Recent content by Petter

You are a freshman. You shouldn't expect to become an expert in a year. You seem to be doing fine.

I'll reply because this thread is a top Google hit: function phi = eulerphi(n) phi = n; for i = 1:length(n) p = unique( factor(n(i)) ); phi(i) = phi(i) * prod(1 - 1./p); end end The for-loop makes the function work with vectors, which is useful for MATLAB commands like...

Here is my (arguably easier solution): Let the breaking poing be x*. The list has n terms. Let the expected number of steps for a certain position x be E(x). For x\leq x^*, E(x)=x-1. For every x>x*, E(x) is constant (=e) (new position independent of old). So for x>x*, E(x) = e =...

Here is my (arguably easier solution): Let the breaking poing be x*. The list has n terms. Let the expected number of steps for a certain position x be E(x). For x\leq x*, E(x)=x-1. For every x>x*, E(x) is constant (=e) (new position independent of old). So for x>x*, E(x) = e =...

That wasn't the Euler number, was it?

He previously proved that 0 < -1, so that's correct.

Assume 1 < 0. Then by (2) 0 < -1. This gives us, using (3), that -1 < 0. Using (2) again we get 0 < 1, a contradiction. Since 0!=1 we are, using (4), done. #