# Solution of Recurrence relation to find a series expression

조회 수: 32(최근 30일)
MINATI 7 Jan 2020
편집: MINATI 7 Jan 2020
syms x k r f(x) g(x) a b beta b1 M L
syms F(k) G(k)
F(0)=0;F(1)=1;F(2)=a/2;G(0)=0;G(1)=1/2;G(2)=b/2;b1=1/beta;
%%%%dnf=diff(f,x,n)
d1f=(k+1)*F(k+1);d2f=(k+1)*(k+2)*F(k+2);d3f=(k+1)*(k+2)*(k+3)*F(k+3);
d1g=(k+1)*G(k+1);d2g=(k+1)*(k+2)*G(k+2);d3g=(k+1)*(k+2)*(k+3)*G(k+3);
fd2f=symsum(((k-r+1)*(k-r+2)*F(r)*F(k-r+2)),r,0,k);%%% f*d2f
gd2g=symsum((k-r+1)*(k-r+2)*G(r)*G(k-r+2),r,0,k);fd2g=symsum((k-r+1)*(k-r+2)*F(r)*G(k-r+2),r,0,k);
gd2f=symsum((k-r+1)*(k-r+2)*G(r)*F(k-r+2),r,0,k); d1fd1f=symsum((k-r+1)*(r+1)*F(r+1)*F(k-r+1),r,0,k); %%(d1f)^2
d1gd1g=symsum((k-r+1)*(r+1)*G(r+1)*G(k-r+1),r,0,k);
%%%%%%%
eqn1=simplify((1+b1)*d3f-d1fd1f+fd2f+gd2f-(M+L)*d1f==0);
eqn2=simplify((1+b1)*d3g-d1gd1g+fd2g+gd2g-(M+L)*d1g);eqns=[eqn1 eqn2];
solve([eqns,{F(k+3),G(k+2)}])
f=sum(x^k*F(k),k,0,inf);g=sum(x^k*G(k),k,0,inf);
%%%%%%%%%
Using the above code (ofcourse after modification), I want to solve the recurrence relations {F(k+3),G(k+2)} which contains series expression
and using given condition (F(0)=0;F(1)=1;F(2)=a/2;G(0)=0;G(1)=1/2;G(2)=b/2;) to find f and g (SERIES FORM)
OR
the attched pdf (similar problem) can be followed
Thanks

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