Error, (in Statistics:-NonlinearFit) invalid input: no implementation of NonlinearFit matches the arguments in call

after tried many times, got this error

x11 := [0.208408965651696e-3, -0.157194487523421e-2, -0.294739401402979e-2, 0.788206708183853e-2, 0.499394753201753e-2, 0.191468321959759e-3, 0.504980449104750e-2, 0.222150494088535e-2, 0.132091821964287e-2, 0.161118434883258e-2, -0.281236534046873e-2, -0.398055875132037e-2, -0.111753680372819e-1, 0.588868146012489e-2, -0.354191562612469e-2, 0.984082837373291e-3, -0.116041186868374e-1, 0.603027845850267e-3, -0.448778128168742e-2, -0.127561485214862e-1, -0.412027655195339e-2, 0.379387381798949e-2, -0.602550446997765e-2, -0.605986284736216e-2, -0.751396992404410e-2, 0.633613424008655e-2, -0.677581832613623e-2]:
y11 := [ -21321.9719565717, 231.709204951251, 1527.92905167191, -32.8508507060675, 54.9408176234139, -99.4222178124229, -675.771433486265, 42.0838668074923, -12559.3183308951, 5.21412214166344*10^5, 1110.50031772203, 3.67149699000155, -108.543878970269, -8.48861069398811, -521.810552387313, 26.4792411876883, -8.32240296737599, -1085.40982521906, -44.1390030597906, -203.891397612798, -56.3746416571417, -218.205643256096, -178.991498697065, -42.2468018350386, .328546922634921, -1883.18308996621, 111.747881085748]:
z11 := [ 1549.88755331800, -329.861725802688, 8.54200301129155, -283.381775745327, -54.5469129127573, 1875.94875597129, -16.2230517860850, 6084.82381954832, 1146.15489803104, -456.460512914647, 104.533252701641, 16.3998365630734, 11.5710907832054, -175.370276462696, 33.8045539958636, 2029.50029336951, 1387.92643570857, 9.54717543291120, -1999.09590358328, 29.7628085078953, 2.58210333216737*10^6, 57.7969622731082, -6.42551196941394, -8549.23677077892, -49.0081775323244, -72.5156360537114, 183.539911458475]:
a1 := Diff(x1(t),t) = k1*x1(t)+ k2*y1(t)+ k3*z1(t);
b1 := Diff(y1(t),t) = k4*x1(t)+ k5*y1(t)+ k6*z1(t);
c1 := Diff(z1(t),t) = k7*x1(t)+ k8*y1(t)+ k9*z1(t);
ICS:=x1(1)=x11[1],y1(1)=y11[1],z1(1)=z11[1];
sol:=dsolve({a1,b1,c1,ICS}, numeric, method=rkf45, parameters=[k1,k2,k3,k4,k5,k6,k7,k8,k9]);
ans:=proc(p1,p2,p3) sol(parameters=[x1=p1,y1=p2,z1=p3]); end proc:
tim := [seq(n, n=1..27)];
ExperimentalData := <<x11>|<y11>|<z11>>;
FitParams:=Statistics:-NonlinearFit(ans, ExperimentalData,initialvalues=[k1=0,k2=0,k3=0,k4=0,k5=0,k6=0,k7=0,k8=0,k9=0], output=parametervalues);

main_screened_Poisso.mw

In the attached Maple program I construct a matrix of which the entries involve numerical integration. Nothing difficult, I think, yet the ReducedRowEchelonForm is taking forever. What am I doing wrong?

expect to export a series of graphs, but no diagram,

then i debug to export one diagram, it is success, why in this case not export

https://drive.google.com/file/d/0B2D69u2pweEvcDZVZ0tsRTc2dTg/edit?usp=sharing

restart;
with(combinat):
list1 := permute([a, b, a, b, a, b], 3);
list1a := subs(b=1,subs(a=0, list1));
n := 3;
list1a := permute([seq(seq(k,k=0..1),k2=1..n)], n);
list2 := permute([a, b, c, d, e, f, g, h, a, b, c, d, e, f, g, h, a, b, c, d, e, f, g, h], 3);
list3 := subs(h=18,subs(g=17,subs(f=16,subs(e=15,subs(d=14,subs(c=13,subs(b=12,subs(a=11,list2))))))));
list3 := permute([seq(seq(k,k=11..18),k2=1..3)], 3);
Iter:= iterstructs(Permutation([seq(seq(k,k=11..(10+nops(list1a))),k2=1..3)]), size=3):
list3b := [];
while not Iter[finished] do
p:= Iter[nextvalue]();
list3b := [p, op(list3b)];
end do:
list5 := Matrix(nops(list1a)*nops(list3), 1);
count := 1;
for n from 1 to nops(list3) do
temp1 := subs(1=list1a[1],list3[n]);
for k from 11 to nops(list1a)+10 do
temp1 := subs(k=list1a[k-10],temp1);
od;
list5[count] := temp1;
count := count + 1;
od;
Lfh := proc(numoflevel, hx, fx, varx)
if numoflevel = 1 then
hello := 0;
for kk from 1 to nops(var) do
hello := hello + diff(hx[kk], varx[kk])*fx[kk];
od;
return hello;
else
hello := 0;
for kk from 1 to nops(var) do
hello := hello + diff(Lfh(numoflevel-1, hx, fx, varx), varx[kk])*fx[kk];
od;
return hello;
end if;
end proc:
CheckRelativeRankZero := proc(h1, f1, g1,variables1,Count)
IsFinish := 0;
Result := 0;
for ii from 1 to 8 do
if IsFinish = 0 then
Lf2h := Lfh(ii,h1,f1,variables1);
Print(“Lf2h=”);
Print(Lf2h);
Lgf2h := Lfh(1,[seq(Lf2h,n=1..nops(variables1))],g1,variables1);
if Lgf2h = 0 then
print(“Lgf2h = 0”)
print(f1);
print(Lf2h);
print("find at ", ii);
IsFinish := 1;
Result :=Lf2h;
end if;
end if;
od;
return Result;
end proc:
IsZeroMatrix := proc(h1)
Iszero := 1;
for ii from 1 to 3 do
for jj from 1 to 3 do
if h1[ii][jj] <> 0 then
Iszero := 0;
end if
od;
od;
return Iszero;
end proc:
with(combstruct):
list6:= convert(list5, list):
list7 := [];
for ii from 1 to nops(list6) do
if list6[ii] <> 0 then
list7 := [list6[ii], op(list7)];
end if;
od;
with(LinearAlgebra):
with(VectorCalculus):
varlist := [x1, x2, x3];
Iter:= iterstructs(Permutation(list7), size=2):
Count := 1;

with(DEtools):
Iter:= iterstructs(Permutation(list7), size=2):
Count := 1;
list8 := [];
while not Iter[finished] do
p:= Iter[nextvalue]();
I1 := 0;
I2 := 0;
if IsZeroMatrix(p[1]) = 0 and IsZeroMatrix(p[2]) = 0 then
group1 := Matrix(p[1]);
for ii from 1 to 3 do
for jj from 1 to 3 do
if group1[ii][jj] = 1 then
I1 := I1 + varlist[ii]*varlist[jj];
end if;
od;
od;
group2 := Matrix(p[2]);
for ii from 1 to 3 do
for jj from 1 to 3 do
if group2[ii][jj] = 1 then
I2 := I2 + varlist[ii]*varlist[jj];
end if;
od;
od;
f2:=[I1, I2];
g2:=[0,-1,1];
h2:=[x1,0,0];
Lf2h := CheckRelativeRankZero(h2,f2, g2, varlist, Count);
print(“Lf2h=”);
print(Lf2h);
RightSide := MatrixMatrixMultiply(Matrix([[0,diff(I2, varlist[3]),-diff(I2,varlist[2])],[-diff(I2, varlist[3]),0,diff(I2, varlist[1])],[diff(I2, varlist[2]),-diff(I2, varlist[1]),0]]), Matrix([[diff(I1, varlist[1])],[diff(I1, varlist[2])],[diff(I1, varlist[3])]]));
print(“RightSide”);
print(RightSide);
Lf2_h := Lfh(1, Lf2h, f2, varlist);
LgLf_h := Lfh(1,Lfh(1,h2,f2,varlist),g2, varlist);
if LgLf_h = 0 then
u:=0;
else
u := -Lf2_h/LgLf_h;
end if;
newsys := [Diff(x1(t),t) = subs(x3=x3(t),subs(x2=x2(t),subs(x1=x1(t),RightSide[1][1]))) + g[1]*u,
Diff(x2(t),t) = subs(x3=x3(t),subs(x2=x2(t),subs(x1=x1(t),RightSide[2][1]))) + g[2]*u,
Diff(x3(t),t) = subs(x3=x3(t),subs(x2=x2(t),subs(x1=x1(t),RightSide[3][1]))) + g[3]*u];
eval(plotsetup):
`plotsetup/devices`[jpeg]:=[jpeg,`plot.jpg`,[],[],``]:
plotsetup(jpeg, plotoutput=cat(cat(`testhamiplot`, Count),`.jpg`),plotoptions=`height=700,width=800`);
DEplot3d(value(newsys), [x1(t), x2(t), x3(t)], t = 0..1,[[x1(0) = 1, x2(0) = 1, x3(0) = 1]]);

Count := Count +1;
end if;
end do:

Say I have the following loops:

for C from 1 to 10 do
    r:=[]:
    for K from 2 to 10 do
    r:=[op(r),2*K+2*C-3];
    end do:
    print(r);
end do:

for C from 1 to 10 do
    r:=[]:
    for K from 2 to 10 do
    r:=[op(r),K*C+K+C-2];
    end do:
    print(r);
end do:

I wonder how could I write a procedure, say use expressions "2*K+2*C-3" and "K*C+K+C-2" as input arguments?

so I can call up like :

myfun(K*C+K+C-2) or myfun("K*C+K+C-2")

myfun(2*K+2*C-3)

I dont care whether the output(s) are lists, tables, or matrices.

My main difficulty is to get the expression to be procedure inputs.

Though if the output can be a  10 by 9 matrix, it's better.

Thanks,

casper

when i got this error, i am confused i guess t is independant variable, x1,y1,z1 are dependant variables

x11 := [0.208408965651696e-3, -0.157194487523421e-2, -0.294739401402979e-2, 0.788206708183853e-2, 0.499394753201753e-2, 0.191468321959759e-3, 0.504980449104750e-2, 0.222150494088535e-2, 0.132091821964287e-2, 0.161118434883258e-2, -0.281236534046873e-2, -0.398055875132037e-2, -0.111753680372819e-1, 0.588868146012489e-2, -0.354191562612469e-2, 0.984082837373291e-3, -0.116041186868374e-1, 0.603027845850267e-3, -0.448778128168742e-2, -0.127561485214862e-1, -0.412027655195339e-2, 0.379387381798949e-2, -0.602550446997765e-2, -0.605986284736216e-2, -0.751396992404410e-2, 0.633613424008655e-2, -0.677581832613623e-2]:
y11 := [ -21321.9719565717, 231.709204951251, 1527.92905167191, -32.8508507060675, 54.9408176234139, -99.4222178124229, -675.771433486265, 42.0838668074923, -12559.3183308951, 5.21412214166344*10^5, 1110.50031772203, 3.67149699000155, -108.543878970269, -8.48861069398811, -521.810552387313, 26.4792411876883, -8.32240296737599, -1085.40982521906, -44.1390030597906, -203.891397612798, -56.3746416571417, -218.205643256096, -178.991498697065, -42.2468018350386, .328546922634921, -1883.18308996621, 111.747881085748]:
z11 := [ 1549.88755331800, -329.861725802688, 8.54200301129155, -283.381775745327, -54.5469129127573, 1875.94875597129, -16.2230517860850, 6084.82381954832, 1146.15489803104, -456.460512914647, 104.533252701641, 16.3998365630734, 11.5710907832054, -175.370276462696, 33.8045539958636, 2029.50029336951, 1387.92643570857, 9.54717543291120, -1999.09590358328, 29.7628085078953, 2.58210333216737*10^6, 57.7969622731082, -6.42551196941394, -8549.23677077892, -49.0081775323244, -72.5156360537114, 183.539911458475]:
a1 := Diff(x1(t),t) = k1*x1(t)+ k2*y1(t)+ k3*z1(t);
b1 := Diff(y1(t),t) = k4*x1(t)+ k5*y1(t)+ k6*z1(t);
c1 := Diff(z1(t),t) = k7*x1(t)+ k8*y1(t)+ k9*z1(t);
ICS:=x1(0)=x11[1],y1(0)=y11[1],z1(0)=z11[1];
sol:=dsolve({a1,b1,c1,ICS}, numeric, method=rkf45, parameters=[k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12]);
ans:=proc(p1,p2,p3) sol(parameters=[x1=p1,y1=p2,z1=p3]); end proc:
tim := [seq(n, n=1..27)];
FitParams:=Statistics:-NonlinearFit(ans, [x11, y11, z11], tim, initialvalues=<0.5,0.5,0.5>, output=parametervalues);

4th_order.mw Is it possible to solve this ODE with perturbation method using maple? If yes, please give the procedure.

Thanks.

Can one solve nonlinear ODE with perturbation method in maple? If yes, please the procedure.

x11 := Vector([0.208408965651696e-3, -0.157194487523421e-2, -0.294739401402979e-2, 0.788206708183853e-2, 0.499394753201753e-2, 0.191468321959759e-3, 0.504980449104750e-2, 0.222150494088535e-2, 0.132091821964287e-2, 0.161118434883258e-2, -0.281236534046873e-2, -0.398055875132037e-2, -0.111753680372819e-1, 0.588868146012489e-2, -0.354191562612469e-2, 0.984082837373291e-3, -0.116041186868374e-1, 0.603027845850267e-3, -0.448778128168742e-2, -0.127561485214862e-1, -0.412027655195339e-2, 0.379387381798949e-2, -0.602550446997765e-2, -0.605986284736216e-2, -0.751396992404410e-2, 0.633613424008655e-2, -0.677581832613623e-2]):
y11 := Vector([ -21321.9719565717, 231.709204951251, 1527.92905167191, -32.8508507060675, 54.9408176234139, -99.4222178124229, -675.771433486265, 42.0838668074923, -12559.3183308951, 5.21412214166344*10^5, 1110.50031772203, 3.67149699000155, -108.543878970269, -8.48861069398811, -521.810552387313, 26.4792411876883, -8.32240296737599, -1085.40982521906, -44.1390030597906, -203.891397612798, -56.3746416571417, -218.205643256096, -178.991498697065, -42.2468018350386, .328546922634921, -1883.18308996621, 111.747881085748]):
z11 := Vector([ 1549.88755331800, -329.861725802688, 8.54200301129155, -283.381775745327, -54.5469129127573, 1875.94875597129, -16.2230517860850, 6084.82381954832, 1146.15489803104, -456.460512914647, 104.533252701641, 16.3998365630734, 11.5710907832054, -175.370276462696, 33.8045539958636, 2029.50029336951, 1387.92643570857, 9.54717543291120, -1999.09590358328, 29.7628085078953, 2.58210333216737*10^6, 57.7969622731082, -6.42551196941394, -8549.23677077892, -49.0081775323244, -72.5156360537114, 183.539911458475]):
a1 := Diff(x1(t),t) = k1*x1(t)+ k2*y1(t)+ k3*z1(t)+k4*u(t);
b1 := Diff(y1(t),t) = k5*x1(t)+ k6*y1(t)+ k7*z1(t)+k8*u(t);
c1 := Diff(z1(t),t) = k9*x1(t)+ k10*y1(t)+ k11*z1(t)+k12*u(t);
ICS:=x1(0)=x11[1],y1(0)=y11[1],z1(0)=z11[1],u(0)=7;
Zt := rhs(dsolve({a1,b1,c1,ICS},[x1(t),y1(t),z1(t),u(t)]));
Params := NonlinearFit(Re(Zt),<seq(k,k=0..N)>, C, [t], parameternames=[k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12], output=parametervalues):
A := eval(a, Params):
B := eval(b, Params):
a = A;
b = B;

i wait for a long time for dsolve still evaluating

we were given a function that counts the number of primes among the arguments after the
rst and returns the result in the rst argument. When calling this, you must make sure
that the rst argument is a name. 

this is it.

cp := proc (YY) local count, i ;
print("nargs=", nargs, "args=", args) ;
count := 0 ;
for i from 2 to nargs do
if isprime(args[i]) then
count := count+1 ;
end if ;
print("i=", i, "count=", count) ;
end do ;
print("count=", count) ;
YY := count ;
end proc ;

EX: cp('noprimes',2,4,5,6,7,9,19)

and this works grand, but then we were given this function with slight adjustments to it and asked to fix it. the hints we were given were to try and forve evaluation at the right places.

This is the function we have to fix...

xcp := proc (count) local i;
print("nargs=", nargs, "args=", args) ;
count := 0 ;
for i from 2 to nargs do
if isprime(args[i]) then
count := count+1 ;
end if ;
print("i=", i, "count=", count) ;
end do ;
print("count=", count) ;
end proc ;

any help is appreciated!

Hi. I'd like to find the solution closest to zero for sum(abs(f(k, m, n)+g(k, m, n)), n = i .. j) , when a < m, n < b . 

Have trouble wrapping my head around how to do that and would appreciate any help.

Even better would be to find a solution where the maximum absolute value of f(k, m, n) + g(k, m, n) is minimized for n = i .. j) and when a < m, n < b , but I'm guessing the sum would be easier, and close enough.

Maybe I'm barking up the wrong tree getting this done with Maple, but I'm hopeful.

Thank you for looking

Dear experts;

How can I solve this problem with maple?

restart:

 X[3](0):=6.3096*10^9;
 c:=0.67;
 d:=3.7877*10^(-8);
 delta:=3.259*d;
 lambda:=(2/3)*10^8*d;
 R[0]:=1.33;
 p:=(c*X[3](0)*delta*R[0])/(lambda*(R[0]-1));
beta:=(d*delta*c*R[0])/(lambda*p);

ode:=diff(x[1](t), t)=(lambda-d*x[1](t)-(1-beta*x[1](t)*x[3](t)*(psi[1](t)-psi[2](t))/A[1])*beta*x[1](t)*x[3](t)),
 diff(x[2](t), t) =((1-beta*x[1](t)*x[3](t)*(psi[1](t)-psi[2](t))/A[1])*beta*x[1](t)*x[3](t)-delta*x[2](t)),
 diff(x[3](t), t) =((1+psi[3](t)*p*x[2](t)/A[2])*p*x[2](t)-c*x[3](t)),diff(psi[1](t), t) =-1+1/A[1]*beta^2*x[1](t)*(x[3](t))^2*(psi[1](t)-psi[2](t))^2-psi[1](t)*(-d+beta^2*(x[3](t))^2*(psi[1](t)-psi[2](t))/A[1]*x[1](t)-(1-beta*x[1](t)*x[3](t)*(psi[1](t)-psi[2](t))/A[1])*beta*x[3](t))-psi[2](t)*(-beta^2*(x[3](t))^2*(psi[1](t)-psi[2](t))/A[1]*x[1](t)+(1-beta*x[1](t)*x[3](t)*(psi[1](t)-psi[2](t))/A[1])*beta*x[3](t)),
 diff(psi[2](t), t) =1/A[2]*psi[3](t)^2*p^2*x[2](t)+psi[2](t)*delta-psi[3](t)*(psi[3](t)*p^2/A[2]*x[2](t)+(1+psi[3](t)*p*x[2](t)/A[2])*p),
 diff(psi[3](t), t) = 1/A[1]*beta^2*(x[1](t))^2*x[3](t)*(psi[1](t)-psi[2](t))^2-psi[1](t)*(beta^2*(x[1](t))^2*(psi[1](t)-psi[2](t))/A[1]*x[3](t)-(1-beta*x[1](t)*x[3](t)*(psi[1](t)-psi[2](t))/A[1])*beta*x[1](t))-psi[2](t)*(-beta^2*(x[1](t))^2*(psi[1](t)-psi[2](t))/A[1]*x[3](t)+(1-beta*x[1](t)*x[3](t)*(psi[1](t)-psi[2](t))/A[1])*beta*x[1](t))+psi[3](t)*c;

ics := x[1](0)=5.5556*10^7, x[2](0)=1.1111*10^7,x[3](0)=6.3096*10^9,psi[1](100)=0,psi[2](100)=0,psi[3](100)=0;

dsolve([ode, ics],numeric);?????????????????????????

Please help me

ode.mws

function [y g] = ques5
% using Simpson Formula to approximate the integration
% input:
% f(x): [a b]


end
% use Euler formula to compute function y
for i = 1:N
    if i ==1

legend('numerical y','exact y','numerical g','exact g')

function g = f2g(a,b)
% f(x) = x
g = (b-a)/6*(a + 4*(a+b)/2 + b);


ican use matlab to solve this problem but not maple
please help

after running the code below, run

f2:=[ x3^2, x3*x2];
g2:=[0,-1,1];
h2:=[x1,0,0];
Lf2h := Lfh(1,h2,f2, varlist);

got

Error, (in Lfh) invalid subscript selector

can not see where is wrong

Code:

restart;

with(combinat):

list1 := permute([a, b, a, b, a, b], 3);

list1a := subs(b=1,subs(a=0, list1));

n := 3;

list1a := permute([seq(seq(k,k=0..1),k2=1..n)], n);

list2 := permute([a, b, c, d, e, f, g, h, a, b, c, d, e, f, g, h, a, b, c, d, e, f, g, h], 3);

list3 := subs(h=18,subs(g=17,subs(f=16,subs(e=15,subs(d=14,subs(c=13,subs(b=12,subs(a=11,list2))))))));

list3 := permute([seq(seq(k,k=11..18),k2=1..3)], 3);

Iter:= iterstructs(Permutation([seq(seq(k,k=11..(10+nops(list1a))),k2=1..3)]), size=3):

list3b := [];

while not Iter[finished] do

     p:= Iter[nextvalue]();

     list3b := [p, op(list3b)];

end do:

list5 := Matrix(nops(list1a)*nops(list3), 1);

count := 1;

for n from 1 to nops(list3) do

        temp1 := subs(1=list1a[1],list3[n]);

        for k from 11 to nops(list1a)+10 do

                temp1 := subs(k=list1a[k-10],temp1);

        od;

        list5[count] := temp1;

        count := count + 1;

od;

Lfh := proc(numoflevel, hx, fx, var)

if numoflevel = 1 then

        hello := 0;

        for kk from 1 to nops(var) do

                hello := hello + diff(hx[kk], var[kk])*fx[kk];

        od;

        return hello;

else

        hello := 0;

        for kk from 1 to nops(var) do

                hello := hello + diff(Lfh(numoflevel-1, hx, fx, var), var[kk])*fx[kk];

        od;

        return hello;

end if;

end proc:

CheckRelativeRankZero := proc(h1, f1, g1,variables1,Count)

IsFinish := 0;

for ii from 1 to 8 do

        if IsFinish = 0 then

        Lf2h := Lfh(ii,h1,f1,variables1);

        Lgf2h := Lfh(1,[seq(Lf2h,n=1..nops(variables1))],g1,variables1);

        if Lgf2h = 0 then

                print(f1);

                print(Lf2h);

                print("find at ", ii);

                IsFinish := 1;

                return Lf2h;

        end if;

        end if;

od;

return 0;

end proc:

IsZeroMatrix := proc(h1)

Iszero := 1;

for ii from 1 to 3 do

for jj from 1 to 3 do

        if h1[ii][jj] <> 0 then

                        Iszero := 0;

        end if

        od;

od;

return Iszero;

end proc:

with(combstruct):

list6:= convert(list5, list):

list7 := [];

for ii from 1 to nops(list6) do

if list6[ii] <> 0 then

        list7 := [list6[ii], op(list7)];

        end if;

od;

with(LinearAlgebra):

with(VectorCalculus):

varlist := [x1, x2, x3];

Iter:= iterstructs(Permutation(list7), size=2):

Count := 1;

I want  to verify the following function expression

is indeed an antiderivative of the function expression

where  A>0 with B^2-4*A*C<0.

I have tried the command

>diff((2),x);simplify(%);

where (2) is the label of the antiderivative expression.  But the result is very awkarward and lengthy. How could I verify the antiderivative expression is indeed an antiderivative of the other function by using Maple 17?