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## Main Question or Discussion Point

The answer doesn't seem obvious to me:

If I set up

[tex]B_0 = 1[/tex] and [tex]B_n = n^2[/tex]

Then let

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

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

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

If I set up

[tex]B_0 = 1[/tex] and [tex]B_n = n^2[/tex]

Then let

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

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

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

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