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 Bold and Disappointing Experiment With Hermite Polynomials!!! JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 01/19/2017, 08:05 PM (This post was last modified: 01/19/2017, 08:28 PM by JmsNxn.) (01/19/2017, 03:14 AM)mike3 Wrote: I'm a little dubious about that differentiation identity: $\frac{d}{dz} \sum_{n=0}^{z-1} f(n) = \sum_{n=0}^{z-1} f'(n)$ . Consider the very simple case $f(z) = z^2$. The derivative is $f'(z) = 2z$. For this simple polynomial we can use Faulhaber's formula and that gives us $\sum_{n=0}^{z-1} f'(n) = \sum_{n=0}^{z-1} 2z = z(z-1) = z^2 - z$. Integrating that, which should give $\sum_{n=0}^{z-1} n^2$, gives $\frac{z^3}{3} - \frac{z^2}{2}$, yet $\sum_{n=0}^{z-1} n^2 = \frac{z^3}{3} - \frac{z^2}{2} + \frac{z}{6}$, and these differ by a non-constant amount. Likewise, differentiating the latter expression for the sum gives $\frac{d}{dz} \sum_{n=0}^{z-1} n^2 = z^2 - z + \frac{1}{6} \ne z^2 - z$. But looking at this, the derivative of the sum does just have a constant shift, so perhaps what we should really say is $\frac{d}{dz} \sum_z f(z) = \sum_z \frac{d}{dz} f(z)$ 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 $(\sum_z f(z))' = \sum_z f'(z) + C$ and drop the C because the solution still works. Using your notation made me forget about that little $C$ 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 $tet''(z) = tet'(z)(\sum_z tet'(z) + C)$ granted we know what $C$ is. Plus, when I work with it I tend to work with the exponential indefinite sum, if $\sum_z = \sum_{j=-\infty}^z$ then the constant is zero. $\frac{d}{dz}\sum_z = \sum_z\frac{d}{dz}$ This is just like how if $\int = \int_{-\infty}^z$ then $\Delta \int = \int \Delta$ where there is no constant error. Of course this definite sum does not really work in this case though, it's really rather restrictive. « Next Oldest | Next Newest »

 Messages In This Thread Bold and Disappointing Experiment With Hermite Polynomials!!! - by mike3 - 01/18/2017, 09:39 AM RE: Bold and Disappointing Experiment With Hermite Polynomials!!! - by JmsNxn - 01/18/2017, 10:20 PM RE: Bold and Disappointing Experiment With Hermite Polynomials!!! - by mike3 - 01/19/2017, 03:14 AM RE: Bold and Disappointing Experiment With Hermite Polynomials!!! - by JmsNxn - 01/19/2017, 08:05 PM

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