Question

I have an example, and then the question.

Answer 1

\[e^x=\sum_{n=0}^{\infty}\frac{1}{n!}x^n=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\frac{1}{24}x^4+~...\]

Answer 2

at first, is what I posted correct ?

Answer 3

then, when I differentiate, I get
\[\left(e^x\right)'=\left(\sum_{n=0}^{\infty}\frac{1}{n!}x^n\right)'=\left( 1 \right)'+\left( x \right)'+\left( \frac{1}{2}x^2 \right)'+\left( \frac{1}{6}x^3 \right)' +\left(\frac{1}{24}x^4 \right)' +~...\]

Answer 4

\[\left(e^x\right)'=\sum_{n=0}^{\infty}\frac{1}{n!}x^n=0+ 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 +\frac{1}{24}x^4 +~...\]

Answer 5

So for every power series, I would need to differentiate the terms to see if the pattern changes or not, and then to write a new pattern/summation.

Answer 6

?

Answer 7

I think so.

Answer 8

But most the time, if we have to expand the summation, we use it. If not, we take (e^x)'= e^x. No need to expand.

Answer 9

yes, derivative of e^x is just e^x..... I was kind of trying to grasp some concept that in my class we didn't go through.

Answer 10

tnx for your verification.

Answer 11

ok