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Bold and Disappointing Experiment With Hermite Polynomials!!!
(01/19/2017, 03:14 AM)mike3 Wrote: I'm a little dubious about that differentiation identity:


Consider the very simple case . The derivative is . For this simple polynomial we can use Faulhaber's formula and that gives us


Integrating that, which should give , gives , yet , and these differ by a non-constant amount. Likewise, differentiating the latter expression for the sum gives .

But looking at this, the derivative of the sum does just have a constant shift, so perhaps what we should really say is

up to a constant, which, when integrated, yields a linear term, where this is just indefinite continuum sum, not definite.
yes yes, yes, I should've been more explicit. I was being too brief. I always tend to just write it and drop the C because the solution still works. Using your notation made me forget about that little and, nonetheless as you can see, it still satisfies the difference equation which was the point I was making. Nonetheless, it does give a much simpler form of your equation

granted we know what is.

Plus, when I work with it I tend to work with the exponential indefinite sum, if

then the constant is zero.

This is just like how if


where there is no constant error. Of course this definite sum does not really work in this case though, it's really rather restrictive.

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RE: Bold and Disappointing Experiment With Hermite Polynomials!!! - by JmsNxn - 01/19/2017, 08:05 PM

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