## 数学代写|组合学代写Combinatorics代考|CS586

2023年3月22日

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## 数学代写|组合学代写Combinatorics代考|The binomial transformation

As we noted during deducing (2.8), the binomial transform of a sequence $a_n$ is defined by
$$b_n=\sum_{k=0}^n\left(\begin{array}{l} n \ k \end{array}\right) a_k$$
We also know that if the exponential generating function of $a_n$ is $f(x)$, then the exponential generating function of $b_n$ is $e^x f(x)$.

We had actually met with a binomial transform at the very beginning. This was the recursion (1.1). This recursion says that for the Bell numbers the binomial transform is $b_n=B_{n+1}$. In other words, the binomial transformation acts on the Bell number sequence as an index shifting.

Let us take another example. The binomial theorem (see the Appendix) says that
$$(x+y)^n=\sum_{k=0}^n\left(\begin{array}{l} n \ k \end{array}\right) x^k y^{n-k} .$$
If here we substitute $y=1$ we have that
$$(x+1)^n=\sum_{k=0}^n\left(\begin{array}{l} n \ k \end{array}\right) x^k .$$
In this case $a_n=x^n$ and $b_n=(x+1)^n$. The binomial transform is naturally applicable on polynomial sequences.

We can invert the transform to get back $a_n$ from $b_n$ in the above example. We just need to substitute $x+1$ in place of $x$ and -1 in place of $y$ :
$$(x+1-1)^n=x^n=\sum_{k=0}^n\left(\begin{array}{l} n \ k \end{array}\right)(x+1)^k(-1)^{n-k} .$$
A similar approach works in general (the one line proof is seen below). If
$$b_n=\sum_{k=0}^n\left(\begin{array}{l} n \ k \end{array}\right) a_k$$
then
$$a_n=\sum_{k=0}^n\left(\begin{array}{l} n \ k \end{array}\right)(-1)^{n-k} b_k \text {. }$$

## 数学代写|组合学代写Combinatorics代考|Applications of the above techniques

We are now endowed with sufficient knowledge to turn back to our combinatorial numbers and study them from a new point of view. Many new identities can be deduced with the generating function technique.

In the first chapter, firstly we met with the Bell numbers. Recall (1.1) and (2.8). Based on these, and letting $f(x)$ be the generating function of $B_n$,
$$f(x) e^x=\sum_{n=0}^{\infty} \frac{x^n}{n !}\left(\sum_{k=0}^n\left(\begin{array}{l} n \ k \end{array}\right) B_k\right)=\sum_{n=0}^{\infty} B_{n+1} \frac{x^n}{n !}$$
We also know that the derivative of $f(x)$ gives the exponential generating function of $B_{n+1}$, that is,
$$f(x) e^x=f^{\prime}(x)$$
Which function can be $f(x)$ ? We can easily see that $f(x)=e^{e^x}$ is a suitable function satisfying the above differential equation. Equation (2.15) can be multiplied by any real number $c, c f(x)$ is also a solution. What distinguishes the actual exponential generating function of the Bell numbers in the whole class $c f(x)$ ? We know from the theory of generating functions that $f(0)$ is the zeroth coefficient in the series expansion of $f(x)$ (see (2.9)). The zeroth coefficient is $B_0=1$ in our case. Since $c e^{e^0}=c e$, we infer that $c=\frac{1}{e}$. We determined the exponential generating function of the Bell numbers:
$$\sum_{n=0}^{\infty} B_n \frac{x^n}{n !}=\frac{1}{e} e^{e^x}$$ This result has a nice and immediate consequence that shows why the generating functions are so useful. This consequence is the Dobiński formula.

# 组合学代考

## 数学代写|组合学代写Combinatorics代考|The binomial transformation

$$b_n=\sum_{k=0}^n(n k) a_k$$

$$(x+y)^n=\sum_{k=0}^n(n k) x^k y^{n-k}$$

$$(x+1)^n=\sum_{k=0}^n(n k) x^k \text {. }$$

$$(x+1-1)^n=x^n=\sum_{k=0}^n(n k)(x+1)^k(-1)^{n-k} \text {. }$$

$$b_n=\sum_{k=0}^n(n k) a_k$$

$$a_n=\sum_{k=0}^n(n k)(-1)^{n-k} b_k .$$

## 数学代写|组合学代写Combinatorics代考|Applications of the above techniques

$$f(x) e^x=\sum_{n=0}^{\infty} \frac{x^n}{n !}\left(\sum_{k=0}^n(n k) B_k\right)=\sum_{n=0}^{\infty} B_{n+1} \frac{x^n}{n !}$$

$$f(x) e^x=f^{\prime}(x)$$

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

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## MATLAB代写

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