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On the existence of rational operators
#11
(12/20/2010, 08:28 PM)sheldonison Wrote:
(12/20/2010, 07:01 PM)JmsNxn Wrote: A taylor series expansion could only work if one also has a slog taylor series expansion. If you give me that I'd be happy to make a graph over domain [0, 2]. Big Grin
Here it is. Taylor series for , which will converge nicely for z in the range [0..2]. If z<0, take before generating slog(z-1)-1. If z>2, iterate , before generating slog(z-1)+n, so that z is in the range [0..2].
Code:
a0=   0.00000000000000000000000000000000
a1=   1.12439780182947880296975296510341
a2=  -0.01233408638319092919966757867732
a3=  -0.15195716580089316798328536602130
a4=   0.01868009944521288546047080416998
a5=   0.03456100685993161409190280063892
a6=  -0.00907417008961111769380973532974
a7=  -0.00882611191544351225979374105298
a8=   0.00382451721437283174576066832193
a9=   0.00228031089800741907723932214202
a10= -0.00151922346582286239408053757523
a11= -0.00055532576725948556607647219741
a12=  0.00058068759568845128571208222568
a13=  0.00011333875202118827889934233768
a14= -0.00021492130432643551427679982642
a15= -0.00001121186618462451210489139936
a16=  0.00007707957627653141354330216317
a17= -0.00000624892419462078938069406186
a18= -0.00002671099173826526447600018191
a19=  0.00000581456717530582001598703419
a20=  0.00000888533730151933862945265998
a21= -0.00000330763773421352876145208923
a22= -0.00000280211888032738989276581338
a23=  0.00000159184385525029832990193555
a24=  0.00000081754010898099012004646318
a25= -0.00000069935182173423145339560199
a26= -0.00000020858066529691830195782405
a27=  0.00000028862436851123339303428296
a28=  0.00000003862884977802212289870391
a29= -0.00000011328157592267567824609526
a30=  0.00000000094217657182114853689258
a31=  0.00000004247233495866309956742421
a32= -0.00000000615478181186900908929094
a33= -0.00000001519770422468823619166244
a34=  0.00000000440788391865597168670175
a35=  0.00000000515674874286180172029316
a36= -0.00000000238020450731772920188547
a37= -0.00000000163450373305911517165825
a38=  0.00000000113080753816867247217337
a39=  0.00000000046806124793717393704457
a40= -0.00000000049617938942328314300887
a41= -0.00000000011074640735137445965583
a42=  0.00000000020520322530692159331762
a43=  0.00000000001419679684872395334262
a44= -0.00000000008069713894112959557995
a45=  0.00000000000566866105603014575315
a46=  0.00000000003024789945770100323766
a47= -0.00000000000635476323844608537350
a48= -0.00000000001077582550414236272297
a49=  0.00000000000393825972677957845029
a50=  0.00000000000361455077455751078451
a51= -0.00000000000202059080878299162503
a52= -0.00000000000111781235812744791900
a53=  0.00000000000093621058575542170278
a54=  0.00000000000030319231965398624547
a55= -0.00000000000040462836021383451852
a56= -0.00000000000006151258021196048097
a57=  0.00000000000016547055506890328371
a58=  0.00000000000000095303168930585396
a59= -0.00000000000006439584830686473714
a60=  0.00000000000000855544017414425242
a61=  0.00000000000002385199982356358626
a62= -0.00000000000000663620464544180588
a63= -0.00000000000000836346245035952446
a64=  0.00000000000000374483888620781880
a65=  0.00000000000000273889585903612029
a66= -0.00000000000000184001327491851237
a67= -0.00000000000000081266892359162900
a68=  0.00000000000000083079153254456575
a69=  0.00000000000000020181213382841263
a70= -0.00000000000000035247402372018293
a71= -0.00000000000000002985852756369760
a72=  0.00000000000000014190787623066308
a73= -0.00000000000000000805476244793420
a74= -0.00000000000000005438469969331962
a75=  0.00000000000000001071535590487842
a76=  0.00000000000000001979662010087425
a77= -0.00000000000000000694989588600789
a78= -0.00000000000000000678704531046374
a79=  0.00000000000000000366768107666540
a80=  0.00000000000000000214963783636005

window screen [xmin = 0, xmax =2, ymin=0, ymax = 10]
[Image: 2q3.png]

I'm not liking the way this looks either (don't worry, I used the same algorithm you defined). Are there any other ways to extend tetration that I can try? Hopefully one of them will make the graph without any angles. If not, I guess it just expresses a very odd connection. The function 2 {x} 3 is defined piecewise, so I guess it's just the way it is :/.
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Messages In This Thread
RE: On the existence of rational operators - by JmsNxn - 12/20/2010, 09:38 PM

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