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.

Is the output of >polarplot command in maple12 is always in circular pattern, we cann't have rectangular boxed plot?

I have  AX=B , I want to find X , A is 3*3 matrix  <.9,-.3,-.6>|<-.2,.6,-.4>|<-.4,-.4,.8>   B=matrix(3, 1, [600, 600, 600]) ? I using  linearslove, but it is not work.