Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
A series involving the Dirichlet Beta function; How to evaluate $\sum_{n=1}^\infty \frac{\beta(n)-1}{n}$? Let the beta and the zeta function be defined as usual:
\begin{align}
& \beta(s) & = & \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s} & = & 1-\frac{1}{3^s}+\frac{1}{5^s}\dots +\frac{(-1)^n}{(2n+1)^s}+\dots
\\ & ... | First of all : $$ \prod_{p=1}^{n}{\left(1+\frac{1}{4p}\right)}\prod_{p=0}^{n}{\left(1-\frac{1}{4p+3}\right)}=\frac{4\Gamma\left(\frac{7}{4}\right)\Gamma\left(n+\frac{5}{4}\right)\Gamma\left(n+\frac{3}{2}\right)}{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)\Gamma\left(n+1\right)\Gamma\left(n+\frac{7}{4}\right)}\underset{n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3639176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of solvable lights out puzzles on m x n rectangle I'm trying to replicate the results on Dr. Brouwer website,
I consulted two of the papers referenced [1] and [2] both of them make use of the same recursively defined polynomial and GCD.
Let $p_n(x)$ be the mod 2 characteristic polynomial of the path of length n... | $$1+x^2 \equiv (1+x)^2 \mod 2$$
$$x(x^3+x^4) = x^4(1+x)$$
So $\gcd(1+x^2,x(x^3+x^4)) = 1+x$ and is also of degree $1$.
Note that if $\gcd(p(x),q(x+1))=r(x)$, then substituting $x=X+1$ gives $\gcd(p(X+1),q(X+2))=r(X+1)$ showing that when working modulo $2$ the degrees of $\gcd(p(x),q(x+1))$ and $\gcd(p(x+1),q(x))$ are e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$.
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.
At first, I tried to evalua... | Suppose that $x_1,x_2,\ldots,x_{n-1}$ are the roots of $z^{n-1}+z^{n-2}+\ldots+z+1=0$. We have
$$\frac{1}{x_j+x_j^{-1}}=\frac{x_j}{1+x_j^2}.$$
If $n$ is odd, then
$$1+z^{2n}=(1+z^2)(1-z^2+z^4-z^6+\ldots+z^{2(n-1)}).$$
Because $x_j^n=1$, we have
$$\frac{1}{1+x_j^2}=\frac{\sum_{k=0}^{n-1}(-1)^kx_j^{2k}}{1+x_j^{2n}}=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3649434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $\iint_{[0,1]^2} \frac{2-4xy}{(9-xy)(8+xy)}dxdy$ I am trying to compute the following double integral:
$$I=\iint_S \frac{2-4xy}{(9-xy)(8+xy)}dxdy$$
with $S=[0,1]\times[0,1].$
What I have tried:
*
*I have written the integral as follows:
$$I=I_1+I_2=-2\iint_S \frac{1}{(9-xy)}dxdy+2\iint_S \frac{1}{(8+xy)}d... | Let's generalize it, in particular the integral in the question is just the case $a=8$.
$$I(a)=\int_0^1\int_0^1\left(\frac{1}{a+xy}-\frac{1}{1+a-xy}\right)dxdy\overset{xy=t}=\int_0^1\int_0^y\frac{1}{y}\left(\frac{1}{a+t}-\frac{1}{1+a-t}\right)dtdy$$
$$=\int_0^1\int_t^1\frac{1}{y}\left(\frac{1}{a+t}-\frac{1}{1+a-t}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why does the difference equation $x_nx_{n+2}=w^5x_n-(w^2+w^3)x_{n+1}+x_{n+2}$ generate cyclic sequences? Let $w$ be a primitive 5th root of unity. Then the difference equation $$x_nx_{n+2}=x_n-(w^2+w^3)x_{n+1}+x_{n+2}$$ generates a cycle of period 5 for general initial values:
$$u,v,\frac{u-(w^2+w^3)v}{u-1},\frac{uv-(... | Let us prove that for all nonzero $a,b\in \mathbb{C}$ the difference equation $$x_nx_{n+2}=ax_n+bx_{n+1}+x_{n+2}$$
never generates a cycle of length $8$.
Let us denote $u=x_0, y=x_1$.
Suppose the contrary. We know that
$$x_{n+2}=\frac{ax_n+bx_{n+1}}{x_n-1}, x_{n-1}=\frac{bx_n+x_{n+1}}{x_{n+1}-a},$$
so for every inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Factorise $Σa^{2}(b^{4}-c^{4})$ I expanded this cyclic expression then found the three factors and assumed third to be constant but after that I got stuck at this question. The options are
(a)$(a-b)^{2}(b-c)^{2}(c-a)^{2}$
(b)$(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)$
(c)$(a+b)^{2}(b+c)^{2}(c+a)^{2}$
(d)None of these
The answer... | Factorizing the difference of two squares comes up a lot here. Let $u=b^4-c^4,\,v=c^4-a^4$ so $a^4-b^4=-u-v$ and the sum is$$\begin{align}a^2u+b^2v-c^2(u+v)&=(a^2-c^2)u+(b^2-c^2)v\\&=(a^2-c^2)(b^2-c^2)(b^2+c^2)+(b^2-c^2)(c^2-a^2)(c^2+a^2)\\&=(b^2-c^2)(c^2-a^2)(-b^2-c^2+c^2+a^2)\\&=(a^2-b^2)(b^2-c^2)(c^2-a^2),\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3654780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
For $a$, $b$, $c$ the sides of a triangle, show $\sum_{cyc}\frac{\sqrt{bc}\,bc}{a(a+b+c)(b+c-a)}\geq1$
Let $a,b,c$ be the side lengths of a triangle. How to prove the following inequality?
$$\frac{\sqrt{bc}\,bc}{a(a+b+c)(b+c-a)}+\frac{\sqrt{ac}\,ac}{b(a+b+c)(a+c-b)}+\frac{\sqrt{ab}\,ab}{c(a+b+c)(a+b-c)}\geq1$$
Via ... | We have to prove that (see @g.kov's result)
$$3\sqrt[3]{\frac{(abc)^2}{(a+b+c)^3(a+b-c)(b+c-a)(c+a-b)}} \ge 1.$$
By using Ravi Substitution $a = x + y, b = y+z, c = z + x$ for $x, y, z > 0$, the inequality becomes
$$3\sqrt[3]{\frac{[(x+y)(y+z)(z+x)]^2}{64(x+y+z)^3xyz}} \ge 1.$$
Let $p = x+y+z, q = xy + yz + zx, r = xyz... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Dividing polynomial $f(x^3)$ by $x^2+x+1$ Given two polynomials with real coefficients $g(x)$ and $h(x)$.
Additionally, $x^2+x+1$ is a factor of $f(x^3)= h(x^3)+xg(x^3)$.
Prove that $x-1|h(x)$ and $x-1|g(x)$
I have tried to solve it by doing this;
$(x-1)(x^2+x+1)|(x-1)(h(x^3)+xg(x^3))$
then;
$(x^3-1)|(x)(h(x^3)-g(x^3)... | I'm not sure what you mean by that separate cases.
But here is how I did it.
Let $a$ and $b$ be roots for $x^2+x+1=0$, then $a^3=b^3=1$ and clearly $a\ne b$ since the discriminat is not $0$.
Since $$x^2+x+1\mid f(x)$$ we see that $f(a)= f(b)=0$
Now put
$$x=a:\;\;\; f(a^3) = h(a^3)+ag(a^3)$$
so $$ 0 = f(1) = h(1)+ag(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Trying to find a general equation for an ellipse given the foci and sum of focal distances I'm trying to find an equation for the ellipse in the form $$Ax^2 + Bxy + Cy^2 +Dx +Ey +F = 0$$ given the foci $(a,b)$ and $(c,d)$ and the sum of focal distances $r$. I started from the definition $$\sqrt{(x-a)^2+(y-b)^2} + \sqrt... | Thanks to User Blue for helping find the errors. Just a couple of negative signs went awry. The correct coefficients are $$A=(a-c)^2-r^2$$ $$B=2(a-c)(b-d)$$ $$C=(b-d)^2-r^2$$ $$D = r^2(a+c)-(a-c)(a^2+b^2-c^2-d^2)$$ $$E=r^2(b+d)-(b-d)(a^2+b^2-c^2-d^2)$$ $$F = \frac{1}{4}(r^4-2r^2(a^2+b^2+c^2+d^2)+(a^2+b^2-c^2-d^2)^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Sequence $\left\{ a_n\right\}$ defined by $a_1=4$ , and $a_{n+1}=\sqrt{a_n+6}$. Prove that $\left| a_{n+1}-3\right|<\frac{1}{3}\left|a_n-3\right|$. As title. Sequence $\left\{a_n\right\}$ defined by $a_1=4$ , and $a_{n+1}=\sqrt{a_n+6}$.
Prove that $\left| a_{n+1}-3\right|<\frac{1}{3}\left|a_n-3\right|$.
I tried the i... | Hint :
$$
a_{n+1}-3=\sqrt{a_n+6}-\sqrt{9}=\frac{a_n+6-9}{\sqrt{a_n+6}+\sqrt{9}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Question about $I = \int x^3 \sqrt{x^2+2} \, dx$ I'm trying to find the result of the following integral:
$$I = \int x^3 \sqrt{x^2+2} \, dx$$
This integral appears in MIT OCW 18.01SC course
I try to make a substitution as follow :
Let $x = \sqrt{2}\tan(u), \quad dx = \sqrt{2}\sec^2(u)\, du$
If we substitute that in our... | You have made an error in simplification after "reverse all previous trig substitution".
\begin{align*}
&\left. \left. 4\sqrt{2}\left( \frac{1}{5}v^5 - \frac{1}{3}v^3 \right) + C \right|_{v = \sec u} \right|_{u = \arctan(x/\sqrt{2})} \\
&= \left. 4\sqrt{2}\left( \frac{1}{5}\sec^5 u - \frac{1}{3}\sec^3 u \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $\omega$ is a cube root of unity $\not = 1$ then find the minimum value of $|a+b\omega +c\omega^2|$, where $a,b,c$ are integers but not all equal. Let $z=a+b\omega + c\omega^2$
$$z=a+b\omega -c (1+\omega)$$
$$z=a-c+\omega (b-c)$$
Therefore $$|a-c+\omega (b-c)| \ge ||a-c|-|b-c||$$
How should I proceed?
| We want to minimize $|x+y\omega|^2=\bigg(x-\frac{y}{2}\bigg)^2+\frac{3}{4}y^2$ where $x$ and $y$ are nonzero integers.
If $y$ is even, we can take $x=\frac{y}{2}$ and hence $|x+y\omega|^2 \geq \frac{3}{4}y^2 \geq 3$.
If $y$ is odd, we can take $x=\frac{y+1}{2}$ and hence $|x+y\omega|^2 \geq \frac{1}{4}+\frac{3}{4}y^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Prove $\sum_{i=1}^{n-1} \left[\frac{n}{i(i+1)} + \frac{n(n-1)}{i(i+1)} (n(H_{n-2} - H_{n-i-1}) - (i-1))) \right] =(n-1)^2$? Apparently the following expression
$$
\sum_{i=1}^{n-1} \Bigg[\frac{n}{i(i+1)} + \frac{n(n-1)}{i(i+1)} (n(H_{n-2} - H_{n-i-1}) - (i-1))) \Bigg] \\
$$
simplifies to $(n-1)^2$, where $H_i$ is... | I got $-(n-1)^2$ and by calculating the first few cases, it seems to be the correct answer. I will assume that $n\geq 2$ and $H_0=0$. First note that
$$
\sum\limits_{i = 1}^{n - 1} {\frac{1}{{i(i + 1)}}} = \sum\limits_{i = 1}^{n - 1} {\left[ {\frac{1}{i} - \frac{1}{{i + 1}}} \right]} = 1 - \frac{1}{n}.
$$
We can deco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate the area of the surface I have to calculate the area of such surface:
$$(x^2+y^2+z^2)^2=x^2-y^2$$
I use the spherical coordinates transform, and get $r^2=\sin^2 \varphi \cos 2\theta$, but I don't know how to use this to calculate the area.
I'll be grateful if there's any help. :)
| First notice that by even symmetry in each of the variables we can reduce the integral to just one in the first octant
$$\iint_S \:dS = 8 \iint_{S\:\cap\:\text{First octant}} \:dS$$
Next we will parametrize the surface like so:
$$x = \sin^2\theta \cos\phi \sqrt{\cos 2\phi}$$
$$y = \sin^2\theta \sin\phi \sqrt{\cos 2\phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Contest Math sum Partition question From IMO 2012 shortlist:
What is the maximum value of $k$ such that the set $S=\{1,2,\cdots,2018\}$ can be partitioned into $k$ disjoint pairs such that the sum of (the two numbers in) each pair is pairwise distinct and none of the sums exceeds $2018$.
I predict that the answer is ... | This problem is in IMO 2012 shortlist, page 20.
The answer is $\left\lfloor \frac{2n-1}{5}\right\rfloor$. In this case $807$.
Original post in the shortlist:
C2. Let $n \geqslant 1$ be an integer. What is the maximum number of disjoint pairs of elements of the set $\{1,2, \ldots, n\}$ such that the sums of the differ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
When $ab/(a+b)$ is an integer, where $a,b$ are positive integers. When $ab/(a+b)$ is an integer, where $a,b$ are positive integers?
clear;
maxn:=30;
for a in [1..maxn] do
for b in [a..maxn] do
q1:=a*b; q2:=a+b;
if q1 mod q2 eq 0 then
print a,b,q1 div q2;
end if;
end for;
end for;
the Magma code given above outputs t... | Let $\gcd(a,b)=d$, $a=du$ and $b=dv$.
Thus, $\gcd(u,v)=1$ and $$\frac{ab}{a+b}=\frac{ab+b^2-b^2}{a+b}=b-\frac{dv^2}{u+v},$$ which says $$d=k(u+v)$$ and we obtain:
$$(a,b)=(k(u^2+uv),k(v^2+uv)),$$ where $u,$ $v$ and $k$ are integers, $k\neq0$ and $\gcd(u,v)=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3668069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving $(1+x)f'(x)=a f(x)$ where $f(x)=\sum_{n=0}^\infty {a\choose n}x^n$ A function $f$ is defined for $-1<x<1$ by $f(x)=\sum_{n=0}^\infty {a\choose n}x^n$. Here $a$ is a real number which is neither zero nor a positive integer. Prove that $(1+x)f'(x)=a f(x)$.
I took the derivative of $f(x)$ which is $\sum _{n=0}^\in... | For problems like these, I like to write out the first few terms in order to really understand what I've got.
We are working with $f$ defined by
$$f(x) = \binom a0 + \binom a1 x + \binom a2 x^2 + \cdots.$$
Your definition of $f'(x)$ is correct, except that the series should start at $n=1$ since the derivative of $\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{n\to\infty} \lvert x_n \rvert ^\frac{1}{n}$ when $x_{n+1}=3x_n + \frac{2}{x_n^2},x_1=1$ $\lim_{n\to\infty} \lvert x_n \rvert ^\frac{1}{n}$ when $x_{n+1}=3x_n + \frac{2}{x_n^2}, x_1=1$
I figured out that $x_n$ increases as n approaches infinity.
However, the power approaches 0 as n approaches infinity.
So, I... | Assuming that
$x_0 > 0$,
$x_{n+1}
=3x_n + \frac{2}{x_n^2}
\gt 3x_n
$
so
$x_n > 3^n x_0$.
Also
$x_{n+1}
=3x_n + \frac{2}{x_n^2}
\lt 3x_n+\frac{2}{(3^nx_0)^2}
= 3x_n+\frac{2}{9^nx_0^2}
$
so
$\dfrac{x_{n+1}}{3^{n+1}}
\lt \dfrac{x_n}{3^n}+\frac{2}{27^nx_0^2}
$.
Letting
$y_n = \dfrac{x_n}{3^n}$,
$y_{n+1}-y_n
\lt \frac{2}{27... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3671525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the area of $\triangle ABC$ with $B$ and $C$ lying on the line $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$
The vertices $B$ and $C$ of a $\triangle ABC$ lie on the line, $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$ such that $BC = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A(1, –... | The problem with your method is that you're not measuring the area of triangle $\triangle ABC$, but the volume of the parallelepiped with vertices $0, A, B, C, A + B, A + C, B + C, A + B + C$. This is what the $3 \times 3$ determinant measures.
Instead, try computing
$$\frac{1}{2}\|(B - A) \times (C - A)\|.$$
The vecto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
continuity of function$f(x,y) = xy\dfrac {x^{2}-y^{2}}{x^{2}+y^{2}} $ in $(0,0)$? I try to find a value for $f(x,y)=xy\dfrac {x^{2}-y^{2}}{x^{2}+y^{2}}$ in (0,0) which f to be continuous.
First we must show that the $\lim_{(x,y) \to (0,0)} xy\dfrac {x^{2}-y^{2}}{x^{2}+y^{2}} $ exits and then do the alignment $f(0,0) = ... | For any given $\varepsilon>0$ we have
$$\begin{align}
\left|\,xy\,\,\frac{x^2-y^2}{x^2+y^2}\right|&\le \frac12|x^2-y^2|\tag1\\\\
&\le \frac12(x^2+y^2)\\\\
&<\varepsilon
\end{align}$$
whenever $\sqrt{x^2+y^2}<\delta=\sqrt{2\varepsilon}$.
Note in arrving at $(1)$ we used the AM-GM inequality $x^2+y^2\ge 2|xy|$.
Alter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3676921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Do we reduce the problem into the known $3^x+4^x=5^x $? I want to solve the following equations: \begin{align*}&(i) \ \ \ \ \ 3^x+4^x=5^x \\ &(ii) \ \ \ \ \ (x+1)^x+(x+2)^x=(x+3)^x, \ x>-1 \\ &(iii) \ \ \ \ \ (x+1)^x+(x+2)^x=(x+4)^x, \ x>-1\end{align*}
$$$$
I have done the following:
For (i):
We have that $$3^x+4^x... | To answer your last question: no this would not work. Finding an appropriate $x$ would mean… that you have solved the equation.
You can try the same "monotonicity trick", but the function is more complicated:
$$\left (\frac{x+1}{x+3}\right )^x+\left (\frac{x+2}{x+3}\right )^x=1$$
Note that a function like
$$\left (\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $\ f(x)=x^{3}$ is continuous at $\ x = -2$ using the delta-epsilon approach Prove that $\ f(x)=x^{3}$ is continuous at $\ x = -2$.
I am really struggling on how to choose an initial $\delta_1$ value, are we able to choose any value or are there specific values to target, in this case I chose $\delta_1=1$
The... | Note that\begin{align}x^2-2x+4&=\bigl((x+2)-2\bigr)^2-2\bigl((x+2)-2\bigr)+4\\&=(x+2)^2-6(x+2)+12\end{align}and that therefore$$|x+2|<1\implies|x^2-2x+4|<1+6+12=19.$$So, take $\delta=\min\left\{1,\frac\varepsilon{19}\right\}$ and then$$|x+2|<\delta\implies\bigl|(x+2)(x^2-2x+4)\bigr|<\frac\varepsilon{19}\times19=\vareps... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3685485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A sequence includes $a_p=\sqrt2$, $a_q=\sqrt3$, $a_r=\sqrt5$ for some $1\leq p
In a sequence $a_1, a_2,\dots$ of real numbers it is observed that $a_p=\sqrt{2}$, $a_q=\sqrt{3}$, and $a_r=\sqrt{5}$, where $1\leq p<q<r$ are positive integers. Then $a_p$, $a_q$, $a_r$ can be terms of
(A) an arithmetic progression
(B) a ha... | Suppose that $x=\sqrt{15}-\sqrt{10}-\sqrt{6}$ were rational.
Then, squaring the number would give the following:
$$\begin{align}
x^2
&=(\sqrt{15}-\sqrt{10}-\sqrt{6})(\sqrt{15}-\sqrt{10}-\sqrt{6})\\
&=15-\sqrt{150}-\sqrt{90}-\sqrt{150}+10+\sqrt{60}-\sqrt{90}+\sqrt{60}+6\\
&=31-2\sqrt{150}-2\sqrt{90}+2\sqrt{60}\\
&=31-2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Taylor Series of $f(x) =\frac{1}{x^2}$ I've found the Taylor Series for $f(x)=\frac{1}{x^2}$ centered at $a=-1$.
$f(-1)=1$, $f'(-1)=2$, $f"(-1)=6$, $f'''(-1)=24$, $f^4(-1)=120$
I used this formula to get each the first coefficients of the terms of the series: $c_n=\frac{f^n(a)}{n!}$
So I got the expansion:
$$f(x)=1+2(x... | You can verify it by computing this sum, because for $|x+1|<1$ this infinite sum is convergent:
$$
\frac{d}{dx}(x+1)\sum_{k=0}^{\infty} (x+1)^{k+1} = \frac{d}{dx}\bigg(-\frac{x+1}{x} - 1 \bigg) = \frac{d}{dx} \bigg( -\frac{1}{x} \bigg) = \frac{1}{x^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$
Prove that
$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$
with $a>0, b>0 , c> 0$ and $d>0.$
My attempt:
$$\begin{align*}\left(\dfrac{b}{a}... | Use AM-GM. $\frac{ad + bc}{2} \ge \sqrt{abcd}$. Squaring both sides, you get the answer. A tiny tip: If everything is positive, and you have an inequality, think about AM-GM once at least.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ . What is the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ ?
I have been stuck on this problem with no direction. I have tried multiplying the sequence with $x$ and trying out $S-Sx$ but have gotten nowhere. Any help?
Thanks.
| Hint:
$$S=2x+6x^2+12x^3+20x^4+\cdots$$
$$S(1-x)=2x+(6-2)x^2+(12-6)x^3+(20-12)x^4+\cdots=V\text{(say)}$$
$$V(1-x)=2x+x^2(4-2)+x^3(6-4)+x^4(8-6)+\cdots=\dfrac{2x}{1-x}\iff|x|<1$$
Can you take it from here?
See : this
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $z+\frac{1}{z}=2\cos\theta,$ where $z\in\Bbb C$, show that $\left|\frac{z^{2 n}-1}{z^{2n}+1}\right|=|\tan n\theta|$ If $z+\frac{1}{z}=2 \cos \theta,$ where $z$ is a complex number, show that
$$
\left|\frac{z^{2 n}-1}{z^{2 n}+1}\right|=|\tan n \theta|
$$
My Approach:
$$
\begin{array}{l}|\sin \theta|=\left|\sqrt{1-\co... | You could express the equation directly as a quadratic:
$z^2 - 2 z \cos \theta + 1 = 0$
and use the quadratic formula, for further insight.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3694013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all values at which the vector is linearly expressed through vectors I need to find all values $\lambda$ at which the vector $(7,-2, \lambda)$ is linearly expressed through vectors $(2,3,5), (3,7,8), (1,-6,1)$
I can't go to the next step after building a system:
$2x_{1} + 3x_{2} + x_{3} = 7$
$3x_{1} + 7x_{2} + 8x_... | Reduce the augmented matrix:
$$
\left[\begin{array}{ccc|c}
2&3&1&7\\
3&7&-6&-2\\
5&8&1&\lambda
\end{array}\right]
\sim
\left[\begin{array}{ccc|c}
1&0&5&11\\ 0&1&-3&-5\\ 0&0&0&15 - \lambda
\end{array}\right].
$$
Therefore, the only possible $\lambda$ that works is $\lambda = 15$, and it does in fact work: if $\lamb... | {
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"timestamp": "2023-03-29T00:00:00",
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Measure of an angle "subtended by each pentagon" in a truncated icosahedron A soccer ball is a truncated icosahedron; it consists of 12 black regular pentagons and 20 white regular hexagons; the edge lengths of the pentagons and hexagons are congruent. What is the measure of the solid angle (in ste-radian) subtended by... | We know that the normal distance $H_p$ of each of 12 regular pentagonal faces from the center of a truncated icosahedron with edge length $a$ is given by following formula
$$H_p=\frac{a(2\sqrt5+1)}{\sqrt{10-2\sqrt5}}=\frac{a}{4}\sqrt{\frac{250+82\sqrt5}{5}}$$
The solid angle subtended by any regular polygonal plane wit... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of ways of arranging 10 tiles in four colors such that any consecutive block of 5 tiles contain all four colors This problem is from Purple Comet High school contest, 2016.
Ten square tiles are placed in a row, and each can be painted with one of the four colors red (R), yellow (Y), blue (B), and white (W). Find... | There are four types of legitimate sequence of length $n$:
Type $A$: the last four colours are distinct,
Type $B$: the second from last colour is repeated later in the sequence,
Type $C$: the third from last colour is repeated later in the sequence,
Type $D$: the fourth from last colour is repeated later in the sequenc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3701810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the volume between the surface $x^2+y^2+z=1$ and $ z=x^2+(y-1)^2$ I'm trying to find the volume between the surface $x^2+y^2+z=1$ and $ z=x^2+(y-1)^2$ but nothing works for me.
I made the plot and it looks like this:
How could you start? Any recommendation?
| Try checking where the two surfaces intersect. Solve the given equations for $z$ and set them equal:
$$\begin{align*}
1-x^2-y^2&=x^2+(y-1)^2\\
0&=2x^2+2y^2-2y\\
0&=x^2+y^2-y\\
0&=x^2+\left(y-\frac12\right)^2-\frac14\\
x^2+\left(y-\frac12\right)^2&=\frac1{2^2}
\end{align*}$$
which corresponds to the cylinder of radius $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3702756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving limits in terms of epsilon delta questions $ f(x)= {x^3}-2x+1. $
We want to show that $ \lim_{x\to 2} ({x^3}-2x+1)= 5 $.
So here,
$ \lvert f(x)-f(2)\rvert = \lvert {x^3}-2x+1-5\rvert$ = $ \lvert {x^3}-2x-4\rvert $.
$ \lvert ({x^3}-2x)+(-4)\rvert \leq \lvert {x^3}-2x\rvert + \lvert -4\rvert = \lvert {x^3}-2x\... | As you have computed, we have
$$|f(x)-f(2)|=|x^3-2x-4|.\tag{0}$$
Now, we have
$$x^3-2x-4=(x-2)(x^2+2x+2)=(x-2)[(x+1)^2+1].\tag{1}$$
If $|x-2|<1$, then $1<x<3$ and we can bound
$$0<(x+1)^2+1\leq 16+1=17.\tag{2}$$
Therefore, for any $\epsilon>0$, we can choose $\delta=\min\{ 1, \epsilon/17\}>0$
such that if $|x-2|<\de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Spivak Calculus chapter 1 problem 13 proof critique Question paraphrased: Prove that the maximum between two numbers $x$ and $y$ is given by:
$$\max(x,y)=\frac{x+y+|y-x|}{2}$$
Proof: Let $x$ and $y$ be two arbitrary numbers. Then, one and only one holds true: $x \ge y$ or $x \le y$
If $x \ge y$, $x=x+(y-y)=(x-y)+y=|x-... | A quicker way to the proof would be to show LHS=RHS. If $x\geq y,$ then LHS$=x$ and $|y-x|=-(y-x)=x-y$ which gives $$\text{RHS }=\frac{x+y+(x-y)}{2}=\frac{x+x}{2}=x.$$
The case $y> x$ is similar.
| {
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Prove that $\exists !c \in \mathbb{R} \exists ! x \in \mathbb{R} (x^2 + 3x + c = 0)$ This is an exercise from Velleman's "How To Prove It". I am struggling with how to finish the final part of the uniqueness proof, so any hints would be appreciated!
*a. Prove that there is a unique real number $c$ such that there ... | Your approach is complicated and I do not recommend it. Instead, consider the quadratic formula applied here:
$$
x = \frac{-3 \pm \sqrt{9 - 4c}}{2}
$$
Let $d$ be another number such that there exists a unique $x$ in which $x^2 + 3x + d = 0$. We note that $d \not> \frac{9}{4}$, as otherwise $x$ it not real (so no such r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Standard limits not working on this particular question I have this question and its been troubling me for so long. I try to use the standard trig. limits but that just fails everytime and I get the answer as $\infty$.
$$\lim_{x \to 0} \frac{2x+x\cos(x)-3\sin(x)}{x^4\sin(x)}$$
note: I have posted the question by alread... | One way you can find this is by using the Taylor series expansions for $\cos x$ and $\sin x$.
Using the first few terms in each,
\begin{align*}
\frac{2x + x \cos x - 3 \sin x}{x^4 \cdot \sin x} &= \frac{2x + x(1 - \frac{x^2}{2!} + \frac{x^4}{4!} + O(x^6)) - 3(x - \frac{x^3}{3!} + \frac{x^5}{5!} + O(x^7))}{x^4(x + O(x^3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ f(x) = f(x+1) $ and $ f''(x) + f(x) = \frac {1} { f( x+ \frac{3}{4}) } $ then $ f(x) = f( x+ \frac{1}{4}) $ I was trying to solve following problem.
Let $ f: \Bbb R \rightarrow \Bbb R $ be a twice differentiable function such that $ f(x) = f(x+1) $ for all $x$. If $$ f''(x) + f(x) = \frac {1} { f( x+ \frac{3}{4})... | Let me try it : We put $x=\ln(y)$ and $y\neq 0$
The equality :$$ f''(x) + f(x) = \frac {1} { f( x+ \frac{3}{4}) } $$
Becomes :
$$\frac{f''(\ln(y))-f'(\ln(y))}{y^2}+f(\ln(y))=\frac{1}{f(\ln(y)+0.75)}$$
Now we assume $$f''(\ln(y))-f'(\ln(y))\neq 0\quad \forall y\neq 0$$
But :
$$\frac{f''(\ln(y))-f'(\ln(y))}{y^2}+f(\ln(y)... | {
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"timestamp": "2023-03-29T00:00:00",
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If real $x$ and $y$ satisfy $x^2+y^2-4x+10y+20=0$, then prove that $y+7-3\sqrt{2}\le x\le y+7+3\sqrt{2}$
Let $x, y \in \mathbb{R}$ such that $x^2 + y^2 - 4x + 10y + 20 = 0$. Prove that
$$y + 7 - 3\sqrt{2} \le x \le y + 7 + 3\sqrt{2}$$
I'm struggling with that problem. I've recognized that the given equation is one... | WLOG $x=2+3\cos t, y=-5+3\sin t$
$$x-y=7+3(\cos t-\sin t)$$
Now $$(\cos t-\sin t)^2+(\cos t+\sin t)^2=2$$
$$\implies(\cos t-\sin t)^2\le2$$
| {
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How do I find the sum of a power series $\underset{n=3}{\overset{\infty}{\sum}}\frac{x^n}{(n+1)!n\,3^{n-2}}$? I have found the area of convergence to be $ x \in (-\infty, \infty)$, and this is how far I had gotten before getting stuck:
$$
\begin{aligned}
\sum_{n=3}^{\infty} \frac{x^{n}}{(n+1) ! n 3^{n-2}} &=\sum_{k=0}^... | $$f(x)= \sum_{n=3}^{\infty} \frac{x^{n}}{(n+1) ! n 3^{n-2}}$$
$$f^{'}(x)= \sum_{n=3}^{\infty} \frac{nx^{n-1}}{(n+1) ! n 3^{n-2}}= \frac{3^3}{x^2}\sum_{n=3}^{\infty} \frac{x^{n+1}}{(n+1) ! 3^{n+1}} = \frac{3^3}{x^2} \left( e^{\frac{x}{3}}-1- \frac{x}{3}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\sqrt{x^2+8x+7}+\sqrt{x^2+3x+2}=\sqrt{6x^2+19x+13}$ I tried squaring both sides but doesn't seem like a good idea.
$$x^2+8x+7+\sqrt{(x^2+8x+7)(x^2+3x+2)}+x^2+3x+2=6x^2+19x+13$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4x^2+8x+4$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$$
Is there a better way for solving this equation?
| After writing our equation in the form $$\sqrt{(x+1)(x+7)}+\sqrt{(x+1)(x+2)}=\sqrt{(x+1)(6x+13)}$$ we obtain the domain:
$$(-\infty-7]\cup[-1,+\infty).$$
1. $x\geq-1.$
We see that $-1$ is a root and it remains to solve here
$$\sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}$$ or
$$\sqrt{(x+7)(x+2)}=2(x+1)$$ or
$$3x^2-x-10=0,$$ whic... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a,b,c>0$ then prove this inequality holds $$\left( \frac{a}{b+c}+\frac{b}{c+a} \right)\left( \frac{b}{c+a}+\frac{c}{a+b} \right)\left( \frac{c}{a+b}+\frac{a}{b+c} \right)\ge 1+\frac{1}{8}{{\left( \frac{a-b}{a+b} \right)}^{2}}{{\left( \frac{b-c}{b+c} \right)}^{2}}{{\left( \frac{c-a}{c+a} \right)}^{2}}$$
I broke the... | Let $\frac{a}{b+c}=\frac{x}{2},$ $\frac{b}{a+c}=\frac{y}{2}$ and $\frac{c}{a+b}=\frac{z}{2}.$
Thus, since $$\frac{x-y}{x+2}=\frac{\frac{2a}{b+c}-\frac{2b}{a+c}}{\frac{2a}{b+c}+2}=\frac{a-b}{a+c},$$ we need to prove that:
$$\prod_{cyc}(x+y)\geq8+\frac{\prod\limits_{cyc}(x-y)^2}{\prod\limits_{cyc}(x+2)^2}.$$
Also, $$2=\s... | {
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Rotating $(3,3)$ by $45^\circ$ about center $(2,1)$
With a two dimensional surface, if we take $(2,1)$ as the center point and consider a transformation with a rotation axis around $45^\circ$, then point $(3,3)$ is transformed into point $(?,?)$
I am a bit stumped on how to do a $45^\circ$ rotation. I'd prefer answer... | While the rotation matrix that rotates $(x, y)$ by angle $\theta$ counterclockwise around a center $(c_x, c_y)$ may look scary,
$$\left[\begin{matrix} x^\prime \\ y^\prime \end{matrix}\right] =
\left[\begin{matrix} c_x \\ c_y \end{matrix}\right] + \left[\begin{matrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\thet... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving $\frac{1}{16} \sum \frac{(b+c)(c+a)}{ab} +\frac{9}{4} \geq 4\sum \frac{ab}{(b+c)(c+a)}$ For $a,b,c>0$. Prove: $$\frac{1}{16} \sum\limits_{cyc} {\frac { \left( b+c \right) \left( c+a \right) }{ba}}+\frac{9}{4} \geq 4\, \sum\limits_{cyc}{
\frac {ba}{ \left( b+c \right) \left( c+a \right) }}$$
My SOS's proof is:
... | We have$:$\begin{align*}\sum\limits_{cyc} {\frac { \left( b+c \right) \left( c+a \right) }{ba}}+36 - 64\, \sum\limits_{cyc}{
\frac {ba}{ \left( b+c \right) \left( c+a \right) }}
=\frac{\prod \,(a+b)\sum\limits_{cyc} a(a-b)(a-c)+\Big[\sum\limits_{cyc} c(a-b)^2\Big]^2}{a\,b\,c\,(a+b)\,(b+c)\,(c+a)}\geq 0\end{align*}
The ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\int_0^{2\pi} \frac{x \cos x}{2 - \cos^2 x} dx$. I have to find the integral:
$$\int_0^{2\pi} \frac{x \cos x}{2 - \cos^ 2 x} dx$$
I rewrote it as:
$$\int_0^{2\pi} \frac{x \cos x}{1 + \sin^ 2 x} dx$$
But nothing further. I plugged it in a calculator and the result was $0$. I can see that the following relation hol... | $$I=\int_0^{2\pi} \frac{x \cos x}{2 - \cos^ 2 x} dx\tag 1$$
$$I=\int_0^{2\pi} \frac{(2\pi-x) \cos x}{2 - \cos^ 2 x} dx\tag 2$$
Adding (1) & (2) $$2I=\int_0^{2\pi} \frac{2\pi \cos x}{2 - \cos^ 2 x} dx$$
$$I=2\pi \int_0^{\pi} \frac{\cos x}{2-\cos^ 2 x} dx\tag 3$$
$$I=2\pi \int_0^{\pi} \frac{\cos (\pi-x)}{2-\cos^ 2(\pi- x... | {
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Simplification of ${0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$ Simplify
$$0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots,$$ where $n \ge 2.$
I think we can write this as the summation $\displaystyle\sum_{i=0}^{n} 2i\binom{n}{2i},$ which simplifies to $\boxed... | Spoilered answer:
$$\sum_{i} 2i \binom{n}{2i}=n\sum_{i} \binom{n-1}{2i-1}= n2^{n-2}$$
To get the second equaltiy note that $\sum_{i}\binom{n-1}{2i-1}=\sum_{i} \binom{n-1}{2i}$ as results from writing the binomial formula for $(1-1)^{n-1}$, and recall that
$$\sum_{i}\binom{n-1}{2i-1}+\sum_{i} \binom{n-1}{2i}= \sum_{i... | {
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Showing this integral is always greater than $\pi/2$ Define $$I(b)= \int_1^\infty {(u+b)^{1/2} du\over (u^2+b)}$$
I want to show that $\mathit I(b)$ is greater than $\pi/2$, for every $\mathit b >0. $
| The following is a more-or-less elementary approach:
Changing variables $u = bv$ in the original $I(b)$ integral, we get
$$ I(b) = \int_{1/b}^{\infty} \frac{b^{1/2} \sqrt{v+1}}{(b^{1/2}v)^2 +1} \, dv = \int_{1/b}^{\infty} \sqrt{v+1} \, \partial_v(\arctan(b^{1/2}v)-\pi/2) \, dv.$$
Integrating by parts, we get
$$I(b) = \... | {
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Using residue theorem to calculate integral $\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}$ - where is my mistake? I am to calculate:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}
$$
We can set $\gamma(t)=e^{it}$ for $t \in [0, 2\pi]$ and then $z = e^{it}$, $\dfrac{dz}{iz}=dt$, $\sin t =\dfrac{1}{2i}(z-\frac{1}{z})$ so th... | The mistake comes from calculating the residue.
We have $Res(f, -\frac{i}{3})=lim_{z\rightarrow\frac{-i}{3}}\frac{z+\frac{i}{3}}{(z+3i)(3z+i)}=\frac{1}{3}lim_{z\rightarrow\frac{-i}{3}}\frac{3z+i}{(z+3i)(3z+i)}=\frac{1}{3}lim_{z\rightarrow\frac{-i}{3}}\frac{1}{(z+3i)}=\frac{1}{3}\frac{1}{\frac{-i}{3}+3i}=\frac{1}{3}\fra... | {
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Solving $(D^2-1)y=e^x(1+x)^2$ I did like this:
$$\text{Let,} y=e^{mx} \text{ be a trial solution of } (D^2-1)y=0$$
$$\therefore \text{The auxiliary equation is } m^2-1=0$$
$$\therefore m=\pm1\\
\text{C.F.} = c_1e^x+c_2e^{-x}$$
$$\begin{align}
\text{P.I.}& =\frac{1}{D^2-1}e^x(1+x)^2\\
& =e^x\frac{1}{(D+1)^2-1}(1+2x+x^2)... | The answers are essentially the same. $-\frac18e^x$ can be absorbed in $c_1e^x$.
| {
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T,N,B and the curvature and the equation of the osculating plane at a given point Given a vector valued function defined by $r(t):=(2t)i+(t^2)j+(\ln t)k$,Find the unit vectors $\vec T,\vec N,\vec B$ at the point $P(2,1,0)$,then find the curvature and the equation of the osculating plane at the point.
Using the definit... | These three vectors are unit vectors, so they must have magnitude $1$. To begin, we will find $\vec{\mathbf{r}}\,'(t)$.
$$\vec{\mathbf{r}}(t)=\left\langle2t,t^2,\ln(t)\right\rangle\implies\vec{\mathbf{r}}\,'(t)=\left\langle2,2t,\frac{1}{t}\right\rangle$$
We are assuming $t\gt0$. Next, we will find the magnitude of $\ve... | {
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Finding the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$
Find the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$.
My attempt: $$\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}+\tan^{-1}\f... | Actually
$\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}=$
$=\pi+\tan^{-1}\left(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}}\right)$
We can notice that
$\frac{\pi}{2}<\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}<\pi$.
I am going to prove that
if $\frac{\pi}{2}<\alpha+\beta<\frac{3\pi}{2}$ then $\alpha+\bet... | {
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Rewrite a term as sum of squares At the moment I am trying to rewrite this term:
$2x^2-2xy+5y^2-4x+2y+2$
as a sum of squares.
So I am trying to find an experession of $2x^2-2xy+5y^2-4x+2y+2=a^2+b^2+c^2$ (for example)
It looks easy, but everything I have tried failed so far. So I wonder if there is such expression. I kn... | The second matrix identity below says
$$ \frac{1}{2} (2x-y-2)^2 + \frac{9}{2} y^2 $$
The method is discussed at
reference for linear algebra books that teach reverse Hermite method for symmetric matrices
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 1 }{ 2 } & 1 & 0 \\
1 & 0 & 1 \\
\end{a... | {
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What is the value of $p^2q + q^2r + r^2p$ for the given cubic equation?
If $p$, $q$, $r$ are the real roots of the equation $x^3-6x^2+3x+1=0$, then find the value of $p^2q + q^2r + r^2p$.
My Attempt:
I tried $(p+q+r)(pq+qr+rp)$ but couldn't really figure out what to do with the extra terms. The roots are also not tri... | As already noted, $A(p,q,r)=p^2 q + q^2 r + r^2 p$ is not a symmetric function of the roots, but if we couple it with $B(p,q,r)=q^2 p + r^2 q + p^2 r$ we have that
$$ A+B = pq(p+q)+qr(q+r)+pr(p+r) = (pq+qr+pr)(p+q+r)-3pqr$$
$$ AB = pqr(p^3+q^3+r^3)+3 p^2 q^2 r^2+\frac{1}{2}(p^3+q^3+r^3)^2-\frac{1}{2}(p^6+q^6+r^6). $$
B... | {
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Evaluate $\int_{-\pi}^{\pi} \frac{x^2}{1+\sin{x}+\sqrt{1+\sin^2{x}}} \mathop{dx}$ I came across this integral:$$\int_{-\pi}^{\pi} \frac{x^2}{1+\sin{x}+\sqrt{1+\sin^2{x}}} \mathop{dx}$$
I tried $u=x+\pi$
$$\int_{-\pi}^{\pi} \frac{(x+\pi)^2}{1-\sin{x}+\sqrt{1+\sin^2{x}}} \mathop{dx}$$
but had no success.
I also tried $u=... | Rationalizing the denominator makes it much easier to proceed as follows
$$\frac{x^2}{1+\sin{x}+\sqrt{1+\sin^2{x}}}=\frac{(1+\sin x-\sqrt{1+\sin^2x})x^2}{(1+\sin{x}+\sqrt{1+\sin^2{x}})(1+\sin{x}-\sqrt{1+\sin^2{x}})}$$
$$=\frac{(1+\sin{x}-\sqrt{1+\sin^2{x}})x^2}{2\sin x}$$
$$\therefore I=\int_{-\pi}^{\pi}\frac{(1+\sin{x... | {
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For real values with $abc\neq0$, if $\frac{xb+(1-x)c}{a}=\frac{xc+(1-x)a}{b}=\frac{xa+(1-x)b}{c}$, show that $x^3 = -1$ and $a=b=c.$
Let $a,b,c$ and $x$ are real numbers such that $abc \neq 0$ and $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c}.$$ Prove that $x^3=-1$ and $a=b=c.$
My attempt $:... | I don't think that's true. Let $x=1/2$. Then:
$$\frac {b+c} {2a} = \frac {c + a}{2b} = \frac {a+b}{2c}.$$
Any $a,b,c\ne 0$ such that $c=-a-b$ and $a+b\ne 0$ are a solution for that:
$$\frac {b-a-b} {2a} = \frac {-a-b + a}{2b} = \frac {a+b}{-2a-2b} \implies \frac {-a} {2a} = \frac {-b}{2b} = \frac {a+b}{-2(a+b)} \implie... | {
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Proving $a^4+2a^3 b+2ab^3+b^4 ≥ 6a^2b^2$ Prove that for any positive real numbers $a$ and $b$,$$a^4 + 2a^3b + 2ab^3 + b^4 ≥ 6a^2b^2.$$I tried using Vieta's formula to show the product of the LHS is greater than the RHS, but I don't think I am correct.
| Rewrite the inequality and employ AM-GM:
$$a^4+a^3b+a^3b+ab^3+ab^3+b^4 \ge 6\sqrt[6]{a^{12}b^{12}} =6a^2b^2$$
| {
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If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $ I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.
EDIT 1. My approach (that I was talking about):
Given: $z^3=x^3+... | Th proof by Bjkjdz. The inequality equivalent to
$${x^2}z + {y^2}z - {z^3} > 6z(z - x)(z - y),$$
or
$${x^2}(z - x) + {y^2}(z - y) > 6z(z - x)(z - y),$$
or
$$\dfrac{{{x^2}}}{{z - y}} + \dfrac{{{y^2}}}{{z - x}} > 6z.$$
But
$$\left\{ \begin{array}{l} {x^2} = \dfrac{{{x^3}}}{x} = \dfrac{{{z^3} - {y^3}}}{x} = \dfrac{{(z - y... | {
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Determine basis for solution space when there are more unknowns than equations Let's say we have
$$3x_1+x_2+x_3+x_4=0$$
$$5x_1-x_2+x_3-x_4=0$$
Then I would have matrix $\begin{bmatrix} 3 &1&1&1\cr8&0&2&0 \end{bmatrix}$ which simplifies to $\begin{bmatrix}-1&1&0&1\cr4&0&1&0\end{bmatrix}$
then $4x_1+x_3=0$ and $-x_1+x_2+... | Take it to reduced row-echelon form. You have
$$\begin{bmatrix}-1&1&0&1\\4&0&1&0\end{bmatrix} \sim \begin{bmatrix}\color{red}1&0&\frac{1}{4}&0\\0&\color{red}1&\frac{1}{4}&1\end{bmatrix}.$$
From here, identify the pivots, coloured in red, and the columns without pivots in them. In this case, we have the third and fourth... | {
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Find affine function to approximate local inverse Let
$$
f(x,y) = (x^2 - y^2, 2xy)
$$
then by the inverse function theorem $f$ is invertible locally at any point $(x,y) \neq (0,0)$ because
$$
\det Df(x,y) = \det \begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix} = 4(x^2+y^2) > 0 \iff (x,y) \neq (0,0)
$$
it's also not gl... | As a function $f:\Bbb{C}\to\Bbb{C}$ it is imply
$$f(z)=z^2$$
with some algebra we can find an inverse
$$f^{-1}(z)=\sqrt{z} = r^{\frac{1}{2}}e^{\frac{i\theta}{2}} = r^{\frac{1}{2}}\left[\cos\left(\frac{\theta}{2}\right)+i\sin\left(\frac{\theta}{2}\right)\right]$$
And using the half angle identities gives us that the inv... | {
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Evaluate $\cos \frac\pi7\cos \frac{3\pi}7\cos\frac{9\pi}7 $ I need help evaluating the expression
$$\cos \frac\pi7\cos \frac{3\pi}7\cos\frac{9\pi}7 $$
Can someone show the steps he or she used to arrive at the answer?
| Note
\begin{align}
& \cos \frac\pi7\cos \frac{3\pi}7\cos\frac{9\pi}7\\
= &-\cos \frac{8\pi}7\cos \frac{4\pi}7\cos\frac{2\pi}7\\
=& -\cos \frac{8\pi}7\cos \frac{4\pi}7\sin\frac{4\pi}7\cdot\frac1{2\sin\frac{2\pi}7}\\
=& -\cos \frac{8\pi}7\sin\frac{8\pi}7\cdot\frac1{4\sin\frac{2\pi}7}\\
=& -\frac{\sin\frac{16\pi}7}{8\sin... | {
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Find all functions which satisfy $f(m^2+n^2)=f(m)^2+f(n)^2$ $\forall\space m,n\in\Bbb{N}$ and $f(1)>0$ Remark. Dear voters, this question is not a duplicate of the following old question. Please refrain from closing it for being a duplicate.
QUESTION: Find all functions $f:\Bbb{N}→\Bbb{N}$ which satisfy
$$f(m^2+n^2)... | The only such function is $f(n)=n$.
The hard part is to show $f(1)=1$. After that, one just follows the answer to
Find all $f$ such that $f\left(m^{2}+n^{2}\right)=f(m)^{2}+f(n)^{2},$ plus some small values of $f(n)=n$ that will be proved below.
Let $f(1)=a$.
The given formula applies directly to any number that is the... | {
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inverse of a funtion I'm trying to calculate the inverse of the function:
$$y= \tan \left( \cos^{-1} \frac{x}{x+2} \right) $$
In my opinion it is $y= \frac{2 \cos ( \tan^{-1} x )}{1-\cos ( \tan^{-1} x )} $ but in my book it is reported :
$f(x)\begin{cases} -2\left( \frac{\sqrt{1+x^2}-1}{x^2} \right) & x < 0\\
-1 &... | Hint
$$y= \tan\big[ \arccos(f(x)) \big]=\frac{\sqrt{1-f(x)^2}}{f(x)}$$ could help you.
| {
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Convergence of series with negative terms Among the series $\sum\limits_{n=1}^{\infty}\frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{n}, \sum\limits_{n=1}^{\infty} \frac{(-1)^{\lfloor\log n\rfloor}}{n}$, and $\sum\limits_{n=1}^{\infty}\frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}$; which are convergent?
The leibniz test fails... | 1. Denote by $S_m = \sum_{n=1}^{m} \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{n}$ the $m$th partial sum. Define $N_0 = 1$ and let $N_k$ be the $k$-th index at which the sign of the sequence $\{(-1)^{\lfloor \sqrt{n}\rfloor}\}_{n=1}^{\infty}$ flips. In fact, we have $N_k = (k+1)^2$. Then $S_m$ lies between $S_{N_{k-1}-1}$ an... | {
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Find the value of $\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}$ given that $\sin\alpha-\cos\alpha=\frac12$ Given that $\sin\alpha-\cos\alpha=\frac12$. What is the value of $$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}?$$
My work:
$$\sin\alpha-\cos\alpha=\frac12$$
$$\sin\alpha\frac1{\sqrt2}-\cos\alpha\frac1{\sqrt2}=... | $$\sin\alpha-\cos\alpha=\frac{1}{2}$$
$$(\sin\alpha-\cos\alpha)^2=\frac{1}{4}$$,
$$\sin\alpha\cos\alpha=\frac38$$
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\left(\frac 1{\sin\alpha}-\frac 1{\cos\alpha}\right)\left(\frac 1{\sin^2\alpha}+\frac 1{\cos^2\alpha}+\frac 1{\sin\alpha \cos\alpha}\right)$$
$$=\left(\frac {... | {
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An inequality involving real numbers Let $x,y,z$ be real numbers such that $xyz=-1$. Prove that
$$\sqrt[3/2]{\frac{3}{2}}\geq E:=\frac{4(x^3+y^3+z^3)}{(x^2+y^2+z^2)^2}$$
I tried to imitate an idea by River Li but it does not work. The idea is to find a function $f$ such that for all $x,y>0$,
$$E\leq f(x+y)$$
and then u... | Let $x+y+z=3u$, $xy+xz+yz=3v^2,$ where $v^2$ can be negative, and $xyz=w^3$.
Thus, we need to prove that:
$$\left(\frac{3}{2}\right)^{\frac{2}{3}}(x^2+y^2+z^2)^2\geq-4\sqrt[3]{xyz}(x^3+y^3+z^3).$$
Now, we can assumme that $x^3+y^3+z^3>0,$ otherwise the inequality is obviously true.
But, if so $$x^3+y^3+z^3-3xyz>0$$... | {
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How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{... | You can use the partial fraction decomposition $$\dfrac{x^3}{(x^2+1)^3}=\dfrac{3i}{16}\left(\dfrac{1}{(x+i)^2}-\dfrac{1}{(x-i)^2}\right)+\dfrac{1}{8}\left(\dfrac{1}{(x+i)^3}+\dfrac{1}{(x-i)^3}\right)$$ to integrate this function. Then simplify the solution to get rid of complex unit $i.$
| {
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"answer_id": 7
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Geometric sequence problem including sum of the numbers Numbers: $a,b,c,d$ generate geometric sequence and $a+b+c+d=-40. $ Find these numbers if $a^2+b^2+c^2+d^2=3280$
I tried this problem and I have system of equations which I can't solve. I think there should be different way to handle this problem.
| Let $\frac{b}{a}=q$ and $1+q^2=2uq.$
Thus, $|u|\geq1$, $$a=-\frac{40}{1+q+q^2+q^3}$$ and $$a^2(1+q^2+q^4+q^6)=3280,$$ which gives
$$\frac{1600}{(1+q)^2(1+q^2)^2}\cdot(1+q^2)(1+q^4)=3280$$ or
$$20(1+q^4)=41(1+q)^2(1+q^2)$$ or
$$10(2u^2-1)=41(u+1)u$$ or
$$21u^2+41u+10=0,$$ which gives $$u=-\frac{5}{3},$$
$$1+q^2=-\frac{1... | {
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How can I find the roots of the polynomial $12x^{4}+2x^3+10x^2+2x-2$? It's clear that I can divide by $2$, but I don't know what can I do with $$6x^{4}+x^3+5x^2+x-1$$
Is there any algorithm for it or a trick? I have found the roots by an online calculator but I don't know how can I calculate them. Thank you for your he... | Here, I try to give a way of factorization, which isn't too hard to be noticed:
$6x^4+x^3+5x^2+x-1$
$=5x^4+x^3+5x^2+x+x^4-1$
$=x^3(5x+1)+x(5x+1)+(x^2+1)(x^2-1)$
$=x(x^2+1)(5x+1)+(x^2+1)(x^2-1)$
$=(x^2+1)(6x^2+x-1)$
$=(x^2+1)(3x-1)(2x+1)$
| {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \t... | Apparently I can evaluate integrals but not powers. To the ones who stumble upon this, when something is raised to a power, it is multiplied to that not added to it, which was the mistake that I was making. So for my problem above
$$(\sec^{2} \theta)^{\frac{3}{2}} = ((\sec \theta)^{2})^{\frac{3}{2}} = (\sec \theta)^{2 ... | {
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"answer_id": 3
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos ... | Here's an approach :
Let :
$$L=\lim_{x\to 0} \frac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{x\tan x}$$
\begin{align}
\lim_{x\to 0} \frac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{x\tan x}&=\lim_{x\to 0}\frac{\bigg(\sqrt{1+x\sin x}-\sqrt{\cos x}\bigg)\bigg(\sqrt{1+x\sin x}+\sqrt{\cos x}\bigg)}{x\tan x\bigg(\sqrt{1+x\sin x}+\sqrt{\cos x}\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Evaluating $\int_{0}^{2\pi}\frac{1}{\cos^2(\theta)+1}\, d\theta$ What would be the values of this definite Integral?
$$\int_{0}^{2\pi}\frac{1}{\cos^2(\theta)+1}\, d\theta$$
So, I have solved this definite integral using the substitution method, taking $u=\tan(\theta)$.
After some simplification, the solution to the def... | Here is an alternate way with contour integration:
$$\begin{aligned}
J= \int_{0}^{2\pi} \frac{d\theta}{\cos^2 \theta+1} &=
\oint_{|z|=1} \frac{\frac{dz}{iz}}{ \left[ \frac{(z+z^{-1})^2}{4}+1 \right]}\\
&= \frac{4}{i}\oint \frac{ z \, dz}{ z^4 + 6 z^2 +1 }\\
\end{aligned}$$
Take the integral in the positive (counte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\... | Substitute $t=x-2$
\begin{align}
\int \dfrac{dx}{(x^2-4x+13)^2}
& =\int \dfrac{dt}{(t^2+9)^2}= \int \frac1{18t}d\left( \frac{t^2}{t^2+9}\right)\\
&= \frac t{18(t^2+9)}+\frac1{18}\int \frac{dt}{t^2+9}\\
&= \frac t{18(t^2+9)}+\frac1{54}\tan^{-1}\frac t3+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 7
} |
Find the matrix $A^{15}$. Let $I=
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}$ and $O=\begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}$.
1.Let $A=\begin{pmatrix}
1 & 3 \\
3 & 5 \\
\end{pmatrix}$ and $ B=\begin{pmatrix}
x & 3 \\
3 & 6 \\
\end{pmatrix}$. Find the value of $x$ which satisfies $AB=BA$.
$AB=\begin{pm... | You may proceed as follows:
From the given condition we have
$$A-A^2 = A(I-A) =I$$
It follows
\begin{eqnarray} A^{15}
& = & A(A^2)^7 \\
& = & A(A-I)^7 \\
& = & -(A-I)^6 \\
& = & -(A^2-2A + I)^3 \\
& = & -(-A)^3 \\
& = & A(A-I) \\
& = & -I
\end{eqnarray}
Another way would be seeing
$$A^3+I = (A+I)(A^2-A+I)=O \Rightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Strange Cube Root Offense in an Inequality I don't know how to tackle the unusual cube root present in this inequality-
$1.$For real numbers $a,b,c > 0$ and $n\le3$ prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c}\right)\ge 3+n$$
Here is another question with the same lesser side (an... | The first inequality for $n=3$.
By AM-GM $$\sum_{cyc}\frac{a}{b}+\frac{9\sqrt[3]{abc}}{a+b+c}=\frac{1}{3}\sum_{cyc}\left(\frac{2a}{b}+\frac{b}{c}\right)+\frac{9\sqrt[3]{abc}}{a+b+c}\geq$$
$$\geq\sum_{cyc}\sqrt[3]{\frac{a^2}{b^2}\cdot\frac{b}{c}}+\frac{9\sqrt[3]{abc}}{a+b+c}=\sum_{cyc}\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
find primes p,q,and r
Find all primes $p, q, r$ such that $pq+r$, $pq+r^2$, $qr+p$, $qr+p^2$, $rp+q$,
$rp+q^2$ are all primes.
| We know that $pq+r>2$. If all of $p,q,r$ are odd, then $pq+r$ would be even, which would make is impossible for the number to be prime. Now, without loss of generality, let $p=2$. We then have $qr+2$ and $qr+4$ to be prime. We know that both these numbers are greater than $3$ and prime, thus cannot be divisible by $3$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Questions about counting the number of triples arranged in geometric progression
Problem: Three tickets are chosen from a set of $100$ tickets numbered $1,2,3,\ldots,100$. Find the number of choices such that the numbers on the three tickets are in geometric progression.
Solution: Let $k, n \in \mathbb Z_+$ s.t. $n... | Your first patterN, with a common factor of $2$, can start with any number from $1$ through $25$. $\lfloor \frac {100}{n^2} \rfloor$ is the number of progressions with common factor $n$. The third number is $n^2$ times the first, so the progression must start with a number small enough that the third term is no great... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Existence of limit for sequence $x_n=\frac12\left(x_{n-1}+\frac8{x_{n-2}}\right)$ with initial values $x_0=5,x_1=10$ Let $x_0=5,x_1=10,$ and for all integers $n\ge2$ let $x_n=\frac12\left(x_{n-1}+\frac8{x_{n-2}}\right).$ By induction, we have $\forall m\in\mathbb Z_{\ge0}\enspace x_m>0,$ so we can avoid division by $0$... | it is easy to show that $|x_{n}-A|<\epsilon$ for all the $x_{n}$ for a given $A$.
such that
$$A-\epsilon<x_{n}<A+\epsilon$$
$$-A-\epsilon<-x_{n}<-A+\epsilon$$
$$\frac{8}{A+\epsilon}<\frac{8}{x_{n}}<\frac{8}{A-\epsilon}$$
use equation $$x_{n}=\frac{1}{2}(x_{n-1}+\frac{8}{x_{n-2}})$$
we have $$x_{n}-x_{n-1}=\frac{1}{2}(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How can I study the convergence of the improper integral $\int_{0}^{ \infty} \frac{\sin(x)}{x+1} \, \mathrm dx\,$? I need to study the convergence of the following improper integral:
$$\int_{0}^{\infty} \dfrac{\sin(x)}{x+1}\, \mathrm dx$$
I did the following:
$$ -1 \leq \sin(x) \leq 1 \\
\implies \dfrac{-1}{x+1} \le... | Notice that
$$\int_0^\infty \frac{\sin x}{x+1}\,dx = \frac{-\cos x}{x+1}\Bigg|_0^\infty - \int_0^\infty \frac{\cos x}{(x+1)^2}\,dx = 1 - \int_0^\infty \frac{\cos x}{(x+1)^2}\,dx$$
and the last integral converges absolutely since
$$\int_0^\infty \frac{\left|\cos x\right|}{(x+1)^2}\,dx \le \int_0^\infty \frac{dx}{(x+1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Find the values of $k$ that satisfy $\gcd(a,b)=5$ Let $a$,$b$ and $k$ be integers such that
$$a=6k+4$$
$$b=11k+4$$
Find $k$ values that satisfy $\gcd(a,b)=5$
Note: $(a,b)$ are solutions to the equation : $11a-6b=20$
My try :
$\gcd(6k+4,11k+4)=\gcd(k-16;20)=5$
Possible values of $\gcd(a,b)$ are ${1,2,5,10,20}$ so we wan... | Use Euclids algorithm
$\gcd( 6k + 4,11k+4) = \gcd(6k+4, 5k)=\gcd(k+4,5k)=\gcd(k+4, 5k-5(k+4))=\gcd(k+4, -20)=\gcd(k+4,20)$
so $5|k+4$ but $k+4$ is odd.
That is $k+4 \equiv 0 \pmod 5$ or $k\equiv 1 \pmod 5$. And $k+4\equiv 1 \pmod 2$ so $k\equiv 1\pmod 2$ and (we know by Chinese remainder theorem that there is one uniq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is the sum of an alternating series satisfies the following inequality
Question if $s$ is sum of the alternating series $\sum (-1)^{n+1}z_n$ and if $s_n$ is the nth partial sum then, $|s-s_{n}|≤z_{n+1}$
My attempt:
$|s-s_n|=|s-s_{n+1}+s_{n+1}-s_n|$
$$≤|s-s_{n+1}|+|s_{n+1}-s_n|$$
$$≤\epsilon + z_{n+1}$$
($s_n\rightarr... | Here, two cases may arrise,one is when " $(z_n)$ is not monotone " and another is "$(z_n)$ is monotone ".
Case $1$, when $(z_n)$ is not monotone, then your statement is wrong.
Take the example, $z_n = \begin{cases}2^{-n} & \text{ n is even} \\ 3^{-n} & \text{ n is odd }\end{cases}$. ,
The sum, $\sum_{n=1}^{\infty} (-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$ The original question is:
Prove that:$$\begin{aligned}\\
\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\neq\int_0^1dy&\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\\
\end{aligned}\\$$
But I can't evaluate the integral $$\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^... | Hint:
$$\begin{align}\int\dfrac{x^2 - y^2}{\left(x^2 + y^2\right)^2}\,\mathrm dy&\equiv \int\dfrac{2x^2}{\left(x^2 + y^2\right)^2} - \dfrac{x^2 + y^2}{\left(x^2 + y^2\right)^2}\,\mathrm dy\\ &= 2x^2\int\dfrac1{\left(x^2 + y^2\right)^2}\,\mathrm dy - \int\dfrac1{x^2 + y^2}\,\mathrm dy\end{align}$$
The first integral ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
System of congurences and the Chinese Remainder Theorem I have the following system of congruences:
\begin{align*}
x &\equiv 1 \pmod{3} \\
x &\equiv 4 \pmod{5} \\
x &\equiv 6 \pmod{7}
\end{align*}
I tried solving this using the Chinese remainder theorem as follows:
We have that $N = 3 \cdot 5 \cdot 7 = 105$ and $... | Here's how I would solve this particular system.
$ x \equiv 4 \pmod{5}$ and $x \equiv 6 \pmod{7}$ means $x\equiv-1\pmod5$ and $x\equiv-1\pmod7$,
which means $x\equiv-1\pmod{35}$, i.e., $x\equiv34\pmod{35}$,
which means $x\equiv34, 69$, or $104\pmod{105}$.
By the Chinese remainder theorem, only one of these satisfies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can I prove that $\sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n}=\frac{13 \sin(7)}{170 - 26 \cos(7)}$? $$\sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n}=\frac{13 \sin(7)}{170 - 26 \cos(7)}$$
Have no clue how to prove it.
Possibly rewrite $\sin(7n)$ as $\frac{1}{2\sin(7)}\left(\cos(7n-7)-\cos(7n+7)\right)$.
But what next?
| Here is a method which does not use complex numbers. Let $x=\sum_{n=0}^\infty\frac{\sin(7n)}{13^n}$ and $y=\sum_{n=0}^\infty\frac{\cos(7n)}{13^n}$.
Then, we have
\begin{align}
x&=\sum_{n=1}^\infty\frac{\sin(7n)}{13^n}=\sum_{n=0}^\infty\frac{\sin(7n+7)}{13^{n+1}}\\
&=\frac1{13}\sum_{n=0}^\infty\frac{\sin(7)\cos(7n)+\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in t... | A polynomial $f(x)$ is a multiple of $x^2-x-1$ (the characteristic polynomial of the Fibonacci sequence) iff $f(\varphi)=f(\bar{\varphi})=0$, with $\varphi=\frac{1+\sqrt{5}}{2}$, $\bar{\varphi}=\frac{1-\sqrt{5}}{2}$ being algebraic conjugates. Since $\varphi^2=\varphi+1$ we have by induction $\varphi^k=F_{k}\varphi+F_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
If $a$, $b$, $c$, $d$ are positive reals so $(a+c)(b+d) = 1$, prove the following inequality would be greater than or equal to $\frac {1}{3}$.
Let $a$, $b$, $c$, $d$ be real positive reals with $(a+c)(b+d) = 1$. Prove that $\frac {a^3}{b + c + d} + \frac {b^3}{a + c + d} + \frac {c^3}{a + b + d} + \frac {d^3}{a + b + ... | Another way.
Since by AM-GM
$$1=(a+c)(b+d)\leq\left(\frac{a+c+b+d}{2}\right)^2,$$ we obtain $$a+b+c+d\geq2.$$
Now, let $a=kx$, $b=ky$, $c=kz$ and $d=kt$ such that $k>0$ and $x+y+z+t=4$.
Thus, $$k(x+y+z+t)\geq2,$$ which gives $$k\geq\frac{1}{2}.$$
But, $$\sum_{cyc}\frac{a^3}{b+c+d}=\sum_{cyc}\frac{k^2x^3}{y+z+t}\geq\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The existence of a point in a geometric configuration Consider $ABC$ a right triangle in $A$ and let $M$ be the middle of the height from the vertex $A$. Construct $B'=f(B)$ and $C'=g(C)$, where $f$ is the reflection about line $CM$ and $g$ is the reflection about line $BM$. Decide when do lines $BC'$ and $CB'$ interse... | Let $D$ be the feet of altitude from $A$ to $BC$. Then $AD=\frac{bc}a$, $DB=\frac{c^2}a$ so $\tan\angle MBC=\frac{b}{2c}$ and hence
$$
\tan\angle C'BC=\frac{4bc}{4c^2-b^2},
$$
and similarly
$$
\tan\angle B'CB=\frac{4bc}{4b^2-c^2}.
$$
The two lines $C'B$ and $B'C$ does not intersects iff $\angle C'BC+\angle B'CB=180^\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the derivative of $f(x)= \int_{\sin x}^{\tan x} \sqrt{t^{2}+t+1}\, \mathrm d t$
Find the derivative of $$f(x)=\int_{\sin x}^{\tan x} \sqrt{t^{2}+t+1}\,
\mathrm d t$$ with respect to $x$
So from may understanding, I need to apply the fundamental theorem of calculus and then differentiate. I think the upper and ... | Let $g(t)$ denote the integrand $\sqrt{t^2+t+1}$. On the one hand, the FTC guarantees
$$
\frac{d}{dx}\int_{\sin x}^{\tan x} g(t)\, dt
$$
$$
=g(\tan(x))\cdot (\tan(x))' - g(\sin(x))\cdot (\sin(x))'
$$
$$
=g(\tan(x))\cdot \sec^2(x) - g(\sin(x))\cdot \cos(x)
$$
$$
=\sqrt{\tan^{2} (x)+\tan(x)+1}\cdot \sec^2(x) - \sqrt{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that by means of the transformation $w=\frac 1z$ the circle $C$ given by $|z-3|=5$ is mapped into the circle $|w+\frac{3}{16}|=\frac{5}{16}$
Show that by means of the transformation $w=\frac 1z$ the circle C given by $|z-3|=5$ is mapped into the circle $\left|w+\frac{3}{16}\right|=\frac{5}{16}$
My try:
$$\begin{... | We need to prove that if $|z-3|=5$ so $$\left|\frac{1}{z}+\frac{3}{16}\right|=\frac{5}{16}.$$
Let $z=x+yi,$ where $x$ and $y$ are reals.
Thus, $$(x-3)^2+y^2=25$$ and we need to prove that:
$$|3z+16|=5|z|$$ or
$$(3x+16)^2+9y^2=25(x^2+y^2)$$ or
$$16(x^2+y^2)=96x+256$$ or
$$x^2-6x+y^2=16$$ or
$$(x-3)^2+y^2=25$$ and we are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Identifying relation between numbers based on equation relating them
If $a^2 + b^2+16c^2=2(3ab+6bc + 4ac)$ , where $a,b,c$ are non zero numbers. Then $a,b,c$ are in __________?
1. Harmonic progression 2. Geometric progression 3. Arithmetic progression 4. None of these
My attempt:
$ a \rightarrow 4a'$
$ b \rightarrow ... | There is a way to make this appear simpler. You are using letters a,b,c. Alright, take any $x,y,z$ you like such that
$2x^2 - 16 y^2 + 81 z^2 = 0.$ This is just a point on a cone. Then still to avoid fractions, let
$$ a = 4x+12y - 11z \; , \; \; b = 4y - 9 z \; , \; \; c = 4z $$
For infinitely many examples, we can in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
what loops and points of numbers are possible when you take the alternating sum of the digits of squared? what loops of numbers are possible when you take the alternating sum of the digits of squared?
I've heard about the happy numbers and the sad numbers. if you don't know the happy numbers are numbers when you add it... | Partial answer: the only points are $0,1$ and $48$.
Let $f()$ be the function that performs one step of your transformation, e.g. $f(125) = |1^2 - 2^2 + 5^2| = 22$ (not $23$ like you said).
If $x$ is a $k$-digit number, we have $f(x) \le \lceil k/2 \rceil \times 9^2$ because half the digits count positively and half co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
Proving $\displaystyle\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
My atempt:
\begin{align*}
\int_0^1 \int_0^1\frac{-x\ln(xy)}{1-x^2y^2} \, dx \, dy &=\int_0^1I_x(y)\,dy\\[6pt]
\text{where }I_x(y)=\int_0^1\frac{-x\ln... | \begin{align}
-\int_0^1\int_0^1\frac{x\ln{(xy)}}{1-x^2y^2}\,\mathrm{d}x\,\mathrm{d}y
&=-\sum_{n=0}^\infty \int_0^1 y^{2n} \int_0^1x^{2n+1} \ln{(xy)} \, \mathrm{d}x \, \mathrm{d}y \\[6pt]
&=-\sum_{n=0}^\infty\int_0^1y^{2n}\left[\frac{x^{2(n+1)}\ln{(xy)}}{2(n+1)}-\frac{x^{2(n+1)}}{4(n+1)^2}\right]_0^1\,\mathrm{d}y\\[6pt]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the greatest integer less than $\frac{1}{\sin^2(\sin(1))}$ without calculator. Find the greatest integer less than $$\frac{1}{\sin^2(\sin(1))}$$
This was on one of my tests. All angles in radians. Here's my work:
$$0<1<\frac{\pi}{3}<\frac{\pi}{2}$$
Since $\sin(x)$ is increasing in the first quadrant,
$$0<\sin(1)<\... | $$1 > \frac{7}{24}\pi \approx 0.916$$
$$\sin(1) > \sin\left(\frac{7}{24}\pi\right) = \frac12\sqrt{2+\sqrt{2-\sqrt{3}}} \approx 0.793 > \frac{\pi}{4} \approx 0.785$$
$$\frac{1}{\sin^2(\sin(1))} < \frac{1}{\sin^2(\frac{\pi}{4})}=2$$
so $$1 < \frac{1}{\sin^2(\sin(1))} < 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to find the determinant of this $6\times 6$ X-matrix? This question was asked in my quiz and i was unable to solve it, so I am asking it here.
Find the value of determinant of this particular matrix .
$$\begin{pmatrix}1&0&0&0&0&2\\0&1&0&0&2&0\\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{pmatrix}$$
I ... | Performing a Laplace expansion along the first column, we get
$$
\begin{vmatrix}1&0&0&0&0&2\\0&1&0&0&2&0\\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{vmatrix}
=
1\cdot
\begin{vmatrix}1&0&0&2&0\\0&1&2&0&0\\0&2&1&0&0\\2&0&0&1&0\\0&0&0&0&1\end{vmatrix}
- 2\cdot
\begin{vmatrix}0&0&0&0&2\\1&0&0&2&0\\0&1&2&0&0\\0&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Proving: $\int_0^1 \int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$
Proving:$$\displaystyle\int_0^1\displaystyle\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$$
I tried using variable switching
$\ln(xy)=t$ But I did not reach any results after the calculation
\begin{align*}
k&=\displaystyl... | \begin{align}J&=\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy\\
&=\int_0^1 \frac{1}{x}\left(\int_0^x \frac{\ln^4 u}{(1+u)^2 du}\right)\\
&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^4 u}{(1+u)^2} du\right)\ln x\right]_0^1-\int_0^1 \frac{\ln^5 x}{(1+x)^2}dx\\
&=-\int_0^1 \frac{\ln^5 x}{(1+x)^2}dx\\
&\overset{\text{IBP}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How can we take the derivative of this function: $y = \frac{x}{x^2+1}$ from first principles (using the limit definition of the derivative)? I was taking the derivative of the function: $y = \frac{x}{x^2+1}$. I know that we can solve it by the quotient rule. But I tried using the limit definition of differentiation. Th... | $$\frac{h-x^2h-xh^2}{x^4h + x^2h^3+2x^3h^2+2x^2h+h^3+2xh^2+h}=\\ \frac{1-x^2-xh}{x^4 + x^2h^2+2x^3h^2+2x^2+h^2+2xh+1} \to \frac{1-x^2}{(1+x^2)^2}$$
when $h \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find limit of $f(x)$ as $x$ tends to $0$ I need some help answering this question:
$$f(x) = \frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}$$
find the limit of $f(x)$ as $x$ tends to $0$ by writing $f(x)$ as a quotient of two powers series.
I have so far:
$$\frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} ... | \begin{align}
& \frac{\left(x\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)\right)-\left(x + \frac{x^3}{3!} + \cdots\right)}{x\left(x + \frac{x^3}{3!}+\cdots\right)} \\ {} \\
= {} & \frac{\left(x + \frac{x^3}{2!} + \frac{x^5}{4!}+ \cdots\right) -\left( x + \frac{x^3}{3!} \right)}{x^2 + \frac{x^4}{3!}+ \cdots} \\ {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Evaluate $\int\limits_{0}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$ Notice that $\frac{n \sin x}{1+n^2x^2}\to 0$ pointwise.
And we have,$$\int\limits_{0}^{\infty}\frac{n \sin x}{1+n^2x^2}dx=\int\limits_{0}^{1}\frac{n \sin x}{1+n^2x^2}dx+\int\limits_{1}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$$
Then for $\int\limits_{1}^{\infty}... | And idea: substitute
$$x=\frac1u\implies dx=-\frac1{u^2}du\implies\int_0^1\frac{n\sin x}{1+n^2x^2}dx=\int_\infty^1-\frac{du}{u^2}\cdot\frac{n\sin\frac1u}{1+\frac{n^2}{u^2}}=$$
$$=\int_1^\infty\frac{n\sin\frac1u}{u^2+n^2}du$$
and now estimate
$$\left|\frac{n\sin\frac1u}{u^2+n^2}\right|\le\frac n{u^2+n^2}\,,\,\,\text{and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Factorize: $x^3 + x + 2$. How do I factorize the term $x^3 + x +2$?
What I have previously tried is the middle term factor method but it didn't work...
$x^3 + x + 2$
$\Rightarrow x^3 + 2x - x + 2$
$\Rightarrow x(x^2 + x) - 2(x - 1)$
This doesn't work.
What should I do?
| Another solution$:$
$$x^3+x+2=(x^3+1)+(x+1)=(x+1)(x^2-x+1)+(x+1)=(x+1)(x^2-x+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Given positive $x,y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $, find minimum $(x+y)$ I am given positive numbers $x, y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $. I need to find the minimum value of $(x+y)$. Here is my try. Using AM-GM inequality for nonnegative numbers, I have
$$
\frac{(x+y)}{... | put $x=r^2{cos}^2a$ and $y=r^2{sin}^2a$ also let $a$ belong to $[0,\frac{\pi}{2}]$
thus we have to find max value of $r^2$
plugging the values in the given equation and simplifying using basic trig formulas we have $r^4(cosa)(sina)(cos2a)=r^2$ or
$ r^2=\frac{4}{sin(4a)} \ge 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Olympiads Number Theory Problem: Show that there are no positive $a,b,c\in\Bbb Z$ such that $\frac{a^2+b^2+c^2}{3(ab+bc+ca)}\in\Bbb Z$ I have been struggling to solve this problem, could someone give me an idea/solution? please!
Show that there are no positive $a,b,c\in\Bbb Z$ such that $$\frac{a^2+b^2+c^2}{3(ab+bc+ca... | I thought that I had a solution using Vieta Jumping but this was not a complete solution so we can use quadratic residues to solve it...
We can choose WLOG $\text{gcd} (a,b,c)=1$.
After letting $\frac{a^2+b^2+c^2}{ab+bc+ca}=3k$, we have $(a+b+c)^2=(3k+2)(ab+bc+ca)$.
Choose a prime $p \equiv 2 \pmod{3}$ with $v_p(3k+2) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding Fourier series and evaluating at a point Ive been asked to find the fourier series of the following
Need help with finishing the question, i have included my work so far below:
and following that i need help with the final part of the question:
| You wrote the correct expression for the coefficients $a_n$, $n\ge1$. However, there must have been an error in your arithmetic for carrying out the integral
$$a_n=\frac1\pi\int_{-\pi}^\pi (x^2-\pi^2)^2\cos(nx)\,dx=\frac{48(-1)^{n-1}}{n^4}$$
Also, we need the term $a_0$, which is
$$a_0=\frac1{2\pi}\int_{-\pi}^\pi (x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Recursive squaring algorithm by reducing to squaring 5 numbers
*(c) Describe a recursive algorithm that squares any $n$-digit number in $O(n^{log_35})$ time, by reducing to squaring only five $(n/3+O(1))$-digit numbers.
[Hint: What is $(a+b+c)^2+(a−b+c)^2$?]
I am currently studying the Toom-Cook algorithm which is hi... | Wikipedia's article on Toom–Cook multiplication has a nice explanation that doesn't involve guessing missing squares; let me simplify it for your case.
If you're writing your number as $10^{2m}a + 10^m b + c$ (essentially, breaking up the digits into three segments $a, b, c$) then you're writing your number as $p(10^m)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.