Find the Power Series in terms of sigma:

Question

Find the Power Series of f(x) = x^5/(4-x^3) in terms of sigma.

Answer 1

x⁵/(4-x³) = (x⁵/4)[1/(1-x³/4)] = (x⁵/4)∑_n=0...∞(x³/4)ⁿ = ∑_n=0...∞x³ⁿ⁺⁵/4ⁿ⁺¹

Answer 2

The first thing you should always do with an improper fraction is to make it proper:

 

f(x)= x⁵/(4-x³)= -x² + 4x²/(4-x³) = -x² + x²/(1-(x³/4))

 

so really we want the power series of 4x²/(4-x³), presumably about x=0 (or else they would have specified a point of expansion). Since the function is undefined at x=4^1/3, the radius of convergence can be no bigger than that. We could use Taylor's formula to come up with the series expansion, but it would take quite a few terms to see a pattern emerging that would allow us to deduce the n-th term expression (which is what we want when they say "in terms of Σ").

 

A quicker way is to reduce the series to one we already know. In this case,

4x²/(4-x³) = x²/(1-(x³/4)) = (-4/3) d/dx (ln(1-(x³/4))

 

The Taylor series of ln(1+u) about u=0 is well known:

ln(1+u) = u - u²/2 + u³/3 -+....+∑_n=1^∞ (-1)ⁿ⁺¹uⁿ/n, which converges for 0<u≤1.

 

So set u=-x³/4 and get

ln (1-(x³/4)) = ∑ (-1)ⁿ⁺¹(-x³/4)ⁿ/n = -∑ x³ⁿ/(4ⁿn)

 

Take the derivative term by term:

d/dx (ln(1-(x³/4))= -3 ∑ x^3n-1/4ⁿ

 

Therefore,

4x²/(4-x³) =4 ∑_n=1^∞ x^3n-1/4ⁿ = x² +∑_n=2^∞ x^3n-1/4ⁿ

where in the last step I separated out the n=2 term, the reason being that the x² will exactly cancel with the -x² from our original function:

f(x) = ∑_n=2^∞ x^3n-1/4ⁿ

 

If you prefer the summation to start with n=1, you can up-shift the indices:

f(x) = ∑_n=1^∞ x³ⁿ⁺²/4ⁿ⁺¹ = x⁵/4 + x⁸/16 + x¹¹/64 + ...

 

The series converges for |x|<4^1/3.