Question
Taylor series:
f(x)=(1/16 +x)^(-1/4)
=2-8x+80x^2 -960x^3.....
and through differentiation g(x) = (1/16 +x)^(-5/4)
=32-640x+11520x^2
but i then need to integrate about 0, h(x)=(1/16 +x)^(3/4)
not really sure where to start out the integration one
f(x)=(1/16 +x)^(-1/4)
=2-8x+80x^2 -960x^3.....
and through differentiation g(x) = (1/16 +x)^(-5/4)
=32-640x+11520x^2
but i then need to integrate about 0, h(x)=(1/16 +x)^(3/4)
not really sure where to start out the integration one
Answers
Reiny
Your Taylor expansion of f(x) is correct
I disagree with your derivative, it should be
(-1/4)((1/16 + x)^(-5/4)
why not just integrate
ʃ(1/16 +x)^(-1/4) dx
to get
(4/3)(1/16 + x)^(3/4) + C
I disagree with your derivative, it should be
(-1/4)((1/16 + x)^(-5/4)
why not just integrate
ʃ(1/16 +x)^(-1/4) dx
to get
(4/3)(1/16 + x)^(3/4) + C