frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} + \ldots$$
The first term of the sum is negative, and since the sum is a strictly alternating sum wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/308909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Solving an equation with fractional powers I was trying to find the maximum value for a function. I took the first derivative and arrived at this horrible expression:
$$ (x^2 + y^2)^\frac{3}{2} - y {\frac{3}{2}}(x^2 + y^2)^{\frac{1}{2}}2y = 0$$
How can I find the extrema by hand?
| Define a new variable:
$$ r = \sqrt{x^2 + y^2} $$
Then it follows:
$$ \begin{array}{rcl}
(x^2 + y^2)^\frac{3}{2} - y \frac{3}{2}(x^2 + y^2)^{\frac{1}{2}}2y & = & 0 \\
(x^2 + y^2)^\frac{3}{2} - 3(x^2 + y^2)^{\frac{1}{2}}y^2 & = & 0 \\
r^3 - 3y^2r & = & 0 \\
r^2 - 3y^2 & = & 0 \\
r^2 & = & 3y^2 \\
x^2 + y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Taylor expansion of $x^x-1$ around 1 How do I find Taylor expansion of(around 1):
$$f(x)=x^x-1$$
The answer should be:
$$(x-1)+(x-1)^2+\frac 12(x-1)^3+\cdots$$
How the answer was obtained?
| Using the power series for $\log(1+x)$, we get
$$
\begin{align}
(1+x)\log(1+x)
&=(1+x)\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots\right)\\
&=x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+\frac{x^6}{30}-\dots\tag{1}
\end{align}
$$
Plugging $(1)$ into the power series for $e^x$ yields
$$
(1+x)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Nested Summation I have the following nested sum :
$$\sum_{i=1}^{n}\sum_{j=1}^{i}\sum_{k=1}^{j}x = x+1$$
I don't have a clue how to solve this one, can somebody help me?
Thanks in advance.
${}{}$
| I will use the rules found on this page.
First, note that $x$ is not in the base or limit of any of the sums, so it can be "pulled out":
$$x\sum_{i=1}^n\sum_{j=1}^i\sum_{k=1}^j 1= x+1$$
Work from inside to outside, simplifying the sum:
$$\sum_{k=1}^j 1 = j$$
$$\sum_{j=1}^ij = \frac{i(i+1)}{2}=\frac{1}{2}\cdot\left(i^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the standard matrix for a linear transformation If $T: \Bbb R^3→ \Bbb R^3$ is a linear transformation such that:
$$
T \Bigg (\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} 5\\ 3 \\ 14 \\ \end{bmatrix}$$ $$T \Bigg (\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix} \Bigg) = \begin{bmatrix}-4 \\ 6... | The standard matrix has columns that are the images of the vectors of the standard basis
$$
T \Bigg (\begin{bmatrix}1\\0\\0\end{bmatrix} \Bigg),
\qquad
T \Bigg (\begin{bmatrix} 0\\1\\0 \end{bmatrix} \Bigg),
\qquad
T\Bigg (\begin{bmatrix}0\\0\\1 \end{bmatrix}\Bigg).
\tag{1}
$$
So one approach would be to solve a syste... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/313798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 0
} |
Value of $\sum_{k=1}^{\infty}\frac{1}{k^2+a^2}$ So my question is to find the value of $\sum_{k=0}^\infty\frac{1}{k^2+1}$ and more generally $\sum_{k=0}^\infty\frac{1}{Q(k)}$
where Q is a quadratic polynomial with no zeroes on the integers.
I can prove it converges, by the comparison test. I think I 've seen such a su... | Using the residue theorem, you can show that
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^2+a^2} = \frac{\pi}{a} \coth{\pi a}$$
This is equivalent to saying that
$$\sum_{n=1}^{\infty} \frac{1}{n^2+a^2} = \frac{1}{2} \left (\frac{\pi}{a} \coth{\pi a} - \frac{1}{a^2}\right )$$
You can also derive this by considering the Maclur... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/314986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
probability proof requiring conditional probability
Consider the experiment where two dice are thrown. Let $A$ be the event that the sum of the two dice is 7. For each $i\in\{1,2,3,4,5,6\}$, let $B_i$ be the event that at least one $i$ is thrown.
(a) Compute $P(A)$ and $P(A\mid B_1)$.
(b) Prove that $P(A\mid B_i)=P(A\... | Hint:
$\qquad\qquad\text{Event }A\\\begin{array}{c|c|c|c|c|c|c|c|}
& 1 & 2 & 3 & 4 & 5 & 6\\\hline
1& & & & & & \bullet \\\hline
2& & & & & \bullet & \\\hline
3& & & & \bullet & & \\\hline
4& & & \bullet & & & \\\hline
5& & \bullet & & & & \\\hline
6& \bullet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/315729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\lim\limits_{n\rightarrow \infty}\int_1^3\frac{nx^{99}+5}{x^3+nx^{66}} d x$ exists and evaluate it. I am trying to show that $\lim_{n\rightarrow \infty}\int_1^3\dfrac{nx^{99}+5}{x^3+nx^{66}} d x$ exists and what its value is. I know that to do this I must show that $\dfrac{nx^{99}+5}{x^3+nx^{66}}\rightarrow... | You are on the good way, and to complete the proof you can by triangle inequality do:
$$\left|\frac{5-x^{36}}{x^3+nx^{66}}\right|\leq \frac{5+x^{36}}{x^3+nx^{66}}\leq\frac{5+3^{36}}{1^3+n1^{66}}= \frac{5+3^{36}}{1+n}\rightarrow0,\quad\forall x\in[1,3]. $$
Hence, $\forall \epsilon>0,$ we can find $N\in\mathbb{N}$ such t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/316666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Calculate $1\times 3\times 5\times \cdots \times 2013$ last three digits. Calculate $1\times 3\times 5\times \cdots \times 2013$ last three digits.
| If $N=1\times 3\times 5\times \cdots \times 2013$
we need to find $N\pmod{1000}$
Now $1000=8\cdot125$
Now, $1\cdot3\cdot5\cdot7\equiv1\pmod 8$ (Find one generalization here )
So, $(1\cdot3\cdot5\cdot7)\cdots (2001\cdot2003\cdot2005\cdot2007)\equiv1\cdot1\cdots\cdot1\cdot1\equiv1\pmod 8 $
So, $N\equiv 1\cdot2009\cdot2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/318386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Range of $z,$ If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$ If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$. Then value of $z$ lie in the interval.
My Try:: Let $x,y,z$ be the roots of the quadratic equation
$t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$
Let $xyz = p, $ Then let $f(t) = t^3-3t^2-9t-... | We have $x+y+z = 3$ and $xy+yz+zx = -9$
$xy+z(3-z)=-9$
$\Rightarrow xy+3z-z^2=-9$
$\Rightarrow xy+3z-z^2+9=0$
$\Rightarrow -xy-3z+z^2-9=0$
$\Rightarrow z^2-3z-9=xy$.....(1)
By A.M G.M inequality we have,
$\displaystyle |xy|\le(\frac{x+y}{2})^2=(\frac{3-z}{2})^2$
Using this in 1 we have,
$-(\frac{3-z}{2})^2\le z^2-3z-9\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/318515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the remainder. Let $a,b$ be positive integers such that $7$ divide $a^2+b^2$ .How to find the remainder when we divide $ab-1$ by $7$
| Answer is 6
a^2+b^2+2=(a+b)^2-2ab+2
2(ab-1)=(a+b)^2-(a^2+b^2)-2
=49d+7k-2
ab-1=7[k+7d]/2-1=7[k+7d]/2-7+6
Note that both a,b are divisible by 7
Assume if not a=7n+1,7n+2,7n+3,b=7m+1,7m+2,7m+3,etc. and check
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Show that the curve $x^2+y^2-3=0$ has no rational points
Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.
Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a... | Suppose $a^2 + b^2 = 3 c^2$
Write $a = 3^p u$, $b = 3^q v$, and $c = 3^r w$,
where $u, v, w$ are all relatively prime to 3.
This does not assume that $a, b, c$ have no common divisor greater than 1.
Assume $p \le q$ (if $p > q$, switch their roles in what follows).
$a^2+b^2 = (3^p u)^2 + (3^q v)^2
= 3^{2p}(u^2+ 3^{2(q-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/319553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
} |
Find the value $\sum_{n=1}^{\infty}(e-(1+\frac{1}{n})^n)$ How to find the following series' value?
$$\sum_{n=1}^{\infty}\bigg(e-\Big(1+\frac{1}{n}\Big)^n\bigg)$$
| Taylor expansion of $\frac {e^{x} - 1} {x}$ around $x = 0$ is $$1 + \frac {x} {2} + \frac {x^2} {3} + \cdots > 1 + \frac {x} {2}.$$ Put $n = 1/x$ and let $n \to \infty$, we have $$n \left(e^{1/n} - \left(1 + \frac {1} {n}\right)\right) > \frac {1} {2n}.$$ By Minkowski inequality, we have $$e - \left(1 + \frac {1} {n}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/319914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
How to determine equation for $\sum_{k=1}^n k^3$ How do you find an algebraic formula for $\sum_{k=1}^n k^3$? I am able to find one for $\sum_{k=1}^n k^2$, but not $k^3$. Any hints would be appreciated.
| As you probably have already realized the formula
$$
\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2
$$
can be proved using induction.
For $n=1$ you have
$$
1^3 = \left(\frac{1\cdot 2}{2}\right)^2.
$$
Say that the formula holds for $n$ and prove that it holds for $n=1$. So you have/get
$$
\sum_{k=1}^{n+1} k^3 = \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/320985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 10,
"answer_id": 5
} |
$x\in [0,1]$ and $n\in N$ $\implies$ $\left| \frac{1-2xn}{n^2} \right| \leq \frac{1}{\sqrt n}$ Let $x\in [0,1]$ and $n\in N$. According to the solutions manual this is true:
$$\left| \frac{1-2xn}{n^2} \right| \leq \frac{1}{\sqrt n}$$
How can I see this is true ?
| Since $0 \le x \le 1$,
$0 \ge 1-2xn \ge 1-2n$.
So, $|1-2xn| \le 2n-1$.
We thus want to show that
$\frac{2n-1}{n^2} \le \sqrt{n}$
or $2n-1 \le n^{2.5}$.
Looking at the $2n-1$, I think of
$(n-1)^2 = n^2-2n+1$.
Since $(n-1)^2 \ge 0$,
$0 \le (n-1)^2 = n^2-2n+1$
or $2n-1 \le n^2$
and since $n^2 \le n^{2.5}$ we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Equal angles formed by the tangent lines to an ellipse and the lines through the foci. Given an ellipse with foci $F_1, F_2$ and a point $P$. Let $T_1, T_2$ the points of tangency on the ellipse determined by the tangent lines through $P$. Show that $\widehat {T_1 P F_1} = \widehat {T_2 P F_2}$.
| Let the ellipse be:
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
Set $P(a,t)$, the tangent line $PT_1$ is $y=k(x-a)+t$, $k=\tan A_1$, since it is tangent line, so put in ellipse equation, we have:
$$ b^2x^2+a^2(kx-ka+t)^2=a^2b^2$$
i.e.,
$$(b^2+a^2k^2)x^2+2a^2k(t-ka)x+a^2(t-ka)^2-a^2b^2=0$$
Solving for the point of tangency,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/323574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Fourier series of function $f(x)=0$ if $-\pi$$f(x) = \begin{cases}0 & \text{if }-\pi<x<0, \\
\sin(x) & \text{if }0<x<\pi.
\end{cases}$$
My attempt:
I went the route of expanding this function with a complex Fourier series.
$$f(x) = \sum_{n=-\infty}^{+\infty} C_{n}e^{inx}$$
$$C_{n} = \frac {1}{2\pi} \int_{0}^{\pi} \frac... | Note that, $C_1$ is a special case and you need to handle separately as
$$ C_1 = \frac {1}{2\pi} \int_{0}^{\pi} \sin(x) e^{-ix} dx \neq 0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Matrix 4x4 calculation Let $$A = \begin{pmatrix}
1 & a+3 & 8 & 2 \\
0 & 0 & 2 & 0 \\
0 & a-2 & 5 & 0 \\
0 & 10 & a & a+3
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z \\
w \\
\end{pmatrix} = \begin{pmatrix}
0 \\
7 \\
-2 \\
0 \\
\end{pmatrix}$$
I'm a bit stuck on this question... I need to find values of a for which t... | $$\begin{align}\det A &= 1\cdot\det\left(\begin{array}{ccc}0 & 2 & 0\\a-2 & 3 & 0\\10 & a & a+3\end{array}\right)\quad (\text{expanded down first column})\\ &= 1\cdot(a+3)\cdot\det\left(\begin{array}{cc}0 & 2\\a-2 & 5\end{array}\right)\quad(\text{expanded down third column})\\ &= 1\cdot(a+3)\cdot\bigl(0-2(a-2)\bigr)\\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/324896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Variable Exponents Question If $a,b$ and $c$ are different positive integers and $2^a\cdot2^b\cdot2^c =64$ then $2^a+2^b+2^c$=?
This is so far my work: I got $2^a\cdot2^b\cdot2^c=2^6$ then $abc=6$ is this so far in the right track?
| Note that
$2^a\cdot2^b\cdot2^c=2^{a+b+c}$ and $64=2^6$.
You know that $2^a\cdot2^b\cdot2^c=64$, then you have that $2^{a+b+c}=2^6$, and from this you conclude that $a+b+c=6$.
Solution $(a,b,c)= (1,2,3)$, therefore $2^a+2^b+2^c= 2^1+2^2+2^3=2+4+8=14$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/325127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Having $A_1=a+b+c$,$A_2=a^2+b^2+c^2$, $A_3=a^3+b^3+c^3$ - how to get $a,b,c$? Perhaps I'm just a bit dense at the moment - I've re-read some of my notes from monthes ago concerning elementary symmetric polynomials, and I find that I've no idea how to approach the "inverse" problem: if you have the -say- three values $A... | As $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca),$
$ A_1^2=A_2+2(ab+bc+ca)\implies ab+bc+ca=\frac{A_1^2-A_2}2$
Again, $a^3+b^3+c^3-3abc=(a+b+c)\{a^2+b^2+c^2-(ab+bc+ca)\}=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}$
$\implies A_3-3abc=A_1\left(A_1^2-3\frac{A_1^2-A_2}2\right)$
Express $abc$ in terms $A_1,A_2,A_3$
So, the equation whose roots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Generating function for the sequence $1,1,3,3,5,5,7,7,9,9,\ldots$ The generating function for the sequence $\left\{1,1,1,1,...\right\}$ is $$1 + x + x^2 + x^3 ... = \frac{1}{1-x}$$
What is the generating function for the sequence $\left\{1,1,3,3,5,5,7,7,9,9,\dots \right\}$?
This is my attempt:
what we want to do is af... | Here is another approach.
The generating function writes $(1+x)(1+3x^2 + 5x^4 + 7x^6 + 9x^8 + \dots)$ where the second term appears as the derivative of $x+x^3 + x^5 + x^7 + x^9 + \dots$
The later being the odd part of $1+x+x^2 + x^3 + x^4 + \dots$, we have:
$$
x + x^3 + x^5 + x^7 + x^9 + \dots = \frac{1}{2}\left(\frac... | {
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$(x-a)(x-b)(x-c)(x-d)=ex$ We can verify that $x=125,162,343$ are the roots of equation $(x-105)(x-210)(x-315)=2584x$.
My question is,Could you find five positive integers $a,b,c,d,e$, which $(x-a)(x-b)(x-c)(x-d)=ex$ has four positive integer roots?
Thanks in advance.
| a. A parametric solution can be given to,
$$(x-a)(x-b)(x-c)(x-d) = ex$$
such that $a,b,c,d$ and all four roots $x_i$ are all positive integers as,
{$a,b,c,d$} = {$2n,\; 2n+1,\; (n+1)(n+3),\;2(n+1)(n+2)$}, and $e = n(n-1)(n+1)(n+2)(2n+3)$
b. In general, this quartic problem can be solved by two quadratics. One must fi... | {
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$\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$ Find the value of $\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$, if $x\in \left(0,\frac{1}{2}\right)$. I know it is equal to $\sqrt{2}+1$, but I don't know how to prove it?
| $x$ can be set to $\sin^22\theta$ as $0<x<\frac 12\implies 0<\sin2\theta<\frac1{\sqrt 2}$
$\implies 0< 2\theta< \frac\pi4$
$1+\sqrt x=1+\sin2\theta=(\cos\theta+\sin\theta)^2\implies \sqrt{1+\sqrt x}=\cos\theta+\sin\theta$
$1-\sqrt x=1-\sin2\theta=(\cos\theta-\sin\theta)^2\implies \sqrt{1+\sqrt x}=\cos\theta-\sin\theta$... | {
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$\textrm{Res}\left(\frac{\log z}{z^3+8}; z_k\right) = \frac{-z_k \log z_k}{24}$ when $z_k$ solves $z^3+8=0$ The problem in the book is
Compute $\int_0^\infty \frac{dx}{x^3+8}$.
I set up the keyhole contour, apply the residue theorem, and go through the tedious algebra. I get stuck in doing so, but looking at the solu... | Things are a lot easier if you choose a contour which is a circular wedge shape encompassing a single pole. I chose a wedge angle of $2 \pi/3$ and used the residue theorem with just the one pole at $z=\pi/3$ and got the correct result, $\pi/(6 \sqrt{3})$. Let me explain.
Consider
$$\oint_C \frac{dz}{z^3+8}$$
where $C... | {
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The roll of the dice Find the probability of rolling a sum of three or less on two dice.
(Yes, I know this is an "easy" question but I second quess myslef and like second opinions.)
| $$
\begin{array}{c|cccccc} X+Y & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \\
1 & \color\green{2} & \color\green{3} & 4 & 5 & 6 & 7 \\
2 & \color\green{3} & 4 & 5 & 6 & 7 & 8 \\
3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\... | {
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$\left(\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\right)^2=\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2$ I'm having trouble proving $$\left(\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\right)^2=\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2$$ where $n\in\mathbb{Z}$ and $\theta\in\mathbb{R}$. Can anyon... | In order to proof
$$\sum_{k=1}^{n}{\sin \left(\varphi + k\alpha \right)}=\frac{\sin\left(\frac{\left(n+1\right)\alpha}{2}\right)\cdot\sin\left(\varphi+\frac{n\alpha}{2}\right)}{\sin \frac{\alpha}{2}}$$
observe
$$\sin \frac{\alpha}{2}\cdot \sin \left(\varphi + k\alpha\right)=\frac{1}{2}\left[\cos \left(\varphi + k\alph... | {
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Abstract algebra polynomial problem
Let $F$ be any field and $a,b\in F,\,\,a\neq b$. Find the greatest common divisor of $f(x) = x + a$ and $g(x) = x + b$.
Since the degree of both is $1$, the gcd is $1$ or $f(x)$ or $g(x)$, since $a\neq b$. So $\gcd(f(x),g(x))=1$.
Am I right for the answer and proving?
| Let $a \sim b$ denote that $a$ is associated with $b.$ The only divisors up to associates of $x+a$ are $1$ and $x+a$, and the only divisors of $x+b$ up to associates are $1$ and $x+b.$ Since $a\neq b$, $x+a$ is not associated to $x+b$ so the greatest common divisor is the only common divisor, $1.$
| {
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Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| A bit late.. but all the calculus and double sums aren't necessary.
$$\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}=\sum_{n=0}^{\infty}\frac{n+1}{2^{n+1}}=\sum_{n=1}^{\infty}\frac{n}{2^{n+1}}+\sum_{n=0}^\infty\frac{1}{2^{n+1}}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{n}{2^{n}}+1$$
Hence $$\displaystyle \sum_{n=1}^{\infty}\f... | {
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Simplify : $\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a}$? Just out of curiosity, is $$\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a},a \gt0\quad?$$
Thanks
| $$\sqrt{\frac{a}{2}} + \sqrt{\frac a2} = 2 \frac{\sqrt a}{\sqrt 2} = \frac {2}{\sqrt 2} \sqrt{a} = \sqrt 2 \sqrt a = \sqrt{2a} > \sqrt a$$
Added: perhaps your question originates from not remembering that $$(a + b)^2 \not\equiv a^2 + b^2\;?$$ If we do square each side of the proposed inequality (which we can do, witho... | {
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$x^4 - y^4 = 2z^2$ has no solution How do I prove that the equation $x^4 - y^4 = 2 z^2$ has no solutions using the fact that the equations $x^4 + y^4 = z^2$ and $x^4 - y^4 = z^2$ have no solutions.
I cant think of a method of reducing the above equation to one of these forms.
| IDENTITY--The identity $(X^4 - Y^4)^2 + 4(XY)^4 = (X^4 + Y^4)^2 $ solves the proposer's question.
The proposer assumes that there is a non-zero triple of integers X,Y,Z
such that $(X^4 - Y^4 ) = 2(Z^2)$. WOLOG we may assume $(X,Y,Z) =1$.
We substitute " $2(Z^2)$ " for $(X^4 -Y^4)$ in the above IDENTITY.
This tr... | {
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Taylor expansion of $\dfrac{e^x}{\cos x}$ Let $g\left(x\right) = \dfrac{e^x}{\cos x}$ and find the first four terms of its taylor polynomial.
The first four terms are: $$1 + x + x^2 + \frac{2x^3}{3} + \frac{x^4}{2}$$
Why do you have to solve this problem term by term? Why does just dividing the terms of the series wo... | I hate denominators.
First write ${e^x\over\cos(x)}=a_0+a_1x+a_2x^2+a_3x^3+\cdots$
Then, since I hate denominators, I consider $e^x=\cos(x)(a_0+a_1x+a_2x^2+a_3x^3+\cdots)$
Let's work from the right hand side, and take an intuitive guess that we only need expand $\cos(x)$ to a third degree polynomial:
$$\begin{align*... | {
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How to draw an ellipse if a center and 3 arbitrary points on it are given? How to draw an ellipse if a center and 3 arbitrary points on it are given?
| The generally centered 2D ellipse equation is
$$ \begin{pmatrix} x\\y\\1 \end{pmatrix}^\top \begin{bmatrix} C_{11} & C_{12} & 0 \\ C_{12} & C_{22} & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{pmatrix} x\\y\\1 \end{pmatrix} = 0 \\ C_{11} x^2 + C_{22} y^2 + 2 C_{12} x y - 1 = 0$$
Using the three points you can find the three c... | {
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If $\left |z-3\right |=\left |z+i\right |$, where $z=x+iy$, prove that $3x+y=4$
If $\left |z-3 \right |=\left |z+i\right |$, where $z=x+iy$, prove that $3x+y=4$.
I have got to the point where I have $\left |z \right |= \sqrt{x^2+(y+1)^2} = \sqrt{(x-3)^2+y^2}$
But really don't know where to go from here or if my start... | $|z-3|=\sqrt {(x-3)^2+y^2}$
$|z-i|= \sqrt {x^2+(y+1)^2}$
Now ,
$(x-3)^2+y^2=x^2+(y+1)^2$
$x^2+9-6x+y^2=x^2+y^2+2y+1$
$8=6x+2y$
| {
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Prove inequality: $\sum \frac{a^4}{a^3+b^3} \ge \frac{a+b+c}{2}$
Prove inequality with $a,b,c >0$ $$\frac{a^4}{a^3+b^3} +
\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3} \ge \frac{a+b+c}{2}$$
I tried the inequality: $\sum \frac {a^4+b^4}{a^3+b^3} \ge \sum \frac{a+b}2=a+b+c$, but seem like it doesn't help
| There is an extensive discussion of this inequality here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=23113
I like the sum of squares method the best:
\begin{align}
& \sum_{cyc}{\frac{a^4}{a^3+b^3}}-\frac{1}{2}\sum_{cyc}{a} \\
& =\frac{(a-b)^2(b-c)^2(c-a)^2(ab+ac+bc)^2}{4(a^3+b^3)(b^3+c^3)(c^3+a^3)} ... | {
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$\sin{\frac{A+B}{2}}+\sin{\frac{B+C}{2}}+\sin{\frac{C+A}{2}} > \sin{A}+\sin{B}+\sin{C}. $ Help me please to prove that:
for any $\triangle ABC$ we have the following inequality: $$\sin{\frac{A+B}{2}}+\sin{\frac{B+C}{2}}+\sin{\frac{C+A}{2}} > \sin{A}+\sin{B}+\sin{C}. $$
It's about convexity ?
thanks :)
| Hint: Eliminate $\angle C$
$A+B+C=180^0 \implies C=180-(C+B)$
Now you just need to show :
$\sin{\frac{A+B}{2}}+\sin{(90-\dfrac{A}{2})}+\sin{(90-\dfrac{B}{2})} \ge \sin{A}+\sin{B}+\sin{(180-(A+B))}$
$\sin{\frac{A+B}{2}} +\cos{\frac{B}{2}}+\cos{\frac{A}{2}} \ge \sin A+ \sin B+ \sin(A+B)$
| {
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Can't find solutions for $\tan{2x} = \tan{x}$ Solving $\tan{2x}=\tan{x}$
Reducing the left side: $$\frac{\sin{2x}}{\cos{2x}} = \frac{2\sin{x}\cos{x}}{2\cos^{2}(x)-1}.$$
Reducing the right side: $$\tan{x} = \frac{\sin{x}}{\cos{x}}.$$
therefore:
$$\frac{2\sin x\cos x}{2\cos^2x-1} = \frac{\sin x}{\cos x}$$
$$\frac{2\cos x... | Recall that
$$\tan(A+B) = \dfrac{\tan(A) + \tan(B)}{1-\tan(A) \tan(B)}$$
Hence,
$$\tan(2x) = \dfrac{2\tan(x)}{1-\tan^2(x)} = \tan(x)$$
This implies $\tan(x) = 0 \text{ or }1-\tan^2(x) = 2$. $1-\tan^2(x) = 2$ is not possible if $x \in \mathbb{R}$. Hence, $$\tan(x) = 0 \implies x = n \pi$$
| {
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Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stu... | Hint: Complete the square in your last expression $2x^2 + 12x + 16$.
Solution:
Completing the square gives
$$2x^2 + 12x + 16 = 2(x^2 + 6x + 4) = 2(x^2 + 2\cdot 3x + 3^2 - 1) = 2(x+3)^2 - 2$$
You already showed that your inequality is equivalent to $2x^2 + 12x + 16 \geq 0$.
Using the result of completing the sqare, we... | {
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Prove: $\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$ How can I show:
$$\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$$
I've tried multiplying with its "conjugate" but that doesn't seem to help that much.
Thanks!
| Probably you meant
$$\lim _{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$$
Which can dealt using the identity
$$x^3-y^3=(x-y)(x^2+y^2+xy)$$
now apply it for $x=x^{\frac{1}{3}}, \,y=(x+1)^{\frac{1}{3}}$
so multiply the numerator and denominator with
$x^\frac{2}{3}+(x+1)^\frac{2}{3}+((x+1)x)^\frac{1}{3}$
and you will get:... | {
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Proving that $a_n$ is an integer for every $n$ For every $k\ge1$ integer number if we define the sequence : $a_1,a_2,a_3,...,$ in the form of :$$a_1=2$$
$$a_{n+1}=ka_n+\sqrt{(k^2-1)(a^2_n-4)}$$
For every $n=1,2,3,....$ how to prove that $a_n$ is an integer for every $n$
| $a_{n+1}=ka_n+\sqrt{(k^2-1)(a^2_n-4)}\Rightarrow a_{n+1}-ka_n=\sqrt{(k^2-1)(a^2_n-4)}$
Squaring both sides we have,
$a_{n+1}^2+k^2a_n^2-2ka_{n+1}a_n=k^2a_n^2-4k^2-a_n^2+4$
$\Rightarrow a_{n+1}^2+k^2a_n^2-2ka_{n+1}a_n-k^2a_n^2+4k^2+a_n^2-4=0 $
$\Rightarrow a_{n+1}^2-2ka_{n+1}a_n+4k^2+a_n^2-4=0 \dots (1)$
Replacing $n+1$... | {
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$f(x - 1) + f(x − 2) $ and the sum of coeficients If $f(x-1)+f(x-2) = 5x^2 - 2x + 9$
and
$f(x)= ax^2 + bx + c$
what would be the value of $a+b+c$?
I was doing
$f(x-1)+f(x-2)= f(x-3)$
then
$f(x)$
a = 5
b = -2
c = 9
$(5-3)+(-2-3)+(9-3)$
But do not think is is correct
What would be correct approach?
| $$f''(x)=2a, 4a=f''(x-1)+f''(x-2)=10, a=2.5$$
$$f'(x)=2ax+b=5x+b, 5(x-1)+5(x-2)+2b$$
$$f'(x-1)+f'(x-2)=10x-2, -15+2b=-2, b=6.5$$
$$f(x)=ax^2+bx+c, 2.5(x-1)^2+6.5(x-1)+2.5(x-2)^2+6.5(x-2)+2c=f(x-1)+f(x-2)=5x^2-2x+9$$
let $x=0$ on both side, $2.5-6.5+10-13+2c=9, c=8$
so $a+b+c=2.5+6.5+8=17$
| {
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Multivariable limit $\lim_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1}$ $$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} $$
According to my textbook the limit equals $2$.
What I have tried:
Using the squeeze theorem:
$$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 ... | The following "trick" can be useful. Let $x=r\cos\theta$, $y=r\sin\theta$. Then we are finding
$$\lim_{r\to 0}\frac{r^2}{\sqrt{r^2+1}-1}.$$
This can be computed using any of the usual one variable techniques.
| {
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"question_score": "7",
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"answer_id": 1
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Why is $\displaystyle\int^{1/2}_{-1/2} \cos(\tfrac{2}{3} \pi x)\, dx= 0$? Why is
$$\int^{1/2}_{-1/2} \cos\left(\frac{2}{3} \pi x\right)\, dx= 0\qquad ?$$
Because mathematically, we have
$$\sin\left( \frac{2}{3}\pi \right) - \sin\left(-\frac{2}{3}\pi\right) = 2\sin\left(\frac{2}{3}\pi\right)$$ instead.
And what about
... | $$\int\limits_{-1/2}^{1/2}\cos\frac{2}{3}\pi x\,dx=\left.\frac{3}{2\pi}\sin\frac{2}{3}\pi x\right|_{-1/2}^{1/2}=\frac{3}{2\pi}\left(\sin\frac{\pi}{3}-\sin\left(-\frac{\pi}{3}\right)\right)=$$
$$=\frac{3}{\pi}\sin\frac{\pi}{3}=\frac{3\sqrt 3}{2\pi}\neq 0$$
| {
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What goes wrong in this derivative? $$ f(x) = \frac{2}{3} x (x^2-1)^{-2/3} $$
and f'(x) is searched.
So, by applying the product rule $ (uv)' = u'v + uv' $ with $ u=(x^2-1)^{-2/3} $ and $ v=\frac{2}{3} x $, so $ u'=-\frac{4}{3} x (x^2-1)^{-5/3} $ and $ v' = \frac{2}{3} $, I obtain
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3... | So
$$\begin{align}
(uv)' = u'v + uv' &= \frac{-4}{3}x(x^2 - 1)^{-5/3}\frac{2}{3}x + (x^2 - 1)^{-2/3}\frac{2}{3} \\
&= \frac{-8}{9}x^2(x^2 - 1)^{-5/3} + \frac{2}{3}(x^2 - 1)^{-2/3}\\
&= \frac{-\color{green}8}{\color{green}9}\color{green}x^\color{green}2(x^2 - 1)^{-5/3} + \frac{\color{red}2}{\color{red}3}(\color{red}x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/346797",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then $3$ divides both $a$ and $b$. If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then show that $3$
divides both $a$ and $b$.
can anyone tell me please how to solve these types of problem oe which formula is required
| As Glen has shown, $9\mid \{(a-b)^2+3ab\}$
$\implies 3\mid \{(a-b)^2+3ab\}\implies 3\mid(a-b)^2$
$\implies 3\mid(a-b)\implies 9\mid(a-b)^2$
$\implies 9\mid3ab\implies 3\mid ab$
But if $3$ divides $a,3$ must divide $b$ and conversely as $3\mid(a-b)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/348964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Showing that a $3^n$ digit number whose digits are all equal is divisible by $3^n$
Let $c$ be a $3^n$ digit number whose digits are all equal. Show that $3^n$ divides
$c$.
I have no idea how to solve these types of problems. Can anybody help me please?
| A number whose digits are all equal and of length $3^n$ is thus of the form $c = \sum\limits_{i=0}^{3^n-1} a \cdot 10^i = a \dfrac {10^{3^n}-1}{10-1}$ by the geometric series.
Since we have to account for the possibility that $a = 1$ we need to show that $3^n \mid \dfrac{10^{3^n}-1}9$, i.e. $3^{n+2} \mid 10^{3^n}-1$.
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/349952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Evaluate $\int\limits_0^\infty \frac{\cos(ax)}{\cos(bx)}\frac{1}{1+x^2}dx$ I would like to show that
$$\text{PV}\int_0^\infty \frac{\cos(ax)}{\cos(bx)}\frac{1}{1+x^2}dx = \frac{\pi}{2}\mathrm{sech}(b)$$
using complex analysis. $a$ and $b$ are real numbers and $a \neq b$.
Please give some hints.
| What the question asking for cannot be right!
At least for $0 < a < b$, we have:
$$\begin{align}\operatorname{PV} \int_0^{\infty} \frac{\cos a x}{\cos b x} \frac{dx}{1+x^2}&= \frac12 \operatorname{PV} \int_{-\infty}^{\infty} \frac{\cos a x}{\cos b x}\frac{dx}{1+x^2}
\\&= \frac12 \lim_{\epsilon\to 0+} \Re\left[\int_{-\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/350017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many intersections has the following two curves at the point $(0,0)$? The curves are as follows:
$y^2+yx^2-x^3$ and $y^2-x^5$
| Let $x=z^2,y=z^5$
$\implies (z^5)^2+z^5\cdot z^2-(z^2)^3=0\implies z^6(z^4+z-1)=0$
So, $z$ has $6$ repeated values for $z=0$,
So, $x$ will have $\frac62=3$ repeated values as one value $t^2$ of $x$ corresponds to two values of $z$ namely, $\pm t$
If $x=0,y^2=0\iff y=0$
So,we will have $3$ intersections.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/350206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that for every positive integer $n$, $1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n$ Base case: n=1. $1/1\le 2-1/1$. So the base case holds.
Let $n=k\ge1$ and assume
$$1/1^2+1/2^2+1/3^2+\cdots+1/k^2\le 2-1/k$$
We want to prove this for $k+1$, i.e.
$$(1/1^2+1/2^2+1/3^2+\cdots+1/k^2)+1/(k+1)^2\le 2-\frac{1}{k+1}$$
This i... | Let's assume it's true for $k$.
Then $\sum_{i=1}^k\dfrac{1}{i^2}\leq2-\dfrac 1 k$
Let's try $k+1$.
$\sum_{i=1}^{k+1}\dfrac{1}{i^2}=\sum_{i=1}^k\dfrac{1}{i^2}+\dfrac{1}{(k+1)^2}\leq2-\dfrac 1 k+\dfrac{1}{(k+1)^2}$
As $\dfrac{1}{(k+1)^2}-\dfrac 1 k<-\dfrac{1}{k+1}$ ,
$\sum_{i=1}^{k+1}\dfrac{1}{i^2}\leq2-\dfrac{1}{(k+1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Geometry Question - Trihedral angles, planar geometry, spherical geometry
Two rays, $OX$ and $OY$, are drawn in the horizontal plane $\pi$, and the third ray, $OZ$, is drawn in space so that the rays $OX$, $OY$, and $OZ$ form a trihedral angle $OXYZ$. The planar angles of this trihedral angle are $a = \angle ZOY$, $b ... | For simplicity, choose $X$, $Y$, $Z$ along their rays such that $|OX|=|OY|=|OZ|=1$. Let $P$ be the foot of the perpendicular dropped from $Z$ into the plane of $\triangle XOY$; then, the angle you seek ---call it "$\theta$"--- is given by
$$\sin\theta=\frac{|PZ|}{|OZ|} = |PZ|$$
We just need a useful expression for $|PZ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/354269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Conjugate Ferrers diagrams
Let $\pi=\langle \pi_1,\pi_2,... \rangle , \ \pi_1\ge\pi_2\ge...,$ be a partition of a number and $\pi'=\langle \pi_1',\pi_2',... \rangle$ be a partition conjugated to $\pi$, which means that Ferrers diagram for $\pi'$ is transposed Ferrers diagram for $\pi$. For example partition conjugated... | HINTS: If you take the first, third, fifth, etc. of a set of $n$ elements, you end up taking $\left\lceil\frac{n}2\right\rceil$ elements; if you take the second, fourth, sixth, etc., you end up taking $\left\lfloor\frac{n}2\right\rfloor$ elements
First identity:
$$\begin{array}{l|l}
\begin{array}{ccc}
\pi\\ \hline
\col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/354797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a parametrization of a hyperbola who has a fixed signature How do I find parametrization of the hyperbola $x^2-y^2=1$ which is the unit sphere of a quadratic form with signature $(1,-1)?$
The only parametrization that comes to mind is $x=\cosh t,y=\sinh t$. What is a quadratic form of a parametrization?
| $$\text {As }(a^2+b^2)^2=(2ab)^2+(a^2-b^2)^2$$
For $ab\ne 0,$
$$\left(\frac{a^2+b^2}{2ab}\right)^2=1+\left(\frac{a^2-b^2}{2ab}\right)^2\implies (x,y)\text{ can be}\left(\frac{a^2+b^2}{2ab},\frac{a^2-b^2}{2ab}\right)$$
For $a^2-b^2\ne 0\implies a\ne \pm b$
$$\left(\frac{a^2+b^2}{a^2-b^2}\right)^2=1+\left(\frac{2ab}{a^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Equivalent modulos In my course notes I came onto two examples for finding remainders of huge numbers with fermat's little theorem and need some helping analyzing them.
1) Find the remainder when 5^183 is divided by 11.
I know that 5^10 ≡ 1 mod(11) so (5^(10*18)) = 5^180*5^3.
That ≡ 1(5^3)mod11 but then the exam... | For the first question, $5^3=5^2\cdot 5$. But $5^2\equiv 3$.
For the second, unless there is a typo, I suspect the question has shape find the remainder when $a^{2^{68}}$ is divided by $19$. For the calculation, if $a$ is not divisible by $19$, we want to use the fact that $a^{18}\equiv 1\pmod{19}$. So we want to begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the number of triples $(a,b,c)$ with $-47 \leq a,b,c \leq 47$ and $a+b+c >0$
Kate is looking for ordered triples $(a,b,c)$ of distinct integers such that $-47 \leq \; a,b,c \; \leq 47$ and $a+b+c>0$. How many such ordered triples can Kate find?
My first step was to let $a_1=a+47, \; b_1=b+47$ and $c_1=c+47$ ... | Let $N = 47$ and $X = \{-N,\ldots,N\}$. We have $|X^3| = |X|^3 = (2N+1)^3$.
Let $\mathscr{N}_{???}$ be the number of solutions for any equation ??? involving $(a,b,c) \in X^3$.
Because of symmetry,
$$\mathscr{N}_{a+b+c>0} = \mathscr{N}_{a+b+c<0} \implies \mathscr{N}_{a+b+c>0} = \frac12 \left((2N+1)^3 - \mathscr{N}_{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/356806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Solving functional equation for generating function Find the functional equation for the generating function whose coefficients satisfy $$
a_n = \sum_{i=1}^{n-1}2^ia_{n-i}, \text{ for } n\ge 2, a_0 = a_1 = 1
$$
This is what I've tried so far:
$$
\begin{align}
g(x) -1 -x &= \sum_{n\ge2} \sum_{i=1}^{n-1} 2^i a_{n-i}\\
&=... | We have
\begin{align}
g(x) & = \sum_{n=0}^{\infty} a_n x^n = 1 + x + \sum_{n=2}^{\infty} \sum_{i=1}^{n-1} 2^i a_{n-i}x^n = 1 + x + \sum_{i=1}^{\infty} \sum_{n=i+1}^{\infty} 2^i a_{n-i} x^n\\
& = 1+x+\sum_{i=1}^{\infty} \sum_{k=n-i=1}^{\infty} 2^i a_{k} x^{k+i} = 1+x + \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} (2x)^i a_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/357888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
limit with $\arctan$ I have to find the limit
and want ask about a hint:
$$\lim_{n \to \infty} n^{\frac{3}{2}}[\arctan((n+1)^{\frac{1}{2}})- \arctan(n^{\frac{1}{2}})]$$
I dont have idea what to do. Derivatives and L'Hôpital's rule are so hard
| Here is another approach, just compute the Taylor series of $\arctan((x+1)^{\frac{1}{2}})- \arctan((x)^{\frac{1}{2}}$ at the point $x=\infty$
$$ \arctan((x+1)^{\frac{1}{2}})- \arctan((x)^{\frac{1}{2}}= \frac{1}{2}\, \frac{1}{x^{3/2}}-\frac{7}{8}\, \frac{1}{x^{5/2}}+O \left( \frac{1}{x^{7/2}} \right). $$
So the above ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/358368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find this $\frac{1}{2m+1}+\frac{1}{2n+1}=\frac{2}{2k+1}$ Let $m,n,k\in \mathbb{N}$, and $m,n,k\ge 1,m\neq n\neq k\neq m $, such that
$$
\dfrac{1}{2m+1}+\dfrac{1}{2n+1}=\dfrac{2}{2k+1}
$$
Is there a solution? Or does this not have any solution?
| $$\frac{1}{2m+1}+\frac{1}{2n+1}=\frac{2n+1+2m+1}{(2m+1)(2n+1)}=\frac{2(n+m+1)}{(2m+1)(2n+1)}$$
So this is of the form $\frac{2}{2k+1}$ if and only if $$(n+m+1)\mid(2m+1)(2n+1)$$
Now $(2n+1)(2m+1)=4nm + 2(n+m) + 1 = 4nm -1 + 2(n+m+1)$. So we want:
$$(n+m+1)\mid 4nm-1$$
Let $d=n+m+1$. Then $n\equiv -1-m\pmod d$ and there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/358479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Test the convergence of $\sum_{n=1}^{n=\infty}\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)$. Test the convergence of
$$
\sum_{n=1}^{\infty}\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)
$$
$$
\frac{u_{n+1}}{u_{n}}=\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}}
$$
$$
\lim_{n\to\infty}\frac{... | A straightforward computation gives
$$
\lim_{n\to\infty}\frac{u_{n+1}}{u_{n}}
=\lim_{n\to\infty}\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}}
=\lim_{n\to\infty}\frac{(n+1)^4+2^{n+1}}{2^{n+1}(n+1)^2}\frac{n^2 2^n}{n^4+2^n}
=\lim_{n\to\infty}\frac{n^2}{(n+1)^2}\frac{2^n}{2^{n+1}}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/359294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find this $a,b,c$ such that $\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$ It is known that$$\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$$
for exactly one set of positive integers $(a,b,c)$ where $0<c<90$
find the value
$$\dfrac{b+c}{a}$$
my idea,$ \sin 50^\circ >\sin 45^\circ >\frac{_5}{^8} $
so$\sqrt{9-8\sin 50^{\cir... | $$\sin c-\cos(2c)$$
$$\implies \cos(90-c)-\cos(2c)$$
$$\implies -2\sin\Bigg(\dfrac{90+c}2\Bigg) \sin \Bigg(\dfrac{90-3c}2\Bigg)=-\sin(50^{\circ})$$
$$2\sin\theta \sin\phi=\sin 50^{\circ}$$
So, one of the solution comes when one of $\sin \theta^{\circ}$ or $\sin\phi$
is equal to $1/2$ and other is $\sin 50^{\circ}$
So, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/359594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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To find the logarithm of $1728$ to the base $2 \sqrt{3}$
Find the logarithm of: $1728$ to base $2\sqrt{3}$.
Let, $\log_{2\sqrt{3}} 1728 = y$, then
$$\begin{align} (2\sqrt{3})^y &= 1728\\
2^y(\sqrt3)^y &= 1728\\2^y(3^\frac12)^y &= 1728\\2^y(3^\frac y2)
&= 1728\\2^y × 3^\frac y2 &= 2^6 × 3^3 \end{align}$$
What should... | $$12^3 = 1728 = (2\sqrt 3)^y = (\sqrt 12)^y = (12^\frac 1 2)^y = 12^\frac y 2 $$
$$ 3 = \frac y 2 $$
$$ y = 6 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/361461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Solve the equation: $x^2+\frac {1}{x^2}=2^{1-y^2}$. I found the following problem interesting but do not know how to proceed:
Solve the equation: $$x^2+\frac {1}{x^2}=2^{1-y^2}.$$
Can someone point me in the right direction?
| We can proceed with the problem in the following way:
We know that for any non-negative $a,b ; \quad a+b \geq 2 \sqrt {ab}$. So, $x^2 + \frac {1}{x^2} \geq 2 \sqrt {x^2. \frac {1}{x^2}}=2$. Hence the min. value of the left hand side of the equation is $2$. On the other hand ,the max. value of the
right hand side is $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/362388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Expressing $\cos\theta - \sqrt{3}\sin\theta = r\sin(\theta - \alpha)$ My book explains that $a\cos\theta + b\sin\theta$ is a sine (or cosine) graph with a particular amplitude/shift (i.e. $r\sin(\theta + \alpha)$) and shows me some steps to solve for $r$ and $\alpha$:
$$r\sin(\theta + \alpha) \equiv a\cos\theta + b\sin... | In your specific case you get either
$$\sin\alpha = \frac{1}{2}\text{ and } \cos\alpha=-\frac{\sqrt{3}}{2}\text{ for } r=2$$
or
$$\sin\alpha = -\frac{1}{2}\text{ and } \cos\alpha=\frac{\sqrt{3}}{2}\text{ for } r=-2$$
In the first case you get $\alpha=\frac{5\pi}{6}$, and in the second case you get $\alpha=-\frac{\pi}{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/363222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find the standard matrix of the transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ that corresponds to the reflection through the line Find the standard matrix of the transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ that corresponds to the reflection through the line $x_2=2x_1$ followed by reflection through the line $x_1=... | First you change coordinates to where your basis is given by $\begin{pmatrix} 1 \\ 2\end{pmatrix}$ and $\begin{pmatrix} 2 \\ -1\end{pmatrix}$.
This is done by multiplying by
$$\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}^{-1} = \frac{1}{5}\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}.$$
Next, you reflect the second coor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/367270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Deriving Closed Form for a Recursion via Generating Functions Consider (1) $a_{n+2} = 2a_{n+1} - a_n + 4n3^n$ with $a_0 = a_1 = 1$.
Using generating functions and setting $A(x) = \sum a_nx^n$ we obtain
$$\begin{align*}&\quad\sum a_{n+2}x^{n+2} = \sum2a_{n+1}x^{n+2} - \sum a_nx^{n+2} + \sum 4n3^nx^{n+2}\\ &\implies [A(x... | This is how I would approach this. Of all the terms $4n3^n$does not depend on the $a_i$, so I'll keep that one apart. The first thing is to view an expression obtained from the remaining terms as a multiple of $A(x)$. So first separate
$$
a_{n+2}-2a_{n-1}+a_n = 4n3^n.
$$
We can recognise the left hand side as the coe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
complex integral with parametrization of an ellipse I'm having trouble with the following complex integral:
$$\int_{C}^{}\frac{dz}{\sqrt{1 - z^{2}}}$$ where C is a positively oriented ellipse $${x^{2}\over a^{2}} +{ y^{2}\over b^{2}} = 1$$ where $$a^{2} - b^{2} = 1$$
I know z(t) = $a\cos(t) + ib\sin(t)$ but I'm not sur... | As you say: $z(t) = a\cos t + ib\sin t$ gives a bijection from $[0,2\pi)$ to the ellipse. Given the definition of $z$, we can see that $\operatorname{d}\!z=(-a\sin t + ib\cos t) \, \operatorname{d}\!t$. Hence:
$$\oint_C \frac{\operatorname{d}\!z}{\sqrt{1-z^2}} = \int_0^{2\pi} \frac{-a\sin t + ib\cos t }{\sqrt{1-(a\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the Laurent series at the given point and its residue
a) $\displaystyle \frac{\sin(z)}{(z-\pi)^2}$ at $z_0=\pi$
b) $\displaystyle \frac{1}{1-\cos(z)}$ at $z_0=0$
I'm having trouble understanding Laurent series, please help!
| A related problem. For the second one, you can advance as
$$ \frac{1}{1-\cos(z)} = \frac{1}{\frac{z^2}{2!}-\frac{z^4}{2!}+\frac{z^6}{6!}-\ldots }= \frac{1}{\frac{z^2}{2!} \left(1-\frac{2! z^2}{4!}+\frac{2! z^4}{6!}-\ldots \right) } $$
$$ =\frac{2}{z^2} \left(1-\left(\frac{2! z^2}{4!}-\frac{2! z^4}{6!}+\ldots\right) \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Express ${x^3 + 2x^2 + 61 \over (x + 3)^2 (x^2 + 4)}$ in partial fractions $$\eqalign{
& {x^3 + 2x^2 + 61 \over (x + 3)^2(x^2 + 4)} \equiv {A \over (x + 3)} + {B \over (x + 3)^2} + {Cx + D \over (x^2 + 4)} \cr
& \equiv {A(x + 3)(x^2 + 4) + B(x^2 + 4) + (Cx + D)(x + 3)^2 \over (x + 3)^2(x^2 + 4)} \cr} $$
so:
$$x^... | You're doing the problem right. I don't think there's a way of avoiding solving a $3\times 3$ system. However, you can try "nice" values for $x$ to make it a nice system, e.g. $x=0, x=1, x=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? In this recent answer to this question by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$
I would like to know if this resu... | Using the fact that: $$x^3+y^3=(x+y)(x^2-xy+y^2)$$ and letting $x=m^{\frac{1}{3}}$ and $y=n^{\frac{1}{3}}$, we get the statement: $$m+n=(m^{\frac{1}{3}}+n^{\frac{1}{3}})(m^\frac{2}{3}-(mn)^{\frac{1}{3}}+n^{\frac{2}{3}})$$Maybe this will help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/374619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 6,
"answer_id": 4
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Verifying a proof that if $x,y,z \geq 0$ and $x+y+z = 1$, then $0 \le xy + yz + zx - 2xyz \le \frac{7}{27}$ I was working some recreational problems from a book (The Art and Craft Of Problem Solving, Zeitz) and came across one from the '84 IMO:
Suppose that $x, y, z$ are non-negative reals, with $x + y + z = 1$. Prove... | For proving the second inequality, take $ x=a+\frac{1}{3}, y=b+\frac{1}{3}, z=c+\frac{1}{3}$. Since $x+y+z=1$, so $a+b+c=0$. Since $x,y,z\geq 0$, so $a,b,c\geq -\frac{1}{3}$.By simple algebraic manipulation we get,
$$xy+yz+zx-2xyz=\frac{2}{3}(ab+bc+ca-3abc)+\frac{7}{27}$$
We just need to show that $ab+bc+ca-3abc\leq 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/374679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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I cannot find the last factor of this expression? I'm supposed to factor $x^8-y^8$ (the exponents are 8 for both if it is too difficult to see) as completely as possible. It is easy to factor this to $(x+y)(x-y)(x^2+y^2)(x^4+y^4)$. However, the book says "(Hint: there are 5 factors. Note that we sat real coefficients, ... | Another approach: if you know that
$$x^4+1=(x^2+\sqrt 2x+1)(x^2-\sqrt 2x+1)$$
then you can do as follows:
$$x^4+y^4=y^4\left(\frac{x^4}{y^4}+1\right)=y^4\left(\frac{x^2}{y^2}+\sqrt 2\frac xy+1\right)\left(\frac{x^2}{y^2}-\sqrt 2\frac xy+1\right)=$$
$$=\left(x^2+\sqrt 2xy+y^2\right)\left(x^2-\sqrt 2xy+y^2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Differential equation with no constants. Is there a way to calculate $\psi$ which is a function of $x$ out of this differential equation:
$$
\frac{d^2 \psi}{d x^2} = x^2 \psi
$$
| $\dfrac{d^2\psi}{dx^2}=x^2\psi$
$\dfrac{d^2\psi}{dx^2}-x^2\psi=0$
Note that this belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0205.pdf.
Let $\psi=e^{-\frac{x^2}{2}}y$ ,
Then $\dfrac{d\psi}{dx}=e^{-\frac{x^2}{2}}\dfrac{dy}{dx}-xe^{-\frac{x^2}{2}}y$
$\dfrac{d^2\psi}{dx^2}=e^{-\frac{x^2}{2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/376181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Polynomial and distinct roots Find all real m such that $x^3-2x^2-2x+m$ has 3 distinct rational roots.
Source: School exam paper. No idea why it seems so hard
I don't think rational root theorem works, since m is not necessarily integer (clearly it is rational though)
Assuming $m=\frac{r}{s}$, we need to check $sx^3-2s... | Let $a, b, c$ be the rational roots. Then we have
$$a+b+c=2,\quad ab+bc+ca=-2, \quad m=-abc.$$
Replace $c=2-a-b$,
$$a^2+b^2+ab-2(a+b)-2=0.$$
(If $a=b$, then $a$ is irrational) Thus,
$$\Delta_b=(a-2)^2-4(a^2-2a-2)=-3a^2+4a+12$$
is a square rational.
The rest is easy.
Added after diner. Since
$$-3a^2+4a+12=-3(a+\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_0^\tau \frac{t\sin(t z)}{z\cos(t z)-\sin(tz)}\text{d}t$ I'm trying to evaluate the following definite integral. Mathematica gives me a complicated expression which I think I can simplify, but I was wondering if there was a "nice" way to evaluate it.
$$\displaystyle\int_0^\tau \frac{t\sin(t z)}{z\cos(t z)... | This integral doesn't seem to have an elementary expression, at least for general $\tau$. At least Mathematica and my own results agree on this, as the final answer involves polylogarithm.
Let's represent the integral as:
$$I=\frac{1}{z^2} \int_0^{\tau z} \frac{y \sin y}{z \cos y-\sin y}dy$$
My attempt uses integration... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the limit of $ x_n = \prod_{j=2}^{n} \left(1 - \frac{2}{j(j+1)}\right)^2$ I am stuck on the following problem:
Let $x_n=(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2 \ldots...(1-\frac{1}{n(n+1)/2})^2, \text{where} \space n \geq 2$. Then $\lim_{n \to \infty}x_n=?$
I see that $x_n^{\frac{1}{2}}=\frac{2}{3... | Let $f(n)=\frac{n-1}{n+1}$. Then show $$1-\frac{1}{n(n+1)/2} = \frac{f(n)}{f(n+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.