How to solve? Find the Taylor series expansion of f(x)=(x-1)e^x about x=1.

Question

anonymous · Answer

What am I supposed to do first? Thanks! :)

anonymous · Answer

(x-1) is already polynomial so leave it as it is

anonymous · Answer

Can you draw it please?

anonymous · Answer

draw what?

anonymous · Answer

in your text book , they might give you taylor expansion for e^x

anonymous · Answer

steps to solve this.

anonymous · Answer

We don't have books. :)) :(

anonymous · Answer

THANKS @imranmeah91 :)

anonymous · Answer

@imranmeah91 what does "about x=1" mean? Were you able to use that?

anonymous · Answer

it mean centered at x=1 =a

so we have to use expanded form
f(a)+f'(a)/1! (x-a)+ f''(a)/2! (x-a)^2

plug in 1

f(x)=e^x
f(1)= e^1

e+ e (x-1)+ e(x-1)^2/2!+e(x-1)^2/3!

anonymous · Answer

so e^x centered at x=1 is

$$e^x=\sum _{n=0}^{\infty } \frac{e(x-1)^n}{n!}$$

anonymous · Answer

multiply that by (x-1)

$$e^x=\sum _{n=0}^{\infty } \frac{e(x-1)^{n+1}}{n!}$$