A weird convolution (recurrence relation)

[rsq_a]

The answer doesn't seem obvious to me:

If I set up

B_0 = 1 and B_n = n^2

Then let

A_n = \sum_{m=0}^n B_m B_{n-m} = 2B_0 B_{n} + 2B_1 B_{n-1} + \ldots

Then I almost expected A_n to grow like n^2. Instead, I'm getting (numerically) that A_n \sim O(n^5)! Why is that?

In general, it seems that if B_n \sim O(n^a) then A_n \sim O(n^{2a+1})

 

[zetafunction]

for big m replace the sum by an integral

\sum_{m=0}^{n}B_{m}B_{n-m}\sym \int_{0}^{x}dtB(t)B(x-t)

the integral \int_{0}^{x}dt(x-t)^{2}t^{2} using the Binomial theorem includes the integral of terms t^{4} which is itself proportinal to x^{5} this is the solution to your paradox

 

[rsq_a]

for big m replace the sum by an integral

\sum_{m=0}^{n}B_{m}B_{n-m}\sym \int_{0}^{x}dtB(t)B(x-t)

the integral \int_{0}^{x}dt(x-t)^{2}t^{2} using the Binomial theorem includes the integral of terms t^{4} which is itself proportinal to x^{5} this is the solution to your paradox

Hmm...I sort of follow it. For large n, we can compare the growth of the sum to the growth of the integral, which is indeed O(n^5).

It's difficult to see how this arises from the direct sum itself, though.