Joe Riel,

I am interested in making some adjustments to the Syrup package to help tailor it to my needs (expanded error descriptions, variable partial differentiations, and others).   A little over a year ago I had some dialog about how to make Syrup changes, but the program I was working on at the time, cancelled the Maple effort...

I need guidance on (1) how the Maple code is structured so I may explore modest changes, (2) what suggestion do you have on the easiest/smartest code development environment, (3) how to compile the changed code, then (4) how to rerun the revised Syrup.

Thank you and happy holidays.
Jeff Belue

Description of program: To try to find 'similar shaped' words.  eg the word 'qwirky' is similar to 'ywivhg' because q, y and g have downstrokes and h, k have an 'upstroke'.  ..at present the 'words' are not words in the dictionary sense.

   I have posed several questions throughout the program, but will focus on the most puzzling which is in the last section of the program, reproduced below.  The variable ii is set at 1 - relating to the first letter, q, of the 'word' qwirky. It correctly outputs the list, listy[4], which contains the letter q.  Similarly if ii is set to 2 it ouputs listy[8] - w.  However when ii is set to 3, it ouputs the number 7, when listy[7] (=[i]) was expected.  Values of ii from 4 to 6 (thelength of 'qwirky') work as expected.  Ideally I would like to dispense with  the ii:=1 statement, and uncomment the for .. do loop:

> for ii from 1 to nops(letters) do
> seq(listy[listsuffix[  Ord(letters[ii] )-96] ][kk], kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) );
> end do:

  Another question was how to incorporate the sequence, seq() statement in a printf statement - but that's enough for now!!:-)

    After reading thris through and seeing theTags, I see I have used a variable 'letters' in my program - perhaps this is causing problems?  I initially thought Maple had a bug in it - but probably a common thought by students!:-)

 ##End part of program  

> ii:=1:   #####Try putting equal to 2, 3, 4, 5, 6
> #ii:=2 returns w - which is correct, but ii:=3 returns 7 when i is expected. Values 4 to 6 work OK
> printf("List of the listy which contains the particular letter, %d  of 'qwirky'  ie the letter %s",ii, letters[ii]);
> printlevel:=5:
> printf("Nops(word)=%d  nops(letters)=%d\n",nops(word),nops(letters));
> #for ii from 1 to nops(letters) do
> seq(listy[listsuffix[  Ord(letters[ii] )-96] ][kk], kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) );
> #end do:

Thanks, David S

START of MAPLE PROGRAM here

# Similar Words - idea is to create words which look similar in shape to other words
> restart:
> with(StringTools):
> #abcdefghijklmnopqrstuvwxyz
> #a  c, e, o, s, u, x, z
> #b  d, h, k,
> #f  l, t
> #g  j, p, q, y
> #i
> #m  n
> #r  v
> #w
>
> #  Take a word eg 'qwirky' & make words which appear similar -
>
> word:="qwirky":
> numLetters:=length(word):
> printf("Length of word is %d",length(word));
> printf("Why is the length of x+2*y equal to %d - as opposed to just 5?",length(x+2*y));
> printf("Is it because x and y each take up 3 'storage locations' - and one each for +2* ?");  
>
> # Help says:For other objects, the length of each operand of expr is computed recursively and added to the number of words used to represent expr. In this way, the measure of the size of expr is returned.
> #...but I'm not much wiser!:-(
> #I ended up using 'nops' instead, but not sure which is better.
> listy[1]:=[a,c,e,o,s,u,x,z]:
> #printf("Number of letters in listy[1] seq(%s, kk=1..nops(listy[1]))   is %d\n",seq(listy[1][kk], kk=1..nops(listy[1])));
> printf("The previous printf statement - commented out - does not work.  The one below does\n - but that's because I've put in eight 'per cents'\n - I'd like the program to work this out.\n");    
>
> printf("Number of letters in listy[1] %s,%s,%s,%s,%s,%s,%s,%s is %d\n",seq(listy[1][kk], kk=1..nops(listy[1])  ),nops(listy[1]) );
> listy[2]:=[b, d, h, k]:
> listy[3]:=[f, l, t]:
> listy[4]:=[g, j, p, q, y]:
> listy[5]:=[m, n]:
> listy[6]:=[r, v]:
> listy[7]:=[i]:
> listy[8]:=[w]:
> #Which list number, for each letter of the 26 letters?
> #Nos below correspond to a, b,c... ..z
> listsuffix:=[1,2,1,2,1,3,4,2,7,4,2,3,5,5,1,4,4,6,1,3,1,6,8,1,4,1]:
> #printf("listsuffix[12]=%d",listsuffix[12]);
> #printf("listy[4][2]=%s",listy[4][2]);
> printf("Ascii number of Second letter in list 4 is %d",Ord(listy[4][2]));
> printf("Second letter in list 4 is %s",Char(Ord(listy[4][2])));
> #printf("listsuffix[listy[4][2]]=%d",listsuffix[listy[4][2]]);
> #Split into a list of letters
> letters:=convert(word, list):
> #Consider the letters close to the letters - go through all combinations and check to see if any match words in the dictionary.
> #printf("1");
> #printf("%d",Ord(letters[2]));
> #printf("2");
> #printf("Suffix number is %d",listsuffix[Ord(letters[2])-71]);
> #printf("3");
> for i from 1 to length(word) do
> #See which list letter belongs to
> #printf("Letter number %d, %A belongs to list %d\n",i,letters[i],listsuffix[Ord(letters[i])-96]);
> printf("Ascii number of letter number %d, %s in list %d is %d\n",i,letters[i],listsuffix[Ord(letters[i])-96],Ord(letters[i]));
>
> #printf("Ascii number of letter number %d, %s is %d in list %d which is \n",i,letters[i],Ord(letters[i]),listsuffix[Ord(letters[i])-96]);
> #printf("Below may not work");
> #printf("Ascii number of letter number %d, %s is %d in list %d which is seq(%s, kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) ))\n",i,letters[i],Ord(letters[i]),listsuffix[Ord(letters[i])-96],seq(listy[listsuffix[  Ord(letters[ii] )-96] ][kk], kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) ) );
>
> #Check out all letters in list, listy[listsuffix[Ord(letters[i])-96]
>
> #seq(  listy[  listsuffix[  Ord(letters[i] )-96]  ], 1..length(listy[                listsuffix[  Ord(letters[i] )-96]);
> #seq(listy[listsuffix[  Ord(letters[ii] )-96] ][kk], kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) )
> end do:
> #seq(listy[2][kk], kk=1..4);
> #List of listy[2]
> printf("List of listy[2] below:");
> seq(listy[2][kk], kk=1..nops(listy[2]));
> #List of the listy which contains the particular letter of 'qwirky'
> ii:=1:   #####Try putting equal to 2, 3, 4, 5, 6
> #ii:=2 returns w - which is correct, but ii:=3 returns 7 when i is expected. Values 4 to 6 work OK
> printf("List of the listy which contains the particular letter, %d  of 'qwirky'  ie the letter %s",ii, letters[ii]);
> printlevel:=5:
> printf("Nops(word)=%d  nops(letters)=%d\n",nops(word),nops(letters));
> #for ii from 1 to nops(letters) do
> seq(listy[listsuffix[  Ord(letters[ii] )-96] ][kk], kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) );
> #end do:

#####  END of program.   Below is output for Maple 7   ##########

Warning, the assigned name Group now has a global binding

Length of word is 6
Why is the length of x+2*y equal to 9 - as opposed to just 5?
Is it because x and y each take up 3 'storage locations' - and one each for +2* ?
The previous printf statement - commented out - does not work.  The one below does
 - but that's because I've put in eight 'per cents'
 - I'd like the program to work this out.
Number of letters in listy[1] a,c,e,o,s,u,x,z is 8
Ascii number of Second letter in list 4 is 106
Second letter in list 4 is j
Ascii number of letter number 1, q in list 4 is 113
Ascii number of letter number 2, w in list 8 is 119
Ascii number of letter number 3, i in list 7 is 105
Ascii number of letter number 4, r in list 6 is 114
Ascii number of letter number 5, k in list 2 is 107
Ascii number of letter number 6, y in list 4 is 121
List of listy[2] below:
                                          b, d, h, k
List of the listy which contains the particular letter, 1  of 'qwirky'  ie the letter q
{--> enter printf, args = "Nops(word)=%d  nops(letters)=%d\n", 1, 6
Nops(word)=1  nops(letters)=6
                                              30
<-- exit printf (now at top level) = }
{--> enter StringTools:-Ord, args = "q"
                                             113
<-- exit StringTools:-Ord (now at top level) = 113}
{--> enter StringTools:-Ord, args = "q"
                                             113
<-- exit StringTools:-Ord (now at top level) = 113}
{--> enter StringTools:-Ord, args = "q"
                                             113
<-- exit StringTools:-Ord (now at top level) = 113}
{--> enter StringTools:-Ord, args = "q"
                                             113
<-- exit StringTools:-Ord (now at top level) = 113}
{--> enter StringTools:-Ord, args = "q"
                                             113
<-- exit StringTools:-Ord (now at top level) = 113}
{--> enter StringTools:-Ord, args = "q"
                                             113
<-- exit StringTools:-Ord (now at top level) = 113}
                                        g, j, p, q, y

My Maple is apparently getting rusty!  Question: how do I recognize a finite constant?

type(Pi, finite) is false (by design, apparently), while type(infinity, constant) is true.  So now I don't know quite what to 'ask' maple!

i use a not good example's polynomials to illustrate the idea

u1:=3*a*c+b^2+c;
u2:=7*a^5*b*c^3;
u3:=a+3*b^2;
T := lexdeg([a,b,c],[e1,e2, e3]);
GB := Basis([e1-u1, e2-u2, e3-u3],T);
result := NormalForm(a*b+b*c, GB, T);

now result is to express a*b+b*c in terms of e1, e2, e3 which represent u1, u2, u3.

is it possible to use preimage to find possible u1,u2,u3 if unknown u1,u2,u3 and given known eqx?

How to use preimage to find possible eqx if given known u1,u2,u3 to find eqx in NormalForm(eqx, GB, T)?

what i confused in code below is that if i know it in terms of -1*e1+2*e2+*e3

it already can be used to find eqx, it seems reasonable to put source1list as unknown to find eqx  or find unknown eq1, eq2, eq3 if given known eqx.

with(RegularChains):
with(ConstructibleSetTools):
source1 := PolynomialRing([e1, e2, e3]);
target1 := PolynomialRing([a, b, c]);
source1list := [...];
target1list := [eq1, eq2, eq3];
cs := PolynomialMapPreimage(target1list, source1list, source1, target1);
Info(cs, source1);
[[e1-e2-e3], [1]]
-1*eq1+2*eq2+*eq3;

Maple 15 does not allow me to do even small changes after executing even the smallest bit of command. Is there a way to overwrite this default setting and force Maple to undo? Do the newer versions of Maple provide this functionality and what is the purpose of undo getting deactivated after a command, at least it should be enabled for the input?

Hi, I am using Maple 18 and struggling with plotting Newton's Method.

I am wanting use the function f(x)=x^3 +cx + 1 where c is a parameter and uses 100 parameter values between -2 and 0, with 100 iterations of each parameter.

Any help would be brilliant.

Thanks in advance,

Neil

deqn4 := {4*eta^4*(diff(f(eta), eta, eta, eta))*(diff(f(eta), eta))-12*eta^3*(diff(f(eta), eta, eta, eta))*f(eta)+12*eta^4*(diff(f(eta), eta, eta))*(diff(f(eta), eta))-36*eta^3*(diff(f(eta), eta, eta))*f(eta)-24*eta^3*(diff(f(eta), eta))^2+12*eta^2*(diff(f(eta), eta))*f(eta)-36*eta*f(eta)^2-4*eta^3*(diff(f(eta), eta, eta))*(diff(f(eta), eta))-12*eta^2*(diff(f(eta), eta, eta))*f(eta)+12*eta^3*f(eta)*(diff(f(eta), eta))+36*eta^2*f(eta)^2-4*eta^3*(diff(f(eta), eta, eta, eta, eta))-12*eta^3*(diff(f(eta), eta, eta, eta))+12*eta^2*(diff(f(eta), eta, eta))-12*eta*(diff(f(eta), eta))+24*f(eta) = 0};
/
| 4 / d / d / d \\\ / d \
< 4 eta |----- |----- |----- f(eta)||| |----- f(eta)|
| \ deta \ deta \ deta /// \ deta /
\

3 / d / d / d \\\
- 12 eta |----- |----- |----- f(eta)||| f(eta)
\ deta \ deta \ deta ///

4 / d / d \\ / d \
+ 12 eta |----- |----- f(eta)|| |----- f(eta)|
\ deta \ deta // \ deta /

3 / d / d \\
- 36 eta |----- |----- f(eta)|| f(eta)
\ deta \ deta //

2
3 / d \ 2 / d \
- 24 eta |----- f(eta)| + 12 eta |----- f(eta)| f(eta)
\ deta / \ deta /

2
- 36 eta f(eta)

3 / d / d \\ / d \
- 4 eta |----- |----- f(eta)|| |----- f(eta)|
\ deta \ deta // \ deta /

2 / d / d \\
- 12 eta |----- |----- f(eta)|| f(eta)
\ deta \ deta //

3 / d \ 2 2
+ 12 eta f(eta) |----- f(eta)| + 36 eta f(eta)
\ deta /

3 / d / d / d / d \\\\
- 4 eta |----- |----- |----- |----- f(eta)||||
\ deta \ deta \ deta \ deta ////

3 / d / d / d \\\
- 12 eta |----- |----- |----- f(eta)|||
\ deta \ deta \ deta ///

2 / d / d \\ / d \
+ 12 eta |----- |----- f(eta)|| - 12 eta |----- f(eta)|
\ deta \ deta // \ deta /

\
|
+ 24 f(eta) = 0 >
|
/
init4 := { {f(0) = 1, (D(f))(0) = 0, ((D@@2)(f))(0) = 0, ((D@@3)(f))(0.1e-2) = 0} };
{{f(0) = 1, D(f)(0) = 0, @@(D, 2)(f)(0) = 0,

@@(D, 3)(f)(0.001) = 0}}
dsol4 := dsolve(deqn4 union init4, numeric, method=rkf45);

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

Hi,

I used to use windows 32 bit, but I have 64 bit windows now. I installed the MapleSim 6.4 and Maple 18 and try to run the simulations that I created with the 32 bit windows. I have this error 'Unable to compile (rc=1), please try again, and if that fails verify your Windows compiller installation'. Could you please let me know what should I do to eliminate this problem?

Best

Onder

Hi,

I have a eight different size lists.

I want to put it into the Array with simplest and clear way as possible respect to first column and containg symbols.
I enclose workseet.

Could you help somone?.
Thank you in advanced

wzel

fill_the_array.mw

i want to plot a arc that starting angle and finishing angle are variable

but it didn't show the arc there is my code

@Markiyan Hirnyk 

how to calculate for module case

M := [[x*y,y,x],[x^2+x,y+x^2,y],[-y,x,y],[x^2,x,y]];

i find maple help using regular chain library, but it has ideal source and ideal target,

when compare with singular, seems different, i am not sure singular code whether correct or not.

so, would like to ask how to do this in maple

if you know singular, i would like to know too.

ring r = 32003, (x,y), lp;
setring r;
ideal Z;
ideal i = x-x2-2*x*y+2*x2*y-2*x*y2+y+y2, 1-x2-2*x*y+2*x2*y-2*x*y2+y2;
map phi = r, i;
ideal i1 = preimage(r,phi,Z);
i1;
ideal i2 = preimage(r,i,i);
i2;
ideal i3 = preimage(r,i,Z);
i3;
ideal i4 = preimage(r,Z,i);
i4;

would like to apply to find Cohen Maculay

While(NewKer <> 0)

Mapping = Basis(M – N) ^ K[M]

NewKer = ker(Mapping, M, N)

N = M

M = NewKer

If IsCM(NewKer) = true then

    NewKer

End if

Do

Hello,

      I would like to solve a system of 9 nonlinear equations, with the constraints on all 9 variables to be that they are nonnegative. How can I do this?

My code is below - I am trying NLPSolve and have tried solve, but am getting stuck.

with(Optimization);

restart; eq1 := 531062-S/(70*365)-(.187*(1/365))*(H+C+C1+C2)*S/N = 0;eq2 := (4/365*(T+C))*S/N-(.187*(1/365))*(H+C+C1+C2)*T/N-(1/(70*365)+1/(5*365))*T = 0; eq3 := (.187*(1/365))*(H+C+C1+C2)*S/N-(4/365)(T+C)*H/N-(1/(70*365)+1/(4*365))*H = 0; eq4 := (.187*(1/365))*(H+C+C1+C2)*T/N+(4/365*(T+C))*H/N-(1/(70*365)+3/(8*365)+.2*(1/365)+.1)*C = 0; eq5 := .1*C-(1/(70*365)+1/(4*365)+1/60+.5)*C1 = 0; eq6 := (1/60)*C1-(1/(70*365)+1/(4*365)+1/210+.5)*C2 = 0; eq7 := .5*C1-(1/(70*365)+1/60+0.1e-2)*CT1 = 0; eq8 := .5*C2-(1/(70*365)+1/210+(1/9)*(0.1e-2*7))*CT2+(1/60)*CT1 = 0; eq9 := N-S-T-H-C-C1-C2-CT1-CT2 = 0; soln := NLPSolve({eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9}, {C, C1, C2, CT1, CT2, H, N, S, T}, assume = nonnegative);

Dear All,

I need your help to plot the phase portrait using DEtools[DEplot]  these are the lines of the code. But when I make RUN, there is an error. I need your help to fix the error. Many thinks.

r1:=1; r2:=1; q2:=2; q1=0.5;  a1:=1;

Sys1 := {diff(N(t),t) = r1*N*(1-N/q1)-P*N/(1+N), diff(P(t),t) = r2*P*(exp(-a1*P)-q2)+P*N/(1+N)};

DEtools[DEplot](Sys1,[N(t),P(t)],t=-10..10,N=0..2,P=0..2);

Many thinks

The problem is The square root of 16-x^2 over the interval [0,-4]  0 being the upper bound, -4 being the lower bound.  I have solved 3/4s of this problem but I don't understand what they mean by "Solve the definite integral exactly by geometry".