Sorry if this is already known, but I haven't found the proper function call.

I just want to "print" this expression without evaluating the boolean calls

restart;
P(X <= 5) = P(-5 <= -X) = P(E(X) - 5 <= E(X) - X);

As it stands, this will evaluate to "false" which I obviously don't want.

I just want to use it as a displayer.

Hi, I'm trying to solve these 2 nonlinear equations in f, g, and x, where x is from 0 to 1 with an increment of 0.1.

I am new to Maple and do not know the basics. Please try and help me. I'd highly appreciate it.

Since this is a nonlinear system, multiple solutions exist, I need to find the first 3 or 5 solutions. Once I solve the system, I would like to plot 2 plots, y-x and z-x.

I currently do extensive modeling that involves Monte Carlo simulations. This kind of analysis seems idially suited to parallel computing. To begin the process I have been following the teachings of chapter 15 of the Programming Guide 2023. As a first excersize I copyied the code on pages 583-584 and attempted to run it on my system. My Maple program is a 2019  Student Version running on anHP laptop using Windows11. The code yields an error code "Error, invalid input: add_range uses a 1st argument, lo, which is missing"

This is an exact copy of the ccode in the guide. Is there posibly a probem with my installation? I am willing to update and upgrade as may be needed. Any suuggestions how to procced?

Thanks

Ray

 I defined the following function L1 and L2 to test, if  Maple is returning the same results. Mathematically they are identical. For all testpoints, L1 returns the correct results (for y := -5 the result is -15).  L2  returns identical results exept for y:=-5. For y:= -5, where you can see on the first glance that the result must be -15,  Maple is returning for L2 a complex number. I am worried about this different treatment of the functions L1 and L2, because I am calculating with functions, where you cannot prove the result as easy as it can be done here. 

L1 := y -> 3*y*((y + 4)^2)^(1/3);
   L1 := proc (y) options operator, arrow, function_assign; 

      3*y*((y+4)^2)^(1/3) end proc

L2 := y -> 3*y*(y + 4)^(2/3);
   L2 := proc (y) options operator, arrow, function_assign; 

      3*y*(y+4)^(2/3) end proc
NULL;
for y from -5 to 0 do
    print("y = ", y, "L1(y)   =  ", L1(1.0*y), "          L2(y)  =,  ", L2(1.0*y));
end do;
  "y =  ", -5, "L1(y)   =  ", -15.0, "          L2(y)  =,  ",     7.500000000 - 12.99038105 I
 "y =  ", -4, "L1(y)   =  ", -0., "          L2(y)  =,  ", -0.
"y =  ", -3, "L1(y)   =  ", -9.0, "          L2(y)  =,  ", -9.0
           "y =  ", -2, "L1(y)   =  ", -9.524406312,              "          L2(y)  =,  ", -9.524406312
           "y =  ", -1, "L1(y)   =  ", -6.240251469,              "          L2(y)  =,  ", -6.240251469
   "y =  ", 0, "L1(y)   =  ", 0., "          L2(y)  =,  ", 0.

Question about using the command „ExtremPoints“

I am using Maple 2019.

Using the command ExtremPoints I got different list when defining the function over a closed intervall piecewise or with f(x), x=a..b.

Maple desciption states:
ExtremePoints(f(x), x = a..b) command returns all extreme points of f(x) in the interval [a,b] as a list of values.

An extreme point is defined as any point which is a local minimum or maximum, which includes any finite end points.

So I expected, Maple returns the same list, independend of how the same function is defined (see example below). Instead: with the piecewise definition Maple returns as extrempoints only the local extrempoints without the finite endpoints.
Defining the same function with f(x),x=a..b Maple returns the list with local minimum or maximum, which includes any finite end points.

Example:

f := x -> piecewise(-1 <= x and x <= 2, x^2, undefined);

Return:  ExtremePoints(f(x));

                              [0] 

g := x -> x^2

ExtremePoints(g(x), x = -1 .. 2);

                           [-1, 0, 2]

restart;
Pr:=0.71: n:=-1:

eta0:=0.0699;

EQ1:=diff(H(x), x ) - x*diff(F(x), x ) ;
 

EQ2:=(1+x^2)*diff(F(x), x$2) + (3*x + x*F(x)-H(x))*diff(F(x), x) + F(x)^2 + G(x)^2 +2*P(x) + x*diff(P(x), x) ;

EQ3:=(1+x^2)*diff(G(x), x$2) + (3*x + x*F(x)-H(x))*diff(G(x), x) ;

EQ4:=(1+x^2)*diff(H(x), x$2) + (3*x + x*F(x)-H(x))*diff(H(x), x) + (1+F(x))*H(x)- diff(P(x), x);

EQ5:=(1+x^2)*diff(theta(x), x$2) + x*(1-2*n)*diff(theta(x), x) + n^2*theta(x) - Pr*( n*F(x)*theta(x) + ( H(x)-x*F(x) )*diff(theta(x), x)  ) ;

EQ:={EQ1=0, EQ2=0,EQ3=0,EQ4=0 ,EQ5=0}:

IC:={ F(0)=0, G(0)=12, H(0)=0, theta(0)= 1, F(eta0)=0, G(eta0)=12, H(eta0)=0, theta(eta0)= 0, P(0)=0};
 

sol:= dsolve(EQ union IC,numeric,output=Array([0,0.0699]));

ques.mw

What will be the range of p and q to get the plot and to get the optimum solution?
If possible get a solution for particular value of p and q.
file attached: q1.mw

Just wanted to ask, what the issue here is:

restart;
Int(1/(1 - x*ln(x)), x);
IntegrationTools:-Change(%,u=1-x*ln(x),u);

doesn't give the proper transformation. It gives

Int(1/u,u)

Solving for x and writing the transformation in terms of LambertW gives something else, if I'm not mistaken.

tortoise_draft2.mw
I am trying to use the function "rt" defined below to solve the equation "eq1", but I get an error "Error, (in PDEtools:-Solve) x is in the equation, and is not solved for". I don't quite understand what prevents me from using this function freely in expressions (for example, to plot it I must use unevaluation, etc.). Could you, please, clarify why this error occurs and how to fix it?

Download tortoise_draft2.mw

Hello,

I have this simple problem which doesn't want to work.

restart;
d := g -> (D@@2)(g) - x^2*g;
((d@@2)(g) assuming x::constant);

The result of the last line is:

(D@@4)(g) - 2*(D@@2)(x)*x*g - 2*D(x)^2*g - 4*D(x)*x*D(g) - x^2*(D@@2)(g) - x^2*((D@@2)(g) - x^2*g)

so Maple doesn't set D(x) to 0. On the other hand if I just write

D(x) assuming x::constant

then Mapel returns 0.

Similarly

D(f^k) assuming k::constant

just returns D(f^k) and not k*D(f)*f^(k-1) as the example in HELP suggests.

I cannot view 3d graphics with my version of Ubuntu 20.04. I've updated all my computer's graphics card drivers and the problem persists. If I run without hardware acceleration, nothing changes; no visualization and no production possible.
Do you have any ideas for solving this problem? Maple uses OpenGL libraries for 3D production and visualization, and these libraries are installed on my computer. Would installing mesa solve the problem, for example?

Thanks in advance.

Isolate function is not working, showing error and hessian matrix is not giving simplified result in readable format. I am attaching the sheet below. 
q1.mw

Hello, 

Ive followed the examples, and yet not the result the book gave. The question before this one, same type, and i did get the right answer there. 

Would you have a clue why it is not going as planned? 

Translation: "Through a resistancewire with the measured resistance R= (1.5+-0.5)Ohm goes an electric current with the measured strength I= (2.5+-0.05)A, During a timespan of t=(10+-0.5)s develops an amount of heat Q=RI^2*t. Give the estimate of the maximum relative deviation in Q."

Here i got the right answer of .42, and yet when doing the exact same thing in assignment 4, i got the wrong answers. 

Translation: "Two bodies with mass M and m have a massmidpointdistance r. For the gravitational force between these bodies counts: F=y*M*m/r^2, at which y is the gravitational constant. Give an estimation of the maximum realative deviation in F on ground of the following measurementdata: M=(7.0+-0.05)kg, m=(2.0+-0.02)kg and r=(5.0)+-0.05)m."

I had the answer -0.0029. The books says its wrong, but i cant see what i did wrong. 

Translation of the example: "According to the law of Ohm the currentstrength in the stream(/current)circle(circuit) with tension(voltage) U and resistance R is equal to I=U/R. For a given currentcircle((part of a)circuit) is given: U=(30.0+-0.05)Volt and R=(2.5+-0.03)Ohm. Determine the maximum absolute and relative deviation in I. 

Solution

The maximum absolute deviation in I is

This gives the currentstrength:

The maximum realtive deviation in I is: .... or about 1.4%"

Could somebody tell me what i am doing wrong?

Thank you!

Greetings,

The Function

Download Mapleprimes_Question_Book_2_Paragraph_5.13_Question_4.mw

The answers from the book dont make any sense after using their solution. Its not producing the expected results. 

This is the translation by the way: "A battery that has a voltage U, is in series with a resistor R, and a condenser with capacity C. The current strength on a given time "t" in this circuit is given by i=i(t)=U/R*e^-(t/R*C). Determine R and C based on the data given in table 5.6, the current is U=100V."

Greetings,

The Function

Download Mapleprimes_Question_Book_2_Paragraph_5.12_Question_5.mw