Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How I find $\gcd(\frac{a^{2m+1}+1}{a+1}, a+1)$? $$\gcd\left(\frac{a^{2m+1}+1}{a+1}, a+1\right)$$
My answer until this moment is:
$$
For: a^{2m+1}+1^{2m+1} = (a + 1)^{m+1} - 2a = (a+1)^m(a+1)-2a
$$ where
\begin{align}
(a + 1)^{m} = a^{2m}+a^{2m-1}+\cdots+1
\end{align}
So,
\begin{align}
a^{2m+1}+1 = [a^{2m}+a^{2m-1}+\... | $$
\begin{align}
\frac{a^{2m+1}+1}{a+1}
&=\frac{((a+1)-1)^{2m+1}+1}{a+1}\tag1\\
&=\sum_{k=1}^{2m+1}(-1)^{k-1}\binom{2m+1}{k}(a+1)^{k-1}\tag2\\[6pt]
&\equiv2m+1\pmod{a+1}\tag3
\end{align}
$$
Explanation:
$(1)$: $a=(a+1)-1$
$(2)$: Binomial Theorem and algebra
$(3)$: the $k=1$ term is the only one without a factor of $a+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Second Method to Find the Volume of a Slice of a Cone
I was looking to formulate a general solution for a vertical slice of a cone. After failing to do this by integrating for the area of an ever shrinking chord segment, I eventually came up with a more geometric solution by subtracting the triangular part from the ar... | A cone of radius $R$ and height $H$
would be:
$z = H - \frac {H}{R} \sqrt {x^2 + y^2}$
And we are going to slice it at some line $y = a$
$v = \int_a^R\int_{-\sqrt{R^2 - x^2}}^{\sqrt {R^2- x^2}} (H - \frac {H}{R} \sqrt {x^2 + y^2}) \ dx \ dy$
Convert to polar:
$x = r\cos \theta\\
y = r\sin \theta\\
z = z\\
dy\ dx = r \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating definite integral $\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$ Question: $$\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$$ is equals to
(a) $-\frac\pi6$
(b) $-\frac{\pi}{12}$
(c) $\frac\pi{12}$
(d) $\frac\pi6$
My attempt: Denoting given integral by $I$ and letting $z=e^{iθ}$ then giv... | COMMENT.-An unorthodox way. The function is continuous and its minimum and maximum are taken in $\dfrac{3\pi}{2}$ and $\dfrac{\pi}{2}$ respectively. Then we have (using areas of rectangles)
$$2\pi\cdot0.0555\approx\dfrac{\pi}{10}\lt\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta\lt 2\pi\cdot\frac 18=\frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to solve the following trigonometrical equation? I have the following equations
\begin{align*}
R_1\cos(\omega T_1-\phi_{1})& =Q_1-R_2\cos(\omega T_2-\phi_{2})\\
R_1\sin(\omega T_1-\phi_{1})& =-Q_2-R_2\sin(\omega T_2-\phi_{2}).
\end{align*}
From these equations how can I obtain the following solution
$$\omega T_2=\p... | Another way to tackle the same problem of solving $a \sin \theta + b \cos \theta + c = 0$ is to use the tan-half-angle substitutions
$$\begin{aligned}
t &= \tan \left( \frac{\theta}{2} \right)\, \rightarrow & \theta = 2 \arctan(t) \\
\sin(\theta) & = \frac{2 t}{1+t^2} \\
\cos(\theta) & = \frac{1-t^2}{1+t^2}
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Convergence of integral with trigonometry
*
*$\displaystyle \int_{-\frac{\pi}{4}}^\frac{\pi}{4} (\frac{\cos x - \sin x}{\cos x + \sin x})^{\frac{1}{3}} dx$
Obviously, problem in $ -\frac{\pi}{4} $ as $ \lim_{x\to -\frac{\pi}{4}^+} = \infty $, integral is positive on whole segment, but I can't use any usual rules t... | Recall that
*
*$\cos x - \sin x=\sqrt 2 \sin \left(\frac{\pi}4-x\right)$
*$\cos x + \sin x=\sqrt 2 \sin \left(\frac{\pi}4+x\right)$
then
$$ =\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)^{\frac{1}{3}} dx
=\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\frac{\sin \left(\frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Non negative integer triplets $(x,y,z)$ in $x^2+2y^2+4z^2+5=2(x+y+xy)+4z(y-1)$
Non negative integer triplets $(x,y,z)$ in $x^2+2y^2+4z^2+5=2(x+y+xy)+4z(y-1)$
Try: Writting equation as $$4z^2-4(y-1)z+x^2+2y^2-2(x+y+xy)+5=0$$
Now if equation has real roots. Then $$z=\frac{(y-1)\pm \sqrt{2xy+2x-x^2-y^2-4}}{2}$$
So for ... | The discriminant was:
$$\begin{align}
2xy+2x-x^2-y^2-4&=k^2\\
\text{let} \qquad y=x \pm &t \\
2x(x \pm t)+2x-x^2-(x \pm t)^2-4&=k^2\\
2x^2 \color{red}{\pm 2tx}+2x-x^2-x^2 \color{red}{\mp 2tx}-t^2-4&=k^2 \\
2x-t^2-4&=k^2 \\
x&=\frac{1}{2}(k^2+t^2+4) \\
y=x\pm t&=\frac{1}{2}(k^2+t^2 \pm 2t+4) \\
&=\frac{1}{2}\left[k^2+(t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
perpendicular distance from $ax-by = 1$ to origin I have a problem from a basic number theory book that asks for the perpendicular distance from the line $ax - by = 1$ to the origin.
My approach was to find the area of the triangle formed by the line and the axes, find the base of the triangle situated at the diagonal... | Minimum Distance
Since
$$
\begin{align}
1
&=\left|\,ax-by\,\right|\\
&=\left|\,(a,-b)\cdot(x,y)\,\right|\\
&\le\left|\,(a,-b)\,\right|\left|\,(x,y)\,\right|
\end{align}
$$
we have
$$
\begin{align}
\left|\,(x,y)\,\right|
&\ge\frac1{\left|\,(a,-b)\,\right|}\\
&=\frac1{\sqrt{a^2+b^2}}
\end{align}
$$
If $(x,y)=\frac{(a,-b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the number of integers $n$ such that the equation $xy^2 + y^2 - x - y= n$ has an infinite number of integer solutions $(x,y)$.
Find the number of integers $n$ such that the equation $xy^2 + y^2 - x - y= n$ has an infinite number of integer solutions $(x,y)$.
Firstly the equation can be rearranged into $(y-1)(y+x... | If $x(y+1)+y=\frac{n}{k}$ for infinitely many values of $x$, then for any two such values $x_0$ and $x_1$ we have
$$(x_0-x_1)(y+1)=0.$$
where $x_0-x_1\neq0$, and so $y+1=0$.
The argument can be put more simply; given any integer $n$ and a solution $(x,y)$ to
$$n=xy^2+y^2-x-y=(y-1)(y+x(y+1)),$$
it follows immediately t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If a,b,c are in harmonic progression value of $\frac{3a+2b}{2a-b} + \frac{3c+2b}{2c-b}$? If a, b and c form a harmonic progressions what is the value of $\frac{3a+2b}{2a-b} + \frac{3c+2b}{2c-b}$?
I tried to do this by substituting $a = \frac{1}{p-q}$, $b=\frac{1}{p}$ and $c=\frac{1}{p+q}$. I got the result $\frac{5p-2q... | Your result is correct.
Notice that $\frac{5p-2q}{p+q}+\frac{5p+2q}{p-q}$ can be written as
$$=\frac{5p^2-2pq-5pq+2q^2+5p^2+2pq+5pq+2q^2}{p^2-q^2}$$
$$=\frac{5p^2+5p^2+2q^2+2q^2-2pq+2pq-5pq+5pq}{p^2-q^2}$$
$$=\frac{10p^2+4q^2}{p^2-q^2}$$
which is nothing but $$10+\frac{14q^2}{p^2-q^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Probability of a $1000 \times 1000$ square matrix over $\mathbb{Z}_2$ having full rank There are only two entries, $0$ and $1$, over $\mathbb{Z}_2$. Thus, only $16$ possible $2\times2$ matrices over $\mathbb{Z}_2$, and $6$ of them have full rank:
$$\begin{pmatrix}0&1\\
1&0\end{pmatrix} \quad
\begin{pmatrix}1&1\\
1... | The general linear group $GL(n,q)$ is the group of invertible $n\times n$ matrices over a field with $q$ elements (note $q=p^k$ for some prime $p$).
The order of $$|GL(n,q)|=\prod_{k=0}^{n-1}(q^n-q^k)$$
So the probability is $$\dfrac{1}{q^{n^2}}\prod_{k=0}^{n-1}(q^n-q^k)$$
For $n=2$ $q=2$ you get $\frac{1}{2^4}(2^2-2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find $\int{\arctan x}\,\mathrm dx$ without substitution 2018-08-15: I'm still looking for an answer that does not rely on
$$\int{f\left[g(x)\right]g'(x)}\,\mathrm dx = F\left[g(x)\right]$$
I'm refreshing my old calculus skills, and the textbook (Kalkulus by Tom Lindstrøm, 3rd edition, a Norwegian book) asks me to fin... | $\newcommand{\dx}{\mathrm dx\,}$The only other way I can see without using a u-substitution, which must I mention, is the easiest way to evaluate this integral, is using the infinite geometric sequence. Hopefully you remember that$$\sum\limits_{n\geq0}(-1)^n\, x^{2n}=\frac 1{1+x^2}$$
Therefore, calling the integral $\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
cross product not associative, outer product associative The cross product is not associative. If $i=(1,0,0)$, $j=(0,1,0)$ and
$k=(0,0,1)$, then
\begin{eqnarray}
i \times (i \times j) = i \times k = -j \\
(i \times i) \times j = 0
\end{eqnarray}
However in Geometric Algebra, if $e_1=(1,0,0)$, $e_2=(0,1,0)$ and
$e_3=... | The basic point is that a difference of associative bilinear products will still be bilinear but need not be associative at all. Still, it's instructive to get an explicit expression for the associator of the cross product.
Let $I := e_1 e_2 e_3$, so that
$$
\forall a,b \in \mathbb{R}^3, \quad a \times b = -I(ab - a \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding value of an algebraic expression. I am given $$2x = a-\frac{1}{a} \;\text{and} \; 2y=b-\frac{1}{b} .$$
I have to find the value of $ xy +\sqrt{(x²+1)(y²+1)}.$
One way to solve this is to simply putting the value of $x$ and $y$ and get the answer, but that's a very lengthy process though lucid. But I am looking ... | I do not think there is any special trick in solving the problem.
$$x=\frac{a}{2}-\frac{1}{2a}$$
$$y=\frac{b}{2}-\frac{1}{2b}$$
$$xy=\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab}$$
$$x^2+1 = \frac{a^2}{4}-\frac{1}{2}+\frac{1}{4a^2}+1=\frac{a^2}{4}+\frac{1}{2}+\frac{1}{4a^2}=(\frac{a}{2}+\frac{1}{2a})^2$$
$$y^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Compute $\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx.$
Evaluate $\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx.$
I tried so many substitutions but none of them led me to the right answer:
$u=\frac 1{\sqrt{x+3}}$, $u=\frac 1{x+3}$, $u=\sqrt{x}$... I even got to something like $\int_0^1 \frac {u^2}{(u^2+3)^{\frac 32}}du$... | If you change $u=\sqrt{x} \Rightarrow u^2=x \Rightarrow 2udu=dx$, then:
$$\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx=\int_0^1\frac {2u^2}{(u^2+3)^{3/2}}du=\int_0^1\frac {2u^2+6-6}{(u^2+3)^{3/2}}du=\\
2\int_0^1\frac {1}{(u^2+3)^{1/2}}du-6\int_0^1\frac {1}{(u^2+3)^{3/2}}du.$$
Both integrals you can evaluate by $u=\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Orthonormal diagonalizable Let:
$$ A = \begin{pmatrix} 5 &2 & -1 \\ 2 & 2 & 2 \\ -1 & 2 & 5 \end{pmatrix} \in Mat_3(\mathbb{R})$$
1) Show that $0$ and $6$ are eigenvalues for $A$ and find the basis for the corresponding eigenspace.
2) Explain why $A$ is orthonormal diagonalizable and find an orthonormal basis for $\m... | Note that $A$ is symetric hence by the spectral theorem it's orthogonal diagonalizable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve the initial value problem $ (x+1)^2 dx+(2xy+x^2-1)dy =0 , \ y(0)=1 \ $ Solve the initial value problem $$ (x+1)^2 dx+(2xy+x^2-1)dy =0 \ , \ \ y(0)=1 $$
Answer:
Consider the above equation with $ \ M(x,y)dx+N(x,y)dy=0 \ $ , then we get
$ M(x,y)=(x+1)^2, \ N(x,y)=2xy+x^2-1 \ $
Since $ \ \frac{\partial M}{\partial ... | Hint:
$(x+1)^2~dx+(2xy+x^2-1)~dy=0$
$(2xy+x^2-1)\dfrac{dy}{dx}=-x^2-2x-1$
$\left(y+\dfrac{x}{2}-\dfrac{1}{2x}\right)\dfrac{dy}{dx}=-\dfrac{x}{2}-\dfrac{1}{2}-\dfrac{1}{2x}$ with $y(0)=1$
This belongs to an Abel equation of the second kind.
Let $u=y+\dfrac{x}{2}-\dfrac{1}{2x}$ ,
Then $y=u-\dfrac{x}{2}+\dfrac{1}{2x}$
$\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof of trig identity using t-formulae Show that $$2\arctan x = \arccos\frac{1-x^2}{1+x^2}$$
if $ x > 0$, and
$$2\arctan x = -\arccos\frac{1-x^2}{1+x^2}$$ if $x <0.$
I have been able to equate the expressions by letting $x=\tan(y/2)$, but I do not know how to show the different signs of the equation for $x>0$ and $x<... | The simplest way to go is probably to calculate $\;\cos(2\arctan x)$.
Set $\theta=\arctan x$ and use the duplication formula:
\begin{align}
\cos 2\theta&=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-x^2}{1+x^2},\\
\text{so }\qquad
2\arctan x&\equiv \pm\arccos\frac{1-x^2}{1+x^2}\mod 2\pi.
\end{align}
Now either $\;0\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the matrix representation of a linear map Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a linear map and consider the matrix representation $$M_{B}^{B} =\begin{pmatrix}
4 & 6\\
6 & 3
\end{pmatrix}$$
with respect to the basis $B=\bigg\{ \begin{pmatrix}
2 \\
2
\end{pmatrix} , \begin{pmatrix}
3\\
... | $$f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) \color{red}{=4e_1 + 6e_2}$$
This is where you are wrong, $4$ and $6$ are coordinates in the basis $B$, so there is one more step to get $e_1$ and $e_2$ into the game. Otherwise your approach is nice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric equation: $2\arcsin \left(\frac{2x}{1+x^2}\right)- \pi x^3 = 0$
The number of solutions of the equation $$2\arcsin \left(\dfrac{2x}{1+x^2}\right)- \pi x^3 = 0$$ is?
Let $x= \tan \theta$
$\implies \sin 2\theta = \sin(\dfrac \pi 2 \tan^3\theta)$
I had to delete the rest of my attempt because it was total... | Let
$$
f(x)=2\arcsin\frac{2x}{1+x^2}-\pi x^3.
$$
Then $f'(x)=-3\pi x^2-\frac{4}{1+x^2}<0$ if $|x|>1$, namely $f(x)$ is decreasing in $(-\infty,-1)$ and $(1,\infty)$. Also
$$ f'(x)=-3\pi x^2+\frac{4}{1+x^2}, |x|<1$$
and
$$ f''(x)=-\frac{2x[4+3\pi(x^2+1)^2]}{(x^2+1)^2}, |x|<1. $$
So $f''(x)>0$ if $x\in(-1,0)$ and $f''(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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The value of $\sum_{1\leq l< m How to solve this summation ? Also, I'm not sure what does $1\leq l< m <n$ supposed to imply in the development of summation form.
$$\sum_{1\leq l< m <n}^{} \frac{1}{5^l3^m2^n}$$
This one is from the Galois-Noether Contest in 2018:
Galois-Contest
| \begin{align*}\sum_{1\leq l< m <n}^{} \frac{1}{5^l3^m2^n}&=\sum_{l=1}^\infty \frac{1}{5^l}\sum_{m=l+1}^\infty \frac{1}{3^m}\sum_{n=m+1}^\infty\frac{1}{2^n}\\&=\sum_{l=1}^\infty \frac{1}{5^l}\sum_{m=l+1}^\infty \frac{1}{3^m}\cdot\frac{1}{2^m}\\&=\sum_{l=1}^\infty \frac{1}{5^l}\cdot\frac{1}{6^l}\cdot\frac15\\&=\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Solving probability problem with complementation approach. Problem
If it is assumed that all $\binom{52}{5}$ poker hands are equally likely, what is the probability of being dealt one pair (i.e. a,a,b,c,d where a,b,c, and d are all distinct)?
I was trying to solve this by complementation.
Let us find number of ways ... | You ignored the part "where $a$, $b$, $c$, and $d$ are all distinct". You counted the hands that contain at least one pair, not the ones that contain exactly one pair.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the pattern to a general matrix Consider the following $(n+1) \times (n+1)$ matrix:
$$\begin{bmatrix} n & 1 & 0 & 0&\cdots &0& 0 & 0 & 0 \\ -n & n-2 & 2 &0& \cdots &0& 0 & 0 & 0 \\ 0 & 1-n & n-4 &3& \cdots &0& 0 & 0 & 0 \\ 0 & 0 & 2-n &n-6& \cdots &0& 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \v... | Add row 1 to row 2
and row 2 is
[0, n-1, 2, 0, ...].
Then add row 2 to row 3
and row 3 is
[0, 0, n-2, 3, 0, ...].
Then add row 3 to row 4
and row 4 is
[0, 0, 0, n-3, 4, 0, ...].
I think that,
with a proper specification
of each row,
this can reduce the matrix
to upper triangular
with the main diagonal being
[n, n-1, n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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find the length of the curve $y= \int_{-2}^x\sqrt{3t^4-1} \, dt$ I'm stuck for a while not sure how to continue, or if there's a mistake that I did that prevent me to continue.
$$y= \int_{-2}^x\sqrt{3t^4-1} \ dt \ , \ -2≤x≤-1$$
$$ y = F(x) - F(-2) $$
$$ y' = f(x) - f(-2) $$
$$ y' = \sqrt{3x^4-1} - \sqrt{47} $$
$$ (y'... | Notice by the FTC
$$ y' = \sqrt{3x^4-1}$$
and Thus,
$$ (y')^2 +1 = 3x^4 $$
it follows that
$$ \mathcal{L} = \int\limits_{-2}^{-1} \sqrt{3}x^2 dx = ... $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Showing that $P(x)$, where $P(x) = P(x+2)-x^2-2$ for all $x$, is a third-degree polynomial I want to show that $P(x)$ is a third degree polymomial:
$P(x) = P(x+2) -x^2-2$ for every real $x$
It is not so difficult, but I am not seeing the proof straight away.
| Let:
$$\begin{align}P(x+2)&=a_0(x+2)^n+a_1(x+2)^{n-1}+a_2(x+2)^{n-2}+\cdots+a_n\\
P(x)&=\color{red}{a_0}x^n+\color{blue}{a_1}x^{n-1}+a_2x^{n-2}+\cdots+a_n \\
x^2+2=P(x+2)-P(x)&=2na_0x^{n-1}+(2n(n-1)a_0+2(n-1)a_1)x^{n-2}+\cdots \Rightarrow \\
x^2&=2na_0x^{n-1} \Rightarrow \\
2&=n-1 \quad \text{and} \quad 2na_0=1\Rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Showing that $f(x,y)=(\lvert x\rvert -\lvert y\rvert)\log\left(2x^2+\lvert y\rvert\right)$ has no local extrema I have an exercise asking to find the local extrema of $$ f(x,y)=(\lvert x\rvert -\lvert y\rvert)\log\left(2x^2+\lvert y\rvert\right)$$if they exist... Wolfram Alpha tells me they don't, but I think I've spen... | In the following I am assuming that $(x,y) \neq (0,0)$.
Since $f(x,y) = f(|x|,|y|)$, we need only examine $x\ge 0, y \ge 0$.
For $x>0,y>0$ we have
$f_x(x,y) = {1 \over 2x^2+y} (4x (x-y) + (2x^2+y) \log(2x^2+y))$,
$f_y(x,y) = {1 \over 2x^2+y} ((x-y) - (2x^2+y) \log(2x^2+y))$.
Suppose $f_x(x,y) = f_y(x,y) = 0$, then the ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to integrate the product of two or more polynomials raised to some powers, not necessarily integral This question is inspired by my own answer to a question which I tried to answer and got stuck at one point.
The question was:
HI DARLING.
USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FO... | Looking at your previous deleted post, one answer suggested to use Euler subtitution
$$\sqrt{x^2-3x+2}=t+x\implies x=\frac{2-t^2}{2t+3}\implies dx=-\frac{2 (t+1) (t+2)}{(2 t+3)^2}\,dt$$ Replacing, we arrive to
$$\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}}=\frac{2 (t+1)^2 \left(3 t^4-4 t^3-2 t^2+56 t+60\right)}{(2 t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prime Numbers and Canonical Factorization. Find $n$ such that $2^{n} \mid 3^{1024}−1$, where $n$ is an integer. I factorized $3^{1024}−1$ as $a^{2}-b^{2}$ and showed $2 \mid 3^{1024}-1$. Help me to get $n$.
| $$2^n||3^{1024}-1$$
We know that $2^{10}=1024$ and $a^2-b^2=(a+b)(a-b)$
Now,
$$3^{2^{10}}=(3^{2^9}+1)(3^{2^9}1)$$
$$=(3^{2^9}+1)(3^{2^8}+1)(3^{2^8}-1)$$
$$=(3^{2^9}+1)(3^{2^8}+1)(3^{2^7}+1)(3^{2^7}-1)$$
$$.$$
$$.$$
$$=(3^{2^9}+1)(3^{2^8}+1)(3^{2^7}+1)....(3^{2^1}+1)(3^{2^0}+1)(3-1)$$
To find the largest $n$, $2^n||3^{1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the PDE: $(xy-x^2)\frac{\partial u}{\partial x} + y^2 \frac{\partial u}{\partial y} + (e^{y/x}+yz)\frac{\partial u}{\partial z} = 0$ As stated in the title, I want to solve the following PDE:
$(xy-x^2)\frac{\partial u}{\partial x} + y^2 \frac{\partial u}{\partial y} + (e^{y/x}+yz)\frac{\partial u}{\partial z} = 0... | $$ \frac{dx}{xy-x^2} = \frac{dy}{y^2} = \frac{dz}{e^{y/x} + yz} \quad \text{is OK.} $$
There is a sign mistake in your first integral. $\psi_{1} = \frac{y}{x} + \ln|y|$ is false and should be :
$$\psi_{1} = \frac{y}{x} -\ln|y|$$
Or, equivalently with $c_1=e^{\psi_1}$ :
$$c_1=\frac{1}{y}e^{y/x}$$
$ \frac{dy}{y^2} = \fr... | {
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Prove $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$ when $a=b+c$ I want to prove this identity:
$$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) \qquad\text{when}\;a=b+c$$
Can somebody give a hint in the easiest way possible? I am debugging this for hours and can't get the left side to be t... | Let $\, X := e^{ia}, Y := e^{ib}, Z := e^{ic}. \,$ Now expand and factor
$$ W \!:=\! \sin^2 a \!+\! \sin^2 b \!+\! \sin^2 c \!-\! 2(1 \!-\! \cos a \cos b \cos c) \!=\!
\frac{(X Y \!-\! Z) (Y Z \!-\! X) (Z X \!-\! Y) (X Y Z \!-\! 1)}{(2 X Y Z)^2}.\,$$
This leads to
$$\, W = 4 \sin\frac{a+b-c}2 \sin\frac{a-b+c}2 \sin\... | {
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Convergence of $\sum_{n=1}^{\infty} 3^n \sin(\frac{1}{4^nx})$ I wish to prove the convergence of:
$$\sum_{n=1}^\infty 3^n \sin\left(\frac 1 {4^nx}\right)$$
for $1\le x \lt \infty$, using Cauchy's criterion.
Here is what I tried:
\begin{align}
|S_{n+p}-S_n| & = \left| 3^{n+1} \sin\left(\frac 1 {4^{n+1}x}\right)
+ \cd... | If you absolutely want to use Cauchy's criterion, you can onserve that, if $n$ is large enough, $0<\frac1{4^n x}<\frac\pi2$, hence
$$0<\sin\frac1{4^n x}<\sin\frac1{4^{n+1}x}<\dotsm<\sin\frac1{4^{n+p}x},$$
and $\;\sin \dfrac1{4^{n+k}x}< \dfrac1{4^{n+k}x}<\dfrac1{4^{n+k}}$, so that, by the triangle inequality,
$$|S_{n+... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$ where $a,b$ are finite natural numbers
Evaluate $$\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$$ where $a,b$ are finite natural numbers
I have spent about a day thinking over this problem. I tried int... | If we assume $a,b>0$ and set
$$ I(a,b)=\int_{0}^{\pi/2}\log(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta $$
we have $I(a,b)=I(b,a)$ from the substitution $\theta\mapsto\frac{\pi}{2}-\theta$. On the other hand $I(a,a)=\pi\log(a)$ is trivial, so $I(a,b)=\pi\log\left(\frac{a+b}{2}\right)$ is a very reasonable conjecture. In... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can't prove an equation using induction I have an equation
$$
\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{n}{2^n} = 2 - \frac{n + 2}{2^n}
$$
Below is what I have already done:
$$
\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{n}{2^n} + \frac{n + 1}{2^{n+1}} = 2 - \frac{n + 2}{2^n} + \frac... | You forgot about the negative sign:
$$
-\frac{n+2}{2^{n}}+\frac{n+1}{2^{n+1}}=\frac{-(2n+4)+n+1}{2^{n+1}}=\frac{-n-3}{2^{n+1}}=-\frac{(n+1)+2}{2^{n+1}}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factoring $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$ I am trying to factor the expressions $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$.
I am rather stuck though. Is there a general method for going about this? I always find myself having to g... | And
$$
(x - a)^3 (b - c)^3 + (x - b)^3 (c - a)^3 + (x - c)^3 (a - b)^3 = 3 (a - b) (a - c) (b - c) (a - x) (b - x) (c - x)
$$
This is easy to conclude because $a,b,c$ are roots
| {
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Determining rational functions that simplify to a simple expression
Question: Is there a way to find, if possible, another way to write $(\sqrt[3]2-1)^2$ and $(\sqrt[3]4-1)^2$, in the form of $\frac {a+b}{c+d}$?
What I meant was that, let's take the second expression $(\sqrt[3]4-1)^2$ as an example. The expression is... | For $\,(\sqrt[3]4-1)^2\,$, using $(a-b)(a^2+ab+b^2)=a^3-b^3$ gives $\,\displaystyle
\left(\sqrt[3]{4}-1\right)\left(\sqrt[3]{4^2}+\sqrt[3]{4}+1\right) = 3
$.
Therefore $\,\displaystyle
\sqrt[3]{4}-1=\frac{3}{2 \sqrt[3]{2}+\sqrt[3]{4}+1} = \frac{3}{\left(\sqrt[3]{2}+1\right)^2}\,$.
Then, using $\,a^2-b^2=(a-b)(a+b)\,$ g... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Induction Problem from Putnam and Beyond I have copied the first step of the solution below in order to save time and space, the problem, with full solution, can be found in the book Putnam and Beyond.
Question:
Let $a_1, a_2, \ldots, a_n, \ldots$ be a sequence of distinct positive integers. Prove that for any positiv... | An inductive approach would use $$a_1{}^2 + \dots + a_n{}^2 \ge \frac{2n + 1}3\bigg(a_1 + \dots + a_n\bigg)$$ to establish $$a_1{}^2 + \dots + a_n{}^2 + a_{n+1}{}^2\ge \frac{2(n+1) + 1}3\bigg(a_1 + \dots + a_n + a_{n+1}\bigg)$$
by using transitivity of $x \ge y \text{ and } y \ge z \text{ implies } x \ge z$. So $x,y,z... | {
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"timestamp": "2023-03-29T00:00:00",
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Given relatively prime positive integers $a,b>1$, how many positive integers are there which are not non-negative integer combination sum of $a,b$? Let $a,b>1$ be relatively prime positive integers. I know that $ab-a-b$ is the largest positive integer which can not be written as a sum of non-negative integer combinatio... | Let $N = ab$, we count the number of integers $0\leq t \leq N$ that can be written as
$$
t = ma+nb
$$
such that $m,n\geq 0$. Let there be $R$ such numbers. Since the largest non-representable (positive)-integer is $ab-a-b$, all non-representable numbers are in $[0,N]$ and hence we know there are $N+1 - R$ of them.
If... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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The ring $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$ is a UFD. Contradiction? Considering the ring $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$, I know that this ring is an euclidian domain and therefore a unique factorization domain.
Now, $23=(3+4\sqrt{2})(3-4\sqrt{2})=(11+7\sqrt{2})(11-7\sqrt{2... | There is just a calculation mistake, since
$$
(3+4\sqrt{2})(3-4\sqrt{2}) = 9 - 32 = -23
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding limit using Euler number $$\lim_{x\rightarrow \infty }\left ( 1+\frac{3}{x+2} \right )^{3x-6}$$
I've tried to factor and simplfy the expression. I got:
$${\left ( 1+\frac{3}{x+2} \right )^{\frac{1}{x+2}}}^{3({x^2-4})}$$
I set $x$ to $1/t$ I get:
$${\left ( 1+\frac{3}{\frac{1}{t}+2} \right )^{\frac{1}{\frac{1}{t... | To get the result notice that $3x-6 = 9 \cdot \frac{x+2}{3} - 12$ then you limit is
$$
\lim_{x \to +\infty} \left[ \left( 1 + \frac{1}{\frac{x+2}{3}} \right)^{\frac{x+2}{3}} \right]^9 \cdot \left( 1 + \frac{3}{x+2} \right)^{-12}.
$$
The second term goes to $1$, while the term inside the brackets goes to $e$. Therefore,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $1+\frac11(1+\frac12(1+\frac13(...(1+\frac1{n-1}(1+\frac1n))...)))=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+...+\frac1{n!}$? I'm trying to prove that
$$1+\frac11(1+\frac12(1+\frac13(...(1+\frac1{n-1}(1+\frac1n))...)))=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+...+\frac1{n!}$$
Using induction, suppose that
$$1+\fr... | You can also start calculating the inner bracket first:
\begin{align}
&1+\frac11\left(1+\frac12\left(1+\cdots + \frac1{n-2}\left(1+\frac1{n-1}\left(1+\frac1n\right)\cdots\right)\cdots\right)\right) \\&= 1+\frac11\left(1+\frac12\left(1+\cdots + \frac1{n-2}\left(1+\frac{n+1}{n(n-1)}\right)\cdots\right)\right) \\&=
1+\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\lim_{n \to \infty} \int_{0}^{1} x^2 \left(1+\frac{x}{n}\right)^n dx$ Evaluate $$I=\lim_{n \to \infty} \int_{0}^{1} x^2 \left(1+\frac{x}{n}\right)^n dx$$
My try:
we have
$$I=\lim_{ n\to \infty} \frac{1}{n^n} \int_{0}^{1} \left((x+n)^2-2n(x+n)+n^2\right)(x+n)^ndx$$ $\implies$
$$I=\lim_{ n\to \infty} \frac{1}{n... | I learned a thing or two about the DCT theorem, so let's give it a try. Observe that if $t \in [0,1] \implies \ln(1+t) \le t$ and substitute $t = \dfrac{x}{n}, 0 \le x \le 1, n \ge 1\implies \ln(1+\dfrac{x}{n}) \le \dfrac{x}{n}\implies 1+\dfrac{x}{n} \le e^{\frac{x}{n}}\implies (1+\dfrac{x}{n})^{n} \le e^x\implies |f_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Integrating $F(x,y,z)=\frac{(0,xz,-xy)}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}$ along a circle
Let $F:\mathbb R^3 - \{0\}\to \mathbb R$ be given by $$F(x,y,z)=\frac{(0,xz,-xy)}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}.$$ Compute $\int_C F\cdot ds$ where $C$ is the unit cirlce on the plane $x+y+z=3$ with the orientation from the point $$\le... | Obviously $D$ is a disk of radius $1$ in the given plane. You just need to find it's origin. The format of the given points on the circle hints that the center is $(1,1,1)$, which you can verify.
I would do a change of variables, to some $X,Y,Z$ such that $X,Y$ are in the plane, then $Z=0$.
| {
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"url": "https://math.stackexchange.com/questions/2831775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Triangles and Similarity. ADEN is a square. BMDF is a square such that F lies on AD and M lies on the extension of ED. C is the point of intersection of AD and BE. If the area of triangle CDE is 6 square units, what is the area of triangle ABC?
I tried solving this as :
tria.CDE is similar to tria.BME is similar to tr... | Let DE = a, DM = b
*
*Area of $\triangle ABF = \frac{1}{2} \cdot b \cdot (a-b)$
*Area of $\triangle CBF = \frac{1}{2} \cdot b \cdot CF$
*Area of $\triangle CDE = \frac{1}{2} \cdot a \cdot CD$
*$CF + CD = b$ and $CF = \frac{b}{a} CD$. Thus, $CD = \frac{ab}{a+b}$ and $CF = \frac{b^2}{a+b}$
Now, from 1, 2 & 4: area... | {
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Proving $x^2+x+1\gt0$ I was doing a question recently, and it came down to proving that $x^2+x+1\gt0$. There are of course many different methods for proving it, and I want to ask the people here for as many ways as you can think of.
My methods:
*
*$x^2+x+1=(x+\frac12)^2+\frac34$, which is always greater than $0$.
... | Put $x=\dfrac{b}{a}$
$x^2+x+1=\dfrac{b^2}{a^2}+\dfrac{b}{a}+1$
$=\dfrac{b^2+ab+a^2}{a^2}=\dfrac{\dfrac{1}{2}(a+b)^2+\dfrac{b^2}{2}}{a^2}+\dfrac{1}{2}>
0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 21,
"answer_id": 20
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Determine Minimum Value.
Find the minimum value of
$$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}$$
for $x>0$.
When $x=1$, $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=6$$
I tried to plot some points on a graph and I observed that the minimum value is $6$.
Any hints would be sufficient.
Thanks... | Hint-$$\frac{(x+\frac1x)^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})} = \frac{(x+\frac{1}{x})^6-(x^3+\frac{1}{x^3})^2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}=(x+\frac{1}{x})^3-(x^3+\frac{1}{x^3})=3(x+1/x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Finding the $LU$ factorization of the matrix
Find the $LU$ factorization of the matrix:
$$\begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix}$$
I am aware that I need to find $L=\begin{bmatrix} 1 & 0 & 0 \\ * & 1 & 0 \\ * & * & 1 \end{bmatrix}$ and $U=\begin{bmatrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 &... | Gaussian Elimination without Pivoting is as follows. The primary purpose of Gaussian elimination if you follow this is to find $\ell_{jk}$ which zeros out the row below. That is why it is the ratio of the two rows and then you subtract them. This continues on and on.
Suppose that
$$ A = \begin{bmatrix} 1 & 1 & 1 \... | {
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"timestamp": "2023-03-29T00:00:00",
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Minimum $e$ where $a,b,c,d,e$ are reals such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ I have a question about this 1978 USAMO problem:
Given that $a,b,c,d,e$ are real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the maximum value that $e$ can attain.
I had the following solution:
Let $a+... | I don't understand, wouldn't the minimum value be $0$? if $$a=b=c=d=2$$ and $$e=0$$ both equations are satisfied. Clearly, zero would also be the smallest positive number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Triples $(x, y, z)$ that satisfy a set of equations
Suppose that $a$ is a fixed (but unknown) real number, with $a^2 \neq 1$. Determine all triples $(x, y, z)$ of real numbers that satisfy the system of equations:
$x + y + z = a$
$xy + yz + xz = -1$
$xyz = -a$
I've tried making substitutions but don't seem to be ab... | We have
\begin{align}
(u-x)(u-y)(u-z) &= u^3 - (x+y+z)u^2 + (xy+yz+xz)u - xyz\\
&= u^3-au^2-u+a\\
&= u^2(u-a)-(u-a)\\
&= (u^2-1)(u-a)\\
&= (u-1)(u+1)(u-a)
\end{align}
so $\{x,y,z\} = \{-1,1,a\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
general solution for a recurrence relation I have the following recurrence relation:
$$x_1=1, x_2=a, x_{n+2}=ax_{n+1}-x_n\hspace{1cm}(*)$$
If we assume that $x_n=r^n$ is a solution for the relation $x_{n+2}=ax_{n+1}-x_n$, then I can deduce that
$r=\frac{a+\sqrt{a^2-4}}{2}$ or $r=\frac{a-\sqrt{a^2-4}}{2}$.
By using the... | A bit of culture. Any two consecutive numbers in your sequence, call them $x_n$ and $x_{n+1},$ satisfy
$$ x_n^2 - a \, x_n \, x_{n+1} + x_{n+1}^2 = 1 $$
Try consecutive values in
$$ 1, \; \; a, \; \; a^2 - 1, \; \; a^3 - 2a, \; \ldots $$
This comes from the matrix
$$
\left(
\begin{array}{cc}
0 & 1 \\
-1 & a
\end{arra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Complex Partial Fraction Decomposition The question I need help with is:
Prove that
$$\sum_{k=0}^{6}\frac{1-z^{2}}{1-2z\cos\left(\frac{2k\pi}{7}\right)+z^{2}}=\frac{7(z^{7}+1)}{1-z^{7}}$$
I have already tried brute forcing this by combining the LHS into a single fraction. While this worked, it is an extremely long proo... | Following the posts that were first to appear we introduce
$\zeta=\exp(2\pi i/n)$ and seek to evaluate
$$S = \sum_{k=0}^{n-1} \frac{1-z^2}{(z-\zeta^k)(z-1/\zeta^k)}.$$
where presumably $z$ is not a power of $\zeta$ and no singularity
appears. Introducing
$$f(v) = \frac{1-z^2}{(z-v)(z-1/v)} \frac{n/v}{v^n-1}
= \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Series Convergence of Harmonic Means
Let $\{x_n\}$ be a sequence of real numbers such that $0< x_1 <x_2$. If $$x_n= \frac{2}{\frac{1}{x_{n-1}}+\frac{1}{x_{n-2}}}, $$then show that
$$\lim_{n\to\infty}x_n=\frac{3x_1x_2}{2x_1+x_2}.$$
| Note that
\begin{align}
& x_n = \frac{2}{\frac{1}{x_{n-1}} + \frac{1}{x_{n-2}}} \\
\iff & \frac{1}{x_{n-1}} + \frac{1}{x_{n-2}} = \frac{2}{x_{n}} \\
\iff & a_n = \frac{1}{2}a_{n-1} + \frac{1}{2}a_{n-2}\\
\iff & a_n-a_{n-1} = -\frac{1}{2}\left(a_{n-1}-a_{n-2}\right)
\end{align}
where $a_i = \frac{1}{x_i}$. Consequently,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find all the solutions of the polynomial equation $x^5+y^3 \equiv 1 \pmod9$. Find all the solutions of the polynomial equation $$x^5+y^3 \equiv 1\pmod9.$$ I have learned to find the solutions of the polynomials like $x^p \equiv n\pmod m$, but never been like two variables. Please help me to solve this. Thanks.
| Guide:
Let $y=3m+r, r=0,1,2$
$y^3=(3m+r)^3=(3m)^3+3(3m)^2r+3(3m)r^2+r^3\equiv r^3 \pmod{9}$
Hence we just have to consider $3$ cases,
Case $1$: $y \equiv 0 \pmod{3}$ and the problem reduces to $x^5 \equiv 1\pmod{9}$.
Case $2$: $y \equiv 1 \pmod{3}$ and the problem reduces to $x^5 \equiv 0\pmod{9}$.
Case $3$: $y \equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$ Solve:
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$
This is taken from one of the TAU entry tests (I have one in 2 weeks :) )
So, I don't really recognize anything speical here ... | With factors following an arithmetic progression, you establish the rule
$$\frac1{a(a+b)}+\frac1{(a+b)(a+2b)}=\frac{2a+2b}{a(a+b)(a+2b)}=\frac2{a(a+2b)}.$$
Applying it three times, the sum reduces to
$$\frac4{(x+1)(x+5)},$$ giving the easy quadratic equation
$$(x+1)(x+5)=\frac4{0.8}$$
or $$x(x+6)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Trigonometry and quadratics : Possible mismatch? There’s this problem I came across, gives me an invalid answer by using general quadratic formula. Wonder why?
$2\sin^2{x} -5\cos{x} -4 =0 $
Here’s what I did:
$2\sin^2{x} -5\cos{x} -4 =0 $
$2(1-\cos^2{x}) - 5 \cos{x} - 4 = 0$
$2 \cos^2{x} + 5 \cos{x} + 2 =0$
This is a q... | Simple factorization..
$$2 \cos^2 x + 5 \cos x + 2 =0$$
$$(2 \cos^2 x + 4 \cos x) +(\cos x + 2) =0$$
$$2 \cos x(\cos x +2)+( \cos x + 2) =0$$
$$(\cos x +2)( 2\cos x + 1) =0$$
$$\implies ( 2\cos x + 1) =0$$
$$\implies \cos x =-\frac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Geometry: Prove that two angles are not equal (This is just a question for fun. I saw a commercial logo today and I was inspired.
I have posted answers for this question and you may post alternative answers!)
Question
In the figure, $\triangle ABC$ is half of a square and $M$ is the midpoint of $BC$. Prove that $\alph... | First assume that $\alpha$ = $\beta$ and that both would be equal to $\frac{\pi}{8}^\circ$ (since $\alpha = \beta$ and $\alpha + \beta = \frac{\pi}{4} \to 2\beta = \frac{\pi}{4} \to \beta = \frac{\pi}{8}$).
If we call side $\overline{AC}$, n, then side $\overline{MC}$ is $\frac{n}{2}$. Next, we solve for $\overline{A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 14,
"answer_id": 9
} |
Calculate $\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$
Calculate:$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$$
Even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.
My own solution
$$\lim_{n\to\i... | Let $1/n=h$
As $n\to\infty,h\to0+,h>0$
$$\sqrt{5n^2+4}=\sqrt{\dfrac{5+4h^2}{h^2}}=\dfrac{\sqrt{5+4h^2}}{\sqrt{h^2}}$$
Now $\sqrt{h^2}=|h|=+h$ for $h>0$
So, we ahve
$$\lim_{h\to0^+}\dfrac{\sqrt{5+4h^2}-\sqrt{5+h}}h=\lim_{h\to0^+}\dfrac1{\sqrt{5+4h^2}+\sqrt{5+h}}\cdot\lim_{h\to0^+}\dfrac{5+4h^2-(5+h)}h=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Series convergence proof verification Show that the series $\sum_{k=0}^{\infty}(-1)^k\frac{1}{2k+1}$ converges to $\pi/4$.
Here is my draft of a proof.
Abel's theorem states that suppose for a nondegenerate $[a,b]$. If $f(x):=\sum_{k=0}^{\infty}a_x(x-x_0)^k$ converges on $[a,b]$, then $f(x)$ is continuous and converg... | This is not entirely correct.
The series $\sum_{k=0}^\infty (-1)^kx^{2k}$ converges to $1/(1+x^2)$ on $(-1,1)$, but the convergence is not uniform. As a power series with convergence radius $1$, it does converge uniformly on any compact interval $[a,b] \subset (-1,1)$ -- a fact you cite but apply incorrectly. Consequ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$
T... | Let us use Roots of unity!
Here's a hint (by the way this question is from the ARML and I recommend you try it out because it is quite nice). We know that if $z\neq1$ is an $z$th root of unity, that $1+z+z^2+z^3+z^4+\cdots+z^{n-1}=0$. Also, note that the $n$th roots of unity form a regular $n$-gon with a vertex at $(1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
The sides of a triangle are in Arithmetic progression If the sides of a triangle are in Arithmetic progression and the greatest and smallest angles are $X$ and $Y$, then show that
$$4(1- \cos X)(1-\cos Y) = \cos X + \cos Y$$
I tried using sine rule but can't solve it.
| Let the sides be $a-d,a,a+d$ (with $a>d)$ be the three sides of the triangle, so $X$ corresponds to the side with length $a-d$ and $Y$ that to with length $a+d$. Using cosine formula
\begin{align*}
\cos X & = \frac{(a+d)^2+a^2-(a-d)^2)}{2a(a+d)}=\frac{a+4d}{2(a+d)}\\
\cos Y & = \frac{(a-d)^2+a^2-(a+d)^2)}{2a(a-d)}=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Help summing the telescoping series $\sum_{n=2}^{\infty}\frac{1}{n^3-n}$. I know a priori that the series $$\sum_{n=2}^{\infty}\frac{1}{n^3-n}$$ converges. However, I am tasked with summing the series by treating it as a telescoping series.
By partial fraction decomposition, the series can be written as: $$\sum_{n=2}^... | $$\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$$
$$=\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{2n}-\frac{1}{2n}\right)$$
$$=1/2\sum_{n=2}^{\infty}\left(\frac{1}{(n+1)}-\frac{1}{n}+\frac{1}{(n-1)}-\frac{1}{n}\right)$$
Do you see two telescoping series?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.
| Note that $$\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{(49-x^2)-(25-x^2)}{\sqrt{49-x^2}-\sqrt{25-x^2}}=\frac{24}{3}=8\,.$$
Together with $\sqrt{49-x^2}-\sqrt{25-x^2}=3$, we conclude that
$$\sqrt{49-x^2}=\frac{3+8}{2}=\frac{11}{2}\,.$$
Therefore, $x^2=\dfrac{75}{4}$, or $x=\pm\dfrac{5\sqrt{3}}{2}$.
P.S.: Oopsie, nextpuzzle al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $... | Simply use Taylor' formula ultimately at order $2$ to find equivalents: near $0$,
$$\cos u=1-\frac{u^2}2+o(u^2),\qquad \sin u=u+o(u)$$
so
\begin{align}
\cos x-\cos 3x&=1-\frac{x^2}2+o(x^2)-\Bigl(1-\frac{9x^2}2+o(x^2)\Bigr)= 4x^2+o(x^2)\\
\sin 3x^2-\sin x^2&=3x^2+o(x^2)-\bigl(\sin x^2+o(x^2)\bigr)=2x^2+o(x^2).
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 3
} |
Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$ Let $a,b,$ and $c$ be real numbers such that
$a+b+c=2 \text{ and } a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{... | Let $a=mc$ and $b=nc$ then
$$m+n=\dfrac{2}{c}-1~~~,~~~m^2+n^2=\dfrac{12}{c^2}-1$$
by Cauchy-Schwarz
$$(m+n)^2\leq2(m^2+n^2)$$
with substitution $-3c^2+4c+20\geq0$ gives $c=-2,\dfrac{10}{3}$ leads us to difference $\dfrac{16}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Application of Chebyschev inequality
I want to prove the inequality above. On the extreme ends we get a clear application of AM-GM, and I want to use the chebyschev inequality for the middle but was having trouble.
My Attempt:
Since Chebyschevs inequalty is
We can square the left hand side of our inequality with the... | Firstly, your inequality is wrong.
Try $a=1$, $b=-1$ and $c=0$.
For non-negatives $a$, $b$ and $c$ we see that $(a,b,c)$ and $(a,b,c)$ they are the same ordered.
This, by Chebyshov we obtain:
$$3(a\cdot a+b\cdot b+c\cdot c)\geq(a+b+c)(a+b+c)$$or
$$3(a^2+b^2+c^2)\geq(a+b+c)^2$$ or
$$3(a^2+b^2+c^2+2(ab+ac+bc))\geq(a+b+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove by definition $\lim_{n\to\infty} \sqrt{\frac{n}{n+1}}=1$
Prove by definition $\lim_{n\to\infty} \sqrt{\frac{n}{n+1}}=1$.
I got to $\left|\sqrt{\frac{n}{n+1}}-1\right|< ε$, but I don't know how to proceed.
| Let me explain Kenny Lau's answer a bit. The first step, he does
$$\left|\sqrt{\frac{n}{n+1}}-1\right|=\left|\frac{\sqrt{n}}{\sqrt{n+1}}-1\right|=\left|\frac{\sqrt{n}}{\sqrt{n+1}}-\frac{\sqrt{n+1}}{\sqrt{n+1}}\right|=\left|\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n+1}}\right|$$
Now he proceeds to multiply both sides of the fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$? On a problem book solution I was faced with the following step:
$$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$$
I identified $\frac{1}{n+1}+\frac{1}{(n+1)^2}...$ as a geometric series so the sum would be $\frac{1}{1-r}$ so that $\frac{1}{1-\frac{1}{n+1}}=1+\frac{1}... | How I recommend approaching these problems:
Use the fact that if $|x| < 1$,
$$1 + x + x^2 + \cdots = \dfrac{1}{1-x}\text{.} \tag{*}$$
Then, factor to rewrite the problem so that it is in terms of the equation (*).
Observe that if I factor out $\dfrac{1}{n+1}$ that
$$\dfrac{1}{n+1} + \dfrac{1}{(n+1)^2} + \cdots = \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Computing the determinant of a $4\times4$ matrix
Compute the determinant of \begin{vmatrix} 2 & -1 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & -1 \\ -1 & 1 & 2 & -2 \end{vmatrix}
My try:
$$\begin{vmatrix} 2 & -1 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & -1 \\ -1 & 1 & 2 & -2 \end{vmatrix}_{R_2\rightarrow2R_2-R_1\\R_3\right... | Because you multiplied the first and the third rows by 2 while making the row operations, hence the determinant you found is 4 times bigger than the determinant of the original matrix. So the answer is $\frac{92*2}{4}$=46.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
What is the domain of the function $f(x)=\sin^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$?
What is the domain of the function
$f(x)=\sin^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$?
I started using the fact $-\frac{\pi}{2}\le f(x) \le \frac{\pi}{2}\implies-1 \le\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\le1$.Now,... | Examine that,
$$\lim_{x\to-\infty}\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=0$$
$$\lim_{x\to\infty}\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=0$$
Also the function is discontinuous at $x=1$.
Now check where it assumes the values $1$ and $-1$.
$$\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=1$$
$$8(3)^{x-2}=1-3^{2(x-1)}$$
$$8(3)^{x-2}+3^{2(x-1)}=1$$
$$3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get i... | We can find the equation must satisfy$x^4+ax^2+bx+c=(x^2+3x-1)(x^2+mx+n)$ and$m=-3,n=-c$. So when$x=1,1+a+b+c=3(1-3-c)\rightarrow a+b+4c+7=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the quotient and the remainder of $(n^6-7)/(n^2+1)$
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.
So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?
| I assume that by "dish" you mean "quotient"; that's not terminology I'm familiar with, but it seems to make sense in context.
When you're dividing polynomials, remember that you are not done until the remainder is of lower degree than the denominator. So you're correct that the first step of dividing $n^2 + 1$ into $n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
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} |
what is the multiplicative order of the following elements in $F_{16}$ For $f(x) = x^4 + x + 1 \in F_2[x]$ . Let $a \in F_{16}$ be a root of $f$.
How can I find the multiplicative order of $a$ and $a^2+a+1$
My attempt:
$f$ clearly has not root in $F_2$ so if it is reducible it must be reduced to polynomials of degree... | Since $x^4+x+1$ is irreducible over $\mathbb{F}_2$ the roots of $x^4+x+1$ in $\mathbb{F}_{16}\simeq \mathbb{F}_2[x]/(x^4+x+1)$ are of the form $a,a^2,a^4,a^8$. Since $a$ is a root of $x^4+x+1$ we have $a^4=-a-1=a+1$. We cannot have $a^3=1$ or $a^5=1$ since $x^3-1$ and $x^5-1$ share no common factor with $x^4+x+1$ over ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4$? Using the Cardano formula, one can show that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}$ is a real root of the depressed cubic
$f(x)=x^3-6x-40$. Actually, one can show by the calculating the determinant that this is the only real root. On the ... | Well, $(2\pm\sqrt2)^3=20\pm14\sqrt2$, that is $\sqrt[3]{20\pm14\sqrt2}
=2\pm\sqrt2$. Therefore
$$\sqrt[3]{2+14\sqrt2}+\sqrt[3]{20-14\sqrt2}=2+\sqrt2+2-\sqrt2=4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Proof by induction for $f^{n+1}(x)=\frac{x}{\sqrt{1+(n+1)x^2}}$ Look at the following function f: $\mathbb{R} \to \mathbb{R}: x \mapsto \frac{x}{\sqrt{1+x^2}}.$
Show with the complete induction that the recursive ( given by $f^1:=f$ and $f^{n+1}:=f\circ f^n$) composition $f^n$ from $f$ has the following explicit illust... | \begin{align}
(f\circ f^n)(x)&= \dfrac{ \dfrac{x}{\sqrt{1+nx^2}} }{\sqrt{1+\dfrac{x^2}{1+nx^2}}}\\
& = \dfrac{ \dfrac{x}{\sqrt{1+nx^2}} }{\sqrt{\dfrac{1+nx^2+x^2}{1+nx^2}}}\\
& =\dfrac{x}{ \sqrt{1+nx^2+x^2}}\\
&=\dfrac{x}{ \sqrt{1+(n+1)x^2}}\\
&= f^{n+1}(x)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Partial fraction of $\frac{2s+12}{ (s^2 + 5s + 6)(s+1)} $ then inverse transform it
Find the inverse Laplace transform of
$$\mathcal{L}^{-1}\left(\frac{2s+12}{ (s^2 + 5s + 6)(s+1)}\right)$$
I recognise I need to use partial fractions to solve it and that is where I got stuck. Here’s my working,
After factoring the... | You can use the methods associated with the "Partial fraction decomposition" (see https://en.wikipedia.org/wiki/Partial_fraction_decomposition). After writing :
$$\frac{2s+12}{ (s+2)(s+3)(s+1)} = \frac{A}{s+2} + \frac{B}{s+3} + \frac{C}{s+1} $$
You multiply and the right and left hand side by $(s+2)$, which gives :
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Roots of unity and large expression
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}.$$
I have tried combining the first and third terms & first and last ter... | Alt. hint: let $\,z=\omega+\dfrac{1}{\omega}\,$ so that $\,z^2=\omega^2+\dfrac{1}{\omega^2}+2\,$, then use that $\,\omega^4=\bar\omega\,$ and $\,\omega^3=\bar\omega^2\,$ so the sum is:
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\bar\omega^2}{1 + \bar\omega^4} + \frac{\bar\omega}{1 + \bar\om... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Write as the sum of a series The question asks to write
$\dfrac{1}{1-i-z}$ as the sum of a series such that $\left|-z-i\right|<1$
but I genuinely have no idea how to do it, or even where to start.
| $$
\frac 1 {1-i-z}
$$
You want the series to converge if $\left| -z-i \right|<1.$ That is the same as $\left|z+i\right|<1,$ and so the same as $\left|z-(-i)\right|<1.$ We therefore want to write this as a sum of powers of $z-(-i),$ i.e. of $z+i.$ The function is
$$
\frac 1 {1 - (z+i)}.
$$
Recall that
$$
\frac 1 {1-r} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Find the Fourier series of the function: $f(x)=\begin{cases}x+1 & -1\leq x< 0,\\1-x &0\leq x< 1\end{cases},\;\;f(x+2)=f(x)$ I want to find the Fourier series of the function. For now, I am clueless on how to handle the function, int that it has $f(x+2)=f(x).$
$$f(x)=\begin{cases}x+1 & -1\leq x< 0,\\1-x &0\leq x< 1\end{... | BACKGROUND
First $f(x+2) = f(x)$ is just a way of saying that you got to repeat your function, i.e. it is periodic of period $2$.
This is needed so that we could derive the fourier series,
\begin{equation}
a_0 + \sum\limits_{k=1}^{\infty}
a_k\cos(k x) + b_k \sin(k x)
\end{equation}
where
\begin{align}
a_0 &= \fra... | {
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"url": "https://math.stackexchange.com/questions/2887826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Solve $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$
Solve
$$8\sin x=\dfrac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$$
My approach is as follow
$8 \sin x-\frac{1}{\sin x}=\frac{\sqrt{3}}{\cos x}$
On squaring we get
$64 \sin^2 x+\frac{1}{\sin^2 x}-16=\frac{3}{\cos^2 x}$
$(64\sin^4 x-16\sin^2 x+1)(1-\sin^2 x)=3 \sin^2 ... | First thing first, if you make the substitution $t=\sin^2x$ the polynomial you get is $$-64t^3+80t^2-20t+1=0$$ Now, to decompose it, you could use Ruffini's rule: first we find a zero of the polynomial that divides the constant term (in our case $\pm 1$). Let us call the polynomial $P(t)$, then $$P(1) =-64+80-20+1 \neq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Expanding product of binomials $(z^k + z^{-k})$ Suppose $z$ is a complex number, and consider the product
$$f_m(z)=\prod_{k=1}^m \left(z^k + \frac 1 {z^k} \right),$$
for $m = 1,2,\dots$ . Of course, one should be able to expand this into a sum of terms that are either powers of $z$ or of $1/z$. It is easy to see that t... | A proof of 1. could go as follows.
We obtain for integral $m\geq 1$
\begin{align*}
\color{blue}{f_m(z)}&=\prod_{k=1}^m\left(z^k+\frac{1}{z^k}\right)\\
&=\prod_{k=1}^m\left(z^{-k}\left(1+z^{2k}\right)\right)\\
&=z^{-\sum_{k=1}^mz^k}\prod_{k=1}^m\left(1+z^{2k}\right)\tag{1}\\
&=z^{-\frac{1}{2}m(m+1)}\sum_{j=0}^{\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Convergence and limit of recursive sequence Consider the sequence $(a_n)_{n \in \mathbb{N}}$ which has startvalue $a_0 > -1$ and recursive relation:
$$a_{n+1} = \frac{a_n}{2} + \frac{1}{1+ a_n}$$
How to prove the convergence and find the limit?
_
I think you need to show the convergence with a Cauchy sequence. It is al... | $$a_0>-1$$
$$a_{n+1}=\dfrac{a_n}{2}+\dfrac{1}{1+a_n}=\dfrac{a_n^2+a_n+2}{2(1+a_n)}$$
We know that $x^2+x+2>0, \forall x\in \mathbb{R}$.
$$a_1=\dfrac{a_0^2+a_0+2}{2(1+a_0)}>0\rightarrow a_{n+1}=\dfrac{a_n^2+a_n+2}{2(1+a_n)}>0\rightarrow \boxed{a_n>0, \forall n\geq 1}$$
We are going to prove by induction that $\boxed{a_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$a+\sqrt{b} = b + \sqrt{a}$$
Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x... | Observe that
$$a-\sqrt a=b-\sqrt b\implies a-b=\sqrt a-\sqrt b\iff\frac{(\sqrt a-\sqrt b)(\sqrt a+\sqrt b)}{\sqrt a-\sqrt b}=1\iff$$
$$\iff \sqrt a+\sqrt b=1\;(\text{ assuming $\,a\neq b\,$)}\implies b=(1-\sqrt a)^2$$
So you need the maximum of $\;f(a):=a+b=a+(1-\sqrt a)^2=2a-2\sqrt a+1\;$ ...can you take it from here?... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
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Disjoint sets in a combinatoral sum (continued) Let $f_S(m)$ be defined as in my previous question:
Let $S = \{1/n^2 : n \in \mathbb{N} \}$. Let $f_S(m)$ be the sum of the products of all $m$-tuples chosen from $S$. That is
$$f_S(m) = \sum_{X \in {S \choose m}} \prod_{x \in X} x $$
I have used inclusion-exclusion... | These may be computed with the unlabeled set operator
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}$$
applied to the Riemann Zeta function
$$\zeta(s) = \sum_{n\ge 1} \frac{1}{n^s}.$$
We use the recurrence by Lovasz for the cycle index $Z(P_n)$ of the
set operator $\textsc{SET}_{=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Critical points and extremum of $f(x,y)=y^2-x^2+x^3+x^2y+\frac{y^3}{3}\;\;\forall\;(x,y)\in\Bbb{R}^2$ Let $f:\Bbb{R}^2\to \Bbb{R}$ be a function defined by
\begin{align}f(x,y)=y^2-x^2+x^3+x^2y+\frac{y^3}{3}\;\;\forall\;(x,y)\in\Bbb{R}^2\end{align}
$i.$ Compute the critical points of $f$
$ii.$ Does $f$ have an extremum?... | \begin{align}\frac{\partial f}{\partial x}=-2x+3x^2+2xy\qquad (1)\end{align}
\begin{align}\frac{\partial f}{\partial y}=2y+x^2+y^2\qquad (2)\end{align}
At \begin{align}\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0\end{align}
we have (1) implies
\begin{align}x(3x+2y-2)=0\end{align}
So there are two cases... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve for $x$ in the equation containing ${\lfloor{x}\rfloor}$ and $\{x\}$
Calculate all possible values of $x$ satisfying,
$$\frac{\lfloor{x}\rfloor}{\lfloor{x-2}\rfloor}-\frac{\lfloor{x-2}\rfloor}{\lfloor{x}\rfloor}=\frac{8\{x\}+12}{\lfloor{x}\rfloor
\lfloor{x-2}\rfloor}$$
where $\{x\}$ stands for fractional part o... | Decompose $x=n+f$. Then
$$\frac n{n-2}-\frac{n-2}n=\frac{8f+12}{n(n-2)}$$ or
$$n^2-(n-2)^2=8f+12$$
or
$$f=\frac{n-4}2.$$
The possible values for $n$ are $4$ and $5$ (so that $0\le f<1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Is $f $ decreasing? Let $f:(0,\infty)\rightarrow (0,\infty) $, $f (x)+f (y)\geq 2f (x+y), \forall x, y>0$.
Is $f $ decreasing?
I need to show that $f (x)+f (y)+f (z)\geq 3f (x+y+z) $ and it's enough to show that $f$ is decreasing.
| We have:
$$f(x)+f(y)+f(z) \ge 2f(x+y)+f(z) = f(x+y) +f(z) + f(x+y)+f(z)-f(z) \ge 2f(x+y+z)+2f(x+y+z)-f(z)=4f(x+y+z)-f(z)$$
In the same manner:
$$f(x)+f(y)+f(z) \ge 2f(x+z)+f(y) = f(x+z) +f(y) + f(x+z)+f(y)-f(y) \ge 2f(x+y+z)+2f(x+y+z)-f(y)=4f(x+y+z)-f(y)$$
And:
$$f(x)+f(y)+f(z) \ge 2f(y+z)+f(x) = f(y+z) +f(x) + f(y+z)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$? This question comes in mind while solving another question.
If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$ ?
I wonder if this question (if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$) helps?
I wondered if AM-GM could help, but the extra square keeps bothering me while solving it... | By Rearrangement inequality we have that
$$(a^2+b^2+c^2)^2\ge a^4+b^4+c^4\ge a^2bc+b^2ac+c^2ab=abc(a+b+c)=a+b+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$x^4-2x^2\sin^2(\frac{\pi x}{2})+1=0\\x^4+1=2x^2\left (1-\cos^2\left(\frac{\pi x}{2}\right)\right )\\(x^2-1)^2=-2x^2\cos^2\left(\frac{\pi x}{2}\right)\\(x^2-1)^2+2x^2\cos^2\left(\frac{\pi x}{2}\right)=0\\x^2-1=0\,\text{and}\, 2x^2\c... | Another way to look:
\begin{eqnarray*}
x^{4}-2 x^{2} \cos^{2}\left(\frac{\pi x}{2}\right) +1 &=& x^{4}-2x^2+1 + 2 x^{2} \cos^{2}\left(\frac{\pi x}{2}\right) \\
&=& \left(\frac{x^{2}-1}{x \sqrt{2}}\right)^{2} + \cos^{2}\left(\frac{\pi x}{2}\right)
\end{eqnarray*}
Now $ \left(\frac{x^{2}-1}{x \sqrt{2}}\right)^{2} + \cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
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Factorization $\cos\left(\tfrac{\pi z}{4}\right)-\sin\left(\tfrac{\pi z}{4}\right)$
Prove that
$$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=\prod_{n=1}^\infty\left(1+\frac{(-1)^nz}{2n-1}\right)$$
My try:
$$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=\cos\left(\frac{\pi ... | $\displaystyle \prod_{n=1}^\infty\left(1+\frac{(-1)^nz}{2n-1}\right) =\prod_{n=0}^\infty\left(1-\frac{z}{4n+1}\right)\prod_{n=1}^\infty\left(1+\frac{z}{4n-1}\right)=$
$\displaystyle =(1-z)\prod_{n=1}^\infty\left(\frac{4n(1-\frac{z-1}{4n})}{4n(1+\frac{1}{4n})}\right)\prod_{n=1}^\infty\left(\frac{4n(1+\frac{z-1}{4n})}{4n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove $\ \frac{z-1}{z+1} $ is imaginary no' iff $\ |z| = 1 $ Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $
Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so
$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \c... | Note that\begin{align}\frac{z-1}{z+1}&=\frac{(z-1)\left(\overline z+1\right)}{(z+1)\left(\overline z+1\right)}\\&=\frac{|z|^2+z-\overline z-1}{|z+1|^2}\\&=\frac{|z|^2-1+2i\operatorname{Im}(z)}{|z+1|^2}\end{align}and that therefore $\frac{z-1}{z+1}$ is purely imaginary if and only $|z|^2-1=0$, which is the same thing as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Linear dependence and independence solution. How was it found? I am reading this text:
I feel like my text skipped a step to find the coefficients/solutions. Did they write a system of linear equations and use Gaussian elimination? How did they do this?
| In order to see whether these vectors are dependent examine the system:
$a(1,3,1)+b(0,1,2)+c(1,0,-5)=(0,0,0)$
which is equivalent with
$$\begin{pmatrix}
1 & 0 & 1\\
3 & 1 & 0\\
1 & 2 & -5
\end{pmatrix}\begin{pmatrix}
a\\
b\\
c
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}$$
then you can use elimination to find... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to get the equation where a circle goes through three points
If I have the equation $ax^2+ay^2+bx+cy+d=0$ how do I get the equation where the circumference goes through the points P = (1,1), Q = (−1,−1) and
R = (−1,1)
I have it in mind to solve it with a matrix, but the instructions seem confusing to me, can I ... | If you need to use matrices/linear algebra, consider this approach: If you move $d$ to the other side of the equation $ax^2 + ay^2 + bx + cy + d = 0$, you get $ax^2 + ay^2 + bx + cy = -d$. We can view the left hand side of this equation as the product of two matrices:
\begin{eqnarray}
\left[ax^2 + ay^2 + bx + cy\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Find all triples $(p,x,y)$ such that $ p^x=y^4+4$ Find all triples $(p,x,y)$ such that $ p^x=y^4+4$, where $ p$ is a prime and $ x$ and $ y$ are natural numbers.
I know that this question has already been asked on another forum, but I want to ask different questions about it.
$ p^x=y^4+4$
$ p^x=(y^2+2)^2 - (2y)^2$
$p^x... | Let $p$ be a prime number, and $x$ and $y$ natural numbers, such that $p^x=y^4+4$. Then as you note
$$p^x=(y^2+2y+2)(y^2-2y+2),$$
and hence both factors are powers of $p$, say
\begin{eqnarray*}
y^2+2y+2&=&p^u,\\
y^2-2y+2&=&p^v.
\end{eqnarray*}
First note that if $v=0$ then $y=1$, and we find that $p=5$ and $x=1$, the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if $q=3$?
Question: If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if
$q=3.$
I am certain the answer to the question is yes, and I believe I have worked out a solution - although I am not certain it ... | If $q \equiv 3 \mod 4$, $q+1$ is divisible by $4$. Say $q+1=4k$ Then
$$ {q+1 \choose 2} = \frac{(q+1)q}{2} = 2 k (4k-1)$$
This is divisible by $2$, and by any primes that divide $k$ or $4k-1$. That makes at least three (not necessarily distinct) primes, unless either $k = 1$ or $4k-1 = 1$ (the latter being impossibl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $\Im(z) > 0$then $\Im (-1/(z+1)) >0$ My work so far:
Letting $z = x+iy$, where $x,y$ are real,
then $\Im(z) = y > 0$
$y+1 > 0$
$1/(y+1) > 0$
But I am not sure how to proceed. I also attempted to multiply $-1/(z+1)$ by its conjugate, but that led to $(-x-1+iy)/((x+1)^2 + y^2)$, which I feel is the wrong way to approa... | $z = x + iy, \; x, y \in \Bbb R, \; y > 0; \tag 1$
$z + 1 = (1 + x) + iy; \tag 2$
$\dfrac{1}{z + 1} = \dfrac{1}{(1 + x) + iy}$
$= \dfrac{(1 + x) - iy}{(1 + x)^2 + y^2} = \dfrac{1 + x}{(1 + x)^2 + y^2} -i\dfrac{y}{(1 + x)^2 + y^2}; \tag 3$
$-\dfrac{1}{z + 1} = -\dfrac{1 + x}{(1 + x)^2 + y^2} +i\dfrac{y}{(1 + x)^2 + y^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find all $n\in\mathbb{N}$ such that the nonzero roots of $(z+1)^n-z^n-1$ are all on the unit circle. I try to use Mathematica to find such $n$, and I believe that only $n=2,3,4,5,6,7$ meet the requirement but I don't know how to prove that for $n\ge 8$, there is always some root lying outside the unit ball by using the... | It is clear that the result is true for $n=2$. Now we consider $n\ge 3$.
Note that $(z+1)^n-z^n-1=\sum_{k=0}^n\binom{n}{k} z^k-z^n-1 = \sum_{k=1}^{n-1}\binom{n}{k} z^k= nz\sum_{k=1}^{n-1}\frac{\binom{n}{k}}{n}z^{k-1}$. Suppose that $\sum_{k=1}^{n-1}\frac{\binom{n}{k}}{n}z^{k-1}=\prod_{k=1}^{n-2}(z-z_k)$. Then we have $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $ The task is to find out if this series is convergent or divergent.
$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$
The solution uses the ratio test and says:
$ \left.\begin{aligned} \frac { a _ { n + 1 } ... | $\sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! }
$
For region of convergence,
just replace $n!$ with
$(n/e)^n$
and then use the
n-th root test.
The additional
$\sqrt{n}$
only affects the boundaries.
In this case
(why the annoying spaces?),
$\dfrac{n!n^{n}}{(2n)!}
$
becomes
$\dfrac{(n/e)^nn^{n}}{(2n/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Prove that $3^{16} -33$ and $3^{15} +5$ is divisible by 4 by means of binomial theorem This is a question that I found in a textbook:
Given that $p=q+1$, $p$ and $q$ are integers, then show that
$p^{2n} - 2nq-1$ is divisible by $q^2$ given that $n$ is a positive integer.
By taking a suitable value of $n$, $p$ and $q$,... | It's better if you use
$$
(1+q)^{2n}=1+2nq+\sum_{k=2}^{2n}\binom{2n}{k}q^k
$$
so
$$
p^{2n}-2nq-1=q^2\sum_{k=2}^{2n}\binom{2n}{k}q^{k-2}
$$
is divisible by $q^2$.
With $q=2$ and $n=8$ you have $p=3$ and $p^{2n}-2nq-1=3^{16}-33$.
For the second case, consider
$$
3^{15}+5=3(3^{14}-29)+3\cdot29+5
$$
and set $q=2$, $n=7$, n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x ... | We could do it using harmonic numbers
$$S_x=\sum_{k=1}^x \frac x {x^2+k}=x\sum_{k=1}^x \frac 1 {x^2+k}=x \left(H_{x^2+x}-H_{x^2}\right)$$
Now, using the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
p^2}+O\left(\frac{1}{p^4}\right)$$ we should get
$$S_x=1-\frac{1}{2 x}-\frac{1}{6 x^2}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Whats wrong in this approach to evaluate $\int_{0}^{\frac{\pi}{2}} \frac{\sin x\cos xdx}{\sin x+\cos x}$ Whats wrong in this approach to evaluate $$I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x\cos xdx}{\sin x+\cos x}$$
Since $f\left(\frac{\pi}{2}-x\right)=f(x)$ we have
$$I=2\int_{0}^{\frac{\pi}{4}} \frac{\sin x\cos xdx}{\si... | We have that
*
*$\sin(\pi/4-x)=\frac{\sqrt 2}2(\cos x-\sin x)$
*$\cos(\pi/4-x)=\frac{\sqrt 2}2(\cos x+\sin x)$
therefore
$$I=2\int_{0}^{\frac{\pi}{4}} \frac{\sin x\cos xdx}{\sin x+\cos x}=\sqrt{2}\int_{0}^{\frac{\pi}{4}}\frac{(\cos x-\sin x)(\cos x+ \sin x)dx }{\cos x-\sin x+\cos x+ \sin x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$. Solve the inequality
$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow$ $\frac{x^2-2}{x^2+2} - \frac{x}{x+4} \leq 0 \Rightarrow$ $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0 \Rightarrow$ $x \geq \frac{1}{2} \pm \sqrt{2.25} ... | You reduced the inequality to $$\frac{4(x+1)(x-2)}{(x^2+2)(x+4)} \ge 0$$
which is equivalent to $$\frac{(x+1)(x-2)}{x+4} \ge 0$$
This is true if and only if an even number of terms $(x+1), (x-2), (x+4)$ are $\le 0$. There are four options:
*
*$x+1 \ge 0$, $x-2 \ge 0$, $x+4 > 0$ which gives $x \in [2, +\infty\rangle$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Solve $a^2 - 2b^2 - 3 c^2 + 6 d^2 =1 $ over integers $a,b,c,d \in \mathbb{Z}$ Are we able to completely solve this variant of Pell equation?
$$ x_1^2 - 2x_2^2 - 3x_3^2 + 6x_4^2 = 1 $$
This has an interpretation as is related to the fundamental unit equation of $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}[x,y]/(x^2 - 2,... | Equation given above is shown below:
$a^2-2b^2-3c^2+6d^2=1$
As shown by "Individ" , what "OP" can do is to take (a,b,c,d) as shown below:
$a=p(6w^2+4w+3)$
$b=p(2w^2+6w+1)$
$c=p(4w^2+4w+2)$
$d=p(2w^2+4w+1)$
Where $(p)= [1/(2w^2-1)]$
For suitable value's of 'w' we get the numerical solutions below:
w=(1), (a,b,c,d)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"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.