janhardo

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11 years, 44 days

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@janhardo 

The FunctionAdvisor in Maple  gives two definitions for arctan

There are two approaches to defining a complex function in Maple.

Method 1. Make  f(x, y)  a function of two real variables  x, y :  f(x,y) 

Method 2. Make  f(z)  a function of the complex variable  z .  :  F(x+ I y)

@Carl Love 

Depends of the functionality of database what forum software can offer on search posts seems to me.
There is some "out the box"  forum software to install if i should start with making one
online forum on a server.
This forum has some base forum progam running what has added extended functionality further inbuilt ?

@Carl Love 

Thanks

In the list of all my question i like to filter out on who had answered my question 

Example: i search for your answers on my  questions 

@Carl Love 

Thanks

I got a answer from Maple , but it was not in a "bookform" , so i did it manually deriving the last answer from Maple to the wanted "bookform" 
Checking this manually derived answer in 2d maple form,  shows now it was the same as the answer in Maple for the derivation of the complex  integral.  
So, its all now solved for the derivation of the complex integral.

@Axel Vogt 

Thanks

I do have a lot of lecture notes about complex analysis and the collection is growing
I studied first the classical example of a two valued function : the logarithmic function
Got a first idea about a branch point and cut from this ln(z) .

Then i went to the arctan (z) function to look at this with the FunctionAdvisor(arctan)
For me important when has een complex function a branch point, if so then it is a multivalued function i think, but not sure about it
Studying the "branch" information Maple gives from arctan(z) is complicated to understand for now, so you need the theoretical background from some books 

restart;

with(plots):

ans := -I*1/2*ln((b + z*I)/(b - z*I))/b;

-((1/2)*I)*ln((b+I*z)/(b-I*z))/b

(1)

 plot3d([Re,Im](arctan(z/b)/b-ans),b=-4..4,z=-4..4,
       color=[red,blue],grid=[100,100],view=-1e-14..1e-14);

 

ans := -I/2*ln((1 + 1*I)/(1 - 1*I))/1; #take 1, for all variables

(1/4)*Pi

(2)

# arctan(z/b)/b-ans = arctan(1/1)/1-ans

FunctionAdvisor(arctan(Pi/4));

 

?FunctionAdvisor

FunctionAdvisor(branch_points, arctan(Pi/4));FunctionAdvisor(branch_cuts, arctan(Pi/4));

[arctan((1/4)*Pi), `in`((1/4)*Pi, [-I, I])]

 

[arctan((1/4)*Pi), "No branch cuts"]

(3)

ans := -I*1/2*ln((b + z*I)/(b - z*I))/b;

-((1/2)*I)*ln((b+z*I)/(b-z*I))/b

(4)

plot3d(arctan(Pi/4));# try getting a complex plot, so Pi/4 must be a complex number  

 

 

I must try to plot th ecomplexfunction with the right notation RE AND Im om axes and make a closer look tot the function for spotting branch cuts ( vertical lines).
Can do a lot with plot options

 plot3d([Re,Im](arctan(z/b)/b-ans),b=-4..4,z=-4..4,
       color=[red,blue],grid=[100,100],view=-1e-14..1e-14);

 

 

Download branch_onderzoek_berekende_complexe_integraal.mw

@Carl Love 

Thanks

Getting a "bookform" expression as last step ( because Maple has give his latest answer and is this is not the bookform) is be done by hand.

?op

restart;

expr1:= -I/2*ln((b + z*I)/(b - z*I))/b;
 

-((1/2)*I)*ln((b+I*z)/(b-I*z))/b

(1)

-I*ln((b+I*z)/(b-I*z))/(2*b)

 (bookform)

-((1/2)*I)*ln((b+I*z)/(b-I*z))/b

(2)

expr2:= -I/(2*b)*ln((b + z*I)/(b - z*I));# convert from 2d input

-((1/2)*I)*ln((b+I*z)/(b-I*z))/b

(3)

%-%;

0

(4)

 

(-I/2*ln((b + z*I)/(b - z*I))/b) - (-I/(2*b)*ln((b + z*I)/(b - z*I)));

0

(5)

 

In 2D input  i transformed expr1 by hand , checked this if it is correct by subtracting expr1 - expr2

Download post-Double_back_quotes_for_hard_parentheses.mw

@Axel Vogt 

Thanks

I try to learn how to examine complex multi-valued function on branches, but  its limited to not so many complex functions in Maple (no,there are more..see branch_cuts in FunctionAdvisor.
I calculated a complex integral , this could be examined for branch points ? 

 plot3d([Re,Im](arctan(z/b)/b-ans),b=-4..4,z=-4..4,
       color=[red,blue],grid=[100,100],view=-1e-14..1e-14);

The most basic case of the multi valued function is  the f(z) = z^1/2  as inverse from f(z)= z^2 as i understand it now.

Another fact about real function opposite complex functions is for example
 e^x  <inverse>  ln x  are one to one functions (bijection) 
e^z   <inverse>  ln z  , e^z  is many to  1  and ln z is  1 to many ( multivalued )

FunctionAdvisor(branch_points, exp);
                  [exp(z), "No branch points"]

@janhardo 

At the moment there are these (multivalued) functions known to Maple for the presence of branches

arcsin
arccos
arctan
arccsc
arcsec
arccot
arcsinh
arccosh
arctanh
arccsch
arcsech
arccoth
ln
LambertW

@vv 
Thanks

There are some more Maple commands to get about handling of complex functions
Knowing more about a branch cut/point ..?

@janhardo 

Now how to use this TS module in other cases?

Here  example: "bookform"

TS:-Mn(2):
TS:-Mn(i):
TS(i/(2*b))*ln(``(b*i+z)/(b*i-z));

Those use of the double back quotes ` ` ...?

@Carl Love 

Thanks

Amazing coding you performed.
It did the job by showing the "bookform"

Is there a general use for your TS module programming , in case there is another calculation in the future made by me what has not a desired "bookform" answer. ?

Otherwise i do have to check this manually a result( imagine i use a computeralgebra program and  must check this answer if it is correct ?)
Its only a last step so  i think it can be that hard (as this integral example proves)

If there is a command possible in Maple in the future  what can split a expression   ?

@janhardo 

For transforming  expression (2) into the bookanswer form :

I could try to divide expression (2) by b for the fraction. 
I could bring the b under  -1/2ln..this is a splitting case of the fraction ( expression (2))

Doing(trying  )handmade transformation is problematic

expr:=-I/2*ln((b + z*I)/(b - z*I))/b;   =  

 denom(expr); =                          numer(expr); =                                          

So Maple does see 2b as denominator , while i consider b as denominator for manipulation for splitting the expr  fraction

Maple is making from expr a only one  fraction what is correct 

@Carl Love 
Thanks

I am curious if the evaluated expression in Maple from the  complex integral calculation can be used in your module code?

restart
:
TS:= module()
uses T= Typesetting;
local
    #Using Typesetting:-_Hold introduces unwanted parentheses. I get
    #around that by using attributes and my own Hold.
    Hold,
    ModuleApply:= e-> setattribute(T:-Typeset(e), Hold(e))
;
export
    #`macro` is needed because Typesetting:-Typeset declares its 1st
    #parameter uneval.
    Mn:= e-> (parse@sprintf)("macro(%a=%a)", ``||e, T:-mn(""||e)),
        
    Value:= e-> subsindets(
        subsindets(
            subs[eval](``= (()-> args), e), attributed,
            e->
            local p, A:= [attributes](e);
                `if`(membertype(specfunc(Hold), A, 'p'), op(A[p]), e)  
        ),
        specfunc({T:-mn, T:-mi}),
        N-> try parse(op(1,N)) catch: N end try
    )
;
end module
:

TS:-Mn(2):
TS(a/(`2`*d)) . ``(b/c);
TS:-Value(%);
lprint(%);

Parse:-ConvertTo1D, "invalid input %1", Typesetting:-mfrac(Typesetting:-mi("a"), Typesetting:-mrow(Typesetting:-mn("2"), Typesetting:-mo("&InvisibleTimes;"), Typesetting:-mi("d"))).``(b/c)

 

(1/2)*a/d.(b/c)

 

1/2*(a/d . (b/c))

 

 

What was the start of this all : the maple prime post: how to perform this complex subtition
https://www.mapleprimes.com/questions/233292-How-To-Perform-This-Complex-Substitution

 

There was een integration of complex integral and the end answer in Maple was not the same as in the book example  

 

 Maple answer was      

 

The bookanswer was :   

 

Maple was not able to do more simplifications in order to get the book answer and both answer are the same.

I  simplified the question by using

                                                  

Evaluated in Maple you get :                                                                  

((1/2)*a*(b/c))/d;

(1/2)*a*b/(c*d)

(1)

-I/2*ln((b + z*I)/(b - z*I))/b;

-((1/2)*I)*ln((b+I*z)/(b-I*z))/b

(2)

 

 If i use your module code it must be then possible  to transform answer( 2)  hereabove into the bookanswer ?

Download Post-how_get_a_wanted_expression-reactie_Carl_met_module_programmering.mw

Seems to be that the earlier posts reactions by @acer on my question are disappaerd, what is happened?

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