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

Question

How to prove that $(1 + x)^\frac{1}{b}$ (where $b$ is an integer) can be expressed as a formal power series without using Binomial theorem?

Answer 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 $\rm\:i>0\:.$

Answer 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.

Answer 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 $a_1=1/b$.

Answer 4

We do it for $b>0$. A similar argument will work for $b<0$. Note that the argument is purely formal.

So we are trying to find $a_0$, $a_1$, $a_2$, and so on such that $$1+x=(a_0+a_1x+a_2x^2+ \cdots +a_nx^n +\cdots)^b$$ where equality is meant in the formal sense. It is clear that we want $a_0=1$ and $a_1=1/b$.

There is a natural proof by induction, if we choose the induction hypothesis correctly.
Let $n \ge 1$. Assume by way of induction hypothesis that there are numbers $a_0, a_1, \dots, a_n$ such that $$(a_0+a_1x+\cdots+a_nx^n)^b$$ formally agrees with $1+x$ in positions $0, 1,\dots,n$. We will show that we can choose $a_{n+1}$ such that
$$(a_0+a_1x+\cdots+a_{n+1}x^{n+1})^b$$ formally agrees with $1+x$ in positions $0, 1,\dots,n+1$.

That is easy. Let $c_{n+1}$ be the coefficient of $x^{n+1}$ in $(a_0+a_1x+ \cdots +a_nx^n)^b$. Define $a_{n+1}$ by $ba_{n+1}=-c_{n+1}$.

Answer 5

We can use Newton's method to find a root to the function

$$ g(t) = t^b - (1+x) $$

by using the iteration

$$t_{n+1} = t_n - \frac{g(t_n)}{g'(t_n)} = t_n - \frac{t_n^b - (1+x)}{b t_n^{b-1}} $$

We can see this converges by observing that if the reciprocal of $t_n$ is a formal power series (i.e. its leading coefficient is nonzero) and we have

$$ g(t_n) \equiv 0 \pmod{x^k} $$

then we must also have have

$$ t_{n+1} \equiv t_n \pmod{x^k} $$

and the Taylor series formula gives

$$ \begin{align*}g(t_{n+1}) &= g(t_n) - g'(t_n) \frac{g(t_n)}{g'(t_n)} \mod{\left(\frac{g(t_n)}{g'(t_n)}\right)^2} \\ &=0 \mod x^{2k} \end{align*} $$

If we set $t_0 = 1$, then it's easy to check that

$$g(t_0) \equiv 0 \pmod x \qquad \qquad t_n \equiv 1 \pmod x $$

and thus the limit of this sequence is a formal power series that is a root of $g(t)$.