If g(x) = sum_{j=0}^{infinity} a_j x^j, where |x|

Question

If g(x) = sum_{j=0}^{infinity} a_j x^j, where |x| < 1, which of the following equals sum_{i=-infinity}^{infinity} g(|i|) x^i? (Assume the series converges.)

A. g(x)
B. i = -1
C. 2g(x)
D. 2^i

Accepted Answer

2g(x) We are summing g(|i|) x^i over all integers i from negative infinity to positive infinity. Because |i| = i for i >= 0 and |-i| = i for i < 0, the positive-index and negative-index halves are symmetric. The i=0 term is g(0)*1 = a_0. For every i > 0 we get a_i x^i (from the positive side) and a_i x^{-i} (from the negative side). But here x^{-i} = 1/x^i, so we need |x| < 1 for convergence. Looking at the answer options, option C = 2g(x) suggests the two halves both contribute g(x).

If g(x) = sum_{j=0}^{infinity} a_j x^j, where |x| < 1, which of the following equals sum_{i=-infinity}^{infinity} g(|i|) x^i? (Assume the series converges.) A. g(x) B. i = -1 C. 2g(x) D. 2^i