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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(f(x)=e^{2x}\) \(f'(x)=2e^{2x}\) \(f''(x)=4e^{2x}\)

perl
 4 years ago
Best ResponseYou've already chosen the best response.1the taylor series is , does it say about x = 0 ? i assume it is

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am assuming that x=0 too

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am not sure why we assume x=0 in this case but whtvr

perl
 4 years ago
Best ResponseYou've already chosen the best response.1sum f^(n) (a) (xa)^n / n! = f(a) + f'(a) (xa) + f '' (a) (xa)^2/2! + ...

perl
 4 years ago
Best ResponseYou've already chosen the best response.1we dont have to, it just makes it much cleaner

perl
 4 years ago
Best ResponseYou've already chosen the best response.1and by cleaner i mean simpler , since x  0 = x , instead of x  1 or x  2 , etc

perl
 4 years ago
Best ResponseYou've already chosen the best response.1so now lets plug in f(0) + f'(0) *x + f ' ' (0) * x^2/2 + ... , since we want just the first three terms

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So our answer wld be 1+2+2?

perl
 4 years ago
Best ResponseYou've already chosen the best response.1e^(2*0) + 2*e^(2*0) * x + 4*e^(2*0) *x^2/2

perl
 4 years ago
Best ResponseYou've already chosen the best response.1, remember a taylor series is approximating a complicated function (here exponential) with a less complicated function (polynomial)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yaaaa I think I get it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I get it lol just need to know how to enter it into a program. Just gotta fiddle around. THANKS

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0THANKKSSSSS :))))))))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok figured it out. Thanks for ur help

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0U got ur medal even b4 i got my answer :P
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