Maple 2018 Questions and Posts

How to save procedures to a library?...

Asked by: Ronan 1022

October 25 2018

1 3

I have asked this before but am still confused. I have a half-dozen procedures I want to save. Don't want them in a module/package. I am using windows 10.

I just can't get the syntax correct on this.

Obviously after saving restart and load to test.

libname;
       "C:\Program Files\Maple 2018\lib", 

         "C:\Users\Ronan\maple\toolbox\CodeBuilder\lib", 

         "C:\Users\Ronan\maple\toolbox\OEIS\lib", 

         "C:\Users\Ronan\maple\toolbox\personal\lib", 

         "C:\Users\Ronan\maple\toolbox\UTF8\lib"
libdir := "C:/Users/Ronan/maple/toolbox/personal/lib";
     libdir := "C:/Users/Ronan/maple/toolbox/personal/lib"
NULL;

LibraryTools:-Save(Pedal, cat(kernelopts(homedir), "/maple/toolbox/personal/lib/Pedal.mpl"));
Error, (in LibraryTools:-Save) could not open `C:\Users\Ronan/maple/toolbox/personal/lib/Pedal.mpl\Pedal.m` for writing

Is the for..do loop in my procedure causing a prob...

Asked by: Ronan 1022

October 22 2018

0 12

I trying to get a proc to work. In the 1st half of the document, I derive four sections for a pedal curve and plot them fine. In the second I try doing it with a procedure called Pedal. Am having problems. I think it might be something to do with the for.. do loop.

C := (x^2+y^2+12*x+9)^2-4*(2*x+3)^3

(1)

[(-x^2-12*x-9+2*(8*x^3+36*x^2+54*x+27)^(1/2))^(1/2), -(-x^2-12*x-9+2*(8*x^3+36*x^2+54*x+27)^(1/2))^(1/2), (-x^2-12*x-9-2*(8*x^3+36*x^2+54*x+27)^(1/2))^(1/2), -(-x^2-12*x-9-2*(8*x^3+36*x^2+54*x+27)^(1/2))^(1/2)]

(2)

X^4+4*X^3*r+2*X^2*Y^2+4*X^2*Y*s+6*X^2*r^2+2*X^2*s^2+4*X*Y^2*r+8*X*Y*r*s+4*X*r^3+4*X*r*s^2+Y^4+4*Y^3*s+2*Y^2*r^2+6*Y^2*s^2+4*Y*r^2*s+4*Y*s^3+r^4+2*r^2*s^2+s^4-8*X^3-24*X^2*r+24*X*Y^2+48*X*Y*s-24*X*r^2+24*X*s^2+24*Y^2*r+48*Y*r*s-8*r^3+24*r*s^2+18*X^2+36*X*r+18*Y^2+36*Y*s+18*r^2+18*s^2-27

(3)

(4)

(5)

xp1 := 1; yp1 := 2; lprp1 := -coeff(ltg1, y)*x+coeff(ltg1, x)*y+K1; for i to nops(sol1) do s := eval(sol1[i], x = r); print("s ", s); K1 := solve(eval(lprp1, [x = xp1, y = yp1]), K1); print("K1 ", K1); sol2 := solve([ltg1, lprp1], [x, y]); Pedal || i := [rhs(sol2[1, 1]), rhs(sol2[1, 2])]; p9 || i := plot([op(Pedal || i), r = -20 .. 20], colour = "MediumSeaGreen"); unassign('K1', 'sol2') end do

"K1 ", -4*(-r^2-12*r-9-2*(8*r^3+36*r^2+54*r+27)^(1/2))^(3/2)-4*r^2*(-r^2-12*r-9-2*(8*r^3+36*r^2+54*r+27)^(1/2))^(1/2)+16*(8*r^3+36*r^2+54*r+27)^(1/2)*r-48*r*(-r^2-12*r-9-2*(8*r^3+36*r^2+54*r+27)^(1/2))^(1/2)+192*r^2+96*(8*r^3+36*r^2+54*r+27)^(1/2)-36*(-r^2-12*r-9-2*(8*r^3+36*r^2+54*r+27)^(1/2))^(1/2)+576*r+432

(6)

(7)

This I can't get to work. All 4 sections of the pedal curve are the same and should not be.

Parse:-ConvertTo1D, "`%1` is not a module or member", RonanRoutines

"K||i", -8*r^3+(-4*(-r^2-12*r-9-2*(8*r^3+36*r^2+54*r+27)^(1/2))^(1/2)+48)*r^2-8*(-(-r^2-12*r-9-2*(8*r^3+36*r^2+54*r+27)^(1/2))^(1/2)-3)^2*r-4*(-r^2-12*r-9-2*(8*r^3+36*r^2+54*r+27)^(1/2))^(3/2)+48*r^2+576*r+432+96*(8*r^3+36*r^2+54*r+27)^(1/2)-36*(-r^2-12*r-9-2*(8*r^3+36*r^2+54*r+27)^(1/2))^(1/2)

(8)

(9)

for i to nops(sol1) do p9 || i := plot([op(Pedal2[i]), r = -20 .. 20], colour = "MediumSeaGreen") end do

Download Proc_for_Pedal_not_working.mw

Evaluate definite integral with infinite sum...

Asked by: s144117 5

October 22 2018

1 7

Hi

I am trying to evaluate a function that contains an infinite sum. I need to evaluate the integral to determine a CDF, and from that hopefully the inverse CDF in order to sample the corresponding PDF.

The sum comes from a 2F1 hypergeometric function, which I have written manually in the attached file.

Can anyone tell me if and how it is possible to evaluate this integral, with respect to x?

I have inserted an image below, but also the entire file :) Marginal2.mw

Thanks in advance.

Maple 2018 hangs after start up...

Asked by: André Devos 5

October 22 2018

1 2

Maple 2018 starts but block and hangs after the first imput.

Even after a clean install on a clean disk where there are no old files of maple.

My specs are: WIN10 pr 64/ I7 /16 GB/ SSD

Who has the same problem and a sollution?

Autonomous first-order nonlinear recurrences with...

by: Carl Love 26488 Product: Maple 2018

October 21 2018

5 8

The attached worksheet shows how to evaluate and graphically analyze an autonomous first-order nonlinear recurrence with two dependent variables and multiple symbolic parameters.

This worksheet shows how a small module that simply encapsulates the given information of a problem combined with some use statements can greatly facilitate the organization of one's work, can encapsulate the setting of parameter values, and can allow one to work with symbolic parameters.

Edit: In the first version of this Post, I forgot to include the qualifier "autonomous". The system being autonomous substantially simplifies its treatment.

Autonomous first-order nonlinear recurrences with parameters and multiple dependent variables

Author: Carl Love <carl.j.love@gmail.com> 20-Oct-2018

The techniques used in this worksheet can be applied to most autonomous first-order nonlinear recurrences with multiple dependent variables and parameters.

This worksheet shows how a small module that simply encapsulates the given information of a problem combined with some use statements

•	can greatly facilitate the organization of one's work,

•	can encapsulate the setting of parameter values,

•	can allow one to work with symbolic parameters.

A Problem from MaplePrimes: A discrete Lottka-Volterra population model is applied to an isolated island with a population of predators (foxes), R, and prey (rabbits), K. [Note that R is the foxes, not the rabbits! Perhaps this problem statement originated in another language.] The change over one time period is given by

K[n+1]:= K[n]*(-b*R[n]+a+1); R[n+1]:= R[n]*(b*e*K[n]-c+1),

where a, b, c, e are parameters of the model. In this problem we will use a= 0.15, b= 0.01, c= 0.02, e= 0.01, when numeric values are needed.

a) Show that there exists an equilibrium (values of K[n] and R[n] such that K[n+1] = K[n] and R[n+1] = R[n]).

b) Write Maple code that solves the recurrence numerically. Assume that if any population is less than 0.5 then it has gone extinct and set the value to 0. Check that your program is idempotent at the equilibrium.

>

restart:

We begin by collecting all the given information (except for specific numeric values) into a module. The ModuleApply lets the user set the numeric values later.

For all two-element vectors used in this worksheet, K is the first value and R is the second value.

>

KandR:= module()
local
   a, b, c, e, #parameters

   #procedure that lets user set parameter values:
   ModuleApply:= proc({
       a::algebraic:= KandR:-a, b::algebraic:= KandR:-b,
       c::algebraic:= KandR:-c, e::algebraic:= KandR:-e
   })
   local k;
      for k to _noptions do thismodule[lhs(_options[k])]:= rhs(_options[k]) od;
      return
   end proc,

   Extinct:= (x::realcons)-> `if`(x < 0.5, 0, x) #force small, insignificant values to 0
;
export
   #Procedure that does one symbolic iteration
   #(Note that this procedure uses Vector input and output.)
   iter_symb:= KR-> KR *~ <-b*KR[2]+a+1, b*e*KR[1]-c+1>,

#Such simple treatment as above is only possible for autonomous
#recurrences.

iter_num:= Extinct~@iter_symb #one numeric iteration
;
end module:

>	#The following expression is the discrete equivalent of the derivative (or gradient). #It represents the change over one time period. P:= <K,R>: OneStep:= KandR:-iter_symb(P) - P

Vector(2, {(1) = K*(-R*b+a+1)-K, (2) = R*(K*b*e-c+1)-R})

>	#An equilibrium occurs when the gradient is 0. Eq:= <K__e, R__e>: Eqs:= solve({seq(eval(OneStep=~ 0, [seq(P=~ Eq)]))}, [seq(Eq)]);

>	#We're only interested here in nonzero solutions. EqSol:= remove(S-> 0 in rhs~(S), Eqs)[];

>	#Set parameters: KandR(a= 0.15, b= 0.01, c= 0.02, e= 0.01);

>	#Show idempotency at equilibrium: use KandR in Eq0:= eval(Eq, EqSol); print(Eq0 = iter_num(Eq0)) end use:

(Vector(2, {(1) = 200.0000000, (2) = 15.00000000})) = (Vector(2, {(1) = 200.0000000, (2) = 15.00000000}))

>

#procedure that fills a Matrix with computed values of a 1st-order recurrence.
#(A more-efficient method than this can be used for linear recurrences.)
#This procedure has no dependence on the module.
Iterate:= proc(n::nonnegint, iter, init::Vector[column])
local M:= Matrix((n+1, numelems(init)), init^+, datatype= hfloat), i;
for i to n do M[i+1,..]:= iter(M[i,..]) od;
M
end proc:

We want to see what happens if the initial conditions deviate slightly from the equilibrium. It turns out that any deviation (as long as the
initial values are still nonnegative!) will cause the same effect. I simply chose the deviation <7,2> because it was the smallest for which

the plot clearly showed what happens using the scale that I wanted to show the plot at. By using a finer scale, it is possible to see the

"outward spiral" efffect from even the tiniest deviation.

>	dev:= <7,2>: use KandR in KR:= Iterate(1000, iter_num, Eq0 + dev) end use:

>

plot(
   [
       KR, #trajectory of population
       KR[[1,1],..], #1st point
       KR[-[1,1],..], #last point,
       <Eq0|Eq0>^+, #equilibrium
       #every 100th point (helps show time scale):
       KR[100*[$1..iquo(numelems(KR[..,1]), 100)-1], ..]
   ],
   #This group of options are all lists, each element of which corresponds
   #to one of the above components of the plot:
   style= [line, point$4],
   symbol= [solidcircle$4, soliddiamond],
   symbolsize= [18$4, 12],
   color= [black, green, red, brown, blue],
   thickness= [0$5],
   legend= [`pop.`, init, final, equilibrium, `100 periods`],

   #This group of options are lists, each element of which corresponds to one
   #coordinate axis (horizontal, then vertical).
   view= [0..max(KR[..,1]), 0..max(KR(..,2))],
   labels= [rabbits, foxes],
   labeldirections= [horizontal,vertical],
   size= [700,700], #measured in pixels

   #options applied to whole plot:
   labelfont= [TIMES, BOLDITALIC, 14],
   title= "Population of foxes and rabbits over time" "\n", titlefont= [TIMES,16],
   caption=
      "\n" "Choosing an initial point near the equilibrium causes"
      "\n" "outward spiraling divergence." "\n",
   gridlines= false
);

A fieldplot helps show what happens for any starting values. An arrow is drawn from each of a 2-D grid of point. The magnitude and direction of the arrow show the gradient (as a vector) in this case.

>

plots:-fieldplot(
   rtable_eval(OneStep),
   K= 0..max(KR[..,1]), R= 0..max(KR[..,2]), grid= [16,16],

   #arrow-specific options:
   anchor= tail, fieldstrength= log, arrows= slim, color= "DarkGreen",

   #other options (same as any 2D plot):
   labels= [rabbits, foxes], labeldirections= [horizontal,vertical],
   labelfont= [TIMES, BOLDITALIC, 14],
   title= "One-step population changes from any point" "\n", titlefont= [TIMES,16],
   caption= "\n" "All trajectories spiral outward from the equilibrium." "\n",
   size= [700,700],
   gridlines= false
);

The above plot is computed only from the symbolic discrete gradient expression OneStep; it does not use the computed population values from the first plot. It only uses the maxima of those computed values to determine the length of the axes.

Conclusion: While this is interesting stuff mathematically, and makes for great plots, divergence from the equilibrium doesn't seem realistic to me.

>

Download FoxesAndRabbits.mw

Defining complex equations using smaller parts...

Asked by: dpananos 10

October 21 2018

2 1

I have a solution to a linear ODE which is very long and complicated. The solition clearly has some parts which are repeated and so it would would be easiest to express those repeated parts as something simpler.

For example, suppose I had

x = (-b + sqrt(b^2 - 4*a*c) ) /2*a

What is the command to take x and do someting like

Z = sqrt(b^2 - 4*a*c)

x = (-b + Z)/2*a

two variable polar graphing...

Asked by: jlb4848 10

October 20 2018

2 3

when plotting a polar function in terms of r and theta, is there a way to animate it?

For instance I want to animate u(r,theta)=rcos(theta) for theta between 0 and 2Pi.

Implicitplot empty, Why?...

Asked by: Ronan 1022

October 19 2018

1 8

Why does the implicit plot return empty?

plots:-implicitplot((x^2+y^2 = 1)^2, x = -3 .. 3, y = -3 .. 3);# plots
   plots:-implicitplot((x^2+y^2-1)^2, x = -3 .. 3, y = -3 .. 3) # empty plot

Putting a metric tensor in maple under polar coord...

Asked by: guru kido 60

October 19 2018

4 2

So I currently have:
with(DifferentialGeometry); with(Tensor);
DGsetup([r, theta], pol);
g1 := evalDG(drdr+r^2d(theta)d(theta))
C1 := Christoffel(g1);
However its coming back saying that g1 is not of metric form, am i missing something? Thanks

pdsolve does not work...

Asked by: torabi 85

October 18 2018

1 2

why pdsolve does not work?

Q and alpha are constant. I want to find f(x,y).

Thanks

Download PLATE.mw

How to do recursion to model population of rats...

Asked by: brian bovril 889

October 17 2018

0 6

Hello

I found this paper.

http://benson-labs.com/uploads/stuff/POW15.pdf

I'm not sure if the population after 1 year is correct.

I wanted to get the formula from rsolve. But i have incorrorectly formulated the sum by day and the total.

Rattus_recursion.mw

Matrix Multiplication with 2D input...

Asked by: tomleslie 13771

October 17 2018

2 6

Whilst trying to debug another problem on this site, this one

https://www.mapleprimes.com/questions/225688-How-To-Compute-Inverse-Of-A-Matrix-For

I came across a Maple bug(?) which was so serious, I thought either

I'm being more than usually dim
it needs it's own entry here, so that others can check

Problem can be summarized as:

When using 2D input, can the '.' character be used for simple matrix multiplication???? Because it isn't working for me!

Entering the '.' character places a bold, midline dot between the two arguments, but executing the result gives "Error unable to parse". Converting to 1-D input, no issues - everything works as expected.I should also point out that

I never normally use 2D-input for my own work
the attached worksheet executes correctly in Maple 2017.3, X86 64 WINDOWS, Sep 13 2017, Build ID 1262472.

If this is reproducible then wtf!

>

restart; with(LinearAlgebra); kernelopts(version)

(1)

>

Matrix(2, 2, {(1, 1) = 44, (1, 2) = -31, (2, 1) = 92, (2, 2) = 67})

Matrix(%id = 18446744074378134462)

(2)

>

Error, unable to parse

>

Matrix(%id = 18446744074378136502)

(3)

>	A:= RandomMatrix(2, 2); B:= RandomMatrix(2, 2); A.B

Matrix(2, 2, {(1, 1) = -32, (1, 2) = -4, (2, 1) = -74, (2, 2) = 27})

Matrix(2, 2, {(1, 1) = -93, (1, 2) = -72, (2, 1) = -76, (2, 2) = -2})

Matrix(%id = 18446744074378132406)

(4)

>

Download matMulProb.mw

Statistics:-Mean fails...

Asked by: sand15 765

October 17 2018

0 4

Hi,

I submit you this strange result:
for somenumerical values of a, b, c (a, b, c real and b > a), Maple 2018 is not able to compute the mean of
c*U where U is a Uniform random variable with support [a, b].

>

restart:

>	with(Statistics):

>	interface(version);

(1)

>	z := .4070716688RandomVariable(Uniform(0.12-0.020.12, 0.12+0.02*0.12)); Mean(z);

Error, (in Statistics:-Mean) the expression does not have a taylor expansion at t = 0

>	z := .4070716688RandomVariable(Uniform(a, b)); Mean(z); subs({a=0.12-0.020.12, b=0.12+0.02*0.12}, %);

(2)

>

Download Mean.mw

PLEASE: Maple still fails if I replace a = 0.12 -0.02*0.12 by its value 0.1174
and b = 0.12+0.02*0.12 by its value 0.1224

This is incomprehensible and could hide a more profound problem.

plot for differential equation in time domain...

Asked by: torabi 85

October 16 2018

0 10

Hi.

How I can Plot voltage (v) versus time (t) in nonlinear differential equation.

if possible please provide a simple example for me. for instance the command that I must use!!

plot.pdf

Can I Insert eqations into a list using equation l...

Asked by: peter2108 95

October 16 2018

0 2

Following creates a list of 10 numbers 1..10:

L := []; for i from 1 to 10 do L := [op(L), i] end do;

Now suppose I have equations labelled (1) ... (10). Can I place these equations into a list by some analogy of the above code using the label references?

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