MHB Maclaurin Series for e^x Example

[ISITIEIW]

Hey Guys!
I'm stick on this question,
I know that the summation of n=0 to infinity for x^n/n! equals e^x

In the question it wants me to come up with a corresponding summation for the function x^2(e^(3x^2) - 1) … I don't know how to manipulate it to get the -1. I know i can substitute x for 3x^2 which would make my summation (3x^(2n))/n!. But I'm stuck after this point.

Thanks,
ISITIEIW

 

[Evgeny.Makarov]

In the question it wants me to come up with a corresponding summation for the function x^2(e^(3x^2) - 1) … I don't know how to manipulate it to get the -1. I know i can substitute x for 3x^2 which would make my summation (3x^(2n))/n!.

It should be $\dfrac{(3x^2)^n}{n!}=\dfrac{3^nx^{2n}}{n!}$. But the first term is still 1...

 

[ISITIEIW]

Right, so now I have e^(3x^2) = (3^n)(x^2n)/n! … what do i do to incorporate this negative 1 in the expression? x^2(e^3x^2 - 1)?

 

[Evgeny.Makarov]

I am saying that the first term of the series is still 1:
\[
e^{3x^2}=1+3x^2+\frac{(3x^2)^2}{2}+\dots
\]
So, in $e^{3x^2}-1$ this 1 cancels.

 

[ISITIEIW]

OH lol,

So then the corresponding series would be (x^2(3^n)(x^2n))/n!
which equals ((x^2n+2)3^n)/n! ?

 

[Evgeny.Makarov]

So then the corresponding series would be (x^2(3^n)(x^2n))/n!
which equals ((x^2n+2)3^n)/n! ?

First, be careful with operation priorities: x^2n+2 is likely to be parsed as $x^{2n}+2$ or even $x^2n+2$. In plain text, $x^{2n+2}$ can be written as x^(2n+2). Second, the series is
\[
\sum_{n=1}^\infty\frac{3^nx^{2n+2}}{n!}
\]
The fact that is starts with $n=1$ and not $n=0$ is due to subtracting 1.