Wikipedia:Reference desk/Archives/Mathematics/2016 September 12

= September 12 =

Factorial and the normalized sinc function
I was playing around with the function $$f(n,x) = \binom{n}{x} = \frac{n!}{(n-x)! x!}$$. For the $$n=0$$ case, I inspected the graph and found that $$ \frac{1}{x!\cdot \left(-x\right)!} = \frac{\sin(\pi x)}{\pi x} $$. I thought this was very interesting! Is there a reason that this is the case? I figured it had something to do with the definition of $$x!$$ in relation to the integral Gamma function, but I wasn't sure how to relate that to the sine function. Thanks for the help. 64.229.38.13 (talk) 00:03, 12 September 2016 (UTC)


 * This is one of the functional identities obeyed by the Gamma function, specifically the one mentioned in Gamma function.  S ławomir  Biały  00:43, 12 September 2016 (UTC)

Cardinality of vector space dimensions
For an infinite-dimensional vector space, do we distinguish between having countably or uncountably infinite dimension? For example, I know that for a Fourier series, we try to represent a function as a sum of countably infinite linearly independent functions. However, we know that Fourier series don't always converge to the target function, and the sums are infinite, which to me means that the trigonometric functions do not span the set of all functions from R to R. Even if we restrict attention to continuously differentiable functions (for which Fourier series always converge to the target), the fact that not every such function has a finite Fourier series seems to imply that we don't have a basis either.

This has led me to wonder, must any basis for the vector space of real-valued functions on the real line be uncountably infinite? Is there a canonical basis like that? To me, the answer to the first question is undoubtably yes if we only allow finite linear combinations, since the set itself is uncountably infinite.--Jasper Deng (talk) 04:55, 12 September 2016 (UTC)
 * Yes, the cardinality of the dimension of a vector space is a well-defined notion that can be of interest. Yes, some (and hence every) basis for the space of real-valued functions has the cardinality of the continuum.  No, there is no canonical basis; it requires the axiom of choice to show that a basis exists at all.
 * The fact that the set is uncountable doesn't on its own mean that it requires an uncountable basis. For example, the real numbers is an uncountable vector space with dimension 1.--2406:E006:2EFE:1:2C15:58BD:2049:412E (talk) 09:57, 12 September 2016 (UTC)
 * it requires the axiom of choice to show that a basis exists at all: For the slow-thinking like me, $$(\delta_x)_{x \in R}, \delta_x(y)=\left(1\ for\ x=y,0\ otherwise\right)$$ is not a base, since a base allows for a representation as a finite sum. Tigraan Click here to contact me 15:59, 12 September 2016 (UTC)
 * More to the point of what you're asking, a Banach space is either finite dimensional or has uncountable dimension as a vector space. For example, the space of continuous functions on [0.1] is of uncountable dimension as a vector space.   Sławomir Biały  (talk) 10:18, 12 September 2016 (UTC)
 * That is interesting. Is there an easy argument to see there can't be a Banach space of countably infinite dimension?  Does every infinite-dimensional separable Banach space have vector-space dimension equal to the cardinality of the continuum?  Or is there maybe some nontrivial cardinal invariant that I'm not familiar with involved here? --Trovatore (talk) 17:52, 12 September 2016 (UTC)
 * It's an easy consequence of the Baire category theorem: a Banach space cannot be written as a countable union of finite dimensional subspaces (these are nowhere dense closed subspaces). Not sure about the cardinal.   Sławomir Biały  (talk) 18:49, 12 September 2016 (UTC)
 * Ah, thanks. So it's at least $$\operatorname{cov}({\mathcal K})$$ or not, see below (see Cichoń's diagram), the smallest number of meager sets whose union can be all the reals.  (For these purposes, every separable Banach space might as well be the reals.) --Trovatore (talk) 19:34, 12 September 2016 (UTC)
 * Actually I should be careful. Apparently $$\operatorname{add}({\mathcal K})$$ (the smallest number of meager sets whose union is nonmeager) can be less than $$\operatorname{cov}({\mathcal K})$$.  I'd have to think about the details of the argument to see if you can get past $$\operatorname{add}({\mathcal K})$$. --Trovatore (talk) 19:50, 12 September 2016 (UTC)
 * I think I was right the first time. The argument for $$\operatorname{cov}({\mathcal K})$$ (or greater; we haven't ruled out that the minimum cardinality is greater) should go through as follows:  Suppose there were a basis of size $$\kappa<\operatorname{cov}({\mathcal K})$$.  Then every point in the space is in a space spanned by some finite collection of basis elements, and there are $$\kappa^{<\omega}=\kappa$$ such subsets.  --Trovatore (talk) 20:17, 12 September 2016 (UTC)
 * There's a simpler argument that the dimension is at least the continuum: every infinite-dimensional Banach space contains a topologically embedded copy of $$\ell^1$$.  Sławomir Biały  (talk) 21:44, 12 September 2016 (UTC)
 * Could you fill in some of the blanks here? Do you mean there's a topological-vector-space embedding?  How does one see that? --Trovatore (talk) 02:39, 13 September 2016 (UTC)
 * Start with an subspace of countable dimension, with basis $$x_i$$ such that $$\|x_i\|=1$$. Then map the sequence $$(a_i)\in\ell^1$$ to $$\sum_i a_ix_i$$.  This is a topological (linear) embedding of $$\ell^1$$.   Sławomir Biały  (talk) 10:52, 13 September 2016 (UTC)