I have the following integral, which is solved. int(14*t*exp(-t/3),t=0..infinity); It equals 126. I now need to find 90%, of the integral. This is just math and the answer is 113.40. The challenge is I need to prove this using maple. So I need to show how to get the upper limit to 11.67, which yeilds the answer of 113.40(rounded).  I just started pluging in numbers for the upper limit till I got my answer, but clearly, this is now what my instructor wants. I need the process for finding the upper limit. Any help would be appreciated.

A bride and groom are very torn on what flavor of wedding cake they should have. They decide to make the cake 1/3 chocolate, 1/3 vanilla, 1/3 strawberry. The bride values Chocolate five times as much as she values Vanilla, but she values Vanilla four times as much as she Values Strawberry. If the cake costs a total of $480, then in the value system of the bride, what is the cost of the entire Vanilla piece.

is there an algorithm to search whether protocol exist for movement of balls in different boxes

for example

assume there are 5 boxes, each box has a variable number of number balls

ball number 1 move from box 1 to box 3 and box 5

then when discover ball number 8 in any box, ball number 1 will disappear

if input data are 5 list of numbers each row

1. [1,2,3],[5,7,12],[12,14,15],[1,5,7,9,15],[12,13,14]

2. [2,3],[5,7,12],[1,12,14,15],[1,5,7,9,15],[1,12,13,14]

....

etc

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]);

When I launch Maple, I have to activate my license, so I type my purchase code and then I fill the form, but I receive the message "not enough activations left on this account". I used and installed Maple on an old computer and sold it because it was really old, now with my new computer I would like to re-install Maple, is there a way I can use my old activation key, instead of buying Maple again ? 

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?

Hi all,

All other programs put it functionallity into Mac's menu line in the top, so your here have the following

"Appel logo" "Program name" File Edit View Inset

but this is not happening with Maple 2017???     anyone who have a solution for this 

best regards

Claus

I would like to plot h over the real interval from 0 to 2*Pi, but

produces an empty plot...  How can I plot successfully?

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

Dear All,

I am actively using Maple for scientific research but I am obliged to use xkill several times a day (~70 times) and restart Maple again since the software freezes and completely stops after a few minutes. I was wondering whether this is an issue with the present verison (Maple 15) and that upgrading to a latest version is inevitable. Your help is highly appreciated.

Thank you

Fede

What i am trying to achieve is to evaluate the sequence as, shown, but from within a try-catch statement that handles and keeps a tally on the number and arguements for which the sequence encounters a division by zero error. ie, instead of haulting evaluation when each error is encountered, i want my code to record the index values at which the error occured, then continue on to the next term.
 

Download PLEASE_HELP_MAPLE.mw

Hi

I have a long column vector containing data in Records.

A:=Vector[column](4, [J_K = `Record(mu = 724.901557888305, sigma = 96.7437910529146)`, I_W = `Record(mu = 775.098442111694, sigma = 96.7437910529198)`, K_J = `Record(mu = 785.098442111694, sigma = 96.7437910529198)`, D_B = `Record(mu = 764.901557888305, sigma = 96.7437910529146)`])

How to I sort this in descending values of mu so I get:

Vector[column](4, [K_J = `Record(mu = 785.098442111694, sigma = 96.7437910529198)`,I_W = `Record(mu = 775.098442111694, sigma = 96.7437910529198)`,D_B = `Record(mu = 764.901557888305, sigma = 96.7437910529146)`,J_K = `Record(mu = 724.901557888305, sigma = 96.7437910529146)`])

Im aware you can extract mu from Records by the rhs(A[1]):-mu

Hello,
I am trying to do the following double integral in tranches as listed here:

P := proc (x, y) options operator, arrow; (1/2)*exp(-(1/2)*(x^2+G*y^2-2*B*x*y)/(-B^2+G))/(Pi*sqrt(-B^2+G)) end proc

Since G and B are constant

((int(x = 0 .. infinity))*(int(y = -infinity .. 0))+(int(x = -infinity .. 0))*(int(y = 0 .. infinity)))*P(x, y)

But does notwork. How do I pass these coordinates to polar?

Regards.

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.

I was told that the following workout was done in Maple.  I have tried to read material about how to do it but I am completely lost.  Can someone indicate me where I can read in oder to do what the image says or give me some tips please?  

where all functions are dependent on the variables (u,v).

Observation: subscripts means partial derivatives of the function while superscripts are just for naming different functions,i.e Gamma^1 and Gamma^2 are two functions.

Sergio

Sergio