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Prove properties of the Bernoulli polynomials

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).

5 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.

  1. 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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