4.2 Formal functions

1. Prove that for any \(\alpha, \beta \in \mathbb{C}\) we have \(e^{\alpha s}e^{\beta s}=e^{(\alpha+\beta)s}\)​

   Proof

   \(\exp(s)=1+\frac{s}{1!}+\frac{s^{2}}{2!}+\frac{s^{3}}{3!}+\cdots\)​

   \(\exp(\alpha s)=1+\frac{\alpha s}{1!}+\frac{(\alpha s)^{2}}{2!}+\frac{(\alpha s)^{3}}{3!}+\cdots\)

   \(\exp(\beta s)=1+\frac{\beta s}{1!}+\frac{(\beta s)^{2}}{2!}+\frac{(\beta s)^{3}}{3!}+\cdots\)​

   And we NTP coefficients at \(s^n\) of \(\exp((\alpha +\beta)s)\) is \(\frac{(\alpha +\beta)^n}{n!}\)

   Since \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\) and \(fg=\sum_{n=0}^{\infty}\left(\sum_{j=0}^{n}f_{j}g_{n-j}\right)s^{n}\), then coefficients at \(s^n\) of \(\exp((\alpha+\beta)s)\) is \(\sum_{j=0}^{n}\frac{\alpha^{j}}{j!}\frac{\beta^{n-j}}{\left(n-j\right)!}\)​

2. Let \(f = f_0 + f_1 s + f_2 s^2 + ... = \sum_{j} f_j s^j \in \mathbb{C}[[s]]\) such that \(f_0 \neq 0\). Prove that there are exactly two formal power series \(g \in \mathbb{C}[[s]]\) such that \(g^2 = f\).

   Assume \(g=g_{0}+g_{1}s+g_{2}s^{2}+...=\sum_{n=0}^{\infty}g_{n}s^{n}\in\mathbb{C}[[s]]\), then \(g^{2}=g_0^2+2g_0g_1s+(2g_0g_2+g_1^2)s^2+(2g_0g_3+2g_1g_2)s^3+...\)​

   Then \(g_0^2=f_0\) two solutions \(\pm\sqrt{f_0}\) (Important \(f_{0}\neq 0\implies g_{0}\neq 0\)

   Because the next equation is \(2g_0g_1=f_1\) is linear equation of the form \(ax+b=c\) for \(a\neq 0\)\, which is unique solution for \(g_1\) for fixed \(g_0\)

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   if the signal is minus, then is the first one since we need to cancel constant

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   Condition for existence of a square root: \(f_0=f_1=...=f_{2n-1}=0,f_{2n}\neq 0\) for some \(n\)​

3. Find the number of subsets of \(\{1, ..., 2000\}\), the sum of whose elements is divisible by 5.

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   We can see here the subsets such that sum of whose elements is divisible by 5 is corresponding to the power of the \(s\)

   Thus it means we just need to solve the sum of coefficients corresponding to \(s^{5},s^{10},...\),

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