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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. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 07/05/2014, 02:11 PM Thats an awesome gadget you got there! I love that. I'm grumpy that my continuum sum using FC won't work here because you're using gaussians. But besides that this is quite beautiful. To me the only thing that's lacking is some rigor and epsilons and deltas, which I'm sure wouldn't take much to put in here. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 07/06/2014, 04:14 AM (This post was last modified: 07/06/2014, 04:21 AM by mike3.) (07/05/2014, 02:11 PM)JmsNxn Wrote: Thats an awesome gadget you got there! I love that. I'm grumpy that my continuum sum using FC won't work here because you're using gaussians. But besides that this is quite beautiful. To me the only thing that's lacking is some rigor and epsilons and deltas, which I'm sure wouldn't take much to put in here. Yes, we need to find a condition for when the continuum sum will converge given a convergent $f(z)$ Hermite series. And I suppose the bit at the beginning about the integral might need to be improved a little -- we could, e.g. show that it is always smaller than a known convergent integral to make the proof complete. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 07/06/2014, 11:03 AM (This post was last modified: 07/06/2014, 11:29 AM by JmsNxn.) (07/06/2014, 04:14 AM)mike3 Wrote: Yes, we need to find a condition for when the continuum sum will converge given a convergent $f(z)$ Hermite series. And I suppose the bit at the beginning about the integral might need to be improved a little -- we could, e.g. show that it is always smaller than a known convergent integral to make the proof complete. Hmm. Can we say that $|f(x+yi)| < C_x e^{\alpha |y|}$ for $0 < \alpha < \pi/2$? for $x$ belonging to the area we want to continuum sum. This will guarantee a converging continuum sum with a triple integral transfrom from FC. I have all this rigorously laid out. If Faulbaher's continuum sum takes $s(s+1)(s+2)\cdots(s+n-1)\to \frac{1}{n}s(s+1)(s+2)\cdots(s+n)$ then My continuum sum may be the same as Faulbaher's. Now as for saying if the representation as continuum summed Hermite polynomials is convergent, I know some techniques from FC again that might work here. But they rely on the above and I'd have to take a closer look. This is really quite interesting, I like this representation. Are you going to show: $e^{\sum_{n=0}^{z-1} f(n)} = \frac{d}{dz} f(z)$ Or do you have a different more convenient pattern for $a_n$ fivexthethird Junior Fellow Posts: 9 Threads: 3 Joined: Nov 2013 07/06/2014, 02:11 PM Hm. Define the Weierstrass transform: $\mathcal{W}\{f\}(x) = \frac{1}{\sqrt{4\pi}}\int_{-\infty}^\infty f(t) \; e^{-\frac{(x-t)^2}{4}} \; dt$ And it's inverse $\mathcal{W}^{-1}\{f\}(x)_{x0} = \frac{1}{\sqrt{4\pi}} \int_{x_0-i\infty}^{x_0+i\infty} f(t)e^{\frac{(x-t)^2}{4}}\;dt$ Where $x_0$ is in the strip of convergence of the forward transform. This transform has the nice property that $\mathcal{W}\{H_n(\frac{x}{2})\}(t) = t^n$ Now let $f(x)$ be a function that is exponentially bounded (i.e. $|f(x+yi)| < Ce^{a|y|}$ on some strip $a < \mathfrak{R}(z) < b$. First, we have that $\mathcal{W}^{-1}\{f\}(x)_{c}$ is convergent for all $x \in \mathbb{R}$ if $a < c < b$. To show this, let $t =a+bi$. Then we have $|f(a+bi)e^{\frac{(x-(a+bi))^2}{4}}| < |Ce^{\frac{(x-(a+bi))^2}{4}+\alpha |b|}|$ For sufficiently large $|b|$ With some simple algebra, we can rearrange that to $|f(a+bi)e^{\frac{(x-(a+bi))^2}{4}}| < Ce^{\alpha |b| +\frac{1}{4}(a^2-b^2-2ax+x^2)}$ This is $o(e^{-|b|})$ as $b \rightarrow \pm\infty$ since $\alpha|b| - \frac{b^2+c}{4} < -|b|$ for sufficiently large $|b|$. Thus the integral converges as the integrand will decay quickly on the line of integration. Now let $a_n$ be an arbitrary sequence of complex numbers and $Ge(a_n,x) = \sum_{k=0}^{\infty} a_k x^k$ $He(a_n,x) = \sum_{k=0}^{\infty} a_k H_k(x)$ If $Ge(a_n,x)$ is exponentially bounded on some strip, we can take it's inverse Weierstrass transform. If $Ge(a_n,x)$ is also entire, we can exchange the integral and sum and use the identity $\mathcal{W}\{H_n(\frac{x}{2})\}(t) = t^n$ to get $\mathcal{W}^{-1}\{ Ge(a_n,t)\}(x)_{c} = \sum_{k=0}^{\infty} a_k H_k(\frac{x}{2})$ Which is just $He(a_n,x)$ Which also means that $\sum_{k=0}^{\infty} a_k H_k(\frac{x}{2})$ converges for all $x \in \mathbb{R}$ So if we let $h_n$ be the nth coefficient in the Hermite series for some function $f(x)$, we get that $-h_0iz + \sum_{n=0}^{iz-1} f(n) = \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} h_n {n \choose k-1} \frac{B_{n-k+1}}{k} (-2i)^{n-k}\right) H_k(z)$. converges for all $x \in \mathbb{R}$ if $g(z) = \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} h_n {n \choose k-1} \frac{B_{n-k+1}}{k} (-2i)^{n-k}\right) z^k$ defines an entire function that is exponentially bounded on some strip $a < \mathfrak{R}(z) < b$ It would probably be easy to weaken the growth bounds to something like $|f(x+iy)| for $0, but the integral would most likely no longer converge for all real x. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 07/06/2014, 11:04 PM (This post was last modified: 07/06/2014, 11:38 PM by mike3.) (07/06/2014, 11:03 AM)JmsNxn Wrote: (07/06/2014, 04:14 AM)mike3 Wrote: Yes, we need to find a condition for when the continuum sum will converge given a convergent $f(z)$ Hermite series. And I suppose the bit at the beginning about the integral might need to be improved a little -- we could, e.g. show that it is always smaller than a known convergent integral to make the proof complete. Hmm. Can we say that $|f(x+yi)| < C_x e^{\alpha |y|}$ for $0 < \alpha < \pi/2$? for $x$ belonging to the area we want to continuum sum. This will guarantee a converging continuum sum with a triple integral transfrom from FC. I have all this rigorously laid out. If Faulbaher's continuum sum takes $s(s+1)(s+2)\cdots(s+n-1)\to \frac{1}{n}s(s+1)(s+2)\cdots(s+n)$ then My continuum sum may be the same as Faulbaher's. Now as for saying if the representation as continuum summed Hermite polynomials is convergent, I know some techniques from FC again that might work here. But they rely on the above and I'd have to take a closer look. This is really quite interesting, I like this representation. Are you going to show: $e^{\sum_{n=0}^{z-1} f(n)} = \frac{d}{dz} f(z)$ Or do you have a different more convenient pattern for $a_n$ I would find that last identity dubious -- that would make the continuum sum just the log of the derivative! So you're suggesting to guarantee the convergence, you should limit $\alpha$ to be less than $\pi/2$? (Which is not a problem for tetration since it's actually bounded on the strip) is this to try and prove equivalence between this and your fractional calculus continuum sum? The coefficients $a_n$ depend on whatever function you're trying to continuum sum -- I don't get what you mean by a "more convenient pattern". It'll depend on the function, just as whether or not there's a "convenient pattern" for the Taylor series coefficients depends on the function. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 07/07/2014, 12:39 PM (This post was last modified: 07/07/2014, 12:42 PM by JmsNxn.) (07/06/2014, 11:04 PM)mike3 Wrote: I would find that last identity dubious -- that would make the continuum sum just the log of the derivative! That's an identity for tetration. If $f$ is tetration its derivative is the continuum product. (at least im pretty certain of that, tommy's always saying it and a little algebra shows it) (07/06/2014, 11:04 PM)mike3 Wrote: So you're suggesting to guarantee the convergence, you should limit $\alpha$ to be less than $\pi/2$? (Which is not a problem for tetration since it's actually bounded on the strip) is this to try and prove equivalence between this and your fractional calculus continuum sum? Well I meant if you have alpha less than pi/2 I can give you a continuum sum. It may not be faulbahers but it probably is (since continuum sums are unique if they are exponentially bounded on the real line and of exponential type less than pi/2 on the imaginary line.) It's a triple integral transform that's a little ugly but it works. (07/06/2014, 11:04 PM)mike3 Wrote: The coefficients $a_n$ depend on whatever function you're trying to continuum sum -- I don't get what you mean by a "more convenient pattern". It'll depend on the function, just as whether or not there's a "convenient pattern" for the Taylor series coefficients depends on the function. I meant, the following. if $f$ is tetration base e, then $e^{\sum_{j=0}^{z-1} f(j)} = \frac{d}{dz}f(z)$ With this relationship you can solve for $a_n$ using what I presume is a lot of summation identities. Is this what you plan to do? Or do you plan to find another way to solve for $a_n$. I.e: "a more convenient pattern." mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 07/08/2014, 06:31 AM (07/07/2014, 12:39 PM)JmsNxn Wrote: (07/06/2014, 11:04 PM)mike3 Wrote: I would find that last identity dubious -- that would make the continuum sum just the log of the derivative! That's an identity for tetration. If $f$ is tetration its derivative is the continuum product. (at least im pretty certain of that, tommy's always saying it and a little algebra shows it) (07/06/2014, 11:04 PM)mike3 Wrote: So you're suggesting to guarantee the convergence, you should limit $\alpha$ to be less than $\pi/2$? (Which is not a problem for tetration since it's actually bounded on the strip) is this to try and prove equivalence between this and your fractional calculus continuum sum? Well I meant if you have alpha less than pi/2 I can give you a continuum sum. It may not be faulbahers but it probably is (since continuum sums are unique if they are exponentially bounded on the real line and of exponential type less than pi/2 on the imaginary line.) It's a triple integral transform that's a little ugly but it works. (07/06/2014, 11:04 PM)mike3 Wrote: The coefficients $a_n$ depend on whatever function you're trying to continuum sum -- I don't get what you mean by a "more convenient pattern". It'll depend on the function, just as whether or not there's a "convenient pattern" for the Taylor series coefficients depends on the function. I meant, the following. if $f$ is tetration base e, then $e^{\sum_{j=0}^{z-1} f(j)} = \frac{d}{dz}f(z)$ With this relationship you can solve for $a_n$ using what I presume is a lot of summation identities. Is this what you plan to do? Or do you plan to find another way to solve for $a_n$. I.e: "a more convenient pattern." OK, so you're talking about tetration specifically -- that's why I was thrown off. I just took "$f$" to mean "arbitrary function". To try to solve these equations, we would need to find the exponential of the Hermite series, and then equate coefficients with the derivative I am not quite sure how to do this. In particular, the formula for an exponential of a power series is well-known, but a Hermite series? Now, the Weierstrass transform provides a transformation from Hermite to power series, as was mentioned here. But it's not clear if it can be used to crack what $\exp(f)$ means when $f$ is represented by a Hermite series. The question is whether or not the system of equations is linear or nonlinear, and if linear, is it infinite linear or not (given that the coefficients of the continuum sum are themselves infinite sums, it seems this is likely to be the case). If it is infinite linear, or nonlinear, it may not have a unique solution (which would be interesting -- what would these alternative-continuum-sum tetrationals be, anyways? Would they be the "alternate fixed point" solutions sheldonison has constructed? Something else altogether?), unless additional constraints can be applied. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 07/08/2014, 12:24 PM @mike the cleanest continuum sum is probably as I mentioned before , an analytic sexp with a parabolic fixpoint at sexp(1) = 1. A system of equations can be solvable even if infinite but that does not garantee convergeance. Im not convinced by all these transforms methods at the moment , though I would be nice. Maybe with additional work. regards tommy1729 « Next Oldest | Next Newest »

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