Recent content by BrownianMan

I though of another way. Induction. Base case is easy. To prove for n=k+1, we have k+1 <= (1+ sqrt(2/k))^(k+1) We have (1+ sqrt(2/k))^k*(1+ sqrt(2/k)) =(1+ sqrt(2/k))^(k+1) Then I can say that (1+ sqrt(2/k))^k <=(1+ sqrt(2/k))^k*(1+ sqrt(2/k)) and since k+1 <= (1+ sqrt(2/k))^k by...

I have a question based on this. Is it true that n^(1/n) - 1 <= sqrt(2/(n-1))? How do you show this? And what is the mini induction you have to prove?

Aren't B(u) and B(v) independent? If so, then the variance of their sum should be the sum of their variance.

B(t) is a standard Brownian Motion. u and v are both => 0. What is the distribution of B(u) + B(v)? The mean is 0. For the variance I get Var(B(u)+B(v)) = u+v. Is this right?

For watches produced by a certain manufacturer, lifetimes follow a single-parameter Pareto distribution with alpha > 1 and theta = 4. The expected lifetime of a watch is 8 years. a) Calculate the probability that the lifetime of a watch is at least 6 years. b) For the same distribution as...

Anyone?

So I found a formula for the number of ways of coloring a shape with 20 triangular faces, 30 edges, and 12 vertices: (1/60)*(k^20+15*k^10+20*k^8+24*k^4). Now I need to find the # of ways of coloring the faces with exactly 5 colors each with each color used exactly 4 times. I know how to find...

I have that B(x)=e^{e^{x}-1} is the generating function for the number of set partitions. Also the Stirling numbers of the second kind are defined by S(0,0)=1, S(n,0)=S(0,n)=0 for n=>1 and S(n,k)=S(n-1, k-1) + kS(n-1, k). Show that e^{u(e^{x}-1)}=1+\sum_{n\geq...

Ok, I wrote a function to do that and then replaced the slow loop with this: vec=sumPowersOf2(N); %vector of elements of powers of 2 that sum to N AN=eye(N_space-1); for i=1:numel(vec) AN=AN*A^vec(i); end But this is actually slower.

Is there a function that allows you to do that in Matlab?

A depends on the inputs of the function, but inside the loop it does not change. For a given set of inputs, it is constant. As far as taking advantage of properties of exponents, N is also an input. N=200 is what I am suppose to use to show that it gives the correct result for that specific N...

Yes, A is constant. I tried that. But the problem is that N is 200. So it actually takes more time to calculate A^N. I was wondering if there are more efficient ways to compute A^N. I tried using [v d]=eigs(A) and v*(d.^N)*v' but it said it wasn't able to find eigenvalues with sufficient accuracy.

A is a sparse matrix. B is a vector. I have the following code: for j=1:N B=A*B; end; This part of the code is inside a function which gets called about 160000 times. I ran the Profiler and this part is the bottleneck. How can I make it more efficient?

Yes, I solved it. Thanks again!

You're right. My mistake. Thanks.