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

x

^{5}/(4-x^{3}) = (x^{5}/4)[1/(1-x^{3}/4)] = (x^{5}/4)∑_{n}_{=0...∞}(x^{3}/4)^{n}= ∑_{n=0...∞}x^{3n+5}/4^{n+1}Find the Power Series of f(x) = x^5/(4-x^3) in terms of sigma.

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Brookline, MA

New Wilmington, PA

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

f(x)= x^{5}/(4-x^{3})= -x^{2} + 4x^{2}/(4-x^{3}) = -x^{2} + x^{2}/(1-(x^{3}/4))

so really we want the power series of 4x^{2}/(4-x^{3}), 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,

4x^{2}/(4-x^{3}) = x^{2}/(1-(x^{3}/4)) = (-4/3) d/dx (ln(1-(x^{3}/4))

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

ln(1+u) = u - u^{2}/2 + u^{3}/3 -+....+∑_{n=1}^{∞} (-1)^{n+1}u^{n}/n, which converges for 0<u≤1.

So set u=-x^{3}/4 and get

ln (1-(x^{3}/4)) = ∑ (-1)^{n+1}(-x^{3}/4)^{n}/n = -∑ x^{3n}/(4^{n}n)

Take the derivative term by term:

d/dx (ln(1-(x^{3}/4))= -3 ∑ x^{3n-1}/4^{n}

Therefore,

4x^{2}/(4-x^{3}) =4 ∑_{n=1}^{∞} x^{3n-1}/4^{n} = x^{2} +∑_{n=2}^{∞} x^{3n-1}/4^{n}

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

f(x) = ∑_{n=2}^{∞} x^{3n-1}/4^{n }

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

f(x) = ∑_{n=1}^{∞} x^{3n+2}/4^{n+1 }= x^{5}/4 + x^{8}/16 + x^{11}/64 + ...

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

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