Maple will multiply two matrices for me perfectly.  

However, when I change the value in one of the cells and try to re-evaluate the product, Maple starts giving me the sum of something in all of my cells.   

Are you simply not allowed to ever change the initial values you set for a matrix?  This doesn't seem to make sense to me.  

I have attached a screen shot to show an example with a random matrix.  

is(k::OrProp(even, odd)) assuming k::integer;
                             false

is(k::Or(posint, nonposint)) assuming k::integer; # a synonym for OrProp
                             false

is(`or`(k::posint, k::negint, k = 0)) assuming k::integer;
                             false

Not including the infix or because that would fail for k=0 for different reasons.

1.

eliminate({-a+x+y+z, -b+x*y+y*z+x*z, -c+x*y*z, -P+x^4+y^4+z^4}, {x, y, z});
    [{x = (sqrt(c^2*(a^2*b^2-2*a*b*c-4*b^3+c^2)/b^4)*b^2+a*b*c-c^2)/(2*b*c),
      y = -(sqrt(c^2*(a^2*b^2-2*a*b*c-4*b^3+c^2)/b^4)*b^2-a*b*c+c^2)/(2*b*c),
      z = c/b}, {-a^4+4*a^2*b-4*a*c-2*b^2+P}],
    [{x = -(sqrt(c^2*(a^2*b^2-2*a*b*c-4*b^3+c^2)/b^4)*b^2-a*b*c+c^2)/(2*b*c),
      y = (sqrt(c^2*(a^2*b^2-2*a*b*c-4*b^3+c^2)/b^4)*b^2+a*b*c-c^2)/(2*b*c),
      z = c/b}, {-a^4+4*a^2*b-4*a*c-2*b^2+P}]

The expression for P is fine, but how are x and y square roots and z=c/b??

2.

eliminate({y*z, x+y-z}, z);
                       [{z = 0}, {x + y}]

Is this supposed to happen? I was expecting y*(x+y):

Groebner:-Basis({y*z, x+y-z}, lexdeg([z], [x, y]));
                     [x*y + y^2, z - x - y]

Although a very general question... does the rope break or not?

A 100 kg weight is hung from the center of a 1/4" polypropelene rope making an angle of 10 degrees at both ends with the horizontal.  According to some characteristics, the breaking strength of the rope is 1125 lbs or 5kN.  The safe load (safety factor of 12) drops the weight to  93.8 lbs or 0.417kN. Find the tension in the rope?  Is it strong enough?

Fairly easy to solve in Maple - just two equations two unknowns making t1 tension on one side and t2 tension on the other side

eq1 := t1*cos(80*Pi*(1/180))+t2*cos(80*Pi*(1/180)) = 100*9.8 # ΣFy=mg

eq2 := t1*sin(80*Pi*(1/180))-t2*sin(80*Pi*(1/180)) = 0 #ΣFx=0 hence basically t1=t2

solve([eq1,eq2])

                 {t1 = 2821.797537, t2 = 2821.797537}

Therefore the tension on the rope t1=t2 is 2821 N  (looks like way beyond the safety factor) so this means the rope will not break, correct?  or is it that the total tension becomes t1+t2=5642 N putting us over the 5kN and the rope breaks?

I am new to maple. how do i segregate different term in long PDE in maple

In the unrelated discussion https://mapleprimes.com/questions/222768-Interrupting-An-Evaluation-Leaves-Mserver-Running
I asked Carl Love if he had an estimate of the number of builtin procedures in Maple.
Carl answered with the code you will find in the link.
His code came up with 316 and 320 in Maple 2016.2 and 2017.2, respectively.
He mentioned that the last one alphabetically was zip.
I happened to look up the help page for type/builtin and found the statement:

"It may be used with anames to list all builtin procedures. For example anames(builtin)."

So I tried:

AB:={anames(builtin)};
nops(AB); # 234 in Maple 2017.2
member(zip,AB); #false
op(3,eval(zip)); # builtin = 589 in Maple 2017.2
showstat(zip); # Works despite being builtin.

In Maple 2016.2 the corresponding numbers were 232 and 585.
So I'm left somewhat confused.
Apparently some procedures are less builtin than others? Does this just mean that the most difficult zipping is builtin, while simple stuff like the following is done by line 16 in the available code for zip?

showstat(zip,16..17);
zip(`=`,[a,b,c], [1,2,3]);

I have saved 3 Maple workbook files on my computer, but only one of them can re-open.
when i try to open the other two a pop-up says

"A problem was ecountered while opening the workbook.

database is not opened" 

and then i get the option to press "ok". What should i do?

If I enter the following I do not get the residue... how do I make the residue command work?

The documentation for the option AllSolutions for int says that the results are always valid for all real parameter values (in the endpoints). That seems like a pretty major claim. Each of these three is already wrong for a=-1/2, b=1/2:

int(1/ln(t), t = a .. b, AllSolutions);
    piecewise(ln(a) < ln(b), piecewise(And(1 < b, a < 1), undefined, piecewise(a = 1, infinity,
    Ei(1, -ln(a)))+piecewise(b = 1, -infinity, -Ei(1, -ln(b)))), ln(b) = ln(a), 0, ln(b) < ln(a),
    -piecewise(And(1 < a, b < 1), undefined, piecewise(b = 1, infinity, Ei(1, -ln(b)))+
    piecewise(a = 1, -infinity, -Ei(1, -ln(a)))))

int(sqrt(t^2-1+I*t), t = a .. b, AllSolutions);
    piecewise(a < b, (1/2)*sqrt(b^2-1+I*b)*b+I*sqrt(b^2-1+I*b)*(1/4)-3*ln(-2*signum(0, -b, 1)^2*
    b^2+2*sqrt(b^4-b^2+1)*signum(0, -b, 1)^2+4*b*sqrt(2*sqrt(b^4-b^2+1)+2*b^2-2)+2*
    signum(0, -b, 1)^2-2*signum(0, -b, 1)*sqrt(2*sqrt(b^4-b^2+1)-2*b^2+2)+6*b^2+2*
    sqrt(b^4-b^2+1)-1)*(1/16)-3*ln((-I*(signum(0, -b, 1)*sqrt(2*sqrt(b^4-b^2+1)-2*b^2+2)-1+I*
    sqrt(2*sqrt(b^4-b^2+1)+2*b^2-2)+(2*I)*b))*(1/sqrt(-2*signum(0, -b, 1)^2*b^2+2*sqrt(b^4-b^2+1)*
    signum(0, -b, 1)^2+4*b*sqrt(2*sqrt(b^4-b^2+1)+2*b^2-2)+2*signum(0, -b, 1)^2-2*signum(0, -b, 1)*
    sqrt(2*sqrt(b^4-b^2+1)-2*b^2+2)+6*b^2+2*sqrt(b^4-b^2+1)-1)))*(1/8)-(1/2)*sqrt(a^2-1+I*a)*a-I*
    sqrt(a^2-1+I*a)*(1/4)+3*ln(-2*signum(0, -a, -1)^2*a^2+2*sqrt(a^4-a^2+1)*signum(0, -a, -1)^2+
    4*a*sqrt(2*sqrt(a^4-a^2+1)+2*a^2-2)+2*signum(0, -a, -1)^2-2*signum(0, -a, -1)*sqrt(2*
    sqrt(a^4-a^2+1)-2*a^2+2)+6*a^2+2*sqrt(a^4-a^2+1)-1)*(1/16)+3*ln((-I*(signum(0, -a, -1)*sqrt(2*
    sqrt(a^4-a^2+1)-2*a^2+2)+I*sqrt(2*sqrt(a^4-a^2+1)+2*a^2-2)-1+(2*I)*a))*(1/sqrt(-2*
    signum(0, -a, -1)^2*a^2+2*sqrt(a^4-a^2+1)*signum(0, -a, -1)^2+4*a*sqrt(2*sqrt(a^4-a^2+1)+2*
    a^2-2)+2*signum(0, -a, -1)^2-2*signum(0, -a, -1)*sqrt(2*sqrt(a^4-a^2+1)-2*a^2+2)+6*a^2+2*
    sqrt(a^4-a^2+1)-1)))*(1/8), b = a, 0, b < a, -(1/2)*sqrt(a^2-1+I*a)*a-I*sqrt(a^2-1+I*a)*(1/4)+
    3*ln(-2*signum(0, -a, 1)^2*a^2+2*signum(0, -a, 1)^2*sqrt(a^4-a^2+1)+4*a*sqrt(2*sqrt(a^4-a^2+1)+
    2*a^2-2)+2*signum(0, -a, 1)^2-2*signum(0, -a, 1)*sqrt(2*sqrt(a^4-a^2+1)-2*a^2+2)+6*a^2+
    2*sqrt(a^4-a^2+1)-1)*(1/16)+3*ln((-I*(signum(0, -a, 1)*sqrt(2*sqrt(a^4-a^2+1)-2*a^2+2)+
    I*sqrt(2*sqrt(a^4-a^2+1)+2*a^2-2)-1+(2*I)*a))*(1/sqrt(-2*signum(0, -a, 1)^2*a^2+
    2*signum(0, -a, 1)^2*sqrt(a^4-a^2+1)+4*a*sqrt(2*sqrt(a^4-a^2+1)+2*a^2-2)+2*signum(0, -a, 1)^2-
    2*signum(0, -a, 1)*sqrt(2*sqrt(a^4-a^2+1)-2*a^2+2)+6*a^2+2*sqrt(a^4-a^2+1)-1)))*(1/8)+(1/2)*
    sqrt(b^2-1+I*b)*b+I*sqrt(b^2-1+I*b)*(1/4)-3*ln(-2*signum(0, -b, -1)^2*b^2+2*sqrt(b^4-b^2+1)*
    signum(0, -b, -1)^2+4*b*sqrt(2*sqrt(b^4-b^2+1)+2*b^2-2)+2*signum(0, -b, -1)^2-
    2*signum(0, -b, -1)*sqrt(2*sqrt(b^4-b^2+1)-2*b^2+2)+6*b^2+2*sqrt(b^4-b^2+1)-1)*(1/16)-
    3*ln(-(I*signum(0, -b, -1)*sqrt(2*sqrt(b^4-b^2+1)-2*b^2+2)-I-sqrt(2*sqrt(b^4-b^2+1)+2*b^2-2)-
    2*b)/sqrt(-2*signum(0, -b, -1)^2*b^2+2*sqrt(b^4-b^2+1)*signum(0, -b, -1)^2+4*b*sqrt(2*
    sqrt(b^4-b^2+1)+2*b^2-2)+2*signum(0, -b, -1)^2-2*signum(0, -b, -1)*sqrt(2*sqrt(b^4-b^2+1)-
    2*b^2+2)+6*b^2+2*sqrt(b^4-b^2+1)-1))*(1/8))

int(arctan(t+2*I), t = a .. b, AllSolutions);
   piecewise(a < b, piecewise(a < 0, I*arctan(4*a/(a^2-3))*(1/2)+(1/4)*ln(a^2+1)+(1/4)*ln(a^2+9)-
   (2*I)*arctan(2*I+a)-arctan(2*I+a)*a+I*Pi*(1/2), a = 0, -I*Pi+3*ln(3)*(1/2), 0 < a, I*arctan(4*a/
   (a^2-3))*(1/2)+(1/4)*ln(a^2+1)+(1/4)*ln(a^2+9)-(2*I)*arctan(2*I+a)-arctan(2*I+a)*a-I*Pi*(1/2))+
   piecewise(b < 0, -I*arctan(4*b/(b^2-3))*(1/2)-(1/4)*ln(b^2+1)-(1/4)*ln(b^2+9)+(2*I)*
   arctan(2*I+b)+arctan(2*I+b)*b-I*Pi*(1/2), b = 0, -I*Pi-3*ln(3)*(1/2), 0 < b, -I*arctan(4*b/
   (b^2-3))*(1/2)-(1/4)*ln(b^2+1)-(1/4)*ln(b^2+9)+(2*I)*arctan(2*I+b)+arctan(2*I+b)*b+I*Pi*(1/2))+
   piecewise(And(0 < b, a < 0), -(2*I)*Pi, 0), b = a, 0, b < a, piecewise(b < 0, -I*arctan(4*b/
   (b^2-3))*(1/2)-(1/4)*ln(b^2+1)-(1/4)*ln(b^2+9)+(2*I)*arctan(2*I+b)+arctan(2*I+b)*b-I*Pi*(1/2),
   b = 0, I*Pi-3*ln(3)*(1/2), 0 < b, -I*arctan(4*b/(b^2-3))*(1/2)-(1/4)*ln(b^2+1)-(1/4)*ln(b^2+9)+
   (2*I)*arctan(2*I+b)+arctan(2*I+b)*b+I*Pi*(1/2))+piecewise(a < 0, I*arctan(4*a/(a^2-3))*(1/2)+
   (1/4)*ln(a^2+1)+(1/4)*ln(a^2+9)-(2*I)*arctan(2*I+a)-arctan(2*I+a)*a+I*Pi*(1/2), a = 0,
   I*Pi+3*ln(3)*(1/2), 0 < a, I*arctan(4*a/(a^2-3))*(1/2)+(1/4)*ln(a^2+1)+(1/4)*ln(a^2+9)-(2*I)*
   arctan(2*I+a)-arctan(2*I+a)*a-I*Pi*(1/2))-piecewise(And(0 < a, b < 0), -(2*I)*Pi, 0))

The first one probably has the correct answer inside, but it has conditions like ln(a)<ln(b), so that case never gets selected when the values are complex.

f := (z, t) -> ln(t)^2/((t^2+1)*(t-z));

int(f(z, t), t = 0 .. infinity) assuming Im(z) > 0;
       int(ln(t)^2/((t^2+1)*(t-z)), t = 0 .. infinity)

int(f(a + I*b, t), t = 0 .. infinity) assuming a::real, b > 0; # 0*infinity
       -sqrt(a^2+b^2-2*b+1)*signum(I*arctan(b, a)-I*arctan(-b, -a)-I*Pi)*
       infinity/((I*b-I+a)*(I*b+I+a))

int(f(z, t), t = 0 .. infinity) assuming Re(z) > 0, Im(z) > 0;
       int(ln(t)^2/((t^2+1)*(t-z)), t = 0 .. infinity)

int(f(a + I*b, t), t = 0 .. infinity) assuming a > 0, b > 0;
      -((3*I)*Pi^3*b+3*Pi^3*a-(16*I)*Pi^2*arctan(b/a)-(6*I)*Pi*ln(a^2+b^2)^2+(24*I)*
      Pi*arctan(b/a)^2+(6*I)*ln(a^2+b^2)^2*arctan(b/a)-(8*I)*arctan(b/a)^3-8*Pi^2*
      ln(a^2+b^2)+24*Pi*ln(a^2+b^2)*arctan(b/a)+ln(a^2+b^2)^3-12*ln(a^2+b^2)*
      arctan(b/a)^2)/(24*(-b^2+(2*I)*a*b+a^2+1))

So it looks like the first three can be made to work as well (and the result in terms of z will be much neater).

I have noticed that I don't receive e-mails anymore when contributions are submitted to my subscriptions.
I used to.
Has this happened to anyone else?

It is embarrasing to have asked somebody a question and gotten a reply you are not made aware of.

What to do about it?
 

Hello every body

I have a plot that contains two curve. I need to chracterize the curves with `min(D_{T})` and `max(D_{E})`. Is it possible to write legend of plot such that when include it in latex file,  T and E be presented in indices?

"The account type or email address or password is incorrect."

None of those things is true. Using those same credentials, I am able to log in to Maple Cloud in my office but not from home. 

Has anyone else had this problem or know of a solution?

Thanks!

Running Maple 2017 in office on Windows 7 and at home on Windows 10.

Office copy is through a site license purchased by my university. Home copy is a personal home-use copy obtained through the Maple Adoption program.

I am attempting to build a text field at the bottom of this worksheet MathApps-ResistorsMark.mw  that asks users for the resistance.

I am hoping to have the value of the text field evaluated using the Module() at the bottom of the startup code, for use as a question within MapleTA.  For some reason I can get the code to work for a slider, but not a textbox.

I have limited knowledge about startup code.

Notice: I will be removing the resistance from the diagram for students after I know it works.

I appreciate any help that can be offered.

Dear Maple users,

I am solving a system of linear equations Ax=b where A is a matrix (243*241) which contains a rational polynomial of unknown "kappa" along with floating numbers. As suggested in some other posts I am using Linear algebra package with the LinearSolve command and option "solve" to find out unknown vector x. b is a vector having entries zero and 1. The system is such that two equations are redundant but it is difficult to recognise which two are redundant and hence for the time being I am keeping them in the matrix. (For a known value of kappa (say kappa=2) I have checked that two equations are redundant.) For the reference, the matrix and the right-hand vector b is attached as text.

There are two issues

1) Maple takes very long time (12 hours and so) to get x=b\A;

2) The result is a long expression i.e.  x[i] is a rational polynomial in kappa; a very long rational polynomial which I am importing as a text. I am not sure if maple exports all terms in the polynomial as for the different value of kappa I see Ax-b >0. 

How can I overcome this problem? Any help will be greatly appreciated.