Recent content by ZioX

This is an interesting question. If f^\prime(f(x))=x for all x in R then f^\prime(x)=f^{-1}(x) so that (f(f(x))^\prime=f^\prime (f(x))=f^\prime(f(x))f^\prime(x)=xf^\prime(x)=xf^{-1}(x) Can you take it from here? Edit: made a mistake, will fix it up

The argument is as follows x_n=\sqrt{x_{n-1}+2} so that if the limit does exist then \lim_{n\to\infty} x_n = \lim_{n\to\infty} \sqrt{x_{n-1}+2} but since sqrt(x) is continuous on R>0 we can move the limit into sqrt() on the RHS so that \lim_{n\to\infty} x_n = \sqrt{\lim_{n\to\infty}...

Hint: 1/x is a positive real too.

If this is for a real analysis course you should probably justify that such a limit must satisfy this equation by mentioning the fact that sqrt(x) is continuous. This is the 'similar' argument you are thinking of when you look at the inference "lim (An + Bn)= lim An + lim Bn then lim (Sn + 1) =...

I don't understand your problem. Maybe you are thinking of this in terms of limits? \mbox{lim}_{k\to\infty}\cap_{n=1}^k\left(0,\frac{1}{n}\right)=\mbox{lim}_{k\to\infty}\left(0,\frac{1}{k}\right)=(0,0)=\emptyset Severe abuse of notation.

What tools do you have available? How was ln(n) derived? There is a nifty theorem that says that if a sequence converges then any subsequence will converge to the same limit. This will enable you to look at ln(x)/x^c in R rather than in N.

By definition, if x is in the intersection then x is in (0,1/n) for every n in N. Equivalently, 0<x<1/n. As n-> infty, 1/n -> 0. Using the N-epsilon definition of limits we know that if we let epsilon=x then there is a natural N such that 1/N<x. Therefore, x is not in A_N. Contradiction...

It is obviously true. There are a lot of ways to prove this. What kind of class is this for? Are you expected to be rigorous in your solutions? N-epsilon proofs and asymptotic behaviour are very different arguments in terms of the development required in an analysis setting.

http://statistics.laerd.com/statistical-guides/normal-distribution-calculations.php

I am puzzled why you have the product of 2x2 matrices yielding a 2x3 matrix? Perhaps this is the source of your problem? If not, then notice that the (dot) product of the 2nd column of the 2nd matrix and the 1st row of the 1st matrix will be the (1,2)-entry in the product of the two matrices...

(a+b)^2=a^2+2ab+b^2

No, it's not. |A_m-A_n|\le |A_m-A_{m-1}|+\dots+|A_{n+1}-A_n|<r^{m-1}+r^{m-2}+\dots+r^{n-1}<(m-n)r^{n-1} Where the last inequality follows from the fact that r^m<r^n for m>n and noting there are m-n terms summed. Read some more Cauchy proofs. Pay particular attention to how m and n are...

You do not know n*r^n goes to zero. If you did you would be done. |Am - Am-1| + |Am-1 - Am-2| +...+ |An-1 -An|< n*r^n This is not true. Also, since r is between 0 and 1 you know there is a natural number, say, a, such that 1/a <= r. You know this because 1/n -> 0.

You're just taking limits in R^n or whatever. What is the definition for f being continuous at v? Is f continuous at v? So does that mean you can slip in the limit?

Same sequence, just at different n. Another way of looking at whether sequences converge is that after some point they become arbitrarily close together. That is, if for any epsilon there is an N such that for n,m > N |x_n-x_m|< epsilon. Such a sequence is called Cauchy sequence.