Maple 2019 Questions and Posts

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

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

#opdracht 3

R := 1.5; A := 2.5; t := 10; Q := R*A^2*t

1.5

 

2.5

 

10

 

93.750

(1)

NULL

restart

Q := R*A^2*t

R*A^2*t

(2)

diff(R*A^2*t, R)

A^2*t

(3)

diff(R*A^2*t, A)

2*R*A*t

(4)

diff(R*A^2*t, t)

R*A^2

(5)

NULL

R := 1.5; A := 2.5; t := 10

1.5

 

2.5

 

10

(6)

.5*A^2*t+2*R*A*t*0.5e-1+.5*R*A^2

39.6875

(7)

Q := R*A^2*t

93.750

(8)

39.6875/(93.750)

.4233333333

(9)

#Opdracht 4

restart

M := 7; m := 2; r := 5; F := y*M*m/r^2

7

 

2

 

5

 

(14/25)*y

(10)

restart

diff(y*M*m/r^2, M)

y*m/r^2

(11)

diff(y*M*m/r^2, m)

y*M/r^2

(12)

diff(y*M*m/r^2, r)

-2*y*M*m/r^3

(13)

M := 7; m := 2; r := 5; y := 9.81

7

 

2

 

5

 

9.81

(14)

0.5e-1*y*m/r^2+0.2e-1*y*M/r^2+(-2*y*M*m/r^3)*0.5e-1

-0.1569600000e-1

(15)

F := y*M*m/r^2

5.493600000

(16)

-0.1569600000e-1/(5.493600000)

-0.2857142857e-2

(17)

NULL

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


 

restart

U := 100

100

(1)

R := 4900

4900

(2)

C := 20*(1/1000)

1/50

(3)

"v(t):=U/(R)*(e)^(-(t/(R*C)))"

proc (t) options operator, arrow, function_assign; U*exp(-t/(R*C))/R end proc

(4)

v(.1)

0.2038734923e-1

(5)

NULL

data := LinearAlgebra:-Transpose(`<,>`(`<|>`(.1, .2, .3, .4, .5), `<|>`(7.36, 2.7, .99, .37, .13)))

Matrix(%id = 18446746712187741238)

(6)

plot(data, style = point)

 

X := data[() .. (), 1]; Y := data[() .. (), 2]

Vector(5, {(1) = .1, (2) = .2, (3) = .3, (4) = .4, (5) = .5})

 

Vector[column](%id = 18446746712187761598)

(7)

sumX := add(X); sumY := add(Y); sumX2 := add(x^2, `in`(x, X)); sumXY := X.Y

1.5

 

11.55

 

.55

 

1.78600000000000003

(8)

eq1 := numelems(X)*a+b*sumX = sumY; eq2 := a*sumX+b*sumX2 = sumXY

5*a+1.5*b = 11.55

 

1.5*a+.55*b = 1.78600000000000003

(9)

solve({eq1, eq2}, {a, b})

{a = 7.347000000, b = -16.79000000}

(10)

``

Download Mapleprimes_Question_Book_2_Paragraph_5.12_Question_5.mw

The following 2D integrals of 0 are seemingly trivial and one would expect them to evaluate to zero, but Maple evaluates them to undefined

int(0, x=0..infinity, y=0..1) # undefined
int(0, x=0..1, y=0..infinity) # undefined

When the 2D integral is split into two 1D integrals, it does evaluate to zero, as the following examples show

int(0,x=0..infinity) # 0
int(int(0,x=0..infinity), y=0..1) # 0
int(int(0,x=0..1),y=0..infinity) # 0

If infinity is replaced by a variable (say 'c'), the first two integrals are also evaluated to zero.

It may be connected by the following

int(a, x=0..infinity, y=0..1) # a*infinity
int(a, x=0..1, y=0..infinity) # a*infinity
int(a,x=0..infinity) # signum(a)*infinity

So for the 1D integrals the signum is applied to 'a' when the interval is infinite, but not for the 2D integrals. I'm not sure about this difference.

I receive the following error when I run maple to solve a diff equation.
Thank you in advance for your help.

eq1:=A__n=[X__n+[(C-z)*(tan(theta1)+tan(theta2))]] * [B + (2*(C-z)*tan(theta3))];
eq2:=P__n=[2*(X__n+B)] + 2 * (C-z) * [(tan(theta1)+tan(theta2)) + (2*tan(theta3))];
eq3:=gamma - [cu*(P__n/A__n)] = diff(sigma__v(z),z);
eq4:=subs([eq1,eq2],eq3);
ics:=sigma__v(0)=sigma__s;
sol:=simplify(dsolve(([eq4,ics]), sigma__v(z)));
NULL

A__n = [X__n+[(C-z)*(tan(theta1)+tan(theta2))]]*[B+2*(C-z)*tan(theta3)]

 

P__n = [2*X__n+2*B]+2*(C-z)*[tan(theta1)+tan(theta2)+2*tan(theta3)]

 

gamma+[-cu*P__n/A__n] = diff(sigma__v(z), z)

 

gamma+[-cu*([2*X__n+2*B]+2*(C-z)*[tan(theta1)+tan(theta2)+2*tan(theta3)])/([X__n+[(C-z)*(tan(theta1)+tan(theta2))]]*[B+2*(C-z)*tan(theta3)])] = diff(sigma__v(z), z)

 

sigma__v(0) = sigma__s

 

Error, (in PDEtools/NumerDenom) invalid input: `PDEtools/NumerDenom` expects its 1st argument, ee, to be of type algebraic, but received [tan(theta1)+tan(theta2)+2*tan(theta3)]

 

``

Download diffeq.mw

Dear power users, I have a probably trivial question. How can I import the attached file in Maple? The file consists of a header part with below the header two columns with data. The file coming from a measuring device has no extension and I could not upload it to the forum as it was considered a null file. In order to solve this I gave it a .txt extension but keep in mind that this normally is not present.

Additional question: if the file would have several headers each time followed by 2 data columns, is there an easy way to separate them in a sort of automatic manner?

Spiking2011622.txt

How to isolate t2 from equation A. Such that t2 can be written in form of i like -> t2 = i * something
Attaching sheetq6m.mw

How should i remove all brackets and conditions, such that i can fetch only equation t2 = something. I am attaching the image below.  

I looking to create a GUI may be from outside say script language or any given language may be 

The main intention is I have my

maple code

And 

A python code

In the same folder say

I create a GUI with two buttons

When I click one button it runs my maple file in that folder and generates the output

Second button runs the python file that folder generates the output

Is such a coding  creatable 

One more import is can we call a maple code to run from python like how we call exe files

And 

Can we call python compiler attached and file from maple and run it 

Dears,
How can I work with a LogNormal distribution represented by a mean and standard deviation?
I see that is with exp(Normal(mu,sigma)), but I dont sure. Would you please help me with a example using LN(mean=20,standard deviation=5)?
I will use in PDF, CDF and RandomVariable. 

How to find G(0)??

M2.mw

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