ray giraffe
Registered User

#2

1 -> 1

2 -> 1x3

3 -> 1x3x5

4 -> 1x3x5x7

5 -> 1x3x5x7x9

6 -> 1x3x5x7x9x11

...

...

n -> 1x3x5x ... x(?)

Hope that helps!

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)^nThe ### 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?