abdulganiy

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These are questions asked by abdulganiy

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);

 

 

Can someone help with the simplification of the result of this code? I am sure the "qs" in the final result should not appear.

Thanking you in anticipation of your positive responses

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

Cc:=eval(<var>, solve(eval({M}),{var}) ):
for i from 1 to 3 do
	a[i-1]:=Cc[i]:
end do:
Cf:=P:
E:=collect(Cf, [y[n], f[n], f[n+1]], recursive):
print():
#y[n+1]=collect(simplify(simplify(expand(eval(Cf,x=q+h^alpha)),size)), [y[n],f[n],f[n+1]], factor);
y[n+1]=simplify(eval(Cf, x=q+h^alpha)):
collect(%, [y[n], f[n], f[n+1]], recursive);

 

Good day. Please can someone kindly help to reduce the result of this code. Thank you and kind regards

restart:
s:=(sum(a[j]*x^j,j=0..3)+sum(a[j]*exp(-(j-3)*x),j=4..7)):
F:=diff(s,x):
p1:=simplify(eval(s,x=q))=y[n]:
p2:=simplify(eval(F,x=q))=f[n]:
p3:=simplify(eval(F,x=q+h/3))=f[n+1/3]:
p4:=simplify(eval(F,x=q+h))=f[n+1]:
p5:=simplify(eval(F,x=q+5*h/3))=f[n+5/3]:
p6:=simplify(eval(F,x=q+2*h))=f[n+2]:
p7:=simplify(eval(F,x=q+7*h/3))=f[n+7/3]:
p8:=simplify(eval(F,x=q+3*h))=f[n+3]:


vars:= seq(a[i],i=0..7):
Cc:=eval(<vars>, solve({p||(1..8)}, {vars})):
for i from 1 to 8 do
	a[i-1]:=Cc[i]:
end do:
Cf:=s:
L:=collect(simplify(simplify(expand(eval(Cf,x=q+3*h)),size)), [y[n],f[n],f[n+1/3],f[n+1],f[n+5/3],f[n+2],f[n+7/3],f[n+3]], factor):
length(L);
H := ee -> collect(numer(ee),[exp],h->simplify(simplify(h),size))/collect(denom(ee),[exp],h->simplify(simplify(h),size)):
M:=y[n+3]=(H@expand)(L);
length(M);

 

In the following codes I am very sure exp(-3*q) is a factor both at numerator and denominator. However, I dont know how to annihilate this factor. Can someone kindly help to ensure it cancels out? Thanks and kind regards.

restart:
s:=(sum(a[j]*x^j,j=0..2)+sum(a[j]*exp(-(j-2)*x),j=3..4)):
F:=diff(s,x):
p1:=simplify(eval(s,x=q))=y[n]:
p2:=simplify(eval(F,x=q))=f[n]:
p3:=simplify(eval(F,x=q+h))=f[n+1]:
p4:=simplify(eval(F,x=q+3*h/2))=f[n+3/2]:
p5:=simplify(eval(F,x=q+2*h))=f[n+2]:

vars:= seq(a[i],i=0..4):
Cc:=eval(<vars>, solve({p||(1..5)}, {vars})):
for i from 1 to 5 do
	a[i-1]:=Cc[i]:
end do:
Cf:=s:
T:y[n+2]=collect(simplify(eval(Cf,x=q+2*h)), [y[n],f[n],f[n+1],f[n+3/2],f[n+2]], recursive);

 

Please I am trying to get a compact result for this code in particular the lines assigne "a" and "b" respectively. I am sure the result shoud not be more than two lines. Can someone be of help? 

restart:
P:=a[0]+(a[1]*x)/(1+(a[2]*x)/(1+(a[3]*x))):
Q:=diff(P,x):
T:=diff(P,x,x):
e1:=simplify(eval(P, x=q))=y[n]:
e2:=simplify(eval(Q,x=q))=f[n]:
e3:=simplify(eval(Q,x=q+h))=f[n+1]:
e4:=simplify(eval(T,x=q+h))=g[n+1]:
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:
a:=y[n+1]=collect(simplify(eval(Cf, x=q+h)),[y[n], f[n], f[n+1],g[n+1]], recursive):
b:=map(eval@allvalues, [a]);

 

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