Example: Bernoulli Polynomials and Sums of Powers

Axiom provides operations for computing definite and indefinite sums. You can compute the sum of the first ten fourth powers by evaluating this. This creates a list whose entries are m^4 as m ranges from 1 to 10, and then computes the sum of the entries of that list.

You can also compute a formula for the sum of the first k fourth powers, where k is an unspecified positive integer.

This formula is valid for any positive integer k. For instance, if we replace k by 10, we obtain the number we computed earlier.

You can compute a formula for the sum of the first k nth powers in a similar fashion. Just replace the 4 in the definition of sum4 by any expression not involving k. Axiom computes these formulas using Bernoulli polynomials; we use the rest of this section to describe this method. First consider this function of t and x.

Since the expressions involved get quite large, we tell Axiom to show us only terms of degree up to 5.

If we look at the Taylor expansion of f(x,t) about t=0, we see that the coefficients of the powers of t are polynomials in x.

In fact, the nth coefficient in this series is essentiall the nth Bernoulli polynomial: the nth coefficient of the series is 1/n!*Bn(x), where Bn(x) is the nth Bernoulli polynomial. Thus, to obtain the nth Bernoulli polynomial, we multiply the nth coefficient of the series ff by n!. For example, the sixth Bernoulli polynomial is this.

We derive some properties of the function f(x,t). First we compute f(x+1,t)-f(x-t).

If we normalize g, we see that it has a particularly simple form.

From this it follows that the nth coefficient in the Taylor expansion of g(x,t) at t=0 is 1/(n-1)!*x^(n-1). If you want to check this, evaluate the next expression.

However, since

      g(x,t)=f(x+1,t)-f(x,t)

it follows that the nth coefficient is

       1/n! * (Bn(x+1) - Bn(x))

Equating coefficients, we see that

       1/(n-1)! * x^(n-1) = 1/n! * (Bn(x+1) - Bn(x))

and, therefore

       x^(n-1) = 1/n * (Bn(x+1) - Bn(x))

Let's apply this formula repeatedly, letting x vary between two integers a and b, with a<b:

           a^(n-1) = 1/n * (Bn(a+1) - Bn(a))
       (a+1)^(n-1) = 1/n * (Bn(a+2) - Bn(a+1))
       (a+2)^(n-1) = 1/n * (Bn(a+3) - Bn(a+2))
                   .
                   .
       (b-1)^(n-1) = 1/n * (Bn(b)   - Bn(b-1))
           b^(n-1) = 1/n * (Bn(b+1) - Bn(b))

When we add these equations we find that the sum of the left-hand sides is

       sum(m=a..b,m^(n-1))

the sum of the (n-1)-st powers from a to b. The sum of the right-hand sides is a "telescoping series". After cancellation, the sum is simply

       1/n*(Bn(b+1)-Bn(a))

Replacing n by n+1, we have shown that

       sum(m=a..b,m^n) = 1/(n+1)*(B<n+1>(b+1)-B<n+1>(a))

Let's use this to obtain the formula for the sum of fourth powers. First we obtain the Bernoulli polynomial B5.

To find the sum of the first k 4th powers, we multiply 1/5 by B5(k+1)-B5(1)

This is the same formula that we obtained via sum(m^4,m=1..k)

At this point you may want to do the same computation, but with an exponent other than 4. For example, you might try to find a formula for the sum of the first k 20th powers.