Riemann Zeta Function

[six_L]

The Zeta(z) Nontrivial Zero looks like this(Attachment).

[TimoVJL]

And what was code for it ?

Long time ago i asked form mathematician, can he show me a mathematical functions for best woman breasts and butts.
I have seen many functions for those and some of them are really good.

For woman breast some GDI functions are very good simulating  those things, but sadly those sources aren't availble right now.

[zedd]

And what was code for it ?

I'll ask an even better question...
"Where is the assembly code for any of it?", even if it is only uasm compatible and not specifically masm compatible (given the board that it is in).

Enquiring minds want to know... 

[six_L]

For me, it's just despatching my interesting time when I have excess energy.

But the question of finding prime is very mysterious.
I guess:
Every Riemann Zeta(z) Nontrivial Zero corresponds to a prime number.
Zeta(0.5+bi) = 0
f(0.5+bi) = p
The larget known maxiprime today is 2^136279841-1. If someone finds the function, we'll find the next maxiprime to be easy.

Long time ago i asked form mathematician, can he show me a mathematical functions for best woman breasts and butts.
I have seen many functions for those and some of them are really good.

For woman breast some GDI functions are very good simulating  those things, but sadly those sources aren't availble right now.

This is a best APP, ought to continue.

[TimoVJL]

I was interest, how Complex numbers are handled in pure masm.

Complex numbers are part of C since C99, not supported by Pelles C anymore.

[FORTRANS]

Hi,

I was interest, how Complex numbers are handled in pure masm.

   I do not think there is any direct support for complex, or imaginary,
arithmetic in MASM.  You would have to implement such actions as macros
or a set of functions.  So you would handle complex numbers with your
own code.

   Of course, someone has probably done this already, and you could use
their implementation.

Regards,

Steve N.

[daydreamer]

And what was code for it ?

Long time ago i asked form mathematician, can he show me a mathematical functions for best woman breasts and butts.
I have seen many functions for those and some of them are really good.

For woman breast some GDI functions are very good simulating  those things, but sadly those sources aren't availble right now.

Seen on TV show a woman named "fern",but there is also exist code for draw a fern plant
Might be possible with combine several curves used for creating meshes
I only have managed to created sphere with trigo curves: 1700 half circles drawing a planet,also made an egg shape

Hourglass body = vertical cosine

[TimoVJL]

In C11 specs _Complex data type is just optional.
So many C programmers are now in same position as asm programmers.

@daydreamer, jokes needs their own topic and this is UAsm topic and i was asking masm way to handle complex numbers.

[daydreamer]

In C11 specs _Complex data type is just optional.
So many C programmers are now in same position as asm programmers.

@daydreamer, jokes needs their own topic and this is UAsm topic and i was asking masm way to handle complex numbers.

I think math expert raymond might have something in his math library ???

[six_L]

I was interest, how Complex numbers are handled in pure masm.

for example:

Code Select Expand

FpuPow proc @Fx:QWORD,@Fy:QWORD
;return: ST(0)=x^y
;x^y =2^[y*log2(x)], x > 0

finit

mov rax,@Fy
fld tbyte ptr[rax] ;y
mov rax,@Fx
fld tbyte ptr[rax] ;x,x11=x11^y
fyl2x ;y*log2(x)
;ST(0)=y*log2(x), ST(1)=zzz
fld  st ;make a second copy
;ST(0)=y*log2(x), ST(1)=y*log2(x), ST(2)=zzz
frndint ;round it to an integer
;ST(0)=int[y*log2(x)], ST(1)=y*log2(x), ST(2)=zzz
fsub st(1),st ;this will leave only a fractional portion in ST(1)
;ST(0)=int[y*log2(x)], ST(1)=y*log2(x)-int[y*log2(x)], ST(2)=zzz
fxch st(1) ;ST(0)=y*log2(x)-int[y*log2(x)], ST(1)=int[y*log2(x)], ST(2)=zzz
f2xm1 ;get the fractional power of 2 (minus 1)
;ST(0)=2ST(0)-1, ST(1)=int[y*log2(x)], ST(2)=zzz
fld1 ;ST(0)=1, ST(1)=2ST(0)-1, ST(2)=int[y*log2(x)], ST(3)=zzz
fadd ;add the 1 to ST(1) and POP ST(0)
;ST(0)=2ST(0), ST(1)=int[y*log2(x)], ST(2)=zzz
fscale ;add the integer in ST(1) to the exponent of ST(0)
;effectively multiplying the content of ST(0) by 2int
;and yielding the final result of x^y
;ST(0)=x^y, ST(1)=int[y*log2(x)], ST(2)=zzz
fstp st(1) ;the content of ST(1) has become useless
;overwrite the content of ST(1) with the result and POP ST(0)
;ST(0)=x^y, ST(1)=zzz
ret

FpuPow endp
;¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤
EComplexNumberPow proc @inRealPart:QWORD,@inImaginaryPart:QWORD,@OutRealPart:QWORD,@OutImaginaryPart:QWORD
LOCAL @e:REAL10

; e^(a+bi)
; i = (-1)^0.5
; e^(a+bi) = e^a * e^[b*ln(e)i] = e^a * [cos(b.ln(e))] + e^a * [sin(b.ln(e))]i
; = e^a * [cos(b)] + e^a * [sin(b)]i

finit
; e^a
fld FP10(2.7182818284590452)
fstp @e
invoke FpuPow, addr @e, @inRealPart ; e^a

mov rax,@inImaginaryPart
fld tbyte ptr [rax] ; b
fsincos ; st0=cos(b),st1=sin(b)
fmul st(0),st(2) ; e^a * cos(b)
mov rax,@OutRealPart
fstp tbyte ptr [rax] ; @OutRealPart = e^a * cos(b)

fmul st(0),st(1) ; e^a * sin(b)
mov rax,@OutImaginaryPart
fstp tbyte ptr [rax] ; @OutImaginaryPart = e^a * sin(b)
ffree st(0)

ret
EComplexNumberPow endp

Code Select Expand

LOCAL ao:REAL10
LOCAL bo:REAL10
LOCAL a:REAL10
LOCAL b:REAL10


fld FP10(0.3)
fstp a
fld FP10(0.4)
fstp b
; e^(0.3+0.4i)
invoke EComplexNumberPow,addr a,addr b,addr ao,addr bo
; Print
; ( 1.243302295069503) + ( 0.525659779196979)i
; ao = 1.243302295069503
; bo = 0.525659779196979


[jack]

Hi six_L
if you need some reference code here's a reasonably complete set of functions complex Riemann Zeta

[TimoVJL]

Thanks six_L 
I used it for test code for poasm.exe V13

[six_L]

Hi,jack
Thanks you.

Code Select Expand

    '  f(k)=(k+nc)^-n

    '  inf           nc              inf                        inf
    '  ====          ====           /                          ==== B
    '  \       1     \       1      [                          \    (2*k)   (2*k-1)
    '   >    ----- =  >    ----- +  I    f(k) dk + B * f(0) -   >   ------ f  (0)
    '  /      n      /      n       ]               1          /    (2*k)!
    '  ====  k       ====  k       /                           ====
    '  k = 1         k = 1          0                          k = 1

I don't understand your Zeta formula. Do you have some detailed documentations about the formula?

1, Maybe some mistakes in calculation.

zeta( 1.0000, -1.0000) =  0.5821580597520036,  0.9268485643308069
zeta( 1.0000, -0.5000) =  0.5784330210993112,  1.9635494964529780
zeta( 1.0000,  0.0000) = -1.#IND000000000000, -1.#IND000000000000
zeta( 1.0000,  0.5000) =  0.5784330210993112, -1.9635494964529780
zeta( 1.0000,  1.0000) =  0.5821580597520036, -0.9268485643308069

zeta( 1.0000, -1.0000) = inf
zeta( 1.0000, -0.5000) = inf
zeta( 1.0000,  0.0000) = inf
zeta( 1.0000,  0.5000) = inf
zeta( 1.0000,  1.0000) = inf

Gamma(s).Gamma(1-s) = Pi/sin(s.Pi)     ; Euler Product.

Zeta(s) = [1/(2.pi.i)].Gamma(1-s).Int_(+inf --> +inf)[(z^(s-1).e^z)/(1-e^z)].dz ; Riemann Product.
i = (-1)^0.5
s = a + bi ; a != 1
pay attention to the (+inf --> +inf):
+inf --> ... --> (0~) --> ... --> +inf , It's not the Int_(0 --> +inf) or Int_(-inf --> +inf)

Zeta(s) = 2.Gamma(1-s).(2Pi)^(s-1).sin(Pi.s/2).Zeta(1-s)

2, some of nontrivial zeros in Riemann Zeta Function:

1, ( 0.5 + 14.1347251 i ) ;  11, ( 0.5 - 14.1347251 i ) 
2, ( 0.5 + 21.0220396 i ) ;  12, ( 0.5 - 21.0220396 i )
3, ( 0.5 + 25.0108575 i ) ;  13, ( 0.5 - 25.0108575 i )
4, ( 0.5 + 30.4248761 i ) ;  14, ( 0.5 - 30.4248761 i )
5, ( 0.5 + 32.9350615 i ) ;  15, ( 0.5 - 32.9350615 i )
6, ( 0.5 + 37.5861781 i ) ;  16, ( 0.5 - 37.5861781 i )
7, ( 0.5 + 40.9187190 i ) ;  17, ( 0.5 - 40.9187190 i )
8, ( 0.5 + 43.3270732 i ) ;  18, ( 0.5 - 43.3270732 i )
9, ( 0.5 + 48.0051508 i ) ;  19, ( 0.5 - 48.0051508 i )
10,( 0.5 + 49.7738324 i ) ;  20, ( 0.5 - 49.7738324 i )

Zeta_1(s) is the aux function of Zeta(s).
Zeta_1(s) = Pi^(-s/2).Gamma(s/2).Zeta(s)
of which:
Pi^(-s/2).Gamma(s/2) is the Euler Factor.

if we want to get the Zero of Zeta(s), we need to get the Zero of Zeta_1(1-s).
for example:
Zeta(0.5+14.1 i) --> Zeta_1[1-(0.5+14.1 i)] --> Zeta_1(0.5-14.1 i)
because getting the Zero of Zeta_1(1-s) value is more easier than the Zero of Zeta(s), and the Zeros of Zeta_1(1-s) is equal the Zeros of Zeta(s).

3, Reflecting Function

Zeta(s) = 2.Gamma(1-s).(2Pi)^(s-1).sin(Pi.s/2).Zeta(1-s)
if we use (1-s) to replace the s. then the result is swaping left and right.

Gamma(s).Gamma(1-s) = Pi/sin(s.Pi)
sin(2a) = 2sin(a).cos(a)
sin(a-b) = sin(a).cos(b) - cos(a).sin(b)

Zeta(1-s) = 2.Gamma(1-(1-s)).(2Pi)^((1-s)-1).sin(Pi.(1-s)/2).Zeta(1-(1-s))
          = 2.Gamma(s).(2Pi)^(-s).sin((Pi/2)-(Pi.s/2)).Zeta(s)                ;sin(Pi.(1-s)/2)
      = 2.Gamma(s).(2Pi)^(-s).[sin(Pi/2).cos(Pi.s/2)-cos(Pi/2).sin(Pi.s/2)].Zeta(s)    ;sin(Pi/2)=1, cos(Pi/2)=0
      = 2.Gamma(s).(2Pi)^(-s).cos(Pi.s/2).Zeta(s)                    ;Gamma(s) = Pi/[sin(s.Pi).Gamma(1-s)]
      = 2.(Pi/[sin(s.Pi).Gamma(1-s)]).(2Pi)^(-s).cos(Pi.s/2).Zeta(s)
      = ((2Pi).(2Pi)^(-s).cos(Pi.s/2)/[sin(s.Pi).Gamma(1-s)]).Zeta(s)        ;(2Pi).(2Pi)^(-s)=(2Pi)^(1-s)
      = ((2Pi)^(1-s).cos(Pi.s/2)/[sin(s.Pi).Gamma(1-s)]).Zeta(s)
      = (cos(Pi.s/2)/[(2Pi)^(s-1).sin(s.Pi).Gamma(1-s)]).Zeta(s)            ;(2Pi)^(1-s)= (2Pi)^[-(s-1)]

Zeta(1-s).(2Pi)^(s-1).sin(s.Pi).Gamma(1-s)/cos(Pi.s/2) = Zeta(s)            ;y=(a/b).x -->(b/a).y=x
[Gamma(1-s).(2Pi)^(s-1).sin(s.Pi)/cos(Pi.s/2)].Zeta(1-s) = Zeta(s)            ;

sin(s.Pi)/cos(Pi.s/2)=sin[2(s.Pi/2)]/cos(Pi.s/2)                    ;sin(2a) = 2sin(a).cos(a)
             =[2.sin(Pi.s/2).cos(Pi.s/2)]/cos(Pi.s/2)
             =2.sin(Pi.s/2)

[Gamma(1-s).(2Pi)^(s-1).sin(s.Pi)/cos(Pi.s/2)].Zeta(1-s) = Zeta(s)
[Gamma(1-s).(2Pi)^(s-1).2.sin(Pi.s/2)].Zeta(1-s) = Zeta(s)
2.Gamma(1-s).(2Pi)^(s-1).sin(Pi.s/2).Zeta(1-s) = Zeta(s)                ;swaped left and right.

4, the progress in prime number researching.
4.1 Adjacent prime numbers
2,3,5,7,11,13,17,...
3-2=1,17-13=4,...
p(n+1)-p(n) < 2460000    ;This means the max gap of any prime numbers less than 2460000.
4.2 there are the 45% nontrivial zeros on 1/2 line.(the attachment paper)

5,the features of Riemann Zeta function's nontrivial zeros
5.1 Count
Zeta(s)=Zeta(a+bi)=0
i = (-1)^0.5
s = a + bi ; a != 1, a = 1/2, real b > 0
CountZeros[-b,b] = 2[(b/2Pi).ln(b/2Pi) - (b/2Pi) + O(ln(b))]

5.2 Distribution
The distribution features of nontrivial zeros is similar to the quantum energy level.

6, more precision at the nontrivial zeros in Riemann Zeta(s) Function
a = 1/2, real b > 0
accurate to over 1000 decimal places.

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  30.424876125859513210311897530584091320181560023715440180962146036993
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