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janhardo

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11 years, 254 days
B. Ed math
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A maplet for complex integration ...

janhardo 860 Maple
complex-analysis interactive
January 12 2022
0 0

This is so useful to see geometrical mapping diagram to visualize Complex analysis

Something that also can be made for Maple 

Mapping Diagram for Cauchy Integral Formula – GeoGebra

Using GeoGebra for visualizing complex variable. (google.com)

I highly encourage everyone interested in complex variable to read Tristan Needham „Visual Complex Analysis” and try to solve problems with or without aid of GeoGebra. I hope that in this workshop we will manage to get a feeling of complex functions and as a final point understand how complex integration works. It is a common misconception that complex integration can't be visualized, and using Tristan Needham's ideas we will try to explore this idea. It's a pity that we don't have a lot of time, thus we will skip a lot of important information and construct only some graphs. 

There is so much experimenting with Geogebra software and doing too this in Maple ?

How to get a prime and subscript notatio...

janhardo 860 Maple
derivative 2dmath typesetting
January 10 2022
1 4

I am studying the help pages for declare() statement in the PDEtool package. 
It can be used also for calculus(prime notation) for 1 variable and two variables (subscript notation) as i understand it now.

There is no input possible in Maple by using a prime notation and indexed functions notation?

Maybe with a alias ? ..remember that I can be chanced I (default) into the i for complex numbers 

How to plot the approximation graph for ...

janhardo 860 Maple
plotting riemann-zeta
January 10 2022
0 1

I must feed the not trivial zeros numbers into this aproximation formula ?

 

Riemann hypothese and staircase of primes

 

> 

restart;
> 

with(NumberTheory)
> 

PrimeCounting(1)

0

 (1)
> 

pi(Pi)

2

 (2)
> 

PrimeCounting(10000)

1229

 (3)
> 

numelems(select(isprime, [seq(1 .. 10000)]))

1229

 (4)

The prime counting function is approximated by Li(x) and x/ln(x).

> 

plot([PrimeCounting(x), Li(x), x/ln(x)], x = 1 .. 500, legend = [pi(x), Li(x), x/ln(x)])

The staircase of primes approximated by two functions
Interesting is the video: How i learned to love and fear the Riemann Hypothesis

https://www.quantamagazine.org/how-i-learned-to-love-and-fear-the-riemann-hypothesis-20210104/

> 

NULL
> 

ps:=Array(1..30):
y:=0:
for n from 1 to 30 do
 if is(n,prime)
     then ps[n]:=plot([[n,y],[n,y+1],[n+1,y+1]]):
     y:=y+1;
     else ps[n]:=plot([[n,y],[n+1,y]]):
 end if ;
od;
with(plots):
display({seq(ps[n],n=1..30)}):  
> 

plot([PrimeCounting(x)] ,x = 1 .. 35, legend = [pi(x)]):
> 

plot([PrimeCounting(x), Li(x), x/ln(x)], x = 1 .. 35, legend = [pi(x), Li(x), x/ln(x)])

 

 

 

 

Prime counting function
What found RIEMANN for the prime counting function in relation to the zeta function after he defined the zeta function?

 

He found further a function what follows exactly the shape of the prime counting function

Final discovery v. Riemann.  

- step in the omhoog in de priemtelfunctie = log(p) (zie video)

 

Using the logarithmic primecount function( from Chebyshev) (approximation)
Further  analyse with this Chebyshev approximation formula in relation to the not trivial zero points from Riemann zeta function ( zeros) gives another real function for approximating the primecounting function what uses the non trivial zeros from Riemanns zeta function  in this function:

 

"(not trivial zeros ) u[k ] = "i*w[k]+v[k]   
Number now all nottrivial zeros in the upperhalfplane from down to bottom,  as u[1], u[2], u[3, () .. ()]

> 

"`ϕ`(x) := x-ln(2Pi)-1/(2 )ln(1-1/(x^(2))) - (∑)2/(|u[k]|) x^(v[k]) cos(w[k] ln(x)-alpha[k])"
> 

NULL

Its only alpha[k] that must be calculated out of the not trivial zeros and i must have a list of  serie of not trivial zeroes from the zeta function. => see Hardy's Z(t) ? from this ..... alpha[k]  can be calculated ?
All not trivial zeros are complex numbers laying on a line ,but  orginating from (0,0) in the complex plane as  vectors to the points    

> 

varphi(x):= x - ln(2*Pi) - 1/2*ln(1 - 1/x^2) - sum(2*x^v[k]*cos(w[k]*ln(x) - alpha[k])/abs(u[k]), k = 1 .. infinity);

x-ln(2*Pi)-(1/2)*ln(1-1/x^2)-(sum(2*x^v[k]*cos(-w[k]*ln(x)+alpha[k])/abs(u[k]), k = 1 .. infinity))

 (5)

This formula seems to be correct .
Now how to make a plot ?
Hardy's Z(t) function shows the not trivial zeros in the upperhalfplane of the critical strip of the Riemann zeta function  as zeros in this Z(t) real function : derived from a alternating serie ?

Download priem_staircase_en_riemann_functie.mw

How to prove this sum is equal to produ...

janhardo 860 Maple
product summation
January 06 2022
2 8

Try to prove those two expressions : a sum and product 
First to know if they are correct defined for some values ?

Eulers productformule in Maple

 
> 

 

Eerst de Euler identiteit in Maple opschrijven
via info over series

 

p-Series

sum(1/n^p, n = 1 .. infinity)

Absolute convergence for p > 1

Diverges for p <= 1 

 

 

> 

restart;

sum(1/n^p, n = 1 .. infinity) = product(1/(1-p^(-s)), i = k .. p)

> 

product(f,i=k..n);

> 

product(1/(1 - p^(-s)), i = 1 .. p);

(1-p^(-s))*(-1/(-1+p^(-s)))^(p+1)

(1.1)
> 

sum(1/n^p, n = 1 .. infinity);

sum(1/n^p, n = 1 .. infinity)

(1.2)
> 

sum(1/n^p, n = 1 .. 4);

1+1/2^p+1/3^p+1/4^p

(1.3)
> 

product(1/(1 - p^(-s)), i = 1 .. 4);
            

1/(1-p^(-s))^4

(1.4)
> 

Product(1/(1 - p^(-s)), i = 1 .. 4)= product(1/(1 - p^(-s)), i = 1 .. 4);

Product(1/(1-p^(-s)), i = 1 .. 4) = 1/(1-p^(-s))^4

(1.5)
> 

sum(1/n^p, n = 1 .. infinity) = product(1/(1 - p^(-s)), i = k .. p);

sum(1/n^p, n = 1 .. infinity) = (-1/(-1+p^(-s)))^(p+1)/(-1/(-1+p^(-s)))^k

(1.6)
> 

 

 How to prove this sum(1/n^p, n = 1 .. infinity) = product(1/(1-p^(-s)), i = k .. p)

For the sum i got some terms , but for the product i don't see factors as i use the commands  

Download the_golden_key_in_Maple.mw

Riemanns formidable formula gives differ...

janhardo 860 Maple
roundoff numeric evalf
January 05 2022
2 5

Some small differences is unavoidable ?

> 

 

Its not symbolic that's why i don't get he same answers ?
Later on i will investigate some values for for this formula  

Riemanns formidabele formule

 
> 

restart:

Volgens het boekje de riemanns hypothese : de functionaal vergelijking  (zonder bewijs)
Zeta(-z) = ((-2*factorial(z))*(1/(2*Pi)^(z+1)))*sin((1/2)*Pi*z)*Zeta(z+1)

Zeta(-z) = -2*factorial(z)*sin((1/2)*Pi*z)*Zeta(z+1)/(2*Pi)^(z+1)

(1.1)
> 

Zeta(-z) = -2*z!*sin(Pi*z/2)*Zeta(z + 1)/(2*Pi)^(z + 1);

Zeta(-z) = -2*factorial(z)*sin((1/2)*Pi*z)*Zeta(z+1)/(2*Pi)^(z+1)

(1.2)
> 

verg:=%;

Zeta(-z) = -2*factorial(z)*sin((1/2)*Pi*z)*Zeta(z+1)/(2*Pi)^(z+1)

(1.3)
> 

verg2:= subs(z=2+3*I,verg);

Zeta(-2-3*I) = -2*factorial(2+3*I)*sin((1+(3/2)*I)*Pi)*Zeta(3+3*I)/(2*Pi)^(3+3*I)

(1.4)
> 

verg3:= eval(verg,z=2.+3*I);

.1329711559-.1230533004*I = (2.450819690-44.87297744*I)/(2*Pi)^(3.+3.*I)

(1.5)
> 

evalf(rhs(verg3));

.1329711558-.1230533005*I

(1.6)
> 

lhs(verg3)=evalf(rhs(verg3));

.1329711559-.1230533004*I = .1329711558-.1230533005*I

(1.7)
> 

lhs(verg3)- evalf(rhs(verg3));

0.1e-9+0.1e-9*I

(1.8)
> 

 

Download post_naar_primes_over_triemanns_geweldige_formule.mw

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