janhardo

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These are replies submitted by janhardo

@acer 

Thanks

I wrote:  I must differentiate z = e^i.phi   in order to get the differential dz

In normal math format i was thinking for the expression here above notated
Indeed it must be  z(phi) = exp(i.phi) for maple and the procedure can differentiate this rightside of the equation! 

dz= z(phi)'. d(phi)     (dz is differential of z )

@janhardo 

The complex function must be "analytic" in order to possible differentiate it 
note: i must look at information for complex functions classes who are proven analytic 

NULL

 

 

restart;

3.2  The Cauchy-Riemann Equations

content

  

a complex analytical function

   

 

NULL

Download reactie_op_complex_differentieren_in_gemaakte_procedure.mw

@acer 

Thanks

"I really hope that you understand that it is impossible for anyone to create a procedure that will produce your as-yet-unstated, desired form of output for all future examples which are not yet known. That is not a Maple thing; it is a logical consequence."

That would be Maplemagical , but i don't believe in fairy tales ( although some people say: it was a fairy tale what they have experienced)

@acer
 Thanks 

You are right ,guessing a intended scope is not workable.

Instead of working all this out in Maple, i show you a  link
Nieuwe pagina 1 (hhofstede.nl) ( in Dutch)
You do see a circle and left of it some math written and above the intergral is written  z= e^i .phi  so,  dz= i.e^i.phi 
dz is differential (already  calculated here) and in general  dy = f '(x) . dx 

I must differentiate z = e^i.phi   in order to get the differential dz
Can the "differentiate procedure" handle complex differentiating maybe? 

Hopefully its clear now i hope, or not and do you want it worked out in Maple the problem description ?

@janhardo 

Don't know how the procedure is behaving for complex differentiating ?

Example : would calculate a differential  for  z= e^i. phi   => dz=  ...d(elta)phi 

@acer 

Thanks

That is a powerful procedure that differentiates all kinds of functions !
For intergration, such a procedure would also be useful and call the indefinite integral: the primitive ( F(x) ). 
The suppression of notation is useful for memorizing certain differentiation rules.

combine(diff(eq, x), power) seems to be also useful for differentation ( what is the advantage here ?) 

@acer 

Thanks

Goal was to derive the powerrule in  textbook form and its easy to type in that form in Maple

 

restart;

interface(typesetting=extended):

Typesetting:-Settings(typesetprime=true):
#Typesetting:-Settings(prime=s):
 

Typesetting:-Settings(typesetprime=true);

true

(1)

f(s)^n=-((1/2)*I)*factorial(n)*(int(f(z)*(z-s)^(-1-n), z))/Pi;
 

f(s)^n = -((1/2)*I)*factorial(n)*(int(f(z)*(z-s)^(-1-n), z))/Pi

(2)

My first goal was to find a general expression for a derivative of a function ( real or complex).

I have succeeded with a intergral formula from Cauchy

 

Now for a function  of type   f(x) = x^n

 

restart;

 

y(x)=x^n;

y(x) = x^n

(3)

map(diff,%,x);# the power rule

diff(y(x), x) = x^n*n/x

(4)

 

In math textbook form it is diff(y(x), x) = n*x^(n-1) for (2) for manual calculating to memorize.
=============================================================

That was my goal: the power rule finding , but it can be typed in as a textbook form

=============================================================

 


I have now done two examples of functions to find a general expression for the derivative.
To do this in Maple differs for both functions and it is impossible for me to have a general approach to do this in Maple for all kinds of functions.

 

Coincidentally, I see this here , but can't recognise here the power rule from a  textbook

diff(f(x)^n,x);# the power rule , compared with (3) ?

f(x)^n*n*(diff(f(x), x))/f(x)

(5)

NULL

The product rule : (f.g)' =f 'g +g'f  (memorize)

h(x)=f(x)*g(x);

h(x) = f(x)*g(x)

(6)

map(diff,%,x);

diff(h(x), x) = (diff(f(x), x))*g(x)+f(x)*(diff(g(x), x))

(7)

restart;

h(x) = f(x)*g(x)

h(x) = f(x)*g(x)

(8)

map(diff, %, x)

diff(h(x), x) = (diff(f(x), x))*g(x)+f(x)*(diff(g(x), x))

(9)

``

 

Download post_maple_primes-algemen_formules_afgeleides.mw

@acer 

Thanks

I needed the other way around as teached in math books for differentiating  

in textbook form it is 

Its impossible to get a general approach to learn for me to get  a general formula for differentiating a (complex) functions

That was the start for this thread, a intergral formula from Caughy in Complex analysis

@acer 

Thanks
It's amazing how you can bend Maple to your will 
This is about getting a particular expression into a desired form but is not really important for the math I am trying to do with Maple, as it is easy to get it into the right form by hand as well 
Probably some advanced users needed this typesetting conversion.

@janhardo 

Ok, i made a mistake by posting a separate question again , about the topic here: a general derative formula and could do it in this thread.

The post is removed and why is it not placed hereunder ( if possible ) ?

Note: i don't care anymore about the answer Maple produces if it is not a textbook form, because some manual elementair math can be performed to get the wanted notebook form.

-------------------------------------------

The additional question was : the general derative for y= x^n   =>  y'= n. x^n-1

 How to get this general formula ?

@acer 

Thanks for the effort

Some manual math...gives the desired formula form!

restart;

-((1/2)*I)*factorial(n)*(int(f(z)*(z-s)^(-1-n), z))/Pi;

-((1/2)*I)*factorial(n)*(int(f(z)*(z-s)^(-1-n), z))/Pi

(1)

manual

 

-I*factorial(n)*(int(f(z)/(z-s)^(1+n), z))/(2*Pi)

 

 

first thinking on a mistake made maybe by the author of source of the integral ?, but  the number i is a special number to handle in complex numbers

 

I^2 = -1
There is a : -I in the formula derived by Maple  it has different forms
-i = -i and -i = i*(1/i^2) and i*(1/i^2) = 1/i

This  formula by Maple derived can be written as

 

-(I*(1/2))*factorial(n)*(int(f(z)*(z-s)^(-1-n), z))/Pi =    "(n!)/(2 Pi I)  (∫(f(z) )/((z-s)^(n+1) ) ⅆz) =  "

So maple gives not always the textbook answer , but doing manual further you get the wanted answer

Download Maple_primes_bvraag_hoger_orde_singulariteit_henk_hofstede-_formule_identiek_in_Maple_.mw

@acer 

Thanks

Your n -the derative of H(s) =   -((1/2)*I)*factorial(n)*(int(f(z)*(z-s)^(-1-n), z))/Pi

Is it the same as this expression below (ask myself) ? 

@Scot Gould 

Thanks

What is the function written symbolically ? : eg.  y(x) = x^2 -6  

I did also ODEStep command for a step solutions for your example, but this type of differential equation is not to get in closed form? ( only numerically solution) 
If it is a numerical solution, then it wll be getting probably hard for Maple to show the function description?

@vv 

Thanks

A spacial complex function what rises out from the complex plane, like the riemann zeta function is, cannot sterographical projected on the riemann sphere.

The projection must be done from the flat complex plane, right?

The complex function as mountain landscape what possible posesses all sorts of zeroes points and sorts of poles what is visible.
More difficult is to do calculation on this. 
The order of  a zero is highest needed derative to get it 0 , or  the pole calculation is with a limit process and with a residu (must study it further, is not complete correct what i write i think) 
Much more not intuiative is intergrating, with a contour and a loops and don't have for this not a general approach yet, especilaly for the loops , what all is involved?

@mmcdara 

Thanks 
Thought it was a help for visualizing complex functions on the Riemann sphere
Probably not used by Riemann himself.

Cartesian to spherical coordinates transformation differs from a  stereographic projection?

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