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These are questions asked by janhardo

How to get a general expression here ?



f(s) is the function value  expressed as a integral around s  (singularity)  of a complex function.

f(s) = (int(f(z)/(z-s), z))/(2*Pi*I)

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


f(s) = int(f(z)/(z - s), z)/((2*Pi)*I):


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



diff(diff(f(s), s), s) = -((1/2)*I)*(int(2*f(z)/(z-s)^3, z))/Pi


"((ⅆ)^(n))/(ⅆn) f(s)   =  "




quote : "Important consequence.

Above it actually says: "If there is a function f(s) that is somewhere analytic, then you can use such an integral as above to make a new function f '(s), which is also analytic there. And from that a new function f '' and so on.
That means something revolutionary for complex numbers:   "



Question: how to get the "((ⅆ)^(n))/(ⅆn) f(s)  "?

Note : again the form of the answers in Maple : Its not possible to force Maple to come up with this form of answer, ex

``  NULL

Download Maple_primes_bvraag_hoger_orde_singulariteit_henk_hofstede.mw

Seems to me informative to see a earth-like surface on a sphere and in particular from the zeta function.
Another simple complex function will do it also.

Note: i saw a  3D picture with a colored sphere , where you can see zeroes and poles on the surface of the sphere
A colored complex function is that hard to make with Maple too?: it are all polar coordinates as complex points in the complex plane. 
The angle is standing for hue and the magnitude is standing for the lightness 

Really , the complex plotting possibilities in Maple are difficult to decipher. 

I tried something, but  in general the visualizing for me is not that easy

Now i must look at the complexplot where i got a circle for  the complexplot(sin(x + I), x = -Pi .. Pi) example ?

Try to calculate the values for the Zeta(z) on the critical line in symbolic form and also as a plot 

Both i did not yet succeed in 



solve(f = 1/2 , z ) 


Alternating serie


sum((-1)^(n+1)/(2*n-1), n = 1 .. infinity)




sum((-1)^(n + 1)/(2*n - 1), n = 1 .. infinity):

sum((-1)^(n+1)/(2*n-1), n = 1 .. 4)



expand(sum((-1)^(n + 1)/(2*n - 1), n = 1 .. 4),symbolic);




series(sum((-1)^(n + 1)/(2*n - 1), n = 1 .. infinity),n=0,5);




Info series


How to get for n= 4  "for (∑)(((-1)^(n+1))/(2 n-1))  =   symbolic term 1+ symbolic term 2+... "



Download onderzoek_reeks_-hoe_krijg_ik_een_partieke_symbolische_som.mw

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