The problem here is, that you use the fact that if p|(n+M) \text{ and } p|n \Rightarrow p|M. But this doesn't cover all the possibilities, namely if p\not|n \text{ and } p\not|M \text{ but } p|(n+M), as micromass pointed out.
For any k\geq 2 choose n to be the product of all prime numbers \leq k. Then \{p\in\mathbb{P} : p < n, p \not |n \}=\emptyset. So M need not exist, if I'm not mistaken.
Yeah, that's the part that caught me as well. I remember seing your function somewhere also, it seems to be the most straight forward. The function sequence I had in mind was
f_n(x) =
\begin{cases}
n\cdot x & \text{if} \; \; 0 \leq x < \frac{1}{n} \\
1 & \text{if} \; \; \frac{1}{n} \leq x...
Under the orthogonality section in the wikipedia article on Legendre polynomials, you find the identity
\displaystyle \frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P(x)\right] = -\lambda P(x)
where the eigenvalue \lambda corresponds to n(n+1).
I suppose P(x)=P_n(x) for any n, but I'm not sure though...
Mayby I'm confused with the notations but I thought that \displaystyle d_{\infty}(f,g)=\sup_{x\in[a,b]}|f(x)-g(x)|<\epsilon meant uniform convergence, since doesn't this mean that
\forall x\in[a,b]:\ |f(x)-g(x)|<\epsilon?
For the last problem, try to find a sequence of continuous functions that converges to a non-continuous function. I have one in mind, but haven't thought of the details yet. Consider piecewise functions...
edit: Sorry, the path I suggested might not lead you anywhere. It seems that if a...
What properties do you know about Legendre Polynomials? If you can use the orthogonal properties that are listed in the article on Legendre polynomials in wikipedia, then integration by parts should do the trick.
I suppose you would prove it by the same reasoning as you'd prove that the sequence \{n\}_{n\in \mathbb{N}} diverges, reductio ad absurdum using the definition of convergence to get a contradiction.
If it is calculated like I suspect it is, then you will get a satisfying answer.
So first we want to find out what f_3^n is when n\geq 3:
f_3^n = \displaystyle \frac{1}{2}\cdot1\cdot\left(\prod_{k=3}^{n-1}\left(\frac{k}{k+1}\right)^a\right)\cdot\left(1-\left(\frac{n}{n+1}\right)^a\right) =...
Yes, it looks like a very tough series to me. I don't know where to go from there. I don't know anything about Markov chains so I don't know how you calculate the f_n^i 's. Did you calculate them by T^{(n)}=T(1)\cdots T(n)?
cause it looks to me like you should have f_3^3 = 1/2 \cdot 1 \cdot...
Aren't consecutive terms at indexes n, n+1 like so:
a_n= \left(\frac{n}{n+1}\right)^{a(n-3)}-\left(\frac{n}{n+1}\right)^{a(n-2)}
a_{n+1}= \left(\frac{n+1}{n+2}\right)^{a(n-2)}-\left(\frac{n+1}{n+2}\right)^{a(n-1)}?
I'm sorry, it doesn't change a thing. But now I noticed that the sum isn't calculated correctly, you should change k back to n-3 so you can see the flaw. The sum is actually
\sum_{n=3}^{\infty}\left(\frac{n}{n+1}\right)^{a(n-3)}-\left(\frac{n}{n+1}\right)^{a(n-2)} so the cancellations are not...
I don't think you haven't copied the problem correctly.
The book i'm looking at also gives a fixed integer k\geq2, and a_n is defined to be the largest integer s.t. \displaystyle \sum_{i=0}^n{\frac{a_i}{k^i}} \leq x. You should be able to prove that \sup\{r_n:n\geq 0\}=x by first proving that...
In these ε-δ-proofs you often find the suitable δ after going through the proof without knowing it, i.e. you find an inequality binding ε and δ, solve for δ and assume this holds from the beginning of your proof. It's hard to explain, but you'll see it in the end.
At this part you should list...
Your indexes k and n are not independent, but it looks like you have "independent" limits and terms in the sums, although I believe you have taken the limit with k = n. If I understood correctly, you should have k = n-3. If you plug this in you should get something else.
Have you encountered the limit of
(1 + 1/n)^n
before? I'm sure it has been mentioned somewhere in your study material. The limit is Euler's number e\ =\ 2.7182....
For the first:
First of all, you need to fiddle around with the expression so that you get something you recognize. Now
n/(n+1) = 1 / ( (n+1)/n ) = 1 / ( (1+1/n) ),
so if you plug this in the original you get
( 1 / ( (1+1/n) ) )^n = 1 /( ( (1+1/n) ) )^n
and now you should recognize the...
Homework Statement
Prove, that a senigruop S is a group if and only if for every x in S there exists a unique x' in S so that xx'x=x.
Homework Equations
The Attempt at a Solution
I am trying to prove that xx'=1S for all x in S. But all I can show is that xx' is in ES={x | x^2=x}...
I think you can assume that the origin is not on the surface of the sphere, so you can differentiate. And then just integrate.
(
dS=r^2\sin{\theta}d\theta d\phi
)
You just do the sub
x=r\cos\theta,\ y=r\sin\theta:
\frac{4x^3}{3(x^4+y^4)^{2/3}}\\
=\frac{4r^3\cos^3\theta}{3(r^4(\cos^4\theta+\sin^4\theta))^{2/3}}\\
=\frac{r^3}{r^{8/3}}\frac{4\cos^3\theta}{3(\cos^4\theta+\sin^4\theta)^{2/3}}\\...