janhardo

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11 years, 44 days

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These are replies submitted by janhardo

@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?

@Scot Gould 
Thanks

This a better presentation than in maple input format , that's my impression.
Also the use of a prime makes that the differential equation is easier for input handling and reading.

Is there a idea what this function y(x) is notated? 

@ecterrab 

Thanks

Enough information and helpful in order for me to make a further start with this Zeta function.

@dharr 
Thanks

I will study it further.

@tomleslie 

Thanks

Looks great this plot and how it is made

The non-trivial zeroes of the Zeta function are all positioned on this critical line 
Knowing a formula for this zeroes positions occur on the 1/2 line, means that you have solved the Riemann Hypothese ( you win a prize ..bingo )
I wonder if there will be ever found a formula what predicts the distribution of primes ?

There is now a calculation going on for years now to find the next mersenne prime

So my question was not possible realizing this afterwards. 
I do want also construct later a 3 d complex plot from the critical line (strip)

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