Contents
The demos for our text book
we tried fern, floatgui, and walker.
ncmgui
Golden ratio phi (1)
notice 1) display format of number, 2) three methods to calculate phi.
phi= (1+sqrt(5))/2 format long phi 1/phi p= [1 -1 -1]; r= roots(p) r= solve('x-1= 1/x') phi= r(1) vpa(phi, 50) % too many digits phi= double(phi) f= @(x) x-1-1/x; figure(2); ezplot(f, [0, 4]) phi= fzero(f, 1) hold on plot(phi, 0, 'o') plot([0, 4], [0, 0])
phi =
1597/987
phi =
1.618033988749895
ans =
0.618033988749895
r =
-0.618033988749895
1.618033988749895
r =
5^(1/2)/2 + 1/2
1/2 - 5^(1/2)/2
phi =
5^(1/2)/2 + 1/2
ans =
1.6180339887498948482045868343656381177203091798058
phi =
1.618033988749895
phi =
1.618033988749895
Golden ratio phi (2)
下面是显示黄金矩形的程序，它是黄金比例名字的由来。 注意看goldrect.m中的plot, text及axis等有关命令。
edit goldrect.m
figure(3)
goldrect
Golden ratio phi (3)
黄金比例也可以由连分数（取无穷项）得到。 goldfract.m是显示有限项连分数的程序，随着项数n增加误差越小吗？
edit goldrect.m goldfract(38) goldfract(39) goldfract(91) % 试试n=91看看，从goldfract.m源码中找原因
p =
1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1))))))))))))))))))))))))))))))))))))))
p =
102334155/63245986
p =
1.618033988749895
err =
2.2204e-16
p =
1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1)))))))))))))))))))))))))))))))))))))))
p =
165580141/102334155
p =
1.618033988749895
err =
0
p =
1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
p =
1.220016e+19/7540113804746346496
p =
1.618033933694774
err =
5.5055e-08
斐波拉契数
看两个函数fibonacci.m和fibnum.m, 后者是递归程序。 递归程序简洁，但运行时间长。 最后，看看相邻斐波拉契数的比值
edit fibonacci.m edit fibnum.m tic, fibonacci(24), toc tic, fibnum(24), toc tic, fibnum(30), toc f= fibonacci(40); f(2:40)./f(1:39) %注意./
ans =
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711
28657
46368
75025
Elapsed time is 0.000195 seconds.
ans =
75025
Elapsed time is 0.832392 seconds.
ans =
1346269
Elapsed time is 11.187684 seconds.
ans =
2.0000
1.5000
1.6667
1.6000
1.6250
1.6154
1.6190
1.6176
1.6182
1.6180
1.6181
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
1.6180
分形蕨
看看复杂的绘图程序，图片文件的读取与显示。
fern edit fern.m F= imread('fern.png'); figure; image(F)
3阶幻方矩阵
熟悉矩阵的一些计算命令
A= magic(3)
sum(A)
sum(A')'
sum(diag(A))
sum(diag(flipud(A)))
det(A)
X= inv(A)
format rat; X
rot90(A,1)
norm(A)
eig(A)
A =
8 1 6
3 5 7
4 9 2
ans =
15 15 15
ans =
15
15
15
ans =
15
ans =
15
ans =
-360
X =
0.1472 -0.1444 0.0639
-0.0611 0.0222 0.1056
-0.0194 0.1889 -0.1028
X =
53/360 -13/90 23/360
-11/180 1/45 19/180
-7/360 17/90 -37/360
ans =
6 7 2
1 5 9
8 3 4
ans =
15
ans =
15
4801/980
-4801/980
4阶幻方矩阵
figure load durer image(X) colormap(map) axis image A= magic(4) inv(A) rank(A) % 更高阶幻方矩阵的秩的规律，看课本
A =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
ans =
* * * *
* * * *
* * * *
* * * *
ans =
3
