Maple 2019 Questions and Posts

These are Posts and Questions associated with the product, Maple 2019


Hello. Here is my question.

 

#GIVEN:

ST := [`162` = Record(mu = 475, sigma = 41), `70` = Record(mu = 480, sigma = 42),
       `168` = Record(mu = 448, sigma = 103)]

[`162` = Record(mu = 475, sigma = 41), `70` = Record(mu = 480, sigma = 42), `168` = Record(mu = 448, sigma = 103)]

(1)

Games:=[[`162`, `70`], [`70`, `168`], [`168`, `162`]]

[[`162`, `70`], [`70`, `168`], [`168`, `162`]]

(2)

#I need Maple commands to follow the order of Games to auto create STO please

STO:=[`162` = Record(mu = 475, sigma = 41),`70` = Record(mu = 480, sigma = 42),
       `70` = Record(mu = 480, sigma = 42),`168` = Record(mu = 448, sigma = 103),
     `168` = Record(mu = 448, sigma = 103),`162` = Record(mu = 475, sigma = 41)]

`Non-fatal error while reading data from kernel.`

(3)

 


 

Download ORDERED.mw


I'm trying to evaluate an integral, but after several hours, MAPLE 2019 is unable to return an answer.  My CPU and memory are not being taxed, and the integrand appears well behaved.  Please see below. Any advice?

restart

istar := y*sinh(Pi*y)*exp(-alpha*y^2)*LegendreP(-1/2+I*y, 1+u)

y*sinh(Pi*y)*exp(-alpha*y^2)*LegendreP(-1/2+I*y, 1+u)

(1)

Istar := `assuming`([int(istar, y = 0 .. infinity)], [alpha > 0, u > 0])

int(y*sinh(Pi*y)*exp(-alpha*y^2)*LegendreP(-1/2+I*y, 1+u), y = 0 .. infinity)

(2)

alpha := 1; u := 2

2

(3)

plot(istar, y = 0 .. 40)

 

evalf(Istar)

``


 

Download Istar.mw

 

How I can substitute Eqs1-16) into Eq (17)?

Should I use subs rule?

Thanks 

 

 

A few seconds after calling up Help starts zucking araound and the whole computer then freezes. Ctrl-Alt-Delete doesn't work, hard reset required. Very funny. Am I alone?

Hi

I'm trying to export a file using the .mw file. without a lot of success. Can someone help?

For 1 game, the output is between (8) and (9).

For 2 games. the output is after (17).

TEST.mw

 

Hey guys, 

Is there any way to remove these imaginary parts of the numbers ?

I tried using the RealDomain package in beginning of code but it starts showing some Float(undefined) numbers and I don't understand why since none of the terms inside the sqrt are negative.

Thanks in advance.

Download Reforço.mw

I'm new to Maple, and i'm learnning Maple with basic commands(the version I use is Maple2019). But after I entered some commands, the results Maple gives me is not what I expected. For example:

When i enter abs(3+5*I), I hope get a $\sqrt{34}$, but Maple gives me a complex number:

abs(3 + 5*I);
                                                         -58                                                     -59  
       1.81847767202745 10    + 7.53238114626421 10    I
sin(0);
                                                        -116                                                    -116  
      2.73949338633639 10     + 2.73949338633639 10     I
sin(pi);
                                                        -116                                                    -116  
      2.73949338633639 10     + 2.73949338633639 10     I
abs(3 + 5*I);
                                                        -58                                                      -59  
       1.81847767202745 10    + 7.53238114626421 10    I

expand(cos(4*x) + 4*cos(2*x) + 3, trig);
                                                                          -231  
                 8. - 3.75241200689042 10     I
Why???

Should I set or define something or what?

Thank you!

Hello! 

I have been using maple not so long ago and I can’t always understand for myself what is the reason for the program not working.
In this case, I need to solve a system of second-order differential equations with respect to four variables. I don’t understand what’s wrong, do not judge strictly.

// 
restart;

m1 := 50;
m2 := 70;
R1 := 0.14;
R2 := 0.17;
O1C1 := 0.3;
alpha := 0.024;
a := 0.28;
b := 0.08;
c := 0.08;
M0 := 12;
k1 := 1.7;
k2 := 0.9;
Zb := -a;
Md := M0 - k1*diff(varphi(t), t);
Mc := -k1*diff(varphi(t), t)^2;

Ixz := 0;
Iyz := m1*O1C1*(b + c) + m2*R2^2/4*alpha;
Izz := m1(R1^2/2 + O1C1^2) + m2*R2^2/4*(alpha^2 + 2);

eqMx := -M*Yc*diff(varphi(t), t $ 2) - M*Xc*diff(varphi(t), t)^2 = Xa + Xb;
eqMy := M*Xc*diff(varphi(t), t $ 2) - M*Yc*diff(varphi(t), t)^2 = Ya + Yb;
eqIy := -Ixz*diff(varphi(t), t $ 2) + Iyz*diff(varphi(t), t)^2 = Zb*Yb;
eqIx := -Iyz*diff(varphi(t), t $ 2) - Ixz*diff(varphi(t), t)^2 = Zb*Xb;
eqMz := Izz*diff(varphi(t), t $ 2) = Md + Mc;
sys := {eqIx, eqIy, eqMx, eqMy};
dsolve({sys, varphi(0) = 0}, Xa, Xb, Ya, Yb);

Error, (in dsolve) invalid arguments; expected an equation, or a set or list of them, received: { and so on }

//

I will be glad if you help me with this problem. if there are certain questions, you - ask, and I'll answer.
Here you can find the file: Курсовой_по_термеху_2.mw

I have an ordinary differential equation to be solved. when I give numeric values to B,H and L__1 (for example B=10,H=10,L__1=4), everything works fine and equations eq37a will be solved.

but when B,H and L__1 are symbolic parameters, dsolve give me this error " Error, (in dsolve) give the main variable as a second argument". and I don't know what it means.

is there any way to solve the ODE with symbolic B,H and L__1??

 

restart;

#B:=10;
#H:=10;
#L__1:=4;

eq0:=R__i=(1/2)*B+L__1;
eq01:=R__f=C+(1/2)*B+L__1;
eq02:=theta__1=arctan((H/2)/(B/2 + L__1));
eq03:=theta__2=Pi - arctan((H/2)/(B/2 - L__1));
eq43:= v__beta(r,theta,beta)= v__m(beta)*(1-(r^2/((R__max(theta,beta))^2)));
eq39:=R(beta)=R__i+2*beta*(R__f-R__i)/Pi;
eq39a:=simplify(subs([eq0,eq01], eq39));
eq38:=R__max(theta,beta)=R__max(theta,0)*(R(beta)/R__i);
eq41:=R__max(theta,0)=piecewise(theta<=rhs(eq02),(B/2+L__1)/cos(theta),theta<=rhs(eq03),(H/2)/sin(theta),rhs(eq03)<theta,(B/2-L__1)/cos(Pi-theta));
eq38a:= simplify(subs([eq41,eq39a,eq0], eq38));
eq44 := v__m(beta) = 1/(2*int((1 - r^2/R__max(theta, beta)^2)*r, [r = 0 .. R__max(theta, beta), theta = 0 .. Pi]));
eq44a:=simplify(subs(eq38a,eq44)) assuming B/L__1 >= 2 , H > 0 , B > 0 , L__1 > 0;
eq44b:= v__m(beta)= rhs(eq44a) * B * H;
eq47:=subs([eq44b,eq38a],eq43);

eq0 := R__i = (1/2)*B+L__1

 

eq01 := R__f = C+(1/2)*B+L__1

 

eq02 := `&theta;__1` = arctan(H/(2*((1/2)*B+L__1)))

 

eq03 := `&theta;__2` = Pi-arctan(H/(2*((1/2)*B-L__1)))

 

eq43 := `v__&beta;`(r, theta, beta) = v__m(beta)*(1-r^2/R__max(theta, beta)^2)

 

eq39 := R(beta) = R__i+2*beta*(R__f-R__i)/Pi

 

eq39a := R(beta) = (1/2)*B+L__1+2*beta*C/Pi

 

eq38 := R__max(theta, beta) = R__max(theta, 0)*R(beta)/R__i

 

eq41 := R__max(theta, 0) = piecewise(theta <= arctan(H/(2*((1/2)*B+L__1))), ((1/2)*B+L__1)/cos(theta), theta <= Pi-arctan(H/(2*((1/2)*B-L__1))), H/(2*sin(theta)), Pi-arctan(H/(2*((1/2)*B-L__1))) < theta, -((1/2)*B-L__1)/cos(theta))

 

eq38a := R__max(theta, beta) = (Pi*(B+2*L__1)+4*beta*C)*piecewise(theta <= arctan(H/(B+2*L__1)), (B+2*L__1)/cos(theta), theta <= Pi-arctan(H/(B-2*L__1)), H/sin(theta), Pi-arctan(H/(B-2*L__1)) < theta, (-B+2*L__1)/cos(theta))/(2*Pi*(B+2*L__1))

 

eq44 := v__m(beta) = 2/(int(R__max(theta, beta)^2, theta = 0 .. Pi))

 

eq44a := v__m(beta) = 2*Pi^2*(B+2*L__1)^2/(H*(B*Pi+4*C*beta+2*L__1*Pi)^2*B)

 

eq44b := v__m(beta) = 2*Pi^2*(B+2*L__1)^2/(B*Pi+4*C*beta+2*L__1*Pi)^2

 

v__beta(r, theta, beta) = 2*Pi^2*(B+2*L__1)^2*(1-4*r^2*Pi^2*(B+2*L__1)^2/((Pi*(B+2*L__1)+4*beta*C)^2*piecewise(theta <= arctan(H/(B+2*L__1)), (B+2*L__1)/cos(theta), theta <= Pi-arctan(H/(B-2*L__1)), H/sin(theta), Pi-arctan(H/(B-2*L__1)) < theta, (-B+2*L__1)/cos(theta))^2))/(B*Pi+4*C*beta+2*L__1*Pi)^2

(1)

eq37:=diff(v__r(r, theta, beta),r)*r*(R__f-r*cos(theta))+v__r(r,theta,beta)*(R__f-2*r*cos(theta))+r*diff(v__beta(r,theta,beta),beta)=0;
eq37a:=simplify(expand(subs(eq47, eq37)));

eq37 := (diff(v__r(r, theta, beta), r))*r*(R__f-r*cos(theta))+v__r(r, theta, beta)*(R__f-2*r*cos(theta))+r*(diff(`v__&beta;`(r, theta, beta), beta)) = 0

 

(-2*(Pi*(B+2*L__1)+4*beta*C)^2*((1/2)*(Pi*(B+2*L__1)+4*beta*C)^3*(r*cos(theta)-R__f)*r*(diff(v__r(r, theta, beta), r))+(Pi*(B+2*L__1)+4*beta*C)^3*(r*cos(theta)-(1/2)*R__f)*v__r(r, theta, beta)+8*C*r*Pi^2*(B+2*L__1)^2)*piecewise(theta <= arctan(H/(B+2*L__1)), (B+2*L__1)/cos(theta), theta <= Pi-arctan(H/(B-2*L__1)), H/sin(theta), Pi-arctan(H/(B-2*L__1)) < theta, (-B+2*L__1)/cos(theta))^2+128*C*r^3*Pi^4*(B+2*L__1)^4)/((Pi*(B+2*L__1)+4*beta*C)^5*piecewise(theta <= arctan(H/(B+2*L__1)), (B+2*L__1)/cos(theta), theta <= Pi-arctan(H/(B-2*L__1)), H/sin(theta), Pi-arctan(H/(B-2*L__1)) < theta, (-B+2*L__1)/cos(theta))^2) = 0

(2)

bc:=v__r(rhs(eq38a),theta,beta)=0;

bc := v__r((Pi*(B+2*L__1)+4*beta*C)*piecewise(theta <= arctan(H/(B+2*L__1)), (B+2*L__1)/cos(theta), theta <= Pi-arctan(H/(B-2*L__1)), H/sin(theta), Pi-arctan(H/(B-2*L__1)) < theta, (-B+2*L__1)/cos(theta))/(2*Pi*(B+2*L__1)), theta, beta) = 0

(3)

sol3:=simplify(dsolve([eq37a,bc], v__r(r, theta,beta)));

Error, (in dsolve) give the main variable as a second argument

 

NULL


 

Download Mechrect.mw

How I can apply the condition or assuming (a>0) in ?

I want to have a plot according to the attached figure.

restart; A5 := 2.4821; A3 := .8417; A4 := -.1384; K := 0.3e-2; `&sigma;&sigma;` := (9*A3*A5*a^2-10*A4^2*a^2)/(24*A3^(3/2))+K/(2*a*A3^(1/2))

2.4821

 

.8417

 

-.1384

 

1.004210973*a^2+0.1634980569e-2/a

(1)

NULL

solve(sigma = 1.004210973*a^2+0.1634980569e-2/a, a)

solve(sigma = 1.004210973*a^2-0.1634980569e-2/a, a)

plot({1.079801433*10^(-25)*(-6.465854665*10^71+1.058832302*10^46*sqrt(-2.057991547*10^56*sigma^3+3.729044048*10^51))^(1/3)+3.074042610*10^24*sigma/(-6.465854665*10^71+1.058832302*10^46*sqrt(-2.057991547*10^56*sigma^3+3.729044048*10^51))^(1/3), 1.079801433*10^(-25)*(6.465854665*10^71+1.058832302*10^46*sqrt(-2.057991547*10^56*sigma^3+3.729044048*10^51))^(1/3)+3.074042610*10^24*sigma/(6.465854665*10^71+1.058832302*10^46*sqrt(-2.057991547*10^56*sigma^3+3.729044048*10^51))^(1/3)}, sigma = -.2 .. .2)

 

``


 

Download plot 

I was writing in a maple document when suddently all the text and equations collapsed into a single line at the top, making it unreadable and unusuable. I was not able to restore a previous backup of the file, however i did manage to salvage a smaller part of the document. Both files are included. 
corrupted.mw 
what_i_managed_to_save.mw

Cursor jumping to end of sheet/page when pressing enter (calculating).

How do I change the settings in maple 2019 so that the cursor doesnt jump to the end of the page/document everytime I make a calculation. 

Thanks!

Jeppe
 

Maple can't solve the following symbolic integration for me and it remains in the state of evaluating when I use simplify and nothing happens after hours.. any solution? 

I was thinking maybe there is something wrong with my maple !!?

 

eq0:=R__i=(1/2)*d+L__1;
eq00:=R__f=c+(1/2)*d+L__1;
eq39:=R(beta)=R__i+2*beta*(R__f-R__i)/Pi;
eq39a:=simplify(subs([eq0,eq00], eq39));
eq38:=R__max(theta,beta)=R__max(theta,0)*(R(beta)/R__i);
eq41:=R__max(theta,0)=((2*L__1*cos(theta)) + sqrt(d^2 - 2*(L__1)^2 + 2*(L__1)^2*cos(2*theta)))/2;
eq38a:= simplify(subs([eq41,eq39a,eq0], eq38));

R__i = L__1+(1/2)*d

 

R__f = c+(1/2)*d+L__1

 

R(beta) = R__i+2*beta*(R__f-R__i)/Pi

 

R(beta) = L__1+(1/2)*d+2*beta*c/Pi

 

R__max(theta, beta) = R__max(theta, 0)*R(beta)/R__i

 

R__max(theta, 0) = L__1*cos(theta)+(1/2)*(d^2-2*L__1^2+2*L__1^2*cos(2*theta))^(1/2)

 

R__max(theta, beta) = (1/2)*(2*L__1*cos(theta)+(d^2-2*L__1^2+2*L__1^2*cos(2*theta))^(1/2))*(2*L__1*Pi+Pi*d+4*beta*c)/(Pi*(2*L__1+d))

(1)

eq44 := v__m(beta) = 1/int((1 - r^2/R__max(theta, beta)^2)*r, [r = 0 .. R__max(theta, beta), theta = 0 .. 2*Pi]);

v__m(beta) = 4/(int(R__max(theta, beta)^2, theta = 0 .. 2*Pi))

(2)

eq44a:=simplify(subs(eq38a,eq44));

``


Download int.mw

 

linkedlist.zip

I am having a problem understanding how to pretty print the results of a procedure incorporated into a module installed as a package using the maple workbook(.maple).  In an effort to understand how this should be done, I installed the LinkedListPackage from the example code using a maple workbook to see if the package works as shown in the help files. The results from this exercise confirmed the problem I am having with a module I am creating. Specifically, as shown in the help file, I assigned the exported name “nil” to L; ie. L := nil; Note that the result is L:= linkedlist:-nil not L:= nil.  Furthermore when I create a linked list the results are as shown below:  

L := ``(1, linkedlist:-nil)
L := ``(2, ``(1, linkedlist:-nil))
L := ``(3, ``(2, ``(1, linkedlist:-nil)))
L := ``(4, ``(3, ``(2, ``(1, linkedlist:-nil))))
L := ``(5, ``(4, ``(3, ``(2, ``(1, linkedlist:-nil)))))
L := ``(6, ``(5, ``(4, ``(3, ``(2, ``(1, linkedlist:-nil))))))
L := ``(7, ``(6, ``(5, ``(4, ``(3, ``(2, ``(1, linkedlist:-nil)))))))
L := ``(8, ``(7, ``(6, ``(5, ``(4, ``(3, ``(2, ``(1, linkedlist:-nil))))))))
L := ``(9, ``(8, ``(7, ``(6, ``(5, ``(4, ``(3, ``(2, ``(1, linkedlist:-nil)))))))))
L := ``(10, ``(9, ``(8, ``(7, ``(6, ``(5, ``(4, ``(3, ``(2, ``(1, linkedlist:-nil))))))))))

I realized that nil is an exported local variable which is being reflected in how it presents in the worksheet, but in an expression of any complexity presenting exported local names in this way makes the expression unreadable. I have been combing help looking for a clue as to how the results from the package can be pretty-printed without a fully-qualified name. Can you suggest how I should proceed.

My other question concerns the macro command which is used in the example code to format the pairs.  As written in the help file, the macro command is intended to replace the name _PAIR with a space.  However, it seems that the macro facility now replaces _PAIR with name quotes instead of a space. Has the macro facility changed since the example was written?

Hey,

Is anyone of you capable of simplifying this expression

f1:=(-3*sin(8*x) + 3*sin(8*x + 2*y) - 3*sin(8*x + 6*y) + 3*sin(8*y + 8*x) + 3*sin(8*y + 6*x) + 3*sin(8*y) - 18*sin(8*y + 4*x) + 3*sin(8*y + 2*x) - 45*sin(6*y + 6*x) + 87*sin(4*y + 6*x) - 3*sin(6*x - 2*y) - 87*sin(6*x + 2*y) + 18*sin(4*x - 4*y) - 93*sin(4*x + 4*y) + 93*sin(4*x + 6*y) - 51*sin(2*x - 4*y) - 342*sin(2*x + 4*y) - 3*sin(-6*y + 2*x) + 51*sin(6*y + 2*x) - 93*sin(-2*y + 4*x) + 342*sin(-2*y + 2*x) + 639*sin(2*x + 2*y) - 639*sin(2*x) + 45*sin(6*x) + 93*sin(4*x) + 231*sin(4*y) - 225*sin(2*y) - 63*sin(6*y) - 57*sqrt(3)*cos(2*x) - 375*sqrt(3)*cos(2*y) + sqrt(3)*cos(8*y + 8*x) - 5*sqrt(3)*cos(8*x + 6*y) - 7*sqrt(3)*cos(8*y + 6*x) + sqrt(3)*cos(8*x) + 192*sqrt(3)*cos(2*y + 4*x) + 43*sqrt(3)*cos(-2*y + 4*x) - 7*sqrt(3)*cos(6*x + 2*y) + 7*sqrt(3)*cos(-6*y + 2*x) - 5*sqrt(3)*cos(6*y) - 149*sqrt(3)*cos(4*x + 4*y) - 149*sqrt(3)*cos(4*x) - 65*sqrt(3)*cos(6*y + 2*x) + 126*sqrt(3)*cos(2*x + 4*y) - 65*sqrt(3)*cos(2*x - 4*y) - 5*sqrt(3)*cos(8*x + 2*y) - sqrt(3)*cos(8*y) + 7*sqrt(3)*cos(8*y + 2*x) + 6*sqrt(3)*cos(8*x + 4*y) - 57*sqrt(3)*cos(2*x + 2*y) + 125*sqrt(3)*cos(4*y) + 126*sqrt(3)*cos(-2*y + 2*x) - 7*sqrt(3)*cos(6*x - 2*y) + 19*sqrt(3)*cos(6*x) + 43*sqrt(3)*cos(4*x + 6*y) + 19*sqrt(3)*cos(6*y + 6*x) - 7*sqrt(3)*cos(4*y + 6*x) + 246*sqrt(3))/(2*(-261*sin(4*x + y) - 297*sin(2*x + 3*y) - 48*sin(5*y + 6*x) + 126*sin(5*y + 2*x) + 9*sin(5*y + 8*x) + 12*sin(7*y + 6*x) - 9*sin(7*y + 4*x) - 36*sin(5*y + 4*x) + 261*sin(3*y + 4*x) + 9*sin(-3*y + 4*x) + 297*sin(-y + 2*x) - 135*sin(3*y) - 21*sin(5*y) - 147*cos(y)*sqrt(3) - 9*sqrt(3)*cos(7*y + 4*x) - 3*sqrt(3)*cos(5*y + 8*x) - 3*sqrt(3)*cos(3*y + 8*x) + 54*sqrt(3)*cos(6*x + 3*y) + 5*sqrt(3)*cos(-5*y + 2*x) + 5*sqrt(3)*cos(7*y + 2*x) - 2*sqrt(3)*cos(6*x - y) - 20*sqrt(3)*cos(6*x + y) - 69*sqrt(3)*cos(4*x + y) + 68*sqrt(3)*cos(4*x - y) + 2*sqrt(3)*cos(8*x + y) + 2*sqrt(3)*cos(7*y + 8*x) - 20*sqrt(3)*cos(5*y + 6*x) - 2*sqrt(3)*cos(7*y + 6*x) + 68*sqrt(3)*cos(5*y + 4*x) - 9*sqrt(3)*cos(-3*y + 4*x) - 69*sqrt(3)*cos(3*y + 4*x) - 171*sqrt(3)*cos(2*x + 3*y) - 35*sqrt(3)*cos(5*y) + 171*sqrt(3)*cos(3*y) - 171*sqrt(3)*cos(-y + 2*x) + 354*sqrt(3)*cos(2*x + y) + sqrt(3)*cos(7*y) + 639*sin(y) - 9*sin(3*y + 8*x) - 12*sin(6*x - y) + 3*sin(7*y) - 9*sin(7*y + 2*x) + 9*sin(-5*y + 2*x) + 48*sin(6*x + y) + 36*sin(4*x - y) - 126*sin(2*x - 3*y)))

 

into

 

cos(y-Pi/3).

 

PS: Actually I managed by expanding the thing out and converting to exp then expanding again and using radnormal. In essence I leave the question, because maybe somebody can explain to me why radnormal seems to be superior (sometimes) to simplify which I thought of as the USEALL choice. Thanks

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