Bernoulli Polynomials &
the Sum-of-Powers Formulas

We used the formulas for the sums of the \(k\)th powers of the first \(n\) integers when \(k = 1, 2,\) and \(3\). Now we’re going to derive these formulas for any \(k\). These formulas were first published in 1713 by the Swiss mathematician James Bernoulli in his book Ars Conjectandi.

  1. The Bernoulli polynomials \(B_n\) are defined by \(B_0(x) = 1\), \(B'_n(x) = B_{n-1}(x)\), and \(\int_0^1 B_n(x) \,\mathrm{d}x = 0\) for \(n = 1, 2,3,\dotsc\). Find an explicit formula for \(B_n(x)\) for \(n = 1, 2, 3\), and \(4\).
  2. Use the Fundamental Theorem of Calculus to show that for \(n \geq 2\) we have \(B_n(0) = B_n(1).\)
  3. If we introduce the Bernoulli numbers \(b_n = n! B_n(0)\), then we can write \[ \begin{align*} B_0(x) &= b_0 \\B_1(x) &= \frac{1}{1!}x + \frac{b_1}{1!} \\B_2(x) &= \frac{1}{2!}x^2 + \frac{b_1}{1!1!}x + \frac{b_2}{2!} \\B_3(x) &= \frac{1}{3!}x^3 + \frac{b_1}{1!2!}x^2 + \frac{b_2}{2!1!}x + \frac{b_3}{3!} \end{align*} \] and in general \[ B_n(x) = \frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k} b_kx^{n-k} \quad\text{where}\quad \binom{n}{k} = \frac{n!}{k!(n-k)!} \,. \] Show that for \(n \geq 2\) that \[ b_n = \sum_{k=0}^{n} \binom{n}{k} b_k \] and therefore \[ b_{n-1} = -\frac{1}{n}\left( \binom{n}{0}b_0 + \binom{n}{1}b_1 + \dotsb + \binom{n}{n-2}b_{n-2} \right) \] This gives an efficient way of computing Bernoulli numbers and therefore the Bernoulli polynomials.
  4. Show that \(B_n(1-x) = (-1)^nB_n(x)\) and deduce that \(b_{2n+1} = 0\) for \(n \gt 1\).
  5. Calculate \(b_6\) and \(b_8\) explicitly, and then calculate the polynomials \(B_5\), \(B_6\), \(B_7\), \(B_8\), and \(B_9\).
  6. Graph the Bernoulli polynomials \(B_1\), \(B_2\), …, \(B_9\) for \(0 \leq x \leq 1\). What pattern do you notice in the graphs?
  7. Use mathematical induction to prove that \(B_{k+1}(x+1) - B_{k+1}(x) = x^k/k!\,.\)
  8. By putting \(x = 0,1,2,\dotsc,n\) in the previous formula, prove that \[\begin{align*} 1^k + 2^k + 3^k + \dotsb + n^k &= k! \left( B_{k+1}(n+1) - B_{k+1}(0) \right) \\&= k! \int\limits_{0}^{n+1} B_k(x) \,\mathrm{d}x \,. \end{align*} \]
  9. Show that the last formula can be written symbolically as \[ 1^k + 2^k + 3^k + \dotsb + n^k = \frac{1}{k+1}\left( (n+1+b)^{k+1} - b^{k+1} \right) \] where the expression \((n + 1 + b)^{k+1}\) is to be expanded formally using the Binomial Theorem and each power \(b^i\) is replaced by the Bernoulli number \(b_i\).
  10. Use this formula to find a closed form for \[1^4 + 2^4 + 3^4 + \dotsb + n^4 \quad\text{and}\quad 1^5 + 2^5 + 3^5 + \dotsb + n^5\] and generally for \(\sum_{i=1}^{n} i^k\) for any positive integer \(k\).