# 数学代写|组合学代写Combinatorics代考|Power sums of arithmetic progressions

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## 数学代写|组合学代写Combinatorics代考|Power sums of arithmetic progressions

The Stirling numbers can be used to calculate sums of powers of consecutive integer numbers. Let us first recall the well-known formulas from elementary mathematics:
$$1+2+3+\cdots+n=\frac{1}{2}\left(n^2+n\right)$$
The sum of the squares of the first $n$ numbers is
$$1^2+2^2+3^2+\cdots+n^2=\frac{1}{6}\left(2 n^3+3 n^2+n\right) .$$
For the third powers we have that
$$1^3+2^3+3^3+\cdots+n^3=\frac{1}{4}\left(n^4+2 n^3+n^2\right) .$$
We can easily recognize a rule: the sums of the powers of the first $n$ positive integers can be expressed as polynomials of $n$. It would be interesting to know what are the coefficients of these polynomials in general. The sums on the left-hand sides will be called power sums.
The Stirling numbers pop up in this problem via the relation (2.46):
$$x^p=\sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} x^{k}$$
On the left-hand side there is the $p$ th power of a real number $x$. Sum on $x$ from 1 to $n-1$. (We shall see later that summing up to $n-1$ in place of $n$ makes the resulting expressions somewhat simpler looking.)
$$1^p+2^p+\cdots+(n-1)^p=\sum_{x=1}^{n-1} \sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} x^{k} .$$
If we could find a simple polynomial expression for the right-hand side we would be ready. What we immediately see is that we should know the sums of the falling factorials, since
$$\sum_{x=1}^{n-1} \sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} x^{k}=\sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} \sum_{x=1}^{n-1} x^{k}$$

## 数学代写|组合学代写Combinatorics代考|The Bernoulli numbers

In the previous section, we have reached our aim and have found an expression for the power sum polynomials. We can go even further because it is possible to express the coefficients of the power sum polynomials in another way using the Bernoulli numbers. These numbers are important in many areas of mathematics and even in physics.

The $B_n$ Bernoulli ${ }^3$ numbers are usually defined by the exponential generating function
$$\sum_{n=0}^{\infty} B_n \frac{x^n}{n !}=\frac{x}{e^x-1}$$

The Bell numbers are also denoted by $B_n$, but the Bell numbers and Bernoulli numbers will not appear together in this book so there will be no risk of confusion during using the notation $B_n$.

We are going to connect our power sum formula to the Bernoulli numbers. We begin with the exponential generating function of
$$\sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} \frac{1}{k+1} \overline{\left[\begin{array}{c} k+1 \ m \end{array}\right]}$$
First determine the generating function of
$$\frac{1}{k+1} \overline{\left[\begin{array}{c} k+1 \ m \end{array}\right]}$$
$$x \sum_{k=0}^{\infty} \frac{1}{k+1} \overline{\left[\begin{array}{c} k+1 \ m \end{array}\right]} \frac{x^k}{k !}=\sum_{k=0}^{\infty} \overline{\left[\begin{array}{c} k+1 \ m \end{array}\right]} \frac{x^{k+1}}{(k+1) !}=\sum_{k=1}^{\infty} \overline{\left[\begin{array}{l} k \ m \end{array}\right]} \frac{x^k}{k !}=\frac{\ln ^m(1+x)}{m !}$$
from which one can infer
$$\sum_{k=0}^{\infty} \frac{1}{k+1} \overline{\left[\begin{array}{c} k+1 \ m \end{array}\right]} \frac{x^k}{k !}=\frac{1}{x} \frac{\ln ^m(1+x)}{m !}$$

# 组合学代考

## 数学代写|组合学代写Combinatorics代考|Power sums of arithmetic progressions

$$1+2+3+\cdots+n=\frac{1}{2}\left(n^2+n\right)$$

$$1^2+2^2+3^2+\cdots+n^2=\frac{1}{6}\left(2 n^3+3 n^2+n\right)$$

$$1^3+2^3+3^3+\cdots+n^3=\frac{1}{4}\left(n^4+2 n^3+n^2\right) .$$

## 数学代写|组合学代写Combinatorics代考|The Bernoulli numbers

$$\sum_{n=0}^{\infty} B_n \frac{x^n}{n !}=\frac{x}{e^x-1}$$

$$\frac{1}{k+1} \overline{[k+1 m]}$$
$$x \sum_{k=0}^{\infty} \frac{1}{k+1} \overline{[k+1 m]} \frac{x^k}{k !}=\sum_{k=0}^{\infty} \overline{[k+1 m]} \frac{x^{k+1}}{(k+1) !}$$从中可以推断
$$\sum_{k=0}^{\infty} \frac{1}{k+1} \overline{[k+1 m]} \frac{x^k}{k !}=\frac{1}{x} \frac{\ln ^m(1+x)}{m !}$$

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