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We know that the set of all polynomials $a + bx + cx^2 +...$ forms a countably infinite dimension vector space.

However, what if we want to form a vector space out of $f(x)=\sum_\alpha c_\alpha x^\alpha$, with $\alpha \in \mathbb R$.

It would be the set of polynomial functions with polynomials that all have exponents within an interval.

We could then define a real function $c(x)$ that gives the values of the coefficients of the polynomials.

For example, if $c(\pi)=6$, then $f(x) = 6\cdot x^\pi + \sum_{\alpha \neq \pi}c_\alpha x^\alpha$, and so forth.

  • Does this work properly?
  • Has this been done before?
  • Are there interesting applications or results from this?
user56834
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  • Your $;f(x);$ is not a polynomial at all in general, unless the exponents opf $;x;$ are all natural numbers. – DonAntonio Mar 28 '17 at 08:54
  • Ok. I used the wrong terminology. I'm not sure what to call it though. – user56834 Mar 28 '17 at 08:55
  • Irrational function, and for such a thing to be well defined all but a finite numbers of $;\alpha,'$ s must be zero, as infinite sums are not well defined in general ... – DonAntonio Mar 28 '17 at 08:57

2 Answers2

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There is the field of (formal) Hahn series. The trick to a fruitful definition is you need to constrain how many nonzero coefficients you have, so that you can define multiplication by

$$ \left( \sum_\alpha c_\alpha x^\alpha \right) \left( \sum_\beta d_\beta x^\beta\right) = \sum_{\mu} \left( \sum_\nu c_\nu d_{\mu - \nu}\right) x^\mu $$

The problem to having a good definition is for all of the sums

$$ \sum_\nu c_\nu d_{\mu - \nu} $$

to be well-defined. The definition of Hahn series arranges it so that you're guaranteed this sum only has finitely many nonzero terms, so that you can define the uncountable sum to simply be the sum of those finitely many terms.

Of course, you can achieve your aims without being so bold — you can instead consider the ring you get by requiring a series to have only finitely many nonzero coefficients.

The result of this is pretty clearly a vector space with a basis indexed by the set of possible exponents. If you take the exponents in the reals as you suggest, it will be uncountably infinitely dimensional.

  • Is it possible to constrain the $c(x)$ function such that still an uncountably infintely many coefficients are non-zero, but that it decreases so fast that $f(x)$ is still well defined? – user56834 Mar 28 '17 at 08:59
  • @Programmer2134: If you're asking about something that is an uncountably infinite sum of (nonzero) real numbers, there is no hope of anything vaguely resembling the approach to infinite sums you see in calculus. –  Mar 28 '17 at 09:06
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This is perfectly alright: Let me make an alternative uncountable dimensional vector space which is more natural (in the sense that the elements are actually familiar real-valued functions that can be evaluated).

Let $V$ be the set of real-valued functions that are defined on the whole real line except (an unspecified) finite subset. (This set where it is undefined is allowed to vary with each function).

Clearly this is a vector space.

For a given real number $\alpha$ consider the function $f(x)= \frac1{x-\alpha}$, it is defined on the whole real line except at $\alpha$. For any set of distinct real numbers $\alpha_1,\alpha_2,\ldots,\alpha_n$, the corresponding functions $1/(x-\alpha_1), 1/(x-\alpha_2),\ldots,1/(x-\alpha_n)$ are easily checked to be linearly independent elements of $V$. Thus $V$ is an uncountable dimensional vector space. (This has been considered by Kaplansky in his proof of of Hilbert's Nullstellensatz. See Artin's ALGEBRA).

P Vanchinathan
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