The calling sequence in ?complex states only the two argument form 'Complex(x, y)', but later in the section "Description", in a bit confusing way, two rules for the single argument form are stated:

I get "hypergeom" with Thomas Calculus exercise 5.6.15 for some unknown reason. the other exercises don't get this. See it at:

http://i36.tinypic.com/258lzqx.jpg

how can i get rid of the "hypergeom" in the answer ?

I need to do the following. Suppose I have an array, c[ ]. I need to be able to specialize the elements of this array, and then later "clear" the elements, making them indeterminates again. I attempted to do this by creating a temporary holding array

i:='i': for i from 1 to 5 do d[i]:=c[i] end do;
i:='i': for i from 1 to 5 do c[i]:=i end do;
i:='i': for i from 1 to 5 do c[i]:=d[i] end do;
 

I expected and want the output to be

c[0] c[1] c[2] c[3] c[4] c[5] 1 2 3 4 5 c[0] c[1] c[2] c[3] c[4] c[5]

Hello!

I have just found this incorrect behaviour of is  (Maple 12):

restart;

Hi all,

> deq := diff(x(t), t) = 3*x(t)/t+(9/2)*t-13;

                             d         3 x(t)   9       
                     deq := --- x(t) = ------ + - t - 13
                             dt          t      2       
> ci := x(3) = 6;

                               ci := x(3) = 6
> p := dsolve({ci, deq}, x(t), numeric);

                        p := proc(x_rkf45)  ...  end;
> plots[odeplot](p, view = [-1 .. 4, -10 .. 10]);

> p(0);

                  [                                     -8]
                  [t = 0., x(t) = 8.65023735199754930 10  ]

but if I do:

> q := dsolve({ci, deq}, x(t), type = numeric, method = taylorseries);

                    q := proc(x_taylorseries)  ...  end;
> plots[odeplot](q, view = [-1 .. 4, -10 .. 10]);


> q(0);

                             [t = 0., x(t) = 0.]
> solex := rhs(dsolve({ci, deq}, x(t)));

                                    9  2   13      3
                         solex := - - t  + -- t + t 
                                    2      2        

But in cases where I don't know the answer, which should I trust?  here is another one

> deq := diff(x(t), t) = 1-t-x(t)/t;

                               d                 x(t)
                       deq := --- x(t) = 1 - t - ----
                               dt                 t  
> ci := x(1) = 0;

                               ci := x(1) = 0
> q := dsolve({ci, deq}, x(t), numeric);

                        q := proc(x_rkf45)  ...  end;
> q(0);

                    [t = 0., x(t) = 1.73003351210698475]
> plots[odeplot](q);


> solex := rhs(dsolve({ci, deq}, x(t)));

                                    1  2   1      1 
                         solex := - - t  + - t - ---
                                    3      2     6 t
> plot(solex, t = -1 .. 1, -100 .. 100);


and for the finish

> r := dsolve({ci, deq}, x(t), numeric, method = taylorseries);

                    r := proc(x_taylorseries)  ...  end;
> plots[odeplot](r);
%;
Warning, could not obtain numerical solution at all points, plot may be incomplete

> r(0);

Error, (in r) cannot continue integration past t=0.585794295977905e-4, step size dropped below minimum

Thanks in advance for any help

Mario

Hi all,

Hi

Just started using the global optimization toolbox. Not sure if I am doing something wrong or if it is genuinely awful. After about 20mins it produces a fit that is dreadful,an alternative cheaper package (scop) produces a great fit in about 2 mins. Can anyone see what I'm doing wrong?

I have the following nonlinear ODE symmetric in the functions A(r) < - > B(r):

eq:= diff(A(r),r)^2/A(r)^2 + diff(A(r),r)*diff(B(r),r)/A(r)/B(r) + diff(B(r),r)^2/B(r)=k;

The task is to solve for A(r) in terms of B(r) and it goes like this: solve the quadratic equation for A'/A in terms of B'/B (or vise versa cause equation is symmetric) and then integrate.

dsolve(eq, A(r));  produces the required solution for A(r) in terms of integral over B and B'.

dsolve(eq, B(r)); returns nothing.

Should it really take 103 seconds to build a list of 95850 floats on a 2GHz machine?

I am using the data from the attached .m file and using the following code (I am not doing anything here but I will be) on the list, angles1_s which is read in from the .m file

Hello,

I need to plot a solid circle on a line. The problem is that you can see the line through the circle:

with(plots):
with(plottools): 
l := plot(x,color=red,thickness=3): 
d:=disk([0.5, 0.5],0.05,color=black): 
display({l, d},view=[0..1,0..1]);

I think I need something which is called in Microsoft Office products "Bring Forward (or to Front)" and "Send Backward (to Back)"

I suppose it's something very simple but I just cannot find it!

Thank you!

I have this function:

Eig1 := proc (R) options operator, arrow; (1/9*(-27*deltaE^3*R^9-10*M^6+3*sqrt(-189*deltaE^4*R^12*M^4-147*deltaE^2*R^6*M^8-27*M^12+60*deltaE^3*R^9*M^6))^(1/3)-(-deltaE^2*R^6-7/9*M^4)/(-27*deltaE^3*R^9-10*M^6+3*sqrt(-189*deltaE^4*R^12*M^4-147*deltaE^2*R^6*M^8-27*M^12+60*deltaE^3*R^9*M^6))^(1/3)+2/3*deltaE*R^3+2/9*M^2)/R^3 end proc

where:

> deltaE := 0.355 , and

M := sqrt(540000.0000)

Hi,

The eigenvalue command gives me a vector like: [ [5] , [4] , [3] ].

Is there a quick way I could grap the first eigenvalue and make that a function without copying and pasting?

Hi I used the "search" option of this forum. I have seen old messages about  the continuation of line "#" of fortran77 that is different from the ones in fortran90. I have thousands lines of fortran code created by codegeneration and I can't use "replace all # with &", because # are at the beginning and "&" at the end of the line. It takes a while to substitute all of them by hand every time I generate the fortran subroutine. Do you know if they are going to introduce the continuation of line with & instead #?

thanks