C_R

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6 years, 96 days

MaplePrimes Activity


These are replies submitted by C_R

I believe most people would prefer a solution where the output appears in a text area after the loop, ideally where repeated output from a loop typically appears. Something like that

restart;

with(DocumentTools):

with(DocumentTools:-Components):

with(DocumentTools:-Layout):

s := "0":

T := TextArea(s, identity = "TextArea0"):
xml := Worksheet(Group(Input(Textfield(T)))):

#insertedname:=InsertContent(xml)[1,1]:

box_printed:=false:
for i to 10 do
if (not(box_printed))then
   insertedname:=InsertContent(xml)[1,1]:
   box_printed:=true:
end if:
Threads:-Sleep(1);

DocumentTools:-SetProperty(insertedname,value,sprintf("%d",i),refresh=true);
end do:

Maplets:-Examples:-Message("Done");

Can't one of the print commands be modified to do this (i.e. avoid a line feed)? This would be much easier than learning DocumentTools.

At least there is a solution that would not have occurred to me. 👍

Download text-area-update-progress_after_loop.mw

L as used in line 3 is of dimension time. In line 4 you want L to be zero when the time is between 1 and 2 (days?).

Otherwise you want L, which is of dimension time, to be C, which is probably of a different dimension.

What you probably want is to define a piecewise function for C:

C:=piecewise(time condition,first concentration dosage 1, concentration dosage 1 + concentration dosage 2) where concentration of dosage 2 is timeshifted.

Concering C: Do you really want C to be a product of a constant and a time depended exponential term?

there is an Error before the plot comand

@tomleslie 

Its a single element list exported from

It looks like I have always been removing brackets from expressions when working manually, without realizing that it is a requirement for using isolate.

Thank you all for the quick response!

@Maxie 

You could try to adopt the following for your problem

https://www.mapleprimes.com/posts/220877-Ball-Bouncing-On-A-Flexible-Beam

 

@sursumCorda 

I was looking for something like that! 👍

 

@ecterrab Could you have a look if the output of Physics:-Substitute in the attached is as intended?

 

Another_way_with_Physics_Substitute.mws

@mmcdara 

I can’t see a better way without referring to advanced statements like applyrule. Thank you!

 That what i get with 2023

As an alternative to dharr's answer, which does the fit in mA, you could divide values in Y by 1000 do get results in A which fit to C in muF.

The book is correct. See the plots in my reply to dharr below

@dharr Just a check that the data is in mA. I have never used units and fits so I stopped after checking that the parameters fit to the data. So maybe your tweak with mF is not required.

restart

U := 100*Unit('V')

100*Units:-Unit(V)

(1)

R := 4900*Unit(Unit('Omega'))

4900*Units:-Unit(`Ω`)

(2)

C := 20*Unit('`μF`')

20*Units:-Unit(`μF`)

(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)

simplify(v(.1*Unit('s')))

0.7356077312e-2*Units:-Unit(A)

(5)

``

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

Matrix(%id = 36893489545658315588)

(6)

p1 := plot(data, style = point)

 

p2 := plot(v(t), t = .1*Unit('s') .. .5*Unit('s'), useunits = [s, mA])

 

plots[display](p1, p2)

 

``

NULL

Download Mapleprimes_Question_Book_2_Paragraph_5.12_Question_5_with_units.mw

@Carl Love 

Maple provides so many solutions, that I rarely need to code (and test) procedures.

In case I have to pass a function I will not try it make work a you said. Thank you for bringing this to my attention.

As for Fortran, I do not mis goto but I have kept a few punch cards that I use as bookmarks for sentimental reasons. I will see if I can find a goto card as physical proof that those statements were in use ;-)

@Kitonum 
Also with my 2023 installation

@acer

This particular value for posint is worth an example in the detaled helppage of fslove.

Infinite thanks!

@dharr 

Thank you for your detialed answer.
It looks like that I have used by wrong intuition a calling sequence of RootOf that is not documented in help(RootOf).

There is one equation too much. You only have 5 unknowns

indets({fa[1], fa[2], fa[3], fa[4], fa[5], fa[6]});

                 {U[0], V[0], W[0], Z[0], r[0]}
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