I propose to give a very simple proof of the following theorem of Weierstrass:
If F(x) is any continuous function in the interval [0,1], it is always possible, regardless how small ε, to determine a polynomial En(x)=a0xn+a1xn−1+⋯+an of degree n high enough such that we have |F(x)−En(x)|<ε for every point in the interval under consideration.
To this end, I consider an event A, whose probability is equal to x. Suppose n experiments are conducted and that is agreed to pay a player the sum F(mn), if the event A occurs m times. Under these conditions, the mathematical expectation En for the player will have the value
En=m=n∑m=0F(mn)⋅Cmn⋅xm⋅(1−x)n−m.
It follows from the continuity of the function F(x) that it is possible to set a number δ, such that the inequality
|x−x0|≤δ
causes
|F(x)−F(x0)|<ε2;
so that, if ¯F(x) is the maximum and F_(x) the minimum of F(x) in the interval (x−δ,x+δ), then
¯F(x)−F(x)<ε2,F(x)−F_(x)<ε2.
Let η be the probability of the inequality |x−mn|>δ and L the maximum of |F(x)| in the interval [0,1].
We then have
F_(x)⋅(1−η)−L⋅η<En<¯F(x)⋅(1−η)+L⋅η.
But by virtue of a theorem of Bernoulli, we can take n large enough to have
η<ε4L.
Inequality (3) will in turn take the form
F(x)+(F_(x)−F(x))−η(L+F_(x))<En<F(x)+(¯F(x)−F(x))+η(L−¯F(x))
and so
F(x)−ε2−2L4Lε<En<F(x)+ε2+2L4Lε;
therefore
|F(x)−En|<ε
En is clearly a polynomial of degree n.
The theorem is therefore proved.
I would only add two points.
The approximate polynomials En(x) are especially convenient, it seems to me, when you know exactly or approximately the values of F(x) for x=mn(m=0,1,⋯n).
Formula (1) and inequality (5) show that, for any continuous function F(x):
F(x)=limn→∞m=n∑m=0F(mn)⋅Cmn⋅xm⋅(1−x)n−m.
Translated by Mike Bertrand (Mar 6, 2015).