A couple weeks ago I wrote about the Weierstrass approximation theorem, the theorem that says every continuous function on a closed finite interval can be approximated as closely as you like by a polynomial.

The post mentioned above uses a proof by Bernstein. And in that post I used the absolute value function as an example. Not only is |*x*| an example, you could go the other way around and use it as a step in the proof. That is, there is a proof of the Weierstrass approximation theorem that starts by proving the special case of |*x*| then use that result to build a proof for the general case.

There have been many proofs of Weierstrass’ theorem, and recently I ran across a proof due to Lebesgue. Here I’ll show how Lebesgue constructed a sequence of polynomials approximating |*x*|. It’s like pulling a rabbit out of a hat.

The staring point is the binomial theorem. If *x* and *y* are real numbers with |*x*| > |*y*| and *r* is any real number, then

.

Now apply the theorem substituting 1 for *x* and *x*² – 1 for *y* above and you get

The partial sums of the right hand side are a sequence of polynomials converging to |*x*| on the interval [−1, 1].

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If you’re puzzled by the binomial coefficient with a top number that isn’t a positive integer, see the general definition of binomial coefficients. The top number can even be complex, and indeed the binomial theorem holds for complex *r*.

You might also be puzzled by the binomial theorem being an infinite sum. Surely if *r* is a positive integer we should get the more familiar binomial theorem which is a *finite* sum. And indeed we do. The general definition of binomial coefficients insures that if *r* is a positive integer, all the binomial coefficients with *k* > *r* are zero.

As an introductory analysis student I found this quite helpful. The only problem I noticed was that the binomial theorem only proves that the power series equals what it’s supposed to on (-1,1), but you assert that it’s true on [-1,1]. It is true that you can prove this for x=+/-1 easily because (1^2-1)^k=0 for all k>0 so the power series at x=+/-1 reduces to nCr(1/2,0), but to realize that this equals 1 you have to know that Π(i for i from m to n)=1 when m>n, so that the 0th falling factorial and therefore the generalized binomial product choosing 0 of any real number is 1. I would suggest that you add this information to make your proof clear.

Actually I misunderstood this before. The binomial theorem proves that this series converges when the absolute value of y in (x+y)^r is less than 1, i.e. |x^2-1|<1, i.e. 0<x<sqrt(2). You still need to demonstrate that the series converges at x=0, where x^2-1=-1, for the proof to work.