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Find the Power Series in terms of sigma:

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

x5/(4-x3) = (x5/4)[1/(1-x3/4)] = (x5/4)∑n=0...∞(x3/4)n = ∑n=0...∞x3n+5/4n+1
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=41/3, 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 ∑ x3n-1/4n
4x2/(4-x3) =4 ∑n=1 x3n-1/4n = x2 +∑n=2 x3n-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 x3n-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|<41/3.