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Prove properties of the Bernoulli polynomials

by RoRi

The Bernoulli polynomials are defined by

    \[ P_0(x) = 1; \qquad P'_n(x) = n P_{n-1} (x) \qquad \text{and} \qquad \int_0^1 P_n (x) \, dx = 0 \qquad \text{if } n \geq 1. \]

  1. Find explicit formulas for the first Bernoulli polynomials in the cases n = 1,2,3,4,5.
  2. Use mathematical induction to prove that P_n(x) is a degree n polynomial in x, where the degree n term is x^n.
  3. For n \geq 2 prove that P_n (0) = P_n (1).
  4. For n \geq 1 prove that

        \[ P_n (x+1) - P_n (x) = nx^{n-1}. \]

  5. Prove that

        \[ \sum_{r=1}^{k-1} r^n = \int_0^k P_n (x) \, dx = \frac{P_{n+1}(k) - P_{n+1}(0)}{n+1} \]

    for n \geq 2.

  6. Prove that for n \geq 1,

        \[ P_n (1-x) = (-1)^n P_n (x). \]

  7. Prove that for n \geq 1,

        \[ P_{2n+1} (0) = 0 \qquad \text{and} \qquad P_{2n-1}\left( \frac{1}{2} \right) = 0. \]

  1. We start with the initial condition P_0(x) = 1. This gives us

        \[ P'_1 (x) = 1 \cdot P_0(x) = 1 \quad \implies \quad P_1 (x) = \int dx = x + C_1. \]

    Now, using the integral condition to find C_1,

        \[ \int_0^1 (x+C_1) \, dx = 0 \quad \implies \quad \left( \frac{1}{2}x^2 + C_1 x \right) \Bigr \rvert_0^1 = 0 \quad \implies \quad C_1 &= -\frac{1}{2}. \]

    Thus,

        \[ P_1 (x) = x -\frac{1}{2}. \]

    Next, using this expression for P_1(x) we have

        \[ P'_2 (x) = 2 \cdot \left( x- \frac{1}{2} \right) = 2x - 1 \quad \implies \quad P_2 (x) = x^2 - x + C_2. \]

    Using the integral condition to find C_2,

        \[ \int_0^1 (x^2 - x + C_2) \, dx = 0 \quad \implies \quad \frac{1}{3} - \frac{1}{2} + C_2 = 0 \quad \implies \quad C_2 = \frac{1}{6}. \]

    Thus,

        \[ P_2 (x) = x^2 - x + \frac{1}{6}. \]

    Next, using this expression for P_2 (x) we have

        \[ P'_3 (x) = 3 \cdot \left( x^2 - x + \frac{1}{6} \right) = 3x^2 - 3x + \frac{1}{2} \quad \implies \quad P_3(x) = x^3 - \frac{3}{2}x^2 + \frac{1}{2}x + C_3. \]

    Using the integral condition to find C_3,

        \[ \int_0^1 \left( x^3 - \frac{3}{2}x^2 + \frac{1}{2}x + C_3\right) \, dx = 0 \quad \implies \quad \frac{1}{4} - \frac{1}{2} + \frac{1}{4} + C_3 = 0 \quad \implies \quad C_3 = 0. \]

    Thus,

        \[ P_3 (x) = x^3 - \frac{3}{2}x^2 + \frac{1}{2} x. \]

    Next, using this expression for P_3 (x) we have

        \[ P'_4(x) = 4 \cdot \left( x^3 - \frac{3}{2}x^2 + \frac{1}{2}x \right) = 4x^3 - 6x^2 + 2x \quad \implies \quad P_4 (x) = x^4 - 2x^3 + x^2 + C_4. \]

    Using the integral condition to find C_4,

        \[ \int_0^1 \left( x^4 - 2x^3 + x^2 + C_4 \right) \, dx = 0 \quad \implies \quad \frac{1}{5} - \frac{1}{2} + \frac{1}{3} + C_4 = 0 \quad \implies \quad C_4 = \frac{1}{30}. \]

    Thus,

        \[ P_4 (x) = x^4 - 2x^3 + x^2 - \frac{1}{30}. \]

    Finally, using this expression for P_4 (x) we have

        \[ P'_5(x) = 5 \cdot \left( x^4 - 2x^3 + x^2 - \frac{1}{30} \right) = 5x^4 - 10x^3 + 5x^2 - \frac{1}{6} \quad \implies \quad P_5 (x) = x^5 - \frac{5}{2} x^4 + \frac{5}{3}x^3 - \frac{1}{6}x + C_5. \]

    Using the integral condition to find C_5,

        \[ \int_0^1 \left( x^5 - \frac{5}{2} x^4 + \frac{5}[3} x^3 - \frac{1}{6} x + C_5 \right) \, dx = \frac{1}{6} - \frac{1}{2} + \frac{5}{12} - \frac{1}{12} + C_5 = 0 \quad \implies \quad C_5 = 0. \]

    Thus,

        \[ P_5 (x) = x^5 - \frac{5}{2}x^4 + \frac{5}{3}x^3 - \frac{1}{6}x. \]

  2. Proof. We have shown in part (a) that this statement is true for n = 0, 1, \ldots, 5. Assume then that the statement is true for some positive integer m, i.e.,

        \[ P_m (x) = x^m + \sum_{k=0}^{m-1} c_k x^k. \]

    Then, by the definition of the Bernoulli polynomials we have,

        \[ P_{m+1}' (x) = (m+1) \cdot \left( x^m + \sum_{k=0}^{m-1} a_k x^k \right) = (m+1)x^m + \sum_{k=0}^{m-1} b_k x^k, \]

    where b_k = (m+1)a_k for k = 1, \ldots, m-1. Then, taking the integral of this expression

        \[ P_{m+1} (x) = \int \left( (m+1)x^m + \sum_{k=0}^{m-1} b_k x^k \right) \, dx = x^{m+1} + \sum_{k=0}^m \frac{b_k}{k+1} x^{k+1}. \]

    Hence, the statement is true for the case m+1; hence, for all positive integers n. \qquad \blacksquare

  3. Proof. From the integral property in the definition of the Bernoulli polynomials we know for n \geq 1,

        \[ \int_0^1 P_n (x) \, dx = 0 \quad \implies \quad \int_0^1 (n+1) P_n (x) \, dx = 0. \]

    Then, using the first part of the definition we have P'_{n+1} (x) = (n+1) P_n (x); therefore,

        \[ 0 = \int_0^1 (n+1) P_n (x) \, dx = \int_0^1 P'_{n+1} (x) \, dx = P_{n+1}(1) - P_{n+1}(0). \]

    Thus, we indeed have

        \[ P_{n+1} (1) = P_{n+1}(0). \qquad \blacksquare \]

  4. Proof. The proof is by induction. For the case n = 1 we have

        \[ P_1 (x) = x - \frac{1}{2}, \qquad \text{and} \qquad P_1 (x+1) = x + \frac{1}{2}. \]

    Therefore,

        \[ P_1 (x+1) - P_1 (x) = 1. \]

    Since nx^{n-1} = 1 \cdot x^0 = 1, the stated difference equation holds for n =1. Assume then that the statement holds for some positive integer m. Then by the fundamental theorem of calculus, we have

        \[ P_{m+1} (x) = \int_0^x P'_{m+1}(t) \, dt = (m+1) \int_0^x P_m (t) \, dt. \]

    Therefore,

        \begin{align*} P_{m+1}(x+1) - P_{m+1}(x) &= (m+1) \left( \int_0^{x+1} P_m (t) \, dt - \int_0^x P_m (t) \, dt \right) \\[9pt] &= (m+1) \left( \int_0^1 P_m (t) \, dt + \int_1^{x+1} P_m (t) \, dt - \int_0^x P_m (t) \, dt \right) \\[9pt] &= (m+1) \left( 0 + \int_0^x P_m (t+1) \, dt - \int_0^x P_m (t) \, dt \right) &( \text{Integral condition})\\[9pt] &= (m+1) \left( \int_0^x (P_m (t+1) - P_m (t)) \, dt\right) \\[9pt] &= (m+1) \left( \int_0^x mt^{m-1} \, dt \right)&(\text{Induction Hypothesis})\\[9pt] &= (m+1) x^m. \end{align*}

    Hence, the statement is true for the case m+1, and so it is true for all positive integers n. \qquad \blacksquare

  5. Proof. (Let’s assume Apostol means for k to be some positive integer.) First, we use the definition of the Bernoulli polynomials to compute the integral,

        \begin{align*} \int_0^k P_n (x) \, dx &= \int_0^k \frac{1}{n+1} P'_{n+1} (x) \, dx \\[9pt] &= \frac{P_{n+1}(k) - P_{n+1}(0)}{n+1}. \end{align*}

    Now, we want to express the numerator as a telescoping sum and use part (d),

        \begin{align*} P_{n+1}(k) - P_{n+1}(0) &= \sum_{r = 1}^{k-1} \left( P_{n+1}(r + 1) - P_{n+1}(r) \right) \\[9pt] &= \sum_{r=0}^{k-1} \left( (n+1)r^n \right) \\ &= (n+1) \sum_{r=1}^{k-1} r^n. \end{align*}

    Thus, we indeed have

        \[ \sum_{r=1}^{k-1} r^n = \int_0^k P_n (x) \, dx = \frac{P_{n+1}(k) - P_{n+1}(0)}{n+1}. \qquad \blacksquare\]

  6. Proof.

Incomplete. I’ll try to fix parts (f) and (g) soon(ish).

7 comments

    • Mihajlo says:

      Actually, while it is probably doable to do f) by induction (e.g. to prove that later odd Bernoulli numbers are 0), the other suggested solution here looks more appropriate.

    • Anonymous says:

      The proof of part (d) is wrong since he assumes that p_m+1(0)=0 which is clearly not true (in general) from part a

    • Mohammad Azad says:

      In part d, he assumes that P_m+1(0)=0 which is clearly wrong from part a, but he got lucky since it got cancelled when he took the difference P_m+1(x+1) – P_m+1(x)

  2. nu creation says:

    For part (f), I found a way to do it online. First, prove that if some polynomials {Qn} satisfy the three conditions above then the polynomial Qn(x) must equal Pn(x) for every n. Then define polynomial Qn(x) := (-1)^n Pn(1-x) and show that this polynomial satisfies the three conditions above. Part (g) follows from part (f) and part (c).

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