symbolic substitution error and convert

syms x
series1(x)=sym(zeros(1));
series2(x)=sym(zeros(1));
U=zeros(1,2,'sym');
syms  x m    z 
alpha=15; 
gamma=0.01;
U(1)=m;
U(2)=0;
A(1)=zeros(1,'sym');
for k=1
    U(k+2)=z;
    A(1)=0;
    for m=1:k
        A(1)=A(1)+U(m)*(k-m+1)*(k-m+2)*U(k-m+3);
    end
    B=k*(k+1)*U(k+2)+alpha*A(1)-gamma*U(k)
    D=simplify(solve(B,z))
    U(k+2)=D
end

B = 

D = 

U = 

disp(U(3))

for k=1:3
    series1(x)=simplify(series1(x)+U(k)*(power(x,k-1)));
end

series1

series1(x) = 

e1=subs(series1,x,1);
e=e1-1;
format long 
accuracy=input('enter the accuracy')

Error using input
Support for user input is required, which is not available on this platform.

f=e(x)
g=inline(f)
g=inline(f)
a=input('enter the ist approximation=')
b=input('enter the 2nd approximation=')
fa=feval(g,a)
fb=feval(g,b)
while fa*fb>0
    a=input('enter the ist approximation=')
    b=input('enter the 2nd approximation=')
    fa=feval(g,a)
    fb=feval(g,b)   
end
for i=1:50
    c=(a+b)/2;
    fc=feval(g,c);
    disp([i   a      fa     b       fb        c       fc     abs(b-a)])
    if fc==accuracy
        fprintf('the root of the equation is %f',c)
        break;
    elseif abs(b-a)<=accuracy
        fprintf('the root of the equation is %f',c) 
        break;
    elseif fa*fc<=0
        b=c;
        fb=fc;
    else  
        a=c;
        fa=fc;
    end
end
fprintf('the value of c=%f', c)
series2(x)=subs(series1,m,c)
toc

the synatx is correct but still the value of m is not substituted in

[Walter Roberson]

In this context, you should be avoiding inline() and feval() . You should either be proceeding symbolically, or else you should be using matlabFunction()

It is unlikely that you really want to input() inside the loop; your should be calculating a and b instead.

    if fc==accuracy

Your code is confused there. First off, you have used exact comparison of binary floating point numbers: those are unlikely to happen to match bit-for-bit. Secondly, suppose the exact root was -0.05 and accuracy was 0.10 and fc happened to come out exactly as 0.10, then you would compare the fc to the accuracy, find an exact match, and declare yourself satisfied, even though fc was not within 0.10 accuracy of -0.05 . You should just get rid of that branch of code; your next line about abs(b-a) is good enough.

[Walter Roberson]

We cannot test your code without knowing the exact values you are providing for the input() statements.

[yogeshwari patel]

0.99999 is the exact value . I am interseted in syntax subs . why the value is not replace by c

[Walter Roberson]

You are saying that you enter 0.99999 in response to

accuracy=input('enter the accuracy')

and as well in response to each of the

a=input('enter the ist approximation=')
b=input('enter the 2nd approximation=')

???

Answer 1

[Broy]

@yogeshwari patel, a possible reason the value of m is not being substituted at the end is that you are overwriting the symbolic variable m with a numeric loop counter inside your first for loop.

[Walter Roberson]

Oh, of course! Good catch !