My answer is not matching with attached file

syms k r  
a=sym('a'); b = sym('b');L=sym('L'); M = sym('M');  b1 = sym('b1');
m=7;  F = sym(zeros(m,1));   F(1)=0;  F(2)=1;    F(3)=a;                
G = sym(zeros(m,1));   G(1)=0;  G(2)=1/2;  G(3)=b;            
for k=1:7
    for r = 1:k
         F3 = F(1)+ F(2)+F(3);    G3 = G(1)+G(2)+G(3);
F(k+3)= ( F3+sum((r+1)*F(r+1)*(k-r+1)*F(k-r+1)) - sum((k-r+1)*(k-r+2)*F(k-r+1)*(F(r)+G(r)))+ (M+L)*(k)*F(k+1))/((1+b1)*(k+1)*(k+2)*(k));
G(k+3) = (G3+ sum((r+1)*G(r+1)*(k-r+1)*G(k-r+1)) - sum((k-r+1)*(k-r+2)*G(k-r+2)*(F(r)+G(r))) + (M+L)*(k)*G(k+1))/((1+b1)*(k+1)*(k+2)*(k));
    end
end
% %%%%%
for N=1:6
disp(F(N))
disp(G(N))
end
f=sum(x^k*F(k),k,0,7)
g=sum(x^k*G(k),k,0,7)

%%%%%%%

Any reply will be greatly appreciated

After getting F(N) and G(N), I neeed to find then f and g

[Walter Roberson]

x is not defined.

That syntax for sum() does not exist.

F is not a function, so F(0) cannot be accessed.

My guess is

syms x
f = sum(x.^(0:7) .* F(1:8));
g = sum(x.^(0:7) .* G(1:8));

Why are you overwriting all of F and G in every iteration of the double nested loop?

[MINATI]

Dear Walter Roberson

All F's and G's are different

i.e, like F(r), F(k-r) and F(k-r+1) are different.(Its a recurrence relation) and same for G also.

But the values I am getting for F(4), F(5),F(6) and G(4, G(5),G(6) etc. are not matching with the actual values (In ATTACHED pdf)

[Walter Roberson]

Look at your code.

for k=1:7
  for r = 1:k
       F3 = F(1)+ F(2)+F(3);    G3 = G(1)+G(2)+G(3);
F(k+3)= ( F3+sum((r+1)*F(r+1)*(k-r+1)*F(k-r+1)) - sum((k-r+1)*(k-r+2)*F(k-r+1)*(F(r)+G(r)))+ (M+L)*(k)*F(k+1))/((1+b1)*(k+1)*(k+2)*(k));
G(k+3) = (G3+ sum((r+1)*G(r+1)*(k-r+1)*G(k-r+1)) - sum((k-r+1)*(k-r+2)*G(k-r+2)*(F(r)+G(r))) + (M+L)*(k)*G(k+1))/((1+b1)*(k+1)*(k+2)*(k));
  end
end

How many times do you set F3 and G3 to the same value?

For any given value of k how often do you write overtop of F(k+3) ?

[MINATI]

If I use

for k=1:7
    for r = 1:k
F(k+3)= ( F(1)+ F(2)+F(3)+sum((r+1)*F(r+1)*(k-r+1)*F(k-r+1)) - sum((k-r+1)*(k-r+2)*F(k-r+1)*(F(r)+G(r)))...
   + (M+L)*(k)*F(k+1))/((1+b1)*(k+1)*(k+2)*(k));
G(k+3) = (G(1)+G(2)+G(3)+ sum((r+1)*G(r+1)*(k-r+1)*G(k-r+1)) - sum((k-r+1)*(k-r+2)*G(k-r+2)*(F(r)+G(r)))...
    + (M+L)*(k)*G(k+1))/((1+b1)*(k+1)*(k+2)*(k));
    end
end

%%%THEN

F(4)=(L + M + a + 1)/(6*b1 + 6)  

and

G(4)=(L/2 + M/2 + b + 1/2)/(6*b1 + 6)

WHICH is not matching

[Walter Roberson]

Okay, now, suppose we have a look at when k = 4, and substituting that value into the code:

    for r = 1:4
F(4+3)= ( F(1)+ F(2)+F(3)+sum((r+1)*F(r+1)*(4-r+1)*F(4-r+1)) - sum((4-r+1)*(4-r+2)*F(4-r+1)*(F(r)+G(r)))...
   + (M+L)*(4)*F(4+1))/((1+b1)*(4+1)*(4+2)*(4));
G(4+3) = (G(1)+G(2)+G(3)+ sum((r+1)*G(r+1)*(4-r+1)*G(4-r+1)) - sum((4-r+1)*(4-r+2)*G(4-r+2)*(F(r)+G(r)))...
    + (M+L)*(4)*G(k+1))/((1+b1)*(4+1)*(4+2)*(4));
    end
    

Now, how many times do you write to F(4+3) ? If you write more than once to F(4+3) then is in the form of adding to a previous result, along the lines of

value = 0;
for r = 1 : 4
    value = value + something involving r
end

Or similarly is it developing the value iteratively, something along the lines of

value = 1;
for r = 1 : 4
    value = some_function(value);
end

or are you instead just assigning new values without reference to the result already stored in the location?

[MINATI]

okay

Thanks for your patience

So what next to modify?

[Walter Roberson]

You have not posted the recurrence formula, so we are restricted to pointing out parts of the code that look suspicious, without being able to make any suggestions as to what code would work.

[MINATI]

for reference

Eqn (3.5) - (3.8) are used to find eqns (3.9)& (3.10) in the attached pdf (GOOD.pdf).