1. ## McLaurin series

Ok, so I've got the following problem:

Find an expression for the following McLaurin-series:

Σ (((n^2) - n +1) / n!) * x^n

(from n = 0 to infinity).

e^x = Σ (x^n) / n!

I then differentiate and get:

e^x = Σ (n*(x^(n-1))) / n!

I differentiate one more time and get:

e^x = Σ (((n^2) - n) / n!) * x^(n-2)

I can't get any further than this though. So what I need now is to get the (+1) in the numerator of the n-term, and x^(n-2) has to be reversed back to x^n. Can I multiply e^x with x^2 to get the x taken care of? And how do I get the (+1) term?

Any tips would be greatly appreciated!

2. Originally Posted by krje1980
Ok, so I've got the following problem:

Find an expression for the following McLaurin-series:

Σ (((n^2) - n +1) / n!) * x^n

(from n = 0 to infinity).

e^x = Σ (x^n) / n!

I then differentiate and get:

e^x = Σ (n*(x^(n-1))) / n!

I differentiate one more time and get:

e^x = Σ (((n^2) - n) / n!) * x^(n-2)

I can't get any further than this though. So what I need now is to get the (+1) in the numerator of the n-term, and x^(n-2) has to be reversed back to x^n. Can I multiply e^x with x^2 to get the x taken care of? And how do I get the (+1) term?

Any tips would be greatly appreciated!

So after differentiating twice you got $e^x=\sum\frac{n^2-n}{n!}x^{n-2} = \sum\frac{x^{n-2}}{(n-2)!}$ .

But now pay attention that $\sum\frac{n^2-n+1}{n!}x^n=\sum\frac{n^2-n}{n!}x^n+\sum\frac{1}{n!}x^n$ , and you can split the series this way because everything converges absolutely here for all x.

Tonio

3. Thanks a lot . So the correct answer will then be:

(x^2)(e^x) + (e^x)

(or (e^x)((x^2) + 1))

?