Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the matrix associated to linear function. Good night, i have a problem when i find the matrix associated to a linear transformation:
$T:P_{1\rightarrow}P_{2}$
$T(p(t))=t(p(t))$
Basis for $P_{1}=\left\{ t,1\right\} $
and for $P_{2}=\left\{ t^{2},t-1,t+1\right\}$
I work in this problem and i made this:
$T(t)=t^{2}=... | As you discovered in the comments, it is easy to solve this one by playing around with linear combinations.
But what about a general strategy?
We know that given a linear transformation $T$ from $V$ to $W$, and two bases $\alpha = (v_1, \ldots, v_n)$ and $\beta = (w_1, \ldots, w_m)$, there is always a unique matrix rep... | {
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"url": "https://math.stackexchange.com/questions/1789739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show that $\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $ Let $a,b,c$ are positive numbers,if $$a+b+c=1$$
show that $$\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $... | $$ f(x) = \frac{1}{1-x}-\frac{2}{1+x} $$
is not a convex function on $(0,1)$, since:
$$ f''(x) = \frac{2}{(1-x)^3}-\frac{4}{(1+x)^3} \geq 0 $$
is equivalent to:
$$ \frac{1+x}{1-x}\geq \sqrt[3]{2} $$
but if we consider that $f'\left(\frac{1}{3}\right)=\frac{27}{8}$, it is not difficult to prove the algebraic inequality
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
} |
How can I prove this equation has no real solution? I have an equation
$$x^2-5 x+10+(x-4) \sqrt{1+x}=0. \tag{1}$$
Now I am trying to prove this equation has no real solution.
I tried. Put $t = \sqrt{1+x}$, then, I got
$$t^4+t^3-7 t^2-5 t+16=0. \tag{2}$$
But I can't prove two equations (1) and (2) have not solution.
H... | $$x^2-5x+10-(4-x)\sqrt{x+1}$$
now note: for roots to exist $-1<x<4$ thus $(4-x)>0$ and $\sqrt{x+1}>0$ now using our favourite AM-GM
$$x^2-5x+10-(4-x)\sqrt{x+1}>x^2-5x+10-\frac{(4-x)^2+(x+1)}{2}=\frac{2x^2-10x+20-16-x^2+8x-x-1}{2}=\frac{x^2-3x+3}{2}$$ now discriminant of the quadratic is negative and first coefficient ... | {
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"url": "https://math.stackexchange.com/questions/1792631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Error in solving $\int \sqrt{1 + e^x} dx$ . I want to solve this integral for $1 + e^x \ge 0$
$$\int \sqrt{1 + e^x} dx$$
I start by parts $$\int \sqrt{1 + e^x} dx = x\sqrt{1 + e^x} - \int x \frac{e^x}{2\sqrt{1 + e^x}} dx $$
Substitute $\sqrt{1 + e^x} = t \implies dt = \frac{e^x}{2\sqrt{1 + e^x}} dx$
So I remain with $... | with $$t=\sqrt{1+e^x}$$ we get $$x=\ln(t^2-1)$$ and from here $$dx=\frac{2t}{t^2-1}dt$$ and our integral will be $$\int\frac{2t^2}{t^2-1}dt$$
| {
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"url": "https://math.stackexchange.com/questions/1793094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $2^n+3^n $ is never a perfect square My attempt :
If $n$ is odd, then the square must be 2 (mod 3), which is not possible.
Hence $n =2m$
$2^{2m}+3^{2m}=(2^m+a)^2$
$a^2+2^{m+1}a=3^{2m}$
$a (a+2^{m+1})=3^{2m} $
By fundamental theorem of arithmetic,
$a=3^x $
$3^x +2^{m+1}=3^y $
$2^{m+1}=3^x (3^{y-x}-1) $
Whic... | This is more convoluted than your way, but I already wrote it so I'll just leave it here :-)
$4^m = (a-3^m)(a+3^m) \implies a-3^m = 2^p, a+3^m = 2^q$ with $p+q = 2m$. Note that $q \ge p$
Then $a = 2^p + 3^m = 2^q - 3^m \iff 2^p (2^{q-p}-1) = 2\cdot 3^m$
But then it must be $p=1$, $q = 2m-1$, and we have
$$2^{2(m-1)}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
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sum of all Distinct solution of the equation $ \sqrt{3}\sec x+\csc x+2(\tan x-\cot x) = 0\;,$
The sum of all Distinct solution of the equation $\displaystyle \sqrt{3}\sec x+\csc x+2(\tan x-\cot x) = 0\;,$
Where $x\in (-\pi,\pi)$ and $\displaystyle x\neq 0,\neq \frac{\pi}{2}.$
$\bf{My\; Try::}$ We can write equation a... | HINT:
$\cos (a+b) \neq \cos a \cos b + \sin a \sin b $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to see $\cos x \leq \exp(-x^2/2)$ on $x \in [0,\pi/2]$? Can anyone help me with the above inequality? I tried looking at the series expansion and I guess the answer indeed lies there, but I fail to see it.
Thanks
| Here is another option to prove the inequality by Taylor expansions as the OP suggested. So we are to prove that
$$
\exp(-x^2/2)\ge\cos x,\quad x\in[0,\pi/2].
$$
By doing Taylor expansions at zero with the Lagrange remainders, we can get the estimations
\begin{eqnarray}
\exp(-t)&=&1-t+\frac{t^2}{2!}-\frac{t^3}{3!}+\u... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Local extrema and minima of the multivariable function $f(x,y) = x^2y+y^2+xy$ Let $f(x,y) = x^2y+y^2+xy$ be a function, I want to find its local extrema an minima.
I easily find that $f$ has 2 critical points: $(x,y)=(0,0)$ and $(x,y) = (-1,0)$.
In order to find its local extrema, I now find the Hessian matrix of the q... | The critical points are points $a \in \mathbb R^2$ such that $\mathrm {grad}\, f(a) = 0$. Now $\mathrm{grad}\, f(x,y) = (2xy + y , x^2 + 2y + x)$. You need to solve $$\begin{cases}2xy + y = 0 \\x^2 + 2y + x = 0\end{cases} \iff \begin{cases}y(2x+ 1) = 0\\x(x + 1) + 2y = 0\end{cases} \iff \begin{cases}y= 0 \,\,\text{or}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Laurent series of $\frac{e^z}{z^2+1}$
I cant figure out the laurent series of the following function.
Let $f(z)= \frac{e^z}{z^2+1} $ and $|z|\gt 1$
$$\frac{1}{z^2+1}=\sum_{n=0}^{\infty}(-1)^nz^{-2n-2}$$
and
$$e^z = \sum_{n=0}^{\infty}\frac{z^{n}}{n!}$$
$$e^z*\frac{1}{z^2+1} =\sum_{n=0}^{\infty}\sum_{k=0}^{n}(-1)^nz^... |
Let
\begin{align*}
f(z)&= \frac{e^z}{z^2+1}
\end{align*}
We assume we need a Laurent expansion for all $|z|>1$ with center $z_0=0$.
We obtain
\begin{align*}
f(z)&=\frac{e^z}{z^2+1}\\
&=\frac{e^z}{z^2}\cdot\frac{1}{1+\frac{1}{z^2}}\\
&=\frac{e^z}{z^2}\sum_{k=0}^\infty(-1)^k\frac{1}{z^{2k}}\\
&=\sum_{j=0}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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QR factorization for least squares This is from my textbook
I don't undertand why small errorr in $A^TA$ can lead to large error in cofficient matrix? Because A=QR, so there should be no difference to use A or QR anyway.Could someone give an example? Thank you very much
| A typical example uses the matrix
$$
\mathbf{A} =
\left(
\begin{array}{cc}
1 & 1 \\
0 & \epsilon \\
\end{array}
\right)
$$
Consider the linear system
$$
\mathbf{A} x= b.
$$
The solution via normal equations is
$$
%
\begin{align}
%
x_{LS} &= \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1}\mathbf{A}^{*}b \\
% x
\left(
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving an algebraic identity Prove:
$$(a + b + c)(ab + bc + ca) - abc = (a + b)(b + c)(c + a)$$
Problem:
I am not sure how to proceed after expanding the brackets on the RHS. I am not sure if I also expanded correctly. My solution is:
| $$(a+b)(b+c)(c+a)$$
$$=(ab+bc+ca+b^2)(c+a)$$
$$=(ab+bc+ca)(a+c)+ab^2+b^2c$$
Adding $abc$ gives:
$$(ab+bc+ca)(a+c)+ab^2+b^2c+abc$$
$$=(ab+bc+ca)(a+c)+b(ab+bc+ac)$$
$$=(ab+bc+ca)(a+b+c)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Find the minimum value of expression involving real numbers
Let $n$ positive integer. Find the minimum value of expression:
$$ E=max(\frac {x_1} {1+x_1},\frac {x_2} {1+x_1+x_2}, ... , \frac {x_n}
{1+x_1+..+x_n})$$
where $x_1,x_2, .. , x_n$ are real not negative so that $x_1+x_2+ .. +x_n=1$
My try
For $x_1=1, x_2=0,... | For $n=2$, the answer is $1-\frac{1}{\sqrt{2}}$, when $x_1=\sqrt{2}-1$ and $x_2=2-\sqrt{2}$.
In general, the answer is $1-\frac{1}{\sqrt[n]{2}}$.
Let's assume $a_0=1$, $a_i=1+x_1+x_2+\dots+x_i$ for $1\leqslant i\leqslant n$. Hence, $a_n=2$ and $a_i\leqslant a_{i+1}$. Then, $\frac{x_i}{1+x_1+x_2+\dots+x_i}=\frac{a_i-a_... | {
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"url": "https://math.stackexchange.com/questions/1802424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a, b, c >0$ prove that $ [(1+a)(1+b)(1+c)]^7 > 7^7a^4b^4c^4 $. I solved it using AM, GM inequalities and reached to $[(1+a)(1+b)(1+c)]^7 > 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.
| Here we have to prove $$(1+a)(1+b)(1+c)>7(abc)^{\frac{4}{7}}$$
So we get $$1+(a+b+c)+(ab+bc+ca)+abc>7(abc)^{\frac{4}{7}}$$
Now Using $\bf{A.M\geq G.M}\;,$ We get $$a+b+c\geq 3(abc)^{\frac{1}{3}}$$
and $$ab+bc+ca\geq 3(abc)^{\frac{2}{3}}$$
So $$1+\sum a+\sum ab+abc\geq 1+3(abc)^{\frac{1}{3}}+3(abc)^{\frac{2}{3}}+abc$$
N... | {
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"url": "https://math.stackexchange.com/questions/1802976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Condition for three points to lie on a Sphere? If A and B are the points (3,4,5) and (-1,3,-7) respectively then the set of the points P such that $PA^2+PB^2 = K^2$ where K is a constant lie on a proper sphere if K = 1 or K^2 <>= 161/2?
The correct answer is K^2 > 161/2. How?
I know that three non - collinear points al... | Note first that the distance between $A$ and $B$ is $\sqrt{161}$. Let $C(1,\frac{7}{2},-1)$ be the midpoint of $AB$ and let $r=\frac{1}{2}\sqrt{2K^2-161}$. Take any point $P(x,y,z)$.
We have $PA^2+PB^2=(x-3)^2+(y-4)^2+(z-5)^2+(x+1)^2+(y-3)^2+(z+7)^2=2x^2+2y^2+2z^2-4x-14y+4z+109=2(x-1)^2+2(y-\frac{7}{2})^2+2(z+1)^2+\fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Switch from $a\cdot \sin(t) + b \cdot \cos(t)$ to $c \cdot \cos(t+f)$ How could I switch from $a\cdot \sin(t) + b \cdot \cos(t)$ to $c \cdot \cos(t+f)$?
Thank you for your time.
| $$
a\sin t+b\cos t=\sqrt{a^2+b^2}\left[\frac{b}{\sqrt{a^2+b^2}}\cos t +\frac{a}{\sqrt{a^2+b^2}}\sin t\right]$$
$$
=\sqrt{a^2+b^2}\left[\cos \alpha \cos t +\sin\alpha \sin t\right]$$
$$ =\sqrt{a^2+b^2} \cos(t-\alpha). $$
$ (c,f) $ have different symbols.
Useful in combining two waves of same frequency but different ampl... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
I think we'll have to use number theory to do it. Simply solving the equations won't do.
If we divide the second equation by the first, we get:
$$x^2 - xy... | A general identity that is nice to know is
$$x^3+y^3=(x+y)(x^2-xy+y^2).$$
Given that $x+y=1-z$ and $x^3+y^3=1-z^2$ we have $z=1-x-y$ and hence
$$x^3+y^3=1-z^2=(1-z)(1+z)=(x+y)(2-x-y).$$
This means that either $x+y=0$, so $y=-x$ and $z=1$, or
$$x^2-xy+y^2=2-x-y.$$
This is a quadratic in $x$, and quadratics are easy. App... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 0
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How is $ \ \frac{1}{1+\frac{1}{x^2}}=\ \frac{x^2}{1+x^2} = \ 1 - \frac{x^2}{1+x^2}$? How did $$ \ \frac{1}{1+\frac{1}{x^2}}$$ become $$\ \frac{x^2}{1+x^2} = \ 1 - \frac{x^2}{1+x^2}$$
I am trying to figure out the equation.
| $$\frac{1}{1+\frac{1}{x^2}}=\frac{x^2}{x^2}\times\frac{1}{1+\frac{1}{x^2}}=\frac{x^2}{x^2(1+\frac{1}{x^2})}=\frac{x^2}{x^2+1}=\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1}$$ So, probably a typo somewhere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $\frac{1}{x_{1}(x_{1}+1)}+\frac{1}{x_{2}(x_{2}+1)}+\cdots+\frac{1}{x_{n}(x_{n}+1)}\ge\frac{n}{2}$ with $x_{1}x_{2}\cdots x_{n}=1$ Let $x_{1},x_{2},\cdots,x_{n}$ be postive real numbers, and such $x_{1}x_{2}\cdots x_{n}=1$, Show that
$$\dfrac{1}{x_{1}(x_{1}+1)}+\dfrac{1}{x_{2}(x_{2}+1)}+\cdots+\dfrac{1}{x_{n}... | Let $x_i=\frac{a_i}{a_{i+1}}$, where $a_i>0$ and $a_{n+1}=a_{1}$. We have
$$\tag{1}\sum_{cyc}\frac{1}{x_i(x_i+1)}=\sum_{cyc}\frac{1}{\frac{a_i}{a_{i+1}}\cdot\left( \frac{a_i}{a_{i+1}}+1\right)}=\sum_{cyc}\frac{a_{i+1}^2}{a_i\cdot\left(a_i+a_{i+1}\right)}$$
Notice that:
$$\sum_{cyc}\frac{a_{i+1}^2}{a_i\cdot (a_i+a_{i+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Can someone help me with this question of finding x as exponent? The equation is:
$$6^{x+1} - 6^x = 3^{x+4} - 3^x$$
I need to find x. I forgot how to use logarithm laws. Help would be appreciated. Thanks.
| This means
$$
6^x(6-1) = 3^x(3^4 - 1) \iff \\
5 \cdot 6^x = 80 \cdot 3^x \iff \\
5 \cdot 2^x \cdot 3^x = 80 \cdot 3^x \iff \\
5 \cdot 2^x = 80 \iff \\
2^x = 16 \Rightarrow \\
x = 4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\int_{0}^{1}{\ln(x) \over 2-x}\mathrm dx={\ln^2(2)-\zeta(2)\over 2}$ $$I=\int_{0}^{1}{\ln(x) \over 2-x}\,\mathrm{d}x={\ln^2(2)-\zeta(2)\over 2}$$
$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}$$
Using binomial series here
$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}={1\over2}+{x\over 4}+{x^2\o... | start with $$\int_0^1 \ln(1-x) x^{n-1}dx = -\frac{H_{n}}{n}$$
change of variable and expand in power series
$$\int_0^1 \frac{\ln(x)}{2-x}dx = \int_0^1 \frac{\ln(1-x)}{1+x} = \sum_{n=0}^\infty (-1)^n \int_0^1\ln(1-x) x^n dx = \sum_{n=1}^\infty (-1)^n \frac{H_{n}}{n}$$
and use a trick found by Jack D'Aurizio
$$\frac{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Differential equation exercise. I am tasked with solving
\begin{cases}
y''(t) &=& \frac{(y(t)')^2}{y} - 2\frac{y'(t)}{y^4(t)} \\
y(0) &=& -1 \\
y'(0) &=& -2
\end{cases}
I proceed by setting $v(s) = y' (y^{-1}(s))$ reducing the problem to
\begin{cases}
v'(s) &=& \frac{v(s)}{s} - \frac{2}{s^4} \\
v(-1) &=& -2
\... | \begin{cases}
y''(t) &=& \frac{(y(t)')^2}{y} - 2\frac{y'(t)}{y^4(t)} \\
y(0) &=& -1 \\
y'(0) &=& -2
\end{cases}
This is an ODE of the autonomeous kind. The usual way to reduce the order is the change of function :
$$\frac{dy}{dt}=F(y) \quad\to\quad \frac{d^2y}{dt^2}=\frac{dF}{dy}\frac{dy}{dt}=F\frac{dF}{dy}$$
$$F... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How can I prove this sharp upper bound? Here :
What are sharp lower and upper bounds of the fast growing hierarachy?
Deedlit mentions that for natural $m,n\ge 2$ and natural $k>n+log_2(n)$ , we have $$2\uparrow^{m-1}n<f_m(n)<2\uparrow^{m-1} k$$
The left inequality can be proven by induction. But what about the right in... | So we will prove that for $m \ge 2, n \ge 3$, $f_m(n) \le 2\uparrow^{m-1}\lceil n + \log_2(n) + 1\rceil$.
Base case $m=2$: $f_2(n) = n 2^n = 2^{n+\log_2(n)} \le 2 \uparrow \lceil n + \log_2(n) + 1\rceil$.
Inductive step:
We assume that $f_m(n) \le 2\uparrow^{m-1}\lceil n + \log_2(n) + 1 \rceil \le 2 \uparrow^{m-1}(2n)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$ without L'hôpital I have this $\lim_{}$.
$$\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$$
Indetermation:
$$\lim_{x\to 0} \frac{\ln(0+4)-\ln(4)}{0}$$
$$\lim_{x\to 0} \frac{0}{0}$$
Then i started solving it:
$$\lim_{x\to 0} \frac{\ln\frac{(x+4)}{4}}{x}$$
$$\lim_{x\to 0} \frac{... | Consider rewriting the expression as
$\dfrac{\ln(x + 4) - \ln(4)}{x} = \dfrac{\ln\Big(\dfrac{x + 4}{4}\Big)}{x} = \ln\Big(\frac{x}{4} + 1\Big)^{1/x} = \frac{1}{4}\ln\Big(\frac{1}{4/x} + 1\Big)^{4/x}$
Then, we find
$\lim_{x \rightarrow 0} \dfrac{\ln(x + 4) - \ln(4)}{x} = \frac{1}{4}\ln\lim_{x \rightarrow 0}\Big(\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Problem based on sum of reciprocal of $n^{th}$ roots of unity
Let $1,x_{1},x_{2},x_{3},\ldots,x_{n-1}$ be the $\bf{n^{th}}$ roots of unity. Find: $$\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+......+\frac{1}{1-x_{n-1}}$$
$\bf{My\; Try::}$ Given $x=(1)^{\frac{1}{n}}\Rightarrow x^n=1\Rightarrow x^n-1=0$
Now Put $\displaystyle ... | Otherwise you could write $x_i=e^{j\frac {i2\pi} {n}}$ ($j^2=-1$)
Then
$$y_i=\frac {e^{-j\frac {i\pi} {n}}} {e^{-j\frac {i\pi} {n}}-e^{j\frac {i\pi} {n}}}=\frac {e^{-j\frac {i\pi} {n}}}{-2j\sin(\frac {i\pi} {n})}=2j\cot(\frac {i\pi} {n})+\frac 1 2$$
Now consider the fact that, if $x_i$ is an $n^{th}$ root of unity, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Evaluate the limit $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$
Evaluate the limit $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$, without using a ... | From the Euler-Maclaurin Summation Formula, we have
$$\begin{align}
\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{k}}&=\frac{1}{\sqrt{n}}\left(1+\int_1^n \frac{1}{\sqrt{x}}\,dx+\frac12\left(\frac{1}{n^{1/2}}-1\right)+O(1)\right)\\\\
&=2+O\left(\frac{1}{n^{1/2}}\right)\\\\
&\to 2\,\,\text{as}\,\,n\to \infty
\end{align}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
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Suppose $-a\sin(s) - b\cos(s) = 0$, then $a^2 + b^2 = 1$?
Suppose $-a\sin(s) - b\cos(s) = 0$ and given that $a^2 + b^2 \le 1$, then $a^2 + b^2 = 1$?
I am having trouble getting the above identity. I vaguely recall that
$$a\sin(s) + b\cos(s)=\sqrt{a^2+b^2}\sin(s+\theta)$$
where $$\tan(\theta) = \frac{b}{a}$$
But I st... | It makes no sense for a mathematical claim to have a condition of the form "$-P-Q = 0$", so maybe the intended claim is:
Given reals $a,b,t$ such that $1 - a \sin(t) - b \cos(t) = 0$ and $a^2+b^2 \le 1$, then $a^2+b^2 = 1$.
And indeed your method proves this.
For fun, here is an alternative proof that uses weighted A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Write the general term of the periodic sequence $1$, $-1$, $-1$, $1$, $-1$, $-1$, $1$, ..., as $(-1)^{g(n)}$ or other closed form How to put mathematically sequence that changes sign like:
$n = 0\quad f = 1$
$n = 1 \quad f = -1$
$n = 2 \quad f = -1$
$n = 3 \quad f = 1$
$n = 4 \quad f = -1$
$n = 5 \quad f = -1$
$n = 6 \... | \begin{align}
a_1 &= 1 \\
a_2 &= -1 \\
a_n &= a_{n-1}a_{n-2}
\end{align}
Using a product recurrence relation, we have
$$
a_n = a_2^{F_{n-1}}a_1^{F_{n-2}}
$$
Where $F_k$ is the $k$'th Fibonacci number. Given a closed form for $F_k$, we can simplify $a_n$.
$$
F_k = \frac{(1 + \sqrt{5})^k - (1 - \sqrt{5})^k}{2^k\sqrt{5}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 1
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Proof of an inequality involving three numbers $(a,b,c)\gt 0$ If $(a,b,c)\gt 0$
the following inequality holds:
$$\dfrac{a^2}{2a^2+(b+c)^2}+\dfrac{b^2}{2b^2+(c+a)^2}+\dfrac{c^2}{2c^2+(a+b)^2}\lt\dfrac{2}{3}$$
I am stuck to find a proof of it.
Can someone help me? Thanks.
| WLOG, assume that $a\ge b \ge c$.
We first show that
$$\dfrac{b^2}{2b^2+(c+a)^2}+\dfrac{c^2}{2c^2+(a+b)^2} \le \frac{(b+c)^2}{2(b+c)^2+a^2}.\qquad (1)$$
Indeed, it is equivalent to each of the following inequalities:
$$\dfrac{c^2}{2c^2+(a+b)^2} \le \frac{(b+c)^2}{2(b+c)^2+a^2} - \dfrac{b^2}{2b^2+(c+a)^2}$$
$$\dfrac{c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\sqrt{1+x}$ can be represented by a power series I need to show that $\sqrt{1+x}$ can be represented as a power series. I need to prove the equality between the function and its Taylor series, not to prove that the Taylor series of the function, $\sum\limits_{n=0}^{\infty}{\frac{1}{2}\choose n}x^n$ converges in ... | Start by simplifying the expression for $R_n(x)$. The large fraction becomes, after taking absolute values:
$$
\frac{1}{(n+1)!} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdots \frac{2n - 3}{2} = \frac{(2n - 3)(2n - 5) \cdots (3)(1)}{2^{n+1}(n+1)!}
$$
Observe that the numerator is bounded above by $(n + 1)!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that $\{a_n\}$ defined by $a_{n+1}=\frac{a_n+2}{a_n+1}$ converges Suppose $a_0$ is an arbitrary positive real number. Define the sequence $\{a_n\}$ by $$a_{n+1}=\frac{a_n+2}{a_n+1}$$ for all $n\geq0$. I have to prove that $\{a_n\}$ converges.
My attempt: If $a=\lim_{n\to\infty}{a_n}$ exists, then it should be a s... | Note: not obvious, to get the matrix for a composition of Möbius transformations you multiply the matrices, in the same order.
The problem is largely about the Pell equations $ p^2 - 2 q^2 = \pm 1.$ You have an integer matrix
$$
M =
\left(
\begin{array}{rr}
1 & 2 \\
1 & 1
\end{array}
\right)
$$ so
$$
M^2 =
\left(
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Does $8a+5$ ever divide $b^2+8$? For natural $a,b$, does $8a+5$ ever divide $b^2+8$ ?
It doesn't for $b$ up to $10^7$.
Couldn't find congruence obstructions for moduli up to $500$.
$b^2+8$ can be even.
| You say you couldn't find a congruence obstruction for moduli up to $500$, but there is such an obstruction to having $-8 \equiv b^2 \bmod 8a+5$, revealed by Jacobi reciprocity. Since $-8$ is a unit modulo $8a+5$, so would be $b$, and therefore using Jacobi symbols we would get $(\frac{-8}{8a+5}) = (\frac{b^2}{8a+5}) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
The expansion of $(a+b+c+d)^{20}$ Let us consider the expansion of $$(a+b+c+d)^{20}.$$
Find:
*
*The coefficients of $a^{11}b^6c^2d$ and $a^{11}b^9$,
*The total number of terms of this expansion,
*The sum of all the coefficients.
Thank you for your help.
| Observe that each term in the expansion of $$(a+b+c+d)^{20}=(a+b+c+d)(a+b+c+d) \cdots (a+b+c+d)$$ is obtained by taking one of the four terms $a,b, c$ or $d$ from each of the $20$ factors. Hence, the coefficient of $a^{11}b^6 c^2 d$ is the number of ways to choose $11$ of the $20$ factors for $a$ (which is ${20 \choos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How would you find the exact roots of $y=x^3+x^2-2x-1$? My friend asked me what the roots of $y=x^3+x^2-2x-1$ was.
I didn't really know and when I graphed it, it had no integer solutions. So I asked him what the answer was, and he said that the $3$ roots were $2\cos\left(\frac {2\pi}{7}\right), 2\cos\left(\frac {4\pi}{... | Consider the equation $$\cos4\theta=\cos3\theta$$ whose roots are $$\theta=n\cdot\frac{2\pi}{7}$$
Representing this as a polynomial in $c=\cos\theta$, we have $$8c^4-4c^3-8c^2+3c+1=0$$
$$\Rightarrow (c-1)(8c^3+4c^2-4c-1)=0$$
Now write $x=2c$ and we see that the polynomial equation $$x^3+x^2-2x-1=0$$ has roots as stated... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Common tangents to circle $x^2+y^2=\frac{1}{2}$ and parabola $y^2=4x$ I'm having trouble with this. What i do is say $\epsilon: y=mx+b$ is the tangent and it meets the circle at $M_1(x_1,y_1)$, i equate the $y$ of the tangent with the circle: $y=\pm \sqrt{1/2-x^2}$ and then the same with the parabola at $M_2(x_2,y_2)$,... | Let $P = (h,k)$ be a point on the circle $x^2 + y^2 = \dfrac 12$.
$\left( \text{Then}\; h^2 + k^2 = \dfrac 12 \right)$. The origin, $O = (0,0)$ is the center of the circle. So the equation of the line
$\overleftrightarrow{OP}$ is $kx - hy = 0$
A line tangent to the line $kx - hy = 0$ must have an equation of the form $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the square of an integer $a$ is of the form $a^2=3k$, or $a^2=3k+1$, where $k\in \mathbb{Z} $ Here's my attempt to prove this. I'm not sure about it, but i hope i'm correct.
Let $a=λ, λ\in \mathbb{Z}$. Then $a^2=λ^2=3\frac{λ^2}{3}$.
When $λ^2$ is divided by 3 there are three possible remainders $0,1,2$. So
... | I don't understand how you eliminated the $3k+2 $ case. It might be easier just to look at the cases $a=3\lambda, a=3\lambda+1, a=3\lambda+2$ and multiply out. as follows:
When $a$ is divided by $3$, there are three possible remainders over an exact multiple of $3$: $0, 1,$ and $2$. Therefore consider the three cases
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Summation of Binomial Coefficient: $\sum\binom{n+k}{2k} \binom{2k}k \frac{(-1)^l}{k+1}$ I am trying to solve this summation problem .
$$\sum\limits_{k = 0}^\infty {\left( {\begin{array}{*{20}{l}}
{n + k}\\
{2k}
\end{array}} \right)} \left( {\begin{array}{*{20}{l}}
{2k}\\
k
\end{array}} \right)\frac{{{{( - 1)}^k}}}{{k ... | Suppose we seek to evaluate
$$\sum_{k=0}^n {n+k\choose 2k} {2k\choose k}\frac{(-1)^k}{k+1}
= \sum_{k=0}^n {n+k\choose n-k} {2k\choose k}\frac{(-1)^k}{k+1}.$$
Introduce
$${n+k\choose n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (1+z)^{n+k}\; dz.$$
This vanishes when $k\gt n$ and we may extend the sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Characterize the elements of a set of polynomials Given $$M = \{x^3 + ax^2 + bx + c \in\mathbb Z_3[x]\}$$ and given $$A=\{f \in M \mid\overline{1} \text{ is a root of }\ f\}.$$
The exercise asks to characterize (tell) what are the elements of $A$ and how many of them there are. My first attempt at solving this exercise... | You want all polynomials $x^3+ax^2+bx+c$ such that $1+a+b+c=0\bmod3\ \ (*)$. Pick any $a$ and any $b$, then $(*)$ holds iff $c=2-a-b\bmod3$. So we have a total of 9 solutions. If you want them explicitly:
$x^3+2,x^3+x+1,x^3+2x$,
$x^3+x^2+1,x^3+x^2+x,x^3+x^2+2x+2$,
$x^3+2x^2,x^3+2x^2+x+2,x^3+2x^2+2x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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how to solve $x^{113}\equiv 2 \pmod{143}$ I need to solve $x^{113} \equiv 2 \pmod{143}$
$$143 = 13 \times 11$$
I know that it equals to $x^{113}\equiv 2 \pmod{13}$ and $x^{113}\equiv 2 \pmod{11}$
By Fermat I got
1) $x^{5} \equiv 2 \pmod{13}$
2) $x^{3} \equiv 2 \pmod{11}$
Now I'm stuck..
| By the Chinese remainder theorem, you have to solve first
$$\begin{cases}x^{113}\equiv2\mod13\\x^{113}\equiv2\mod11\end{cases}$$
Now Little Fermat says for any $x\not\equiv 0\mod 13\enspace(\text{resp. }11)$, one has $x^{12}\equiv 1\mod 13$, resp. $x^{10}\equiv 1\mod 11$. Hence the system of congruences is equivalent t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How to obtain this factorization of $x^4+4$? $x^4 + 4 = (x^2 + 2x +2)(x^2 - 2x +2)$
I am curious how would one obtain this factorization?
Clearly, once the factorization is known it is routine to verify it, however the hard part is how to find the factorization in the first place?
Thanks!
I observed that $(A+B)(A-B)=A... | This is pretty straight forward if you employ complex numbers and this formula: $$(a+b)(a-b)=a^2-b^2$$
Here we go:
\begin{align}
x^4 + 4 & = (x^2)^2 - (-4) \\
& = (x^2)^2 - (2i)^2 \\
& = (x^2 + 2i) \ (x^2 - 2i) \\
& = (x+ i\sqrt{2i}) \ (x- i\sqrt{2i}) \ (x+ \sqrt{2i}) \ (x- \sqrt{2i})
\end{align}
We know that $\sqrt{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Finding the action of T on a general polynomial given a basis I am given a question as follows:
Suppose $T: P_{2} \rightarrow M_{2,2}$ is a linear transformation whose action on a basis for $P_{2}$ is
$$T(2x^2+2x+2)=\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix} \\T(4x^2+2x+2)=\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix} \... |
Hint:
Let $p_1 = 2x^2 + 2x + 2$, $p_2 = 4x^2 + 2x + 2$, $p_3 = 1$. Then
$$
-\frac{1}{2}p_1 + \frac{1}{2}p_2 = x^2, \\
p_1 -\frac{1}{2}p_2 - p_3 = x, \\
p_3 = 1.
$$
Thus
$$
ax^2 + bx + c = \left(-\frac{a}{2} + b\right)p_1 + \frac{a-b}{2}p_2 + (c - b)p_3
$$
and as $T$ is a linear operator we get
$$
T(ax^2 + bx + c) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluation of Irrational Integral
Evaluation of $$\int\frac{x^4}{(1-x^4)^{\frac{3}{2}}}dx$$
$\bf{My\; Try::}$ Let $$I = \int\frac{x^4}{(1-x^4)^{\frac{3}{2}}}dx = -\frac{1}{4}\int x\cdot \frac{-4x^3}{(1-x^{4})^{\frac{3}{2}}}dx$$
Using Integration by parts, We get
$$I =\frac{x}{2(1-x^4)^{\frac{1}{2}}}-\int\frac{1}{(1-x... | An elliptic integral (of the first kind) is one of the form
$$
F(t\mid k^2)=\int\frac{1}{\sqrt{1-k^2\sin^2t}}\,dt
$$
or by substituting $x=\sin t$
$$
G(x\mid k^2)=\int\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx
$$
In our case, the integral
$$
\int\frac{1}{\sqrt{1-x^4}}\,dx
$$
is $G(x\mid-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$then.. If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$ then find the value of $f(0) + f'(0) + f''(0)$.
I tried differentiating the given. But it is getting too long and complicated. So there must be a way... | Try to simplify the denominator of f(x):
Note that by the identity $\cos(2t)=2\cos^2t-1$
$$
\cos(3t)+6\cos(2t)+15\cos t+10=4\cos^3t+12\cos^2t+12\cos t+4=4(\cos t+1)^3
$$
so
$$
\cos(6x)+6\cos(4x)+15\cos(2x)+10=4(\cos(2x)+1)^3
$$
Therefore
$$
f(x)=\frac{1}{4}\cdot\overbrace{\frac{\cos x}{(\cos(2x)+1)^3}}^{(1)}+\frac{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
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Prove $\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0$ without $\varepsilon - \delta$. Unlike Multivariable Delta Epsilon Proof $\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}$ --- looking for a hint I would like to avoid the $\varepsilon - \delta$ criterium.
Prove $$\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0 ... | Hint 1: $$\frac{x^2 y^3}{x^4+y^4}=\frac{y}{(x/y)^2+(y/x)^2}. $$
Hint 2: $f(u)=u^2+u^{-2}$ has an absolute minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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If $x= m-m^2-2$ then find $x^4+3x^3+2x^2-11x+6$ where m is a cube root of unity If $$x= m-m^2-2$$ then find $$x^4+3x^3+2x^2-11x+6$$ where $m$ is a cube root of unity.
My try:
Since $ m+ m^2+1=0$ the value of $x$ is $-1$.
Let $f(x)=x^4+3x^3+2x^2-11x+6$
then $ f(-1)=5$ So the answer is $5$.
Am I correct? In my book it sa... | $$x=m-m^2-2=-(m+m^2+1)+2m-1=2m-1$$
And $2m-1 \neq1$, here is your mistake. Then we can continue our calculations :
$$f(x)=x^4+3x^3+2x^2-11x+6\\=16m^4-32m^3+24m^2-8m+1+24 m^3-36 m^2+18 m-3+8 m^2-8 m+2-22m+11+6\\=16 m^4-8 m^3-4 m^2-20 m+17\\=16m^2(m^2+m+1)-24m^3-20m^2-20m+17\\=-24m^3-20m^2-20m+17=-24m(m^2+m+1)+4m^2+4m+17... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the maximum value of $n$ What's the maximum value for $n$ for which there is a set of distinct positive integers $k_{1}, k_{2}, ... , k_{n}$ for which $k_{1}^{2} + k_{2}^{2} + ... + k_{n}^{2} = 2002$
| $1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+14^2+16^2+18^2+24^2=2002$ which uses 16 distinct squares.
$1^2+\dots+18^2=2109>2002$, so any 18 distinct squares have too large a sum. So the only question is whether it can be done with 17.
Note that $1^2+2^2+\dots+17^2=1785=2002-217$ and $18^2-10^2=224>217$, so we m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For which $a$ and $b$ is this matrix diagonalizable? For which $a$ and $b$ is this matrix diagonalizable?
$$A=\begin{pmatrix} a & 0 & b \\ 0 & b & 0 \\ b & 0 & a \end{pmatrix}$$
How to get those $a$ and $b$? I calculated eigenvalues and eigenvectors, but don't know what to do next?
| Calculating the eigenvalues, $b, b+a, a-b,$ one can then easily calculate their respective eigenvectors. For eigenvalue $b,$
\begin{equation}
\begin{pmatrix} a-b&0&b\\0&0&0\\b&0&a-b \end{pmatrix}\begin{pmatrix} 0\\1\\0\end{pmatrix} = \begin{pmatrix} 0\\0\\0\end{pmatrix},
\end{equation}
There may be special cases, such... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Roots of $y=x^3+x^2-6x-7$ I'm wondering if there is a mathematical way of finding the roots of $y=x^3+x^2-6x-7$?
Supposedly, the roots are $2\cos\left(\frac {4\pi}{19}\right)+2\cos\left(\frac {6\pi}{19}\right)+2\cos\left(\frac {10\pi}{19}\right)$, $2\cos\left(\frac {2\pi}{19}\right)+2\cos\left(\frac {14\pi}{19}\right)+... | We may take $\xi$ as a primitive $19$-th root of unity, $\xi=\exp\left(\frac{2\pi i}{19}\right)$, then check that the elementary symmetric functions of
$$ a = \xi^2+\xi^3+\xi^5+\xi^{14}+\xi^{16}+\xi^{17} $$
$$ b = \xi^4+\xi^6+\xi^9+\xi^{10}+\xi^{13}+\xi^{15} $$
$$ c = \xi^1 + \xi^7 + \xi^8+\xi^{11}+\xi^{12}+\xi^{18} $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $h$ if $x^2 + y^2 = h$
Consider equation $x^2 + y^2 = h$ that touches the line $y=3x+2$ at some point $P$. Find the value of $h$
I know that $x^2 + y^2 = h$ is a circle with radius $\sqrt{h}$.
Also, since $y = 3x + 2 $ is a tangent, we know that the slope of the radius perpendicular to the tangent i... | So we have equation $x^2 + y^2 = r^2$, who’s geometrical representation would be a circle with the radius $r$. Now we also have an equation of a line represented by: $y = mx + c$ that touches the circle at point $P$. To find this point we will use the method of solving simultaneous equation where one is a quadratic and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 10,
"answer_id": 1
} |
Proving Gerretsen's Inequality Today in class we were shown Gerretsen's inequality:
$$16Rr-5r^2\leq s^2 \leq 4R^2+4Rr+3r^2$$
Where $R$, $r$, and $s$ are the circumradius, in radius, and semiperimeter of a triangle. After some research, I found that this inequality was proven by showing the following:$$IH^2=4R^2+4Rr+3r... | Your question is very interesting. Let me give my idea.
Firstly, we will compute $IG$. Note that we have $$\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}.$$ Then, $$3\vec{IG} = \vec{IA} + \vec{IB} + \vec{IC}.$$
$$9IG^2 = \sum_{\circlearrowleft} IA^2 + 2\sum_{\circlearrowleft}\vec{IA}\vec{IB} $$
Recall that $IA^2 = r^2 + (s-a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Expressing Tornheim sums in terms of Riemann's Zeta If
$$T(a,b,c)=\sum_{r\geq1}\sum_{s\geq1} \frac{1}{r^as^b(r+s)^c}$$
How to prove that :
$$T(3,1,2)=-\frac13 \zeta(6)+\frac{\zeta^2(3)}{2}$$
I tried some algebraic manipulations but did not work. Can you please help me ?
Any solution will be appreciated
| The easiest way to do it is to use partial fraction decomposition. Let us choose the variable for the decomposition to be $s$. Then we have:
\begin{equation}
\frac{1}{r^a s^b (s+r)^c} =
\sum\limits_{l_1=1}^b \binom{b+c-1-l_1}{c-1} \frac{(-1)^{b-l_1}}{s^{l_1} r^{b+b+c-l_1}} +
\sum\limits_{l_1-1}^c \binom{b+c-1-l_1}{b-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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finding the angle between two vectors If $a,b$ and $c$ be three vectors such that $|\, a \,|=3$, $|\, b \,|=5$ and $|\, c \,|=7$ and $a+b+c=0$. Then find the angle between $a$ and $b$.
I tried by taking $a=-(b+c)$ and $b=-(c+a)$. But couldn't proceed further. Please help.
| \begin{align*}
\mathbf{a+b+c} &=\mathbf{0} \\
\mathbf{c} &=-(\mathbf{a+b}) \\
\mathbf{c}\cdot \mathbf{c} &= (\mathbf{a+b})\cdot (\mathbf{a+b}) \\
c^2 &= a^2+b^2+2\mathbf{a\cdot b} \\
2\mathbf{a\cdot b} &= c^2-a^2-b^2 \\
\cos \theta_{ab} &= \frac{c^2-a^2-b^2}{2ab} \\
&= \frac{7^2-3^2-5^2}{2(3)(5)} \\
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\theta$ I''m stuck in a exercise in complex analysis concerning integration of rational trigonometric functions. I have the solution but I don't understand a specific part. Here it goes:
We want to find $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\t... | $$\begin{align*} 3z^2 + 10z + 3
&= 3 \left(z^2 + \frac{10}{3}z + 1\right) \\
&= 3\left(z^2 + 3z + \frac{1}{3}z + 1\right) \\
&= 3\left(z(z + 3) + \frac{1}{3}(z + 3)\right) \\
&= 3\left(z + \frac{1}{3}\right)(z+3). \end{align*}$$
You are missing a factor of $2$ when you go from step $3$ to step $4$. The denominator s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Smallest number whose $\sin(x)$ in radian and degrees is equal Question:
What is the smallest positive real number $x$ with the property that the sine of $x$ degrees is equal to the sine of $x$ radians?
My try: 0. But zero isn't a positive number. How do I even begin to solve it? I tried taking inverse on both sides... | We have
$$\sin (x) = \sin (\beta x)$$
where $\beta := \frac{\pi}{180}$. Using
$$\sin (\alpha x) = \frac{e^{i \alpha x} - e^{-i \alpha x}}{2i}$$
we conclude that the equation $\sin (x) = \sin (\beta x)$ can be rewritten as follows
$$2 \, \sin \left( \left(\frac{1-\beta}{2}\right) x\right) \, \cos \left( \left(\frac{1+\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Finding Laurent and Taylor series I need to find both a Laurent and a Taylor expansion.
$$f(z)=\frac{z}{(z-1)(z-2)} = \frac{-1}{(z-1)}+\frac{2}{(z-2)}$$
If I choose $z_0=0$
$$f(z)=\frac{1}{(1 + z)} - \frac{4}{\left(1 - \frac{z}{4}\right)}$$
$$f(z)=\sum_{n}^{\infty}(-1)^n{z^n} - 4\sum_{n}^{\infty}(\frac{z}{4})^n$$
Whi... | It is sufficient to consider a Laurent expansion around $z=0$. Depending on the region which you then choose you obtain a principal part of a Laurent expansion respectively an expansion as Taylor series.
The function
\begin{align*}
f(z)&= \frac{-1}{z-1}+\frac{2}{z-2}\\
\end{align*}
has two simple poles at $1$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Sum and product of analytic functions that is not analytic The function
$$f(x) = \frac{2 + \cos x}{3} (2π - x) + \sin x$$
is the sum/product of analytic functions ($\cos(x)$,$\sin(x)$, linear), but all it's derivatives at $2\pi$ are $0$ ($f^n(2\pi)=0$).
I simply don't understand why.
Thanks,
| Set $x-2\pi=t$, so $x=t+2\pi$ and $\cos x=\cos t$, $\sin x=\sin t$. Thus you can consider
$$
g(t)=-\frac{2}{3}t-\frac{1}{3}t\cos t+\sin t
$$
and the Taylor coefficients at $0$ of $g$ are the same as the Taylor coefficients at $2\pi$ of $f$. We have
\begin{align}
g(t)&=-\frac{2}{3}t-\frac{1}{3}t\left(1-\frac{1}{2}t^2+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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solve $x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$ Help with this excercise.. :)
$$x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$$
the book says it is an exact differential equation, but how?
$$x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$$
$$x(x^2+y^2)^{-1/2}dx+y(x^2+2y^2)dy=0$$
$M=x(x^2+y^2)^{-1/2}$
$N=y(x^2+2y^2)$
$$\frac{\partial ... | Hint:
$x(x^2+y^2)^{-1/2}+yy'(x^2+2y^2)=0$
$y\dfrac{dy}{dx}(x^2+2y^2)=-\dfrac{x}{\sqrt{x^2+y^2}}$
$\dfrac{dx}{dy}=-\dfrac{y(x^2+2y^2)\sqrt{x^2+y^2}}{x}$
Let $t=y^2$ ,
Then $\dfrac{dx}{dy}=\dfrac{dx}{dt}\dfrac{dt}{dy}=2y\dfrac{dx}{dt}$
$\therefore2y\dfrac{dx}{dt}=-\dfrac{y(x^2+2y^2)\sqrt{x^2+y^2}}{x}$
$\dfrac{dx}{dt}=-\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Attempt:
$$\gcd(4n-5,60-12n)=(4n-5,-8n+55)=(4n-5,-4n+50)=(4n-5,45)$$
$$n=1: (4-5,45)=1\quad \checkmark\\
n=2: (3,45)=3\quad \times\\
n=3: (7,45)=1\quad \checkmark\\
n=4... |
I want to verify that my solution is correct
Your solution is not correct.
For $m\not=1$,
$$\frac{4\times\color{red}{5m}-5}{60-12\times\color{red}{5m}}=\frac{5(4m-1)}{5(12-12m)}$$
can be reduced.
If $n$ is of the form $3k+2$, then
$$\frac{4(3k+2)-5}{60-12(3k+2)}=\frac{3(4k+1)}{3(-12k+12)}$$
can be reduced.
If $n$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Proving that $1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } $ by induction
Prove that
$$1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 }. $$
I can get to $1/3(k+1)(k+2) + (k+1)(k+2)$ but then finishing off an... | Let $ 2 = \frac{1 \cdot 2 \cdot 3}{3} $ be the basis. The inductive step consists of simple distribution of multiplication:
$$
\begin{align}&\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) \\&= \tfrac{1}{3}n^3+2n^2+\tfrac{11}{3}n+2 \\&= \frac{(n + 1)\Big((n+1)+1\Big)\Big((n + 1)+2\Big)}{3}\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Maximize $k=x^2+y^2$ Subject to $x^2-4x+y^2+3=0$ Question
Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. Find the maximum and minimum values of $x^2+y^2$.
My work
Let $k=x^2+y^2$
Therefore, $x^2-4x+y^2+3=0$ ---> $k-4x+3=0$ .
What do I do next? How do I find an expression in terms of $k$ tha... | Although this kind of problem is usually solved using Lagrange multipliers, this one can be solved using methods from precalculus.
We wish to maximize the value of $x^2+y^2$ on the circle
\begin{equation}
(x-2)^2+y^2=1
\end{equation}
Which is defined on the interval $[1,3]$.
Since $x^2+y^2=4x-3$ the maximum and minimum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simplify this:$\sqrt{5\sqrt{3}+6\sqrt{2}}$? How to simplyfy this:$\sqrt{5\sqrt{3}+6\sqrt{2}}$.
I know I should use nested radicals formula but which one is $A$ and $B$.Using the fact $A>B^2$ you can find $A$ and $B$.
But $C^2=A-B^2$ isn't a rational number then we have again a nested radical.
What to do?
| $$\begin{align}
\sqrt{5\sqrt{3}+6\sqrt{2}}
&= \sqrt{5\sqrt{3}+\left(2\sqrt{6}\right)\sqrt{3}} \\
&= \sqrt[4]{3}\cdot\sqrt{5+2\sqrt{6}} \\
&= \sqrt[4]{3}\cdot\sqrt{2+2\sqrt{6} + 3} \\
&= \sqrt[4]{3}\cdot\sqrt{\frac{4+4\sqrt{6}+\left(\sqrt{6}\right)^2}{2}} \\
&= \sqrt[4]{3}\cdot\sqrt{\frac{\left(2+\sqrt{6}\right)^2}{2}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove $\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$ This series represents the sum of reciprocal pentagonal numbers (multiplied by $3$). I got its value from Wolfram Alpha:
$$\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$$
My attempt: Turn the series into an integral:
$$\sum_{n=1}^\infty \frac{x... | We can also continue your way. We have $$-\int\frac{\log\left(1-x^{3}\right)}{x^{2}}dx=\frac{\log\left(1-x^{3}\right)}{x}-\int\frac{3x}{1-x^{3}}dx
$$ $$=\frac{\log\left(1-x^{3}\right)}{x}-\int\frac{1-x}{x^{2}+x+1}dx-\int\frac{1}{x-1}dx
$$ $$=\frac{\log\left(1-x^{3}\right)}{x}-\log\left(x-1\right)-\frac{1}{2}\int\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Showing that $2^6$ divides $3^{2264}-3^{104}$
Show that $3^{2264}-3^{104}$ is divisible by $2^6$.
My attempt: Let $n=2263$. Since $a^{\phi(n)}\equiv 1 \pmod n$ and
$$\phi(n)=(31-1)(73-1)=2264 -104$$
we conclude that $3^{2264}-3^{104}$ is divisible by $2263$.
I have no idea how to show divisibility by $2^6$.
| Another approach:
$$\text{Euler's formula: }\quad a^{\phi(n)}\equiv 1 \pmod{n} \text{ when} \gcd(a,n)=1$$
$\implies a^{32}\equiv 1 \pmod{2^6}$
We want $3^{2264}-3^{104}\equiv 0 \pmod{2^6}$
$$\phi(2^6)=2^5=32$$
$$ 3^{2264}-3^{104}=3^{2240}3^{24}-3^{96}3^{8}=(3^{70})^{32}3^{24}-(3^3)^{32}3^8\equiv 3^{24}-3^8\equiv 17^6-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 4
} |
Find the highest point of intersection
Find the highest point of intersection of the sphere $x^2+y^2+z^2=30$ and the cone $x^2+2y^2-z^2=0$.
Am I supposed to use the Lagrange multiplier for this?
EDIT:
So this is what I've tried...
$z^2=-x^2-y^2+30$ and $z^2=x^2+2y^2$. Then $-x^2-y^2+30 = x^2+2y^2$.
This gives us $3y... | By using Lagrange multipliers, we need to solve system $$F'_x=0,$$ $$F'_y=0,$$ $$F'_z=0,$$ where is $F(x,y,z,\alpha,\beta)=z+\alpha(x^2+y^2+z^2-30)+\beta(x^2+2y^2-z^2)$,
with conditions $$x^2+y^2+z^2-30=0,$$ $$x^2+2y^2-z^2=0.$$
Stationary points are ($\pm \sqrt{15}$,$0$,$\pm \sqrt{15}$) and ($0$,$\pm\sqrt{10}$, $\pm\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the probability that $1984!$ is divisible by $n$
Let $a,b,c,d$ be a permutation of the numbers $1,9,8,4$ and let $n = (10a+b)^{10c+d}$. Find the probability that $1984!$ is divisible by $n$.
I was told this could be solved by casework on $a$ and using Fermat's Little Theorem. For example, if $a = 1$, there are $... | The following PARI/GP program determines the permutations for which
$1984!$ divides the given number :
? q=0;x=[1,9,8,4];for(j=1,24,p=numtoperm(4,j);a=x[p[1]];b=x[p[2]];c=x[p[3]];d=x[
p[4]];if(Mod(1984!,(10*a+b)^(10*c+d))==0,q=q+1;print(q," ",a," ",b,"
",c," ",d)))
1 1 9 4 8
2 1 8 9 4
3 ... | {
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"url": "https://math.stackexchange.com/questions/1859250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form of function $f(n) = \frac1n \sum\limits _{k=1}^{n-1} f(k)$? Could anyone help me get to the closed form of the function? Here $\mathbb{N}=\{1,2,3,\ldots\}$.
Find all functions $f:\mathbb{N}\to\mathbb{R}$ such that $f(1)=1$ and
$$f(n) = \frac 1 n \sum _{k = 1}^{n-1}f(k)$$
for every integer $n>1$.
So far, ... | Start by computing the first few terms to gain an intuition into the problem, we can see that $f(2) = \frac{1}{2} f(1) = \frac{1}{2}$. And then, it follows that $f(3) = \frac{1}{3} \left(1 + \frac{1}{2}\right) = \frac{1}{2} $, etc...
Claim: $f(n) = \frac{1}{2}$ for all $n \geq 2$.
Proof: Assume the above holds, then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show : $(-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \right)}=\tfrac{(-1)^{n}}{n}+\mathcal{O}\left(\tfrac{\ln(n)}{n^{2}} \right)$
I would like to show that :
$$(-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)}=\dfrac{(-1)^{n}}{n}+\mathcal{O}\left(\dfrac{\ln(n)}{n^{2}} \right)$$
My proof:
Note that :... | First, we note that $\tan\left(\frac{\pi}{4}+\frac1n\right)$ can be expanded as
$$\begin{align}
\tan\left(\frac{\pi}{4}+\frac1n\right)&=\frac{1+\tan(1/n)}{1-\tan(1/n)}\\\\
&=1+\frac2n +\frac2{n^2}+O\left(\frac{1}{n^3}\right)
\end{align}$$
Then,
$$\begin{align}
e^{-\tan\left(\frac{\pi}{4}+\frac1n\right)\log(n)}&=e^{-\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Calculate limit for, $\lim\limits_{x\to 0}\frac{1-cos(x^6)}{x^{12}}$, but in there have suprize. Let's think about this function, $\quad \to f(x)=\dfrac{x^2-1}{x-1}$,
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=0/0$ ,
First Solution :
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=\lim\limits_{x\to 1}\dfrac{x+1}{1}$
$=\lim\limits... | Hint:
$$
\begin{align}
\frac{1-\cos\left(x^6\right)}{x^{12}}
&=\frac{1-\cos\left(x^6\right)}{\sin^2\left(x^6\right)}\left(\frac{\sin\left(x^6\right)}{x^6}\right)^2\\
&=\frac{1-\cos\left(x^6\right)}{1-\cos^2\left(x^6\right)}\left(\frac{\sin\left(x^6\right)}{x^6}\right)^2\\
&=\frac1{1+\cos\left(x^6\right)}\left(\frac{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
show that $b,c \in \Bbb N$ for $b=c^{\frac{1}{c-1}}$ $\iff$ c = 2. I am trying to show that $b,c \in \Bbb N$ for $b=c^{\frac{1}{c-1}}$ $\iff$ c = 2. The reverse implication is easy, but the forward implication is tricky. It seems essentially to state the the (n-1)th root of a natural number n is not an integer for n>2.... | Define $f(x) = \dfrac{\ln x}{x-1}$. Then, $f'(x) = -\dfrac{1}{(x-1)^2}\ln x + \dfrac{1}{x-1} \cdot \dfrac{1}{x} = -\dfrac{x(\ln x - 1)+1}{x(x-1)^2}$.
For $x \ge 3$, we have $f'(x) < 0$, so $f$ is decreasing on $[3,\infty)$.
Hence, for all $c \ge 3$, we have $f(c) \le f(3)$, i.e. $\dfrac{\ln c}{c-1} \le \dfrac{\ln 3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ How can I prove the inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ ?
The derivative of $f(x):=\sqrt{\cos x}-\cos(\sin x)$ is very unpleasant, so the standard method is probably not be the right choice...
| Both sides being positive, we can raise each to the fourth power, to get
$$ \cos^2 x > \cos^4 \sin x $$
and then (as suggested by @Bacon), change variable to $u=\sin x$, which gives us
$$ 1-u^2 > \cos^4 u, \qquad u\in(0,1/\sqrt 2) $$
By Taylor's theorem $\cos u = 1 - \frac12 u^2 + \frac{\cos \xi}{4!}u^4$ for some $\xi\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Rewriting product to a binomial I'm currently researching Wigner matrices. I wanted to calculate the moments of its spectral density. The probability density is
$$\frac{1}{2\pi} \sqrt{4-x^2} \text{ for } x \in [-2,2] $$
I have found an expression for the $k^{th}$ moment by integrating
$$ \int_{-2}^2 \frac{x^k}{2\pi} \... | Since we have to consider $k$ even, we set $k=2l$ and show
The following is valid for $l\geq 1$:
\begin{align*}
\frac{1}{2l+1}\prod_{i=1}^l\frac{8l-8i+12}{2l-2i+4}=\frac{1}{l+1}\binom{2l}{l}
\end{align*}
with $\frac{1}{l+1}\binom{2l}{l}$ the Catalan numbers.
We obtain
\begin{align*}
\frac{1}{2l+1}\prod_{i=1}^l\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proof about Pythagorean triples Show that if $(x,y,z)$ is a Pythagorean triple, then $10\mid xyz$
Proof
First, if $x$, $y$, $z$ are all odd, then so are $x^2$, $y^2$, $z^2$, so $x^2+y^2$ is even, which means that $x^2+y^2 \neq z^2 $. Hence, at least one of $x$, $y$, $z$ is even, so $2\mid xyz$ (clear).
Next, for any... | No problem with divisibility by $2$, good.
Suppose $5\nmid xyz$, so $5$ divides none of the numbers $x$, $y$ and $z$.
Then $x^2\equiv\pm1\pmod{5}$ and the same for $y$ and $z$. However, it can't be $x^2\equiv1$ and $y^2\equiv-1$, otherwise $z^2\equiv0\pmod{5}$. Similarly it's impossible that $x^2\equiv-1$ and $y^2\equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
nature of the series $\sum (-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \right)}$ I would like to study the nature of the following serie:
$$\sum_{n\geq 0}\ (-1)^{n}n^{-\tan\left(\dfrac{\pi}{4}+\dfrac{1}{n} \right)} $$
we can use simply this question :
Show : $(-1)^{n}n^{-\tan\left(\tfrac{\pi}{4}+\tfrac{1}{n} \rig... | Without digging into asymptotics, the original series is simply convergent by Dirichlet's test, since $\{(-1)^n\}$ has bounded partial sums and
$$ n^{-\tan\left(\frac{\pi}{4}+\frac{1}{n}\right)} = \frac{1}{n^{\frac{1+\tan(1/n)}{1-\tan(1/n)}}}$$
is decreasing towards zero for any $n\geq 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find $ \tan \left(\frac{x}{2}\right) $ knowing that $\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $ Good evening to everyone. I don't know how to find $ \tan \left(\frac{x}{2}\right) $ knowing that $$\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $$ and x$\in (0,\frac{\pi}{3})$ Here's what I've tried... | Another approach:
$$\sqrt{2}\cos\left(x + \frac{\pi}{4}\right) = \cos(x) + \sin(x)$$
Thus,
$$\cos(x) + \sin(x) = \frac{7}{5} \rightarrow \sqrt{2}\cos\left(x + \frac{\pi}{4}\right) = \frac{7}{5}$$
And hence
$$x + \frac{\pi}{4} = \arccos\left(\frac{7}{5\sqrt{2}}\right) \rightarrow x = \arccos\left(\frac{7}{5\sqrt{2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
At which points is the following function continuous? From Royden and Fitzpatrick's Real Analysis, Fourth Edition (Chapter 1, Problem 48):
Let $f$ be the function defined by
$$
f(x) = \begin{cases}
x & \text{if $x$ is irrational} \\
p \cdot \sin \frac{1}{q} & \text{if $x = \frac{p}{q}$ in lowest terms}
\end{cases}
$$
... | Let us first show that $f$ is continuous at 0.
By construction, $f(0) = 0$.
Now pick any $\epsilon > 0$.
Since $\lim_{n \to \infty} n \sin \dfrac{1}{n} = 1$, there exists an index $N$ such that
$$
n \sin \dfrac{1}{n} < 1+\epsilon
$$
for all $n \geq N$.
Let $\delta = \min\left\{\dfrac{1}{N},\dfrac{\epsilon}{1+\epsilon}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Another way to evaluate $\int\frac{\cos5x+\cos4x}{1-2\cos3x}{dx}$? What I've done is this:$$\int\dfrac{\cos5x+\cos4x}{1-2\cos3x}{dx}$$
$$\int \dfrac{\sin 3x}{\sin 3x}\left[\dfrac{\cos5x+\cos4x}{1-2\cos3x}\right]{dx}$$
$$\dfrac {1}{2}\int\dfrac{\sin 8x -\sin 2x +\sin 7x -\sin x}{\sin 3x - \sin 6x}$$
$$-\dfrac {1}{2}\int... |
Is there any other way to evaluate it ?
Hint. One may observe that
$$
\frac{\cos(5x)+\cos(4x)}{1-2\cos(3x)}=-\cos(2x)-\cos (x)
$$ then the evaluation is easier.
Edit. Here is a way to obtain such a simplification. One may set $u=e^{ix}$ then using De moivre's formula for $\cos(\cdot)$ one gets
$$
\begin{align}
\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to prove by induction that $3^{3n}+1$ is divisible by $3^n+1$ for $(n=1,2,...)$ So this is what I've tried:
Checked the statement for $n=1$ - it's valid.
Assume that $3^{3n}+1=k(3^n+1)$ where $k$ is a whole number (for some n). Proving for $n+1$: $$3^{3n+3}+1=3^33^{3n}+1=3^3(3^{3n}+1)-26=3^3k(3^n+1)-26=3^3k(3^n+1)-... | As an interesting alternative, note that $x^3 +1 = (x+1)(x^2 - x + 1)$, so setting $x = 3^n$ gives $3^{3n} + 1 = (3^n + 1)(3^{2n} - 3^n + 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$
*
*$\sin x + \sin y = 1$
*$\cos x + \cos y = 0$
Any valid pair of $(x, y)$ is fine, as the restrictions on the board in the image below are obscured.
I got the question from chapter 26 of a comic called Yamada-kun.
How can I solve this equa... | HINT
Squaring both equations you get
$$
\sin^2 x + \sin^2 y + 2\sin x \sin y = 1\\
\cos^2 x + \cos^2 y + 2\cos x \cos y = 0
$$
Now add them together to get
$$
2 + 2 \sin x \sin y + 2 \cos x \cos y = 1
$$
or in other words
$$
\frac{-1}{2} = \cos x \cos y + \sin x \sin y = \cos (x-y)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 2
} |
Compute $1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \cdots + n \cdot \frac {1}{2^n} + \cdots $ I have tried to compute the first few terms to try to find a pattern but I got
$$\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}$$
but I still don't see any obvious pattern... | Let
$$ L = \sum_{n=1}^\infty \frac{n}{2^n} $$
Then,
$$ L = \frac{1}{2} + \sum_{n=2}^\infty \frac{n}{2^n} \\
L = \frac{1}{2} + \sum_{n=1}^\infty \frac{n+1}{2^{n+1}} \\
L = \frac{1}{2} + \frac{1}{2} \sum_{n=1}^\infty \frac{n}{2^n} + \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \\
L = \frac{1}{2} + \frac{L}{2} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Trigonometry Olympiad problem: Evaluate $1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$
Find the value of
$$1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$$
My attempt
I converted the $\sin$ functions which have arguments greater than $90^\circ$ to $\cos$... | As a complement to other answers, you should know that the following identities are valid
$$ \begin{align}
2 \sin \frac{1}{2} x \sum_{k=1}^{n} \cos kx &= +\sin(n+\frac{1}{2})x - \sin\frac{1}{2}x \\
2\sin \frac{1}{2} x \sum_{k=1}^{n} \sin kx &= -\cos(n+\frac{1}{2})x + \cos\frac{1}{2}x
\end{align}$$
Thsese formulas can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Partial fractions and using values not in domain I'm studying partial fraction decomposition of rational expression. In this video the guy decompose this rational expression:
$$ \frac{3x-8}{x^2-4x-5}$$
this becomes:
$$\frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1} $$
$$[(x-5)(x+1)]\times \frac{3x-8}{(x-5)(x+1)... | Observe that the equation
\begin{eqnarray}
\frac{3x-8}{(x+1)(x-5)}=\frac{A}{x-5}+\frac{B}{x+1}
\end{eqnarray}
is valid only for $x\in \mathbb{R}\backslash \{-1,5\}$, as you have rightly pointed out. However, the equation
\begin{eqnarray}
3x-8=A(x+1)+B(x-5)
\end{eqnarray}
is valid for all $x\in \mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Compute trigonometric limit without use of de L'Hospital's rule
$$ \lim_{x\to 0} \frac{(x+c)\sin(x^2)}{1-\cos(x)}, c \in \mathbb{R^+} $$
Using de L'Hospital's rule twice it is possible to show that this limit equals $2c$. However, without the use of de L'Hospital's rule I'm lost with the trigonometric identities.
I c... | Using Taylor series make life "easy". Start with $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ $$\sin(x^2)=x^2-\frac{x^6}{6}+O\left(x^8\right)$$ $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ So,$$\frac{(x+c)\sin(x^2)}{1-\cos(x)}=\frac{(x+c)\left(x^2-\frac{x^6}{6}+O\left(x^8\right) \right)}{\frac{x^2}{2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 6
} |
Evaluation of $\int^{\pi/2}_{0} \frac{x \tan(x)}{\sec(x)+\tan(x)}dx$ Evaluate the given integral:
$$\int^{\pi/2}_0 \frac{x \tan(x)}{\sec(x)+\tan(x)}dx$$
I multiplied and divided by $\sec(x)+\tan(x)$ to get denominator as $1$ but In calculation of integral, $x$ is creating problem. Is there any way to eliminate $x$ her... | There might be an easier solution exploiting symmetry but the one follows is what I came up:
\begin{align*}
\int_{0}^{\pi/2} \frac{x\tan x}{\sec x + \tan x} \, {\rm d}x &= \int_{0}^{\pi/2} \frac{x \tan x}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \, {\rm d}x \\
&= \int_{0}^{\pi/2} \frac{x \cos x \frac{\sin x}{\cos x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Linear programming exercise I want some help for the following exercise
I know that the leaving variable is the basic variable associated with the smallest nonnegative ratio with the strictly positive denominator. I can't understand how the current basic variables ($x_5$, $x_6$, $x_7$, $x_8$) are matched with the eq... | Let $x_r$ and $x_j$ denote the leaving and entering variables respectively. ($r$ stands for "row"; $j$ runs through columns.)
Write the simplex tableaux.
$$
\require{enclose}
% inner horizontal array of arrays
\begin{array}{cc}
% inner array of minimum values
\begin{array}{r|r|r|r}
\text{basis} & a_j & b & b/a_j \\
\h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Convergence/Divergence of $\sum_{n=1}^{\infty}\frac{n+n^2+\cdots+n^n}{n^{n+2}}$
$$\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}$$
$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \su... | Easy with equivalents (this requires the general term to have a constant sign) and some computation:
$$n+n^2+\dots+n^n=\frac{n(n^n-1)}{n-1}\sim_\infty n^n$$
so
$$\frac{n+n^2+\cdots+n^n}{n^{n+2}}\sim_\infty\frac{n^n}{n^{n+2}}=\frac1{n^2},$$
which converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Formula for $1^k+2^k+3^k...n^k$ for $n,k \in \mathbb{N}$ So I've been looking for a formula where I can input the parameter $k$ and it will give me a formula for $1^k+2^k+3^k...+ n^k$ with $n,k \in \mathbb{N}$. The result is always a polynomial with $k+1$ as highest power. I've taken the time to calculate the polynomes... | Another way to derive a formula for $$S(k,n)=\sum_{s=1}^{n}{s^k}$$ is to use the binomial expansion. To do this, you can start by changing the summation index $s$ to $t+1$. That is:$$S(k,n)=\sum_{t=0}^{n-1}{(t+1)^k}=1+\sum_{t=1}^{n-1}{(t+1)^k}$$Using binomial expansion we have:$$S(k,n)=1+\sum_{t=1}^{n-1}{\sum_{j=0}^{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Can this systems of equations be solved? There is a system of equations as follow:
$$\left[\begin{matrix}x\\ y\\ \end{matrix} \right] = \left[\begin{matrix}a& b\\b & c\\\end{matrix} \right]\left[\begin{matrix}\frac{1}{x}\\ \frac{1}{y}\\ \end{matrix} \right]$$
Given $ \left[\begin{matrix}a& b\\b & c\\\end{matrix} \right... | After pie314271's answer
The equation $$b^2(x^2-a)+c(x^2-a)^2=b^2x^2$$ reduces to $$a\left(a c- b^2\right)-2 a c x^2+c x^4=0$$ which is just a quadratic in $x^2$ and the solutions are given by $$x=\pm\sqrt{a\pm\frac{\sqrt{a}}{\sqrt{c}}\,b}$$ So, the whole set of solutions is
$$\left\{x= -\sqrt{a-\frac{\sqrt{a} b}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1881765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Computing $\int \frac{1}{(1+x^3)^3}dx$ I tried various methods like trying to break it into partial fractions after factorizing, applying substitutions but couldn't think of any. How will we integrate this?
| Hint. Another path. Consider a fixed real number $a>0$. From
$$
a^3+x^3=(a+x)(a^2-ax+x^2)
$$ one gets
$$
\begin{align}
\frac{1}{a^3+x^3}&=\frac{1}{3 a^2 (a+x)}+\frac{2 a-x}{3 a^2 \left(a^2-a x+x^2\right)}
\end{align}
$$ integrate the preceding identity to get
$$
\int\frac{dx}{a^3+x^3}=\frac1{3a^2}\ln(a+x)-\frac1{6a^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integration of trigonometric function $\int\frac{\sin(2x)}{\sin(x)-\cos(x)}dx$
$$\int\frac{\sin(2x)}{\sin(x)-\cos(x)}dx$$
My attempt: Firstly, $\sin(2x)=2\sin(x)\cos(x)$.
After that, eliminate the $\cos(x)$ seen in both the numerator and denominator to get
$$2\int\frac{\sin(x)}{\tan(x)-1}\ dx.$$
From here onwards, sh... | Hint:
$$\int\frac{\sin(2x)}{\sin(x)-\cos(x)}dx = \int\frac{2\sin(x)\cos(x) }{\sin(x)-\cos(x)}dx = \int\frac{\sin(x)\cos(x)+\cos(x)\sin(x) }{\sin(x)-\cos(x)}dx\\
=\int\frac{\sin(x)\cos(x)-\sin^2(x)+\cos(x)\sin(x)-\cos^2(x)+1 }{\sin(x)-\cos(x)}dx \\=\int -\sin(x)dx+\int \cos(x)dx + \int \frac{1}{\sin(x) -\cos(x)}dx \\= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Better way to evaluate $\int \frac{dx}{\left (a +b\cos x \right)^2}$ $$\int \frac{dx}{\left (a +b\cos x \right)^2}$$
$$u=\frac{b +a \cos x}{a +b\cos x }$$
$$du=\frac{\sin x\left(b^2 -a^2\right)}{ \left (a +b.\cos x \right)^2}$$
$$\frac{du}{\sin x\left(b^2 -a^2\right)}=\frac{dx}{\left (a +b\cos x \right)^2}$$
$$\cos x=\... | HINT
Put $$A=\frac{\ sin x}{a+b\cos x} $$now differentiate both sides and after simplification integrate it up to get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What is $\sqrt{-4}\sqrt{-9}$? I assumed that since $a^c \cdot b^c = (ab)^{c}$, then something like $\sqrt{-4} \cdot \sqrt{-9}$ would be $\sqrt{-4 \cdot -9} = \sqrt{36} = \pm 6$ but according to Wolfram Alpha, it's $-6$?
| The property $a^c \cdot b^c = (ab)^{c}$ that you mention only holds for integer exponents and nonzero bases. Since $\sqrt{-4} = (-4)^{1/2}$, you cannot use this property here.
Instead, use imaginary numbers to evaluate your expression:
$$
\begin{align*}
\sqrt{-4} \cdot \sqrt{-9} &= (2i)(3i) \\
&= 6i^2 \\
&= \boxed{-6}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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How to solve this determinant equation in a simpler way
Question Statement:-
Solve the following equation
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$$
My Solution:-
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=
\begin{vmatrix}
x+5 & 2 & 3 \\
x+5 & x & 1 ... | Just expand the determinant! We want
$$
5x^2+2x+24-2x-40-3x^2=0
$$
which simplifies to $2x^2-16=0$.
(In general, finding determinants via row/column relations is faster when a matrix is large. But $3 \times 3$ matrices aren't that large yet...)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Integration: $\int\frac{dx}{\sqrt{x(1-x)}}$ My teacher wrote this on the blackboard: $$\int_0^z \frac{1}{\pi \sqrt{x(1-x)}}dx= \frac{2}{\pi} \arcsin(\sqrt{z})$$
But when I try to calculate the integral:
\begin{align}
\int \frac{1}{\pi \sqrt{x(1-x)}}dx&=\int \frac{1}{\pi \sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}}dx\\
&= ... | if you make the substitution $x=u^2$ the integral becomes:
$$
\int_0^{\sqrt{z}} \frac2{\pi \sqrt{1-u^2}}du
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\int \frac{1+x}{1+\sqrt x} dx$ I picked this exercice from B. Demidovitch and i started solving, but when it's complete, i get only the half of the solution (the first two fractions and the ln, doesn't show up).
$$\int \frac{1+x}{1+\sqrt x} dx$$
The Solution is:
$$2 \left[\frac{\sqrt{x^3}}{3}-\frac{x}{... | By setting $\color{red}{x=z^2}$ we just have to compute
$$ I=\int \frac{2z(1+z^2)}{1+z}\,dz $$
and by polynomial division we have $2z(1+z^2)=(2z^2-2z+4)(1+z)-4$, hence
$$ I = \int (2z^2-2z+4)\,dz -4\int\frac{dz}{1+z} = \color{red}{\frac{z^3}{3}-z^2+4-4\log(1+z)+C}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Let a, b, c be positive real numbers. Prove that Let a,b,c be positive real numbers. Prove that
$$\frac{a^3+b^3+c^3}{3}\geq\frac{a^2+bc}{b+c}\cdot\frac{b^2+ca}{c+a}\cdot\frac{c^2+ab}{a+b}\geq abc$$
I will post what I had solved originally, however it is unfortunately incorrect. Please help solve and/or aid in findi... |
This is what I have although it is incorrect any help would be greatly appreciated.!
$\mathbf{1.)}$ Consider now the expression $\frac{a^3+b^3+c^3}{3}$
The expressions has minimum value given by:
$a\geq b \geq c > 0 \implies a^3 \geq b^3 \geq c^3$
Thus $\frac{a^3+b^3+c^3}{3}$ has minimum value $c^3$
$\mathbf{2.)}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $a +b $,= 5 and $ab = 6$ . Find $\frac{1}{a}$ + $\frac{1}{b}$ $a +b $ = 5 and $ab = 6$ . Find $\cfrac{1}{a}$ + $\cfrac{1}{b}$
Any Ideas on how to begin?
Many Thanks
| $$\frac { 1 }{ a } +\frac { 1 }{ b } =\frac { a+b }{ ab } =\frac { 5 }{ 6 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although:
$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$
for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might... |
I thought it might be instructive to present a brute force approach. To that end we proceed.
Let $S_n=\sum_{k=1}^n (-1)^{k-1}\sqrt{k}$ be the sequence of interest. Then, we can write the even and odd terms, respectively by
$$\begin{align}
S_{2n}&=\sum_{k=1}^n \left(\sqrt{2k-1}-\sqrt{2k}\right)\\\\
S_{2n+1}&=1+\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
Solve for x and find an approximate value $15 = \dfrac{((x+3)+(2x-3))h}{2}$
h = $(2x-3) -(x+3) $
& it is also given that $\sqrt{19}$ = 4.36
How can I simplify this & can you help me by explaining the steps
| $$30 = ((2x-3)+(x+3))((2x-3)-(x+3))=(2x-3)^2-(x+3)^2=3x^2-18x$$
$$x^2-6x=10$$
$$x^2-6x+9=19$$
$$x=3\pm\sqrt {19}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$, what can we say about $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$?
Suppose that $a,b,c$ are three real numbers such that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$. What are the possible values for $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$?
Aft... | Assuming
$$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}=1$$
we also have
$$\frac{a^{2}}{b+c}+\frac {ab}{c+a}+\frac {ac}{a+b}= a$$
as well as
$$\frac{ab}{b+c}+\frac {b^2}{c+a}+\frac {bc}{a+b}= b$$
and
$$\frac{ac}{b+c}+\frac {bc}{c+a}+\frac {c^2}{a+b}= c$$
These three sum together as but all terms without $(.)^2$ on the other ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Find the last digit of $7^{802} -3^{683}$
Find the last digit of $$7^{802} -3^{683}$$
$$7^{802} -3^{683}=(50 -1)^{401} -3(10 -1)^{341}$$
$$=50k -1 -3(10m -1) \space \text{where k,m are fixed integer } $$
$$=50k -30m +2 $$
So,the last digit will be $2$
Someone just added this question, but deleted it. I solved this qu... | Okay, after reading the comments I realize my answer is completely out of the experience level of the OP. But read ahead for a taste of exciting things to come.
=====
Very nice!
I want to introduce you to the idea of Fermat's Little Thereom however.
If $\gcd(a, m) = 1$. And $\phi(m) = $ the number of natural numbers ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Generating function for cubes of Harmonic numbers By generalizing the approach in Integral involving a dilogarithm versus an Euler sum. meaning by using the integral representation of the harmonic numbers and by computing a three dimensional integral over a unit cube analytically we have found the generating function ... | Here we provide a closed form for another related sum. We have:
\begin{eqnarray}
&&\sum\limits_{n=1}^\infty \frac{H_n^3}{n} \cdot x^n =\\
&&3 \zeta(4)-3 \text{Li}_4(1-x)+3 \text{Li}_3(1-x) \log (1-x)+\log (x) \log ^3(1-x)+\\
&&\text{Li}_2(x){}^2-2 \text{Li}_4(x)-3 \text{Li}_4\left(\frac{x}{x-1}\right)+\frac{3}{2} \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"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.