Simple application of generating function (Derivation of general term formula of Cattelan number and Fibonacci sequence)

Solving Recurrence Relations Using Generating Functions

simple example

Generating function is a kind of infinite sequence $f_0,f_1,f_2,f_3,...$ Representation of coefficients expressed as a power sequence:

$F(x)=f_0+f_{1}x+f_2{x}^2+f_3{x}^3+...$

For example, the sequence $1,1,1,...$ , we use the generating function to express:

$G(x)=1+x+x^2+x^3+...$

We can calculate its simple expression:

$\begin{array}{r}G(x)=1+x+x^2+x^3+...\\ -xG(x)=-x-x^2-x^3-...\end{array}\}\Rightarrow G\left(x\right)-xG(x)=1\Rightarrow G\left(x\right)=\sum _{i=0}^{\infty }{x}^i=\frac{1}{1-x}$

Another example is the sequence of positive integers $1,2,3,4,...$ , expressed as a generating function:

$N(x)=1+2x\_+3x{\_}^2+4x{\_}^3+...+(n+1)x^n$

We use the same method to obtain its simple expression:

$N\left(x\right)-xN(x)=1+x+x^2+x^3+...=G(x)=\frac{1}{1-x}$

therefore:

$N\left(x\right)=\sum _{i=0}^{\infty }(i+1)x^i=\frac{1}{(1-x{)}^2}$

Commonly used generating functions

$\frac{1}{1-ax}=\sum _{i=0}^{\infty }{a}^i{x}^i=1+ax+a^{2x}{\_}^2+a^{3x}{\_}^3+...$

$\frac{1}{1-x^r}=\sum _{i=0}^{\infty }{x}^{ri}=1+x^r+x^{2r}+x^{3r}+...$

$\frac{1}{(1-x{)}^n}=\sum _{i=0}^{\infty }\left(\begin{array}{c}n+i-1\\ i\end{array}\right)x^i=1+\left(\begin{array}{c}n\\ 1\end{array}\right)x+\left(\begin{array}{c}n+1\\ 2\end{array}\right)x^2+\left(\begin{array}{c}n+2\\ 3\end{array}\right)x^3+...$

arithmetic progression

Generating functions can be used to find general term formulas for sequences of numbers. For example, $a_0$as the first item, $d$ is the tolerance arithmetic sequence$a_0,a_1,a_2,a_3,...$ . Let's first write its generating function:

$A(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+...$

Then misplaced and subtracted:

$A\left(x\right)-xA\left(x\right)=a_0+d(x+x^2+x^3+...)=a_0+\frac{xd}{1-x}=a_0-d+\frac{d}{1-x}$

so:

$A(x)=\frac{a_0-d}{1-x}+\frac{d}{(1-x{)}^2}=\sum _{i=0}^{\infty }((a_0-d)+(i+1)d)x^i=\sum _{i=0}^{\infty }(a_0+id)x^i$

From this we can derive the general term formula of the arithmetic sequence as:

$a_{n+1}=a_0+nd$

Fibonacci sequence

The general term formula of the Fibonacci sequence can also be obtained by this method. We know the recurrence relationship of the Fibonacci sequence:

$f_0=0$

$f_1=1$

$f_n=f_{n-1}+f_{n-2},n⩽2$

Its generating function is denoted as:

$F(x)=f_0+f_{1}x+f_2{x}^2+f_3{x}^3+...+f_n{x}^n+...$

Dislocation subtraction:

$F\left(x\right)-xF(x)-x^{2}F(x)$

$=f_0+(f_1-f_0)x+(f_2-f_1-f_0)x^2+...+(f_n-f_{n-1}-f_{n-2})x^n+...$

$=0+1x+0x{\_}^2+...+0x{\_}^n+...$

$=x$

then:

$F(x)=\frac{x}{1-x-x^2}$

This is a partial fraction. We know that partial fractions can be decomposed. The theorem is as follows:

Let $p\left(x\right)$ is the degree less thannnPolynomials of$n$$a_1,a_2,...,a_n$are different non-zero numbers, then there are constants $c_1,c_2,...,c_n$make

$\frac{p(x)}{(1-a_{1}x)(1-a_{2}x)...(1-a_{n}x)}=\frac{c_1}{1-a_{1}x}+\frac{c_2}{1-a_{2}x}+...+\frac{c_n}{1-a_{n}x}$

For $\frac{x}{1-x-x^2}$, we assume it can be decomposed into $\frac{c_1}{1-a_{1}x}+\frac{c_2}{1-a_{2}x}$。由$1-x-x^2=(1-a_{1}x)(1-a_{2}x)$ , can solve$a_1=\frac{1+\sqrt{5}}{2},a_2=\frac{1-\sqrt{5}}{2}$。又由$\frac{x}{1-x-x^2}=\frac{c_1}{1-a_{1}x}+\frac{c_2}{1-a_{2}x}$The general score is $x=c_1(1-a_{2}x)+c_2(1-a_{1}x)$ . Substitute into$a_1$and $a_2$Get $c_1=\frac{1}{\sqrt{5}},c_2=-\frac{1}{\sqrt{5}}$。所以$\frac{x}{1-x-x^2}=\frac{1}{\sqrt{5}}(\frac{1}{1-a_{1}x}-\frac{1}{1-a_{2}x})$。

Root generalized generating function $\frac{1}{1-ax}=\sum _{i=0}^{\infty }{a}^i{x}^i=1+ax+a^{2x}{\_}^2+a^{3x}{\_}^3+...$ , we can write the above fraction as:

$\frac{x}{1-x-x^2}=\frac{1}{\sqrt{5}}((1+a_{1}x+a_1^2{x}^2+a_1^3{x}^3+...)-(1+a_{2}x+a_2^2{x}^2+a_2^3{x}^3+...))=\sum _{i=0}^{\infty }\frac{a_1^i-a_2^i}{\sqrt{5}}{x}^i$

So we solve the general term formula of the Fibonacci sequence as:

$f_n=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2}{)}^n-(\frac{1-\sqrt{5}}{2}{)}^n)$

Other examples of linear recurrence with constant coefficients

Next, we use generating functions to solve a practical mathematical problem:

Topic: For numbers from $0\sim A\; code\; string\; composed\; of\; 9$ , it must contain an even number of 0, the length isnnHow many code strings of$n\; are\; there?$

answer:

We set the length to be $The\; code\; string\; of\; n$ has$a_n$indivual.

when $n=1$ , the code string "$1$ ", code string "$2$ ", ..., code string "$9$ "are symbolic conditions. So$a_1=9$。

when $n=When\; k$ , the legal code string has$a_k$, add a non-zero number behind it, and get a legal code string; the illegal code string has $10^k-a_k$, add a 0 behind it to get a legal code string. Then we get the recursive relation:

$a_{k+1}=9a{\_}_k+(10^k-a_k)=8a_k+10^k$

Our order $G(x)=\sum _{i=0}^{\infty }{a}_i{x}^i$ generates a function for it, then there are:

$G\left(x\right)-1$

$=\sum _{i=1}^{\infty }{a}_i{x}^i$

$=\sum _{i=0}^{\infty }{a}_{i+1}{x}^{i+1}$

$=\sum _{i=0}^{\infty }8a_i{x}^{i+1}+\sum _{i=0}^{\infty }10^i{x}^{i+1}$

$=8x\sum _{i=0}^{\infty }{a}_i{x}^i+x\sum _{i=0}^{\infty }10^i{x}^i$

$=8xG(x)+\frac{x}{1-10x}$

From this we get:

$G(x)=\frac{1}{2}(\frac{1}{1-8x\_\_\_}+\frac{1}{1-10x})$

so:

$G\left(x\right)=\sum _{i=0}^{\infty }\frac{8^i+10^i}{2}{x}^i$

$a_n$The general formula for is:

$a_n=\frac{1}{2}(8^n+10^n)$

Cattelan number

We can use generating functions to solve some problems with more complex recurrence relations. For example, the Catalan number we will discuss below. We introduce it from a practical problem.

Railway platforms are often designed as a last in first out (LIFO) stack structure. As shown below:

------------------------------------------------------------


<-出站               <------+     +---------         <-进站
                            |    |
                            |    | 
------------------------+   |    |   +----------------------
                        |   |    |   |
                        |   |    V   |
                        |            |
                        |            |
                        |     站     |
                        |            |
                        |     台     |
                        |            |

When the trains numbered 1, 2, 3, and 4 enter the station in sequence (only entering and leaving), the situation on the platform is as follows:

--------------------------------------------------------------


<-出站               <------+        +---------       <-进站
                            |       |
                            |       | 
------------------------+   |       |   +-------------------
                        |   |       |   |
                        |   |  站   V   |
                        |      台       |
                        |               |
                        | +-----------+ |
                        | |           | |
                        | |     4     | |
                        | |           | |
                        | +-----------+ |
                        |               |
                        | +-----------+ |
                        | |           | |
                        | |     3     | |
                        | |           | |
                        | +-----------+ |
                        |               |
                        | +-----------+ |
                        | |           | |
                        | |     2     | |
                        | |           | |
                        | +-----------+ |
                        |               |
                        | +-----------+ |
                        | |           | |
                        | |     1     | |
                        | |           | |
                        | +-----------+ |

Now these trains will leave the station, obviously only the train 4 closest to the outside of the platform can leave the station. Other trains can only leave the station after train 4 leaves the station. Because train 4 is the last to enter the station and must be the first to exit, the platform model is a last-in-first-out stack structure.

now have $n$ trains, numbered$1,2,...,n$ . They enter an empty platform in sequence (for$i<j$ , train$i$ must be greater than train$j$ advances to the station), and then leaves the station (as long as there is a train in the station, the train closest to the station can choose to leave the station, even at this time$n$ trains have not all entered the station). For example, when dispatching a train for the first time, the station is empty at this time, and no train can leave the station, so only train$1$ pit stop. When dispatching trains for the second time, you can choose train$1$ out of the station, you can also choose the train$2$ pit stops. If the second dispatch selects train$2$ enters the station, then when dispatching the train for the third time, you can choose train$2$ out of the station, you can also choose the train$3$ pit stops. And so on.

We define a sequence called "outbound sequence", which is obtained by arranging the numbers of outbound trains in the order of outbound trains. With $n=3$ o'clock for example. First let 3 trains enter the station in sequence, and then let the 3 trains in the station leave the station in sequence, then the order of the trains leaving the station is: train$3$ , train$2$ , train$1$ . The outbound sequence for this scheduling method is$321$ . Another example is this scheduling: train$1$ into the station, train$2$ into the station, train$2$ exits, train$3$ into the station, train$3$ exits, train$1$ outbound. Its outbound sequence is$231$ . In fact, each time the train is dispatched, either the train with the smallest number among the trains that have not entered the station is selected to enter the station, or the train with the largest number in the station is selected to leave the station. So we describe the scheduling process like this: inbound, inbound, outbound, inbound, outbound, outbound. This also determines the outbound sequence as$231$。

Now we ask the question: for $How\; many\; different\; outbound\; sequences\; are\; there\; for\; n$ incoming trains?

Answer: We remember that for $There\; are\; n$ trains to enter the station, a total of$c_n$different outbound sequences.

Obviously $c_1=1$。

For $1⩽k⩽n$ , we assume train$k$ is the last to leave the station. Because the train with the largest number in the station leaves the station every time it leaves the station, so in the train$Before\; k$ enters the station, all trains with numbers smaller than it have left the station. For this$k-1$ car, with$c_{k-1}$different outbound sequences. And for numbers than kkFor trains with large$k$$Stop\; after\; k$ . And because they will be on the train$i$ leaves the station before leaving the station, so these trains have$c_{n-k}$outbound sequence. We might as well let $c_0=1$ , then there is a recursive formula:

$c_n=\sum _{k=1}^n{c}_{k-1}{c}_{n-k}=\sum _{k=0}^{n-1}{c}_k{c}_{n-k-1}$

This $c_n$It is the Cattelan number. Now we want to use its recursive formula to find its general term formula.

We remember its generating function as $C(x)=\sum _{i=0}^{\infty }{c}_i{x}^i$ , and square it to get:

$C^2(x)$

$=c_0^2+(c_0{c}_1+c_1{c}_0)x+(c_0{c}_2+c_1{c}_1+c_2{c}_0)x^2+...$

$=\sum _{i=0}^{\infty }(\sum _{k=0}^i{c}_k{c}_{i-k})x^i$

$=\sum _{i=0}^{\infty }{c}_{i+1}{x}^i$

Dislocation subtraction:

$C(x)-x{C}^2(x)=c_0=1$

Solving the equation gives:

$C\left(x\right)=\frac{1\pm \sqrt{1-4x}}{2x\_}$

We are now going to drop a root. From C ( x ) C(x)The definition of$C$$($$x$$)$$\underset{x\to 0}{\lim }C(x)=c_0=1$

而$\underset{x\to 0}{\lim }C\left(x\right)=\underset{x\to 0}{\lim }\frac{1\pm \sqrt{1-4x}}{2x\_}=\underset{x\to 0}{\lim }\frac{2}{1\mp \sqrt{1-4x}}$

So discard the solution with positive sign and get $C(x)$：

$C\left(x\right)=\frac{1-\sqrt{1-4x}}{2x\_}$

Next continue to calculate $C\left(x\right)$ , we need to use the binomial theorem:

$(1+x{)}^a=1+\sum _{i=1}^{\infty }\left(\begin{array}{r}a\\ i\end{array}\right)x^i,a\in R$

Let $a=\frac{1}{2}$, substitute into the above formula to calculate $C(x)$：

$C(x)$

$=\frac{1-\sqrt{1-4x}}{2x\_}$

$=\frac{1}{2x\_}(1-(1+\sum _{i=1}^{\infty }\left(\begin{array}{c}\frac{1}{2}\\ i\end{array}\right)(-4x)\_{}^i))$

$=2\sum _{i=1}^{\infty }\left(\begin{array}{c}\frac{1}{2}\\ i\end{array}\right)(-4x)\_{}^{i-1}$

$=2\sum _{i=1}^{\infty }\frac{\frac{1}{2}\cdot (-\frac{1}{2})\cdot (-\frac{3}{2})\cdot ...\cdot (\frac{3}{2}-i)}{i!}(-4x)\_{}^{i-1}$

$=1+\sum _{i=2}^{\infty }\frac{\frac{1}{2}\cdot \frac{3}{2}\cdot ...\cdot (i-\frac{3}{2})}{i!}(4x)\_{}^{i-1}$

$=1+\sum _{i=2}^{\infty }\frac{(2i-3)!}{i!}(2x)\_{}^{i-1}$

$=1+\sum _{i=2}^{\infty }\frac{(2i-3)!}{i!}\cdot \frac{(2i-2)!}{(i-1)!}{x}^{i-1}$

$=1+\sum _{i=2}^{\infty }\frac{(2i-2)}{i!(i-1)!}{x}^{i-1}$

$=\sum _{i=1}^{\infty }\frac{(2i-2)}{i!(i-1)!}{x}^{i-1}$

$=\sum _{i=0}^{\infty }\frac{(2i)}{(i+1)!i!}{x}^i$

$=\sum _{i=0}^{\infty }\frac{\left(\begin{array}{c}2i\_\\ i\end{array}\right)}{i+1}{x}^i$

所以$c_n=\frac{\left(\begin{array}{c}2n\\ n\end{array}\right)}{n+1}$, the outbound sequence has $\frac{\left(\begin{array}{c}2n\\ n\end{array}\right)}{n+1}$kind.

$c_n=\frac{\left(\begin{array}{c}2n\\ n\end{array}\right)}{n+1}$, which is the general term formula of Cattelan number.

reference article

Part of Cattelan numbers: Deriving the general term formula of Cattelan numbers with generating functions