[Combinatorics/Computer Mathematics] Homework Chapter 3 Recursion Relations

1

Draw on the plane$n$ infinite straight lines, each of which intersects at different points. The number of infinite regions they constitute is recorded as$f(n)$，求$f\left(n\right)$Complete feedthrough system .
$f(1)=2,f(2)=4,f(3)=6$
$$\{\begin{array}{l}f(n)=f(n-1)+2\\ f\left(1\right)=2\end{array}$$
Special expedition process:$x_1-1=0$，即$x_1=1$，1是$P(x)=Single\; root\; of\; 0$.
Special solution: $an{1}^n=an$. Substitution system: $an=a(n-1)+2$得$a=2$。
通解为：$c_1+c_2\times 1^n$，ie$f\left(n\right)=c_1+c_2+2n$，以$f\left(1\right)=2$, because$c_1=c_2=0$.
Transfer system: $f\left(n\right)=2n$

2

$In\; the\; n$ ternary number, the number of sequences in which 1 does not appear to the right of any 2 is recorded as$f\left(n\right)$，求$f\left(n\right)$'s recommendation system< /span>

记$f(n-1)$为$n-A\; sequence\; that\; satisfies\; the\; condition\; in\; 1$ digit ternary number. When introducing a new number at the leftmost position: based on $n-For\; a\; qualified\; sequence\; of\; 1$ bits, additional 0s and 1s can be introduced without affecting subsequent sequences. There is$2f(n-1)$ situations. If 2 is forcibly introduced on the left, then only 0 and 2 can be selected later, there are $2^{n-1}$different circumstances. Note:$f\left(n\right)=2f(n-1)+2^{n-1}$

初值：$f\left(1\right)=3,f\left(2\right)=8$。

Use high school students’ method to homogenize the recurrence expression:
$$\begin{gathered} f\left(n\right)=2f(n-1)+2^{n-1} \\ 2f(n)=4f(n-1)+2^n \\ 2f(n-1)=4f(n-2)+2^{n-1} \end{gathered}$$
Type 1-Type 3 arrangement:
$$f(n)-4f(n-1)+4f(n-2)=0$$
Special expedition route:
$$x^2-4x+4=0$$
Obtained double root: $x_1=x_2=2$
则$f\left(n\right)=(b_1+b_{2}n)2^n$
将$f(1)=3,f(2)=8$Assignment, solution$b_1=1,b_2=0.5$

即：$f\left(n\right)=(1+n/2)2^n$

3

$In\; the\; n$ digit quaternary number, the number of sequences in which 2 and 3 appear an even number of times is recorded as$f(n)$，求$The\; recurrence\; relation\; satisfied\; by\; f\left(n\right)$ .
Let the number that satisfies the condition be $f(n)$
则对满足条件的$f(n-1)$, No. 1 new intake 0 or 1, Yes$2f(n-1)$species.
记$g\left(n\right)$ means there is an even number of 2 An odd number of 3's. Because of symmetry, $g\left(n\right)$ can also mean that there are odd numbers An even number of 2s and a sequence of 3s.
Note $h\left(n\right)$even number 2 sequence quantity.
Present$n-In\; the\; 1$ quaternary number, there is an even number of sequences of 2 plus a non-2 number, totaling$3h(n-1)$Sequence.
Present$n-In\; 1$ quaternary number, there is an odd number of 2 sequences with a 2 added, and there are$4^{n-1}-h(n-1)$Sequence.
$h(n)=3h(n-1)+4^{n-1}-h(n-1)=4^{n-1}+2h(n-1)$，$h(1)=3$
The number of sequences with an even number of 2 minus the number of sequences with an even number of 2 and odd number of 3 is:
$f\left(n\right)=h\left(n\right)-g\left(n\right)$
Even number length $(n-1)$ 0/1, yes$2f(n-1)$ species. One of the numbers 2 and 3 is odd and the other is even, there is $2g(n-1)$种
$f(n)=2f(n-1)+2g(n-1)$

求解$h(n)=2h(n-1)+4^{n-1}$: Special expedition$x-2=0$，$x=2$ is a double root.
This is a non-homogeneous linear equation.
The special solution is: $b_0\cdot 4^n$. General substitution$h\left(n\right)=2h(n-1)+4^{n-1}$，得$b_0\cdot 4^n=2b_0\cdot 4^{n-1}+4^{n-1}$，解得$b_0=\frac{1}{2}$
The general solution of the homogeneous recurrence relation is $c_1\cdot 2^n$
$h\left(n\right)=c_1\cdot 2^n+\frac{1}{2}\cdot 4^n$, first assignment$h(1)=3$，解得$c_1=\frac{1}{2}$
即：$h\left(n\right)=\frac{1}{2}\cdot 2^n+\frac{1}{2}\cdot 4^n$

General$h(n)$代入
$$\{\begin{array}{l}f\left(n\right)=h\left(n\right)-g(n)\\ f\left(n\right)=2f(n-1)+2g(n-1)\end{array}$$
Use the method of high school students:
$f(n-1)=h(n-1)-g(n-1)$，$2f(n-1)=2h(n-1)-2g(n-1)$
$f\left(n\right)=2h(n-1)-2g(n-1)+2g(n-1)=2h(n-1)=2^{n-1}+4^{n-1}$

即：$f\left(n\right)=2^{n-1}+4^{n-1}$

4

Find the value that satisfies the difference between adjacent bits being 0$n$Number of binary sequences$f\left(n\right)$
Last first place 1 , any second place in the number, yes$f(n-1)$ species.
The last digit is 0, the second to last digit is only 1, and the third to last digit is arbitrary, there are $f(n-2)$种。
初值：$f(1)=2,f(2)=3,f(3)=5,f(4)=8$
$$\{\begin{array}{l}f(n)=f(n-1)+f(n-2)\\ f\left(1\right)=2,f\left(2\right)=3\end{array}$$
This is a constant coefficient linear homogeneous recurrence relationship, and the characteristic equation is $x^2-x-1=0$，Special expedition base$x_1=\frac{1+\sqrt{5}}{2},x_2=\frac{1-\sqrt{5}}{2}$, the general solution is$f\left(n\right)=c_1(\frac{1+\sqrt{5}}{2}{)}^n+c_2(\frac{1-\sqrt{5}}{2}{)}^n$
Root$f\left(1\right)=2,f\left(2\right)=3$, 反推$f(0)=1$,General$0,1$代入$f(n)$
$$\{\begin{array}{l}{c}_1+c_2=1\\ \frac{1+\sqrt{5}}{2}{c}_1+\frac{1-\sqrt{5}}{2}{c}_2=2\end{array}$$
解得：$c_1=\frac{\sqrt{5}+3}{2\sqrt{5}},c_2=\frac{\sqrt{5}-3}{2\sqrt{5}}$

$f\left(n\right)=\frac{\sqrt{5}+3}{2\sqrt{5}}(\frac{1+\sqrt{5}}{2}{)}^n+\frac{\sqrt{5}-3}{2\sqrt{5}}(\frac{1-\sqrt{5}}{2}{)}^n$

This sequence is similar to the Fibonacci sequence (the recursive formula is exactly the same as the Fibonacci sequence, but the initial values ​​are different) and it looks like a lot$\sqrt{5}$, but it can actually be converted into integers.

8

The length of the three letters a, b, c is transmitted on the channel: $A\; string\; of\; n$, if there are two $If\; a$ appears continuously, the channel cannot transmit. Let$f\left(n\right)$Display credit path available as well as long为$The\; number\; of\; strings\; in\; n$, find$f\left(n\right)$Complete feedthrough system .

$f\left(1\right)=3$,$f(2)$可为ab,ac,ba,bb,bc,ca,cb,cc,$f\left(2\right)=8$
令$f\left(n\right)$为长为$The\; number\; of\; strings\; in\; n$. When the leftmost character is $b$or$When\; c$, the remaining characters are arbitrary. Total$2f(n-1)$ species. When the leftmost character is a, the second character is b or c, and the remaining characters are arbitrary. Total$2f(n-2)$species. Note:$f(n)=2f(n-1)+2f(n-2)$
This is a homogeneous recurrence relationship with constant coefficients, and the characteristic equation is $x^2-2x-2=0$,Special base$x_1=1+\sqrt{3},x_2=1-\sqrt{3}$
Root$f(1)$和$f\left(2\right)$Anti-inference$f(0)=1$
$$\{\begin{array}{l}{c}_1+c_2=1\\ (1+\sqrt{3})c_1+(1-\sqrt{3})c_2=3\end{array}$$
解得$c_1=\frac{1}{2}+\frac{\sqrt{3}}{3},c_2=\frac{1}{2}-\frac{\sqrt{3}}{3}$

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

9(1)

See 8

9(2)

$$\{\begin{array}{l}f\left(n\right)=4f(n-1)-4f(n-2)\\ f\left(0\right)=1,f(1)=3\end{array}$$

Special expedition$x^2-4x+4=0$，解得$x_1=x_2=2$, double root$x=2$
$f\left(n\right)=(b_1+b_{2}n)2^n$
代入
$$\{\begin{array}{l}1=(b_1)2^0\\ 3=(b_1+b_2)2^1\end{array}$$
解得$b_1=1,b_2=\frac{1}{2}$
即：$f\left(n\right)=(1+\frac{1}{2}n)2^n$

9(3)

$$\{\begin{array}{l}f\left(n\right)=-f(n-1)+3f(n-2)+5f(n-3)+2f(n-4)\\ f\left(0\right)=1,f\left(1\right)=0,f(2)=1,f(3)=2\end{array}$$

Special expedition course$x^4+x^3-3x^2-5x-2=0$,解得$x_1=x_2=x_3=-1,x_4=2$
inside, $-1$This is a triple root,$2$This is 1 layer root.
$f(n)=c_1(-1{)}^n+c_2(-1{)}^{n}n+c_3(-1{)}^n{n}^2+c_4{2}^n$
代入初值$f(0)=1,f(1)=0,f(2)=1,f(3)=2$
$$\{\begin{array}{l}{c}_1+c_4=1\\ -c_1-c_2-c_3+2c_4=0\\ c_1+2c_2+4c_3+4c_4=1\\ -c_1-3c_2-9c_3+8c_4=2\end{array}$$
解得$c_1=\frac{7}{9},c_2=-\frac{1}{3},c_3=0,c_4=\frac{2}{9}$

$f\left(n\right)=\frac{7}{9}(-1{)}^n-\frac{1}{3}(-1{)}^{n}n+\frac{2}{9}{2}^n$

9(4)

$$\{\begin{array}{l}f\left(n\right)-4f(n-1)+4f(n-2)=n\cdot 2^n\\ f\left(0\right)=0,f\left(1\right)=1\end{array}$$

Homogeneous characteristic equation$x^2-4x+4=0$,$x=2$ is the 2-fold root of the characteristic equation.

Then the non-homogeneous special solution is$n^2(b_{1}n+b_0)2^n$. General 1,0,-1,-2 assignment special solution:

$f\left(1\right)=2(b_1+b_0),f\left(0\right)=0,f(-1)=\frac{-b_1+b_0}{2},f(-2)=(-2b_1+b_0)$

Substitute 0 and 1 into the original equation:
$$\{\begin{array}{l}0-4\times \frac{-b_1+b_0}{2}+4(-2b_1+b_0)=0\\ 2(b_1+b_0)-0+4\frac{-b_1+b_0}{2}=2\end{array}$$
解得$b_0=\frac{1}{2},b_1=\frac{1}{6}$, then the non-homogeneous special solution is$n^2(\frac{1}{6}n+\frac{1}{2})2^n$
The general solution of the homogeneous equation is: $(c_{1}n+c_0)2^n$.
Add the general solution and the specific solution: $f(n)=(c_{1}n+c_0)2^n+n^2(\frac{1}{6}n+\frac{1}{2})2^n$

Substitute the general solution into the original equation:
$$\{\begin{array}{l}{c}_0{2}^0+0=0\\ (c_1+c_0)\times 2^1+1(\frac{1}{6}+\frac{1}{2})\times 2=1\end{array}$$
解得$c_0=0,c_1=-\frac{1}{6}$

$f\left(n\right)=(c_{1}n+c_0)2^n+n^2(\frac{1}{6}n+\frac{1}{2})2^n$

9(5)

$$\{\begin{array}{l}f\left(n\right)=nf(n-1)+n!(n\ge 1)\\ f(0)=2\end{array}$$

Use an iterative method (a high school student method from beginning to end).

Two connections$n!$
$$\frac{f(n)}{n!}=\frac{f(n-1)}{(n-1)!}+1$$
设$g(n)=\frac{f\left(n\right)}{n!}$

$g\left(n\right)=g(n-1)+1$，$g(0)=2$，则$g\left(n\right)=n+2$，则$\frac{f\left(n\right)}{n!}=n+2$，

$f\left(n\right)=(n+2)n!$

9(6)

$$\{\begin{array}{l}f\left(n\right)=(n+2)f(n-1)(n\ge 1)\\ f\left(0\right)=1\end{array}$$

Use the iterative method (the method used by high school students from beginning to end).
$$\begin{gathered} f\left(n\right)=(n+2)f(n-1)=(n+2)(n+1)f(n-2) \\ =... \\ =(n+2)(n+1)...(n+i)f(n+i-3) \\ =(n+2)(n+1)...(3)f(0) \\ =(n+2)!/2 \end{gathered}$$

11

设有$n$ elliptic curves, any two intersect at two points, any three elliptic curves do not intersect at one point. Ask this question: $How\; many\; parts\; do\; n$ ellipses divide the plane into?
Note$n-1$ ellipses divide the plane into $f(n-1)$ parts, then the new ellipse will pass through the original $n-1$Place of arrival$2(n-1)$ new planar areas.
$$\begin{gathered} f(n)=f(n-1)+2(n-1) \\ f(1)=2,f(2)=4 \end{gathered}$$

Solution: Use the method of high school students

$$\begin{gathered} f\left(n\right)=f(n-1)+2(n-1) \\ =f(n-2)+2(n-2)+2(n-1) \\ =f(n-i)+2(n-i)...2(n-1) \\ =f\left(1\right)+2+4+...2(n-1) \\ =2+2\frac{(1+n-1)(n-1)}{2} \\ =2+n(n-1) \end{gathered}$$

12

request$n$The number of positive decimal integers with an even number of 5's
$f\left(1\right)=8,f\left(2\right)=81$,记$f(n)$为$The\; number\; of\; even\; 5\; occurrences\; in\; n$ decimal number.
$9\times f(n-1)$ 电影$f(n-1)$An even number of 5 appears, but the newly introduced lowest bit is not 5.
$9\times 10^{n-2}-f(n-1)$Before display$n-An\; odd\; number\; of\; 5s\; appear\; in\; the\; 1$ bit, and 5 is introduced in the lowest bit. $9\times 10^{n-2}$Display: 对于一个$n-1$ digit number, middle $n-The\; 2$ bit is optional (0 to 9, a total of 10 selection methods), but the highest bit cannot be 0 (1 to 9, a total of 9 selection methods). Therefore, $9\times 10^{n-2}-f(n-1)$display$n-The\; total\; number\; of\; 1$ digits is subtracted from the number of even numbers of 5, leaving an odd number of 5s.

$f(n)=9\times f(n-1)+9\times 10^{n-2}-f(n-1)=9\times 10^{n-2}+8f(n-1)$

Using the high school method,
$f(n+1)=9\times 10^{n-1}+8f(n)$
$10f(n)=9\times 10^{n-1}+80f(n-1)$

Identity:$f\left(n\right)=18f(n-1)-80f(n-2)$
Special expedition course:$x^2-18x+80=0$，解得$x_1=10,x_2=8$
则$f\left(n\right)=c_1\times 10^n+c_2\times 8^n$

Initial value: Because 0 is a 0-digit number, there are 8 types of 1-digit numbers, 1/2/3/4/6/7/8/9, $f(1)=8$; in 2nd place,$\overline{1x}$/$\overline{2x}$/$\overline{3x}$/$\overline{4x}$/$\overline{6x}$/$\overline{7x}$/$\overline{8x}$/$\overline{9x}$有$8\times 9=72$, and another 55, a total of 73 with an even number of 5s. $f(2)=73$
$$\{\begin{array}{l}10c_1+8c_2=8\\ 100c_1+64c_2=73\end{array}$$
解得$c_1=\frac{9}{20},c_2=\frac{7}{16}$

则$f\left(n\right)=\frac{9}{20}\times 10^n+\frac{7}{16}\times 8^n$