janhardo

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11 years, 47 days

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These are replies submitted by janhardo

@vv 
Thanks

With this correct sum now Maple knows perhaps the Euler prorduct formula as closed formula?

@tomleslie 

Thanks

Your definition of the sum and product seems to be correct: they are equal , if look to the comparison
I am unsure if i defined this correct in my worksheet too
Is there a way in Maple doing this direct without filling in first some values in the sum and product?

@vv 

Thanks 
informative the wiki 

At least then : sum(1/n^p, n = 1 .. infinity)

Proving for all numbers in Maple is impossible , so how i can check  this symbolically ?
In order to check these sum and product if they are equal then i must fill in some values then. 
note: i think i must chance p in the sum into s  ( is real number >1 , or the same as  Re s>1) after looked in a book and seen in the @tom leslie code

@tomleslie 
Thanks

It is not that important for examing the graph of the zeta function i think?

@Carl Love 

Thanks

Those definitions can  be tricky

f = u +i v form   f = u(x,y) + i(x,y)  form see example 2.1

Question : use of `  hereunder ?

`f(x,y) ` = f(x,y); ` `;

About the one-argument and two argument form of arctan : for real arctan versus complex arctan ?

restart;

I got this information from this course and thought it can be used for the function definition via the FunctionAdvisor of arctan

FunctionAdvisor(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 .

Method 2. Make  f(z)  a function of the complex variable  z .

 

Example 2.1.  Write  f(z) = z^4  in the  f = i*v+u  form.  

 

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

 

f:='f': x:='x': y:='y': z:='z':
f := proc(x,y)
  local z,w;
  z := x + I*y;
  w := expand(z^4);
end:
`f(z) ` = z^4;
`f(x,y) ` = f(x,y); ` `;
`At  z = 1 + 2i: `;
`f(1,2) ` = f(1,2);

`f(z) ` = z^4

`f(x,y) ` = x^4+(4*I)*x^3*y-6*x^2*y^2-(4*I)*x*y^3+y^4

` `

`At  z = 1 + 2i: `

`f(1,2) ` = -7-24*I

 

Method 2. Make  f(z)  a function of  z .

 

F:='F': x:='x': y:='y': z:='z':
F := proc(z)
  local w;
  w := expand(z^4);
end:
`F(z) ` = F(z);
`F(x + I y) ` = F(x + I*y); ` `;
`At  z = 1 + 2i: `;
`F(1 + I 2) ` = F(1 + I*2);

`F(z) ` = z^4

`F(x + I y) ` = x^4+(4*I)*x^3*y-6*x^2*y^2-(4*I)*x*y^3+y^4

` `

`At  z = 1 + 2i: `

`F(1 + I 2) ` = -7-24*I

restart;

# arctan(x + y*I) = arctan(y, x); ?

arctan(y, x)# also defined (see functionadvisor) as complex function in u(x,y),v(x,y) form, but defined in Maple as # arctan(x,y) its a a real function

arctan(2. + 5*I);# complex arctan function

1.499847799+.1732867951*I

(1)

arctan(5., 2); # real arctan function

1.190289950

(2)

# arctan(Pi);

arctan(Pi)

(3)

arctan(-sqrt(3)/3);

-(1/6)*Pi

(4)

arctan(-1);

-(1/4)*Pi

(5)

arctan(y, x) = -I*ln((x + y*I)/sqrt(x^2 + y^2));

 

Download reaction_functionadvisor_arctan.mw

@The function 

I don't think it's a shame that the book does not come up with such answers you got here on the Maple Primes forum.
Here some people on the MaplePrimes  forum are really specialists in the Maple language.

@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

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