Let's forget the factor of 5, and just expand
f(x) = 1/(x^2+4)
f' = -2x/(x^2+4)^2
f" = 8(3x^2-4)/(x^2+4)^3
f(3) = -96x(x^2-4)/(x^2+4)^4
and then just evaluate the Taylor Series at x=0
f(x) = 1/4 - x^2/16 + x^4/64 - ...
Expand 4/(x^2+4) as a power series centered at 0.
3 answers
ohh that makes sense...but is there a way of solving it without using Taylor series?
what else is a power series?
I guess you could use the binomial theorem, and expand
(4+x^2)^(-1) = 4^-1 + (-1)/1! 4^-2 x^2 + (-1)(-2)/2! 4^-3 x^4 + ...
= 1/4 - x^2/16 + x^4/64 - ...
∞
= ∑ (-1)^k x^(2k)/4^(k+1)
k=0
I guess you could use the binomial theorem, and expand
(4+x^2)^(-1) = 4^-1 + (-1)/1! 4^-2 x^2 + (-1)(-2)/2! 4^-3 x^4 + ...
= 1/4 - x^2/16 + x^4/64 - ...
∞
= ∑ (-1)^k x^(2k)/4^(k+1)
k=0
1 answer