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Derivative, endpoint, parity, and boundedness identities for Bernoulli polynomials

Statement

For the polynomials of Bernoulli polynomials by a normalized coefficient recursion and every natural $m \geq 1$ , $B_{m}^{'} (x) = ι (m) B_{m - 1} (x), \int_{0}^{1} B_{m} (x) d x = 0.$ In particular, $B_{0} (x) = 1$ and $B_{1} (x) = x - \frac{1}{2}$ . Consequently $B_{m} (1) = B_{m} (0)$ for $m \geq 2$ . The reflection identity $B_{m} (1 - x) = (- 1)^{m} B_{m} (x)$ holds for every $m \geq 0$ , so $B_{m} (0) = 0$ for odd $m \geq 3$ . Each $B_{m}$ is bounded on $[0, 1]$ .

Facts & Assumptions

Given: The recursive coefficient definition of $B_{m}$ and an integer index $m$ .

[A1]

$m$ is a natural index; derivative and mean identities are asserted for $m \geq 1$ .

[F1]

Bernoulli polynomials by a normalized coefficient recursion: if $B_{m - 1} (x) = \sum_{q = 0}^{m - 1} β_{m - 1, q} x^{q}$ , then $B_{m} (x) = ι (m) \sum_{q = 0}^{m - 1} β_{m - 1, q} (x^{q + 1} / ι (q + 1) - 1 / (ι (q + 1) ι (q + 2)))$ .

[F2]

For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term and Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : differentiating a finite polynomial sum termwise gives the sum of the power derivatives.

[F3]

The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : the integral of a continuous derivative on $[0, 1]$ is the endpoint difference of its primitive.

[F4]

The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ : the derivative of $B_{m} (1 - x)$ is $- B_{m}^{'} (1 - x)$ .

[F5]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion and For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent: polynomials are integrable and bounded on $[0, 1]$ .

[F6]

Monotone change of variable for Riemann-integrable functions: the decreasing affine map $x \mapsto 1 - x$ preserves the Riemann integral after multiplication by its absolute derivative $1$ .

[F7]

The principle of mathematical induction: induction over the natural polynomial index is valid.

Proof

1.1givenA1F1F2F3algebra

For the nonnegative natural index $m$ of [A1], the case $m = 1$ gives $B_{1} (x) = x - \frac{1}{2}$ by [F1]. For every $m \geq 1$ , [F1]–[F2] give $B_{m}^{'} (x) = ι (m) \sum_{q = 0}^{m - 1} β_{m - 1, q} x^{q} = ι (m) B_{m - 1} (x)$ . Integrating each monomial in [F1] over $[0, 1]$ , the integral of $x^{q + 1} / ι (q + 1)$ is $1 / (ι (q + 1) ι (q + 2))$ , which cancels the subtracted constant term; hence $\int_{0}^{1} B_{m} = 0$ .

1.2givenF1basealgebra

The reflection formula holds at $m = 0$ because $B_{0} (x) = 1$ by [F1], so $B_{0} (1 - x) = B_{0} (x) = (- 1)^{0} B_{0} (x)$ .

2.1givenF1F3step 1.1algebra

For $m \geq 2$ , [F3] and step 1.1 applied to $B_{m - 1}$ give $B_{m} (1) - B_{m} (0) = ι (m) \int_{0}^{1} B_{m - 1} = 0$ . Thus the endpoints agree for every such $m$ .

2.2givenA1F2F3F4F6F7step 1.1IHinductionalgebra

Let $m \geq 1$ and assume $B_{m - 1} (1 - x) = (- 1)^{m - 1} B_{m - 1} (x)$ for all $x$ . By [F2], [F4], and step 1.1, the derivative of $D_{m} (x) : = (- 1)^{m} B_{m} (1 - x) - B_{m} (x)$ is $D_{m}^{'} (x) = (- 1)^{m + 1} ι (m) B_{m - 1} (1 - x) - ι (m) B_{m - 1} (x) = 0$ , using the induction hypothesis. Hence $D_{m}$ is constant by [F3]; its integral on $[0, 1]$ is zero because step 1.1 gives $\int_{0}^{1} B_{m} = 0$ and [F6] gives $\int_{0}^{1} B_{m} (1 - x) d x = \int_{0}^{1} B_{m} (x) d x$ . A constant with zero integral on an interval of length one is zero, so reflection holds at $m$ .

3.1A1F7step 1.2step 2.2induction

The base case in step 1.2 and the induction step in step 2.2 prove the reflection identity for every natural $m$ in [A1] by [F7].

4.1givenA1F5step 2.1step 3.1inductiondischarge-inductionalgebra∎

For the index $m$ in [A1], if it is odd and at least $3$ , step 3.1 gives $B_{m} (1) = - B_{m} (0)$ while step 2.1 gives $B_{m} (1) = B_{m} (0)$ ; therefore $B_{m} (0) = 0$ . Finally each $B_{m}$ is a polynomial and hence continuous, so [F5] bounds it on the compact interval $[0, 1]$ .

Depends on

Used by

Finite Euler–Maclaurin formula for the composite trapezoid Theorem

Dependency tree · two levels

69 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

L. Ridgway Scott, Numerical Analysis, 2nd ed., §13.3 (standard reference, not scraped)
NIST Digital Library of Mathematical Functions, Chapter 24 §24.2(i), Bernoulli Numbers and Polynomials (standard reference, not scraped)