# New proof of the polynomial theorem

The article provides a new proof of the beautiful and difficult theorem of mathematical analysis, stated in such a way that it is accessible to high school students of specialized mathematical schools.

Let be  - an infinitely many times differentiable real function, and for each point  there is a natural  such that . Then polynomial.

# Evidence

We need the Baire theorem on a system of closed sets:

1. Let and  closed subsets of the line, and  and . Then in there is a point that is contained in one of along with its surroundings. More precisely, there is a point, natural  and  such that .

Indeed (by contradiction), we choose a point  and surround it with surroundings where . We hypothesized that the statement of Baire’s theorem is not true. Means. Choose in a point . Surround interval  such that the ends of this interval are points  and  lie in , a . By assumption. This allows you to choose in some point  Continuing the process, we construct a nested contracting sequence of intervals  It's clear that

, (1)
(2)

Since each gapthen , and from (1) and (2) it follows that  for everybody . So we found the point, but not lying in any of the sets
.
We say that a point on a real line is regular if, in a neighborhood of this point, the functionIs a polynomial. The set of all regular points is denoted by. A bunch ofadditional to  denote by and call it the set of irregular points. (We say that ifthen  - wrong point).
2. If each point of the segment correct then narrowing  on the  Is a polynomial.

Indeed, for every point  there is an interval such that the narrowing on this interval is a polynomial. Those. for each point there is an interval and some natural, what equal to zero in this interval.

From the compactness of the segment it follows that there is such a natural , what  everywhere on , Consequently  Is a polynomial.

3. If each point of the half-intervalcorrect then
narrowing on the  Is a polynomial.

Evidence. Consider the increasing sequence such that  converges to . As proved in the previous paragraph, on each of the segments narrowing Is a polynomial. Let be - polynomial coinciding with  on the segment . It's clear that for all  therefore  coincides with  on the , then at the point . (Recall that and  continuous everywhere on )

Similarly to the previous one, it is easy to prove that:

4. If each point of the half-interval or interval  Is correct then  - polynomial on .

We proceed to the study of irregular points, i.e. set points.

5. Many does not contain isolated dots.

Really. Let beIs an isolated point. Then for some and consist of the right points. Means narrowing on the  and on polynomials. It is clear that for sufficiently large ( there must be more degrees of each of these polynomials)  will be zero everywhere on . Those. is the right point.

6. Let the setwrong points are not empty. Put. It's clear that and each closed. It follows from Baire’s theorem (see 1.) that there is an interval such that  and  lies in one of .

Consider the function . This function is zero at every point.. Since each wrong point is the limit for the setthen  for all integers  and all .

Let us prove that is equal to  everywhere on . Not so. Then there is such that . Since many is not empty and closed, then we find a point in it closest to . For definiteness, we put. Function differentiable infinitely many times on  and all derivatives . Becausethen by the Lagrange finite increment theorem  cannot be zero everywhere on  for no natural .

Slobodnik Semyon Grigoryevich ,
developer of content for the application “Tutor: mathematics” (see article on Habré ), candidate of physical and mathematical sciences, teacher of mathematics, school 179, Moscow