How do I use msolve to solve y^2 + y - 11=0 in Zp for all primes p with 41=< p =<107 ?

Also, using the results make a conjecture describing the primes p for which there are solutions to y^2 + y - 11 = 0 in Zp

This was what I did.

41<=p<=107

msolve(y^2 + y - 11=0, p)

but I received this error, no implementation of msolve matches the arguments in call, msolve(y^2 + y - 11=0, p)

Any help is appreciated. Thanks

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

I have a coefficient matrix A and the constant matrix b.

I want to find the unknowns x_1,x_2,...,x_8. When I put my code and run it in Maple I get an error. Any ideas as to why this is happening? The code I tried is as follows:

with(Student[NumericalAnalysis]):

A := Matrix([[-1, 0, 0, (1/2)*sqrt(2), 1, 0, 0, 0], [0, -1, 0, (1/2)*sqrt(2), 0, 0, 0, 0], [0, 0, -1, 0, 0, 0, 1/2, 0], [0, 0, 0, -(1/2)*sqrt(2), 0, -1, -1/2, 0], [0, 0, 0, 0, -1, 0, 0, 1], [0, 0, 0, 0, 0, 1, 0, 0], [0, 0, 0, -(1/2)*sqrt(2), 0, 0, (1/2)*sqrt(3), 0], [0, 0, 0, 0, 0, 0, -(1/2)*sqrt(3), -1]]);

b := Vector([0, 0, 0, 0, 0, 10000, 0, 0]);

LinearSolve(A, b, maxiterations = 300, method = SOR(1.25), tolerance = 10^(-2));

I also tried it with a SOR variable of 1.5 instead of 1.25, but that gives me the same error.

Hello,

Is it possible in Maple to select summand with some command?

For example:

eq := a+b = 0;
                           a + b = 0
eq-b;
                             a = -b

How to generalize second line?(in this case I explicitly told to subtract b, but I want to pick b by some command)

And I am not talking about solving, isolation or anything like that in this case. I just want to transfer something from lhs to rhs by selecting it beforehand.

Thank you.

Hello,

How can one simplify following expression:

After applying 'simplify' command I am getting this:

Powers are not distributed between bases.

How to force Maple simplify it further to

Thank you.

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.

I have a DEplot :

DEplot(sys, [x(t), y(t)], t = 0 .. 30, [[x(0) = 0.8, y(0) = 0]], x = -1.25 .. 1.25, y = -1.25 .. 1.25, numpoints = 300, axes = boxed);

I would like to draw on this plot two circles centered in the origin with different radiuses.

How can I proceed?

How can I plot a vertical line say x = a in Maple 15 worksheet?

Thanks

I have about a year of experience in Maple, but as we know, we often come accross very simple problems that are difficult to find solutions to. Part of the problem is that I have not used conditional tests within for loops before.

I have a very large matrix, but until I solidify the algorithm, I'm practicing on a small matrix.

I created a 3 by 3 fibonacci matrix A:=<21,5,1|13,3,1|6,2,0>

I am trying to change the 1s to 0s without specific index reference. Here is my psuedocode that I hope to work on after some tips:

• for i from 1 to 9
  if A[k] = 1 then
  A[k]=0
  end if
  end do

I get an error message saying "unterminated for loop," which is confusing me.

I have 4 retangular equations:

z0:=sqrt(3)+I
z1:=-1+I*sqrt(3)
z2:=-sqrt(3)-I
z3:=1-I*sqrt(3)

on polar form they apear as:

polar(z0)=(2,Pi/6)
polar(z1)=(2,2*Pi/3)
polar(z2)=(2,-5*Pi/6)
polar(z3)=(2,-Pi/3)

I want those polar expressions plottet in a polarplot as points(x,y), how is that possible ?

I want to integrate the following cumulative distribution function with variable bounds but maple returns the integral.

>with(Statistics);

>N := RandomVariable(Normal(0, 1)):

>a := (ln(98.53*(1/95))+(0.1e-1+.5*x^2)*.5)/(x*sqrt(.5))

>y:= CDF(N, a, inert = true);

How do i make maple return a function of x instead of an integral?

I'm trying to write the cumulative distribution function into maple but I seem to get an error. The code is:

>with(Statistics):

>N:=RandomVariable(Normal(0,1)):

>CDF(N,t,inert=true);

I get an error message "Error, (in Statistics:-CDF) unexpected parameters: inert = true".

It used to work, maybe because I was using a newer version of Maple.

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