Registered User
#1

Im trying to find the taylor series approx of 1/sqrt x, centered at a = 9

So here are my derivatives

f'(x) = -1/2 . x ^-3/2
f''(x) = 3/4 . x ^-5/2
f'''(x) = -15/8 . x ^-7/2
f''''(x) = 105/16 . x ^-9/2

f'(9) = -1/2 . 1/(3^3)
f''(9) = 3/4 . 1/(3^5)
f'''(9) = -15/8 . 1/(3^7)
f''''(9) = 105/16 . 1/(3^9)

So in trying to find a sigma expression for the taylor series I have the following -

Σ (-1)^n . ###/2^n . 1/3^(2n+1) . (x-9)^n

The ### is the part I cant get - As you can see it is coming out like this -
n = 1 ===> 1
n = 2 ===> 3
n = 3 ===> 15
n = 4 ===> 105
n = 5 ===> 945

How can I put this in terms of n so I can complete my taylor series representation?

Registered User
#2

irishdude11 said:

n = 1 ===> 1
n = 2 ===> 3
n = 3 ===> 15
n = 4 ===> 105
n = 5 ===> 945

How can I put this in terms of n so I can complete my taylor series representation?

1 -> 1
2 -> 1x3
3 -> 1x3x5
4 -> 1x3x5x7
5 -> 1x3x5x7x9
6 -> 1x3x5x7x9x11
...
...
n -> 1x3x5x ... x(?)

Hope that helps!