Solution of Recurrence relation to find a series expression
3 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
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
0 commentaires
Réponses (0)
Voir également
Catégories
En savoir plus sur Calculus dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!