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These are replies submitted by janhardo

I do read here more messages about  deleted questions
While ago is also deleted the answering of my post 

Isn't one of the moderators the culprit here , who removes post just like that?

To get idea how to doing a proof by induction

Its here solved with Mathematica 

Mathematica Tutorial 12 - Proof by Induction - Bing video

In this case its a sum ,so you need a another solutionstrategy

I don't see my thread here anymore ?

What is happened?

Is it removed or something malfunctioning on the website ?


The presentation of this format of a complex function, here in this example is not accepted by the procedure :  

What is representation of f(z)  = with variable z ?



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 )


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 





3.2  The Cauchy-Riemann Equations



a complex analytical function




Download reactie_op_complex_differentieren_in_gemaakte_procedure.mw



"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)


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 ?


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 



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 ?) 



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








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


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





y(x) = x^n


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

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



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)



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


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



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



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

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


map(diff, %, x)

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




Download post_maple_primes-algemen_formules_afgeleides.mw



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


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.


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 ?


Thanks for the effort

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


-((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




-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

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