MaplePrimes Questions

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.
 

restart

with(Statistics):

``

F := (-2*n[1]*n[3]-4*n[2]*n[1]*n[3]+4*n[2]*n[3]+4*n[3]*n[4]*n[1]+n[1])/(-n[1]-2*n[2]*n[1]+2*n[2]+2*n[4]*n[1]):

S := proc (N, M, G, L) options operator, arrow; [seq(seq(seq(seq(F, n[1] = N .. N), n[2] = M .. M), n[3] = G .. G), n[4] = L .. L)] end proc

proc (N, M, G, L) options operator, arrow; [seq(seq(seq(seq(F, n[1] = N .. N), n[2] = M .. M), n[3] = G .. G), n[4] = L .. L)] end proc

(1)

A := {}; -1; U := 0; -1; K := 0; -1; E := 0

0

(2)

J := 0:

H := 1

1

(3)

K[1] := J+H:

try S(K[1], K[2], K[3], K[4]) catch "numeric exception: division by zero": E := E+1; `union`({[K[1], K[2], K[3], K[4]]}, A) end try

[-1]

(4)

J := 0:

H := 1

1

(5)

K[1] := J:

try S(K[1], K[2], K[3], K[4]) catch "numeric exception: division by zero": E := E+1; `union`({[K[1], K[2], K[3], K[4]]}, A) end try

[0]

(6)

J := 0:

H := 1

1

(7)

K[1] := J:

try S(K[1], K[2], K[3], K[4]) catch "numeric exception: division by zero": E := E+1; `union`({[K[1], K[2], K[3], K[4]]}, A) end try

1

 

{[0, 0, 1, 0]}

(8)

J := 0:

H := 1

1

(9)

K[1] := J:

try S(K[1], K[2], K[3], K[4]) catch "numeric exception: division by zero": E := E+1; `union`({[K[1], K[2], K[3], K[4]]}, A) end try

2

 

{[0, 0, 0, 1]}

(10)

``


 

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

 

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).

 

the Eigenvalues are showing with I and am not expecting a complex eigenvalues so what is that I stand for? Can you please help? Thank You

Hi

I've got this list:

L:=[[TC,DB], [], [TD,JK], [IW,CM], [], [KJ,DJ]]

What command to remove the 'null sets', leaving :=[[TC,DB],[TD,JK], [IW,CM], [KJ,DJ]]

this doesn't work:

remove(has, L, 0)

a::real and b::real;
Error, type `real` does not exist

`and`(a::real, b::real);
                       a::real and b::real

Since a::real by itself is fine, why does the conjunction give an error?

These two are evaluated differently as well:

Im(a) > 0 and a <> 0;
                           0 < Im(a)
`and`(Im(a) > 0, a <> 0);
                      0 < Im(a) and a <> 0

The consequence is that these two will work differently:

is(a <> 0) assuming Im(a) > 0 and a <> 0;
                             FAIL

is(a <> 0) assuming `and`(Im(a) > 0, a <> 0);
                             true

is(a <> 0) assuming Im(a) > 0; # sadly, just Im(a) > 0 is not enough
                             FAIL

It looks like (a and b) and `and`(a, b) just do completely different things:

x := proc() local r; r := rand(); print(r); r end proc;

a > x() and a > x() and a = a;
                          395718860534
                        395718860534 < a

`and`(a > x(), a > x(), a = a);
                          193139816415
                          22424170465
         193139816415 < a and 22424170465 < a and a = a

That is, (a and b) first simplifed the expression and then evaluated a>x() once, but `and`(a, b) evaluated the arguments without doing any simplifications.

Also, should this work (it works with `real`, which supposedly doesn't exist):

Re(a+b) assuming a::realcons, b::realcons;
                           Re(a + b)

 

I've compiled a program with Maple 18. unfortunately it gives me the message "Large sortie de plus de 1000000 noeuds". If any one have any idea to resolve this problem. Thanks you very mutch for your support. 

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