Question about pi. [Archive]

TobiasTheCommie

10th July 2005, 05:55 PM

hmm.. haven't heard that one.. but

Some amusing tidbits are on these pages - for example, when the Simpsons writers contacted NASA for the 40,000th digit of pi, NASA actually sent them a printout of all 40,000 digits."

btw.. it's 1.

Unnamed

10th July 2005, 06:26 PM

Originally posted by Hawk one
Is it true or an urban legend that the nine million nine hundred and ninety-nine thousand, nine hundred and ninety-ninth decimal place of pi is nine?

It's true (http://www.angio.net/pi/bigpi.cgi?UsrQuery=1&startpos=9999999&querytype=substr).

Marvel Frozen

10th July 2005, 06:30 PM

It's true.

http://www.angio.net/pi/piquery

Hawk one

10th July 2005, 06:55 PM

Nice, unnamed and Marvel Frozen.

Now, I am as many knows, rather illiterate in math, but what would happen if we had a 9-digit system going on (i.e. from 0 to 8)? Would the digit before the 10 millionth digit then be 8? :D

Dr Adequate

10th July 2005, 07:33 PM

Originally posted by Hawk one
Now, I am as many knows, rather illiterate in math, but what would happen if we had a 9-digit system going on (i.e. from 0 to 8)? Would the digit before the 10 millionth digit then be 8? :D Wow, reasoning by false analogy.

No, there's no particular reason why it should be.

Iconoclast

10th July 2005, 08:34 PM

Originally posted by Hawk one
Is it true or an urban legend that the nine million nine hundred and ninety-nine thousand, nine hundred and ninety-ninth decimal place of pi is nine? It's true, so all i can say is WOW! What are the chances of that?!?!? Must be a bizillion to one.... or 1 in 10.

boooeee

10th July 2005, 10:13 PM

Originally posted by Iconoclast
It's true, so all i can say is WOW! What are the chances of that?!?!? Must be a bizillion to one.... or 1 in 10.
Or 1 to 1, for this type of "coincidence". If the 9,999,999 digit wasn't 9, then just look at the 99,999,999 digit, and so on.

Hawk one

11th July 2005, 04:37 AM

Originally posted by Dr Adequate
Wow, reasoning by false analogy.

No, there's no particular reason why it should be.

Reasoning? You misunderstand me here, A. I was merely wondering if the coincidence would stretch that far. There's of course no particular reason it would be so, but there is a theoretical chance of being so, isn't it?

CurtC

11th July 2005, 05:14 PM

As boooeee pointed out, you might also say "wow!" if the 999th digit were 9; but it isn't. Someone looking for this kind of coincidence then looked at 9,999. But it isn't. So they looked at 99,999, but it isn't either. On and on, until finally, wow, the 9,999,999th digit is 9.

By the way, the 77,777th digit of pi is 7. Wow.

Manny

11th July 2005, 06:27 PM

Also, the first decimal is 1. :D

geni

11th July 2005, 06:47 PM

Originally posted by CurtC
As boooeee pointed out, you might also say "wow!" if the 999th digit were 9; but it isn't. Someone looking for this kind of coincidence then looked at 9,999. But it isn't. So they looked at 99,999, but it isn't either. On and on, until finally, wow, the 9,999,999th digit is 9.

By the way, the 77,777th digit of pi is 7. Wow.

you have too much time on your hands.

Diogenes

12th July 2005, 08:12 AM

An interesting spin.

Calculating Pi in binary is illegal..

http://www.verdandi.co.nz/blog/PermaLink.aspx?guid=95876d31-49aa-41a0-acb0-006721096aa0

WARNING: Do NOT calculate Pi in binary. It is conjectured that this number is normal, meaning that it contains ALL finite bit strings.

If you compute it, you will be guilty of:

Copyright infringement (of all books, all short stories, all newspapers, all magazines, all web sites, all music, all movies, and all software, including the complete Windows source code) ................. (more)

Just thinking

12th July 2005, 08:22 AM

Originally posted by Diogenes
Calculating Pi in binary is illegal..

Also, Pi in base Pi is 10.

Ziggurat

12th July 2005, 09:29 AM

Originally posted by Just thinking
Also, Pi in base Pi is 10.

There are 10 kinds of people in the world: those who can read binary, and those who can't.

Badly Shaved Monkey

12th July 2005, 02:39 PM

rppa

12th July 2005, 08:43 PM

Originally posted by Badly Shaved Monkey
"normal" in that context was not really explained by the definition that followed it. Why is "normal" the word to use?

Because it's the mathematical term for a number in which all digit sequences of any given length appear, on average, equally often.

That's stronger than the statement given, which is that in a normal number all sequences will appear at least once (including the contents of every copyrighted work).

Badly Shaved Monkey

13th July 2005, 12:12 AM

Originally posted by rppa
Because it's the mathematical term for a number in which all digit sequences of any given length appear, on average, equally often.

That's stronger than the statement given, which is that in a normal number all sequences will appear at least once (including the contents of every copyrighted work).

OK. I can see why "normal" could end up being the word chosen to represent that property. In the hierarchy of number classifications is it exactly equivalent to anything else, e.g. are all irrational numbers normal and all normal numbers irrational? Or, does being normal overlap other categories?

MRC_Hans

13th July 2005, 01:08 AM

Just to sum up on the original question: Since (the decimals of) pi is a virtually random sequence of digits, all digits are equally represented, so the chance that the nnn,nnn,nnn th decimal is n is 1/10. If you were using base 8, the chance would be 1/8. SO these things are not coincidences, they are simply the result of searching till it fits. Since there is no restriction on how many decimals you choose to use, you can find this for any digit you want, just keep searching and you will find a match.

Hans

rppa

13th July 2005, 05:00 AM

Originally posted by Badly Shaved Monkey
OK. I can see why "normal" could end up being the word chosen to represent that property. In the hierarchy of number classifications is it exactly equivalent to anything else, e.g. are all irrational numbers normal and all normal numbers irrational? Or, does being normal overlap other categories?

Congratulations for suspecting a connection between normal and irrational numbers. As I think you are guessing (correctly), a normal number is irrational since it does not terminate or repeat.

However, very little is known about them. This (http://pi314.at/math/normal.html) is a good writeup. One thing I see I forgot is that "normal" means "normal in all bases". If every binary sequence occurs equally often, it's "normal in base 2" but not necessarily "normal".

In that article, the statement that "the set of non-normal numbers has measure zero" means that most numbers are in fact normal. Yet despite that, not only do we not know if pi is normal (in *any* base), we don't know any normal numbers. Not one.

I hope that reduced rather than added to the confusion. Number theory can be a very strange place.

rppa

13th July 2005, 05:03 AM

Originally posted by MRC_Hans
Just to sum up on the original question: Since (the decimals of) pi is a virtually random sequence of digits, all digits are equally represented, so the chance that the nnn,nnn,nnn th decimal is n is 1/10. If you were using base 8, the chance would be 1/8. SO these things are not coincidences, they are simply the result of searching till it fits. Since there is no restriction on how many decimals you choose to use, you can find this for any digit you want, just keep searching and you will find a match.

Hans

It's the numerologist's dream. I think Randi has written on numerology, showing how given a couple of collections of facts, names or numbers (say, the biographies of Abraham Lincoln and John Kennedy), you can always find coincidences if you look hard enough.

But with this pi stuff, you can dream up specific KINDS of coincidences and then find examples.

Badly Shaved Monkey

13th July 2005, 09:07 AM

Originally posted by rppa
Number theory can be a very strange place.

Thanks for the link.

A bit more revision please.

Integer-rational-irrational-transcendental.

All are real?

Is there another category 'beyond' transcendental that is yet more abtruse or do those 4 categories comprise all the reals?

Badly Shaved Monkey

13th July 2005, 09:12 AM

Sorry to be a nuisance, but could someone please remind me of that stuff about pi being a better-behaved number in some base other than 10. I know that's badly phrased, but if you know the answer, I expect you'll understand the question!

CurtC

13th July 2005, 09:57 AM

IANAM, but I think pi is not "well behaved" in any number base. It's always transcendental, and I think it's normal, but also I think that it's not formally proved to be normal (yet).

Except, as someone has already mentioned, in base pi, pi has the value 10.

phildonnia

13th July 2005, 11:22 AM

Originally posted by Diogenes
An interesting spin.

Calculating Pi in binary is illegal..

http://www.verdandi.co.nz/blog/PermaLink.aspx?guid=95876d31-49aa-41a0-acb0-006721096aa0

I remember reading something similar about Graham's number: that written in binary, it contained the complete works of shakespeare, translated into all the world's languages. The argument is different of course, Graham's number is an integer, and is therefore not "normal" as pi is conjectured to be. There may be something inherently non-shakespearean in its construction.

Pi is a turing-computable number. Put another way, even though it continues infinitely, it contains finite information. Most numbers are not turing-computable.

jan

13th July 2005, 12:08 PM

Originally posted by Badly Shaved Monkey
A bit more revision please.

Integer-rational-irrational-transcendental.

All are real?

Is there another category 'beyond' transcendental that is yet more abtruse or do those 4 categories comprise all the reals?

The real numbers can be divided into the rationals and the irrationals. The irrational real numbers can be further subdivided into the algebraic irrational numbers and the transcendental numbers (all rational numbers are also algebraic). Almost all real numbers are transcendental.

Is there a subset of the transcendentals being even more abstruse? There are as many as you are able to imagine. phildonnia just mentioned one: the set of those real numbers that are not turing-computable. Almost all real numbers are not turing-computable.

How do the normal numbers fit in here? Not very well. All rational numbers are non-normal numbers, but that's almost all that can be said. There are transcendental numbers that are non-normal, and it could be possible that some irrational algebraic numbers are normal. Or not normal. Almost all real numbers are, of course, transcendental, normal, and not turing-computable.

ETA: Added a missing "algebraic".

drkitten

13th July 2005, 12:18 PM

Originally posted by Badly Shaved Monkey
Sorry to be a nuisance, but could someone please remind me of that stuff about pi being a better-behaved number in some base other than 10. I know that's badly phrased, but if you know the answer, I expect you'll understand the question!

See this page (http://www.math.toronto.edu/mathnet/questionCorner/pidigs.html)

Basically, the formula for calculating pi in base-16 is somewhat easier than (it is known to be) in other bases; one can calculate the zillionth base-16 digit directly without needing to calculate the zillion-1 preceeding digits. SImilarly, one doesn't need to store the preceeding zillion digits, so the memory requirements are much less.

jan

13th July 2005, 01:05 PM

It could be that there are other "digit-extraction" algorithms for π that work for base 10 — but if such an algorithm should exist, it would have to be completely different from the one discovered, and it is possible that there isn't one, so perhaps base 16 and π really have a special relationship.

There are similair algorithms for other famous numbers, some of them listed here (http://mathworld.wolfram.com/BBP-TypeFormula.html).

Badly Shaved Monkey

13th July 2005, 04:10 PM

Originally posted by new drkitten
See this page (http://www.math.toronto.edu/mathnet/questionCorner/pidigs.html)

Basically, the formula for calculating pi in base-16 is somewhat easier than (it is known to be) in other bases; one can calculate the zillionth base-16 digit directly without needing to calculate the zillion-1 preceeding digits. SImilarly, one doesn't need to store the preceeding zillion digits, so the memory requirements are much less.

Thanks. It was the base-16 connection that I had remembered.

Is it productive to think what might connect the number 16 to pi?

Hawk one

13th July 2005, 04:34 PM

Gee, I only wanted to find out if the claim in the OP was true or not (because a friend told me about it), and basically leave it at that, since it's nothing more than a cute li'l piece of trivia. Then all you guys go pi-serk... :p

jan

13th July 2005, 05:49 PM

Originally posted by Badly Shaved Monkey
Is it productive to think what might connect the number 16 to pi?

God did it.

phildonnia

13th July 2005, 06:34 PM

Originally posted by Hawk one
Gee, I only wanted to find out if the claim in the OP was true or not (because a friend told me about it), and basically leave it at that, since it's nothing more than a cute li'l piece of trivia. Then all you guys go pi-serk... :p

No need to be irrational...

Just thinking

13th July 2005, 08:15 PM

Originally posted by Badly Shaved Monkey
Integer-rational-irrational-transcendental.

All are real?

Is there another category 'beyond' transcendental that is yet more abtruse or do those 4 categories comprise all the reals?

Actually, there are (sometimes) included other categories.

They goes like this ...

Natural (1 - 2 - 3 - 4 - 5 ->)

Whole (0 -1 - 2 - 3 - 4 - 5 ->)

Integer (<- -3 - -2 - -1 - 0 - 1 - 2 - 3 - 4 - 5 ->)

Rational (Integer + Fractions)

Irrational (Rational + Non-perfect roots + Non-repeating non-terminating decimals + Transcendentals)

All of the above are Real numbers. As you can see, transcendentals are a subset of the irrationals.

Just thinking

13th July 2005, 10:38 PM

Originally posted by Just thinking
Irrational (Rational + Non-perfect roots + Non-repeating non-terminating decimals + Transcendentals)

Whoops!

Sorry -- please take out 'Rational' from Irrational description; all others however do apply.

Then we have ...

Real (Rationals + Irrationals)

After this we go to Immaginary and Complex.

Hawk one

14th July 2005, 07:35 AM

Originally posted by phildonnia
No need to be irrational...

I think you're going in circles now...

jan

14th July 2005, 10:23 AM

Originally posted by Just thinking
Irrational (Rational + Non-perfect roots + Non-repeating non-terminating decimals + Transcendentals)

Irrational (Non-perfect roots + Transcendentals) = Non-repeating non-terminating decimals

Just thinking

14th July 2005, 03:56 PM

Originally posted by jan
Irrational (Non-perfect roots + Transcendentals) = Non-repeating non-terminating decimals

Ah .... not quite.

Here is an example of a non-ending non-repeating decimal that is not a non-perfect root and is not transcendental.

1.010010001000010000010000001 ... and on and on.

It is not the root of 2 or 3 or 5 or whatever (and since you have no final decimal you cannot tell me the square of this number or what exactly it is the root of).

And it's not one of the handful of transcendental numbers.

phildonnia

14th July 2005, 06:32 PM

Originally posted by Just thinking
...And it's not one of the handful of transcendental numbers.

Handful?

Just thinking

14th July 2005, 08:22 PM

Originally posted by phildonnia
Handful?

I meant known ones ... relatively speaking.

drkitten

15th July 2005, 07:58 AM

Originally posted by Just thinking

Here is an example of a non-ending non-repeating decimal that is not a non-perfect root and is not transcendental.

1.010010001000010000010000001 ... and on and on.

It is not the root of 2 or 3 or 5 or whatever (and since you have no final decimal you cannot tell me the square of this number or what exactly it is the root of).

And it's not one of the handful of transcendental numbers.

Um, no. It's a transcendental number. It's merely one of the handful of transcendental numbers that are not normal.

And given that the transcendental numbers have measure 1 over the reals, I'm rather surprised that you consider there to be only a handful of them. Could I possibly borrow a fistful of dollars from you?

jan

15th July 2005, 10:22 AM

Originally posted by Just thinking
I meant known ones ... relatively speaking.

What is the mathematical difference between a "known" and an "unknown" transcendental? I assume you don't have something in mind like the ability of a turing machine to compute the successive digits, but rather something like "some human being has mentioned this number". But that's not a mathematical distinction.

Even if we would restrict the meaning of "transcendental" to "known transcendental", your definitions seems to me to be still misleading, since the "+" should be read as "union of disjoint sets". But then it should be called "unknown transcendentals", not "Non-repeating non-terminating decimals": the algebraic irrationals and the "known" transcendentals don't repeat either.

jan

15th July 2005, 10:45 AM

It just occurred to me that even if we restrict the meaning of "transcendental" to "known transcendental", the number 1.010010001000010000010000001... is still transcendental, since we know it, since you just defined it.

By the way, your "proof" that this number is not algebraic could be improved.

Take this one:
0.618033988749894848204586834...

It is not the root of 2 or 3 or 5 or any natural number, and you have no final decimal. But nevertheless, it is still a pretty algebraic number.

jan

15th July 2005, 01:11 PM

By the way — another method to split the reals is this: if you start with a line of length 1, some lengths can be constructed using only ruler and compass, and some not. All numbers that can be constructed with this method are basically the numbers that can be constructed by starting with the rationals and taking square roots. That means, that the golden ratio can be constructed, but 2^(1/3) can't (therefor, the delian problem can't be solved with ruler and compass, although 2^(1/3) is not transcendental). Some algebraic irrationals are constructible, some are not. All rationals can be constructed, but no transcendental number (therefore, the squaring of the circle is impossible).

jan

15th July 2005, 01:30 PM

To sum it up, so far we have

Natural ⊂ Whole ⊂ Integer ⊂ Rationals ⊂ Ruler-And-Compass-Constructible ⊂ Algebraic ⊂ Turing-Computable Digit Sequences ⊂ Real

Just thinking

15th July 2005, 01:58 PM

Originally posted by jan
Some algebraic irrationals are constructible, some are not. All rationals can be constructed, but no transcendental number (therefore, the squaring of the circle is impossible).

True, but I believe that before any number is given the title of 'Transcendental' it must be proven to be so. Both e and Pi are transcendental -- but it is not yet proven (I believe) that (e + Pi) is transcendental. So what does that make it? Since we don't know it to be algebraic or Transcendental for sure, I simply describe it as non-ending non-repeating. Yes, yes ... I know it must be one or the other, but in the meantime ....

phildonnia

15th July 2005, 03:02 PM

Originally posted by jan
To sum it up, so far we have

Natural ⊂ Whole ⊂ Integer ⊂ Rationals ⊂ Ruler-And-Compass-Constructible ⊂ Algebraic ⊂ Turing-Computable Digit Sequences ⊂ Real

You might also throw in the Algebraic Integers (http://mathworld.wolfram.com/AlgebraicInteger.html)

...Constructible Numbers ⊂ Algebraic Integers ⊂ Algebraic Numbers...

jan

15th July 2005, 03:14 PM

To: Just thinking

Well. Since there are more "transcendental" numbers (in the usual, non-Just thinking-meaning) than can be counted, you can prove that for almost all of them, it is not possible to prove that they are transcendental (since there can be only countable many different proofs). This means that almost all "Numbers-Formerly-Known-As-Transcendental" are not "Provably-Transcendentals".

Of course you are free to define the term "transcendental" as it pleases you, but I am afraid that "almost all" mathematicians will be unable to communicate with you if you do.

Another example: it is known that there are irrational numbers a and b with a^b being rational. Occasionally, it can be quite tricky to decide whether a^b is rational or irrational. So assume that you don't know whether a^b is rational or irrational. How would you call it?

i) "Either rational or irrational, but I don't know which."
ii) "Neither rational nor irrational, but something else."

I would prefer i).

Perhaps even better would be iii) "real" ;)

η: Added a "To: Just thinking", since I agree with everything phildonnia says.

Just thinking

15th July 2005, 03:19 PM

Correction field

jan

15th July 2005, 03:21 PM

x = p/q

Then x is the solution of

qx - p = 0

Sounds pretty algebraic, doesn't it?

η: Damn! Just thinking learns faster than I am able to type. This post refers to a deleted post. Move on, move on, nothing to see here.

phildonnia

15th July 2005, 03:36 PM

Originally posted by jan
To: Just thinking
Well. Since there are more "transcendental" numbers (in the usual, non-Just thinking-meaning) than can be counted, you can prove that for almost all of them, it is not possible to prove that they are transcendental (since there can be only countable many different proofs). This means that almost all "Numbers-Formerly-Known-As-Transcendental" are not "Provably-Transcendentals".

Oh, now that's just making my head spin. I started wondering if "provably-trancendental"+ "algebraic" = computable?

Of course you are free to define the term "transcendental" as it pleases you, but I am afraid that "almost all" mathematicians will be unable to communicate with you if you do.

If you do, please use the spelling "trankendetal" to indicate your own made-up definition.

jan

15th July 2005, 03:52 PM

Originally posted by phildonnia
Oh, now that's just making my head spin. I started wondering if "provably-trancendental"+ "algebraic" = computable?

I don't have a proof for this, but I guess that the answer is no.

To be computable you just need a program that spawns the digits. The program might be quite arbitrary. That doesn't mean that you know whether its output is irrational or transcendental. Perhaps it is transcendental, but there is no proof for this.

On the other hand, it is not completely inconceivable that for any of the countable many computable transcendentals, there is also a proof of its transcendentality. I don't know. Interesting question.

Just thinking

15th July 2005, 08:06 PM

Originally posted by phildonnia
Oh, now that's just making my head spin. I started wondering if "provably-trancendental"+ "algebraic" = computable?

What of rational computable numbers?

For example, has anyone computed an exponential ladder of seven 9's? Does anyone know what the first digit is or to what power of 10 the answer is (in scientific notation)? My point being, is such a value (rational and whole) ever able to be determined due to its size?

69dodge

15th July 2005, 08:35 PM

Originally posted by Just thinking
What of rational computable numbers?

For example, has anyone computed an exponential ladder of seven 9's? Does anyone know what the first digit is or to what power of 10 the answer is (in scientific notation)? My point being, is such a value (rational and whole) ever able to be determined due to its size?"Computable" doesn't mean "computable in a reasonable amount of time by a real, physical computer". It means "computable by a theoretical computer that is allowed to use as much time and memory as it needs".

Iamme

17th July 2005, 06:00 PM

jan

18th July 2005, 09:05 AM

More about the problem of computable and provably transcendental numbers:

The term "provable" is not a mathematical, but a metamathematical one. To be able to talk about it within mathematics, we would have to replace it with something like "provable within the formal system S", or "there is a turing machine U which is able to determine whether a real-number-spawning turing machine T produces an algebraic or transcendental number". That means, whether or not a computable transcendental number is also a provably transcendental number or not would depend on the formal system S or the decision machine U that we use.

So the "provably transcendental" real numbers would be either identical with the computable transcendental numbers, or (which I guess to be more likely), the extension of this set of numbers would be rather arbitrary, artificial, unnatural, inelegant. Therefore I tend to doubt that it would be a useful concept, and would prefer to exclude it from the "one true hierarchy of real numbers".