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

#1**+1 **

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**+1 **

We can do this with something known as the Maclaurin Series

Each term is given by:

[f ^{n }(0) / n!] * x^{n}

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)x^{2}

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

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

etc.....

CPhill Jun 11, 2017