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I'm sure the answer is negative, they are both called "basis" after all, but there is a "paradox" that I can't wrap my head around.

We can write the dirac delta function in the fourier basis as $$ \delta(x) = \frac{1}{2\pi} + \frac{1}{\pi}\sum_{n=1}^\infty \cos nx $$ yet since the delta function is not a "function" per se, it seems that it can't have a taylor series. Does this mean that it can't be expressed in the polynomial basis?

I guess there are 2 possible answers to this question:

  1. It can still be expressed in the polynomial basis, just by coefficients that are different from those given by a taylor series
  2. Since the dirac delta is not a function, "basis" theory loses its meaning, and so it does make sense that it can only be expressed by sins and coses. In this sense, is the fourier basis "more powerful/expressive" than the polynomial basis (ie it can also express things such as distributions)
samlaf
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  • Any basis can express the space it spans. If you choose a polynomial basis that spans the same space then it will be able to span the same space, but I don't see any specification on the polynomial basis, nor which inner product is used. – mathreadler May 25 '17 at 21:42
  • I think Fourier Transform is usually first defined on the quite restrictive Schwarz class of infinitely continously differentiable functions ($\mathbf C^\infty$) with compact support, which is so odd that is usually required to show the students a proof that it's in fact not empty. Then we can find families of functions which we can use to help expand the transform to other spaces of functions. – mathreadler May 25 '17 at 21:56
  • @mathreadler Quibble: this is Laurent Schwartz with a "t"... also, "the Schwartz space" of functions is nowadays often the collection of smooth functions so that they and all derivatives are rapidly decreasing, while the space of smooth, compactly-supported functions is (Schwartz') test functions. – paul garrett May 25 '17 at 22:09
  • @paulgarrett: yes you are right maybe I risked messing up the wrong Schwar(t)z:es, it is late over here. Is the one I spelled the one with the famous inequality in linear algebra? Yes maybe the space of test functions, but that's what distributions are defined for after all and the question regards the Dirac distribution which works on such test functions from that space so it would only make sense to talk about the most restrictive of the two spaces, right? – mathreadler May 25 '17 at 22:18
  • @mathreadler, heh, yes, the linear-algebra inequality-Schwarz is Herman Amandus (or similar) Schwarz. And, yes, there is ambiguity in "Schwartz' space...", though/and the dual to (what I call) Schwartz' space, a.k.a. tempered distributions, is smaller than the dual of the smaller space of (what I call) test functions. But of course these are just descriptors... :) – paul garrett May 25 '17 at 22:21
  • After pondering this for more than ten minutes, I'm starting to have some guesses as to just what question is being asked here. My best guess so far is that the functions $n\mapsto e^{inx}$ for $n\in\mathbb Z$ are being thought of as a basis of a linear space and the functions $x\mapsto x^n$ are thought of as another basis, and they should span the same space. But this seems to be an attempt to apply ideas outside the contexts in which they have been shown to make sense. When ${ x\mapsto e^{inx}:n\in\mathbb Z}$ is thought of as a "basis" of a vector space, sometimes it means$,\ldots\qquad$ – Michael Hardy May 25 '17 at 22:30
  • $\ldots,$that it's an "orthonormal basis" of the Hilbert space of square-integrable periodic functions. But when that is done, then the Fourier coefficients $c_n$ of a square-integrable function satisfy $\sum_{n\in\mathbb Z} |c_n|^2 < \infty.$ The Dirac delta function is not a member of that space and the sum of squares of norms of its Fourier coefficients is not finite. That space includes functions that have no smoothness in them and even have discontinuities. Functions to which their own Taylor series converge are very smooth within their regions of convergence. – Michael Hardy May 25 '17 at 22:40
  • So these are different spaces, neither of which contains the delta function. – Michael Hardy May 25 '17 at 22:41
  • $\ldots,$and BTW one should remember that the function $\displaystyle \delta(x) = \frac 1 {2\pi} + \frac 1 {\pi}\sum_{n=1}^\infty \cos nx$ is a periodic delta function with period $2\pi. \qquad$ – Michael Hardy May 25 '17 at 22:47

2 Answers2

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It's not a Taylor series, but you can also write the Dirac delta as a sum (convergent in the sense of distributions on the interval $(-1,1)$) of Chebyshev polynomials $T_n(x)$:

$$ \delta(x) = \frac{1}{\pi} + \frac{2}{\pi} \sum_{k=1}^\infty (-1)^k T_{2k}(x) $$

Robert Israel
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To write a function, or a "generalized function" (distribution? hyperfunction?) as an infinite sum of simpler functions (rigorously, as opposed to heuristically... even though many physicists have made tremendous use of physically intuitive heuristics! E.g., Dirac!) is probably best done when/if one has a suitable (!) topology on a relevant space of functions (or its dual, or its extension, or both simultaneously, most usefully, as in the case of distributions, tempered distributions, etc.)

Somewhat ironically, the vector space of polynomials on a closed interval is a bit clumsy, insofar as polynomials are not eigenfunctions for any convenient self-adjoint (bounded or not) operator. That is, despite their intuitive appeal, they are not as well adapted to subtler situations as, for example, Fourier series of various sorts (where exponentials or sines and cosines are indeed eigenfunctions for Laplacians...)

A related example where things turn out somewhat better is the case of the "quantum harmonic oscillator" $-\Delta+x^2$ on $\mathbb R$, which does have an orthonormal basis of eigenfunctions in $L^2(\mathbb R)$, and whose subtler theory does allow expression of tempered distributions as infinite sums of terms which are constant multiples of $n$th Hermite polynomial times suitable Gaussian.

paul garrett
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