iteration using for loop is not working

a(1)=0;
b(1)=1;
c(1)=0;


n=1;

for n=1:5
    
    
    a(2)=0.5*(a(1)-b(1));
    b(2)=0.5*((a(1))-(b(1))-(a(1)*c(1)));
    c(2)=0.5*((a(1)*b(1)-c(1)));
    
    a(1)=a(2);
    b(1)=b(2);
    c(1)=c(2);
    
    n=n+1;
    
    
    
    
    
    
    
end

disp(a(2))

how to run this program to give from old value to new value using iteration a(2) giving more than five value

Answer 1

[Image Analyst]

Try it like this:

numElements = 5;
a = zeros(1, numElements);
b = ones(1, numElements);
c = zeros(1, numElements);
for n = 2 :  numElements
    a(n) = 0.5*(a(n-1)-b(n-1));
    b(n) = 0.5*((a(n-1))-(b(n-1))-(a(n-1)*c(n-1)));
    c(n) = 0.5*((a(n-1)*b(n-1)-c(n-1)));
end
a

a = 1×5

0 -0.5000 0 0 0

b

b = 1×5

1.0000 -0.5000 0 0 0

c

c = 1×5

0 0 0.1250 -0.0625 0.0312

[shiv gaur]

why the initial value is not included

[Torsten]

It is included. a and c are initialized as zeros, b is initialized as ones.

This especially means a(1)=c(1)=0 and b(1)=1.

[shiv gaur]

if this work why this eq is not working

numElements = 20;
a = zeros(1, numElements);
b = ones(1, numElements);
c = zeros(1, numElements);
for n = 2 :  numElements
    a(n) = (1/n)*10*(a(n-1)-b(n-1));
    b(n) = (1/n)*((28*a(n-1))-(b(n-1))-(a(n-1)*c(n-1)));
    c(n) = (1/n)*((a(n-1)*b(n-1)-3.3*c(n-1)));
end
a
b
c

1.0e+199 *

Columns 1 through 11

0.0000 -0.0000 -0.0000 -0.0000 -0.0000 0.0000 0.0000 0.0000 -0.0000 -0.0000 -0.0000

Columns 12 through 20

0.0000 0.0000 2.0928 -Inf NaN NaN NaN NaN NaN

c =

1.0e+199 *

Columns 1 through 11

0 0 0.0000 0.0000 -0.0000 -0.0000 0.0000 -0.0000 -0.0000 0.0000 -0.0000

Columns 12 through 20

-0.0000 0.0000 -3.0100 -Inf Inf NaN NaN NaN NaN

>>

[Torsten]

It is working, but the recursion gives high values for a, b and c for numElements being a large number.

[shiv gaur]

this is the high value is not possible 10^199

[Torsten]

Then your recursion is wrong.

[shiv gaur]

this is your last problem you answer but value is not right

[Image Analyst]

We did not evaluate the "correctness" of your formula. I simply used it as is. If your formula is wrong, then correct it. I really have no idea what the "next" value might be based on the prior values, so all I can do is to assume you wrote down the formulas correctly. If you didn't, only you know what the correct formulas should be, not us.

[Walter Roberson]

numElements = 20;
a = sym(zeros(1, numElements));
b = sym(ones(1, numElements));
c = sym(zeros(1, numElements));
for n = 2 :  numElements
    a(n) = (1/sym(n))*10*(a(n-1)-b(n-1));
    b(n) = (1/sym(n))*((28*a(n-1))-(b(n-1))-(a(n-1)*c(n-1)));
    c(n) = (1/sym(n))*((a(n-1)*b(n-1)-3.3*c(n-1)));
end
a

a = 

b

b = 

c

c = 

digits(10)
vpa(a)

ans = 

vpa(b)

ans = 

vpa(c)

ans = 

[Walter Roberson]

numElements = 10;
syms a [1 numElements]
syms b [1 numElements]
syms c [1 numElements]
for n = 2 :  numElements
    a(n) = (1/sym(n))*10*(a(n-1)-b(n-1));
    b(n) = (1/sym(n))*((28*a(n-1))-(b(n-1))-(a(n-1)*c(n-1)));
    c(n) = (1/sym(n))*((a(n-1)*b(n-1)-3.3*c(n-1)));
end

Ea = expand(a);
Ca  = vpa(findSymType(Ea, 'number'),5);
[~, idx] = sort(abs(Ca), 'desc');
format long g
double(reshape(Ca(idx(1:10)),[],1))

ans = 10×1

1.0e+00 * 2.14539347488294e+24 2.1388215038686e+24 -1.97683468274523e+24 -1.66919049501844e+24 1.57009953308725e+24 -1.53553980415839e+24 1.25698231464319e+24 -1.12800014100977e+24 -1.03005272271465e+24 1.02716449593526e+24

So by the time you get to 10, you are working with coefficients as large as 2*10^24

Ca0 = vpa(findSymType(expand(subs(Ea, [sym('a1'), sym('a3')], [0,0])),'number'),5);
[~, idx] = sort(abs(Ca0), 'desc');
double(reshape(Ca0(1:10), [], 1))

ans = 10×1

1.0e+00 * -9.35288582447091e+18 -6.04096409393064e+18 -1.99711770767288e+18 -1.57887744010591e+18 -1.23085737821287e+18 -3.24385496525373e+17 -2.04394672654123e+17 -5.32546405724897e+16 -4.96159656693842e+16 -4.67790559622922e+16

And after you take into account that a and c start out as 0, you are still working with coefficients in the 10^18 range by the time you get to numElements = 10.

It isn't MATLAB messing up: your formulas are not working out for b as large as 1. If b were less than 1, then you might get very different results.