I have a Maple result presented in the first (top) formula which I would like to format as in the second (bottom) formula:

Can this be done? This is Maple 2016.1

Hello

I have experienced that maple does not save all of the varibles. But some it does.

I calculate with units, could that be the reason?

I have allso been thinking that it has something to do with saving the document online in onenote. But that works like the file is saved on the Pc's harddrive.

Are there anybody else that has experienced this?

I calculate with units, but as the varible does not appear in the calculation with units, I make one varible with the same result, to get the next calculation to work.   

Regards

Heide

Why won't this procedure Compile? All I get is a cryptic and ungrammatical error message.

    step:= proc(
          n::integer[4],
          XYZ::Matrix(datatype= float[8]),
          E::Vector(datatype=integer[4]), F::Vector(datatype=integer[4]),
          W::Matrix(datatype= float[8]), #3x2 scratch matrix
          mu::integer[4]
     )
     option autocompile;
     local
          i::integer[4], j::integer[4],
          ed::float[8],
          fd::float[8],
          p::float[8], t::float[8]       
     ;
          to mu do
               for i to n do
                    ed:= 0;  fd:= 0;
                    for j to 3 do
                          p:= XYZ[i,j];
                          t:= XYZ[E[i],j] - p;
                          ed:= ed+t^2;
                          W[j,1]:= t;
                          t:= XYZ[F[i],j] - p;
                          fd:= fd+t^2;
                          W[j,2]:= t
                    od;
                    ed:= sqrt(ed) + .01;  fd:= sqrt(fd) + .01;
                    for j to 3 do
                         XYZ[i,j]:= 0.995*XYZ[i,j] - 0.01*W[j,1]/ed + 0.02*W[j,2]/fd
                    od
               od
          od
     end proc;
Compiler:-Compile(step);

Error, (in Compiler:-Compile1) In memory compilation failed

Hello guys,

I was just playing around with differential equations, when I noticed that symbolic solution is  different from the numerical.What is the reason for this strange behavior?

ODE := (diff(y(x), x))*(ln(y(x))+x) = 1

sol := dsolve({ODE, y(1) = 1}, y(x))

a := plot(op(2, sol), x = .75 .. 2, color = "Red");
sol2 := dsolve([ODE, y(1) = 1], numeric, range = .75 .. 2);

with(plots);
b := odeplot(sol2, .75 .. 2, thickness = 4);
display({a, b});

Strange_issue.mw

Mariusz Iwaniuk

Hello,

I need to crate a function to be evaluated in a range of values, and this function i would to use in other expression, example:

cel1      "seq(i,i=0.001..2,0.001)"

cel2      "A:=&1";cel1

cel3      "f:=x->diff(KelvinBei(0,x),x)"

cel4      ""B:=map(x->f(x),[A])"

This is ok with a lot of function but with diff(KelvinBei(0,x),x) in cel4 show this error "Error,(in f) invalid input:.1e-2, which is not valid for its 2nd argument.

Why??? How can I do??

The first example is very slow compared to the second.  The difference is in the definition of f().

restart;
y := int(1/(-0.4016e-1*m^(2/3)-0.211e-3*m^(5/3)), m);
f:= unapply(abs(y), m):
n := 500: ## sample size
M := <seq(2*idx/n,idx=1..n)>; ## m
Y := f~(M)+~Statistics:-Sample(Normal(0,3), n)^+; ## signal + noise

restart;
y := int(1/(-0.4016e-1*m^(2/3)-0.211e-3*m^(5/3)), m);
f:= (x) -> abs(subs(m=x,y));
n := 500: ## sample size
M := <seq(2*idx/n,idx=1..n)>; ## m
Y := f~(M)+~Statistics:-Sample(Normal(0,3), n)^+; ## signal + noise

The code below projects a torus illuminated by a point light source located at the origin onto an ellipsoid as the torus's centre moves along the x axis from [-0.8,0,0] to [0.8,0,0].

Ellipsoid := (1/32)*x^2+(1/18)*y^2+(1/12)*z^2 = 1:
EllipsoidTitle := Typesetting:-Typeset((1/32)*x^2+(1/18)*y^2+(1/12)*z^2 = 1):
DispEllipsoid := implicitplot3d(Ellipsoid, x = -10 .. 10, y = -10 .. 10, z = 1.25 .. 5, style = surface, color = yellow, grid = [30, 30, 30]):
TorusVorig := Vector([(1+.25*cos(v))*cos(u), (1+.25*cos(v))*sin(u), .6+.25*sin(v)]):

i := 0;

for x from -.8 by .1 to .81 do

i := i+1;

TorusV := `<,>`(x, 0, 0)+TorusVorig:

DispTorusV := plot3d(TorusV, u = 0 .. 2*Pi, v = 0 .. 2*Pi, shading = xyz):

NormTorusV := Normalize(TorusV, Euclidean):

Vadj := solve((1/32)*w^2*NormTorusV[1]^2+(1/18)*w^2*NormTorusV[2]^2+(1/12)*w^2*NormTorusV[3]^2 = 1, w):

if evalf(eval(Vadj[1]*NormTorusV[3], [u = Pi, v = Pi])) > 0 then

DispTorusVproj := plot3d(Vadj[1]*NormTorusV, u = 0 .. 2*Pi, v = 0 .. 2*Pi):

else

DispTorusVproj := plot3d(Vadj[2]*NormTorusV, u = 0 .. 2*Pi, v = 0 .. 2*Pi):

end if:

Disp[i] := display(DispTorusV, DispEllipsoid, DispTorusVproj, view = [-10 .. 10, -10 .. 10, 0 .. 5], scaling = constrained, Plot3Daxes, title = typeset("Project %1 onto %2 from a light at the origin", TorusV, EllipsoidTitle), titlefont = [Courier, bold, 14]):

end do:

display(seq(Disp[j], j = 1 .. i), insequence = true);

How can I plot the unmoving torus centred on the z axis projected onto the ellipsoid illuminated by a point light source moving on the x axis from [-0.8,0,0] to [0.8,0,0]?

Is there an elegant way to plot a surface with three given parameters, such as
x=(5+w\cos v)\cos u, y=(5+w\cos v)\sin u, z=w\sin v

with u,v between 0 and 2Pi and w between 0 and 3?

I would like the C-Text style in 14pt Times Roman while the C-2D-Math style is 12pt.

My use case is that I am typing in a single execution block.

I use Format >> Styles and select the style for C-Text and set it to 14pt Times. I click OK to close the dialog. Next I repeat this for the C-2D-Math style but this time set the font size to 12pt and, for testing, the colour to blue. The effect is to give me 14pt Times for both styles, though the C-text is black and the 2-D-Math is blue.

Can the effect I want be achieved, or is this a bug/feature?

This is Maple 2016.1 on Windows 10 64bit

 Thanks for any help

With a somewhat complicated equation for a line, draw fails.

with(geometry):

point(P1,[47+(38+22/60)/60, -(122+(43+4/60)/60)]);
point(P2,coordinates(P1) +~ [cos(30*Pi/180),sin(30*Pi/180)]);
line(L1,[P1,P2]);
Equation(L1);
draw(L1); ## no line

point(P1,[0,0]);
point(P2,[7,9]);
line(L1,[P1,P2]);
draw(L1);  ## works

Tom Dean

I have been working on a general solution to motion analysis and seem to be going backwards.  I have an numerical solution in Octave I use for comparison.  I have reduced the problem to a small example that exhibits the problem.

I posted a question similar to this, but, without a set of known values.

I am doing something wrong, but, what?

Tom Dean

## bearing.mpl, solve the target motion problem with bearings only.
##
## Consider a sensor platform moving through points (x,y) at times
## t[1..4] with the target bearings, Brg[1..4] taken at times t[1..4]
## with the target proceeding along a constant course and speed.
##
## time t, bearing line slope m, sensor position (x,y) are known
## values.
##
## Since this is a generated problem the target position at time t is
## provided to compare with the results.
##
#########################################################################
##
restart;
##
genKnownValues := proc()
    description "set the known values",
    "t - relative time",
    "x - sensor x location at time t[i]",
    "y - sensor y location at time t[i]",
    "m - slope of the bearing lines at time t[i]",
    "tgtPosit - target position at time t[i]";
    global t, m, x, y, tgtPosit;
    local dt, Cse, Spd, Brg, A, B, C, R, X;
    local tgtX, tgtY, tgtRange, tgtCse, tgtSpd;
## relative and delta time
    t := [0, 1+1/2, 3, 3+1/2];
    dt := [0, seq(t[idx]-t[idx-1],idx=2..4)];
## sensor motion
    Cse := [90, 90, 90, 50] *~ Pi/180; ## true heading
    Spd := [15, 15, 15, 22];  ## knots
## bearings to the target at time t
    Brg := [10, 358, 340, 330] *~ (Pi/180);
## slope of the bearing lines
    m:=map(tan,Brg);
## calculate the sensor position vs time
    x := ListTools[PartialSums](dt *~ Spd *~ map(cos, Cse));
    y := ListTools[PartialSums](dt *~ Spd *~ map(sin, Cse));
## target values  start the target at a known (x,y) position at a
## constant course and speed
    tgtRange := 95+25/32; ## miles at t1, match octave value...
    tgtCse := 170 * Pi/180; ## course
    tgtSpd := 10; ## knots
    tgtX := tgtRange*cos(Brg[1]);
    tgtX := tgtX +~ ListTools[PartialSums](dt *~ tgtSpd *~ cos(tgtCse));
    tgtY := tgtRange*sin(Brg[1]);
    tgtY := tgtY +~ ListTools[PartialSums](dt *~ tgtSpd *~ sin(tgtCse));
## return target position vs time as a matrix
    tgtPosit:=Matrix(4,2,[seq([tgtX[idx],tgtY[idx]],idx=1..4)]);
end proc:
##
#########################################################################
## t[], m[], x[], and y[] are known values
##
## equation of the bearing lines
eq1 := tgtY[1] - y[1]    = m[1]*(tgtX[1]-x[1]):
eq2 := tgtY[2] - y[2]    = m[2]*(tgtX[2]-x[2]):
eq3 := tgtY[3] - y[3]    = m[3]*(tgtX[3]-x[3]):
eq4 := tgtY[4] - y[4]    = m[4]*(tgtX[4]-x[4]):
## target X motion along the target line
eq5 := tgtX[2] - tgtX[1] = tgtVx*(t[2]-t[1]):
eq6 := tgtX[3] - tgtX[2] = tgtVx*(t[3]-t[2]):
eq7 := tgtX[4] - tgtX[3] = tgtVx*(t[4]-t[3]):
## target Y motion along the target line
eq8 := tgtY[2] - tgtY[1] = tgtVy*(t[2]-t[1]):
eq9 := tgtY[3] - tgtY[2] = tgtVy*(t[3]-t[2]):
eq10:= tgtY[4] - tgtY[3] = tgtVy*(t[4]-t[3]):
##
#########################################################################
##
## solve the equations
eqs  := {eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10}:

Sol:= solve(eqs, {tgtVx, tgtVy, seq([tgtX[k], tgtY[k]][], k= 1..4)}):
##

genKnownValues():
## these values are very close to Octave
evalf(t);evalf(m);evalf(x);evalf(y);evalf(tgtPosit);
## The value of tgtX[] and tgtY[] should equal the respective tgtPosit values
seq(evalf(eval([tgtX[idx],tgtY[idx]], Sol)),idx=1..4);

I accidentally came across a nice Mma animation. Unfortunately, I am able to present only few frames of it in MaplePrimes. See two inconsecutive frames below

I find this animation very deep. I don't remember something similar. It looks like an iterative
map shown in its dynamics. Not being an expert in Mathematica, I don't understand the machinery of the generating code.
n = 1000;
r := RandomInteger[{1, n}];
f := (#/(.01 + Sqrt[#.#])) & /@ (x[[#]] - x) &;
s := With[{r1 = r}, p[[r1]] = r; q[[r1]] = r];
x = RandomReal[{-1, 1}, {n, 2}];
{p, q} = RandomInteger[{1, n}, {2, n}];
Graphics[{PointSize[0.007], Dynamic[If[r < 100, s];
Point[x = 0.995 x + 0.02 f[p] - 0.01 f[q]]]}, PlotRange -> 2]
Here is its fragment translated into Maple:
>with(MmaTranslator):
>FromMma(" (#/(.01 + Sqrt[#.#])) & /@ (x[[#]] - x) &;");
map(unapply(_Z1/(0.1e-1+sqrt(_Z1 . _Z1)), _Z1), unapply(x(_Z1)-x, _Z1))
To my regret,
>FromMma(" n = 1000;
r := RandomInteger[{1, n}];
f := (#/(.01 + Sqrt[#.#])) & /@ (x[[#]] - x) &;
s := With[{r1 = r}, p[[r1]] = r; q[[r1]] = r];
x = RandomReal[{-1, 1}, {n, 2}];
{p, q} = RandomInteger[{1, n}, {2, n}];
Graphics[{PointSize[0.007], Dynamic[If[r < 100, s];
Point[x = 0.995 x + 0.02 f[p] - 0.01 f[q]]]}, PlotRange -> 2]");
Error, (in MmaTranslator:-FromMma) incorrect syntax (at position 11) in last character of "...0)
r"

Hi all,

Using the Physics package, I have defined a density operator, shown in the image below:

(I did assume(s ≠i), assume(v≠h).

I then operate on it with the Bracket as shown in the next image:

I think this should evaluate to (|Y,s,h><Y,s,h|)/2; or at least that's the result I want. However, Maple evaluates this to zero.

Is there a where to convince Maple to evaluate this to what I want/expect?

Thanks in advance,

Kevin

I would prefer that all the polynomials generated in my workbook by MAPLE were in expanded form.  For instance, it the elements of a matrix are polynomials, I want to see the expanded form for all of them.  What do I type into a workbook to make this happen.  (I am a new user of MAPLE.) 

12pt type is difficult for me to read.  I would like to fix a workbook so that everything yet to appear will be in 18pt type, without my having to enlarge things one at a time.   (I am a new MAPLE user.)