Question

Find the Maclaurin expansion for the interval from 0 to x (e^t - 1)/t dt

Answer 1

\[\int\limits_{0}^{x} (e^t - 1)/t dt\]

Answer 2

first write the expansion for \(e^t\)
then subtract 1
then divide by t
then integrate term by term

Answer 3

you got that?

Answer 4

\[ e^{t} = \sum_{n=0}^{\infty} t^{n}/n!\]

Answer 5

I use this to expand e^t?

Answer 6

\[e^{t}-1 = \sum_{n=0}^{\infty} t^{n}/n! -1\]
so
\[e^{t}-1/t = \(1/t)sum_{n=0}^{\infty} t^{n}/n! -1\]

Answer 7

Then I integrate term by term ?

Answer 8

\[e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+...=\sum\frac{t^n}{n!}\]

Answer 9

\[e^t-1=t+\frac{t^2}{2}+\frac{t^3}{3!}+..=\sum_{n=1}^{\infty}\frac{t^n}{n!}\]

Answer 10

divide by \(t\) and get
\[\frac{e^t-1}{t}=1+\frac{t}{2}+\frac{t^2}{3!}+...=\sum_{n=0}^{\infty}\frac{t^n}{(n+1)!}\]

Answer 11

then integrate term by term

Answer 12

\[t+\frac{t^2}{4}+\frac{t^3}{3\times 3!}+...\] make up a nice formula and you are done

Answer 13

^ this guy = #winning

Answer 14

I have to make up a formula from this or I'm done at this step?

Answer 15

best make up a formula
of course i don't mean "make up" i mean find it

Answer 16

How do I do that :/