# Maclaurin Series calculation

1. Nov 21, 2007

### chrono210

For the function f(x) = (10x^2) e^(-2x), I calculated the first term of the Maclaurin series to be 0.

However, for the second term, I also calculated it to be 0, but apparently this is wrong. Shouldn't the second term be f'(x) = 10x^2 * -2e^(-2x) + e^(-2x) * 20x? Or what am I doing incorrectly here?

2. Nov 21, 2007

### HallsofIvy

Staff Emeritus
No, that is not wrong, apparently or not! If $f(x)= 10x^2e^{-2x}$ then, yes, f(0)= 0. $f'(x)= 20xe^{-2x}- 20x^2e^{-x}= 20(x-x^2)e^{-x}$ which is also 0 at x= 0. You don't get a non-zero term until the second derivative: $f"(x)= 20(1- 2x)e^{-x}- 40(x-x^2)e^{-x}$ which is 20 when x= 0. The first non-zero term of the series is (20/2)x2.

Actually, you don't need to do any differentiations for this. Presumably you know that the MacLaurin series for ex is $\Sum x^n/n!$ so the MacLaurin series for e-2x is $\Sum (-2x)^n/n!= \Sum (-2)^nx^n/n!$ and so the MacLaurin series for $10x^2e^{-2x}$ is $\Sum 10(-2)^n x^{n+2}/n!$.

3. Nov 21, 2007

### EnumaElish

Is x the point at which the series is evaluated? Or is it the argument of the series, e.g. the x in Taylor series f(a) + f'(a)(x - a) + ...?

4. Nov 21, 2007

### chrono210

Interesting. Granted, this is an online-based homework program, so I wouldn't be surprised if it had an error in it (although it's been fine so far this semester). When I put in 0 for the first term, it indicated "correct", but when I put in 0 for the second term, it indicated "incorrect". Thanks for the help!
Not sure exactly what you mean by this, but the question just asks to find the first five terms of the Maclaurin series for the function.

EDIT: I figured out the problem. Apparently if a term is 0, it wants you to just skip to the next one...but it still wants the first term at 0? *wishes this thing would make sense*

Last edited: Nov 21, 2007