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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

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 our answer wld be 1+2+2?

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

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