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

Hi every one,

Q1:

I tried to get the max $ min of a following function:

l:=1:alpha:=1:b:=100:k:=20:

eq1 := (alpha+(l+alpha)*u+alpha*k*u^2)*a =
u*(alpha+(l+alpha)*u+alpha*k*u^2)*(1+l*alpha*b/((alpha+(l+alpha)*u+alpha*k*u^2))):

I did this code but it seems it didnt work for this equation

maximize(eq, u=1..12, location);

minimize(eq, u=1..12, location);

Also, I think about solving the cubic i feel i'm so close to the solve but couldn't

factor((rhs-lhs)(eq1));

eq:=collect(%,u);

Q:=(a,u)->eq;sol:=evalf(solve(Q(a,u),u)): S:=array([],1..3): S[1]:=sol[1]:S[2]:=sol[2]:S[3]:=sol[3]:

Q2:

the same thing wanted to get the maximum and the minimum of the function v

here the code

restart;
eq1:=(alpha+(l+alpha)*u+alpha*k*u^2)*a=
u*(alpha+(l+alpha)*u+alpha*k*u^2)*(1+l*alpha*b/((alpha+(l+alpha)*u+alpha*k*u^2)));
eq2:=v=alpha*b*(1+u+k*u^2)/(alpha+(l+alpha)*u+alpha*k*u^2);
factor((rhs-lhs)(eq1));
eq1:=collect(%,u);
params:={l=10,alpha=0.5,b=100,k=20};
U:=[solve(eval(eq1,params),u)]; #3 solutions for u
#plots:-complexplot(U,a=0..20,style=point); #plot in the complex u-plane
vua:=eval(solve(eq2,v),params): #v expressed in terms of u and a
V:=eval~(vua,u=~U): #the 3 solutions for v in terms of a

## PLOT the function V
plot(V,a=0..75,v=0..100,color=black,labels=[a,v],axes=boxed,numpoints=90,linestyle=1,font=[1,1,18],thickness=2,tickmarks=[4,4],view=[0..65,25..100]);

I do appricaited any advises

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

Minimize doesn't work with dsolve porcedure?

experiment_real.mw

Download experiment_real.mw

with trool procedure minimize dosent work .... and its make me realy sad, couse i need to optimize alpha and leng in other (big one) porcedure with same dsolve.

get this errors:
"Warning, The use of global variables in numerical ODE problems is deprecated, and will be removed in a future release. Use the 'parameters' argument instead (see ?dsolve,numeric,parameters)"
"Error, (in XX) parameter 'alpha' must be assigned a numeric value before obtaining a solution"

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;

http://homepages.lboro.ac.uk/~makk/MathRev_Lie.pdf

ode1 := Diff(f(x),x$2)+2*Diff(f(x),x)+f(x);
with(DEtools):
with(PDETools):
gen1 := symgen(ode1);
with(PDEtools):
DepVars := ([f])(t);
NewVars := ([g])(r);
SymmetryTransformation(gen1, DepVars, NewVars);

Error, invalid input: too many and/or wrong type of arguments passed to PDEtools:-SymmetryTransformation; first unused argument is [_xi = -x, _eta = f*x]

generator1 := rhs(sym1[3][1])*Diff(g, x)+ rhs(sym1[3][2])*Diff(g, b)

what is X1 and X2 so that [X1, X2] = X1*X2 - X2*X1 = (X1(e2)-X2(e1))*Diff(g, z) ?

is it possible to use lie group to represent a differential equation, and convert this group back to differential equation ? how do it do?

how to find symmetry z + 2*t*a, when you do not know before taylor calcaulation?

fza := z + 2*t*a;
fza := x;
fza := z + subs(a=0, diff(fza,a))*a;

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? 

 

How can I've shaded region with non-linear inequality, I have some potential function like V(r)=(1-2Me^-2lr+q²e^-4lr) and I want to shade area between this & x-axis,

when I'm doing it with inequlityplot, it says that inequality should be linear.

Hello,

First of all, I've searched already a bit around, but couldn't find a similar topic, so I thought I'd open a new one. Also, English isn't my main language, so terminology may be wrong, but I hope you'll still understand what I want to say.

So, I have this procedure:

restart;
functions:= proc(n)
local L, list, p, f, sum, i, part, g, normg, x:
L:=1/sqrt(2):
list:=[L]:

for p from 2 to n do
f := x**(p-1):
sum := 0:

for i from 1 to (p-1) do
part:= int(list[i]*f*(x+1),x=-1..1)*list[i]:
sum := sum + part:
end do:

g:= f-sum:
normg:= sqrt(int(g^2*(x+1),x = -1..1)):
L:=g/normg:
list := [op(list), unapply(L,x)]:
end do:
list;
end proc:

What this procedure does, is calculating n orthonormal functions (but that doesn't really matter here). The result is a list of functions, or should be. What I get when I enter e.g. functions(5), is a list of very weird functions with 'nexpr' and more, instead of some polynomials.
When I replace the 'unapply(L,x)' in the 4th last line by just 'L', I do get the correct expressions, but I can't manage to calculate the function values for those. The expression just gets returned. By the way, I can't do that in the situation before the edit either.

So what I eventually want to do, is calculating some function values for each function in the list, or (if this isn't the right terminology) in Maple code e.g.:

f:= x -> x^2 + 3*x;
f(3);
(The result should be 18 in this case)

Could someone help me? :)

Jeroen

Two questions:

The algortihms that Groebner[Basis] uses at each step computes some "tentative" or "pseudo-basis". The "tentative" basis is not a Groebner basis but it is in the ideal generated by the original system of polynomial eq.

1) Is this correct ? Provided this is correct, then

2) How can one retrive the last "tentative" basis?
 If I just use timelimit I can abort the computations but how can one retrive the last computation?

Please respond me by email, thanks.

wingwatson7@gmail.com

Good evening, dear experts. 

I want to solve the system of PDEs, but i get a mistake:

"Error, (in pdsolve/numeric/xprofile) unable to compute solution for t>HFloat(0.0):

solution becomes undefined, problem may be ill posed or method may be ill suited to solution"

How  can I determine the method for pdsolve?
Thanks for answers.

It's my file with the functions:
restart;
alias(X = x(t, tau), Y = y(t, tau));
xt, yt := map(diff, [X, Y], t)[];
xtau, ytau := map(diff, [X, Y], tau)[];
M := (xtau*(-Y+.1*X)+ytau*X)/(xtau^2+ytau^2);
pde1 := xt = -Y+.1*X-xtau*M;
pde2 := yt = -M*ytau+X;

cond := {x(0, tau) = 1, x(t, 0) = 1, y(0, tau) = 1, y(t, 0) = 1};
Sol := pdsolve({pde1, pde2}, cond, numeric, time = t, range = 0 .. 1);
Sol:-value(t = 1);
%;
Error, (in pdsolve/numeric/xprofile) unable to compute solution for t>HFloat(0.0):
solution becomes undefined, problem may be ill posed or method may be ill suited to solution

I was answering this question on another platform, and wanted to compare with Maple. On Matlab, this is the result

format long
916.536 + 3.3

      9.198359999999999e+02

When using Maple I get

evalhf(916.536 + 3.3);
     919.836000000000013

I am on a 64 bit intel PC, and the OS is windows 7, 64 bit, and I my Maple is 64 bit version also.

Using Mathematica on my PC, which is 64 bit also, I get same as Matlab:

FullForm[916.536 + 3.3]
    919.8359999999999`

and help on evalhf says:
"A call to evalhf evaluates an expression to a numerical value using the floating-point 
hardware of the underlying system. The evaluation is done in double precision."

So, the question is, why I am not gettting the same result as those others shown above?
May be it has to do with this:

"The evalhf function converts all its arguments to hardware floats, computes the answer 
and converts the answer to a Maple float result."

?

Here is the original question http://www.mapleprimes.com/ViewTemp.ashx?f=21095_1386318320/screen06.12.13.docx , replaced by the questioner.  She/he must not do such things.

The differential equation dy/dt = t / (2-y), y(0)=1 fails the tests in section 5.1 at y=2. [ f(t,y) is undefined at y=2 and the y-partial derivative of f(t,y) is also undefined there. ] If a solution stays away from y=2, there is no problem at all. Try a few different initial conditions and summarize your findings. Use the Runge-Kutta order 4 method with a fixed step size.

Hint: You may find Maple's solution of the differential equation helpful:

s1 := dsolve({diff(y(t),t)= t/(2-y(t))}, y(t)); 

In the solution _C1 is a constant to be determined using the initial conditions.