Posts for OmnipotentEntity

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Posted: 7/16/2017 5:42 PM
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OmnipotentEntity
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Joined: 9/11/2004
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Plush, you do good work. But there's a number of very good reasons for the migration from Mupen. I understand it's the standard for Mario 64 TASers. But it's not really being maintained anymore, and the timing, while slightly better in an absolute scale, is far too generous, and results in a better time than possible on console. BizHawk is undeniably a more accurate emulator overall. And it's much easier for other people, like judges, encoders, and publishers, to work with. No one is saying you can't continue to personally use Mupen, or that you can't continue to use it as a benchmark for your old files. Just that if you'd like to submit the file as a movie it needs to be in the BizHawk format. I know there's a lot of drama in the SM64 TAS community. Like... all of the time. But this, at least, isn't personal. It's simply applying uniform standards, to try to make the site a better place.
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Posted: 7/9/2017 7:58 PM
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OmnipotentEntity
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(1) n is the smallest m for which p divides Fm, and p2 also divides Fn.
Does this imply that n = m? That seems nonsensical, so I must be misunderstanding. What is the relationship between n and m?
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Posted: 7/8/2017 12:15 AM
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OmnipotentEntity
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As noted by thatguy, Fibonacci numbers have the property that Fmn is divisible by Fm and Fn (but not necessarily FmFn, (F3)2 ∤ F9 is a simple counterexample). Of course, because Fibonacci numbers have exponential growth rates, Fmn is much larger than Fm and Fn and has other factors. If we consider the set of divisors of some positive integer n (not including n itself), we can generate several numbers which are divisors of Fn. If we find the LCM of these numbers, we have a number comprised of all of the "Fibonacci factors" of Fn. Dividing Fn by this number we have a number comprised of the "left Over factors." Let's call this number On. With the exception of O6 = 4 these numbers seem to be square free. Are they? A short python program and a sample run can be found at https://pastebin.com/iKjBq13C
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 7/5/2017 5:46 AM
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OmnipotentEntity
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Wouldn't master also mean the AI rubber bands more tightly, so finishing in second is faster?
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 6/30/2017 7:38 PM
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OmnipotentEntity
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Warp wrote:
This is a very easy problem mathematically speaking, but that's not the challenge. The problem is: What is the area of the red square compared to the area of the green square? The challenge is not finding the answer. The challenge is to find a geometric proof (or, could it perhaps be more accurate to say "geometric argument"?) that requires no math at all. Perhaps put your answer in spoiler tags, if other people would want to think about it as well.
The red square is half the area of the green square, you can cut the red square in half down a diagonal and place the hypotenuses on two edges of the green square to see geometrically that it covers half.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 6/12/2017 3:30 PM
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OmnipotentEntity
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No one cares about rerecord count.
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Posted: 6/9/2017 5:23 PM
(Edited: 7/7/2017 8:33 PM)
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OmnipotentEntity
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I thought it might be interesting to examine what other divisors Fmn have if we remove all of the divisors from Fm and Fn 
3 : {"2": 1}
4 : {"3": 1}
5 : {"5": 1}
6 : {"2": 2}
7 : {"13": 1}
8 : {"7": 1}
9 : {"17": 1}
10 : {"11": 1}
11 : {"89": 1}
12 : {"2": 1, "3": 1}
13 : {"233": 1}
14 : {"29": 1}
15 : {"61": 1}
16 : {"47": 1}
17 : {"1597": 1}
18 : {"19": 1}
19 : {"37": 1, "113": 1}
20 : {"41": 1}
21 : {"421": 1}
22 : {"199": 1}
23 : {"28657": 1}
24 : {"2": 1, "23": 1}
25 : {"5": 1, "3001": 1}
26 : {"521": 1}
27 : {"53": 1, "109": 1}
28 : {"281": 1}
29 : {"514229": 1}
30 : {"31": 1}
31 : {"557": 1, "2417": 1}
32 : {"2207": 1}
33 : {"19801": 1}
34 : {"3571": 1}
35 : {"141961": 1}
36 : {"3": 1, "107": 1}
37 : {"73": 1, "149": 1, "2221": 1}
38 : {"9349": 1}
39 : {"135721": 1}
40 : {"2161": 1}
41 : {"2789": 1, "59369": 1}
42 : {"211": 1}
43 : {"433494437": 1}
44 : {"43": 1, "307": 1}
45 : {"109441": 1}
46 : {"139": 1, "461": 1}
47 : {"2971215073": 1}
48 : {"2": 1, "1103": 1}
49 : {"97": 1, "6168709": 1}
50 : {"101": 1, "151": 1}
51 : {"6376021": 1}
52 : {"90481": 1}
53 : {"953": 1, "55945741": 1}
54 : {"5779": 1}
55 : {"661": 1, "474541": 1}
56 : {"7": 1, "14503": 1}
57 : {"797": 1, "54833": 1}
58 : {"59": 1, "19489": 1}
59 : {"353": 1, "2710260697": 1}
60 : {"2521": 1}
61 : {"4513": 1, "555003497": 1}
62 : {"3010349": 1}
63 : {"35239681": 1}
64 : {"1087": 1, "4481": 1}
65 : {"14736206161": 1}
66 : {"9901": 1}
67 : {"269": 1, "116849": 1, "1429913": 1}
68 : {"67": 1, "63443": 1}
69 : {"137": 1, "829": 1, "18077": 1}
70 : {"71": 1, "911": 1}
71 : {"6673": 1, "46165371073": 1}
72 : {"103681": 1}
73 : {"9375829": 1, "86020717": 1}
74 : {"54018521": 1}
75 : {"230686501": 1}
76 : {"29134601": 1}
77 : {"988681": 1, "4832521": 1}
78 : {"79": 1, "859": 1}
79 : {"157": 1, "92180471494753": 1}
80 : {"1601": 1, "3041": 1}
81 : {"2269": 1, "4373": 1, "19441": 1}
82 : {"370248451": 1}
83 : {"99194853094755497": 1}
84 : {"83": 1, "1427": 1}
85 : {"9521": 1, "3415914041": 1}
86 : {"6709": 1, "144481": 1}
87 : {"173": 1, "3821263937": 1}
88 : {"263": 1, "881": 1, "967": 1}
89 : {"1069": 1, "1665088321800481": 1}
90 : {"181": 1, "541": 1}
91 : {"13": 1, "741469": 1, "159607993": 1}
92 : {"4969": 1, "275449": 1}
93 : {"4531100550901": 1}
94 : {"6643838879": 1}
95 : {"761": 1, "29641": 1, "67735001": 1}
96 : {"2": 1, "769": 1, "3167": 1}
97 : {"193": 1, "389": 1, "3084989": 1, "361040209": 1}
98 : {"599786069": 1}
99 : {"197": 1, "18546805133": 1}
100 : {"401": 1, "570601": 1}
101 : {"743519377": 1, "770857978613": 1}
102 : {"919": 1, "3469": 1}
103 : {"519121": 1, "5644193": 1, "512119709": 1}
104 : {"103": 1, "102193207": 1}
105 : {"8288823481": 1}
106 : {"119218851371": 1}
107 : {"1247833": 1, "8242065050061761": 1}
108 : {"3": 1, "11128427": 1}
109 : {"827728777": 1, "32529675488417": 1}
110 : {"11": 1, "331": 1, "39161": 1}
111 : {"1459000305513721": 1}
112 : {"10745088481": 1}
113 : {"677": 1, "272602401466814027129": 1}
114 : {"229": 1, "95419": 1}
115 : {"1381": 1, "2441738887963981": 1}
116 : {"347": 1, "1270083883": 1}
117 : {"29717": 1, "39589685693": 1}
118 : {"709": 1, "8969": 1, "336419": 1}
119 : {"159512939815855788121": 1}
120 : {"241": 1, "20641": 1}
121 : {"97415813466381445596089": 1}
122 : {"5600748293801": 1}
123 : {"68541957733949701": 1}
124 : {"3020733700601": 1}
125 : {"5": 1, "158414167964045700001": 1}
126 : {"1009": 1, "31249": 1}
127 : {"27941": 1, "5568053048227732210073": 1}
128 : {"127": 1, "186812208641": 1}
129 : {"257": 1, "5417": 1, "8513": 1, "39639893": 1}
130 : {"131": 1, "2081": 1, "24571": 1}
131 : {"1066340417491710595814572169": 1}
132 : {"261399601": 1}
133 : {"3457": 1, "42293": 1, "351301301942501": 1}
134 : {"4021": 1, "24994118449": 1}
135 : {"1114769954367361": 1}
136 : {"23230657239121": 1}
137 : {"19134702400093278081449423917": 1}
138 : {"691": 1, "1485571": 1}
139 : {"277": 1, "2114537501": 1, "85526722937689093": 1}
140 : {"12317523121": 1}
141 : {"108289": 1, "1435097": 1, "142017737": 1}
142 : {"688846502588399": 1}
143 : {"8581": 1, "1929584153756850496621": 1}
144 : {"10749957121": 1}
145 : {"349619996930737079890201": 1}
146 : {"151549": 1, "11899937029": 1}
147 : {"293": 1, "3529": 1, "347502052673": 1}
148 : {"11987": 1, "81143477963": 1}
149 : {"110557": 1, "162709": 1, "4000949": 1, "85607646594577": 1}
150 : {"12301": 1, "18451": 1}
151 : {"5737": 1, "2811666624525811646469915877": 1}
152 : {"1091346396980401": 1}
153 : {"17": 1, "7175323114950564593": 1}
154 : {"229769": 1, "9321929": 1}
155 : {"21701": 1, "12370533881": 1, "61182778621": 1}
156 : {"12280217041": 1}
157 : {"313": 1, "11617": 1, "7636481": 1, "10424204306491346737": 1}
158 : {"32361122672259149": 1}
159 : {"317": 1, "97639037": 1, "229602768949": 1}
160 : {"23725145626561": 1}
161 : {"8693": 1, "612606107755058997065597": 1}
162 : {"3079": 1, "62650261": 1}
163 : {"977": 1, "4892609": 1, "33365519393": 1, "32566223208133": 1}
164 : {"163": 1, "800483": 1, "350207569": 1}
165 : {"86461": 1, "518101": 1, "900241": 1}
166 : {"35761381": 1, "6202401259": 1}
167 : {"18104700793": 1, "1966344318693345608565721": 1}
168 : {"167": 1, "65740583": 1}
169 : {"337": 1, "89909": 1, "104600155609": 1, "126213229732669": 1}
170 : {"1158551": 1, "12760031": 1}
171 : {"6841": 1, "5741461760879844361": 1}
172 : {"313195711516578281": 1}
173 : {"1639343785721": 1, "389678749007629271532733": 1}
174 : {"349": 1, "947104099": 1}
175 : {"701": 1, "17231203730201189308301": 1}
176 : {"93058241": 1, "562418561": 1}
177 : {"2191261": 1, "805134061": 1, "1297027681": 1}
178 : {"179": 1, "22235502640988369": 1}
179 : {"21481": 1, "156089": 1, "3418816640903898929534613769": 1}
180 : {"10783342081": 1}
181 : {"8689": 1, "422453": 1, "8175789237238547574551461093": 1}
182 : {"689667151970161": 1}
183 : {"1097": 1, "14297347971975757800833": 1}
184 : {"253367": 1, "9506372193863": 1}
185 : {"1702945513191305556907097618161": 1}
186 : {"63799": 1, "35510749": 1}
187 : {"373": 1, "10157807305963434099105034917037": 1}
188 : {"563": 1, "5641": 1, "4632894751907": 1}
189 : {"38933": 1, "955921950316735037": 1}
190 : {"191": 1, "41611": 1, "87382901": 1}
191 : {"4870723671313": 1, "757810806256989128439975793": 1}
192 : {"2": 1, "11862575248703": 1}
193 : {"9465278929": 1, "1020930432032326933976826008497": 1}
194 : {"3299": 1, "56678557502141579": 1}
195 : {"88999250837499877681": 1}
196 : {"5881": 1, "61025309469041": 1}
197 : {"15761": 1, "25795969": 1, "227150265697": 1, "717185107125886549": 1}
198 : {"991": 1, "2179": 1, "1513909": 1}
199 : {"397": 1, "436782169201002048261171378550055269633": 1}
200 : {"9125201": 1, "5738108801": 1}
201 : {"5050260704396247169315999021": 1}
202 : {"809": 1, "7879": 1, "201062946718741": 1}
203 : {"1217": 1, "56470541": 1, "2586982700656733994659533": 1}
204 : {"409": 1, "66265118449": 1}
205 : {"821": 1, "125598581": 1, "36448117857891321536401": 1}
206 : {"619": 1, "1031": 1, "5257480026438961": 1}
207 : {"4072353155773627601222196481": 1}
208 : {"3329": 1, "106513889": 1, "325759201": 1}
209 : {"57314120955051297736679165379998262001": 1}
210 : {"21211": 1, "767131": 1}
211 : {"22504837": 1, "38490197": 1, "800972881": 1, "80475423858449593021": 1}
212 : {"1483": 1, "2969": 1, "1076012367720403": 1}
213 : {"1277": 1, "185790722054921374395775013": 1}
214 : {"47927441": 1, "479836483312919": 1}
215 : {"2607553541": 1, "67712817361580804952011621": 1}
216 : {"6263": 1, "177962167367": 1}
217 : {"433": 1, "44269": 1, "217221773": 1, "2191174861": 1, "6274653314021": 1}
218 : {"128621": 1, "788071": 1, "593985111211": 1}
219 : {"123953": 1, "4139537": 1, "3169251245945843761": 1}
220 : {"59996854928656801": 1}
221 : {"203572412497": 1, "90657498718024645326392940193": 1}
222 : {"4441": 1, "146521": 1, "1121101": 1}
223 : {"4013": 1, "108377": 1, "251534189": 1, "164344610046410138896156070813": 1}
224 : {"223": 1, "449": 1, "1154149773784223": 1}
225 : {"11981661982050957053616001": 1}
226 : {"412670427844921037470771": 1}
227 : {"23609": 1, "5219534137983025159078847113619467285727377": 1}
228 : {"227": 1, "26449": 1, "212067587": 1}
229 : {"457": 1, "2749": 1, "40487201": 1, "132605449901": 1, "47831560297620361798553": 1}
230 : {"1151": 1, "5981": 1, "324301": 1, "686551": 1}
231 : {"9164259601748159235188401": 1}
232 : {"299281": 1, "834428410879506721": 1}
I dunno. There do seem to be more very large primes than I expected. Something I found surprising while writing this program. Fm2 is divisible by Fm but not necessarily by Fm2. The first counter example is F9.
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Posted: 6/6/2017 12:52 PM
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OmnipotentEntity
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thatguy wrote:
Here's a more interesting twin prime problem: prove that the only Fibonacci numbers that are also twin primes are 3, 5 and 13. (I can remember the result, and I have been shown a proof, but I can't remember the details and Google isn't helping. But I think it relies pretty heavily on recurrence formulas for the Fibonacci numbers, and for Fn+-2, that enable you to factorise them.)
The solution is behind a paywall.
Warp wrote:
Btw, in a related note, is it hard to prove how many Fibonacci numbers are prime (as in, are there infinitely many of them)?
Hard enough that it hasn't been done yet.
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Posted: 6/2/2017 4:28 PM
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OmnipotentEntity
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Bumping to add, v1.50 came out recently and added/changed a ton of stuff. So if anyone is interesting in the surely mindnumbing work that updating this TAS would be, there you go.
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Posted: 5/31/2017 11:59 PM
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OmnipotentEntity
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Warp wrote:
In other words is the derivative of "(1/2)*x*f(x) + (1/4)*sinh-1(2x)" "f(x)"?
No, absolutely not.
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Posted: 5/31/2017 8:27 PM
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OmnipotentEntity
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So, the movie seems technically sound, to the extent of my knowledge of the game. And I gave it a yes vote. I supposed I'll lead this controversy off... Is this a secret warp intended to be found in normal play or a debugging code leftover from final boss testing? And if it is a debugging code, does it run afoul with the rule, or is this a case where this rule is "not strict" and does not necessarily apply.
Rules.html wrote:
Cheats, debugging codes, and arcade continues are not allowed This includes any input sequences such as the Konami Code, as well as immediately accessible hidden menus. Note that, if the button sequence is mentioned in the manual as a normal means of playing, such as level restart shortcuts in the Legend of Zelda or Metroid, it is usually allowed. Additionally, buying continues with coins in arcade games is considered to be a cheat-like practice, as it provides the player with a free and virtually unlimited power resource, and as such goes against the typical concept of a TAS. These rules are not strict, but are motivated by the same concept as the guideline that says you should play on the hardest difficulty. As such, you can use a code to unlock the hardest difficulty, although it's better to first ask on the forums if this is a good idea. Indeed, sometimes it may make the movie worse!
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Posted: 5/31/2017 8:17 PM
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OmnipotentEntity
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I'm not sure what you're asking.
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Posted: 5/31/2017 3:31 PM
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OmnipotentEntity
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Warp wrote:
FractalFusion wrote:
f(x) = (1/2)*x*sqrt(4x2 + 1) + (1/4)*sinh-1(2x), f'(x) = sqrt(4x2 + 1)
The second expression is the derivative of the first one? That seems surprising.
My very first post was taking the anti-derivative of sqrt(4x2 + 1) and getting (1/2)*x*sqrt(4x2 + 1) + (1/4)*sinh-1(2x). Just remarking that it didn't seem surprising to you then. ;)
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Posted: 5/29/2017 6:11 PM
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OmnipotentEntity
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If you need to perform a walk in the range [-1, 1] on the original f(x) = x^2 function, then unfortunately you'll need to find a piecewise estimate to this function. For clarity, the arclength from 0 to 1 along the parabola is approx 1.47894285754... Meaning you need convergence on the domain [0, 1.47894285754...] for the inverse function in order to go from arclength traveled to x coordinate. The radius of convergence of the Taylor series is almost certainly smaller than this value, so piecewise is the only way.
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Posted: 5/29/2017 5:00 PM
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OmnipotentEntity
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More important is on what domain you want it to converge on. A Taylor series will only be accurate around a point, and will get less and less accurate around other points. Further, I think that there might be a radius of convergence with this particular function (because the coefficients seem to be increasing with each iteration). Meaning it doesn't matter how many terms you use, you won't get good estimates outside of a certain interval. The Taylor series around x=0 is x-(2 x^3)/3+(26 x^5)/15-(1972 x^7)/315+O[x]^9, which is only accurate in the domain -0.5 < x < 0.5
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Posted: 5/28/2017 7:52 PM
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OmnipotentEntity
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This one is actually pretty easy. The arc length of a curve from x in range [a,b] is given by Integral from a to b of sqrt(1 + (dy/dx)^2) dx In the case of the parabola the antiderivative of this actually does have a closed form: integral sqrt(1 + ((dx^2)/(dx))^2) dx = 1/4 (2 sqrt(4 x^2 + 1) x + sinh^(-1)(2 x)) + constant However, this function does not seem to be invertible. So, you would need to approximate it to model it and calculate the evenly spaced points.
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Posted: 5/27/2017 11:37 PM
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OmnipotentEntity
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I was looking into continued fractions, and I found this OEIS wiki page on the subject. It lists the basic continued fraction [0, 1, 2, 3, 4, 5]... as I_1(2)/I_0(2) where I_n is the modified Bessel Function of the 1st kind. This is the last place I expected to see the Bessel function crop up. Any insight on why this is? I have been unable to find a proof.
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Posted: 5/25/2017 12:48 PM
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OmnipotentEntity
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I asked about this on Math Stackexchange: https://math.stackexchange.com/questions/2294892/very-slowly-converging-formulas-for-pi/2295488#2295488
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Posted: 5/24/2017 12:38 PM
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OmnipotentEntity
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I was able to find a French paper that provides this formula as: "The slowest and heaviest formula imaginable to access pi, developed to verify a hypothesis on the volume of the sphere" I don't know how slowly it converges. It might be sub-logarithmic. A formula the converges approximately as quickly as Leibniz formula is the Wallis Product formula. This formula also has logarithmic convergence. 2/1 * 2/3 * 4/3 * 4/5 * 6/5 ... Hope this helps.
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Posted: 5/18/2017 1:38 AM
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OmnipotentEntity
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Well, x^(1/x) = e^(log(x)/x) in the domain of interest. Using a similar approach, this satisfies the differential equation y' = x^-2(1 - log(x))y The form of this is not so nice. But the x^-2 bit shouldn't affect the roots, even under repeated differentiation (it will however ensure that there will always be a log and a constant, so there will always be exactly one root from each term). So... that's the outline of a proof, but the only problem, is I can't seem to figure out a way to prove that roots will never repeat.
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Posted: 5/17/2017 12:56 PM
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OmnipotentEntity
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(d^n x^(1/x))/(dx^n) = x^(-n + 1/x) (1 - n + 1/x)_n for (n element Z and n>=0 and x!=1/x) Where (x)_n is the Pochhammer symbol and represents (x)(x+1)(x+2)...(x+n-1) This formula is polynomial of 1/x of degree n which can be easily shown to have positive zeroes (when 1/x = 1-n+m), multiplied by a term which is zero at no point while x > 0. However, one of the zeroes is at 1/x = 0. I'd look into this more to figure out where the final zero is coming from, but I have to drive.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 5/8/2017 6:03 PM
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OmnipotentEntity
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SPACKlick wrote:
GIM crashes on N64 but not on Virtual Console. I don't know if there's a way of testing runs on VC or not.
I'm highlighting this because it implies that the run might not be accepted even if it were submitted due to GIM. I'm honestly rather happy it didn't get submitted because that would be a solid MONTH of drama.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 5/7/2017 4:33 PM
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OmnipotentEntity
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Would anything by the Cernettes count? http://cernettes.wixsite.com/cernettes https://www.youtube.com/channel/UCAGLwiOVLh6ztaKBL84vtUA
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 4/26/2017 6:13 PM
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OmnipotentEntity
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You probably already have acapella science, but if you don't: https://www.youtube.com/user/acapellascience Two of my favorites: Link to video Link to video Also if you could provide a link to your playlist, that would help us not suggest duplicates.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 4/23/2017 3:01 AM
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OmnipotentEntity
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Perhaps you're seeking something analogous to a logarithmic interpolation, https://www.cmu.edu/biolphys/deserno/pdf/log_interpol.pdf
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
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