Question

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

Answer 1

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

Answer 2

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

Answer 3

Can you draw it please?

Answer 4

draw what?

Answer 5

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

Answer 6

steps to solve this.

Answer 7

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

Answer 8

THANKS @imranmeah91 :)

Answer 9

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

Answer 10

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!

Answer 11

so e^x centered at x=1 is

\[e^x=\sum _{n=0}^{\infty } \frac{e(x-1)^n}{n!}\]

Answer 12

multiply that by (x-1)

\[e^x=\sum _{n=0}^{\infty } \frac{e(x-1)^{n+1}}{n!}\]