hi,

I am trying to implement some data-intensive algorithms like: comuting similarity scores, clustering etc. How efficient Maple is in doing these operations? I mean is it comparable to MATLAB in these operations? What toolboxes are useful in these? I am trying to build a recommender system.

I am using Maple18 and not sure if Maple has any advantage in implementing these type of algorithms over MATALB??

thanks

siba

Please consider this code:

restart;

with(DEtools):

test:=(diff(x(t),t,t))+(diff(a(t),t,t))=0;

FirstOrderSys := convertsys(test, [], x(t), t, y, yp );

When it is executed Maple says: Error, (in is/internal) too many levels of recursion

Now if i change just the letter a to, say, p (a(t)->p(t)) like this:

restart;

with(DEtools):

test:=(diff(x(t),t,t))+(diff(p(t),t,t))=0;

FirstOrderSys := convertsys(test, [], x(t), t, y, yp );

Lo and Behold! Suddenly Maple gives the answer:

/ d / d \\ / d / d \\ |--- |--- x(t)|| + |--- |--- p(t)|| = 0 \ dt \ dt // \ dt \ dt // [[ d / d \] [[yp[1] = y[2], yp[2] = ---- |--- p(t)|], [[ dt \ dt /] [ d ] ] [y[1] = x(t), y[2] = --- x(t)], undefined, []] [ dt ] ]

Why is that so? I don't see how one letter makes this difference. I have learned Maple on my own, so maybe I have missed something?

Please, i need help USING ODE1 and ODE2 with given BCS and Pr=0.714

it is needed to to generates   

                   [-0.2], [0.51553], [0.4000]
                  [-0.1], [0.57000], [0.4371]
                   [0.], [0.62756], [0.4764]
                  [0.1], [0.68811], [0.5176]
                  [0.2], [0.75153], [0.5609]

but it is generarting

                  [-0.2], [0.51553], [0.42342]
                  [-0.1], [0.57000], [0.46114]
                   [0.], [0.62756], [0.50088]
                  [0.1], [0.68811], [0.54261]
                  [0.2], [0.75153], [0.58628]

the values of D(theata)(0) is wrong. Please i need HELP. this the code below that i use:
>restart;
>with (plots):ode1:=diff(f(eta),eta,eta,eta)+f(eta)*diff(f(eta),eta,eta)-M*diff(f(eta),eta)=0:

>ode2:=diff(theta(eta),eta,eta)+Pr*f(eta)*diff(theta(eta),eta)=0:

>bcs1:= f(0)=w,D(f)(0)=1,D(f)(10)=0:
>bcs2theta(10)=0,theta(0)=1:
>fixedparameter1:=[M=0.0]:
>ode3:=eval(ode1,fixedparameter1):
>fixedparameter2:=[Pr=0.714]:
>ode4:=eval(ode2,fixedparameter2):

>G:=[-0.2,-0.1,0.0,0.1,0.2]:
>for ode3 and ode4:
  for k from 1 to 5 do
  sol_All:=dsolve(eval({ode3,ode4,bcs1},w=G[k]),    [f(eta),theta(eta)],numeric,output=listprocedure);
Y_sol||k:= -rhs(sol_All[4]);
YP_sol||k:=-rhs(sol_All[6]);
end do:
>Digits:=5:

>for k from 1 to 5 do
evalf([G[k]]),evalf([(Y_sol||k(0))]),evalf([YP_sol||k(0)]);    
od;
                  [-0.2], [0.51553], [0.42342]
                  [-0.1], [0.57000], [0.46114]
                   [0.], [0.62756], [0.50088]
                  [0.1], [0.68811], [0.54261]
                  [0.2], [0.75153], [0.58628]

Hi all.

I am using Maple2015.

I typed in as input y=x/sqrt(1-x^2).

I hit enter.  The output is:

I know the 2 answers are equivalent.

My question is why did Maple swap 1-x^2 to -x^2+1???

Any advice to swap it back would be greatly appreciated.

How to find asymptotic behaviour of a function.

For example at infinity

sinh(x) behaves as 1/2*exp(x)

1/sinh(x)  behaves as 2*exp(-x)

exp(-x)*(exp(-x)+1) behaves as exp(-x)

so that it works with a more complex expression.

when solving a system of equations, I want to get rid of all the absolute functions.

for example, |y-2|=x，I don't want maple solve this equation directly...because maple may has difficulties when dealing with absolute values. Instead, I want to transform this equation by squaring the both sides at the same time which end up this equation: (y-2)^2=x^2.

The example I provide is kind of simple...what if there are multiple absolute term in the equations? Is there a general way to get what I need? Or Is it practical to use Maple to achive such thing?

thanks in advance.

When I use fsolve with equation 

-x^2 + 2*x + 5 + (x^2 + 2*x - 1)* sqrt(2 - x^2)=0

I got only one solution.

fsolve(-x^2 + 2*x + 5 + (x^2 + 2*x - 1)* sqrt(2 - x^2)=0,x);

In fact, it have two reals solutions.  

I posted at here

http://mathematica.stackexchange.com/questions/83985/does-the-equation-have-two-roots/83991#83991

Hello Everyone

I have an expression which I wish to integrate. I would be grateful if you could please help me with it. I have uploaded the maple file for your refrence.

Thanks a lot for your time.

IntegrationExample.mw

restart;
pp:=-55471918776960000*tanh((1/3220)*sqrt(10368400-cp^2)*Pi*x/cp)+5350094400*tanh((1/3220)*sqrt(10368400-cp^2)*Pi*x/cp)*cp^2-129*tanh((1/3220)*sqrt(10368400-cp^2)*Pi*x/cp)*cp^4+2670899840*tanh((1/6450)*sqrt(41602500-cp^2)*Pi*x/cp)*sqrt(41602500-cp^2)*sqrt(10368400-cp^2);
Student[Calculus1]:-Roots(subs(x=8000,pp),cp=1..3220,numeric);
p1:=proc(v)
option hfloat;
local a;
a:=Student[Calculus1]:-Roots(subs(x=v,pp),cp=1..3220,numeric);
if nops(a)>=1 then seq([v,a[i]],i=1..nops(a));
end if;
end proc:
SS1:=[seq(p1(i),i=3500..20000,200)]:
plot(SS1,style=point,gridlines);

The final figure is different between maple12 and maple17.

On 17, unwanted points apprear.

is it a bug?

Hello,

I don't why I can no longer see my multibody model in the 3D construction mode.

I probably should desactive something.

I would like to observe my system with the kinematic constraints not enforced.

The idea is that my model is computing but the computation is very long. I should have done a mistake in my initial conditions.

Do you have an idea of what to do to reactivate/ show again the 3d contruction mode ?

Thank you for your help.

Dear Colleges

I have a problem with the following code. As you can see, procedure Q1 converges but I couldn't get the resutls from Q2.

I would be most grateful if you could help me on this problem.

Sincerely yours

Amir

restart;

Eq1:=diff(f(x),x$3)+diff(f(x),x$2)*f(x)+b^2*sqrt(2*reynolds)*diff(diff(f(x),x$2)^2*x^2,x$1);
Eq2:=diff(g(x),x$3)+diff(g(x),x$2)*g(x)+c*a^2*sqrt(2*reynolds)*diff(diff(g(x),x$2)^2*x,x$1);
eq1:=isolate(Eq1,diff(f(x),x,x,x));
eq2:=subs(g=f,isolate(Eq2,diff(g(x),x,x,x)));
EQ:=diff(f(x),x,x,x)=piecewise(x<c*0.1,rhs(eq1),rhs(eq2));
Eq11:=diff(theta(x),x$2)+pr*diff(theta(x),x$1)*f(x)+pr/prt*b^2*sqrt(2*reynolds)*diff(diff(f(x),x$2)*diff(theta(x),x$1)*x^2,x$1);
Eq22:=diff(g(x),x$2)+pr*diff(g(x),x$1)*f(x)+pr/prt*a^2*c*sqrt(2*reynolds)*diff(diff(f(x),x$2)*diff(g(x),x$1)*x^1,x$1);
eq11:=isolate(Eq11,diff(theta(x),x,x));
eq22:=subs(g=theta,isolate(Eq22,diff(g(x),x,x)));
EQT:=diff(theta(x),x,x)=piecewise(x<c*0.1,rhs(eq11),rhs(eq22));
EQT1a:=eval(EQT,EQ):
EQT2:=eval(EQT1a,{f(x)=G0(x),diff(f(x),x)=G1(x),diff(f(x),x,x)=G2(x)}):
bd:=c;
a:=0.13:
b:=0.41:
pr:=1;
prt:=0.86;
reynolds:=12734151.135786774055543653356602;     #10^6;   #1.125*10^8:

c:=88.419896050808975395120916434619:
;
Q:=proc(pp2) local res,F0,F1,F2;
print(pp2);
if not type(pp2,numeric) then return 'procname(_passed)' end if:
res:=dsolve({EQ,f(0)=0,D(f)(0)=0,(D@@2)(f)(0)=pp2},numeric,output=listprocedure);
F0,F1,F2:=op(subs(subs(res),[f(x),diff(f(x),x),diff(f(x),x,x)])):
F1(bd)-1;
end proc;
fsolve(Q(pp2)=0,pp2=(0..1002));
se:=%;
res2:=dsolve({EQ,f(0)=0,D(f)(0)=0,(D@@2)(f)(0)=se},numeric,output=listprocedure):
G0,G1,G2:=op(subs(subs(res2),[f(x),diff(f(x),x),diff(f(x),x,x)])):
plots:-odeplot(res2,[seq([x,diff(f(x),[x$i])],i=1..1)],0..c);



Q2:=proc(rr2) local solT,T0,T1;
print(rr2);
if not type(rr2,numeric) then return 'procname(_passed)' end if:
solT:=dsolve({EQT2,theta(0)=1,D(theta)(0)=-rr2},numeric,known=[G0,G1,G2],output=listprocedure):
T0,T1:=op(subs(subs(res),[theta(x),diff(theta(x),x)])):
T0(bd);
end proc;
fsolve(Q2(rr2)=0,rr2=(0..100));


shib:=%;
sol:=dsolve({EQT2,theta(0)=1,D(theta)(0)=-shib},numeric,known=[G0,G1,G2],output=listprocedure):
plots:-odeplot(sol,[x,theta(x)],0..c);
#fsolve(Q2(pp3)=0,pp3=-2..2):

Amir

could you help me about maple
i try to calculating using chevypade rational approximating and the answer for cos(x) xe is(-.221091073962959*T(1, x-1)+.7710737338*T(0, x-1)-0.4212446689e-1*T(2, x-1))/(0.836360586596837e-1*T(1, x-1)+T(0, x-1)+0.3360079945e-1*T(2, x-1)) i can not to convert to rational form as x^^n .maple is not very friendship
Thanks

> restart;
> with*plots;
> Eq1 := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2 = 0; 

> N := 1;

> blt := 10;
> Eq2 := (diff(theta(eta), eta, eta))/Pr+f(eta)*(diff(theta(eta), eta)) = 0; 
> bcs1 := f(0) = 0, (D(f))(0) = 1, (D(f))(blt) = 0;
> bcs2 := (D(theta))(0) = -N*(1+theta(0)), theta(blt) = 0;
> L := [2.5, 3, 5, 7, 10];
> for k to 5 do R := dsolve(eval({Eq1, Eq2, bcs1, bcs2}, Pr = L[k]), [f(eta), theta(eta)], numeric, output = listprocedure); X1 || k := rhs(R[3]); X2 || k := rhs(R[4]); Y1 || k := rhs(R[5]); Y2 || k := -rhs(R[6]) end do;

 how I will draw the graph for Pr against theta   for Pr=2.5 until 7  taking rest of the parameter fix

Problem: I have two polynomials with arbitrary coefficients. I set them both to 0 and I used the 'map' , as well as 'coeffs' command to make the coefficients equal to 0.

Now for some reason, Maple does not print in order for one of the polynomials and it does for the other.

Note that 'order' refers to the coefficients attached to the powers of the variable.

Quick example: (this one actually works on Maple, but just not the one I have)

e1:= ax + (b + c)x^2 = 0

e2:= (c + d)x + (a + c)x^3 = 0

After applying map and coeff, one expects it to output

a = 0, (c + d) = 0, (b + c)=0, (a + c) = 0

instead I got

a = 0, (c + d) = 0, (a + c) = 0,(b + c)=0

Here is the problematic file

OutOfOrder.mw

I wish to evaluate the expression

knowing that

where a is a constant.  It is not hard to see, assuming enough differentiability,  that the expression evaluates to

I know how to do this when all the derivatives are expressed in terms of the diff() operator.  Here it is:

eq := diff(u(x,t),t) = a^2*diff(u(x,t),x,x);
expr := diff(u(x,t),t,t);
eval['recurse'](expr,[eq]);

However, I would prefer to do the computations when all derivatives are expressed in terms of the D operator but cannot get that to work.  What is the trick?