1. ## Taylor Series Question

Got a taylor series question here, just want to see if I'm on the right track

f(x) = x^(-1/2) at a = 9

Took the derivative a bunch..

f'(x) = (-1/2)x^-(3/2)
f''(x) = (-1/2)(-3/2)x^-(5/2)
f'''(x) = (-1/2)(-3/2)(-5/2)x^-(7/2)
f4(f) = (-1/2)(-3/2)(-5/2)(-7/2)x^-(9/2)

so the taylor series at a = 9 would be..

[(-1)^n][(x-9)^(-1/2 - n)][2n-1]/[(n!)(2^n)]

or am I way off?

2. Originally Posted by coolguy99
Got a taylor series question here, just want to see if I'm on the right track

f(x) = x^(-1/2) at a = 9

Took the derivative a bunch..

f'(x) = (-1/2)x^-(3/2)
f''(x) = (-1/2)(-3/2)x^-(5/2)
f'''(x) = (-1/2)(-3/2)(-5/2)x^-(7/2)
f4(f) = (-1/2)(-3/2)(-5/2)(-7/2)x^-(9/2)

so the taylor series at a = 9 would be..

[(-1)^n][(x-9)^(-1/2 - n)][2n-1]/[(n!)(2^n)]

or am I way off?
Do you mean that is a coefficient? What you give is not a series and, since there is no "x" certainly not a Taylor's series.

3. The question asked to find a taylor series for the equation f(x) = x^(-1/2) at a = 9... that's all the info it said, and I quoted it directly hahah