Question

What are the first 3 terms of the Taylor expansion for the function \(f(x)=e^{2x}\)

Answer 1

\(f(x)=e^{2x}\)
\(f'(x)=2e^{2x}\)
\(f''(x)=4e^{2x}\)

Answer 2

the taylor series is , does it say about x = 0 ? i assume it is

Answer 3

I am assuming that x=0 too

Answer 4

I am not sure why we assume x=0 in this case but whtvr

Answer 5

sum f^(n) (a) (x-a)^n / n!
= f(a) + f'(a) (x-a) + f '' (a) (x-a)^2/2! + ...

Answer 6

we dont have to, it just makes it much cleaner

Answer 7

and by cleaner i mean simpler , since x - 0 = x , instead of x - 1 or x - 2 , etc

Answer 8

so now lets plug in
f(0) + f'(0) *x + f ' ' (0) * x^2/2 + ... , since we want just the first three terms

Answer 9

So our answer wld be 1+2+2?

Answer 10

e^(2*0) + 2*e^(2*0) * x + 4*e^(2*0) *x^2/2

Answer 11

, remember a taylor series is approximating a complicated function (here exponential) with a less complicated function (polynomial)

Answer 12

1 , 2x , 4x^2/2

Answer 13

ok?

Answer 14

Yaaaa I think I get it

Answer 15

I get it lol just need to know how to enter it into a program. Just gotta fiddle around. THANKS

Answer 16

THANKKSSSSS :))))))))

Answer 17

Ok figured it out. Thanks for ur help

Answer 18

:D

Answer 19

where is my medal?

Answer 20

thanks

Answer 21

U got ur medal even b4 i got my answer :P