## Monday Math 13

Consider the sum of the m-th powers of the first n integers:

Suppose that the sum is a polynomial in n: . Show that  is an m+1 degree polynomial with a leading coefficient of .

Consider the integral . We know that the value of this integral is
.
However, we can of course express the integral as the limit of a Riemann Sum (Riemann Integral). If we divide the interval into n equally sized partitions, and choose the right side of the partition for the height, then the k-th partition has width  and height . Then the Riemann Sum is:



But the limit of this sum as  is the integral, and so:
. For  a polynomial in n, this means the highest power term in n must be , and so  is an m+1 degree polynomial with a leading coefficient of .

We can also see from the n=0 case that  and so the constant term in  must be zero. Further, for n=1, we have  and so the sum of the coefficients of  must be unity.

The first few of these polynomials are known to many:


.

These polynomials can be found by Faulhaber’s Formula, which gives them in terms of the Bernoulli numbers (or related Bernoulli polynomials).

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### 3 Responses to “Monday Math 13”

1. Monday Math 26 « Twisted One 151’s Weblog Says:

[…] any positive even integer. More specifically, , where the are the Bernoulli numbers, which I have mentioned earlier in the context of Faulhaber’s Formula. Possibly related posts: (automatically […]

2. Monday Math 56 « Twisted One 151’s Weblog Says:

[…] for the sum of the nth power of the first m integers in terms of the Bernoulli polynomials (see here). The values of the Bernoulli polynomials at x=0 (the constant term of the polynomial) are known […]

3. Monday Math 138 « Twisted One 151's Weblog Says:

[…] first proof is algebraic: we find an explicit expression in terms of n for both sides. So, we use Faulhaber’s formula. We recall and Thus, for the left-hand side, we have . Now, for the right-hand side, we have , […]