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This question comes from physics but I think it fits better here. On page 488 of Sakurai's Modern Quantum Mechanics he says that in order to find an infinite series of higher order derivatives of a function, i.e. $$ i\frac{\partial}{\partial t}\psi(x,t) = \left[m - \frac{1}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{8m^3}\frac{\partial^4}{\partial x^4}-\cdots\right]\psi(x,t), $$ you would have to specify the function at "farther and farther away from the origin" (or whatever point you are evaluating at).

The problem (in the physics context) is that this implies nonlocality. Ideally, you should be able to specify the time derivative and spacial derivatives at a point, solving the equation. However, I think Sakurai means to imply that knowing all derivatives at a point is equivalent to knowing the value of the function everywhere.

My question is is it possible to take all derivatives $\frac{\partial^n}{\partial x^n}\psi$ for $n \in \mathbb{N}$ only knowing the value of the function on a finite interval? (thanks to Roland for the wording). If so, can't we do the same thing we do with the first derivative (making the interval smaller and smaller) in order to make the interval vanishingly small? Could we then recover locality?

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    I don't see a connection between taking a derivative at a point $x_0$ (which is calculating a limit with $x \rightarrow x_0)$ and the size of the interval $x_0$ is in Can you clarify this? I also don't get the 'farther way from the origin' part. – 2017-01-27
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    I think as long as the function is defined on a dense set around a point, then you can take the infinity-th derivative, if it exists. – 2017-01-27
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    @Roland I think he's thinking that if you have a finite amount of points, the derivative is estimated by two points near a point (since this comes from a physics context), so with each derivative, you have less points to work with. – 2017-01-27
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    Another noteworthy point: I think it's more important to take all derivatives $\frac{\partial^n}{\partial x^n}\psi$ for $n\in \mathbb{N}$, because the infiniy-th derivative is rather [boring](http://math.stackexchange.com/questions/362873/is-there-any-meaning-to-an-infinite-derivative) – 2017-01-27
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    Perhaps you can try to understand the series in terms of the Fourier transform of the wave function. The series of derivatives is then transformed into a series in the wave number which in turn is just a function of the wave number. – 2017-01-27
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    Physics has limitations that do not exist in pure mathematics--the Heisenberg Uncertainty Principle, for example. I think you really should put this question to physicists, not mathematicians, after all. – 2017-01-28
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    I downloaded [this version of the book](http://www.fisica.net/quantica/Sakurai%20-%20Modern%20Quantum%20Mechanics.pdf) but the page numbering is different. Which chapter is it in? – 2018-10-17
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    It's not in that edition, sadly. It's in the second edition and that is the revised edition. The second edition contains a chapter on Relativistic Quantum Mechanics. – 2018-10-18

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Think of the limit definition of the derivative in the form ...

$$ f(x)=\lim_{h \to 0}\frac1h\left[ f(x+h)-f(x) \right] $$

this can be generalized to ... $$ f^{(n)}(x)=\lim_{h \to 0}\frac1h \displaystyle \sum_{k=0}^{n}(-1)^{n-k}\binom nk f(x+kh) $$

The crucial point being that $f$ needs to be defined on the interval $[x,x+nh]$ , this is not an issue for finite $n$ because the interval has zero length when you take $h\to 0$ but if you take the limit as $n \to \infty$ you get an interval of undefined length.

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    There is a factor $\binom{n}{k}$ missing in the summation. – 2017-01-28
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    Interesting argument. Is it necessary to use this definition of the $n^{th}$ derivative instead of $f^{(n)}(x)=\lim_{h \to 0}[f^{(n-1)}(x+h)-f^{(n-1)}(x)]$ – 2017-01-28
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    I think it comes down to the same thing once you make substitutions like $$f^{(n-1)}(x+h)=\lim_{h \to 0}\frac 1h[f^{(n-2)}(x+2h)-f^{(n-2)}(x+h)] $$ – 2017-01-28
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    It makes me think that perhaps my formula for $f^{(n)}(x)$ should have $(\frac 1h)^n$ instead of $\frac 1h$ – 2017-01-28
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    Where is it written that you must use the same values of $h$ for every value of $n$? If your interval has width $d,$ take $h – 2017-01-28
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    @WW1 for some simple functions like $x^{-1}$ we can get an expression like $f^{(n)}(x) = (-1)^nn!x^{-n}$ which is valid for any finite $n$. I'm not sure if it can be extended to an infinite derivative – 2017-01-28
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    Ok, so you have an interval of undefined length. Is it possible to take the limits correctly so that that length becomes defined? Zero? – 2017-02-03