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# Series expansion

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Expand the following for at least 5 terms: (1 + x)^(1/5). Please explain the method used for the expansion. Thank you for any help.

Guest Jun 11, 2017
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#1
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Taylor series expansion is used for such series. In general, we have: (1 + x)^(n/m) = [1 +(n/m)]x - [n(m - n)/(2!m^2)]x^2 + [n(m -n)(2m - n)/(3!m^3)]x^3 - .......etc.
So, ( 1+ x)^1/5 = 1 + x/5 - 2x^2/25 + 6x^3/125 - 21x^4/625 + 399x^5/15,625 - 1,596x^6/78,125 + .......etc.

Guest Jun 11, 2017
#2
+75344
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We can do this with something known as the Maclaurin Series

Each term is given by:

[f (0) / n!]  * xn

Where  f n   is the nth derivative  of  (1 + x)1/5       with n begiining at 0

1st term (n = 0) =    [ (1 + 0)(1/5)/ 0! ] * x(0) =   1

2nd term (n = 1) =  [ (1/5)(1 + 0)(-4/5)/ 1! ] * x(1)   =  x / 5

3rd term  (n = 2)  =  [(-4/25)(1 + 0)(-9/5)/ 2! ] * x(2) =   ( -2 / 25)x2

4th term (n = 3)  =  [ (36/125)(1 + 0) (-14/5) / 3! ] * x(3)  =  ( 6 / 125)x3

5th term ( n = 4)  = [ (-504/3125) (1 + 0)(-19/5) / 4!] * x(4) =  ( -21 / 3125)x4

etc.....

CPhill  Jun 11, 2017
edited by CPhill  Jun 11, 2017

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