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 Hermite Polynomials mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 07/05/2014, 06:07 AM (This post was last modified: 07/05/2014, 06:30 AM by mike3.) Hi. I've just started exploring a new possibility based around what are known as "Hermite Polynomials". See here: http://en.wikipedia.org/wiki/Hermite_polynomials You may ask what this is for. Well, I'm thinking about the possibility of it providing yet another tetration method based on the continuum sum. In this post, I show how the polynomials can be used to make a continuum sum for a kind of function that should include Tetration. ------------------------------------------------------------------------------------------------------------ The key property of the Hermite polynomials that we are interested in is that they form what is known as an "orthogonal basis" of the space $L^2(\mathbb{R}, e^{-x^2})$ (and so $L^2(\mathbb{C}, e^{-x^2})$) (for the polynomials of the second type mentioned, denoted $H_n$), which is the space of all Lebesgue-integrable functions $f: \mathbb{R} \rightarrow \mathbb{R}$ (and by extension $f: \mathbb{R} \rightarrow \mathbb{C}$) satisfying $\int_{-\infty}^{\infty} |f(x)|^2 e^{-x^2} dx < \infty$. In particular, every function which is exponentially bounded, i.e. for which $|f(x)| < Ce^{a|x|}$ (and also satisfies the Lebesgue integrability requirement) belongs to this space. This is easy to see, since we then have $|f(x)|^2 < C^2 e^{2a|x|}$ and then $|f(x)|^2 e^{-x^2} < C^2 e^{2a|x| - x^2}$. This is $o(e^{-|x|})$ as $x \rightarrow \pm i\infty$ because $2a|x| - x^2 < -|x|$ for sufficiently large $|x|$ (take $|x| > 2a + 1$). Therefore the integral converges as the integrand will always decay quickly to 0. So, suppose $f: \mathbb{C} \rightarrow \mathbb{C}$ now is a complex function satisfying: 1. $f(z)$ is holomorphic for $\Re(z) > -2$ 2. $f(z)$ is exponentially bounded in the strip $-1 < \Re(z) < 1$, that is, $|f(z)| < Ce^{a|\Im(z)|}$ in that strip. Then $f(ix)$, $x \in \mathbb{R}$, will meet the requirements for membership in that space and so we can write it as a "Hermite series": $f(ix) = \sum_{n=0}^{\infty} a_n H_n(-ix)$. Of course, we are interested in the case where $f(z) = \mathrm{tet}(z)$. Hermite Transform This series formula leads to the following notion. Given a sequence $a_n$, we could define a "Hermite Transform" by $\mathcal{H}\{a\}(x) = \sum_{n=0}^{\infty} a_n H_n(x)$. The inverse transform can be found via the orthogonality properties of the polynomials, namely $\int_{-\infty}^{\infty} H_n(x) H_m(x) e^{-x^2} dx = \sqrt{\pi} 2^n n! \delta_{n,m}$ where $\delta_{n,m}$ is the Kronecker delta. This allows us to integrate a function against the Hermite polynomial and so extract a coefficient of the sequence $a_n$, giving $\mathcal{H}^{-1}\{f\}_n = \frac{1}{\sqrt{\pi} 2^n n!} \int_{-\infty}^{\infty} f(x) H_n(x) e^{-x^2} dx$. In particular, we can use the inverse Hermite to obtain the coefficients for a function. Continuum sum of Hermite series We now turn to making a continuum sum of the Hermite polynomials, or a Hermite series. We get $\sum_{n=0}^{z-1} \sum_{k=0}^{\infty} a_k H_k(-in) = \sum_{k=0}^{\infty} \sum_{n=0}^{z-1} a_k H_k(-in) = \sum_{k=0}^{\infty} a_k \sum_{n=0}^{z-1} H_k(-in) = \sum_{k=0}^{\infty}a_k Hs_k(z)$ where we define the "Hermite sum polynomials" to be $Hs_n(z) = \sum_{k=0}^{z-1} H_n(-iz)$. The continuum sum operator we use here is the Faulhaber's formula one, which sends polynomials to other polynomials (and every polynomial has a unique continuum sum which is a polynomial, given by this operator). To obtain expressions for the sum polynomials $Hs_n$, we can proceed as follows. First, there is a way to obtain a recurrence formula, and second, a way to obtain an explicit formula in terms of the original Hermite polynomials. Recurrence formula Start with the identity $H_n(x + y) = \sum_{k=0}^{n} {n \choose k} H_k(x) (2y)^{n-k}$. Now take $x = -iz$ and $y = -i$ (so $x + y = -iz - i = -i(z + 1)$) and subtract $H_n(-iz)$ to get $\Delta H_n(-iz) = H_n(-i(z + 1)) - H_n(-iz) = \left(\sum_{k=0}^{n} {n \choose k} H_k(-iz) (-2i)^({n-k}\right) - H_n(-iz) = \sum_{k=0}^{n-1} {n \choose k} H_k(-iz) (-2i)^{n-k}$ where the last equality follows from $H_0(x) = 1$. Summing each side with the Faulhaber operator gives $H_n(-iz) = \sum_{l=0}^{z-1} \sum_{k=0}^{n-1} {n \choose k} H_k(-il) (-2i)^{n-k} = \sum_{k=0}^{n-1} {n \choose k} (-2i)^{n-k} \sum_{l=0}^{z-1} H_k(-il) = \sum_{k=0}^{n-1} {n \choose k} (-2i)^{n-k} Hs_k(z)$. Taking the last term off the sum, we get $-2in Hs_{n-1}(z) = H_n(-iz) - \sum_{k=0}^{n-2} {n \choose k} (-2i)^{n-k} Hs_k(z)$ or $-2i(n+1) Hs_n(z) = H_{n+1}(-iz) - \sum_{k=0}^{n-1} {{n+1} \choose {k}} (-2i)^{n-k+1} Hs_k(z)$. Starting with $Hs_0(z) = z$, we have a full recurrence formula. Explicit formula Now for the explicit formula. To do this, we start with the generating function of the Hermite polynomials: $\exp(2xt - t^2} = \sum_{n=0}^{\infty} H_n(x) \frac{t^n}{n!}$ or, for the imaginary axis, $\exp(-2izx - x^2) = \sum_{n=0}^{\infty} H_n(-iz) \frac{x^n}{n!}$. Continuum summing with the Faulhaber operator (which is valid analytically for $|x| < \pi$ (and so extends elsewhere by analytic continuation) and valid formally), we get $\sum_{l=0}^{z-1} \exp(-2ilx - x^2) = \sum_{l=0}^{z-1} \sum_{n=0}^{\infty} H_n(-il) \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{x^n}{n!} \sum_{l=0}^{z-1} H_n(-il) = \sum_{n=0}^{\infty} Hs_n(z) \frac{x^n}{n!}$. Now the left hand sum sums to $\frac{\exp(-2izx - x^2) - 1}{\exp(-2ix) - 1}$, so we have the generating function for the Hermite sum polynomials as $\frac{\exp(-2izx - x^2) - 1}{\exp(-2ix) - 1} = \sum_{n=0}^{\infty} Hs_n(z) \frac{x^n}{n!}$. We can now obtain an explicit solution for $Hs_n(z)$. Rewrite the left side as follows: $\frac{\exp(-2izx - x^2) - 1}{\exp(-2ix) - 1} = \frac{\exp(-2izx - x^2) - 1}{\exp(-2ix) - 1} \frac{(-2ix)}{(-2ix)} = (\exp(-2izx - x^2) - 1)\frac{(-2ix)}{(-2ix)}\frac{1}{\exp(-2ix) - 1} = \frac{\exp(-2izx - x^2) - 1}{(-2ix)} \frac{(-2ix)}{\exp(-2ix) - 1}$. Now we have the generating function as a product of two functions which have no poles. Using that $H_0(x) = 1$, we see that the constant term of $\exp(-2izx - x^2) - 1$ is 0, and carrying out the division, we get $\frac{\exp(-2izx - x^2) - 1}{(-2ix)} = \frac{1}{(-2ix)} \sum_{n=1}^{\infty} H_n(-iz) \frac{x^n}{n!} = \frac{1}{(-2i)} \sum_{n=1}^{\infty} H_n(-iz) \frac{x^{n-1}}{n!} = \frac{1}{(-2i)} \sum_{n=0}^{\infty} H_{n+1}(-iz) \frac{x^n}{(n+1)!} = \frac{1}{(-2i)} \sum_{n=0}^{\infty} \frac{H_{n+1}(-iz)}{n+1} \frac{x^n}{n!}$. Then, for the other function, we recognize that $\frac{x}{\exp(x) - 1}$ is the generating function for the Bernoulli numbers, and so we get $\frac{(-2ix)}{\exp(-2ix) - 1} = \sum_{n=0}^{\infty} B_n (-2i)^n \frac{x^n}{n!}$. Finally, multiplying together both of these functions and applying the binomial convolution to the series, we get \begin{align}\frac{\exp(-2izx - x^2) - 1}{\exp(-2ix) - 1} &= \left(\frac{1}{(-2i)} \sum_{n=0}^{\infty} \frac{H_{n+1}(-iz)}{n+1} \frac{x^n}{n!}\right) \left(\sum_{n=0}^{\infty} B_n (-2i)^n \frac{x^n}{n!}\right) \\ &= \sum_{n=0}^{\infty} \left(\sum_{k=0}^{n} {n \choose k} \frac{H_{k+1}(-iz)}{k+1} B_{n-k} (-2i)^{n-k-1}\right) \frac{x^n}{n!}\end{align} and therefore $Hs_n(z) = \sum_{k=0}^{n} {n \choose k} \frac{H_{k+1}(-iz)}{k+1} B_{n-k} (-2i)^{n-k-1}$. Then we have, by rearranging the order of summation in the equation for the continuum sum of the function, $\sum_{n=0}^{z-1} f(n) = a_0 z + \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} a_n {n \choose k-1} \frac{B_{n-k+1}}{k} (-2i)^{n-k}\right) H_k(-iz)$. ------------------------------------------------------------------------------------------------------------ What do you think of this approach? Note that we still need to be able to take the exponential of a Hermite series and we need to know when the continuum sum will converge. This is just a starting post to put the idea out. « Next Oldest | Next Newest »

 Messages In This Thread Hermite Polynomials - by mike3 - 07/05/2014, 06:07 AM RE: Hermite Polynomials - by JmsNxn - 07/05/2014, 02:11 PM RE: Hermite Polynomials - by mike3 - 07/06/2014, 04:14 AM RE: Hermite Polynomials - by JmsNxn - 07/06/2014, 11:03 AM RE: Hermite Polynomials - by mike3 - 07/06/2014, 11:04 PM RE: Hermite Polynomials - by JmsNxn - 07/07/2014, 12:39 PM RE: Hermite Polynomials - by mike3 - 07/08/2014, 06:31 AM RE: Hermite Polynomials - by fivexthethird - 07/06/2014, 02:11 PM RE: Hermite Polynomials - by tommy1729 - 07/08/2014, 12:24 PM

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