diff(a(t), t) = diff(a(t), t);
diff(b(t),t) = 0;
diff(c(t),t) = -b(t)/a(t);

[diff(rhs(sol[1][2]), a(t)),diff(rhs(sol[1][2]), b(t)),diff(rhs(sol[1][2]), c(t))];
Error, (in VectorCalculus:-diff) invalid input: diff received a(t), which is not valid for its 2nd argument

hey guys Im new client in maple and today I was about check out the resualt of my mathematic quastion with maple.

I need a step by step solution and exact command to give me true resualts 

for example 

how can I expand a factorization like (x^2-y^2) to (x-y)(x+y)

in a little more  complicated form (cd-1)^2-(c-d)^2/(d^2)(c-1)=5 the value of c=?

for solve this problem I need to expand (cd-1)^2-(c-d)^2 than other expands & in the end value of c

I dont have anymore time for my mathemathic exam so know that how maple works in basic and intermadiate mathematic level is important to me

thank you guys

eq2 := b(t)*(diff(c(t), t))*(diff(a(t), t))+b(t)*(diff(a(t), t))+a(t)*(diff(c(t), t));
eq3 := a(t)*(diff(b(t), t))(diff(a(t), t))+b(t)*(diff(b(t), t))*(diff(c(t), t));
eq4 := b(t)*(diff(c(t), t))(diff(b(t), t))+a(t)*(diff(b(t), t))+b(t)*(diff(c(t), t));
dfieldplot([eq2,eq3,eq4],[t,x],t=0..5,a=-5..5,b=-5..5,c=-5..5);
dfieldplot([eq2,eq3],[t,x],t=0..5,a=-5..5,b=-5..5);
eq2a := eval(subs(c(t)=exp(t), eq2));
eq3a := eval(subs(c(t)=exp(t), eq3));
eq4a := eval(subs(c(t)=exp(t), eq4));
dfieldplot([eq2a,eq3a], [a(t), b(t)], t = -5 .. 5, a = -5 .. 5, b = -5 .. 5, arrows = SLIM, color = black, dirfield = [10, 10]);

Hey,

I think I found a bug concerning the useage of assume and alias:

Commenting out the alias command produces correct results. I am on linux with build 922027.

alias_bug.mw

I want to solve a system of equations using f-solve (two unknowns) and exporting the solutions to a matrix where the solutions are in seperate columns. How do I do this?

I have tried:

for i from 1 to 937 do

AP[i,1]:=fsolve(x=KL[i,1].y=KL[i,2],x=..8,y=0..15))

end if

end do

But this returns the solutions for X and Y in the same column. Also, for the values that are not possible to solve, it returns the entire expression instead of e.g. 0 or "undefined".

Thank you.

 from determinant's polynomial?                                                                                                       

Dear All

I have updated my Maple 18, I am surprised to see that ordinary "solve" do not work and return error massage like;

"Error, (in coulditbe) invalid input: `coulditbe/internal` uses a 1st argument, obj, which is missing"

or like;

"Error, (in solve) invalid input: hastype expects 2 arguments, but received 1"

I am totally confused !!!

Can anybody help me out please !!!!

Download Solve_Command.mw

Regards

Dear All

I have third party Maple package saved along path E:/Maple work/General Maple Workout/TWS.mpl, but after using march command for other package, Maple is reporting error like "unable to read; E:/Maple work/General Maple Workout/TWS.mpl". What could be possible reason for this?

Moreover when I type "currentdir()" it shows me "C:\WINDOWS\system32" which is right path from where Maple is working

Regards

How do I write a procedure to find a root of f(x)=0 in the vicinity of a given value x0. The procedure should initially use the rearrangement method to produce a linearly convergent sequence of values, and should, when appropriate, switch to Aitkin's Method. The input for the procedure should be the re-arranged function and the velue for x0. The output should be the root and the number of iteration taken. The procedure should check that re-arrangement will converge. This program should do in Maple V Release5.

Thank you for your help.

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