Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 8 / d

For either exponent, the recurrence from part (c) impliesfor every finite complex $x$. The ratio test therefore givesThe factor multiplying $x^{\sigma }$ is an entire function in each case. For $\sigma =1/2$, this does not make the full Frobenius solution single-valued at zero: its square-root prefactor still requires a branch. The infinite radius refers to $\sum a_n{x}^n$, not to a Taylor expansion of the branched full solution.