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My answer is not matching with attached file
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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 Commenti
Walter Roberson
il 14 Gen 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.
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