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

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

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\(f(x)=e^{2x}\) \(f'(x)=2e^{2x}\) \(f''(x)=4e^{2x}\)
the taylor series is , does it say about x = 0 ? i assume it is
I am assuming that x=0 too

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I am not sure why we assume x=0 in this case but whtvr
sum f^(n) (a) (x-a)^n / n! = f(a) + f'(a) (x-a) + f '' (a) (x-a)^2/2! + ...
we dont have to, it just makes it much cleaner
and by cleaner i mean simpler , since x - 0 = x , instead of x - 1 or x - 2 , etc
so now lets plug in f(0) + f'(0) *x + f ' ' (0) * x^2/2 + ... , since we want just the first three terms
So our answer wld be 1+2+2?
e^(2*0) + 2*e^(2*0) * x + 4*e^(2*0) *x^2/2
, remember a taylor series is approximating a complicated function (here exponential) with a less complicated function (polynomial)
1 , 2x , 4x^2/2
ok?
Yaaaa I think I get it
I get it lol just need to know how to enter it into a program. Just gotta fiddle around. THANKS
THANKKSSSSS :))))))))
Ok figured it out. Thanks for ur help
:D
where is my medal?
thanks
U got ur medal even b4 i got my answer :P

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