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)
(and so )
) (for the polynomials of the second type mentioned, denoted 
), which is the space of all Lebesgue-integrable functions 
(and by extension 
) satisfying
|^2 e^{-x^2} dx < \infty)
.
In particular, every function which is exponentially bounded, i.e. for which| < Ce^{a|x|})
(and also satisfies the Lebesgue integrability requirement) belongs to this space. This is easy to see, since we then have |^2 < C^2 e^{2a|x|})
and then |^2 e^{-x^2} < C^2 e^{2a|x| - x^2})
. This is )
as 
because 
for sufficiently large 
(take 
). Therefore the integral converges as the integrand will always decay quickly to 0.
So, suppose
now is a complex function satisfying:
1.)
is holomorphic for  > -2)
2. is exponentially bounded in the strip  < 1)
, that is, | < Ce^{a|\Im(z)|})
in that strip.
Then)
, 
, will meet the requirements for membership in that space and so we can write it as a "Hermite series":
 = \sum_{n=0}^{\infty} a_n H_n(-ix))
.
Of course, we are interested in the case where = \mathrm{tet}(z))
.
Hermite Transform
This series formula leads to the following notion. Given a sequence
, we could define a "Hermite Transform" by
 = \sum_{n=0}^{\infty} a_n H_n(x))
.
The inverse transform can be found via the orthogonality properties of the polynomials, namely
 H_m(x) e^{-x^2} dx = \sqrt{\pi} 2^n n! \delta_{n,m})
where
is the Kronecker delta. This allows us to integrate a function against the Hermite polynomial and so extract a coefficient of the sequence , giving
 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_{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 = \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
, 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
 = \sum_{k=0}^{n} {n \choose k} H_k(x) (2y)^{n-k})
.
Now take
and 
(so )
) and subtract )
to get
 = 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 = 1)
.
Summing each side with the Faulhaber operator gives
 = \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
 = H_n(-iz) - \sum_{k=0}^{n-2} {n \choose k} (-2i)^{n-k} Hs_k(z))
or
 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 = 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:
 \frac{t^n}{n!})
or, for the imaginary axis,
 = \sum_{n=0}^{\infty} H_n(-iz) \frac{x^n}{n!})
.
Continuum summing with the Faulhaber operator (which is valid analytically for
(and so extends elsewhere by analytic continuation) and valid formally), we get
 = \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 - 1}{\exp(-2ix) - 1})
, so we have the generating function for the Hermite sum polynomials as
 - 1}{\exp(-2ix) - 1} = \sum_{n=0}^{\infty} Hs_n(z) \frac{x^n}{n!})
.
We can now obtain an explicit solution for)
. Rewrite the left side as follows:
 - 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, we see that the constant term of  - 1)
is 0, and carrying out the division, we get
 - 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 - 1})
is the generating function for the Bernoulli numbers, and so we get
}{\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
 - 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
 = \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,
 = 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
In particular, every function which is exponentially bounded, i.e. for which
So, suppose
1.
2.
Then
Of course, we are interested in the case where
Hermite Transform
This series formula leads to the following notion. Given a sequence
The inverse transform can be found via the orthogonality properties of the polynomials, namely
where
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
where we define the "Hermite sum polynomials" to be
To obtain expressions for the sum polynomials
Recurrence formula
Start with the identity
Now take
where the last equality follows from
Summing each side with the Faulhaber operator gives
Taking the last term off the sum, we get
or
Starting with
Explicit formula
Now for the explicit formula. To do this, we start with the generating function of the Hermite polynomials:
or, for the imaginary axis,
Continuum summing with the Faulhaber operator (which is valid analytically for
Now the left hand sum sums to
We can now obtain an explicit solution for
Now we have the generating function as a product of two functions which have no poles. Using that
Then, for the other function, we recognize that
Finally, multiplying together both of these functions and applying the binomial convolution to the series, we get
and therefore
Then we have, by rearranging the order of summation in the equation for the continuum sum of the function,
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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.
