The Bernstein Convergence Theorem. Binomial Series

Let $f(x)$ be a function defined for $a\le x\le b$; suppose also that for every integer $n\ge 0,f^{(n)}(x)$ exists and $f^{(n)}(x)\ge 0$ for $a\le x\le b$. Let $x_1$ be some point of the interval and choose some fixed positive integer $n$. Then consider Taylor's formula for $f(x)$ about $x_1$, where the polynomial of the formula has more than $n$ terms, say $m$ terms. In particular, consider what this gives us for $f(b)$, when we use the Cauchy remainder:

\begin{aligned} f(b) = f(x_1) &+ \frac{(b-x_1)}{1!} f'(x_1) + \dots + \frac{(b-x_1)^n}{n!} f^{(n)}(x_1) + \dots \\ &+ \frac{(b-x_1)^{m-1}}{(m-1)!} f^{(m-1)}(x_1) + \frac{(b-x_1)(b-\xi)^{m-1}}{(m-1)!} f^{(m)}(\xi), \end{aligned}

where . Now all terms of the sum on the right are non-negative; therefore $f(b)$ is greater than or equal to any of them. In particular,

\frac{(b-x_1)^n}{n!} f^{(n)}(x_1) \le f(b), \tag{1}

and this is true for any $n$.

Now consider the function f''(x). Since the derivatives of f''(x) are also derivatives of $f(x)$, they all are non-negative in $a\le x\le b$. Therefore we can use f''(x) in place of $f(x)$ in (1), obtaining

\frac{(b-x_1)^n}{n!} f^{(n+2)}(x_1) \le f''(b), \quad \text{or}f^{(n+2)}(x_1) \le \frac{n!}{(b-x_1)^n} f''(b) \cdot \tag{2}

This last inequality also holds for any $n$; in particular, it holds for $n-2$, where $n$ is as in (1). Then it becomes

f^{(n)}(x_1) \le \frac{(n-2)!}{(b-x_1)^{n-2}} f''(b) \cdot \tag{3}

This is true for any point $x_1$ of the interval.

Now let $x_0$ be some point of the interval, and consider the Taylor expansion about $x_0$ for $f(x)$, where $x$ is also in the interval. The Cauchy remainder after $n$ terms has absolute value

$$\frac{|x-x_0||x-\xi {|}^{n-1}}{(n-1)!}{f}^{(n)}(\xi ),$$

where $\xi$ is between $x$ and $x_0$. By (3), this is less than or equal to

\begin{aligned} &\frac{|x-x_0|\ |x-\xi|^{n-1}}{(n-1)!} \cdot \frac{(n-2)!}{(b-\xi)^{n-2}} f''(b) \\ &= \frac{|x-x_0|\ |x-\xi| \cdot f''(b)}{n-1} \left( \frac{|x-\xi|}{b-\xi} \right)^{n-2}. \end{aligned}

This will approach $0$ as $n$ becomes large if $|x-\xi |\le b-\xi$. If , then $|x-\xi |=x-\xi \le b-\xi$. If , and if $|x-x_0|\le b-x_0$, then $|x-\xi |\le |x-x_0|\le b-x_0\le b-\xi$. Therefore we know that the remainder approaches zero whenever $|x-x_0|\le b-x_0$. Let us summarize the result we have proved, which is the Bernstein convergence theorem.

For instance, this theorem tells us that the Taylor's series for $e^x$ formed about any point at all converges to $e^x$ for all $x$. The result can be extended to a considerably larger class of functions if we make a few observations. First suppose there is some number $M$ such that f(x) \ge M, f'(x) \ge M, f''(x) \ge M, \dots whenever $a\le x\le b$. If $M\ge 0$, then the Taylor's series of $f(x)$ converges to $f(x)$ by the Bernstein theorem. If , let

$$g(x)=f(x)-M{e}^{(x-a)}\cdot$$

Then $g(x)\ge M-M{e}^{x-a}=M(1-e^{x-a})\ge 0$, since both factors are less than or equal to zero. Likewise, g'(x) = f'(x) - M e^{x-a} \ge 0, g''(x) \ge 0, \dots. Therefore the Taylor's series for $g(x)$ converges to $g(x)$. As we have remarked, the Taylor's series for $e^{x-a}$ converges to $e^{x-a}(=e^{-a}\cdot e^x)$. Since the operation of forming the Taylor's series is a linear one, the Taylor's series for $f(x)$ is that for $g(x)$ plus $M$ times that for $e^{x-a}$; the remainder in each of these goes to zero, and therefore the remainder for $f(x)$, which is the sum of these remainders, goes to zero. Thus we see that the Taylor's series for $f(x)$ converges to $f(x)$ if all derivatives of $f(x)$ are bounded from below.

Next suppose $f^{(n)}(x)\ge 0$ for $a\le x\le b$, whenever for some fixed $N$. Then let $M$ be the smallest of the numbers

0, \min f(x), \min f'(x), \dots, \min f^{(N)}(x) \quad \text{on } a \le x \le b \cdot

Then $f^{(i)}(x)\ge M$ for all $i\ge 0$, and we see as above that the Taylor's series for $f(x)$ converges to $f(x)$. Since the Taylor's series for $-f(x)$ has as terms the negatives of those in the Taylor's series for $f(x)$, we can use $-f(x)$ to prove that the Taylor's series for $f(x)$ converges to $f(x)$ whenever all $f^{(n)}(x)$ are bounded from above, in particular if all $f^{(n)}(x)\le 0$.

Now let f(x) \ge 0, f'(x) \le 0, f''(x) \ge 0, f'''(x) \le 0, \dots; in other words, the derivatives of $f(x)$ alternate in sign. Let . Then , and if $-b\le \xi \le -a$, let $g(\xi )=f(-\xi )$. Then

g(\xi) \ge 0, g'(\xi) = -f'(-\xi) \ge 0, g''(\xi) \ge 0, \dots ,

or $g^{(n)}(\xi )\ge 0$ for all $n$ whenever $-b\le \xi \le -a$. Thus we know by the Bernstein theorem that the Taylor's series for $g(\xi )$ about $-x_0$ converges to $g(\xi )$ whenever $|\xi -(-x_0)|\le -a-(-x_0)$, or whenever $|\xi +x_0|\le x_0-a$. The Taylor's series for $g(\xi )=f(-\xi )$ about $-x_0$ is:

\begin{aligned} g(-x_0) + \frac{(\xi+x_0)}{1!} g'(-x_0) + \frac{(\xi+x_0)^2}{2!} g''(-x_0) + \dots \\ = f(x_0) - \frac{(\xi+x_0)}{1!} f'(x_0) + \frac{(\xi+x_0)^2}{2!} f''(x_0) - \dots \cdot \end{aligned}

Now let $x=-\xi ,\xi =-x$. Then whenever $|-x+x_0|\le x_0-a$, or $|x-x_0|\le x_0-a$, we have, in the limit,

\begin{aligned} f(x) = g(\xi) &= f(x_0) - \frac{(-x+x_0)}{1!} f'(x_0) + \frac{(-x+x_0)^2}{2!} f''(x_0) - \dots \\ &= f(x_0) + \frac{(x-x_0)}{1!} f'(x_0) + \frac{(x-x_0)^2}{2!} f''(x_0) + \dots \cdot \end{aligned}

But this is exactly the Taylor's series for $f(x)$. Note that in this case it converges to $f(x)$ whenever $x$ is closer to $x_0$ than $x_0$ is to $a$, while in the original theorem we had convergence to $f(x)$ whenever $x$ was closer to $x_0$ than $x_0$ was to $b$. Observe also that as before it is enough that the signs alternate from some $N$ on.

We shall apply this last observation to investigate the function $f(x)=x^{\alpha }$ for , $\alpha$ any number. Then

f'(x) = \alpha x^{\alpha-1},\quad f''(x) = \alpha(\alpha - 1)x^{\alpha-2},\quad \dots ,

and eventually

Then $f^{(N+1)}(x)=\alpha (\alpha -1)\dots (\alpha -N+1)(\alpha -N)x^{\alpha -N-1}$ has the opposite sign from $f^{(N)}(x)$ for all , since has the opposite sign from $f^{(N+1)}(x)$, and the signs of $f^{(n)}(x)$ alternate for . Therefore if $ϵ$ is any small positive number and , the Taylor's series for $f(x)$ about $x_0$ converges to $f(x)$ provided $|x-x_0|\le x_0-ϵ$. If , then we are sure that $|x-x_0|\le x_0-ϵ$ for some positive $ϵ$. Therefore the series converges to $f(x)$ provided .

Now let us change notation a little and write $x_0=a,x=a+b$. Then we expand $f(x)=(a+b{)}^{\alpha }$ in powers of $(x-x_0)=b$. (Here .) Thus

\begin{aligned} x^\alpha = (a+b)^\alpha = a^\alpha + \alpha a^{\alpha-1}b + \frac{\alpha(\alpha-1)}{2} a^{\alpha-2}b^2 + \cdots\\ + \frac{\alpha(\alpha-1) \dots (\alpha-n)}{n!} a^{\alpha-n}b^n + \cdots \end{aligned}

in the sense that the series converges to $(a+b{)}^{\alpha }$ whenever , or whenever . This is the so-called binomial series, and the coefficients

$$\frac{\alpha (\alpha -1)\dots (\alpha -n)}{n!}=(\genfrac{}{}{0ex}{}{\alpha }{n})$$

are called the binomial coefficients. Observe that if $\alpha$ is a positive integer, all the coefficients are $0$ from some point on, and we have the ordinary binomial theorem.

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