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Please expand the following at x = 0:  sqrt(1 + x)^ -1. Thank you for help.

Guest Jul 20, 2017
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 #1
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 (1 + x)^ -1/2  at x = 0

 

Using the Maclaurin Series, we have

 

f(0) + f ' (0)x  + f ''(0) x^2 / 2!  +   f  '''(0) x^3 / 3! + ......  + f(n)(0) x^n / n!

 

f(0)  =  1

 

f ' (x) =  ( -1/2) (1 + x)^( -3/2)   and   f ' (0)  =  (-1.2)  ( 1 + 0)^-(3/2)=   (-1/2)

 

f '' (x)  =  (3/4) (1 + x) ^(-5/2)  and   f '' (0)  = (3/4)  (1 + 0)^(-5/2)  =   (3/4)

 

f ''' (x)  =   (-15 / 8 ) ( 1 + x)^(-7/2)   and  f''' (0)  = (-15/8) (1 + 0)^(-7/2)  = ( - 15 / 8)

 

 

So we have the first four terms

 

1 - (1/2)x + (3/4)x^2/2   - (15/8)x^3 / 6 + ..... +  f(n)(0) x^n / n!  =

 

1 - x/2  + 3x^2/8  - 15x^3/48 + .....+  f(n)(0) x^n / n!  =

 

1 - x/2  + 3x^2/8  - 5x^3/16 + ...... +  f(n)(0) x^n / n!

 

 

 

cool cool cool

CPhill  Jul 20, 2017
 #2
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sqrt(1 + x)^ -1 =1 - 1/2x + [1.3/2.4]x^2 - [1.3.5/2.4.6]x^3 + [1.3.5.7/2.4.6.8]x^4.........etc.

Guest Jul 20, 2017

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