Section 6.12

Further Maximum-Minimum Problems on Polynomials

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Problem 6.103

Let P(x)P(x) be an arbitrary polynomial of nnth degree with real coefficients satisfying

∫−11[P(x)]2 dx=1. \int _{-1}^{1} [P(x)]^2\, dx = 1.

Then for −1≦x≦1-1 \leqq x \leqq 1 we have

∣P(x)∣≦n+12. \left|P(x)\right| \leqq \frac{n+1}{\sqrt{2}}.

The equality sign here holds if and only if either

P(x)=±2n+1Sn(x)(problem 88)andx=1 P(x) = \pm \frac{\sqrt{2}}{n+1}S_n(x) \quad \text{(problem 88)} \quad \text{and} \quad x=1

or

P(x)=±2n+1Sn(−x)andx=−1 P(x) = \pm \frac{\sqrt{2}}{n+1}S_n(-x) \quad \text{and} \quad x=-1

(n>0n>0).

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Remark.
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The integral of [P(x)]2[P(x)]^2 is a quadratic form in n+1n+1 variables. Choose these variables in such a manner that the quadratic form becomes a sum of n+1n+1 squares.

Problem 6.104

Let P(x)P(x) be an arbitrary polynomial of nnth degree with real coefficients satisfying

∫−11(1−x)[P(x)]2 dx=1. \int _{-1}^{1} (1-x)[P(x)]^2\, dx = 1.

Then we have

∣P(1)∣≦12(n+22),∣P(−1)∣≦12(n+22). \left|P(1)\right| \leqq \frac{1}{\sqrt{2}}\binom {n+2}{2}, \qquad \left|P(-1)\right| \leqq \frac{1}{\sqrt{2}}\sqrt{\binom {n+2}{2}}.

These bounds cannot be replaced by any smaller bounds.

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Problem 6.105

Let α\alpha and β\beta be constants, α>−1\alpha >-1, β>−1\beta >-1, and P(x)P(x) an arbitrary polynomial of nnth degree with real coefficients satisfying

∫−11(1−x)α(1+x)β[P(x)]2 dx=1. \int _{-1}^{1} (1-x)^\alpha (1+x)^\beta [P(x)]^2\, dx = 1.

Determine the maximum of ∣P(1)∣\left|P(1)\right| and ∣P(−1)∣\left|P(-1)\right|, as P(x)P(x) ranges through the set of all polynomials of the specified kind. How do these maximum values behave for large values of nn?

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Problem 6.106

Let α\alpha be a constant, α>−1\alpha >-1. Determine the maximum of ∣P(0)∣\left|P(0)\right| for all polynomials P(x)P(x) of nnth degree which satisfy the condition

∫0∞e−xxα[P(x)]2 dx=1. \int _0^{\infty } e^{-x}x^\alpha [P(x)]^2\, dx = 1.

How does this maximum behave for large values of nn?

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Problem 6.107

Determine the maximum of ∣P(0)∣\left|P(0)\right| for all polynomials P(x)P(x) of nnth degree, which satisfy the condition

∫−∞∞e−x22[P(x)]2 dx=1. \int _{-\infty }^{\infty } e^{-\frac{x^2}{2}}[P(x)]^2\, dx = 1.

How does this maximum behave for large values of nn?

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Problem 6.108

Let P(x)P(x) be a polynomial of nnth degree which assumes only non-negative values in the interval −1≦x≦1-1 \leqq x \leqq 1, and which satisfies

∫−11P(x) dx=1. \int _{-1}^{1} P(x)\, dx = 1.

Then we have

P(1)≦{q(q+1)2for odd n;n=2q−1,]6pt](q+1)22for even n,n=2q. P(1) \leqq \begin{cases} \dfrac {q(q+1)}{2} & \text{for odd } n; \quad n=2q-1, \\ ]6pt] \dfrac {(q+1)^2}{2} & \text{for even } n, \quad n=2q. \end{cases}

The same estimate holds for P(−1)P(-1). These bounds cannot be replaced by any smaller bounds.

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Problem 6.109

The first mean-value theorem of integral calculus may be strengthened for polynomials of nnth degree as follows: Let P(x)P(x) be a polynomial of nnth degree and mm the minimum, MM the maximum of P(x)P(x) in the interval a≦x≦ba \leqq x \leqq b. Then we have

m+M−mαn≦1b−a∫abP(x) dx≦M−M−mαn, m + \frac{M-m}{\alpha _n} \leqq \frac{1}{b-a}\int _a^b P(x)\, dx \leqq M - \frac{M-m}{\alpha _n},

where αn=q(q+1)\alpha_n = q(q+1) for odd nn, n=2q−1n=2q-1, and αn=(q+1)2\alpha_n = (q+1)^2 for even nn, n=2qn=2q.

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Problem 6.110

Let α\alpha and β\beta be constants, α>−1\alpha >-1, β>−1\beta >-1, and P(x)P(x) a polynomial of nnth degree which assumes only non-negative values in the interval −1≦x≦1-1 \leqq x \leqq 1 and which satisfies

∫−11(1−x)α(1−x)βP(x) dx=1. \int _{-1}^{1} (1-x)^\alpha (1-x)^\beta P(x)\, dx = 1.

Then we have

P(1) \leqq \begin{cases} \dfrac {1}{2^{\alpha +\beta +1}}\dfrac {\Gamma (q+\alpha +1)\Gamma (q+\alpha +\beta +2)}{\Gamma (\alpha +1)\Gamma (\alpha +2)\Gamma (q)\Gamma (q+\beta +1)} & \text{for odd } n, \quad n=2q-1, \\ ]8pt] \dfrac {1}{2^{\alpha +\beta +1}}\dfrac {\Gamma (q+\alpha +2)\Gamma (q+\alpha +\beta +2)}{\Gamma (\alpha +1)\Gamma (\alpha +2)\Gamma (q+1)\Gamma (q+\beta +1)} & \text{for even } n, \quad n=2q. \end{cases} \[ These bounds cannot be replaced by any smaller bounds. The corresponding bounds for $P(-1)$ are obtained by interchanging $\alpha $ and $\beta $. \]
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Problem 6.111

Let α\alpha be a constant, α>−1\alpha >-1, and P(x)P(x) a polynomial of nnth degree which assumes only non-negative values for non-negative values of xx, and which satisfies

∫0∞e−xxαP(x) dx=1. \int _0^{\infty } e^{-x}x^\alpha P(x)\, dx = 1.

Then we have

P(0)≦Γ(p+α+2)Γ(α+1)Γ(α+2)Γ(p+1),p=[n2]. P(0) \leqq \frac{\Gamma (p+\alpha +2)}{\Gamma (\alpha +1)\Gamma (\alpha +2)\Gamma (p+1)}, \qquad p = \left[\frac{n}{2}\right].

This bound cannot be replaced by any smaller bound.

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Problem 6.112

Let P(x)P(x) be a polynomial of nnth degree, which is non-negative for non-negative values of xx, and which satisfies

∫0∞e−xP(x) dx=1. \int _0^{\infty } e^{-x}P(x)\, dx = 1.

Then we have

P(0)≦[n2]+1. P(0) \leqq \left[\frac{n}{2}\right] + 1.
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Problem 6.113

(Continuation of problem 112.) Let ξ\xi be an arbitrary non-negative number. Then we have

e−ξP(ξ)≦[n2]+1. e^{-\xi }P(\xi ) \leqq \left[\frac{n}{2}\right]+1.
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