Hello,

This might sound silly, but 

how do you find a tangent line equation at an exact point of a curve?

>TangentLine(curve,var);

I know this method but how can i put the exact point like (0,1) for an example?

Some of the examples would be very helpful! Thank you

I have a set of formulas I saved in a matrix and the exported the matrix to Excel. However when I open the file in Excel the equations are in prefix notation (I think).  That's not exactly human readable friendly.  If I copy and paste straight into excel they are readable.

It there a way to make the Maple export similar?

EDIT:-  I exported as a.csv file because if export as .xlsx    all the equation change into "#NUM!" in the spreadsheed.

Download test_eqn_export.mw

I am trying to plot the following kind of sequence. I was expecting a curve for one set of [] inside plot command, but I am getting twin curve there. How can I get a single curve for one single parametric command there. Please reply asap.

plot([[(1.428571429*(r^2+.49-4*r*(r^2+.49-2*exp(-.1/r))/(2*r-.2*exp(-.1/r)/r^2)))*csc((1/3)*Pi), sqrt(16*r^2*(r^2+.49-2*exp(-.1/r))/(2*r-.2*exp(-.1/r)/r^2)^2+.49*cos((1/3)*Pi)^2-2.040816327*(r^2+.49-4*r*(r^2+.49-2*exp(-.1/r))/(2*r-.2*exp(-.1/r)/r^2))^2*cot((1/3)*Pi)^2), r = -10 .. 10], [(1.428571429*(r^2+.49-4*r*(r^2+.49-2*exp(-.1/r))/(2*r-.2*exp(-.1/r)/r^2)))*csc((1/3)*Pi), -sqrt(16*r^2*(r^2+.49-2*exp(-.1/r))/(2*r-.2*exp(-.1/r)/r^2)^2+.49*cos((1/3)*Pi)^2-2.040816327*(r^2+.49-4*r*(r^2+.49-2*exp(-.1/r))/(2*r-.2*exp(-.1/r)/r^2))^2*cot((1/3)*Pi)^2), r = -10 .. 10], [(5.*(r^2+0.4e-1-4*r*(r^2+0.4e-1-2*exp(-.2/r))/(2*r-.2*exp(-.2/r)/r^2)))*csc((1/3)*Pi), sqrt(16*r^2*(r^2+0.4e-1-2*exp(-.2/r))/(2*r-.2*exp(-.2/r)/r^2)^2+0.1e-1*cos((1/3)*Pi)^2-25.*(r^2+0.4e-1-4*r*(r^2+0.4e-1-2*exp(-.2/r))/(2*r-.2*exp(-.2/r)/r^2))^2*cot((1/3)*Pi)^2), r = -10 .. 10], [(5.*(r^2+0.4e-1-4*r*(r^2+0.4e-1-2*exp(-.2/r))/(2*r-.2*exp(-.2/r)/r^2)))*csc((1/3)*Pi), -sqrt(16*r^2*(r^2+0.4e-1-2*exp(-.2/r))/(2*r-.2*exp(-.2/r)/r^2)^2+0.1e-1*cos((1/3)*Pi)^2-25.*(r^2+0.4e-1-4*r*(r^2+0.4e-1-2*exp(-.2/r))/(2*r-.2*exp(-.2/r)/r^2))^2*cot((1/3)*Pi)^2), r = -10 .. 10]], -13 .. 5, -5 .. 5)

I noticed in object constructor I had to write  _self:-name  to refer to object own variable called name.  But inside a another proc I can use just name without having to add _self:- to it. It also works when adding _self:-

Is this becuase constructor is special proc, and the object is not yet full constructed?  Here is a MWE

restart;
person:=module()
    option object;
    local name::string:="";

    export ModuleCopy::static:= proc( _self::person, proto::person, the_name::string, $ ) 
            print("Enter  constructor");
           _self:-name:= the_name;

           #Why this fails here, but not in process proc? Is it because of special
           #case since done inside constructor?
           #print(name); 

           print(_self:-name);
    end proc;
 
   export process::static :=proc(_self,$)
     #here both cases work
     print(_self:-name);
     print(name);
   end proc;

end module;

#and now

p:=Object(person,"me")

The above works. But if I uncomment #print(name); inside the constructor, maple gives error

p:=Object(person,"me")
Error, static procedure `ModuleCopy` refers to non-static local or export `name::string` in surrounding scope

But this works with no error

p:-process()

                              "me"
                              "me"

But no such error when doing same thing inside a local static proc inside same module.

Is this special just for the constructor than one must use :-self ? Just wanted to make sure.

Maple 2022.1

Hi.

What wrong could be there with the color line?

Download Legend_Inside.mw

How can we export image with minimum size? I plot the solution and then export the image as an .eps file. But when I try to generate the generate the pdf file on Latex, the file size is too large. But when I plot the solutions on Matlab and export the images as .eps file, then Latex generate pdf file with samll size. Why Maple generate images of large sizes?

solution.mw

I want to plot the solutions of the equation (x-y)^2+(1-z)^2=0.

However, implicitplot3d is not able to plot them, at least using the default arguments. Any recommendations?

I know a priori that it is going to be a curve contained in a plane in case that makes this task easier.

The final solution is the sum of 2 terms and I am doing it individually for 1st term. I think I am doing it wrong because answer did not match when comparing with textbook answer. Can anyone teach me or hint me compute the final series for m=1 to 10. An example of final series for m=1 would be helpful. 

Download ODE_sol.mw

Hello,

Attached is a simple worksheet dealing with units and angular motion.   I need to reverse calculate from an acceleration to the uniform motion (rad/sec or RPM) which is used in forces and FEA simulation.

The initial units work fine and get to a velocity (feet/sec) but then fail to convert to rad/sec or RPM.   The attached shows the results of units Hz (1/sec) and the problems with resolving to RPM or rad/sec.   the unit (length / length (radius)) as an answer stil confounds me.

Uniform_circular_motion_units_issue.mw

Any help greatly appreciated,
Bill

Hello, I have the following (nested) summation:

add(binomial(196, j)*0.5^(32.1+j)*add(binomial(109, l)*(-1)^(j+l)*add(0.5^(3.1*h)*GAMMA(-4.1+h)*GAMMA(9.3170731707317073170731707317073170731707317073170+.24390243902439024390243902439024390243902439024390*l+.24390243902439024390243902439024390243902439024390*j+.75609756097560975609756097560975609756097560975610*h)/(factorial(h)*(10.354838709677419354838709677419354838709677419355+.32258064516129032258064516129032258064516129032258*j+h)*GAMMA(.24390243902439024390243902439024390243902439024390*j+.24390243902439024390243902439024390243902439024390*l+15.417073170731707317073170731707317073170731707317+.75609756097560975609756097560975609756097560975610*h)), h = 0 .. infinity), l = 0 .. 109), j = 0 .. 196)

But it takes ages to sum h up to 80, for example. I was wondering whether we could improve to gain time. any ideas?

CorrBasis := {<sqrt(3),2,0>, <0,1/5,1>};
CorrMat := Matrix([op(CorrBasis)]);

Basis := {<3^(0.5), 2., 0.>, <0.,0.2,1.>}; #float
Basis := remove(i -> is(LinearAlgebra:-Rank(<CorrMat|i>) <> nops(CorrBasis)),  Basis);

I need to use matrices/ vectors with different datatypes. Do I need to turn them all into floats in order for Basis to remain the same after calling

Basis := remove(i -> is(LinearAlgebra:-Rank(<CorrMat|i>) <> nops(CorrBasis)),  Basis);

Is there any better way to manage this?

How do you stop Maple during this code execution:

proc()
x:
print(1);
goto(x);
end():

     Hello everyone !

     I have a problem asking for help:

     In the Oxy coordinate plane, for rectangles are limited by straight lines: x=1, x=7, y=1, y=9 and there are 63 points distinguished from coordinates that are integers located on this rectangle.

     These include:

• 7 black points with coordinates are listed in the list:

[[1,1], [2,1], [3,1], [4,1], [5,1], [6,1], [7,1]].

• 7 red points with coordinates are listed in the list:

[[1,2], [2,2], [3,2], [4,2], [5,2], [6,2], [7,2]].

• 8 yellow points with coordinates are listed in the list:

[[1,3], [4,3], [5,3], [7,3], [1,4], [4,4], [5,4], [7,4] ].

• 6 pink points with coordinates are listed in the list:

[[2,3], [3,3], [6,3], [2,4], [3,4], [6,4]].

• 8 brown points with coordinates are listed in the list:

[[1,5], [3,5], [5,5], [7,5], [1,6], [3,6], [5,6], [7,6]].

• 6 purple points with coordinates are listed:

[[2,5], [4,5], [6,5], [2,6], [4,6], [6,6]].

• 9 blue points with coordinates are listed in the list:

[[1,7], [2,7], [7,7], [1,8], [2,8], [7,8], [1,9], [2,9], [7,9]].

• 6 green points with coordinates are listed:

[[3,7], [5,7], [3,8], [5,8], [3,9], [5,9]].

• 6 orange points with coordinates are listed in the list:

[[4,7], [6,7], [4,8], [6,8], [4,9], [6,9]].

     Help me find the integer coordinates of the 63 points when arranging them on the rectangle knowing that their HorizontalCoord has not changed, and the VerticalCoord of the points of the same color is always different with the Maple command.

     Thank you so much for your help!

I would show you what I have but I actually don't know where to start this problem..

If n people (numbered 1 to n) stand in a circle and someone starts going around the circle and eliminating every other person till only one person is left, the number J(n) of the person left at the end is given by 

    J(n) = 1                           if n = 1
    J(n) = 2*J(n/2) - 1          if n > 1 and n is even
    J(n) = 2*J((n-1)/2) + 1   if  n > 1 and n is odd

(i) Write a recursive procedure to compute J. [As a check the first 16 values (starting with 1) of J(n) are 1,1,3,1,3,5,7,1,3,5,7,9,11,13,15,1]. 
(ii)Compute the value of J(10000). 
(iii) Can you explain why this is so much faster than our recursive procedure to compute the n-th Fibonacci number?

Why this error message ?
with(geometry);
vartheta := (2*Pi)/17;
x || k := cos(k*theta);
y || k := sin(k*theta);
 xk := cos(k theta): yk := sin(k theta)
.M||k:=(point,x||k,y||k);
 with(plots):
Points := pointplot([seq(M || k, k = 1 .. 16)], symbol = solidcircle, color = red, symbolsize = 10):
Error, (in plots:-pointplot) cannot convert data to an Array of datatype float[8]; Thank you for your answer..