Question

Find MacLurin Series for f(x)..The process to solve is attached. I do not understand how you get from step, one to step two. Step three and four I understand.

Answer 1

Hmm I would have gone about this in a different way...

In any case, I suppose you understand how
\[\large e^{-t^2-1}=\sum_{n=0}^\infty\frac{(-t^2)^n}{n!}-1=\sum_{n=0}^\infty\frac{(-1)^nt^{2n}}{n!}-1\]
right?

Answer 2

What they do is make the first term of the series disappear:
\[\begin{align*}
\sum_{n=0}^\infty\frac{(-1)^nt^{2n}}{n!}&=\frac{(-1)t^0}{0!}+\sum_{n=1}^\infty\frac{(-1)^nt^{2n}}{n!}&\text{extract the first term}\\\\
&=1+\sum_{n=1}^\infty\frac{(-1)^nt^{2n}}{n!}\\\\
\sum_{n=0}^\infty\frac{(-1)^nt^{2n}}{n!}-1&=\left(1+\sum_{n=1}^\infty\frac{(-1)^nt^{2n}}{n!}\right)-1&\text{subtract 1 from both sides}\\\\
e^{-t^2}-1&=\sum_{n=1}^\infty\frac{(-1)^nt^{2n}}{n!}\\\\
&=\sum_{n=0}^\infty\frac{(-1)^{n+1}t^{2(n+1)}}{(n+1)!}&\text{shift the index}\\&&\text{back down to 0}\\\\
&=\sum_{n=0}^\infty\frac{(-1)^{n+1}t^{2n+2}}{(n+1)!}
\end{align*}\]
And from here they integrate over \([0,x]\).

Answer 3

Thank you

Answer 4

yw

Answer 5

Which way would you have gone about doing it?

Answer 6

When you say the first term what they are doing is getting rid of the 1 in th equation correct?