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.
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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,
\( \sum_{n=0}^{z-1} f(n) = a_0 z + \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} a_n {n \choose k-1} \frac{B_{n-k+1}}{k} (-2i)^{n-k}\right) H_k(-iz) \).
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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.
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,
\( \sum_{n=0}^{z-1} f(n) = a_0 z + \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} a_n {n \choose k-1} \frac{B_{n-k+1}}{k} (-2i)^{n-k}\right) H_k(-iz) \).
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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.