Kvant Math Problem 806

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 13m10s
Source on kvant.digital

Problem

Prove that if $$a_1+\dfrac{a_2}{2}+\dfrac{a_3}{3}+\ldots+\dfrac{a_n}{n}=0,$$ then the polynomial $a_nx^{n-1}+a_{n-1}x^{n-2}+\ldots+a_2x+a_1$ has a root between 0 and 1.
Prove that if for some $p\gt0$, $$\dfrac{a_1}{p+1}+\dfrac{a_2}{p+2}+\dfrac{a_3}{p+3}+\ldots+\dfrac{a_n}{p+n}=0,$$, then this polynomial also has a root between 0 and 1.

A. Gokhberg, 10th-grade student (Donetsk), M. Ovetsky, 10th-grade student (Donetsk)

Exploration

Let

$$P(x)=a_nx^{,n-1}+a_{n-1}x^{,n-2}+\cdots+a_2x+a_1.$$

The condition in part 1 is

$$a_1+\frac{a_2}{2}+\cdots+\frac{a_n}{n}=0.$$

The coefficients appear with denominators matching the exponents that arise when integrating powers of $x$ on $[0,1]$. Indeed,

$$\int_0^1 P(x),dx =\sum_{k=1}^n a_k\int_0^1 x^{k-1},dx =\sum_{k=1}^n\frac{a_k}{k}.$$

Thus the hypothesis is exactly $\int_0^1P(x),dx=0$.

If $P$ were positive on all of $[0,1]$, the integral would be positive. If $P$ were negative on all of $[0,1]$, the integral would be negative. Hence either $P\equiv0$ or $P$ changes sign, and continuity then gives a zero in $(0,1)$. This settles part 1.

For part 2, the denominators are $p+k$. Since

$$\int_0^1 x^{p+k-1},dx=\frac1{p+k},$$

multiplying $P(x)$ by $x^p$ gives

$$\int_0^1 x^pP(x),dx =\sum_{k=1}^n\frac{a_k}{p+k}.$$

The hypothesis becomes $\int_0^1 x^pP(x),dx=0$. Because $p>0$, the weight $x^p$ is strictly positive on $(0,1)$. If $P$ had no zero in $(0,1)$, continuity would force $P$ to keep a constant sign there, making $x^pP(x)$ keep the same sign and giving a nonzero integral. Hence $P$ must vanish at some point of $(0,1)$.

The step most likely to hide an error is the passage from a zero weighted integral to the existence of a root. One must use continuity and the fact that the weight $x^p$ is positive on $(0,1)$.

Problem Understanding

We are given a polynomial

$$P(x)=a_nx^{n-1}+a_{n-1}x^{n-2}+\cdots+a_2x+a_1.$$

In part 1 we must prove that the condition

$$\sum_{k=1}^n\frac{a_k}{k}=0$$

forces $P$ to have a root between $0$ and $1$.

In part 2 we must prove the analogous statement under the more general condition

$$\sum_{k=1}^n\frac{a_k}{p+k}=0, \qquad p>0.$$

This is a Type B problem. The task is to prove the stated implication.

The core difficulty is recognizing the given sums as definite integrals of $P$ or of $x^pP(x)$, and then showing that a continuous function with zero integral against a strictly positive weight must vanish inside the interval.

Proof Architecture

First lemma: for the polynomial $P$,

$$\int_0^1P(x),dx=\sum_{k=1}^n\frac{a_k}{k}.$$

This follows by integrating each monomial separately.

Second lemma: for every $p>0$,

$$\int_0^1x^pP(x),dx=\sum_{k=1}^n\frac{a_k}{p+k}.$$

This follows from $\int_0^1x^{p+k-1},dx=\frac1{p+k}$.

Third lemma: if a continuous function $f$ on $[0,1]$ satisfies $\int_0^1w(x)f(x),dx=0$, where $w(x)>0$ for all $x\in(0,1)$, then either $f\equiv0$ or $f$ has a zero in $(0,1)$.

The reason is that a continuous function with no zero in $(0,1)$ has a constant sign there, and multiplication by a positive weight preserves that sign, forcing the integral to be nonzero.

The most delicate point is the third lemma, because one must exclude the possibility that a sign-definite function could still have weighted integral $0$.

Solution

Let

$$P(x)=a_nx^{n-1}+a_{n-1}x^{n-2}+\cdots+a_2x+a_1.$$

For part $1$, integrating term by term yields

$$\int_0^1P(x),dx = \sum_{k=1}^na_k\int_0^1x^{k-1},dx = \sum_{k=1}^n\frac{a_k}{k}.$$

By the hypothesis,

$$\int_0^1P(x),dx=0.$$

Assume that $P$ has no root in $(0,1)$. Since $P$ is continuous, it cannot change sign on $(0,1)$. Hence either $P(x)>0$ for all $x\in(0,1)$, or $P(x)<0$ for all $x\in(0,1)$, or $P\equiv0$.

If $P(x)>0$ on $(0,1)$, then

$$\int_0^1P(x),dx>0.$$

If $P(x)<0$ on $(0,1)$, then

$$\int_0^1P(x),dx<0.$$

Both conclusions contradict $\int_0^1P(x),dx=0$. Thus the only remaining possibilities are that $P\equiv0$, in which case every point is a root, or that $P$ has a root in $(0,1)$.

Hence the polynomial has a root between $0$ and $1$.

For part $2$, let $p>0$. Again integrating term by term,

$$\int_0^1x^pP(x),dx = \sum_{k=1}^na_k\int_0^1x^{p+k-1},dx = \sum_{k=1}^n\frac{a_k}{p+k}.$$

The hypothesis gives

$$\int_0^1x^pP(x),dx=0.$$

Assume that $P$ has no root in $(0,1)$. Since $P$ is continuous, it has a constant sign on $(0,1)$.

Because $p>0$,

$$x^p>0 \qquad\text{for all }x\in(0,1).$$

Hence $x^pP(x)$ has the same sign as $P(x)$ throughout $(0,1)$.

If $P(x)>0$ on $(0,1)$, then $x^pP(x)>0$ on $(0,1)$, and therefore

$$\int_0^1x^pP(x),dx>0.$$

If $P(x)<0$ on $(0,1)$, then $x^pP(x)<0$ on $(0,1)$, and therefore

$$\int_0^1x^pP(x),dx<0.$$

Both conclusions contradict

$$\int_0^1x^pP(x),dx=0.$$

Thus either $P\equiv0$ or $P$ possesses a root in $(0,1)$. In either case the polynomial has a root between $0$ and $1$.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the identity

$$\int_0^1P(x),dx=\sum_{k=1}^n\frac{a_k}{k}.$$

Writing

$$P(x)=\sum_{k=1}^na_kx^{k-1},$$

we obtain

$$\int_0^1P(x),dx = \sum_{k=1}^na_k\int_0^1x^{k-1},dx = \sum_{k=1}^na_k\Bigl[\frac{x^k}{k}\Bigr]0^1 = \sum{k=1}^n\frac{a_k}{k}.$$

No index shift is involved, so there is no risk of replacing $k$ by $k-1$.

The second delicate step is proving that a continuous function with zero weighted integral must vanish. Suppose $P$ has no zero in $(0,1)$. Continuity implies that $P$ cannot take both positive and negative values there, because a sign change would produce a zero by the intermediate value theorem. Hence $P$ is strictly positive or strictly negative on $(0,1)$. Since the weight $x^p$ is strictly positive on $(0,1)$, the product $x^pP(x)$ has the same sign everywhere on that interval. The integral of a continuous function that is strictly positive on an interval is positive, and the integral of one that is strictly negative is negative. Thus the integral cannot be $0$.

A careless argument might overlook the necessity that the weight be positive. The proof works because $x^p>0$ on $(0,1)$.

Alternative Approaches

Part $1$ can be viewed as an application of the integral mean value theorem. Since $P$ is continuous,

$$\int_0^1P(x),dx=P(\xi)$$

for some $\xi\in(0,1)$. The hypothesis makes the left-hand side equal to $0$, hence $P(\xi)=0$. This immediately gives a root in $(0,1)$.

Part $2$ admits the same idea with the positive weight $x^p$. The weighted mean value theorem yields

$$\int_0^1x^pP(x),dx = P(\xi)\int_0^1x^p,dx$$

for some $\xi\in(0,1)$. Since the first factor on the left is $0$ and the second integral is positive, $P(\xi)=0$.

The integral-sign argument used in the main proof is preferable because it treats both parts uniformly and requires only continuity and elementary properties of definite integrals.