Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How many ordered triples of integers which are between 0 and 10 inclusive do we actually have if $a * (b+c) = a * b +c$ How many ordered triples of integers $(a,b,c)$ which are between 0 and 10 inclusive do we have if:
$a * (b+c) = a * b +c$
| We have that
$$a(b+c)=ab+c \Rightarrow ab+ac=ab+c.$$
If $a=1$, then we have $b+c=b+c$, and if $c=0$ then $ab=ab$.
When $a=1$ there are $11 \cdot 11=121$ ways to write two of the eleven numbers from $0$ to $10$ inclusive for $b,c$ with repeat, and when $c=0$, there are again $11 \cdot 11=121$ ways to write two of the eleven numbers from $0$ to $10$ inclusive for $a,b$ with repeat.
These two sets have eleven duplicates when $a=1$ and $c=0$, since $b$ is one of the same eleven values in both cases.
Hence there are $2 ( 11 \cdot 11)-11=231$ ordered triples $(a,b,c)$ that satisfy your equation.
| {
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"url": "https://math.stackexchange.com/questions/584276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to integrate $\int\frac{\sqrt{1-x}}{\sqrt{x}}\ \mathrm dx$ I'm having a bit of trouble solving this integral: $$\int\frac{\sqrt{1-x}}{\sqrt{x}}dx$$
Here is my attempt at a solution:
I multiplied the numerator and the denominator of $\frac{\sqrt{1-x}}{\sqrt{x}}$ by $\sqrt{x}$, yielding $$\int\frac{\sqrt{x-x^2}}{x}dx.$$
Further simplification resulted in $$\int\frac{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}{x}dx.$$
Using trigonometric substitution, I set $$x-\frac{1}{2}=\frac{1}{2}\sin\theta$$ and solving for the differential $dx$ got $$dx=\frac{1}{2}\cos\theta.$$ Substituting this all back into $\int\frac{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}{x}dx$ (and some simplification later) yielded $$\frac{1}{2}\int{\frac{\cos^2\theta}{\sin\theta+1}}d\theta.$$ By substituting $1-\sin^2\theta$ for $\cos^2\theta$ I obtained $$\frac{1}{2}\int{\frac{1}{\sin\theta+1}-\frac{\sin^2\theta}{\sin\theta+1}d\theta}.$$ The issue I'm having is trying to solve this resultant integral. If there is an easier method to solve the problem, that would be graciously accepted.
| Set $\sqrt{x} = \sin(t)$. We then have $x = \sin^2(t)$. Hence, $1-x = \cos^2(t)$. This gives us
\begin{align}
\int \dfrac{\sqrt{1-x}}{\sqrt{x}}dx & = \int \dfrac{\cos(t)}{\sin(t)} 2 \sin(t) \cos(t) dt = 2\int \cos^2(t) dt\\
& = \int(1+\cos(2t))dt = t + \dfrac{\sin(2t)}2 + c\\
& = \arcsin(\sqrt{x}) + \sqrt{x}\sqrt{1-x} + c
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/586320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Equations with unknowns and powers $a + b +c = 17$
$a^2 + b^2 + c^2 = 101$
$a^3 + b^3 + c^3 = 623$
How does one go about solving this?
Thanks
| my weaknesses show up particularly when doing tedious calculations, so as a penance, just this once, i will bludgeon my way through this, probably generating more heat than light in the process!
we have: $$a+b=17-c$$
thus
$$(a+b)^2 = 289 -34c+c^2$$
and
$$(a+b)^3 = 4913 - 867c+51c^2 -c^3$$
but $$(a+b)^2 = a^2+b^2 + 2ab = 101 - c^2 +2ab$$
giving
$$ab = 94 -17c +c^2$$
likewise $$(a+b)^3 = a^3+b^3 +3ab(a+b) \\
= 623 -c^3 +3(94-17c+c^2)(17-c) \\
=5417 -1149c +102c^2 -4c^3$$
so we have
$$ 5417 -1149c +102c^2 -4c^3 = = 4913 -867c+51c^2 -c^3$$
i.e.
$$504 -282c+51c^2 -3c^3 =0$$
or
$$c^3-17c^2+94c-168 =0$$
or
$$(c-4)(c-6)(c-7) = 0$$
| {
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"url": "https://math.stackexchange.com/questions/589568",
"timestamp": "2023-03-29T00:00:00",
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Let a,b,c be positive real number, proof. Let a,b,c be positive real number, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that :
$\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2} \leq \frac{3}{16}$
Can anyone help me how to deal with it?
| Since $$\displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c}=a+b+c$$
$$\Rightarrow \frac{ab+bc+ca}{abc \left( a+b+c \right)} = 1 $$
$$\Rightarrow (ab+bc+ca)^2=3abc(a+b+c)$$
$$\Rightarrow \frac{3}{ab+bc+ca} = \frac{ab+bc+ca}{abc \left( a+b+c \right)} = 1$$
Or $$\frac{9}{16 \left( ab+bc+ca \right)} = \frac{3}{16}$$
Need proof $$\dfrac{1}{\left( 2a+b+c \right)^2}+\dfrac{1}{\left( 2b+c+a \right)^2}+ \dfrac{1}{\left( 2c+a+b \right)^2} \le \frac{9}{16 \left( ab+bc+ca \right)} $$
Use Am-Gm have $$\left( 2a+b+c \right)^2 = \left( a+b + a+c \right)^2 \ge 4 \left( a+b \right) \left( a+c \right)$$
So $$\dfrac{1}{\left( 2a+b+c \right)^2}+\dfrac{1}{\left( 2b+c+a \right)^2}+ \dfrac{1}{\left( 2c+a+b \right)^2} \le \frac{a+b+c}{ 2 \left( a+b \right) \left( b+c \right) \left( c+a \right)} \le \frac{9}{16 \left( ab+bc+ca \right)}$$
$$ \Leftrightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2 \ge 6abc$$. This is true follow to inequality AM-GM
| {
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"url": "https://math.stackexchange.com/questions/591134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Does a matrix $A$ of $\pm1$'s of order $11$ exist with $\det A >4000$? How to prove or disprove that statement: there exists a square matrix of size 11 having all entries $ \pm 1$ and its determinant greater than 4000?
| I found such matrix, having executed a few times the Maple code $$interface(rtablesize = 15):with(LinearAlgebra):with(Statistics): $$ $$P := Array([-1, 1]):X := RandomVariable(EmpiricalDistribution(P)): $$ $$M := convert(Matrix(11, 11, convert(Sample(X, 121), list)),rational); Determinant(M);$$ which outputed $$ \left[ \begin {array}{ccccccccccc} - 1&- 1& 1&- 1& 1&- 1&
1& 1& 1& 1& 1\\- 1& 1&- 1&- 1&-
1& 1&- 1& 1& 1& 1& 1\\ - 1& 1&
1& 1&- 1& 1& 1& 1&- 1&- 1&- 1\\
1& 1& 1&- 1& 1& 1& 1&- 1&- 1& 1& 1
\\ 1&- 1&- 1& 1& 1&- 1&- 1&- 1& 1
& 1& 1\\ 1&- 1& 1& 1&- 1&- 1&- 1&
- 1& 1& 1& 1\\ 1-& 1& 1&- 1&- 1&-
1&- 1& 1&- 1& 1& 1\\ 1&- 1& 1&
1& 1&- 1& 1&- 1&- 1& 1&- 1\\ - 1&
- 1& 1&- 1&- 1& 1&- 1& 1& 1& 1&- 1
\\ 1&- 1& 1&- 1& 1&- 1& 1& 1&- 1&
- 1&- 1\\ 1& 1&- 1& 1& 1&- 1&- 1&
- 1& 1&- 1&- 1\end {array} \right]
$$ and $4096. $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 3
} |
How to Integerate $\frac{x}{x^2-4x+8}$? I want to know how to integrate this ?
$$\frac{x}{x^2-4x+8}$$
Thanks all
| Well, you can rewrite this as $$\frac{x}{x^2-4x+4+4}=\frac{x}{(x-2)^2+4}=\frac{x}{(x-2)^2+2^2}.$$ Try the substitution $$x-2=2\tan\theta.$$ This means that $x=2+2\tan\theta$ and $\frac{dx}{d\theta}=2\sec^2\theta,$ and we also have $$(x-2)^2+2^2=2^2(\tan^2\theta+1)=2^2\sec^2\theta.$$ Can you take it from there?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\frac{(n+1)^4}{4}+(n+1)^3\le\frac{(n+2)^4}{4}$ for all $n \ge 1$.
$$\frac{(n+1)^4}{4}+(n+1)^3\le\frac{(n+2)^4}{4}$$
For all $n\ge 1$. I thought that I could get rid of the denominators like this:
$$(n+1)^4+4(n+1)^3\le(n+2)^4$$
Then, maybe, take $(n+1)^3$ as common factor:
$$(n+1)^3\cdot((n+1)+4)\le(n+2)^4$$
$$(n+1)^3\cdot(n+5)\le(n+2)^4$$
But I get the feeling that I'm just getting stuck. How can I further prove this?
This came up because I was doing an induction exercise. I needed to prove for all $n \ge 1$:
$$1^3+2^3+3^3+...+n^3\le \frac{(n+1)^4}{4}$$
By the hypothesis I know that
$$1^3+2^3+3^3+...+n^3+(n+1)^3\le \frac{(n+1)^4}{4}+(n+1)^3$$
So what I finally need to prove is that
$$\frac{(n+1)^4}{4}+(n+1)^3 \le \frac{(n+2)^4}{4}$$
| It is equivalent to see that $$4(n+1)^3\leq(n+2)^4-(n+1)^4=\bigg((n+1)^2+(n+2)^2\bigg)\big(2n+3 \big)\leq \bigg((n+1)^2+(n+2)^2\bigg)\big(3n+3)\leq 3(n+1)^3+3(n+1)(n+2)^2,$$
which is always true, where we have used the identity $a^4-b^4=(a^2+b^2)(a^2-b^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/593427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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A trigonometic integral with complex technicals Let $a,b \in \mathbb{R}^+$. Show that $$\int_0^{2\pi}\frac{1}{a^2\cos^2(t)+b^2\sin^2(t)}dt=\frac{2\pi}{ab}$$ Help please!
Thanks.
| Use this to conclude what you want:
$$\int_0^{2 \pi} \dfrac{dt}{1+r \sin^2(t)} = \sum_{k=0}^{\infty} (-r)^k \int_0^{2\pi} \sin^{2k}(t)dt = 2 \pi \sum_{k=0}^{\infty}(-r/4)^k \dbinom{2k}k = \dfrac{2\pi}{\sqrt{1+r}}$$
where we used the fact that
$$\int_0^{2\pi} \sin^{2k}(t) dt = \dfrac{2\pi}{4^k}\dbinom{2k}k$$
and
$$\dfrac1{\sqrt{1+r}} = \sum_{k=0}^{\infty}(-r/4)^k \dbinom{2k}k$$
We have
$$a^2 \cos^2(t) + b^2 \sin^2(t) = a^2 + (b^2-a^2)\sin^2(t) = a^2\left(1+\left(\left(\dfrac{b}a\right)^2 - 1\right)\sin^2(t)\right)$$
Hence, in our case, we have $r = \left(\dfrac{b}a \right)^2 - 1$ and hence $\sqrt{1+r} = \dfrac{b}a$.
$$I = \int_0^{2\pi} \dfrac{dt}{a^2 \cos^2(t) + b^2 \sin^2(t)} = \dfrac1{a^2} \times \dfrac{2\pi}{b/a} = \dfrac{2\pi}{ab}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/597125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find an accurate value of $f(x)=\sqrt{4x^2+x}-2x$ for large values of x. Calculate $\lim_{x\to\infty}f(x)$ My works:
$x^2$ can be very large if x is large, thus the function has lose-of-significance error and we need to reformulate it. $$ f(x)=\sqrt{4x^2+x}-2x=\sqrt{x(4x+1)}-2\sqrt{x}\sqrt{x}=\sqrt{x}(\sqrt{4x+1}-2\sqrt{x})$$ The latter will not have any lose-of-significance error in the evaluation.
I am not sure if I did this correctly and I do not know how to compute the limit...
| Note that
$$\sqrt{4x^2+x}-2x = \dfrac{\sqrt{4x^2+x}+2x}{\sqrt{4x^2+x}+2x} \cdot \sqrt{4x^2+x}-2x = \dfrac{4x^2+x-4x^2}{\sqrt{4x^2+x}+2x} = \dfrac{x}{\sqrt{4x^2+x}+2x}$$
Hence, use $\dfrac{x}{\sqrt{4x^2+x}+2x}$ for computation purposes, especially when $x$ is large. There is no catastrophic cancellation of digits as opposed to evaluating $\sqrt{4x^2+x}-2x$ directly.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $\sum ^{\infty }_{n=2}\dfrac {2+4+6+\ldots +2n}{1+a_{n}}$ Define $a_{n}=1!\left( 1^{1}+1+1\right) +2!\left( 2^{2}+2+1\right) +3!\left( 3^{2}+3+1\right) +\ldots +n!\left( n^{2}+n+1\right) $ , Find $\sum ^{\infty }_{n=2}\dfrac {2+4+6+\ldots +2n}{1+a_{n}}$
| You can show that $1+a_n = (n+1) \cdot (n+1)!$ by induction on $n$.
Since $2+4+6+ ... + 2n = n(n+1)$, your sum becomes $$\sum_{n=2}^{\infty} \frac{n}{(n+1)!} = \sum_{n=2}^{\infty} \frac{1}{n!} - \sum_{n=2}^{\infty} \frac{1}{(n+1)!} = \frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/602496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of a Sequence Defined Recursively Fix $r > 1$ and $a_1$ = 1 and define the sequence recursively as $a_{n+1}$ = $\frac{1}{r}$$(a_n + r + 1)$. Find the limit of the sequence.
| This can also be done using induction:
$$a_1=1\implies a_2=\dfrac{2+r}{r}\implies a_3=\dfrac{2+2r+r^2}{r^2}\implies a_4=\dfrac{2+2r+2r^2+r^3}{r^3}\implies\dots\dots$$
$$$$
$$\implies a_n=\dfrac{2+2r+2r^2+\dots+2r^{n-2}+r^{n-1}}{r^{n-1}}=1+\dfrac{2}{r}+\dfrac{2}{r^{2}}+\dfrac{2}{r^{3}}+\dots\dots+\dfrac{2}{r^{n-1}}$$
$$\implies a_n=1+2\sum_{i=1}^{n-1}\dfrac{1}{r^i}$$
$$\implies a_n\stackrel{n\to\infty}{\longrightarrow}1+2\dfrac{\dfrac{1}{r}}{1-\dfrac{1}{r}}=\dfrac{r+1}{r-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/603102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I find the fifth root of unity? I need to find fifth root of unity in the form $x+iy$.
I'm new to this topic and would appreciate a detailed "dummies guide to..." explanation!
I understand the formula, whereby for this question I would write:
$1^{1/5} = r^{1/5}e^{2ki\pi/5}$.
However, I don't know what to do next.
Any help is appreciated.
| Let's do it the hard way. We want to solve the equation $$x^5-1=(x-1)(x^4+x^3+x^2+x+1)=0.$$ Then we are interested in solving $x^4+x^3+x^2+x+1=0$. Note the symmetry: if $r$ is a root, $1/r$ is also a root. Why? Thereafter, we have two ways out. First, we divide everything by $x^2$ to get $$x^2+x+1+\frac 1x+\frac{1}{x^2}=\left(x+\frac1x\right) +\left(x+\frac1x\right)^2-1=u^2+u-1=0.$$ What did we do? We noted $\left(x+x^{-1}\right)^2-2=x^2+x^{-2}$ and substituted $u=x + \frac{1}{x}$. The solutions for $u^2+u=1$ are $-\varphi$ and $\varphi-1$, $\varphi$ being the golden ratio. We now have two equations: $$x+\frac1x=-\varphi\implies x^2+\varphi x+1=0\implies x=\frac{\pm\sqrt{\varphi-3}-\varphi}{2}$$
$$x+\frac{1}{x}=\varphi-1\implies x^2+(1-\varphi)x+1=0\implies x=\frac{\pm\sqrt{-\varphi-2}+\varphi-1}{2}.$$
You can now manipulate it to get a $a+bi$ form. Alternatively, you can multiply $(x-r)(x-r^{-1})$, divide $x^4+x^3+x^2+x+1$ by it and discover which $r$ makes remainder zero. This would be somewhat long, so I won't explictly do it, but here is the idea.
| {
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Find a circle orthogonal to two other circles Given two circles
$x^2+(y-4)^2=4^2$
and
$(x-8)^2+(y+2)^2=2^2$
I need to find the circle that's orthogonal to both the above circles, and contains the point $P=(8,4).$
Is there an algebraic way to find this circle? Or is a geometric way easier?
Any advice will be greatly appreciated.
Thanks in advance!
| Consider $\bigcirc A$ with radius $a$ and $\bigcirc B$ of radius $b$, and $\bigcirc R$ of radius $r$ orthogonal to both.
That $\bigcirc R$ is orthogonal to $\bigcirc A$ means that either point of intersection determines a right triangle with legs $r$ and $a$ and hypotenuse $|AR|$. Likewise for $\bigcirc B$. Thus,
$$a^2 + r^2 = |AR|^2 \qquad\qquad b^2 + r^2 = |BR|^2$$
If point $P$ is on $\bigcirc R$, then also
$$r^2 = |PR|^2$$
Writing $R(h,k)$, the above equations make a non-linear system in unknowns $h$, $k$, $r$. The non-linearness is daunting, but it can be overcome. In fact, the specific nature of your problem makes it easier than it would be in general.
In your problem, we have $A = (0, 4)$, $B = (8,-2)$, $P = (8,4)$, and $a = 4$, $b = 2$. The above equations become
$$\begin{align}
16 + r^2 &= ( h- 0 )^2 + ( k - 4 )^2 \\
4 + r^2 &= ( h - 8 )^2 + ( k + 2 )^2 \\
r^2 &= ( h - 8 )^2 + ( k - 4 )^2
\end{align}$$
where I won't multiply-out the terms. Instead, I'll use the last equation to replace $r^2$ in the first two and get some convenient cancellation (which is what makes your specific problem easier):
$$\begin{align}
16 + ( h - 8 )^2 + ( k - 4 )^2 &= ( h- 0 )^2 + ( k - 4 )^2 \quad \to \quad 16 + ( h - 8 )^2 = h^2 \\
4 + ( h - 8 )^2 + ( k - 4 )^2 &= ( h - 8 )^2 + ( k + 2 )^2 \quad \to \quad \phantom{1}4 + ( k - 4 )^2 = ( k + 2 )^2\\
\end{align}$$
Now, multiplying-out, we see that the quadratic terms $h^2$ and $k^2$ cancel in their respective equations, and we solve to get
$$h = 5 \qquad k = \frac{4}{3} \qquad \text{and, thus} \qquad r^2 = (5-8)^2 + \left(\frac{4}{3} - 4 \right)^2 = \frac{145}{9}$$
so that the equation of $\bigcirc R$ is
$$\left( x - 5 \right)^2 + \left( y - \frac{4}{3} \right)^2 = \frac{145}{9}$$
This agrees with @Apurv's answer.
(If we hadn't gotten the convenient cancellation, we would still have been able to cancel $h^2$ and $k^2$ from the equation pair. This would leave a linear system in $h$ and $k$ that could be solved.)
Note: One can prove @Apurv's orthogonality condition by observing that
$$x^2 + y^2 + 2 m x + 2 n y + c = 0 \quad\text{and}\quad x^2 + y^2 + 2 p x + 2 q y + d = 0$$
represent circles about $U(-m,-n)$ and $V(-p,-q)$, with respective radii $u$ and $v$ satisfying $u^2 = m^2+n^2-c$ and $v^2 = p^2+q^2-d$. Thus,
$$\begin{align}
u^2 + v^2 = |UV|^2 \quad&\implies\quad m^2 + n^2 + p^2 + q^2 - (c + d) = (m-p)^2 + (n-q)^2 \\
&\implies\quad c + d = 2 m p + 2 n q
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculation of $\int\frac{x^2 \left(\sin 2x - \cos 2x\right)}{\left(1+\sin 2x\right)\cos^2 x}dx$ Calculation of $\displaystyle \int\frac{x^2 \left(\sin 2x - \cos 2x\right)}{\left(1+\sin 2x\right)\cos^2 x}dx$
$\bf{My\; Try}::$ We can write the given Integral as $\displaystyle \int\frac{x^2 \left\{(1+\sin 2x)-(1+\cos 2x)\right\}}{(1+\sin 2x)\cos^2 x}dx$
$\displaystyle = \int\frac{x^2}{\cos^2 x}dx - \int \frac{2x^2}{1+\sin 2x}dx$
$\displaystyle = \int x^2 \sec^2 xdx - \int\frac{x^2}{\cos^2 \left(\frac{\pi}{4}-x\right)}dx$
$\displaystyle = \int x^2 \sec^2 xdx - \int x^2 \sec^2 \left(\frac{\pi}{4}-x\right)dx$
Now I did not understand how can i solve that
Help Required
Thanks
| My hint
$$\int x^2d(\tan x)+\int x^2d(\tan(\frac{\pi}{4}-x))=x^2(\tan x+\tan(\frac{\pi}{4}-x))-2\int x(\tan x+\tan(\frac{\pi}{4}-x))dx$$
$$\int x(\tan x+\tan(\frac{\pi}{4}-x))dx=\int \frac{x}{\cos x(\sin x+\cos x)}dx$$
This is the result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this inequality: $\sum{\frac{1}{a^2\sqrt{a^2+2ab}}}\ge\frac{\sqrt{3}}{abc}$ Let $a,b,c>0$. Prove this inquality:
$\frac{1}{a^2\sqrt{a^2+2ab}}+\frac{1}{b^2\sqrt{b^2+2bc}}+\frac{1}{c^2\sqrt{c^2+2ac}}\ge\frac{\sqrt{3}}{abc}$
| hint:
$\sqrt{a^2+2ab}=\sqrt{\dfrac{3a(a+2b)}{3}}\le \dfrac{2a+b}{\sqrt{3}}$
then the inequality becomes: $\sum_{cyc}\dfrac{bc}{a(2a+b)} \ge1$ which is easier .
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate: $\int \frac{1}{(x+a)(x+b)}$ Evaluate:
$$\int \frac{1}{(x+a)(x+b)}$$
My attempt:
$$\int \frac{1}{(x+a)(x+b)} = \frac{A}{x+a} + \frac{B}{x+b}$$
$$1 = A(x+b) + B(x+a)$$
$$x = -b$$
$$1 = A(-b + b) + B(-b + a)$$
$$1 = B(-b + a)$$
$$x = -a$$
$$1 = A(-a + b) + B(-a + a)$$
$$1 = A(-a + b)$$
At this point I have no idea how to proceed. Can someone help me with this? Please.
| First, it is easy to see there are two cases,(the problem has implied that $x\neq -a,\,-b$)
Case 1, if $a=b$, then the integration is $\int \frac{1}{(x+a)^2}=-\frac{1}{x+a}$.
Case 2, if $a\neq b$, then $\int \frac{1}{(x+a)(x+b)}=\int \frac{1}{b-a}\left(\frac{1}{x+a}-\frac{1}{x+b}\right)=\frac{1}{b-a}\left(\int \frac{1}{x+a}-\int \frac{1}{x+b}\right)=\frac{1}{b-a}\ln \left|\frac{x+a}{x+b}\right|+C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/608601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate the integral $I=\int_{0}^{\infty}\frac{\ln^3{x}}{(1+x^2)(1+x)^2}dx$ Find this integral
$$I=\int_{0}^{\infty}\dfrac{\ln^3{x}}{(1+x^2)(1+x)^2}dx$$
My try: let $x=\tan{t}$
then
$$I=\int_{0}^{\frac{\pi}{2}}\dfrac{\ln^3{\tan{t}}}{(1+\tan{t})^2}dt$$
I am unable to simplify after this. This problem is from QQ.
| Filling in the steps of Mhenni Benghorbal's solution we have
The Mellin transform
\begin{equation}
\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x
\label{eq:160813a2}
\tag{2}
\end{equation}
where
\begin{equation}
f(x) = \frac{1}{(1+x^{2})(1+x)^{2}} = -\frac{1}{2}\frac{x}{x^{2}+1} + \frac{1}{2}\frac{1}{x+1} + \frac{1}{2}\frac{1}{(x+1)^{2}}
\label{eq:160813a3}
\tag{3}
\end{equation}
via partial fraction expansion.
Applying the Mellin transform, yields
\begin{align}
\mathcal{M}[f(x)](s) & = \int\limits_{0}^{\infty} \frac{x^{s-1}}{(1+x^{2})(1+x)^{2}} \mathrm{d} x \\
& = -\frac{1}{2}\left[\frac{1}{2}\pi\sec\left(\frac{\pi}{2}s\right)\right] + \frac{1}{2}\pi\csc(\pi s) + \frac{1}{2} \mathrm{B}(s,2-s)
\label{eq:160813a4}
\tag{4}
\end{align}
Taking the 3rd derivative of equation \eqref{eq:160813a4} with respect to s and then taking $\lim s \to 1$ yields
\begin{align}
\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x & = -\frac{1}{2}\left( -\frac{7}{960} \pi^{4} \right) + \frac{1}{2} \left( -\frac{7}{60} \pi^{4} \right) + \frac{1}{2}0 \\
& = -\frac{7}{128} \pi^{4}
\label{eq:160813a5}
\tag{5}
\end{align}
Let us fill in the details. Handling the beta function first, we have
\begin{equation}
\mathrm{B}(s,2-s) = \frac{\Gamma(s)\Gamma(2-s)}{\Gamma(2)} = \Gamma(s)\Gamma(2-s)
\label{eq:160813a6}
\tag{6}
\end{equation}
To take derivatives, we note that
\begin{equation}
\frac{\mathrm d}{\mathrm d s} \Gamma(s) = \Gamma(s) \psi^{(0)}(s) \quad \mathrm{and} \quad \psi^{(n)}(s) = \frac{\mathrm{d}^{n}}{\mathrm{d} s^{n}} \psi^{(0)}(s)
\end{equation}
Where $\psi^{(n)}(s)$ is the polygamma function.
Taking the third derivative of equation \eqref{eq:160813a6} and letting $\lim s \to 1$ equals 0. Here we used
\begin{equation}
\psi^{(0)}(1) = -\gamma \quad \mathrm{and} \quad \psi^{(1)}(1) = \frac{\pi^{2}}{6}
\end{equation}
and fortunately $\Gamma^{(3)}(s) = -\Gamma^{(3)}(2-s)$ which leads to some cancellations.
Doing the same for the first two terms on the right hand side of equation \eqref{eq:160813a4}, we have to be careful. Each of them individually goes to $\infty$ as $\lim s \to 1$ but the $(s-1)^{-4}$ terms in the Laurent expansions about $s=1$ cancel. Here I used Wolfram Alpha. Doing it with the two terms combined yielded our final answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/610435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Nonhomogeneous Linear ODE $$
x'
=\left(\begin{array}{rr}4 & 8 \\ -2 & -4\end{array}\right)x + \left(\begin{array}{rr}t^{-3} \\ -t^{-2}\end{array}\right), t>0
$$
To find the general solution of the given system above, my solution is:
$det(\left(\begin{array}{rr}4-r & 8 \\ -2 & -4-r\end{array}\right)) = r^2$
, eigenvalues are $r_{1,2}=0$
$\left(\begin{array}{rr}4 & 8 \\ -2 & -4\end{array}\right)\left(\begin{array}{rr}v_{11} \\ v_{12}\end{array}\right) = \left(\begin{array}{rr}0 \\ 0\end{array}\right)$ Therefore $v_{1} =\left(\begin{array}{rr}-2 \\ 1\end{array}\right)$
$\left(\begin{array}{rr}4 & 8 \\ -2 & -4\end{array}\right)\left(\begin{array}{rr}v_{21} \\ v_{22}\end{array}\right) = \left(\begin{array}{rr}-2 \\ 1\end{array}\right)$ From there, I have and equation $2v_{21}+4v_{22}=-1$
What should I do now? Thanks.
| Since $\begin{pmatrix}
4 & 8\\
-2 & -4
\end{pmatrix}\cdot \binom{-1/2}{0} = \binom{-2}{1}$ works, we get a homogeneous solution to the ODE of:
$c_{1}\binom{-2}{1}+c_{2}(t\binom{-2}{1}+\binom{-1/2}{0})$. Now it suffices to find a particular solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/613432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $\sum_{n=1}^{\infty}\arctan{\left(\frac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\frac{\sqrt{2}-1}{2}}\right)\cdots\right)}$
show that:
$$\sum_{n=1}^{\infty}\arctan{\left(\dfrac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\dfrac{\sqrt{2}-1}{2}}\right)\cdot\dfrac{e^{\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}+e^{-\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}}{e^{\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}-e^{-\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}}\right)}-\dfrac{\pi}{8}$$
This relust is nice.(maybe have some wrong)
becasue I know this famous problem
$$\sum_{n=1}^{\infty}\arctan{\dfrac{2}{n^2}}=\dfrac{3\pi}{4}$$
and follow AMM( E3375) problem
$$\sum_{n=1}^{\infty}\arctan{\dfrac{1}{n^2}}=\arctan{\left(\dfrac{\tan{\frac{\pi}{\sqrt{2}}}-th{\dfrac{\pi}{\sqrt{2}}}}{\tan{\dfrac{\pi}{\sqrt{2}}}+th{\dfrac{\pi}{\sqrt{2}}}}\right)}$$
Follow is AMM solution:
My try: my problem I want use this methods,But at last failure it.
Thank you for you help.
This problem is similar this:we konw this
$$\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$$
and then little hard problem:
$$\sum_{n=1}^{\infty}\dfrac{1}{n^2+1}=\dfrac{1}{2}(\pi\coth{\pi}-1)$$
| For any $\alpha > 0$, let $u + iv = \pi\sqrt{\alpha^2 + i}$, we have
$$\begin{align}
& \tan\left\{\sum_{n=1}^\infty \tan^{-1}\left(\frac{1}{n^2+\alpha^2}\right)\right\}
=\tan\left\{\sum_{n=1}^\infty\arg\left(1+\frac{i}{n^2+\alpha^2}\right)\right\}\\
= &\tan\left\{\arg\prod_{n=1}^\infty \left(1+\frac{i}{n^2+\alpha^2}\right)\right\}
=\tan\left\{\arg\prod_{n=1}^\infty \left(
\frac{1+\frac{\alpha^2+i}{n^2}}{1+\frac{\alpha^2}{n^2}}
\right)\right\}\\
= &\tan\left\{\arg\prod_{n=1}^\infty \left(1+\frac{\alpha^2+i}{n^2}\right)\right\}
=\tan\left\{\arg\prod_{n=1}^\infty \left(1+\frac{(u+iv)^2}{n^2\pi^2}\right)\right\}\\
= & \tan\left\{\arg\left(\frac{\sinh(u+iv)}{u+iv}\right)\right\}
=\tan\left\{\arg\left(\frac{\tanh u + i\tan v}{u+iv}\right)\right\}\\
= & \tan\left\{\arg(\tanh u + i\tan v) - \arg(u+iv)\right\}\\
= & \tan\left\{\tan^{-1}\left(\frac{\tan v}{\tanh u}\right)-\frac12\arg(\alpha^2 + i)\right\}\\
= & \tan\left\{\tan^{-1}\left(\frac{\tan v}{\tanh u}\right)-\frac12\tan^{-1}\left(\frac{1}{\alpha^2}\right)\right\}
\end{align}$$
For $\alpha = 1$, we have $u + iv = \pi\sqrt{\frac{\sqrt{2}+1}{2}} + i\pi\sqrt{\frac{\sqrt{2}-1}{2}}$, so
$$\sum_{n=1}^\infty \tan^{-1}\left(\frac{1}{n^2+1}\right) =
\tan^{-1}\left(\frac{\tan\left(\pi\sqrt{\frac{\sqrt{2}-1}{2}}\right)}{\tanh\left(\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)}\right)- \frac{\pi}{8} + N \pi$$
for some integer $N$ to be determined. Numerically, the RHS excluding the unknown term $N\pi$ is about $1.0373$. On the LHS, we know it is a number around $1$. So the unknown constant $N$ is $0$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/613670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Does the series $\sum^\infty_{n=1}\frac{n!}{\sqrt{(2n)!}}$ converge/diverge?
Does the series $\displaystyle\sum^\infty_{n=1}\frac{n!}{\sqrt{(2n)!}}$ converge/diverge?
I used the ratio test but I'm not sure:
$\begin{align} \frac{\frac{(n+1)!}{\sqrt{(2n+2)!}}}{\frac{n!}{\sqrt{(2n)!}}}
&=\frac{n+1}{\sqrt{(2n+1)(2n+2)}}\\
&=\frac{n+1}{4(n+1)^2\sqrt{2n+1}}\\
&=\frac{1}{4\sqrt{2n+1}} \end{align}$
The limit of that is smaller than $1$ so the series does converge. Is it correct ?
| This is in response to your question here. It's too long for a comment so I'm posting it as an answer.
No, you can't take them out of the root like that. One way to handle the denominator is by writing it like
$$
\begin{align}
\sqrt{(2n+1)\cdot(2n+2)} &= \sqrt{(2n+1)\cdot 2\cdot (n+1)} \\
&= \sqrt{2n+1} \sqrt{2} \sqrt{n+1},
\end{align}
$$
so that
$$
\frac{n+1}{\sqrt{(2n+1)(2n+2)}} = \frac{n+1}{\sqrt{2} \sqrt{2n+1} \sqrt{n+1}}.
$$
Now
$$
\frac{n+1}{\sqrt{n+1}} = \sqrt{n+1},
$$
so
$$
\frac{n+1}{\sqrt{2} \sqrt{2n+1} \sqrt{n+1}} = \frac{\sqrt{n+1}}{\sqrt{2}\sqrt{2n+1}}.
$$
To calculate the limit of this as $n \to \infty$, divide the numerator and the denominator by $\sqrt{n}$ to get
$$
\begin{align}
\frac{\frac{1}{\sqrt{n}}\sqrt{n+1}}{\frac{\sqrt{2}}{\sqrt{n}}\sqrt{2n+1}} &= \frac{\sqrt{1+\frac{1}{n}}}{\sqrt{2}\sqrt{2 + \frac{1}{n}}} \\
&\overset{n\to\infty}{\longrightarrow} \frac{\sqrt{1+0}}{\sqrt{2}\sqrt{2 + 0}} \\
&= \frac{1}{2}.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Rational roots of a cubic polynomial Find all distinct non-zero rational numbers $a$, $b$ and $c$ such that $x (x+a) (x+b) +c$ has 3 distinct non-zero rational roots.
What I have so far:
Let the polynomial have factorization $(x+x_1)(x+x_2)(x+x_3)$ where $x_i$'s are distinct non-zero
rational number. Equate coefficients of like power,
$$(x+x_1)(x+x_2)(x+x_3) = x (x+a) (x+b) +c,$$
after some algebraic manipulation we get:
$$
x_3^2 - (a+b - x_1) x_3 + (a \cdot b - x_1 (a+b - x_1)) = 0
$$
Since $x_3$ is assumed to be rational, the discriminate must be rational
$$
r = \sqrt{(a+b - x_1)^2 - 4 (a \cdot b - x_1 (a+b - x_1))}
$$
Assuming we can choose $a$, $b$ and $x_1$ such that $r$ is rational,
\begin{align*}
x_3 & = \frac{a + b - x_1 \pm r}{2} \\
x_2 &= a + b - x_1 - x_3 \\
c &= x_1 x_2 x_3
\end{align*}
This meets all the requirements, if $x_1 \ne x_2 \ne x_3 \ne 0$.
These conditions can be written as
\begin{align*}
r &\ne 0 \\
x_1 & \ne \frac{a+b\pm r}{3} \\
x_1 &\ne a+b\pm r \\
\end{align*}
So the problem is equivalent to finding all $a$, $b$ and $x_1$ such that $r$ is rational and
the 3 conditions are met. But I cannot show if such an $r$ exists, or if does,
enumerate some values of $a$, $b$, and $x_1$, let alone all of them.
Edit: corrected an algebraic mistake.
| Actually, your condition,
$(a+b - x_1)^2 - 4 (a \cdot b - x_1 (a+b - x_1))=r^2\tag{1}$
is quite easy to solve. Collecting powers of $x_1$, we get the equivalent,
$(a-b)^2+2(a+b)x_1-3x_1^2=r^2\tag{2}$
Assume $r=nx_1+(a-b)$ for some free variable $n$, and (2) reduces to,
$x_1(-2 a - 2 b + 2 a n - 2 b n + 3 x_1 + n^2 x_1) = 0\tag{3}$
You can then rationally solve for $x_1$ in terms of free variables $a,b,n$ (though choose them to avoid the 3 conditions you mentioned).
P.S. I forgot to add that (3) is the complete rational solution to (2). For any given $a,b,x_1$ with $x_1\neq0$ such that (2) is a square, one can always find rational n as $n = (r-(a-b))/x_1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Intersection of Trig Functions The questions asks to find the intersections of
$$f(x) = 2 \sin(x-7) + 6$$ and $$g(x) = \cos(2x-10) + 8$$
within the interval $[6,14]$.
So my general strategy was, 1) equate the functions, 2) get all the $X$s on one side and 3) convert to the same trig function.
So
$$2 \sin(x-7) + 6 = \cos(2x-10) + 8$$
I recognized the double angle in the cosine function, so
$$2 \sin(x-7) + 6 = \cos[ 2 (x-5) ] + 8$$
then
$$2 \sin(x-7) + 6 = \cos^2(x-5) - \sin^2(x-5) + 8$$
$\cos^2$ can be replaced with an identity, so
$$2 \sin(x-7) + 6 = 1 - \sin^2(x-5) - \sin^2(x-5) + 8$$
Group like terms and move then around,
$$2 \sin(x-7) + 2 \sin^2(x-5) = 3$$
Extracting the $2$ from the left side.
$$\sin(x-7) + \sin^2(x-5) = \frac 3 2$$
So here is where I hit a mental wall.
I could use the sine addition formula, but that would reintroduce cosine.
I can't simplify the terms any further since the angles are different.
Where would I go from here? Or is my approach off completely?
| $$\sin(x-7)+\sin^2(x-5)=\frac32\iff\sin(t-2)+\sin^2t=\frac32$$
$$\sin t\cdot\cos(-2)+\cos t\cdot\sin(-2)+\sin^2t=\frac32$$
$$\sin^2t+\cos2\cdot\sin t-\sin2\cdot\cos t=\frac32\iff y^2+\cos2\cdot y-\sin2\cdot\bigg(\!\!\pm\sqrt{1-y^2}\bigg)=\frac32$$
$$y^2+\cos2\cdot y-\frac32=\sin2\cdot\bigg(\!\!\pm\sqrt{1-y^2}\bigg)\iff\bigg(y^2+\cos2\cdot y-\frac32\bigg)^2=\sin^22\cdot(1-y^2)$$
Now you're left with solving a quartic equation in $y=\sin t=\sin(x-5)\iff x=5+\arcsin y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$. If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$$
Without using calculus.
There are a few ways I've tried solving this:
$1)$ We could try using the Cauchy-Schwarz inequality: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge \frac{(x+y+z)^2}{(x-1)^2+(y-1)^2+(z-1)^2}$$
But it's apparent that nothing's useful here.
$2)$ We could use AM-GM as well: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 3\sqrt[3]{\frac{x^2y^2z^2}{(x-1)^2(y-1)^2(z-1)^2}}=\frac{3}{\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}}$$
So we only have to prove that: $$\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}\le 3$$
We could raise both sides to the power of 3: $$(x-1)^2(y-1)^2(z-1)^2\le 27$$
But this inequality doesn't hold.
$3)$ We could try cleaning the denominators by multiplying both sides by $(x-1)^2(y-1)^2(z-1)^2$. After a bunch of expanding and simplifying we get that: $$x^2y^2+y^2z^2+x^2z^2-6(xy+yz+xz)+2(x+y+z)+9\ge 0$$
I can't tell so easily whether the inequality is true or not. You could help me out on this one. Just don't forget that $x,y,z\in\mathbb R\setminus\{1\}$ and so we can't just simply use AM-GM, unless we're using it for squares that have to be non-negative.
| Proof:
Use Cauchy-Schwarz inequality,we have
$$\left[\sum_{cyc}\left(\dfrac{a}{a-b}\right)^2\right]\left[\sum_{cyc}(a-b)^2(a-c)^2\right]\ge\left(\sum_{cyc}a^2-\sum_{cyc}ab\right)^2$$
and since
\begin{align*}
\sum_{cyc}(a-b)^2(a-c)^2&=\sum_{cyc}(a-b)^2(a-c)^2+2\sum_{cyc}(a-b)(a-c)(b-c)(b-a)\\
&=\left[\sum_{cyc}(a-b)(a-c)\right]^2=\left(\sum_{cyc}a^2-\sum_{cyc}ab\right)^2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Prove that $n(n^2 - 1)(n + 2)$ is divisible by $4$ for any integer $n$
Prove that $n(n^2 - 1)(n + 2)$ is divisible by $4$ for any integer $n$
I can not understand how to prove it. Please help me.
| The first very obvious thing to note is that $n(n^2 - 1)(n+2)$ can be written as $n(n - 1)(n+1)(n+2)$ which means that they are four consecutive integers. Obviously one of these must be divisible by $4$.
The proof:
Any number $n$ is either of the form $2m$ or $2m+1$.
Case I: $n = 2m$
$n(n^2 - 1)(n+2)$ = $2m(4m^2 - 1)(2m + 2) = 2m(4m^2 - 1)*2(m+1) = 4x$
Case II: $n = 2m+1$
$n(n^2 - 1)(n+2) = (2m+1)((2m+1)^2 - 1)(2m+1 + 2) = (2m+1)((4m^2 + 4m +1) - 1)(2m+3) = 4x$
Hence proved.
The link provided by @lab bhattacharjee is quite useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
$\sin (A-B) = \pm\frac{1}{3}.$ if....... please help me, how can I approach. I can not understand what to do. How can I start. Help me. If $\sqrt{2}\cos A =\cos B +\cos^3 B$ and $\sqrt{2} \sin A = \sin B - \sin^3 B$ show that $\sin (A-B) =\pm\frac{1}{3}$.
| \begin{align}
\sqrt 2 \sin A \cos B = \sin B \cos B \left ( 1-\sin^2B\right ) \tag 1 \\
\sqrt 2 \cos A \sin B = \sin B \cos B \left( 1 + \cos^2 B\right ) \tag 2
\end{align}
Now $(1) - (2)$ and use $\sin(A-B) = \sin A \cos B - \cos A \sin B$
\begin{align}
\sqrt 2 \sin (A-B) &= \sin B \cos B \left( 1-\sin^2B - 1 - \cos^2B\right ) = -\sin B \cos B \\
\sin(A-B) &= -\frac {\sin B\cos B}{\sqrt 2} \tag 3
\end{align}
\begin{align}
\sqrt 2 \cos A \cos B &= \cos^2B \left( 1+\cos^2 B\right ) \tag 4\\
\sqrt 2 \sin A \sin B &= \sin^2B \left( 1-\sin^2B\right ) = \sin^2 B \cos^2B\tag 5
\end{align}
Now, $(4) + (5)$ and use $\cos(A-B) = \cos A \cos B - \sin A \sin B$
\begin{align}
\sqrt 2 \cos(A-B) &= \cos^2B \left( 1+\cos^2B + \sin^2B\right ) = 2\cos^2B \\
\cos(A-B) &= \sqrt 2 \cos^2B \tag 6
\end{align}
Now use $(3)^2+(6)^2 \equiv 1$, so
$$
\frac {\sin^2B \cos^2B}2 + 2 \cos^4B = 1 \\
\cos^2B \left( \sin^2B + 4 \cos^2B\right ) = 2 \\
\cos^2B \left ( 1 + 3 \cos^2B\right ) = 2 \\
3\cos^4B + \cos^2B - 2 = 0 \\
\cos^2B = \frac {-1 \pm \sqrt{1 + 24}}6 = \frac {-1 \pm 5}6 = \frac 23
$$
Therefore
$$
\sin^2B = 1 - \cos^2B = \frac 13
$$
and
$$
\sin B \cos B = \pm \sqrt{\sin^ 2B \cos^2B} = \pm \frac {\sqrt 2}3
$$
Finally
$$
\sin(A-B) = -\frac 1{\sqrt2} \sin B \cos B = \mp \frac 13
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Calculate the sum using Fourier series of $\big|\cos(\frac{x}{2})\big|$. I have been given task to evaluate the sum $\dfrac{(-1)^{n-1}}{4n^2-1}$ using Fourier series for function $|\cos{\frac{x}{2}}|$.
I used the interval $(-\pi, +\pi)$ for evaluation of the sum and I noticed that $|\cos{\frac{x}{2}}|$ = $\cos{\frac{x}{2}}$ on this interval which made my life a lot easier. :) And also that $a_n$ does not exist in this Fourier series (except $a_0$), but when I calculate $b_n$, I get $\dfrac{1}{\pi} \cdot (-1)^{n-1} \cdot \dfrac{2}{1-n^2}$ which doesn't help me very much with this task.
Can somebody give me some advice how to solve this, and more generally how to find an interval of integration to find some sum?
Thanks in advance.
| This is a classic application of Fourier series to computation of series.
First, cosine is $2\pi$-periodic, thus $f(x)=|\cos x/2|$ is $2\pi$-periodic. Since it's even, the sine coefficient will vanish, I will only consider the cosine coefficients.
With $T=2\pi$, you have (I write $S(x)$ for the series, not $f$ since you have also to prove $S=f$, but it follows from the fact that $f$ is continuous and piecewise $C^1$, and Dirichlet theorem).
$$S(x)=\frac{a_0}2+\sum_{n=1}^\infty a_n \cos \left(\frac{2\pi}{T}nx\right)$$
$$a_n=\frac 2T\int_{-T/2}^{T/2} f(x) \cos \left(\frac{2\pi}{T}nx \right)\mathrm{d}x $$
Since $\cos (x/2)\geq 0$ for $x\in[-\pi,\pi]$,
$$a_n=\frac 1{\pi}\int_{-\pi}^{\pi} \cos\left(\frac x2\right) \cos \left(nx \right)\mathrm{d}x $$
$$a_n=\frac{2}{\pi}\int_0^{\pi} \cos\left(\frac x2\right) \cos \left(\frac{2nx}2 \right)\mathrm{d}x $$
And since $2\cos a\cos b=\cos(a+b)+\cos(a-b)$,
$$a_n=\frac{1}{\pi}\int_0^{\pi} \cos\left(\frac {2n+1}2x\right) +\cos\left(\frac {2n-1}2x\right)\mathrm{d}x $$
$$a_n=\frac{1}{\pi}\left[\frac 2{2n+1}\sin \left(\frac {2n+1}2x\right) + \frac 2{2n-1}\sin \left(\frac {2n-1}2x\right)\right]_0^\pi$$
$$a_n=\frac{1}{\pi} \left(\frac 2{2n+1}\sin \left(\frac {2n+1}2\pi\right) + \frac 2{2n-1}\sin \left(\frac {2n-1}2\pi\right)\right)$$
And $\sin \left(\frac {2n+1}2\pi\right)=\sin \left(n\pi+\frac {\pi} 2\right)=(-1)^n$, so
$$a_n=\frac{1}{\pi} \left(\frac 2{2n+1}(-1)^n - \frac 2{2n-1}(-1)^n\right)$$
$$a_n=\frac{2(-1)^n}{\pi}\left(\frac 1{2n+1} - \frac 1{2n-1}\right)=\frac{2(-1)^n}{\pi}\frac{-2}{4n^2-1}$$
$$a_n=\frac{4}{\pi}\frac{(-1)^{n-1}}{4n^2-1}$$
And $a_0=\frac 4{\pi}$
Now, your function is continuous and piecewise $C^1$, so $f=S$ and
$$\left|\cos \frac x2\right|=\frac{2}{\pi}+\frac{4}{\pi}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{4n^2-1} \cos nx$$
And for $x=0$,
$$ 1=\frac{2}{\pi}+\frac{4}{\pi}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{4n^2-1}$$
And finally,
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{4n^2-1}=\frac {\pi}4-\frac{1}{2}$$
| {
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Smallest value that a certain variable can take in a system of equations.
Consider the solutions $(x,y,z,u)$ of the system of equations:
$$\begin{cases}
x+y=3(z+u)\\
x+z=4(y+u)\\
x+u=5(y+z)\\
\end{cases}$$
where $x,y,z \text{ and } u$ are positive integers.
What is the smallest value that $x$ can have?
What I have done:
$$\begin{pmatrix}
1&1&-3\\
1&-4&1\\
1&-5&-5\\
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z\\
\end{pmatrix}
=\begin{pmatrix}
3\\
4\\
-1\\
\end{pmatrix}\ u
$$
$$ \left( \begin{array}{lcr|c}
1&1&-3&3\\
1&-4&1&4\\
1&-5&-5&-1\\
\end{array}
\right)
\equiv
\left( \begin{array}{lcr|c}
17&0&0&83\\
0&17&0&7\\
0&0&17&13\\
\end{array}
\right)
$$
Hence
$$\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z\\
\end{pmatrix}
=\frac{1}{17}\begin{pmatrix}
83\\
7\\
13\\
\end{pmatrix}\ u
$$
$$x=\frac{83}{17}u,\; y=\frac{7}{17}u,\; z=\frac{13}{17}u$$
Since $x,y,z,u \in \mathbb{Z^+},x_{min}=83 \quad $ Is that right?
| Given that the variables have to be positive integers, $83$ is correct for the minimum value of $x$. You have done well.
| {
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If two polynomials both of n degree have n identical real roots, are they equal? Proof? CORRECTION: The polynomials don't have to be equal, but one has to be a constant multiple of the other.
I ask the question because I saw this fact used in this solution to a problem:
Problem:
Given that
$ \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 $
$ \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 $
$ \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 $
$ \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 $
find the value of $ w^2+x^2+y^2+z^2 $
Here's the solution
http://www.isinj.com/aime/AIME-Solutions-1983-2011.pdf
Starts at the bottom of page 22, problem #15.
On page 23, the solution compares the LHS of eq (2) to the LHS of eq (3) because they are both 4th degree polynomials with 4 identical roots. I am trying to prove that this must be true for all polynomials. Can someone help me prove it?
| Clarification: To come to the conclusions in the answers we must assume the multiplicity of each root matches. For example, It seems all the answers rule this possibility out. For example, one can make a case saying these cubics have the same roots of 1 and -1. However, they are obviously not the same polynomial.
$$f(x)=(x-1)(x-1)(x+1)$$ and
$$g(x)=(x-1)(x+1)(x+1)$$
Essentially, add the words "identical, including multiplicity, real roots" to rule out any possibility of confusion.
| {
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how do i write $y = 2\sin2(x + \frac{\pi}{4}) - \cos2(x + \frac{\pi}{4} )$ in format $y = A\sin(kx) + B\cos(kx)$ The problem is the double angles. I tried to simplify them and change them around but no luck,
$$\begin{align} y&=A\sin(kx)+B\cos(ky)\\
y&=2\sin2(x+\pi/4)-\cos2(x+\pi/4)\\
&=2\left(2\sin(x+\pi/4)\cos(x+\pi/4)\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=4\left(\sin x\cos\frac\pi 4+\cos x\sin\frac\pi4\right)\left(\cos x\cos\frac\pi 4-\sin x\sin\frac\pi4\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=4\left(\sin x\frac1{\sqrt2}+\cos x\frac1{\sqrt2}\right)\left(\cos x\frac1{\sqrt2}-\sin x\frac1{\sqrt2}\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=\frac 4{\sqrt 2}\left(\sin x+\cos x\right)\left(\cos x-\sin x\right)-\cos^2(x+\pi/4)+\sin^2(x+\pi/4)\\
&=2\sqrt 2(\cos^2x-\sin^2x)+\left(\sin x\frac1{\sqrt2}+\cos x\frac1{\sqrt2}\right)^2-\left(\cos x\frac1{\sqrt2}-\sin x\frac1{\sqrt2}\right)^2\\
&=2\sqrt 2(\cos^2x-\sin^2x)+4\left(\frac1{\sqrt2}\sin x\frac1{\sqrt2}\cos x\right)\\
&=2\sqrt 2(\cos^2x-\sin^2x+\sin x\cos x) \end{align}$$
| First, notice that $2(x+\tfrac{\pi}{4}) \equiv 2x+\tfrac{\pi}{2}$. Using this fact, we have
$$y= 2\sin\left(x+\tfrac{\pi}{2}\right) - \cos\left(x+\tfrac{\pi}{2}\right).$$
There are two standard formulae:
$$\begin{eqnarray*}
\sin(A+B) &\equiv& \sin A \cos B + \sin B \cos A \\ \\
\cos(A+B) &\equiv& \cos A \cos B - \sin A \sin B
\end{eqnarray*}$$
Applying these two formulae with $A=2x$ and $B=\tfrac{\pi}{2}$ gives:
$$
\color{blue}{y = 2\left(\sin 2x \cos \tfrac{\pi}{2} + \sin \tfrac{\pi}{2} \cos 2x\right) - \left( \cos 2x \cos \tfrac{\pi}{2} - \sin 2x \sin \tfrac{\pi}{2}\right) \equiv 2\cos 2x + \sin 2x}$$
Moreover, we can simplify $2\cos 2x + \sin 2x$ even further. Consider
$$R\cos(2x-\alpha) \equiv 2\cos 2x + \sin 2x$$
Expanding the left hand side gives
$$(R\cos\alpha)\cos 2x + (R\sin\alpha)\sin 2x \equiv 2\cos 2x + \sin 2x$$
Solving $R\cos\alpha=2$ and $R\sin\alpha=1$ gives $R=\sqrt{5}$ and $\alpha=\arctan(1/2)$. Hence:
$$\color{blue}{2\sin\left[2\left(x+\tfrac{\pi}{4}\right)\right] - \cos\left[2\left(x+\tfrac{\pi}{4}\right)\right] \equiv \sqrt{5}\cos\left(2x-\arctan\tfrac{1}{2}\right)} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Contour Integral of Exponential I want to show the following for $a > 0$:
$$e^{-a} = \int_{0}^{\infty}{\frac{e^{-x}}{\sqrt{x}}e^{-a^{2}/(4x)}dx}.$$
| $$\int_{0}^{\infty} \frac{e^{-(x+a^2/4x)}}{\sqrt{x}} \ dx = 2 \int_{0}^{\infty} e^{-(t^{2}+a^{2}/4t^{2})} \ dt$$
In general,
$$ \int^{\infty}_{0} e^{-ax^{2}-b/x^{2}} \ dx = \int_{0}^{\infty} \exp \left[ -a \left(x^{2}+\frac{b}{a}\frac{1}{x^{2}} \right) \right] \ dx $$
$$ = \exp\left[ -a\left(x^{2}+\frac{b}{a} \frac{1}{x^{2}} - 2 \frac{\sqrt{b}}{\sqrt{a}} \right) - 2 \sqrt{ab}\right] \ dx= e^{-2 \sqrt{ab}} \int_{0}^{\infty} \exp \left[ -a\left(x- \frac{\sqrt{b}}{\sqrt{a}} \frac{1}{x} \right)^{2} \right] \ dx $$
$$ = \frac{\sqrt{b}}{\sqrt{a}} e^{-2\sqrt{ab}} \int_{0}^{\infty} \exp \left[ -a\left(\frac{\sqrt{b}}{\sqrt{a}} \frac{1}{u}- u \right)^{2} \right] \frac{du}{u^{2}}$$
$$ \implies \int_{0}^{\infty} e^{-ax^{2}-b/x^{2}} \ dx = \frac{1}{2} e^{-2 \sqrt{ab}} \int_{0}^{\infty} \left( 1+ \frac{\sqrt{b}}{\sqrt{a}} \frac{1}{x^{2}}\right) \exp \left[-a \left(x- \frac{\sqrt{b}}{\sqrt{a}} \frac{1}{x} \right)^{2} \right] \ dx$$
$$ = \frac{1}{2} e^{- 2\sqrt{ab}} \int_{-\infty}^{\infty} e^{-at^{2}} \ dt = \frac{1}{2} \sqrt{\frac{\pi}{a}} e^{- 2 \sqrt{ab}}$$
Therefore,
$$ \int_{0}^{\infty} \frac{e^{-(x+a^2/4x)}}{\sqrt{x}} \ dx = \sqrt{\pi} \ e^{-2 \sqrt{a^{2}/4}} = \sqrt{\pi} e^{-a}$$
| {
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How to solve $(x^2-2x)^2+4(x^2-2x)+3=0$? How to solve $(x^2-2x)^2+4(x^2-2x)+3=0\,$? I tried
$$(x^2-2x)(x^2-2x+4)+3=0 \Longrightarrow (x^2-2x)(x^2-2x+4)=-3,$$ but couldn't continue. So that is why I came here to ask for some help, please. Thank you.
Lorie Kirchner
| Due to Viëta we have $x^2-2x=-1$ or $x^2-2x=-3$. Clearly $x=1$ is the only (real) solution.
| {
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How to calculate $ \int \frac{x^3+x^2}{x^3-1} \mathrm{d}x$ I was trying to solve this integral:
$$ \int \frac{x^3+x^2}{x^3-1} \mathrm{d}x$$
I made the following steps:
*
*Polynomial division: $$ \frac{x^3+x^2}{x^3-1} = \left(1+\frac{x^2+1}{x^3-1}\right)$$
*Hermite polynomials: $$ \frac{x^2+1}{x^3-1} = \frac{2}{3}\frac{1}{x-1}+\frac{\frac{x}{3}-\frac{1}{3}}{x^2+x+1}$$
So, now the integral becomes: $$x+\frac{2}{3}\log{|x-1|}+\int \dfrac{\frac{x}{3}-\frac{1}{3}}{x^2+x+1} \mathrm{d}x$$
But i think there's something wrong, i can not continue. Any ideas?
| What you have done so far is correct, so
$$
\int\frac{x^3+x^2}{x^3-1}dx=
x+\frac{2}{3}\log\left|x-1\right|+\frac{1}{3}\int\frac{x-1}{x^2+x+1}dx.
$$
Now, to determine the last integral, we want to have the derivative of the denominator in the numerator, so we write
$$
\int\frac{x-1}{x^2+x+1}dx=
\int\frac{\frac{1}{2}(2x+1)-\frac{3}{2}}{x^2+x+1}dx=
\frac{1}{2}\log\left|x^2+x+1\right|-\frac{3}{2}\int\frac{dx}{x^2+x+1}.
$$
Can you take it from here?
| {
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Finding $y^{\prime \prime}$ of $2x^2+3y^2=4$
Find $y^{\prime}$ and $y^{\prime \prime}$ of $2x^2+3y^2=4$
$$y^{\prime}=\dfrac{d}{dx}(2x^2)+\dfrac{d}{dx}(3y^2)=\dfrac{d}{dx}(4)$$
$$4x+6yy^{\prime}=0$$
$$y^{\prime}=\dfrac{-2x}{3y}$$
This is how I started finding $y^{\prime\prime}$:
$$y^{\prime\prime}=\dfrac{3y \dfrac{d}{dx}(-2x) - (-2x)\dfrac{d}{dx}(3y)}{(3y^2)}$$
$$\dfrac{3y(-2)-[-2x(3y^{\prime})]}{9y^2}$$
$$\dfrac{-6y-[-6xy^{\prime}]}{9y^2}$$
This isn't right since the correct answer is $\dfrac{-6y^2+4x^2}{9y^3}$
Can you please show how to find $y^{\prime\prime}?$ Thank you.
| Note that
$$6x \cdot \frac{-2x}{3y} = \frac{-12x^{2}}{3y} = \frac{-4x^{2}}{y}$$
Therefore, we have that
$$y'' = \frac{-6y + \frac{-4x^{2}}{y}}{9y^{2}}$$
Multiplying by $\frac{y}{y} = 1$, we find:
$$y'' = \frac{-6y^{2} - \frac{4x^{2}}{y}\cdot y}{9y^{3}} = \frac{-6y^{2}-4x^{2}}{9y^{3}}$$
| {
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Find max: $M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}$ For $a,b,c>0$ and $abc=1$, find the maximum of
$$M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}.$$
| the Max should be $1$, so we need to prove:
$\dfrac{a}{b^2+c^2+a}+\dfrac{b}{c^2+a^2+b}+\dfrac{c}{a^2+b^2+c} \le 1 \iff \dfrac{b^2+c^2}{b^2+c^2+a}+\dfrac{c^2+a^2}{c^2+a^2+b}+\dfrac{a^2+b^2}{a^2+b^2+c} \ge 2$
$\dfrac{b^2+c^2}{b^2+c^2+a} \ge \dfrac{2bc}{2bc+a} =\dfrac{2}{2+a^2} \implies \sum \dfrac{1}{2+a^2} \ge 1 \iff a^2b^2+b^2c^2+c^2a^2 \ge3$
| {
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minimal polynomial of $\alpha+\alpha^2$ where $\alpha^5=1$ and $\alpha\neq 1$ I have to find the minimal polynomial of of $\alpha+\alpha^2$ where $\alpha^5=1$ and $\alpha\neq 1$ in $\mathbb{Q}$. First I found the minimal polynomial of $\alpha$: $X^5-1=(X-1)(X^4+X^3+X^2+X+1)$ so $X^4+X^3+X^2+X+1$ must be the minimal polynomial of $\alpha$ since its irreducible and has $\alpha$ as a root. I didn't come much further than this. I really need help. Thanks.
| Let $\beta=\alpha+\alpha^2$.
Then
$$ \beta^2=\alpha^4+2\alpha^3+\alpha^2=\alpha^3-\alpha-1$$
$$ \beta^3=\alpha^6+3\alpha^5+3\alpha^4+\alpha^3=3\alpha^4+\alpha^3+\alpha+3=-2\alpha^3-3\alpha^2-2\alpha$$
$$ \beta^4=\alpha^8+4\alpha^7+6\alpha^6+4\alpha^5+\alpha^4=\alpha^4+\alpha^3+4\alpha^2+6\alpha+4=3\alpha^2+5\alpha+3$$
and of course $\beta^0=1$. This gives us five vectors expressed as linear combinations of $1,\alpha,\alpha^2,\alpha^3$, which are therefore linearly dependent. That is you can find nontrivial coefficients $a_4,a_3,a_2,a_1,a_0\in\mathbb Q$ such that $a_4\beta^4+a_3\beta^3+a_2\beta^2+a_1\beta+a_0=0$, that is some nonzero polynomial that has $\beta$ as root.
| {
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Proving that $\frac{3}{2} \sum_{k=1}^{\infty} \frac{4}{k^3+k^2} = \pi^2-6$ I'm trying to prove that:
$$\frac{3}{2} \sum_{k=1}^{\infty} \frac{4}{k^3+k^2} = \pi^2-6$$
I've tried looking at the partial sums, but no luck there. I just have no idea where to begin.
Knowing that $\frac{4}{k^3+k^2} = \frac{4}{k^2 (k+1)} \implies \frac{4}{k^2}+\frac{4}{(k+1)}-\frac{4}{k}$ seemed to help, but I get stuck no matter how I look at it.
Any help would be greatly appreciated.
| Hint:
$$
\frac{4}{k^3+k^2} = \frac{4}{k^2(k+1)} = \frac{a_1}{k^2} + \frac{a_2}{\vphantom{k^2}k} + \frac{a_3}{\vphantom{k^2}k+1}
$$
for some constants $a_1$, $a_2$, and $a_3$, and
$$
\zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}.
$$
| {
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Find a closed form for the generating function for each of these sequences. Find a closed form for the generating function for each
of these sequences. (Assume a general form for the terms
of the sequence, using the most obvious choice of such a
sequence.)
a) 0, 1, −2, 4, −8, 16, −32, 64, ...
b) 1, 0, 1, 0, 1, 0, 1, 0, ...
actually I don't know how should I solve this kind of problems. can anyone help?
| For the first one, we obtain:
\begin{align*}
f(x) &= 0x^0 + 1x^1 + (-2)x^2 + 4x^3 + (-8)x^4 + 16x^5 + (-32)x^6 + 64x^7+\cdots \\
&= x - 2x^2 + 4x^3 - 8x^4 + 16x^5 - 32x^6 + 64x^7+\cdots
\end{align*}
This is an infinite geometric series with initial term $a = x$ and common ratio $r = -2x$, so we obtain:
$$
f(x) = \frac{a}{1 - r} = \frac{x}{1+2x}
$$
For the second one, we obtain:
\begin{align*}
g(x) &= 1x^0 + 0x^1 + 1x^2 + 0x^3 + 1x^4 + 0x^5 + 1x^6 + 0x^7+ 1x^8 + \cdots \\
&= 1 + x^2 + x^4 + x^6 + x^8 + \cdots
\end{align*}
This is an infinite geometric series with initial term $a = 1$ and common ratio $r = x^2$, so we obtain:
$$
g(x) = \frac{a}{1 - r} = \frac{1}{1-x^2}
$$
| {
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Differentiation d^2y/dx^2 of trigonometric function We have a function $x \sin y = y^2$. Let $P$ be $\displaystyle (\frac{\pi^2}{4}, \frac{\pi}{2})$. Evaluate $\displaystyle \frac{dx}{dy}$ and $\displaystyle \frac{d^2y}{dx^2}$ at $P$.
For $\displaystyle \frac{dx}{dy}$:
$\displaystyle \frac{\pi^2}{4} \sin \left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2$
$\displaystyle D_x \left[\frac{\pi^2}{4}\right] \sin \left(\frac{\pi}{2}\right) + \frac{\pi^2}{4} D_x \left[\sin \frac{\pi}{2}\right] = D_x \left[\left(\frac{\pi}{2}\right)^2\right]$
$\displaystyle \frac{\pi}{2} \sin \left(\frac{\pi}{2}\right) + \frac{\pi^2}{4} \cos \left(\frac{\pi}{2}\right) = \pi$
I'm wondering if I could directly substitute the points to the equation or solving first for $\frac{dx}{dy}$.
$\displaystyle x \sin y = y^2 \rightarrow x' \sin y + x D_x [\sin y] = D_x [y^2] \rightarrow \sin y + x \cos \frac{dx}{dy} = 2y$
I tried to use implicit differentiation. The left side by product rule and right by power rule.
| We have $x \sin y = y^2$.
Taking the derivative with respect to $y$, we get $\displaystyle x \cos y + \frac{dx}{dy} \sin y = 2y$, and from this we get $\displaystyle \frac{dx}{dy} = \frac{2y - x \cos y}{\sin y}$.
Taking a derivative of the original equation with respect to $x$, we get $\sin y + x \cos y \frac{dy}{dx} = 2y \frac{dy}{dx}$, which means that $\displaystyle (2y - x \cos y) \frac{dy}{dx} = \sin y$, or $\displaystyle \frac{dy}{dx} = \frac{\sin y}{2y - x \cos y}$. Taking a derivative with respect to $x$ again, we have $\displaystyle \frac{d^2y}{dx^2} = \frac{\cos y \frac{dy}{dx} \cdot (2y - x \cos y) - \sin y (2 + x \sin y \frac{dy}{dx})}{(2y - x \cos y)^2}$. I'll let you do the substitution and simplification on your own.
| {
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Finding the limit $\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$
Find the following limit: $$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$
Well I tried to do the $\exp\left(\frac{ \ln\frac{\sin x}{x}}{x^2}\right)$ then apply LHR but I seem to get to endless dervivations...
There's got to be a more simple approach.
| Using Taylor Series ,
$$\sin x=x-\frac{x^3}{3!}+O(x^5)\implies \frac{\sin x}x=1-\frac{x^2}{3!}+O(x^4)$$
$$\implies \lim_{x\to0}\left(\frac{\sin x}x\right)^{\frac1{x^2}}=\lim_{x\to0} \left(1-\frac{x^2}{3!}+O(x^4)\right)^{\frac1{x^2}}$$
$$=\left(\lim_{x\to0}\left(1-\frac{x^2}{3!}+O(x^4)\right)^{\frac1{-\frac{x^2}{3!}+O(x^4)}}\right)^{\lim_{x\to0}\frac{-\frac{x^2}{3!}+O(x^4)}{x^2}}$$
Now if we set $\displaystyle -\frac{x^2}{3!}+O(x^4)=-\frac1u,$ the inner limit reduces to $\displaystyle\lim_{u\to\infty}\left(1+\frac1u\right)^u=e$
For the exponent, $\displaystyle\lim_{x\to0}\frac{-\frac{x^2}{3!}+O(x^4)}{x^2}=\lim_{x\to0}\left({-\frac1{3!}+O(x^2)}\right)$ as $x\ne0$ as $x\to0$
$\displaystyle\implies\lim_{x\to0}\frac{x^2}{-\frac{x^2}{3!}+O(x^4)}=-\frac16$
| {
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Using substitution to evaluate indefinite integral $ \int{x\sqrt{4x+1}}dx$
Evaluate this indefinite integral.
$$I= \int{x\sqrt{4x+1}}dx$$
Let $u=4x+1$
$$\frac{du}{dx}=4\rightarrow{dx=\frac{du}{4}}$$
$$I=\int{x}\sqrt{u}\frac{1}{4}du=\frac{1}{4}\int{x}\sqrt{u}du$$
Then I got stuck at this point.
|
$$I= \int{x\sqrt{4x+1}}dx$$
Using Euler Substitution
$$\sqrt{4x+1}=t\iff x=\frac{t^2-1}{4}\iff dx=\frac{t}{2}dt$$
$$\begin{align}
I&= \int{x\sqrt{4x+1}}dx\\
&=\int \left(\frac{t^2-1}{4}\right)\frac{t^2}{2}dt\\
&=\frac{t^5}{40}-\frac{t^3}{24}+C\\
&I=\frac{(4x+1)^{5/2}}{40}-\frac{(4x+1)^{3/2}}{24}+C\\
\end{align}$$
$$\int{x\sqrt{4x+1}}dx=\frac{3(4x+1)^{5/2}-5(4x+1)^{3/2}}{120}+C$$
| {
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trigonometry identity I have some problem with proving this identity:
$$2\left(1+\cos\alpha \right)-\sin^2\alpha=4\cos^4\frac{\alpha}{2}$$
I tried to start from the right side rewritting it to $(2\cos^2(2\frac{\alpha}{4}))^2$ but it's not working.
| We have the double angle identities $sin(x)=2sin(\frac{x}{2})cos(\frac{x}{2})$ and $cos(x)=2cos^2(\frac{x}{2})-1$
Substituting into the left hand side of your expression, we have
$2(1+2cos^2(\frac{\alpha}{2})-1) - (2sin(\frac{x}{2})cos(\frac{x}{2}))^2$
$=4cos^2(\frac{\alpha}{2})-4sin^2(\frac{\alpha}{2})cos^2(\frac{\alpha}{2})$
$=4cos^2(\frac{\alpha}{2})[1-sin^2(\frac{\alpha}{2})]$
$=4cos^2(\frac{\alpha}{2})[cos^2(\frac{\alpha}{2})]$
$=4cos^4(\frac{\alpha}{2})$
| {
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Sum of odd numbers always gives a perfect square. $1 + 3 = 4$ (or $2$ squared)
$1+3+5 = 9$ (or $3$ squared)
$1+3+5+7 = 16$ (or $4$ squared)
$1+3+5+7+9 = 25$ (or $5$ squared)
$1+3+5+7+9+11 = 36$ (or $6$ squared)
you can go on like this as far as you want, and as long as you continue to add odd numbers in order like that, your answer is always going to be a perfect square.
But how to prove it?
| Because
$$\begin{align}1+3+5+\cdots+(2n-1)&=\sum_{k=1}^{n}(2k-1)\\&=2\sum_{k=1}^{n}k-\sum_{k=1}^{n}1\\&=2\cdot\frac{n(n+1)}{2}-n\\&=n(n+1)-n\\&=n^2.\end{align}$$
| {
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Evaluation of a particular type of integral involving logs and trigonometric function Is there any closed form for
$$ \int _0 ^{\infty}\int _0 ^{\infty}\int _0 ^{\infty} \log(x)\log(y)\log(z)\cos(x^2+y^2+z^2)dzdydx$$
if yes then how to prove it?
| One way to attack this is to exploit the symmetry of the integral. Start by expanding the cosine term into individual pieces, i.e.,
$$\begin{align}\cos{(x^2+y^2+z^2)} &= \cos{x^2} \cos{(y^2+z^2)} - \sin{x^2} \sin{(y^2+z^2)}\\ &= \cos{x^2} \cos{y^2} \cos{z^2} - \cos{x^2} \sin{y^2} \sin{z^2} \\ &\quad -\sin{x^2} \sin{y^2} \cos{z^2} - \sin{x^2} \cos{y^2} \sin{z^2} \end{align}$$
Next define
$$C = \int_0^{\infty} dx \, \cos{x^2} \: \log{x} $$
$$S = \int_0^{\infty} dx \, \sin{x^2} \: \log{x} $$
Then the triple integral is equal to $C^3-3 C S^2$.
We obtain $C$ and $S$ using Cauchy's theorem to find $C+i S$. We do this by considering the integral
$$\oint_{\eta} dz \, e^{i z^2} \log{z}$$
where $\eta$ is a circular wedge of radius $R$ and angle $\pi/4$ in the upper half plane with base along the positive real axis. One may show that the integral over the circular arc vanishes as $\pi \log{R}/(4 R)$ as $R \to \infty$. Thus, we have
$$C+i S = e^{i \pi/4} \int_0^{\infty} dt \, e^{-t^2} \left (\log{t} + i \frac{\pi}{4} \right ) $$
One may derive the value of this integral by noting that
$$\int_0^{\infty} dt \, e^{-t^2} \log{t} = -\frac{\sqrt{\pi}}{4} (\gamma + 2 \log{2})$$
This value is derived from using the value of
$$\frac12 \left [\frac{d}{d\alpha} \Gamma \left (\frac{\alpha+1}{2} \right ) \right ]_{\alpha=0} = \frac14 \Gamma \left (\frac12 \right ) \psi \left ( \frac12 \right )$$
We also use
$$\int_0^{\infty} dt \, e^{-t^2} = \frac{\sqrt{\pi}}{2}$$
to obtain
$$C = -\frac18 \sqrt{\frac{\pi}{2}} (\pi + 2 \gamma + 4 \log{2})$$
$$S = \frac18 \sqrt{\frac{\pi}{2}} (\pi - 2 \gamma - 4 \log{2})$$
and therefore the triple integral is
$$C^3-3 C S^2 = \frac{\pi ^{3/2} \left(4 \gamma ^2-8 \gamma \pi +\pi ^2+16 \log{2} \,(\gamma
-\pi +\log{2})\right) (2 \gamma +\pi +4 \log{2})}{512 \sqrt{2}}$$
| {
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Integration by substitution fail for $\int \frac{1}{(1+\sin x)}\, \mathrm dx$. We tried to do an integral with the universal trigonometric substitution $$\int \frac{1}{(1+\sin x)}\, \mathrm dx$$
Meaning, we substituted: $ t = \tan \frac{x}{2} \Rightarrow$
$$\int \frac{1}{(1+\sin x)}\, \mathrm dx = \int \frac{\frac{2}{1+t^2}}{1+\frac{2t}{1+t^2}}\, \mathrm dt = \int \frac{2}{(1+t)^2}\, \mathrm dt = \frac{-2}{1+t} = \frac{-2}{1+\tan \frac{x}{2}} + C$$
But the answer is:
$$ \tan x - \frac{1}{\cos x} + C $$
What did we do wrong?
| Another answer is
$$
\int\frac{1}{1+\sin x} dx=-\frac{\cos x}{1+\sin x}+c
$$
To show this we write
$$
\tan x-\sec x=\frac{\sin x-1}{\cos x}=
\frac{(\sin x-1)\cos x}{\cos^2 x}=
\frac{(\sin x-1)\cos x}{1-\sin^2 x}=
\frac{(\sin x-1)\cos x}{(1-\sin x)(1+\sin x)}=
-\frac{\cos x}{1+\sin x}+c
$$
| {
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Mathematical Induction of $\frac{n(n-1)(n-2)}{6}$
Show that the number of triples that can be chosen from $n$ items is precisely $$\frac{n(n−1)(n−2)}{6}.$$
Suppose n = k+1,
We want $\frac{(k+1)k(k-1)}{6}$ therfore,
$\frac{k(k-1)(k-2)}{6}$ + (k+1)
and then solve the rest.
What am I doing wrong here?
| I will show how to prove $$1+3+6+\cdots+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{6}.$$
For $n=1$ we have $1=\frac{1\cdot2\cdot3}{6}=1$, so the base case holds.
Assume for any $n-1$ we have that the sum holds: $$1+3+6+\cdots+\frac{(n-1)n}{2}=\frac{(n-1)n(n+1)}{6}.$$
We need to show that by adding the next term $\frac{n(n+1)}{2}$ to the left and right hand sides the equality still holds. For the LHS it's obvious. For the RHS, in detail:
$$\frac{(n-1)n(n+1)}{6}+\frac{n(n+1)}{2}=\frac{(n-1)n(n+1)+3n(n+1)}{6}=\frac{(n-1+3)n(n+1)}{6}=\frac{n(n+1)(n+2)}{6}.$$
Hopefully you see how to apply this to the proof you need. (Hint: subtract 2 from each $n$.)
| {
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Discuss the convergence of the series $\sum_{n=1}^{\infty}\frac{1/2+(-1)^n}{n}$
Discuss the convergence of the series $\displaystyle\sum_{n=1}^{\infty}\frac{\frac{1}{2}+(-1)^n}{n}$.
Attempt: Since the given series is alternating series, I am using Leibniz test.
The terms of the series are $\frac{-1}{2\cdot 1},\frac{3}{2\cdot 2},\frac{-1}{2\cdot 3},\frac{3}{2\cdot 4},\frac{-1}{2\cdot 5},\ldots$
That is equal to $\frac{1}{2}(-1,\frac{3}{2},\frac{-1}{3},\frac{3}{4},\frac{-1}{5},\ldots)$.
Hence $a_n= \frac{-1}{2n-1}$ if $n$ is odd and $a_n=\frac{3}{n}$ if $n$ is even.
$a_n$ goes to zero. But $a_n$ not decreasing. Hence the series divergent.This argument will work or not please help me!
| I would look at creating a sequence that consists of summing 2 subsequent terms of the original sequence.
$b_{n} = a_{2n+1} + a_{2n+2} = \frac{-\frac{1}{2}}{2n + 1} + \frac{\frac{1}{2}}{2n+ 2}
= \frac{1}{2}(\frac{1}{2n+2} - \frac{1}{2n + 1}) = \frac{1}{2}\frac{(2n+1)-(2n+2)}{(2n+1)(2n+2)} = -\frac{1}{2}\frac{1}{(2n+1)(2n+2)}$
I'm not sure about the actual value but if I am correct the series of $b_n$ converges and therefor the series of $a_n$ converges.
Edit: Sorry, I made an error. My the series of my sequence $b_n$ as constructed above is convergent but is not the same as yours.
The correct one is:
$b_n = a_{2n+1} + a_{2n+2} = \frac{\frac{1}{2}}{2n+1} + \frac{\frac{3}{2}}{2n+2} = \frac{1}{2}\frac{4n+1}{(2n+1)(2n+2)}$
Which is divergent.
| {
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How prove this $f(n)\le f(n+1)$ where $f(n)=\sum_{k=1}^{n}\frac{n}{n^2+k^2}$ let $$f(n)=\sum_{k=1}^{n}\dfrac{n}{n^2+k^2}$$
prove or disprove
$$f(n)\le f(n+1)$$
this inequality is found when I deal this follow limit:
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{1+(k/n)^2}=\int_{0}^{1}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{4}$$
But I can't prove
$$f(n)\le f(n+1)$$
since
$$f(n+1)=\dfrac{n+1}{(n+1)^2+1}+\dfrac{n+1}{(n+1)^2+2^2}+\cdots+\dfrac{n+1}{(n+1)^2+n^2}+\dfrac{n+1}{(n+1)^2+(n+1)^2}$$
$$f(n)=\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2^2}+\cdots+\dfrac{n}{n^2+n^2}$$
so
$$f(n+1)-f(n)=\left(\dfrac{n+1}{(n+1)^2+1}-\dfrac{n}{n^2+1}\right)+\left(\dfrac{n+1}{(n+1)^2+2^2}-\dfrac{n}{n^2+2^2}\right)+\cdots+\left(\dfrac{n+1}{(n+1)^2+n^2}-\dfrac{n}{n^2+n^2}\right)+\dfrac{1}{2(n+1)}$$
so
$$f(n+1)-f(n)=\sum_{k=1}^{n}\dfrac{k^2-n^2-n}{(k^2+n^2)((n+1)^2+k^2)}+\dfrac{1}{2(n+1)}$$
This problem is my found it,can you help to solve this problem?
| This might lead to a solution, but at the very least, demonstrates just how delicate the estimate is.
We want to get a grip on
$$
S_n = \sum_{k = 1}^n \frac{n^2 + n - k^2}{(n^2 + k^2) ((n+1)^2 + k^2)}
$$
and show that it dies faster than $\frac{1}{2n}$. Let's be cavalier and drop the '$n+1$' business in the denominator, to make our lives simpler. Then, we're bounding
$$
S_n' = \sum_{k = 1}^n \frac{n^2 + n - k^2}{(n^2 + k^2)^2}
= \frac{1}{n^2} \left(\sum_{k = 1}^n \frac{1-(k/n)^2}{(1 + (k/n)^2)^2} + \frac{1}{n} \frac{1}{(1 + (k/n)^2)^2} \right)
$$
We can realize this as a pair of Riemann sums: the first (dominant) term is the right-endpoint Riemann sum of the decreasing function $g_1(x) = \frac{1-x^2}{(1 + x^2)^2}$, and the second term is the right-endpoint Riemann sum of $g_2(x) = \frac{1}{(1+x^2)^2}$. Both of these Riemann sums will be strictly smaller than their destined integrals. Now the fun part:
$$
\int_0^1 g_1(x) dx = \frac{1}{2}
$$
and so for large $n$, the dominant term is just just barely small enough to conclude. To make this a proof, one would have to find a lower bound for the error in the Riemann sum of $g_1$ along $[0,1]$, and compare this with the $g_2$ term. Thankfully, $g_1$ is actually concave on $[0,\sqrt{2}-1]$, and so there is some hope of taking a small subinterval in there and cooking up a lower bound for this error that defeats the $g_2$ term.
I did out the following. It didn't work the way I hoped, but I figured I would record the method here, if someone can improve it.
WolframAlpha computes the integral $\int_0^1 g_2(x) dx = \frac{2 + \pi}{8} \leq \frac{7}{10}$ and $\int_{2/5}^1 g_1(x) dx = 9/58, \int_{0}^{1/5} g_1(x) dx = \frac{5}{26}$. So,
$$
S_n \leq S_n' \leq \frac{1}{n^2} \sum_{k = \lfloor \frac{n}{10} \rfloor }^{ \lceil \frac{2 n}{5}\rceil } g_1(\frac{k}{n}) + \frac{C}{n} + \frac{7}{10 n^2}
$$
with $C = \frac{9}{58} + \frac{5}{26}$. (the above might be off by a term or two in the remaining sum) We truncate the Riemann sum for $g_1$ from $2/5$ to $1$ because $g_1$ isn't concave there, and we truncate from $0$ to $1/5$ so that we have a positive lower bound on $g_1'$ on $[1/5, 2/5]$.
Now, $g_1'(x) \leq - 1$ for $x \in [1/5,2/5]$; each of the vertical strip segments in the remaining part of the Riemann sum will underestimate the true integral of $g_1$ there by at least $\frac{1}{n^2}$. In total, the underestimation contribution to the total Riemann sum from such terms is at least $\frac{n }{5} \cdot \frac{1}{ n^2} = \frac{1}{5 n}$, and so we have shown that
$$
S_n' \leq \frac{1}{n}\left(\frac{1}{2} - \frac{1}{5 n} + \frac{7}{10 n}\right)
$$
Unfortunately this isn't strong enough. I don't think adjusting the endpoints will work: one probably has to make a more careful estimate using additionally the (nonconcave) segment $[2/5,1]$.
| {
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Unitary similarity transformation I have a matrix:
$ A= \dfrac{i}{3} \begin{bmatrix} 1&-2&1\\-2&1&1\\1&1&-2\end{bmatrix} $
Could someone explain me how to find a corresponding diagonal matrix for a diagonalizable matrix or linear map by unitary similarity transformation, please?(in the simplest way, if it's possible)
| @Amzoti: Could you tell me where is my mistake: $A= \frac{i}{3} \begin{bmatrix} 1&-2&1\\-2&1&1\\1&1&-2\end{bmatrix} = \begin{bmatrix} \frac{i}{3}&\frac{-2i}{3}&\frac{i}{3}\\\frac{-2i}{3}&\frac{i}{3}&\frac{i}{3}\\\frac{i}{3}&\frac{i}{3}&\frac{-2i}{3}\end{bmatrix} \\ A_{\lambda} = \begin{bmatrix} \frac{i}{3}-\lambda&\frac{-2i}{3}&\frac{i}{3}\\ \frac{-2i}{3}&\frac{i}{3}-\lambda&\frac{i}{3}\\ \frac{i}{3}&\frac{i}{3}&\frac{-2i}{3}-\lambda\end{bmatrix}$
After transformations($C_{1}-C_{2}$ and $W_{2}+W_{1}$), I have:
$A_{\lambda} = \begin{bmatrix} i-\frac{\lambda}{3}&\frac{-2i}{3}&\frac{i}{3}\\ 0&\frac{-i-\lambda}{3}&\frac{2i}{3}\\ 0&\frac{i}{3}&\frac{-2i-\lambda}{3}\end{bmatrix}$
So then I wanted to develop it with the first column:
$(i-\frac{\lambda}{3})(-1)^{2} \begin{bmatrix} \frac{-1-\lambda}{3}&\frac{2i}{3}\\\frac{i}{3}&\frac{-2i-\lambda}{3} \end{bmatrix}$
And the CP: $\lambda(i-\frac{\lambda}{3})(i+\frac{\lambda}{3})=0$
| {
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Mathematical Induction (summation): $\sum^n_{k=1} k2^k =(n-1)(2^{n+1})+2$ I am stuck on this question from the IB Cambridge HL math text book about Mathematical induction. I am sorry about the bad formatting I am new and have no idea how to write the summation sign.
Using mathematical induction prove that the
$$\sum^n_{k=1} k2^k =(n-1)(2^{n+1})+2$$
[correction made]
I tried solving it and got stuck on the let $n=k+1$ part
So first I made $n=1$ and both sides equaled to $2$
then assume $n=k$ and got an expression which I don't know how to write here because of the formatting
then $n=K+1$
Thanks again
| $\sum_{k=1}^{n}k2^{k}=2+(4+4)+(8+8+8)+...+(2^{n}+2^{n}+...+2^{n})$
$=(2+4+8+...+2^{n-1}+2^{n})+(4+8+...+2^{n-1}+2^{n})+(8+...+2^{n-1}+2^{n})+...+(2^{n-2}+2^{n-1}+2^{n})+(2^{n-1}+2^{n})+2^{n}$
$2(1+2+4+...+2^{n-2}+2^{n-1})+4(1+2+4+...+2^{n-3}+2^{n-2})+8(1+...+2^{n-3})+...+2^{n-2}(1+2+4)+2^{n-1}(1+2)+2^{n}$
$=2(2^{n}-1)+4(2^{n-1}-1)+8(2^{n-2}-1)+...+2^{n-2}(2^{3}-1)+2^{n-1}(2^{2}-1)+2^{n}$
$=(n-1)2^{n+1}+2^{n}-2-4-...-2^{n-1}=(n-1)2^{n+1}+2^{n}-2(2^{n-1}-1)$
$=(n-1)2^{n+1}+2$
This should be the formula. Now we prove this by induction. Both sides are $2$ at $n=1$. Assume it is true for $n\ge1$ and we show it for $n+1$.
$\sum_{k=1}^{n+1}k2^{k}=\sum_{k=1}^{n}k2^{k}+(n+1)2^{n+1}=(n-1)2^{n+1}+2+(n+1)2^{n+1}$
$=(2n)2^{n+1}+2=n2^{n+2}+2=((n+1)-1)2^{(n+1)+1}+2$.
| {
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Show that $\int_0^{2\pi}\frac{R^2-r^2}{R^2 - 2Rr\cos (\varphi-\vartheta) + r^2}d\vartheta$ is independent of $R>r>0$, using only real numbers. The poisson kernel is sometimes written as
$$
\frac{1}{2\pi}\int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\varphi-\vartheta) + r^2} \mathrm{d}\vartheta = 1 \ , \ \ R>r>0
$$
Where $\varphi$ is some arbitary angle.
This much used in complex analysis amongst other fields. Is there some basic, elementary way of showing that the integral is independent on $R$ and $r$?
Splitting the integral at $\int_0^\pi + \int_\pi^{2\pi}$ and using the Weierstrass substitution seems like somewhat ugly and a messy way to approach the problem.
| Step I. Using integration by parts one can easily show that
$$
J_n=\int_0^{2\pi} \cos^{2n}x\,dx=\frac{2\pi}{4^n}\binom{2n}{n},
$$
as for the cosine integral we make use of a recurrence,
\begin{align}
J_{n+1}&=\int_0^{2\pi} \cos^{2(n+1)}x\,dx=\int_0^{2\pi} (\cos^{2n+1}x)(\sin x)'\,dx =0+(2n+1)\int_0^{2\pi} \cos^{2n}x\,\sin^2 x\,dx \\&=
(2n+1)\int_0^{2\pi} \cos^{2n}x\,(1-\cos^2 x)\,dx=(2n+1)J_n-(2n+1)J_{n+1},
\end{align}
and thus $J_{n+1}=\frac{2n+1}{2n+2}J_n=2^{-n-1}\prod_{j=0}^{n+1}\frac{2j-1}{j+1}J_0=\frac{2\pi}{4^{n+1}}\frac{(2n+2)!}{(n+1)!(n+1)!}$.
Step II. If $|a|<1$, then
$$
\frac{1}{\sqrt{1-a^2}}=(1-a^2)^{-1/2}=\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}a^{2n}=\cdots=\sum_{n=0}^\infty \binom{2n}{n}\frac{a^{2n}}{2^{2n}}.
$$
Step III. Then
$$
I=\frac{1}{2\pi}\int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\phi-\theta) + r^2} \mathrm{d}\theta
=\frac{R^2-r^2}{2\pi(R^2+r^2)}\int_0^{2\pi}\frac{dx}{1-a\cos x},
$$
where $a=\dfrac{2rR}{r^2+R^2}.$
Also
$$
\int_0^{2\pi}\frac{dx}{1-a\cos x}=\sum_{n=0}^\infty\int_0^{2\pi} a^{2n}\cos^{2n}x\,dx=\sum_{n=0}^\infty \frac{2\pi\,a^{2n}}{4^n}\binom{2n}{n}=\frac{2\pi}{\sqrt{1-a^2}}
$$
Step IV. Finally,
\begin{align}
I&=\frac{1}{2\pi}\int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\phi-\theta) + r^2} \mathrm{d}\theta
=\frac{R^2-r^2}{2\pi(R^2+r^2)}\int_0^{2\pi}\frac{dx}{1-a\cos x} \\
&=\frac{R^2-r^2}{2\pi(R^2+r^2)}\cdot
\frac{2\pi}{\sqrt{1-\left(\frac{2rR}{r^2+R^2}\right)^2}}
=\cdots=1.
\end{align}
Note. I prefer the proof of T.A.E, but the question was asking of a proof without the use of complex numbers!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/650846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
If $a,b,c,d>0$ and $a+b+c+d=4$, then $\frac{1}{11+a^2}+\frac{1}{11+b^2}+\frac{1}{11+c^2}+\frac{1}{11+d^2} \leq \frac {1}{3}$ Prove if $a,b,c,d>0$ and $a+b+c+d=4$, then $$\dfrac{1}{11+a^2}+\dfrac{1}{11+b^2}+\dfrac{1}{11+c^2}+\dfrac{1}{11+d^2} \leq \dfrac {1}{3}$$
This was an Inequality Olympiad Problem1.
I proved by using Lagrange Multipliers method. Can you do without calculus?
| oh sorry, there is some detail in my post ! i omit it and i write it here :
$\Longrightarrow\dfrac{1}{11+a^2}+\dfrac{1}{11+b^2}+\dfrac{1}{11+c^2}+\dfrac{1}{11+d^2} \leq \\\dfrac{1}{11+d^2}+\dfrac{1}{11+d^2\cdot\frac{(4-b)^2}{(4-d)^2}}+\dfrac{1}{11+d^2\cdot\frac{(4-c)^2}{(4-d)^2}}+\dfrac{1}{11+d^2\cdot\frac{(4-a)^2}{(4-d)^2}} $
this inequality holds, that is because if we assume :
$a=\frac{1}{x_{1}}$$,b=\frac{1}{x_{2}}$$,c=\frac{1}{x_{3}}$$,d=\frac{1}{x_{4}}$
then, our inequality transform to :
$\Longrightarrow\dfrac{x_{1}^2}{11x_{1}^2+1}+\dfrac{x_{2}^2}{11x_{2}^2+1}+\dfrac{x_{3}^2}{11x_{3}^2+1}+\dfrac{x_{4}^2}{11x_{4}^2+1}\le{\dfrac{4x_{4}^2}{11x_{4}^2+1}}\le\frac{1}{3}$
and $f(x)=\dfrac{x^2}{11x^2+1}$ monotone increase in its domain !
the hint is :
your inequality reach its upper bound as :
$a=b=c=d$ and $d$ can be replaced by $a$ !
so our inequality seems feasible !
| {
"language": "en",
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"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Does $\sum_{n=2}^{\infty}\frac{\cos\left(\frac{\pi n^2}{n+1}\right)}{(\log n)^2}$ converge? I'm pretty sure the sum converges by Abel or Dirichelet, I just have no idea how to tackle the numerator. Any tips would be appreciated!
| Note that
$$
\frac{n^2}{n+1}= \frac{(n^2 + n) - (n+1) + 1}{n+1} = n-1+\frac{1}{n+1}
$$
From there, we note that
$$
\begin{align}
\cos\left(\frac{\pi n^2}{n+1}\right) &=
\cos\left(\pi(n-1) + \frac{\pi}{n+1}\right)
\\&=
\cos\left(\pi(n-1)\right)\cos\left(\frac{\pi}{n+1}\right)
- \sin\left(\pi(n-1)\right)\sin\left(\frac{\pi}{n+1}\right)
\\&=
(-1)^n\cos\left(\frac{\pi}{n+1}\right)
\end{align}
$$
Noting that for all $n \geq 2$, $0 < \frac{\pi}{n+1}< \pi/2$ so that $\cos(\frac{\pi}{n+1})$ is strictly positive and less than $1$. It follows that
$$
0<\frac{\cos(\frac{\pi}{n+1})}{(\log(n))^2} < \frac{1}{(\log(n))^2}
$$
Now, in order to apply the alternating series test, we would need to show that $\frac{\cos(\frac{\pi}{n+1})}{(\log(n))^2}$ is decreasing for sufficiently large $n$.
In order to do so, note that
$$
\frac{d}{dx} \frac{\cos(\frac{\pi}{x+1})}{(\log(x))^2} =
\frac{\pi \sin(\frac{\pi}{x+1})}{(1+x^2)(\log(x))^2}
-\frac{2 \cos(\frac{\pi}{x+1})}{x(\log(x))^3}
$$
If we may find an $x_0$ so that the above is negative for all $x>x_0$, then we will have shown that the sequence eventually decreases, as is required.
As David Mitra points out in the comment below, $x_0 = 2$ works perfectly well.
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve $\cos(z)=\frac{3}{4}+\frac{i}{4}$ I tried solving this using the definition of $cos(z)=\frac{e^{iz}+e^{-iz}}{2}$ and equating it to $\frac{3}{4}+\frac{i}{4}$ and converting it to a complex quadratic equation through a substitution $t=e^{iz}$ and finding roots via the complex quadratic formula but it didn't seem to work. I would prefer solutions via elementary methods.
Here is my attempt:
By definition we have $\frac{e^{iz}+e^{-iz}}{2}=\frac{3}{4}+\frac{i}{4} \implies e^{iz}+e^{-iz}=\frac{3}{2}+\frac{i}{2}$. Let $t=e^{iz}$ so then we have $t+\frac{1}{t}=\frac{3}{2}+\frac{i}{2}$ and if we multiply both sides by $t$ we have $t^2+1=(\frac{3}{2}+\frac{i}{2})t$ and hence $t^2+(\frac{3}{2}+\frac{i}{2})t+1=0$ By the quadratic formula for complex numbers we have, $a=1, b=\frac{3}{2}+\frac{i}{2}, c=1 \implies z=\frac{-(\frac{3}{2}+\frac{i}{2}) \pm \sqrt{(\frac{3}{2}+\frac{i}{2})^2-4(1)(1)}}{2(1)}$. Simplyifing we have $z=\frac{-\frac{3}{2}-(\frac{1}{2})i \pm \sqrt{-2+(\frac{3}{2})i}}{2}$ We wish to express $-2+(\frac{3}{3})i$ in polar form so we have $|-2+(\frac{3}{2})i|=\frac{5}{2}$. Now equating the real and imaginary parts we have $\frac{5}{2}\cos(\theta)=-2 \implies \cos(\theta)=-\frac{4}{5}$ and $\frac{5}{2}\sin(\theta)=\frac{3}{2} \implies \sin(\theta)=\frac{3}{5}$. From this we have $\tan(\theta)=-\frac{3}{4} \implies \theta=\arctan(-\frac{3}{4}) \approx -.6435$ rad. So we have $w=-2+(\frac{3}{2})i=\frac{5}{4}(\cos(\theta)+i\sin(\theta))=\frac{5}{4}e^{i\theta}$. By Proposition 1.3.12 we have $\sqrt{w}=\sqrt{\frac{5}{4}}e^{\frac{i\theta}{2}}=\frac{\sqrt{5}}{2}e^{\frac{i\theta}{2}}$. Similarily for $-\frac{3}{2}-(\frac{1}{2})=\frac{\sqrt{10}}{2}e^{i\varphi}$ Where $\varphi=\arctan(\frac{1}{3})$. So finally we have $z=\frac{-(\frac{3}{2}+\frac{i}{2}) \pm \sqrt{(\frac{3}{2}+\frac{i}{2})^2-4(1)(1)}}{2(1)}=\frac{\sqrt{10}e^{i\varphi} \pm \sqrt{5}e^{\frac{i\theta}{2}}}{4}$ as solutions to $\cos(z)=\frac{3}{4}+\frac{i}{4}$.
|
Two solutions of the system above: (Pi /4 , - ln(2)/2) and (- Pi/4 , ln(2)/2), among others.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Mysterious subleading corrections to sums with internal dependence on limit Is there a standard method for finding expansions in $N$ of sums like
$$S(N)=\sum_{n=0}^N \sqrt{N^2-n^2}$$
beyond the first term?
It is easy to compute here that $$S(N)=N^2 \int_0 ^1 \sqrt{1-x^2} \mathrm
d x + \mathcal O(N) = \pi N^2/4 + \mathcal O(N)$$
but finding $S(N) \approx \pi N^2 /4 + N/2$ is harder (I can do this by an unverified exchange of limits which fails at higher order), and at the moment I find the fact that the next correction seems to be a multiple of $\sqrt N$ pretty perplexing, and don't have an exact form for the coefficient. (Numerically I find $S \sim \pi N^2/4 + N/2 -0.2939955\ldots\sqrt N$, though this may not in fact be the correct form; I went up to $N=3,\!000,\!000$ plotting the ratio with $\sqrt N$).
Please note that I'm interested in generally applicable techniques for such a computation rather than this specific summation.
| Usually one can use the Euler-Maclaurin sum formula to get information like this. Its first iteration takes the form
$$
\begin{align}
&\frac{1}{n} \sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) \\
&\qquad = \int_0^1 f(x)\,dx + \frac{f(0)-f(1)}{2n} - \frac{1}{n} \int_0^1 \left(-nx - \lceil -nx\rceil - \frac{1}{2}\right)f'(x)\,dx,
\end{align}
$$
and when applied to your sum we get
$$
\frac{S(n)}{n^2} = \frac{\pi}{4} + \frac{1}{2n} + \frac{1}{n} \int_0^1 \left(-nx - \lceil -nx\rceil - \frac{1}{2}\right)\frac{x}{\sqrt{1-x^2}}\,dx.
$$
Now it just remains to estimate the integral. To do this, we first expand the periodic factor in Fourier series as
$$
-nx - \lceil -nx\rceil - \frac{1}{2} = \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{\sin(2\pi k n x)}{k},
$$
so that
$$
\begin{align}
&\int_0^1 \left(-nx - \lceil -nx\rceil - \frac{1}{2}\right)\frac{x}{\sqrt{1-x^2}}\,dx \\
&\qquad = \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{1}{k} \int_0^1 \sin(2\pi k n x) \frac{x}{\sqrt{1-x^2}}\,dx \\
&\qquad = \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{1}{k} \left[\left. -\sin(2\pi k n x) \sqrt{1-x^2} \right|_0^1 + 2\pi k n \int_0^1 \cos(2\pi k n x) \sqrt{1-x^2}\,dx\right] \\
&\qquad = \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{1}{k} \cdot 2\pi k n \int_0^1 \cos(2\pi k n x) \sqrt{1-x^2}\,dx \\
&\qquad = \frac{1}{2} \sum_{k=1}^{\infty} \frac{J_1(2\pi k n)}{k},
\end{align}
$$
where $J_1$ is the Bessel function of the first kind of order $1$. This Bessel function has the asymptotic form
$$
J_1(x) = - \sqrt{\frac{2}{\pi x}} \cos\left(x+\frac{\pi}{4}\right) + O\left(x^{-3/2}\right)
$$
as $x \to \infty$, and since $\cos(2\pi k n + \pi/4) = 1/\sqrt{2}$ we get
$$
J_1(2\pi k n) = - \frac{1}{\pi \sqrt{2 k n}} + O\left((kn)^{-3/2}\right).
$$
It follows that
$$
\begin{align}
\sum_{k=1}^{\infty} \frac{J_1(2\pi k n)}{k} &= -\frac{1}{\pi \sqrt{2n}} \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} + O\left(n^{-3/2}\right) \\
& = -\frac{\zeta(3/2)}{\pi\sqrt{2n}} + O\left(n^{-3/2}\right)
\end{align}
$$
and hence that
$$
\int_0^1 \left(\lceil nx\rceil - nx - \frac{1}{2}\right)\frac{x}{\sqrt{1-x^2}}\,dx = -\frac{\zeta(3/2)}{\pi2^{3/2}\sqrt{n}} + O\left(n^{-3/2}\right).
$$
Therefore
$$
\frac{S(n)}{n^2} = \frac{\pi}{4} + \frac{1}{2n} - \frac{\zeta(3/2)}{\pi 2^{3/2} n^{3/2}} + O\left(n^{-5/2}\right).
$$
or
$$
S(n) = \frac{\pi}{4}n^2 + \frac{1}{2}n - \frac{\zeta(3/2)}{\pi 2^{3/2}} n^{1/2} + O\left(n^{-1/2}\right).
$$
Note that the constant you estimated on the $n^{1/2}$ term is
$$
- \frac{\zeta(3/2)}{\pi 2^{3/2}} \approx -0.293\ 995\ 518\ 793\ 519\ 291.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/653707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why is it trivial that $\left(1+\frac{2\ln3}{3}\right)^{-3/2}\leq\frac{2}{3}$? Can someone tell me why $$\left(1+\dfrac{2\ln3}{3}\right)^{-3/2}\leq\dfrac{2}{3}$$ is trivial because for me its not and I will need to do the calculation to see it.
| We can see everything is positive, so just take reciprocal and square both sides:
$\left(1+ \frac{2 \ln 3}{3}\right) ^ {\frac{3}{2}} \ge \frac{3}{2}$
$\left(1+ \frac{2 \ln 3}{3}\right) ^ 3 \ge \frac{9}{4}$
Bring down the three inside the bracket (first step of expansion), since everything is positive the extra terms will be positive:
$1 + 2 \ln 3 + \text{stuff} \ge 1+ \frac{5}{4}$
Which is clearly true since $\ln 3 > 1$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
How prove this $\sum_{k=0}^n \binom{n}{k} \binom{(p-1)n}{k} \binom{pn+k}{k} = \binom{pn}{n}^2 $ I think the following equality is true ($p\in \mathbb{N},p\ge 2$):
$$\sum_{k=0}^n \binom{n}{k} \binom{(p-1)n}{k} \binom{pn+k}{k} = \binom{pn}{n}^2 $$
when $p=2$, then
$$\sum_{k=0}^{n}\binom{n}{k}^2\binom{2n+k}{k}=\binom{2n}{n}^2$$
But I can't prove this
and I can't prove this $p\ge 3$ ?
Thank you for your help
| Suppose we seek to verify that
$$\sum_{k=0}^n {n\choose k}
{pn-n\choose k} {pn+k\choose k} = {pn\choose n}^2.$$
We use the integrals
$${pn-n\choose k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z^{k+1}} \; dz$$
and
$${pn+k\choose k} =
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn+k}}{w^{k+1}} \; dw.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn}}{w}
\sum_{k=0}^n {n\choose k} \frac{(1+w)^k}{z^k w^k}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn}}{w}
\left(1+\frac{1+w}{zw}\right)^n
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn}}{w^{n+1}}
(1+w+zw)^n
\; dw \; dz.$$
Expanding the binomial in the inner sum we get
$$\sum_{q=0}^n {n\choose q} w^q (1+z)^q$$
which yields
$$\sum_{q=0}^n {n\choose q}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n+q}}{z^{n+1}}
{pn\choose n-q} \; dz
\\ = \sum_{q=0}^n {n\choose q}
{pn-n+q\choose n}
{pn\choose n-q}.$$
The inner term is
$${n\choose q}
{pn-n+q\choose n}
{pn\choose pn-n+q}
\\ =
\frac{(pn)!}{q!\times (n-q)! \times
(pn-2n+q)! \times
(n-q)!}
\\ = {pn\choose n}
\frac{n! \times (pn-n)!}{q!\times (n-q)! \times
(pn-2n+q)! \times
(n-q)!}
\\ = {pn\choose n}
{n\choose q} {pn-n\choose n-q}.$$
Thus it remains to show that
$$\sum_{q=0}^n {n\choose q} {pn-n\choose n-q} = {pn\choose n}.$$
This can be done combinatorially or using the integral
$$\frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{pn-n}}{v^{n+1}}
\sum_{q=0}^n {n\choose q} v^q \; dv
\\ = \frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{pn-n}}{v^{n+1}}
(v+1)^n \; dv
\\ = \frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{pn}}{v^{n+1}}
= {pn\choose n}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Help with the algebra in for this number theory proof? For all $n\geq 1$, prove with mathematical induction
$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$
So far.. I have substituted 1 and saw that the statement is true and I have plugged in n+1 to show that the proof is true for all integers but I don't know how to go about the simplification.. right now I have
LHS: $2-\frac{1}{k}+\frac{1}{(k+1)^2} \leq 2-\frac{1}{k+1}$
Should I try to find common denominators for the left? Step by step explanation please!
| you simply have
$$2-\frac{1}{k}+\frac{1}{(k+1)^2} \\
=2-\{\frac{1}{k}-\frac{1}{(k+1)^2}\} \\
=2-\{\frac{(k+1)^2-k}{k(k+1)^2}\}\\
=2-\{\frac{k^2+k+1}{k(k+1)^2}\}\\
\leq 2-\frac{k(k+1)}{k(k+1)^2}$$
Since
$$k^2+k+1 \gt k^2+k\\
\frac{k^2+k+1}{k(k+1)^2} \gt \frac{k^2+k}{k(k+1)^2}\\
-\frac{k^2+k+1}{k(k+1)^2} \le -\frac{k^2+k}{k(k+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Two answers for the same limit I have a limit:
$$\lim_{x \to \infty} \left(\sqrt{x^4+5x^2+1} - x^2\right)$$
I know the answer is $\frac{5}{2}$, and you can get it by multiplying and dividing both sides by the conjugate ($\sqrt{x^4+5x^2+1} + x^2$) to get
$$ \lim_{x \to \infty} \left(\frac{x^4+5x^2+1-x^4}{\sqrt{x^4+5x^2+1} + x^2}\right) = \lim_{x \to \infty} \left(\frac{5x^2+1}{x^2 \sqrt{1+\frac{5}{x^2}+\frac{1}{x^4}} + x^2}\right) $$
It is then pretty easy to see that the stuff under the square root in the denominator will go to 1, and we will get $\frac{5x^2}{2x^2}$ = $\frac{5}{2}$. This seems to be the correct answer according to WolframAlpha as well. However, I got to thinking that there seems to be another completely legitimate way to do this limit, as follows:
$$ \lim_{x \to \infty} \left(\sqrt{x^4+5x^2+1} - x^2\right) = \lim_{x \to \infty} \left(x^2 \left(\sqrt{1+\frac{5}{x^2}+\frac{1}{x^4}}\right) - x^2\right) $$
Again, the stuff inside the square root will go to 1, so we will get
$$ = \lim_{x \to \infty} \left( x^2 - x^2\right ) = 0$$
Obviously this second way is wrong, but can someone explain why?
| The second method is wrong as it results in an indeterminate form. If I write the last term as
\begin{equation}
x^2\left(\sqrt{1 + \frac{5}{x^2} + \frac{1}{x^4}} - 1\right).
\end{equation} After putting the limiting value of $x$ I get $x^2$((a term that tends to $1$ but is greater than $1)-1)$ or $ \infty $*(a number tending to $0)$. Hence an indeterminate form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/659179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
A cubic diophantine Equation While reading Diophantine equations I came across the following equation $$x^3+cy^3-3yx=0$$
Is there any known method to solve this equation for any $c$?
| First note that the solutions involving zero are $x=y=0$ or $c=0,x=3t,y=3t^2$. Otherwise, we can give full solutions in terms of the (rare) solutions of much simpler equations.
For $c\ne0$, the equation $x^3+cy^3=3xy$ is solvable in non-negative integers if and only if there is a solution in non-negative integers of the equation $$a^2u^3+bv^3=1\text{ or }3,\text{ where }c=ab.$$
Proof
Let $x=tz,y=tv$, where $z$ and $v$ are corime. Then $$t(z^3+cv^3)=3zv.$$
$z$ is a factor of $tcv^3$ and therefore of $tc$. Let $c=ab$ and $z=au$, where $b$ and $u$ are coprime and where we can choose $b$ to be positive. Then $$t(a^2u^3+bv^3)=3uv.$$
Then $uv$ is coprime to $a^2u^3+bv^3$ and so $t=suv$ and $$s(a^2u^3+bv^3)=3.$$
If this equation is solvable then the original equation has solution $x=asu^2v,y=suv^2.$ Note that switching the signs of $s,u$ and $v$ leaves $x$ and $y$ unchanged so we can suppose $s$ is positive.
Example 1 Solve $x^3+y^3=3xy.$
The equation $u^3+v^3=1$ has no non-zero solutions (Fermat) and the equation $u^3+v^3=3$ is impossible modulo $9$. So $x^3+y^3=3xy$ has no non-zero solutions.
Example 2 Solve $x^3+6y^3=3xy.$
The equation $4u^3+3v^3=1 (s=3)$, for example, has solution $u=1,v=-1$, giving $x=-6,y=3.$
In general one can use PARI/GP to obtain all solutions. For example see the solution by @YongHaoNg to the following post:-
Solutions of $ax^3+by^3=1$
| {
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For $n \geq 2$, prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$ For $n \geq 2$ prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$
We need to use induction.
The Principle of Mathematical Induction, Theorem 4.2.1, states that $n_0 \in \mathbb{Z}$. For each integer $n \geq n_o$, let $P(n)$ be a statement about $n$. Suppose that the following two statements are true:
*
*$P(n_o)$
*$( \forall n \geq n_o)[P(n) \rightarrow P(n+1)]$
Then, for all integers $n \geq n_o$, that statement $P(n)$ is true.
Suppose,
$(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$, the equation, is true for some natural number $k$.
$\frac{1}{2^2})*...*(1- \frac{1}{k^2}) = \frac{k+1}{2k}$
The induction hypothesis $P(k) \rightarrow P(k+1)$ must hold for any natural number $k, P(k),$ from the basis and inductive step.
$(\frac{1}{2^2})*...*(1- \frac{1}{k^2})*(1- \frac{1}{(k+1)^2}) = \frac{k+1}{2k}(1- \frac{1}{(k+1)^2})$
$\leftrightarrow$ $\frac{k+1}{2k}(\frac{(k+1)^2}{(k+1)^2}- \frac{1}{(k+1)^2})$
$\leftrightarrow$ $(\frac{(k+1)}{2k})( \frac{(k+1)^2-1}{(k+1)^2})$
$\leftrightarrow$ $( \frac{(k^2+2k+1-1}{(2k)(k+1)})$
$\leftrightarrow$ $( \frac{(k^2+2k}{(2k)(k+1)})$
$\leftrightarrow$ $( \frac{k+2}{(2)(k+1)})$
We need to demonstrate that
$(\frac{1}{2^2})*...*(1- \frac{1}{n^2}) = \frac{n+1}{2n}$
holds true for $n \geq 2$
The question is do I just substitute a number that is greater than two after I did the induction?
| Here is an answer that uses logarithmic functions.
Let
$$
P_n=\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)
$$
and
$$
S_n=\log P_n.
$$
Then
\begin{eqnarray}
S_n&=&\log\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)\\
&=&\sum_{k=2}^{n}\log\left(1-\frac{1}{k^2}\right)\\
&=&\sum_{k=2}^{n}\log\left(\frac{k^2-1}{k^2}\right)\\
&=&\sum_{k=2}^{n}\log\left[\left(\frac{k-1}{k}\right)\left(\frac{k+1}{k}\right)\right]\\
&=&\sum_{k=2}^{n}\left[\log\left(\frac{k-1}{k}\right)+\log\left(\frac{k+1}{k}\right)\right]\\
&=&\sum_{k=2}^{n}\left[\log(k-1)-\log(k)\right]+\sum_{k=2}^n\left[\log(k+1)-\log(k)\right]\\
&=&\sum_{k=1}^{n-1}\log(k)-\sum_{k=2}^{n}\log(k)+\sum_{k=3}^{n+1}\log(k)-\sum_{k=2}^{n}\log(k)\\
&=&\log(1)+\sum_{k=2}^{n-1}[\log(k)-\log(k)]-\log(n)+\log(n+1)+\sum_{k=3}^{n}[\log(k)-\log(k)]-\log(2)\\
&=&0+0-\log(n)+\log(n+1)+0-\log(2)\\
&=&\log\left(\frac{n+1}{2n}\right).
\end{eqnarray}
Since
$$
S_n=\log P_n=\log\left(\frac{n+1}{2n}\right),
$$
it follows that
$$
P_n=\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right) =\frac{n+1}{2n}
$$
| {
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Prove that, $\int_0^{\pi/2}\frac{\sin (2n+1)x}{\sin x}~dx=\frac{\pi}{2}$ Prove that, $$\int_0^{\pi/2}\dfrac{\sin (2n+1)x}{\sin x}~dx=\dfrac{\pi}{2}$$ For every integer $n\ge 0$
I tried using induction without any help. Possibly my method of application was not right. Please help me!
| First, note that for any positive integer $n$, $$ \int_{x=0}^{\pi/2} \cos 2nx \, dx = \left[ \frac{\sin 2nx}{2n} \right]_{x=0}^{\pi/2} = 0.$$ Now let $$ \begin{align*} f_n(x) &= \frac{\sin(2n+1)x}{\sin x} = \frac{\sin(2n-1)x \cos 2x + \cos(2n-1)x \sin 2x}{\sin x} \\ &= f_{n-1}(x) (1 - 2 \sin^2 x) + 2 \cos(2n-1)x \cos x \\ &= f_{n-1}(x) + 2(\cos(2n-1)x \cos x - \sin(2n-1)x \sin x) \\ &= f_{n-1}(x) + 2 \cos 2nx. \end{align*}$$ Therefore, $$I_n = \int_{x=0}^{\pi/2} f_n(x) \, dx = \int_{x=0}^{\pi/2} f_{n-1}(x) \, dx + 2 \int_{x=0}^{\pi/2} \cos 2nx \, dx = I_{n-1}. $$ Since $I_0 = \frac{\pi}{2}$ trivially, the result immediately follows.
| {
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Can you prove $(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x)$? Show that $$(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x)$$
This can be shown through expansion but there is a more elegant solution
I cannot discover anything I would consider elegant. Can anyone help?
| $$a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc$$
$$=(a+b)^3+c^3-3ab(a+b)-3abc$$
$$=[(a+b)+c][(a+b)^2-(a+b)c+c^2]-3ab[(a+b)+c]$$
$$=(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
Though not required in the current case, $$a^2+b^2+c^2-ab-bc-ca=\frac{(a-b)^2+(b-c)^2+(c-a)^2}2$$ which will be zero iff $\displaystyle a-b=b-c=c-a=0\implies ?$
| {
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Find the least nonnegative residue Find the least nonnegative residue of $5^{18} \mod 11$.
To do this I took $5^2 \equiv 3 \mod 11$. Then I did $(5^2)^5 \equiv 3^5 \mod 11$. And $3^5 \equiv 1 \mod 11$.
So now I have $5^{10} \equiv 1 \mod 11$. Then I multiplied both sides by $5^8$ to get $5^{18} \equiv 5^8 \mod 11$. So I believe $5^8$ is the least nonnegative residue but I am not entirely sure. Can someone please confirm that this is correct?
| As you mentioned, we have $5^2 \equiv 3$ mod 11. However, I find it easier to proceed by showing,
$$5^8 \equiv (5^2)^4 \equiv 3^4 \equiv 4 \text{ mod }11, $$
so that $$5^{16} \equiv (5^8)^2 \equiv 4^2 \equiv 5 \text{ mod }11,$$ and then finally,
$$ 5^{18} \equiv 5^2\cdot5^{16} \equiv 3 \cdot 5 \equiv 4 \text{ mod } 11.$$
Also, this should be a much more satisfying answer since the least nonnegative residue of $n$ will always be in the interval $[0,n]$.
| {
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Proving $A^{-1}=-\frac{1}{6}A^{2}+\frac{2}{3}A-\frac{1}{6}I$ I have a matrix A
\begin{bmatrix}
1 & 1 & 2\\
0& 2 &0 \\
2&3 & 1
\end{bmatrix}
I have calculated the $A^{-1}$
\begin{bmatrix}
-\frac{1}{2} & -\frac{5}{4} & 2\\
0& \frac{3}{4} &0 \\
1&\frac{1}{4} & -\frac{1}{2}
\end{bmatrix}
And I want to prove that $A^{-1}=-\frac{1}{6}A^{2}+\frac{2}{3}A-\frac{1}{6}I$
I tried doing all the calculations (not really productive, I know) but the result wasn't correct, or I made some calculation mistake. Is there any more productive way to prove this?
I have also found the characteristic polynomial, and it's $-\lambda ^{3}+4\lambda ^{2}-\lambda-6$
I would prefer a little tip to help me rather than the solution itself
| Note that $\operatorname{char}_A(\lambda)=-\lambda^3+4\lambda^2-\lambda-6$. The Cayley-Hamilton Theorem then implies
$$
-A^3+4A^2-A-6I=\mathbf 0
$$
so that
$$
A\left(-A^2+4A-I\right)=6I
$$
That is,
$$
A^{-1}=
-\frac{1}{6}A^2+\frac{2}{3}A-\frac{1}{6}I
$$
as advertised.
| {
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Chromatic polynomial for a bipartite graph I need to get the chromatic polynomial for the complete bipartite graph: $K_{2,3}$
Im using the Fundamental Reduction Theorem, and the picture below shows mi attempt to it.
I omitted vertex names because is just a small graph.
With this procedure i get: $P(K_{2,3},x)=\frac{C_4.K_2}{K_1}-\frac{K_3.K_3}{K_2}$
Here $C_4$ is a cycle lenght 4 joined to a complete graph lenght 2 just by one vertex. And is well known that: $P(C_4,x)=x(x-1)(x^2-3x+3)$.
I think im doing well, but the final result is: $x (-3 x^3+12 x^2-16 x+7)$ and is not correct.
The correct result is supposed to be: $x (x-1) (x^3-5 x^2+10 x-7)$
I think im close, but evidently im failing at some point.
| The correct answer is $x (x-1) (x^3-5 x^2+10 x-7)$.
Here is another argument to prove it:
In any coloring of $K_{2,3}$ the two vertices in the first part will have either the same color or two different colors. So, in the first case, there are $x$ colors to choose the same color and $x-1$ colors to choose from to color the remaining three vertices. Thus there are $x(x-1)^3$ colorings of this first kind.
In the second case, there $x(x-1)$ ways to color the two vertices with different colors and we are left with $x-2$ colors to choose from to color the remaining three vertices. Thus, there are $x(x-1)(x-2)^3$ colorings of this second kind.
So, in total we have: $x(x-1)^3 + x(x-1)(x-2)^3$ colorings. After simplification you get your answer.
** Edit:
Also your method wields the same result:
So $P(K_{2,3},x)=\frac{P(C_4, x).P(K_2, x)}{P(K_1, x)}-\frac{P(K_3, x).P(K_3, x)}{P(K_2, x)}$
and we have:
$P(C_4,x)=x(x-1)(x^2-3x+3)$,
$P(K_3, x)= x(x-1)(x-2)$,
$P(K_2, x)= x(x-1)$
and $P(K_1, x)= x$
So, $P(K_{2,3},x)=\frac{x(x-1)(x^2-3x+3).x(x-1)}{x}-\frac{(x(x-1)(x-2))^2}{x(x-1)}$
$P(K_{2,3},x)=x(x-1)^2(x^2-3x+3)- x(x-1)(x-2)^2$
$P(K_{2,3},x)=x(x-1)[(x-1)(x^2-3x+3)- (x-2)^2 ]$
$P(K_{2,3},x)=x(x-1)[(x-1)(x^2-3x+3)- (x-2)^2 ]$
This will give you the correct answer.
| {
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Factoring the quintic polynomial $x^5+4x^3+x^2+4=0$ I am trying to factor
$$x^5+4x^3+x^2+4=0$$
I've used Ruffini's rule to get
$$(x+1)(x^4-x^3+5x^2-4x+4)=0$$
But I don't know what to do next.
The solution is $(x+1) (x^2+4) (x^2-x+1) = 0$. I've tried using the completing square method but with no result. Could you give me hints?
| Or, alternatively, note that $$x^4-x^3+5x^2-4x+4=x^2(x^2-x+1)+4x^2-4x+4$$ and factor $4$ from the last three terms.
| {
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Integral of $\int^\sqrt2_1\frac{1}{1+\sqrt{x^2 - 1}}dx$ by substitution? In a maths question I have the question:
$$\int^\sqrt2_1\frac{1}{1+\sqrt{x^2 - 1}}dx$$ by substitution?
All other questions have been by trigonometric substitution so I assume that is how to solve.
| Another substitution that is useful here is $x=\cosh{t}$; the integral is thus equal to
$$\int_0^{\log{(1+\sqrt{2})}} dt \frac{\sinh{t}}{1+\sinh{t}} = \log{(1+\sqrt{2})} - \int_0^{\log{(1+\sqrt{2})}} \frac{dt}{1+\sinh{t}}$$
Now use $\sinh{t} = (e^t-e^{-t})/2$ and sub $y=e^t$ to get that the integral on the right is
$$2 \int_1^{\sqrt{2}+1} \frac{dy}{y^2+2 y-1} = 2 \int_1^{\sqrt{2}+1} \frac{dy}{(y+1)^2-2} = 2 \int_2^{2+\sqrt{2}} \frac{du}{u^2-2}$$
This integral is equal to
$$\frac1{\sqrt{2}} \left [\log{\left (\frac{u-\sqrt{2}}{u+\sqrt{2}} \right )} \right ]_2^{2+\sqrt{2}} = \frac1{\sqrt{2}} \log{\left ( 1+\sqrt{2}\right )}$$
Thus the original integral is
$$\int_1^{\sqrt{2}} \frac{dx}{1+\sqrt{x^2-1}} = \left ( 1-\frac1{\sqrt{2}}\right ) \log{\left (1+\sqrt{2}\right )}$$
| {
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How to find the sum of series $\sum_{i=1}^{\infty}\frac{i}{2^i}$? I am learning about series of numbers at the moment. In the book there is an exercise in which I need to find the sum of :
$$\sum_{i=1}^{\infty}\frac{i}{2^i}$$
I know it is equal to $2$. But how do I get to that result?
Are there any general ways of finding the sums of series? In the books I am using, there is a lot about series, their convergence etc. but almost no examples.
| Let
$$S := \sum_{k=1}^\infty \frac{k}{2^k}.$$
Since all of its elements are positive and the series is absolutely convergent, we can do the following:
\begin{align*}
S &= \sum_{k=1}^\infty \frac{k}{2^k} = \sum_{k=1}^\infty \frac{k-1}{2^k} + \sum_{k=1}^\infty \frac{1}{2^k} = \frac{1}{2} \sum_{k=1}^\infty \frac{k-1}{2^{k-1}} + 1 = \frac{1}{2} \sum_{k=2}^\infty \frac{k-1}{2^{k-1}} + 1 = \frac{1}{2} \sum_{k=1}^\infty \frac{k}{2^k} + 1 \\
&= \frac{1}{2}S + 1,
\end{align*}
so $S = 2$.
| {
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Problem with logarithms Problem:
Solve:
$$\frac{1}{2^x} = \frac{5}{8^{x+2}}$$
My attempt:
$$\frac{1}{2^x} = \frac{5}{8^{x+2}}$$
$$\Rightarrow 5 \cdot 2^x = 8^{x+2}$$
$$\Rightarrow 2^{\log_2 5+x} = 8^{x+2} $$
$$\Rightarrow (\log_2 5 + x)(\log_a 2) = (x+2)(\log_a 8)$$
And then just keep going like this, but I'm obviously wrong already as the answer is:
$$ x = \frac{\ln 5 - 9 \ln 2}{2 \ln 2}$$
What am I doing wrong?
EDIT: If someone could enlarge the latex for me that would be great.
| With a slight modification, we have
$$5\cdot 2^x=8^{x+2}=(2^3)^{x+2}=2^{3x+6}$$
which means that
$$5=2^{2x+6}\\
\log_2 5=2x+6\\
x=\frac{\log_2 (5)-6}2$$
Transforming to the natural logarithm would look like
$$5=2^{2x+6}\\
\ln 5=(2x+6)\ln 2\\
x=\frac{\ln 5-6\ln 2}{2\ln 2}$$
| {
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Let $p$ be prime and $\left(\frac{-3}p\right)=1$. Prove that $p$ is of the form $p=a^2+3b^2$
Let $p$ be prime and $\left(\frac{-3}p\right)=1$, where $\left(\frac{-3}p\right)$ is Legendre symbol. Prove that $p$ is of the form $p=a^2+3b^2$.
My progress:
$\left(\frac{-3}p\right)=1 \Rightarrow$ $\left(\frac{-3}p\right)=\left(\frac{-1}p\right)\left(\frac{3}p\right)=(-1)^{\frac{p-1}2}(-1)^{\left\lfloor\frac{p+1}6\right\rfloor}=1 \Rightarrow$ $\frac{p-1}2+\left\lfloor\frac{p+1}6\right\rfloor=2k$ I'm stuck here. This is probably not the way to prove that.
Also tried this way:
$\left(\frac{-3}p\right)=1$, thus $-3\equiv x^2\pmod{p} \Rightarrow$ $p|x^2+3 \Rightarrow$ $x^2+3=p\cdot k$
stuck here too.
Any help would be appreciated.
| $(\frac{p}{3})=(\frac{-3}{p})(\frac{p}{3})=(\frac{-1}{p})(\frac{3}{p})(\frac{p}{3})=(-1)^{\frac{p-1}{2}}(\frac{3}{p})(\frac{p}{3})=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p-1}{2}\frac{3-1}{2}} = 1$
Hence,$(\frac{-3}{p})=1$ iff, $p\equiv1\mod3$.
Since, there is $u \in \mathbb{Z}$ such that, $-3\equiv u^2\pmod{p}$
Consider the lattice defined by $L=\{(a,b)\in\mathbb{Z}^2\, : \,a\equiv ub\pmod p\}$ generated by $(u,1)$ and $(0,p)$. $L$ has index $p$ in $\mathbb{Z}^2$, and area of its fundamental domain is $p$. Now, consider an ellipse $E_n$ defined by $x^2+3y^2=n$, then the area of $E_n=\frac{\pi n}{\sqrt3}>1.8n$
Choose, $n=2.3 p$, then Area of $E_{n}>4p$ and $E_n\cap L$ has a non zero point $(a,b)$.
Now, $a^2+3b^2\equiv(ub)^2+3b^2\equiv b^2(u^2+3)\equiv0 \pmod p$.
Since, $(a,b)\in E_n \implies a^2+3b^2<2.3p$ we have $a^2+3b^2=p,2p$.
But, $a^2+3b^2=2p \implies a^2\equiv 2p \pmod 3 \equiv 2 \pmod 3$ contradiction !!
Therefore, $a^2+3b^2=p$.
| {
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How to solve this recurrence $T(n)=2T(n/2)+n/\log n$ How can I solve the recurrence relation $$T(n)=2T\left(\frac n2\right)+\frac{n}{\log n}$$?
I am stuck up after few steps..
I arrive till
$$T(n) = 2^k T(1) + \sum_{i=0}^{\log(n-1)} \left(\frac{n}{\log n} - i\right)$$
How to simplify this log summation?
| Your summation is wrong, and you should have replaced $k$ with $log(n)$ in your last expression.
Here are detailed steps of this recurrence relation:
$$
\\
T(n) = 2T(\frac{n}{2}) + \frac{n}{\log n}
\\
T(n) = 2(2T(\frac{n}{4}) + \frac{\frac{n}{2}}{\log \frac{n}{2}}) + \frac{n}{\log n} = 2^2T(\frac{n}{2^2}) + \frac{n}{\log(n) - 1} + \frac{n}{\log n}
\\
T(n) = 2(2(2T(\frac{n}{8}) + \frac{\frac{n}{4}}{\log \frac{n}{4}}) + \frac{\frac{n}{2}}{\log \frac{n}{2}}) + \frac{n}{\log n} = 2^3T(\frac{n}{2^3}) + \frac{n}{\log(n) - 2} + \frac{n}{\log(n) - 1} + \frac{n}{\log n}
\\
...
\\
T(n) = 2^kT(\frac{n}{2^k}) + \sum_{i = 0}^{ k - 1}\frac{n}{\log(n) - i}
\\
\text{When } \frac{n}{2^k} = 1 \Rightarrow n = 2^k \Rightarrow k = \log_2(n)
\\
T(n) = 2^{\log_2(n)}T(1) + \sum_{i = 0}^{ \log_2(n) - 1}\frac{n}{\log(n) - i}
\\
T(n) = \Theta (n) + \sum_{i = 0}^{ \log_2(n) - 1}\frac{n}{\log_2(n) - i}
\\
T(n) \approx \Theta (n) + \sum_{j = 1}^{ \log_2(n)}\frac{n}{j}
\\
T(n) \approx \Theta (n) + nH_{\log_2(n)}
\\
T(n) \approx \Theta (n) + n \ln(\log_2(n))
\\
\text{And finally }
T(n)\ \in \Theta(n \ln(\log_2(n)))
$$
| {
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Probability of transfers Probability of transferring materials between containers.
| Define the random variable: $X=$ is the number of green marbles drawn, which has the hypergeometric distribution with parameters $N=8+4=12$, $K=4$ and $n=3$. Then you have that the probability mass function of $X$ is equal to: $$P(X=k)=\frac{\dbinom{4}{k}\dbinom{8}{3-k}}{\dbinom{12}{3}}$$ for $k=0,1,2,3$. So, by substitung in the above formula we find that: $$P(X=0)=\frac{14}{55}, P(X=1)=\frac{28}{55}, P(X=2)=\frac{12}{55}P(X=0)=\frac{1}{55}$$
1st Question. Define the event $G$ the marble drawn from the second jar. According to the Total Probability Law you have (by conditioning on X) that:
$$\begin{align*}P(G)&=P(G|X=0)P(X=0)+P(G|X=1)P(X=1)+P(G|X=2)P(X=2)\\&\quad+P(G|X=3)P(X=3)=\\&=0P(X=0)
+\frac{1}{3}P(X=1)+\frac{2}{3}P(X=2)+1P(X=3)=0+\frac{1}{3}\frac{28}{55}+\frac{2}{3}\frac{12}{55}+\frac{1}{55}\\&=\frac{1}{3}\end{align*}$$
2nd Question. You want to calculate the probability: $P(X=1|G)$. According to Bayes theorem you have that: $$P(X=1|G)=\frac{P(G|X=1)\cdot P(X=1)}{P(G)}=\frac{\frac{1}{3}\frac{28}{55}}{\frac{1}{3}}=\frac{28}{55}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/691297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding out the coeffcient next to $x^2$ in $(\cdots(x-2)^2-2)^2\cdots-2)^2$. In need to find out the coefficient next to $x^2$ in polynomial $(\cdots(x-2)^2-2)^2\cdots-2)^2$, where we nest the expression $(x-2)^2$ n times.
Meaning that for $n=1$ we get $(x-2)^2$, for $n=2$ we get $((x-2)^2-2)^2=(x^2-4x+2)^2$
Which I won't even try to expand, since already I see that no trivial recurrence applies...
| After $n$ compositions, the constant term is (clearly?) $4$.
The linear coefficient is $-4$ for $n=1$, and the linear coefficient after $n$ compositions (call it $a_n$) satisfies the recurrence $$a_n=2a_{n-1}(4)$$ so $a_n=-4\cdot8^{n-1}$ where the $4$ is the constant term form the previous composition.
The quadratic coefficient is $1$ for $n=1$ and the quadratic coefficient after $n$ compositions (call it $b_n$) satisfies the recurrence $$b_n=a_{n-1}^2+2b_{n-1}(4)$$ which means $$b_n=2^{6n-8}+8b_{n-1}$$ If we introduce $c_n$ defined by $b_n=2^{6n-8}c_n$, we have
$$2^{6n-8}c_n=2^{6n-8}+8\cdot2^{6n-14}c_{n-1}$$
$$c_n=1+\frac18c_{n-1}$$
Now $c_n$ has a nonhomogeneous linear recurrence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/691748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k = 1}^{n}{f(\frac{k}{n}) }$ $=\int_0^1 f(x)dx.$ Question:
Let $f$ be continuous on $[0,1]$. Prove that
$\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k = 1}^{n}{f(\frac{k}{n}) }$ $=\int_0^1 f(x)dx.$
where $k=0,1,...,n.$
Attempt:
I don't even know where to start. It makes sense reading the sum, as $k\rightarrow n$, and dividing it by the number of partitions, I should reach the definition of the integral. Hoping for a little push to get started.
| $ f(x) =f(0)+\frac{f'(0)x}{1!}+\frac{f''(0)x^2}{2!}+.....=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n $
$$\int _0^x {f(t) dt}=\int _0^x(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} t^n)dt=\sum_{n=0}^{\infty} (\frac{f^{(n)}(0)}{n!}\int _0^x t^n dt)=\sum_{n=0}^{\infty} (\frac{f^{(n)}(0)}{n!}\frac{x^{n+1}}{n+1})$$
$$\int _0^x {f(t) dt}=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^{n+1}}{(n+1)!}$$ $$(1)$$
$$f(\frac{kx}{n})=\sum_{m=0}^{\infty} \frac{f^{(m)}(0)}{m!} (\frac{kx}{n})^m$$
$$\sum \limits_{k=1}^{n} k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$ where $a_j$ are constants.
More information about summation http://en.wikipedia.org/wiki/Summation
$$\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})=\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n \sum_{m=0}^{\infty} \frac{f^{(m)}(0)}{m!} (\frac{kx}{n})^m=\lim_{n\to\infty} \frac{x}{n}\sum_{m=0}^{\infty} \frac{x^m}{n^m} \frac{f^{(m)}(0)}{m!} \sum \limits_{k=1}^n k^m=\lim_{n\to\infty} \frac{x}{n}[f(0)n+\frac{f'(0)x}{n 1!}(\frac{n^2}{2}+\frac{n}{2})+ \frac{f''(0)x^2}{n^2 2!}(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})+\frac{f'''(0)x^3}{n^3 3!}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+\frac{f^{(4)}(0)x^4}{n^4 4!}(\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30})+...... ]=
\lim_{n\to\infty} [f(0)x+\frac{f'(0)x^2}{n^2 1!}(\frac{n^2}{2}+\frac{n}{2})+ \frac{f''(0)x^3}{n^3 2!}(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})+\frac{f'''(0)x^4}{n^4 3!}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+\frac{f^{(4)}(0)x^5}{n^5 4!}(\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30})+...... ]= [f(0)x+\frac{f'(0)x^2}{ 2!}+ \frac{f''(0)x^3}{ 3!}+\frac{f'''(0)x^4}{ 4!}+\frac{f^{(4)}(0)x^5}{ 5!}+...... ]$$
$$\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})=\sum_{m=0}^{\infty} \frac{f^{(m)}(0)x^{m+1}}{(m+1)!}$$
$$(2)$$
Equation $(1)$ and equation $(2)$ are equal to each other. Thus
$$\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})=\int _0^x {f(t) dt}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/692088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
How can I prove the trigonometric Problem? How can I show the following trigonometric problem :
$$\frac{1}{3}\leq \frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^2
\theta}\leq 3$$
I have tried in the following way :
$$
\begin{align}
& \phantom{\Rightarrow}
\frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{1}{\sec^2\theta+\tan^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{\cos^2\theta}{1+\sin^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{1-\sin^2\theta}{1+\sin^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{2-4\sin^2\theta}{3(1+\sin^2
\theta)} \\[8pt]
& \Rightarrow \frac{2\cos 2\theta}{3(1+\sin^2
\theta)}
\end{align}
$$
How can I show $ \dfrac{2\cos 2\theta}{3(1+\sin^2
\theta)}\geq 0$ ?
| We can't show $$ \dfrac{2\cos 2\theta}{3(1+\sin^2\theta)}\ge\iff\cos 2\theta\ge0$$ which can be $<0$ if $\displaystyle2n\pi+\frac\pi2<2\theta<2n\pi+\pi+\frac\pi2$ where $n$ is an integer
Why don't we adapt the following method:
$$ \frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^2\theta}=\frac1{2\tan^2\theta+1}$$
Now, $$0\le\tan^2\theta\le\infty\iff0\le2\tan^2\theta\le\infty\iff1\le1+2\tan^2\theta\le\infty$$
$$\implies1\ge\frac1{1+2\tan^2\theta}\ge0\iff 0\le\frac1{1+2\tan^2\theta}\le1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/694637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find this ODE solution $(y-x)\sqrt{x^2+1}\dfrac{dy}{dx}=(1+y^2)^{\frac{3}{2}}$ Find all solutions $y$ to the ODE
$$(y-x)\sqrt{x^2+1}\dfrac{dy}{dx}=(1+y^2)^{\frac{3}{2}}$$
My try:
$$(y-x)\dfrac{d(\sqrt{1+y^2})}{y}=\dfrac{(1+y^2)dx}{\sqrt{x^2+1}}$$
then
$$\dfrac{(y-x)\cdot d(\sqrt{1+y^2})}{y(1+y^2)}=\dfrac{d(\sqrt{1+x^2})}{x}$$
then I can't.Thank you for you help
| Maple says:
$$
\arctan \left( x \right) -\int ^{-\arctan \left( x \right) +\arctan
\left( y \left( x \right) \right) }\!{\frac {2-\cos \left( 4\,t+4
\right) -2\,\cos \left( 2\,t \right) +2\,\cos \left( 2\,t+4 \right) -
\cos \left( 4 \right) +\sqrt {4\,\cos \left( 4\,t+8 \right) -2\,\cos
\left( 8+2\,t \right) -2\,\cos \left( 6\,t+8 \right) -8\,\cos \left(
4 \right) +16\,\cos \left( 2\,t+4 \right) -12\,\cos \left( 2\,t
\right) -8\,\cos \left( 4\,t+4 \right) +12}}{2+\cos \left( 4\,t+4
\right) +2\,\cos \left( 2\,t \right) +2\,\cos \left( 2\,t+4 \right) +
\cos \left( 4 \right) }}{dt}-C=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove convergence of a series little with Direct comparison test I have the following series
$$\sum_{k=1}^{\infty} \log\left(1+\frac{1}{k^2}\right)$$
This series should converge but when I apply the Direct comparison test it diverges
$$\left|\sum_{k=1}^{\infty} \log(1+\frac{1}{k^2})\right| \le \sum_{k=1}^{\infty} \log\left|1+\frac{1}{k^2}\right| \le \sum_{k=1}^{\infty} 1+\frac{1}{k^2} = \sum_{k=1}^{\infty} 1 + \sum_{k=1}^{\infty}\frac{1}{k^2}$$
so we know that $\sum_{k=1}^{\infty} \left(1+\frac{1}{k^2}\right) = \sum_{k=1}^{\infty} 1 + \sum_{k=1}^{\infty}\frac{1}{k^2}$ diverges because $\sum 1$ diverges, so the series should diverges.
what am I doing wrong?
Should it be in this way :
$$\left|\sum_{k=1}^{\infty} \log(1+\frac{1}{k^2})\right| \le \sum_{k=1}^{\infty} \log\left|1+\frac{1}{k^2}\right| \le \sum_{k=1}^{\infty} \frac{1}{k^2}$$
so it will converges because $\sum \frac{1}{k^2}$ converges ? if yes, why do we ignore $\sum 1$
| Since $\log(1+x)\le x$ for all $x\gt-1$, we have
$$
\begin{align}
\sum_{k=1}^\infty\log\left(1+\frac1{k^2}\right)
&\le\sum_{k=1}^\infty\frac1{k^2}\\
&=\frac{\pi^2}6\tag{1}
\end{align}
$$
so the sum converges by comparison.
Since $\log(1+x)\ge\frac{x}{1+x}$ for all $x\gt-1$, we have
$$
\begin{align}
\sum_{k=1}^\infty\log\left(1+\frac1{k^2}\right)
&\ge\sum_{k=1}^\infty\frac1{k^2+1}\\
&=\frac\pi2\coth(\pi)-\frac12\tag{2}
\end{align}
$$
In fact,
$$
\begin{align}
\prod_{k=1}^\infty\frac{k^2+1}{k^2}
&=\lim_{n\to\infty}\prod_{k=1}^n\frac{k+i}{k}\frac{k-i}{k}\\
&=\lim_{n\to\infty}\frac{\Gamma(n+1+i)}{\Gamma(1+i)}\frac{\Gamma(n+1-i)}{\Gamma(1-i)}\frac1{\Gamma(n+1)^2}\\
&=\frac1{i\,\Gamma(i)\,\Gamma(1-i)}\lim_{n\to\infty}\frac{\Gamma(n+1+i)\Gamma(n+1-i)}{\Gamma(n+1)^2}\\
&=\frac{\sin(\pi i)}{\pi i}\cdot1\\[3pt]
&=\frac{\sinh(\pi)}\pi\tag{3}
\end{align}
$$
Therefore,
$$
\sum_{k=1}^\infty\log\left(1+\frac1{k^2}\right)
=\log\left(\frac{\sinh(\pi)}\pi\right)\tag{4}
$$
As shown in $(1)$ and $(2)$,
$$
\frac\pi2\coth(\pi)-\frac12
\lt\log\left(\frac{\sinh(\pi)}\pi\right)
\lt\frac{\pi^2}6\tag{5}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/696255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of a+b+c If $x+1$ is a factor of $ax^4 + bx^2 + c$, find the value of $a + b + c$?
I know that it is equal to zero, but I have to know How to do it.
| As a shortcut for the long division, note that if $x+1$ is a factor of $ax^4+bx^2+c=0$, then $x-1$ is also a factor. Divide $ax^4+bx^2+c$ by $(x+1)(x-1)=x^2-1$ to get the remainder $a+b+c$, which must be equal to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/696646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
Proving the limit $\lim_{x\to 1} \frac{100}{x} = 100$ using epsilon-delta definition $$\lim_{x\to 1} \frac{100}{x} = 100$$
I'm trying to proving the limit using epsilon-delta definition and here's what I've come up with so
My attempt:
$\forall ε>0, \exists δ>0$, such that if for x, $0<|x-1|<δ$ then $|\frac{100}{x} - 100| < ε$
$$100*|\frac{1}{x} - 1| < ε$$
$$100*\frac{1}{|x|} * |x-1| < ε$$
δ : $$100*\frac{1}{|x|} * |x-1| < ε$$
$$|x-1| < δ$$, choose δ = 1/2
then,
$$|x-1| < \frac{1}{2} \Rightarrow 0 < x < \frac{3}{2}$$
From here I'm not too sure how to proceed on. I was wondering if anyone can help me on what to do next.
| You want $$\left|\frac{100}{x} - 100\right| < \varepsilon$$
whenever $\left|x-1\right| < \delta$.
Your task is to find a $\delta$ that makes this true for a given $\varepsilon.$
Let $\varepsilon > 0$, we have
\begin{align}
\left|\frac{100}{x} - 100\right| &=
100\cdot\left|\frac{1}{x} - 1\right| =
100\cdot\left|\frac{1-x}{x}\right| =
100\cdot\frac{\left|1-x\right|}{\left|x\right|}.
\end{align}
If we choose $\delta\le\frac 1 2$ we have $\left|x-1\right|<\frac 1 2$ which says
$$-\frac 1 2 < x - 1 < \frac 1 2$$
or equivalently
$$ \frac 1 2 < x < \frac 3 2.$$
In particular $\left|x\right|=x > \frac 1 2$, so we continue with
\begin{align}
\left|\frac{100}{x} - 100\right| &=
100\cdot\frac{\left|1-x\right|}{\left|x\right|} =
100\cdot\frac{\left|1-x\right|}{\left|x\right|} \\&<
100\cdot\frac{\left|1-x\right|}{1/2} = 200 \cdot \left|1-x\right| < 200\cdot \delta.
\end{align}
For this to be less than or equal to $\varepsilon$ we need $200\cdot\delta\le\varepsilon$, which is equivalent to $\delta \le \frac{\varepsilon}{200}$.
So if we have both $\delta\le\frac 1 2$ and $\delta<\frac{\varepsilon}{200}$, which can be achieved by picking
$$ \delta = \min \left\{\frac 1 2, \frac{\varepsilon}{200}\right\},$$
we get
\begin{align}
\left|\frac{100}{x} - 100\right| &=
100\cdot\frac{\left|1-x\right|}{\left|x\right|} < 200\cdot \delta \le 200\cdot \frac{\varepsilon}{200} = \varepsilon.
\end{align}
You get this when you look at $$\frac{\left|1-x\right|}{\left|x\right|},$$ and realize that to get this fraction small you need $\left|1-x\right|$ to be small, which can be achievied directly by choosing $\delta$ small, but you also want $\left|x\right|$ to don't be small, since a small denominator gives a large fraction. So you need a way to make $\left|x\right|>c>0$ by making $\left|x-1\right|$ small. I chose to make $\left|x-1\right|<\frac 1 2$ which gave us $\left|x\right|>\frac 1 2$ so we can finish the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/696809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determine the value of the integral $I=\int_{0}^{1}\frac{\ln\left(1-a^2x^2\right)}{\sqrt{1-x^2}}dx$ Determine the value of the integral $$I(a)=\int_{0}^{1}\frac{\ln\left(1-a^2x^2\right)}{\sqrt{1-x^2}}dx, \: |a|\leq 1$$
My try:
$\to I'(a)=\int_{0}^{1}\frac{-2ax^2}{\left(1-a^2x^2\right)\sqrt{1-x^2}}dx$
Set $x=\cos t\to dx=-\sin tdt$
Hence $I'(a)=\int_{0}^{\frac{\pi}{2}}\frac{-2a\cos^2t}{1-a^2\cos^2t}dt=\pi\left(\frac{1}{a}-\frac{1}{\sqrt{1-a^2}}\right)\to I(a)=\pi\left(\ln|a|-\arcsin a\right)+C$
Question: Find C?
| $$\begin{align}I(a) &= -\sum_{k=1}^{\infty} \frac{a^{2 k}}{k} \int_0^1 dx \frac{x^{2 k}}{\sqrt{1-x^2}}\\ &= -\frac{\pi}{2}\sum_{k=1}^{\infty} \frac1{k} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k} \end{align} $$
$$I'(a) = -\frac{\pi}{2} \sum_{k=1}^{\infty} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k-1} $$
$$a I'(a) = -\pi \sum_{k=1}^{\infty} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k} = -\pi \left (\frac1{\sqrt{1-a^2}} -1\right )$$
$$\implies \begin{align}I(a) &= -\pi \int da \left (\frac1{a\sqrt{1-a^2}}-\frac1{a} \right )\\ &= \pi \log{\left (1+\sqrt{1-a^2}\right )} +C \end{align}$$
$$I(0)=0 \implies C=-\pi \log{2}$$
$$\therefore I(a) = \pi \log{\left (\frac{1+\sqrt{1-a^2}}{2} \right )}$$
To derive the result for the integral in the first line, see here.
| {
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"url": "https://math.stackexchange.com/questions/699675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Calculating $e^{3x} \pmod{27}$ I am following some notes that say that $\exp(3x) = 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3 = 1 +3x+18x^2+18x^3 \pmod{27}$
I can't understand this. Firstly why do we stop the expansion after 4 terms? And secondly I don't see why $ \frac{9}{2}x^2 + \frac{9}{2}x^3 = 18x^2+18x^3 \pmod{27}$
Could someone please explain it to me? I wish to follow the same method for $\log(1+5x) \pmod{5^4} $
Thanks in advance
| $$e^{3x}=\sum_{k=0}^\infty\frac{(3x)^k}{k!}=1+3x+\frac{9x^2}2+\underbrace{\frac{9x^3}2}_{=\frac{27}{3!}}+\underbrace{27\left(\frac{x^4}{8}+\frac{3x^5}{40}+\ldots\right)}_{=0\pmod{27}}$$
and now check that
$$\frac12=2^{-1}=14\pmod{27}\implies \frac92=9\cdot14=18\pmod{27}$$
so we finally get
$$e^{3x}=1+3x+18x^2+18x^3\pmod{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/700276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Abstract Algebra Computation
Compute:
$3^{47}$ mod $23$
$3^{49}$ mod $7$
$2^{2^{17}}$ mod $23\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ hint: compute first $2^{17} + 1$ mod $19$
This is the first time I've had to compute elements, so I would appreciate any help.
For the first problem, I got $3^{47} \equiv 3^{22} \cdot 3^{22} \cdot 3^{3} \equiv 4$
For the second, I got $3^{49} \equiv (3^7)^7 \equiv 3^7 \equiv 3$
For the third, I'm not really sure where to begin. I don't understand how computing $2^17$ mod $18$ would help. Using Euler's theorem, though, I got that $2^{17}$ mod $18 \equiv 12$ since $\phi(18) = 6$.
EDIT: I'm sorry, I mistyped the third problem. I will ask it in another question.
| Even for the second, $\displaystyle3^3=27\equiv-1\pmod7,$
$\displaystyle\implies3^{49}=(3^3)^{16}\cdot3\equiv(-1)^{16}\cdot3\pmod7 $
For the third,
$\displaystyle2^{\left(2^{17}\right)}\equiv2^{\left(2^{17}\pmod{\phi(23)}\right)}\pmod{23}$
Let us find $\displaystyle2^{17}\pmod{22}$ as $\phi(23)=22$
As $\displaystyle(2^{17},22)=2,$ we shall try finding $\displaystyle2^{16}\pmod{11}$
As $\displaystyle2^5=32\equiv-1\pmod{11},2^{16}=2\cdot(2^5)^3\equiv2\cdot(-1)^3\pmod{11}\equiv-2\equiv9$
$\displaystyle\implies2^{17}\equiv2\cdot9\pmod{2\cdot11}\equiv18$
$\displaystyle\implies2^{\left(2^{17}\right)}\equiv2^{18}\pmod{23}$
Now, $\displaystyle2^6=64\equiv-5\pmod{23}$
$\displaystyle\implies2^{18}\equiv(-5)^3\equiv-125\equiv13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/700483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $(a-1),a,(a+1)$ are consecutive positive integers, $ (a+1)^3 \neq a^3 + (a-1)^3$ I had to prove the following statement:
If $(a-1),a,(a+1)$ are consecutive positive integers, $(a+1)^3 \neq a^3 + (a-1)^3$
My attempt at the solution was to first expand each side to get
$$a^3 + 3a^2 + 3a + 1 \neq 2a^3 - 3a^2 +3a - 1\\
0 \neq a^3 - 6a^2 - 2$$
However, $a^3 - 6a^2 - 2$ does hit the $x$-axis at $a = 6.0546$.
Does that mean that the statement is incorrect?
| $a^3−6a^2−2$ does hit the $x$-axis at $a=6.0546$, so, after placing the value of $a$ in the equation, you get the value of $a^3−6a^2−2$ to be $0.001536$ which is not equal to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/701166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Integral $\int_0^1\frac{\log(1-x)}{\sqrt{x-x^3}}dx$ I have a trouble with this integral
$$I=\int_0^1\frac{\log(1-x)}{\sqrt{x-x^3}}dx.$$
Could you suggest how to evaluate it?
| Here is a proof of Cleo's answer.
Rewrite the integral as
$$
\begin{align*}
I &= \int_0^1 \frac{\log(1-x)}{\sqrt{x}\sqrt{1-x^2}}dx \\
&= \int_0^1 \frac{\log(1-x^2)-\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\
&= \int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx
\end{align*}
$$
The first integral can be computed by calculating a derivative of the beta function.
$$
\begin{align*}
&\;\int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx \\
&\stackrel{y=x^2}{=}\frac{1}{2}\int_0^1 \frac{\log(1-y)}{y^{\frac{3}{4}}\sqrt{1-y}}dy \\
&= \frac{1}{2}\frac{\partial}{\partial \alpha}\left\{B\left(\frac{1}{4},\alpha \right)\right\}_{\alpha=\frac{1}{2}}\\
&= \frac{\sqrt{\pi}\Gamma\left(\frac{1}{4}\right)}{2\Gamma\left(\frac{3}{4} \right)}\left\{ \psi_0\left(\frac{1}{2} \right)-\psi_0\left(\frac{3}{4} \right)\right\} \\
&= \frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(2\log 2-\pi \right)
\end{align*}
$$
The other integral can be evaluated by using equation $(3.22)$ of this paper.
$$
\begin{align*}
&\;\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\
&\stackrel{x=\sin^2 t}{=}2\int_0^{\frac{\pi}{2}}\frac{\log(1+\sin^2 t)}{\sqrt{1+\sin^2 t}}dt\\
&= \log(2) K(\sqrt{-1}) \\
&= \log(2)\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}
\end{align*}
$$
where $K(k)$ is the complete elliptic integral of the first kind. After combining everything, one gets
$$I=\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(\log 2-\pi \right)\approx -3.20998$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/702681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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$a,b,c$ are positive real numbers such that, $a+b+c\ge abc$. Prove that $a^2+b^2+c^2\ge \sqrt{3}abc$ $a,b,c$ are positive real numbers such that, $a+b+c\ge abc$. Prove that $a^2+b^2+c^2\ge \sqrt{3}abc$
My work:
I tried using Cauchy-Schwarz inequality to find that,
$(a^2+b^2+c^2)(1^2+1^2+1^2)\ge (a+b+c)^2$
$(a^2+b^2+c^2)\ge \dfrac13(a+b+c)^2$
$\sqrt{(a^2+b^2+c^2)}\ge \dfrac{1}{\sqrt{3}}(a+b+c)\ge \dfrac{1}{\sqrt{3}}abc$
which is not what I need and neither I can use it to prove the required inequality. Please help.
| We have $a^2 + b^2 + c^2 \geq \frac{(a+b+c)^2}{3} \geq \frac{(abc)^2}{3}$ and $a^2+b^2+c^2 \geq 3 \sqrt[3]{a^2b^2c^2}$ by AM-GM. Take the $1/4$-th power of the first inequality and the $3/4$-th power of the second inequality, and multiply (this is allowed since everything is positive). The result is $a^2+b^2+c^2 \geq \sqrt{3}{abc}$.
| {
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"source": "stackexchange",
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If $a\neq 1$, find $(a+1)(a^2+1)(a^4+1)\ldots(a^{2^n}+1)$.
If $a\neq 1$, find $$(a+1)(a^2+1)(a^4+1)\ldots(a^{2^n}+1)$$
Or i.e.
If $a\neq 1$, find $\prod_{i=0}^n(a^{2^i}+1)$.
It really does seem like $$(a+1)(a^2+1)(a^4+1)\ldots(a^{2^n}+1)=a^{2^n+2^{n-1}+\ldots+2+1}+a^{2^n+2^{n-1}+\ldots+2+1-1}+a^{2^n+2^{n-1}+\ldots+2+1-2}+\ldots+a^2+a+1$$
Or i.e. $$\prod_{i=0}^n(a^{2^i}+1)=\sum_{i=0}^{2^n+2^{n-1}+\ldots+2+1}a^i$$
Is there an interesting proof of this? I'd appreciate any help.
| The "generating function" interpretation of this is that every number from $0$ to $2^n-1$ can be written in exactly one way in base $2$ number of $n$ digits or less.
So the base 3 way of writing this is that $$(1+x+x^2)(1+x^3+x^6)(1+x^9+x^{18})\dots(1+x^{3^n}+x^{2\cdot 3^n}) = \frac{x^{3^{n+1}}-1}{x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/703077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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a simple problem on limit made me confused I tried to solve this problem : $$\lim_{x \to \frac{\pi}{2}} (1+\cos x)^{\tan x}$$
if $y=(1+\cos x)^{\tan x} \implies \log y=\tan x \log(1+\cos x)$
$$\lim_{x \to \frac{\pi}{2}}\log y=\lim_{x \to \frac{\pi}{2}}[\tan x\log(1+\cos x)]=\lim_{x \to \frac{\pi}{2}}\frac{\log(1+\cos x)}{\cos x} \times \lim_{x \to \frac{\pi}{2}} \sin x$$$$=\lim_{x \to \frac{\pi}{2}} \frac{(-\sin x)}{(1+\cos x)(-\sin x)}=\lim_{x \to \frac{\pi}{2}} \frac{1}{1+\cos x}=1$$
$$\implies\log\lim_{x \to \frac{\pi}{2}}(1+\cos x)^{\tan x}=1$$
$$\implies\lim_{x \to \frac{\pi}{2}}(1+\cos x)^{\tan x}=e$$
I got answer $"e"$ but in my book it is $"e^{-1}"$. So I am confused whether I am right , please solve this, it will be definitely appreciable
| Built around $x=\frac{\pi }{2}$ , the Taylor series of $(1+\cos x)^{\tan x}$ is $$e+\frac{1}{2} e \left(x-\frac{\pi }{2}\right)-\frac{1}{24} e \left(x-\frac{\pi
}{2}\right)^2-\frac{7}{48} e \left(x-\frac{\pi
}{2}\right)^3+O\left(\left(x-\frac{\pi }{2}\right)^4\right)$$
This come from the fact that the Taylor series (built at $x=\frac{\pi }{2}$ are, respectively for $(1+\cos x)$ and $\tan x$,
$$A=1-\left(x-\frac{\pi }{2}\right)+\frac{1}{6} \left(x-\frac{\pi
}{2}\right)^3-\frac{1}{120} \left(x-\frac{\pi
}{2}\right)^5+O\left(\left(x-\frac{\pi }{2}\right)^7\right)$$
$$B=-\frac{1}{x-\frac{\pi }{2}}+\frac{1}{3} \left(x-\frac{\pi }{2}\right)+\frac{1}{45}
\left(x-\frac{\pi }{2}\right)^3+\frac{2}{945} \left(x-\frac{\pi
}{2}\right)^5+O\left(\left(x-\frac{\pi }{2}\right)^7\right)$$ Now, compose the series for $A^B$ (using logarithms make things slightly easier).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometry Identity: Prove that $\sin(a-b)=\sin a \cos b - \cos a \sin b$ First, I do not want a proof using $\sin(a+b)=\sin a \cos b + \cos a \sin b$. Second, I suspect that it has something to do with Euler's formula; $e^{ix}=\cos x + i\sin x$, but I am not sure. Can anyone give me some direction? Thanks in advance.
P.S. I apologize if this question is a duplicate. I did not find any when scrolling down the "Questions that may already have your answer" list.
| I searched for an alternate way to deriving sum of angles relations, and by congruence came up with this alternate derivation. Perhaps it is acceptable.
Take a coordinate pair say $(x,y)$ with angle $A$. We have the definitions $\sin(A)= y/(x^2+y^2)^.5 $, and $\cos(A)= x/(x^2+y^2)^.5$.
Now rotate, counterclockwise, the coordinate pair $(x,y)$ by an angle $B$ such that
$\angle A + \angle B = \angle C$
After rotation $(x,y)$ becomes $(x\cos B-y\sin B, x\sin B+y\cos B)$. (I'm leaving out a derivation for rotation since it is everywhere, See Rotate the graph of a function?). Since the angle of the rotated pair $(x,y)$ is $\angle C$ the following equations follow:
$$\cos(C)=\frac{x\cos B-y\sin B}{((x\cos B-y\sin B)^2+(x\sin B+y\cos B)^2)^.5}$$
Notice that:
$$(x\cos B-y\sin B)^2+(x\sin B+y\cos B)^2)= (x^2+y^2)$$
Hence,
$$\cos(C)=\frac{x\cos B-y\sin B}{(x^2+y^2)^.5}$$
And
$$\cos(A+B)= {\cos(A)\cos(B)}-{\sin(A)\sin(B)}$$
Other Relations--------------
$$\sin(C)=\frac{x\sin B+y\cos B}{(x^2+y^2)^.5} = {\cos(A)\sin(B)}+{\sin(A)\cos(B)}$$
$$\sin(A+B)= {\cos(A)\sin(B)}+{\sin(A)\cos(B)}$$
We also have
$$A\tan(y/x)= \angle A$$
and
$$A\tan\frac{x\sin B+y\cos B}{x\cos B-y\sin B}= \angle C$$
$$ \angle B +A\tan(y/x)= A\tan\frac{x\sin B+y\cos B}{x\cos B-y\sin B}$$
$$ \angle B +A\tan(y/x)= A\tan\frac{\tan B+y/x}{1-(y/x)\tan B}$$
Or, for example, by setting $Tan B = u$ and $ y/x = v$
$$ A\tan(u) +A\tan(v)= A\tan\frac{u+v}{1-uv}$$
For the tangent is straight forward.
$$tan(A + B) = \frac{x\sin B+y\cos B}{x\cos B-y\sin B}$$
$$tan(A + B) = (tan B + tan A) / (1 − tan A *tan B)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Calculating the determinant gives $(a^2+b^2+c^2+d^2)^2$? I need to calculate the following determinant in order to prove the following equality:
$$\det\begin{pmatrix} a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{pmatrix} = (a^2+b^2+c^2+d^2)^2.$$
I tried using Gauss-algorithm to get an easier matrix, but I'm not sure if I did it correctly.
Calling the $4$ lines $I$, $II$, $III$ and $IV$, I did:
(1) $II \cdot a$
(2) $III \cdot a$
(3) $IV \cdot a$
After this I did:
(4) $II' + I \cdot b$
(5) $III' + I \cdot c$
(6) $IV' + I \cdot d$
So finally I got the following matrix:
$$\begin{pmatrix} a & b & c & d \\ 0 & a^2+b^2 & bc-ad & ac+bd \\ 0 & ad+bc & a^2+c^2 & cd-ab \\ 0 & bd-ac & ab-cd & a^2+d^2 \end{pmatrix}.$$
I thought this would make the determinant a bit easier, unfortunately I must have done something wrong. Is multiplication with single lines allowed as I have done it?
Thank you very much.
| Notice that the rows are perpendicular to each other and the length of the rows are all the same.If you call the matrix $M$ then
$\det(M^tM)=\det(M^t)\det(M)=\det(M)^2$ But $$
M^tM=
(a^2+b^2+c^2+d^2)\left(
\begin{array}{cccc}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{array}
\right)$$
Therefor $\det(M^tM)=(a^2+b^2+c^2+d^2)^4\Rightarrow \det(M)=\pm(a^2+b^2+c^2+d^2)^2$
Set $N=\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}M$ then all the rows of $N$ have length $1$. Further $N$ is completely determined by its first row. Now we have $\det(N)=\pm 1$
. Notice that the determinant gives a smooth function, $$\det:S^3\to \{-1,1\} $$ By sending $(a,b,c,d)$ to our matrix and then to the determinant of this matrix. (This is smooth because it is a polynomial function). But the point $(1,0,0,0)\in S^3$ is mapped to $1$. Now we can conclude that because $S^3$ is connected, $\{-1,1\}$ is discrete, and our function is continuous, that the entire $S^3$ is mapped to $1$. Thus $\det(N)=1$ and also $\det(M)=(a^2+b^2+c^2+d^2)^2$
| {
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"source": "stackexchange",
"question_score": "7",
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Operations with surds (Pre-calculus) I have tried multiple ways to solve this:
$$ \left (2\sqrt 8 - 3\sqrt 5 \right)^2 $$
(original screenshot)
and my answer was way off. Can someone show me their method? Thanks.
| Remember the formula:
$$(a-b)^2=a^2-2ab+b^2$$
Apply it to your expression $(2\sqrt{8}-3\sqrt{5})^2$. But first, simplify $2\sqrt{8}$ to $4\sqrt{2}$. Now we have to find the value of:
$$(4\sqrt{2}-3\sqrt{5})^2$$
Use the $(a-b)^2=a^2-2ab+b^2$ formula.
$$a=4\sqrt{2}$$
$$b=3\sqrt{5}$$
$$(4\sqrt{2}-3\sqrt{5})^2=(4\sqrt{2})^2-2(4\sqrt{2})(3\sqrt{5})+(3\sqrt{5})^2$$
Remember, $(ab)^2=a^2b^2$.
$$(4\sqrt{2})^2-2(4\sqrt{2})(3\sqrt{5})+(3\sqrt{5})^2=4^2(\sqrt{2})^2-2(4\sqrt{2})(3\sqrt{5})+3^2(\sqrt{5})^2$$
$$4^2(\sqrt{2})^2-2(4\sqrt{2})(3\sqrt{5})+3^2(\sqrt{5})^2=16(2)-2(4\sqrt{2})(3\sqrt{5})+9(5)$$
$$16(2)-2(4\sqrt{2})(3\sqrt{5})+9(5)=32-2(4\sqrt{2})(3\sqrt{5})+45$$
$$32-2(4\sqrt{2})(3\sqrt{5})+45=77-2(4\sqrt{2})(3\sqrt{5})$$
Now we will simplify $2(4\sqrt{2})(3\sqrt{5})$.
$$77-2(4\sqrt{2})(3\sqrt{5})=77-2(4)(3)(\sqrt{2})(\sqrt{5})$$
Rememeber. $\sqrt{a}\cdot\sqrt{b}$.
$$77-2(4)(3)(\sqrt{2})(\sqrt{5})=77-24\sqrt{2\cdot 5}$$
$$77-24\sqrt{2\cdot 5}=77-24\sqrt{10}$$
$$\displaystyle \boxed{\therefore \left(2\sqrt{8}-3\sqrt{5}\right)^2=77-24\sqrt{10}}$$
| {
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"url": "https://math.stackexchange.com/questions/706678",
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"source": "stackexchange",
"question_score": "2",
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Limits of Indeterminate Quotient: $\lim\limits_{x \to 0^+} \frac{\ln(e^x - 1)}{\ln(x)}$ and $\lim\limits_{x \to -1}(\frac1{x+1} - \frac3{x^3+1})$ I was preparing for my exam and found myself struggling with finding limits of indeterminate quotient.
$$\lim\limits_{x \to 0^+} \dfrac{\ln(e^x - 1)}{\ln(x)}$$
I have tried using L'Hopital's Rule to reduce it to:
$\lim\limits_{x \to 0^+} \dfrac{xe^x}{e^x-1}$
but still does not solve the problem.
Another problem that I've faced:
$$\lim\limits_{x \to -1}(\frac{1}{x+1} - \frac{3}{x^3+1})$$
I have tried to combine it into 1 term:
$\lim\limits_{x \to -1}(\dfrac{x^3-3x-2}{x^4+x^3+x+1})$
and applied L'Hopital's Rule but still got an Indeterminate Quotient.
Any advice on the 2 above qns is really much appreciated!
| For the second limit you can also apply algebra.
$\displaystyle
\begin{align}
\lim_{x \to -1}\left(\frac{1}{x+1} - \frac{3}{x^3+1}\right) &= \lim_{x \to -1}\left(\frac{1}{x+1} - \frac{3}{(x+1)(x^2-x+1)}\right) \\
&= \lim_{x \to -1}\left(\frac{1}{x+1} - \left(\frac{1}{x+1}+\frac{2-x}{x^2-x+1}\right)\right) \\
&= \lim_{x \to -1}\left(\frac{x-2}{x^2-x+1}\right)
\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the sum of $i(i+1)\cdots(i+k)$ for fixed $k$ between $i = 1$ and $n$? I learned that $$\sum \limits_{i=1}^n i(i+1) = \frac{n(n+1)(n+2)}{3}$$ or in general $$\sum \limits_{i = 1}^n i(i+1)(i+2) \dots (i + k) = \frac{n(n+1)\dots (n+k+1)}{k+2}$$ From a mathematical standpoint why is this true? I'm not asking for inductive proof. I am asking if you only given the left hand side, how would you go about writing a closed form expression for the sum?
| $$
S=\sum \limits_{i = 1}^n i(i+1)(i+2) \dots (i + k) = \sum_{i=1}^{n}\frac{(i+k)!}{(i-1)!}
$$
$$
\frac{S}{(k+1)!}=\sum_{i=1}^{n}\frac{(i+k)!}{(i-1)!(k+1)!}=\sum_{i=0}^{n-1}\binom{i+k+1}{i}
$$
$$
\frac{S}{(k+1)!}=\binom{k+1}{0} + \binom{k+2}{1} + \dots + \binom{n+k}{n-1} \\
=\binom{k+2}{0} + \binom{k+2}{1} + \dots + \binom{n+k}{n-1} \\
=\binom{k+3}{1} + \binom{k+3}{2} + \dots + \binom{n+k}{n-1} \\
=\binom{n+k+1}{n-1}
$$
the above uses $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$
$$
S=\binom{n+k+1}{n-1}*(k+1)!=\frac{(n+k+1)!(k+1)!}{(n-1)!(k+2)!}= \frac{n(n+1)\dots (n+k+1)}{k+2}
$$
| {
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How does $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$? Good afternoon my wonderful friends!
Whenever I do this equation I set it up using the difference of two cubes, which is as follows:
$(a+b)^3 = (a+b)(a^2-ab+b^2) = a^3 + b^3$
Whenever I try to use this formula I always get:
$(x+h) x^2 - xh + h^2$
Simplify:
$x^3 - x^2h + h^2x + x^2h - xh^2 + h^3$
Simplify:
x^3 + h^3$
I don't understand were the three's come from in the final answer: $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$
| $$(x+h)^3=(x+h)(x+h)^2=(x+h)(x^2+2hx+h^2)=$$
$$x^3+2hx^2+h^2x+hx^2+2h^2x+h^3=$$$$x^3+3hx^2+3h^2x+h^3$$
There are quicker ways to do this - the binomial theorem, for example - but I thought that longhand might help you to see a bit better what is going on. Another view: In
$$(x+h)^3=(x+h)(x+h)(x+h)$$ The eventual term in $h$ can get the $h-$factor from any of the three brackets - hence the factor $3$. There is only one way to take an $x$ from each bracket, hence the coefficient of $x^3$ is $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I get an exact value for the trigonometric expression? I'm trying find an exact value for $$\cos\left(\frac{1}{3}\arctan\left(\frac{-10}{9\sqrt{3}}\right)\right)$$
Evaluating $\cos(\arctan(\frac{-10}{9\sqrt{3}}))$ is straighforward, but I'm having trouble with the above expression. I plugged it into Wolfram Alpha and it returned $3/2 \sqrt{\frac{3}{7}}$.
Any help would be appreciated.
| Recall the inverse tangent identity $$\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x+y}{1-xy}.$$ So in particular, $$2 \tan^{-1}x = \tan^{-1} \frac{2x}{1-x^2},$$ and $$3 \tan^{-1} x = \tan^{-1} \frac{x(3-x^2)}{1-3x^2}.$$ Consequently, we wish to find $x$ such that $$\frac{x(3-x^2)}{1-3x^2} = -\frac{10}{9 \sqrt{3}},$$ which is equivalent to $$0 = 9 \sqrt{3} x^3 + 30 x^2 - 27 \sqrt{3} x - 10.$$ This equation happens to factor, giving the solutions $$x = -\frac{5}{\sqrt{3}}, -\frac{1}{3 \sqrt{3}}, \frac{2}{\sqrt{3}}.$$ Consequently, $$\cos \Bigl( \frac{1}{3} \tan^{-1} \frac{-10}{9 \sqrt{3}} \Bigr) = \frac{1}{2} \sqrt{\frac{3}{7}}, \frac{3}{2} \sqrt{\frac{3}{7}}, \sqrt{\frac{3}{7}}.$$ Which of these are correct? The second one is correct if we take $\tan^{-1} x \in (-\tfrac{\pi}{2}, \tfrac{\pi}{2})$. The first is correct if we take $\tan^{-1} x \in (-\tfrac{3\pi}{2}, -\tfrac{\pi}{2})$. The third is correct if we take $\tan^{-1} x \in (\tfrac{\pi}{2}, \tfrac{3\pi}{2})$. In other words, all three are correct values, but they correspond to three different branches of the inverse tangent function: this arises from the $\frac{1}{3}$ term.
| {
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"timestamp": "2023-03-29T00:00:00",
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Value of this definite integral $\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $ So I came across this question on brilliant.org and didn't know how to go about it:
$$\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $$
I tried to complete the squares in the denominator and then use a trigonometric substitution (with tan) but I didn't get anything beyond that.
How do I do it? Do we need to differentiate under the integral sign here?
| As mentioned above, use a keyhole contour in the complex plane, which we'll call $C$. Here though you should consider the contour integral
$$\oint_C dx \frac{\log^2{z}}{z^2+2 z+4} $$
I will leave it to the reader to show that the integral vanishes over the circular portions of $C$; we are thus left with the contour integral equaling
$$\int_0^{\infty} dx \frac{\log^2{x}-(\log{x}+i 2 \pi)^2}{x^2+2 x+4} = -i 4 \pi \int_0^{\infty} dx\frac{\log{x}}{x^2+2 x+4} + 4 \pi^2 \int_0^{\infty} \frac{dx}{x^2+2 x+4}$$
Now, the integral on the right we need to know as well. We may evaluate that any way we wish, but let's stick to this formalism. This integral may be evaluated using the same contour and the residue theorem. Let
$$R_n = \sum_k \operatorname*{Res}_{z=z_k} \frac{\log^{n+1}{z}}{z^2+2 z+4}$$
for $n=0,1$. That is,
$$\oint_C dz \frac{\log^{n+1}{z}}{z^2+2 z+4} = i 2 \pi R_n$$
Then from the above, you may show that
$$\int_0^{\infty} dx\frac{\log{x}}{x^2+2 x+4} = i \pi R_0 - \frac12 R_1$$
Note that the poles of the integrands are at $z_1=2 e^{i 2 \pi/3}$ and $z_2 = 2 e^{i 4 \pi/3}$. Note that we must use these arguments because of how we defined the contour.
We mat deduce that
$$R_0 = \frac{\log{2}+i 2 \pi/3}{i 2 \sqrt{3}} + \frac{\log{2}+i 4 \pi/3}{-i 2 \sqrt{3}} = -\frac{\pi}{3 \sqrt{3}}$$
$$R_1 = \frac{(\log{2}+i 2 \pi/3)^2}{i 2 \sqrt{3}} + \frac{(\log{2}+i 4 \pi/3)^2}{-i 2 \sqrt{3}} = -\frac{2 \pi \log{2}}{3 \sqrt{3}} - i \frac{2 \pi^2}{3 \sqrt{3}}$$
Putting it all together, the imaginary pieces cancel and we get
$$\int_0^{\infty} dx\frac{\log{x}}{x^2+2 x+4} = \frac{\pi \log{2}}{3 \sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/709516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
} |
Find a second degree polynomial that goes through 3 points I am having trouble calculating the quadratic curve $f(x)$ that goes through 3 points;
for example a curve that goes through $A(1,3), B(-1,-5), and C(-2,12)$.
I can only guess that the curve is upwards and that I may create the system:
$$
y_1 = ax^2_1 + bx_1 + c\\
y_2 = ax^2_2 + bx_2 + c\\
y_3 = ax^2_3 + bx_3 + c
$$
assuming that the points are in the format $A(x,y)$ and from this point what do I do?
Do I build a matrix and use the Gaussian eliminations?
EDIT: I also know that $f(4) = 120$
| Each of the points (1,3), (-1,-5) and (-2,12) satisfies the equation $y = ax^2 + bx + c$ for some unknown a,b,c. The task is to find a,b and c. Start by substituting each of the points into the equation, we have
$$
\begin{align}
3 &= a(1)^2 + b(1) + c \\
-5 &= a(-1)^2 + b(-1) + c \\
12 &= a(-2)^2 + b(-2) + c
\end{align}$$
We can write this more compactly as a matrix equation
$$
\begin{bmatrix} 1 & 1 & 1\\ 1 & -1 & 1\\ 4 & -2 & 1
\end{bmatrix} \begin{bmatrix}a\\ b\\ c \end{bmatrix} = \begin{bmatrix} 3\\ -5\\ 12\end{bmatrix}
$$
Write the augmented matrix and do elementary row operations
$$
\begin{bmatrix} 1 & 1 & 1 & 3\\ 1 & -1 & 1& -5\\ 4 & -2 & 1 & 12
\end{bmatrix} \Rightarrow \begin{bmatrix} 1 & 1 & 1 & 3\\ 0 & -2 & 0 & -8\\ 0 & -6 & -3 & 0
\end{bmatrix} \Rightarrow \begin{bmatrix} 1 & 1 & 1 & 3\\ 0 & -2 & 0 & -8\\ 0 & 0 & -3 & 24
\end{bmatrix}
$$
and now, back substitution.
starting with the last row,
$$\begin{align}-3c &= 24 \\ c &= -8 \end{align} $$
and then the second row
$$\begin{align} -2b &= -8 \\ b &= 4 \end{align}$$
and finally back substituting these into the first row
$$\begin{align}a + b + c &= 3 \\ a + (4) + (-8) &= 3 \\ a &= 7\end{align}$$
So, I think the equation is:
$$ y = 7x^2 + 4x -8 $$
Please check my work, I did this in a hurry.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/710750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to algebraically prove $\binom{n+m}{2} = nm + \binom{n}{2} + \binom{m}{2}$? Need help trying to prove this problem algebraically.
$$\binom{n+m}{2} = nm + \binom{n}{2} + \binom{m}{2}$$
The farthest I've got is simplifying the RHS to $$nm + \frac{n(n-1)}{2!} + \frac{m(m-1)}{2!}$$
but not sure what to do after that.
| By definition,
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
Hence,
\begin{align*}
nm + \binom{n}{2} + \binom{m}{2} & = nm + \frac{n!}{2!(n - 2)!} + \frac{m!}{2!(m - 2)!}\\
& = nm + \frac{n(n - 1)(n - 2)!}{2 \cdot 1 \cdot (n - 2)!} + \frac{m(m - 1)(m - 2)!}{2 \cdot 1 \cdot (m - 2)!}\\
& = nm + \frac{n(n - 1)}{2} + \frac{m(m - 1)}{2}\\
& = \frac{2nm + n(n - 1) + m(m - 1)}{2}\\
& = \frac{2nm + n^2 - n + m^2 - m}{2}\\
& = \frac{n^2 + 2nm + m^2 - n - m}{2}\\
& = \frac{(n + m)^2 - (n + m)}{2}\\
& = \frac{(n + m)(n + m - 1)}{2}\\
& = \frac{(n + m)(n + m - 1)(n + m - 2)!}{2(n + m - 2)!}\\
& = \frac{(n + m)!}{2!(n + m - 2)!}\\
& = \binom{n + m}{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/711311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Derivation of factorization of $a^n-b^n$ How does one prove that:
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\dots+a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$$
Better yet, why is $a^n-b^n$ divisible by $a-b$? I would very much appreciate some help on this. Thanks
| I'm not sure why this never occurred to me before, but looking at your general formula written out in full, I thought, "Where have we seen something like this before?" We can write
$$ a^n \ - \ b^n \ = \ a^n \ \left[ \ 1 \ - \ \left( \frac{b}{a} \right)^n \ \right] $$
$$ = \ \left[ \ 1 \ - \ \left( \frac{b}{a} \right) \ \right] \ \sum_{k=0}^{n-1} \ a^n \cdot \left( \frac{b}{a} \right)^k $$
$$ = \ \left[ \ 1 \ - \ \left( \frac{b}{a} \right) \ \right] \ \cdot \ a \ \cdot \ a^{n-1} \ \cdot \ \left( \ 1 \ + \ \frac{b}{a} \ + \ \frac{b^2}{a^2} \ + \ \ldots \ + \ \frac{b^{n-1}}{a^{n-1}} \ \right) $$
$$ = \ (a - b) \ ( \ a^{n-1} \ + \ a^{n-2}b \ + \ a^{n-3}b^2 \ + \ \ldots \ + \ b^{n-1} \ ) \ \ . $$
This also suggests a variant proof analogous to the usual one for the "finite geometric series formula":
$$ s \ = \ 1 \ + \ \frac{b}{a} \ + \ \frac{b^2}{a^2} \ + \ \ldots \ + \ \frac{b^{n-1}}{a^{n-1}} $$
$$ \underline{- \ \ \left(\frac{b}{a} \right) \ s \ = \ \frac{b}{a} \ + \ \frac{b^2}{a^2} \ + \ \ldots \ + \ \frac{b^{n-1}}{a^{n-1}}\ + \ \frac{b^n}{a^n} }$$
$$ \left[ \ 1 \ - \ \left(\frac{b}{a} \right) \ \right] \ s \ = \ 1 \ - \ \frac{b^n}{a^n} \ , $$
with multiplication of both sides by $ \ a^n \ $ carried through in the same manner as in the derivation above:
$$ a \ \left[ \ 1 \ - \ \left(\frac{b}{a} \right) \ \right] \ \cdot \ a^{n-1} \ \cdot \ s \ = \ a^n \ - \ b^n \ \ . $$
[This is related to the expression Git Gud uses, but this is a proof without the use of induction.]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/712758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 4
} |
e is irrational Prove that e is an irrational number.
Recall that $\,\mathrm{e}=\displaystyle\sum_{n=0}^\infty\frac{1}{n!},\,\,$ and assume $\,\mathrm{e}\,$ is rational, then
$$\sum\limits_{k=0}^\infty \frac{1}{k!} = \frac{a}{b},\quad \text{for some positive integers}\,\,\, a,b.$$
so
$$b\sum\limits_{k=0}^\infty \frac{1}{k!} =a$$
or
$$
b\left(1+1+\frac{1}{2} + \frac{1}{6} +\cdots \right)= a.
$$
Where can I go from here?
| Here is yet another one, which is one of my favorite irrationality/transcendence proofs :
The confluent hypergeometreic series
$$_{0}F_{1}(k; z) = \sum_{n = 0}^\infty \frac1{(k)_n} \frac{z^n}{n!}$$
Satisfies the more-or-less easily verifiable identity
$$_0F_1(k-1;z) - {}_0F_1(k; z) = \frac{z}{k(k-1)}{}_0F_1(k+1;z)$$
Iterating this, one ends up with the continued fraction
$$\frac{{}_0F_1(k+1;z)}{k{}_0F_1(k;z)} = \frac1{k+\cfrac{z}{k+1+\cfrac{z}{k+2+\cfrac{z}{k+3+\cdots}}}}$$
Now note that $_0F_1(3/2;x^2/4) = \cosh(x)$ and $x \,{}_0F_1(3/2;x^2/4) = \sinh(x)$, hence applying the above one has the pretty well-known continued fraction
$$\tanh(x) = \cfrac{x/2}{\frac{1}{2} + \cfrac{\frac{x^2}{4}}{\frac{3}{2} + \cfrac{\frac{x^2}{4}}{\frac{5}{2} + \cfrac{\frac{x^2}{4}}{\frac{7}{2} + \cdots}}}} = \cfrac{x}{1 + \cfrac{x^2}{3 + \cfrac{x^2}{5 + \cfrac{x^2}{7 + \cdots}}}}$$
Note, however, that $$\tanh(x) = \frac{\exp(x) - \exp(-x)}{\exp(x) + \exp(-x)}$$ but since the continued fraction above is not finite, $\tanh(1)$ is not rational. Hence $e$ is not rational either.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/713467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 3
} |
prove that there are no reals such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{x+y}$ I'm going through an introduction to complex analysis and there are two problems
that I'm having problems with.
A) Prove that there are no reals $x$ and $y$ such that $\displaystyle\frac{1}{x} + \frac{1}{y} =
\frac{1}{x+y}$.
B) Prove that there are complex numbers $z$ and $w$ such that $\displaystyle\frac{1}{z} + \frac{1}{w} = \frac{1}{z+w}$.
Note that this is not homework. I'm just trying to get a better understanding of how to solve these types of problems. Thanks.
| A somewhat (though perhaps only slightly) different approach:
I assume $x, y, x + y \ne 0$, so that the division operations are all well-defined. Then the equation
$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{x + y} \tag{1}$
becomes, upon multiplication by $x + y$,
$\dfrac{x + y}{x} + \dfrac{x + y}{y} = 1, \tag{2}$
or, after a little turning of the algebra crank,
$2 + \dfrac{x}{y} + \dfrac{y}{x} = 1, \tag{3}$
or
$\dfrac{x}{y} + \dfrac{y}{x} + 1 = 0. \tag{4}$
Now set
$\omega = \dfrac{x}{y} \tag{5}$
so that (4) becomes
$\omega + \omega^{-1} + 1 = 0 \tag{6}$
or
$\omega^2 + \omega + 1 = 0. \tag{7}$
The roots of (7) are
$\omega = \dfrac{1}{2}(-1 \pm i\sqrt 3); \tag{8}$
by (5), $x = \omega y$, so $x$ and $y$ cannot both be real. Thus is item (A) resolved. As for item (B), let $0 \ne z \in \Bbb C$ and set $w = \omega z$. Then from (6),
$\dfrac{w}{z} + \dfrac{z}{w} + 1 = 0 \tag{9}$
or
$\dfrac{w}{z} + \dfrac{z}{w} + 2 = 1 \tag{10}$
or
$(\dfrac{w}{z} + 1) + (\dfrac{z}{w} +1) = 1 \tag{11}$
or
$\dfrac{w + z}{z} + \dfrac{w + z}{w} = 1. \tag{12}$
Note that
$w + z = \omega z + z = (1 + \omega)z \ne 0, \tag{13}$
since $1 + \omega \ne 0$; we can thus divide (12) by $w + z$, yielding
$\dfrac{1}{z} + \dfrac{1}{w} = \dfrac{1}{z + w}. \tag{14}$
And that takes care of item (B). QED.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
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