Taylor series of i[x]
#9
Okey, I programmed it and I got the next graph for y = Y(1;x)*(1 xor x):
https://www.dropbox.com/s/81rxif99w7j50u...D.bmp?dl=0

And this is my PARI/gp code:
Code:
twist(p,q)={
L=ceil(1.4426950408889634*log((max(q+1,p+1))));
res=1;sta=1;
if(round(p)!=p||round(q)!=q,p*=2^60;q*=2^60;L+=60);
p=round(p);q=round(q);
T=[1,2,3,4;2,2,5,6;3,6,3,5;4,5,6,4;5,6,6,6;6,5,5,5];
for(i=0,L,j=L-i-1;k=2*bittest(p,j)+bittest(q,j)+1;sta=T[sta,k]);
if(sta==4||sta==6,return(-1),return(1));
}
xor(x,y)={return(bitxor(round(x*2^60),round(y*2^60))/2^60)}


I checked it in a really littleĀ interval (from 0 to 10^-8 ) and it does not look like something which is differentialable. So should I say it has not a derivative and Taylor series?
Anyway, can anyone determine the integral of it from 0 to x? It would be really interesting, I think so.
Xorter Unizo
Reply


Messages In This Thread
Taylor series of i[x] - by Xorter - 01/12/2017, 04:50 PM
RE: Taylor series of i[x] - by sheldonison - 01/12/2017, 08:50 PM
RE: Taylor series of i[x] - by Xorter - 01/13/2017, 05:26 PM
RE: Taylor series of i[x] - by Xorter - 01/13/2017, 07:13 PM
RE: Taylor series of i[x] - by Xorter - 01/14/2017, 10:14 AM
RE: Taylor series of i[x] - by mike3 - 01/23/2017, 07:38 AM
RE: Taylor series of i[x] - by Xorter - 02/26/2017, 11:10 AM
RE: Taylor series of i[x] - by Xorter - 03/01/2017, 03:06 PM
RE: Taylor series of i[x] - by Xorter - 03/04/2017, 09:40 AM
RE: Taylor series of i[x] - by Xorter - 04/06/2017, 03:42 PM
RE: Taylor series of i[x] - by Xorter - 04/11/2017, 12:18 PM
RE: Taylor series of i[x] - by Xorter - 07/10/2017, 04:07 PM
RE: Taylor series of i[x] - by Xorter - 02/20/2018, 09:55 PM

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