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September 7

Denumerable sets of trigonometric polynomials

Hi! I've been puzzled by a problem for some time now:

1. A real number is algebraic if it is a root of a polynomial, integer coefficients etc. 2. How is it provable that the set of roots for trigonometric polynomials is countable? It seems any proof I can imagine depends on the definition of hyperbolic trig. functions. But, I feel like there must be a possibility of proving this that does not rely on that.

```` — Preceding unsigned comment added by 76.102.205.17 (talk) 01:59, 7 September 2014 (UTC)[reply]

I'm not sure why you brought up the definition of algebraic real. Perhaps you meant to ask a second question?

As far as roots of trigonometric polynomials, trigonometric polynomials are analytic, and in fact any analytic non-zero function must have only countably many zeros. The reasoning is as follows: since the real line (and indeed the complex plane) is second-countable, any uncountable set contains an accumulation point. By unique analytic continuation, only the identically zero function can have a zero-set that contains an accumulation point.--80.109.106.3 (talk) 07:53, 7 September 2014 (UTC)[reply]

What's the full pi number

With all the numbers jacobroozie@gmail.com 65.175.250.157 (talk) 20:18, 7 September 2014 (UTC)[reply]

See Pi. The numbers never come to an end. Never. They've worked out the first 12 trillion digits, literally, and they haven't even scratched the surface. -- Jack of Oz ^{[pleasantries]} 20:28, 7 September 2014 (UTC)[reply]

You're thinking too decimal. I can satisfy the request very easily. The full pi number is pi. That's all the numbers in pi. Hope this helps. --Trovatore (talk) 20:43, 7 September 2014 (UTC)[reply]

10, in base pi. Double sharp (talk) 06:51, 9 September 2014 (UTC)[reply]

For what it's worth, you can download the 12-trillion-digit approximation here (split into 120000 zip files), though you need some serious storage space for the whole thing (12 terabytes for the ASCII form, natch).--Link ^(t•c•m) 21:20, 7 September 2014 (UTC)[reply]

Even better than that are the clever spigot algorithms for pi that allow you to compute any digit of pi in a reasonable amount of time. If you want every digit of pi, you will need to run a calculation that will continue for an infinite amount of time. But if you want any specific digit, no matter how far "down the line" it is, it can be calculated with such an algorithm. Nimur (talk) 16:25, 9 September 2014 (UTC)[reply]

But keep in mind that such spigot algorithms aren't available for every base. The ones given for pi are typically in binary or hexadecimal or some similar power-of-two base. I'm unaware of a spigot algorithm for pi which can be done in decimal, and attempting to convert from binary (or hexadecimal) to decimal requires you know all of the preceding digits, effectively "unspigotting" your algorithm. -- 160.129.138.186 (talk) 17:21, 10 September 2014 (UTC)[reply]

Not an issue with pi.

The pi algorithm I know is pi = 2 + (1/3) (2 + (2/5) (2 + (3/7) (2 + (4/9) ( ... )), and that can be evaluated using a quite short array of integers, of which all except the first are small:

1 |1/3 2/5 3/7 4/9

2 | 2 | 2 | 2 | 2 Multiply by 10 (you can multiply by 2 or 8 for a binary approximation here): 20 |20 |20 |20 |20 Resolve carries 20 |20 |20 |28 | 2 (9 in the last place equals 4 in the second-to-last; we do this twice) 20 |20 |32 | 0 | 2 (7 in the 4th column become 3 in the 3rd, etc) 20 |32 | 2 | 0 | 2 30 | 2 | 2 | 0 | 2 Multiply by 10 again: 300|20 |20 | 0 |20 Resolve carries 300|20 |20 | 8 | 2 300|20 |23 | 1 | 2 300|28 | 3 | 1 | 2 309| 1 | 3 | 1 | 2

So, 309 is our approximation to 100 pi. This sucks, but only because the array is so short. Another entry will about halve the truncation error. 10 more entries mean about 3 more decimals.

There can be unresolved carries in the leftmost column (for example, the initial return was pi > 2 and the next step returned 10pi > 30) but they are quite low (usually 0 or 1); it's an approximation to a true spigot algorithm. - ¡Ouch! (^{hurt me} / _{more pain}) 08:47, 11 September 2014 (UTC)[reply]

64 digits of Pi is all you ever need for home renovation. 202.177.218.59 (talk) 01:59, 8 September 2014 (UTC)[reply]

$\pi =4\times \left[1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-{\frac {1}{11}}+\cdots \right]$

Count Iblis (talk) 17:26, 10 September 2014 (UTC)[reply]

Which is infamous for extremely slow convergence. - ¡Ouch! (^{hurt me} / _{more pain}) 08:47, 11 September 2014 (UTC)[reply]

For a tongue-in-cheek answer, see Feynman point. Sławomir Biały (talk) 16:13, 12 September 2014 (UTC)[reply]