Your series would just be . I don't know why you're subtracting it from .
Hi,
The question asks to find a Maclaurin series for the function
f(x) = (x-sinx)/x^3 when x does not = 0
and
f(x) = 1/6 when x = 0
Since sinx = x - x^3/3! + x^5/5! - ...
x-sinx = x^3/3! - x^5/5! + x^7/7! - ...
and (x-sinx)/x^3 = 1/3! - x^2/5! + x^4/7! - ... = 1/6 - x^2/5! + x^4/7! - ...
I noticed that I can make a series out of the second term onward, and in doing so would make f(x) = 1/6 when x = 0. This gives me 1/6 - series(n=0,infinity) ((-1)^(x+1) * x^(2n+2))/(2n+3)!
Would that be my answer? Just curious, because I've never done a problem like this before.
Oh thanks, I was thinking 0^0 = 0 or undefined; I couldn't really find a definite answer online, does 0^0 = 1? That's why I was confused.
0^0 - Wolfram|Alpha
Would that be my answerReplica Handbags,Replica Wallets For Sale
? Just curious, because I've never www.louisvuittonend.com
done a problem like this before.