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Oct-14-17
 | | al wazir: <john barleycorn: Hint #1>. If you have an answer, post it. |
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| Oct-14-17 | | john barleycorn: https://en.wikipedia.org/wiki/Appro... The continued fraction representation of π can be used to generate successive best rational approximations. These approximations are the best possible rational approximations of π relative to the size of their denominators. Here is a list of the first thirteen of these:[53][54] 3 1 , 22 7 , 333 106 , 355 113 , 103993 33102 , 104348 33215 , 208341 66317 , 312689 99532 , 833719 265381 , 1146408 364913 , 4272943 1360120 , 5419351 1725033 {\displaystyle {\frac 31},{\frac 227},{\frac 333106},{\frac 355113},{\frac 10399333102},{\frac 10434833215},{\frac 20834166317},{\frac 31268999532},{\frac 833719265381},{\frac 1146408364913},{\frac 42729431360120},{\frac 54193511725033}} {\frac 31},{\frac 227},{\frac 333106},{\frac 355113},{\frac 10399333102},{\frac 10434833215},{\frac 20834166317},{\frac 31268999532},{\frac 833719265381},{\frac 1146408364913},{\frac 42729431360120},{\frac 54193511725033}
Of all of these, 355 113 {\displaystyle {\frac 355113}} {\frac 355113} is the only fraction in this sequence that gives more exact digits of π (i.e. 7) than the number of digits needed to approximate it (i.e. 6). The accuracy can be improved by using other fractions with larger numerators and denominators, but, for most such fractions, more digits are required in the approximation than correct significant figures achieved in the result. |
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| Oct-14-17 | | john barleycorn: As I mentioned in a previous post it is hard to re-invent problems. You mastered this to perfection. Not saying that you adorn yourself with borrowed plumes but it smells funny. However, this wiki article can be found no time when searching for "approximations of pi". |
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Oct-14-17
| | al wazir: <john barleycorn: As I mentioned in a previous post it is hard to re-invent problems.> No, it's all too easy. What's hard is inventing *new* problems. Thanks for the link -- but as you point out, it doesn't add much to what I found. |
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| Oct-14-17 | | john barleycorn: <al wazir: <john barleycorn: As I mentioned in a previous post it is hard to re-invent problems.> No, it's all too easy. What's hard is inventing *new* problems.> Haven't you noticed the irony?
<Thanks for the link -- but as you point out, it doesn't add much to what I found.> <al wazir: I found 355/133 myself ...>
Too bad that 355/133=2,669...,
500 AD the Chinese knew that 355/113 is the better approximation. But in fact, the article does add if you would understand it, eventually. |
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Oct-14-17
 | | alexmagnus: So, in terms of "efficiency" as defined by al wazir 355/113 is the best? Can it be proven strictly? |
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Oct-14-17
| | alexmagnus: 355/113 is so good because of the large term in the continued fraction that follows next (3 7 15 1 <292>). Now, the continued fraction of pi has even larger terms (though it is still an open problem whether they get arbitrarily large and if yes whether it contains all natural numbers). Maybe somewhere we come across such a big term that the resulting fraction from before the term is more efficient than 355/113 |
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Oct-14-17
| | alexmagnus: Talking about large terms in contintued fraction, a funny case of such is the Champernowne constant 0.12345678910111213... etc. It's continued fraction begins 0 8 9 1 <149083> 1 1 1 4 1 1 1 3 1 1 1 15 and the next term is the mindboggling <457540111391031076483646628242956118599603939710-
457555000662004393090262652563149379532077471286-
563138641209375503552094607183089984575801469863-
14883392141783010987>. It continued this behaviour, with occasionally randomly popping large (ever larger) numbers all throughout. Interestingly, the constant built by concatenation of all <primes> (instead of all integers) exhibits no such erratic behavior. |
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Oct-14-17
| | alexmagnus: Haha, that large number was so large that Chessgames decided to include a couple of out-of-place hyphens. |
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Oct-14-17
| | al wazir: <john barleycorn: Too bad that 355/133=2,669...> I told you I'm a lousy typist. |
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| Oct-14-17 | | john barleycorn: <al wazir: <john barleycorn: Too bad that 355/133=2,669...> I told you I'm a lousy typist.> No, you did not tell me that. Actually, your typing is pretty good compared to your mathematics. |
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| Oct-14-17 | | john barleycorn: <alexmagnus> the approximation of pi by rational numbers p/q has fundamental restrictions. https://link.springer.com/article/1... |
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Oct-14-17
| | al wazir: <john barleycorn: No, you did not tell me that.> A dozen posts up on this page. Actually, my typing is about as good as your reading. |
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| Oct-14-17 | | john barleycorn: <al wazir: <john barleycorn: No, you did not tell me that.> A dozen posts up on this page....> I am not Count Wedgemore, by the way.
<al wazir: <Count Wedgemore: Digit no.14 after the decimal point is a 9, not an 8.> Sorry. I typed it from memory instead of copying it. My memory is pretty good, but I'm a lousy typist.> |
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| Oct-14-17 | | john barleycorn: <al wazir: ...
Actually, my typing is about as good as your reading.> I acknowledged that your typing is pretty good. It is youre basic mathematics that has plenty of room for improvement. |
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Oct-14-17
| | al wazir: <john barleycorn: It is youre basic mathematics that has plenty of room for improvement.> My math is better than your reading. You still haven't gone through the proof I posted that a^3 + b^3 = 2c^3 has no non-trivial solutions in the natural numbers. |
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| Oct-14-17 | | john barleycorn: <al wazir:...
You still haven't gone through the proof I posted that a^3 + b^3 = 2c^3 has no non-trivial solutions in the natural numbers.>True, I have better ways to waste my time than checking whether you are able to adjust and copy an existing proof. The more so as I showed you that this utter triviality of an exercise can be solved in a less than 5 lines. |
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Oct-14-17
| | al wazir: <john barleycorn: True, I have better ways to waste my time than checking whether you are able to adjust and copy an existing proof. The more so as I showed you that this utter triviality of an exercise can be solved in a less than 5 lines.> If you had trouble understanding it, I'll be glad to answer your questions. Your exercise in "less than 5 lines" expanded to more than 100 once I pressed you on the contradictions and omissions -- and you still haven't filled in the gaps. Your M.O. is like that of any fake: you wave your arms and say "it's trivial," and when I ask questions you call me stupid and other insulting names. Real mathematicians and scientists argue about content and often disagree, but they let the *facts* decide the issue. They don't resort to name-calling. |
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| Oct-14-17 | | john barleycorn: <al wazir: ...
Your M.O. is like that of any fake: you wave your arms and say "it's trivial," and when I ask questions you call me stupid and other insulting names.Real mathematicians and scientists argue about content and often disagree, but they let the *facts* decide the issue. ...> Real mathematicians? How come you know? hahaha it's not name calling when saying what an ignorant person you are. It is the truth. simple like that. You made points where no points are for a mathematician. you played your game and failed as seen by other users as well. |
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Oct-15-17
| | al wazir: Weisstein (p. 2231) gives some characteristically non-intuitive pi approximations due to Ramanujan. In terms of "efficiency," the champion is (97 + 9/22)^(1/4),
which is accurate to 14 (!) significant figures while using only seven digits. And it can be written with just five digits: √[√(97 + 9/22)].
But that's really a cheat. To be fair I suppose you should count the √ symbols too, and maybe the +, /, and ^ as well. |
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Oct-15-17
 | | beatgiant: <al wazir>
I seem to recall this came up before. As I already pointed out, to define efficiency, you must first fix the set of symbols allowed and the cost of each symbol.If you allow square roots of numbers that are not perfect squares, you need to specify how you are approximating *them* as rationals too. At first glance, the expression you gave above is also an irrational. From where are you getting the 14 significant figures? Since we are talking about approximation, I assume you don't allow trig functions, natural logs, complex numbers, integration signs, the infinity sign, etc. which could give us short exact formulae for pi but beg the question of how they are approximated. More realistically, do you allow "looping" symbols such as a summation sign? And if you do, how much cost do you charge for them in the efficiency calculation? |
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Oct-15-17
| | al wazir: <beatgiant: I seem to recall this came up before.> I posted essentially the same problem five or six years ago. <As I already pointed out, to define efficiency, you must first fix the set of symbols allowed and the cost of each symbol.> I agree. Otherwise I could define a new symbol (notated, say, by #), that stands for ".14159265358979." Then 3# would be a two-symbol pi approximation accurate to 15 digits. <If you allow square roots of numbers that are not perfect squares, you need to specify how you are approximating *them* as rationals too. At first glance, the expression you gave above is also an irrational.> Sure it is. What's wrong with that? (When I suggested trying expressions of the form a^b, I was opening the door to irrationals.) <From where are you getting the 14 significant figures?> I was quoting Weisstein. I checked it, but I can only get to about 12 decimal places or so on my calculator or using double precision arithmetic in a code. (There are ways to extend the precision arbitrarily by using arrays instead of single words to represent numbers, or with a symbolic manipulation language like Mathematica, but that would have been too much trouble.) <Since we are talking about approximation, I assume you don't allow trig functions, natural logs, complex numbers, integration signs, the infinity sign, etc. which could give us short exact formulae for pi but beg the question of how they are approximated.> True. I was tacitly accepted all elementary arithmetic operations, but only those. |
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Oct-15-17
| | beatgiant: <al wazir>
<Sure it is. What's wrong with that?> Nothing. You posed the problem, so you can make the rules. But you need to make clear what those rules are.As far as I understand it currently, the rules are the following: We are allowed to use rationals, plus, minus, times, divide, and raise to power. If raising to a power yields an irrational, we are given an "oracle machine" that yields an arbitrary precision approximation of it. We are not allowed to use a looping sign such as summation or repeated product. The efficiency of the expression is defined as the number of significant figures of accuracy per number of digits used. Did I get that right? |
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Oct-15-17
| | beatgiant: <al wazir>
And by the rules I posted above, the two most efficient posted so far are: 355/115: 7 digits accuracy, 6 digits in expression => efficiency 1.167. (97 + 9/22)^(1/4): 14 digits accuracy, 7 digits in expression => efficiency 2.0. Can the efficiency be improved? I'd bet my bottom chessbuck that it can. What's the ultimate efficiency limit? Great open problem for a real mathematician. |
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Oct-15-17
| | al wazir: <beatgiant: Did I get that right?> Well, so far neither of us has mentioned parentheses or decimal points. I used parentheses above, and in my searches for an exponential approximation, which yielded for example 1.423^3.245 = 3.141592857...,
I used decimal points.
<Great open problem for a real mathematician.> I'm awed by Ramanujan's (97 + 9/22)^(1/4), and I can't imagine how he could have found it without an electronic calculator or computer. |
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