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Searching for an asymptotic to exp[0.5]
(10/08/2015, 08:41 AM)tommy1729 Wrote:
(10/08/2015, 03:41 AM)sheldonison Wrote:




Yes but this exp ( - n^2 / 4 ) is far from Jay's 2 ^ ( - n (n-1) ) / n !

Its a different base ; exp(- 1/4) =\= 2^(-1).

So this is the worst fit , rather then the best ?

It seems to disprove the conjectures ?!

Or do i need less or more medication ?

Regards

Tommy1729

It is a different base; I interpolated as per your earlier post. I leave it as a home work problem for you to generate the fake function for , which should be closer to Jay's series, in the sense that I think as x gets arbitrarily large, for J(x). But this is a very rough order of magnitude approximation of J(x). If you like, I can post the results for the fake function for Jay's series exactly, but it would be a numerical Taylor series, not a nice closed form function.

Ok, here are the numerical approximations for the ratio of the Gaussian approximation for fake_a_n/a_n, for J(x). I would expect that as n gets arbitrarily large, this would go to a constant, but not get to exactly 1, unlike the case for the fake function for exp(z) or exp^{0.5}(z), where the ratio gets arbitrarily close to 1. The error term is going to be the ratio of the Gaussian approximation over the exact integral, which is complicated, but for this function, we think g'' goes to a constant, instead of getting arbitrarily small. So the error term involves the integral going from -infinity to infinity instead of the exact value of the integral from -pi to pi. Also the exact a_n integral includes all of the derivatives of g(z), not just the g'' approximation with all higher derivatives zero.

My "guess" for the ratio as n goes to infinity: 1.00000000002528325425; this is the limiting ratio as n goes to infinity for the fake function for . Now take the entire function, fake(x)=~f(x); and generate fake2(x) from fake(x), and that is how you get this limiting ratio. More later, only if there is interest.

(10/08/2015, 12:26 PM)tommy1729 Wrote: Sheldon , in your link you apparantly considered similar things.

But what is that about Laurent series ?
You mention Laurent and then you drop the negative terms ??
If one includes the full Laurent series with negative terms, then there is a closed form for fake(x)-f(x), which might help with proofs. The negative Laurent series terms, terms cause a singularity at zero, but otherwise quickly become insignificant; elsewhere the full Laurent series converges. One might be able to prove the "guessed ratio" above.
Code:
n  ratio of fake_a_n/a_n for J(x)
1 1.048770528303
2 1.013850038715
3 1.005968616357
4 1.003144747621
5 1.001868901379
6 1.001204340672
7 1.000822844603
8 1.000587713520
9 1.000434687401
10 1.000330708190
11 1.000257534636
12 1.000204519322
13 1.000165153630
14 1.000135302082
15 1.000112249218
16 1.000094160750
17 1.000079766664
18 1.000068168339
19 1.000058717592
20 1.000050938776
21 1.000044477428
22 1.000039065682
23 1.000034498530
24 1.000030617249
25 1.000027297641
26 1.000024441575
27 1.000021970809
28 1.000019822446
29 1.000017945527
30 1.000016298466
31 1.000014847101
32 1.000013563191
33 1.000012423249
34 1.000011407636
35 1.000010499843
36 1.000009685927
37 1.000008954052
38 1.000008294132
39 1.000007697532
40 1.000007156832
41 1.000006665634
42 1.000006218397
43 1.000005810309
44 1.000005437176
45 1.000005095334
46 1.000004781568
47 1.000004493054
48 1.000004227300
49 1.000003982105
50 1.000003755518
51 1.000003545807
52 1.000003351428
53 1.000003171005
54 1.000003003307
55 1.000002847233
56 1.000002701791
57 1.000002566091
58 1.000002439331
59 1.000002320786
60 1.000002209802
61 1.000002105785
62 1.000002008196
63 1.000001916547
64 1.000001830392
65 1.000001749326
66 1.000001672978
67 1.000001601010
68 1.000001533112
69 1.000001469001
70 1.000001408415
71 1.000001351117
72 1.000001296886
73 1.000001245519
74 1.000001196830
75 1.000001150646
76 1.000001106809
77 1.000001065171
78 1.000001025596
79 1.000000987958
80 1.000000952139
81 1.000000918031
82 1.000000885534
83 1.000000854553
84 1.000000825000
85 1.000000796795
86 1.000000769861
87 1.000000744128
88 1.000000719529
89 1.000000696003
90 1.000000673491
91 1.000000651940
92 1.000000631299
93 1.000000611520
94 1.000000592559
95 1.000000574374
96 1.000000556926
97 1.000000540177
98 1.000000524094
99 1.000000508642
100 1.000000493793
101 1.000000479516
102 1.000000465784
103 1.000000452571
104 1.000000439854
105 1.000000427608
106 1.000000415814
107 1.000000404449
108 1.000000393494
109 1.000000382932
110 1.000000372744
111 1.000000362915
112 1.000000353428
113 1.000000344269
114 1.000000335424
115 1.000000326879
116 1.000000318622
117 1.000000310641
118 1.000000302925
119 1.000000295462
120 1.000000288242
121 1.000000281255
122 1.000000274493
123 1.000000267945
124 1.000000261604
125 1.000000255462
126 1.000000249511
127 1.000000243743
128 1.000000238151
129 1.000000232729
130 1.000000227471
131 1.000000222370
132 1.000000217420
133 1.000000212616
134 1.000000207953
135 1.000000203425
136 1.000000199028
137 1.000000194756
138 1.000000190606
139 1.000000186573
140 1.000000182653
141 1.000000178842
142 1.000000175136
143 1.000000171532
144 1.000000168026
145 1.000000164616
146 1.000000161296
147 1.000000158066
148 1.000000154921
149 1.000000151859
150 1.000000148878
151 1.000000145973
152 1.000000143144
153 1.000000140388
154 1.000000137702
155 1.000000135084
156 1.000000132532
157 1.000000130044
158 1.000000127618
159 1.000000125251
160 1.000000122943
161 1.000000120692
162 1.000000118495
163 1.000000116351
164 1.000000114258
165 1.000000112215
166 1.000000110221
167 1.000000108274
168 1.000000106372
169 1.000000104515
170 1.000000102701
171 1.000000100928
172 1.000000099196
173 1.000000097503
174 1.000000095849
175 1.000000094232
176 1.000000092651
177 1.000000091105
178 1.000000089594
179 1.000000088116
180 1.000000086670
181 1.000000085255
182 1.000000083871
183 1.000000082517
184 1.000000081192
185 1.000000079896
186 1.000000078626
187 1.000000077384
188 1.000000076167
189 1.000000074976
190 1.000000073810
191 1.000000072667
192 1.000000071549
193 1.000000070453
194 1.000000069379
195 1.000000068327
196 1.000000067296
197 1.000000066286
198 1.000000065296
199 1.000000064325
200 1.000000063374
201 1.000000062441
202 1.000000061527
203 1.000000060630
204 1.000000059751
205 1.000000058889
206 1.000000058043
207 1.000000057213
208 1.000000056399
209 1.000000055601
210 1.000000054817
211 1.000000054048
212 1.000000053294
213 1.000000052553
214 1.000000051826
215 1.000000051113
216 1.000000050412
217 1.000000049724
218 1.000000049049
219 1.000000048386
220 1.000000047735
221 1.000000047095
222 1.000000046467
223 1.000000045850
224 1.000000045244
225 1.000000044648
226 1.000000044063
227 1.000000043488
228 1.000000042923
229 1.000000042368
230 1.000000041822
231 1.000000041286
232 1.000000040759
233 1.000000040240
234 1.000000039731
235 1.000000039230
236 1.000000038737
237 1.000000038253
238 1.000000037777
239 1.000000037308
240 1.000000036848
241 1.000000036394
242 1.000000035949
243 1.000000035510
244 1.000000035079
245 1.000000034655
246 1.000000034237
247 1.000000033826
248 1.000000033422
249 1.000000033024
250 1.000000032632
251 1.000000032247
252 1.000000031867
253 1.000000031494
254 1.000000031126
255 1.000000030764
256 1.000000030408
257 1.000000030057
258 1.000000029712
259 1.000000029372
260 1.000000029037
261 1.000000028707
262 1.000000028382
263 1.000000028062
264 1.000000027746
265 1.000000027436
266 1.000000027130
267 1.000000026828
268 1.000000026532
269 1.000000026239
270 1.000000025951
271 1.000000025667
272 1.000000025387
273 1.000000025111
274 1.000000024839
275 1.000000024571
276 1.000000024307
277 1.000000024047
278 1.000000023790
279 1.000000023537
280 1.000000023287
281 1.000000023042
282 1.000000022799
283 1.000000022560
284 1.000000022324
285 1.000000022092
286 1.000000021863
287 1.000000021637
288 1.000000021414
289 1.000000021194
290 1.000000020977
291 1.000000020763
292 1.000000020552
293 1.000000020344
294 1.000000020138
295 1.000000019936
296 1.000000019736
297 1.000000019539
298 1.000000019344
299 1.000000019152
300 1.000000018963
301 1.000000018776
302 1.000000018591
303 1.000000018409
304 1.000000018229
305 1.000000018052
306 1.000000017877
307 1.000000017704
308 1.000000017533
309 1.000000017365
310 1.000000017198
311 1.000000017034
312 1.000000016872
313 1.000000016712
314 1.000000016554
315 1.000000016398
316 1.000000016244
317 1.000000016092
318 1.000000015942
319 1.000000015793
320 1.000000015647
321 1.000000015502
322 1.000000015359
323 1.000000015218
324 1.000000015078
325 1.000000014941
326 1.000000014804
327 1.000000014670
328 1.000000014537
329 1.000000014406
330 1.000000014276
331 1.000000014148
332 1.000000014022
333 1.000000013897
334 1.000000013773
335 1.000000013651
336 1.000000013530
337 1.000000013411
338 1.000000013293
339 1.000000013177
340 1.000000013062
341 1.000000012948
342 1.000000012836
343 1.000000012725
344 1.000000012615
345 1.000000012506
346 1.000000012399
347 1.000000012293
348 1.000000012188
349 1.000000012084
350 1.000000011982
351 1.000000011880
352 1.000000011780
353 1.000000011681
354 1.000000011583
355 1.000000011486
356 1.000000011390
357 1.000000011295
358 1.000000011202
359 1.000000011109
360 1.000000011017
361 1.000000010927
362 1.000000010837
363 1.000000010748
364 1.000000010661
365 1.000000010574
366 1.000000010488
367 1.000000010403
368 1.000000010319
369 1.000000010236
370 1.000000010154
371 1.000000010073
372 1.000000009993
373 1.000000009913
374 1.000000009834
375 1.000000009756
376 1.000000009679
377 1.000000009603
378 1.000000009528
379 1.000000009453
380 1.000000009379
381 1.000000009306
382 1.000000009234
383 1.000000009162
384 1.000000009091
385 1.000000009021
386 1.000000008952
387 1.000000008883
388 1.000000008815
389 1.000000008748
390 1.000000008681
391 1.000000008615
392 1.000000008550
393 1.000000008485
394 1.000000008421
395 1.000000008358
396 1.000000008295
397 1.000000008233
398 1.000000008172
399 1.000000008111
400 1.000000008051

- Sheldon
Reply
Thanks for the reply.

I Will Try to finish my homework today but im only at 2% of my potential.

Homework sounds a bit ... Oh well.

Anyway exp( (ln(x)^2) / 2 ln(2) ) is indeed a better estimate.

Why are you so carefull with the exact integral ?
Cant we simply express it with erf or similar special functions ?
I havent investigated it yet, but you seem to hold back or so.
Not sure about the how and why here.

How you computed your limit is a mystery to me.

Something to do with iterating the gaussian method ?

Reminds me of the Tommy-Sheldon iterations.

I Will now take my medication and do my other homework first ( number theory ).

Im not a student though.
There is no PhD in fake function theory yet.
And your not my professor.

But thanks for the post.

Regards

Tommy1729

Reply
I edited post 150 , where I mentioned the tommy-sheldon iterations.

Typo's , mistakes and confusion should be gone.

Although the convergeance conjecture disagrees with sheldon's recent not 1 ratio ... Maybe ...

Things should be clear now.


Regards

Tommy1729
Reply
Dont you just hate it when you Try to prove

D < 1.

And on day 1 you prove D < ln(n).

Day 2 : D < ln^[2] (n)

Day 3 : D < ln^[3] (n)

Etc

Oh Well


Regards

Tommy1729
Reply
I was thinking about g ' ' (h_n).

To avoid it being too small for the gaussian method , I invented the Tommy-Sheldon iterations.

However perhaps computing a fake g works better.

Although computing a fake g with the gaussian could result in " the same problem with g_2 ".
( the Tommy-Sheldon iterations do not have that problem )


Or maybe combining those.

Hmm.

Regards

Tommy1729
Reply
(10/08/2015, 11:08 PM)tommy1729 Wrote: I edited post 150 , where I mentioned the tommy-sheldon iterations.

Typo's , mistakes and confusion should be gone.

Although the convergeance conjecture disagrees with sheldon's recent not 1 ratio ... Maybe ...

Things should be clear now.


Regards

Tommy1729

Start with

g''(x) for f(x) is conjectured to approach exactly g''(x) for J(x), as x gets arbitrarily large, which is the initial reason for choosing this particular f(x). But f(x) is interesting on its own.

I generate the fake function for f(x) using the Gaussian method. I assume that this is what Tommy means in post#150 by F_2(x).
> F_2(x) = F_1(x) • sqrt( 2 pi G_1 '' (h_n) )

where for f(x) optionally
Is this the same as Tommy's F_2? Tommy has the g'' in the numerator which is a typo. It looks like Tommy's F_3 would be the fake function for F_2(x)? I'm not sure if that's what Tommy intended or not.

f2(x) is the fake Gaussian approximation for f(x)

Now, starting with let's generate
From there, we can generate a new set of values, which should be nearly identical to the original set of h_n values, and a new set of values which should be nearly identical to the values.

I did this numerically. I think I might be able to generate a closed form equation for this new ratio result, using the equations from post#85. I guess this ratio not going to a limiting value of 1 is a contradiction for your conjecture. Its actually rather interesting, especially if you consider the ratio for non integer values of a_n; n=20.5; vs the equation above. I can post more later if interested; it turns out we have a sine wave oscillating around Anyway, assuming g''(x) for f(x) is asymptotically the same as g''(x) for J(x) as x gets arbitrarily large, I expect the limiting ratio for J(x) to be the same as the limiting ratio below, as n gets arbitrarily large.
Code:
ratio of b_n over a_n where f2(x) is the function from above
1 1.13160761703913345046
2 1.03756115378093045262
3 1.00584054797835817399
4 1.00042570875240853058
5 1.00001412678446418263
6 1.00000021835990293497
7 1.00000000163026326669
8 1.00000000003096900943
9 1.00000000002529304393
10 1.00000000002528326249
11 1.00000000002528325425
12 1.00000000002528325425
13 1.00000000002528325425
14 1.00000000002528325425
15 1.00000000002528325425
16 1.00000000002528325425
17 1.00000000002528325425
18 1.00000000002528325425
19 1.00000000002528325425
20 1.00000000002528325425
21 1.00000000002528325425
22 1.00000000002528325425
23 1.00000000002528325425
24 1.00000000002528325425
25 1.00000000002528325425
26 1.00000000002528325425
27 1.00000000002528325425
28 1.00000000002528325425
29 1.00000000002528325425
30 1.00000000002528325425
- Sheldon
Reply
(10/09/2015, 08:15 AM)sheldonison Wrote:
(10/08/2015, 11:08 PM)tommy1729 Wrote: I edited post 150 , where I mentioned the tommy-sheldon iterations.

Typo's , mistakes and confusion should be gone.

Although the convergeance conjecture disagrees with sheldon's recent not 1 ratio ... Maybe ...

Things should be clear now.


Regards

Tommy1729

Start with

g''(x) for f(x) is conjectured to approach exactly g''(x) for J(x), as x gets arbitrarily large, which is the initial reason for choosing this particular f(x). But f(x) is interesting on its own.

I generate the fake function for f(x) using the Gaussian method. I assume that this is what Tommy means in post#150 by F_2(x).
> F_2(x) = F_1(x) • sqrt( 2 pi G_1 '' (h_n) )

where for f(x) optionally
Is this the same as Tommy's F_2? Tommy has the g'' in the numerator which is a typo. It looks like Tommy's F_3 would be the fake function for F_2(x)? I'm not sure if that's what Tommy intended or not.

f2(x) is the fake Gaussian approximation for f(x)

Now, starting with let's generate
From there, we can generate a new set of values, which should be nearly identical to the original set of h_n values, and a new set of values which should be nearly identical to the values.

I did this numerically. I think I might be able to generate a closed form equation for this new ratio result, using the equations from post#85. I guess this ratio not going to a limiting value of 1 is a contradiction for your conjecture. Its actually rather interesting, especially if you consider the ratio for non integer values of a_n; n=20.5; vs the equation above. I can post more later if interested; it turns out we have a sine wave oscillating around Anyway, assuming g''(x) for f(x) is asymptotically the same as g''(x) for J(x) as x gets arbitrarily large, I expect the limiting ratio for J(x) to be the same as the limiting ratio below, as n gets arbitrarily large.
Code:
ratio of b_n over a_n where f2(x) is the function from above
1 1.13160761703913345046
2 1.03756115378093045262
3 1.00584054797835817399
4 1.00042570875240853058
5 1.00001412678446418263
6 1.00000021835990293497
7 1.00000000163026326669
8 1.00000000003096900943
9 1.00000000002529304393
10 1.00000000002528326249
11 1.00000000002528325425
12 1.00000000002528325425
13 1.00000000002528325425
14 1.00000000002528325425
15 1.00000000002528325425
16 1.00000000002528325425
17 1.00000000002528325425
18 1.00000000002528325425
19 1.00000000002528325425
20 1.00000000002528325425
21 1.00000000002528325425
22 1.00000000002528325425
23 1.00000000002528325425
24 1.00000000002528325425
25 1.00000000002528325425
26 1.00000000002528325425
27 1.00000000002528325425
28 1.00000000002528325425
29 1.00000000002528325425
30 1.00000000002528325425

Wait.

If a_n ~~ exp(g(h_n) - n h_n)

Is An underestimate ( as you say )

And the gaussian is beter then

Gaussian > exp( g(h_n) - n h_n).

Right ?

So gaussian can not be exp ... / sqrt( 2 pi g" (h_n)).

It has to be exp( g(h_n) - n h_n) * sqrt( 2 pi g " (h_n) ).

Or am I crazy today ?

Regards

Tommy1729
Reply
(10/09/2015, 11:56 AM)tommy1729 Wrote: So gaussian can not be exp ... / sqrt( 2 pi g" (h_n)).

It has to be exp( g(h_n) - n h_n) * sqrt( 2 pi g " (h_n) ).

Start with

I think your just missing a little bit of algebra. We are interested an approximation for
This is an exact equation if

see post#16 for more details on getting to this step







All of the numerical approximations I have posted, use this version of the Gaussian approximation, and get excellent results. In most cases, the Gaussian approximation is also an over-approximation of an entire function with all positive derivatives.
- Sheldon
Reply
(10/10/2015, 03:08 AM)sheldonison Wrote:
(10/09/2015, 11:56 AM)tommy1729 Wrote: So gaussian can not be exp ... / sqrt( 2 pi g" (h_n)).

It has to be exp( g(h_n) - n h_n) * sqrt( 2 pi g " (h_n) ).

Start with

I think your just missing a little bit of algebra. We are interested an approximation for
This is an exact equation if

see post#16 for more details on getting to this step







All of the numerical approximations I have posted, use this version of the Gaussian approximation, and get excellent results. In most cases, the Gaussian approximation is also an over-approximation of an entire function with all positive derivatives.

Hmm.

The reason I got confused is

1) i incorrectly thought 1/n! ~ sqrt(2 pi n) (e/n)^n.
See the fake exp.


2) I assumed the gaussian was bigger then S9 because

2 a) the Gaussian is better then S9.

2 b) You Said in post 9 that the method ( S9 ) always gives An underestimate.

So apparantly 2 b) is false.

This was not mentioned before !

Im sorry. Big misunderstanding.

Did I understand my misunderstanding Well ? Smile

Since 2 b) is false, does this imply there exist functions equal to their fake , independent of the method for fake ?
I guess so.
So we could solve S9( f(x) ) = f(x) ?

Intresting.

I need to read the entire thread again.

Regards

Tommy1729


Reply
Despite my recent confusions and mistakes , I edited the Tommy-Sheldon equations.

I still believe they are the best so far.

Regards

Tommy1729
Reply


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