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.
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.
+128474
We can do this with something known as the Maclaurin Series
Each term is given by:
[f n (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.....
