Hello dear!

Hope everyone fine, I am facing to solve the attached problem please find and fix it. I am waiting your kind response. 

naveed.mw

y := int(1/(-0.4016e-1*m^(2/3)-0.211e-3*m^(5/3)), m)

I have to  draw  plot m against y. I can draw plot y against m but i don't can draw plot m against y. Please help me.

I have entered three functions into Maple and I would like to create a set of all possible two-function and three-function compositions involving the three functions. For instance, for my three functions f(x), g(x) and h(x), the set would contain f(g(x)), g(f(x)), f(g(h(x))), f(f(f(x))), etc. 

I'm also looking for a method that will be generalisable to larger numbers of initial functions than just three.

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

I have used solve to find the solution to an equation that has two solutions, and I want to give each solution a label so that I can use each individually in subsequent manipulations. How do I label each solution separately?

I am having trouble to display a 3d and a 2d plots in a same figure. I tried with the display command but no luck.

f1:=exp(x*y);

plot3d(f1,x=0..1,y=0..1);

p1:=%:

f2:=exp(x);

plot(f2,x=0..1);

p2:=%:

To combine both I used display

display(p1,p2);

I ends up with a structure error

The second question is how to plot f(x,y,z)=exp(x+y+z)?

Thanks

I have the following integration:

Int(c-sqrt(a+b*(v*x+u)^2), x = -(b*u+sqrt(b*c^2-a*b))/(b*v) .. (-b*u+sqrt(b*c^2-a*b))/(b*v));

This integration is equivalent to the following integration (between dashed lines):

-----------------------------

NumericEventHandler(invalid_operation = `Heaviside/EventHandler`(value_at_zero = 1)):

Heaviside(0)

Heaviside(x) = convert(Heaviside(x), piecewise)

Int(Heaviside(x+(b*u+sqrt(b*c^2-a*b))/(b*v))*(c-sqrt(a+b*(v*x+u)^2)), x = -infinity .. (-b*u+sqrt(b*c^2-a*b))/(b*v));

-----------------------------

In this form the function, which is to be integrated, is taken as equal to zero for x < -(b*u+sqrt(b*c^2-a*b))/(b*v).

Now I want to write a form such that the function is also taken as equal to zero for x > (-b*u+sqrt(b*c^2-a*b))/(b*v).

How should I do this? Can this also be done with Heaviside?

Download Forcing_Maple_Output.mw

Regards

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

Hi,

I am having trouble in plotting the following surface (it is a quite complicated expression, but should be fine).

OwnSurface := [-Re(arctan(exp(I*Pi*(1/4))*(u^2+v^2)^(1/2)))-(1/2)*ln((1+(u^2+v^2)^2)^(1/2)), -arctan(1/((2*(u^2+v^2))^(1/2)-1))+7*Pi*(1/4), (1/8)*ln(u^2+v^2-(2*(u^2+v^2))^(1/2)+1)-(1/8)*ln(u^2+v^2+(2*(u^2+v^2))^(1/2)+1)-(1/4)*arctan(1/((2*(u^2+v^2))^(1/2)+1))+(1/4)*arctan(1/((2*(u^2+v^2))^(1/2)-1))];

plot3d(OwnSurface, u = -.4 .. .4, v = -.4 .. .4, labels = [x1, x2, x3]);

The only thing maple does is plotting a box with a diagonal line. How can I fix this?

Hello people in mapleprimes,

I cannot obtain a proper result from the following code.

a:=int(((beta/beta[1,2])^(-theta/(1-theta))-kappa[1]^(-theta/(1-theta)))*m*beta^(m-1),beta=0 .. kappa[1]*beta[1,2]);

Please tell me if you know how to have maple calculate it.

Thanks in advance.

taro

Hi all,

Let $A$ be a 0-1 square matrix of order $n$. I want to obtain a matrix $D$ from powers $A$, $A^2$, $\dots$, $A^{n-1}$, where the ($i,j$)-element of $D$ is the smallest $k$ for which the ($i,j$)-element of $A^k$ is nonzero. I only consider the non-diagonal elements of $D$.For example, if the matrix $A$ is Matrix( [[0, 1, 0, 0, 0], [1, 0, 1, 0, 0], [0, 1, 0, 1, 0], [0, 0, 1, 0, 1], [0, 0, 0, 1, 0]]), then $D$ shoule be Matrix([[0, 1, 2, 3, 4], [1, 0, 1, 2, 3], [2, 1, 0, 1, 2], [3, 2, 1, 0, 1], [4, 3, 2, 1, 0]]). However, I cannot obtain this result.

The code is as follows.

p := proc ()

   local A, B, D, m, n, k, r;

   A := Matrix([[0, 1, 0, 0, 0], [1, 0, 1, 0, 0], [0, 1, 0, 1, 0], [0, 0, 1, 0, 1], [0, 0, 0, 1, 0]]);

   r := LinearAlgebra:-RowDimension(A);

   D := A;

   for k from 2 by 1 to r-1 do

       B := A^k;

       for m from 1 by 1 to r do

           for n from 1 by 1 to r do

               if m <> n and B[m, n] <> 0 and D[m, n] = 0 then

                    D[m, n] := k

                end if;

           end do;

        end do;

   end do;

   D;

end proc;

By executing this procedure, I obtain D=Matrix([[0,1,2,3,3],[1,0,1,2,3],[2,1,0,1,2],[3,2,1,0,1],[3,3,2,1,0]]), which is not I want.

Thanks.

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"

Hello Dear!

I want to solve the system of linear equation but facing some problem please see the attachmen. I am waiting your positive response 

1_(1).mw