August 27

Define the set of bounded functions as the set of functions for which there exists ${\displaystyle c\in \mathbb {R} }$ such that ${\displaystyle |f(x)|<c}$ for all real x. Anyway I guess the approach is to show that, given any arbitrary finite linearly independent subset of this set of functions, show that their exists another bounded function which is not in the span of them. How do you do that, or is there a better approach? Widener (talk) 03:38, 27 August 2011 (UTC)[reply]

Hint: Think about the characteristic functions of singletons — functions that are zero everywhere except at one point, where they take the value 1. Given a finite set of such functions, can you see a way to find a bounded function not in their span? --Trovatore (talk) 03:46, 27 August 2011 (UTC)[reply]

(or more to the point, a function that is in the span of all characteristic functions of singletons, but not in the span of the given subset) --Trovatore (talk) 03:49, 27 August 2011 (UTC)[reply]

Obviously any other characteristic function of singletons will do. But that only shows that the set of characteristic functions of singletons is infinite dimensional, it does not show that all bounded functions are infinite dimensional. Widener (talk) 03:55, 27 August 2011 (UTC)[reply]

The span of the set of characteristic functions of singletons is a subspace of the space of all bounded functions. --Trovatore (talk) 04:02, 27 August 2011 (UTC)[reply]

I'm guessing that if A is a subspace of B and A is infinite dimensional, that implies B is infinite dimensional. Can you prove that? Widener (talk) 04:08, 27 August 2011 (UTC)[reply]

Try it the other direction. Take a subspace of a finite-dimensional space. Can you show it's finite-dimensional? --Trovatore (talk) 04:10, 27 August 2011 (UTC)[reply]

Okay thanks. Widener (talk) 04:23, 27 August 2011 (UTC)[reply]

Suppose in the problem above, instead of a fair die you use a weighted die, so the probability of getting i is p_i (i>0). We don't need to assume the number of sides is 6 or even finite. The argument given above shows that if limit of the probability (call this q_n) that position n will be reached exists, then that limit is 1/E(i), provided of course that E(i) exists. This won't always be the case though, for example if p₂=1 and all other p_i are 0, then the probability of reaching n is 0 if n is odd and 1 if n is even, so there is no limit. There are also distributions with no expected value. But if you add conditions to allow for these possibilities then is it true that the limit always exists? Specifically, prove or disprove: If gcf{i: p_i>0}=1 then lim(n→∞) q_n = 1/E(i) if E(i) exists and lim(n→∞) q_n = 0 if E(i) does not exist.--RDBury (talk) 09:07, 27 August 2011 (UTC)[reply]

See above. If

${\displaystyle \phi (x)=\sum _{n=1}^{\infty }p_{n}x^{n))$

and

${\displaystyle F(x)=\sum _{n=0}^{\infty }q_{n}x^{n))$, then

${\displaystyle F(x)={\frac {1}{1-\phi (x)))}$

If φ(x) = 1 has no solutions with absolute value 1 other than x = 1, (which corresponds to the gcf/gcd equation above), then (1-x)F(x) has radius of convergence greater than 1 about x = 0, and

${\displaystyle F(x)-{\frac {1}{\phi '(1))){\frac {1}{1-x))}$ is bounded on the unit disk, and has radius of convergence greater than 1.

It follows that ${\displaystyle \lim _{n\to \infty }\left(q_{n}-{\frac {1}{\phi '(1)))\right)=0}$, and

${\displaystyle \lim _{n\to \infty }q_{n}={\frac {1}{\phi '(1)))={\frac {1}{E(i))).}$ — Arthur Rubin (talk) 00:53, 29 August 2011 (UTC)[reply]

I'm with except for how to get that there are no solutions to φ(x) = 1 when the gcd equations holds. If you factor the 1-x it comes down to proving that if 1 ≥ a₁ ≥ a₂ ≥ a₃ ... ≥ 0, with lim(n→∞) a_n =0, then the zeros of 1+a₁z+a₂z²... have absolute value ≥1, with equality only if gcf{i: a_i-1 < a_i}=1. The ≥1 part can be shown from the original problem; if not then the q_n's would be unbounded, contradicting that they are probabilities. I can't get the "equality only if" part though. The idea of finding a bound on the real part on the unit circle I used above fails in general since it turns out the real part can be negative.--RDBury (talk) 12:15, 29 August 2011 (UTC)[reply]

If:

1. ${\displaystyle |z|\leq 1,}$
2. All ${\displaystyle p_{n}\geq 0,}$ and
3. ${\displaystyle \sum _{n=1}^{\infty }p_{n}=1,}$ then

${\displaystyle \left|\sum _{n=1}^{\infty }p_{n}z^{n}\right|\leq \sum _{n=1}^{\infty }p_{n}|z|^{n}\leq \sum _{n=1}^{\infty }p_{n}=1,}$

where equality only occurs if |z| = 1 and zⁿ = z^m whenever p_n and p_m are nonzero. If

${\displaystyle \sum _{n=1}^{\infty }p_{n}z^{n}=1,}$

then, in addition, zⁿ = 1 whenever p_n is nonzero. If, in addition, z ≠ 1, then

${\displaystyle z^{\gcd\{n|p_{n}\neq 0\))=1,}$ so

${\displaystyle \gcd\{n|p_{n}\neq 0\}>1.}$ — Arthur Rubin (talk) 05:11, 31 August 2011 (UTC)[reply]

Very nice.--RDBury (talk) 07:25, 31 August 2011 (UTC)[reply]

What is 12/20 - 15/20 — Preceding unsigned comment added by Lightylight (talk • contribs) 20:28, 27 August 2011 (UTC)[reply]

Yes I can solve it :) Did you try it out, what is 12-15 dollars for instance? How about if the dollars are replaced by nickels (5 cents or a twentieth of a dollar) Dmcq (talk) 20:34, 27 August 2011 (UTC)[reply]

Holy shit, I am blown away!

You have presented the problem whose solution will solve... ALL OF MATHEMATICS. --COVIZAPIBETEFOKY (talk) 21:06, 27 August 2011 (UTC)[reply]

Is the sarcasm really necessary? It's a perfectly valid question, although it does seem a little strange age-group wise. Grandiose (me, talk, contribs) 21:08, 27 August 2011 (UTC)[reply]

Yes. --COVIZAPIBETEFOKY (talk) 00:51, 28 August 2011 (UTC)[reply]

The only possible source of confusion on this problem is the order of operations. The standard order of operations specifies that the division be done before the subtraction:

(12/20) - (15/20) = (12-15)/20

You could get different answers, though, if you used a different order of operations:

12/(20-15)
    20

But, assuming you meant to use the standard order of operations, just complete the first one, noting that you will get a negative fraction. StuRat (talk) 21:58, 27 August 2011 (UTC)[reply]

Its a bit deeper than that. Problems like this are one of the biggest stumbling blocks in mathematical development, I guess about half the population, many adults included, would not be able to do it. Conceptually it represents the first generalisation of number after the integers, it involves a whole new techniques for manipulating numbers which follow counter intuitive rules. Why do you only subtract the tops not the bottoms as well? Why are the add/subtract rules so different to multiply rules? How come two different fractions are equal to each other. And to make matters worse the sums going to have a negative result as well, what the heck does minus half an apple mean?

So to try and give some meaning, take a sheet of paper, fold it in four one way, fold it in five the other way. Cut along the fold so you get 20 bits. All the bits together make one whole, each bit is 1/20th. Try the easier sum 15/20 - 12/20. Count out 15 bits take away 12, how many is left, whats that as a fraction?. Play with 1/2, 1/4, 1/5 all of which can be made from the bits, add and subtract them and see what happens. --Salix (talk): 23:22, 27 August 2011 (UTC)[reply]

Is it okay to use commas to separate digits in numbers that are not in base ten? For example, instead of writing 10001000100011100011001011011010000100111011101100110000000011100010110010010₂, could one write 10,001,000,100,011,100,011,001,011,011,010,000,100,111,011,101,100,110,000,000,011,100,010,110,010,010₂? -Metroman (talk) 23:54, 27 August 2011 (UTC)[reply]

While the comma is a commonly used digit group separator, there are many options, and your choice should be based on context. The comma is certainly a permissible choice for your own use, but a space or an underscore ("_") is more commonly used with binary numbers. The space or thin space is more likely to be seen in general texts, and the underscore in computer science. The underscore is a permissible digit group separator symbol for constants (of any base, including decimal) in many computer languages because the comma is reserved to separate values.

Also note that you need not group only by threes as you are used to in decimal (although several cultures use different groupings of two or four in decimal numbers, and may even vary the group size within a single number as with Indians writing 7,00,00,00,000 where you may be used to 7,000,000,000 -- see lakh). Grouping a binary number in three suggests that while you are writing it in binary, you are representing it in support of octal. These days it is more common to group in fours, suggesting support of hexadecimal, or eights, marking byte (or octet) boundaries. -- 110.49.242.14 (talk) 01:16, 28 August 2011 (UTC)[reply]

It's problematic. See Digit group separator. Groups of 3 hints at decimal numbers. I wouldn't use binary groups of 3 except in special circumstances where octal numbers are very common. PrimeHunter (talk) 01:18, 28 August 2011 (UTC)[reply]

Thanks for all your answers. You've answered my question nicely. -- Metroman (talk) 06:06, 1 September 2011 (UTC)[reply]