Hi,

I am using the solve command to solve an equation of the form "linear over quadratic is equal to a constant" where the constant is assumed to be nonzero. This is easily solved by hand, of course, but I to use the solution in other computations. So I asked maple to solve it for me. But when I check maple's solution (i.e. just plug the two solutions in on the left hand side and simplify) maple does not return the original constant. Can anyone help me understand what is going wrong?

Construction of arabesques of melodic line BACH

Elena, Liya "Construction of arabesques of melodic line BACH", Kazan, Russia, school#57
       
> restart:
> with(plots):with(plottools):

      The setting and visualization of line BACH: B - note b-flat, A - note la, C - note do, H - note si.
> p0:=plot([[0,1],[2,0],[4,1.5],[6,1]],thickness=4,color=cyan,scaling=constrained);
>
>   p0 := PLOT(
>
>         CURVES([[0, 1.], [2., 0], [4., 1.500000000000000], [6., 1.]])
>
>         , SCALING(CONSTRAINED), THICKNESS(4), AXESLABELS( ,  ),
>
>         COLOUR(RGB, 0, 1.00000000, 1.00000000),
>
>         VIEW(DEFAULT, DEFAULT))
>
> plots[display](p0);
> r_i:=seq(rotate(p0,i*Pi/4),i=1..8):
> p1:=display(r_i,p0):plots[display](p1,scaling=constrained);

> c1:=circle([0,0],6,color=blue,thickness=2):
> plots[display](c1,p1,scaling=constrained);
> p_c:=plots[display](c1,p1,scaling=constrained):

> pt_i_2:=seq(translate(p1,0,2*6*i),i=0..4):
> plots[display](pt_i_2,scaling=constrained);
> pt_i_22:=seq(translate(p1,0,6*i),i=0..4):
> plots[display](pt_i_22,scaling=constrained);
> pt_i_222:=seq(translate(p1,0,1/2*6*i),i=0..4):
> plots[display](pt_i_222,scaling=constrained);

> pr:=rotate(p1,Pi/8):
> plots[display](pr,scaling=constrained);
> plots[display](p1,pr,scaling=constrained);
> pr_i:=seq(rotate(p1,Pi/16*i),i=0..8):
> plots[display](pr_i,scaling=constrained);


> pt_1:=translate(p1,0,2*6):
> pr_1_i:=seq(rotate(pt_1,Pi/3.5*i),i=0..6):
> plots[display](pr_1_i,scaling=constrained);
> pr_11_i:=seq(rotate(pt_1,Pi/5*i),i=0..10):
> plots[display](pr_11_i,scaling=constrained);
> pr_111_i:=seq(rotate(pt_1,Pi/6.5*i),i=0..12):
> plots[display](pr_111_i,scaling=constrained);

Elena, Liya "Designing of islamic arabesques", Kazan, Russia, school #57

> restart:
      At the theorem of cosines  c^2 = a^2+b^2-2*a*b*cos(phi);
      In our case  c=a0 ,  a=1 ,  a=b , phi; - acute angle of a rhombus (the tip of the kalam).
      s0 calculated at theorem of  Pythagoras.
     (а0 - horizontal diagonal of a  rhombus, s0 - vertical diagonal of a  rhombus)
> a:=1:phi:=Pi/4:
> a0:=sqrt(a^2+a^2-2*a^2*cos(phi));

                       a0 := sqrt(2 - sqrt(2))

> solve((s0^2)/4=a^2-(a0^2)/4,s0);

                sqrt(2 + sqrt(2)), -sqrt(2 + sqrt(2))


      The setting of initial parameters : the size of the tip of the pen-kalam and  depending on its - the main module size - point
       (а0 - horizontal diagonal of a  rhombus, s0 - vertical diagonal of a  rhombus)
> a0:=sqrt(2-sqrt(2)):
> s0:=sqrt(2+sqrt(2)):
      Connection the graphical libraries Maple
> with(plots):with(plottools):
      Construction of unit of measure (point) - rhombus - the tip of the kalam
> p0:=plot([[0,0],[a0/2,s0/2],[0,s0],[-a0/2,s0/2],[0,0]],scaling=constrained,color=gold,thickness=3):
> plots[display](p0);

The setting and construction of altitude of alif - the basis of the rules compilation of the proportions      Example, on style naskh altitude of alif amount five points
> p_i:=seq(plot([[0,0+s0*i],[a0/2,s0/2+s0*i],[0,s0+s0*i],[-a0/2,s0/2+s0*i],[0,0+s0*i]],scaling=constrained,color=black),i=0..4):
> pi:=display(p_i):
> plots[display](p_i);
The setting of appropriate circle of diameter, amount altitude of alifd0:=s0+s0*i:
> i:=4:
> d0:=d0:
> c0:=circle([0,d0/2],d0/2,color=blue):
> plots[display](p_i,c0);

Construction of flower by turning "point"r_i:=seq(rotate(p0,i*Pi/4),i=1..8):
> p1:=display(r_i,p0):plots[display](p1,scaling=constrained);

 The setting of circumscribed circlec1:=circle([0,0],s0,color=blue,thickness=2):
      Construction and the setting of flower inscribed in a circle
> plots[display](c1,p1,scaling=constrained);
> p_c:=plots[display](c1,p1,scaling=constrained):

The setting and construction of arabesque by horizontal parallel transport original flower with different stepspt_i_1:=seq(translate(p1,5*a0*i,0),i=0..4):
> plots[display](pt_i_1);
> pt_i_11:=seq(translate(p1,2*a0*i,0),i=0..4):
> plots[display](pt_i_11);
> pt_i_111:=seq(translate(p1,a0*7*i,0),i=0..4):
> plots[display](pt_i_111);

 The setting and construction of arabesque by vertical parallel transport original flower with different stepspt_i_2:=seq(translate(p1,0,2*s0*i),i=0..4):
> plots[display](pt_i_2);
> pt_i_22:=seq(translate(p1,0,s0*i),i=0..4):
> plots[display](pt_i_22);
> pt_i_222:=seq(translate(p1,0,1/2*s0*i),i=0..4):
> plots[display](pt_i_222);
 Getting arabesques by turning original flower on different anglespr:=rotate(p1,Pi/8):
> plots[display](pr);
> plots[display](p1,pr);

> pr_i:=seq(rotate(p1,Pi/16*i),i=0..8):
> plots[display](pr_i);


> pt_1:=translate(p1,0,2*s0):
> pr_1_i:=seq(rotate(pt_1,Pi/3.5*i),i=0..6):
> plots[display](pr_1_i);
> pr_11_i:=seq(rotate(pt_1,Pi/5*i),i=0..10):
> plots[display](pr_11_i);
> pr_111_i:=seq(rotate(pt_1,Pi/6.5*i),i=0..12):
> plots[display](pr_111_i);

Construction of standard quadrilaterals

      Muchametshina Liya,  8th class,  school № 57, Kazan, Russia


                   Square

                  Rectangle     
                  
                  Rhombus        
 
                  Parallelogram

                   Trapeze

Construction of square

> restart:
> with(plottools):
       Сoordinates (x;y) of the lower left corner of the square and the side "а"
> x:=0;y:=3;a:=6;

                                x := 0


                                y := 3


                                a := 6

      Construction of the square
> P1:=plot([[x,y],[x,y+a],[x+a,y+a],[x+a,y],[x,y]],color=green,thickness=4):
> plots[display](P1,scaling=CONSTRAINED);

The setting of the second square wich moved relative to the first on the vector (2;-3) (vector can be changed) and with side "а-1" (the length of a side can be changed)P2:=plot([[x+2,y-3],[x+2,y-3+a-1],[x+2+a-1,y-3+a-1],[x+2+a-1,y-3],[x+2,y-3]],color=black,thickness=4):
> plots[display](P1,P2,scaling=CONSTRAINED);

Construction of rectangle

> restart:
> with(plottools):
        Сoordinates (x;y) of the lower left corner of the square and the "а" and "b" sides
> x:=0;y:=2;a:=3;b:=9;
>

                                x := 0


                                y := 2


                                a := 3


                                b := 9

       The rectangle is specified by the sequence of vertices with given the lengths "a" and "b"
> l:=plot([[x,y],[x,y+a],[x+b,y+a],[x+b,y],[x,y]]):
> plots[display](l,scaling=CONSTRAINED,thickness=4);
Construction of rhombus

> restart:
> with(plottools):
      The coordinates (x;y) of the initial vertex of the rhombus and the half of the diagonals "a" and "b"
> x:=0;y:=2;a:=3;b:=4;

                                x := 0


                                y := 2


                                a := 3


                                b := 4

       Rhombus is specified by the sequence of vertices with the values "a" and "b"
> ll:=plot([[x,y],[x+a,y+b],[x+a+a,y],[x+a,y-b],[x,y]]):
> plots[display](ll,scaling=CONSTRAINED,thickness=4);

Construction of parallelogram

> restart:
> with(plottools):
      (х;у) - the starting point, (i;j) - the displacement vector of starting point, "а" - the base of the parallelogram
> x:=0;y:=0;i:=4;j:=5;a:=10;

                                x := 0


                                y := 0


                                i := 4


                                j := 5


                               a := 10

     The parallelogram is defined by the sequence of vertices
> P1:=plot([[x,y],[x+i,y+j],[x+i+a,y+j],[x+a,y],[x,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);
 If  i= 0  it turns out the rectangleget.
       If  j= а  it turns out the  square.
       If  a := sqrt(i^2+j^2) it turns out the rhombus. a:=sqrt(i^2+j^2):

Construction of trapeze

Trapeze general form
> restart:
> with(plottools):
>
        (х;у) - the starting point, (i;j) - the displacement vector of starting point, а - the larger base of the trapezoid
> x:=0;y:=2;i:=1;j:=5;a:=11;

                                x := 0


                                y := 2


                                i := 1


                                j := 5


                               a := 11

         The trapez is defined by the sequence of vertices     
> P1:=plot([[x,y],[x+i,y+j],[x+i+j,y+j],[x+i+a,y],[x,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);
Rectangular trapezoid
> restsrt:
> with(plottools):
> x:=0;y:=2;i:=0;j:=6;a:=11;

                                x := 0


                                y := 2


                                i := 0


                                j := 6


                               a := 11

> P1:=plot([[x,y],[x,y+j],[x+j,y+j],[x+a,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);
Isosceles trapezoid
> restart:
> with(plottools):
> x:=0;y:=2;i:=4;j:=6;a:=15;

                                x := 0


                                y := 2


                                i := 4


                                j := 6


                               a := 15

> P1:=plot([[x,y],[x+i,y+j],[x+j+i,y+j],[x+a,y],[x,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);

Dear Forum, 

I am a new Maple user, and its symbolic prowess is really amazing. So we are trying to interface it with a C library. I want to generate some C code through Maple, and am trying the CodeGeneration package. 

But the default conversion of C(a, b) is b = C language equivalent of expression a.

Now this should be fine for most purposes, but the C library that we are working with, "ACADOToolkit" in this case, requires the equations to be formatted in a certain way. So, I need the following equation in C:

f << dot(v) == (u-0.2*v*v)/m

Now the LHS part of == is to be hard-coded, but we want to generate the equation on the right using maple. Even if I define an equation as 

eq1:=  and then use C(rhs(eq1)), I get the result in the form of cg = u - 0.2 ...., whereas I want this to be assigned to something else, in this case - "f << dot(v)= ".

How can I achieve this ?

Thanks 

Chintan Pathak 

Research Scholar, 

University of Washington

hello , 

how i can exract value from pdsolve ,i need to use dU(x,R)/dR 

thank you 

Download U(R)_numériqueg2.mw

Dear all

I have a confusion between these symbol

Sum , add and sum

If we consider u(n) is a sequence and n integer

and what is the difference between 

sum( u(n),n=0..infinity)

Sum(u(n),n=0..infinity)

and sum('u(n)', n=0..infinity)

Many thanks

HI all,

I have 

> sol2 := dsolve({odesys, H(0) = 4995, R(0) = 65000, W(0) = 102000, l(0) = 96000}, numeric, method = rosenbrock);
print(`output redirected...`); 
proc(x_rosenbrock) ... end;

I want to have a list of my solutions, t, H(t), R(t), W(t), l(t) that I can put into a spreadsheet (.csv, .txt, etc.), for 600 timesteps. 

There are some answers out there, but I am confused by them, and have not been able to make it work.

Thanks!

How can I simplify $\sqrt{1−r^2\exp(2i\theta)}$ in Maple. I could do it by hand but I need this type of simplification later for far more complicated expressions.  I allready tried to enter this as a complex number using II, but simplify(...,'symbolic') didn't simplify this expression. Any suggestion?

I have a nonlinear function Q(a,b,c,d,x,y) and I'd like to get the optimum (x*,y*) for different values of (a,b,c,d). The usual sintax:

NLPSolve(Q(10, 1, 5, 2, x,y), x= 0 .. 50, y = 0 .. 50, initialpoint = {x = 2,y= .5}, assume = nonnegative) does not work when Q contains numerical integration, that is evalf (Int). I have no problem with the definite integral evalf(int). The problem is that most of the cases required numerical integration so I need the former expression.

I'd appreciate very much if someone could help me.

Hi,

how can I check in maple if my variable P is positive (always or only for some certain conditions)

with assumptions

a>0,b>0,c>0,d>0,e>0,f>0,g>0 and a>c,e,g

I need to prove that P is always positive with that assumptions, how?

While performing integration of some expessions I bumped into a strange problem. My expession consists of quite a lot of terms, but here I present the susbset of only 2. If I integrate it as a whole maple does not want to solve it, and leaves it as there was no closed expression to this integral. But if I split this sum and integrate the parts, it all works fine. What is happening here? What do I miss? I used simplify and allvalues but didn't change a thing...

Is there a way to split my terms into list, integrate one by one and they recreate the solution by summing the parts? Its a bit of a workaround, but surely better that doing it manually (I have around 50 terms). I use Maple 2015

Thanks,

Jeremi

Good morning everybody.

I have tried Explore to draw plots in an interactive way.

The function I want to draw (plot3d) is automatically constructed by a specific code, as well as the names of the parameters it depends on. These are of the form P||1, P||2, ...P||N (N is an integer determined during the construction)

... but I did not succeed !

After some investigations I was able to reduced the problem to the following one

N  := 2:
F  := add(P||k * x^k, k=0..N):
G := [seq(P||k=-1.0 .. 1.0, k=0..N)]:
Explore(plot(F, x=0..1), parameters=G)

The result is the classical imbedded window with a plot of -1-x_x^2 and three sliders below.
But as soon as I push anu of them the plot disappears and connot be recoverd.

But the following sequence seems to work as expected

N  := 2:
F  := add(P__k * x^k, k=0..N):
G := [seq(P__k=-1.0 .. 1.0, k=0..N)]:
Explore(plot(F, x=0..1), parameters=G)


Before changing my code, could you please confirm me that parameter names such that P||k are not correctly accounted for by Explore ?


Auxiliary question : more generally, are there points which I have to count carefully when I use the P||k construction ?

Thanks in advance

Hello,

a=number      b=number

=maple("Qm:=x->(diff(KelvinBei(0,x),x)*psi2(x)-(diff(KelvinBer(0,x),x)*psi1(x)))/&1";B11)

=maple("Qv:=x->(&1*psi2(x)-(&2*psi1(x)))/(&3*&4)";B6;B7;B2;B11)

=maple("Fm:=x->(Qv(x)+(&1*Qm(x)))/2";B3)

I need abs(max(Fm(x))) and abs(min(Fm(x))) values of function Fm(x), locals, for a<x<b in excel.

Now I use a vector to do this, but I need an exact values not an approximation of a fuction evaluated with n values of x.

what I do:

=maple("seq(i,i=&1..&2,&3)";N2;N3;N4)

=maple("A:=&1";N5)

=maple("G:=map(g->evalf(eval(Fm(x),x=g)),[A])")

=maple("max(abs~(G))")

Someone can help me??

Hello,

I tried to plot the problem presented below:

restart; with(plots); C := setcolors(); with(LinearAlgebra);

formula1 := 2.6*BodyWeight*abs(sin(4*Pi*t));
2.6 BodyWeight |sin(4 Pi t)|
BodyWeight := 80*9.81;
plot(formula1, t = 0 .. 2);

eq2 := formula1-SpringConstant*y(t)-DampConstant*(diff(y(t), t)) = Mass*(diff(y(t), `$`(t, 2)));
2040.480 |sin(4 Pi t)| - SpringConstant y(t)

/ d \ / d / d \\
- DampConstant |--- y(t)| = Mass |--- |--- y(t)||
\ dt / \ dt \ dt //
DampConstant := 50;
50
Mass := .200;
Springt := 200;
200
SpringConstant := Youngsmodulus*Surface/DeltaLength;
DeltaLength := 0.2e-1-y(t);
Surface := .15;
Youngsmodulus := 6.5*10^6/(t+1)+6.5*10^6;
plot(Youngsmodulus, t = 0 .. 10000);

eq2;
2040.480 |sin(4 Pi t)|

/ 6 \
|6.5000000 10 6|
0.15 |------------- + 6.5000000 10 | y(t)
\ t + 1 / / d \
- ----------------------------------------- - 50 |--- y(t)| =
0.02 - y(t) \ dt /

/ d / d \\
0.200 |--- |--- y(t)||
\ dt \ dt //

incs := y(0) = 0, (D(y))(0) = 0;
eq4 := dsolve({eq2, incs}, y(t), type = numeric, method = lsode[backfull], maxfun = 0);
proc(x_lsode) ... end;

plots:-odeplot(eq4, [t, y(t)], 0 .. 5);

 When I try to plot it beyond t=5, Maple gives the following error:

Warning, could not obtain numerical solution at all points, plot may be incomplete

Does anyone know how to plot it even further?