Question: How do I obtain Taylors series expansion of fractional differential equation?

Good day everyone.

Please can help me with this code on the taylor series expansion involving Fractional Differential Equation (FDE)? Particularlly, the lines highlighted blue and green respectively in relation to FDE.

Thank you and kind regards

#k=2
restart:q:=n*h:
P:=sum((a[k]*x^(k))/GAMMA(k+1-alpha), k=0..3):
assume(alpha>0,alpha < 1):
Q:=fracdiff(P,x,alpha):
e1:=simplify(eval(P, x=q+h))=y[n+1]:
e2:=simplify(eval(Q,x=q))=f[n]:
e3:=simplify(eval(Q,x=q+h))=f[n+1]:
e4:=simplify(eval(Q,x=q+2*h))=f[n+2]:
var:=seq(a[i], i=0..3):
M:=e||(1..4):

Cc:=eval(<var>, solve(eval({M}),{var}) ):
for i from 1 to 4 do
	a[i-1]:=Cc[i]:
end do:
Cf:=P:
E:=collect(Cf, [y[n+1], f[n], f[n+1],f[n+2]], recursive):
print():
s2:=y[n+2]=simplify(eval(Cf, x=q+2*h)):
collect(%, [y[n+1], f[n], f[n+1],f[n+2]], recursive):

s1:=y[n]=simplify(eval(Cf, x=q)):
collect(%, [y[n+1], f[n], f[n+1],f[n+2]], recursive):

Y[n+1]:=convert(taylor(y(x+h),h=0,12),polynom):
F[n]:=convert(taylor((D(y))(x), h = 0,12), polynom):
F[n+1]:=convert(taylor((D(y))(x+h), h = 0,12), polynom):
F[n+2]:=convert(taylor((D(y))(x+2*h), h = 0,12), polynom):


W:=asympt(expand(eval(rhs(s2),[y[n+1]=Y[n+1],f[n]=F[n],f[n+1]=F[n+1],f[n+2]=F[n+2]])),h,6);
X:=convert(taylor(y(x+2*h),h=0,12),polynom)-W;
lte:=convert(asympt(X,h,8),polynom);

 

 

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