prove that $(1 + x)^\frac{1}{b}$ is a formal power series

Solution 1:

The coefficient recurrence arises from the obvious first-order differential equation, namely

$$\rm\ \frac{y'}y\ =\ (log\ y)'\:=\ \bigg(\frac{log(1+x)}b\bigg)'\:=\ \frac{1}{b\ (1+x)}\ \ \ \Rightarrow\ \ \ y\: =\ b\ (1+x)\ y'$$

Therefore $\displaystyle\rm\ \ y\ =\ \sum_{k\ge 0}\ a_k x^k\ =\ b\ (1+x)\ \sum_{k\ge 0}\ (k+1)\ a_{k+1}\ x^k\:,\ \: $ which, after algebra, yields

$$\rm a_{k+1}\ =\ \frac{1-b\:k}{b\:(k+1)}\ a_k,\ \ \ a_0 = 1$$

As a check, note that it yields the binomial formula for $\rm\ b = 1/n,\ y = (1+x)^n\ $ since then

$$\rm \frac{a_{k+1}}{a_k}\ =\ \frac{n-k}{k+1}\ =\ \frac{n\choose k+1\:}{n\choose k\:}$$

and $\rm\ a_k = 0\ $ for $\rm\ k > n\ $ since the above implies $\rm\ a_{n+1} = 0\ $ hence $\rm\ a_{n+i} = 0\ $ for all $\rm\:i>0\:.$

Solution 2:

I don't think you mean it is (literally) a formal power series, because it isn't. It can be expressed as a formal power series (in $x$), because it is an analytic function of $x$ in a neighbourhood of $x=0$. You could use the Inverse Function Theorem for that.

Solution 3:

We are looking for a power series $f=1+a_1x+a_2x^2+\cdots$ such that $f^b=1+x$. Writing out $f^b$, you get $1+ba_1x+g_2(a_1,a_2)x^2+g_3(a_1,a_2,a_3)x^3+\cdots$, where $g_i(a_1,a_2,\ldots,a_i)$ is a linear function of the coefficients. Then for $f^b=1+x$, we need to solve a system of equations $$ba_1=1$$ $$g_2(a_1,a_2)=0$$ $$g_3(a_1,a_2,a_3)=0$$ $$\vdots$$ This gives us an infinite system of equations which can be solved iteratively, starting with $a_1=1/b$.