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My answer is not matching with attached file

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MINATI
MINATI 12 Jan 2020
댓글: MINATI 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
  댓글 수: 8
MINATI
MINATI 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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