Why would a maclauren series for f(x) = x^2*e^x^2 be 0 at f^(7)(0) and f^(9)(0) but 8! /3! at f^(8)(0)? i know how to find the series (x^2e^2^2 = x^2 + x^4 + x^6/2! + x^8/3! + you basically take the maclauren series for e^x, modify it for e^x^2, and then multiply that by x^2). i'm just confused why it's 8! /3! at f^(8)(0). shouldn't it be 0 as well since f^(8)(0) = x^18/8! and if x = 0 then 0^18/8! = 0? what am i missing?

By definition, the Maclaurin series for is



It sounds to me like you're asking why is zero for some terms, and non-zero for others. Take the derivative to see why this is the case.






and so on. There's a pattern here. Every odd-order derivative involves an odd-degree polynomial multiplied by with no constant term. So when is odd. Meanwhile, every even-order derivative consists of an even-degree polynomial multiplied by and *does* have a constant term, which is why when is even.

This would explain the absence of odd powers of in the series expansion for .

But you can see why this must happen just by referring to the manipulated series for :


(both even and odd powers)


(even powers only)


(again, even powers only)

So when or , the corresponding term in the series is 0, but this is not the case for when or any other even number.