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Dear all;

I open this good discussion, and hope can get a nice and strong idea in this domain of approximation of Hankel funciton and order truncation of infinite series. Thanks for all idea, can improve the discussion. 

Using the asymptotics of Hankel function for large argument and large orders ( both together) and
find   an order of truncation N of the obove series so that we can ensure an error bound  of epsilon( epsilon very small given).  abs(sum ( c[m]*HankelH1(m,x)*exp(I*m*theta), m=-infinity..infinity)-sum ( c[m]*HankelH1(m,x)*exp(I*m*theta), m=-N..N))<epsilon.

A first idea come in mind: the series converge, so that the general terms of this series converge to zero, and in particularity,  abs(c[m]* HankelH1(m,x))<1:
then abs(c[m])<1/ abs(HankelH1(m,x)).
So we can ensure an error bound of epsilon on the coefficient c[m] by imposing  (HankelH1(m,x))<epsilon  this lead to abs(c[m])<epsilon.
I consider the case where m and x are very large, I can suppose for example m=x*(1+zeta), with 0<zeta<1. zeta parameter. So that our truncation N depend on zeta.
And then how can I find and approximation of the inverse of Hankel function for large argument and large order. using m=x*(1+zeta). I think this give us N the truncation order.
I hope get a good discussion in this subjet.
Of course maybe there are other strong idea to compute the truncation series.
I get the following error in the code:

Error, (in MultiSeries:-multiseries) unable to expand with respect to parameter

 

 

 

###### Code### and error
restart:

with(MultiSeries):
assume(0 <= x);

assume(0 <= zeta<1);

HankelH1(v,x):

sum(c[m]*HankelH1(m, x)*exp(I*m*theta),m=-infinity..infinity);

1/HankelH1(x*(1+zeta),x);

MultiSeries:-asympt(%,x, 4);

eval(%, O=0);

convert(%,exp);

simplify(%);

truncated_series.mw

The issue Type check of parameters was resolved using the depends modifier. As far as I can tell, this modifier is not allowed for expected or keyword parameters, though. Thus the issue seems to reemerge for these types of parameters. Consider the following test example:

createModule := proc(V::Vector)
   local dim := LinearAlgebra:-Dimension(V);
   module()
      export f,g,h;
      f := proc( x::depends('Vector'(dim))              ) x end proc;
      g := proc( x::expects('Vector'(dim)) := something ) x end proc;
      h := proc({x::        'Vector'(dim)  := something}) x end proc;
   end module
end proc:
createModule(Vector(4)):-f(    Vector(4));
createModule(Vector(4)):-g(    Vector(4));
createModule(Vector(4)):-h(x = Vector(4));

The function f is just a restatement of the already resolved issue, compare the above link, while the functions g and h are for the expected and keyword parameter cases, respectively. The problem remains the same: the variable dim is not evaluated for g and h. What to do? Does there exist a solution equally satisfactory as the one for f?

Hello dears! Hope all you are fine. I am facing problem to integrate some expression please see the attachment and fix the problem. I am very thankful for your favour. 

Integration.mw

Mob #: 0086-13001903838

I want to solve the equation

sqrt(x)+sqrt(-x^2+1) = sqrt(-4*x^2-3*x+2)

in Real domain. I tried

RealDomain:-solve(sqrt(x)+sqrt(-x^2+1) = sqrt(-4*x^2-3*x+2), x);

and I got -5/9+(1/9)*sqrt(34).

But, with Mathematica, I posted my question at http://mathematica.stackexchange.com/questions/51316/how-can-i-get-the-exact-real-solution-of-this-equation

Mathematica had two solutions 

x ==-1-Sqrt[2]|| x ==1/9(-5+Sqrt[34])

If I understand correctly, when Maple solve in RealDomain of this equation, the solution of equation must satisfy conditions x>=0 and -x^2+1 >=0 and -4*x^2-3*x+2 >=0. Therefore, the number

x ==-1-Sqrt[2] 

is not a solution. My question is the given equation has one solution (Maple) or two solutions (Mathematica)?

I get the maple result as a product and sum off terms

 

x= (a+b*c)/d

i want to convert it to

x= a/d+b*c/d

i try to use convert(x,?)  

 

I am trying to find a general solution to the 1D-wave equation

Eq1:=diff(u(x,t),t$2)=diff(u(x,t),x$2);

pdsolve(Eq1,HINT=f(x)*g(t)); # Hinting pdsolve gives general solution using separation of variables

pdsolve({Eq1,u(x,0)=f(x),D[2](u)(x,0)=g(x)}); # without HINT and using intial conditions, I get travelling wave solution

pdsolve({Eq1,u(x,0)=f(x),D[2](u)(x,0)=g(x)},HINT=f(x)*g(t)); # Now when I try to use hint and ICs both, pdsolve return nothing.

I want to use separation of variables to find solution to the wave equation.

Any comment?

Thanks

 

It returns unevaluated.  The solution is x=-ln(3),y=0.  In fact it doesn't give a solution even if the solution is provided as the initial point.  The value of Digits doesn't seem to make a difference.

(Tested Maple 2015.2 Macintosh and Maple 2015.1 Linux)

Salut j'ai fait un programme avec maple   pour minimiser une fonction à deux variable avec la méthode de fletcher reevers car la fonction minimize ne marche pas avec les fonction plus compliqué.
SP veuillez voir mon programmme et essayer de le corriger car la boucle while ne marche pas:
test.mwtest.mw
Cordialement:

Does any one know if you can extract the linear graph from a Maple Sim model?  And by linear graph I mean the alternative to a a bond graph, not a type of plot.

the program is not recognized kthe program is not recognized the k

I have a set of differential equations on 3 variables, B[1],B[2] and C. Its not physically meaningful for B[1]+B[2]>0.5 so i would ideally like to replace the cube that the solutions are displayed on (the axis take the limits B[1]=0...0.5,B[2]=0...0.5,C=0...100 ) with a triangular prism (the axis take the limits B[1]=0...0.5,B[2]=0...0.5,B[1]+B[2]<0.5,C=0...100 ).

Failing that i'd like the plot to display a plane showing where the meaningful values for the variables end.

here is the code I use to put the ODEplot together

Model := [diff(B[1](t), t) = k[a1]*C(t)*(R-B[1](t)-B[2](t))-k[d1]*B[1](t), diff(B[2](t), t) = k[a2]*C(t)*(R-B[1](t)-B[2](t))-k[d2]*B[2](t), diff(C(t), t) = (-(k[a1]+k[a2])*C(t)*(R-B[1](t)-B[2](t))+k[d1]*B[1](t)+k[d2]*B[2](t)+k[m]*((I)(t)-C(t)))/h];
DissMod := subs((I)(t) = 0, Model);
AssMod := subs((I)(t) = C[T], Model);

Pars1a := [k[a1] = 6*10^(-4), k[d1] = 7*10^(-3), k[a2] = 6*10^(-4), k[d2] = (7/5)*10^(-3), R = .5, k[m] = 10^(-4), C[T] = 100, h = 10^(-6)];

Pars := Pars1a; thing11 := subs(Pars, AssMod[1]), subs(Pars, AssMod[2]); thing12 := diff(C(t), t) = piecewise(t <= 100, subs(Pars, rhs(AssMod[3])), subs(Pars, rhs(DissMod[3]))); sol := dsolve({thing11, thing12, C(0) = 0, B[1](0) = 0, B[2](0) = 0}, {C(t), B[1](t), B[2](t)}, numeric, output = listprocedure, maxstep = 2, maxfun = 1000000); ParsPlot1a := odeplot(sol, [B[1](t), B[2](t), C(t)], t = 0 .. 700, color = blue, view = [0 .. .5, 0 .. .5, 0 .. 100], tickmarks = [[0 = 0, .5 = R], [0 = 0, .5 = R], [0 = 0, 25 = (1/4)*C[T], 50 = (1/2)*C[T], 100 = C[T]]]);


But I can't see a way of either making the plane or making the ODEplot axis into something other than a cube.

Could someone explain, in layman terms, what dilog and polylog are?

I don't seem to understand them from the Help and I cannot find anything on the web.

 

Thank you

I have the mathematica code given below. I am using Wolfram Mathematica Online.
I am very new to Mathematica. When I put this code, I do not get any output.
It just does not show me anything.




\[Alpha] = 3; F[s_] := Exp[-A*s^(2/\[Alpha])]; integral = Re[Assuming[{A > 0, t > 0, {t, A} \[Element] Reals}, Integrate[F[s]*Exp[s*t] /. s -> I*y, {y, 0, Infinity}]/Pi]]

Can some help me?

Dear all

I have problem related to collection of coefficient of differtials in differential expression containing multiple dependent variables and we want to collect coefficient wrt to selected dependent variables. Please see attached Maple file for details.

 


with(PDEtools):

DepVars := [u(x, t), v(x, t), a[1](t), a[2](t), a[3](t), b[1](t), b[2](t), b[3](t), r(x, t), s[1](x, t), p[1](x, t), s[2](x, t), p[2](x, t)]

[u(x, t), v(x, t), a[1](t), a[2](t), a[3](t), b[1](t), b[2](t), b[3](t), r(x, t), s[1](x, t), p[1](x, t), s[2](x, t), p[2](x, t)]

(1)

alias(u = u(x, t), v = v(x, t), a[1] = a[1](t), a[2] = a[2](t), a[3] = a[3](t), b[1] = b[1](t), b[2] = b[2](t), b[3] = b[3](t), r = r(x, t), s[1] = s[1](x, t), p[1] = p[1](x, t), s[2] = s[2](x, t), p[2] = p[2](x, t))

u, v, a[1], a[2], a[3], b[1], b[2], b[3], r, s[1], p[1], s[2], p[2]

(2)

Suppose we differential expression like:

a[1]*(diff(u, x))*s[1]*u-2*a[1]*u*(diff(r, x))*(diff(u, x))+2*a[2]*(diff(v, x))*s[2]*v-2*a[2]*v*(diff(r, x))*(diff(v, x))-(diff(a[3], t))*r*(diff(u, x))/a[3]+diff(p[1], t)+a[3]*(diff(p[1], x, x, x))+(diff(r, t))*(diff(u, x))+(diff(s[1], t))*u-(diff(a[3], t))*s[1]*u/a[3]-s[1]*a[2]*v*(diff(v, x))-(diff(a[3], t))*a[1]*u*(diff(u, x))/a[3]-(diff(a[3], t))*a[2]*v*(diff(v, x))/a[3]-3*(diff(r, x))*p[1]+(diff(a[1], t))*u*(diff(u, x))+(diff(a[2], t))*v*(diff(v, x))+a[2]*(diff(v, x))*p[2]+a[2]*v^2*(diff(s[2], x))+a[2]*v*(diff(p[2], x))+a[1]*u*(diff(p[1], x))+a[1]*(diff(u, x))*p[1]+a[1]*u^2*(diff(s[1], x))+3*a[3]*(diff(s[1], x))*(diff(u, x, x))+3*a[3]*(diff(s[1], x, x))*(diff(u, x))+a[3]*(diff(r, x, x, x))*(diff(u, x))-(diff(a[3], t))*p[1]/a[3]-3*r*(diff(r, x))*(diff(u, x))-3*(diff(r, x))*s[1]*u+a[3]*(diff(s[1], x, x, x))*u+3*a[3]*(diff(r, x, x))*(diff(u, x, x)) = 0

3*a[3]*(diff(diff(r, x), x))*(diff(diff(u, x), x))+3*a[3]*(diff(s[1], x))*(diff(diff(u, x), x))+3*a[3]*(diff(diff(s[1], x), x))*(diff(u, x))+a[3]*(diff(diff(diff(r, x), x), x))*(diff(u, x))+a[3]*(diff(diff(diff(s[1], x), x), x))*u+diff(p[1], t)+(diff(r, t))*(diff(u, x))+(diff(s[1], t))*u-3*(diff(r, x))*p[1]+a[3]*(diff(diff(diff(p[1], x), x), x))-(diff(a[3], t))*a[1]*u*(diff(u, x))/a[3]-(diff(a[3], t))*a[2]*v*(diff(v, x))/a[3]+a[1]*(diff(u, x))*s[1]*u-2*a[1]*u*(diff(r, x))*(diff(u, x))+2*a[2]*(diff(v, x))*s[2]*v-2*a[2]*v*(diff(r, x))*(diff(v, x))-(diff(a[3], t))*r*(diff(u, x))/a[3]-(diff(a[3], t))*s[1]*u/a[3]-s[1]*a[2]*v*(diff(v, x))+(diff(a[1], t))*u*(diff(u, x))+a[1]*u*(diff(p[1], x))+a[2]*v*(diff(p[2], x))+a[2]*v^2*(diff(s[2], x))+a[2]*(diff(v, x))*p[2]+a[1]*(diff(u, x))*p[1]+a[1]*u^2*(diff(s[1], x))-(diff(a[3], t))*p[1]/a[3]-3*r*(diff(r, x))*(diff(u, x))-3*(diff(r, x))*s[1]*u+(diff(a[2], t))*v*(diff(v, x)) = 0

(3)

We can collect coefficients of differential like u[x], u[x, x], v[x], u, vin following manner:

The Procedure

   

 

 

Now how can we collect coefficents with respect to u[x], u[x, x], v[x], u, vso that differential expression (3) appear as
"(......)*u+(.......)*v+(......)*u[x]+(........)*uu[x]+(.........)vv[x]+(........)u[xx]  =0....................."????????""

``


Download Collecting_Coefficients_in_Differential_Expression.mw

Regards

We are running Maple16 in a virtual lab setup using Citrix, when exporting to PDF and saving the file to the local machine the PDF comes out as a 0kb file and the PDF is corrupt.  Has anyone else had an issue like this before?

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