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A standard representation of the $\delta $-distribution is $$ \pi \delta (x) = \lim_{\epsilon \searrow 0} \frac{\sin (x/\epsilon )}{x} $$ Is there a sense in which this could be seen as the leading term in an expansion of $\sin (x/\epsilon ) /x$ in positive powers of $\epsilon $, presumably with distributions as coefficients? If so, is it possible to give the expansion explicitly?

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  • $\begingroup$ Do you see the sign of the expression depends on the sign of $\epsilon$, so the direction of the limit is important? . $\endgroup$ – Anixx Jan 18 at 17:40
  • $\begingroup$ For what it worth, at $x\to0$ the function becomes $\frac1\epsilon$. $\endgroup$ – Anixx Jan 18 at 17:43
  • $\begingroup$ Of course, I had in mind the usual $\epsilon \searrow 0$. I've edited the question. $\endgroup$ – Michael Engelhardt Jan 18 at 17:44
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    $\begingroup$ Your distribution is the Dirichlet kernel, so the discrepancy $f(0)-(D_{1/\epsilon}*f)(0) = \int_{|t|>1/\epsilon} \widehat{f}(t)\, dt$ goes to zero faster than any power of $\epsilon$, thanks to the rapid decay of $\widehat{f}$. $\endgroup$ – Christian Remling Jan 18 at 17:57
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    $\begingroup$ @ChristianRemling - ah yes, I see what you're saying, this provides some clarity. The behavior is manifestly determined by the behavior of the Fourier transform of the test function at infinity, and strictly speaking, that should decay faster than any power. Now, I might get physicist notions like applying this on less well-behaved spaces than bona-fide test functions - so the Fourier transform might only decay as a power, and then I'd get nonvanishing coefficients in the expansion in powers of $\epsilon $. I wonder how this would look in $x$-space, that's another way of putting the question. $\endgroup$ – Michael Engelhardt Jan 18 at 19:06
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If you are interested in Laurent series, the expansion is

$$\frac{\sin (x/\epsilon)}x=\frac{1}{\epsilon }-\frac{x^2}{6 \epsilon ^3}+\frac{x^4}{120 \epsilon ^5}-\frac{x^6}{5040 \epsilon ^7}+\frac{x^8}{362880 \epsilon ^9}+O\left(\frac{1}{\epsilon^{11} }\right)=\sum_{n=0}^\infty \frac {(-1)^{2n}x^{2n}}{(2n+1)!\epsilon^{2n+1}}$$

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    $\begingroup$ Ah, I see I've again not been specific enough in the formulation of the question. No, by power series I meant positive powers, not the Laurent series, which is of course not very useful. I've once more edited the question. $\endgroup$ – Michael Engelhardt Jan 18 at 18:40
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The expansion around point $a$ in positive powers exists, but what is the general form of the term and whether it is convergent requires further research.

$$\frac{\sin (x/\epsilon)}x=\frac{\sin \left(\frac{x}{a}\right)}{x}-\frac{(\epsilon -a) \cos \left(\frac{x}{a}\right)}{a^2}+\frac{(\epsilon -a)^2 \left(2 a \cos \left(\frac{x}{a}\right)-x \sin \left(\frac{x}{a}\right)\right)}{2 a^4}$$ $$+\frac{(\epsilon -a)^3 \left(-6 a^2 \cos \left(\frac{x}{a}\right)+x^2 \cos \left(\frac{x}{a}\right)+6 a x \sin \left(\frac{x}{a}\right)\right)}{6 a^6}$$ $$+\frac{(\epsilon -a)^4 \left(24 a^3 \cos \left(\frac{x}{a}\right)-36 a^2 x \sin \left(\frac{x}{a}\right)+x^3 \sin \left(\frac{x}{a}\right)-12 a x^2 \cos \left(\frac{x}{a}\right)\right)}{24 a^8}$$ $$+\frac{(\epsilon -a)^5 \left(-120 a^4 \cos \left(\frac{x}{a}\right)+240 a^3 x \sin \left(\frac{x}{a}\right)+120 a^2 x^2 \cos \left(\frac{x}{a}\right)-x^4 \cos \left(\frac{x}{a}\right)-20 a x^3 \sin \left(\frac{x}{a}\right)\right)}{120 a^{10}}$$ $$+O\left((\epsilon -a)^6\right)$$

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    $\begingroup$ This of course works for $a\neq 0$. Can we make sense of this as $a\searrow 0$? Do the coefficients perhaps converge in a suitable (distribution?) sense to give a series convergent to $\sin(x/\epsilon)/x$? $\endgroup$ – Wojowu Jan 18 at 19:11

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