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Calculus Question - change order of integration Can We always change the order of integration in double integrals ?
| Here is a counterexample that shows the value might not coincide.
Consider the function
$$
\frac{x^2-y^2}{(x^2+y^2)^2}
$$
A $y$-primitive for this on $[1,\infty)$ is
$$
\frac{y}{x^2+y^2}
$$
which simplifies to $$-\frac{1}{1+x^2},$$ when evaluated from $1$ to $\infty$.
Knowing this, and the fact that $$\int \frac{1}{1+x^2}=\arctan(x)+C,$$ we get
$$
\int_1^\infty \int_1^\infty \frac{x^2-y^2}{(x^2+y^2)^2} dydx=\int_1^\infty -\frac{1}{1+x^2}=-\frac{\pi}{4}.
$$
Now changing the order, we get
$$
\int_1^\infty \int_1^\infty \frac{x^2-y^2}{(x^2+y^2)^2} dxdy=-\int_1^\infty \int_1^\infty \frac{-x^2+y^2}{(x^2+y^2)^2} dxdy=\int_1^\infty \frac{1}{1+y^2}=\frac{\pi}{4}.
$$
There are theorem that can garantee the change will provide the same answer, namely Fubini's. As Glebovg pointed out, you can find this example and some other theorems here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/244092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Function expansion help I want to write a function, $f(k,a,b)$, I made, in terms of combinations of the fractional part function, $$ j\left\{\frac{c \ }{d}k\right\},$$ where $c,d,$ and $j$ are any integers.
The function is as follows: $f(k,a,b)=1$ if $k\equiv b$ mod a
and $f(k,a,b)=0$, if it is not
I need a general method for finding the expansion of this function in terms of the fractional part function for any given coprime integers $a,b$.
An example of one is $$ f(k,6,1)= -\{k/6 \}+\{k/2\}+\{2k/3\} $$
Such that if, $k\equiv 1$ mod 6 , then $f(k,6,1)=1$, if not $f(k,6,1)=0$.
So I need a general method for finding the expansion of $f(k,a,b)$ in terms of the fractional part operator.
| Let C be a (m-1)*(m-1) matrix of numbers $\large\ c_{ij}=\left\{\Large\frac{i\cdot j}{m}\right\}$
Let $D=C^{-1}$
Then
$\large f(k,m,1)=\sum_{i=1}^{m-1}d_{i1}\Large\{\frac{i}{m}\cdot\large k\}$
$\large f(k,m,r)=f(k-r+1,m,1)$
Example: $\ \ \ \ \ m=6$
$$
C=\frac{1}{6}\cdot\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 0 & 2 & 4 \\
3 & 0 & 3 & 0 & 3 \\
4 & 2 & 0 & 4 & 2 \\
5 & 4 & 3 & 2 & 1
\end{pmatrix}$$
$$D=C^{-1}=\begin{pmatrix}
-1 & 0 & 1 & 1 & 0 \\
0 & 1 & -1 & -1 & 1 \\
1 & -1 & 0 & -1 & 1 \\
1 & -1 & -1 & 1 & 0 \\
0 & 1 & 1 & 0 & -1
\end{pmatrix}$$
$$\large f(k,6,1)=(-1)\cdot\{\frac{1}{6}\cdot k\}+1\cdot\{\frac{3}{6}\cdot k\}+1\cdot\{\frac{4}{6}\cdot k\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/244648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why are these two expressions different in this induction problem? Prove with $n \ge 1$:
$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot4}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$$
First, I prove it for $n=1$:
$$\left(\frac{1+2}{1(1+1)2^1} = 1-\frac{1}{(1+1)2^1}\right) \implies \left(\frac{3}{4} = 1- \frac{1}{4}\right) \implies \left(\frac{3}{4} = \frac{3}{4}\right)$$
Which is true.
So I will now assume this:
$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$$
And I want to prove it for $n+1$, i.e:
$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} = 1 - \frac{1}{((n+1)+1)2^{n+1}}$$
This is how I tried to prove it:
$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} =$$
$$ 1 - \frac{1}{(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}}$$
Before continuing, I usually like to grab a calculator, give a value to $n$ and evaluate my current expression with the expression I want to reach. If both values are equal, it means I'm doing okay.
So I took $n=5$ and evaluated the expression I want to reach:
$$1 - \frac{1}{(5+2)\cdot2^5+1} = \frac{447}{448}$$
Then, still with $n = 5$ I evaluated my current expression:
$$1 - \frac{1}{(5+1)\cdot2^5}+\frac{5+3}{(5+1)(5+1)\cdot2^{5+1}} = \frac{575}{576}$$
So I got $\frac{447}{448}$ for the expression I want to reach and $\frac{575}{576}$ for what I got so far. Something went wrong.
My problem with this is that I haven't done any calculations yet. All my steps so far were rather mechanical - things I always do with mathematical induction.
Maybe I simply evaluated them wrongly. But I can't see it - I've tested it many times already.
Why are both expressions different? They should be the same.
| Elevating comment to answer, at suggestion of OP:
In the last term of the last displayed equation, there is a $5+1$ where there should be a $5+2$. Jonas Meyer notes that this correction gets rid of the discrepancy.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral by partial fractions
$$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x$$ find the value of the constant when the antiderivative passes threw (6,0)
factor out the 5, and use partial fraction
$$ 5 \left[\int \frac{A}{x-5} + \frac{B}{\left(x-5\right)^2}\, \mathrm{d}x \right] $$
Solve for $A$ and $B$.
$A\left(x-5\right) + B = x$ Then $B-5A$ has to be zero and $A$ has to be 1.
Resulting in
$$ 5 \left[\int \frac{1}{x-5} + \frac{5}{\left(x-5\right)^2}\, \mathrm{d}x \right]$$
$$ \Rightarrow 5 \left[ \ln \vert x - 5 \vert -\frac{5}{x-5}\right] + C$$
However, this approach doesn't give the answer in the book.
Book's Answer
$$ \frac{5}{x-5} \left(\left(x-5\right) \ln \vert x - 5 \vert - x \right) + C $$
The value should be 30, according to the book.
| Your solution is correct, but books solution is also. Differentiate the solutions and you will see, that both of them are Antiderivatives.
Moreover it is:
$$ \frac{5}{x-5} \left(\left(x-5\right) \ln \vert x - 5 \vert - x \right)
= 5 \left(\ln \vert x - 5 \vert - \frac{x-5+5}{x-5}\right) = 5 \left(\ln \vert x - 5 \vert - \frac{5}{x-5}\right) +\mathcal{Const}$$
Optional way to get your solution:
$$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x= \frac{5}{2}\int\frac{2x-10}{\left(x-5\right)^2}+\frac{10}{\left(x-5\right)^2}\,\mathrm{d}x=\frac{5}{2}\left(2\ln\vert x-5\vert-\frac{10}{x-5}\right)+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/247178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Area of an ellipse An ellipse has equation :
$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0$$
Can you provide an optimum method to find it's area?
| When rotating conics in implicit form
$$
Ax^2+Bxy+Cy^2+Dx+Ey+F=0\tag{1}
$$
around the origin there are 5 invariants:
$$
\begin{array}{rl}
I_1&=A+C\\
I_2&=(A-C)^2+B^2\\
I_3&=D^2+E^2\\
I_4&=(A-C)(D^2-E^2)+2DEB\\
I_5&=F\tag{2}
\end{array}
$$
Assuming that we have rotated to eliminate $B$, we have
$$
Ax^2+Cy^2+Dx+Ey+F=0\tag{3}
$$
Translating the center to the origin gives
$$
Ax^2+Cy^2+F-\frac{D^2}{4A}-\frac{E^2}{4C}=0\tag{4}
$$
which is the same as
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\tag{5}
$$
if we set
$$
a=\sqrt{\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{A}}\quad\text{and}\quad
b=\sqrt{\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{C}}\tag{6}
$$
Using $\text{Area}=\pi ab$ and $(4)$ and rewriting in terms of the invariants to remove the rotation, we get
$$
\begin{align}
\text{Area}
&=\pi\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{\sqrt{AC}}\\
&=\pi\frac{\color{#C00000}{2CD^2+2AE^2}-\color{#00A000}{8ACF}}{\color{#0000FF}{8(AC)^{3/2}}}\\
&=\pi\frac{\color{#C00000}{I_1I_3-I_4}-\color{#00A000}{2(I_1^2-I_2)I_5}}{\color{#0000FF}{(I_1^2-I_2)^{3/2}}}\tag{7}
\end{align}
$$
Therefore, $(2)$ and $(7)$ give the area in terms of the coefficients in $(1)$.
Invariants Under Rotation and Translation
$I_1$ and $I_2$ are invariant under rotation and translation, but there is one more: the constant coefficient when the center is translated to the origin. Writing $F-\dfrac{D^2}{4A}-\dfrac{E^2}{4C}$ in terms of the rotational invariants and expanding yields
$$
I_6=F-\frac{AE^2-BDE+CD^2}{4AC-B^2}\tag{8}
$$
Thus, the conic satisfying $(1)$, when rotated and translated becomes
$$
(I_1-\sqrt{I_2})x^2+(I_1+\sqrt{I_2})y^2+2I_6=0\tag{9}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/247332",
"timestamp": "2023-03-29T00:00:00",
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Find parametrization of a manifold Find parametrization of $x^2y^2+y^2z^2+z^2x^2=xyz$, $x,y,z>0$.
I've checked it's manifold but I can't find parametrization(s).
| So, $$\frac {xy}z+\frac{yz}x+\frac{zx}y=1$$
As $x,y,z>0$
we can write $\frac {xy}z+\frac {yz}x=\cos^2\theta$ and $\frac{zx}y=\sin^2\theta$
So, $\frac {xy}{z\cos^2\theta}+\frac {yz}{x\cos^2\theta}=1$
So, we can write $\frac {xy}{z\cos^2\theta}=\cos^2\phi$ and $\frac {yz}{x\cos^2\theta}=\sin^2\phi$ as $\frac {xy}{z\cos^2\theta},\frac {yz}{x\cos^2\theta}>0$
So, $\frac {xy}z=\cos^2\theta\cos^2\phi--->(1)$
$\frac {yz}x=\cos^2\theta\sin^2\phi--->(2)$
$\frac{zx}y=\sin^2\theta--->(3)$
On multiplication, $xyz=\cos^4\theta\cos^2\phi \sin^2\phi \sin^2\theta-->(4) $
$(4)/(1),z^2=\cos^2\theta\sin^2\phi \sin^2\theta\implies z=\cos\theta\sin\phi \sin\theta=\frac{\sin\phi \sin2\theta}2$
Similarly, $(4)/(2)$ and $(4)/(3)$ will supply
$x=\frac{\cos\phi \sin2\theta}2$ and $y=\cos\phi\sin\phi \cos^2\theta$
As $x,y,z>0$
either $\sin2\theta, \sin\phi,\cos\phi>0\implies 0<\phi<\frac{\pi}2$ and $0<2\theta<\pi$ with $\theta\ne \frac{\pi}2$
or, $\sin2\theta, \sin\phi,\cos\phi<0\implies \pi<\phi<\frac{3\pi}2$ and $\pi<2\theta<2\pi$ with $\theta\ne \frac{3\pi}2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find Jordan-Normal form of a given matrix I am trying to write the matrix $A=\begin{pmatrix} 2 & 2 & -5 \\ 3 & 7 & -15 \\ 1 & 2 & -4 \end{pmatrix}$ in Jordan-Normal Form.
It's characteristic polynomial is $|A-xI|=x^3-5x^2+7x-3=(x-1)^2(x-3)=0$
So, it has eigenvalues 1 and 3. I calculated three eigenvectors $v_1=\left(\begin{smallmatrix} 1 \\ 3 \\ 1 \end{smallmatrix}\right), v_2=\left(\begin{smallmatrix}5\\0\\1\end{smallmatrix}\right), v_3=\left(\begin{smallmatrix}-2\\1\\0\end{smallmatrix}\right)$.
So $Z=\begin{pmatrix} 1&5&-2\\3&0&1\\1&1&0\end{pmatrix},\text{ }Z^{-1}=-\frac{1}{2}\begin{pmatrix}-1&-2&5\\1&2&-7\\3&4&-15\end{pmatrix}$
$Z^{-1}AZ=\begin{pmatrix}3&0&0\\0&1&0\\0&0&1\end{pmatrix}$, but shouldn't the answer be
$Z^{-1}AZ=\begin{pmatrix}3&0&0\\0&1&1\\0&0&1\end{pmatrix}$?
Thanks.
| Everything you have calculated is correct. A Jordan-Normal-Form for your matrix is $$\pmatrix{3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}$$
Another solution would be for example $$\pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3}$$
The Jordan-Normal-Form do not have to be unique and depends on how you arange the vectors in your matrix $\mathcal{Z}$
But $$\pmatrix{3 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1}$$ cannot be a Jordan-Normal-Form of your matrix. For the eigenvalue $1$ the dimension of the corresponding eigenspace is $2$ so your matrix is even diagonalizable!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/249948",
"timestamp": "2023-03-29T00:00:00",
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How to simplify $\frac{4 + 2\sqrt6}{\sqrt{5 + 2\sqrt{6}}}$? I was tackling through an olympiad practice book when I saw one of these problems:
If $x = 5 + 2\sqrt6$, evaluate $\Large{x \ - \ 1 \over\sqrt{x}}$?
The answer written is $2\sqrt2$, but I can't figure my way out through the manipulations. I just know that I have the following:$${4+ 2\sqrt6} \over {\sqrt{5 + 2\sqrt6}}$$
| $$\frac{x-2}{\sqrt x}=\frac{4+ 2\sqrt6}{\sqrt{5 + 2\sqrt6}}=\frac{4+ 2\sqrt6}{\sqrt{5 + 2\sqrt6}}\cdot\frac{\sqrt{5-2\sqrt 6}}{\sqrt{5-2\sqrt 6}}=\left(4+2\sqrt 6\right)\sqrt{5-2\sqrt 6}\Longrightarrow$$
$$(x-2)^2=x(40+16\sqrt 6)(5-2\sqrt 6)=x(8)\Longrightarrow $$
$$x^2-12x+4=0\ldots...$$
Take it from here
| {
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"timestamp": "2023-03-29T00:00:00",
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Computation of a certain integral I would like to compute the following integral. This is for a complex analysis course but I managed to around some other integrals using real analysis methodologies. Hopefully one might be able to do for this one too.
$$\int_{0}^{2\pi} \frac{1}{a-\cos(x)}dx, \text{ with } a > 1.$$
Any suggestion will be greatly appreciated.
| $$\cos x = \frac{e^{i x}+e^{-ix}}{2}$$
Thus
$$\int_0^{2\pi} \frac{dx}{a-\frac{e^{i x}+e^{-ix}}{2}} = \oint_C \frac{1}{a-\frac{z+z^{-1}}{2}} \frac{dz}{iz} = - i\oint_C \frac{dz}{az-\frac{z^2+1}{2}} = 2i\oint_C \frac{dz}{z^2-2az+1}$$
where $C$ describes the unit circle $|z|=1$, centred at the origin, parametrized by $e^{iz}$ where $0\le z\le 2\pi$.
Letting $f(z)=\frac{1}{z^2-2az+1}$, we find that the poles of $f$ are at $z=a\pm\sqrt{a^2-1}$. Noting that $a>1$, the only pole in $C$ is the one with the negative sign. Then
$$\operatorname*{Res}_{z = a-\sqrt{a^2-1}}f(z)= \lim_{z\to a-\sqrt{a^2-1}}\frac{z-a+\sqrt{s^2+1}}{z^2-2az+1}= \lim_{z\to a-\sqrt{a^2-1}}\frac{1}{2z-2a}=- \frac{1}{2\sqrt{a^2-1}}$$
And thus we wrap up:
$$\int_0^{2\pi} \frac{dx}{a-\cos x} = 2i\oint_C \frac{dz}{z^2-2az+1} = 2i\left(-\frac{2\pi i }{2\sqrt{a^2-1}}\right) = \frac{2\pi}{\sqrt{a^2-1}}$$
$\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/250250",
"timestamp": "2023-03-29T00:00:00",
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Problem when integrating $e^x / x$. I made up some integrals to do for fun, and I had a real problem with this one. I've since found out that there's no solution in terms of elementary functions, but when I attempt to integrate it, I end up with infinite values. Could somebody point out where I go wrong?
So, I'm trying to determine: $$ \int{\frac{e^x}{x}} \, dx $$
Integrate by parts, where $u = 1/x$, and $v \, ' = e^x$. Then $u \, ' = - 1/x^2$, and $v=e^x$. So,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \int{\frac{e^x}{x^2}} \, dx$$
Integrate by parts again, $u = 1/x^2$, $v \, ' = e^x$, so that $u \, ' = -2/x^3$ and $v=e^x$. So,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2\int{\frac{e^x}{x^3}} \, dx$$
Repeat this process ad infinitum to get,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2 \left( \frac{e^x}{x^3} + 3 \left( \frac{e^x}{x^4} + 4 \left( \frac{e^x}{x^5} + \, \cdots \right) \right) \right) $$
Expanding this gives,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + \frac{2e^x}{x^3} + \frac{6 e^x}{x^4} + \frac{24 e^x}{x^5} + \cdots $$
And factoring that gives,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( 1 + \frac{1}{x} + \frac{2}{x^2} + \frac{6}{x^3} + \frac{24}{x^4} + \cdots \right) $$
Now, considering the series itself, the ratio between the $n^{th}$ term and the $(n-1)^{th}$ term = $\Large \frac{n}{x}$. Eventually, $n$ will be larger than $x$, so the ratio between successive terms will be positive, so (assuming $x$ is positive), the series diverges, meaning (and I'm sure everybody will cringe upon seeing notation used like this), that:
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( \infty \right) = \infty $$
| This part looks right:
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + \frac{2e^x}{x^3} + \frac{6 e^x}{x^4} + \frac{24 e^x}{x^5} + \cdots+ \frac{n!e^x}{x^{n+1}}+(n+1)!\int \frac{e^x}{x^{n+1}}$$
When you say "repeating to infinity" you want to take the limit of that...in order for your equality to hold, you need
$$\lim_n (n+1)!\int \frac{e^x}{x^{n+1}}=0 \,.$$
because that is your error in you partial sum "approximation". But not only the above limit is not 0, it actually makes no sense (an integral is a family of functions, what happens with the constant???).
That's why formally, whenever you use a process like this, you need to prove that the difference between your n-th term and the limit goes to 0...
Your idea is similar to the following
\begin{eqnarray}
1&=&1+1-1\\
&=&1+1+1-2 \\
&=&1+1+1+1-3\\
&=&....
\end{eqnarray}
Taking limit to infinity you get
$$1=1+1+1+...+1+...= \infty$$
On this example you can see immediately that the "errror" in our appoximations don't go to 0, so our approximations are not approximations.
| {
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"url": "https://math.stackexchange.com/questions/251795",
"timestamp": "2023-03-29T00:00:00",
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Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that:
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried :
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I want to try to prove that for example $$a+\frac{bc}{a} \geq a+b$$
but I don't know if is is a good idea.
Thanks:)
| $$LHS-RHS=abc\sum_{sym} a(a-b)(a-c)+\frac{1}{2}(ab+bc+ca)\left[c^2(a-b)^2+a^2(b-c)^2+b^2(c-a)^2\right]\ge 0$$
True by Schur 3 deg
| {
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"url": "https://math.stackexchange.com/questions/253015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
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How to find these integrals? How to find the following two integrals?
$$\int_{0}^{1}{\sqrt{{{x}^{3}}-{{x}^{4}}}dx}$$
and
$$\int_{0}^{1}{x\sqrt{{{x}^{3}}-{{x}^{4}}}dx}$$
| $$I:=\int_0^1 x\sqrt {x^3-x^4}\,dx=\int_0^1 x^2\sqrt{x-x^2}\,dx=\int_0^1x^2\sqrt{\frac{1}{4}-\left(\frac{1}{2}-x\right)^2}\,dx=$$
$$=\frac{1}{2}\int_0^1x^2\sqrt{1-\left(1-2x\right)^2}\,dx $$
Substitute now
$$\sin u=1-2x\Longrightarrow \cos u\,du=-2\,dx\,,\,x=0\Longrightarrow u=\frac{\pi}{2}\;\;,\;x=1\Longrightarrow u=\frac{3\pi}{2}\,\,,\,\text{thus:}$$
$$I=-\frac{1}{4}\int_{\pi/2}^{3\pi/2}\left(\frac{1-\sin u}{2}\right)^2\sqrt{1-\sin^2u}\cos u\,du=$$
$$-\frac{1}{16}\int_{\pi/2}^{3\pi/2}\left(\cos^2u-2\sin u\cos^2u+\cos^2u\sin^2u\right)du=\ldots\,\,etc.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that the following series is convergent: $\sum\limits_{i=1}^\infty \left({n^2+1\over n^2+n+1}\right)^{n^2}$ Can someone please help me prove that this series is convergent?
$$ \sum_{i=1}^\infty \left({n^2+1\over n^2+n+1}\right)^{n^2} $$
I guess I'm supposed to show that the limit of the sequence is an "e" limit, that means something of the kind: $$ \left( 1\pm{1 \over a} \right)^a $$
But how? I came to this state by now and that's where I'm stuck:
$$ \left( {n^2+n+1-n \over {n^2+n+1}} \right)^{n^2} = \left( 1+{n \over {n^2+n+1}} \right)^{n^2} $$
| Note that $${\left( \dfrac{n^2+n+1}{n^2+1}\right)^{n^2}}={\left(1+ \dfrac{n}{n^2+1}\right)^{\frac{n^2+1}{n}\cdot\frac{n^3}{n^2+1}}}\geqslant 2^{\frac{n}{2}},$$ because $2<\left(1+ \dfrac{n}{n^2+1}\right)^{\frac{n^2+1}{n}}<3$ and $\frac{n^3}{n^2+1}\geqslant \frac{n}{2}$ for $n\geqslant 1.$ Therefore,
$$\left( \dfrac{n^2+1}{n^2+n+1}\right)^{n^2}\leqslant {\frac{1}{2^{\frac{n}{2}}}}=\left(\frac{1}{\sqrt{2}}\right)^n$$
and the series
$$\sum_{n=1}^\infty \left({n^2+1\over n^2+n+1}\right)^{n^2}$$
converges by comparison test.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Number of ordered sets of integers How many ordered sets of integers $(x,y,z)$ satisfying $$x,y,z \in [-10,10]$$ are solutions to the following system of equations:
$$x^2y^2+y^2z^2=5xyz$$
$$y^2z^2+z^2x^2=17xyz$$
$$z^2x^2+x^2y^2=20xyz$$
By a quite cumbersome method, which is too long for me to type down here, I arrived at the answer 5: the ordered sets being $(0,0,0), (8,2,4), (-8,-2,4), (-8,2,-4)$ and $(8,-2,-4)$. But this does not seem to be the right answer. What I basically did was divide the system of equations with $(xyz)$, then by a process of elimination of the variables, found the possible values of x, y and z. However, it seems I've somehow missed out other solutions as well. Can anyone help me out with this?
Thanks!
| If any two of $x,y,z$ are $0$, all the equations say $0 = 0$.
On the other hand, if only one is $0$, it's easy to see that another must also be $0$.
So now let's assume none of them are $0$.
Then we can cancel a $y$, $z$, or $x$ respectively from each equation, leaving
$$ \eqalign{(x^2 + z^2) y &= 5 x z\cr
(x^2 + y^2) z &= 17 x y\cr
(y^2 + z^2) x &= 20 y z\cr}$$
Next, notice that changing signs of any two of the variables preserves all three equations. Thus we may assume, say, $x > 0$ and $y > 0$, and get the other solutions by symmetry.
Now $x z \le (x^2 + y^2)/2$ (because $x^2 - 2 x z + y^2 = (x-y)^2 \ge 0$), and similarly $-x z \le (x^2 + y^2)/2$, i.e $|x z| \le (x^2 + y^2)/2$.
So from the first equation we get
$$|y| = \frac{5 |x z|}{x^2 + y^2} \le \frac{5}{2}$$
Since at this point $y$ is a positive integer, the only possibilities are $y= 1$ and $y = 2$.
If $y = 1$, the second equation says $z (x^2 + 1) = 17 x$. Since $x^2 + 1$ and $x$ are relatively prime, this implies that $x$ divides $z$. So let $z = m x$. We then
have $m (x^2 + 1) = 17$. Since $17$ is prime, one of $m$ and $x^2 + 1$ must be $1$ and the other $17$. But $x \ne 0$ so it must be $x^2 + 1 = 17$ (i.e. $x=4$), and $m= 1$ so $z=4$. But $x=4,y=1,z=4$ doesn't satisfy the first or third equation.
The remaining possibility is $y=2$. Then the second equation says $z (x^2 + 4) = 34 x$. If $x$ is odd, so is $x^2 + 4$, and that is relatively prime to $x$ so it divides $34$. The only odd factors of $34$ are $1$ and $17$, so we'd need $x^2 + 4 = 17$, but that doesn't work since $17-4=13$ is not a square. So $x$ must be even: say $x = 2 t$. Now the second equation, after dividing by $4$, becomes $z (t^2 + 1) = 17 t$. As in the previous paragraph, we conclude that $t=4$ (so $x=8$) and $z=4$. Thus we have $(x,y,z) = (8,2,4)$, which is a solution. By symmetry (changing any two signs) we also have solutions $(-8,-2,4), (-8,2,-4)$ and $(8,-2,-4)$.
Summing up, we have the solutions $(x,y,z) = (0,0,z)$, $(0,y,0)$, $(x,0,0)$, $(8,2,4)$, $(-8,-2,4), (-8,2,-4)$ and $(8,-2,-4)$.
| {
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How do you integrate$\int_{-\infty}^{\infty} dw \frac{1}{(\alpha^2-w^2)^2+ w^2\beta^2} $ I am having trouble integrating the following expression appearing in a mechanical problem:
$$\int_{-\infty}^{\infty} dw \frac{1}{(\alpha^2-w^2)^2+ w^2\beta^2} $$
I tried using the residue theorem, but having a polynom of degree 4 in the denominator causes me trouble. Mathematica doesn't seem to take it as I type it.
| Here is how we might go about it while minimizing the use of computer algebra tools. The idea is to compute the integral on a contour consisting of the line segment from $-R$ to $R$ on the real axis and a semicircle of radius $R$ in the upper half plane, letting $R$ go to infinity. Since our function $$f(w) = \frac{1}{(\alpha^2-w^2)^2 + \beta^2 w^2}$$ is $O(1/R^4)$ on the semicircle its contribution is $O(1/R^3)$ and vanishes in the limit. So to calculate the integral we only need to find the sum of the residues at the poles in the upper half plane.
First, calculate the location of the poles:
$$ (\alpha^2 - w^2)^2 = - \beta^2 w^2 \\
\alpha^2 - w^2 = \pm i \beta w $$ so that
$$ w^2 \pm i \beta w - \alpha^2 = 0$$ or
$$ \rho_{0, 1, 2, 3} = \pm \frac{1}{2} i\beta \pm \frac{1}{2} \sqrt{4\alpha^2-\beta^2}.$$ We assume that $\alpha$ and $\beta$ are positive and that $2\alpha>\beta$ and take the poles in the upper half plane which correspond to
$$\alpha^2 - w^2 = - i \beta w $$
and have the values
$$ \rho_{0, 1} = \frac{1}{2} i\beta \pm \frac{1}{2} \sqrt{4\alpha^2-\beta^2}.$$
We now need to find the residues at these poles. Take $\rho$ as a placeholder that will be instantiated to $\rho_{0,1}$ later on. The poles are simple and the residues are
$$\lim_{w\to \rho} \frac{w-\rho}{(\alpha^2-w^2)^2 + \beta^2 w^2} =
\lim_{w\to \rho} \frac{1}{2(\alpha^2-w^2)(-2w) + 2\beta^2 w} =
\lim_{w\to \rho} \frac{1}{2(-i\beta w)(-2w) + 2\beta^2 w}$$
or
$$\frac{1}{\beta}\lim_{w\to \rho} \frac{1}{4iw^2 + 2\beta w} =
\frac{1}{\beta}\lim_{w\to \rho} \frac{1}{4i(\alpha^2+i\beta w) + 2\beta w}
= \frac{1}{\beta}\lim_{w\to \rho} \frac{1}{4i\alpha^2 -2\beta w}$$
or $$q = \operatorname{Res}_{w=\rho} f(w) =
\frac{1}{\beta} \frac{1}{4i\alpha^2 -2\beta \rho}.$$
Observe that
$$\frac{1}{x+iy} + \frac{1}{-x+iy} = - \frac{2iy}{x^2+y^2}$$ and apply this to the second factor in $q_0$ and $q_1$ to get
$$ x^2 + y^2 = 4\beta^2 \frac{1}{4} (4\alpha^2-\beta^2) + (4\alpha^2-\beta^2)^2
= (4\alpha^2-\beta^2) (4\alpha^2-\beta^2 + \beta^2) = (4\alpha^2-\beta^2) 4\alpha^2$$
and $$- \frac{2iy}{x^2+y^2} = - \frac{2i(4\alpha^2-\beta^2)}{(4\alpha^2-\beta^2)4\alpha^2}=
-\frac{i}{2\alpha^2}.$$
It follows that the end result is
$$2\pi i (q_0+q_1) = - 2\pi i \frac{1}{\beta} \frac{i}{2\alpha^2} =
\frac{\pi}{\alpha^2\beta}.$$
| {
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How to compute $7^{7^{7^{100}}} \bmod 100$? How to compute $7^{7^{7^{100}}} \bmod 100$? Is $$7^{7^{7^{100}}} \equiv7^{7^{\left(7^{100} \bmod 100\right)}} \bmod 100?$$
Thank you very much.
| First note that $$7^4 \equiv 1 \pmod{100}$$
Hence, we get that $$7^n \equiv \begin{cases}1 \pmod{100} & n \equiv 0 \pmod4\\ 7 \pmod{100} & n \equiv 1 \pmod4\\ 49 \pmod{100} & n \equiv 2 \pmod4\\ 43 \pmod{100} & n \equiv 3 \pmod4 \end{cases}$$
Hence, all we need to figure out is $7^{7^{100}} \pmod 4$. Since $7 \equiv (-1) \pmod4$, we have that $$7^{7^{100}} \equiv -1^{\text{odd}} \pmod 4 \equiv -1 \pmod 4 \equiv 3 \pmod 4$$
Hence, $$7^{7^{7^{100}}} \equiv 43 \pmod{100}$$
| {
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compute the following integral $\int^{a}_{-a} \sqrt{a^2-x^2}dx$ I have to compute the following integral:
$\int^{a}_{-a} \sqrt{a^2-x^2}dx$
I did the substitution: $x=a\sin\theta$ so $dx=a\cos\theta d\theta$. The boundaries becomes $\pi/2+2k\pi$ and $-\pi/2-2k\pi$. So:
$\int^{\pi/2+2k\pi}_{-\pi/2-2k\pi} a^2\cos^2\theta d\theta$ becomes the integral.
My question is whether I am doing it correct. Thanks.
| Let $I=\int\sqrt{a^2-x^2}dx=\sqrt{a^2-x^2}\int dx-\int\left(\frac{\sqrt{a^2-x^2}}{dx}\int dx\right)dx$
$=x\sqrt{a^2-x^2}-\int\left(\frac{(-2x)x}{2\sqrt{a^2-x^2}}\right)dx$
$=x\sqrt{a^2-x^2}-\int\left(\frac{(a^2-x^2-a^2)}{\sqrt{a^2-x^2}}\right)dx$
$=x\sqrt{a^2-x^2}-\int \sqrt{a^2-x^2}dx +a^2\int\frac{dx}{\sqrt{a^2-x^2}}$
$=x\sqrt{a^2-x^2}-I +a|a|\arcsin\frac xa+C$ where $C$ is the indefinite constant for indefinite integration.
Or, $I=\frac{x\sqrt{a^2-x^2}}2+\frac{a|a|}2\arcsin\frac xa+\frac C2$
So, $$\int^{a}_{-a} \sqrt{a^2-x^2}dx=(\frac{x\sqrt{a^2-x^2}}2+\frac{a|a|}2\arcsin\frac xa+C)\mid_{-a}^a=\frac{a|a|}2\{ \arcsin1-\arcsin(-1)\}$$
$$=\frac{a|a|}2\{\frac\pi2-(-\frac\pi2)\}=\frac{\pi a|a|}2$$
Alternatively,
taking $a>0,x=a\sin \theta,dx=a\cos \theta d\theta$
and $\sqrt{a^2-x^2}=|a|\cos \theta=a\cos \theta$ as $a>0$
$$\int^{a}_{-a} \sqrt{a^2-x^2}dx=\int^{\pi/2+2k\pi}_{-\pi/2+2k\pi} (a\cos\theta)^2 d\theta=\frac{a^2}2\int^{\pi/2+2k\pi}_{-\pi/2+2k\pi}(1+\cos2\theta)d\theta$$
$$=\frac{a^2}2(\theta+\frac{\sin2\theta}2)\mid_{-\pi/2+2k\pi}^{\pi/2+2k\pi}$$
$$=\frac{a^2}2\{\frac \pi 2+2k\pi-(-\frac \pi2+2k\pi)\}+\frac{a^2}4(\sin(4k+1)\pi-\sin(4k-1)\pi)=\frac{\pi a^2}2$$
| {
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$2^{10 - x} \cdot 2^{10 - x} = 4^{10-x}$ $$2^{10 - x} \cdot 2^{10 - x} = 4^{10-x}$$
Is that correct?
I would've done
$$
2^{10 - x} \cdot 2^{10 - x}\;\; = \;\;
(2)^{10 - x + 10 - x} \; = \; (2)^{2 \cdot (10 - x)} \;=\; 4^{10 - x}\tag{1}
$$
Is that allowed?
If so, can I say that
$$
\frac{4^x}{2^y} = 2^{x - y} \tag{2}
$$
| First statement is correct, but $4^x/2^y = 2^{2x-y}$.
| {
"language": "en",
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Help Solving Trigonometry Equation I am having difficulties solving the following equation:
$$4\cos^2x=5-4\sin x$$
Hints on how to solve this equation would be helpful.
|
$\cos^2(x) = 5-4 \sin(x)$
Move everything to the left hand side.
$\cos^2(x)-5+4 \sin(x) = 0$
Write in terms of $sin(x)$ using the identity $\cos^2(x) = 1-\sin^2(x)$:
$4 \sin(x)-4-\sin^2(x) = 0$
Factor constant terms from the left hand side and write the remainder as a square:
$-(\sin(x)-2)^2 = 0$
Multiply both sides by -1:
$(\sin(x)-2)^2 = 0$
Take the square root of both sides:
$\sin(x)-2 = 0$
Add 2 to both sides:
$\sin(x) = 2$
Your edited your post:
$$4 cos^2(x) = 5-4 sin(x)$$
Subtract $5-4 sin(x)$ from both sides:
$$4 cos^2(x)-5+4 sin(x) = 0$$
Using the identity $cos^2(x) = 1-sin^2(x):$
$$4 sin(x)-1-4 sin^2(x) = 0$$
Factor constant terms from the left hand side and write the remainder as a square:
$$-(2 sin(x)-1)^2 = 0$$
Multiply both sides by -1:
$$(2 sin(x)-1)^2 = 0$$
Take the square root of both sides:
$$2 sin(x)-1 = 0$$
Add 1 to both sides:
$$2 sin(x) = 1$$
Divide both sides by 2:
$$sin(x) = 1/2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\int_0^{2\pi} \frac{d\theta}{A+B\cos\theta}$? I'm having a trouble with this integral expression:
$$\int_0^{2\pi} \frac{d\theta}{A+B \cos\theta}$$
I've done this substitution: $t= \tan(\theta/2)$
and get: $\displaystyle \cos\theta= \frac{1-t^2}{1+t^2}$ and $\displaystyle d\theta=\frac{2}{1+t^2}dt$ where $\displaystyle \cos^2\theta/2=\frac{1}{1+t^2}$
then the integral becomes:
$$\int\frac{2 \, dt}{(A-B)t^2+(A+B)}= \sqrt\frac{A+B}{A-B} \arctan \left(\left(\sqrt\frac{A+B}{A-B} \right) t\right)$$
However, I'm not sure about the new limits since $\tan$ has period $\pi$ so what I have to do at this point to decide the new limits? And of course if you find some mistake in what I've done before, please let me know!
| To avoid confusion with limits, note that
$$\int_0^{\pi} \dfrac{dx}{a+b \cos(x)} = \int_{\pi}^{2\pi} \dfrac{dx}{a+b \cos(x)}$$
Hence, we have
$$I = \int_0^{2\pi} \dfrac{dx}{a+b \cos(x)} = 2\int_0^{\pi} \dfrac{dx}{a+b \cos(x)}$$
Now use your substitution $t = \tan(x/2)$ and note that $t$ goes from $0$ to $\infty$ as $x$ goes from $0$ to $\pi$.
EDIT
An easier way if you are familiar with a little bit of complex analysis is as follows.
Let $z = e^{ix}$. Then we have that $dz = iz dx$. Hence,
\begin{align}
I & = \int_0^{2 \pi} \dfrac{dx}{a+b \cos(x)} = \oint_{\vert z \vert = 1} \dfrac{dz}{iz \left( a + \dfrac{b}2 \left(z+ \dfrac1z \right)\right)}\\
& = \dfrac1i \oint_{\vert z \vert = 1} \dfrac{dz}{az + \dfrac{b}2 \left(z^2+1 \right)} = \dfrac2{ib} \oint_{\vert z \vert = 1}\dfrac{dz}{z^2 + \dfrac{2a}b z + 1}\\
& = \dfrac2{ib} \oint_{\vert z \vert = 1} \dfrac{dz}{\left(z + \dfrac{a}b \right)^2 + 1 - \dfrac{a^2}{b^2}} = \dfrac2{ib} \oint_{\vert z \vert = 1} \dfrac{dz}{\left(z + r + \sqrt{r^2-1} \right)\left(z + r - \sqrt{r^2-1} \right)}
\end{align}
where $r = \dfrac{a}b \in \mathbb{R}$.
First note that if $\vert r \vert \leq 1$, then $\left \vert r \pm \sqrt{r^2-1} \right \vert = 1$. Hence, the integral doesn't exist.
If $r > 1$, then $ \left \vert r + \sqrt{r^2 - 1} \right \vert > 1$ and $\left \vert r - \sqrt{r^2-1} \right \vert < 1$. Hence, by Cauchy integral formula, we have that
$$I = \dfrac{2}{ib} 2 \pi i \dfrac1{2 \sqrt{r^2-1}} = \dfrac{2 \pi \, \text{sign}(a)}{\sqrt{a^2 - b^2}}$$
If $r < -1$, then $\left \vert r - \sqrt{r^2 - 1} \right \vert > 1$ and $\left \vert r + \sqrt{r^2-1} \right \vert < 1$. Hence, by Cauchy integral formula, we have that
$$I = \dfrac{2}{ib} 2 \pi i \dfrac{-1}{2 \sqrt{r^2-1}} = \dfrac{2 \pi \, \text{sign}(a)}{\sqrt{a^2 - b^2}}$$
Hence, to summarize,
$$I = \begin{cases} \dfrac{2 \pi \, \text{sign}(a)}{\sqrt{a^2 - b^2}} & \text{if } \vert a \vert > \vert b\\ \text{Does not exists} & \text{if } \vert a \vert \leq \vert b\end{cases}$$
| {
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Can someone please explain to me how I did this summation formula wrong? I was trying to show that
$\sum \limits_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ but instead I got this $[\frac{n(n+1)}{2}]^2$ which from my understanding I basically proved another summation formula which is $\sum \limits_{k=1}^n k^3$. Obviously I must have done something wrong. So I am going to show you how I got the summation formula wrong.
$s_n = 1^2 + 2^2 + ... + (n-1)^2 + n^2 $
In reverse order
$s_n = n^2 + (n-1)^2 + ... + 2^2 + 1^2 $
I decided to square root the partial sum which probably what led to the wrong answer. But I do not know, I am clumsy when I write on paper.
$\sqrt{s_n} = 1 + 2 + ... + (n-1) + n$
In reverse order
$\sqrt{s_n} = n + (n-1) + ... + 2 + 1$
Adding the two partial sums
$\sqrt{s_n} + \sqrt{s_n} = 2\sqrt{s_n}$
$2\sqrt{s_n} = (n+1) + (n+1) + ...$
$2\sqrt{s_n} = n(n+1)$
$\sqrt{s_n} = \frac{n(n+1)}{2}$
$(\sqrt{s_n})^2 = [\frac{n(n+1)}{2}]^2$
Now my question what did I do wrong. Can somebody show me the correct way. I am pretty sure this a fake proof, or a minor error. Thank you.
P.S. I am no latex expert and this not homework just for practice.
| $\sqrt{a^2+b^2}\ne a+b$ in general unless at least one of $a,b$ is $0$
If $s_n=1^2+2^n+\cdots+(n-1)^2+n^2,$
how can you write $s_n=1+2+\cdots+(n-1)+n?$
(1)One way to proof is :
$ (r+1)^3-r^3=3r^2+3r+1$
Put $r=0,1,2,\cdots,n-1,n$ and add to get
$(n+1)^3=3S_n+3(1+2+3+\cdots+n)+n=3S_n+3\frac{n(n+1)}2+n$
So, $S_n=...$
(2)We can use induction too:
Let $S(m)= \frac{m(m+1)(2m+1)}6$
$S(1)=\frac{1\cdot\cdot3}6=1$ which is true.
So, $S(m+1)=S(m)+m+1$
$=\frac{m(m+1)(2m+1)}6+(m+1)^2=\frac{(m+1)\{(m+1)+1\}\{2(m+1)+1\}}6$
So the proposition holds true for $n=m+1$ if it is true for $n=m.$
| {
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Distinct natural numbers such that $ab=cd=a+b+c+d-3$ Find the distinct natural numbers $a,b,c,d$ who satisfying $ab=cd=a+b+c+d-3$.
| Assume $a$ is the largest number among $a,b,c,d$; then $(a-1)b=a+c+d-3$
$$b=(a+c+d-3)/(a-1)<(a+a+a-3)/(a-1)=3$$
Hence, $b=1$ or $b=2$.
If $b=1$, then $a=a+1+c+d-3$. This implies $c+d=2$. Not the ideal pair.
If $b=2$, then $2a=a+2+c+d-3$. This implies $a=c+d-1$.
$$cd=ab=(c+d-1) \times 2$$
Hence, $$(c-2)(d-2)=2$$
Note that $c$ and $d$ are natural numbers.
Hence, $c-2 d-2$ is either $(-1,-2)$ or $(1,2)$. Only $(1,2)$ is the pair we want.
Hence, putting all this together, we get that
$$a = 6, b=2, c=3, d= 4$$
| {
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for a $3 \times 3$ matrix A ,value of $ A^{50} $ is I f
$$A= \begin{pmatrix}1& 0 & 0 \\
1 & 0 & 1\\
0 & 1 & 0 \end{pmatrix}$$
then $ A^{50} $ is
*
*$$ \begin{pmatrix}1& 0 & 0 \\
50 & 1 & 0\\
50 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0 & 0 \\
48 & 1 & 0\\
48 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0 & 0 \\
25 & 1 & 0\\
25 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0 & 0 \\
24 & 1 & 0\\
24 & 0 & 1\end{pmatrix}$$
I am stuck on this problem. Can anyone help me please...............
| The answer is 3. $$\begin{pmatrix}1& 0 & 0 \\
25 & 1 & 0\\
25 & 0 & 1 \end{pmatrix}$$
Just compute $A^2$ , $A^3$ , $A^4$ and $A^5$ and you will understand the repeated pattern.
| {
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Calculate a double integral I would like to ask a pretty easy question (at least I believe so). I know that:
$$\phi_{11}(k) = \frac{E(k)}{4\pi k^4}(k^2 - k_1^2)$$
$$E(k) = \alpha \epsilon^{\frac{2}{3}}L^{\frac{5}{3}}\frac{k^4}{(1 + k^2)^{\frac{17}{6}}}$$
therefore, substituting the expression of $E(k)$ in $\phi_{11}(k)$:
$$\phi_{11}(k) = \frac{\alpha\epsilon^{\frac{2}{3}}L^{\frac{5}{3}}}{4\pi}\frac{k^2 - k_1^2}{(1 + k^2)^{\frac{17}{6}}}.$$
Furthermore
$$k = \sqrt{k_1^2 + k_2^2 + k_3^2}$$
hence the above expression becomes
$$\phi_{11}(k_1,k_2,k_3) = \frac{\alpha\epsilon^{\frac{2}{3}}L^{\frac{5}{3}}}{4\pi}\frac{k_2^2 + k_3^2 }{(1 + k_1^2 + k_2^2 + k_3^2)^{\frac{17}{6}}}.$$
The question is how to manually compute the function
$$F_{11}(k_1) = \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty \phi_{11}\, \mathrm dk_2 \mathrm dk_3$$
$F_{11}$ is therefore the double integral of $\phi_{11}$ over unlimited range.
Of course, the first step is to write
$$F_{11}(k_1) = 2 \cdot 2 \cdot \int\limits_0^\infty\int\limits_0^\infty \phi_{11}\, \mathrm dk_2\mathrm dk_3$$
| The simplest way to approach this is to note that the integrand has radial symmetry: it depends only on $r^2\equiv k_2^2 + k_3^2$. (In other words, use $k^2=k_1^2+k_{\perp}^2$ in the first place, with the appropriate area element.) So the general form here is
$$
\int_{-\infty}^\infty \int_{-\infty}^\infty f(x^2+y^2) \, dx \, dy=2\pi\int_{0}^{\infty}rf(r^2) \, dr.
$$
In your case, this results in
$$
F_{11}(k_1)=\frac{1}{2}\alpha\epsilon^{2/3}L^{5/3}\int_{0}^{\infty}\frac{r^3 dr}{\left(1+k_1^2+r^2\right)^{17/6}}
$$
This you can integrate by parts. Note that even if this integral were intractable, you could rescale using $r\rightarrow r\sqrt{1+k_1^2}$ to find the functional dependence on $k_1$ (which in many cases is all you need):
$$
\begin{eqnarray}
F_{11}(k_1)&=&\frac{1}{2}\alpha\epsilon^{2/3}L^{5/3}\left(1+k_1^2\right)^{-5/6}\int_0^\infty \frac{r^3 \, dr}{\left(1+r^2\right)^{17/6}} \\
&=& \frac{9}{55}\alpha\epsilon^{2/3}L^{5/3}\left(1+k_1^2\right)^{-5/6}.
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/267784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$3x^2 ≡ 9 \pmod{13}$ What is $3x^2 ≡ 9 \pmod{13}$?
By simplifying the expression as $x^2 ≡ 3 \pmod{13}$ and applying brute force I can show that the answers are 4 and 9, but how to approach this in a more efficient way?
I tried by stating that what the expression above says essentially means $13|(3x²-9)$, which only gives me more variables ($3x²-9=13k, k \in \mathbb{Z}$)
| We have $13\mid 3(x^2-3)\iff 13\mid(x^2-3)$ as $(3,13)=1$ so, $x^2\equiv3\pmod {13}$.
Now, any number $x$ can be $\equiv 0,\pm1,\pm2,\pm3,\pm4,\pm5,\pm6 \pmod {13}$
So, $x^2\equiv 0,1,4,9,16(\equiv3),25(\equiv 12\equiv-1),36(\equiv10\equiv-3)\pmod {13}$
So, $x\equiv\pm4\pmod {13}$
For a larger prime, we can use Quadratic Reciprocity Theorem, to check the solutions exists or not before the trial as follows:
$$\left(\frac 3{13}\right)\left(\frac{13}3\right)=(-1)^\frac{(13-1)(3-1)}4=1$$
Now, $\left(\frac{13}3\right)=\left(\frac13\right)$ and $y\equiv\pm1\iff y^2\equiv1\pmod 3\implies \left(\frac{13}3\right)=1\implies \left(\frac{13}3\right)=1$ hence $3$ is a quadratic residue of $13$ and the given equation is solvable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/269147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^1{\frac{y}{\sqrt{y(1-y)}}dy}$ I'm a little rusty with my integrals, how may I evaluate the following:
$$
\int_0^1{\frac{y}{\sqrt{y(1-y)}}dy}
$$
I've tried:
$$
\int_0^1{\frac{y}{\sqrt{y(1-y)}}dy} = \int_0^1{\sqrt{\frac{y}{1-y}} dy}
$$
Make the substitution z = 1-y
$$
= \int_0^1{\sqrt{\frac{1-z}{z}} dz} = \int_0^1{\sqrt{\frac{1}{z} - 1} dz}
$$
Which seems simpler but I have not managed to evaluate it. All help appreciated (just a few pointers would be appreciated just as much as the full solution).
Thank you
EDIT:
Solution following Ng Hong Wai suggestion:
At
$$
\int_0^1{\sqrt{\frac{y}{1-y}} dy}
$$
We make the substitution $y = sin^2(\theta)$, so $dy = 2sin(\theta)cos(\theta)d\theta$. This results in:
$$
\int_0^\frac{\pi}{2}{2 sin^2(\theta)d\theta} = \int_0^\frac{\pi}{2}{[1-cos(2\theta)]d\theta} = [x-\frac{1}{2}sin(2\theta)]_0^\frac{\pi}{2} = \frac{\pi}{2}
$$
Solution according to M. Strochyk's suggestion:
$$
\int_0^1{\sqrt{\frac{y}{1-y}} dy} = 2 \int_0^\infty{\frac{u^2}{(1+u^2)^2}du}
$$
By decomposing into partial fractions this becomes:
$$
2 \int_0^\infty{\frac{1}{1+u^2}du} - 2 \int_0^\infty{\frac{1}{(1+u^2)^2}du}
$$
The first term integrates directly to $arctan(u)$ while the second can be evaluated by making the substitution $u=tan(x)$ and then applying the double-angle formula for cosine. The result is (unsurprisingly):
$$
2[arctan[u]]_0^\infty -[x + \frac{1}{2}sin(2x)]_0^\frac{\pi}{2} = \frac{\pi}{2}
$$
| Substitute $u=\sqrt{\dfrac{y}{1-y}}.$ Then
$$\dfrac{y}{1-y}=\dfrac{y-1+1}{1-y}=-1+\dfrac{1}{1-y}=u^2, \\
( {0}< {y} <{1} \Leftrightarrow {0}< {u}<{+\infty}),\\
\dfrac{1}{1-y}=1+u^2, \\
1-y=\dfrac{1}{1+u^2}, \\
y=1-\dfrac{1}{1+u^2}, \\
dy=\dfrac{2u}{(1+u^2)^2}\, du,
$$
so
$$\int\limits_0^1{\sqrt{\dfrac{y}{1-y}} dy}=\int\limits_0^{+\infty}{\dfrac{2u^2}{(1+u^2)^2}}\, du=2\int\limits_0^{+\infty}{\dfrac{u^2}{(1+u^2)^2}}\, du$$
Last integral can be calculated by decomposition into partial fractions.
| {
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"url": "https://math.stackexchange.com/questions/272183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}\right)$ has a limit $$x_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}$$
How can we prove that the sequence $(x_n)$ has a limit? I have to use the fact that an increasing sequence has a limit iff it is bounded from above. No more "advanced" tools can be used. It's obvious that this sequence is increasing, but I am having trouble finding a bound.
| Alternately you can use
$$x_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!} \leq x_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{(n-1)\cdot n}$$
and
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{(n-1)\cdot n}$$ is telescopic, since
$$\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1} \,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/272245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How many ways to write one million as a product of three positive integers? In how many ways can the number 1;000;000 (one million) be written as the product
of three positive integers $a, b, c,$ where $a \le b \le c$?
(A) 139
(B) 196
(C) 219
(D) 784
(E) None of the above
This is my working out so far:
$1000000 = 10^{6} = 2^{6} \cdot 5^{6}$ and then to consider this as the product of three factors i.e.
$10^6 = 2^6 \cdot 5^6$
$= 2^a 5^p \cdot 2^b5^q \cdot 2^c 5^r$ (where $a+b+c = 6$ and $p+q+r = 6$).
However there are repetitions here because $2^3\,5^3 \cdot 2^2\,5^2 \cdot 2^1\,5^1$ is the product of the same three factors as $2^2\,5^2 \cdot 2^3\,5^3 \cdot 2^1\,5^1$.
I think there are 139 such factors.
| Hint: Solve the problem for $abc = 10^6$. This has $ {8 \choose 2}^2=784$ solutions.
Count the number of solutions where $a=b=c$.
Count the number of solutions where $a=b$ or $b=c$ or $c=a$.
Count the number of solutions where $a, b, c$ are pairwise distinct.
Account for your repeated counting above, to find the cases where $a \leq b \leq c$.
Motivation: Bars and stripes, Orbit Stabilizier Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/273137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$(\tan^2(18^\circ))(\tan^2(54^\circ))$ is a rational number Assuming $$\cos(36^\circ)=\frac{1}{4}+\frac{1}{4}\sqrt{5}$$ How to prove that $$\tan^2(18^\circ)\tan^2(54^\circ)$$ is a rational number? Thanks!
| Use the fact that
$$ \tan^2{18^{\circ}} = \frac{1-\cos{36^{\circ}}}{1+\cos{36^{\circ}}} = 1-\frac{2}{5} \sqrt{5} $$
Then use the fact that
$$ \tan^2{54^{\circ}} = \frac{1}{\tan^2{36^{\circ}}} $$
so that
$$ \tan^2{18^{\circ}} \tan^2{54^{\circ}} = \frac{\tan^2{18^{\circ}}}{\tan^2{36^{\circ}}} = \frac{1}{4} (1 -\tan^2{18^{\circ}})^2 = \frac{1}{5} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Calculate $\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$ Please help me calculate this:
$$\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$$
Here I've tried multiplying by $\sqrt[4]{x+9}+2$ and few other method.
Thanks in advance for solution / hints using simple methods.
Edit
Please don't use l'Hosplital rule. We are before derivatives, don't know how to use it correctly yet. Thanks!
| $$\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$$
$$=\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}.\frac{x-7}{x-7}$$
$$=\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{x-7}.\underset{x\rightarrow7}{\lim}\frac{x-7}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}$$
$$=\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}-3+3}{x-7}.\underset{x\rightarrow7}{\lim}\frac{x+9-16}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}$$
$$=[\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt9}{x-7}-\underset{x\rightarrow7}{\lim}\frac{\sqrt[3]{x+20}-\sqrt[3]27}{x-7}].\underset{x\rightarrow7}{\lim}\frac{(x+9)-16}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}$$
$$= \frac{112}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
$\displaystyle\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} = -\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}$ Please help me, to prove that
$$
\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} =
-\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}.
$$
| Maybe we wanna use the fact that $\displaystyle \sum_{n=1}^{\infty}\frac{1}{ns^n}=\ln\frac{s}{s-1}, \space s>1$. Then
$$\sum_{n=2}^{\infty} \frac {2}{(n^3-n)3^n}=\frac{1}{3}\sum_{n=2}^\infty \frac {1}{(n-1)3^{n-1}} - 2\sum_{n=2}^\infty \frac {1}{n3^n} + 3\sum_{n=2}^\infty \frac {1}{(n+1)3^{n+1}}=\frac{4}{3} \log\frac{3}{2}-\frac{1}{2}$$
Chris. (the auxiliary limit is straightforward by Taylor series)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
} |
Proof of tangent half identity
Prove the following: $$\tan \left(\frac{x}{2}\right) = \frac{1 + \sin (x) - \cos (x)}{1 + \sin (x) + \cos (x)}$$
I was unable to find any proofs of the above formula online. Thanks!
| You could always use Euler's formula, setting $$ z = e^{i x} $$ which gives for the left
$$ \tan\left(\frac{x}{2}\right) = \frac{1}{i} \frac{z^{1/2}-z^{-1/2}}{z^{1/2}+z^{-1/2}}
= \frac{1}{i} \frac{z-1}{z+1}$$
and for the right
$$ \frac{1+\sin(x)-\cos(x)}{1+\sin(x)+\cos(x)} =
\frac{1+ 1/(2i) z - 1/(2i) 1/z - (1/2) z - (1/2) 1/z}
{1+ 1/(2i) z - 1/(2i) 1/z + (1/2) z + (1/2) 1/z} =
\frac{z + 1/(2i) z^2 - 1/(2i) - (1/2) z^2 - (1/2)}
{z + 1/(2i) z^2 - 1/(2i) + (1/2) z^2 + (1/2)} $$
which is
$$ \frac{-1/2(1+i)(z-1)(z+i)}{1/2(1-i)(z+1)(z+i)} =
-\frac{1+i}{1-i} \frac{z-1}{z+1} = \frac{1}{i} \frac{z-1}{z+1}.$$
While this may not be the most intuitive approach it does show that these trigonometric identities are amenable to automated theorem proving -- just make the substitution from Euler's formula, factorize LHS and RHS, done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/277106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Finding $\frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3}+\cdots+\frac{1}{S_{2013}}$ Assume $S_1=1 ,S_2=1+2,S=1+2+3+,\ldots,S_n=1+2+3+\cdots+n$
How to find :
$$\frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3}+\cdots+\frac{1}{S_{2013}}$$
| Hint 1:
$$ \sum_{k=1}^n k = \frac{n (n+1)}{2} $$
Hint 2:
$$ \frac{1}{n (n+1)} = \frac{1}{n} - \frac{1}{n+1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/281318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The number of ways of completing this partial Latin square If we want to fill the empty squares by the numbers $1$, $2$, $3$, $4$, $5$, $6$ so that all the numbers appear in each row and column, how can we find the number of ways to do that?
$$\begin{array}{|c|c|c|c|c|c|}
\hline
\;1\strut\;& \;2\; & \;3\; & \;4\; & \;5\; &\;6\;\\\hline
2 \strut& & & & & 5\\\hline
3 \strut& & & & & 4\\\hline
4 \strut& & & & & 3\\\hline
5 \strut& & & & & 2\\\hline
6\strut & 5& 4& 3& 2& 1\\\hline
\end{array}$$
(original image)
| The second row must contain a $3$ and a $4$, neither of which can be in either the third or the fourth column; thus, they must be in the second and fifth columns, in either order, and the missing $1$ and $6$ can then be filled in in either of two ways. That is, there are $2^2=4$ ways to fill in the second row acceptably.
Now proceed to the third row. It must contain a $2$ and a $5$, and they must go in the middle two columns; they can still go there in either order. The missing $1$ and $6$ will then fill the second and fifth columns, and they can appear in either order, so there are $4\cdot2^2=16$ ways to fill the second and third rows acceptably.
In the fourth row the missing $2$ and $5$ must occupy the middle two columns, just as in the third row. Thus, there is only one possible way to place them. The $1$ and $6$ must occupy the second and fifth columns, again just as in the third row, so they must be placed in the order not used in the third row. Thus, the layout of the fourth row is completely determined by the layout of the third row.
Similarly, the layout of the fifth row is completely determined by the layout of the second row, and the total number of acceptable arrangements is just $16$.
The array below shows the possibilities. Numbers separated by a slash are alternatives for that cell. The four red cells are free choices, but once they’ve been chosen everything else is completely determined.
$$\begin{array}{|c|c|c|c|c|c|} \hline
1&2&3&4&5&6\\ \hline
2&\color{red}{3/4}&\color{red}{1/6}&6/1&4/3&5\\ \hline
3&\color{red}{1/6}&\color{red}{2/5}&5/2&6/1&4\\ \hline
4&6/1&5/2&2/5&1/6&3\\ \hline
5&4/3&6/1&1/6&3/4&2\\ \hline
6&5&4&3&2&1\\ \hline
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/282136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a transformation in specified basis My task is to find a matrix of linear transformation $\varphi$ in basis $A,B$
$\varphi:\mathbb{R}^{2}\to\mathbb{R}^{4}
\varphi((x_{1},x_{2}))=(3x_{1}+x_{2},x_{1}+5x_{2},-x_{1}+4x_{2},2x_{1}+x_{2})$
$\mathcal{A}=\{(3,1),(4,2)\}
\mathcal{B}=\{(1,0,1,0),(0,1,1,1),(0,1,2,3),(0,0,0,1)\}$
How I've started:
$M_{st}^{st}(\varphi)=\left[\begin{matrix}3 & 1\\
1 & 5\\
-1 & 4\\
2 & 1
\end{matrix}\ \right] $
$M_{B}^{st}(id)=\left[\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 1 & 0\\
1 & 1 & 2 & 0\\
0 & 1 & 3 & 1
\end{array}\right]
$
$M_{A}^{st}(id)=\left[\begin{matrix}3 & 4\\
1 & 2
\end{matrix}\right]
$
$M_{st}^{A}(id)=(M_{A}^{st}(id))^{-1}
$
$\left[\begin{matrix}3 & 4 & 1\\
1 & 2 & & 1
\end{matrix}\right]\sim\left[\begin{matrix}1 & 0 & 1 & -2\\
1 & 2 & & 1
\end{matrix}\right]\sim\left[\begin{matrix}1 & 0 & 1 & -2\\
0 & 2 & -1 & 3
\end{matrix}\right]\sim\left[\begin{matrix}1 & 0 & 1 & -2\\
0 & 1 & -\frac{1}{2} & \frac{3}{2}
\end{matrix}\right]
$
$M_{st}^{A}(id)=\left[\begin{matrix}1 & -2\\
-\frac{1}{2} & \frac{3}{2}
\end{matrix}\right]
$
$M(id)_{st}^{A}\cdot M_{st}^{st}(\varphi)=M_{st}^{A}(\varphi)$
$M_{st}^{A}(\varphi)=\left[\begin{matrix}1 & -2\\
-\frac{1}{2} & \frac{3}{2}
\end{matrix}\right]\cdot\left[\begin{matrix}3 & 1\\
1 & 5\\
-1 & 4\\
2 & 1
\end{matrix}\right] = ???$
I was doing everything with my algorithm. But I did something wrong. Could someone point me where and how to fix it?
Thanks in advance!
| Assuming that vectors in $\mathbb{R}^2$ and $\mathbb{R}^4$ are represented by column vectors, you should find $M_B^{st}(id)^{-1}M_{st}^{st}(\varphi)M_A^{st}(id)$ instead. If you adopt a row vector convention, just transpose the resulting matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/282390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $a,b$ and $c$ be real numbers.evaluate the following determinant: |$b^2c^2 ,bc, b+c;c^2a^2,ca,c+a;a^2b^2,ab,a+b$| Let $a,b$ and $c$ be real numbers. Evaluate the following determinant:
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}$$
after long calculation I get that the answer will be $0$. Is there any short processs? Please help someone thank you.
| If $b=0,$
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix} =\begin{vmatrix}0 &0&c\cr c^2a^2&ca&c+a\cr 0&0&a\cr\end{vmatrix}$$
Now, if $a=0,$ $$\text{the determinant becomes }\begin{vmatrix}0 &0&c\cr 0&0&c\cr 0&0&0\cr\end{vmatrix}=0$$
else for $ca\ne 0$ $$\text{the determinant becomes }c^3a^3\begin{vmatrix}0 &0&c\cr 1&1&c\cr 0&0&0\cr\end{vmatrix}=0$$
So, if $abc\ne 0,$
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}$$
$$=\frac1{abc}\begin{vmatrix}a(b^2c^2) &abc& a(b+c)\cr b(c^2a^2)&cab&b(c+a)\cr c(a^2b^2)&abc&c(a+b)\cr\end{vmatrix} \text{ applying } R_1'=aR_1, R_2'=bR_2, R_3'=cR_3$$
$$=abc\begin{vmatrix}bc &1& a(b+c)\cr ca&1&b(c+a)\cr ab&1&c(a+b)\cr\end{vmatrix}\text{ applying } C_1'=\frac{C_1}{abc} \text{ and } C_2'=\frac{C_2}{abc}$$.
$$=abc\begin{vmatrix}bc &1& a(b+c)+bc\cr ca&1&b(c+a)+ca\cr ab&1&c(a+b)+ab\cr\end{vmatrix} \text{ applying } C_3'=C_3+C_1$$
If $ab+bc+ca=0,$ $$\text{the determinant becomes } abc\begin{vmatrix}bc &1& 0\cr ca&1&0\cr ab&1&0\cr\end{vmatrix}=0 $$
Else
$$=abc(ab+bc+ca)\begin{vmatrix}bc &1& 1\cr ca&1&1\cr ab&1&1\cr\end{vmatrix} \text{
applying } C_3'=\frac{C_3}{ab+bc+ca}$$
$$=abc(ab+bc+ca)\cdot 0 \text { as the 2nd & the last columns are identical.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/282655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
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Problem related to a square matrix Let $A$ be an $n\times n$ matrix with real entries such that $A^{2}+I=\mathbf{0}$. Then:
(A) $n$ is an odd integer.
(B) $n$ is an even integer.
(C) $n$ has to be $2$
(D) $n$ could be any positive integer.
I was thinking about the problem.I noticed for a $2\times 2$ matrix $A$ of the form
$$\begin{pmatrix}
1 &-2 \\
1& -1
\end{pmatrix},$$ the given condition holds good.So option (C) is a possibility.But I am not sure about other options. Is there any convenient way to tackle it?With regards..
| Note:
$n=1\,$ is ruled out, since e.g., $A$ consists of the single scalar entry $1$: $A = [1],\; I = I_1,\;, A^2 + I = 2.\;$
Indeed there is no real scalar $\,k\,\neq 0\,$ (in the case $n = 1, A = k\,$) such that $\,k^2 = -1.\,$
So option (A) is ruled out, since $\,n = 1\,$ is odd, and option (D) is ruled out, since $\,n = 1 >0\, n \in \mathbb{Z}^+$.
What remains is to decide between (B) and (C).
You know for $n = 2\,$, the equality is satisfied (hence "(C)" is in the "running") but $n = 2\,$ is also even: so "(B)" has a chance. You can rule out (C) if there exists any $\,n= 2k,\; k\in \mathbb{Z}^+$, $n > 2,\,$ such that $\,A_{n\times n}^2 + I_n\, = 0$.
*
*Hint: try $n = 4$: construct a $4\times 4$ matrix made of $4$-square block $2 \times 2$ matrices, using your matrix for each of the two block entries on the diagonal, and zero blocks off the diagonal.)
Let $A =
\begin{pmatrix}
1 &-2 \\
1& -1
\end{pmatrix}.
\quad$ So using $A^2 + I = 0$, construct
$A_{4 \times 4} =
\begin{pmatrix}
A & 0 \\
0 & A
\end{pmatrix}
$
$$
A_{4\times 4}^2 + I_4 =
\begin{pmatrix}
A &0 \\
0& A
\end{pmatrix}
\cdot
\begin{pmatrix}
A & 0 \\
0 & A
\end{pmatrix}
+ I_4 =
\begin{pmatrix}
A^2 & 0 \\
0& A^2
\end{pmatrix} +
\begin{pmatrix}
I_2 & 0\\
0 & I_2
\end{pmatrix}
$$
$$
=
\begin{pmatrix}
A^2 + I_2 & 0\\
0 & A^2 + I_2
\end{pmatrix}= 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$? Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$ ?
$f(x)=x^4+x^3+x^2+x+1$
$f(x^5)=x^{20}+x^{15}+x^{10}+x^5+1$
| Let $x^{20}+x^{15}+x^{10}+x^5+1=(x^4+x^3+x^2+x+1)Q(x)+R(x)$.
$x^4+x^3+x^2+x+1=0$ has $4$ complex roots $a_1,a_2,a_3,a_4$. And these are also roots of $x^5=1$, so when $x=a1,a2,a3,a4$, the above equation becomes
$5=R(a_i)$ ($i=1,2,3,4$)
This is true when $R(x)=5$ for every $x$, and it is easy to show
that a polynomial of degree $3$ like this is unique. So the remainder is just $5$ (constant).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$ How can I find the formula for the following equation?
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$$
More importantly, how would you approach finding the formula? I have found that every time, the denominator number seems to go up by $n+2$, but that's about as far as I have been able to get:
$\frac12 + \frac16 + \frac1{12} + \frac1{20} + \frac1{30}...$ the denominator increases by $4,6,8,10,12,\ldots$ etc.
So how should I approach finding the formula? Thanks!
| Hint: Use the fact that $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and find $S_n=\sum_1^n\left(\frac{1}{k}-\frac{1}{k+1}\right)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
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Proving that $ f(1)=\frac{1-\sqrt{5}}{2}$ for this function
Let $f:(0,+\infty)\mapsto R$ be a strictly increasing function such that $\forall x\ge0,$ $$f(x)+\frac{1}{x}\ge0, \qquad f(x)f\left(f(x)+\frac{1}{x}\right)=1.$$
Show that $$f(1)=\frac{1-\sqrt{5}}{2}.$$
Please give an example that satisfies these conditions.
Thanks in advance.
| $f(x)f(f(x)+\frac{1}{x})=1$
$x=1 $ gives $ f(1)f(f(1)+1)=1$
$x=f(1)+1\ge0$ gives $ f(f(1)+1)f(f(f(1)+1)+\frac{1}{f(1)+1})=1$
By replacing $f(f(1)+1)$ by $\frac{1}{f(1)}$ you get $\frac{1}{f(1)}f(\frac{1}{f(1)}+\frac{1}{f(1)+1})=1$
Multiply both sides by $f(1)$ to get $f(\frac{1}{f(1)}+\frac{1}{f(1)+1})=f(1)$
Then you know that the function is strictly increasing and therefore injective.
So $\frac{1}{f(1)}+\frac{1}{f(1)+1}=1$
And then solve for $f(1)$.
Let $x=f(1)$
$\frac{1}{x}+\frac{1}{x+1}=1$
Multiply by $x$ and $x+1$ to get $x+1+x=x(x+1)$
$2x+1=x^2+x$
$0=x^2-x-1$
$x=\cfrac{1\pm\sqrt{5}}{2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Equations over permutations Let $\sigma=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix} \in S_4$ and $\theta=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \end{pmatrix} \in S_4$.
Solve the following equations $(x \in S_4)$:
a) $x \sigma = \sigma x$;
b) $x^2 = \sigma$;
c) $x^2 = \theta$.
I'm writing here my work. I want to know if something goes wrong in my proof.
We denote the number of inversions in $σ$ with $N(\sigma)$ and the sign of permutation $\sigma$ with $\text{sgn}(\sigma)$.
b) We have
$\text{sgn}(\sigma)=(-1)^{N(\sigma)}=(-1)^3=-1$, and $\text{sgn}(x^2)=\text{sgn}(x) \cdot \text{sgn}(x)=(-1)^{2 \cdot N(x)}=1$.
Relation $\text{sgn}(\sigma) \ne \text{sgn}(x^2)$ imply that the equation $x^2 = \sigma$ has no solutions in $S_4$.
Ideas for the rest of the exercise, please?
| a) Let $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ a & b & c & d \end{pmatrix} \in S_4$. Then
$x\sigma=\begin{pmatrix} 1 & 2 & 3 & 4 \\ a & b & c & d \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix}=\begin{pmatrix} 1 & 2 & 3 & 4 \\ c & a & d & b \end{pmatrix}$.
We conclude that $x(\sigma(1))=c$, $x(\sigma(2))=a$, $x(\sigma(3))=d$, and $x(\sigma(4))=b$.
Case 1. If $a=1$, then $x(1)=1$, which implies that $\sigma(x(1))=\sigma(1)=3$. On the other hand, $\sigma(x(1))=c$. Hence $c=3$.
Now, $x(3)=3$, which implies that $\sigma(x(3))=\sigma(3)=4$. On the other hand, $x(\sigma(3))=d$. Hence $d=4$.
Now, $x(4)=4$, which implies that $\sigma(x(4))=\sigma(4)=2$. On the other hand, $x(\sigma(4))=b$. Hence $b=2$.
In this case we obtain the solution $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{pmatrix}$.
Case 2. If $a=2$, then $x(1)=2$, which implies that $\sigma(x(1))=\sigma(2)=1$. On the other hand, $\sigma(x(1))=c$. Hence $c=1$.
Now, $x(3)=1$, which implies that $\sigma(x(3))=\sigma(1)=3$. On the other hand, $x(\sigma(3))=d$. Hence $d=3$.
Now, $x(4)=3$, which implies that $\sigma(x(4))=\sigma(3)=4$. On the other hand, $x(\sigma(4))=b$. Hence $b=4$.
In this case we obtain the solution $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 1 & 3 \end{pmatrix}$.
Case 3. If $a=3$, then $x(1)=3$, which implies that $\sigma(x(1))=\sigma(3)=4$. On the other hand, $\sigma(x(1))=c$. Hence $c=4$.
Now, $x(3)=4$, which implies that $\sigma(x(3))=\sigma(4)=2$. On the other hand, $x(\sigma(3))=d$. Hence $d=2$.
Now, $x(4)=2$, which implies that $\sigma(x(4))=\sigma(2)=1$. On the other hand, $x(\sigma(4))=b$. Hence $b=1$.
In this case we obtain the solution $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix}$.
Case 4. If $a=4$, then $x(1)=4$, which implies that $\sigma(x(1))=\sigma(4)=2$. On the other hand, $\sigma(x(1))=c$. Hence $c=2$.
Now, $x(3)=2$, which implies that $\sigma(x(3))=\sigma(2)=1$. On the other hand, $x(\sigma(3))=d$. Hence $d=1$.
Now, $x(4)=1$, which implies that $\sigma(x(4))=\sigma(1)=3$. On the other hand, $x(\sigma(4))=b$. Hence $b=3$.
In this case we obtain the solution $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1 \end{pmatrix}$.
Conclusion: The equation $x\sigma=\sigma x$ has $4$ solutions in $S_4$:
$x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{pmatrix}$, $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 1 & 3 \end{pmatrix}$, $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix}$, and $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1 \end{pmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/287339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find five positive integers whose reciprocals sum to $1$ Find a positive integer solution $(x,y,z,a,b)$ for which
$$\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} + \frac{1}{a} + \frac{1}{b} = 1\;.$$
Is your answer the only solution? If so, show why.
I was surprised that a teacher would assign this kind of problem to a 5th grade child. (I'm a college student tutor) This girl goes to a private school in a wealthy neighborhood.
Please avoid the trivial $x=y=z=a=b=5$. Try looking for a solution where $ x \neq y \neq z \neq a \neq b$ or if not, look for one where one variable equals to another, but explain your reasoning. The girl was covering "unit fractions" in her class.
| Note that $\frac{4!}{4!}=1$. Now write $4!=24$ as $1+2+3+6+12$ (basically the divisors of $24$) Then we have $$1=\frac{4!}{4!}=\frac{1+2+3+6+12}{24}=\frac{1}{24}+\frac{1}{12}+\frac{1}{8}+\frac{1}{4}+\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "373",
"answer_count": 16,
"answer_id": 14
} |
Compute the limit $\displaystyle \lim_{x\rightarrow 0}\frac{n!.x^n-\sin (x).\sin (2x).\sin (3x).......\sin (nx)}{x^{n+2}}\;\;,$ How can i calculate the Given limit
$\displaystyle \lim_{x\rightarrow 0}\frac{n!x^n-\sin (x)\sin (2x)\sin (3x)\dots\sin (nx)}{x^{n+2}}\;\;,$ where $n\in\mathbb{N}$
| Since $\sin x = x - x^3/6 +O(x^5)$ as $x\to 0$, we get
$$\begin{array}
. & &\frac{n!x^n-\sin (x)\sin (2x)\sin (3x)\cdots\sin (nx)}{x^{n+2}}
\\&=&\frac{n!x^n - (x-x^3/6+O(x^5))\cdots(nx-(nx)^3/6+O(x^5))}{x^{n+2}}
\\&=& \frac{\frac{1}{6}x^{n+2}n! (1^2+2^2+\cdots+n^2)+O(x^{n+4})}{x^{n+2}}
\end{array}$$
as $x\to0$.
So desired limit is $\frac{1}{36}n!\cdot n(n+1)(2n+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/291087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Computing the value of $\operatorname{Li}_{3}\left(\frac{1}{2} \right) $
How to prove the following identity
$$ \operatorname{Li}_{3}\left(\frac{1}{2} \right) = \sum_{n=1}^{\infty}\frac{1}{2^n n^3}= \frac{1}{24} \left( 21\zeta(3)+4\ln^3 (2)-2\pi^2 \ln2\right)\,?$$
Where $\operatorname{Li}_3 (x)$ is the trilogarithm also the result from above can be found here in $(2)$.
In particular for $x=\frac12$ we have:$$\operatorname{Li}_3\left(\frac12\right)=\sum_{n=1}^\infty \frac{1}{n^3}\frac{1}{2^n}=\frac14 \sum_{n=1}^\infty \frac{1}{2^{n-1}} \int_0^\frac12 x^{n-1}\ln^2 xdx=\frac14 \int_0^1 \frac{\ln^2 x}{1-\frac{x}{2}}dx$$
| We can find the value similarly as seen here for $\operatorname{Li}_2\left(\frac12\right)=\frac{\pi^2}{12}-\frac{\ln^2 2}{2}$.
From the question we have:
$$\operatorname{Li}_3\left(\frac12\right)=\sum_{n=1}^\infty \frac{1}{n^3}\frac{1}{2^n}=\frac12 \int_0^1 \frac{\ln^2 x}{2-x}dx\overset{\large \frac{x}{2-x}=t}=\frac12\int_0^1 \ln^2\left(\frac{2t}{1+t}\right)\frac{dt}{1+t}$$
$$=\frac12 \int_0^1\frac{\ln^2 (2t)-2\ln(2t)\ln(1+t)+\ln^2(1+t)}{1+t}dt$$
We can rewrite the first two terms as:
$$\ln^2(2t)-2\ln(2t)\ln(1+t)=\color{blue}{\ln^2 2-2\ln 2 \ln(1+t)} +2\ln 2 \color{red}{\ln t}+\color{green}{\ln^2 t}-\color{orange}{2\ln t\ln(1+t)}$$
$$\small \color{blue}{\int_0^1 \frac{\ln^2 2-2\ln 2\ln(1+t)}{1+t}dt}=\ln 2 \int_0^1 \frac{\ln\left(\frac{2}{(1+t)^2}\right)}{1+t}dt\overset{\large t\to \frac{1-t}{1+t}}=-\ln 2 \int_0^1 \frac{\ln\left(\frac{2}{(1+t)^2}\right)}{1+t}dt=0$$
$$\color{red}{\int_0^1 \frac{\ln t}{1+t}dt}= \sum_{n=1}^\infty (-1)^{n-1} \int_0^1 t^{n-1} \ln t \, dt=-\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}=-\frac{\pi^2}{12}$$
$$\color{green}{\int_0^1 \frac{\ln^2 t}{1+t}dt}=\sum_{n=1}^\infty (-1)^{n-1} \int_0^1 t^{n-1} \ln^2 t \, dt=2\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3}=\frac32 \zeta(3)$$
$$\color{orange}{\int_0^1 \frac{2\ln t \ln(1+t)}{1+t}dt}\overset{IBP}=-\int_0^1 \frac{\ln^2(1+t)}{t}dt=-\frac{\zeta(3)}{4}$$
For the latter integral look here or just let $m=1,n=0,q=1,p=0$ in the following relation:
$$\small \int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx=\left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)$$
The last term is easily found:
$$\int_0^1 \frac{\ln^2(1+t)}{1+t}dt=\frac{\ln^3(1+t)}{3}\bigg|_0^1 =\frac{\ln^3 2}{3}$$
$$\Rightarrow \operatorname{Li}_3\left(\frac12\right)=\frac12 \left(-\frac{\pi^2}{6}\ln 2+\frac32\zeta(3)+\frac{\zeta(3)}{4}+\frac{\ln^3 2}{3}\right)=\boxed{\frac78\zeta(3)+\frac{\ln^3 2}{6}-\frac{\pi^2}{12}\ln 2}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "9",
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If $A\equiv 1\pmod{3}$, then $4p=A^2+27B^2$ uniquely determines $A$. If $p\equiv 1\pmod{3}$, it's well know that $p$ can be expressed as
$$
p=\frac{1}{4}(A^2+27B^2).
$$
In this letter by Von Neumann, he mentions that Kummer determined that $A$ is in fact uniquely determined by the additional condition $A\equiv 1\pmod{3}$.
The reference that is listed is Kummer's De residuis cubicis disquisitiones nonnullae analyticae, but I'm having a hard time getting my hands on the original source. Is there an elelementary proof of why $A$ is uniquely determined under these two conditions? Thanks.
| It looks like you're doing calculation in the number field $\mathbb{Q}(\sqrt{-3})$. In particular, the typical algebraic integer is of the form
$$ u + v \frac{1 + \sqrt{-3}}{2} $$
or put differently, as
$$ \frac{A + B \sqrt{-3}}{2} $$
where $A$ and $B$ have the same parity. The norm of this element is
$$ N = \frac{A + B \sqrt{-3}}{2} \frac{A - B \sqrt{-3}}{2} = \frac{1}{4}(A^2 + 3 B^2)$$
The unit group of (the ring of integers of) this number field is simply the set of sixth roots of unity.
Assuming $N$ is prime, there are only 12 elements whose norm is $N$: the six multiples of this element by the sixth roots of unity, and their complex conjugates:
*
*$\frac{1}{2} (A + B \sqrt{-3}) $
*$\frac{1}{2} (\frac{A-3B}{2} + \frac{A+B}{2} \sqrt{-3}) $
*$\frac{1}{2} (\frac{-A-3B}{2} + \frac{A-B}{2} \sqrt{-3}) $
*$\frac{1}{2} (-A - B \sqrt{-3}) $
*$\frac{1}{2} (\frac{-A+3B}{2} + \frac{-A-B}{2} \sqrt{-3}) $
*$\frac{1}{2} (\frac{A+3B}{2} + \frac{-A+B}{2} \sqrt{-3}) $
*$\frac{1}{2} (A - B \sqrt{-3}) $
*$\frac{1}{2} (\frac{A-3B}{2} + \frac{-A-B}{2} \sqrt{-3}) $
*$\frac{1}{2} (\frac{-A-3B}{2} + \frac{-A+B}{2} \sqrt{-3}) $
*$\frac{1}{2} (-A + B \sqrt{-3}) $
*$\frac{1}{2} (\frac{-A+3B}{2} + \frac{A+B}{2} \sqrt{-3}) $
*$\frac{1}{2} (\frac{A+3B}{2} + \frac{A-B}{2} \sqrt{-3}) $
If $N$ is a prime integer, then $(A + B \sqrt{-3})/2$ is actually a prime element of $\mathbb{Z}[(1 + \sqrt{-3})/2]$. If we also require $N \neq 3$, then $A,B$ are both odd and $A$ is not divisible by $3$.
It's easy to check that exactly four of these elements have a coefficient on $\sqrt{-3}$ equivalent to $0$ modulo $3$. WLOG, we may assume it is the four $(\pm A \pm B \sqrt{-3})/2$. Among these, exactly two have the first term equivalent to $1$ modulo $3$. WLOG assume it is the two $(A \pm B \sqrt{-3})/2$. Finally, WLOG we may assume $B$ is positive, and we're down to a single possibility.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Gaussian-Jordan Elimination question? I have the linear system
$$
\begin{align*}
2x-y-z+v&=0 \\
x-2y-z+5u-v&=1 \\
2x-z+v&=1
\end{align*}$$
Very well. I form the matrix
$$
\left[
\begin{array}{@{}ccccc|c@{}}
2&-1&-1 & 0 & 1 &0 \\
1&-2&-1 & 5 & -1 &1 \\
2&0&-1 & 0&1&1 \\
\end{array}
\right]
$$
So I thought about exchanging the first row with the second one,and the first column with the last column. The first row is $a1$,the second $a$2 and the third $a3$. So what I do is find $a3-a2$ ,then multiply $a1$ by $-2$, then do $a2+a1$ , then $a3-a1$, this way I will have a $scaled$ $matrix$ and it will look like
$$
\left[
\begin{array}{@{}ccccc|c@{}}
-2&2&2 & -10 & -4 &-2 \\
0&-3&1 &-10 & -5 &-2 \\
0&0&0& 0&1&1 \\
\end{array}
\right]
$$
When I proceed with the Gaussian method, I don't know what to multiply or divide to get some of the numbers in the matrix zero. Help me ? Please.
| Assuming that you've done everything right so far
$$
\left[
\begin{array}{@{}ccccc|c@{}}
-2&2&2 & -10 & -4 &-2 \\
0&-3&1 &-10 & -5 &-2 \\
0&0&0& 0&1&1 \\
\end{array}
\right]
$$
Now, let the first row be L1, the second L2, the third L3.
L1$\times$(-0.5) and then, L1-2L2.
You got
$$
\left[
\begin{array}{@{}ccccc|c@{}}
1&-1&-1 & 5 & 0&-1 \\
0&-3&1 &-10 & -5 &-2 \\
0&0&0& 0&1&1 \\
\end{array}
\right]
$$
next, you do L2$\times$(-1/3)-(5/3)L3. You got
$$
\left[
\begin{array}{@{}ccccc|c@{}}
1&-1&-1 & 5 & 0&-1 \\
0&1&-1/3 &10/3 & 0 &-2/3 \\
0&0&0& 0&1&1 \\
\end{array}
\right]
$$
Finally, L1+L2, you have:
$$
\left[
\begin{array}{@{}ccccc|c@{}}
1& 0& -4/3 & 25/3 & 0&-2 \\
0&1&-1/3 &10/3 & 0 &-1 \\
0&0&0& 0&1&1 \\
\end{array}
\right]
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding Binomial expansion of a radical I am having trouble finding the correct binomial expansion for $\dfrac{1}{\sqrt{1-4x}}$:
Simplifying the radical I get: $(1-4x)^{-\frac{1}{2}}$
Now I want to find ${n\choose k} = {\frac{-1}{2}\choose k}$
\begin{align}
{\frac{-1}{2}\choose k} &= \dfrac{\frac{-1}{2}(\frac{-1}{2}-1)(\frac{-1}{2}-2)\ldots(\frac{-1}{2}-k+1)}{k!} \\
&= (-1)^k\dfrac{\frac{1}{2}(\frac{1}{2}+1)(\frac{1}{2}+2)\ldots(\frac{1}{2}+k-1)}{k!} \\
&=(-1)^k{\frac{1}{2}+k-1\choose k}
\end{align}
Now, applying Newton's Generalized Binomial Theorem I get:
\begin{equation}\sum\limits_{k=0}^\infty(-1)^k{\frac{1}{2}+k-1\choose k}(4x)^k
\end{equation}
Is this correct? it seems odd to me since the top number of the combination is always less than the bottom number.
Furthermore, I would like to write $\left(\dfrac{1}{\sqrt{1-4x}}\right)^2$ as an infinite sum like:
\begin{equation}\sum\limits_{k=0}^\infty\sum\limits_{i=0}^k(expression)x^k
\end{equation}
I should be able to square the entire sum for what I derived above, and get what I know from calculus:
\begin{equation}
\left(\dfrac{1}{\sqrt{1-4x}}\right)^2 = \dfrac{1}{1-4x} = \sum\limits_{k=0}^\infty(4x)^k
\end{equation}
The trouble is finding the expression for the double sum.
All help is greatly appreciated!
EDIT: With the help from below we should get the double sum to be: $\sum\limits_{k=0}^\infty\sum\limits_{i=0}^k{k\choose i}^2x^k$
| Also, generalizing what Andre Nicolas wrote,
\begin{align}
{\frac{-1}{2}\choose k} &= \dfrac{\frac{-1}{2}(\frac{-1}{2}-1)(\frac{-1}{2}-2)\ldots(\frac{-1}{2}-k+1)}{k!} \\
&= (-1)^k\dfrac{\frac{1}{2}(\frac{1}{2}+1)(\frac{1}{2}+2)\ldots(\frac{1}{2}+k-1)}{k!} \\
&=(-1)^k \frac{1\ 3\ 5 ... (2k-1)}{2^k k!} \\
&=(-1)^k \frac{1\ 2\ 3\ 4\ 5 ... (2k-1)(2k)}{2^k k!(2\ 4\ ...(2k))} \\
&=(-1)^k \frac{(2k)!}{2^k k!2^k(1\ 2\ ...k)} \\
&=(-1)^k \frac{(2k)!}{4^k k!^2} \\
&= (-1)^k {2k \choose k} \big/4^{k} \\
\end{align}
This is a standard bit of manipulation
that, combined with Stirling's factorial approximation,
allows good approximations for this expression
to be derived.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/299421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defined a function $f(x)$ such that
$$
f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}}
$$
We can see that
$$\begin{align}
f(0) &= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\
&= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
\end{align}$$
And we begin solving by
$$\begin{align}
f(x) &= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\
f(x)^2 &= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\
&= 2^x + f(x + 1) \\
f(x + 1) &= f(x)^2 - 2^x
\end{align}$$
At this point I find myself stuck, as I have little experience with recurrence relations.
How would this recurrence relation be solved? Would the method extend easily to
$$\begin{align}
f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\
f_n(x)^2 &= n^x + f_n(x + 1)~\text ?
\end{align}$$
| Since you have recurrence relation $(f(x))^2=2^x+f(x+1)$ you could find an approximate solution by approximating $f(x+1)\approx f(x)$ and then you get quadratic equation for the function $f$ and in doing so you can find an approximate value $f(x)$ for every value of $x$. It is clear from the recurrence relation that the exact expression for $f(x)$ may not be expressible as some combination of the functions that have been studied to this day, but maybe I am wrong.
If you still want to search for closed forms for certain values of $f$ maybe it is better to study expressions of this type that are finite, such as $f(2,x,n) =\sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} +\sqrt {\ldots+\sqrt{2^{x+n}}}}}}}$ and then let $n\to\infty$ to get your $f$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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"answer_id": 3
} |
Find all pairs of positive whole numbers Find all pairs of positive whole numbers x and y which are a solution for $ \dfrac{2}{x} + \dfrac {3}{y} = 1 $.
I don't really understand how to tackle this question. I rewrote $ \dfrac{2}{x} + \dfrac {3}{y} = 1 $ as $2y+ 3x =xy$ but that's it..
| If we multiply both sides the original equation by $xy$, we ge $2y+3x=xy$. We can rewrite this as $2y+3x-xy=0$. We now perform a little trick.
Note that what we have is very much like $(3-y)(x-2)=3x+2y-xy-6$. If we subtract $6$ from both sides, we get $2y+3x-xy-6=-6$ and $(3-y)(x-2)=-6$. Multiplying through by $-1$ gives $(y-3)(x-2)=6$. You can now search for pairs of small $x$ and $y$ that satisfy the condition. (Hint: Use the prime factorization of $6$ to help you.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to solve a system of 3 equations with Cramer's Rule? I am given the following system of 3 simultaneous equations:
$$
\begin{align*}
4a+c &= 4\\
19a + b - 3c &= 3\\
7a + b &= 1\end{align*}
$$
How do I solve using Cramers' rule?
For one, I do know that by putting as a matrix the LHS
$$\begin{pmatrix}
4&0&1\\19&1&-3\\7&1&0
\end{pmatrix}$$
and then computing its determinant to be $24$ is of some use...but could somebody give more ideas?
| $$\begin{pmatrix}
4&0&1\\19&1&-3\\7&1&0
\end{pmatrix}=\frac{1}{a}\begin{pmatrix}
4a&0&1\\19a&1&-3\\7a&1&0
\end{pmatrix}$$
$$\Rightarrow \frac{1}{a}\begin{pmatrix}
(4a+0.b+1.c)&0&1\\(19a+b-3c)&1&-3\\(7a+b+0.c)&1&0
\end{pmatrix}$$
$$\Rightarrow \frac{1}{a}\begin{pmatrix}
4&0&1\\3&1&-3\\1&1&0
\end{pmatrix}$$
$$\Rightarrow \begin{pmatrix}
4&0&1\\19&1&-3\\7&1&0
\end{pmatrix}=\frac{1}{a}\begin{pmatrix}
4&0&1\\3&1&-3\\1&1&0
\end{pmatrix}$$
$$\Rightarrow \frac{det(\begin{pmatrix}
4&0&1\\3&1&-3\\1&1&0
\end{pmatrix})}{det \begin{pmatrix}
4&0&1\\19&1&-3\\7&1&0
\end{pmatrix} }=a$$
Similarly for others.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$ \log_{\frac 32x_{1}}\left(\frac{1}{2}-\frac{1}{36x_{2}^{2}}\right)+\cdots+ \log_{\frac 32x_{n}}\left(\frac{1}{2}-\frac{1}{36x_{1}^{2}}\right).$ Let $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ be $n$ real numbers in $\left(\frac{1}{4},\frac{2}{3}\right)$. Find the minimal value of the expression: $ \log_{\frac 32x_{1}}\left(\frac{1}{2}-\frac{1}{36x_{2}^{2}}\right)+\log_{\frac 32x_{2}}\left(\frac{1}{2}-\frac{1}{36x_{3}^{2}}\right)+\cdots+ \log_{\frac 32x_{n}}\left(\frac{1}{2}-\frac{1}{36x_{1}^{2}}\right). $
| The expression equals to $2n$ when all $x_i=\frac13$.
The expression cannot be less than $2n$ because of AM-GM inequality due to the following:
$$\frac{\ln\left(\frac12-\frac1{36x^2}\right)}{\ln\left(\frac32x\right)}=2\frac{\ln\left(\frac94x^2-\left(\frac32x-\frac1{6x}\right)^2\right)}{\ln\left(\frac94x^2\right)}\ge2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{\pi}{4}\le\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n+1}-\sqrt{n}}{n+1}\right)$ Prove that
$$\frac{\pi}{4}\le\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n+1}-\sqrt{n}}{n+1}\right)$$
EDIT: inspired by Michael Hardy's suggestion I got that
$$\arcsin \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}=\arcsin\frac{1}{\sqrt{n+1}}-\arcsin\frac{1}{\sqrt{n+2}}$$
and then
$$\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n+1}-\sqrt{n}}{n+1}\right)\ge\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}\right)\rightarrow\frac{\pi}{4}$$
because $\sum_{n=1}^{\infty} \left(\arcsin\frac{1}{\sqrt{n+1}}-\arcsin\frac{1}{\sqrt{n+2}}\right)=\arcsin \frac{\sqrt{2}}{2}=\frac{\pi}{4}$
Sis & Chris.
| Since $\arcsin x\ge \arctan x$ for $x \in [0,1]$, thus we shall have
$$\arcsin(\frac{\sqrt{n+1}-\sqrt{n}}{n+1})\ge \arctan(\frac{\sqrt{n+1}-\sqrt{n}}{n+1})\ge \arctan\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}+1} $$ $$= \arctan{\sqrt{n+1}}-\arctan{\sqrt{n}}.$$
(The last equality uses $\arctan(x) - \arctan(y) = \arctan(\frac{x-y}{1 + xy})$.)
Done.
P.S.
$\sum \arcsin(\frac{\sqrt{n+1}-\sqrt{n}}{n+1}) \ge \sum(\arctan{\sqrt{n+1}}-\arctan{\sqrt{n}}) = \pi/2-\arctan(1)=\pi/4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Integral of type $\displaystyle \int\frac{1}{\sqrt[4]{x^4+1}}dx$ How can I solve integral of types
(1) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4+1}}dx$
(2) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4-1}}dx$
| Looks like the variable substitution in Lab Bhattacharjee's answer can be generalized to handle indefinite integrals of the form:
$$ \int \frac{dx}{\sqrt[n]{x^{n}+1}}$$
Let $y = \frac{x}{\sqrt[n]{x^n+1}}$, we have:
$$n y^{n-1} dy = \frac{n x^{n-1} dx}{(1+x^n)^2} \implies \frac{dy}{y} = \frac{dx}{x(1+x^n)} = ( 1 - y^n)\frac{dx}{x} \implies \frac{dx}{\sqrt[n]{x^n+1}} = \frac{dy}{1-y^n}$$
Let $\zeta$ be a $n^{th}$ primitive root of $1$, we have:
$$\int\frac{dy}{1 - y^n} = -\int\left[\frac{1}{n} \sum_{k=0}^{n-1} \frac{\zeta^k}{y - \zeta^k}\right]dy = -\frac{1}{n}\sum_{k=0}^{n-1}\zeta^k \log(1-\frac{y}{\zeta^k}) + C \tag{*}$$
For $n = 4$, R.H.S of (*) reduces to
$$\begin{align}
&-\frac14 \left( \log(1-y) - \log(1+y) + i \left[\log(1-\frac{y}{i}) - \log(1 + \frac{y}{i})\right]\right) + C \\
=&
\frac12 \left[ \frac12 \log(\frac{1+y}{1-y}) + \frac{1}{2i}\log(\frac{1+yi}{1-yi}) \right] + C\\
=&
\frac12 \left[ \frac12 \log(\frac{1+y}{1-y}) + \arctan(y) \right] + C
\end{align}$$
For other $n$, let $c_k = \cos(\frac{2k\pi}{n})$ and $s_k = \sin(\frac{2k\pi}{n})$.
When $n$ is even, (*) reduces to:
$$ \frac{1}{n} \log(\frac{1+y}{1-y}) - \frac{2}{n} \sum_{k=1}^{\frac{n}{2}-1}
\left[ c_k \log( 1 + y^2 - 2 c_k y) - s_k \arctan(\frac{s_k y}{1 - c_k y})\right] + C
$$
When $n$ is odd, (*) reduces to:
$$ -\frac{1}{n} \log(1-y) - \frac{2}{n} \sum_{k=1}^{\frac{n-1}{2}}
\left[ c_k \log( 1 + y^2 - 2 c_k y) - s_k \arctan(\frac{s_k y}{1 - c_k y})\right] + C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$ The following question comes from Some integral with sine post
$$\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$$
but now I'd be curious to know how to deal with it by methods of complex analysis.
Some suggestions, hints? Thanks!!!
Sis.
| I have a generalized elementary method for this problem，If f (x) is an even function, and the period is $\pi$,we have:
$$\int_{0}^\infty f(x)\frac{\sin^nx}{x^n}dx=\int_{0}^\frac{\pi}{2}f(x)g_n(x)\sin^nxdx \qquad (1)$$
Where the $g_n(x)$ in (1) is as follows
$$g_n(x)=\begin{cases}\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left(\csc x\right),& \text{for n is odd $n\in\Bbb N$ and}\\[2ex]
\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left(\cot x\right),& \text{ for n is even .}
\end{cases}$$
——————————————————————————————————————————————————
Proof:
\begin{align}
\int_{0}^\infty f(x)\frac{\sin^nx}{x^n}dx&=\sum_{k=0}^\infty\int_{k\pi}^{(2k+1)\frac{\pi}{2}}f(x)\left(\frac{\sin x}{x}\right)^ndx+\sum_{k=1}^\infty\int_{(2k-1)\frac{\pi}{2}}^{k\pi}f(x)\left(\frac{\sin x}{x}\right)^ndx\\
&=\sum_{k=0}^\infty\int_{0}^{\frac{\pi}{2}}f(x+k\pi)\left(\frac{\sin (x+k\pi)}{x+k\pi}\right)^ndx+\sum_{k=1}^\infty\int_{-\frac{\pi}{2}}^{0}f(x+k\pi)\left(\frac{\sin (x+k\pi)}{x+k\pi}\right)^ndx\\
&=\sum_{k=0}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(x)\left(\frac{\sin x}{x+k\pi}\right)^ndx+\sum_{k=1}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(-x)\left(\frac{\sin x}{x-k\pi}\right)^ndx\\
&=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nx\left(\frac{1}{x^n}+\sum_{k=1}^\infty(-1)^{nk}\left[\frac{1}{(x+k\pi)^n}+\frac{1}{(x-k\pi)^n}\right]\right)dx\\
&=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nxg_n(x)dx
\end{align}
We know by the Fourier series
\begin{align}
\csc x&=\frac{1}{x}+\sum_{k=1}^\infty(-1)^k\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right)\\
\end{align}
and
\begin{align}
\cot x&=\frac{1}{x}+\sum_{k=1}^\infty\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right)
\end{align}
Take the n-1 order derivative,thus we obtain $g_n(x)$.
——————————————————————————————————————————————————
Example:
\begin{align}
(1.)\qquad\int_{0}^{\infty}\frac{\sin^3x}{x}dx&=\int_{0}^{\frac{\pi}{2}}\sin^2xg_1(x)\sin xdx\\
&=\int_{0}^{\frac{\pi}{2}}\sin^2x\frac{1}{\sin x}\sin xdx\\
&=\int_{0}^{\frac{\pi}{2}}\sin^2xdx\\
&=\frac{\pi}{4}\\
\end{align}
\begin{align}
(2.)
\int_{0}^{\infty}(1+\cos^2x)\frac{\sin^2x}{x^2}dx
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)g_2(x)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(-\frac{d}{dx}\cot x\right)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(\frac{1}{\sin^2x}\right)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)dx\\
&=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}\\
\end{align}
\begin{align}
(3.)
\int_{0}^{\infty}\frac{1}{(1+\cos^2x)}\frac{\sin^3x}{x^3}dx
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}g_3(x)dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\left(\frac{1}{2}\frac{d^2}{dx^2}(\csc x)\right)dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\frac{(1+\cos^2x)}{2\sin^3x}dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1}{2}dx=\frac{\pi}{4}\\
(4.)
\int_{0}^{\infty}\frac{1}{3+\cos2x}\frac{\sin^2x}{x^2}dx
&=\int_{0}^{\frac{\pi}{2}}\frac{1}{3+\cos2x}dx
=\frac{\pi}{4\sqrt{2}}\\
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/307510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
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Need help using De Moivre's theorem to write $\cos 4\theta$ & $\sin 4\theta$ as terms of $\sin\theta$ and $\cos\theta$ I need help with the following question:
"Use De Moivre's theorem to write $\cos 4\theta$ & $\sin 4\theta$ as terms of $\sin\theta$ and $\cos\theta$"
You could write the problem as:
$(\cos\theta+i\sin\theta)^4 - i\sin 4\theta$ = $\cos 4\theta$
I don't think it is ment that you should develop the $exp 4$ parenthesis, so there must be a simpler way? What would be your strategy with approaching this problem?
Thank you
| Use the binomial theorem for $(x+i y)^4$:
$$\begin{align}(x+i y)^4 &= x^4 + 4 i x^3 y +6 i^2 x^2 y^2+4 i^3 x y^3 + i^4 y^4\\&= x^4-6 x^2 y^2 + y^4 + i (4 x^3 y - 4 x y^3) \end{align}$$
using $i^2=-1$, etc. Now let $x=\cos{\theta}$, $y=\sin{\theta}$:
$$e^{i 4 \theta} = \cos{4 \theta} + i \sin{4 \theta} = (\cos{\theta} + i \sin{\theta})^4 $$
so that
$$\cos{4 \theta} = \cos^4{\theta} - 6 \cos^2{\theta} \sin^2{\theta} + \sin^4{\theta}$$
$$\sin{4 \theta} = 4 \cos{\theta} \sin{\theta} (\cos^2{\theta} - \sin^2{\theta}) = 2 \sin{2 \theta} \cos{2 \theta}$$
Note also that $\cos{4 \theta}$ could be derived from
$$\cos{4 \theta} = \cos^2{2 \theta} - \sin^2{2 \theta}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$ Converges? $$\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$$
Do you have an idea about this serie? If it converges what is the sum?
| This is
$$\sum_{k=1}^{\infty} (-1)^{k+1} \frac{2 k+1}{k (k+1)}$$
For large $k$, the summand behaves as $2(-1)^k/k$. By comparison to
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \log{2}$$
the sum converges.
Note that the sum may be expressed as
$$\sum_{k=1}^{\infty} (-1)^{k+1} \left ( \frac{1}{k} + \frac{1}{k+1} \right ) = \log{2} + (1-\log{2}) = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding the derivative of $x$ to the power something that is a function of $x$ if $y = x^{(x+1)^\frac12}$
then how can I get the first derivative of $y$?
| Another way to solve it is to take $\ln$ of both sides, and apply implicit differentiation:
$$y=x^{(x+1)^{\frac {1}{2}}}$$
$$\ln y=\ln x^{(x+1)^{\frac {1}{2}}}$$
Rewriting using $\log$ properties:
$$\ln y=(x+1)^{\frac {1}{2}} \cdot \ln x$$
Now take the derivative implicitly:
$$\frac{1}{y}\cdot y' = \frac{1}{2}(x+1)^{-\frac{1}{2}} \ln x + (x+1)^{\frac {1}{2}} \cdot \frac{1}{x}$$
We want $y'$, so we multiply $y$ on both sides to isolate $y'$.
$$ y' = y \cdot \left( \frac{1}{2}(x+1)^{-\frac{1}{2}} \ln x + (x+1)^{\frac {1}{2}} \cdot \frac{1}{x} \right)$$
We want the right hand side to be in terms of $x$. The original problem says $y=x^{(x+1)^{\frac {1}{2}}}$, so
$$ y' = \big( x^{(x+1)^{\frac {1}{2}}} \big) \cdot \left( \frac{1}{2}(x+1)^{-\frac{1}{2}} \ln x + (x+1)^{\frac {1}{2}} \cdot \frac{1}{x} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/310822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does this sum go to $0$? http://www.math.chalmers.se/Math/Grundutb/CTH/tma401/0304/handinsolutions.pdf
In problem (2), at the very end it says
$$\left(\sum_{k = n+1}^{\infty} \frac{1}{k^2}\right)^{1/2} \to 0$$
I don't see how that is accomplished. I understand the sequence might, but how does the sum $$\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$?
Is one allowed to do the following?
$$\lim_{n \to \infty}\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$
$$=\left (\lim_{n \to \infty}\frac{1}{(n+1)^2}+ \lim_{n \to \infty}\frac{1}{(n+2)^2} + \lim_{n \to \infty}\frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$
$$(0 + 0 + 0 + \dots)^{1/2} = 0$$
| Since the sum $\sum_{k=1}^\infty \frac{1}{k^2}$ is convergent, you can make the "tail" of the sum ($\sum_{k=n+1}^\infty \frac{1}{k^2}$) as small as you want.
Explicitly, since $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$, for any $\epsilon>0$, you can find an $n$ such that $\sum_{k=n+1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}-\sum_{k=1}^n \frac{1}{k^2}<\epsilon$.
In that case, the tail sums form a strictly decreasing sequence of positive numbers, which can be made smaller than any $\epsilon>0$.
That is convergence to 0! :)
Naturally if the expression converges to zero, so does the square root of the expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/311528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 4
} |
Maclaurin series for $\frac{x}{e^x-1}$ Maclaurin series for
$$\frac{x}{e^x-1}$$
The answer is
$$1-\frac x2 + \frac {x^2}{12} - \frac {x^4}{720} + \cdots$$
How can i get that answer?
| One way is to write $e^x-1 $ as $1 + x + x^2/2 + ... - 1$ and then factor out $x$ and cancel up the top and expand it as geometric series and collect the coefficients of like powers.
$\displaystyle
\begin{align*}
e^x - 1 &= x + \frac{x^2}{2!} + \frac{x^3}{3!} + o(x^4)\\
\frac{x}{e^x - 1} &= \frac{1}{1 + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )} \\
&= 1 - \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right ) + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )^2 - \left( \frac x 2 - \frac{x^2}{6} + o(x^3) \right )^3 + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )^4... \\
&= 1 - \frac{x}{2} + x^2 \left( \frac 1 4 - \frac 1 6 \right ) + x^3 \left(-\frac{1}{4!} + 2 \cdot \frac 1 2 \cdot \frac 1 6 - \frac{1}{2^3} \right ) + x^4 \left(-\frac{1}{5!} + \frac{1}{6^2} + 2 \cdot \frac 12 \cdot \frac{1}{4!} + \frac{1}{2^4}\right )+o(x^5)
\end{align*} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/311817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
} |
Determining a point's coordinates on a circle So I have a circle (I know its center's coordinates and radius) and a point on the circle (I know its coordinates) and I have to determine the coordinates of another point on the circle which is exactly at the distance L from the first point.
| If the equation of the circle be $(x-h)^2+(y-k)^2=r^2$ any point on the circle can be parametrized as $(r\cos\theta+h,r\sin\theta+k)$ and if the known point on the circle be $S(p,q)$
So, we need to know the two values of $\theta$ from
$(r\cos\theta+h-p)^2+(r\sin\theta+k-q)=L^2$
$\implies 2r(h-p)\cos\theta+2r(k-q)\sin\theta=L^2-r^2-(h-p)^2-(k-q)^2$
Now, we know $-\sqrt{c^2+d^2}\le c\cos A+d\sin A\le \sqrt{c^2+d^2}$(Proof below)
There will be two or zero solutions according as $c\cos A+d\sin A$ lies within or outside the range.
[
Proof: Let $c=R\cos B,d=R\sin B$ where real $R\ge0$
So, $c^2+d^2=R^2\implies R=\sqrt{c^2+d^2}$ and $\tan B=\frac dc$
and $c\cos A+d\sin A=\sqrt{c^2+d^2}\cos(A-B)$
As $-1\le\cos(A-B)\le1, -\sqrt{c^2+d^2}\le\sqrt{c^2+d^2}\cos(A-B)\le \sqrt{c^2+d^2}$
$\implies -\sqrt{c^2+d^2}\le \cos A+d\sin A\le \sqrt{c^2+d^2}$
]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/312704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Rolling 2 fair, 6 sided dice find P(sum=12, 2times in 36 rolls) I have a dice problem.
If we roll 2 fair dice, and the sum is 12 then our test is a pass, otherwise its a fail.
What is the probability that the number of passes in 36 tests is greater then 1.
S={1,2,3,4,5,6}
(S,S) = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
event
Let e = sum of both dice is 12
e in s = {(6,6)}
P(e)=1/36
Then, i guess i would need to take the inverse of the failures
1 - (35/36)$*$(35/36)$*$(35/36)$*$ ... $*$(35/36)
1 - .3267 = .6372
If im looking at this right, then i have just calculated the probability that my test passes at least once. How do i modify this to calculate at least 2 times?
EDIT:
I think i should look at it like this:
P(eefff...f) = 1/36^2 * 35/36^34
| Note that there is only way way to achieve a $12$ when rolling two dice (whose outcomes we will denote by the random variables $X_{1}$ and $X_{2}$) and taking their sum, which is when both dice show $6$, which are both independent events with probability of $\frac{1}{6}$, therefore:
$$P(X_{1}+X_{2}=12)=P(X_{1}=6 \cap X_{2}=6)=P(X_{1}=6)P(X_{2}=6)=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$$
We now know the outcome of each Bernoulli trial is a success with probability $\frac{1}{36}$, therefore, we can model the probability of the number of successes in $36$ trials using a Binomial distributions $Y\sim B(36,\frac{1}{36})$. Therefore, the probability of achieving exactly $2$ successes is:
$$P(Y=2)={36\choose 2}\left(\frac{1}{36}\right)^{2}\left(\frac{35}{36}\right)^{34}\approx0.1865$$
EDIT: Your question asks for the probability that there are at least $2$ successes (i.e. $P(Y\geq2)$), which is:
$$P(Y\geq2)=1-P(Y\leq1)=1-P(Y=0\cup Y=1)=1-\left(\left(\frac{35}{36}\right)^{36}+{36 \choose 1}\left(\frac{1}{35}\right)\left(\frac{35}{36}\right)^{35}\right)\approx1-0.2642\approx0.7358$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/314223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple AM-GM inequality Let $a,b,c$ positive real numbers such that $a+b+c=3$, using only AM-GM inequalities show that
$$
a+b+c \geq ab+bc+ca
$$
I was able to prove that
$$
\begin{align}
a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\
&\ge \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\frac{2\sqrt{a^2c^2}}{2} \\
&= ab+bc+ca
\end{align}
$$
but now I am stuck. I don't know how to use the fact that $a+b+c=3$ to prove the inequality. Anybody can give me a hint?
| $9 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac) \geq 3(ab+bc + ac)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/315699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Two finite fields are isomorphic. Let $F = \Bbb{Z}_2$. Given the irreducible polynomials $f(x)= x^3 + x + 1$, and $g(y) = y^3 + y^2 + 1$, form the fields $K = F[x]/(f(x))$ and $E = F[y] / (g(y))$. These are fields of order 8 (given), so they must be isomorphic. Is the map $[x] \mapsto [y + 1]$ an isomorphism? It's clearly onto, and and it's one-one since $F$ and $E$ both have 8 elements.
| An element of $K$ has a unique representation as $p + (f)$, where $\deg p \leq 2$ and an element of $E$ has a unique representation as $q + (g)$, where $\deg q\leq 2$. Take $A = ax^2 + bx + c + (f)$, $B = a'x^2 + b'x + c + (f)$.
\begin{align*}
\phi(A + B) &= \phi(ax^2 + bx + c + (f) + a'x^2 + b'x + c + (f))\\
&= \phi((a + a')x^2 + (b + b')x + c + c' + (f))\\
&= (a + a')(y + 1)^2 + (b + b')(y + 1) + c + c' + (g)\\
&= (ay^2 + a + by + b + c + (g)) + (a'y^2 + a' + b'y + b' + c' + (g))\\
&= \phi(A) + \phi(B)
\end{align*}
Then if you can show $\phi(AB) = \phi(A)\phi(B)$ using a similar calculation, $\phi$ is a bijective homomorphism, and hence an isomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/316305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving the inequality $\arctan\frac{\pi}{2}\ge1$ Do you see any nice way to prove that
$$\arctan\frac{\pi}{2}\ge1 ?$$
Thanks!
Sis.
| I'm not sure if this would be considered nice, but anyway:
It suffices to show that $\tan{1} \leq \frac{\pi}{2}$, or equivalently $\sin{1} \leq \frac{\pi}{2}\cos{1}$.
$$\sin{x}=(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!})+\sum_{n=2}^{\infty}{\left(\frac{x^{4n+1}}{(4n+1)!}-\frac{x^{4n+3}}{(4n+3)!}\right)}$$
$$\cos{x}=(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!})+\sum_{n=2}^{\infty}{\left(\frac{x^{4n}}{(4n)!}-\frac{x^{4n+2}}{(4n+2)!}\right)}$$
$$(1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!})<\frac{\pi}{2}(1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!})$$
reduces to $3.1+\frac{40.7}{2723}=\frac{8482}{2723}<\pi$, which is true.
$$\left(\frac{1}{(4n+1)!}-\frac{1}{(4n+3)!}\right)<\frac{\pi}{2}\left(\frac{1}{(4n)!}-\frac{1}{(4n+2)!}\right)$$
reduces to $(4n+2)(4n+3)-1<\frac{\pi}{2}((4n+1)(4n+2)(4n+3)-(4n+3))$, which is true.
Thus $\sin{1} \leq \frac{\pi}{2}\cos{1}$, so we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Problem involving permutation matrices from Michael Artin's book. Let $p$ be the permutation $(3 4 2 1)$ of the four indices. The permutation matrix associated with it is
$$ P =
\begin{bmatrix}
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0
\end{bmatrix}
$$
This is the matrix that permutes the components of a column vector.
The problems asks you to decompose $p$ into transpositions and show that the associated matrix product equals the above matrix. However, I'm not getting that it does and I've run through it several times. Here are my calculations:
$p = (12)(14)(13)$
$$
P_{(12)} = \begin{bmatrix}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix},
P_{(14)} =\begin{bmatrix}
0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0
\end{bmatrix},
P_{(13)}=\begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
However,
$$P_{(12)} (P_{(14)} P_{(13)}) =
\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0
\end{bmatrix}
\neq P$$
I don't see where I've made a mistake.
| The problem is that you are multiplying cycles from left to right, but matrices multiply from right to left.
Let's take this out of cycle notation for a second. $$(3421)=\left(\begin{array}{cccc}1&2&3&4\\3&1&4&2\end{array}\right)$$
You want to say that $$(3421)=(12)(14)(13).$$ If you are multiplying right to left, it works out like this.
$$\begin{eqnarray*}(13)&=&\left(\begin{array}{cccc}1&2&3&4\\ \color{red}{3}&2&\color{red}{1}&4\end{array}\right)\\
(14)(13)&=&\left(\begin{array}{cccc}1&2&3&4\\\color{red}{4}&2&1&\color{red}{3}\end{array}\right)\\
(12)(14)(13)&=&\left(\begin{array}{cccc}1&2&3&4\\ \color{red}{2}&\color{red}{4}&1&3\end{array}\right)\end{eqnarray*}$$
That didn't work. Now let's try multiplying left to right.
$$\begin{eqnarray*}(12)&=&\left(\begin{array}{cccc}1&2&3&4\\ \color{red}{2}&\color{red}{1}&3&4\end{array}\right)\\
(12)(14)&=&\left(\begin{array}{cccc}1&2&3&4\\ \color{red}{4}&1&3&\color{red}{2}\end{array}\right)\\
(12)(14)(13)&=&\left(\begin{array}{cccc}1&2&3&4\\ \color{red}{3}&1&\color{red}{4}&2\end{array}\right)\end{eqnarray*}$$
Now, there are competing conventions in group theory about whether to multiply permutations left to right or right to left. You've likely been taught to multiply them left to right; if so, you found the proper decomposition. In general, however, matrices are always multiplied right to left. So when you convert your transpositions to matrices and multiply them, you have to reverse their order. In fact,
$$\left(
\begin{array}{cccc}
0 & 0 & \color{red}{1} & 0 \\
0 & 1 & 0 & 0 \\
\color{red}{1} & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{array}
\right)\left(
\begin{array}{cccc}
0 & 0 & 0 & \color{red}{1} \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\color{red}{1} & 0 & 0 & 0
\end{array}
\right)\left(
\begin{array}{cccc}
0 & \color{red}{1} & 0 & 0 \\
\color{red}{1} & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}
\right)=\left(
\begin{array}{cccc}
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0
\end{array}
\right)$$
So, to sum it all up, if you have matrices $P_a P_b$ associated with cycles $a$ and $b$, if you use the left to right convention of multiplying cycles, you'll want $P_{ab}=P_bP_a$; if you use the right to left convention, use $P_{ab}=P_aP_b$.
Remark. I should mention that some group theorists actually get so frustrated with this that they write all function composition in the left to right convention. Some even go as far as to write functions as $(x)f$, rather than $f(x)$, so that when they write $(x)fg$, it's clear that $f$ acts first, then $g$, whereas usually when we write $fg(x)$, we have that $g$ acts first, then $f$ (like with matrix multiplication). To me this seems kind of silly, but I must admit this consistency does fix issues like the one you were having.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
$2005|(a^3+b^3) , 2005|(a^4+b^4 ) \implies2005|a^5+b^5$ How can I show that if $$2005|a^3+b^3 , 2005|a^4+b^4$$ then $$2005|a^5+b^5$$
I'm trying to solve them from $a^{2k+1} + b^{2k+1}=...$ but I'm not getting anywhere.
Can you please point in me the correct direction?
Thanks in advance
| Actually you don't need the $a^4 + b^4$.
$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. It turns out that $a^2 - ab + b^2 \equiv 0$ has no solutions except $(0,0)$ mod either $5$ or $401$, so the only way to have $a^3 + b^3 \equiv 0 \mod 2005$ is
$a + b \equiv 0$, and then you also have $a^k + b^k \equiv 0 \mod 2005$ for every odd positive integer $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
An Inequality question I have the following question. I have to find a $\delta>0$ such that for all complex numbers $x,y$ the following holds true - \begin{equation}
\frac{1}{2\pi}\int_0^{2\pi}|x+e^{it}y|\,dt \ge (|x|^2+\delta|y|^2)^{1/2}.
\end{equation}
I have proceeded in the following way. Clearly, if $x=0$, then the requirement on $\delta$ is that $\delta\le 1$. So assume that $x\ne 0$. Then we write $x=re^{i\varphi}$ and $y=se^{i\tau}$ \begin{eqnarray} LHS &=& \frac{1}{2\pi}\int_0^{2\pi}|x+e^{it}y|\,dt \\ &=& \frac{r}{2\pi}\int_{\tau-\varphi}^{2\pi+\tau-\varphi}|1+e^{it}r^{-1}s|\,dt \\ &=& \frac{r}{2\pi}\int_{\tau-\varphi}^{2\pi+\tau-\varphi}(|1+e^{it}r^{-1}s|^2)^{1/2}\,dt \\ &=& \frac{r}{2\pi}\int_{\tau-\varphi}^{2\pi+\tau-\varphi}(1+r^{-2}s^2+2r^{-1}s\cos t)^{1/2}\,dt
\end{eqnarray}
I am stuck at this point. What inequality should I use to get the desired inequality.
| I think the inequality is not correct if we are looking at 2-uniform PL convex (see http://poincare.matf.bg.ac.rs/~pavlovic/BLAsco.PDF page 750) .
The case if a power of 2 should be on the integrad and no square root on the RHS.
Note if $y=0$ the equality holds for all $\delta \in \mathbb{R}$. Hence let $y \neq 0$ and let
\begin{align*}
x &=re^{i\theta} \\
y &=Re^{i\phi}
\end{align*}
First note
\begin{align*}
\left| x+e^{it}y \right|^2
&= \left( x+e^{it}y \right)\overline{\left( x+e^{it}y \right)} \\
&= \left| x \right|+2\text{Re} \left(\overline{x}ye^{it} \right) +\left| y \right|
\end{align*}
and
\begin{align*}
\text{Re}\left(\overline{x}ye^{it}\right)
&= \text{Re}\left(rR\; e^{i\left(t+\phi-\theta \right)} \right) \\
&= rR \cos{ \left( t+\phi-\theta \right)}
\end{align*}
Hence
\begin{align*}
\frac{1}{2 \pi} \int_{0}^{2\pi}\!{\left|x+e^{it}y \right|^2}\,\text{d}t
&= \frac{1}{2 \pi} \int_{0}^{2\pi} \! {\left|x\right|^2+ 2rR \cos{ \left( t+\phi-\theta \right)}+\left|y\right|^2}\,\text{d}t \\
&= \left|x\right|^2+\left|y\right|^2 + \frac{rR}{\pi}\int_{0}^{2\pi}\! {\cos{ \left( t+\phi-\theta \right)}} \,\text{d}t\\
&= \left|x\right|^2+\left|y\right|^2 \\
\end{align*}
Hence
\begin{align*}
\left|x\right|^2+\left|y\right|^2
\geq \left|x\right|^2+ \delta \left|y\right|^2
\; \iff \; \delta \leq 1
\end{align*}
The case if I am wrong.
An idea for a way to proceed:
\begin{align*}
\text{Re}\left(\overline{x}ye^{it}\right)
&= \text{Re}\left(rR\; e^{i\left(t+\phi-\theta \right)} \right) \\
&= rR \cos{ \left( t+\phi-\theta \right)} \\
&\geq -rR
\end{align*}
Since $-1 \leq \cos{ \left( t+\phi-\theta \right)} \leq 1$.
Hence
\begin{align*}
\frac{1}{2 \pi} \int_{0}^{2\pi}\!{\left|x+e^{it}y \right|}\,\text{d}t
&= \frac{1}{2 \pi} \int_{0}^{2\pi} \! \sqrt{\left|x\right|^2+ 2rR \cos{ \left( t+\phi-\theta \right)}+\left|y\right|^2}\,\text{d}t \\
&\geq \frac{1}{2 \pi} \int_{0}^{2\pi} \!\sqrt{\left|x\right|^2 - 2rR + \left|y\right|^2}\,\text{d}t \\
&=\frac{1}{2 \pi} \int_{0}^{2\pi} \!\sqrt{\left(\left|x\right| - \left|y\right| \right)^2}\,\text{d}t \\
&= \left|x\right|-\left|y\right|
\end{align*}
Hence
\begin{align*}
\left|x\right|-\left|y\right|
& \geq \left( \left|x\right|^2+ \delta \left|y\right|^2 \right)^{1/2}
\end{align*}
But this will lead to a contradiction if $\delta>0$. So I must of lost too much information in $\text{Re}\left(\overline{x}ye^{it}\right)\geq -rR$.
Alternatively you can try to expand the square root form of the integrand by the Binomial expansion to some order and then see where integration leads you to.
Visualise the problem and think of the vector $y$ added to $x$ giving a vector $x+y$. Now the introduction of $e^{it}$ turns the $y$ vector such that the point $z=x+y$ rotates about a circle of radius $|y|$ centred at $x$. Hence if you plot $|z|$ against $t$ you will get a periodic graph of period $2\pi$. So the integral will be the area of this graph between $0$ and $2\pi$ ("averaged" by $1/2\pi$). Consequently, this should be bigger than the area represented by the RHS. According to your inequality.
Hope this helps.
| {
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prove $ \frac{1}{13}<\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}<\frac{1}{12} $ Without the aid of a computer，how to prove
$$ \frac{1}{13}<\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}<\frac{1}{12} $$
| Another approach, perhaps simpler:
Let $X = \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{99}{100}$
We need to show $\dfrac{1}{13} < X < \dfrac{1}{12}$, which is equivalent to showing $144 < \dfrac{1}{X^2} < 169$, and we shall proceed to do this.
For the lower bound, note that we need to show: $$ \frac{2^2}{1} \cdot \frac{4^2}{3^2}\cdot \frac{6^2}{5^2} \cdots \frac{100^2}{99^2} > 144$$
$$\Leftrightarrow \frac{2^2}{1 \cdot 3} \cdot \frac{4^2}{3 \cdot 5}\cdot \frac{6^2}{5 \cdot 7} \cdots \frac{100^2}{99 \cdot 101} > \frac{144}{101} $$
$$\Leftrightarrow \prod_{k=1}^{50} {\frac{(2k)^2}{(2k)^2-1}} > \frac{144}{101} $$
Now all the factors in the LHS are higher than $1$, and converge fairly fast to $1$, so taking the first three terms, we have $\dfrac{4}{3} \cdot \dfrac{16}{15} \cdot \dfrac{36}{35} = \dfrac {256}{175}$ which is easily verified to be higher than $\dfrac{144}{101}$. Hence $X < \dfrac{1}{12}$ is established.
For the other part viz. $\dfrac{1}{X^2} < 169$ the approach is exactly similar, you should get an equivalent inequality of form:
$$ \prod_{k=1}^{49} \frac{2k (2k+2)}{(2k+1)^2} < \frac{169}{200}$$
Similarly, each factor now is less than $1$ and convergence is rapid. So taking the first three terms, we have $\dfrac{1}{X^2} < \dfrac{1024}{1225}$ which is verifiably less than $\dfrac{169}{200} = 0.845$. Hence $\dfrac{1}{13} < X$ is also established.
| {
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Prove that only parallelogram satisfies these conditions
Sum of distances between middle points of two opposite sides of a quadrilateral is equal to its semiperimeter. Prove that the quadrilateral has to be parallelogram.
I have no idea where to start. I tried with using middle line of triangle, messed around with areas, drew parallel lines though vertices, tried some inequalities...
| Hint: Use vector notation. Write out both sides in terms of the position vectors of $A, B, C, D$.
Hint: Apply the triangle inequality to show that equality must occur.
Hint: Equality only holds when the 3 vertices of a triangle are in a straight line, giving us the parallel condition that we desire.
(Abusing vector notation because I'm lazy to draw the arrows)
Setup: Let the vectors be $A, B, C, D$. The midpoints of 2 opposite sides are $\frac {A+B}{2} $ and $\frac {C+D}{2}$ . So the distance between them is $ | \frac {A+B}{2} - \frac {C+D}{2} | $. Likewise, the other opposite sides are $\frac {A+D}{2} $ and $\frac {B+C}{2}$, so the distance between them is $| \frac {A+D}{2} - \frac {B+C} {2}|$. So, we are given that
$$ | \frac {A+B}{2} - \frac {C+D}{2} | + | \frac {A+D}{2} - \frac {B+C} {2}| = |\frac {A-B}{2} | + | \frac {B-C} {2} | + | \frac {C-D}{2} | + | \frac {D-A} {2} | $$
Proof: By the triangle inequality, we have
$$ | \frac {A+B}{2} - \frac {C+D}{2} | \leq | \frac {A-D} {2} | + | \frac {B-C}{2} | \\
| \frac {A+D}{2} - \frac {B+C} {2}| \leq | \frac {A-B}{2} | + | \frac {D-C}{2} | \\$$
Hence, the condition given in the question may only hold when both triangle inequalities hold. For each triangle inequality to hold, we know that all the vectors involved must be parallel to each other. Writing this out, we get
$$ A-D \parallel B-C, A-B \parallel D-C $$
Hence, $A, B, C, D$ define a parallelogram.
| {
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Is there an easy way to calculate $\lim\limits_{k \to \infty} \frac{(k+1)^5(2^k+3^k)}{k^5(2^{k+1} + 3^{k+1})}$? Is there an easy way to calculate $$\lim_{k \to \infty} \frac{(k+1)^5(2^k+3^k)}{k^5(2^{k+1} + 3^{k+1})}$$
Without using L'Hôpital's rule 5000 times?
Thanks!
| It is a product of the following two expressions
$\frac{(k+1)^5}{k^5}=\left(1+\frac1k\right)^5$
$\frac{2^k+3^k}{2^{k+1}+3^{k+1}}= \frac{3^k(1+(2/3)^k)}{3^{k+1}(1+(2/3)^{k+1})}= \frac13\cdot$ $\frac{1+(2/3)^k}{1+(2/3)^{k+1}}$
| {
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Need help finding unknowns in simplex tableau. I need help with this homework problem.
The objective is to maximize $2x_1 - 4x_2$, and the slack variables are $x_3$ and $x_4$. The constraints are $\le$ type.
Tableau
$\begin{matrix}z & x_1 & x_2 & x_3 & x_4 & \text{RHS}\\
1 & b & 1 & f & g & 8\\
0 & c & 0 & 1 & 1\over5 & 4\\
0 & d & e & 0 & 2 & a\end{matrix}$
a) Find the unknowns $a$ through $g$.
b) Find $B^{-1}$.
c) Is the tableau optimal?
I can't figure out which columns make up the basis. Can anyone please help me get started?
| Additional notations
*
*$A \in M_{m\times n}(\Bbb R)$ ($m\le n$) has rank $n$ and basis matrix $B$.
*$x_B$ denotes the basic solution.
*$c_B$ denotes the reduced objective function so that $c^T x=c_B^T x_B$. (So the order/arrangement of basic variables is very important.)
Unknown entries in the tableau
*
*From the $x_3$ column, we know that $x_3 = 4$ is a basic variable, so $f=0$.
*The current objective value is $8$. Since $x_i$'s are nonnegative, this forces $x_1 = 4$, $x_2 = 0$, $a = 4$, $b = 0$, $c = 0$, $d = 1$.
*As OP's comment suggests, $B^{-1}$ can be directly read under the $x_3,x_4$ columns because the initial tableau has the form
\begin{array}{cc|c}
-c^T & 0 & 0 \\ \hline
A & I & b
\end{array}
Multiplying $B^{-1}$ on both sides gives
\begin{array}{cc|c}
c_B^T B^{-1}A-c^T & c_B^T B^{-1} & c_B^T B^{-1}b \\ \hline
B^{-1}A & B^{-1} & B^{-1}b \tag1 \label1
\end{array}
It's easy to (mentally) calculate $B = \begin{bmatrix} 1 & -1/10 \\ 0 & 1/2 \end{bmatrix}$.
The simplex tableau becomes
\begin{array}{r|rrrr|r}
z & x_1 & x_2 & x_3 & x_4 & \text{RHS} \\ \hline
1 & 0 & 1 & 0 & g & 8 \\ \hline
0 & 0 & 0 & 1 & 1/5 & 4 \\
0 & 1 & e & 0 & 2 & 4
\end{array}
$c^T = (2,-4)$ is given, and $c_B^T = (0,2)$ and $x_B^T = (4,4)^T$ since $x_3,x_1$ are current basic variables. (Note the arrangement of $x_3,x_1$ in the above tableau.) This allows us to (mentally) calculate $g$ using \eqref{1}
$$(0,g) = c_B^T B^{-1} = (0,2) \begin{bmatrix} 1 & 1/5 \\ 0 & 2 \end{bmatrix} = (0,4). $$
Therefore, the current solution is optimal.
Finally, to find $e$ \eqref{1}, we focus on the $x_2$ column.
\begin{align}
1 &= c_B^T B^{-1} {\bf a}_2 - c_2 \\
-3 &= (0,4){\bf a}_2 \\
4a_{22} &= -3 \\
a_{22} &= -\frac34
\end{align}
Calculate $B^{-1}{\bf a}_2$ in \eqref{1}.
\begin{align}
B^{-1}{\bf a}_2 &= (0,e)^T \\
{\bf a}_2 &= B (0,e)^T \\
(\star, a_{22})^T &= (\clubsuit,e/2)^T \\
e &= 2a_{22} = 2\left( -\frac34 \right) = -\frac32
\end{align}
Hence the optimal tableau is
\begin{array}{r|rrrr|r}
z & x_1 & x_2 & x_3 & x_4 & \text{RHS} \\ \hline
1 & 0 & 1 & 0 & 4 & 8 \\ \hline
0 & 0 & 0 & 1 & 1/5 & 4 \\
0 & 1 & -3/2 & 0 & 2 & 4
\end{array}
| {
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How do I transform the left side into the right side of this equation? How does one transform the left side into the right side?
$$
(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2
$$
| $$(a^2+b^2)(c^2+d^2)$$
$$=a^2.c^2+a^2.d^2+b^2.c^2+b^2.d^2$$
$$=a^2.c^2+b^2.d^2+a^2.d^2+b^2.c^2$$
$$=a^2.c^2-2a.b.c.d+b^2.d^2+a^2.d^2+2a.b.c.d+b^2.c^2$$
$$=(ac)^2 - 2.(ac).(bd)+(bd)^2 + (ad)^2 + 2(ad)(bc) + (bc)^2$$
$$=(ac - bd)^2+(ad+bc)^2$$
$$=R.H.S$$
| {
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Half-symmetric, homogeneous inequality Let $x,y,z$ be three positive numbers. Can anybode prove the follwing inequality :
$(x^2y^2+z^4)^3 \leq (x^3+y^3+z^3)^4$ (or find a counterexample, or find a reference ...)
| We need to show $(x^2y^2 + z^4)^3 \le (x^3+y^3+z^3)^4.$
By AM-GM, we have $(x^3+y^3+z^3)^4 \ge \left(2(xy)^{\frac{3}{2}} + z^3\right)^4$
Let $a = \sqrt{xy} > 0$. Then it is sufficient to show that
$(2a^3 + z^3)^4 \ge (a^4 + z^4)^3$
Let $t = \frac{a}{z} > 0$, then we need to show
$f(t) = (2t^3 + 1)^4 - (t^4 + 1)^3 \ge 0$ for $t > 0$.
or $f(t) = t^3 (15 t^9+32 t^6-3 t^5+24 t^3-3t+8) > 0$
Now, note that for $t \ge 1$,
$3t^6 - 3t^5 = 3t^5(t-1)$ and $3t^3 - 3t = 3t(t^2 - 1)$
so $f(t) = t^3 [15t^9 + 29t^6 + 3t^5(t-1) + 21t^3 + 3t(t^2-1) + 8] > 0$.
Similarly, when $0 < t < 1$,
$- 3t^5 + 3t^3 = 3t^3(1 - t^2)$ and $-3t + 3 = 3(1-t)$
so $f(t) = t^3 [15t^9 + 32t^6 + 3t^3(1-t^2) + 21t^3 + 3(1-t) + 5] > 0.$
Thus $f(t) > 0$ when $t > 0$, and hence the inequality holds.
| {
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Simplify $ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $ Please help me find the sum
$$
\frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}}
$$
| Adding the terms together, you should get:
$$
\sum_{\text{all terms}} = \frac{(32 x^{31})}{(x^{32}-y^{32})}
$$
This result is obtained by using the LCD, (Least Common Denominator).
LCD:
$$
(x-y) (x+y) (x^2+y^2) (x^4+y^4) (x^8+y^8) (x^{16}+y^{16}) =\\ (x^2-y^2)(x^2+y^2)(x^4+y^4) (x^8+y^8) (x^{16}+y^{16})=\\
(x^4-y^4)(x^4+y^4)(x^8+y^8) (x^{16}+y^{16})=\\
(x^8-y^8)(x^8+y^8)(x^{16}+y^{16})=\\
(x^{16}-y^{16})(x^{16}+y^{16})=\\
(x^{32}-y^{32}) =\\
\text{LCD(LCD(LCD(LCD(LCD( $x-y$, $x+y$), $x^2+y^2$), $x^4+y^4$), $x^8 + y^8$), $x^{16}+y^{16}$)}
$$
Note
It would be preferable to move from left to right simplifying the pairwise additions.
I included the LCD so you could see the general pattern in the denominators as you progress in your simplification from left to right.
| {
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Showing divergence of the series. I am having hard time trying to show that the following series is divergent. Can someone help me please?
$$\sum \frac{(-1)^n}{\log n} b_n $$
where $b_n=\frac{1}{\log n} $ if n is even
and $b_n=\frac{1}{2^n}$ if n is odd.
This would be a great help!
| Your series is:
$$
\frac{1}{\log{2}}\frac{1}{\log{2}} - \frac{1}{\log{3}}\frac{1}{2^3} + \frac{1}{\log{4}}\frac{1}{\log{4}}-\frac{1}{\log{5}}\frac{1}{2^5}+-\ldots
$$
Which is:
$$
\sum_{2}^{\infty}\left(\frac{1}{\log{2n}}\right)^2 - \sum_{2}^{\infty} \frac{1}{\log{(2n+1)}2^{2n+1}}
$$
$$
\sum_{2}^{\infty} \frac{1}{\log{(2n+1)}2^{2n+1}} < \sum_{2}^{\infty} \frac{1}{2^{2n+1}} = \frac{1}{24}
$$
So
$$
\sum_{2}^{\infty}\left(\frac{1}{\log{2n}}\right)^2 - \sum_{2}^{\infty} \frac{1}{\log{(2n+1)}2^{2n+1}} > \left(\sum_{2}^{\infty}\left(\frac{1}{\log{2n}}\right)^2\right) - \frac{1}{24} =\\
\infty - \frac{1}{24} = \infty
$$
| {
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Logarithm simplification Simplify: $\log_4(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})$
Can we use the formula to solve this: $\sqrt{a+\sqrt{b}}= \sqrt{\frac{{a+\sqrt{a^2-b}}}{2}}$
Therefore first term will become: $\sqrt{\frac{3}{2}}$ + $\sqrt{\frac{1}{2}}$
$\log_4$ can be written as $\frac{1}{2}\log_2$
Please guide further..
| Hint:
$(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})=\sqrt k$
Your $k=6$, Now its just $\frac{1}{2} \log_46$
| {
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Derivative of $\ln\left(\frac{2x}{1+x}\right)$ I know that the derivative of
$$f(x)=\ln(x) \ ,\ x>0$$
is just simply
$$f'(x)=\frac{dx}{x}$$
But how do you find the derivative for the function:
$$g(x)=\ln\left(\frac{2x}{1+x}\right)\ , \ x>0$$
| The trick hinted at by Peter Tamaroff in another answer is a clever one, but in general you use the chain rule to find the derivative of a composition:
$$(f(g(x)))' = f'(g(x))\cdot g'(x).$$
In your case, $f(x)=\ln x$ and $g(x)=2x/(1+x)$. Let us first find the derivatives of $f$ and $g$:
\begin{align*}
f'(x) &= (\ln x)' = \frac{1}{x},\\
g'(x) &= \left(\frac{2x}{1+x}\right)' = \frac{2(1+x) - 2x\cdot 1}{(1+x)^2} = \frac{2}{(1+x)^2}.
\end{align*}
Now we can apply the chain rule:
$$(f(g(x)))'=\frac{1}{\frac{2x}{1+x}} \cdot \frac{2}{(1+x)^2} = \frac{1+x}{2x}\cdot \frac{2}{(1+x)^2}= \frac{1}{x(1+x)}.$$
Note that $\frac{1}{x(1+x)}=\frac{1}{x} - \frac{1}{1+x},$ which is the answer you would get from Peter's answer.
| {
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Calculating $\lim_{x \rightarrow 1}(\frac{23}{1-x^{23}} - \frac{31}{1-x^{31}})$ How to calculate following limit?
$$\lim_{x \rightarrow 1}\left(\frac{23}{1-x^{23}} - \frac{31}{1-x^{31}}\right)$$
| Let $x=1+y$, then
$$
\begin{align}
\frac{23}{1-x^{23}}-\frac{31}{1-x^{31}}
&=\frac{23}{1-(1+23y+\frac{23\cdot22}{2\cdot1}y^2+O(y^3))}\\
&-\frac{31}{1-(1+31y+\frac{31\cdot30}{2\cdot1}y^2+O(y^3))}\\
&=-\frac1y(1-\tfrac{22}{2\cdot1}y+O(y^2))\\
&\hphantom{=}+\frac1y(1-\tfrac{30}{2\cdot1}y+O(y^2))\\
&=(\tfrac{22}{2\cdot1}-\tfrac{30}{2\cdot1})+O(y)\\
&=-4+O(y)
\end{align}
$$
Therefore,
$$
\lim_{x\to1}\frac{23}{1-x^{23}}-\frac{31}{1-x^{31}}=-4
$$
| {
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Show that $x_n \leq \frac{1}{\sqrt{3n+1}}$ Let $$x_n=\frac{1}{2} \frac{3}{4}\frac{5}{6}\cdots\frac{2n-1}{2n}$$
Then show that $$x_n \leq \frac{1}{\sqrt{3n+1}}$$ for all $n=1,2,3,\dots$
I try induction but unable to solve this equality.
| If (induction)
$$
x_n \leq \frac{1}{\sqrt{3n +1}}
$$
Then
$$
x_{n+1} = \frac{1}{2} \frac{3}{4}\frac{5}{6}\cdots\frac{2n-1}{2n}\frac{2n+1}{2n+2} \leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}
$$
You want now to prove that
$$
\frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2} \leq \frac{1}{\sqrt{3n+4}}.
$$
That is:
$$
\sqrt{3n+4}(2n+1) \leq (2n+2)(\sqrt{3n+1})
$$
Everything is positive, so the above is true if and only if it is true with the square on both sides. When you so that you get
$$
(3n + 4)(4n^2+4n + 1)\leq (4n^2 + 8n + 4)(3n+1).
$$
Now just expand both sides.
| {
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Find a closed form of the series $\sum_{n=0}^{\infty} n^2x^n$ The question I've been given is this:
Using both sides of this equation:
$$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$
Find an expression for $$\sum_{n=0}^{\infty} n^2x^n$$
Then use that to find an expression for
$$\sum_{n=0}^{\infty}\frac{n^2}{2^n}$$
This is as close as I've gotten:
\begin{align*}
\frac{1}{1-x} & = \sum_{n=0}^{\infty} x^n \\
\frac{-2}{(x-1)^3} & = \frac{d^2}{dx^2} \sum_{n=0}^{\infty} x^n \\
\frac{-2}{(x-1)^3} & = \sum_{n=2}^{\infty} n(n-1)x^{n-2} \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n(n-1)\frac{x^n}{x}(x+1) \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} (n^2x + n^2 - nx - n)\frac{x^n}{x} \\
\frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n^2x^n + n^2\frac{x^n}{x} - nx^n - n\frac{x^n}{x} \\
\end{align*}
Any help is appreciated, thanks :)
| there is a wrong sign in the first differentiation. I'd say:
$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$
Differentiating (and multiplying with $x$)we have,
$\displaystyle \frac{x}{(1-x)^2}=\sum_{n=0}^{\infty}nx^n$
Differentiating(and multiplying with $x$) we have,
$\displaystyle \frac{x(1+x)}{(1-x)^3}=\sum_{n=0}^{\infty}n^2x^n$
| {
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$x^2\dfrac{\partial u}{\partial x}+y^2\dfrac{\partial u}{\partial y}=u^2$ Help me please to solve the following PDE equation:
$x^2\dfrac{\partial u}{\partial x}+y^2\dfrac{\partial u}{\partial y}=u^2,\; \: u(x,2x)=1$
Thanks a lot!
| Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x^2$ , letting $x(1)=-1$ , we have $-\dfrac{1}{x}=t$
$\dfrac{dy}{dt}=y^2$ , we have $-\dfrac{1}{y}=t+y_0=-\dfrac{1}{x}+y_0$
$\dfrac{du}{dt}=u^2$ , we have $\dfrac{1}{u}=-t+f(y_0)=\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$ , i.e. $u(x,y)=\dfrac{1}{\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)}$
$u(x,2x)=1$ :
$\dfrac{1}{\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{2x}\right)}=1$
$\dfrac{1}{x}+f\left(\dfrac{1}{2x}\right)=1$
$f\left(\dfrac{1}{2x}\right)=1-\dfrac{1}{x}$
$f(x)=1-\dfrac{1}{\dfrac{1}{2x}}=1-2x$
$\therefore u(x,y)=\dfrac{1}{\dfrac{1}{x}+1-2\left(\dfrac{1}{x}-\dfrac{1}{y}\right)}=\dfrac{1}{1+\dfrac{2}{y}-\dfrac{1}{x}}=\dfrac{xy}{xy+2x-y}$
| {
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Proof that $t^8+2t^6+4t^4+t^2+1$ is reducible in $\mathbb{F}_p$ Prove that the polynomial $t^8+2t^6+4t^4+t^2+1$ is reducible in $\Bbb F_p$, for all $p\in \Bbb P$.
Here are some examples:
$t^8+2t^6+4t^4+t^2+1=(1 + t + t^4)^2\pmod{2}$
$t^8+2t^6+4t^4+t^2+1=(1 + t) (2 + t) (1 + t^2) (2 + 2 t^2 + t^4)\pmod{3}$
$t^8+2t^6+4t^4+t^2+1=(2 + t^2) (3 + 4 t^2 + t^6)\pmod{5}$
$t^8+2t^6+4t^4+t^2+1=(10 + 9 t + 3 t^2 + 2 t^3 + t^4) (10 + 2 t + 3 t^2 + 9 t^3 + t^4)\pmod{11}$
Thanks in advance.
| According to Polynomials irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$
you should calculate galois group of this polynomial over $\mathbb Q$.
If you ask Magma:
P:=PolynomialAlgebra(Rationals()); f:=x^8+2*x^6+4*x^4+x^2+1; G:=GaloisGroup(f); print G;
then you have
Permutation group G acting on a set of cardinality 8
Order = 192 = 2^6 * 3
(1, 2, 3, 4)(5, 6, 7, 8)
(1, 2)(5, 6)
(1, 6)(2, 5)
(1, 6, 7, 8)(2, 3, 4, 5)
Now you should check that there are no 8-cycles in this group.
Most simple way is to generate all 192 elements and check.
| {
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When are both fractions integers? The sum of absolute values of all real numbers $x$, such that both of the fractions $\displaystyle \frac{x^2+4x−17}{x^2−6x−5}$ and $\displaystyle \frac{1−x}{1+x}$ are integers, can be written as $\displaystyle \frac{a}{b}$, where $a$ and $b$ are coprime positive integers. What is the value of $a+b$?
| Hints:
$$\text{I}\;\;\;\;\;\;\;\frac{x^2+4x−17}{x^2−6x−5}=1+\frac{10x-12}{x^2-6x-5}=1+2\frac{5x-6}{(x-3)^2-14}$$
$$\text{II}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1-x}{1+x}=\frac{2}{1+x}-1$$
Look at II: the first fraction on the RHS must be an integer, so... The values you found with this substitute in I and do then some mathematics...
| {
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"url": "https://math.stackexchange.com/questions/339042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$x^4-4y^4=z^2$ has no solution in positive integers $x$, $y$, $z$. How do I prove that the diophantine equation $x^4-4y^4=z^2$ has no solution in positive integers $x$, $y$, $z$.
| Let show that the equation in question is equivalent to another one:$x^4+y^4=z^2$, called as $\Gamma$-equation.
If $p$ and $q$ and $r$ satisfy $(\Gamma)$, then, upon setting $x=r$, $y=pq$, and $z=p^4-q^4$, we obtain a solution of $x^4=4y^4+z^2$. Conversely, if $x$, $y$, and $z$ satisfy the equation in question, then there are $m$ and $n$ such that $x^2=m^2+n^2$, $2y^2=2mn$, and $z=m^2-n^2$. Hence we find that there are $p$ and $q$ with $m=p^2$ and $n=q^2$. Then $x^2=p^4+q^4$. Therefore the two equations are indeed equivalent, speaking in integers.
Now the transformed equation $(\Gamma)$ is easy to show its insolvability, and here we offer two methods:
I. Here have I proved that there is no pythagorean triple $(a,b,c)$ with both $a$ and $b$ squares, nor both $a$ and $c$ squares. If $(\Gamma)$ has a solution $(x,y,z)$ then $(x^2,y^2,z)$ is a pythagorean triple contradiction what is just said. So there is no solution at all.
II. Here is a proof contained in Diophantine Equation by Mordell.
Suppose $(x,y,z)$ is a solution with $(x,y)=1$. Since $x$ and $y$ cannot both be odd by considerations of $2$-powers, we assume that $y$ is even, and $x$ is odd. Now by Euclid's lemma, we know that there are $a$ and $b$ such that $x^2=a^2-b^2$, $y^2=2ab$, and $z=a^2+b^2$, with $a, b$ coprime. By dint of the same lemma again, we find $p$ and $q$ such that $x=p^2-q^2$, $b=2pq$, $a=p^2+q^2$, with $p, q$ coprime. Since $p$, $q$ and $p^2+q^2$ are pairwise prime to each other, and for $y^2=2ab=4pq(p^2+q^2)$, there are coprime $r, s$ and $t$ such that $p=r^2$, $q=s^2$, and $p^2+q^2=t^2$. So $r^4+s^4=t^2$, while $z=a^2+b^2=p^4+q^4+6p^2q^2=r^8+s^8+6r^4s^4$. Hence $z>(r^4+s^4)^2=t^4$, namely, $t$ is strictly less than $z$. So this completes the proof "by descent".
Though the length is still a pain, I hope that, this time, length brings not only pain, but the result, the richness of the subject, and its depth, as well. And tell me if something occurs that is inappropriate here. Thanks in advance.
| {
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Prove $ax^2+bx+c=0$ has no rational roots if $a,b,c$ are odd If $a,b,c$ are odd, how can we prove that $ax^2+bx+c=0$ has no rational roots?
I was unable to proceed beyond this: Roots are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
and rational numbers are of the form $\frac pq$.
| Consider a quadratic equation of the form $a\cdot x^2 + b\cdot x + c = 0$. The only way, it can have rational roots IFF there exist two integers $\alpha$ and $\beta$ such that
$$\alpha \cdot \beta = a\cdot c\tag1$$
$$\alpha + \beta = b\tag2$$
$$
Explanation\left\{ \begin{align}
if\,\alpha\cdot \beta &= a\cdot c,\\
\frac{\alpha}{a} &= \frac{c}{\beta}\\
a\cdot x^2 + b\cdot x + c& = a\cdot x^2 + (\alpha + \beta)\cdot x + c\\
& = a\cdot x^2 + \alpha\cdot x + \beta\cdot x + c\\
& =a\cdot x(x + \frac{\alpha}{a}) + \beta\cdot (x + \frac{c}{\beta}))\\
& =a\cdot x(x + \frac{\alpha}{a}) + \beta\cdot (x + \frac{\alpha}{a}))\\
& =(x+\frac{\alpha}{a})\cdot (a\cdot x + \beta)\\
&\text {As a Quadratic equation has only two roots,}\\
&\text {there would be no other way to factorize the equation}
\end{align}\right.
$$
$$\text{Reason }\alpha,\beta\in\mathbb{Z}\begin{cases}
\text{Given a Ring R, with two operations }\left\{⋅,+\right\},\text{ on }\mathbb{Q}\\
\text{and if } \alpha,\beta \in \mathbb{Q},\alpha\cdot \beta \in \mathbb{Z},\alpha+\beta\in\mathbb{Z}\\
\Rightarrow\alpha,\beta \in \mathbb{Z}\\
\text{ where } \mathbb{Z}\subset\mathbb{Q}
\end{cases}
$$
If $(a,b,c)$ are odd then $\alpha \cdot \beta$ is odd and $\alpha + \beta$ is odd, but you cannot have two integers whose product and sum are odd.
So by contradiction we prove that the equation cannot have rational roots
| {
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"source": "stackexchange",
"question_score": "9",
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How many ways are there to distribute 5 balls into 3 boxes, under additional conditions? How many ways are there to distribute 5 balls into 3 boxes if:
*
*both the boxes and balls are labeled
*the balls are labeled but the boxes are not
*the balls are unlabeled but the boxes are labeled
*both the balls and boxes are unlabeled
My way of doing these is as follows:
*
*$3^5$?
*$C(5,5)$?
*$3!5! - C(5,5)$?
*$C(3,5)$?
Where $C(k,n)$ means $\binom{n}{k}$.
| (2) Since the boxes are indistinguishable, there are 5 different cases for arrangements of the number of balls in each box: $(5,0,0)$, $(4,1,0)$, $(3,2,0)$, $(3,1,1)$, or $(2,2,1)$.
$(5,0,0)$: There is only $1$ way to put all 5 balls in one box.
$(4,1,0)$: There are $\binom{5}{4} = 5$ choices for the 4 balls in one of the boxes.
$(3,2,0)$: There are $\binom{5}{3} = 10$ choices for the 3 balls in one of the boxes.
$(3,1,1)$: There are $\binom{5}{3} = 10$ choices for the 3 balls in one of the boxes, and we simply split the last two among the other indistinguishable boxes.
$(2,2,1)$: There are $\binom{5}{2} = 10$ options for one of the boxes with two balls, then $\binom{3}{2} = 3$ options for the second box with two balls, and one option remaining for the third. However since the boxes with two balls are indistinguishable, we are counting each pair of balls twice, and must divide by two. So there are $\dfrac{10 \times 3}{2} = 15$ arrangements of balls as $(2,2,1)$.
Thus the total number of arrangements for 3 indistinguishable boxes and 5 distinguishable balls is $1 + 5 + 10 + 10 + 15 = \boxed{41}$.
$\textbf{Alternate solution:}$ There are $3^5 = 243$ arrangements to put 5 distinguishable balls in 3 distinguishable boxes. Among these 243 arrangements, there is one case in our problem that is counted three times: if all 5 balls are placed in one box and the other two boxes both contain nothing. This leaves 240 other arrangements.
For every other case, the contents of each box is different, and so these cases are each counted $3! = 6$ times. Therefore there must be 40 of these cases, and we have $\boxed{41}$ cases total.
| {
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"question_score": "10",
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find $n$ so $n/k$ is a $k$th power, $k=2,3,5$. Find a natural number n, in canonical form, such that:
$n/2=a^2$
$n/3=b^3$
$n/5= c^5$
for some a,b and c (natural numbers).
| Attempt an $n$ of the form $n = 2^x 3^y 5^z$.
You have $n/2 = 2^{x-1} 3^y 5^z$ as a square, so $x \equiv 1 (\bmod 2),\; y \equiv z \equiv 0 (\bmod 2).$
Analogously, $x \equiv z \equiv 0 (\bmod 3)$, $y \equiv 1 (\bmod 3)$ and
$x \equiv y \equiv 0 (\bmod 5)$, $z \equiv 1 (\bmod 5)$.
The Chinese remainder theorem guarantees that you will find such natural numbers $x,y,z$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ $$\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$$
I have attempted this question by expanding the left side using the cosine sum and difference formulas and then multiplying, and then simplifying till I replicated the identity on the right. I am not stuck. What's bothering me is that the way I went about this question seemed like a rather "clunky" method. I'm just curious if I've missed some underlying pattern that could have made it easier to reproduce the identity on the right.
The way I did it:
$$\begin{array}{l} \cos (A + B)\cos (A - B)\\ \equiv (\cos A\cos B - \sin A\sin B)(\cos A\cos B + \sin A\sin B)\\ \equiv {\cos ^2}A{\cos ^2}B - {\sin ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A(1 - {\sin ^2}B) - (1 - {\cos ^2}A){\sin ^2}B\\ \equiv {\cos ^2}A - {\cos ^2}A{\sin ^2}B - {\sin ^2}B + {\cos ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A - {\sin ^2}B\end{array}$$
| You can use the addition theorem which states that
$$\cos(\alpha+\beta)=\cos(\alpha)\cdot \cos(\beta) -\sin(\alpha)\sin(\beta)$$
$$\cos(\alpha + \beta) \cdot \cos(\alpha -\beta)= ( \cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta))\cdot (\cos(\alpha) \cdot \cos(-\beta)-\sin(\alpha)\cdot \sin(-\beta))$$
As $\cos(x)=\cos(-x)$ and $\sin(x)=-\sin(-x)$ this is equal to
$$(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta))\cdot (\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta))$$
With the third binom you get
$$\cos^2(\alpha) \cos^2(\beta) -\sin^2(\alpha)\sin^2(\beta)$$
As $\cos^2(\beta)+\sin^2(\beta)=1$ we have
$$\cos^2(\alpha)(1-\sin^2(\beta))-(1-\cos^2(\alpha))\sin^2(\beta)$$
Multplying it out gives you
$$\cos^2(\alpha) -\sin^2(\beta)\cos^2(\alpha)-\sin^2(\beta)+\cos^2(\alpha)\sin^2 (\beta)$$
And this is
$$\cos^2(\alpha)-\sin^2(\beta)$$
Another way:
$$2\cos(\alpha)\cdot \cos(\beta)=\cos(\alpha+\beta) + \cos(\alpha-\beta)$$
Using this in your formula gives us
$$\cos(A+B)\cdot \cos(A-B)=\frac{1}{2}\left( \cos(2A) + \cos(-2B)\right)=\frac{1}{2} \left(\cos(2 A)+ \cos(2 B)\right)$$
As $\cos(2 A)=1-2\sin^2(A)$ and $\cos(2 B) = 1-2 \sin^2 (B)$ this is equal to
$$1-\sin^2(A) -\sin^2 (B)=\cos^2(A)-\sin^2(B)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Simplifying $\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt {5 +\cdots}}}}$ How to simplify the expression:
$$\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\cdots}}}}.$$
If I could at least know what kind of reference there is that would explain these type of expressions that would be very helpful.
Thank you.
| Let $x = 2\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{...}}}}$. Then (if this converges) $x = 2\sqrt{5+x}$. Solving, $x = 2(1+\sqrt6)$, so the answer to your original question is $1+\sqrt{6}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$ Is there an easy way to prove the identity?
$$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3} \right )$$
While solving one question, I am stuck, which looks obvious but without any feasible way to approach.
Few observations, not sure if it would help
$$
\begin{align}
\dfrac{\dfrac{\pi}{7}+\dfrac{3\pi}{7}}{2} &= \dfrac{2\pi}{7}\\\\
\dfrac{\pi}{7} + \dfrac{3\pi}{7} + \dfrac{2\pi}{7} &= \pi - \dfrac{\pi}{7}
\end{align}
$$
| Hint
*
*$$ \cos A + \cos B = 2 \cos \left( \dfrac{A + B}{2} \right) \cos \left( \dfrac{A - B}{2} \right) $$
*$$ \cos A - \cos B = - 2 \sin \left( \dfrac{A + B}{2} \right) \cos \left( \dfrac{A - B}{2} \right) $$
*$$ \cos \left( \dfrac{2 \pi}{7} \right) = \cos { \left( 2 \theta \right) } \tag{ $ \theta = \dfrac{\pi}{7}$ } $$
*$$ - \cos \left( \dfrac{2 \pi}{7} \right) = \cos { \left( \pi -\dfrac{2\pi}{7} \right) } = \cos { \left( \dfrac{5\pi}{7} \right) } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/347112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
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Proving $\tan \left(\frac{\pi }{4} - x\right) = \frac{{1 - \sin 2x}}{{\cos 2x}}$ How do I prove the identity:
$$\tan \left(\frac{\pi }{4} - x\right) = \frac{{1 - \sin 2x}}{{\cos 2x}}$$
Any common strategies on solving other identities would also be appreciated.
I chose to expand the left hand side of the equation and got stuck here:
$$\frac{\cos x-\sin x}{\cos x+\sin x}$$
| As @ChristopherErnst suggests in a comment, some things become "more obvious after experience". Here are two alternative approaches to your problem that bear this out.
If you find yourself working with double- and half-angle arguments often, you might get the immediate sense that the right-hand side would be better if the roles of sine and cosine were reversed, since
$$\frac{1-\cos 2\theta}{\sin 2\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta \qquad\qquad(*)$$
As it turns out, reversing those roles is as easy as replacing the arguments with their complements:
$$\cos x = \sin\left(\frac{\pi}{2}-x\right) \qquad \sin x = \cos\left(\frac{\pi}{2}-x\right)$$
So, writing $2x= \frac{\pi}{2}-2\theta$ (equivalently, $2\theta = \frac{\pi}{2}-2x$) turns your identity into $(*)$.
Another approach immediately recognizes "$1-\sin 2x$" as a perfect square:
$$1-\sin 2x = (\cos^2 x + \sin^2 x) - 2 \cos x \sin x = \left( \cos x - \sin x \right)^2$$
(This isn't something I've seen exploited all that often, but it has come up with unusual frequency in some trig manipulations in my current research, so it kinda jumps out at me.) It's a convenient counterpart to the difference-of-squares version of the cosine double-angle identity:
$$\cos 2x = \cos^2 x - \sin^2 x = \left( \cos x - \sin x \right)\left( \cos x + \sin x \right)$$
Thus, the right-hand side of your identity reduces nicely ...
$$\frac{\left( \cos x - \sin x \right)^2}{\left( \cos x - \sin x \right)\left( \cos x + \sin x \right)} = \frac{\cos x - \sin x}{\cos x + \sin x}$$
... to the expression you have already shown to be equal to $\tan\left(\frac{\pi}{4}-x\right)$.
| {
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Probability: Permutations
Consider the experiment of picking a random permutation $\pi$ on $\{1,2,...,n\}$, and define the associated random variable $f(\pi)$ as the number of fixed points of $\pi$, i.e, the number of $i$ such that $f(i)=i$.
I know that a permutation of $X=\{1,2,\ldots ,n\}$ is a one-to-one function $\pi: X \rightarrow X$. Any two functions, $\pi_1, \pi_2$ can be composed and the resulting function is also a permutation. How can I find $E(F)$ and what do we know about $E(f(\pi^2))$ and $ E(f( \pi^k))$?
| These can be done using generating functions. First, consider $E[F].$ The exponential generating function of the set of permutations by sets of cycles where fixed points are marked is
$$ G(z, u) = \exp\left(uz - z + \log \frac{1}{1-z}\right) =
\frac{1}{1-z} \exp(uz-z).$$
Now to get $E[F]$ compute
$$ \left.\frac{d}{du} G(z, u)\right|_{u=1} =
\left. \frac{u^0 z}{1-z} \exp(uz-z) \right|_{u=1} = \frac{z}{1-z}.$$
This means that $E[F] = 1,$ there is one fixed point on average.
Let $E[F_2]$ denote the expectation of the number of fixed points in $\sigma^2$, where $\sigma$ is a random permutation. Now we need to mark both fixed points and two-cycles, since the latter turn into two fixed points under squaring. We get
$$ G(z, u) = \exp\left(uz - z + u^2 \frac{z^2}{2} - \frac{z^2}{2}
+ \log \frac{1}{1-z}\right) =
\frac{1}{1-z} \exp\left(uz - z + u^2 \frac{z^2}{2} - \frac{z^2}{2}\right).$$
Continuing as before, we find
$$ \left.\frac{d}{du} G(z, u)\right|_{u=1} =
\left. \frac{u^0 z+ u^1 z^2}{1-z}
\exp\left(uz - z + u^2 \frac{z^2}{2} - \frac{z^2}{2}\right) \right|_{u=1} =
\frac{z+z^2}{1-z}.$$
The conclusion is that $E[F_2] = 2$ for $n\ge 2$ and there are two fixed points on average.
The pattern should now be readily apparent. For every divisor $d$ of $k$ a cycle of length $d$ splits into $d$ fixed points when raised to the power $k.$ Hence we need to mark these cycles with $u^d.$ To illustrate this consider $E[F_6].$
We get
$$ G(z, u) = \exp\left(uz - z + u^2 \frac{z^2}{2} - \frac{z^2}{2} +
u^3 \frac{z^3}{3} - \frac{z^3}{3} + u^6 \frac{z^6}{6} - \frac{z^6}{6}
+ \log \frac{1}{1-z}\right) \\ =
\frac{1}{1-z} \exp\left(uz - z + u^2 \frac{z^2}{2} - \frac{z^2}{2} +
u^3 \frac{z^3}{3} - \frac{z^3}{3} + u^6 \frac{z^6}{6} - \frac{z^6}{6}\right).$$
Once more continuing as before, we find
$$ \left.\frac{d}{du} G(z, u)\right|_{u=1} \\ =
\left. \frac{u^0 z+ u^1 z^2+ u^2 z^3+u^5 z^6}{1-z}
\exp\left(uz - z + u^2 \frac{z^2}{2} - \frac{z^2}{2} +
u^3 \frac{z^3}{3} - \frac{z^3}{3} + u^6 \frac{z^6}{6} - \frac{z^6}{6}\right) \right|_{u=1} =
\frac{z+z^2+z^3+z^6}{1-z}.$$
The conclusion is that $E[F_6] = 4$ for $n\ge 6$ and there are four fixed points on average.
The general procedure is
$$ G(z, u) = \exp\left(\sum_{d\mid k}
\left(u^d \frac{z^d}{d} - \frac{z^d}{d}\right) +
\log \frac{1}{1-z} \right)=
\frac{1}{1-z} \exp\left(\sum_{d\mid k}
\left(u^d \frac{z^d}{d} - \frac{z^d}{d}\right)\right).$$
Once more continuing as before, we find
$$ \left.\frac{d}{du} G(z, u)\right|_{u=1} =
\left. \frac{\sum_{d\mid k} u^{d-1} z^d}{1-z}
\exp\left(\sum_{d\mid k}
\left(u^d \frac{z^d}{d} - \frac{z^d}{d}\right)\right) \right|_{u=1} =
\frac{\sum_{d\mid k} z^d}{1-z}.$$
We have shown that the value of $E[F_k]$ is equal to $\tau(k)$ (the number of divisors of $k$) as soon as $n\ge k.$ It starts out at $1$ for $n=1$ and increases by one every time $n$ hits a divisor of $k$ up to and including $k$ itself.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral solutions of hyperboloid $x^2+y^2-z^2=1$ Are there integral solutions to the equation $x^2+y^2-z^2=1$?
| The advanced approach is to rewrite this as $x^2+y^2= z^2+1$ and use unique factorization in $\mathbb Z[i]$. Assuming you disallow the obvious answers where $x$ or $y$ is $\pm 1$, you get that $x+yi=(a+bi)(c+di)$ and $z+i=(a+bi)(c-di)$ for some $a,b,c,d$. That yields a condition on $a,b,c,d$ and an explicit formulas form $x,y,z$ in terms of $a,b,c,d.$
Specifically, if $bc-da=1$ then $x=ac-bd, y=bc+ad, z=ac+bd$ is a solution to your equation. I think this gives all of them, but there is some concern about units.
This doesn't give us all ordered triples, since $y=bc+ad=bc-ad+2ad=1+2ad$, and hence this only gives us $y$ odd. However, it is clearly true that one of $x,y$ has to be odd, so this might sill give all un-ordered pairs $x,y$.
For example, $(a,b,c,d)=(7,2,4,1)$ then $x=26, y=15, z=30$ is the solution.
A more elementary solution is to assume $y=2n+1$ is odd (one of $x,y$ must be odd.) Then $x$ and $z$ have to have the same parity, and we get:
$$4n(n+1)=y^2-1 = z^2-x^2=(z-x)(z+x)$$
Dividing by $4$:
$$n(n+1)=\frac{z-x}{2}\frac{z+x}{2}$$
So we just need a factorization, $n(n+1)=UV$ with $U<V$. Then $z=V+U$ and $x=V-U$.
The most basic answer is $V=n(n+1)$ and $U=1$. Then $x=n(n+1)-1, y=2n+1, z=n(n+1)+1$.
The "obvious" answer, $U=n$, $V=n+1$ just gets you $x=1, y=2n+1, z=2n+1$.
Since $n(n+1)$ is even, we can write $U=2$, $V=\frac{n(n+1)}{2}=T_n$. Then $$x=T_n-2, y=2n+1, z=T_n+2$$
When $n>0$ we see that the number of distinct positive solutions $(x,2n+1,z)$ is $$\frac{\tau(n(n+1))}{2}=\frac{\tau(n)\tau(n+1)}2$$
If $UV=n(n+1)$ we can get $a,b,c,d$ from the first solution by defining:
$$a=(V,n), b=(U,n+1), c=(V,n+1), d=(U,n)$$
then $$bc=n+1, ad=n, ac=V, bd=U$$
So $bc-ad=1$, $bc+ad=2n+1=y$, $V-U=ac-bd=x$ and $V+U=ac+bd=z$.
So any result from the second method is a result from the first method, which means the first method gets all positive solutions, too.
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How many solutions does $x^2 \equiv {-1} \pmod {365}$ have?
How many solutions does $x^2 \equiv {-1} \pmod {365}$ have?
My thought:
$365 = 5 \times 73$ where $5$ and $73$ are prime numbers.
So we can obtain $x^2 \equiv {-1} \pmod 5$ and $x^2 \equiv {-1} \pmod {73}$.
For $x^2 \equiv {-1} \pmod 5$,
we checked $5 \equiv 1 \pmod 4$, therefore it is solvable.
Using the Euler criterion, $(-1)^{\frac{5-1}{2}} \equiv 1 \pmod 5$.
Thus it has two solutions.
For $x^2 \equiv {-1} \pmod {73}$,
we checked $73 \equiv 1 \pmod 4$, therefore it is solvable.
Using the Euler criterion, $(-1)^{\frac{73-1}{2}} \equiv 1 \pmod{73}$.
Thus it has two solutions.
So, there are four solutions in total.
| Working modulo $\,73\,$ :
$$2^6=-9\implies 2^{12}=(-9)^2=8=2^3\implies 2^9=1$$
$$3^4=8=2^3\implies3^{12}=(2^3)^3=2^9=1$$
$$5^6=3\implies 5^{36}=3^6=8\cdot3^2=72=-1$$
Thus, we've found a primitive root modulo $\,73\,$ (namely $\,5\,$), and from here we get two solutions to $\,x^2=-1\pmod{73}\,$ : $\;\;5^{18}=(5^6)^3=3^3=27\;,\;-27=46\;$ , and the two solutions to $\,x^2=-1\pmod 5 \;$ are $\,\;2,3\;$
So we have the solutions modulo $\,365\,$ : $\;173\;,\;-173=192\;,\;338\ldots$ and I think that's all, folks: four different solutions: $\;27,\; 46,\; 173,\; 339\pmod{365}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining series formula Is there any procedure to follow when determining the function of a series? This seems simple but for I can't figure it out.
$$ \frac15 + \frac18 + \frac1{11} +\frac1{14} + \frac1{17}+\cdots$$
| Look at the denominators: $5, 8, 11, 14, 17, \ldots$ Can you find an expression for these?
Start by looking at the differences: $8-5=3$, $11-8=3$, $14-11=3$ and $17-14=3$. To get from one term to the next, we simply add $3$; the term-to-term rule is $+3$. That means that the sequence $5,8,11,14,17,\ldots$ is like the three times table, i.e. it resembles $3n$.
However, the three times table goes $3,6,9,12,15,\ldots$ Our sequence, $5,8,11,14,17,\ldots$, is always two bigger than the three times table. Hence the $n$-th term rule is $3n+2$.
Let's check: $3\times 1 + 2 = 5$, $3\times 2 + 2 = 8$, $3\times 3 + 2 = 11$, $3 \times 4 + 2=14$ and $3\times 5 + 2 = 17$.
If the sequence $5,8,11,14,17,\ldots$ is given by $3n+2$ then the sequence $\frac{1}{5},\frac{1}{8},\frac{1}{11},\frac{1}{14},\frac{1}{17},\ldots$ is given by:
$$\frac{1}{3n+2}$$
The partials sums are then given by
$$\sum_{n=1}^p \frac{1}{3n+2}$$
Sadly, as $p \to \infty$, we don't get a sensible answer. Notice that $5n \ge 3n+2$ for all $n \ge 1$ and so
$$\frac{1}{5n} \le \frac{1}{3n+2}$$
for all $n \ge 1$. If we can show that $\sum_{n \ge 0} 1/5n$ diverges then clearly our series will diverge. Well:
$$\sum_{n \ge 1} \frac{1}{5n} = \frac{1}{5}\sum_{n \ge 1} \frac{1}{n}$$
The sum $\sum_{n \ge 1} 1/n$ is called the Harmonic Series and is well-known to diverge. Since each and every term of $\sum_{n \ge 1}1/5n$ is less than or equal to the corresponding term in the series $\sum_{n \ge 1}1/(3n+2)$, and the series $\sum_{n \ge 1}1/5n$ diverges, it follows that the series $\sum_{n \ge 1}1/(3n+2)$ also diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/354236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Triangle proof using law of sines
In triangle $ABC$, suppose that angle $C$ is twice angle $A$. Use the law of sines to show that $ab= c^2 - a^2$.
| Put $\,\angle C=2w\;,\;\;\angle A=w\;,\;\;\angle B=180- 3w\;$ , then
$$\frac{b}{\sin 3w}=\frac{c}{\sin 2w}=\frac{a}{\sin w}\implies ab=\frac{b^2\sin w}{\sin 3w}{}\;,\;\;a^2-c^2=\frac{b^2\sin^22w}{\sin^33w}-\frac{b^2\sin^2w}{\sin^23w}\implies$$
$$ab=c^2-a^2\iff \frac{b^2}{\sin 3w}\sin w=\frac{b^2}{\sin 3w}\left(\frac{\sin^22w}{\sin 3w}-\frac{\sin^2w}{\sin 3w}\right)\iff$$
$$\iff \sin w=\frac{\sin^22w-\sin^2w}{\sin 3w}=\frac{4\sin^2w\cos^2w-\sin^2w}{\sin 3w}\iff$$
$$\iff \frac{4\sin w\cos^2 w-\sin w}{\sin 3w}=1$$
But
$$(1)\;\;\;4\sin w\cos^2w-\sin w=\sin w(4\cos^2w-1)$$
$$(2)\;\;\;\sin 3w=\sin w\cos 2w+\sin 2w\cos w=\sin w(2\cos^2w-1)+2\sin w\cos^2w$$
Now just check that $\,(1)=(2)\;$...
| {
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"timestamp": "2023-03-29T00:00:00",
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How to factor $x^4 +3x -2$? I have figured out there is two roots between $0$ and $1 ,-1$ and $-2$ for $x^4 +3x -2 = 0$.
Therefore there should be two factors $(x + a)$ and $(y - b)$ where $a,b \in R^+$.
But how to find these $a$ and $b$?
When they found I can find the next factor in $ax^2+bx+c$ form and can check for further factors easily.
| For it to have some "nice" linear factors, the roots must be one of $\pm 1,\pm 2$ (this is due to the rational root theorem). You can quickly check that these are not the roots. The next bet is quadratic factors, i.e.,
$$(x^4+3x-2) = (x^2+ax+b)(x^2+cx+d)$$
Expanding the right hand side gives us
\begin{align}
a+c & = 0\\
b + d + ac & = 0\\
ad+bc & = 3\\
bd & = - 2
\end{align}
This gives us the factors to be
$$(x^2-x+2) \text{ and } (x^2+x-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/356105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the sum of all of the products of $3$ of the digits $1, \dots, 9$? Consider the numbers $1, 2, 3, \dots, 9$. Take the product of any three of them. What is the sum of all such products?
In other words, calculate $1 \cdot 2 \cdot3 + 1 \cdot 2 \cdot 4 + 1 \cdot 2 \cdot 5 + \dots + 7 \cdot 8 \cdot9$.
If we consider products of four numbers or more, what's the answer?
| Use the Principle of Inclusion and exclusion.
Your sum is equal to
$$ \left[ (1 + 2 + \ldots + 9) ^3 - 3 \times ( 1^2 + 2^2 + \ldots + 9^2) \times (1 +2 + \ldots + 9 ) + 2 \times (1^3 + 2^3 + \ldots + 9^3) \right] \div 6$$
As an explanation, you can see if that $a\neq b, b\neq c, c\neq a$, then it will appear 6 times in the first time, 0 times in the second, 0 times in the third, hence appear $6/6 =1 $ time.
Terms of the form $aab$ will appear thrice in the first term, thrice in the second (which is subtracted), 0 times in the third, hence appear $(3 - 3)/6 = 0 $ times.
Terms of the form $aaa$ will appear once in the first term, thrice in the second, twice in the third, for a total of $(1 - 3 + 2)/6 = 0 $ times.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/357193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The minimum of a function Could anyone possibly give me any help with finding the minimum of this function? I believe the result to be $2\pi |n|$ from page 619 of this paper by W. G. C. Boyd.
\begin{equation}
\frac{\frac{1}{2}(1+\zeta(m))\Gamma(m)}{(2\pi)^{m+1}|n|^m}
\end{equation}
Thanks!
| When $m$ is large, we have
$$
a_m : = \frac{{\frac{1}{2}\left( {1 + \zeta \left( m \right)} \right)\Gamma \left( m \right)}}{{\left( {2\pi } \right)^{m + 1} \left| n \right|^m }} \sim \frac{{\Gamma \left( m \right)}}{{\left( {2\pi } \right)^{m + 1} \left| n \right|^m }} \sim \frac{{m^{m - 1/2} e^{ - m} \sqrt {2\pi } }}{{\left( {2\pi } \right)^{m + 1} \left| n \right|^m }} = \left( {\frac{m}{{2\pi e\left| n \right|}}} \right)^m \frac{1}{{\sqrt {2\pi m} }}.
$$
Whence
$$
\frac{a_m}{a_{m + 1}} \sim \frac{\left( \frac{m}{2\pi e\left| n \right|} \right)^m \frac{1}{{\sqrt {2\pi m} }}}{\left( \frac{{m + 1}}{{2\pi e\left| n \right|}} \right)^{m + 1} \frac{1}{\sqrt {2\pi \left( m + 1 \right)} }} \sim \frac{{\left( {\frac{m}{{2\pi e\left| n \right|}}} \right)^m }}{{\left( {\frac{{m + 1}}{{2\pi e\left| n \right|}}} \right)^{m + 1} }} = \frac{2\pi e\left| n \right|}{m + 1}\frac{1}{\left( 1 + \frac{1}{m} \right)^m } \sim \frac{2\pi e\left| n \right|}{m}\frac{1}{e} = \frac{2\pi \left| n \right|}{m}.
$$
So for large $m$, your sequence attends its minimum at $m=2\pi \left| n \right|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/357500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"Incremental" hypergeometric distribution? Box contains $5$ yellow and $3$ red balls. $4$ balls are drawn without replacement. Let $X$ be the number of yellow balls appearing in the first two draws, and let $Y$ be the number of yellow balls appearing in total. Give the joint probability distribution of $X$ and $Y$.
| We want to find, for all relevant $x$ and $y$, the probability that $X=x$ and $Y=y$. Call this number $f_{X,Y}(x,y)$, or, for simplicity, $f(x,y)$.
There are not many possible values of $x$ and $y$, so we can find $f(x,y)$ for each combination. (Alternately, we can find a general formula for $f(x,y)$. For these small numbers, that is probably not worthwhile.)
We show how to find $f(0,y)$ for all possible values of $y$. The probability that $X=0$ (no yellow on the first two trials) is $\frac{3}{8}\cdot\frac{2}{7}$.
Given that there were no yellow on the first two trials, the probability that $Y=0$ is $0$, since there must be at least $1$ yellow in the next two trials. Thus $f(0,0)=0$.
Suppose there were no yellow on the first two. Then the probability of $1$ yellow overall is $\frac{1}{6}+\frac{5}{6}\cdot\frac{1}{5}=\frac{2}{6}$. Thus $f(0,1)=\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{2}{6}$. This can, if we wish, be simplified. Now we have found $f(0,1)$.
Suppose there were no yellow on the first two trials. Then the probability of $2$ yellow in the next two trials is $\frac{5}{6}\frac{4}{5}=\frac{4}{6}$. Thus $f(0,2)=\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{4}{6}$.
We may want to note that $f(0,3)=f(0,4)=0$.
Now you can do the same sort of calculations for $f(1,y)$ and $f(2,y)$.
Added: We develop a formula. There are $\binom{8}{2}\binom{4}{2}$ equally likely ways to choose $2$ balls and then $2$ mre. There are all equally likely. (Think of the balls as having invisible ID numbers.) This will be the denominator in all our probabilities.
For the numerator, we find the number of ways to have $x$ yellow in the first two, and $y$ yellow overall, so $y-x$ in the next two.
The $x$ yellow can be chosen in $\binom{5}{x}$ ways. The accompanying $2-x$ red can be chosen in $\binom{3}{2-x}$ ways. Now the $y-x$ yellow can be chosen from the $5-x$ remaining yellows in $\binom{5-x}{y-x}$ ways. And the $2-(y-x)$ red can be chosen from the $3-(2-x)$ remaining reds in $\binom{3-(2-x)}{2-(y-x)}$ ways. That gives a total of
$$ \binom{5}{x}\binom{3}{2-x}\binom{5-x}{y-x} \binom{3-(2-x)}{2-(y-x)} .$$
(In the formula, we use the convention that $\binom{a}{b}=0$ if $a\lt b$.)
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"timestamp": "2023-03-29T00:00:00",
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