Prove this f'(x) = f(x2), f(0) = 1 The Unique Entire Solution — Synthesis function S(x) S(x) = 1 + summ=1^infinity x 2^(m−1)k=1^m-1(2^\,k+1-1)

To prove that the given series $\mathcal{S}(x)$ is the unique entire solution to the differential equation $f'(x) = f(x^2)$ with the initial condition $f(0) = 1$, we need to show two things:

1. $\mathcal{S}(0) = 1$
2. $\mathcal{S}'(x) = \mathcal{S}(x^2)$

The given series is: $$\mathcal{S}(x) = 1 + \sum_{m=1}^{\infty} \frac{x^{2^m-1}}{\displaystyle\prod_{k=1}^{m-1}\bigl(2^{,k+1}-1\bigr)}$