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My answer is not matching with attached file

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MINATI
MINATI on 12 Jan 2020
Commented: MINATI on 14 Jan 2020
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

  8 Comments

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MINATI
MINATI on 14 Jan 2020
okay
Thanks for your patience
So what next to modify?
Walter Roberson
Walter Roberson on 14 Jan 2020
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
MINATI on 14 Jan 2020
for reference
Eqn (3.5) - (3.8) are used to find eqns (3.9)& (3.10) in the attached pdf (GOOD.pdf).

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