If Pi has infinite numbers following the decimal, at same point those numbers must fit into a sequence that has already been done, for whatever length of numbers. Mathemetecians need only figure out how many numbers are repeated and for how long and in what places of the sequence.
>>11 is false. The guy in the cartoon ASSUMES Pi is a normal number, meaning its decimal expansion is all "random" digits. Nobody has proven Pi is normal.
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Anonymous2008-02-06 23:30
And yeah it becomes kind of obvious >>9 is a corollary of Pi being normal.
Wow, I never thought that 4chan would have something like this on it. Very nice!
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Anonymous2008-02-07 17:27
If you use successively higher bases ad infinitum, you will get less and less frequent occurrences of repeated digits. Using base pi, in fact, you get one digit: pi. There you go.
>>23
Actually, troll, as >>18, I can say that I am not >>19/>>20. And I am right. You may check.
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Anonymous2008-02-08 18:35
Speaking out in a peaceful protest that will be full of high school faggots is stupid, yes.
Half of you wearing masks and the other half not is stupid, yes. It's fucking stupid.
Coagulating the sidewalk outside of the Co$ is going to what exactly?
Every single person that isn't a fucking retard knows they are scammers.
The actual Scifags themselves don't give a FUCK what you think.
What exactly are you doing handing out fliers?
Mail them some fucking anthrax, hang yourself in front of the church, SET THE NEW HIGH SCORE.
Don't hand out FUCKING FLIERS.
This is the only chance /b/ will EVER have of curing the cancer. I don't give a fuck if it's done through pissbombing some underage b& or making fun of and dismantling a church full of retards.
All I know is that the fate of /b/ WILL be decided on the 10th for better or worse.
>>30
You know what to do next. Hint: It rhymes with "fap".
>>31
You don't get that you were TROLLED. Do a barrel roll. NAO!
>>32
I was speaking about ABSOLUTE VALUE, fucktard. Shit, do I have to stoop to drawing you a fucking PICTURE, or something? Is there crayon in your "math book", asstard?
The roots of such a polynomial would be roots of some other polynomial with integer coefficients.
For example, the roots of x^2 - (sqrt(2))x + 1 are also roots of x^4 + 1.
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Anonymous2008-02-12 23:30
>>51
It's not random in any rigorously defined mathematical sense. It's completely deterministic; there's dozens of compact summation formulas that will give you pi to any precision you want.
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Anonymous2008-02-13 1:14
"How do we know pi isn't some fraction of a square root, like 20sqrt(2)/9 ?"
Wtf?
Why are people with no knowledge of math posting on a math/science message board?
That is a good question. Pi is not the root of any "polynomial" equation where the polynomials have powers of x that are fractions, either positive or negative.
For radical exponents, I believe the question is open. That means we don't know the answer.
If you allow *arbitrary* exponents, pi is a root of a zillion different equations. For example, pi is a root of the equation x^q - 2 = 0, where q = log(2)/log(pi).
If you allow the "polynomials" to have infinitely many terms, you can find equations of which pi is a root. For example, it is the smallest positive root of
x - x^3 / 6 + x^5 / 120 - x^7 / 5040 + ... . But this isn't a polynomial.
I hope this does more to clear up the situation than it does to confuse it.
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Anonymous2008-02-16 10:51
>>47
almost - it's not a solution to any non-zero polynomial with coefficients in Q. but it's essentially with integer coefficients, since you can multiply through by something to get all integers.
That's obvious, but I'd imagine it'd not be a solution to any finite polynomial with radical co-efficients, but proving that would be presumably harder.
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Anonymous2008-02-17 3:03
>>65
If you accept that pi is transcendental, it isn't hard at all. The radicals are a subset of the algebraic numbers, which are (by definition) an algebraically closed field. If there was a polynomial with algebraic number coefficients which had pi as a root, then there are two possibilities:
(1) Pi is an algebraic number; contradiction, as pi is transcendental
(2) The algebraic numbers have a nontrivial algebraic extension; also a contradiction, as they are algebraically closed.
Of course, the hard part is the proof that pi is transcendental; if I recall correctly the idea is to show that i*pi (and therefore pi itself) must be transcendental because e is transcendental and e^(i*pi) = -1 is rational.
>>66
Ah, one more thing: it has to be shown that the algebraic numbers only contain elements which are in fact algebraic over the rationals. This is necessary for the "algebraic number or transcendental number" dichotomy used in (1).
Yes, it should follow from the Gelfond-Schneider theorem and the transcendentality of e. But I believe the way Lindemann showed it was by some variation of Liouville's method, by showing that there are rational approximations of pi that are too accurate.