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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 个评论
Walter Roberson
2020-1-14
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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