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 gap$$then $$, 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 function$$Is a polynomial. The set of all regular points is denoted by$$. A bunch of$$additional to $$ denote by $$and call it the set of irregular points. (We say that if$$then $$ - 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-interval$$correct then
narrowing$$ 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 be$$Is an isolated point. Then for some$$ 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 set$$wrong points are not empty. Put$$. 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 set$$then $$ 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 $$ and all derivatives $$. Because$$then by the Lagrange finite increment theorem $$ cannot be zero everywhere on $$ for no natural $$.

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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