Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to prove that for all natural $n$, $133|11^{n+2}+12^{2n+1}$
How to prove that for every natural $n$, the number
$$11^{n+2} +12^{2n+1}$$
is divisible by $133$?
I tried Induction method, so assuming that
$$
n=k \implies
A = (11^{k+2} +12^{2k+1} ) \pmod{133} \equiv 0
$$
Then trying to prove that
$$
n=k+1 \impli... | Given that $11^{k+2}+12^{2k+1}=133\lambda$, you can infer that $11^{k+2}=133\lambda-12^{2k+1}$.
Thus you have that $$11^{k+3}+12^{2k+3}= 11^{k+2}\cdot 11 +144 \cdot 12^{2k+1}=11\cdot 133\lambda -11\cdot 12^{2k+1} +144 \cdot 12^{2k+1} =11\cdot 133\lambda +12^{2k+1} \cdot (144-11)=11\cdot 133\lambda +133\cdot 12^{k+1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"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} ... | It might be useful to simplify the inequality before trying to solve:
$$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow x^2-2 \leq \frac{x^3+2x}{x+4}$$
Now, note that (using polynomial division)
*
*$\frac{x^3+2x}{x+4} = x^2 - 4x+18 -\frac{72}{x+4}$
So, the inequality turns into
$$x^2-2 \leq x^2 - 4x+18 -\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Prove that for $f(z)=\sum_{n=0}^{\infty} \frac{z^{n}}{2^n+1}$
Prove that for $f(z)=\sum_{n=0}^{\infty} \frac{z^{n}}{2^n+1}$,
$$\lim_{z \to2} f(z) (2-z) = 2$$
My approach :
I was thinking of applying Abel's limit theorem to compute the limit but since it is valid only for $z \to 1 $ I am totally clueless
I also trie... | Just note that for any $z \in (-2,2)$, because of absolute convergence in $[-\frac{2+z}{2},\frac{2+z}{2}]$, we have $$f(z)(2-z) = \sum\limits_{n=0}^{+\infty} \frac{2z^n}{2^n+1} - \sum\limits_{n=0}^{+\infty} \frac{z^{n+1}}{2^n+1} = 1 + \sum\limits_{n=0}^{+\infty} z^{n+1} \cdot \frac{2(2^n+1)-(2^{n+1}+1)}{(2^n+1)(2^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$n^3+n<3^n$ for $n \geq4$ by induction. To prove that $n^3+n<3^n$ for $n \geq4$ by induction.
I have proved the fact but it became very long and I have to use two more induction proof within the proof.
Can someone give a better solution by induction?
Thank You.
| $n^3 + n < 3^n \tag 1$
certainly holds for $n = 4$; so this will be the base case; assuming (1) binds for some $n$, then
$3(n^3 + n) < 3^{n + 1}; \tag 2$
now,
$(n + 1)^3 + (n + 1) = n^3 + 3n^2 + 3n + 1 + n + 1 = n^3 + 3n^2 + 4n + 2, \tag 3$
and
$3(n^3 + n) = 3n^3 + 3n, \tag 4$
so what we need is
$n^3 + 3n^2 + 4n + 2 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Given that $m$ is a real number not less than $-1$ The question is:
Given that $m$ is a real number not less than $-1$, such that the equation in $x$ is $x^2+2(m-2)x+m^2-3m+3=0$ has two distinct roots $r$ and $s$. If $r^2+s^2=6$, find the value of $m$.
Here's what I've tried:.
Using Vieta's formulas:
$rs = m^2-3m+3... | The posted solution is correct up to that point, though it remains to be verified that $\,r \ne s\,$ for those values of $\,m\,$.
Alt. hint: $\color{blue}{r+s}=-2m+4$, and each root satisfies $\,x^2= -\big(2(m-2)x+m^2-3m+3\big)\,$, so:
$$
\begin{align}
6 = r^2+s^2 &= -2(m-2)(\color{blue}{r+s})- 2(m^2-3m+3) \\
&= 2(m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using Gram-Schmidt to find the QR decomposition I'm having problems doing a QR decomposition of a matrix...
Let $A=\begin{bmatrix}
{1} & {1} & {0} \\
{0} &{1} &{1} \\
{1} & {0} &{1}
\end{bmatrix}$
Find the QR decomposition for A
Here's what I've been doing:
I choose this basis, $B=\left \{(1,0,1), (1,1,0), ... | $u_1 $ is correct, your mistake is in $u_2$
\begin{equation}
u_2 =(1,1,0) - \frac{<(1,0,1), (1,1,0)>}{<(1,0,1), (1,0,1)>}(1,0,1)
\end{equation}
which is
\begin{equation}
u_2 = (1,1,0) - (\frac{1}{2})(1,0,1)
=
\frac{1}{2}(1,1,-1)
=
(\frac{1}{2}, \frac{1}{2},-\frac{1}{2})
\end{equation}
Now,
\begin{equation}
u_3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Area of ellipse using double integral I am trying to find the area of a quadrant of an ellipse by double integrating polar coordinates but the answer I'm getting is incorrect.
ellipse : $ x^2/a^2 + y^2/b^2 =1 $
Any point on ellipse : $ ( a\cos(\theta), b\sin(\theta)) $
At $ \theta$, taking $ d\theta $ segment, Thus $ ... | You can also use green's theorem.
Set x=acos($\theta$), y=bsin($\theta$). Now we can set up the intergal.
$A=\frac{1}{2}\int_{c} xdy-ydx. \\
dy=bcos(\theta), dx=-asin(\theta) \\
A=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}abcos^2(\theta)+absin^2(\theta)\text{ }d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}ab \text{ }d\theta=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\iiint_R\left(1-\frac{x}{\sqrt{(x^2+y^2)}}\right)\,dx\,dy\,dz$ given the region $R$
Evaluate $$\iiint_R\left(1-\frac{x}{\sqrt{(x^2+y^2)}}\right)\,dx\,dy\,dz$$, where $R$ is the region bounded by $z=x^2+y^2$ and $z=1-x^2-y^2$.
Here is what I did:
Let: $x=r\cos\theta$, $y=r\sin\theta$, $z=z$.
First, con... | Your subsitution is fine, cylindrical coordinates. But you messed up with the limits. Obviously $\theta \in (0,2\pi)$ . We have that z goes from the bottom surface to the top one, so you have to first identify them. In this case $z=r^2$ is the one below and $z=1-r^2$ the one above, so $z \in (r^2,1-r^2)$. Then we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the generating function for sequence $1,2,4,0,8,24,120,184,312,56,568,1592,...$ I'm having trouble finding a generating function for the sequence that has a closed form. The sequence can be deduced using two powers, with alternating negative and multiples of three as shown:
first term:
$1=1$
second term:
$2=1+2^{0... | Begin by finding the generating function for the sequence $2^0, 2^1, -2^2, 2^3, 2^4, 3 \cdot 2^5, 2^6, 2^7, -2^8, \dots$. This is the sum of
\begin{align}
\frac{1}{1-2x} &= 2^0 + 2^1x + 2^2x^2 + 2^3x^3 + 2^4x^4 + 2^5x^5 + \dots \\
-2\cdot\frac{(2x)^2}{1-(2x)^6} &= -2 \cdot 2^2x^2 - 2 \cdot 2^8x^8 - 2\cdot 2^{14}x^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Determinant of Symmetric Matrix $\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T$ To give the close-form of $\det(\mathbf{G})$, where $\mathbf{G}$ is
\begin{align}
\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T
\end{align}
in which $a$ and $b$ are constant, and $\boldsymbol{e}$ is a column vector with all elements being $1$. In ... | Note that $G$ is a circulant matrix, so we know the eigenvectors are $(1,\zeta,\zeta^2,\dots,\zeta^{u-1})$ where $\zeta^u=1$. Its eigenvalue is
$$ a+b\sum_{j=0}^{u-1} \zeta^j=
\begin{cases}
a & \text{if }\zeta\neq 1\\
a+bu & \text{if }\zeta=1
\end{cases} $$
and so the determinant is $a^{u-1}(a+bu)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Is the sequence $a_{n} = \frac {1}{2^2} + \frac{2}{3^2} + .... +\frac{n}{(n+1)^2}$ Cauchy? First:
I tried substituting natural numbers for $n$ to calculate the consecutive terms of the sequence and then see the difference between their values and I found that the difference is decreasing for large values of $n$ (not ve... | "I tried substituting natural numbers for n to calculate the consecutive terms of the sequence and ...found that the difference is decreasing"
It doesn't matter if the difference between consecutive terms get smaller (and converge to zero). The terms $a_n$ and $a_{n+1}$ may get minisculely close together but in all th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
A ring such that $(a+b)^2=a^2+b^2$ and $(a+b)^3=a^3+b^3$
Let $(A,+,\cdot)$ be a ring such that there are $a,b \in A$ which satisfy $$(a+b)^2=a^2+b^2, \quad (a+b)^3=a^3+b^3$$
Prove that $(a+b)^n=a^n+b^n,$ for all positive integers $n.$
I have found the following solution, but I am not quite satisfied with it.
From t... | This is the same, but maybe rearranged to see an idea of norming (non-commutative) monomials in $a,b$. As noticed in the OP, from $a^2+b^2=(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2$ we get the "supercommutativity relation"
$$
ab= -ba\ .
$$
The other relation, extended as $a^3+b^3=(a+b)^3=(a+b)^2(a+b)=(a^2+b^2)(a+b)=a^3+a^2b+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2920135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 2,
"answer_id": 0
} |
Coefficient of a generating function I have the following generating function,
$$f(x)=(1+x+x^2+x^3)^4$$
I want to find the $[x^5]$ coefficient, to do so I wrote,
$$[x^5]f(x)=[x^5](1+x+x^2+x^3)^4=[x^5]\left(\frac{1-x^4}{1-x}\right)^4=
[x^5](1-x^4)^4(1-x)^{-4}
$$
I am wondering how to proceed from here?
| Write the two binomials in terms of series:
$$[x^5](1-x^4)^4(1-x)^{-4}=[x^5]\sum_{i=0}^{4} {4\choose i}(-x^4)^i \cdot \sum_{j=0}^{\infty}{-4\choose j}(-x)^j=\\
[x^5]{4\choose 0}(-x^4)^\color{green}{0}\cdot \color{red}{{-4\choose 5}}(-x)^\color{green}5+{4\choose 1}(-x^\color{green}4)^\color{green}1\cdot \color{blue}{{-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Existence proof of infinite solutions of a diophantine equation Prove by construction that $x(x+1)=2 y^2$ has an infinite number of solutions where $x$ and $y$ are both positive whole numbers. Do not use an iterative procedure like that for solutions of Pell's equation in your proof.
Background: I came up with a proof ... | If you have one solution to
$x^2-dy^2 = 1$,
by using
$\begin{array}\\
(x^2-dy^2)(u^2-dv^2)
&=x^2u^2-d(x^2v^2+y^2u^2)+d^2y^2v^2\\
&=x^2u^2+d^2y^2v^2\pm 2dxuyv-d(x^2v^2+y^2u^2\pm 2xuyv)\\
&=(xu\pm dyv)^2-d(xv\pm yu)^2\\
\end{array}
$
you can construct
an arbitrary number of them.
When $d=8$,
since
$3^2-8\cdot1^2 = 1$,
f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the fact that $x^2=(x-1)(x+1)+1$ have a name? Just curious about this pattern
$$x^2 = (x-1)(x+1) +1$$
So:
$$\begin{align}
1^2 &= \phantom{1}0\cdot\phantom{1}2+1 = 1 \\
2^2 &= \phantom{1}1\cdot\phantom{1}3+1 = 4 \\
3^2 &= \phantom{1}2\cdot\phantom{1}4+1 = 9 \\
4^2 &= \phantom{1}3\cdot\phantom{1}5+1 = 16... | Trickery;
$x^2= x^2 -1 +1 = (x^2-1) +1=$
$ (x+1)(x-1)+1$.
Used: $x^2-1=(x+1)(x-1)$.
P.S. Example :
$19^2= (20)(18)+1 = 360 +1=$
$ 361$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\ \int_{|z|=2}\frac{z^2+1}{(z-3)(z^2-1)}\ dz$
I am trying to solve $$\ \int_{|z|=2}\frac{z^2+1}{(z-3)(z^2-1)}\ dz,$$
using Cauchy's Integral Formula.
My attempt:
$$I=\int_{|z|=2}\frac{z^2+1}{(z-3)(z^2-1)}\ dz=\int_{|z|=2}\frac{z^2+1}{(z-3)(z-1)(z+1)}\ dz$$
Now, $I$ has singularities $\pm 1, 3$, but only ... | *
*Method 1: Using Partial fractions:
$$ \int_{|z|=2}\frac{z^2+1}{(z-3)(z^2-1)} dz = \int_{|z|=2}\frac{-\frac12}{z-1} + \frac{\frac14}{z+1} + \frac{\frac54}{z-3} dz$$
$$ = \int_{|z|=2}\frac{-\frac12}{z-1}dz + \int_{|z|=2}\frac{\frac14}{z+1}dz + \int_{|z|=2}\frac{\frac54}{z-3} dz$$
$$ = -\frac12 (2\pi i) + \frac14 (2\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving $\sqrt{8-x^2}-\sqrt{25-x^2}\geq x$
I would like to find the solution of
$$\sqrt{8-x^2}-\sqrt{25-x^2}\geq x.$$
My try:
First I used the hint of this answer.
$$ \frac{8-x^2-25+x^2}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x \leftrightarrow \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x.$$
Then the solution can be found by... | Unfortunately the approach I did there is not useful here. However, here is another approach:
Write the inequality as
$$\sqrt{8-x^2}-x\geq\sqrt{25-x^2}. \tag{1}$$
Now use the Cauchy–Bunyakovsky–Schwarz inequality,
$$(x_1y_1+x_2y_2+\cdots+x_ny_n)^2\leq\left(x_1^2+x_2^2+\cdots+x_n^2\right)\left(y_1^2+y_2^2+\cdots+y_n^2\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Positive Definite Matrix (Block Matrix) Let $B$ be an $m\times n$ matrix. Is $$ A=\begin{pmatrix} I & B \\ B^T & I+B^TB \end{pmatrix} $$ positive definite?
Attempt:
Let $\mathbf{z}=\begin{pmatrix} \mathbf{x} \\ \mathbf{y} \end{pmatrix}$. To show that $A$ is positive definite, $\mathbf{z}^TA\mathbf{z}>0$. Expanding $\ma... | One hint could be to try to rewrite with sum of self-outer products:
$$A = \begin{bmatrix}I&B\\B^T&I+B^TB\end{bmatrix}= \cdots\\=\begin{bmatrix}I\\B^T\end{bmatrix} \begin{bmatrix}I&B\end{bmatrix} + \begin{bmatrix}0\\I\end{bmatrix} \begin{bmatrix}0&I\end{bmatrix}$$
What can we say about it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
First order differential equation solution check. I have an equation:
$\frac{dy}{y}=(0.0001x^2+0.005x)dx$
$y=ce^{(0.00003x+0.0025)x^2}$
Now I try to substitute $y$ in the euqation and I am getting
$(\frac{1}{ce^{(0.00003x+0.0025)x^2}})'=0.0001x^2+0.005x$
$(\frac{1}{ce^{(0.00003x+0.0025)x^2}})'=(\frac{1}{c}e^{(-0.00003x... | The equation is of the form:
$$ \frac{\mathrm{d}y}{y} = (ax^2+bx)\mathrm{d}x $$
where $a=0.0001$ and $b=0.005$. Integrating both sides gives:
$$ \int\frac{\mathrm{d}y}{y} = \int(ax^2+bx)\mathrm{d}x $$
The solution is
$$ \ln y = \frac{a}{3}x^3+\frac{b}{2}x^2+c$$
Therefore, in terms of $y$:
$$ y = C\mathrm{e}^{(\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Does the elliptic curve $y^2 = 4 x^3 -6075$ have any integer points? Let $E$ be the elliptic curve $y^2 = 4 x^3 -6075$. I ran the following Mathematica code, which searches naively for integer solutions to $E$ but it did not find any solutions $(x, y) \in E(\mathbb{Z})$ satisfying $0 \leq x \leq 10^6$.
T = Table[ z =... | No, $y^2=4x^3-6075$ has no integer solution.
An elementary observation: $3\nmid x$: because $$3\mid x \implies 27\mid y^2 \implies 9\mid y \implies
9\mid x$$ but $y^2+6075$ has no solution in $\mathbb{Z}/3^6\mathbb{Z}$.
The ring of integer of $\mathbb{Q}(\sqrt{-3})$ is a UFD. The equation can be written as
$$\left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Find the relationship of the length of triangle's sides.
Denote the three sides of $\triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22 $$
Now determine what kind of triangle $\triangle ABC$ is.
A.Isosceles triangle which its leg and base is not equal.
B.equilateral triangle
... | We have
$$|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22-a^2-b\tag1$$
from which
$$10a+2\sqrt{b-4}-22-a^2-b\ge 0\tag2$$
follows.
$(2)$ is equivalent to
$$a^2-10a+b-2\sqrt{b-4}+22\le 0,$$
i.e.
$$(a-5)^2-25+(b-4)+4-2\sqrt{b-4}+22\le 0,$$
i.e.
$$(a-5)^2+(\sqrt{b-4}-1)^2\le 0$$
from which
$$a-5=\sqrt{b-4}-1=0$$
i.e.
$$a=b=5$$
follows.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Divergence of infinite series $(\frac{3k-2}{4k+2})^{2k-3}$ I would like to ask if my solution for testing the divergence of the infinite series below is correct.
$$\sum_{k=1}^{\infty} \left(\frac{3k-2}{4k+2}\right)^{2k-3}$$
I used the Cauchy ratio test.
$$
\lim_{k \to \infty} \left(\frac{3k-2}{4k+2}\right)^{2k-3} =
\l... | For this problem, the ratio test requires careful calculations with which you made some mistakes. The easiest approach is to note the $k$th term is positive but less than $(3/4)^{2k-3}$, so the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determine $(n-1)$ eigenvectors of $f$ associated to $0$?
$$A=\begin{pmatrix}1&1&\cdots&1\\1&1&\cdots&1\\\vdots&\vdots&\ddots&1\\1&1&\cdots&1\end{pmatrix}\in\mathcal M_n(\mathbb R).$$
*
*Prove that $0$ and $n$ are eigenvalue of $A$.
*Determine the characteristic polynomial. (Hint: You can use the basis $(e_1, e_2 - ... | Let $f_1 = e_1$, $f_2 = e_2 - e_1$, ..., $f_n = e_n - e_1$ as in the hint. Note that $\sum f_i = (\sum e_i) - (n-1) e_1$.
Then $A$ annihilates $f_2$, $f_3$, ..., and $f_n$. Meanwhile,
\begin{align*}
Af_1 &= \sum_i {e_i} \\
&= \sum_i f_i + (n-1) e_1 \\
&= nf_1 + \sum_{i \neq 1} f_i,
\end{align*}
so the matrix for $A$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find the slant asymptote for $y=\frac{x^2\arctan\left(x\right)}{3x+3}$ I got stuck at the end and wonder if you can give some guide on how to proceed to find the slant asymptote?
$$y=\frac{x^2\arctan\left(x\right)}{3x+3}$$
$$y= kx+m\\
\frac{f(x)}{x} \rightarrow k\\
f(x) - kx \rightarrow m $$
Solution:
$$\lim ... | hint
Put $y=\frac 1x$ and use the fact that
for $y>0$, we have
$$\arctan(\frac 1y)=\frac {\pi}{2}-\arctan(y)$$
use L'Hopital rule to compute the limit when $y\to 0^+$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is $15^{15} + 16^{16} + 17^{17} + 18^{18} + 19^{19} + 20^{20} \pmod{7}$? I am trying to evaluate $15^{15} + 16^{16} + 17^{17} + 18^{18} + 19^{19} + 20^{20} \pmod{7}$. I have found that $15^{15} \equiv 1 \pmod{7}$ and that $16^{16} \equiv 2 \pmod{7}$.
To evaluate $15^{15} \pmod{17}$, I did the following:
$$15 = 2 \... | You can go with the same process I think:
$$17^{17} \equiv 3^{17} \mod{7}$$
and we have
$$3^1 \equiv 3 \mod{7}$$
$$3^2 \equiv 2 \mod{7}$$
$$3^3 \equiv 6 \mod{7}$$
$$3^4 \equiv 4 \mod{7}$$
$$3^5 \equiv 5 \mod{7}$$
$$3^6 \equiv 1 \mod{7} \text{ (We could have this by using Fermat's Little Theorem as well)}$$
From here, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
placing balls into compartments
Suppose we have $n$ balls that are randomly distributed into $N$
compartments. Find the probability that $m$ balls will fall into the
first two compartments. Assume all $N^n$ arrangements are equally
likely.
Try.
First, we select $m$ balls out of $n$, this we can do in ${n \choose m}$... | There are $\binom{n}{m}$ ways to select which $m$ of the $n$ balls will be in the first two compartment and $2^m$ ways to distribute them to those two compartments. The remaining $n - m$ balls can be distributed to the remaining $N - 2$ compartments in $(N - 2)^{n - m}$ ways. Hence, the number of favorable cases is
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify the Dot Product in terms of $a$ and $b$ Where $a$ and $b$ are arbitrary vectors
$(a+2b) \cdot (2a-b)$
$$a\cdot(2a-b)+2b\cdot(2a-b) = 2(a\cdot a)-a\cdot b+1(b\cdot a)-2(b\cdot b)$$
$$=2(a)-ab+4ab-2(b)^2$$
$$=2a^2-2b^2$$
$$=2(a^2-b^2)$$
Where did i go wrong in simplifying this?
| I marked the mistakes in red and corrected these
\begin{align}
(a+2b) * (2a-b)& =a*(2a-b)+2b*(2a-b)\\
& = 2(a*a)-a*b+\color{red}{4}(b*a)-2(b*b)\\
& = 2(a)\color{red}{^2}\underbrace{-ab+4ab}_{\color{red}{3ab}}-2(b)^2\\
& = 2a^2-2b^2+\color{red}{3ab}\\
& = 2(a^2-b^2)+\color{red}{3ab}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proving $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers. I would like to verify whether my proof is correct. The answer sheet used a much more intuitive and logical approach but I think mine is correct also.
To prove: $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers
Proof: We proceed by induction and s... | It is correct, but:
*
*When you wrote that “We rewrite $13\cdot 6^{m}+8\cdot 13^{m}$ as $6(13\cdot 6^m)+13(8\cdot 13^m)$”, what you meant was “We rewrite $13\cdot 6^{m+1}+8\cdot 13^{m+1}$ as $6(13\cdot 6^m)+13(8\cdot 13^m)$”.
*You don't need to compute $6\times8$ and $13\times8$. Note that\begin{align}6(7x-8\times1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Maclaurin series of $\frac{1}{1-x}\ln\frac{1}{1-x}$ Find the Maclaurin series expansion (power law series) of the function:
$$f(x) = \frac{1}{1-x}\ln\frac{1}{1-x}.$$
I could find the expansion terms, up to the 5-th term, and is:
$$F(x) = x + \frac{3}{2!}x^2 + \frac{11}{3!}x^3 + \frac{50}{4!}x^4 + \frac{274}{5!}x^5 + \d... | $$ 1/(1-x)ln(1/(1-x)) = $$
$$ (1+x+x^2+x^3+x^4+...)(x+x^2/2+x^3/3+x^4/4+...) = $$
$$ x+(1+1/2)x^2+(1+1/2+1/3)x^3+(1+1/2+1/3+1/4)x^4+...= $$
$$ x + 3x^2/2 + 11x^3/6+25x^4/12+...=$$
$$ \sum_{a=1..\infty } (\sum_{b=1..a} 1/b) x^a =$$
$$ \sum_{a=1..\infty} H_a x^a$$
Numerators of harmonic numbers $H_n$ : https://oeis.org/A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_0^1\ln(1+x^2)\ln(x^2+x^3)\frac{dx}{1+x^2}$
How to evaluate $$I=\int_0^1\ln(1+x^2)\ln(x^2+x^3)\frac{dx}{1+x^2}?$$
It equals $\frac5{64}\pi^3-\frac92G\ln2+\frac14\pi\ln^22$ according to Mathematica, where $G$ denotes Catalan's constant.
Attempt
$$I=\frac d{ds}\int_0^1\ln(x^2+x^3)\frac{dx}{(1+x^2)^{1-s}... | $$I=\int_0^1\frac{\ln(1+x^2)(2\ln(x)+\ln(1+x))}{1+x^2}dx$$
$$=2\int_0^1\frac{\ln(x)\ln(1+x^2)}{1+x^2}dx+\int_0^1\frac{\ln(1+x)\ln(1+x^2)}{1+x^2}dx$$
$$=2I_1+I_2$$
$I_1$ is calculated here:
$$\boxed{I_1=-2\,\Im\operatorname{Li_3}(1+i)+\frac{3\pi^3}{32}+\frac{\pi}8\ln^2(2)-\ln(2)G}$$
For $I_2$, let $x\to (1-x)/(1+x)$
$$I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 1,
"answer_id": 0
} |
Is my particular solution of $ {(3xy + y^2)}\ dx + {(x^2 + xy)}\ dy = 0 $ for $x=1$ and $y=1$ correct?
Find the particular solution to
$$
{(3xy + y^2)}\ dx + {(x^2 + xy)}\ dy = 0
$$
for $x=1$ and $y=1$.
My Solution
$$
\frac{dy}{dx}=-\frac{3xy+y^2}{x^2+xy}
$$
Substituting $y=vx$ , $\frac{dy}{dx}=v+x\frac{dv}{dx}$
... | Neither you nor the book is correct. Both are partially correct, but not correct entirely. The false step is the belief that
$$\int\,\frac{1}{t}\,\text{d}t=\ln|t|+\text{constant}\,.$$
Since $t=0$ is not in the domain of the function $t\mapsto\dfrac1t$, what is correct is saying that
$$\int\,\frac1t\,\text{d}t=\ln|t|+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How evaluate $ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$ How prove
$$ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}=\frac{\pi^2\ln2}{6}-\frac{\ln^32}{3}-\frac{3\zeta(3)}{4} $$
$\mathbf {My\,Attempt:}$
... | The second round of integration is completed with use of
\begin{align}
I_{1} &= \int_{0}^{1} \mathrm{Li}_2 \left(\frac{1-x}{1+x}\right) \, \frac{dx}{1+x} = \zeta(2) \, \ln 2 - \frac{5}{8} \, \zeta(3) \\
I_{2} &= \int_{0}^{1} \mathrm{Li}_2 \left(\frac{1}{1+x}\right) \, \frac{dx}{1+x} = \frac{\zeta(3)}{8} + \frac{\zeta(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Euclidean GCD calculation and mod
Calculate $6/87 \pmod{137}$
I do not understand the Euclidean GCD algorithm. If someone can please explain the overall logic of this it would be much appreciated.I have posted what the solution is supposed to be below:
| $$ \frac{ 137 }{ 87 } = 1 + \frac{ 50 }{ 87 } $$
$$ \frac{ 87 }{ 50 } = 1 + \frac{ 37 }{ 50 } $$
$$ \frac{ 50 }{ 37 } = 1 + \frac{ 13 }{ 37 } $$
$$ \frac{ 37 }{ 13 } = 2 + \frac{ 11 }{ 13 } $$
$$ \frac{ 13 }{ 11 } = 1 + \frac{ 2 }{ 11 } $$
$$ \frac{ 11 }{ 2 } = 5 + \frac{ 1 }{ 2 } $$
$$ \frac{ 2 }{ 1 } = 2 + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x>0, y>0,x+y=\frac{\pi}{3}$ then maximum value of $\tan x\tan y$
If $x>0, y>0$ and $x+y=\frac{\pi}{3}$, then find the maximum value of $\tan x\tan y$
My Attempt
$x>0, y>0, x+y=\frac{\pi}{3}\implies x, y$ in $1^\text{st}$ quadrant.
$\tan x, \tan y>0, \tan(x+y)=\sqrt{3}, \tan x\tan y>0$
$$
\tan x+\tan y\geq2\sqrt{\... | Hint: One way is finding the extrimum of $f(x,y)=\tan x\tan y$ with condition $x+y=\dfrac{\pi}{3}$, then with
$$F(x,y)=\tan x\tan y-\lambda\left(x+y-\dfrac{\pi}{3}\right)$$
we have
$$\tan y=\lambda\cos^2x~~~,~~~\tan x=\lambda\cos^2y$$
then
$$\tan y\lambda\cos^2y=\tan x\lambda\cos^2x$$
shows $\sin x\cos x=\sin y\cos y$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How did I obtain incorrect results for $P_{B\leftarrow C}$ and $P_{C\leftarrow B}$ geometrically? I'm given two bases in $\mathbb{R}^2$:
$B = \{\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} -1 \\ 2 \end{pmatrix}\}$ and $C = \{\begin{pmatrix} -4 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 5 \end{pmatrix}\}$ and I'm... | First a bit of review. The coordinates of a vector are the coefficients of its unique expression as a linear combination of basis vectors. That is, if we have the ordered basis $\mathcal B = \left(\mathbf b_1,\mathbf b_2\right)$, then $[\mathbf v]_{\mathcal B} = (v_1,v_2)^T$ means that $\mathbf v = v_1\mathbf b_1+v_2\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$ Prove that $$\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$$
Hence, show that $$(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5})^5+i(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5})^5=0.$$
In the first ... | I write
$i\cos \theta + \sin \theta = i(\cos \theta -i\sin \theta) = i(\cos (-\theta) + i\sin (-\theta)) = ie^{-i\theta}; \tag 1$
then
$\sin \theta -i\cos \theta = -i(\cos \theta + i\sin \theta) = -ie^{i \theta}; \tag 2$
we have
$(1 + \sin \theta - i\cos \theta)(\sin \theta + i\cos \theta) = (1 - ie^{i \theta})ie^{-i\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Simplifying $\sin^{-1}(\cos x)$ I have a question on $$\sin^{-1}(\cos x)$$
Since $\cos(x)=\sin(\frac \pi 2 \pm x)$, the above expression can simplify to either $\frac \pi 2 + x$ or $\frac \pi 2 - x$. This seems like a contradiction.
What's the problem here?
| Let $x=\pm\frac{\pi}{2}\mp(y+2\pi n),y\in [-\frac{\pi}{2},\frac{\pi}{2}]$.
Then:
$$\begin{align}\arcsin(\cos x)&=\arcsin(\cos [\pm\frac{\pi}{2}\mp(y+2\pi n)])=\\
&=\arcsin(\sin (y+2\pi n))=\\
&=\arcsin(\sin y)=\\
&=y.\end{align}$$
Example 1: If $x=300^\circ$, then $300^\circ=-90^\circ+(30^\circ+360^\circ)$ and:
$$\arcs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
find $\cos\theta$ if $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$
If $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$ then find $\cos\theta$
The solution is given in my reference as $\frac{1}{6}$.
$$
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}
$$
$$
\cos... | The problem is actually incorrect. Here is why:
Consider a right triangle and angle $\theta$ in that triangle such that $\sin\theta=\dfrac{3}{4}.$
That means that the right triangle has one leg $3$ and hypotenuse $4$. This implies that the other leg is $\sqrt{4^2-3^2}=\sqrt{7}.$
This implies that $\cos \theta=\dfrac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sum of series $\sum_{n=1}^{\infty}\sin(\frac{1}{2^n})\cos(\frac{3}{2^n})$ How can we get sum of this series? $$\sum_{n=1}^{\infty}\sin(\frac{1}{2^n})\cos(\frac{3}{2^n})$$
I think we must apply this theorem.
if in $\sum_{n=1}^\infty a_{n}$, $a_n = b_{n+1}-b_n$ and $\lim_{n\to\infty} b_n = b$; $ \implies $ $\sum_{n=1}... | The identity $\sin x\cos 3x=\frac{1}{2}(\sin 4x-\sin 2x)$ gives$$\sum_{n\ge 1}\sin\frac{1}{2^n}\cos\frac{4}{2^n}=\frac{1}{2}\sum_{n\ge 1}(\sin\frac{4}{2^n}-\sin\frac{2}{2^n})=\frac{1}{2}\sin 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Showing that two random variables are independent I have the following problem:
Given two independent standard normal random variables, call them $X$ and $Y$, how can I show that $Z = X^2 + Y^2$ and $W=\frac{X}{Y}$ are also independent?
I know that because $X$ and $Y$ are standard normal I can write their distribution... | The joint distribution for $(X^2,Y^2)$ is
$$\frac{x^{-1/2}e^{-x/2}}{\sqrt{2}\Gamma(1/2)}1_{(0,+\infty)}(x)\times \frac{y^{-1/2}e^{-y/2}}{\sqrt{2}\Gamma(1/2)}1_{(0,+\infty)}(x)=\frac{(xy)^{-1/2}e^{-(x+y)/2}}{2(\Gamma(1/2))^2}1_{(0,+\infty)}(x)1_{(0,+\infty)}(y)$$
Since $X^2=Z-\frac{Z}{W^2+1}=\frac{ZW^2}{W^2+1},Y^2=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that X$^3$+X+1|X$^7$+1 This is a part of the question. Which states Let K= $\mathbb{Z}$/2$\mathbb{Z}$[X]/(d)$\mathbb{Z}$/2$\mathbb{Z}$[X], where
d=X$^3$+X+1 and let a be the class of X modulo d. This is the only part that i do not know how to solve. I have tried calculating the quotient and the remainder but i do... | Here are the steps for polynomial long division $\mod 2$, where the notation
$\dfrac{p(X)}{q(X)} \longrightarrow X^k \tag 0$
indicates the quotient of the leading terms is $X^k$:
$\dfrac{X^7 + 1}{X^3 + X + 1} \longrightarrow X^4; \tag 1$
$X^7 + 1 - X^4(X^3 + X + 1) = X^7 + 1 - X^7 - X^5 - X^4 = X^5 + X^4 + 1; \tag 2$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How do I evaluate the integral $\int_0^1\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$. stuck on this integral
$$\int_0^1\dfrac{(x^2+x+1)}{(x^4+x^3+x^2+x+1)}\ dx$$
I was attempting to evaluate the infinity sum S = $ 1- \frac{1}{4} + \frac {1}{6} - \frac {1}{9} + \frac {1}{11} -\frac {1}{14}+ ........ $
what I then did was def... | The hint:
Use $$x^4+x^3+x^2+x+1=\left(x^2+\frac{1}{2}x+1\right)^2-\left(\frac{\sqrt5}{2}x\right)^2=$$
$$=\left(x^2+\frac{1-\sqrt5}{2}x+1\right)\left(x^2+\frac{1+\sqrt5}{2}x+1\right).$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sum_{1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} \lt 2 $
Prove that $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} \lt 2$$
I have found that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} < \pi / 2 $ with integrating from $1$ to infinity but integral isn’t allowed.
| Let us go with creative telescoping. As $n\to +\infty$,
$$ \frac{1}{n+n\sqrt{n}}=\frac{1}{n\sqrt{n}}-\frac{1}{n^2}+\frac{1}{n^2\sqrt{n}}+O\left(\frac{1}{n^3}\right) $$
while
$$ \frac{2}{\sqrt{n-\frac{7}{6}}}-\frac{2}{\sqrt{n-\frac{1}{6}}}=\frac{1}{n\sqrt{n}}+\frac{1}{n^2\sqrt{n}}+\frac{95}{96 n^2\sqrt{n}}+o\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding minimum value of $ \frac{x^2 +y^2}{y} $
Finding the minimum value of $\displaystyle \frac{x^2 +y^2}{y}.$ where $x,y$ are
real
numbers satisfying $7x^2 + 3xy + 3y^2 = 1$
Try: Equation $7x^2+3xy+3y^2=1$ represent Ellipse
with center is at origin.
So substitute $x=r\cos \alpha $ and $y=r\sin \alpha$
in $7... | Lagrange function is
$$L=k\, \left( 3 {{y}^{2}}+3 x y+7 {{x}^{2}}-1\right) +\frac{{{y}^{2}}+{{x}^{2}}}{y}$$
Solve system:
$$L'_x=0,\quad L'_y=0,\quad L'_k=0.$$
We get two solutions
$$k=1/4,y=-2/5,x=-1/5$$ $$k=-1/4,y=2/5,x=1/5$$
Answer:
$$f_{min}=f\left(-\frac{1}{5},-\frac{2}{5}\right)=-\frac{1}{2}$$
with CAS Maxima:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How do I use the definition of partial derivative to get $f_x(x,y) = \frac{y(y^2-x^2)}{(x^2 + y^2)^2}$? $\\f(x,y) =
\begin{cases}
\frac{xy}{x^2 + y^2}, & \text{if $(x,y) \ne (0,0)$} \\
0, & \text{if $(x,y) = (0,0)$}
\end{cases}$
Using the definition of partial derivatives:
\begin{align*}
f_x(x,y) &= \lim_{h\to 0}... | Continuing your calculation, we get
$$f_x(x,y)=\lim_{h\rightarrow 0}\frac{hy(y^2-x^2-hx)}{h(x^2+2hx+h^2+y^2)(x^2+y^2)}.$$
Or
$$f_x(x,y)=\lim_{h\rightarrow 0}\frac{y(y^2-x^2-hx)}{(x^2+2hx+h^2+y^2)(x^2+y^2)}.$$
Can you finish from here?
| {
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"url": "https://math.stackexchange.com/questions/2966379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Finding the $2n$ th derivative of $\frac{1}{1+x^2y^2}$ with repect to $x$ We know that with repect to $x$,
\begin{align}
\left( \frac{1}{1-x^2 y^2} \right)^{(2n)} =
\frac{(2n)!y^{2n}}{2 } \left( \frac{1}{(1+xy)^{2n+1}} +\frac{1}{(1-xy)^{2n+1}} \right)
\end{align}
What about
\begin{align}
\left( \frac{1}{1+x^2 y^2} \r... | The first thing to do is make this substitution $y = ik$, we can do this freely because the derivative is with respect to x not to y, and it is only a contsant.Then after this substitution we get
$$\left( \frac{1}{1+x^2 y^2} \right)^{(2n)}=\left( \frac{1}{1-x^2 k^2} \right)^{(2n)}
$$
Which, as you said, is equal to
$$\... | {
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"answer_count": 1,
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Help with induction proof I need help with the following induction proof which I am not sure if I am doing correctly.
$$\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$$
I check for $n=1$ (Base case)
$$\frac{1}{1\cdot3}=\frac13$$
$$\frac{1}{2\cdot1+1}=\frac13$$
Now, is this the correct nex... | $$
\\\frac{1}{1\cdot3}+\frac{1}{1\cdot5}+...+\frac{1}{(2n-1)(2n+1)}=
\\\frac{1}{2}\cdot((\frac{1}{1}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+\cdots+(\frac{1}{2n-1}-\frac{1}{2n+1}))=
\\\frac{1}{2}\cdot(1-\frac{1}{2n+1})=\frac{n}{2n+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Show that $\int_0^{2 \pi} \dfrac{d \theta}{5+4 \sin\theta}=\dfrac{2 \pi}{3}.$ By change of variable and letting $C_0$ be the unit circle about the origin, we have:
\begin{align*}
\int_0^{2 \pi} \dfrac{d \theta}{5+4 \sin\theta} &= \int_{C_0} \dfrac{dz/zi}{5+4(z-z^{-1})/2i}\\
& = \int_{C_0} \dfrac{dz}{2z^2+5iz-2}\\
& = ... | $\oint_{|z|=1} \frac {1}{i(2z^2 + 5iz -2} \ dz\\
\oint_{|z|=1} \frac {1}{2(z +2i)(z + \frac 12i)} \ dz$
You dropped a factor of $\frac 12$
$2\pi i\frac {1}{2(-\frac 12 i + 2i)} = \frac {2\pi}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sequence Inequality question from RMO 2018 Define a sequence {$a_n$} of real numbers by $a_1 = 2$ and $a_{n+1} = \dfrac{a_n^2+1}{2}$, for $n\ge 1$,
Prove that for every natural number $N, \sum_{j=1}^{N} \frac{1}{1+a_j} \lt 1$
I tried mathematical induction after coming to the step where
$Sum_n = \frac{1}{2}\left(\fra... | To get $\dfrac{1}{a_n+1}$, Firse subtract $1$ from both sides
$$a_{n+1}-1=\frac{a_{n}^2-1}{2}=\frac{(a_n+1)(a_n-1)}{2}.$$
Then take the multiplicative inverse ( assume $a_n\neq \pm1$, as we'll see later)
$$\frac{1}{a_{n+1}-1}=\frac{2}{(a_n+1)(a_n-1)}$$
where the right side can be written as
$$\frac{1}{a_n-1}-\frac{1}{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that $10|n+3n^3+7n^7+9n^9$ Prove that $10|n+3n^3+7n^7+9n^9$ for every $n\in \mathbb N$
Only what i see that 10=5*2 and both number is free numbers, and if I show that $5|n+3n^3+7n^7+9n^9$ and $2|n+3n^3+7n37+9n^9$ that I prove, since 5 and 2 is free numbers i can use Fermat's little theorem such that $5|n^5-n$ an... | $\begin{align} \bmod 10\!:\ f_n & = \ n+9n^9 +\ 3n^3\,+ 7n^7\\
&\equiv n(1\!-\!n^8) + 3n^3(1\!-\!n^4)\ \ \ {\rm by}\ \ \ 9\equiv -1,\,\ 7\equiv -3\\
&\equiv \color{#c00}{n(1\!-\!n^4)}\,g_n
\end{align}$
Notice that $\,2,5\mid \color{#c00}{n\,-\,n^5}\ $ by Fermat.
Or, equivalently, $\,\color{#c00}{n^5\equiv n}\,\Rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
What does the minimum x that satify $x=24n+12=15m+6=11k+2$? My attempt so far was: let $x=24n+12=15m+6=11k+2$
find $x$ as the form $x=24*15*a+24*11*b+15*11*c$
$$2\equiv 24\cdot 15\cdot a\quad (mod\quad 11)\quad \Rightarrow 1\equiv 4\cdot a\Rightarrow \quad a=3\\ 6\equiv 24\cdot 11\cdot b\quad (mod\quad 15)\quad \Right... | $\!x\equiv 12\pmod{\!24}\!\iff x/3\,\equiv\ \ \ 4\,\pmod{\! 8}$
$\!\!\left.\begin{align}
&x\equiv \ 6\!\!\pmod{\!15}\!\iff x/3\,\equiv\ \ \ 2\!\!\!\pmod{ 5}\\
&x\equiv \ 2\!\!\pmod{\!11}\!\iff x/3\,\equiv -3\!\!\!\pmod{\!\!11}\\
\end{align}\right\}\!\!\!\iff\!\dfrac{x}3\equiv -3\pmod{\!55}\!\iff\! \dfrac{x}3 = \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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ّFind $x$ such that $ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$.
ّFind $x$ such that
$$ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}\,.$$
My attempt:
After clearing the denominators, I obtain this quartic equation
$$104 x^{4} -624 x^{3} +886 x^{2} +150x-225=0.$$
I don't know how to proceed from here.
| Let $u=x$ and $v=3-x$. Then we have
$$
u+v = 3,
\qquad
\frac{1}{u^2} + \frac{1}{v^2} = \frac{104}{25}
$$
But
$$
\frac{1}{u^2} + \frac{1}{v^2}
=
\frac{u^2+v^2}{(uv)^2}
=
\frac{(u+v)^2-2uv}{(uv)^2}
=
\frac{9-2uv}{(uv)^2}
$$
Thus, $uv$ is a root of
$$
\frac{9-2z}{z^2} = \frac{104}{25}
$$
a quadratic equation. Once you kno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$. I have no idea how to do this question.
I'm given $\int^{\infty}_{-\infty}\frac{1}{x^2+2ax+b^2}dx=\frac{\pi}{\sqrt{b^2-a^2}}$ if $b>|a|$ and I'm asked to prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$.
What I've tr... | Differentiate with respect to $b$ gives
$$\dfrac{d}{db}\int^{\infty}_{-\infty}\frac{1}{x^2+2ax+b^2}dx=\dfrac{d}{db}\frac{\pi}{\sqrt{b^2-a^2}}$$
$$\int^{\infty}_{-\infty}\frac{-2b}{(x^2+2ax+b^2)^2}dx=\frac{-2b\pi}{2\sqrt{(b^2-a^2)^3}}$$
or
$$\int^{\infty}_{-\infty}\frac{1}{(x^2+2ax+b^2)^2}dx=\frac{\pi}{2\sqrt{(b^2-a^2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Show $x_{n+1} = {1\over 2}x_n^2 - 1$ is bounded below and unbounded above and $x_n$ is increasing.
Let:
$$
\begin{cases}
x_{n+1} = {1\over 2}x_n^2 - 1\\
x_1 = 3\\
n\in \mathbb N
\end{cases}
$$
Show that the sequence $x_n$ is bounded only below and is increasing.
I've started with the following:
$$
x_1 = 3 \\
x_2... | $x_{n+1} - x_n = \dfrac{x_n^2}{2}-1 - x_n= \dfrac{x_n^2 - 2x_n - 1}{2}=\dfrac{(x_n-1)^2-2}{2}$. Thus we need to show: $x_n \ge 3, \forall n \ge 1$. We show this by induction on $n\ge 1$. We have $x_1 = 3$, assume $x_n \ge 3$, we have $x_{n+1} = \dfrac{x_n^2}{2} - 1 \ge \dfrac{3^2}{2}-1= 4.5-1 = 3.5 > 3$. Thus $x_{n+1} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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How to eliminate vector from equation This is probably a super simple question.
Let's say I have an equation
$$
\mathbf{v}^T(\mathbf{a} + \mathbf{b}) = \mathbf{v}^T(\mathbf{c}).
$$
Does this imply that
$$
(\mathbf{a} + \mathbf{b}) = \mathbf{c}
$$?
Intuitively, I would say that it does. On the other hand, there is no le... | Consider $\mathbf{v} = \begin{bmatrix}1\\0\\0\end{bmatrix}$, $\mathbf{x} = \begin{bmatrix}0\\1\\0\end{bmatrix}$, $\mathbf{y} = \begin{bmatrix}0\\0\\1\end{bmatrix}$.
Then $\mathbf{v}^T\mathbf{x} = \mathbf{v}^T\mathbf{y} = 0$ but $\mathbf{x} \neq \mathbf{y}$.
In general, $\mathbf{u}^T\mathbf{v}$ is the dot product of $\m... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$? If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?
I tried with Tchebyshev inequality on sets $\{a, b, c\}$ and $\{a^2, b^2 , c^2\... | Okay, the question requires knowledge of the AM-GM inequality, which you do have.
You need to apply it twice; first on $a^2,b^2,c^2$ and obtain an inequality on the product $abc$; then on $a^3,b^3,c^3$ and substitute $abc$.
The answer is $81$.
Hope this helps.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solution of complex differential equation involving $2$ variables
Solve the differential equation
$\displaystyle \frac{dy}{dx} = \sqrt{\frac{1}{2}+\int^{-\sin^4
\theta}_{\cos^4 \theta}\frac{\sqrt{f(\phi)}d\phi}{\sqrt{f(\theta)}+\sqrt{f(\cos 2 \theta-\phi)}}},$
where $\theta = x+y$ and $\displaystyle x+y\in \bigg(\f... | I'll assume that you meant to have $f(\phi)$ and not $f(\theta)$ in the denominator of the integral
You made a sign error
$$ I = \frac12 \left(-\sin^4 \theta - \cos^4 \theta \right) = -\frac12 +\sin^2\theta\cos^2\theta $$
Then
$$ \frac{dy}{dx} = \sqrt{\frac12 + I} = \big\vert\sin\theta\cos\theta\vert $$
You also have
$... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Infinite sequence $2^{n}-3 (n=2,3,...)$ contains no term divisible by 65 Show that the infinite sequence $2^{n}-3 (n=2,3,...)$ contains infinitely many terms which are divisible by $5$ and infinitely many terms which are divisible by $13$, but no terms which are divisible by $65$
My attempt at this:-
By Fermat's theore... | Hint:
Write $12k+4$ as $12n+4$ which needs to be equal $4k+3$ which is odd unlike the former and $k,n$ are arbitrary integers
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find $k$ in Maclaurin series expansion of $\frac{dy}{dx}=-\frac{1}{2}+\frac{1}{4}x+kx^2+...$ where $y=\ln\Bigl(\frac{e^{-x}+1}{2}\Bigl)$
Given that $y=\ln\Bigl(\frac{e^{-x}+1}{2}\Bigl)$, show $\frac{dy}{dx}=\frac{1}{2}e^{-y}-1$. Show that the series expansion of $\frac{dy}{dx}$ in ascending powers of $x$, up to and in... | Yes, you are correct $k=0$. This is an alternative solution where we use the expansions of $e^t$ and $(1+t)^{-1}$ at $t=0$:
$$\begin{align}
\frac{dy}{dx}&=\frac{2}{e^{-x}+1}\cdot \frac{-e^{-x}}{2} =-\frac{1}{1+e^x}\\
&=-\frac{1}{1+1+x+\frac{x^2}{2}+o(x^2)}\\
&=-\frac{1}{2}\left(1+\frac{x}{2}+\frac{x^2}{4}+o(x^2)\right... | {
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"url": "https://math.stackexchange.com/questions/3007897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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General method to find sum of binomial Specifically I want to ask the method to solve
$$\sum_{k=0}^n \binom{4n+b}{4k};\ b=[0,1,2,3]$$
And how to solve series of type
$$\sum_{k=0}^n \binom{an+b}{ak};\ a=[1,2,3,...],\ b=[0,1,2,...,a-1]$$
| I am sure that you will enjoy the result !
If $$S_n=\sum_{k=0}^n \binom{4n+b}{4k}$$ then a CAS gives
$$4S_n=2^{\frac{4 n+b}{2} } \left((-1)^b\, 2^{\frac{4 n+b}{2} } \cos (\pi b)+\cos
\left(\frac{ (4 n+b)\pi}{4} \right)+(-1)^b \cos \left(\frac{3(4
n+b)\pi}{4} \right)\right)-$$ $$4 \binom{4 n+b}{4 n+4} \,
_5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008839",
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"source": "stackexchange",
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Proving $(n+1)^2+(n+2)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$ by induction My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$
My workings
*
*LHS=$2^2$ =$4$ RHS= $\frac{24}{6} =4 $
*$(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= \frac{k(2k+1)(7k+1)}{6}$
*LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^... | Assuming the step for $n=k$ $\:$For
$n=k+1$ $(k+2)^2+\cdots+(2k+2)^2=\dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=\dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=\dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=\dfrac{(k+1)(2k+3)(7k+8)}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$ Hey so I'm having a bit of a hard time understanding this one.
$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$
1) $x+\sqrt{x^2-3x}$ * $(\frac{x-\sqrt{x^2-3x}}{x-\sqrt{x^2-3x}})$
2) $\frac{x^2-(x^2-3x)}{x-\sqrt{x^2-3x}}$
3) $\frac{3x}{x-\sqrt{x^2(1-\frac{3}{x}})}$
4) $\frac{3x}{x-\sqrt{x^2}*... | We are near $-\infty$, so $x<0$ and $|x|=-x$.
thus
$$x+\sqrt{x^2-3x}=x+\sqrt{x^2(1-\frac 3x)}$$
$$=x+|x|\sqrt{1-\frac 3x}$$
$$=x(1-\sqrt{1-\frac 3x})$$
$$=x\frac{\frac 3x}{1+\sqrt{1-\frac 3x}}$$
the limit is $\frac 32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Use $\cos 5\theta$ to find the roots of $x(16x^4 - 20x^2 + 5) = 0$ I used $\cos(3\theta + 2\theta)$ to prove the first part, but I don't know how to the $2$nd part.
Show that $\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta,$ and hence show that
$$\text{the roots of }x(16x^4 - 20x^2 + 5) \text{ are: } 0,\cos\fra... |
Here the value of cosine will repeat after point.
Final values wil be only 5
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How to get from $x^{p-1}-1$ to $(x-1)(x^{p-2}+x^{p-3}+\cdots+x+1)$? How would I get from $x^{p-1}-1$ to $(x-1)(x^{p-2}+x^{p-3}+\cdots+x+1)$?
It make sense to me logically. When one multiplies it out, it would condense to $x^{p-1}-1$. But it's just not clicking. What is the arithmetic between these steps?
| I'm not sure if this fully answers your question, however, when I am factoring higher degree polynomials (degree greater than 2) I often like to use the following trick that I will demonstrate below:
On polynomials like the one you have there, it is easy to see that $x=1$ is a zero of the polynomial $x^{p-1}-1$. Then y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Produce formula Let $a_{i,j}$ any positive integers wher $1\leq i,j \leq 3$.
I want to write closed form for following summation:
$$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$
$\textbf{My attempt:}$
\begin{... | $$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$
$$=(a_{1,1}+a_{1,2}+a_{1,3})(a_{2,1}+a_{2,2}+a_{2,3})(a_{3,1}+a_{3,2}+a_{3,3})-(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})(a_{3,1}+a_{3,2})$$
$$=\prod_{i=1}^3\sum_{j=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is the inequality $\sum_{n=1}^{\infty} \frac{1}{n^2} \leq 1 + \int_1^{\infty} \frac{1}{x^2}$ true? $$\sum_{n=1}^{\infty} \frac{1}{n^2} \leq 1 + \int_1^{\infty} \frac{1}{x^2}dx$$
I'm having trouble figuring out why the inequality above is true. I understand the following inequality:
$$\int_1^{\infty} \frac{1}{x^2}dx... | The right endpoint sums for the integral have the form:
$$
\sum_{n = 2}^\infty \frac{1}{n^2}
$$
and we know:
$$
\sum_{n = 2}^\infty \frac{1}{n^2} \leq \int_1^\infty \frac{1}{x^2} \leq \sum_{n = 1}^\infty \frac{1}{n^2}
$$
Subtracting the RHS, we have:
$$
-1 \leq \int_1^\infty \frac{1}{x^2} - \sum_{n = 1}^\infty
\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3016140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Number of solutions in quadratic congruence I use an example to explain my question:
How many solutions are there to $x^2+3x+18\equiv 0$ (mod $28$).
Usually I will encounter these kind of problems. My first step must transform this equation to become 2 equations:
$$x^2+3x+18\equiv 0 \quad (\text{mod }7) $$
$$x^2+3x+1... | $x\equiv2\pmod4,x\equiv2\pmod7\implies$lcm$(4,7)|(x-2)\implies x\equiv2\pmod{28}$
For the second, $$7a+2=4b+3\iff7a=4b+8-7\iff\dfrac{7(a+1)}4=b+2$$ which is an integer
$\implies4|7(a+1)\iff4|(a+1)$ as $(4,7)=1$
$\implies a+1=4c$
$\implies x=7a+2=7(4c-1)+2=28c-5\equiv-5\pmod{28}\equiv-5+28$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3019723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the Diophantine equation $y^2 = 4x^3 + 1$ for $x,y \in \mathbb{Z}$
I want to solve the Diophantine equation $y^2 = 4x^3 + 1$ for $x,y \in \mathbb{Z}$.
Note that $y$ is odd, since $y$ even would give a contradiction $\mod{2}$. Hence $\frac{y-1}{2}, \frac{y+1}{2} \in \mathbb{Z}$.
So we can rewrite the equation ... | Of course you can, from $f^3-e^3=1$ we get $$(f-e)(f^2+fe+e^2)=1$$
so $f-e=1$ and $f^2+fe+e^2 = 1$ or $f-e=-1$ and $f^2+fe+e^2 = -1$
So you have 2 cases. You can express $f$ with $e$ and plug in the second equation...
Else, I would introduce $y= 2k+1$ for some $k$, then you would have $k(k+1)=x^3$ which is a little e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3020596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Laurent Series expansion about the point $z_0 = i$ of $\frac{z}{z^2+1}$ I am trying to construct the Laurent series expansion of $f(z) = \frac{z}{z^2+1}$ about $z_0 = i$ in the region $\{z \in \mathbb{C}: 0 < |z - i| < 2\}$ but I am stuck.
We can re-write $f(z) = \dfrac{z}{z^2+1} = \dfrac{z}{(z-i)(z+i)}$
This allows us... | \begin{align}&f(z)=\frac{z}{z^2+1}=\frac{A}{z-i}+\frac{B}{z+i}\\&=\frac{1}{2}\frac{1}{z-i}+\frac{1}{2}\frac{1}{z+i}\end{align}
\begin{align}&A=Res[f,i]=\lim\limits_{z\rightarrow i}\frac{z}{z+i}=\frac{1}{2}
\\&B=Res[f,-i]=\lim\limits_{z\rightarrow -i}\frac{z}{z-i}=\frac{1}{2}\end{align}\begin{align}f(z)&= \frac{1}{2}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\sqrt{2}\sec x+\tan x=1$
Solve $\sqrt{2}\sec x+\tan x=1$
I understand it can be very easily solved by expanding in terms of $\sin x$ and $\cos x$, gives $x=2n\pi-\frac{\pi}{4}$. But, what if I do the following:
$$
\sqrt{2}\sec x+\tan x=1\\
\text{Differentiating}\implies\sqrt{2}\sec x\tan x+\sec^2 x=0\implies\s... | HINT
As noticed we can't use differentiation to obtain the result indeed in general
$$f'(x)=g'(x) \not \Rightarrow f(x)=g(x)$$
consider for example the simple case
$$2x+3=1 \to 2=0$$
I suggest to use tangent half-angle identities by $t = \tan \frac x2$ to obtain
$$\sqrt{2}\sec x+\tan x=1 \iff \sqrt{2}\frac{1+t^2}{1-t^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Radius of convergence of $\sum\limits_{n=1}^\infty((\frac{1}{4})^n+(\frac{1}{3})^n)x^n$ according to wolfram alpha the radius is 3. I'm struggeling with the proof and would be very glad if someone could take a short look. Here my approach using the root test:
$a_n:=\sqrt[n]{(\frac{1}{4})^n+(\frac{1}{3})^n}<\sqrt[n]{(\f... | We have that
$$|x|\cdot\sqrt[n]{\left(\frac{1}{4}\right)^n+\left(\frac{1}{3}\right)^n}=|x|\cdot\frac13\sqrt[n]{\left(\frac{3}{4}\right)^n+1} \to|x|\cdot\frac13\cdot 1 = |x|\cdot\frac13<1$$
and therefore the radius of convergence is $3$.
Moreover note that
*
*for $x=3 \implies \left(\left(\frac{1}{4}\right)^n+\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3026242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Seeking Methods to solve $\int_{0}^{\frac{\pi}{2}} \ln\left|\sec^2(x) + \tan^4(x) \right|\:dx $ After weeks of going back and forth I've been able to solve the following definite integral:
$$I = \int_{0}^{\frac{\pi}{2}} \ln\left|\sec^2(x) + \tan^4(x) \right|\:dx $$
To solve this I employ Feynman's Trick with Glasser's ... | $$I = \int_{0}^{\frac{\pi}{2}} \ln\left(\sec^2(x) + \tan^4(x) \right)dx=\int_0^\infty \frac{\ln(1+x^2+x^4)}{1+x^2}dx$$Consider: $$I(a)=\int_0^\infty \frac{\ln((1+x^2)a+x^4)}{1+x^2}dx$$
Derivating under the integral sign with respect to $a$ gives: $$I'(a)=\int_0^\infty \frac{1+x^2}{(1+x^2)a+x^4}\frac{dx}{1+x^2}=\int_0^\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Limit Question: doubting about my answer! A limit where $x \to \infty$ Hey guys so I have this limit:
$$\lim_{x \to ∞} f(x) = \frac{(3x+1)^3(x-1)}{(x-2)^4}$$
and I got $27$ as final answer, just wondering if you guys can check;
I expanded the numerator and denominator and then divided everything by $x^4$
| Note that
$$\frac{(3x+1)^3(x-1)}{(x-2)^4}=27\frac{(x+1/3)^3(x-1)}{(x-2)^4}=27\frac{\frac{(x+1/3)^3}{x^3}\frac{(x-1)}x}{\frac{(x-2)^4}{x^4}}\\=27\frac{(1+1/x)^3(1-1/x)}{(1-2/x)^4}\to27\frac{(1+0)^3(1-0)}{(1-0)^4}=27$$
as $x\to\infty$, so your result is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Joint PDF P[X+Y<=0.5] I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:
Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.
I need to find the $P(x+y\leq 0.5)$.
For the double integration, I have the following bounds:
*
*Outer... | Your made the sign error in step 2 (as indicated by KaviRamaMurthy).
As an alternative, you can simplify in step 2 before integrating:
$$\int_0^{0.5}\left(\frac{x}{2} -x^2+\frac{(0.5-x)^2}{2}\right) dx=
\int_0^{0.5}\left(\require{cancel}\cancel{\frac{x}{2}} -x^2+\frac{1}{8}-\cancel{\frac x2}+\frac{x^2}{2}\right) dx=\\
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integer equations I have $2$ following problems. Find integer roots of
$$\begin{align}
&1)~\frac{x+y}{x^2-xy+y^2}=\frac3z \\
&2)~x^3y^3-4xy^3+y^2+x^2-2y-3=0
\end{align}$$
I have no idea to solve them. I try to guess roots of the second, they are $\left( -2, 1\right), \left( 0, -1\right), \left( 0, 3\right), \left( 2, 1... | Question 1:
\begin{align}
& \frac{x+y}{x^2-xy+y^2} = \frac 3z \\
\iff & \frac{3(x^2-xy+y^2)}{x+y} = z\\
\iff & 3(x+y) - \frac{9xy}{x+y} = z \\
\iff & \frac{9xy}{x+y} = 3x+3y-z \\
\implies & \frac{9}{\frac 1x + \frac 1y} = 3x+3y-z
\end{align}
From the LHS we see that $\frac 1x + \frac 1y$ must be equal to $\pm 1, \pm 3,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the Jordan form and basis for a matrix I have this matrix $A=\left[ {\begin{array}{ccc}
-3&3&-2\\
-7&6&-3\\
1&-1&2\\
\end{array} } \right]$
I computed the characteristic polynomial $C_A(t)=-(t-2)^2(t-1)$
When I go to try and find the eigenvectors and generalized eigenvectors I compute $A-2I=\left[ {\begin{array... | Because you are looking for a Jordan canonical basis, $Ax = 2x+v$, where $x$ is the generalized eigenvector you're looking for. Equivalently, $(A-2I)x = v$. Applying $A-2I$ one more time to each side will give you the following statement: $(A-2I)^{2}x = 0$.
So, your generalized eigenvector is in the nullspace of $(A-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve for positive relatively prime numbers? When finding the relative prime for large numbers like 360 (that has a prime factorization with multiples of the same prime factor), how would it be solved?
ie. How many numbers between 1 and 360 are relatively prime to 360?
I would think either
*
*N = 360 [prime ... | What primes do you need to worry about? The primes that divide $360$ are $2,3,5$, so you really want to find those numbers which are not divisible by $2, 3$, or $5$.
Now
*
*$360/2 = 180$ numbers up to $360$ are divisible by $2$.
*$360/3 = 120$ numbers up to $360$ are divisible by $3$.
*$360/5 = 72$ numbers up to ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A functional equation (another) I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $x\in [1/2,1]$
$$f(x)= \frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right).$$
I am actually not sure about existence of a solution.
| $f(x)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2x}{1+x}\right)$
$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2\times\dfrac{2^x}{2^x-1}}{1+\dfrac{2^x}{2^x-1}}\right)$
$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{\dfrac{2^{x+1}}{2^x-1}}{\dfrac{2^{x+1}-1}{2^x-1}}\right... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution for $(a\cdot x+b)\cdot \sin(x)+c\cdot \cos(x)=0$? As part of an engineering problem, I've been trying for a generic solution for $(a\cdot x+b)\cdot \sin(x)+c\cdot \cos(x)=0$
Here's something I tried:
$(a⋅x+b)⋅\tan(x)+c=0$, when $\cos(x)\neq 0$
$$\tan(x)=\frac{-c}{a⋅x+b}$$
It sort of looking like this, when, sa... | Besides numerical solutions for particular parameter values, you could try series solutions. Thus when $c$ is small compared to $a$ and $b$, the solution near $x=0$ is
$$ -\frac{b}{a} + \cot\left(\frac{b}{a}\right) \frac{c}{a} + \cot\left(\frac{b}{a}\right) \csc^2 \left(\frac{b}{a}\right) \frac{c^2}{a^2}
+ \left(\cot... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x=y+\sqrt{y^2+1}$
Then how to write $y$ in terms of $x$? If $$x=y+\sqrt{y^2+1}$$
Then how to write $y$ in terms of $x$? Please tell me some hints
| I assume that
$x, y \in \Bbb R. \tag 0$
We are given
$x = y + \sqrt{y^2 + 1}. \tag 1$
We observe that this equation implies
$x > 0, \forall y \in \Bbb R, \tag{1.2}$
since
$y^2 < y^2 + 1 \Longrightarrow \vert y \vert < \sqrt{y^2 + 1}; \tag{1.5}$
we proceed:
$x - y = \sqrt{y^2 + 1}; \tag 2$
$x^2 - 2xy + y^2 = (x - y)^2 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An inequality for positive definite matrix with trace 1 Given a positive definite matrix $A \in \mathbb R^{n \times n}$. If $\operatorname{trace}(A) = 1$, for $n \geq 3$, prove that $$\text{det}(A) \leq \frac{n^n}{(n-1)^{2n}} \text{det}(I -A)^2$$ and the equation only holds when $A = \frac{1}{n}I$.
What I have tried:
F... | In order to investigate behavior of the function $f$, at the first we consider the following auxiliary problem.
Let $0<x,y,a$ and $x+y+a=1$. If $a$ is fixed then
$$h(x,y)=\frac{(1-x)^2(1-y)^2}{xy}=\frac{(a+xy)^2}{xy}=\frac{a^2}{xy}+2a+xy.$$
A function $\frac{a^2}{t}+t$ has a derivative $-\frac{a^2}{t^2}+1$, so it dec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Calculate the area of trapezoid. In an isosceles trapezoid $ABCD$ the leg $AB$ and the smaller base $BC$ are 2 cm long, and $BD$ is perpendicular to $AB$.Calculate the area of trapezoid.
Let $\angle BAD=\theta,$
$BD=h$,
$\angle ABD=90^\circ$
$\angle CBD=90^\circ-\theta$
$CD=2$because trapezoid is isosceles
Apply cosin... | you are correct. The answer should be $3\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\int\limits_{-\infty}^\infty \frac{1}{x^8+1}dx$ through Glasser's Master Theorem Trying to find a way to solve
$$\int_{-\infty}^\infty \frac{1}{x^8+1}dx$$
through Glasser's Master Theorem, more specifically the Cauchy–Schlömilch substitution. Preferably, I'm looking for the closed form solution, and I am a... | I don't know what is GMT. I looked it up on Wikipedia and it states $u = x - 1/x$ as Cauchy–Schlömilch substitution. Here is my solution that splits the integral into 4 integrals, each one of them uses $u = x - 1/x$. I will happily delete the answer if it is useless or wrong :)
$$\int^{\infty}_{-\infty} \dfrac{1}{1+ x... | {
"language": "en",
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"source": "stackexchange",
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Can any improvements be made to my proof that "$\sqrt{3} $ is irrational"? Suppose, for contradiction that $\sqrt{3}$ is rational. Then there exists $a,b \in \mathbb{Z}$ such that $$\frac{a}{b}= \sqrt{3},$$
where $a/b$ is in its simplest form. Then the above equation implies $$a^2=3b^2.$$ If $b$ is even, then $a$ is ev... | Your proof looks correct to me. Instead of doing the algebra at the end, you could reduce the equation $a^2 = 3b^2$ modulo $4$. If $a$ and $b$ are both odd, then $$a^2 \equiv 1 \mod 4,$$
and
\begin{align*}
b^2 &\equiv 1\mod 4\\
3b^2 &\equiv 3\mod 4,
\end{align*}
contradiction.
Another approach is to note that in the p... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve the differential equation $\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$ Solve the differential equation $$\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$$
My try:
we can write the equation as:
$$\frac{dy}{dx}=\frac{1}{y^3}\frac{\left(1+2y^4\right)}{1+\frac{4x}{y^4}}$$
Multiplying both sides with $\frac{1}{y^5}$ we get:
$$\frac{1}{y... | This is an "almost" exact equation. Write it in $A(x,y)\text dx + B(x,y)\text dy=0$ form
$$(-y-2y^5)\text dx + (4x+y^4)\text dy = 0$$
Then, look for a function $\mu(y)$ such that
$$ (-y-2y^5)\mu(y)\text dx + (4x+y^4)\mu(y)\text dy = 0 $$
Let $\mu(y) = y^{-5}$. Then, the equation becomes
$$ (-y^{-4}-2)\text dx + (4xy^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^... | Make the substitution $x = r\cos(\theta)$ and $y = r\sin(\theta)$ in order to get
\begin{cases}
r^{2}\cos(\theta) + 3\cos(\theta) - \sin(\theta) = 3r\\
r^{2}\sin(\theta) - \cos(\theta) - 3\sin(\theta) = 0\\
\end{cases}
From the second equation, it results that $\cos(\theta) = (r^{2}-3)\sin(\theta)$. Therefore we get
\b... | {
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"url": "https://math.stackexchange.com/questions/3049686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\iiint x^2\,{\rm d}x{\rm d}y{\rm d}z$ over the volume bounded by $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2 }= 1$
Finding the value of $$\iiint x^2\,\mathrm dx\mathrm dy\mathrm dz$$ over the volume bounded by the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}= 1.$$
I am taking limit... | First, rewrite the integral as
$$\int_{-a}^a x^2 \int\int_{E_x}1\ dy\ dz\ dx $$
where $E_x$ is the ellipse in the $yz$-plane satisfying $\frac{y^2}{b^2} + \frac{z^2}{c^2} = (1-\frac{x^2}{a^2})$. By arranging the formula of the ellipse into the more familiar form $\frac{y^2}{b^2\left(1-\frac{x^2}{a^2}\right)} + \frac{z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049891",
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"source": "stackexchange",
"question_score": "3",
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Proof verification. $\{x_n\}$ is a sequence such that $|x_{n+1} - x_n| \le C\alpha^n$ for $\alpha\in (0, 1), n\in\Bbb N$. Prove $x_n$ converges.
Let $\{x_n\}, n\in \Bbb N$ denote a sequence such that:
$$
\begin{cases}
|x_{n+1} - x_n| \le C\alpha^n \\
0 < \alpha < 1
\end{cases}
$$
Prove $\{x_n\}$ converges.
Given ... | Seems fine.
We have $C \ge 0$.
In the event that $C=0$, we have a constant sequence and hence it converges.
| {
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"url": "https://math.stackexchange.com/questions/3051454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Compute $\sum\limits_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right)$ where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$ As part of solving:
\begin{equation}
I_m = \int_0^1 \ln\left(1 + x^{2m}\right)\:dx.
\end{equation}
where $m \in \mathbb{N}$. I found an unresolved component that I'm unsur... | This does not answer the question as asked in the post.
Consider
$$J_m=\int \log(1+x^{2m})\,dx$$ One integration by parts gives
$$J_m=x \log \left(1+x^{2 m}\right)-2m\int \frac{ x^{2 m}+1-1}{x^{2 m}+1}\,dx=x \log \left(1+x^{2 m}\right)-2mx+2m\int \frac{dx}{x^{2 m}+1}$$ and
$$\int \frac{dx}{x^{2 m}+1}=x \, _2F_1\left(1,... | {
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"url": "https://math.stackexchange.com/questions/3053596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Factoring $3^6-3^3 +1$ $3^6-3^3 +1$ factors?, 37 and 19, but how to do it using factoring, $3^3(3^3-1)+1$, can't somehow put the 1 inside
| $x^2-x+1$ factorises as $(x-\omega)(x+\omega^2)$ where $\omega^3=-1$.
Here $x=27$ and working modulo $27$ the cubes are $1,8,0,10,17,0,19,26, 0, 1, 8 \dots$ so $8^3\equiv -1$, and we can take $\omega = 8, \omega^2=64\equiv 10$ and obtain the factorisation $$(27-8)(27+10)=19\times 37$$
Since we have a cube root involved... | {
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"url": "https://math.stackexchange.com/questions/3053975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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A Series For the Golden Ratio
Question: Can we show that $$\phi=\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2} $$; where $\phi={1+\sqrt{5} \above 1.5pt 2}$ is the golden ratio ?
Some background and motivation:
Wikipedia only provides one series for the golden ratio - see also the link in the ... | Using Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
$$\dfrac{(2n)!}{5^{3n+1}(n!)^2}=\dfrac{2^n\cdot1\cdot3\cdot5\cdots(2n-3)(2n-1)}{5^{3n+1}n!}=\dfrac{-\dfrac12\left(-\dfrac12-1\right)\cdots\left(-\dfrac12-(n-1)\right)}{n!\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
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Calculate $\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})$ I know that $$\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})\\=\lim\limits_{ x\to + \infty}x\cdot \sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right).$$ If $x \rightarrow + \infty$, then $\sin\left(\frac{1}... | \begin{align}
x \sin \left( \sqrt{x^2+3}-\sqrt{x^2+2}\right)
&=x \sin \left( \frac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)\\
&=\left(\dfrac{x}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)
\dfrac{\sin \left( \dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}
{\left(\dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$
Evaluate
$$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$$
So in a previous part of the question I calculated that
$$\sum_{r=1}^{n} \frac{2-r}{r(r+1)(r+2)} = \sum_{r=1}^{n}\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\frac{n}{(n+1)(n+2)}$$
So my questio... | If $H_{n}$ is the $n$-th harmonic number, we have $$\sum_{r=2}^{n}\left(\frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\left(H_{n}-1\right)-3\left(H_{n}-1-\frac{1}{2}\right)+2\left(H_{n}-1-\frac{1}{2}-\frac{1}{3}\right)$$ $$=-1+\frac{3}{2}-\frac{2}{3}=-\frac{1}{6}.$$ Now take the limit as $n\rightarrow+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Laplace transform of square root of a trigonometric function Need help with this question from my university paper.
My question : Find Laplace Transform of the following: $\sqrt{1 + \sin(4t)}$
I do know how to solve $\sqrt{1 + \sin(t)}$
By taking $1 = \sin^2 \left(\frac{t}{2}\right) + \cos^2 \left(\frac{t}{2}\ri... | The function $\sqrt{1+\sin at}$ is periodic with period $T = 2\pi/a$ and the first zero at $3\pi/2a$. Following your steps, we can write this as
$$ \sqrt{1+\sin at} = \sqrt{\cos^{2}\frac{at}{2} + \sin^{2}\frac{at}{2} + 2\sin\frac{at}{2}\cos\frac{at}{2}} = \left|\cos\frac{at}{2}+\sin\frac{at}{2}\right|. $$
Using the Lap... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial
$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
Ho... | $15!=2^{11}\cdot 5^3\cdot 7^2\cdot 11\cdot 13=(1000)X$ where $X=2^8\cdot 3^6\cdot 7^2\cdot 11\cdot 13.$
The last digits of $2^8(=16^2), 3^6 (=9^3),7^2, 11,13$ are, respectively $6,9,9,1,3 .$
Modulo $10$ we have $6\cdot 9\cdot 9 \cdot 1\cdot 3\equiv 6\cdot(-1)^2\cdot 3\equiv 18\equiv 8$. So the last digit of $X$ is an $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
Show that if $a \equiv b\pmod{kn}$, then $a^k\equiv b^k\pmod{k^2n}$
Problem
Show that if $a \equiv b\pmod{2n}$, then $a^2\equiv b^2\pmod{4n}$. More generally, show that if $a \equiv b\pmod{kn}$, then $a^k\equiv b^k\pmod{k^2n}$. [Introduction to Higher Arithmetic, ex. 2.1]
Proof of the first statement
Note that $2n \m... | Note that $a\equiv b\pmod{kn}\implies a=b+kln$ for some integer $l$. Then $$a^k-b^k=(b+kln)^k-b^k=\binom k1b^{k-1}kln+\binom k2b^{k-2}(kln)^2+\cdots+(kln)^k$$ Every term has a factor of $k^2n$ from $(kln)^t$ for $1<t\le k$ but $\dbinom k1=k$ so $k^2n\mid (a^k-b^k)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For integers $a,b$ and prime $p$, are there infinitely many solutions to $p = (a-b)(a+b)$? I was thinking on this problem: suppose, for a prime $p$ and integers $a,b$, we have
$$p = (a-b)(a+b)$$
This implies immediately that $p \mid (a-b)$ or $p \mid (a+b)$. Then, since $p$ is prime and thus its only factors are $1$ an... | Of course not!
If $p$ is prime so its only factors are $\pm 1$ and $\pm p$ so so if $p = (a-b)(a+b)$ then $a+b$ and $a-b$ can only be $\pm 1$ and/or $\pm p$.
So either $a-b = 1$ and $a+b = p$ so $a=\frac {p+1}2; b = \frac {p-1}2 = a-1$.
Or $a-b = p$ and $a + b = 1$ so $a= \frac {p+1}2; b = -\frac {p-1}2= 1-a$
Or $a-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving integral using residues I'm trying to find the value of the integral $\int_0^{2\pi}\frac{\cos^2u}{2-\sin u}du$ using the substitution: $\cos u =\frac{1}{2}(z+\frac{1}{z})$ and $\sin u = \frac{1}{2i}(z-\frac{1}{z})$. Making the substitution, we obtain $\int_C(z^2+1)^2/(2z^2(z^2+4iz-1)dz$ where C denotes the unit... | Let $\tan\frac{u}{2}=x$.
Thus, $$dx=\frac{1}{2\cos^2\frac{u}{2}}du=\frac{1}{2}(1+x^2)du,$$ which gives
$$du=\frac{2}{1+x^2}dx.$$
Hence, $$\int\frac{\cos^2u}{2-\sin{u}}du=\int\left(2+\sin{u}-\frac{3}{2-\sin{u}}\right)du=$$
$$=2u-\cos{u}-3\int\frac{1}{2-\frac{2x}{1+x^2}}\frac{2}{1+x^2}dx=$$
$$=2u-\cos{u}-3\int\frac{1}{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
On Basak's "Bounds On Factors Of Odd Perfect Numbers" Let $N = q^k n^2$ be an odd perfect number given in Eulerian form, i.e. $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. In what follows, we denote the abundancy index of $x \in \mathbb{N}$ by $I(x)=\sigma(x)/x$, where $\si... | We get $$\frac{16q^3}{7\zeta(3)(q^3 - 1)} < I(q)I(n^2)$$ from which we have
$$I(n^2)\gt\frac{16}{7\zeta(3)}f(q)$$
where
$$f(q)=\frac{q^4}{(q^3-1)(q+1)}$$
Now, we have, for $q\ge 5$,
$$f'(q)=\frac{ q^3 (q(q^2-3)-4)}{(q + 1)^2 (q^3 - 1)^2}\gt 0$$
from which we have
$$I(n^2)\gt\frac{16}{7\zeta(3)}f(q)\ge \frac{16}{7\zet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find minimum value of $a^2+b^2$ Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.
I began this way: Let the polynomial be factorized as $(x^2+\alpha x + 1)(x^2+\beta x +1)$. Then expanding and comparing coeffi... | I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.