(Deceptively?) Simple question about Taylor series expansions

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SUMMARY

The discussion centers on the conditions under which the Taylor series expansion of a function u(x) in L²(-∞, ∞) can be accurately represented as u(x-t) = u(x) - (du/dx)t + (1/2)(d²u/dx²)t² - ... = ∑(n=0)∞ (u^(n)(x)/n!)(-t)ⁿ. It is established that the function must be analytic for this representation to hold. The conversation highlights that mere infinite differentiability is insufficient, as demonstrated by bump functions, which do not converge to their Taylor series. Additionally, the discussion notes that while complex functions have clearer criteria for analyticity, they are rarely found in L² spaces.

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Under what circumstances is it correct to say of the function [itex]u(x) \in L^2(-\infty,\infty)[/itex] that

[tex] u(x-t) = u(x) - \frac{du}{dx}t + \frac 12 \frac{d^2u}{dx^2}t^2 - \cdots = \sum_{n=0}^\infty \frac{u^{(n)}(x)}{n!}(-t)^n.[/tex]
 
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We can do that if the function is analytic. I don't really know if there are specific conditions which I could give.

Some notes:
1) Saying that u is differentiable an infinite number of times is not enough. A bump function is a counterexample. http://en.wikipedia.org/wiki/Bump_function These functions have the property that theTaylor expansion doesn't converge to the function.

2) If we were talking about complex functions instead of real functions then the answer is simpler. A complex function is analytic if and only if it is differentiable. However, these functions will rarely be in [itex]L^2[/itex].
 

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