Recent content by aodesky

Thanks for the replies, everyone. To Mute: This should explain where I got the recursive formula. $$\sigma_z(k) = \sum_{n=1}^\infty n^k z^n = z + 2^k z^2 + 3^k z^3 + ... \\ = (z + z^2 + z^3 + ...) + ((2^k - 1)z^2 + (3^k -1)z^3 + ... ) \\ = \sum_{n=1}^\infty z^n + z\left[ \sum_{n=1}^\infty...

Consider the infinite sum: \sum_{n=1}^\infty \frac{n^2}{2^n} For the impatient of you, the answer is here. Anyways, I'm trying to generalize this result, so let me state a definition: \sigma_\alpha(k) = \sum_{n=1}^\infty n^k \alpha ^ n This sum converges so long as the magnitude...

Thanks for the responses. To micromass: I don't quite see how the set's being nonempty necessarily implies that its infimum is finite. And to foxjwill: the set's being closed has no pertinence (at least I don't think it does) to the part of the question I asked here, but there are two other...

Suppose (X,d) is a metric space and A, a subset of X, is closed and nonempty. For x in X, define d(x,A) = infa in A{d(x,a)} Show that d(x,A) < infinity. I really don't have much of an idea on how to show it must be finite. An obvious thought comes to mind, namely that a metric is...

Hello, The qualities of the medium in which a wave is traveling basically determines the qualities of the wave. When the air is warmer, the molecules are all hitting each other more often then they would on a colder day, and hence they are not as easily shifted as when colder. This is related...

The levi-civitia symbol sets repeated indices to zero, so it would be the opposite of what you say, in the sense that when k or j do equal i, the element goes to zero. I think you might have mistaken the levi-civitia symbol for the kronecker delta symbol... That goes for what you're doing...