Taylor series approximation of 1/sqrt x - boards.ie
Boards.ie uses cookies. By continuing to browse this site you are agreeing to our use of cookies. Click here to find out more x
Post Reply  
 
Thread Tools Search this Thread
30-11-2011, 19:53   #1
irishdude11
Registered User
 
Join Date: Mar 2011
Location: Dublin
Posts: 301
Taylor series approximation of 1/sqrt x

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?
irishdude11 is offline  
Advertisement
01-12-2011, 18:19   #2
ray giraffe
Registered User
 
ray giraffe's Avatar
 
Join Date: Jun 2007
Location: Dublin
Posts: 277
Quote:
Originally Posted by irishdude11 View Post
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!
ray giraffe is offline  
Post Reply

Quick Reply
Message:
Remove Text Formatting
Bold
Italic
Underline

Insert Image
Wrap [QUOTE] tags around selected text
 
Decrease Size
Increase Size
Please sign up or log in to join the discussion

Thread Tools Search this Thread
Search this Thread:

Advanced Search



Share Tweet