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

1. swissgirl

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

2. perl

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

3. swissgirl

I am assuming that x=0 too

4. swissgirl

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

5. perl

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

6. perl

we dont have to, it just makes it much cleaner

7. perl

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

8. perl

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

9. swissgirl

So our answer wld be 1+2+2?

10. perl

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

11. perl

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

12. perl

1 , 2x , 4x^2/2

13. perl

ok?

14. swissgirl

Yaaaa I think I get it

15. swissgirl

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

16. swissgirl

THANKKSSSSS :))))))))

17. swissgirl

Ok figured it out. Thanks for ur help

18. perl

:D

19. perl

where is my medal?

20. perl

thanks

21. swissgirl

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