Maple 2024 Questions and Posts

These are Posts and Questions associated with the product, Maple 2024

A student wakes up at the end of the lecture and just catches the professor saying:
"... and the roots form an arithmetic sequence."

On the board there is a 5th degree polynomial as homework, but unfortunately the student only manages to write down
x^5 - 5x^4 - 35x^3 +
before the professor wipes the board.

But the student still finds all the roots of the polynomial.

And the roots now have to be calculated.

I have and input parameter to a procedure called shift:=05. It sets the position of test between to points.

this particular procedure has two potential sets of text one in Q and the other in vec.
Can Q and vec have the own individual values for shift?

I have shown the input to the procedure and what it outputs. The procedure itself to too long and akward to the post.

Also given is the input definiton of the procedure and how I could possibly find a shift in a list.

 

restart

#with( RationalTrigonometry)

P1:=[-1,3]:P2:=[4,5]:

# shift operates as such mp:=P1+shift*(P2-P1). It's default value 0.5 It moves
#   the text along between P1 and P2
# shift and shift are optional inputs to move individual test pieces.
#If Q contains shift then use it else use shift
  qp:=P1+shift*(P2-P1) or qp:=P1+shift*(P2-P1)
#Similarily for vec

Qdim([P1,typeset("P1=",P1),align=[left],colour=red] ,
     [P2,typeset("P2=",P2),align=[right],colour=purple] ,
      Q=[Q_12,align =[below,right],shift=0.6] ,
      point=[symbolsize=12,symbol=solidcircle,colour=blue,shift=0.4] ,
      vec=[typeset("v[12]=",P2-P1),align=[below,right],shift=0.1],
      shift=0.7):  

#The definition input of the procedure is

 

#proc(P1, P2,
#    {Q:=[ NULL,:-align={':-left'}]},
#        {corp::{"cart","proj"}:="cart"},
#        {shift:=0.5},
#        {vec::list:=[]},
#        {scale::{list,integer}:=1},
#        {leader::{-1,0,1}:= 1},
#        {dimoffset::{1/6,1/5,1/4,3/4,2/3,1/2,1/3,1,3/2,2,3,4,5,6}:=1},
#        {point::list:=['color' = ':-blue', symbol = ':-solidcircle', 'symbolsize' = 8]},
#        {line::list:=['thickness'=3]},
#        {plopts::list:=[]})

restart

Q:=[Q_12,shift=0.6,align =[below,right]]  
      
      

 

[Q_12, shift = .6, align = [below, right]]

(1)

nops(Q)

3

(2)

for i to nops(Q) do
 if has([op(Q[i])],`shift`) then
 qp:=rhs(Q[i]);
 print(i,qp);
 break;
end if;
end do;

2, .6

(3)

 


 

Download 2024-09-18_Select_name_and_Value_from_List.mw

In this post, I would like to share some exercises that I recently taught to an undergraduate student using Maple. These exercises aimed to deepen their understanding of mathematical concepts through computational exploration and visualization. With its powerful symbolic computation capabilities, Maple proved to be an excellent tool for this purpose. Below, I present a few of the exercises and the insights they provided. Interestingly, the student found Maple to be more user-friendly and efficient compared to the software he usually uses for his studies. Below, I present a few of the exercises and the insights they provided.

One of the first topics we tackled was the Fourier series. We used Maple to illustrate how the Fourier series approximates a given function as more terms are added. We explored this through both static plots and interactive animations.

To help the student understand the behavior of different types of functions, we defined piecewise functions using Maple's piecewise command. This allowed us to model functions that behave differently over various intervals, such as the following cubic function exercise

Maple's Explore command was an effective tool for creating an interactive learning environment. We used it to create sliders that allowed the student to vary parameters, such as the number of terms in a Fourier series, and see the immediate impact on the plot.

restart; with(plots)

" F(x):={[[-1,-1<x<0],[1,0<x<1]];  "

proc (x) options operator, arrow, function_assign; piecewise(-1 < x and x < 0, -1, 0 < x and x < 1, 1) end proc

(1)

p1 := plot(piecewise(-3 < x and x < -1, F(x+2), -1 < x and x < 1, F(x), 1 < x and x < 3, F(x-2)), x = -3 .. 3, color = blue)

 

L := 2; a__0 := (int(F(x), x = -(1/2)*L .. (1/2)*L))/L

0

(2)

a__n := proc (n) options operator, arrow; 2*(int(F(x)*cos(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

proc (n) options operator, arrow; 2*(int(F(x)*cos(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

(3)

b__n := proc (n) options operator, arrow; 2*(int(F(x)*sin(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

proc (n) options operator, arrow; 2*(int(F(x)*sin(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

(4)

" #` Fourier series function`  `F__fourier`(x,N):=`a__0`+(&sum;)(`a__n`(n)&lowast;cos(2 * n * Pi * x / L) +`b__n`(n)&lowast;sin(2* n * Pi * x / L));"

proc (x, N) options operator, arrow, function_assign; a__0+sum(a__n(n)*cos(2*n*Pi*x/L)+b__n(n)*sin(2*n*Pi*x/L), n = 1 .. N) end proc

(5)

p2 := plot([F__fourier(x, 40)], x = -3 .. 3, numpoints = 200, color = [purple])

display([p1, p2])

 

Explore(plot([piecewise(-3 < x and x < -1, F(x+2), -1 < x and x < 1, F(x), 1 < x and x < 3, F(x-2)), F__fourier(x, N)], x = -3 .. 3, color = [blue, purple], numpoints = 200), N = 1 .. 40, title = "Fourier Series Approximation with N Terms")

restart; with(plots)

" #` Define the piecewise function`  F(x):={[[0,-1<x<0],[x^(2),0<x<1]];  "

proc (x) options operator, arrow, function_assign; piecewise(-1 < x and x < 0, 0, 0 < x and x < 1, x^2) end proc

(6)

p3 := plot(piecewise(-3 < x and x < -1, F(x+2), -1 < x and x < 1, F(x), 1 < x and x < 3, F(x-2)), x = -3 .. 3, color = blue)

 

L := 2; a__0 := (int(F(x), x = -(1/2)*L .. (1/2)*L))/L

1/6

(7)

a__n := proc (n) options operator, arrow; 2*(int(F(x)*cos(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

proc (n) options operator, arrow; 2*(int(F(x)*cos(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

(8)

b__n := proc (n) options operator, arrow; 2*(int(F(x)*sin(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

proc (n) options operator, arrow; 2*(int(F(x)*sin(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

(9)

" #` Fourier series function`  `F__fourier`(x,N):=`a__0`+(&sum;)(`a__n`(n)&lowast;cos(2 * n * Pi * x / L) +`b__n`(n)&lowast;sin(2* n * Pi * x / L));"

proc (x, N) options operator, arrow, function_assign; a__0+sum(a__n(n)*cos(2*n*Pi*x/L)+b__n(n)*sin(2*n*Pi*x/L), n = 1 .. N) end proc

(10)

p4 := plot([F__fourier(x, 40)], x = -3 .. 3, numpoints = 200, color = [purple])

display([p3, p4])

 

Explore(plot([piecewise(-3 < x and x < -1, F(x+2), -1 < x and x < 1, F(x), 1 < x and x < 3, F(x-2)), F__fourier(x, N)], x = -3 .. 3, color = [blue, purple], numpoints = 200), N = 1 .. 40, title = "Fourier Series Approximation with N Terms")

restart; with(plots)

" #` Define the piecewise function`  F(x):={[[x+2,-2<x<0],[2-2 x,0<x<2]]; "

proc (x) options operator, arrow, function_assign; piecewise(-2 < x and x < 0, x+2, 0 < x and x < 2, 2-2*x) end proc

(11)

p5 := plot(piecewise(-3 < x and x < -1, F(x+2), -1 < x and x < 1, F(x), 1 < x and x < 3, F(x-2)), x = -3 .. 3, color = blue)

 

L := 2; a__0 := (int(F(x), x = -(1/2)*L .. (1/2)*L))/L

5/4

(12)

a__n := proc (n) options operator, arrow; 2*(int(F(x)*cos(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

proc (n) options operator, arrow; 2*(int(F(x)*cos(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

(13)

b__n := proc (n) options operator, arrow; 2*(int(F(x)*sin(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

proc (n) options operator, arrow; 2*(int(F(x)*sin(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

(14)

" #` Fourier series function`  `F__fourier`(x,N):=`a__0`+(&sum;)(`a__n`(n)&lowast;cos(2 * n * Pi * x / L) +`b__n`(n)&lowast;sin(2* n * Pi * x / L));"

proc (x, N) options operator, arrow, function_assign; a__0+sum(a__n(n)*cos(2*n*Pi*x/L)+b__n(n)*sin(2*n*Pi*x/L), n = 1 .. N) end proc

(15)

p6 := plot([F__fourier(x, 40)], x = -3 .. 3, numpoints = 200, color = [purple])

display([p5, p6])

 

Explore(plot([piecewise(-3 < x and x < -1, F(x+2), -1 < x and x < 1, F(x), 1 < x and x < 3, F(x-2)), F__fourier(x, N)], x = -3 .. 3, color = [blue, purple], numpoints = 200), N = 1 .. 40, title = "Fourier Series Approximation with N Terms")

restart; with(plots)

F := proc (x) options operator, arrow; piecewise(-1 < x and x < 1, x-x^3, 0) end proc

proc (x) options operator, arrow; piecewise(-1 < x and x < 1, x-x^3, 0) end proc

(16)

p7 := plot(piecewise(-3 < x and x < -1, F(x+2), -1 < x and x < 1, F(x), 1 < x and x < 3, F(x-2)), x = -3 .. 3, color = blue)

 

L := 2; a__0 := (int(F(x), x = -(1/2)*L .. (1/2)*L))/L

0

(17)

a__n := proc (n) options operator, arrow; 2*(int(F(x)*cos(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

proc (n) options operator, arrow; 2*(int(F(x)*cos(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

(18)

b__n := proc (n) options operator, arrow; 2*(int(F(x)*sin(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

proc (n) options operator, arrow; 2*(int(F(x)*sin(2*n*Pi*x/L), x = -(1/2)*L .. (1/2)*L))/L end proc

(19)

b__n(n)

-4*(n^2*Pi^2*sin(n*Pi)+3*cos(n*Pi)*Pi*n-3*sin(n*Pi))/(n^4*Pi^4)

(20)

" #` Fourier series function`  `F__fourier`(x,N):=`a__0`+(&sum;)(`a__n`(n)&lowast;cos(2 * n * Pi * x / L) +`b__n`(n)&lowast;sin(2* n * Pi * x / L));      #` Plot the Fourier series approximation`  p8:=plot([`F__fourier`(x,40)],x = -3.. 3 ,numpoints=200, color=[red]) :"

proc (x, N) options operator, arrow, function_assign; a__0+sum(a__n(n)*cos(2*n*Pi*x/L)+b__n(n)*sin(2*n*Pi*x/L), n = 1 .. N) end proc

(21)

display([p7, p8])

 

Explore(plot([piecewise(-3 < x and x < -1, F(x+2), -1 < x and x < 1, F(x), 1 < x and x < 3, F(x-2)), F__fourier(x, N)], x = -3 .. 3, color = [blue, purple], numpoints = 200), N = 1 .. 40, title = "Fourier Series Approximation with N Terms")

NULL

Download Fourier_Series.mw

The first time they are evaluated, some aborts can occur; the second time they are evaluated, no exception is thrown: 
 

Physics:-Version()

`The "Physics Updates" version in the MapleCloud is 1806 and is the same as the version installed in this computer, created 2024, September 11, 11:27 hours Pacific Time.`

(1)

restart;

RootOf(
        9*x1-5+RootOf(8*_Z**2+_Z-43, 41629632769253767815/18446744073709551616 .. 20814816384626883921/9223372036854775808), x1, 171590466306199/562949953421312 .. 343180932612401/1125899906842624
);

Error, (in property/ProbablyNonZero) cannot determine if this expression is true or false: ln(.1e11*abs(-.304805898398896+1.*RealRange(.304805898398895,.304805898398897)))/ln(10) < -6

 

RootOf(
        9*x1-5+RootOf(8*_Z**2+_Z-43, 41629632769253767815/18446744073709551616 .. 20814816384626883921/9223372036854775808), x1, 171590466306199/562949953421312 .. 343180932612401/1125899906842624
);

5/9-(1/9)*RootOf(8*_Z^2+_Z-43, 41629632769253767815/18446744073709551616 .. 20814816384626883921/9223372036854775808)

(2)

RootOf(
        2*x1-3+RootOf(_Z**2+2*_Z-11, 181818607464242035159/73786976294838206464 .. 363637214928484070345/147573952589676412928), x1, 301683970796757/1125899906842624 .. 150841985398379/562949953421312
);

Error, (in property/ProbablyNonZero) cannot determine if this expression is true or false: ln(2500000000.*abs(-.267949192431123+1.*RealRange(.267949192431122,.267949192431123)))/ln(10) < -6

 

RootOf(
        2*x1-3+RootOf(_Z**2+2*_Z-11, 181818607464242035159/73786976294838206464 .. 363637214928484070345/147573952589676412928), x1, 301683970796757/1125899906842624 .. 150841985398379/562949953421312
);

3/2-(1/2)*RootOf(_Z^2+2*_Z-11, 181818607464242035159/73786976294838206464 .. 363637214928484070345/147573952589676412928)

(3)


 

Download run__twice.mw

Why does (the outer) RootOf have to be evaluated twice? 
Note. There are other similar examples, but they are less concise (so they are omitted here).

If you draw a chord in any curve, when the latter becomes infinitely small, the ratio of the surface segment to the triangle formed by the chord and the associated tangents is 2:3.

(Source: Archive of Mathematics and Physics, editor Johann August Grunert, 31st part of 1858, pp. 449-453, "On a remarkable general theorem on curves",
Author: Andreas Völler)

(The curve may be assumed to be sufficiently differentiable.)

I have a quite complex expression (where and are real numbers): 

expr := Or(And(-p^2 + 3*q < 0, p < 0, p^2 - 4*q < 0, Or(And(p < 0, -q < 0), p < 0, q < 0), Or(And(-2*p^2 + 3*q < 0, -q^2 < 0), And(p <= 0, Or(-2*p^2 + 3*q < 0, q^2 < 0))), Or(And(Or(And(p < 0, -q < 0), p < 0, q < 0), Or(And(-2*p^2 + 3*q < 0, -q^2 < 0), And(p <= 0, Or(-2*p^2 + 3*q < 0, q^2 < 0)))), And(p < 0, -q < 0), p < 0, q < 0, And(2*p^2 - 3*q < 0, -q^2 < 0), And(-p <= 0, Or(2*p^2 - 3*q < 0, q^2 < 0))), -2*p^5 + 15*p^3*q - 27*p*q^2 <= 0, p^2*q^2 - 4*q^3 = 0), And(p^2 - 3*q = 0, p < 0, -2*p^2 + 3*q < 0, Or(And(p < 0, -2*p^2 + 3*q < 0), p < 0, 2*p^2 - 3*q < 0), 2*p^3 - 9*p*q = 0), And(-p^2 + 3*q < 0, Or(And(p < 0, p^2 - 4*q < 0), p < 0, -p^2 + 4*q < 0), p < 0, -q < 0, Or(And(-2*p^2 + 3*q < 0, -q^2 < 0), And(-p <= 0, Or(-2*p^2 + 3*q < 0, q^2 < 0))), Or(And(p < 0, -q < 0, Or(And(-2*p^2 + 3*q < 0, -q^2 < 0), And(-p <= 0, Or(-2*p^2 + 3*q < 0, q^2 < 0)))), And(p < 0, -q < 0), And(2*p^2 - 3*q < 0, -q^2 < 0), And(p <= 0, Or(2*p^2 - 3*q < 0, q^2 < 0))), 2*p^5 - 15*p^3*q + 27*p*q^2 <= 0, p^2*q^2 - 4*q^3 = 0)):

According to coulditbe, is satisfiable: 

_EnvTry := 'hard':
coulditbe(expr) assuming real;
 = 
                              true

But according to SMTLIB:-Satisfiable, is not satisfiable: 

SMTLIB:-Satisfiable(expr) assuming real;
 = 
                             false

Why are the two results opposite

For reference, below is the output from RealDomain:-solve

RealDomain:-solve(expr);
 = 
               /           1  2\                 
              { p = p, q = - p  }, {p = p, q = 0}
               \           4   /                 

I also tried using RealDomain:-simplify, yet the output remains almost unchanged (Why?). 

Bernoulli first order ode has form as show in wikipedia  and also on Maple own site as

Notice that it is P(x)*y above and not P(x)* y^(-1) so the y(x) must be linear in that term.   But when I give Maple this ode

ode:=diff(y(x),x) + x*y(x)^(-1)= y(x)^(-1);

Which is clearly not of the form above, it solves it as Bernoulli.  In the above ode, P(x) is x and Q(x) is 1 and n is -1.

The ode advisor correctly said it is separable. But trace shows it used Bernoulli. Also when asking it to solve it as Bernoulli, it does.

What Am I missing here?  Is it not wrong for Maple to use Bernoulli method on this ode which is not Bernoulli?

Worksheet below

interface(version);

`Standard Worksheet Interface, Maple 2024.1, Windows 10, June 25 2024 Build ID 1835466`

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1805 and is the same as the version installed in this computer, created 2024, September 3, 11:35 hours Pacific Time.`

libname;

"C:\Users\Owner\maple\toolbox\2024\Physics Updates\lib", "C:\Program Files\Maple 2024\lib"

restart;

ode:=diff(y(x),x) + x*y(x)^(-1)= y(x)^(-1);
IC:=y(1) = 0;
DEtools:-odeadvisor(ode);

diff(y(x), x)+x/y(x) = 1/y(x)

y(1) = 0

[_separable]

infolevel[dsolve]:=5;

5

dsolve(ode,y(x));  #why this says it solved it as Bernoulli ?

Methods for first order ODEs:

--- Trying classification methods ---

trying a quadrature

trying 1st order linear

trying Bernoulli

<- Bernoulli successful

y(x) = (-x^2+c__1+2*x)^(1/2), y(x) = -(-x^2+c__1+2*x)^(1/2)

dsolve(ode,y(x),[Bernoulli])

Classification methods on request

Methods to be used are: [Bernoulli]

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

* Tackling ODE using method: Bernoulli

--- Trying classification methods ---

trying Bernoulli

<- Bernoulli successful

y(x) = (-x^2+c__1+2*x)^(1/2), y(x) = -(-x^2+c__1+2*x)^(1/2)

 

 

Download why_this_ode_bernullli_sept_15_2024.mw

Dear Maple community,

I was going to export my math assignment about vector functions as an PDF, when I noticed that it butchered every column vector printed out in blue in the document as math-output. I only have this issue when exporting my work.

How do I make Maple display both coordinates of vector in the blue output field?

I produced the below image as an illustration of my problem:

Thank you in advance.

A classic result states that the equation x3px2qxr=0 with real coefficients p, q, r has positive roots iff p<0, q>0, r<0 and -27r2 - 2p(2p2 - 9q)r + q2(p2 - 4q) ⩾ 0 (see for example this question). 
However, Maple appears unable to find the condition: 

a, b, c := allvalues(RootOf(x^3 + p*x^2 + q*x + r, x), 'implicit'):
RealDomain:-solve({a, b, c} >~ 0, [p, q, r]);
 = 
Warning, solutions may have been lost
                               []

Is there a way to get the above conditions in Maple with as little human intervention as possible (I mean, without a priori knowledge of the theory of polynomials)? 

Edit. An interesting problem is when these three positive roots can further be the lengths of sides of a triangle. For reference, here are some (unenlightening) results from some other software: 

I'm trying to solve a system of coupled differential equations numerically, but I'm getting the following error

Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations

The error occurs at the dsolve step, despite trying to ensure that all equations and conditions are in the correct form (sets/lists).

Could someone help me identify what I'm missing here?

Thanks in advance!

L := 200; K := 99; kappa := 1; omegaD := 1; beta := 1; delta := 0.5e-1; j := 2; tmax := 3000; h := L/(K+1); nsp := [`$`(-(1/2)*L+h*i, i = 0 .. L/h)]; km := nops(nsp); omegaD2 := h^2*omegaD^2; deltaHat := h*delta; a := 2; var := seq(x[i](t), i = 1 .. km); initialPositions := seq(x = a*sin(j*h*Pi*nsp[i]/L), i = 1 .. km); initialVelocities := seq((D(x[i]))(0) = 0, i = 2 .. km-1)

boundaryConditions := [x[1](t) = 0, x[km](t) = 0]

equations := seq(diff(x[n](t), t, t)-kappa*(x[n+1](t)-2*x[n](t)+x[n-1](t))+deltaHat*(diff(x[n](t), t))-omegaD2^2*(x[n](t)-beta*x[n](t)^3) = 0, n = 2 .. km-1)

sol := dsolve({equations, boundaryConditions, initialPositions, initialVelocities}, var, numeric, method = rkf45, range = 0 .. tmax)

Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations

 

NULL

Download dsolve_error.mw

I wanted to change  eq:= 1/2 * sqrt(-2*lambda)  to 1/2 %* sqrt(-2*lambda)  using a rule.

It works outside of rule ofcourse. But when I put %* in the RHS of the rule, maple hangs. It seems it is going into infinite loop.

I tried the trick of using '%*' but this gives syntax error.

Same problem happens when using %. and not just %*

Is there a workaround?

Attached worksheet. Make sure to save all work before trying it.
 

interface(version);

`Standard Worksheet Interface, Maple 2024.1, Windows 10, June 25 2024 Build ID 1835466`

Physics:-Version()

`The "Physics Updates" version in the MapleCloud is 1804. The version installed in this computer is 1802 created 2024, September 3, 11:35 hours Pacific Time, found in the directory C:\Users\Owner\maple\toolbox\2024\Physics Updates\lib\`

libname;

"C:\Users\Owner\maple\toolbox\2024\Physics Updates\lib", "C:\Program Files\Maple 2024\lib"

restart;

eq:= 1/2 * sqrt(-2*lambda)

(1/2)*(-2*lambda)^(1/2)

eq:= 1/2 %. sqrt(-2*lambda); #no problem

`%.`(1/2, (-2*lambda)^(1/2))

eq:= 1/2 %* sqrt(-2*lambda); #no problem

`%*`(1/2, (-2*lambda)^(1/2))

restart;

eq:= 1/2 * sqrt(-2*lambda)

(1/2)*(-2*lambda)^(1/2)

applyrule(sqrt(x::anything)/y::anything = 1/y %. sqrt(x),eq); #why this hangs?

 

restart;

eq:= 1/2 * sqrt(-2*lambda)

(1/2)*(-2*lambda)^(1/2)

applyrule(sqrt(x::anything)/y::anything = 1/y %* sqrt(x),eq); #why this hangs?

 


 

Download maples_hangs_applyrule_sept_10_2024.mw

Hello!

I present a simple work-up of rolling a plane curve along a fixed plane curve in 2d space. Maple sources are attached.

Kind regards!

Source.zip

Hello,

Please help me solve this type of equations, I have more of them, but there is only one in the attached file. fsolve does not work. Please help.

Best regards!

tg_Test.mw

The help page of interface('longdelim') states: 

If true, Maple control structures such as if, do, proc, and so on are displayed with the newer-style long ending delimiters such as end if, end do, end proc, and so on. If false, ending delimiters are displayed as fi, od, end, and so on

If I set interface('longdelim' = false):, the Maple Input

try f:=0 catch:end try;

can be converted into 

via , but why is

use f=0 in f+1 end use;

still converted into

 instead of “use f = 0 in f + 1 end”? 

With the attached task I would like to learn how Maple handles polynomials and plots. For this I have chosen a task with an interesting history from 1593. Adriaan van Roomen posed it to F. Vieta, who solved it after a short consideration.

My questions are:
1.) How are very long terms entered and displayed in Maple in an appropriate manner?
2.) How are the graphs of several functions displayed in the same coordinate system?
3.) Which graphics settings must be selected for differentiable functions in order to obtain nice, rounded curves rather than angular ones from the numerical result?
4.) Can the apparently random oscillations of the curve at the end of the interval be suppressed?
5.) The curves for p(x) and sinnx(x) are theoretically identical, since p(x) is the trigonometric expansion of sinnx(x). Their graphs must therefore be identical. How can this be displayed?
6.) The polynomial has many real zeros. How can the zeros be clearly presented in a table?

AF_20240909.mw

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