How would one in Maple solve this, which is an inequality equation in some variables, which can be nonlinear, with constraints on range of each variable. I.e. I want to find conditions on the variables to make the inequality satisfied.

In Mathematica, I use the Reduce command

Clear[x, y];
eq = 1/2 - x + x^2 - y + y^2;
Reduce[{eq > 0, 0 < x < 1 && 0 < y < 1}, {x, y}, Reals]

How would one do the same in Maple? I tried solve, but can't give constraints.

restart;
eq:=1/2 -x+x^2-y+y^2:
solve(eq>0 , {x1, x2});

So I need to do the same as in the Mathematica command, but in Maple. I do not want numerical solution, but algebraic as shown above.

Using Maple 18.2 on windows.

Hello people in mapleprimes,

I want to ask a question about modification of expression.

Basically, what I want to do is to change the expression of sqrt((-a)^2) to sqrt(a^2).

The expression I want to modify is this:

aa:=phi[n, j] = a[j, D]-a-sqrt((-a[j, D]+a+tau[n, j])^2+4*gamma*f[c, n, j]/L[j])

I want to change this to 

phi[n, j] = a[j, D]-a-sqrt((a[j, D]-a-tau[n, j])^2+4*gamma*f[c, n, j]/L[j])

To do this, I wrote as 

subs(sqrt((-a[j, D]+a+tau[n, j])^2+4*gamma*f[c, n, j]/L[j]) = sqrt((a[j, D]-a-tau[n, j])^2+4*gamma*f[c, n, j]/L[j]), phi[n, j] = a[j, D]-a-((-a[j, D]+a+tau[n, j])^2+4*gamma*f[c, n, j]/L[j])^(1/2))

Or, I wrote as 

subs(op([2,3,2,1,1],aa)=(a[j, D]-a-tau[n, j])^2,aa)

If there is better ways, please tell me them.

Best wishes.

taro

I would be grateful if anyone would comment on the compatibility of Maple 18 and MacOS 10.10 (Yosemite).

A 37 foot ladder is placed against a wall that is 9 feet away from its base. Will the top of the ladder reach a window ledge that is 35 feet above ground? Explain.

How do I get values out of solved augmented matric in 2D math input? I cannot simply assign the solutions to a value by x:=A[1,5}, because it's an augmented matrix.

I have a new question here about using maple, welcome to answer

http://math.stackexchange.com/questions/1049655/how-to-calculate-a-in-this-description

As title , there has 2 balls(A and B) which I can change Quality 、Radius and Position for two dimensions(x and y axis),

A has a velocity and B quiescence , let A strike to B .

As shown below

I did a sample ,but i can't make it accurately .

test.mw

I want to let my sample like the below sample

http://www.mapleprimes.com/questions/202467-How-Does-Maple-Make-A-Rutherford-Scattering-

I have couple questions on displaying/printing items I see from ShowContents(LibLocation). Here is an example

restart;
with(LibraryTools):
LibLocation:=cat(kernelopts(mapledir),"/lib/maple.mla");
c:=ShowContents(LibLocation):

First quesionNow, when I do
c[1];
then I see

When is "%?.m" there? Is this suppoed to be an actual function one can print?

second question

c[30]; gives

But when I do:

interface(verboseproc=3);
print(AiryBiZeros);

I get listing that ends up calling

---------------------------
proc(n::{algebraic, algebraic .. algebraic})
local fn;
   1   if nargs <> 1 then
   2     error "wrong number of arguments"
       end if;
   3   fn := traperror(evalf(n));
   4   if type(fn,'numeric') and not type(n,'posint') or type(fn,('numeric') .. ('anything')) and not type(op(1,n),'posint') or type(fn,('anything') .. ('numeric')) and not type(op(2,n),'posint') then
   5     error "expecting positive integers in argument"
       elif type(n,('integer') .. ('integer')) and op(2,n) < op(1,n) then
   6     NULL
       else
   7     ('AiryBiZeros')(n)   <----- Is this C function/compiled that is why it does not show?
       end if
end proc
--------------------------------------

Which is the same name I printed. So it seems to be another internal procedure with same name? How can I print it as well?

Is there a better way to print Maple procedures/command than what I am doing above so one gets full listing?
I tried Browse(); command in LibraryTools, but found it very cluncky to use (keeps losing listing and screen become blank. Very buggy)

if given a prime ideal p, is it possible to find back possible ideal A which output this prime ideal

such that p*A = 0

Hello,

Is it possible to hide the geometry linked to the center of gravity and masses in the implicit geometry and keep only the kinematic joint.

If yes, may you explain how ? For the moment, I only manage to show/hide all the element of the implicit geometry. 

I use MapleSim 6.4.

Thank you for your help.

http://uknowledge.uky.edu/cgi/viewcontent.cgi?article=1001&context=math_etds

Example 3.23 in page 24

degree(2*z^2 + z + 2); #2

how to convert rational function to into (something) - (something) + (something) like in the example

the example 's rational function is 2*z^2+z+2

if rational function is has denominator how to calculate?

2z^2 + z + 2 =

z+2
3


−

z−2
3


+

z+1
2


−

z−4
2


+

z
1


−

z−8
1

when calculate delta p1, what is bi? how to calculate like the example's one?

(z+i-1 i+1-j) - (z+i-bi-1 i+1-1)

sum(d,i=j) (z+i-1 i+1-j) - (z+i-bi-1 i+1-1)

3.23
(z 1) - (z-bi 1)

how to do the expansion in step 3, as i use right shift that can not calculate the same result as in the example.

x0*x1/x0 = x1; # not same as in the example for I

x1^4*x2/x1 = x2*x1^3;  # not same as in the example for I

finally would like to apply to hilbert series 1/((1+z)*(1+z^2))

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

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;

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