Let a be som expression like
a:=t^2

Now how do I define a function using a.

f:=t->a
doesnt work

I've got the following piecewise function :

(x^2+y^2)^(alpha).arcsin(y/x) if (x,y) are in [-pi/2,pi/2]

0, (x,y)=(0,0)

1. How do I plot this function taking the alpha variable and the piecewise construct into account?

2. How can I check for points of discontinuity, indifferentiability from the plot/function itself?

Hello everybody,

I'm an engineer and i want to solve the following ODE:

EJ*diff(w(x),x,x,x,x) - (Nr/2*(1+x)+Nl/2*(1-x))*diff(w(x),x,x) = q(x)

constants: EJ, Nr, Nl

Is it possible to solve this ODE in Maple? I tried dsolve with coeffs=rational but it doesn't work!

Thx!!!

I am attempting to maximize Planck's intesnity function for any given temperature T with respect to lambda. The function is as follows:

My initial attempt was to simply take the derivative of this function with respect to labda, set it equal to zero, and then solve for lamda. Theoretically this should give me the value of lambda that maximizes the intesity with respect to the constants / values h, c, k, and T (where T is temperature).

However doing this in maple just leads to the result "warning, solutions may have been lost".

Does anyone know how I can go about maximizing this?

Thanks!

When I try and evaluate the minimize command the result is just echoed.

If I try Maples example function, they echo such as:

minimize(x^2+y^2-3*x+3*y+3);

         minimize(x^2 + y^2 - 3 x + 3 y + 3)

Any suggestions? 

Thanks.

Hi everyone,

I am using the following code (by dr. Corless) to animate a Riemann surface:

restart;
with(plots);
B := 1.5;
u2 := r*cos(th); v2 := r*sin(th);
w1 := u1+I*v1; w2 := u2+I*v2;
z1 := evalc(w1^2); z2 := evalc(w2^2);
x1 := evalc(Re(z1)); x2 := evalc(Re(z2));
y1 := evalc(Im(z1)); y2 := evalc(Im(z2));
f1 := proc (theta) options operator, arrow; plot3d([-x1, -y1, v1], u1 = -6 .. 1, v1 = -B .. B, grid = [50, 50], orientation = [theta, 80], color = u1) end proc;
f2 := proc (theta) options operator, arrow; plot3d([-x2, -y2, v2], r = 1 .. 1, th = -Pi .. Pi, grid = [50, 50], orientation = [theta, 80], color = black) end proc;
display(seq([f1(10*k), f2(10*k)], k = -17 .. 18), insequence = true, axes = box, view = [-1 .. 1, -1 .. 1, -B .. B]);

f1 is the Riemann surface and f2 is the winding curve.

The animation works fine on the individual plots, i.e. when I do:

display(seq(f1(10*k), k = -17 .. 18), insequence = true, axes = box, view = [-1 .. 1, -1 .. 1, -B .. B]);

or

display(seq(f2(10*k), k = -17 .. 18), insequence = true, axes = box, view = [-1 .. 1, -1 .. 1, -B .. B]);

but with both f's (as in my modification)  the animation does not combine correctly the two plots and shows them separately. Is there any way I can combine the plots so that display produces a smooth animation with each rotated frame containing its winding curve?

Many thanks,

Yiannis

Hello there, could someone tell me how to input this differential equation?

d²x/dt² - (3)dx/dt + 2x = 2t -3 

x=2, t=0, dx/dt=4

So it looks like radians work.

Why does degrees fail and how do I get degrees work?

How do you guys like to access pi? Do you keep a symbol of it around in a random document to open? 

http://i.imgur.com/07o1dyH.png

Let A(x , 2 , -1) ,B(0, y , 1) ,C(3 , 4 , -2) , AB=AC and ABC is isosceles right triangle. Find x and y ?

How would you insert a label-coordinate in this command:

textplot3d([subs(E[11]=0,x[11]),subs(E[11]=0,y[11]),subs(E[11]=0,z[11]),"Planet"]):

What is the command for filling the space between two circular spacecurves? Something like a Saturnian ring?

Hi there,

I'm trying to get the values from the output of a dsolve command.

I have a system of differential equations:

de1 := diff(V(t), t) = V(t)-(1/3)*V(t)^3-W(t)+Ie;
de2 := diff(W(t), t) = 0.8e-1*(V(t)+.7-.8*W(t))

For a range of the independent variable t and for some given values of the parameter Ie, I would like to get the value of the dependent variable V, as well as its minimum/maximum values for each Ie.

Can anybody suggest a solution please?

This is the worksheet: MaplePrimes_dsolve_min-max.mw

Thanks,

jon

Hello all,

I'm experiencing an aggravating issue when substituting numerical values into a symbolic expression.  I'm using Maple 14.  I believe the issue may be related to floating point precision.  Either way, it's ruining my life.  If anyone has a solution or a work-around, I'd be most grateful.  The following code reproducibly produces the anomolay on my machine:

vx := Vector( 4, symbol=x ):

# A numerator and a denominator. They're equal.
num := (x[1]*x[3]+x[1]*x[4]+x[2]*x[3]+x[2]*x[4])^2:
den := ((x[1]+x[2])^2*(x[3]+x[4])^2):

# Prints true...
evalb(num=den);

# This equals zero
y := 0.1 - 0.1*num/den;

# Prints 0...
simplify(y);

# Prints 0...
simplify(subs(x[3]=1,x[4]=3,0.1-0.1*num/den));

# Prints 0...
simplify(subs(x[3]=1,x[4]=3,y));

# Prints 0.1 x 10^-10 !?!?
simplify(subs(x[3]=1,x[4]=2,y));

Hi,

I need you help to understand this problem.

Let alpha=0.1.;

When I do :  (-0.2)^alpha I get a complex number.

I must find a real number.

What's the problem

Thanks for your idea.

Hello,

I have a little problem. I have a system of ODEs, I can solve (with numeric solve) it but I can't plot it...
Any idea to help ?

Thank you !!

Here is my code:

with(DEtools);
beta := 1; lambda := 5; nu := 1; Delta := 1;


sys := {C(0) = 0, In(0) = .2, diff(C(u), u) = beta*In(u)*s(u)-C(u)/nu, diff(In(u), u) = C(u)/nu-In(u)/lambda+Delta*(diff(diff(In(u), u), u)), diff(s(u), u) = -beta*In(u)*s(u), s(0) = .8, (D(In))(0) = .2}

dsys := dsolve(sys, numeric, [In(u), s(u), C(u)])

plot(dsys[1], u = 0 .. .5, axes = boxed)

And I have this error:

Warning, expecting only range variable u in expression dsys[1] to be plotted but found name dsys[1]


PS: Maple solves the system without any problem:

 dsys(.5);

        [                                              
        [u = 0.5, In(u) = HFloat(0.33305297276372425),
        [                                              

           d                                       
          --- In(u) = HFloat(0.33950703755414946),
           du                                      

          s(u) = HFloat(0.7022184797811994),

                                           ]
          C(u) = HFloat(0.0781621551198558)]
                                           ]