Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Check proof that $\prod_{k=1}^n(1+{1\over a_k})$ is bounded if $a_{n+1} = (n+1)(a_n + 1)$ and $a_1 = 1$, $n\in \mathbb N$
Let $n \in \mathbb N$ and:
$$
\begin{cases}
a_1 = 1 \\
a_{n+1} = (n+1)(a_n + 1)
\end{cases}
$$
Prove that
$$
x_n = \prod_{k=1}^n\left(1+{1\over a_k}\right)
$$
is a bounded sequence.
Obviously $x_n > 0$. So suppose $x_n$ has an upper bound $M$:
$$
0 < x_n < M
$$
Inspecting $a_n$ one may see that the sequence is divergent and $a_n \le n$, so:
$$
\begin{align}
a_{n+1} &> a_n \iff \\
\iff {1\over a_{n+1}} &< {1 \over a_n}
\end{align}
$$
and
$$
\begin{align}
a_n &\ge n \iff \\
\iff {1\over a_n} &\le {1\over n}
\end{align}
$$
Therefore:
$$
1 + {1 \over a_n} \le 1 + {1\over n}
$$
Now lets compare the products:
$$
\prod_{k=1}^n\left(1 + {1 \over a_k}\right) \le \prod_{k=1}^n\left(1 + {1 \over n}\right)
$$
RHS is the definition of $e$ which may be proved to be bounded in various ways so i will not put it here.
So finally we have:
$$
0 < \prod_{k=1}^n\left(1 + {1 \over a_k}\right) \le \prod_{k=1}^n\left(1 + {1 \over n}\right) < 3 \\
0 < x_n < 3
$$
Which proves that $x_n$ is bounded.
Is the above a valid proof?
|
Observe that $b_{n+1} = \dfrac{a_{n+1}}{(n+1)!}= \dfrac{a_n}{n!} + \dfrac{1}{n!}= b_n+\dfrac{1}{n!}\implies b_n = (b_n - b_{n-1}) + (b_{n-1} - b_{n-2})+\cdots+(b_2 - b_1)+ b_1 = \dfrac{1}{(n-1)!}+\dfrac{1}{(n-2)!}+\cdots + \dfrac{1}{1!} + 1\implies a_n
= n!b_n = n + n(n-1) + n(n-1)(n-2)\cdots + n(n-1)(n-2)\cdot\cdot\cdot1+1\implies x_n = \dfrac{a_{n+1}}{(n+1)!} = b_{n+1}= \dfrac{1}{n!}+\dfrac{1}{(n-1)!}+\cdots+\dfrac{1}{1!}+1< \dfrac{1}{n(n-1)}+\dfrac{1}{(n-1)(n-2)}+\cdots + 1 = \dfrac{1}{n-1}- \dfrac{1}{n}+\dfrac{1}{n-2}-\dfrac{1}{n-1}+\dfrac{1}{n-3}-\dfrac{1}{n-2} +\cdots+\dfrac{1}{2} - \dfrac{1}{3}+\dfrac{1}{1} - \dfrac{1}{2}+1 = 2 - \dfrac{1}{n} < 2$. Thus $x_n < 2, \forall n \ge 1$, hence is bounded.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2959680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Partial Fraction of $\int \frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)}$ If
$$\int\frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)} = \int \frac{A \ \mathrm{d}x}{(\sin x)} + B \int\frac{\sin x \ \mathrm{d}x}{ 1 + \sin^2 x} + C \int \frac{\mathrm{d}x}{1 + \sin^2 x}$$
then, which of the following is correct?
*
*$A+B+C=4$
*$A+B+C=2$
*$A+BC=1$
*$A+B+C=5$
My approach is as follows:
$$\int\frac{\left( \cos x + \sin 2x \right) \ \mathrm{d}x}{(2 - \cos^2 x)(\sin x)}= \int \frac{\left( \cos x + \sin 2x \right) \ \mathrm{d}x}{(1 + \sin^2 x)(\sin x)} =$$
$$= \int \frac{\cos x \ \mathrm{d}x}{( 1 + \sin^2 x)(\sin x)} + \int \frac{\sin 2x \ \mathrm{d}x}{(1 + \sin^2 x )(\sin x)} =$$
$$= \int \frac{ \cos x \ \mathrm{d}x}{(1 + \sin^2 x )(\sin x)} + \int \frac{ 2 \cos x \ \mathrm{d}x}{(1 + \sin^2 x )} =$$
$$= \int\frac{- \sin \cos x \ \mathrm{d}x}{ 1 + \sin^2 x } + \int\frac{\cos x \ \mathrm{d}x}{ \sin x } + \int \frac{2 \cos x \ \mathrm{d}x}{1 + \sin^2 x}$$
After this step I am not able to approach
|
The integrand can be written
$$
\frac{\cos x+\sin2x}{(1+\sin^2x)\sin x}
$$
Let's try and determine $A$, $B$ and $C$ so that
$$
\frac{A}{\sin x}+\frac{B\sin x}{1+\sin^2x}+\frac{C}{1+\sin^2x}
$$
is the same. Removing the denominators this forces
$$
A(1+\sin^2x)+B\sin^2x+C\sin x=\cos x+\sin2x
$$
If we set $x=0$, we get $A=1$; with $x=\pi/2$, we get $2A+B+C=0$; with $x=-\pi/2$ we get $2A+B-C=0$.
The linear system
\begin{cases}
A=1\\
2A+B+C=0\\
2A+B-C=0
\end{cases}
has the solution $A=1$, $B=-2$, $C=0$.
Hence $A+BC=1$. However, the two functions are not equal as you can check at $\pi/4$.
In order to compute the integral, one has rather to find $A$, $B$ and $C$ such that
$$
\frac{1+2\sin x}{(1+\sin^2x)\sin x}=
\frac{A}{\sin x}+\frac{B\sin x}{1+\sin^2x}+\frac{C}{1+\sin^2x}
$$
which is possible and equivalent to solving for partial fractions
$$
\frac{1+2u}{u(1+u^2)}=\frac{A}{u}+\frac{Bu+C}{1+u^2}
$$
This certainly has a solution and allows to compute your integral with $u=\sin x$.
This translates into $A+Au^2+Bu^2+Cu=1+2u$, so
\begin{cases}
A+B=0\\
C=2\\
A=1
\end{cases}
so $A=1$, $B=-1$ and $C=2$.
With this fix, the right answer would be $A+B+C=2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2960467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Integral $\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$ A while ago I encountered this integral $$I=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$$ To be fair I spent some time with it and solved it in a heuristic way, I want to avoid that way so I won't show that approach, but the result I got is $\frac{\pi}{6} \ln(2+\sqrt 3)$ so I bet it can be shown in a nice way.
Solving other integrals I also encountered this one: $$J=\int_0^\infty \frac{\ln(1+x^2+x^4)}{1+x^2}dx=\pi \ln(2+\sqrt 3)$$ Which is pretty easy to compute, so most of my time I tried to show that $J=6I$, however without explictly evaluating them I had no luck.
Also I tried to use partial fractions: $$I=\frac12 \left(\int_0^1 \frac{\ln(1+x)}{x^2+\sqrt 3x +1}dx - \int_0^1 \frac{\ln(1+x)}{x^2-\sqrt 3 x+1}dx\right) $$
Considering: $$K(t) =\int_0^1 \frac{\ln(1+x)}{x^2-2\cos(t)x+1}dx$$
We have that $I=\frac12 \left(K\left(\frac{5\pi}{6}\right)-K\left(\frac{\pi}{6}\right)\right) $ and since: $$\frac{\sin t}{x^2-2x\cos t+1}=\frac{1}{2i}\left(\frac{e^{it}}{1-xe^{it}}-\frac{e^{-it}}{1-xe^{-it}}\right)=\Im \left(\frac{e^{it}}{1-xe^{it}}\right)=$$
$$=\sum_{n=0}^{\infty} \Im\left(x^n e^{i(n+1)t}\right)=\sum_{n=0}^\infty x^n\sin((n+1)t)$$
$$\small \Rightarrow K(t)=\frac12 \left(\frac{1}{\sin \left(\frac{5\pi}{6}\right)}\sum_{n=0}^\infty \sin\left(\frac{5\pi}{6} (n+1)t \right) + \frac{1}{\sin \left(\frac{\pi}{6}\right)}\sum_{n=0}^\infty \sin\left(\frac{\pi}{6} (n+1)t \right)\right)\int_0^1 x^n \ln(1+x)dx$$
$$=2\sum_{n=0}^\infty \sin\left(\frac{\pi}{6} (n+1)t \right)\int_0^1 x^n \ln(1+x)dx$$
Now I don't know how to deal with the integral and the sum combined.
|
Integrate by parts
\begin{align}
I&=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx
= \int_0^1 \frac{1}{1+x} \cot^{-1}\frac x{1-x^2}dx
\end{align}
Let $J(a) =\int_0^1 \frac{1}{1+x} \cot^{-1}\frac {2x\sin a}{1-x^2}\ dx$
\begin{align}
J’(a) &= \int_0^1 \frac{2\cos a \ (x^2-x)}{(x^2+1)^2-(2x\cos a)^2}dx
=\frac12\left( a \ln\tan\frac a2\right)’
- \frac\pi4\tan\frac a2
\end{align}
Then, with $J(0)= \frac\pi2 \int_0^{1}\frac {1}{1+x}dx=\frac\pi2\ln2$
\begin{align}
I&= J\left(\frac\pi6\right)= J(0)+\int_0^{\frac\pi6}J’(a)da \\
&= \frac\pi2 \ln2-\frac\pi4 \int_0^{\frac\pi6}\tan\frac a2 da+\frac12\int_0^{\frac\pi6} d\left( a\ln\tan\frac a2\right)\\
&=\frac\pi2\ln2+\frac\pi2\ln\cos\frac\pi{12}+\frac\pi{12}\ln\tan\frac\pi{12}
= \frac\pi6\ln(2+\sqrt3)
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2961277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
}
|
Is this induction proof mathematically correct? Proof that $\frac{n^3}{3} < 3n-3$ is true for $n=2$ but false for every other $ n \in \mathbb{N}$.
Idea is to proof that $\frac{n^3}{3} \geq 3n-3$.
Let $n$ be $n=3$
$\frac{27}{3} \geq 9-3 $.
That means $\exists n \in \mathbb{N}$ such that $\frac{n^3}{3} \geq 3n-3$ and therefore the same for $n = n+1$.
We have to proof that $\frac{n^3}{3} >= 3n-3$ for $n=n+1$ which means $\frac{(n+1)^3}{3} \geq 3(n+1)-3 = \frac{(n+1)^3}{3} \geq 3n$
We have $\frac{(n+1)^3}{3} = \frac{n^3+3n^2+3n+1}{3} = \frac{n^3}{3} + \frac{3n^2+3n+1}{3}$
Now we use that $\frac{n^3}{3} \geq 3n-3$.
$\frac{3n-3}{3} + \frac{3n^2+3n+1}{3} = \frac{3n^2+6n-2}{3} = n^2+2n-\frac23$
Can I now just say that $\forall n > 3; n \in \mathbb{N}$
$n^2+2n-\frac23 > 3n$
is true und therefore our statement $\frac{n^3}{3} < 3n-3$ is proven?
Thanks to everyone who takes some time to help me!
|
Dont write $n=n+1$ you can write $n\mapsto n+1$.
The typical way to estimate from $n^2+2n-\frac23$ is to use, that $n\geq 3$.
This helps dealing with the quadratic term. The rest is simple.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2961411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$ Prove that $$\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$$
Hence, show that $$(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5})^5+i(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5})^5=0.$$
In the first part, I tried realising LHS so I got $$LHS=\frac{(1+\sin\theta+i\cos\theta)^2}{(1+\sin\theta)^2+\cos^2\theta}=\frac{1+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta+\sin^2\theta+i^2\cos^2\theta}{2+2\sin\theta}.$$
but now I am stuck :( . Any help would be greatly appreciated, thanks!
|
Replace $\theta$ with $90^\circ-2y$
$$F=\dfrac{1+\cos2y+i\sin2y}{1+\cos2y-i\sin2y}$$
Using Double angle formula,
$$F=\dfrac{2\cos^2y+2i\sin y\cos y}{2\cos^2y-2i\sin y\cos y}$$
If $\cos y\ne0$
$$F=\dfrac{\cos y+i\sin y}{\cos y-i\sin y}=(\cos y+i\sin y)^2=\cos2y+i\sin2y$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2961689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Help with this proof by induction with inequalities. Show that mathematical induction can be used to
prove the stronger inequality $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n + 1}}$ for all integers greater than 1, which, together with a verification for the case where n = 1, establishes the weaker inequality we originally tried to prove using mathematical induction.
Base Case: P(2)
\begin{aligned}
\frac{1}{2}\cdot\frac{2(2)-1}{2(2)} &< \frac{1}{\sqrt{3(2) + 1}}\\
\frac{3}{8} &< \frac{1}{\sqrt{7}}\\
\frac{1}{8} &< \frac{1}{3\sqrt{7}}\\
\end{aligned}
This is true as $8 > 3\sqrt{7}$.
Inductive Hypothesis: $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n+1}}$
In the inductive step, we want to show that $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} \cdot \frac{2n+1}{2n+2} < \frac{1}{\sqrt{3n+4}}$.
Using the inductive hypothesis, we can get to the following:
\begin{aligned}
\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} \cdot \frac{2n+1}{2n+2} &< \frac{1}{\sqrt{3n+1}}\cdot \frac{2n+1}{2n+2}\\
&< \frac{1}{\sqrt{3n+1}}\cdot 1\\
\end{aligned}
I am not sure how to get to $< \frac{1}{\sqrt{3n+4}}$ from here because i know that if the denominator would get bigger by adding 3 to it so the inequality wouldn't follow...
|
You need to show that $$\frac{1}{\sqrt{3n+1}}\cdot \frac{2n+1}{2n+2}\\
< \frac{1}{\sqrt{3n+4}}\\$$Square both sides and cross multiply to get $$ (2n+1)^2 (3n+4)<(3n+1)(2n+2)^2$$
Multiply and cancel equal terms to get $$ 12n^2+12n+3 <12n^2 +13n+3 $$
Which is true for all positive integers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2965485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Proving Sums or Products by Induction I am currently teaching Extension 1 Mathematics and seem to be coming unstuck on a few induction questions.
$$\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}$$
First step is to prove the equation works for "1" - which it does. The second step is to prove the function works for $k+1$ with my working set out as follows:
$$LHS=\sum_{k=1}^n{\frac{1}{(2k-1)(2k+1)}}+\frac{1}{(2n+1)(2n+3)}$$
$$=\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}$$
$$=\frac{n(2n+3)+1}{(2n+1)(2n+3)}$$
After setting out these steps - my text tells me the answer i'm looking for is $\frac{n+1}{2n+3}$ and I cant seem to find how to get from the final step of my working to the final answer! What am I missing here? (this is the second question I have come unstuck on a step before the final answer)
|
Fro the induction step we need to prove that
$$\sum_{k=1}^{n+1}\frac{1}{(2k-1)(2k+1)}=\frac{1}{(2n+1)(2n+3)}+\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}\stackrel{Ind. Hyp.}=\frac{1}{(2n+1)(2n+3)}+\frac{n}{2n+1}\stackrel{?}=\frac{n+1}{2n+3}$$
that is
$$\frac{1}{(2n+1)(2n+3)}+\frac{n}{2n+1}\stackrel{?}=\frac{n+1}{2n+3}$$
$$\frac{1}{(2n+1)(2n+3)}=\frac{n+1}{2n+3}-\frac{n}{2n+1}$$
$$\frac{1}{(2n+1)(2n+3)}=\frac{(2n+1)(n+1)-n(2n+3)}{(2n+1)(2n+3)}$$
$$\frac{1}{(2n+1)(2n+3)}=\frac{\color{red}{2n^2+3n}+1\color{red}{-2n^2-3n}}{(2n+1)(2n+3)}=\frac{1}{(2n+1)(2n+3)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2965663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Finding complex roots of $(z-i)^n-(z+i)^n =0$ I have a problem. First part of it was to find $z$ such that $$\left|\frac{z-i}{z+i}\right| = 1$$ and I've quickly figured out, that $z \in \Bbb R$.
How ever, now I have to (using previous part of the problem) find all roots of the following equation: $$(z-i)^n - (z+i)^n = 0$$
I have no idea how to do this. I tried adding the second part to the right side, then dividing by it, but with no success.
|
$$(z-1)^n-(z+1)^n=0\ (n\ \epsilon\ \mathbb{Z}^+)$$
$$\Rightarrow (z-1)^n=(z+1)^n \Rightarrow \left(\frac{z-1}{z+1}\right)^n=1=1\angle 2\pi k\ (k=0,1,2,...n-1)$$
$$\Rightarrow \frac{z-1}{z+1}=1\angle \frac{2\pi k}{n}\Rightarrow 1 -\frac{2}{z+1}=1\angle \frac{2\pi k}{n}$$
$$\frac{2\pi k}{n}=\theta \Rightarrow 1 -\frac{2}{z+1}=1\angle \theta \Rightarrow z=\frac{1+1\angle \theta}{1-1\angle \theta}$$
$$\Rightarrow z = \frac{1+i\sin \theta + \cos \theta}{1-i\sin \theta - \cos \theta} = \frac{(1+i\sin \theta + \cos \theta)^2}{1-(i\sin \theta + \cos \theta)^2}$$
$$\Rightarrow z= \frac{2\cos^2 \theta + 2i\sin \theta \cos \theta +2i\sin \theta + 2\cos \theta}{2\sin^2 \theta -2i\sin \theta \cos \theta}$$
$$\Rightarrow z=\frac{2(i\sin \theta + \cos \theta)(\cos \theta + 1)}{-2i\sin \theta (i\sin \theta + \cos \theta)}=\frac{i(\cos \theta +1)}{\sin \theta}$$
$$\Rightarrow z = \csc \theta + i\cot \theta$$
$$\Rightarrow z = \csc \frac{2\pi k}{n} + i\cot \frac{2\pi k}{n}\ (k=0,1,2,...n-1)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2967442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$?
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$ then $\frac{d}{dx}=?$
Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivative of tan inverse and the following with chain rule and i got my answer as ($-\frac{1}{2x\sqrt{x^2-1}}$), which is not same as the above correct answer. 2nd method is that you can substitute $x=\sec\left(\theta\right)$ and while solving in last step we will get $\sec^{-1}\left(\theta\right)$ whose derivative contains $\left|x\right|$, but still i searched and don't know why its derivative has $\left|x\right|$
Here's my attempt stepwise
$\displaystyle\frac{dy}{dx}=\frac{1}{1+\left(\sqrt{\frac{x+1}{x-1}}\right)^2}\cdot\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\cdot\frac{\left(x-1\right)-\left(x+1\right)}{\left(x-1\right)^2}$
$\displaystyle=\frac{\left(x-1\right)}{\left(x-1\right)+\left(x+1\right)}\cdot\frac{1\sqrt{x-1}}{2\sqrt{x+1}}\cdot-\frac{2}{\left(x-1\right)^2}$
$\displaystyle=-\frac{1}{2x}\cdot\frac{\left(x-1\right)\sqrt{x-1}}{\left(x-1\right)^2}\cdot\frac{1}{\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x-1}\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x^2-1}}$
Can you tell what i am doing wrong in my 1st attempt?
|
Let $x = \sec \theta, \ 0 \le \theta \le \pi, \ \theta \ne \frac{\pi}{2}$
Then $\sqrt {\dfrac {x+1}{x-1}} =\sqrt {\dfrac {\sec \theta+1}{\sec \theta-1}} = \sqrt {\dfrac {1+\cos \theta}{1-\cos \theta}} = \tan \dfrac{\theta}{2} $ (positive root as $\dfrac{\theta}{2}$ is in the 1st Quadrant.
$\therefore y = \tan^{-1} \tan \dfrac{\theta}{2} = \dfrac{\theta}{2}$
Hence $\dfrac{dy}{dx} = \dfrac{1}{2 \sec \theta \tan \theta}= \dfrac{1}{2|x| \sqrt{x^2-1}}$ (as $\sec \theta \tan \theta \ge 0$ in the first two quadrants)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2970280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
}
|
find the sum to n term of $\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $ $$\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $$
$$=\sum \limits_{k=1}^{n} \frac{2k-1}{k\cdot(k+1)\cdot(k+2)}$$ $$= \sum \limits_{k=1}^{n} - \frac{1}{2}\cdot k + \sum \limits_{k=1}^{n} \frac{3}{k+1} - \sum \limits_{k=1}^{n}\frac{5}{2\cdot(k+2)} $$
I do not know how to get a telescoping series from here to cancel terms.
|
Let the fractions be $\frac{a}{k}$, $\frac{b}{k+1}$, and $\frac{c}{k+2}$.
$\frac{a}{k}+\frac{b}{k+1}+\frac{c}{k+2}=\frac{a(k+1)(k+2)+bk(k+2)+ck(k+1)}{k(k+1)(k+2)}=\frac{2k-1}{k(k+1)(k+2)}$
We want the following
$a+b+c=0$
$3a+2b+c=2$
$2a=-1$
Solve, $a=-\frac{1}{2}$, $b=3$, and $c=-\frac{5}{2}$.
The rest is standard.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2972505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
}
|
Probability that tossing $10$ coins has $4$ consecutive heads I'd like to compute the probability of having a string of four consecutive heads after $10$ coin tosses.
I tried to solve this problem by considering $10$ blank spaces. To have $4$ consecutive heads, we can consider $7$ blank spaces, where the $7^{\text{th}}$ space is a string consisting of $4$ heads.
Then, there are $7!$ ways to arrange the $7$ spaces, and there are $4!$ internal permutations of the $4$ heads. Also, there are $2^{6}$ choices for the $6$ non-head coins.
This is all out of $10!$ arrangements and $2^{10}$ choices for the coins. So my probability is
$$\frac{7! \cdot 4! \cdot 2^6}{10! \cdot 2^{10}} = 0.00208333333,$$
which is wrong. I want to know what's wrong with my reasoning. I can't figure it out. I know there's a similar post, but I don't know what's wrong with my approach.
|
The total number of possible outcomes is $2^{10}$
Using the counting method for the numerator .......
The number of outcomes with a single exact maximum $HHHH$ is
$HHHHT----- \Rightarrow 2^5 - 3$
$THHHHT---- \Rightarrow 2^4 - 1$
$-THHHHT--- \Rightarrow 2^4$
$--THHHHT-- \Rightarrow 2^4$
$---THHHHT- \Rightarrow 2^4$
$----THHHHT \Rightarrow 2^4 - 1$
$-----THHHH \Rightarrow 2^5 - 3$
$$P(4H) = \frac{2(2)^5 + 5(2)^5 - 8}{2^{10}} = .132813$$
Where two $HHHH$ are allowed......
$$P(4H+) = \frac{2(2)^5 + 5(2)^5 - 2}{2^{10}} = .138672$$
The number of outcomes with at least one run of at least $4$ consecutive heads is $2^6 + 6(2)^5 - 5$
$HHHH------ \Rightarrow 2^6$
$THHHH----- \Rightarrow 2^5$
$-THHHH---- \Rightarrow 2^5$
$--THHHH--- \Rightarrow 2^5$
$---THHHH-- \Rightarrow 2^5$
$----THHHH- \Rightarrow 2^5 - 2$
$-----THHHH \Rightarrow 2^5 - 3$
$$P(\ge4H) = \frac{2^6+6(2)^5-5}{2^{10}} = .245117$$
Does anyone have a more formal/elegant method to check these answers?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2975429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Power Series (Cambridge Tripos 1900) If
$$a/(a+bz+cz^2)=1+p_1z+p_2z^2
+\dots$$
then
$$1+p_1^2z+p_2^2z^2+\dots=\frac{a+cz}{a-cz}\frac{a^2}{
a^2-(b^2-2ac)z+c^2z^2}$$
A tricky problem from G.H.Hardy's "A Course in Pure Mathematics", or maybe I'm missing the obvious.Any help will be greatly appreciated.
|
If we start with the original function, we can reformulate is conveniently as
$$
\frac{a}{a+bz+cz^2} = \frac{1}{1+(\frac{b}{a})z+(\frac{c}{a})z^2} = \frac{1}{(1-\frac{z}{z_+})(1-\frac{z}{z_-})} = \frac{z_+}{(z_+-z_-)} \frac{1}{(1-\frac{z}{z_-})} - \frac{z_-}{(z_+-z_-)} \frac{1}{(1-\frac{z}{z_+})}
$$
with $z_\pm$ the roots of the denominator, i.e.,
$$
z_\pm = \frac{- b \pm \sqrt{b^2 - 4 a c}}{2 c}
$$
If we expand the two expressions on the left as a series in $z$, we get
$$
\frac{z_+}{(z_+-z_-)} \frac{1}{(1-\frac{z}{z_-})} = \sum_{n\geq0} l_n z^n \quad \text{ with } \quad l_n = \frac{z_+}{(z_+-z_-)} \frac{1}{z_-^n}
$$
$$
\frac{z_-}{(z_+-z_-)} \frac{1}{(1-\frac{z}{z_+})} = \sum_{n\geq0} r_n z^n \quad \text{ with } \quad r_n = \frac{z_-}{(z_+-z_-)} \frac{1}{z_+^n}
$$
and hence we identify $p_n=l_n-r_n$ for $n\geq0$. Now let us construct a series with coefficients $p_n^2 = (l_n-r_n)^2$. It is relatively easy to recognise that we can make the following identifications
$$
\sum_{n\geq0} l_n^2 z^n = \frac{z_+^2}{(z_+-z_-)^2} \frac{1}{(1-\frac{z}{z_-^2})}
$$
$$
\sum_{n\geq0} r_n^2 z^n = \frac{z_-^2}{(z_+-z_-)^2} \frac{1}{(1-\frac{z}{z_+^2})}
$$
$$
\sum_{n\geq0} l_n r_n z^n = \frac{z_+ z_-}{(z_+-z_-)^2} \frac{1}{(1-\frac{z}{z_+ z_-})}
$$
and hence that the required series
$$
\sum_{n\geq0} p_n^2 z^n = \frac{1}{(z_+-z_-)^2} \left(\frac{z_+^2}{(1-\frac{z}{z_-^2})} + \frac{z_-^2}{(1-\frac{z}{z_+^2})} - \frac{2 z_+ z_-}{(1-\frac{z}{z_+ z_-})}\right) =
\frac{z_+^2 z_-^2 (z_+ z_- + z)}{(z_+ z_- -z)(z_-^2 - z)(z_+^2 - z)}
$$
Using that $z_+ z_- = a/c$ and $z_+^2+z_-^2=(b^2-2 ac)/c^2$ it then follows that
$$
\sum_{n\geq0} p_n^2 z^n = \frac{a+c z}{a - c z} \frac{a^2}{a^2 - (b^2-2ac)z + c^2 z^2}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2977675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
}
|
Evaluating $\int_1^2\frac{\arctan(x+1)}{x}\,dx$
Evaluate the following integral
$$\int_1^2\frac{\arctan(x+1)}{x}\,dx$$
with $0\leq\arctan(x)<\pi/2$ for $0\leq x<\infty.$
I proceeded the following way
$$\begin{aligned}
&\int_1^2\frac{\arctan(x+1)}{x}\,dx\to {\small{\begin{bmatrix}&u=x+1&\\&du=dx&\end{bmatrix}}}
\to\int_2^3\frac{\arctan(u)}{u-1}\,du=\\
&\ln(2)\arctan(3)-\int_2^3\frac{\ln(u-1)}{u^2+1}\,du\to {\small{\begin{bmatrix}&u=\tan(\theta)&\\&du=\sec^2(\theta)d\theta&\end{bmatrix}}}
\to\\
&\ln(2)\arctan(3)-\int_\alpha^\beta\ln\left(\tan(\theta)-1\right)\,d\theta=\ln(2)\arctan(3)-\int_\alpha^\beta\ln\left(\sin(\theta)-\cos(\theta)\right)\,d\theta+\\
&+\int_\alpha^\beta\ln\left(\cos(\theta)\right)\,d\theta.
\end{aligned}$$
But
$$\int_\alpha^\beta\ln\left(\sin(\theta)-\cos(\theta)\right)\,d\theta\to {\small{\begin{bmatrix}&\theta=s+3\pi/4&\\&d\theta=ds&\end{bmatrix}}}
\to\int_{\alpha-3\pi/4}^{\beta-3\pi/4}\ln\left(\sqrt2\cos(s)\right)\,ds$$
so
$$\begin{aligned}\int_1^2\frac{\arctan(x+1)}{x}\,dx&=\ln(2)\arctan(3)+\ln(\sqrt{2})(\alpha-\beta)\\
&\phantom{aaaaa}-\int_{\alpha-3\pi/4}^{\beta-3\pi/4}\ln\left(\cos(s)\right)\,ds+\int_\alpha^\beta\ln\left(\cos(s)\right)\,ds.
\end{aligned}$$
Here $\alpha=\arctan(2)$ and $\beta=\arctan(3)$.
The problem here is that I am not able to find a way to cancel the last two integrals on the RHS of the latter equality.
ADDENDUM
Using Mathematica 11.3 I found that the answers is $\frac{3}{8} \pi \ln(2)\approx0.81659478386385079894.$
In my equality, if we assume the integrals that involve cosines cancel, we have that the result of the integral is
$\frac{1}{2} \ln (2) \left(\arctan(2)-\arctan(3)\right)+\ln (2) \arctan(3)\approx 0.81659478386385079894$.
Which are exactly equal up to $20$ decimal places! How would I go about canceling the integrals involving cosines (if they actually do cancel)?
|
Related Integral
First, let me tackle another integral which will come in useful later:
$$J=\int_{\arctan 2}^{\arctan 3}\ln(\tan u-1)du$$
$$\tag 1=\int_{\frac{3\pi}4-\arctan 2}^{\frac{3\pi}4-\arctan 3}\ln\Biggl(\tan\left(\frac{3\pi}4-u\right)-1\Biggr)(-du)$$
$$\tag 2=-\int_{\arctan 3}^{\arctan 2}\ln\left(\frac{\tan \frac{3\pi}4-\tan u}{1+\tan\left(\frac{3\pi}4\right)\tan u}-1\right)du$$
$$\tag 3=\int_{\arctan 2}^{\arctan 3}\ln\left(\frac{-1-\tan u}{1-\tan u}-1\right)du$$
$$=\int_{\arctan 2}^{\arctan 3}\ln\left(\frac{-1-\tan u-(1-\tan u)}{1-\tan u}\right)du$$
$$=\int_{\arctan 2}^{\arctan 3}\ln\left(\frac{-2}{1-\tan u}\right)du$$
$$=\int_{\arctan 2}^{\arctan 3}\ln\left(\frac{2}{\tan u-1}\right)du$$
$$=\int_{\arctan 2}^{\arctan 3}\ln 2 - \ln(\tan u-1)du$$
$$\tag 4=\ln 2\int_{\arctan 2}^{\arctan 3}du-\int_{\arctan 2}^{\arctan 3}\ln(\tan u-1)du$$
$$=(\arctan 3-\arctan 2)\ln 2-J$$
$$\tag 5=(2\arctan 3-\frac{3\pi}4)\ln 2-J$$
$$\therefore2J=\left(2\arctan 3-\frac{3\pi}4\right)\ln 2$$
$$\boxed{J=\int_{\arctan 2}^{\arctan 3}\ln(\tan u-1)du=\arctan 3\ln 2 - \frac{3\pi\ln2}8}$$
Main Integral
Now, onto the main integral:
$$I=\int_1^2 \frac{\arctan(x+1)}x dx$$
$$\tag 6=\int_2^3 \frac{\arctan x}{x-1}dx$$
$$\tag 7=\int_{\arctan 2}^{\arctan 3} \frac{u\sec^2u}{\tan u-1}du$$
$$\tag 8=[u\ln(\tan u-1)]_{\arctan 2}^{\arctan 3}-\int_{\arctan 2}^{\arctan 3}\ln(\tan u-1)du$$
$$=\arctan3\ln2-\int_{\arctan 2}^{\arctan 3}\ln(\tan u-1)du$$
$$=\arctan3\ln2-\left(\arctan 3\ln 2 - \frac{3\pi\ln2}8\right)$$
$$\boxed{I=\frac{3\pi\ln2}8}$$
Elaboration
Elaboration for numbered equations:
(1) Substitute $u\rightarrow\frac{3\pi}4-u, du\rightarrow -du$
(2), (5) $\arctan 2+\arctan 3=\frac{3\pi}4 \because\tan(\arctan 2+\arctan 3)=\frac{2+3}{1-2*3}=-1$
(3) $\tan\frac{3\pi}4=-1$ and $-\int_b^a=\int_a^b$
(4) $\ln\frac{a}b=\ln a-\ln b$
(6) Substitute $x\rightarrow x-1, dx\rightarrow dx$
(7) Substitute $x=\tan u -1, dx=\sec^2 udu$
(8) Integration by parts
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2979918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
}
|
Solution to equation using permutation I am hoping I am on the right track for my homework in Discrete Math.
How many solutions does the equation $$a + b + c + d + e + f = 30$$ have if each variable must be a non-negative integer and $a ≤ 3, b ≤ 7$ and $d ≥ 8$?
I started my solution using the formula: ${m+n-1 \choose n-1}$ where m is the number of non-distinct objects and n is the number of distinct objects.
d ≥ 8 where ${22+9-1 \choose 9-1}$ = ${30 \choose 8}$
Then, a ≤ 3 which can be represented as a ≥ 4
${17+6-1 \choose 6-1}$ = ${21 \choose 5}$
Finally, b ≤ 7 which can be represented as b ≥ 8
${9+6-1 \choose 6-1}$ = ${14 \choose 5}$
During this process we have removed solutions for a ≤ 3 and b ≤ 7 twice so we have to add the solution back in once. We removed 25 - (3+7) = 15 so we must add back ${15+6-1 \choose 6-1}$ = ${20 \choose 5}$
Therefore, our solution is as follows:
${30 \choose 8}$ - ${21 \choose 5}$ - ${14 \choose 5}$+ ${20 \choose 5}$
|
The number of all 6-couples in our universe, which is set of all solutions of the equation $a+b+c+d'+e+f =22$ where $d'=d-8\geq 0$ is $${27\choose 5}$$
Let $A$ be a set of all 6-couples where $a\geq 4$ and $B$ be a set of all 6-couples where $b\geq 8$.
So $$|A| = {23\choose 5}\;\;\;{\rm and}\;\;\;|B| = {19\choose 5}$$
Since $$|A\cap B| = {15\choose 5}$$ the finaly answer is (using formula of inclusion and exclusion) $$ {27\choose 5}- {23\choose 5}- {19\choose 5}+ {15\choose 5}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2982158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
If $y=\frac25+\frac{1\cdot3}{2!}\left(\frac25\right)^2 +\frac{1\cdot3\cdot5}{3!}\left(\frac25\right)^3+\ldots$ then what is the value of $y^2+2y$? If
$$y=\frac25+\frac{1\cdot3}{2!}\left(\frac25\right)^2 +\frac{1\cdot3\cdot5}{3!}\left(\frac25\right)^3+\ldots$$
then what is the value of $y^2+2y$?
This is a question from my coaching material in which binomial theorem, multinomial theorem and binomial theorem with fractional and negative indices are covered. How do I approach this problem? What is the pattern in it?
|
Recall the binomial series $(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}+\dots$
Let $x=\frac{-4}{5}$ and $\alpha=\frac{-1}{2}$. Then
$$
\sqrt{5} = \left(1-\frac{4}{5}\right)^{\frac{-1}{2}}
= 1 +\frac{2}{5} + \frac{1\cdot3}{2!}\cdot\left(\frac{-1}{2}\right)^2\cdot\left(\frac{-4}{5}\right)^2+\dots =y+1
$$
so $y^2+2y=(y+1)^2-1=4$
How to arrive at this: in your question you stated that you want to use the binomial theorem. The thing stopping you is that the numerators are decreasing by two each time, rather than by one. This suggests that the exponent is a half-integer. Once you have this it's just a matter of fiddling to come up with the values of $x$ and $\alpha$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2982713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Prove convergence of sequence given by $a_{1}=1$ and $a_{n+1}= \frac{1}{a_1+a_2+\ldots+a_n}$ For sequence given by $a_{1}=1$ and $a_{n+1}= \frac{1}{a_1+a_2+\ldots+a_n}$ I have to prove that it converges to some number and find this number.
I tried toshow that it's monotonic by calculating
$$
\frac{a_{n+1}}{a_{n}} = \frac{1}{a_{n}(a_1+a_2+\ldots+a_n)}
$$
but I cannot say anything about the denominator. How can I try to find it's limit?
|
Let $s_n = \sum\limits_{k=1}^n a_k$. We can rewrite the recurrence relation as
$$s_{n+1} - s_n = a_{n+1} = \frac{1}{s_n} \quad\implies s_{n+1} = s_n + \frac{1}{s_n}$$
This implies
$$s_{n+1}^2 = s_n^2 + 2 + \frac{1}{s_n^2} \ge s_n^2 + 2$$
So for all $n > 1$, we have
$$s_n^2 = s_1^2 + \sum_{k=1}^{n-1} (s_{k+1}^2 - s_k^2) \ge 1 + \sum_{k=1}^{n-1} 2 = 2n - 1$$
Since all $a_n$ is clearly positive, we have $\displaystyle\;0 < a_n = \frac{1}{s_{n-1}} \le \frac{1}{\sqrt{2n-3}}$.
By squeezing, $a_n$ converges to $0$ as $n\to\infty$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2982842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Combinatorial proof of $\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3$?
Give a combinatorial proof of the following identity: $$\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3.$$
I've been working on this proof for hours, however I'm not able to show LHS = RHS-
I completely understand binomial theorem and few combinatorial proofs but not able to succeed this one.
Help would be appreciated.
|
Let $A,B,C$ be the $3$ groups of $n$ students each with grades (marks) $A,B,C$, respectively. You need to select $3$ students.
The LHS is simply the combination of $3n$ students chosen $3$ at a time.
The RHS is to choose $3,2,1$ or $0$ students from $A,B$ and $C$, respectively:
$${n\choose 3}{n\choose 0}{n\choose 0}+{n\choose 2}{n\choose 1}{n\choose 0}+{n\choose 2}{n\choose 0}{n\choose 1}+{n\choose 1}{n\choose 2}{n\choose 0}+{n\choose 1}{n\choose 1}{n\choose 1}+\\
{n\choose 1}{n\choose 0}{n\choose 2}+{n\choose 0}{n\choose 3}{n\choose 0}+{n\choose 0}{n\choose 2}{n\choose 1}+{n\choose 0}{n\choose 1}{n\choose 2}+{n\choose 0}{n\choose 0}{n\choose 3}=\\
{n\choose 3}\cdot 1\cdot 1+{n\choose 2}\cdot n\cdot 1+{n\choose 2}\cdot 1\cdot n+n\cdot {n\choose 2}\cdot 1+n\cdot n\cdot n+\\
n\cdot 1\cdot{n\choose 2}+1\cdot{n\choose 3}\cdot1+1\cdot{n\choose 2}\cdot n+1\cdot n\cdot{n\choose 2}+1\cdot 1\cdot {n\choose 3}=\\
3{n\choose 3}+6n{n\choose 2}+n^3.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2983705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
}
|
Find the value of $(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ Find the value of
$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ where $\alpha,\beta,\gamma$ are roots of the equation $x^3+2x^2+3x+3=0$.
I tried to use the formula which is wrong
$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3=((\frac{\alpha}{\alpha +1})+(\frac{\beta}{\beta +1})+(\frac{\gamma}{\gamma +1}))^3-3((\frac{\alpha}{\alpha +1})(\frac{\beta}{\beta +1})(\frac{\gamma}{\gamma +1}))$
And then I broke the terms to get
$(\frac{13}{5})^3-\frac{36}{5}$
but this is $\ne 44$, which should be the answer.
I know where I was wrong, the formula is $(a+b+c)^3=a^3+b^3+c^3-3(a+b)(b+c)(c+a)$
|
For simplicity, let me use $\alpha=a, \beta=b,\gamma=c$.
We can find:
$$\begin{cases}a+b+c=-2 \\ ab+bc+ca=3\\ abc=-3\end{cases} \Rightarrow \\
\begin{cases}\color{red}{a^2+b^2+c^2}=4-2(ab+bc+ca)=\color{red}{-2} \\ \color{blue}{a^2b^2+b^2c^2+c^2a^2}=9-2abc(a+b+c)=\color{blue}{-3}\end{cases}$$
We can express:
$$x^3+2x^2+3x+3=0 \Rightarrow (x+1)^3=x^3+3x^2+3x+1=x^2-2$$
Hence:
$$\left(\frac{a}{a +1}\right)^3+\left(\frac{b}{b +1}\right)^3+\left(\frac{c}{c +1}\right)^3=\\
\frac{a^3}{a^2-2}+\frac{b^3}{b^2-2}+\frac{c^3}{c^2-2}=\\
a+\frac{2a}{a^2-2}+b+\frac{2b}{b^2-2}+c+\frac{2c}{c^2-2}=\\
-2+2\cdot \frac{a(b^2-2)(c^2-2)+b(a^2-2)(c^2-2)+c(a^2-2)(b^2-2)}{(a^2-2)(b^2-2)(c^2-2)}=\\
-2+2\cdot \frac{abc(ab+bc+ca)-2(a^2b+ab^2)-2(a^2c+ac^2)-2(b^2c+bc^2)+4(a+b+c)}{a^2b^2c^2-2(\color{blue}{a^2b^2+b^2c^2+c^2a^2})+4(\color{red}{a^2+b^2+c^2})-8}=\\
-2+2\cdot \frac{-9-2ab(-2-c)-2ac(-2-b)-2bc(-2-a)-8}{9+6-8-8}=\\
-2+2\cdot \frac{-9+4(ab+bc+ca)+6abc-8}{-1}=\\
-2+2\cdot \frac{-9+12-18-8}{-1}=44.\\
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2983790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
}
|
Is $x(1 - 2x) \le \frac{1}{8}$ and further, is $x(1 - ax) \le \frac{1}{4a}$ It is clear that $x(1-x) \le \frac{1}{4}$
Does it likewise follow that $x(1-2x) \le \frac{1}{8}$?
Here's my reasoning:
(1) For $x < \frac{1}{4}$, $x(1-2x) < \frac{1}{8}$
(2) For $\frac{1}{4} < x < \frac{1}{2}$, $x(1-2x) < \frac{1}{8}$
(3) For $\frac{1}{2} < x$, $x(1-2x) < 0$
Further, can this be generalized to $x(1-ax) \le \frac{1}{4a}$
Since:
(1) For $x < \frac{1}{2a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(2) For $\frac{1}{2a} < x < \frac{1}{a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(3) For $\frac{1}{a} < x$, $x(1-ax) < 0$
Are both of these observations correct? Is only one correct? Is there an exception that I am missing?
|
Consider the function
$$f(x)=x(1-ax) \implies f'(x)=1-2ax\implies f''(x)=-2a$$ The first derivative cancels at $x_*=\frac 1 {2a}$ and $$f\left(\frac{1}{2 a}\right)=\frac{1}{4 a}$$ The point $x_*$ is a maximum if $a>0$ by the second derivative test and a minimum otherwise.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2983897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Proving that for $n \equiv 0 \pmod{2}$, we get ${n \choose 0} < {n \choose 1} <\ldots< {n \choose n/2-1}<{n\choose n/2}$ etc. How can one prove that for $n \equiv 0$ mod $2$ we have
$${n \choose 0} < {n \choose 1} <\ldots< {n \choose n/2-1}<{n \choose n/2}>{n\choose n/2+1}>\ldots>{n\choose n-1}>{n\choose n}\,?$$
Can I say that for fixed $n$, the binomial coefficients $n \choose k$ increase with $k$ for $k < n/2$? If n is even (like in our case), then the central binomial coefficient $n \choose n/2$ is the largest one.
So $n \choose k+1$ is greater than, equal to, or less than $n \choose k$ according as $n-k$ is greater than, equal to, or less than $k+1$, that is according as $k$ is less than, equal to, or greater than $(n−1)/2$.
Is that correct?
|
Let's try using induction on $m$ where $n=2m$. For $m=0$, there is nothing to be proven, for $m=1$, then ${2\choose 0} = {2\choose 2} = 1 < 2 = {2\choose 1}$.
To get the induction step, observe that ${n+2\choose k} = {n+1\choose k-1} + {n+1 \choose k} = {n\choose k-2} + 2 {n\choose k-1} + {n\choose k}$. Suppose that $k<m+1 = \frac{n+2}{2}$, then
\begin{align*}
{2(m+1)\choose k} - {2(m+1)\choose k+1} &= {2m \choose k-2}+{2m\choose k-1}-{2m \choose k}-{2m\choose k+1}\\
&= \left[ {2m \choose k-2} - {2m \choose k} \right] + \left[ {2m\choose k-1} -{2m\choose k+1}\right]
\end{align*}
The first term is directly strictly negative by induction, the second term is too if $k<m$, if $k=m$ then ${2m\choose m-1}={2m\choose m+1}$ and the second term is $0$. This proves that ${2n+2\choose k}<{2n+2\choose k+1}$. The same can be done when $k>m+1$ to obtain ${2n+2\choose k}<{2n+2\choose k-1}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2984998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
If $a=b+c$ prove that $S=a^4+b^4+c^4$ is twice the square of a positive integer
If $a=b+c$, and $a$,$b$,$c\in \Bbb N$, prove that $S=a^4+b^4+c^4$ is twice the square of a positive integer.
Source: a list of problems used in the preparation to math contests.
My attempt:
By making the substitution $a=b+c$ in $S$ and developing $(b+c)^4$, it is easy to show that
$$S=2(b^4+c^4+bc(2b^2+3bc+2c^2))$$
an expression with the form $S=2K$. The problem now is how to prove that K is a square of a positive integer. I also tried to use Newton Identities but with no luck (Note: later, after a hint, I found a way to solve using this approach, see below).
Hints and answers are welcomed. Sorry if this is a dup.
|
$(b+c)^4=b^4+4b^3c+6b^2c^2+4bc^3+c^4=(b^4+c^4)+4bc(b^2+c^2)+6b^2c^2=>$
$$
S=a^4+b^4+c^4=a^4+a^4-4bc(b^2+c^2)-6b^2c^2=2a^4-4bc(b+c)^2+2b^2c^2=2a^4-4bca^2+2b^2c^2=2(a^4-2bca^2+b^2c^2)=2(a^2-bc)^2
$$
$=>S=2(a^2-bc)^2$
$$
Q.E.D
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2985553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
Simultaneous real solution of $x^3+y^3+1+6xy=0$ & $xy^2+y+x^2=0$ I am trying to solve the following system of non-linear equations in real numbers:
$x^3+y^3+1+6xy=0$ & $xy^2+y+x^2=0$, with $x,y$ real.
I can only see that $xy\ne 0$.
I have no clue whether a solution exists or not and how to find any solution. I cannot seem to be able to separate $x,y$ or write it in a good parametric form.
Please help.
|
The given set of equations has three real solutions:
$$(x,y) =
\left(\frac{1}{y_2},y_1\right),
\left(\frac{1}{y_3},y_2\right),
\left(\frac{1}{y_1},y_3\right)
$$
where $y_1 = 2\cos\frac{\pi}{9}$, $y_2 = 2\cos\frac{7\pi}{9}$ and $y_3 = 2\cos\frac{13\pi}{9}$.
To derive them, we will exploit a symmetry hidden underneath the set of equations.
Introduce variables $(X,Y,Z)$ and let $(x,y) = \left(\frac{X}{Z},\frac{Y}{Z}\right)$,
we have
$$\begin{align}
x^3 + y^3 + 1 + 6xy = 0
& \iff X^3 + Y^3 + Z^3 + 6XYZ = 0 \tag{*1a}\\
xy^2 + y + z^2 = 0
&\iff XY^2 + YZ^2 + X^2Z = 0\tag{*1b}
\end{align}
$$
Let $u = \frac{Y}{Z}, v = \frac{Z}{X}, w = \frac{X}{Y}$, we have $uvw = 1$.
Divide RHS(*1a) by $XYZ$, we obtain
$$\frac{w}{v} + \frac{u}{w} + \frac{v}{u} + 6 = \frac{X^2}{YZ} + \frac{Y^2}{XZ} + \frac{Z^2}{XY} + 6 = 0$$
Multiply both sides by $uvw = 1$, we obtain
$$w^2 u + u^2v + v^2w + 6 = 0 \tag{*2}$$
Similarly, divide RHS(*1b) by $XYZ$ give us
$$u + v + w = \frac{Y}{Z} + \frac{Z}{X} + \frac{X}{Y} = 0$$
Let $\omega = e^{\frac{2\pi}{3}i}$ be the cubic root of unity. Since $u+v+w = 0$, we can find
two complex numbers $a, b$ such that
$$\begin{cases}
u &= a + b\\
v &= a\omega + b\omega^2\\
w &= a\omega^2 + b\omega
\end{cases}\tag{*3}$$
In terms of $a,b$, the condition $uvw = 1$ reduces to
$$a^3 + b^3 = (a+b)(a\omega + b\omega^2)(a\omega^2+b\omega) = uvw = 1$$
Substitute the parameterization (*3) into (*2) and simplify, we obtain
$$3a^3\omega + 3b^3\omega^2 + 6 = 0 \iff a^3\omega + b^3\omega^2 = - 2$$
Solving this last two equations give us
$$(a^3, b^3 ) = ( -\omega^2, -\omega ) = \left( e^{\frac{\pi}{3}i}, e^{-\frac{\pi}{3}i}\right)$$
We are looking for real solutions for $(x,y)$, we need $u = a + b$ and $v = a\omega + b\bar{\omega}$ to be real. This forces $b = \bar{a}$ and there are three possibilities:
$$a = \bar{b} = e^{\frac{\pi}{9}i}, e^{\frac{7\pi}{9}i}, e^{\frac{13\pi}{9}i}
\implies (u,v,w) = (y_1,y_2,y_3), (y_2,y_3,y_1), (y_3,y_1,y_2)$$
Notice $(x,y) = (v^{-1},u)$, the three real solutions of the problem follow.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2988031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
If $11z^8+20iz^7+10iz-22=0$,then show that $1<|z|<2$ If $$11z^8+20iz^7+10iz-22=0$$then show that $$1<|z|<2$$
My Attempt:
If $z=x+iy$; then$$z^7=\frac{22z-10iz}{11z+20i}$$
$$|z|^7=\sqrt{\frac{400+440y+100(x^2+y^2)+84}{400+440y+100(x^2+y^2)+21(x^2+y^2)}}$$
If $x^2+y^2=4$ i.e. $|z|=2$
then $|z|^7=1$
which is clearly a contradiction.
Not able to proceed from here.
|
Let $p(z) = 11z^8+20iz^7 + 10iz - 22$.
The part for $|z|<2$ is already in the other answer, but I include it for the sake of completeness. When $|z|=2$,
$$|20iz^7 + 10iz - 22| < 20(2^7)+10(2)+22 \\
< 21(2^7) < 22(2^7) = 11(2^8) = |11z^8|.$$
$11z^8$ has eight zeros (counted as many times as its multiplicity) in the open disk $|z| < 2$, so by Rouché's Theorem, $p$ has eight roots in $|z| < 2$. Since $\deg(p) = 8$, all roots of $p$ lies inside $|z| < 2$.
To show that all roots of $p$ lies outside $|z| < 1$, apply Rouché's Theorem on $f(z) := 20iz^7 - 22$ and $g(z) := 11z^8 + 10iz$. Observe that $p = f+g$, $f(z) = 2(10iz^7-11)$ and $g(z) = z(11z^7+10i)$, and that all roots of $f$ lies outside $|z| \le 1$.
$$f(z) = 0 \iff 20iz^7=22 \implies |z^7| = 1.1,$$
so on $|z| = 1$, $f(z) \ne 0$ and
$$|g(z)| = |11z^7+10i| = |11z^{-7}-10i| \\ = |11-10iz^7| = \frac{|f(z)|}{2} < |f(z)|,\tag{*}\label{*}$$
so all roots $z$ of $p$ satisfy $1 \le |z| < 2$, and it remains to show that $|z| \ne 1$.
Suppose $|z|=1$ and $p(z) = 0$. $p = f + g$ by construction, so
$|f(z)| = |g(z)|$. But $|f(z)| = 2|g(z)|$ from \eqref{*}, so $|f(z)| = |g(z)| = 0$. When $f(z) = 0$, $z^7 = -11i/10$, so $1 = |z^7| = 11/10$, which is absurd. Therfore, the unit circle $|z| = 1$ does not contain any root of $p$.
Hence, all roots of $p$ lie in the annular region $1 < |z| < 2$.
OP's reformulation $$z^7=\frac{22z-10iz}{11z+20i} \tag1\label1$$ can be hardly applied to solve this problem. Let $z = x+yi$ with $x,y \in \Bbb{R}$.
$$\begin{aligned}
|z|^7 &= \left\lvert\frac{22z-10iz}{11z+20i}\right\rvert \\
1 &= \left\lvert\frac{(22x+10y)+(22y-10x)i}{11x + (11y+20)i} \right\rvert \\
&= \sqrt{\frac{584(x^2+y^2)}{121(x^2+y^2)+440y+400}} \\
&= \sqrt{\frac{584}{521+440y}} \\
y &= \frac{63}{440}, x = \pm\sqrt{1-y^2}
\end{aligned}$$
Check that $x+yi$ does not verify \eqref{1} to conclude.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2990850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
What value does $\frac{1}{n} \sqrt[n]{n(n+1)(n+2)\cdots(2n)}$ tend to? I need to find where this sequence tends to:
$$\frac{1}{n} \sqrt[n]{n(n+1)(n+2)...(2n)}$$
My answer is $2$, by using a trick for the $n^{th}$ root: I take the inside and find where $\dfrac{A_{n+1}}{A_n}$ goes, and this is where the original sequence goes.
|
$$ a_n
= \frac{1}{n} \sqrt[n]{n(n+1)(n+2)\cdots (2n)}
= \sqrt[n]{1(1+\frac{1}{n})(1+\frac{2}{n})\cdots (1+\frac{n}{n})}.
$$
$$
\log a_n
= \frac{1}{n} \{ \log 1 + \log (1+\frac{1}{n}) + \log (1+\frac{2}{n})
+ \cdots + \log (1+\frac{n}{n}) \}.
$$
We know the above equations. Hence,
$$
\lim\limits_{n \rightarrow \infty} \log a_n = \int_1^2 \log x dx
= [x\log x -x ]_1^2 = 2\log 2 - 1 = \log \frac{4}{e}.
$$
Thus, the limit of $a_n$ is $\frac{4}{e}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2992724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Polynomial division with remainder
If the polynomial
$$x^4-x^3+ax^2+bx+c$$
divided by the polynomial
$$x^3+2x^2-3x+1$$
gives the remainder $$3x^2-2x+1$$
then how much is (a+b)c?
So what I know, and how I solved these problems before, I can write this down like this:
$x^4-x^3+ax^2+bx+c=Q(x)(x^3+2x^2-3x+1)+3x^2-2x+1$
And from here I should get the roots of $x^3+2x^2-3x+1$ so I could use the remainder only since I don't know $Q(x)$
$x^3+2x^2-3x+1=0$
but I can't find the roots of this by hand, so I used WolframAlpha to get the roots, but they are not whole numbers so I couldn't write this in the form of $(x-x_1)(x-x_2)(x-x_3)$ and then just use $P(x_1)$, $P(x_2)$, $P(x_3)$ to get $a,b,$ and $c$
I tried regular dividing on these polynomials to get $Q(x)$ but I couldn't divide them because of the $a,b$ and $c$ present
|
Let (original) polynomial $P(x) = x^4 - x^3 + ax^2 + bx + c$ and divisor polynomial $D(x) = x^3 + 2x^2 - 3x + 1$. The remainder polynomial $R(x) = 3x^2 - 2x + 1$. We can also define quotient polynomial $Q(x) = \alpha x + \beta$.
We have that $P(x) = Q(x)\cdot D(x) + R(x)$.
By inspection of the coefficient of $x^4$ on both sides, $\alpha = 1$, i.e. $Q(x)$ is monic.
Considering the coefficient of $x^3$ on both sides, we have that $-x^3 = 2x^3 + \beta x^3$, giving $\beta = -3$. So $Q(x) = (x-3)$.
By now substituting convenient values for $x$, we can solve for $a, b, c$ quite easily. But we don't have to go all the way on that, since we only need to find $(a+b)c$.
We can find $c$ by putting $x = 0$, giving $c = (-3)(1) + 1 = -2$.
We can find $a+b$ by putting $x = 1$, giving $1-1 + a + b - 2 = (-2)(1+2-3+1) + 3-2+1$, giving $a+b = 2$.
So $(a+b)c = 2(-2) = -4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2993694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
$M$ is a point in an equaliateral $ABC$ of area $S$. $S'$ is the area of the triangle with sides $MA,MB,MC$. Prove that $S'\leq \frac{1}{3}S$.
$M$ is a point in an equilateral triangle $ABC$ with the area $S$. Prove that $MA, MB, MC$ are the lengths of three sides of a triangle which has area $$S'\leq \frac{1}{3}S$$
|
Choose a coordinate system so that triangle $ABC$ is lying on the unit circle centered at origin and $A$ on the $x$-axis. Let $a = AM$, $b = BM$, $c = CM$ and $S'$ be the area of a triangle with sides $a,b,c$. In this coordinate system, $S = \frac{3\sqrt{3}}{4}$, we want to show $S' \le \frac{\sqrt{3}}{4}$. Using Heron's formula, this is equivalent to
$$16S'^2 = (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) \stackrel{?}{\le} 3$$
Identify euclidean plane with the complex plane. The vertices $A,B,C$ corresponds to $1, \omega, \omega^2 \in \mathbb{C}$ where $\omega = e^{\frac{2\pi}{3}i}$ is the cubic root of unity. Let $z$ be the complex number corresponds to $M$ and $\rho = |z|$, we have
$$
\begin{cases}
a^2 = |z-1|^2 = \rho^2 + 1 - (z + \bar{z})\\
b^2 = |z-\omega|^2 = \rho^2 + 1 - (z\omega + \bar{z}\omega^2)\\
c^2 = |z-\omega^2|^2 = \rho^2 + 1 - (z\omega^2 + \bar{z}\omega)
\end{cases}
\implies
a^2 + b^2 + c^2 = 3(\rho^2 + 1)
$$
Thanks to the identity $\omega^2 + \omega + 1 = 0$, all cross terms involving $\omega$ explicitly get canceled out.
Doing the same thing to $a^4 + b^4 + c^4$, we get
$$\begin{align}a^4 + b^4 + c^4
&= \sum_{k=0}^2 (\rho^2 + 1 + (z\omega^k + \bar{z}\omega^{-k}))^2\\
&= \sum_{k=0}^2\left[ (\rho^2 + 1)^2 + (z\omega^k + \bar{z}\omega^{-k})^2\right]\\
&= 3(\rho^2 + 1)^2 + 6\rho^2\end{align}$$
Combine these, we obtain
$$16S'^2 = 3(\rho^2+1)^2 - 12\rho^2 = 3(1 - \rho^2)^2$$
Since $M$ is inside triangle $ABC$, we have $\rho^2 \le 1$. As a result,
$$S' = \frac{\sqrt{3}}{4}(1-\rho^2) \le \frac{\sqrt{3}}{4} = \frac13 S$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2993965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$. I have no idea how to do this question.
I'm given $\int^{\infty}_{-\infty}\frac{1}{x^2+2ax+b^2}dx=\frac{\pi}{\sqrt{b^2-a^2}}$ if $b>|a|$ and I'm asked to prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$.
What I've tried:
$\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)}dx=\frac{2\pi}{\sqrt3}$. I've tried setting a variable as the power, ie: $I(r)=\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^r}dx$ but differenciating the inside w.r.t to $r$ doesn't seem to form any differential equation that can be solved.
Answers and hints appreciated!
|
$$I(r)=\int_{-\infty}^{\infty} \frac {dx}{\left(\left(x+\frac 12\right)^2+\frac 34\right)^r}=\int_{-\frac 12}^{\infty} \frac {dx}{\left(\left(x+\frac 12\right)^2+\frac 34\right)^r}+\int_{-\infty}^{-\frac 12} \frac {dx}{\left(\left(x+\frac 12\right)^2+\frac 34\right)^r}$$
Substitute $x+\frac 12=u$
$$I(r)=\int_0^{\infty} \frac {du}{\left(u^2+\frac 34\right)^r}+\int_{-\infty}^0 \frac {du}{\left(u^2+\frac 34\right)^r}$$
Since the integrand is even we get $$I(r)=2\int_0^{\infty} \frac {du}{\left(u^2+\frac34\right)^r}=\frac {2\cdot 4^r}{3^r} \int_0^{\infty} \frac {du}{\left(\frac {4u^2}{3}+ 1\right)^r}$$
On substituting $\frac {4u^2}{3}=t$ we get $$I(r)=\frac {2\cdot 4^{r-1}}{3^{r-\frac12}} \int_0^{\infty} \frac {t^{-\frac 12} dt}{(1+t)^r}$$
Now using the definition of Beta function and it's relation with Gamma function we get $$I(r)=\frac {2\cdot 4^{r-1}}{3^{r-\frac12}} B\left(\frac 12,r-\frac 12\right) =\frac {2\cdot 4^{r-1}}{3^{r-\frac12}} \frac {\Gamma\left(\frac 12\right)\Gamma\left(r-\frac 12\right)}{\Gamma(r)}=\frac {2\cdot 4^{r-1}}{3^{r-\frac12}} \frac {\sqrt {\pi}\cdot \Gamma\left(r-\frac 12\right)}{\Gamma(r)}$$
In general it can be proved that $$I(a,b,r)=\int_{-\infty}^{\infty} \frac {dx}{(x^2+2ax+b^2)^r} = \frac {\sqrt {\pi}\cdot\Gamma\left(r-\frac 12\right)}{\Gamma(r)} \left(\frac {1}{b^2-a^2}\right)^{r-\frac 12}$$ if $\vert b\vert \gt \vert a\vert$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2994751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
Is there a $\gamma>1$ such that $\frac{\binom{4p}{2p-1}}{\binom{4p}{0}+\binom{4p}{1}+\cdots+\binom{4p}{p-1}}>\gamma^{p}$ In a proof of the Larman-Rogers conjecture (there is $\gamma>1$ such that $\chi(\mathbb{R}^{d})>\gamma^d) $ they used that there is a $\gamma>1$ such that $\frac{\binom{4p}{2p-1}}{\binom{4p}{0}+\binom{4p}{1}+\cdots+\binom{4p}{p-1}}>\gamma^{p}$. I do not see why this is true. I tried manipulating both parts of the fraction to get the desired result. I also tried to use Stirling's approximation but without success.
EDIT: using $(\frac{n}{k})^k$ $\geq$ $\binom{n}{k}$ $\geq$ $(\frac{en}{k})^k$ seems to help a lot but it is still not getting quite right.
|
By induction on $n$, using Pascal's rule:
$$
\binom{n}{r} = \binom{n - 1}{r} + \binom{n - 1}{r - 1} \quad (n \geqslant r \geqslant 1),
$$
we get, for $n > r \geqslant 0$:
\begin{equation}
\tag{$*$}\label{eq:id}
\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{r} = \binom{n - 1}{r} + 2\binom{n - 2}{r - 1} + \cdots + 2^{r - 1}\binom{n - r}{1} + 2^r.
\end{equation}
Suppose:
$$
\frac{r}{n - 1} \leqslant \frac{1}{4}.
$$
Then:
$$
\frac{r - k}{n - k - 1} \leqslant \frac{1}{4} \quad (k = 0, 1, \ldots, r),
$$
whence:
\begin{gather*}
\binom{n - k - 1}{r - k} = \left(\frac{r}{n - 1}\right)\left(\frac{r - 1}{n - 2}\right)\cdots\left(\frac{r - k + 1}{n - k}\right)\binom{n - 1}{r} \\
\leqslant 4^{-k}\binom{n - 1}{r} \quad (k = 0, 1, \ldots, r).
\end{gather*}
From \eqref{eq:id}, therefore:
\begin{gather*}
\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{r} \leqslant \left(1 + 2^{-1} + \cdots + 2^{-r+1} + 2^{-r}\right)\binom{n - 1}{r} \\
< 2\binom{n - 1}{r} \quad \left(0 \leqslant r \leqslant \left\lfloor\frac{n - 1}{4}\right\rfloor\right).
\end{gather*}
In particular:
$$\binom{4p}{0} + \binom{4p}{1} + \cdots + \binom{4p}{p - 1} < 2\binom{4p - 1}{p - 1}
\quad (p \geqslant 1).
$$
Therefore:
\begin{gather*}
\frac{\binom{4p}{2p-1}}{\binom{4p}{0}+\binom{4p}{1}+…+\binom{4p}{p-1}}
> \frac{\binom{4p}{2p - 1}}{2\binom{4p - 1}{p - 1}}
= \frac{2p(p - 1)!(3p)!}{(2p - 1)!(2p + 1)!} \\
= 2\left(\frac{2p + 2}{p + 1}\right)
\left(\frac{2p + 3}{p + 2}\right)\cdots
\left(\frac{3p}{2p - 1}\right) \\
\geqslant 2\left(\frac{3}{2}\right)^{p - 1}
\geqslant \left(\frac{3}{2}\right)^p\quad (p \geqslant 1).
\end{gather*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2995177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$? If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?
I tried with Tchebyshev inequality on sets $\{a, b, c\}$ and $\{a^2, b^2 , c^2\}$ but was blocked. I also tried with mean of $m-$the power with $m=3$ but again could not proceed further.
I only know elementary inequalities like Cauchy-Schwarz and AM-GM apart from above mentioned inequalities. Please help.
|
For $a=b=c=3$ we obtain a value $81$.
We'll prove that it's a minimal value.
Indeed, we need to prove that $$a^3+b^3+c^3\geq81,$$ which is true because
$$a^3+b^3+c^3-81=\sum_{cyc}(a^3-27)=$$
$$=\sum_{cyc}\left(a^3-27-\frac{9}{2}(a^2-9)\right)=\frac{1}{2}\sum_{cyc}(a-3)^2(2a+3)\geq0.$$
Also, Holder helps:
$$a^3+b^3+c^3=\sqrt{\frac{(a^3+b^3+c^3)^2(1+1+1)}{3}}\geq\sqrt{\frac{(a^2+b^2+c^2)^3}{3}}=81.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3001046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Evaluating $\int_0^1\frac{\ln(1+x-x^2)}xdx$ without using polylogarithms.
Evaluate $$I=\int_0^1\frac{\ln(1+x-x^2)}xdx$$ without using polylogarithm functions.
This integral can be easily solved by factorizing $1+x-x^2$ and using the values of dilogarithm at some special points.
The motivation of writing this post is someone said that this integral cannot be solved without using special functions.
Another alternative solutions will be appreciated.
|
Put
\begin{equation*}
I=\int_{0}^1\dfrac{\ln(1+x-x^2)}{x}\,\mathrm{d}x = \int_{0}^1\dfrac{\ln(1+x(1-x))}{x}\mathrm{d}x.
\end{equation*}
If we change $x$ to $ 1-x $ we get
\begin{equation*}
I=\int_{0}^1\dfrac{\ln(1+x-x^2)}{1-x}\,\mathrm{d}x.
\end{equation*}
Consequently
\begin{equation*}
2I = \int_{0}^1\ln(1+x-x^2)\left(\dfrac{1}{x}+\dfrac{1}{1-x}\right)\,\mathrm{d}x.
\end{equation*}
The next step will be integration by parts.
\begin{equation*}
2I= \underbrace{\left[\ln(1+x-x^2)\ln\dfrac{x}{1-x}\right]_{0}^{1}}_{=0} -\int_{0}^1\dfrac{1-2x}{1+x-x^2}\ln\dfrac{x}{1-x}\, \mathrm{d}x
\end{equation*}
Then\begin{equation*}
I=\dfrac{1}{2}\int_{0}^1\dfrac{2x-1}{1+x-x^2}\ln\dfrac{x}{1-x}\, \mathrm{d}x.
\end{equation*}
If we substitute $ z=\dfrac{x}{1-x} $ we get
\begin{equation*}
I = \int_{0}^{\infty}\dfrac{(z-1)\ln z}{2(z+1)(z^2+3z+1)}\,\mathrm{d}z.
\end{equation*}
In order to evaluate this integral we integrate $\displaystyle \dfrac{(z-1)\log^2 z}{2(z+1)(z^2+3z+1)}$ along a keyhole contour and use residue calculus. We get that
\begin{equation*}
I = 2\ln^2\varphi
\end{equation*}
where $ \varphi = \dfrac{1+\sqrt{5}}{2}. $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3002309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
How do I simplify this expression? How do I simplify:
$$d=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+2+x^2-y^2-2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
If simplification is possible, it should be possible with elementary algebra, but I'm completely lost as to how to go about it.
What I've done so far:
$$d=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+2+x^2-y^2-2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
$$=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+x^2-y^2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}=\frac{-α^2 x^2 - 2 α x^2 + α^2 y^2 + 2 α y^2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
$$\implies d^2=\frac{(-α^2 x^2 - 2 α x^2 + α^2 y^2 + 2 α y^2 - 2)^2}{\alpha^2x^2+\alpha^2y^2}$$
|
$$\frac{-\alpha^2x^2+\alpha^2y^2-2\alpha x^2+2\alpha y^2}{\sqrt{\alpha^2x^2+\alpha^2y^2}}=\frac{\alpha^2(y^2-x^2)+2\alpha(y^2-x^2)}{\alpha\sqrt{x^2+y^2}}$$
$$\frac{\alpha^2(y^2-x^2)}{\alpha\sqrt{x^2+y^2}}+\frac{2\alpha(y^2-x^2)}{\alpha\sqrt{x^2+y^2}}=\alpha\frac{(y^2-x^2)}{\sqrt{x^2+y^2}}+2\frac{y^2-x^2}{\sqrt{x^2+y^2}}$$
Thanks: @Timmay & @David Diaz
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3003860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Binomial Theorem with Three Terms $(x^2 + 2 + \frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
|
The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$\left((x^2+2)+\frac{1}{x}\right)^7=\sum _{k=0}^7 \binom{7}{k} \left(x^2+2\right)^k x^{k-7}$$
Second step
$$\left(x^2+2\right)^k=\sum _{m=0}^k 2^{k-m} x^{2 m} \binom{k}{m}$$
Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$
$$\sum _{m=0}^7 2^{15-3 m} \binom{7}{15-2 m} \binom{15-2 m}{m}$$
Since, for $n, m = 0,1,2,...$ the binomial coefficient $\binom{n}{m}$ is zero unless $n\ge m$ we find $7\ge 15-2m \to m \ge 4$ and $15-2m\ge m \to m\le 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3004752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
}
|
Find $\alpha$,$\beta$ if $\lim_{x→∞}[ \sqrt {ax^2+2bx+c} - \alpha x -\beta] = 0$ Here is my approach.
Consider;
$$ax^2 + 2bx + c = 0$$
or;
$$ x_{±} = \frac {-b±\sqrt {b^2-ac}}{a} = \frac {-b±\sqrt D}{a}$$
Hence;
$$\sqrt {ax^2+2bx+c} = \sqrt {(x+\frac {b-√D}{a})(x+\frac {b+√D}{a})}$$
$$=x\sqrt {(1+\frac {b-√D}{ax})(1+\frac {b+√D}{ax})}$$
For large values of $x$ we may apply the binomial approximation, so that;
$$\sqrt {ax^2+2bx+c} ≈ x(1+\frac {b-√D}{2ax})(1+\frac {b+√D}{2ax})$$
$$=x + \frac {b}{a} + \frac {c}{4ax}$$
As $x→∞$ the final term in the above expression vanishes. Hence;
$$\lim_{x→∞}[ \sqrt {ax^2+2bx+c} - \alpha x - \beta ] = 0,$$
gives;
$$x+\frac {b}{a} - \alpha x - \beta = 0,$$
or;
$$(1-\alpha)x + (\frac {b}{a} - \beta) = 0$$
As $x→∞$, $1-\alpha$ must be $0$ for the former term to vanish, hence,
$$\alpha = 1, \beta = \frac {b}{a}$$
But I doubt it is hardly correct. Is there any better method for the problem?
|
Another approach could be Taylor series
$$\sqrt{a x^2+2 b x+c}=x \sqrt{a+\frac{2 b}{x}+\frac{c}{x^2} }$$ So, for large $x$,
$$\sqrt{a x^2+2 b x+c}=\sqrt{a} x+\frac{b}{\sqrt{a}}+\frac{a c-b^2}{2 a^{3/2}
x}+O\left(\frac{1}{x^2}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3007306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Number of integral solutions $x_1 + x_2 + x_3 = 10$ $x_1 + x_2 + x_3 = 10, \ \ 0 \leq x_1 \leq 10 , \ 0 \leq x_2 \leq 6 , \ 0 \leq x_3 \leq 2 $
$[x^{10}] (1 + x^1 + x^2 + ... + x^{10})(1 + x^1 + x^2 + ... + x^{6})(1 + x + x^2)$
$[x^{10}] \Large (\frac{1 - x^{11}}{1-x})(\frac{1 - x^{7}}{1-x})(\frac{1 - x^{3}}{1-x})$
$[x^{10}] \Large( \frac{(1 - x^7 - x^{11} + x^{18}) (1-x^3)}{(1-x)^3})$
$[x^{10}] \Large( \frac{1 - x^3 - x^7 + x^{10}}{(1-x)^3})$
$[x^{10}] (1 - x^3 - x^7 + x^{10})(1-x)^{-3}$
how to proceed further?
|
Using
$$(1-x)^{-n}=1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+\dotsb+\frac{n(n+1)\dotsb(n+k-1)}{k!}x^k+\dotsb$$
We get
$$(1-x)^{-3}=1+3x+\frac{12}{2}x^2+\frac{60}{6}x^3+\dotsb +\frac{3\cdot 4\cdot \dotsb (3+k-1)}{k!}x^k+\dotsb$$
Now you want coefficient of $x^{10}$ in $(1 - x^3 - x^7 + x^{10})\color{blue}{(1-x)^{-3}}$. This can be obtained by collecting coefficient of $x^{10}$ obtained by multiplying the various terms. What I mean by that is we will obtain $x^{10}$ by multiplying $1$ with $\color{blue}{x^{10}}$, $x^3$ with $\color{blue}{x^7}$, $x^7$ with $\color{blue}{x^3}$ and finally $x^{10}$ with $\color{blue}{1}$. Thus the coefficient of $x^{10}$ obtained will be
\begin{align*}
&=(1)\color{blue}{\left(\frac{3\cdot 4 \dotsb 12}{10!}\right)}+(-1)\color{blue}{\left(\frac{3\cdot 4 \dotsb 9}{7!}\right)}+(-1)\color{blue}{\left(\frac{3\cdot 4 \cdot 5}{3!}\right)}+(1)\color{blue}{(1)}\\
&=66-36-10+1\\
&=21.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3007431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find the function equation $f(x+f(y)+yf(x))=y+f(x)+xf(y)$ find all functions $f:\Bbb R\to \Bbb R$ and such for any $x,y\in\Bbb R$,and such
$$f(x+f(y)+yf(x))=y+f(x)+xf(y)$$
I have proven that $$f(f(x))=x$$
proof:Let $y=0$,we have
$$f(x+f(0))=f(x)+xf(0)$$
let $x=0,y\to x$,we have
$$f(f(x)+xf(0))=x+f(0)$$
so we have
$$f(f(x+f(0))=x+f(0)$$
then $$f(f(x))=x$$
and $f(x)=x$ is one solution but I can't argue this to be the only solution.
It doesn't seem easy.Thank for you help!
|
This is only a partial result (EDIT: full result is given below), but I'll post it anyway. Let us write $(x,y)$ if $y = f(x)$ holds. Since $f(f(x)) = x$, $(x,y)$ is equivalent to $(y,x)$ and $(x,y)\wedge (x,y')$ implies $y= y'$. What we can directly observe is that $0= f(f(0)) = f(0)$ by letting $x=y=0$. What is less trivial is $f(1) = 1$. Let $\gamma =f(1)$. Then by letting $y = 1$ and $y= \gamma$, we get
$$
(x+ \gamma + f(x),1+ f(x) +\gamma x)\wedge \;(x + 1+\gamma f(x), \gamma + f(x) +x).$$
Hence, $1+ f(x) +\gamma x = x + 1+\gamma f(x)$ for all $x$, and if $\gamma \neq 1$, then $f(x) = x$, leading to contradiction.
We next show that $$(-\frac{f(x) - f(x')}{x-x'}, -\frac{x-x'}{f(x) - f(x')}),\quad \forall x\neq x'\;\cdots(*).$$ Assume $(a,b)\wedge (x,y)\wedge (x',y').$ Then, by the FE,
$$
(a+x+by, b+y +ax)\wedge (a+x'+by', b+y' +ax').
$$ We can equate $b+y +ax$ and $ b+y' +ax'$ by letting $a= -\frac{y-y'}{x-x'}$. Then $b$ should satisfy $a+x+by = a+x'+by'$, that is, $ b= -\frac{x-x'}{y-y'}.$ Because $y = f(x), y' = f(x')$, this proves the claim.
From the previous claim, we know that $(-1,-1)$, i.e. $f(-1) = -1$. By letting $x' = 0, - 1$, we also know that $(x,y)$ implies
$$(-\frac{y}{x}, -\frac{x}{y})\wedge (-\frac{y+ 1}{x + 1}, -\frac{x+ 1}{y + 1}).
$$If we write $-\frac{y-y'}{x-x'}= k$, then $(-\frac{1/k+ 1}{k + 1}, -\frac{k+ 1}{1/k + 1}) = (-\frac{1}{k}, -k)$ also shows that $(\frac{y-y'}{x-x'}, \frac{x-x'}{y-y'})$ whenever $x\neq x'$. Next, by putting $y=-1$ in the original functional equation, we also have $(x,y)$ implies
$$(x-y-1, y-x-1).$$ Iterating this $n$ times we also have
$$(2^n(x-y)-1, 2^n(y-x)-1) \;\cdots (**).
$$
Our next claim is that if $(x,x) \vee (x, x+2^{1-N})$, then $(x+ 2^{-N}, x+2^{-N})$ ($N\geq 0$.) To show this, let us write $(x+ 2^{-N}, \alpha)$ and derive an equation about $\alpha$. If $(x,x)$, $(*)$ implies $( 2^N(x-\alpha ), \frac{1}{2^N(x-\alpha )})$. By applying $(**)$ to $(x+ 2^{-N}, \alpha)$ we also have $(2^N(x-\alpha), 2^N(\alpha-x)-2)$. Hence, $2^N(\alpha-x)-2 = \frac{1}{2^N(x-\alpha )}$ and this implies $2^N(x-\alpha )=-1$ as desired. In case $(x, x+2^{1-N})$ can be dealt with similarly by noting that $(-2^N(x-\alpha )-2, - \frac{1}{2^N(x-\alpha )+2})$.
A very similar argument can also prove that
$$(x,x) \vee (x, x-2^{1-N})\Rightarrow (x-2^{-N}, x-2^{-N}).$$
So far, all the tedious arguments show that $f(x) = x$ holds for every dyadic rationals $x=\frac{j}{2^n}$, that is, the set $F$ of fixed points of $f$ contains every dyadic rational. The following are some other facts about $F$: (i) If $(x,y)$, then $x+y+xy$ and $x+y+1$ belongs to $F$.
(ii) If $x,y \in F$, then $x+y+xy \in F$.
(iii) If $x \in F$, then $x\pm \frac{j}{2^n}\in F$. (Above claim)
And for some $(x,f(x))$ such that $x\notin F$, this generates so many non-fixed pairs
$$(\pm \left[\frac{f(x) - q}{x-q}\right]^r,\pm \left[\frac{x- q}{f(x)-q}\right]^r), \quad (x+q+qf(x),f(x)+q+qx)_{q \neq 1},$$ for all dyadic rational $q$ and $r\in \mathbf{N}$.I tried but failed to get further ideas about how $F$ and $\mathbf{R}\setminus F$ looks like. But I wish this helps somehow.
(Note: Actually in the above proposition, $(x, x\pm 2^{1-N})$ cannot occur since their difference cannot be in $F$.)
Following yesterday's post, I've completed proof: $f = id$ is the only solution of the equation.
I'll briefly review some facts already proved.
(i) $F + \mathbf{Q}_{dyad} = F.$
(ii) If $x,y \in F$, then $xy \in F$, or equivalently, $F\cdot F = F$. (Since $x-1, y-1 \in F$ implies $xy -1\in F$ and thus $xy \in F$.)
(iii) If $(x,y)$, then $x+y, \;xy + x+ y \in F$.
Our claim starts with: If $0\neq x \in F$, then $\frac{1}{x} \in F$. Proof is simple. Let $(\frac{1}{x},\gamma).$ Then $\gamma +\frac{1}{x}\in F$. Since $x\in F$, $\gamma x + 1 \in F$, and $\gamma x \in F$. Note that $(\gamma x, \frac{1}{\gamma x}).$(slope of $(\frac{1}{x},\gamma),(0,0).$) This shows $\gamma = \pm \frac{1}{x}.$ If it were that $(\frac{1}{x},-\frac{1}{x})$, then $\frac{1}{x^2} \in F$. This implies that $x\cdot\frac{1}{x^2}=\frac{1}{x} \in F$, as desired.
Our next claim is that $(x,y)$ implies $xy \in F$. We start from $x+y, \;x+y+xy \in F$. If $x+y =0$, it's already done. Otherwise, $\frac{x+y+xy}{x+y} = 1+ \frac{xy}{x+y} \in F$, hence $\frac{xy}{x+y}\in F$. Then this implies $$(x+y)\cdot \frac{xy}{x+y} = xy \in F.$$
Our almost final claim is that $0< y \in F$ implies $\sqrt{y} \in F$. Suppose $(\sqrt[4]{y}, \gamma)$. Then $\gamma^2 \sqrt{y} \in F$. Since $\frac{1}{y} \in F$, we have $\frac{\gamma^2}{\sqrt{y}} \in F.$ Notice that $(\frac{\gamma}{\sqrt[4]{y}}, \frac{\sqrt[4]{y}}{\gamma})$ and hence $(\frac{\gamma^2}{\sqrt{y}}, \frac{\sqrt{y}}{\gamma^2})$ Thus we have $\gamma^4 = y$, $\gamma = \pm \sqrt[4]{y}$. Suppose $\gamma = \sqrt[4]{y}$. Then, $\sqrt[4]{y} \in F$ implies $\sqrt{y} \in F$. Otherwise, $(\sqrt[4]{y}, -\sqrt[4]{y})$ implies $-\sqrt{y} \in F$. Thus $\sqrt{y} \in F.$
Finally we are ready to prove that $(x,y)$ implies $ x-y=d =0$. Assume to the contrary that $d>0$. Then, as I showed in yesterday's post, $(d-1, -d-1).$ But this implies $(-1+d)\cdot(-1-d) = 1-d^2 \in F$, and thus $d^2 \in F.$ So $d$ must be in $F$ and $d-1$ also. This leads to $d-1 = -d-1$, contradicting $d>0$. So, the only solution of the functional equation is $f(x) = x$!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3007645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Contour intergal of a rational trigonometric function, can't find my mistake. This is the integral :
$$I = \int_{0}^{2\pi} \frac{dx}{(5+4\cos x)^2}\ $$
Which, according to wolfram alpha, should evaluate to $\frac{10\pi}{27} $,
but the value i find is $\frac{20\pi}{27} $.
These are my calculations :
Using complex form of cosine i get $ \cos x = \frac{t^2+1}{2t}$
where $ t = e^{ix}$ and $ dx = \frac{dt}{it}$.
If $\partial{D} $ is the unit circle
$$ I = \int_{\partial{D}} \frac{-i}{t(\frac{5t+2t^2+2}{t})^2} dt =
-i \int_{\partial{D}}\frac{t}{(t+2)^2(t+1/2)^2}dt = -i I_a$$
$I_a = \pi i $ times the residue in -$\frac{1}{2}$
The residue is equal to
\begin{align}
\lim_{z\to-\frac{1}{2}} \frac{d}{dz} \left( \frac{t(t+1/2)^2}{(t+2)^2(t+1/2)^2}\right)
&= \lim_{z\to-\frac{1}{2}} \frac{d}{dz} \frac{t}{(t+2)^2}
\\&= \lim_{z\to-\frac{1}{2}} \frac{(t+2)^2-2t(t+2)}{(t+2)^4}
\\&= \lim_{z\to-\frac{1}{2}} \frac{2-t}{(t+2)^3}
= \frac{2+1/2}{(-1/2+2)^3}
\\&= \frac{5/2}{27/9} = \frac{20}{27}
\end{align}
so $ I_a = i\pi\frac{20}{27}$
So our original integral $$I = -i I_a = -i^2 \pi\frac{20}{27} = \frac{20\pi}{27}$$
|
You made an error in calculating the factorization of the denominator. You should get $$5t+2t^2+2=(2+t)(1+2t)=2(t+2)(t+1/2),$$ where you missed to transfer the leading factor $2$ to the factorized expression. The correction leads to $I=-\frac i4 I_a$ and the additional division by $4$ corrects the residuum that is computed with the correct factor $2\pi i$.
You could also use the geometric/binomial series to get the correct residual coefficient. Then, using $u=t+\frac12$,
$$
\frac{t}{(t+2)^2(t+\frac12)^2}=\frac{u-\tfrac12}{(u+\frac32)^2u^2}
=\frac49(u-\tfrac12)\sum_{k=0}^\infty (k+1)\left(-\frac23\right)^ku^{k-2}
$$
so that the coefficient for the power $u^{-1}=(t+\frac12)^{-2}$ is $\frac49(1+\frac12\cdot2\cdot\frac23)=\frac{20}{27}$. The integration over the unit circle adds the factor $2\pi i$, so that in the final result $I=\frac{10\pi}{27}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3008258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Condition in terms of b and a if $ax^2+bx+c=0$ has two consecutive odd positive integers as roots
The roots of the equation$$ax^2+bx+c=0$$, where $a \geq 0$, are two consecutive odd positive integers, then
(A) $|b|\leq 4a$
(B) $|b|\geq 4a$
(C) $|b|=2a$
(D) None of these
My attempt
Let p and q be the roots then if they are consecutive positive integers (q>p) then $$ pq=\frac{c}{a} \geq 0$$
So, $$c \geq 0$$ and $$q-p=2$$
So, $$\frac{\sqrt{b^2-4ac}}{a}=2$$
So,$$|b|>2a$$
$(Since, a>0,c>0)$
But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it.
Any hints and suggestions are welcome!
|
Since we are given that there are two distinct roots and that $a \ge 0$, we must have
$a > 0, \tag 0$
since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.
Let
$n \ge 0, \tag 1$
$r = 2n + 1, \tag 2$
$s = 2n + 3; \tag 3$
suppose for the moment
$a = 1; \tag 4$
then
$(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; \tag 5$
here we have
$b = -(4n + 4), \tag 6$
whence
$\vert b \vert = 4n + 4 \ge 4 = 4a; \tag 7$
thus (B) binds when $a = 1$; now if
$a \ne 1, \tag 8$
the quadratic of the form
$ax^2 + bx + c = a(x^2 + \dfrac{b}{a} x + \dfrac{c}{a}) \tag 9$
has zeroes $2n + 1$, $2n + 3$ provided
$(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = x^2 + \dfrac{b}{a} x + \dfrac{c}{a}; \tag{10}$
thus,
$\dfrac{b}{a} = -(4n + 4), \tag{11}$
or
$b = -(4n + 4)a, \tag{12}$
whence, with $a > 0$,
$\vert b \vert = (4n + 4)a \ge 4a, \tag{13}$
and we see that the correct choice is (B) here as well.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3008382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$ If $x \ge 5$, is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$
I believe the answer is yes.
Here is my thinking:
(1) $\log_2{5} > 2.32 > 2.284 > 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$
(2) Assume up to $x$ that $\log_2 x > \sum\limits_{i \le x}\frac{1}{i}$
(3) $(x+1)^{x+1} > {{x+1}\choose{x+1}}x^{x+1} + {{x+1}\choose{x}}x^x = x^{x+1} + (x+1)x^x = 2x^{x+1} + x^x > 2(x^{x+1})$
(4) $(x+1)\log_2\left(\frac{x+1}{x}\right) > 1$
(5) $\log_2(x+1) - \log_2(x) > \frac{1}{x+1}$
(6) $\log_2(x+1) - \log_2(x) + \log_2(x) = \log_2(x+1) > \sum\limits_{i \le x+1}\frac{1}{i}$
|
We have that
$$1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x \iff 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \frac{\ln x}{\ln 2}$$
$$\ln 2\left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} \right)< \ln x$$
then recall that by harmonic series
$$\sum_{k=1}^x \frac1k \sim\ln x+\gamma+\frac1{2x}$$
then
$$\ln 2\left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} \right) \sim \ln 2\ln x+\gamma\ln 2+\frac{\ln 2}{2x}\stackrel{?}<\ln x$$
$$(1-\ln 2)\ln x> \gamma\ln 2+\frac{\ln 2}{2x}$$
which is true for $x$ sufficiently large.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3010094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
How do I simplify $\frac{\sin(2x)}{1-\cos (2x)}$? $$\dfrac{\sin(2x)}{1-\cos (2x)}$$
How do I simplify given expression?
My attempt:
We know the double-angle identity for $\sin(2x)$ and $\cos(2x)$ as shown below
$$\sin(2x) = 2\sin (x)\cos(x)$$
$$\cos(2x) = 2\cos(x)-1$$
So we have that
$$\dfrac{2\sin (x)\cos(x)}{1-2\cos(x)-1} = \dfrac{2\sin(x)\cos(x)}{-2\cos^2(x)} = \dfrac{2\sin(x)}{-2\cos(x)} = -\tan (x)$$
I believe I've gone wrong somewhere.
|
You got one of the signs mixed up.
$$\frac{\sin(2x)}{1-\cos (2x)} = \dfrac{\sin(2x)}{1-(2\cos^2x-1)} = \frac{\sin(2x)}{1-2\cos^2x+1}$$
Instead, you accidentally used
$$\frac{\sin(2x)}{1-\cos (2x)} \color{red}{\neq \frac{\sin(2x)}{1-2\cos^2x-1}}$$
As for the simplification itself,
$$\frac{\sin(2x)}{1-\cos (2x)} = \frac{2\sin x\cos x}{1-(1-2\sin^2x)} = \frac{2\sin x\cos x}{2\sin^2 x} = \frac{\cos x}{\sin x} = \cot x$$
Also, to point out, you used $\cos(2x) = 2\cos^2x-1$, which doesn’t actually get you anywhere since you get two $+1$’s. Using $\cos^2x = 1-2\sin^2x$ is the way to go.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3010796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Recurrence k-th pattern I am trying to solve this recurrence $T(n) = 6 T(\frac{n}{3}) + n$.
1st recurrence: $6^2T(\frac{n}{3^2}) + \frac{6n}{3} + n$
2nd: $6^3T(\frac{n}{3^3}) + \frac{6^3n}{3^2} + n$
3rd: $6^4T(\frac{n}{3^4}) + \frac{6^6n}{3^3} + n$
I am having trouble describing the general pattern after the k-th iteration.
|
If you are trying to solve the recurrence, try finding the initial pattern and the number of terms - this will help solve the equation in most of the cases.
In the first step, we have $n$ work to do, we do $n$ work and have 6 problems of the same type but of size $\frac{n}{3}$ this time.
In the second step, we have $\frac{n}{3}$ work to do, we do $\frac{n}{3}$ work and have 6 problems of the same type but of size $\frac{n}{9}$ this time. We have 6 problems of this type.
In the third step, we have $\frac{n}{9}$ work to do, we do $\frac{n}{9}$ work and have 6 problems of the same type but of size $\frac{n}{27}$ this time. We have 6 problems of this type.
This gives the equation,
$T(n) = n + 6(\frac{n}{3}) + 36 (\frac{n}{9}) + 216(\frac{n}{27}) + ... + $
$T(n) = n + 2n + 4n + 8n + ... $
You will have $log_{3}n$ such terms, as you are cutting down the problem size by a factor of three all the time. This is a sum of a geometric series, given the number of terms and common ratio, it is easy to sum.
$T(n) = n + 2n + 4n + 8n + ... + 2^{log_{3}n}n $
Kth term will be of the form $2^{k}n$.
The solution should be $\Theta(n^{log_{3}6})$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3011002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Sum of $\{n\sqrt{2}\}$ I would like to prove (rigorously, not intuitively) that
$$\sum_{n=1}^N \{n\sqrt{2}\}=\frac{N}{2}+\mathcal{O}(\sqrt{N})$$
where $\{\}$ is the "fractional part" function. I understand intuitively why this is true, and that's how I came up with this claim - $\{n\sqrt{2}\}$ behaves like a random variable uniformly distributed in $(0,1)$, and treating it as a random variable makes the sum analogous to a random walk, and the $\mathcal{O}(\sqrt{N})$ bound can be shown using expected values.
However, just saying that $\{n\sqrt{2}\}$ behaves "like a random variable" is highly non-rigorous. Can someone show me how to justify this in a more airtight way (ideally without using any theorems that require specialized background knowledge)?
Thanks!
|
Writing
\begin{align}
&\quad \sum_{n=1}^N \left( n\sqrt{2} - \lfloor n\sqrt{2} \rfloor \right) \\
&=\int_1^N \left( n\sqrt{2} - \lfloor n\sqrt{2} \rfloor \right) {\rm d}n + {\cal O}(1) \tag{1} \\
&=\frac{1}{\sqrt{2}} \int_\sqrt{2}^{\sqrt{2}N} \left(x- \lfloor x \rfloor \right) {\rm d}x + {\cal O}(1) \\
&=\frac{1}{\sqrt{2}} \Bigg( N^2 - 1 - \Bigg[ \frac{\sqrt{2}N \left(\sqrt{2}N-1\right)}{2} + \frac{\left\{\sqrt{2}N\right\} \left(1-\left\{\sqrt{2}N\right\}\right)}{2} \\
&\quad - \frac{\sqrt{2} \left(\sqrt{2}-1\right)}{2} - \frac{\left\{\sqrt{2}\right\} \left(1-\left\{\sqrt{2}\right\}\right)}{2} \Bigg] \Bigg) + {\cal O}(1) \\
&=\frac{N}{2} + {\cal O}(1)
\end{align}
where we used
$$
\int_0^x \lfloor t \rfloor \, {\rm d}t = \frac{x(x-1)}{2} + \frac{\left\{x\right\}\left(1-\left\{x\right\}\right)}{2} \, .
$$
The order follows from the fact that
$$
\frac{\left( \sqrt{2}N - \lfloor \sqrt{2}N \rfloor \right) + \left( \sqrt{2} - \lfloor \sqrt{2} \rfloor \right)}{2} = {\cal O}(1) \tag{2}
$$
and
$$\int_1^{N} \left( n\sqrt{2} - \lfloor n\sqrt{2} \rfloor \right)'' {\rm d}n \tag{3} \\
= \left( \sqrt{2}n - \lfloor \sqrt{2}n \rfloor \right)'\big|_{n=N} - \left( \sqrt{2}n - \lfloor \sqrt{2}n \rfloor \right)'\big|_{n=1} \\
=\sqrt{2} \sum_{k=-\infty}^{\infty}\left[ \delta\left( \sqrt{2} - k \right) - \delta\left( \sqrt{2}N - k \right) \right]
$$
but this requires Euler-Maclaurin.
Due to the harsh critic about my somewhat heuristic argument I want to correct my approach as far as possible.
Set $$f(x)=x-\lfloor x \rfloor$$ and $$f_n(x)=\frac{1}{2} - \frac{1}{\pi} \sum_{k=1}^n \frac{\sin(2\pi k x)}{k} \, ,$$such that
$$
\lim_{n\rightarrow \infty} f_n(x) = f(x) \, .
$$
Since $f_n$ is differentiable we can use Euler-Maclaurin to calculate the sum
$$
\sum_{k=1}^N f_n(ak)
$$
with some $a$. The integral in (1) does not create much of an issue in the limit $n \rightarrow \infty$, since the limit is piecewise continuous and the integral can be splitted accordingly and then integrated. Also the limit of (2) is of ${\cal O}(1)$. So the problematic term which needs to be examined is the remainder $R_2$ ((3) was very heuristic) which can be written as
$$
R_2=\int_1^N B_2\left(t-\lfloor t \rfloor\right) \frac{\rm d}{{\rm d}t} f_n'(at) \, {\rm d}t
$$
neglecting unnecessary constants and $B_2$ is the second Bernoulli polynomial. We can express
\begin{align}
f_n'(at) &= 1-\sum_{k=-n}^{n} {\rm e}^{i2\pi k at} = 1 - \frac{\sin\left((2n+1)\pi at\right)}{\sin\left(\pi at\right)} \\
B_2\left(t-\lfloor t \rfloor\right) &= \left(t-\lfloor t \rfloor\right)(\left(t-\lfloor t \rfloor - 1\right) + \frac{1}{6} = \lim_{M \rightarrow \infty} \sum_{k=1}^M \frac{\cos(2\pi kt)}{\pi^2 k^2}
\end{align}
and integrate by parts
$$
R_2=-B_2\left(t-\lfloor t \rfloor\right) \frac{\sin\left((2n+1)\pi at\right)}{\sin\left(\pi at\right)} \Bigg|_{1}^{N} - 2 \sum_{k=1}^{M} \int_1^N \frac{\sin(2\pi kt)}{\pi k} \frac{\sin\left((2n+1)\pi at\right)}{\sin\left(\pi at\right)} \, {\rm d}t \tag{4} \, .
$$
For $a$ not an integer, the first term is bounded and ${\cal O}(1)$ as $n \rightarrow \infty$. The integral can be viewed as a functional for $n \rightarrow \infty$ in which case the Dirichlet kernel acts as a periodic delta-distribution $\sum_{m=-\infty}^{\infty} \delta(at-m)$
$$
\lim_{n \rightarrow \infty} \int_1^N \frac{\sin(2\pi kt)}{\pi k} \frac{\sin\left((2n+1)\pi at\right)}{\sin\left(\pi at\right)} \, {\rm d}t = \sum_{m=\lceil a \rceil}^{\lfloor Na \rfloor} \frac{\sin\left(\frac{2\pi km}{a}\right)}{\pi k a} \, .
$$
Evaluating
$$
\sum_{m=\lceil a \rceil}^{\lfloor Na \rfloor} \lim_{M\rightarrow\infty} -2\sum_{k=1}^{M} \frac{\sin\left(\frac{2\pi km}{a}\right)}{\pi k a} = \sum_{m=\lceil a \rceil}^{\lfloor Na \rfloor} \frac{2\{m/a\}-1}{a} \tag{5}
$$
and using $\sum_{n=1}^{N}\{an\} = \frac{N}{2} + {\cal O}(?)$
this becomes ${\cal o}(N)$. So it is actually true ${\cal O}(1)$ does not follow.
We continue with the integral in (4) for $N$ integer
\begin{align}
&\quad -2\sum_{k=1}^\infty \int_1^N \frac{\sin(2\pi kt)}{\pi k} \frac{\sin\left((2n+1)\pi at\right)}{\sin\left(\pi at\right)} \, {\rm d}t \\
&=-4\sum_{m=1}^n \sum_{k=1}^\infty \int_1^N \frac{\sin(2\pi kt)\cos(2\pi m a t)}{\pi k} \, {\rm d}t \\
&=\frac{4}{\pi^2} \sum_{m=1}^n \sum_{k=1}^\infty \frac{\cos^2(N\pi m a)-\cos^2(\pi ma)}{k^2-m^2 a^2} \\
&=2\sum_{m=1}^n \left[ \frac{\cos^2(N\pi m a)-\cos^2(\pi ma)}{\pi^2 m^2 a^2} - \frac{\cot(\pi ma)\left(\cos^2(N\pi ma) - \cos^2(\pi ma)\right)}{\pi ma} \right]
\end{align}
where $a$ must be an irrational number now. The first term is ${\cal O}(1)$ for $n\rightarrow \infty$.
Any idea for the second?
It can be rewritten as
\begin{align}
&\quad \, \, \sum_{m=1}^n \frac{\cot(\pi ma)\left(\cos^2(N\pi m a) - \cos^2(\pi ma)\right)}{\pi ma} \\
&= \sum_{m=1}^n \frac{\cot(\pi ma)\left(\cos(N2\pi ma) - \cos(2\pi ma)\right)}{2\pi ma} \\
&= - \sum_{m=1}^{n}\cos(m\pi a) \, \frac{\sin\left((N+1)m\pi a\right)}{m\pi a} \, \frac{\sin\left((N-1)m\pi a\right)}{\sin(m\pi a)} \\
&= - \sum_{m=1}^{n} \frac{\sin\left((N+2)m\pi a\right)}{m\pi a} \, \frac{\sin\left((N-1)m\pi a\right)}{\sin(m\pi a)} + \sum_{m=1}^n \frac{ \sin\left(N2\pi ma\right) - \sin\left(2\pi ma\right) }{2\pi ma}
\end{align}
so the second sum is bounded again $\forall N$ and $n \rightarrow \infty$.
Not sure if it helps, but I have the following two identities for the sines
$$
\frac{\sin\left((N-1)nx\right)}{\sin(nx)} = \sum_{l=2}^N \cos\left((N-l)nx\right) \cos^{l-2}(nx)
$$
and
$$
\frac{\sin\left((N-1)nx\right)}{\sin(nx)} = 1+2\sum_{l=1}^{\frac{N}{2}-1} \cos\left(l2nx\right) \, ,
$$
but evaluating this feels as if I'm running in circles.
I added a Figure of the RHS of (5) up to $N=10^6$ which doesn't look anything like $\log$, so either the numbers are just too small or I dont why it has to be $\log$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3014473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
}
|
Finding a mean of biased sample An RV $X$ follows a uniform distribution over $[0,1]$.
Suppose that we cannot observe the realization $x$ of $X$.
Instead, we can observe a value $y$, meaning that $x\in\{y,\frac{1}{2}+\frac{y}{2},\frac{3}{4}+\frac{y}{2}\}$. That is, the true $x$ is one of the three value.
If so, what is the conditional expectation of $X$ given an obersvation of $y\in[0,\frac{1}{2}]$?
I initially thought it should be
$$\frac{1}{3}y+\frac{1}{3}\left(\frac{1}{2}+\frac{y}{2}\right)+\frac{1}{3}\left(\frac{3}{4}+\frac{y}{2}\right)$$
but it looks to be wrong somewow and I can't figure out why..
Any idea about the conditional probabilities and mean?
|
First off, the problem is far less confusing if you state given the value $x$, what values of $y$ can you observe. For example, if $y=1$, then as you put it, $x$ is 1, 1, or 5/4. The last is impossible, and what does two equivalent possibilities mean?
Next, the problem is underdetermined unless you know the probabilities of observing the three different types. You're using 1/3 for each, but it is an unstated assumption.
To have a solvable problem, suppose that the three cases $X=Y$, $X=\frac{1}{2}+\frac{Y}{2}$, and $X=\frac{3}{4}+\frac{Y}{2}$ happen with 1/3 probability each independent of $X$.
First invert these relations to have $Y_1=X$, $Y_2=2X-1$, and $Y_3=2X-\frac{3}{2}$. We have the mixture distribution
$$f_Y(y)=\Pr(X=Y)f_{Y_1}(y)+\Pr\left(X=\frac{1}{2}+\frac{Y}{2}\right)f_{Y_2}(y)+\Pr\left(X=\frac{3}{4}+\frac{Y}{2}\right)f_{Y_3}(y)\text{.}$$
Since $Y_1=X$, it is clear that $f_{Y_1}(y)=1$ for $0\leq y\leq1$. The monotonic change of variables formula gives
$$f_{Y_2}(y)=\left|\frac{dx}{dy}\right|f_X(x)$$
when $x=\frac{1}{2}+\frac{y}{2}$, so $f_{Y_2}(y)=\frac{1}{2}$ for $-1\leq y\leq1$. Similarly $f_{Y_3}(y)=\frac{1}{2}$ for $-\frac{3}{2}\leq y\leq\frac{1}{2}$.
Writing down the conditional expectation formally is a pain because of the mixed discrete/continuous distributions, so just consider all the possibilities. When $0\leq y\leq\frac{1}{2}$, all 3 cases are applicable, so $f_Y(y)=2/3$.
With relative probability $\Pr(X=Y)f_{Y_1}(y)=1/3$ we are in the $X=Y$ case (so $X=y$), and with relative probability $\Pr\left(X=\frac{1}{2}+\frac{Y}{2}\right)f_{Y_2}(y)=1/6$ we are in the $X=\frac{1}{2}+\frac{Y}{2}$ case (so $X=\frac{1}{2}+\frac{y}{2}$); similarly for the third.
Thus the conditional expectation is
$$\frac{\frac{1}{3}(y)+\frac{1}{6}(\frac{1}{2}+\frac{y}{2})+\frac{1}{6}(\frac{3}{4}+\frac{y}{2})}{\frac{2}{3}}=\frac{1}{2}y+\frac{1}{4}\left(\frac{1}{2}+\frac{y}{2}\right)+\frac{1}{4}\left(\frac{3}{4}+\frac{y}{2}\right)$$
Recap: your comment has the right answer, but your reasoning on the intervals is backwards. The first possibility $X=Y$ has $Y$ in an interval of length 1; the second $X=\frac{1}{2}+\frac{Y}{2}$ requires $Y$ to spread over an interval of length 2 so each range of values of $Y$ is half as likely, in the weighted average that is the conditional expectation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3018353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Inequality for $\sin(20°)$ Prove that $$\frac{1}{3} < \sin{20°} < \frac{7}{20}$$
Attempt
$$\sin60°=3\sin20°-4\sin^{3}(20°)$$
Taking $\sin20°$=x
I got the the equation as
$$8x^3-6x+\sqrt{3} =0$$
But from here I am not able to do anything. Any suggestions?
Thanks!
Edit-graph of p(x)
|
Mostly for fun, here's a way to show that
$$6\left(1\over3\right)-8\left(1\over3\right)^3\lt\sqrt3\lt6\left(7\over20\right)-8\left(7\over20\right)^3$$
with only a small amount of multi-digit arithmetic:
$$6\left(1\over3\right)-8\left(1\over3\right)^3=2-{8\over27}\lt2-{8\over28}=2-{2\over7}={12\over7}$$
and $12^2=144\lt147=3\cdot7^2$, which gives the first inequality, while
$$6\left(7\over20\right)-8\left(7\over20\right)^3={7\over10}\left(3-\left(7\over10\right)^2 \right)={7\over10}\left(300-49\over100 \right)\gt{7\over10}\cdot{250\over100}={7\over4}$$
and $7^2=49\gt48=3\cdot4^2$ gives the second inequality.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3024090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
identities and binomial coefficients
I'm having some problems proving this identity. I tried using some formulas I found on the internet so I can turn that $2$ base number into something else but i'm not really sure how to do that. I would be really thankful if someone could help!
|
We have by inspection that
$$\sum_{k=0}^n {m+k\choose k} 2^{n-k}
= 2^n [z^n] \frac{1}{1-z} \frac{1}{(1-z/2)^{m+1}}.$$
This is
$$2^n \times \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{1-z} \frac{1}{(1-z/2)^{m+1}}
\\ = - 2^n \times \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{z-1} \frac{2^{m+1}}{(2-z)^{m+1}}
\\ = 2^{n+m+1} (-1)^{m} \times \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{z-1} \frac{1}{(z-2)^{m+1}}.$$
With
$$f(z) = 2^{n+m+1} (-1)^{m} \frac{1}{z^{n+1}}
\frac{1}{z-1} \frac{1}{(z-2)^{m+1}}$$
we will be using the fact that residues sum to zero i.e.
$$\mathrm{Res}_{z=0} f(z)
+ \mathrm{Res}_{z=1} f(z)
+ \mathrm{Res}_{z=2} f(z)
+ \mathrm{Res}_{z=\infty} f(z) = 0.$$
The residue at infinity is zero since $\lim_{R\to\infty} 2\pi R /
R^{n+1} / R / R^{m+1} = 0.$
The residue at one is
$$2^{n+m+1} (-1)^{m} \times (-1)^{m+1} = - 2^{n+m+1}.$$
For the residue at two we use the Leibniz rule:
$$\frac{1}{m!} \left( \frac{1}{z^{n+1}}
\frac{1}{z-1} \right)^{(m)}
\\ = \frac{1}{m!} \sum_{k=0}^m {m\choose k}
(-1)^k \frac{(n+k)!}{n!} \frac{1}{z^{n+1+k}}
(-1)^{m-k} \frac{(m-k)!}{(z-1)^{m-k+1}}
\\ = (-1)^m \sum_{k=0}^m {n+k\choose k}
\frac{1}{z^{n+1+k}} \frac{1}{(z-1)^{m-k+1}}.$$
Restore factor in front and evaluate at $z=2$:
$$2^{n+m+1} (-1)^{m} \times
(-1)^m \sum_{k=0}^m {n+k\choose k}
\frac{1}{2^{n+1+k}}
= \sum_{k=0}^m {n+k\choose k} 2^{m-k}.$$
Summing the residues we have shown that
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n {m+k\choose k} 2^{n-k}
+ \sum_{k=0}^m {n+k\choose k} 2^{m-k}
- 2^{n+m+1} = 0} $$
which is the claim.
Remark. This is a special case of an identity by Gosper which
was proved at the following MSE
link (set $x=1/2.$)
Addendum. The initial step may be done using an Iverson
bracket. We have
$$\sum_{k=0}^n {m+k\choose k} 2^{n-k}
= \sum_{k\ge 0} {m+k\choose k} 2^{n-k} [[0\le k\le n]]
\\ = \sum_{k\ge 0} {m+k\choose k} 2^{n-k}
[z^n] \frac{z^k}{1-z}
= [z^n] \frac{1}{1-z} \sum_{k\ge 0} {m+k\choose k} 2^{n-k} z^k
\\ = 2^n [z^n] \frac{1}{1-z} \sum_{k\ge 0} {m+k\choose k} 2^{-k} z^k
= 2^n [z^n] \frac{1}{1-z} \frac{1}{(1-z/2)^{m+1}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3024722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Finding determinant of $3\times3$ matrix $$A = \left(\begin{matrix}
\lambda - 1 & -1 & -1 \\
1 & \lambda - 3 & 1 \\
-3 & 1 & \lambda + 1 \\
\end{matrix}\right)$$
$$\det A = \begin{vmatrix}A\end{vmatrix} =
(\lambda - 1) \begin{vmatrix}
\lambda - 3 & 1 \\
1 & \lambda + 1 \\
\end{vmatrix} + 1\begin{vmatrix}
1 & 1 \\
-3 & \lambda + 1 \\
\end{vmatrix} -
1\begin{vmatrix}
1 & \lambda - 3 \\
-3 & 1 \\
\end{vmatrix}\\
= ((\lambda - 1)(\lambda - 3)(\lambda + 1)) - (1 + (\lambda + 1) + 3) + (1 +3\lambda - 9) \\
= (\lambda - 1)(\lambda - 3)(\lambda + 1) + (\lambda + 1) + 2(\lambda - 3)$$
The solution says that it is $(\lambda - 2)(\lambda + 2)(\lambda - 3)$. I feel like I am so close, but I don't get what I am supposed to do to get to the solution.
|
HINT
We can simplify a little bit summing the third to the first row
$$\begin{vmatrix}
\lambda - 1 & -1 & -1 \\
1 & \lambda - 3 & 1 \\
-3 & 1 & \lambda + 1 \\
\end{vmatrix}= \begin{vmatrix}
\lambda -4 & 0 & \lambda \\
1 & \lambda - 3 & 1 \\
-3 & 1 & \lambda + 1 \\
\end{vmatrix}$$
and using Laplace expansion in the first row to obtain
$$\begin{vmatrix}
\lambda -4 & 0 & \lambda \\
1 & \lambda - 3 & 1 \\
-3 & 1 & \lambda + 1 \\
\end{vmatrix}
=(\lambda -4) \cdot \begin{vmatrix}
\lambda - 3 & 1 \\
1 & \lambda + 1 \\
\end{vmatrix}
+\lambda\cdot \begin{vmatrix}
1 & \lambda - 3 \\
-3 & 1 \\
\end{vmatrix}$$
Can you proceed from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3027647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Prove that for all $n$ $\gcd(n, x_n)=1$, given $x_{n+1}=2(x_n)^2-1$ and $x_1=2$ I have a sequence $x_{n+1} = 2(x_n)^2-1$; first values are $2, 7, 97, 18817,\dots$
I noticed that if prime $p$ divides $x_n$, then $x_{n+1} \equiv -1\pmod p$ and for all $k>n+1$, $x_k\equiv 1\pmod p$. But I have no idea what to do next.
|
I'll do my usual playing around
and see what happens.
tl;dr - Nothing but dead ends.
If
$x_{n+1}
= 2(x_n)^2-1
$
then
$x_{n+1} -1
= 2(x_n)^2-2
= 2(x_n^2-1)
= 2(x_n-1)(x_n+1)
$.
Therefore,
if $y_n =x_n-1$ then
$y_{n+1}
=2y_n(y_n+2)
$.
Since
$x_1 = 2$,
$y_1 = 1$.
Therefore
$y_2 = 2\cdot 1\cdot 3 = 6$,
$y_3 = 2\cdot 6\cdot 8= 96$,
$y_4 = 2\cdot 96\cdot 98 = 18616$,
and
$y_5 = 2\cdot 18616\cdot 18618 = 693185376$.
If we can show that
$n | y_n$,
then
$(n, x_n) = 1$
since,
if $d|n$ and $d|x_n$
and $d \ge 2$
then
$d | n | y_n$
so
$d | (x_n-1)$
implies
$d | 1$.
Unfortunately,
this doesn't work
as $y_5$ shows.
Next try:
see if can find
$a_n, b_n$ such that
$na_n-x_nb_n = 1$.
This will show that
$(n, x_n) = 1$.
If
$na_n-x_nb_n = 1$
and
$(n+1)a_{n+1}-x_{n+1}b_{n+1} = 1$
then
$\begin{array}\\
1
&=(n+1)a_{n+1}-(2x_n^2-1)b_{n+1}\\
&=na_{n+1}+a_{n+1}-2x_n^2b_{n+1}+b_{n+1}\\
&=na_{n}+n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1})+b_{n+1}\\
&=x_nb_n+1+n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1})+b_{n+1}\\
&=1+n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1}+b_{n})+b_{n+1}\\
\text{so}\\
0
&=n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1}+b_{n})+b_{n+1}\\
\end{array}
$
and I don't know where to go from here.
Hope someone else can make use of these.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3030131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
Conjecture: if $a$, $b$ and $c$ have no common factors, dividing each of them by their sum yields at least one irreducible fraction
Let $a$, $b$ and $c$ be $3$ integers with no common factors.
I conjecture that at least one of the three fractions:
$$\frac{a}{a+b+c},\quad\frac{b}{a+b+c},\quad\frac{c}{a+b+c}$$
is irreducible.
I know that they are not necessarily all irreducible.
For instance, taking $(a,b,c)=(2,7,15)$, it is verified that $2$, $7$ and $15$ have no common factors.
We have $a+b+c=24$ so
$$\dfrac{2}{2+7+15}=\dfrac{2}{24}\quad(\text{reducible)}$$
$$\dfrac{7}{2+7+15}=\dfrac{7}{24}\quad(\text{irreducible)}$$
$$\dfrac{15}{2+7+15}=\dfrac{15}{24}\quad(\text{reducible)}$$
I have yet to find any counter-example to the initial statement so I'm pretty certain it must be true. I believe I managed to prove the analogous statement for $2$ integers but I would like to generalize it to $3$, and eventually $n$, integers.
Proof for two integers $a$ and $b$ with no common factors
Suppose $k$ is a factor of $a$ and that $a'$ is an integer such that $a=ka'$. Then:
$$\frac{a}{a+b}=\frac{ka'}{ka'+b}=\frac{ka'}{k\left(a'+\dfrac{b}{k}\right)}=\dfrac{a'}{a'+\dfrac{b}{k}}$$
But $k$ is a factor of $a$ and $a$ and $b$ have no common factors by hypothesis, so $k$ does not divide $b$. So $\dfrac{b}{k}$ is irreducible. Since $a'$ is an integer and $\dfrac{b}{k}$ isn't (assuming $k\neq 1$), then $\dfrac{a'}{a'+\dfrac{b}{k}}$ is not a 'fraction' in the sense that it is not a ratio of two integers, making the original fraction $\dfrac{a}{a+b}$ irreducible.
This proves (correct me if I'm wrong) that at least one of the the fractions $\dfrac{a}{a+b}$ or $\dfrac{b}{a+b}$ is irreducible.
For the case with $3$ integers, a similar argument doesn't seem to work. Taking $(4,5,11)$ as an example:
$$\frac{4}{4+5+11}=\frac{4}{4\left(1+\dfrac{5}{4}+\dfrac{11}{4}\right)}=\frac{1}{1+\dfrac{16}{4}}=\frac{1}{5}$$
the fact that there are now $3$ numbers allows for some fractions to combine into an integer. Because of this, I'm having difficulty proving the statement further.
I'm not particularly good with number theory proofs and I apologize if this is not as elegant as it should be (or worse, if there are mistakes). I'm welcoming any tips and possible improvements. Thanks in advance!
|
Counterexample: $5,7,58$.
\begin{align*}\frac{5}{5+7+58} &= \frac{5}{70} = \frac1{14}\\
\frac{7}{5+7+58} &= \frac{7}{70} = \frac1{10}\\
\frac{58}{5+7+58} &= \frac{58}{70} = \frac{29}{35}\end{align*}
How did I find it? Choose $a$ and $b$ odd and relatively prime, then let $c=2ab-a-b$. This $c$ must be even, which makes $\frac{c}{a+b+c}$ reducible.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3030448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Prove: $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2} < \infty$ without L'Hôpital's Given $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}$, prove that it converges.
I tried to use the Ratio test.
I got a terrible algebraic expression: $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left(\frac{n^4 +2n^3 +3n^2 + 2n +2}{n^4 + 4n^3 + 7n^2 + 6n +3}\right)\cdot \left(\frac{n^2 + 2n + 2}{n^2 + 3n + 3}\right)^{2n +1}$$
Now problem is I'm not allowed to use L'Hoptial and I don't really know what to do with this disaster.
|
Try the root criterion.
Prove that
$$\sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}}=\left(1-\frac{n}{n^2+n+1}\right)^n$$
and that this tends to $L<1$ as $n\to \infty$. Then, the criterion implies that the sum converges.
To find the limit, remember that if $a_n\to 0$ then
$$(1+a_n)^{\frac1{a_n}}\to e.$$
So
$$\left(1-\frac{n}{n^2+n+1}\right)^n=\left(1-\frac{n}{n^2+n+1}\right)^{\left(-\frac{n^2+n+1}{n}\right)\cdot \left(-\frac{n}{n^2+n+1}\right)\cdot n}=$$
$$=\left[\left(1-\frac{n}{n^2+n+1}\right)^{-\frac{n^2+n+1}{n}}\right]^{ \left(-\frac{n}{n^2+n+1}\right)\cdot n}.$$
But the bracketed expression is exactly of the form
$$(1+a_n)^{\frac1{a_n}},\quad a_n\to 0,$$
so it tends to $e$. And the expression in the exponent clearly goes to $-1$, so the limit is $e^{-1}<1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3031251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
How do we write $f(x+2)$ in terms of $f(x)$? $$f : R^+ \rightarrow R^+$$
$$f(x) = \dfrac{x}{x+1} $$
How do we write $f(x+2)$ in terms of $f(x)$? This is a general question that I wondered how to algebraically write it.
Regards
|
A more general method:
$$f(x) = \frac{x}{x+1}$$
$$f(x)(x+1) = x$$
$$(f(x)-1)x=-f(x)$$
$$x=\frac{f(x)}{1-f(x)}$$
Hence,
$$f(x+2)=\frac{x+2}{x+3}=\frac{\frac{f(x)}{1-f(x)}+2}{\frac{f(x)}{1-f(x)}+3}=\frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=\frac{2-f(x)}{3-2f(x)}$$
Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=\frac{3}{4}$ by the original expression, and indeed $f(1+2)=\frac{2-f(1)}{3-2f(1)}=\frac{2-\frac{1}{2}}{3-2(\frac{1}{2})}=\frac{\frac{3}{2}}{2}=\frac{3}{4}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3033503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Is right the inequality: $|a|+|b| \leq 2|a+ib|$
I want know if is right this inequality: $\forall a,b \in \mathbb{R}$,
$$|a|+|b| \leq 2|a+ib|$$
|
For real numbers $a$ and $b$ we have
\begin{align*}
&&a^2&\le a^2+b^2\\
&\implies&\sqrt{a^2}&\le\sqrt{a^2+b^2}\\
&&|a|&\le\sqrt{a^2+b^2}
\end{align*}
In the same way we can prove $|b|\le\sqrt{a^2+b^2}$. Since $\sqrt{a^2+b^2}=|a+ib|$, the asked inequality follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3034285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Compute the limit $\lim_{n\to\infty} I_n(a)$ where $ I_n(a) :=\int_0^a \frac{x^n}{x^n+1}\,\mathrm{d}x, n\in N$. For $a>0$ we define
$$\space I_n(a)=\int_0^a\frac{x^n}{x^n+1}\,\mathrm{d}x , n\in N.$$
*
*Prove that $0\le I_n(1) \le \frac{1}{n+1}$
*Compute $\lim_{n\to\infty} I_n(a)$
My attempt:
*
*I regard $I_n(1)=\int_0^1\frac{x^n}{x^n+1}$. If $x\in (0,1)$ then $x^n\in(0,1)$ and $x^n+1\in(1,2)$.
$$x^n>0 \Rightarrow x^n+1>1 \Rightarrow 1>\frac{1}{1+x^n }\Rightarrow x^n>\frac{x^n}{x^n+1}\Rightarrow \int_0^1\frac{x^n }{x^n+1}dx<\int_o^1 x^n \mathrm{d}x\\ \Rightarrow \int_0^1\frac{x^n }{x^n+1}dx<\frac{1}{n+1} \\ 0\le\frac{x^n}{x^n+1} \\ \text{In concusion } 0\le I_n(1) \le \frac{1}{n+1}.$$
*first case $a\in(0,1) \Rightarrow \lim_{n\to\infty} I_n(a) =0$. $I_n(a)\le\frac{1}{n+1})\text{case 2 . }a\in(1,\infty) \Rightarrow$ ???????
I don't believe the limit is $\infty$ because $\frac{x^n }{x^n+1}\le 1$.
I would appreciate some hints.
|
Lets consider the interval $(1, a)$. We have
$|\frac{x^n}{1 + x^n} - 1| = |\frac{1}{1+x^n}|$
By Bernoulli's Inequality, $x^n = (1 + (x - 1))^n \geq 1 + n(x - 1)$. This implies that
$|\frac{x^n}{1 + x^n} - 1| \leq \frac{1}{2 + n(x-1)}$
Now, fix $\epsilon > 0$ small enough and choose $n$ big enough such that $\frac{1}{2 + n(x-1)} < \frac{\epsilon}{2(a - 1)}$ for every $ x \in (1 + \frac{\epsilon}{2}, a)$. We then have
$\int_1^a |\frac{x^n}{1 + x^n} - 1| dx = \int_1^a |\frac{1}{1+x^n}| dx = $
$\int_{1}^{1 + \frac{\epsilon}{2}} |\frac{1}{1+x^n}| dx + \int_{1 +\frac{\epsilon}{2}}^{a} |\frac{1}{1+x^n}| dx \leq $
$\int_{1}^{1 + \frac{\epsilon}{2}} 1 dx + \int_{1 +\frac{\epsilon}{2}}^{a} \frac{1}{2 + n(x - 1)} dx < \epsilon $
Now separate the original integral in $(0, 1)$ and $(1, a)$, and conclude.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3042601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Proof verification: For $a$, $b$, $c$ positive with $abc=1$, show $\sum_{\text{cyc}}\frac{1}{a^3(b+c)}\geq \frac32$ I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:
Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)}\geq \frac32$$
First, substitute $\frac1a = x, \frac1b = y, \frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$\frac{x^2}{z+y} + \frac{y^2}{x+z} + \frac{z^2}{x+y} \geq \frac32$$ Now, let $f(x)=\dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=\dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,z\in \mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$\begin{alignat*}{2}\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y} & =\frac{x^2}{S-x} + \frac{y^2}{S-y} + \frac{z^2}{S-z} \\
&\geq 3f\left(\frac{x+y+z}{3} \right) \\
&= 3\frac{\left(\frac{x+y+z}{3}\right)^2}{S-\frac{x+y+z}{3}} = \frac{1}{3}\frac{(x+y+z)^2}{\frac{2x+2y+2z}{3}} = \frac13 \frac{(x+y+z)^2}{\frac23 (x+y+z)} \\
&= \frac12 (x+y+z) \geq \frac32\end{alignat*}$$
The final inequality follows by AM-GM ($x+y+z\geq 3\sqrt[3]{xyz}=3$)
|
You make a little mistake i think: $$f''(x)=\frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3043074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
integration question $\int_{-1}^{1}\left(\frac{1-x}{1+x}\right)^a\frac{dx}{(x-b)^2}$ How to integrate this integral $\displaystyle\int_{-1}^{1}\bigg(\frac{1-x}{1+x}\bigg)^a\frac{dx}{(x-b)^2}$ where $0<a<1$ and $b>1$
Answer. I put this into online integral calculator, but it said it cann't do this integration.
|
First, substitute $t = (1-x)/(1+x)$.
$$\begin{aligned} I = \int_{-1}^{1}\left(\frac{1-x}{1+x}\right)^{a}\frac{\mathrm{d}x}{(x-b)^{2}} &= \frac{1}{2}\int_{0}^{\infty}t^{a}\left(\frac{x+1}{x-b}\right)^{2}\mathrm{d}t \\
&= 2\int_{0}^{\infty}\frac{t^{a}\,\mathrm{d}t}{(1-t-b(1+t))^{2}} \\
&= 2\int_{0}^{\infty}\frac{t^{a}\,\mathrm{d}t}{((1-b) - (1+b)t)^{2}} \\
&= 2\int_{0}^{\infty}\frac{t^{a}\,\mathrm{d}t}{((b+1)t + (b-1))^{2}}\end{aligned}$$
Let $b_{\pm} = b\pm 1$. Then after $u = b_{+}t/b_{-}$,
$$\begin{aligned} I = 2\int_{0}^{\infty}\frac{t^{a}\,\mathrm{d}t}{(b_{+}t + b_{-})^{2}} &= \frac{2}{b_{-}^{2}}\left(\frac{b_{-}}{b_{+}}\right)^{a+1}\int_{0}^{\infty}\frac{u^{a}\,\mathrm{d}u}{(u+1)^{2}}.\end{aligned}$$
Using the beta function
$$ \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \int_{0}^{\infty}\frac{z^{y-1}\,\mathrm{d}z}{(z+1)^{x+y}},$$
we identify $x=1-a$ and $y=1+a$. Then
$$ I = \frac{2}{b_{-}^{2}}\left(\frac{b_{-}}{b_{+}}\right)^{a+1}\frac{\Gamma(1-a)\Gamma(1+a)}{\Gamma(2)}.$$
Using the reflection identity $\Gamma(z)\Gamma(1-z) = \pi/\sin\pi z$ and recursion $\Gamma(1+a) = a\Gamma(a)$, we have
$$ \frac{\Gamma(1-a)\Gamma(1+a)}{\Gamma(2)} = \Gamma(1-a)a\Gamma(a) = \frac{\pi a}{\sin\pi a}.$$
The answer is then
$$ I = \frac{2}{(b-1)^{2}}\left(\frac{b-1}{b+1}\right)^{a}\frac{b-1}{b+1}\frac{\pi a}{\sin\pi a} = \boxed{\frac{2}{b^{2}-1}\left(\frac{b-1}{b+1}\right)^{a}\frac{\pi a}{\sin\pi a}.}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3048211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given condition is equivalent
$$\lvert z^2+1\rvert=\lvert z\rvert$$
and if $z=a+bi$,
\begin{align*}
\lvert z\rvert&=\lvert a^2-b^2+1+2ab i\rvert=\sqrt{(a^2-b^2+1)^2+4 a^2b^2}\\
&=\sqrt{(a^2-b^2)^2+2(a^2-b^2)+1+4a^2b^2}\\
&=\sqrt{(a^2+b^2)^2+2(a^2-b^2)+1}\\
&=\sqrt{\lvert z\rvert^4+2(a^2-b^2)+1}
\end{align*}
I think it is not leading to something useful... the approach I followed is probably not useful.
Hints and answers are welcomed.
|
Since $$|z| = |z^2-(-1)|\geq |z|^2-|-1|$$
Write $r=|z|$ and we have $$r^2-r-1\leq 0\implies r\in [0,r_2]$$ where $$r_{1,2} ={1\pm \sqrt{5}\over 2} $$
So $r\leq \displaystyle{1+ \sqrt{5}\over 2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3048864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
}
|
Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^2+2)x+y^3-3y^2-4y=0.$$
But I have no idea how to solve this problem. How can I solve it?
|
If $x+3y=0$ so from the second equation we obtain $y=0$, which gives $x=0$, which is impossible.
Thus, $x+3y\neq0$ and since $$\frac{(x^2+y^2)y}{x+3y}=1,$$ from the first equation we obtain:
$$x+\frac{3x-y}{x^2+y^2}\cdot\frac{(x^2+y^2)y}{x+3y}=3\sqrt{\frac{(x^2+y^2)y}{x+3y}}$$ or
$$x^2+6xy-y^2=3\sqrt{y(x^2+y^2)(x+3y)}$$ or with $x^2+6xy-y^2\geq0$
$$(x^2+6xy-y^2)^2=9y(x^2+y^2)(x+3y)$$ or
$$(x+y)(x-2y)(x^2+4xy+9y^2)=0,$$ which gives $y=-x$ or $x=2y$.
Can you end it now?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3049686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
}
|
Prove $x^3+3y^3+9z^3-9xyz=1$ has infinity integers solutions! A question from Alibaba Global Mathematics Competition (number theory)
Prove $\displaystyle x^{3} +3y^{3} +9z^{3} -9xyz=1$ has infinitely many integer solutions.
The hint for the question is to transform the left side to a complex polynomial.
I found that:
Set $\lambda =e^{j2\pi /3}$ then the equation can be transformed to:
$$ \left( x+3^{1/3} y+3^{2/3} z\right)\left( x+3^{1/3} \lambda y+3^{2/3} \lambda ^{2} z\right)\left( x+3^{1/3} \lambda ^{2} y+3^{2/3} \lambda z\right)=1$$
but I don't know how to continue.
|
this is my anser
first suppose we have an answer x,y,z , for simplicity,
set $\displaystyle a=x,b=3^{1/3} y\ ,c=3^{2/3} z$
and take both sides pow3 ,we get
$\displaystyle ( a+b+c)^{3}\left( a+\lambda b+\lambda ^{2} c\right)^{3}\left( b+\lambda ^{2} b+\lambda c\right)^{3}$=1
⟹
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\left( a^{3} +b^{3} +c^{3} +3a^{2} b+3ab^{2} +3a^{2} c+3ac^{2} +3b^{2} c+3bc^{2} +6abc\right) *\\
\left( a^{3} +b^{3} +c^{3} +3\lambda a^{2} b+3\lambda ^{2} ab^{2} +3\lambda ^{2} a^{2} c+3\lambda ac^{2} +3\lambda b^{2} c+3\lambda ^{2} bc^{2} +6abc\right) *\\
\left( a^{3} +b^{3} +c^{3} +3\lambda ^{2} a^{2} b+3\lambda ab^{2} +3\lambda a^{2} c+3\lambda ^{2} ac^{2} +3\lambda ^{2} b^{2} c+3\lambda bc^{2} +6abc\right) \ =1
\end{array}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
set\ A=a^{3} +b^{3} +c^{3} +6abc\ ,\ B=3a^{2} b+3b^{2} c+3ac^{2} \ ,C=3ab^{2} +3a^{2} c+3bc^{2} ,\ we\ get\\
( A+B+C)\left( A+\lambda B+\lambda ^{2} C\right)\left( A+\lambda ^{2} B+\lambda C\right) =1\\
set\ A'=A\ ,\ \ B'=B/\left( 3^{1/3}\right) ,\ \ \ C’=C/\left( 3^{2/3}\right)\\
we\ get\\
\\
\ \ \left( A'+3^{1/3} B'+3^{2/3} C'\right)\left( A'+3^{1/3} \lambda B'+3^{2/3} \lambda C'\right)\left( A'+3^{1/3} \lambda ^{2} B'+3^{2/3} \lambda C'\right) =1\\
\\
then:\\
A'\ =a^{3} +b^{3} +c^{3} +6abc\ \ is\ interger\\
B'=B/\left( 3^{1/3}\right) =\left( 3a^{2} b+3b^{2} c+3ac^{2}\right) /3^{1/3} =3x^{2} y+9y^{2} z+9xz^{2} \ is\ interger\\
C'=\left( 3ab^{2} +3a^{2} c+3bc^{2}\right) /3^{2/3} =3xy^{2} +3x^{2} z+9yz^{2} \ is\ interger\\
\\
so\ we\ get\ \ A^{\prime 3} +3B^{\prime 3} +9C^{\prime 3} -9A'B'C'=1\ such\ that\ A' >a,B' >b,C' >c\\
we\ can\ repeat\ the\ process\ to\ get\ infinity\ solutions!
\end{array}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3050178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
The volume of solid of revolution rotated about the line $y=x$ Find the volume of solid of revolution of region between curves $y=\sqrt x$ and $y=x^2$ in $xy-$plane about the line $y=x$. I know the answer, $\pi/30\sqrt 2$, but how we can obtain it? Should we rotate axis?
|
$x + w/\sqrt{2} = \sqrt{x - w/\sqrt{2}}$
means $w = \frac{1}{2} \left(\sqrt{2} \sqrt{8 x+1}-\sqrt{2} (2 x+1)\right)$.
Volume:
$$\int\limits_{x=0}^1 \sqrt{2} \pi \left(\sqrt{2} \sqrt{8 x+1}-\sqrt{2} (2 x+1)\right)^2\ dx = {\pi \over 30 \sqrt{2}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3052413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Find all integers $a$ that satisfy $c \equiv a \pmod{ab+1}$ Let $a,b,c$ positive integers with $b|c$ I am looking for triplets that satisfy $c \equiv a \pmod{ab+1}$.
What I found by now is that given a solution, we can write $c=q(ab+1)+a=a(qb+1)+q$
So we get $c \equiv q \pmod{qb+1}$ where q is the whole part of $\frac{c}{ab+1}$.
Another thing I observed is that both $a$ and $c$ are coprime to $ab+1$. This means they are both elements of the multiplicative group modulo $ab+1$.
It seems like something can be done to get more information, but I am stuck.
Thanks
|
First, we have $b \mid c$, which means that we can replace $c = bk$. Now, we have $a \equiv bk \pmod{ab+1}$. As you noted, we can see that $\gcd(b,ab+1)=1$. Thus, we can instead write $k \equiv \frac{a}{b} \pmod{ab+1}$.
We can also note that- $$ab \equiv -1 \pmod{ab+1} \implies \frac{1}{b} \equiv -a \pmod{ab+1} \implies k \equiv \frac{a}{b} \equiv -a^2 \pmod{ab+1}$$
Thus, $k = (ab+1)q - a^2$ . Substituting this back, we get our solutions:
$$\{a,b,c\} = \{a \space ,\space b \space, \space b ((ab+1)q-a^2)\}$$
where $q$ is any integer.
If you want $a$ instead in terms of $b,c$ , we can see that $c = -a^2+b^2a+bq$
$$a^2-(b^2)a+(c-bq)=0$$
This provides us a quadratic equation.
$$a = \frac{b^2 \pm \sqrt{b^4+4bq-4c}}{2}$$
where $q$ is a suitable integer such that $b^4+4bq-4c=x^2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3053097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $\sin a + \sin b + \sin(a+b) = 4 \sin\frac12(a+b) \cos \frac12a \cos\frac12b$ Helping my son with his trigonometry review. We know that
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$
We also know that
$$\sin a \cos b = \frac12 \left(\sin(a-b) + \sin(a+b)\right)$$
And we have
$$\sin a + \sin b = 2 \sin\frac12(a+b)\cos\frac12(a-b)$$
From there, we seem to be missing how to get to the right-hand side of the equation.
We first expand
$$\sin a + \sin b = 2 \sin\frac12(a+b) \cos\frac12(a-b)$$
Then we add $\sin(a+b)$, which is $\sin a \cos b + \cos a \sin b$.
We now have:
$$2 \sin\frac12(a+b) \cos\frac12(a-b) + \frac12 \left(\sin(a-b) + \sin (a+b)\right) + \frac12 \left( \sin(b-a) + \sin (a+b)\right)$$
From there, we can't see how to obtain the right-hand side of the equation which is
$$4 \sin\frac12(a+b) \cos\frac12a \cos\frac12b$$
|
Following Thomas Shelby and Dr. Sonnhard Graubner, I provide you a full proof:
\begin{align}\sin(a)+\sin(b)+\sin(a+b)&=2\cos\left(\frac{a-b}{2}\right)\sin\left(\frac{a+b}{2}\right)+2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a+b}{2}\right)\\&=2\sin\left(\frac{a+b}{2}\right)\left[\cos\left(\frac{a-b}{2}\right)+\cos\left(\frac{a+b}{2}\right)\right]\end{align}
Since
\begin{align}\cos\left(a-b\right)+\cos\left(a+b\right)=2\cos(a)\cos(b)\end{align}
[see: last page of https://services.math.duke.edu/~leili/teaching/uwm/math222s11/problems/quizzes/trig.pdf or search for An elementary proof of two formulas in trigonometry on google] then,
\begin{align}\cos\left(\frac{a-b}{2}\right)+\cos\left(\frac{a+b}{2}\right)=2\cos\left(\frac{a}{2}\right)\cos\left(\frac{b}{2}\right)\end{align}
Hence,
\begin{align}2\sin\left(\frac{a+b}{2}\right)\left[\cos\left(\frac{a-b}{2}\right)+\cos\left(\frac{a+b}{2}\right)\right]=&2\sin\left(\frac{a+b}{2}\right)\times 2\cos\left(\frac{a}{2}\right)\cos\left(\frac{b}{2}\right)\\=&4\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a}{2}\right)\cos\left(\frac{b}{2}\right)\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3053978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
}
|
Proof of polynomial divisibility without using complex numbers? My question is the same as polynomial of degree n and its divisor except I want a solution that does not make use of complex numbers
Problem: Find all positive integers $n$ such that $x^2+x+1\mid (x+1)^n+x^n+1$
Using wolframalpha, I can see that a number of solutions work, such as $2,4,7$. I tried substituting various values of $x$ in and then working in specific mod cases, but this doesn't really seem to work because it only gives possible values, not 'actual' values. Polynomial long division here is also quite unwieldy...
|
You can use a form of polynomial modular arithmetic here, working modulo $x^2+x+1$
We have $x^2+x+1\equiv 0$ so that $x+1\equiv -x^2$
also $x^3+x^2+x\equiv 0$ so that $x^3\equiv -x^2-x\equiv 1$
So $p_n(x)=(x+1)^n+x^n+1\equiv (-1)^nx^{2n}+x^n+1$
Now since $x^3\equiv 1$ and you have a $(-1)^n$ there you can work modulo $6$, because extra multiples of $6$ in the powers change nothing.
We have $$p_0(x)\equiv 3$$ $$p_1(x)\equiv -x^2+x+1\equiv 2x+2$$$$p_2(x)\equiv x^4+x^2+1\equiv x+x^2+1\equiv 0$$$$p_3(x)\equiv -x^6+x^3+1\equiv 1$$$$p_4(x)\equiv x^8+x^4+1\equiv x^2+x+1\equiv 0$$$$p_5(x)=-x^{10}+x^5+1\equiv-x+x^2+1\equiv-2x$$
So the division works for $n\equiv 2,4 \bmod 6$ and you have the remainders on polynomial division for the other residue classes.
Note that because $x^2+x+1$ is an irreducible polynomial, this method is algebraically equivalent to replacing $x$ by a root of the equation $x^2+x+1$, but the computations all take place in a polynomial ring which isn't identified with any particular known ring or field.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3054769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
find the smallest $N$ that $\varphi(n)\ge5$ for every $n\ge N$ I know that the solution is with Euler function. I could not understand how to show this. thanks.
|
The smallest $N$ where $\phi(N) \ge 5$ for every $n \ge N$ would be $M + 1$ where $M$ is the largest $M$ where $\phi(M) \le 4$.
Now if $\gcd(n,k) = 1$ then $\phi(nk) = \phi(n)\phi(k)$ and for any $p^m$ for a prime $p$, then $\phi (p^m) = (p-1)p^{m-1} \ge p-1$. So if $\phi(M) \le 4$ then $p \le 5$ so the only possible prime factors of $M$ are $2,3$ or $5$.
So $M = 2^a3^b5^c$ and $\phi(M) = \phi (2^a)\phi(3^b)\phi(5^c)$.
Case 1: Now if $c \ge 1$ then $\phi(5^c) = 4*5^{c-1}$ so $c=1$ and $\phi(M) = 4$. And $\phi(3^b) = \phi(2^a) =1$.
If $b\ge 1$ then $\phi(3^b) = (3-1)3^{b-1} = 2*3^{b-1}$ that's impossible so $b = 0$.
If $a \ge 1$ then $\phi(2^a) = (2-1)2^{a-1} = 2^{a-1}$. That is possible only if $a =1$ so if $M=10$, $\phi(M) =4$ and that's the largest such number that is a multiple of $5$.
Case 2: $c=0$ and If $b\ge 1$: we can have $\phi(3^b) = 2*3^{b-1}$. If this is to be less than or equal to $4$ we must have $b=1$ and $\phi(3) = 2$ so $M= 2^a*3$ and $\phi(M) = \phi (2^a)\phi 3 = 2\phi(2^a)$.
SO $\phi(2^a) \le 2$ and as $\phi(2^a)= 2^{a-1}$ we may have $a = 2$ and $M = 12$ and $\phi(12) = 4$
And that's the largest such number that is a multiple of $3$.
Case 3: $c=0; b=0$ and $M = 2^k$ so $\phi(M) = \phi(2^{k})=2^{k-1}\le 4$. This means $k \le 3$ and $M \le 8$. If $M = 8$ then $\phi 8 = 4$.
And there we have it.
$M =12$ is the largest number so that $\phi(M) \le 4$ and $N = 13$ is the smallest number so that for all $n\ge N$, $\phi(n) \ge 5$.
=======
Worth noting the only time $\phi(k)$ is odd is if $k=1,2$ so $\phi(k) =5$ is impossible. (If $p$ is an odd factor of $k$ then $p-1|\phi(k)$. If $k = 2^k$ then $\phi(2^k) = 2^{k-1}$.)
$\phi(k) \ge 5; k \le 12$ occur with $\phi(7) = 6; \phi(9)=6; \phi(11)=10$
$\phi(k) = 4$ occur with $\phi(5)=\phi(8)= \phi(10)=\phi(12)= 4$
And the rest $\phi(3)=\phi(4)=\phi(6) = 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3055862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Evaluate $\frac{2}{\sqrt{2}}\cdot \frac{2}{\sqrt{2+\sqrt{2}}}\cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$ Problem
Evaluate the infinite product
$$\frac{2}{\sqrt{2}}\cdot \frac{2}{\sqrt{2+\sqrt{2}}}\cdot
\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$$
Attempt
For convenience，let's rewrite the limit. Denote$$x_1=\sqrt{2},~~~x_{n+1}=\sqrt{2+x_n}(n=1,2,\cdots),$$
then
$$\lim_{n \to \infty}\left(\frac{2}{x_1}\cdot \frac{2}{x_2}\cdots \frac{2}{x_n}\right)$$ is what we want.
It's easy to obtain that，$\{x_n\}$ is increasing with a greater $n$，and convergent to $2$. Hence $$x_1<x_2<\cdots x_n<2.$$Therefore$$\frac{2}{x_1}>\frac{2}{x_2}>\cdots>\frac{2}{x_n}>1.$$
Can we go on from here?
|
Use $\cos2A=2\cos^2A-1,$
$$\dfrac2{\sqrt2}=\dfrac2{2\cos\dfrac\pi4}\text{ and }2+\sqrt2=2\left(1+\cos\dfrac\pi4\right)=4\cos^2\dfrac\pi8$$
$$\implies\dfrac2{\sqrt{2+\sqrt2}}=\dfrac2{2\cos\dfrac\pi8}$$
$\dfrac1{\cos\dfrac\pi4}\dfrac1{\cos\dfrac\pi8}=\dfrac{2\sin\dfrac\pi8}{\cos\dfrac\pi4\sin\dfrac\pi4}=\dfrac{4\sin\dfrac\pi8}{\sin\dfrac\pi2}=?$
$\dfrac1{\cos\dfrac\pi4}\dfrac1{\cos\dfrac\pi8}\dfrac1{\cos\dfrac\pi{16}}=2^3\sin\dfrac\pi{2^4}$
$$\implies\prod_{r=2}^n\cos\dfrac\pi{2^r}=2^{n-1}\sin\dfrac\pi{2^n}$$
$$\implies\lim_{n\to\infty}\prod_{r=2}^n\cos\dfrac\pi{2^r}=\dfrac\pi2\cdot\lim_{n\to\infty}\dfrac{\sin\dfrac\pi{2^n}}{\dfrac\pi{2^n}}=?$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3056715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
Generalizing $\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)}=2$ I was looking at this paper on section [17],
$$\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)}=2\tag1$$
Let generalize $(1)$
$$\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)\cdots [2n-(2k+1)]}\tag2$$
Where $k\ge 0$
I conjectured the closed form for $(2)$ to be
$$\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)\cdots [2n-(2k+1)]}=\frac{2(-1)^k}{(2k+1)!!(2k+1)}\tag3$$
Here are a first few values of $k=1,2$ and $3$
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)}&=-\frac{2}{9}\tag4\\
\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)}&=\frac{2}{75}\tag5\\
\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)(2n-7)}&=-\frac{2}{735}\tag6
\end{align}$$
How do we go about to prove this conjecture $(3)?$
|
Using Lemma $(13)$ in this paper provided by the OP
$$\sum_{n=1}^{\infty}H_n{2n \choose n}x^n=\frac{2}{\sqrt{1-4x}}\ln\left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\right)$$
Relpace $x$ with $\displaystyle \frac{x^2}{4}$, we get
$$\sum_{n=1}^{\infty}\frac{H_n}{4^n}{2n \choose n}x^{2n}=\frac{2}{\sqrt{1-x^2}}\ln\left(\frac{1+\sqrt{1-x^2}}{2\sqrt{1-x^2}}\right)$$
Now divide both sides by $x^2$ and integrate from $x=0$ to $x=1$, we get
\begin{align}
S&=\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{4^n(2n-1)}=\int_0^1\frac{2}{x^2\sqrt{1-x^2}}\ln\left(\frac{1+\sqrt{1-x^2}}{2\sqrt{1-x^2}}\right)\ dx,\quad \color{blue}{x=\sin\theta}\\
&=2\int_0^{\pi/2}\csc^2\theta\ln\left(\frac{1+\cos\theta}{2\cos\theta}\right)\ d\theta\overset{IBP}{=}2\int_0^{\pi/2}\frac{d\theta}{1+\cos\theta}=2(1)=2
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3059744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
}
|
Find the $\frac mn$ if $T=\sin 5°+\sin10°+\sin 15°+\cdots+\sin175°=\tan \frac mn$ It's really embarrassing to be able to doesn't solve this simple-looking trigonometry question.
$$T=\sin(5^\circ) +\sin(10^\circ) + \sin(15^\circ) + \cdots +\sin(175^\circ) =\tan \frac mn$$
Find the $\frac mn=?$, where $m$ and $n$ are positive integer numbers.
Attepmts:
$$T=2\Big(\sin(5°)+\sin(10°) + \cdots + \sin(85°)\Big) + 1 = 2\Big((\sin(5°) + \cos(5°))+(\sin(10°)+ \cos(10°))+\cdots + (\sin(40°)+\cos(40°))\Big)+1 = 2\Big(\sqrt 2((\sin50°+\sin55°)+\cdots+\sin(80°))\Big)+1.$$
and then I can not see an any way...
|
Not exactly the most elementary approach, but with complex numbers we could write
$$
T = \operatorname{Im}\left(1 + e^{\frac{\pi}{36}i} + e^{2\frac{\pi}{36}i} + \cdots + e^{35\frac{\pi}{36}i}\right) \\
=\operatorname{Im}\left(1 + z + z^2 + \cdots + z^{35}\right) \quad \left(z = e^{\frac{\pi}{36}i}\right)\\
= \operatorname{Im}\left(\frac{1 - z^{36}}{1 - z}\right) = \operatorname{Im}\left(\frac{2}{(1 - \cos(5^\circ)) - i\sin(5^\circ)}\right)\\
= \operatorname{Im}\left(\frac{2}{[1 - \cos(5^\circ)]^2 + \sin^2(5^\circ)}[(1 - \cos(5^\circ)) + i\sin(5^\circ)]\right)
\\ = \frac{2 \sin(5^\circ)}{[1 - \cos(5^\circ)]^2 + \sin^2(5^\circ)}
\\ = \frac{2 \sin(5^\circ)}{2 - 2\cos(5^\circ)} = \frac{\sin(5^\circ)}{1 - \cos(5^\circ)}
$$
As the commenter below points out, we have $\displaystyle\frac{\sin x}{1 - \cos x} = \tan\left(90^\circ - \frac{x}{2}\right)$. It follows that our answer is
$$
\frac{\sin(5^\circ)}{1 - \cos(5^\circ)} = \tan\left(90^\circ - \frac{5^\circ}{2}\right) = \tan(87.5^\circ) = \tan\left(\left[\frac{175}{2}\right]^\circ\right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3059906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Find every $z$s that fit $\cos(z) = -2$ I couldn't find any, I tried to write $\cos(z)$ as $\cos(x)\cos(iy)-\sin(x)\sin(iy)$ which then gave me
$\cos(x)\cosh(y) - i\sin(x)\sinh(y) = -2$
$\sin(x)=0$ so that imaginary part become $0$
now we have to find $\cosh(y) = -2$ which is not true for no $y$.
is it right or i made a mistake in my substitutions?
|
Be:
$$
\cos z=\frac{e^{iz}+e^{-iz}}{2}=-2
$$
Then:
\begin{eqnarray}
\frac{e^{iz}+e^{-iz}}{2} &=& -2 \\
e^{iz}+e^{-iz} &=& -4 \\
e^{2iz} + 1 &=& -4e^{iz} \\
\left(e^{iz}\right)^2 + 4\left(e^{iz}\right) + 1 &=& 0 \\
e^{iz} &=& \frac{-4 \pm \sqrt{16-4}}{2} \\
e^{iz} &=& -2 \pm \sqrt{3} \\
\end{eqnarray}
Then since $e^{iz}=\cos(z)+i\sin(z)$, $e^{iz}=e^{i(z+2k\pi)}$ with $k\in\mathbb{Z}$:
\begin{eqnarray}
e^{i(z+2k_1\pi)} &=& -2 \pm \sqrt{3} \\
i(z +2k_1\pi) &=& \ln\left(-2 \pm \sqrt{3}\right) \\
z &=& 2k_1\pi - i\ln\left(-2 \pm \sqrt{3}\right) \\
\end{eqnarray}
There are two set solution to differents solutions:
\begin{eqnarray}
z_{+} &=& 2k_1\pi - i\ln\left(-2 + \sqrt{3}\right) \\
z_{-} &=& 2k_1\pi - i\ln\left(-2 - \sqrt{3}\right) \\
\end{eqnarray}
For $z_{-}$:
\begin{eqnarray}
z_{-} &=& 2k_1\pi - i\ln\left(\left(2 + \sqrt{3}\right)e^{-i(1+2k_2)\pi}\right) \\
z_{-} &=& 2k_1\pi - i\left[\ln\left(2 + \sqrt{3}\right)-i(1+2k_2)\pi\right] \\
z_{-} &=& 2k_1\pi - (1+2k_2)\pi - i\ln\left(2 + \sqrt{3}\right) \\
z_{-} &=& (1+2k)\pi - i\ln\left(2 + \sqrt{3}\right) \\
\end{eqnarray}
Then, with $k\in\mathbb{Z}$ the solutions are:
\begin{eqnarray}
z_{+} &=& 2k\pi - i\ln\left(-2 + \sqrt{3}\right) \\
z_{-} &=& (1+2k)\pi - i\ln\left(2 + \sqrt{3}\right) \\
\end{eqnarray}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3059995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Find the maximum and minimum (there are four in total) that the function $f(x,y)=3xy$ achieves when $(x,y)$ travel the ellipse $x^2+y^2+xy=3$ Find the maximum and minimum (there are four in total) that the function $f(x,y)=3xy$ achieves when $(x,y)$ travel the ellipse $x^2+y^2+xy=3$
I have thought about doing the following but I do not know if I am doing the right thing:
I'm using Lagrange multipliers and I start with $\bigtriangledown f=\lambda\bigtriangledown g$, so from this I get $(3y,3x)=\lambda(2x+y,2y+x)$ so $\lambda=\frac{3y}{2x+y}$ and $\lambda=\frac{3x}{2y+x}$ so $\frac{3y}{2x+y}=\frac{3x}{2y+x}$ so $x^2=y^2$ so $y=\pm x$. So, I get to that $(0,0), (1,1), (-1,-1), (\sqrt{3},-\sqrt{3})$ and $(-\sqrt{3},\sqrt{3})$ are critical points, the problem is that I do not know how to determine if they are maximum or minimum, how can I do this? Thank you.
|
1) $(x+y)^2- xy= 3;$
$xy = (x+y)^2 -3;$
$f(x,y)=3xy = 3(x+y)^2-9.$
$f_{min}=-9$, at $x=-y$;
$-3x^2= -9; x=\pm √3; y=\mp √3$;
2) $(x-y)^2+3xy= 3;$
$f(x,y)= 3xy= 3-(x-y)^2$;
$f_{max}= 3$, at $x=y$.
$x^2=1$; $x =\pm 1$, $y=\pm 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3061300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
}
|
In $\Delta ABC$, find $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ if $b+c=3a$
If in a triangle ABC, $b+c=3a$, then $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ is equal to ?
My reference gives the solution $2$, but I have no clue of where to start ?
My Attempt
$$
\cot\dfrac{B}{2}.\cot\dfrac{C}{2}=\frac{\cos\frac{B}{2}.\cos\frac{C}{2}}{\sin\frac{B}{2}.\sin\frac{C}{2}}=\frac{\cos(\frac{A-B}{2})+\cos(\frac{A+B}{2})}{\cos(\frac{A-B}{2})-\cos(\frac{A+B}{2})}
$$
|
Hint:
$$b+c=3a\implies\sin B+\sin C=3\sin A$$
$$\iff2\sin\dfrac{B+C}2\cos\dfrac{B-C}2=6\sin\dfrac A2\cos\dfrac A2$$
Now use $\dfrac{B+C}2=\dfrac\pi2-\dfrac A2,\cos\dfrac{B+C}2=?$
As $0<A<\pi,\sin\sin\dfrac A2\ne0$
$\implies\cos\dfrac{B-C}2=3\sin\dfrac A2=3\cos\dfrac{B+C}2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3061601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
prove $a,b,c$ in A.P if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$
In $\Delta ABC$, if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
\sin A=\frac{2.5}{6}.\frac{36}{61}=\frac{60}{61}\\
\sin C=\frac{2.2}{5}.\frac{25}{29}=\frac{20}{29}\\
$$
it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?
|
$$\dfrac1{\tan\dfrac B2}=\cot B/2=\tan(A/2+C/2)=?$$
Use
$$\sin2x=\dfrac{2\tan x}{1+\tan^2x}=?$$ for $2x=A,B,C$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3061807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Finding the Jordan Form of a matrix... I know that this type of question has been asked on here before but I am still having a hard understanding what is going on. The text that I am learning from is "Linear Algebra Done Right by Sheldon Axler" and I don't think it covers this topic very well. Does anyone have any suggestions? I am studying for my linear comp at the end of January.
Given the matrix
$$A = \left(\begin{matrix}0&-1&-1\\-3&-1&-2\\7&5&6\end{matrix}\right).$$
Find the Jordan form $J$ and an invertible matrix $Q$ such that $A = QJQ^{-1}$.
I know that I want to start by finding the eigenvalues of this matrix. I ended up getting det$(A - \lambda I) = (\lambda + 2)(\lambda62 + 3\lambda - 2) \implies \lambda = -2, \frac{-3 -\sqrt{17}}{2}, \frac{\sqrt{17} - 3}{2}$.
Now, from what I have read on this website, I know now that I want to find the null space of $A - \lambda I$ for each eigenvalue. Then my invertible matrix $Q$ will have entries whose columns are these vectors. Why does one do this? Is there a "nicer" way to go about doing this other than direct computation?
This problem is from a previous linear comp at my university, so I can't imagine that they would have wanted me to the direct calculation.
Hopefully, my questions and explanation of the problem were clear. Thanks for any help in advance.
|
The characteristic polynomial is
$$
p(\lambda)=\lambda^3-5\lambda^2+8\lambda-4=(\lambda-1)(\lambda-2)^2.
$$
The vectors involved in the Jordan blocks associated with eigenvalue $2$ is the column space of
$$
A-I = \begin{pmatrix}-1 & -1 & -1 \\
-3 & -2 & -2 \\
7 & 5 & 5\end{pmatrix}.
$$
The column space of $A-I$ is two-dimensional, which gives $\mbox{dim}(\mbox{ker}(A-I))=1$, and it is obvious that $\mbox{ker}(A-I)$ is spanned by
$$
\begin{pmatrix}0 \\ 1 \\ -1\end{pmatrix}.
$$
(It is obvious because the last two columns of $A-I$ are identical.) Then
\begin{align}
(A-2I)(A-I)&=\begin{pmatrix}-2 & -1 & -1 \\
-3 & -3 & -2 \\
7 & 5 & 4\end{pmatrix}.
\begin{pmatrix}-1 & -1 & -1 \\
-3 & -2 & -2 \\
7 & 5 & 5\end{pmatrix} \\
&= \begin{pmatrix}-2 & -1 & -1 \\
-2 & -1 & -1 \\
6 & 3 & 3\end{pmatrix}
\end{align}
Because $(A-2I)^2(A-I)=0$, you have
\begin{align}
(A-2I)\begin{pmatrix} -1 \\ -2 \\ 5\end{pmatrix}&=\begin{pmatrix}-1 \\ -1 \\ 3\end{pmatrix} \\
(A-2I)\begin{pmatrix}-1 \\ -1 \\ 3\end{pmatrix} &= 0.
\end{align}
The Jordan form is
$$
J = \begin{pmatrix} 1 & 0 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2\end{pmatrix}
$$
and the transition matrix $Q$ is
$$
Q = \begin{pmatrix} 0 & -1 & -1 \\
1 & -1 & -2 \\
-1 & 3 & 5
\end{pmatrix}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3062207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find all parameters $a,b,c,d \in \mathbb R$ for which the function $f: \mathbb R \rightarrow \mathbb R$ a) has a limit at point 0 b) is continuous My function $f: \mathbb R \rightarrow \mathbb R$ it is given a pattern:
$$f(x)=\begin{cases} \frac{2^{7^{x}-1}-a}{\ln(1-x)},& x<0\\ b,& x=0 \\ \frac{\sin\left(c \sqrt{x^{2}+d^{2}x}\right) }{x},& x>0 \end{cases}$$ for $$a,b,c,d \in \mathbb R$$
I have a problem with this task because function patterns for $x</>0$ have they reach zero in the denominator for $x=0.$
That is why I think that I should cut a denominator with a numerator but I cannot do it because I have logarithm and sine so I need other way to this task.
I thought about doing some substitution: for example $y=\frac{1}{x}$, then for $x \rightarrow 0^\mp$ I have $y \rightarrow \pm \infty$ and but I still can not solve this.
Can I count on any tips?
|
The first thing to note is that as $x\nearrow 0,$ the denominator $\ln(1-x)$ vanishes, so the only way to avoid $\bigl|f(x)\bigr|\to+\infty$ in that case is to make sure that the numerator vanishes as well. Since $$2^{7^x-1}-a\to2^{7^0-1}-a=2^{1-1}-a=2^0-a=1-a,$$ this means we must put $a=1$ to accomplish this.
If you're familiar with the result that $$\lim_{t\to 0}\frac{\sin(t)}t=1,\tag{1}$$ then we can tackle the right-hand limit as follows. Since $\lim_{x\searrow0}c\sqrt{x^2+d^2x}=0,$ then by $(1)$ we have $$\lim_{x\searrow0}\frac{\sin\left(c\sqrt{x^2+d^2x}\right)}{c\sqrt{x^2+d^2x}}=1,$$ so, since $$\frac{\sin\left(c\sqrt{x^2+d^2x}\right)}{x}=\frac{\sin\left(c\sqrt{x^2+d^2x}\right)}{c\sqrt{x^2+d^2x}}\cdot\frac{c\sqrt{x^2+d^2x}}{x}=\frac{\sin\left(c\sqrt{x^2+d^2x}\right)}{c\sqrt{x^2+d^2x}}\cdot c\sqrt{1+\frac{d^2}x}$$ whenever $x>0,$ then $\lim_{x\searrow0}f(x)=+\infty$ unless $d=0,$ in which case the limit is $c.$ Thus, for the right-hand limit to exist, we require $d=0.$ Moreover, for the two-sided limit to exist, we need $a=1,$ $d=0,$ and $c=\lim_{x\nearrow0}f(x),$ assuming the left-hand limit exists.
Unfortunately, I'm not familiar with any elementary ways to proceed further. Using L'Hôpital's rule allows us to find that $$\begin{eqnarray}\lim_{x\nearrow0}\frac{2^{7^x-1}-1}{\ln(1-x)} &=& \lim_{x\nearrow0}\cfrac{\ln(2)\ln(7)7^{x}2^{7x-1}}{-\frac1{1-x}}\\ &=& \lim_{x\nearrow0}(x-1)\ln(2)\ln(7)7^{x}2^{7x-1}\\ &=& (0-1)\ln(2)\ln(7)7^{0}2^{0-1}\\ &=& -\frac{\ln(2)\ln(7)}2,\end{eqnarray}$$ so for the two-sided limit to exist, we require $a=1,$ $c=\frac{\ln(2)\ln(7)}2,$ and $d=0.$ For continuity, we additionally require $b=\frac{\ln(2)\ln(7)}2.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3062828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Super hard system of equations
Solve the system of equation for real numbers
\begin{split}
(a+b) &(c+d) &= 1 & \qquad (1)\\
(a^2+b^2)&(c^2+d^2) &= 9 & \qquad (2)\\
(a^3+b^3)&(c^3+d^3) &= 7 & \qquad (3)\\
(a^4+b^4)&(c^4+d^4) &=25 & \qquad (4)\\
\end{split}
First I used the identity
$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(bc+ad)^2$$
Use this identity to (4) too
and simplify (3),
we obtain $$(a^2+b^2-ab)(c^2+d^2-cd)=7$$
And suppose $x=abcd$ use $ac=x /bd , bc=x/ad$
But got stuck...
|
Not really complete, but an interesting result using simple algebraic manipulations.
Write:
$$\begin{align}
a^2+b^2&=(a+b)^2-2ab\\
a^3+b^3&=(a+b)(a^2+b^2-ab)=(a+b)((a+b)^2-3ab)\\
a^4+b^4&=((a^2)^2+(b^2)^2)=\cdots=(a+b)^4+2a^2b^2-4ab(a+b)^2\\
\vdots
\end{align}$$
From $(2)$, we have:
$$\begin{align}
(a^2+b^2)(c^2+d^2)&=((a+b)^2-2ab)((c+d)^2-2cd)=9\\
&=\color{red}{(a+b)^2(c+d)^2}-2ab(c+d)^2-2cd(a+b)^2-4abcd=9
\end{align}$$
But from $(1)$, we know that $(a+b)(c+d)=1$, then the text in $\color{red}{\text{red}}$ is also equal to $1$, so the result above becomes:
$$2abcd-ab(c+d)^2-cd(a+b)^2=4\tag{1*}$$
From $(3)$, we have:
$$\begin{align}
(a^3+b^3)(c^3+d^3)&=\color{red}{(a+b)}((a+b)^2-3ab)\color{red}{(c+d)}((c+d)^2-3cd)=7\\
&=((a+b)^2-3ab)((c+d)^2-3cd)=7\\
&\qquad\vdots\\
&=3abcd-ab(c+d)^2-cd(a+b)^2=2\tag{2*}
\end{align}$$
Adding $(1*)$ and $(2*)$, we get $abcd=-2$ and that $\color{pink}{ab(c+d)^2+cd(a+b)^2=-8}$.
Now $(4)$ is really tricky, but you can write it as:
$$\begin{align}
\left((a+b)^4+2a^2b^2-4ab(a+b)^2\right)\left((c+d)^4+2c^2d^2-4cd(c+d)^2\right)&=25
\end{align}$$
Expanding, and we can eliminate $(a+b)^4(c+d)^4$ since it is equal to $1$. Then we have:
$$\begin{align}
a^2b^2(c+d)^4-2ab(c+d)^2+c^2d^2(a+b)^4+2(abcd)^2-4abc^2d^2(a+b)^2-2cd(a+b)^2-4a^2b^2cd(c+d)^2-4abcd=12
\end{align}$$
Using the fact that $abcd=-2$, then we can shorten the equation above into:
$$\color{red}{a^2b^2(c+d)^4+c^2d^2(a+b)^4}+5ab(c+d)^2-10cd(a+b)^2+16=12\tag{3*}$$
However, you can see that the text in $\color{red}{\text{red}}$ looks very close to the square of two sums:
$$\color{red}{a^2b^2(c+d)^4+c^2d^2(a+b)^4}=(ab(c+d)^2+cd(a+b)^2)^2-2\color{blue}{abcd(a+b)^2(c+d)^2}$$
However we already know the value of the part in $\color{blue}{\text{blue}}$ to be $-2
\cdot 1$.
Now we can write $(3*)$ as:
$$(ab(c+d)^2+cd(a+b)^2)^2+6ab(c+d)^2-10cd(a+b)^2=-8$$
From here, you can substitute $x=ab(c+d)^2$ and $y=cd(a+b)^2$, which gives two systems of equation:
$$(x+y)^2+6x-10y=-8\\
x+y=-8$$
This has one solution:
$$x=-\frac{19}2\,\,y=\frac32$$
You can try working from here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3063839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
"answer_id": 0
}
|
What is $\int_0^{\pi/2}\sin^7(\theta)\cos^5(\theta)d\theta$ I have to integrate the following:
$\int_0^\limits\frac{\pi}{2}\sin^7(\theta)\cos^5(\theta)d\theta$
I decided to use a $u$ substitution of $u=\sin^2(\theta)$, and $\frac{du}{2}=\sin(\theta)\cos(\theta)$
and arrived at this integral
$\int_\limits{0}^{1}u^3(1-u)^2du$
From here I decided to use integration by parts using $g=u^3$ and $dv=(1-u)^2du$
I get the following:
$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\int_\limits{0}^{1}(u^2*(1-u)^3)du$$
Repeated again $g=u^2$, and $dv=(1-u)^3du$
$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\biggl[\frac{u^2*(1-u)^4}{4}\biggr]_0^1+\frac{1}{2}\int_\limits{0}^{1}u(1-u)^4$$
Repeating again $g=u$, and $dv=(1-u)^4$
$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\biggl[\frac{u^2*(1-u)^4}{4}\biggr]_0^1+\frac{1}{2}\biggl[\frac{u(1-u)^5}{5}\biggr]_0^1-\frac{1}5\int_\limits{0}^{1}(1-u)^5$$
and I get
$$\biggl[\frac{u^3*(1-u)^3}{3}\biggr]_0^1-\biggl[\frac{u^2*(1-u)^4}{4}\biggr]_0^1+\frac{1}{2}\biggl[\frac{u(1-u)^5}{5}\biggr]_0^1-\frac{1}{30}\biggl[(1-u)^6\biggr]_0^1$$
|
To lower the exponents a bit, notice that substitution $\theta \mapsto \frac\pi2-\theta$ yields
$$\int_0^{\frac\pi2} \cos^7\theta\sin^5\theta\,d\theta = \int_0^{\frac\pi2} \sin^7\theta\cos^5\theta\,d\theta$$
so we have
$$2I = \int_0^{\frac\pi2}\sin^5\theta\cos^5\theta(\cos^2\theta+\sin^2\theta)\,d\theta = \int_0^{\frac\pi2}\sin^5\theta\cos^5\theta\,d\theta = \int_0^{\frac\pi2}\sin^5\theta(1-\sin^2\theta)^2\cos\theta\,d\theta$$
Now setting $u = \sin\theta$ yields
$$I = \frac12 \int_0^1u^5(1-u^2)^2\,du = \frac1{120}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3064190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
}
|
Integral over recurrence relationship I'm interested in evaluating the following definite integral
\begin{equation}
I_n = \int_0^{\gamma} F_n(x)\:dx
\end{equation}
Where $\gamma \gt 0$ and $F_n(x)$ is based on the recurrence relationship:
\begin{equation}
F_{n + 1}(x) = \frac{1}{1 + F_n(x)}
\end{equation}
Here $F_0(x) = f(x)$ where $f$ is a continuous function on $\left[0,\gamma\right]$. My first task was to find a general solution for $F_n(x)$ and this is where I've become unstuck.
I started by letting $F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)}$. Applying it to the recurrence relationship we have:
\begin{equation}
F_{n + 1}(x) = \frac{\alpha_{n+1}(x)}{\beta_{n+1}(x)} = \frac{\beta_n(x)}{\alpha_n(x) + \beta_n(x)}
\end{equation}
And so we have a recurrence relationship over both $\alpha_n(x)$ and $\beta_n(x)$ with $F_0(x) = f(x) = \frac{\alpha_0(x)}{\beta_0(x)}$.
To begin with I'm focused on $f(x) = \sec(x)$ with $\alpha_0(x) = 1$ and $\beta_0(x) = \cos(x)$ and $\gamma = \frac{\pi}{2}$.
With a few iterations via Wolframalpha the pattern that emerges is that $F_n(x)$ takes the form:
\begin{equation}
F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)} = \frac{a_n\cos(x) + b_n}{c_n \cos(x) + d_n}
\end{equation}
Where $a_n, b_n, c_n, d_n \in \mathbb{N}$ Here I would like to be able to solve for each but I'm not sure how to start.
Does anyone have any good starting points/references that I can use to begin?
Edit: Changed the definition of $I_n$ to have generalised upper limit of $\gamma$.
Update Thanks to hypernova's comment's below, it can be seen that $\beta_n(x)$ follows a Fibonacci Sequence:
\begin{equation}
\beta_{n + 1}(x) = \beta_n(x) + \alpha_n(x) = \beta_n(x) + \beta_{n - 1}(x)
\end{equation}
And so we can now represent $F_n(x)$ as:
\begin{equation}
F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)} = \frac{\beta_{n-1}(x)}{\beta_n(x)}
\end{equation}
for $n \geq 2$.
For the specific example above we have:
\begin{equation}
F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)} = \frac{a_{n-1}\cos(x) + b_{n-1}}{a_n \cos(x) + b_n}
\end{equation}
Where $a_n$ and $b_n$ are Fibonacci Sequences with $b_0 = 0$, $b_1 = 1$ and $a_0 = 1$, $a_1 = 1$. We see that $a_n \gt b_n$ (this will be important later)
So, we may now evaluate the integral
\begin{equation}
I_n = \int_0^{\tfrac{\pi}{2}}\frac{a_{n-1}\cos(x) + b_{n-1}}{a_n \cos(x) + b_n}\:dx
\end{equation}
I will here employ the Weierstrass substitution $t = \tan\left(\frac{x}{2} \right)$:
\begin{align}
I_n &= \int_0^{\tfrac{\pi}{2}}\frac{a_{n-1}\cos(x) + b_{n-1}}{a_n \cos(x) + b_n}\:dx = \int_0^1 \frac{a_{n-1}\frac{1 - t^2}{1 + t^2} + b_{n-1}}{a_n \frac{1 - t^2}{1 + t^2} + b_n}\frac{2\:dt}{1 + t^2} \\
&= 2\int_0^1 \frac{a_{n - 1}\left(1 - t^2\right) + b_{n - 1}\left(1 + t^2\right)}{\left(1 + t^2\right)\left(a_n\left(1 - t^2\right) + b_n\left(1 + t^2\right)\right)}\:dt \\
&= 2\int_0^1 \frac{\left(b_{n - 1} - a_{n-1}\right)t^2 + \left(b_{n-1} + a_{n-1}\right)}{\left(1 + t^2\right)\left(\left(b_n - a_n\right)t^2 + \left(b_n + a_n\right)\right)}\:dt \\
&= 2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right) \int_0^1 \frac{t^2 - \theta_{n - 1}}{\left(1 + t^2\right)\left(t^2 + \theta_n\right)}\:dt
\end{align}
Where
\begin{equation}
\theta_n = \frac{b_n + a_n}{b_n - a_n}
\end{equation}
As $a_n \gt b_n \geq 0$ we see that $\theta_n \lt 0$. Hence:
\begin{align}
I_n &= 2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right) \int_0^1 \frac{t^2 - \left|\theta_{n - 1}\right|}{\left(1 + t^2\right)\left(t^2 - \left| \theta_n\right)\right|}\:dt \\
&= 2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right)\left[ \frac{1}{\left|\theta_n\right| + 1} \left[\left(\left|\theta_{n-1}\right| + 1 \right) \arctan(x) + \frac{\left|\theta_{n-1}\right| - \left|\theta_{n}\right|}{\sqrt{\left|\theta_n\right|}}\operatorname{arctanh}\left(\frac{x}{\sqrt{\left|\theta_n\right|}} \right)\right]\right]_0^1 \\
&=2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right)\left(\frac{1}{\left|\theta_n\right| + 1} \right)\left[\left(\left|\theta_{n-1}\right| + 1 \right) \frac{\pi}{4} + \frac{\left|\theta_{n-1}\right| - \left|\theta_{n}\right|}{\sqrt{\left|\theta_n\right|}}\operatorname{arctanh}\left(\frac{1}{\sqrt{\left|\theta_n\right|}} \right)\right]
\end{align}
Note $b_n = a_{n - 1}$ and thus:
\begin{align}
I_n &=2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right)\left(\frac{1}{\left|\theta_n\right| + 1} \right)\left[\left(\left|\theta_{n-1}\right| + 1 \right) \frac{\pi}{4} + \frac{\left|\theta_{n-1}\right| - \left|\theta_{n}\right|}{\sqrt{\left|\theta_n\right|}}\operatorname{arctanh}\left(\frac{1}{\sqrt{\left|\theta_n\right|}} \right)\right] \\
&= \frac{a_{n - 1}}{a_n}\frac{\pi}{2} + \left[1 - \frac{a_{n + 1}\left(a_{n - 1} - a_{n - 2} \right)}{a_n\left(a_n - a_{n - 1} \right)} \right]\sqrt{\frac{a_n - a_{n - 1}}{a_{n + 1}}}\operatorname{arccoth}\left(\sqrt{\frac{a_{n + 1}}{a_n - a_{n - 1}}} \right)
\end{align}
|
Since the recurrence relation is fully nonlinear, it might be better to try the first few terms with the initial condition $F_0=f$. By Mathematica, it is easy to find the general expression should be of the form
$$
F_n=\frac{a_n+b_nf}{a_n+b_n+a_nf}
$$
for $n\ge 1$, where $a_n$ and $b_n$ are constants. By the recurrence relation, these constants must follow
$$
\left(
\begin{array}{c}
a_{n+1}\\
b_{n+1}
\end{array}
\right)=\left(
\begin{array}{cc}
1&1\\
1&0
\end{array}
\right)\left(
\begin{array}{c}
a_n\\
b_n
\end{array}
\right),
$$
with
$$
\left(
\begin{array}{c}
a_1\\
b_1
\end{array}
\right)=\left(
\begin{array}{c}
1\\
0
\end{array}
\right).
$$
Thanks to linear algebra, you may solve this Fibonacci-like $a_n$ and $b_n$ immediately.
Hope this could be somewhat helpful for you.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3064562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Maximizing $f$ in $\mathbb{R}^3$
Find the domain and the maximum value that the function
$$f(x,y,z)=\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}$$
may attain in its domain.
I have found the domain of the function to be $\mathbb{R^3\backslash\mathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having
$$f_x=\frac{-2 x y-3 x z+y^2+z^2}{\left(x^2+y^2+z^2\right)^{3/2}},\quad f_y=\frac{2 x^2-x y+z (2 z-3 y)}{\left(x^2+y^2+z^2\right)^{3/2}},\quad f_z=\frac{3 \left(x^2+y^2\right)-z (x+2 y)}{\left(x^2+y^2+z^2\right)^{3/2}}$$
But to solve the system $f_x=0,f_y=0$ and $f_z=0$ is rather hard. What are the plausible values of $x,y,z$?
|
Consider the vectors $\,\vec{u}=(x,y,z)\,$ and $\,\vec{v}=(1,2,3)$, then we can write
$$\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}=\frac{\vec{u}\boldsymbol{\cdot}\vec{v}}{\Vert\vec{u}\Vert}=\frac{\Vert\vec{u}\Vert\Vert\vec{v}\Vert\cos(\alpha)}{\Vert\vec{u}\Vert}=\Vert\vec{v}\Vert\cos(\alpha)=\sqrt{1^2+2^2+3^2}\cos(\alpha
)=\sqrt{14}\cos(\alpha)$$
When is the last expression maximized? When $\alpha=k\pi,\,k\in\mathbb{Z}$, thus the maximum value of $f$ is $\sqrt{14}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3065678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Solve the limit using Taylor seris with Big-O notation I have a limit
$$
\lim_{x \to 0} \frac{1}{\sin x} - \frac{1}{x^2}
$$
I've tried to solve it like this:
\begin{align}
&\lim_{x \to 0} \frac{1}{\sin x} - \frac{1}{x^2} =\\
&\lim_{x \to 0} \frac{x^2 - \sin x}{x^2\sin x} = \\
&\lim_{x \to 0} \frac{ x^2 - x + \frac{x^3}{6} - \frac{x^5}{120} - \mathcal{O}\left(x^7\right) }{x^3 - \frac{x^5}{6} + \frac{x^7}{120} + x^2\mathcal{O}(x^7)}
\end{align}
My questions:
*
*How am I supposed to evaluate $x^2\mathcal{O}(x^7)$ in the denominator? I'd say it will be $\mathcal{O}(x^9)$, but I'm not sure about it.
*How am I supposed to evaluate the whole limit with respect to 0 considering Big-O notation? Could you, please, provide some intuitive explanation?
|
Intuitively. You understand that it is the indeterminate form $\infty-\infty$. To avoid it take $\frac1x$ out of brackets:
$$\frac1{\sin x}-\frac1{x^2}=\frac1x\left(\frac x{\sin x}-\frac1x\right).$$
When $x\to 0^+$, the limit is $(+\infty)(-\infty)=-\infty$. When $x\to 0^-$, the limit is $(-\infty)(+\infty)=-\infty$. Because for $x\to 0$, $x^2$ approaches zero faster than $\sin x\sim x$.
When you use Taylor expansion:
$$\sin x=x-\frac{x^3}{6}+O\left(x^5\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right),\\
\lim_{x \to 0} \frac{ x^2 - x + \frac{x^3}{6} - \frac{x^5}{120} - O\left(x^7\right) }{x^3 - \frac{x^5}{6} + \frac{x^7}{120} + \underbrace{x^2O(x^7)}_{O(x^9)}}=\\
\lim_{x \to 0} \frac{\require{cancel}\cancel{x}\left( x - 1 + \frac{x^2}{6} - \frac{x^4}{120} - O\left(x^6\right)\right)}{\cancel{x}\left(x^2 - \frac{x^4}{6} + \frac{x^6}{120} + O(x^8)\right)}=-\infty.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3066897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Find all integer solutions to $x^2+xy+y^2=((x+y)/3 +1)^3$
Find all ordered pairs of integers $(x,y)$, that satisfy the following relation:
$$x^2+xy+y^2=((x+y)/3 +1)^3$$
I tried bashing at first. Then I assumed $x+y = 3k$ for some integer $k$ so that $y=3k-x$, substituted in the given equation and got a cubic polynomial involving $k$ and $x$ if I had solved it right. But I dont know how to proceed further. Please help. Any other solution is also welcome.
|
Bashing seems like a good approach. Setting $x+y=3k$ as you suggest, the equation becomes $$x^2-3kx+9k^2=(k+1)^3.$$ For a fixed integer $k$, we can consider this as a quadratic in $x$, and it has integer roots iff the discriminant $$(-3k)^2-4(9k^2-(k+1)^3))=-27k^2+4(k+1)^3$$ is a perfect square. Serendipitously, this discriminant factors as $$(k-2)^2(4k+1),$$ so it is a perfect square iff either $k=2$ or $4k+1$ is a perfect square. When $k=2$, $4k+1$ is in fact also a perfect square, so $4k+1$ is a perfect square in all cases.
We can thus write $4k+1=(2a+1)^2$ for some integer $a$, so $k=a^2+a$ and the discriminant is $(a^2+a-2)^2(2a+1)^2$. The quadratic formula then gives
$$x=\frac{3a^2+3a\pm(a^2+a-2)(2a+1)}{2}$$ and $$y=3a^2+3a-x=\frac{3a^2+3a\mp(a^2+a-2)(2a+1)}{2}$$ which are a solution for any integer $a$. Or, simplified, we can say $x$ and $y$ are $a^3+3a^2-1$ and $-a^3+3a+1$ in some order.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3068288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
What is the maximum value of $(a+ b+c)$ if $(a^n + b^n + c^n)$ is divisible by $(a+ b+c)$ where the remainder is 0?
The ‘energy’ of an ordered triple $(a, b, c)$ formed by three
positive integers $a$, $b$ and $c$ is said to be n if the following $c$
$\ge b\geq a$, gcd$(a, b, c) = 1$, and $(a^n + b^n + c^n)$ is
divisible (remainder is 0) by $(a +b+ c) $. There are some
possible ordered triple whose ‘energy’ can be of all values of $n \ge$
$1$. In this case, for which ordered triple, the value of $(a+b+c)$ is
maximum?
Second part (of the original problem) Determine all triples $(a,b,c)$ that are simultaneously $2004$-good and $2005$-good, but not $2007$-good.
Source: Bangladesh Math Olympiad 2017 junior category (Originally from Canada, 2005).
I can't understand the first line of this question. Any 3 consecutive numbers have a gcd of 1. Moreover if $n=1$, then $(a^n + b^n + c^n) = (a +b+ c) $.
|
The answer is indeed $6$. Here is a complete solution.
First, take a prime $p$ such that, $p\mid a+b+c$. Note that, if $p$ divides two of $a$ and $b$, then it must divide the third, contradicting with $(a,b,c)=1$. Thus, $p$ either divides none of them, or only one.
*
*If $p\nmid a,b,c$, then by taking $n=p-1$, we have, by Fermat's theorem that $a^{p-1}\equiv b^{p-1}\equiv c^{p-1}\equiv 1 \pmod{p}$. Thus, $p\mid 3$, hence $p=3$.
*If $p\mid a$, and $p\nmid b,c$, then we obtain $p\mid 2$, by taking $n=p-1$.
Therefore, only prime divisors of $a+b+c$ are $2$ or $3$. Now, for $n=2$, we get using $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ that $a+b+c\mid 2(ab+bc+ca)$. Also, for $n=3$, using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we have $a+b+c\mid 3abc$. Next, note that, if $9\mid a+b+c$, then $3\mid abc$. Hence, either $a$ or $b$ or $c$ is divisible by $3$. Now, if $3\mid a$, then using $a+b+c\mid 2(ab+bc+ca)$, we obtain that $3\mid bc$, hence $3\mid b$ or $3\mid c$. However, this, together with $3\mid a+b+c$ contradicts with $(a,b,c)=1$. Hence, $9\nmid a+b+c$.
Similarly, if $4\mid a+b+c$, then note that among $a,b,c$ exactly one is even, suppose it is $a$, that is, $4\mid a$. But this gives, $4\mid 2(ab+bc+ca)$, yielding $2\mid (ab+bc+ca)$, yielding $2\mid bc$. However, since $b$ and $c$ are odd, this is a clear contradiction.
Thus, $a+b+c=3$ or $a+b+c=6$. In former, we get $(1,1,1)$, which is really $n$-good for any $n$. For the latter, we have $(3,2,1)$ or $(4,1,1)$. For the former, $(3,2,1)$ is not $n-$ good for any even $n$, using modulo $3$. For the latter, it is easy to see that it is $n$-good for any $n$.
Hence, we are done.
Note (Alternative) There is an alternative way of proving that $a+b+c$ can only admit $2$ or $3$ as its prime divisors. To see this, suppose $p>3$ divides $a+b+c$. Then, $p\mid 3abc$ implies $p\mid abc$. Using $(a,b,c)=1$,
we see that exactly one of $a,b,c$ is divisible by $p$. Suppose, it is $a$. Then, $p\mid ab+bc+ca$, together with $p\mid ab+ac$ implies $p\mid bc$, which clearly is a contradiction. From here, one can finish in exact same way as in above proof, i.e., prove $4,9\nmid a+b+c$, and finish.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3069401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Find a matrix and analytic formula for a linear map $f: \mathbb R^{4} \rightarrow \mathbb R^{3}$ This linear map $f: \mathbb R^{4} \rightarrow \mathbb R^{3}$ meets the conditions:
\begin{align}\newcommand{\ran}{\operatorname{ran}}
f(1,2,3,1) &=(1,3,1) \\
(1,5,4,1) &\in \ker f\\
(1,1,2) &\in \mathrm{im} f \\
(7,5,0) &\in \mathrm{im} f
\end{align}
I think to do this task firstly, I should create a matrix, so I have:
$$
\begin{bmatrix}
1 & 2 & 3 & 1 &
| & 1 & 3 & 1 \\
1 & 5 & 4 & 1 &
| & 0 & 0 & 0\\
& & & & | & 1 & 1 & 2\\
& & & & | & 7 & 5 & 0
\end{bmatrix}
$$
However, I don't know how to use information about $\mathrm{im} f$ so my matrix is incomplete.
|
$\newcommand{\ran}{\operatorname{ran}}\newcommand{\R}{\mathbb{R}}$I guess you want a matrix which correspond with the linear map with respect to the standard basis.
First of all, let me tell you that you cannot find a unique solution to this exercise. So at some point you have to choose some properties of your solution.
Let $v_1 = \left(\begin{smallmatrix}1 \\ 2\\3 \\ 1\end{smallmatrix}\right)$ and $v_2 = \left(\begin{smallmatrix}1\\5\\4\\1\end{smallmatrix}\right)$.
You have to choose two more vectors $v_3,v_4$, such that $\{v_1,v_2,v_3,v_4\}$ is a basis of $\R^{4}$. One possible choice would be $v_3 = \left(\begin{smallmatrix}0\\0\\1\\ 0\end{smallmatrix}\right)$ and $v_4 = \left(\begin{smallmatrix}0\\0\\0\\ 1\end{smallmatrix}\right)$. Then you choose that $f$ maps
$$
v_3 \mapsto \left(\begin{smallmatrix}1\\1\\2\end{smallmatrix}\right), \quad
v_4 \mapsto \left(\begin{smallmatrix}7\\5\\0\end{smallmatrix}\right).
$$
Now you just have to calculate the image of $e_1 (=\left(\begin{smallmatrix}1\\0\\0\\0\end{smallmatrix}\right))$ and $e_2 (=\left(\begin{smallmatrix}0\\1\\0\\0\end{smallmatrix}\right))$.
\begin{align}
\left(\begin{matrix}
1 & 1 & 0 & 0 \\
2 & 5 & 0 & 0 \\
3 & 4 & 1 & 0 \\
1 & 1 & 0 & 1 \\
\hline
1 & 0 & 1 & 7 \\
3 & 0 & 1 & 5 \\
1 & 0 & 2 & 0 \\
\end{matrix}\right)
\rightsquigarrow \text{many elementary steps} \rightsquigarrow
\left(\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\hline
-\frac{23}{3} & -\frac{2}{3} & 1 & 7 \\
-\frac{7}{3} & -\frac{4}{3} & 1 & 5 \\
-3 & -1 & 2 & 0 \\
\end{matrix}\right)
\end{align}
Therefore, the matrix which represents your linear mapping $f$ is
$$
\left(
\begin{matrix}
-\frac{23}{3} & -\frac{2}{3} & 1 & 7 \\
-\frac{7}{3} & -\frac{4}{3} & 1 & 5 \\
-3 & -1 & 2 & 0 \\
\end{matrix}\right).
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3071388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Using the R method for finding all solutions to $\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$. My solution differs from official answer.
How many solutions does
$$\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$$
have between $-90^\circ$ and $90^\circ$?
I used the R method and got
$$2a-45^\circ = \arcsin\left(\frac{\sqrt{3}}{2}\right).$$
Since $a$ is between $-90^\circ$ and $90^\circ$, then $2a$ is between $-180^\circ$ and $180^\circ$. The RHS can be $60^\circ$, $120^\circ$, $-240^\circ$, and $-300^\circ$. Only $60^\circ$ and $120^\circ$ fit the criteria, but the answer is 4 solutions.
Where did I go wrong?
|
Hint:
$$2a-45^\circ=180^\circ n+(-1)^n\arcsin\dfrac{\sqrt3}2$$
If $n$ is odd$=2m+1$(say)
$$2a-45=360m+180-60$$
But $-180-45\le2a-45\le180-45$
$-225\le360m+120\le135$
$?\le m\le?$
What if $n$ is even $=2m$(say)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3072784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Evaluate $\lim\limits_{x \to 1}\dfrac{x-x^x}{1-x+\ln x}$. Problem
Evaluate $$\lim\limits_{x \to 1}\frac{x-x^x}{1-x+\ln x}$$.
Solution
Consider using Taylor's formula. Expand $x^x $ and $\ln x$ at $x=1$. We obtain
$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$\ln x=(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
\begin{align*}
\lim_{x \to 1}
\frac{x-x^x}{1-x+\ln x}&=\lim_{x \to 1}\frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2)]}\\
&=\lim_{x \to 1}\frac{-(x-1)^2-o((x-1)^2)}{-\frac{1}{2}(x-1)^2+o((x-1)^2)}\\
&=2.
\end{align*}
Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?
|
$1-x=h\implies$
$$\lim_{h\to0}(1-h)\cdot\dfrac{1-(1-h)^{-h}}{h+\ln(1-h)}=\lim_{h\to0}\dfrac{1-\left(1+(-h)(-h)+\dfrac{(-h)(-h-1)}2\cdot h^2+O(h^3)\right)}{h-\left(h+\dfrac{h^2}2+O(h^3)\right)}=\dfrac1{\dfrac12}$$
So, you are correct
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3075787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Limit distribution of a reducible Markov chain I had a simple question yesterday when I was trying to solve an exercise on a reducible,aperiodic Markov Chain. The state spase S was $$S=\{1,...,7\}$$ and we could partition it into two closed classes $\{5,6,7\} \bigcup \{3,4\}$ but the class $\{1,2\}$ was open (thus transient). The exercise was giving the transition matrix and was asking to find the stationary distributions. At the end it was also asking to find the limit distribution, given that the INITIAL distribution was $$P[X_0=1]=P[X_0=2]=\frac{1}{2}$$ I only know what to do in case the initial probability gives mass 1 to some state (I find the absorption probabilities in each closed class, etc..) but when I have initial distribution which gives mass to both of these transient states I do not know what to do. Any helpful ideas/theorems about this??
Thanks a lot !
$$ \begin{pmatrix}
1/3 & 1/6 & 1/3 & 0 & 0 & 1/6 \\
3/5 & 0 & 0 & 1/5 & 1/5 & 0 & 0 \\
0 & 0 & 1/2 & 1/2 & 0 & 0 & 0 \\
0 & 0 & 1/4 & 3/4 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1/4 & 3/4 \\
0 & 0 & 0 & 0 & 1/2 & 0 & 1/2 \end{pmatrix} $$
|
As dan_fulea noticed above, it suffices to calculate the case when $ P[X_0=1]=1 $ and $ P[X_0=2]=1 $ . The first is already treated in the notes: it gives the limit vector $$ \begin{pmatrix} 0, & 0, & \frac{11}{51}, & \frac{22}{51}, & \frac{18}{221}, & \frac{24}{221}, & \frac{36}{221} \end{pmatrix} $$
For the case $ P[X_0=2]=1 $ , since $q_{2,E_1}=\frac{10}{17}$ and $ q_{2,E_2}=\frac{7}{17}$, we get the limit vector : $$\begin{pmatrix} 0, & 0, & \frac{10}{51}, & \frac{20}{51}, & \frac{21}{221}, & \frac{28}{221}, & \frac{42}{221} \end{pmatrix} $$ Hence, the final answer to our question is : $$ \frac{1}{2}\begin{pmatrix} 0, & 0, & \frac{11}{51}, & \frac{22}{51}, & \frac{18}{221}, & \frac{24}{221}, & \frac{36}{221} \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 0, & 0, & \frac{10}{51}, & \frac{20}{51}, & \frac{21}{221}, & \frac{28}{221}, & \frac{42}{221} \end{pmatrix} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3078555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$Ax^2+Bxy+Cy^2=1$ with $A=C=1$ and $B=2$ should be a parabola (because $B^2 = 4 AC$). Instead, it represents parallel lines. What went wrong? My calculus book says
The equation $Ax^2+Bxy+Cy^2=1$ produces a hyperbola if $B^2>4AC$ and
an ellipse if $B^2<4AC$. A parabola has $B^2=4AC$.
When I set $B=2,A=C=1$, the equation becomes $x^2+2xy+y^2=1$, which is equivalent to $(x+y)^2=1$ or $y=-x\pm1$. This is not a parabola but two parallel lines. What went wrong?
|
You are right: a nondegenerate parabola doesn't admit an equation of the form $Ax^2+2Bxy+Cy^2=1$.
Indeed, if $B^2=AC\ne0$, we can multiply all coefficients by $A$, finding
$$
A^2x^2+2ABxy+B^2y^2=B
$$
that becomes $(Ax+By)^2=B$. If $B>0$ this factors as
$$
(Ax+By-\sqrt{B})(Ax+By+\sqrt{B})=0
$$
which is a pair of parallel lines.
If $B<0$, the factorization is over the complex numbers and we get no real point.
If $B=0$, then either $A=0$ or $C=0$, so the equation becomes either $Cy^2=1$ or $Ax^2=1$, again a situation as before.
Only nondegenerate conics with a center admit an equation of the given form, which is obtained by translating the center of the conic in the origin.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3079164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Is there always exactly one solution to $a \cos\left(\frac{x}{2}\right)- b \sin\left(\frac{x}{2}\right) = 0$ in the interval $0\leq x\leq 2\pi$? I have a probably simple question.
If I have an equation like
\begin{align}
a \cos \left(\frac{x}{2}\right) - b \sin \left(\frac{x}{2}\right) &= 0\\
\frac ab \cos \left(\frac{x}{2}\right) - \sin \left(\frac{x}{2}\right) &= 0
\end{align}
with $a, b > 0$. Is it true that there is always precisely one solution in the interval $0 \leq x \leq 2 \pi$.
I think it is obvious, but can I use this in a proof without proving it?
|
Simply note the definition of tangent: $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$.
$$a\cos\left(\frac{x}{2}\right)-b\sin\left(\frac{x}{2}\right) = 0 \iff a\cos\left(\frac{x}{2}\right) = b\sin\left(\frac{x}{2}\right) \iff \\ \frac{a}{b} = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} \iff \frac{a}{b} = \tan\left(\frac{x}{2}\right)$$
Since the range of $\tan(x)$ includes all real numbers, there is a solution in the interval $x \in [0, 2\pi]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3079433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I solve limits like this by property $ \left(a-b\right)\left(a+b\right)=a^2-b^2 $
I made this far:
$$
\lim _{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right)
=n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}=\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)} = \frac{-n^8+n^6\sqrt{n^4+1}}{n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}}$$
I will appreciate every help. Thank you
|
Like Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $,
$$F=\lim _{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4}+1}}-\sqrt{2}n\right)\right)
=\lim_{y\to0^+}\dfrac{\sqrt{1+\sqrt{1+y}}-\sqrt2}y$$
Set $\sqrt{1+\sqrt{1+y}}-\sqrt2=u\implies u\to0$
and $1+\sqrt{1+y}=(u+\sqrt2)^2=2+2\sqrt2u+u^2$
$\implies y=(1+2\sqrt2u+u^2)^2-1=4\sqrt2u+O(u^2)$
$$F=\lim_{u\to0^+}\dfrac u{4\sqrt2u+O(u^2)}=?$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3080173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
}
|
$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted.
Now how do I proceed?
|
Let $\alpha=a+b=c+d$, $\beta=a^3+b^3=c^3+d^3$.
Then $$ \alpha^3=(a+b)^3=a^3+3a^2b+3ab^2+b^3=\beta+\alpha ab.$$
It follows that $a,b$ (and likewise $c,d$) are the two solutions of
$$ X^2-\alpha X+\frac{\alpha^3-\beta}{\alpha}=0,$$
i.e., $\{a,b\}=\{c,d\}$ and hence $$a^n+b^n=c^n+d^n $$
for arbitrary integers $n$. this argument fails only when we cannot divide by $\alpha$, i.e., when $b=-a$. However, in that case also $d=-c$ and so
$$ a^n+b^n=0=c^n+d^n$$
for arbitrary odd integer $n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3080393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
}
|
Differential Equations - Is the solution to $(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$, $y=\ln(-5(e^x+1)^5+c)$? This is the problem. It is question 4. I typed the answer in explicit form, but my professor had said it is incorrect. Could someone check over my work, because I think I did have it correct. My Work
Because someone asked for my work due to bad handwriting:
$$(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$$
Trying to seperate variables:
$$(e^y+1)^2(e^{-y})dx=-(e^x+1)^6(e^{-x})dy$$
$$\frac{-e^x}{(e^x+1)^6}dx=\frac{e^y}{(e^y+1)^2}dy$$
$$\int{\frac{-e^x}{(e^x+1)^6}dx}=\int{\frac{e^y}{(e^y+1)^2}dy}$$
Flip:
$$\int{\frac{e^y}{(e^y+1)^2}dy}=\int{\frac{-e^x}{(e^x+1)^6}dx}$$
where $u = (e^y+1), du =e^ydy$ and $v = (e^x+1), dv=e^xdx$
$$\int{u^{-2}dy}=-\int{v^{-6}}dx$$
$$-u^{-1} + c_1=\frac{1}{5}v^{-5}+c_2$$
Now, substitute:
$$-\frac{1}{e^y+1}+c_1=\frac{1}{5(e^x+1)^5}+c_2$$
The $c$ gets absorbed (also, this is the implicit solution)
$$-\frac{1}{e^y+1}=\frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
But $-c_3$ is the same as $+c_3$
$$-5(e^x+1)^5+c_3=e^y+1$$
$$-5(e^x+1)^5+c_3=e^y$$
Take the natural log of both sides:
$$\ln(-5(e^x+1)^5+c_3)=\ln(e^y)$$
$$\ln\left(-5(e^x+1)^5+c_3\right)=y$$
This is the answer I got. It took long to convert this to Latex, so there might be some discontinuities in the work, but the last answer is for sure what I got.
|
The wrong step is here :
$$-\frac{1}{e^y+1}=\frac{1}{5(e^x+1)^5}+c_3$$
Multiply out:
$$-5(e^x+1)^5-c_3=e^y+1$$
The correct steps are :
Multiply both left-hand and right-hand terms by $(e^y+1)(5(e^x+1)^5)$
$$\left(-\frac{1}{e^y+1}\right)(e^y+1)\big(5(e^x+1)^5\big)=\left(\frac{1}{5(e^x+1)^5}+c_3\right)(e^y+1)\big(5(e^x+1)^5\big)$$
Simplify :
$$-\big(5(e^x+1)^5\big)=\left(1+c_3 5(e^x+1)^5\right)(e^y+1)$$
$$ e^y+1=\frac{-5(e^x+1)^5}{1+ 5c_3(e^x+1)^5} $$
$$y=\ln\left(\frac{-5(e^x+1)^5}{1+c_3 5(e^x+1)^5}-1 \right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3081139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Evaluating $\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)\,dx$
How to prove $$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx = \frac43G + \frac13\pi\ln\left(2+\sqrt3\right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $\Im\operatorname{Li}_2\left(2\pm\sqrt3\right)$.
Attempt
$$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx \\ =\int_0^\infty\frac{\operatorname{arcsinh}(2x)}{1+x^2} \, dx\\
=2\int_0^\infty\frac{x\cosh x}{4+\sinh^2x} \, dx\\
=2\int_0^\infty\frac{x\cosh x}{3+\cosh^2x} \, dx\\
=2\int_0^\infty\sum_{n=0}^\infty x(-3)^n\cosh^{-2n-1}(x) \, dx$$
I failed to integrate $x\cosh^{-2n-1}(x)$. Mathematica returns a hypergeometric term while integrating it.
|
I will begin with the same approach as in @DavidG's answer, but will settle in a different argument. Let
$$ J(t) = \int_{0}^{\infty} \frac{\operatorname{arsinh}(tx)}{1+x^2} \, \mathrm{d}x. $$
Our goal is to compute $J(2)$ using Feynman's trick. Differentiating $J(t)$ and substituting $x=\sqrt{u^{-2}-1}$, we obtain
\begin{align*}
J'(t)
= \int_{0}^{\infty} \frac{x}{(1+x^2)\sqrt{1+t^2x^2}} \, \mathrm{d}x
= \int_{0}^{1} \frac{1}{\sqrt{1 - (1-t^2)(1-u^2)}} \, \mathrm{d}u.
\end{align*}
So it follows that
\begin{align*}
J(2)
&= \int_{0}^{2} \int_{0}^{1} \frac{1}{\sqrt{1 - (1-t^2)(1-u^2)}} \, \mathrm{d}u\mathrm{d}t \\
&= \int_{0}^{1} \int_{0}^{1} \frac{1}{\sqrt{1 - (1-t^2)(1-u^2)}} \, \mathrm{d}u\mathrm{d}t
+ \int_{1}^{2} \int_{0}^{1} \frac{1}{\sqrt{1 + (t^2-1)(1-u^2)}} \, \mathrm{d}u\mathrm{d}t.
\end{align*}
The inner integral is easily computed, yielding
\begin{align*}
J(2)
&= \int_{0}^{1} \frac{\operatorname{artanh}\left( \sqrt{1 - t^2} \right)}{\sqrt{1 - t^2}} \, \mathrm{d}t
+ \int_{1}^{2} \frac{\arctan\left(\sqrt{t^2-1}\right)}{\sqrt{t^2 - 1}} \, \mathrm{d}t.
\end{align*}
Now we substitute $t = \operatorname{sech} \varphi$ for the first integral and $t = \sec \theta$ for the second integral. This yields
\begin{align*}
J(2)
&= \int_{0}^{\infty} \frac{\varphi}{\cosh\varphi} \, \mathrm{d}\varphi
+ \int_{0}^{\frac{\pi}{3}} \frac{\theta}{\cos\theta} \, \mathrm{d}\theta.
\end{align*}
These integrals can be computed as follows:
*
*Using $ \operatorname{sech}\varphi = \frac{2e^{-\varphi}}{1 + e^{-2\varphi}} = 2 \sum_{n=0}^{\infty} (-1)^n e^{-(2n+1)\varphi} $, we obtain
$$ \int_{0}^{\infty} \frac{\varphi}{\cosh\varphi} \, \mathrm{d}\varphi
= 2 \sum_{n=0}^{\infty} (-1)^n \int_{0}^{\infty} \varphi e^{-(2n+1)\varphi} \, \mathrm{d}\varphi
= 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}
= 2G. $$
*Taking integration by parts,
\begin{align*}
\int_{0}^{\frac{\pi}{3}} \frac{\theta}{\cos\theta} \, \mathrm{d}\theta
&= \left[ - \theta \log \left( \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right) \right]_{0}^{\frac{\pi}{3}} + \int_{0}^{\frac{\pi}{3}} \log \left( \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right) \, \mathrm{d}\theta \\
&= \frac{\pi}{3} \log \left( 2 + \sqrt{3} \right) + 2 \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \log \left( \tan \theta \right) \, \mathrm{d}\theta.
\end{align*}
This can be computed by using the Fourier series $\log \left( \tan \theta \right) = - 2 \sum_{n=0}^{\infty} \frac{\cos(4n+2)\theta}{2n+1} $ to yield
\begin{align*}
\int_{0}^{\frac{\pi}{3}} \frac{\theta}{\cos\theta} \, \mathrm{d}\theta
&= \frac{\pi}{3} \log \left( 2 + \sqrt{3} \right) - 2 \sum_{n=0}^{\infty} \frac{\sin\left( \frac{\pi}{2} (2n+1) \right) - \sin\left( \frac{\pi}{6} (2n+1) \right)}{(2n+1)^2} \\
&= \frac{\pi}{3} \log \left( 2 + \sqrt{3} \right) - \frac{2}{3}G.
\end{align*}
Combining two result, we obtain the desired answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3082875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 3
}
|
How to Find the integral $\int_{S}−(xy^2)~dydz + (2x ^2 y)~dzdx − (zy^2)~dxdy$ where is the portion of the sphere $x^2 + y^2 + z^2 = 1$
Find the integral
$$
\int_{S}−(xy^2)~dydz + (2x^2y)~dzdx − (zy^2)~dxdy
$$
where $S$ is the portion of the sphere $x^2 + y^2 + z^2 = 1$ above the plane $z=\frac{1}{2}$. Choose the direction of the normal to be outside the sphere.
Can i get some help? I got that the flux is
$$
\huge \phi = \huge 0.
$$
and I am not sure if thats right.
\begin{split}
\vec F &= (-xy^2,2x^2y,-zy^2)\\
\vec n_\mathrm{sphere} &= (2x,2y,2z)\\
\vec{F}\cdot{}\vec{n} = 2y^2(x^2 - z^2) &=^{{~(z^2 = 1-x^2-y^2)}}~4x^2y +2y^3 -2y\\
\int\int 4x^2y +2y^3 - 2ydydx &=
\end{split}
$$\int_{r=0}^{r=\sqrt{\frac{3}{4}}}\int_{\theta=0}^{\theta=2\pi}4r^4\cos^2\theta \sin\theta + 2r^4\sin^3\theta - 2r^2\sin\theta d\theta dr = \boxed{0}\\
$$
|
Let $$
\Omega: \frac{1}{2}\le z\le \sqrt{1-x^2-y^2}
$$ be the region enclosed by $S\bigcup S_1$ where $S_1 :z=\frac{1}{2}, x^2+y^2\le\frac{3}{4}$. We obtain
$$
\iiint_\Omega 2(x^2-y^2)\ \mathrm{d}x\mathrm{d}y\mathrm{d}z =\iint_{S\bigcup S_1}\vec{F}\cdot \vec{n}\ \mathrm{d}S
$$ by divergence theorem. Note that
$$
\iiint_\Omega x^2\ \mathrm{d}x\mathrm{d}y\mathrm{d}z=\iiint_\Omega y^2\ \mathrm{d}x\mathrm{d}y\mathrm{d}z
$$ by the rotational symmetry, hence the left-hand side equals $0$. This gives
$$\iint_{S\bigcup S_1}\vec{F}\cdot \vec{n}\ \mathrm{d}S=0.
$$ Since outward unit normal vector $\vec{n}=(0,0,-1)^T$ on $S_1$, it follows
$$\begin{eqnarray}
\iint_{S}\vec{F}\cdot \vec{n}\ \mathrm{d}S&=&-\iint_{S_1}\vec{F}\cdot \vec{n}\ \mathrm{d}S\\&=&-\frac{1}{2}\iint_{x^2+y^2\le \frac{3}{4}} y^2\ \mathrm{d}x\mathrm{d}y\\
&=&-\frac{1}{4}\iint_{x^2+y^2\le \frac{3}{4}} (x^2+y^2)\ \mathrm{d}x\mathrm{d}y\\
&=&-\frac{1}{4}\int_0^{2\pi}\int_{0}^{\frac{\sqrt{3}}{2}}r^3\ \mathrm{d}r\mathrm{d}\theta=-\frac{9\pi}{128}.
\end{eqnarray}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3083747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
pdf of transformed variable Given pdf $f_X(x)=\frac{x+2}{18}$ where $-2 < x < 4$, I wanted to find another r.v. $Y = \frac{12}{|X|}$. I think the support of $Y$ would be $3 < y < \infty$ but I wasn't super sure. I found the cdf of $X$, which was $F_X(x) = \int_{-2}^x\frac{x+2}{18}dx=\frac{x^2}{36}+\frac{x}{9}+\frac{1}{9}$, $-2 < x < 4$. I tried using the cdf of $X$ to find the cdf of $Y$ and consequently find the pdf of $Y$ but have been struggling to do so. Could anyone help me with the derivation?
\begin{align*}
F_Y(y)&=P(Y \leq y) \\
&=P(\frac{12}{|X|}\leq y) \\
&= P(|X| \geq \frac{12}{y}) \\
&= 1 - P(|X| \leq \frac{12}{y})\\
&=1- P(-\frac{12}{y} \leq X \leq \frac{12}{y}) \\
&= 1 - F_X(\frac{12}{y}) + F_X(-\frac{12}{y})\\
&= 1 - \frac{8}{3y}
\end{align*}
Then I just differentiate to get the pdf:
\begin{align*}
f_Y(y)&=\frac{d}{dy}F_Y(y) \\
&= \frac{8}{3y^2}
\end{align*}
But this pdf doesn't integrate to $1$ so I'm not sure what's wrong.
|
Take care with the absolute signage and the supports for the functions.
Note that $12/\lvert X\rvert$ folds both the domains $(-2;0)$ and $(0;2)$ onto $(6;\infty)$, and the domain $[2;4)$ onto $(3;6]$.
More clearly, only when $Y>6$ it is mapped to by a positive and negative value for $X$. When $3<Y\leq 6$ then $Y$ is mapped to by only a positive value of $X$ (on $[2;4)$).
$$\begin{split}\mathsf P(Y\leq y) &=\mathsf P(\lvert X\rvert\leq 12/y)\mathbf 1_{6<y}+\mathsf P(\lvert X\rvert\leq 12/y)\mathbf 1_{3<y\leq 6}\\ &= \mathsf P(-12/y\leq X\leq 12/y)\mathbf 1_{6<y}+\mathsf P(0\leq X\leq 12/y)\mathbf 1_{3<y\leq 6}\\ &= -F_X(-12/y)\mathbf 1_{6<y}-F_X(0)\mathbf 1_{3<y\leq 6}+F_X(12/y)\mathbf 1_{3<y}\end{split}$$
So
$$\begin{split}f_Y(y)&=\begin{vmatrix}\dfrac{\partial (-12/y)}{\partial y}\end{vmatrix}f_X(-12/y)\mathbf 1_{y\in(6;\infty)}+\begin{vmatrix}\dfrac{\partial (12/y)}{\partial y}\end{vmatrix}f_X(12/y)\mathbf 1_{y\in(3;\infty)}\\ &=\dfrac{(-24+4y)}{3y^3}\mathbf 1_{6<y}+\dfrac{(24+4y)}{3y^3}\mathbf 1_{3<y}\\ &= \dfrac{8}{3y^2}\mathbf 1_{6<y}+\dfrac{(24+4y)}{3y^3}\mathbf 1_{3<y\leq 6}\end{split}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3086587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Transforming integral to polar coordinates By transforming to polar coordinates, show that
$$\int_{0}^{1} \int_{0}^{x}\frac{1}{(1+x^2)(1+y^2)} \,dy\,dx$$
Is equal to
$$ \int_{0}^{\pi/4}\frac{\log(\sqrt{2}\cos(\theta))}{\cos(2\theta)} d\theta$$
I have tried the standard $x=r\cos(\theta)$ etc. And can easily see that the limits for theta is $0,\pi/4$ with a sketch. However, I'm struggling to compute the integral that comes out, which ranges $r$ from $r=0$ to $r=\frac{1}{\cos{\theta}}$. Also, I can integrate this explicitly using tan substitution but does that help?
|
You have correctly converted the region of integration and thus we have:
\begin{align}
I &= \int_0^{\frac{\pi}{4}}\int_1^{\sec(\theta)} \frac{1}{\left(1 + r^2\cos^2(\theta)\right)\left(1 + r^2\sin^2(\theta)\right)} \cdot r\:dr\:d\theta \\
\end{align}
Applying a partial fraction decomposition we arrive at:
\begin{align}
I &= \int_0^{\frac{\pi}{4}}\int_1^{\sec(\theta)} \frac{1}{\left(1 + r^2\cos^2(\theta)\right)\left(1 + r^2\sin^2(\theta)\right)} \cdot r\:dr\:d\theta \\
&= \int_0^{\frac{\pi}{4}}\int_1^{\sec(\theta)} \frac{1}{\cos^2(\theta) - \sin^2(\theta)}\left[\frac{\cos^2(\theta)}{1 + r^2\cos^2(\theta)} - \frac{\sin^2(\theta)}{1 + r^2\sin^2(\theta)} \right] r\:dr\:d\theta \\
&= \int_0^{\frac{\pi}{4}}\frac{1}{\cos^2(\theta) - \sin^2(\theta)}\int_1^{\sec(\theta)}\left[\frac{r\cos^2(\theta)}{1 + r^2\cos^2(\theta)} - \frac{r\sin^2(\theta)}{1 + r^2\sin^2(\theta)} \right]\:dr\:d\theta \\
&=\int_0^{\frac{\pi}{4}}\frac{1}{\cos^2(\theta) - \sin^2(\theta)}\cdot\frac{1}{2}\bigg[\ln\left|1 + r^2\cos^2(\theta) \right| + \ln\left|1 + r^2\sin^2(\theta) \right| \bigg]_1^{\sec(\theta)} \:d\theta
\end{align}
You okay from here?
Edit: The rest of the solution:
\begin{align}
I &= \int_0^{\frac{\pi}{4}}\frac{1}{\cos^2(\theta) - \sin^2(\theta)}\cdot\frac{1}{2}\bigg[\ln\left|1 + r^2\cos^2(\theta) \right| + \ln\left|1 + r^2\sin^2(\theta) \right| \bigg]_1^{\sec(\theta)} \:d\theta \\
&= \int_0^{\frac{\pi}{4}}\frac{1}{\cos^2(\theta) - \sin^2(\theta)}\cdot\frac{1}{2}\bigg[\ln(2) + \ln\left|\sec^2(\theta) \right| \bigg] \:d\theta \\
&= \int_0^{\frac{\pi}{4}}\frac{1}{\cos(2\theta)}\cdot\frac{1}{2}\ln\left|2\sec^2(\theta)\right|\:d\theta\int_0^{\frac{\pi}{4}}\frac{\ln\left| \sqrt{2}\sec(\theta)\right|}{\cos(2\theta)}\:d\theta
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3086741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
The Integral $\int \frac {dx}{(x^2-2ax+b)^n}$ Recently I came across this general integral,
$$\int \frac {dx}{(x^2-2ax+b)^n}$$
Putting $x^2-2ax+b=0$ we have,
$$x = a±\sqrt {a^2-b} = a±\sqrt {∆}$$
Hence the integrand can be written as,
$$
\frac {1}{(x^2-2ax+b)^n}
=
\frac {1}{(x-a-\sqrt ∆)^n(x-a+\sqrt ∆)^n}
$$
Resolving into partial fractions we have,
$$
\frac {1}{(x^2-2ax+b)^n}
=
\sum \frac {A_r}{(x-a-\sqrt ∆)^r} + \sum \frac {B_r}{(x-a+\sqrt ∆)^r}
$$
Putting $-\frac {1}{2\sqrt ∆} = D$ , I could produce a table of the coefficients $A$ and $B$ for different $n$.
\par
For $n=1$,
$$A_1=-D , B_1=D$$
For $n=2$,
$$A_1=2D^3 , B_1=-2D^3$$
$$A_2=D^2 , B_2 = D^2$$
For $n=3$,
$$A_1=-6D^5 , B_1=6D^5$$
$$A_2=-3D^4 , B_2 = -3D^4$$
$$A_3=-D^3, B_3=D^3$$
For $n=4$,
$$A_1=20D^7, B_1=-20D^7$$
$$A_2=10D^6 , B_2 = 10D^6$$
$$A_3=4D^5, B_3=-4D^5$$
$$A_4=D^4, B_4=D^4$$
For $n=5$,
$$A_1=-70D^9, B_1=70D^9$$
$$A_2=-35D^8, B_2 = -35D^8$$
$$A_3=-15D^7, B_3=15D^7$$
$$A_4=-5D^6, B_4=-5D^6$$
$$A_5=-D^5, B_4=D^5$$
Yet I am unable to deduce a general formula for the coefficients. If I have the coefficients, the integral is almost solved , for then I shall have a logarithmic term and a rational function in $x$. More directly, I seek a result of the form,
$$\kappa \log \left( \frac {x-a-\sqrt ∆}{x-a+\sqrt ∆}\right) + \frac {P(x)}{Q(x)}$$
Any help would be greatly appreciated.
Conjecture 1(Proved below)
$$A(n,r)= (-1)^n \binom {2n-r-1}{n-1} D^{2n-r}$$
$$B(n,r)= (-1)^{n-r} \binom {2n-r-1}{n-1} D^{2n-r}$$
|
Let $a$, $b$ and $x$ be real and $n$ be a positive integer. Let $\Delta:= a^2-b$.
then the following formula holds:
\begin{eqnarray}
\frac{1}{(x^2-2 a x+b)^n} &=&
\sum\limits_{l_1=1}^n \binom{2 n-1-l_1}{n-1} \frac{(-1)^n}{(x-a-\sqrt{\Delta})^{l_1}} \cdot \frac{1}{(-2 \sqrt{\Delta})^{2 n-l_1}} + \\
&& \sum\limits_{l_1=1}^n \binom{2 n-1-l_1}{n-1} \frac{(-1)^n}{(x-a+\sqrt{\Delta})^{l_1}} \cdot \frac{1}{(+2 \sqrt{\Delta})^{2 n-l_1}}
\end{eqnarray}
The result follows from the second formula from the top in my answer to How to quickly solve partial fractions equation? .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3086878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
How to convert $\frac{1}{(1-x)(1-x^3)}$ into a sum of multiple fractions? What I mean is, that one can convert $\frac{1}{(1-x)^2(1-x^2)}$ into the following sum:
$\frac{1}{8}(\frac{1}{1+x}+\frac{1}{1-x} +\frac{2}{(1-x)^2} + \frac{4}{(1-x^3)})$
But I can't seem to do the same here, because when I try to simplify $1-x^3$ I get $(1-x)(1+x+x^2)$, and the second term can't be further simplified.
For context, this is part of another problem in combinatorics, which I'm trying to solve using generating functions.
|
Since you know that$$\frac1{(x-1)(x^3-1)}=\frac1{(x-1)^2(x^2+x+1)},$$use this to deduce that$$\frac1{(x-1)(x^3-1)}=\frac{x+1}{3\left(x^2+x+1\right)}-\frac1{3(x-1)}+\frac1{3 (x-1)^2},$$by partial fraction decomposition.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3087102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Extracting two balls with the same color vs distinct color I am trying to find which has higher probability when extracting from a box without replacement: two balls of the same color, or two balls with different colors.
Let's assume that in the box are $a$ white balls and $b$ black balls.
$$P(\text{distinct color})=P(\text{first white})P(\text{second black/first white})+P(\text{first black})P(\text{second white/first black})$$
$$=\frac{a}{a+b}\frac{b}{a+b-1}+\frac{b}{a+b}\frac{a}{a+b-1}=\frac{2ab}{(a+b)(a+b-1)}$$
$$P(\text{same color})=P(\text{first white})P(\text{second white/first white})+P(\text{first black})P(\text{second black/first black})$$
$$=\frac{a}{a+b}\frac{a-1}{a+b-1}+\frac{b}{a+b}\frac{b-1}{a+b-1}=\frac{a^2+b^2-(a+b)}{(a+b)(a+b-1)}$$
But there is still needed to find which is more probably to happen, the case which I don't know how to proceed. Do I have to solve (I don't know how) for which domain of $a$ and $b$ does the following inequality hold true?
$$a^2+b^2-(a+b)< 2ab \equiv (a-b)^2<a+b $$
This is like in the below answer, so I did it correctly, but still that doesn't help me to conclude an answer..
|
Your inequality
\begin{align*}
(a-b)^2< a + b
\end{align*}
can be expressed as
\begin{align*}
a - \sqrt{2a + \frac{1}{4}}+\frac{1}{2} < b < a + \sqrt{2a + \frac{1}{4}}+\frac{1}{2}
\end{align*}
or
\begin{align*}
\left|b - a - \frac{1}{2}\right| < \sqrt{2a + \frac{1}{4}}
\end{align*}
This domain isn't pretty in any way, but we can make statements about its behavior as $a,b \rightarrow \infty$. For example, dividing both sides by $a$,
\begin{align*}
\left|\frac{b}{a} - 1 - \frac{1}{2a}\right| < \sqrt{\frac{2}{a} + \frac{1}{4a^2}}
\end{align*}
Therefore, in order for the probability of distinct colors to be greater than probability of the same, we need $b = a$ for sufficiently large values of $a$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3087772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Proving irrationality of $\sqrt[3]{3}+\sqrt[3]{9}$ I need to prove $$\sqrt[3]{3}+\sqrt[3]{9}$$
is irrational, I assumed
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m}{n}$$
I cubed both sides and got
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m^3-12n^2}{9n^3}$$
I tried setting $$\frac{m^3-12n^2}{9n^3} = \frac{m}{n}$$ but that led me nowhere. so what can I do?
|
Let $a=\sqrt[3]{3}$ and $b=\sqrt[3]{9}$, and suppose that $a+b=r$ is rational. Then $$r^3 =(a+b)^3 = a^3+3ab(a+b)+b^3 = 12+9(a+b)=12+9r$$
So $r$ is a solution of equation $x^3-9x-12=0$ which has no rational root.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3090626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
}
|
Evaluate $\lim\limits_{n\to \infty} \left( \cos(1/n)-\sin(1/n) \right) ^n $? How to calculate
$\lim\limits_{n\to \infty} \left( \cos(1/n)-\sin(1/n) \right) ^n $?
Since $\lim\limits_{n\to \infty} \frac {\cos(1/n)-\sin(1/n) }{1-1/n} = 1 $, I guess that the limit above is $\frac{1}{e}$, but since the form $(\to 1)^{\to \infty}$ is indeterminate, I don't know how to prove it formally.
Thanks!
|
We have :
$\begin{align*}
\lim\limits_{n\to \infty} \left( \cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right) \right) ^n&=\lim\limits_{n\to \infty}\left(1+\cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right)-1\right)^n\\&=\lim\limits_{n\to \infty}\left[\bigg(1+\cos(\frac{1}{n})-\sin(\frac{1}{n})-1\bigg)^{\frac{1}{\cos(\frac{1}{n})-\sin(\frac{1}{n})=1}}\right]^{n(\cos(\frac{1}{n})-\sin(\frac{1}{n})-1)}\\&=e^{\lim\limits_{n\to \infty}n(\cos(\frac{1}{n})-\sin(\frac{1}{n})-1)}\end{align*}$
But
$\begin{align*}\lim\limits_{n\to \infty}n\left(\cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right)-1\right)&=\lim\limits_{n\to \infty}\frac{\cos(\frac{1}{n})-\sin(\frac{1}{n})-1}{\frac{1}{n}}\\&=\lim\limits_{x\to 0}\frac{\cos x-\sin x-1}{x}\\&=\lim\limits_{x\to 0}(-\sin x-\cos x)\\&=-1,\end{align*}$
so your limit equals $\frac{1}{e}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3093466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
}
|
Evaluate $\sum_{n=1}^{\infty}\arctan(\frac{2}{n^2})$? $$\begin{align}
\sum_{n=1}^{\infty}\arctan\left(\frac{2}{n^2}\right)&=\sum_{n=1}^{\infty}\arctan\left(\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}\right)\\
&=\sum_{n=1}^{\infty}\arctan(n+1)-\arctan(n-1)\\
&= \frac\pi4
\end{align}$$
Is this correct?
|
The series telescopes. The partial sum $S_N$ is equal to
\begin{align}
S_N &= [\arctan (2) - \arctan (0)] + [\arctan (3) - \arctan (1)] + \cdots\\
& \qquad \cdots + [\arctan (N) - \arctan (N - 2)] + [\arctan(N + 1) - \arctan (N - 1)]\\
&= - \arctan (0) - \arctan (1) + \arctan (N) + \arctan (N + 1)\\
&= -\frac{\pi}{4} + \arctan (N) + \arctan (N + 1)
\end{align}
So for the sum $S$ we have
\begin{align}
S &= \lim_{N \to \infty} S_N\\
&= \lim_{N \to \infty} \left [- \frac{\pi}{4} + \arctan(N) + \arctan(N + 1) \right ]\\
&= -\frac{\pi}{4} + \frac{\pi}{2} + \frac{\pi}{2}\\
&= \frac{3 \pi}{4}.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3094394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Inverse of Roto-Translation Matrix I'm trying to implement two JavaScript functions to calculate
*
*A 2D coordinate transformation from the reference frame $O$ to $O'$;
*The inverse transformation from $O'$ to $O$;
The first step it's very simple for me and it works. My mathematical process is this:
$$
\begin{pmatrix}
x'\\
y'\\
1
\end{pmatrix}
=
\begin{pmatrix}
R & u_0\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
u\\
0
\end{pmatrix}
=
\begin{pmatrix}
cos\theta & -sin\theta & x_0\\
sin\theta & cos\theta & y_0\\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
1
\end{pmatrix}
$$
Where $R$ is the rotation matrix, $u_0$ is the $O'$ origin in the $O$ frame reference and $u$ is a point in $O$. At the end I get this two equations:
$$
\left\{
\begin{array}{c}
x' = xcos\theta - ysin\theta + x_0\\
y' = xsin\theta + ycos\theta + y_0\\
\end{array}
\right.
$$
The problem is how to calculate the inverse transformation. I'm doing something like this but it doesn't work:
$$
\begin{pmatrix}
x\\
y\\
1
\end{pmatrix}
=
\begin{pmatrix}
R^{-1} & -R^{-1}u_0\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
u'\\
0
\end{pmatrix}
=
\begin{pmatrix}
\frac{1}{cos\theta} & sin\theta & -\frac{x_0}{cos\theta}-y_0sin\theta\\
-sin\theta & \frac{1}{cos\theta} & {x_0}{sin\theta}-\frac{y_0}{cos\theta}\\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
x'\\
y'\\
1
\end{pmatrix}
$$
The final equations are:
$$
\left\{
\begin{array}{c}
x = x'\frac{1}{cos\theta} + y'sin\theta -\frac{x_0}{cos\theta}-y_0cos\theta\\
y = -x'sin\theta + y'\frac{1}{cos\theta} + {x_0}{sin\theta}-\frac{y_0}{cos\theta}\\
\end{array}
\right.
$$
Can you tell me where I wrong?
|
Pay attention to what transformation you want to represent, whether active or passive.
Anyway, if $R$ is the matrix that represents a rotation of angle $\theta$ in a plane, its inverse is the matrix that represents a rotation of the opposite angle $- \theta$, around the same origin, and it is easy to see that it is represented by the transpose matrix: $R^{-1}=R^T$ (it seems that this is your mistake, when you use $\frac{1}{\cos \theta}$ in the inverse matrix.).
From this it is also easy to see that the inverse of a rigid tranformation $T$ that , in homogeneous coordinates, is represented by the matrix
$$
T=\begin{pmatrix}
R&\vec u\\
0&1
\end{pmatrix}
$$
is represented by the matrix
$$
T^{-1}=\begin{pmatrix}
R^T&-R^T\vec u\\
0&1
\end{pmatrix}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3094716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
How do I show $\sum\limits_{j=0}^{\infty}e^{-wj}=\frac{1}{w}+\frac{1}{2}+\frac{w}{12}-\frac{w^3}{720}+\frac{w^5}{30240}+\cdots$? Since $$\sum_{k=0}^{\infty}x^k=\frac{1}{1-x},$$we have
$$\sum_{j=0}^{\infty}e^{-wj}=\sum_{j=0}^{\infty} (e^{-w})^j=\frac{1}{1-e^{-w}}=\frac{e^w}{e^w-1}$$
Also, since $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},$$we have
$$\sum_{j=0}^{\infty}e^{-wj}=\frac{1+w+\frac{w^2}{2}\cdots}{w+\frac{w^2}{2}+\frac{w^3}{6}+\cdots}$$
How can I obtain $$\sum_{j=0}^{\infty}e^{-wj}=\frac{1}{w}+\frac{1}{2}+\frac{w}{12}-\frac{w^3}{720}+\frac{w^5}{30240}+\cdots?$$
|
The Bernoulli numbers are defined by
$$\frac{t}{e^t-1}=\sum_{n=0}^{\infty} B_n \frac{t^n}{n!}$$
We have
$$\begin{split}
\sum_{j=0}^{\infty}e^{-wj} &=\frac{1}{1-e^{-w}}\\
&=\frac 1 w \frac{(-w)}{e^{-w}-1}\\
&= \frac 1 w \sum_{n=0}^{\infty} B_n \frac{(-w)^n}{n!}\\
&= \frac 1 w + \sum_{n=0}^{\infty} \frac{(-1)^{n+1}B_{n+1}}{(n+1)!}w^n \text{ (as } B_0=1)
\end{split}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3095477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Simplify $\frac{x^3-x}{x^2+xy+x+y}$ $$\frac{x^3-x}{x^2+xy+x+y}$$
What I did:
$$\frac{x}{xy+x+y}$$
through simplifying the $x$'s.
But it's not right.
What did I do wrong?
|
A good way of checking your working is to put some numbers in. Let's take $x=4,y=2$. Then what you have said is that $$\frac{64-4}{16+4\times2+4+2}=\frac{4}{4\times 2+4+2}$$
but we can see that the first fraction is: \begin{align}\frac{64-4}{16+4\times2+4+2}&=\frac{60}{30}\\
&= 2\end{align} and the second fraction is \begin{align}\frac{4}{4\times2+4+2}&=\frac{4}{14}\\
&= \frac 27\end{align} and so we can very quickly notice that we have done something wrong.
You will probably have been told that we can't cancel out when there is a $+$, so we can't say that $$\frac {x+y}x = y$$ Well this is the same principle, we just have more $+$'s in the denominator.
You may remember that, when we have $+$'s in the numerator or the denominator, that we have to cancel the same thing from all terms, so we could say $$\frac{x+xy}{x} = \frac{\color{red}x+\color{red}xy}{\color{red}x} = \frac{1+y}1 = 1+y$$ because we have cancelled an $x$ out of every term in the fraction.
We can apply the same principle here. First, we must factorise the top and the bottom of the fraction, as we cannot cancel the same thing from every term yet.
The top becomes \begin{align}x^3-x &= x(x^2-1)\\&= x(x-1)(x+1)\end{align} (we do the second step by remembering the difference of 2 squares formula) and the bottom becomes \begin{align}x^2+xy+x+y &= (x^2+xy)+(x+y)\\
&= (x(x+y)) + 1(x+y)\\
&= (1+x)(x+y) \end{align}
Therefore, we have $$\frac{x^3-x}{x^2+xy+x+y} = \frac{x(x-1)(x+1)}{(x+1)(x+y)}$$
Now we can say \begin{align}\frac{x(x-1)(x+1)}{(x+1)(x+y)} &= \frac{x(x-1)\color{red}{(x+1)}}{\color{red}{(x+1)}(x+y)}\\
&= \frac{x(x-1)}{(x+y)}\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3096139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
value of $x$ in Trigonometric equation Find real $x(0<x<180^\circ)$ in
$\tan(x+100^\circ)=\tan(x-50^\circ)+\tan(x)+\tan(x+50^\circ)$
what i try
$\displaystyle \tan(x+100^\circ)-\tan(x)=\tan(x+50^\circ)+\tan(x-50^\circ)$
$\displaystyle \frac{\sin(100^\circ)}{\cos(x+100^\circ)\cos x}=\frac{\sin(2x)}{\cos(x+50^\circ)\cos(x-50^\circ)}$
$\displaystyle \frac{\sin(100^\circ)}{\cos(2x+100^\circ)+\cos(100^\circ)}=\frac{\sin(2x)}{\cos(2x)+\cos(100^\circ)}$
How do i solve further
Help me please
|
Let $\cos100^\circ=a$ and $\sin100^\circ=b$; let $\cos2x=X$ and $\sin2x=Y$. The final equation can be written
$$
\frac{b}{aX-bY+a}=\frac{Y}{X+a}
$$
that is,
$$
bX+ab=aXY-bY^2+aY
$$
together with $X^2+Y^2=1$. This is the intersection between a hyperbola and a circle, so generally a degree 4 equation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3096262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.