Does a 'general extrapolation formula' exist?...

**chisigma** · March 27th 2009, 05:13 AM

Ladies and Gentleman…
… at the end of this thread…

http://www.mathhelpforum.com/math-he...variables.html

… Opalg kindly invited me to ask you some questions regarding a problem about which I’m working now. The problem is to examine the possibility to full reconstruction of a function $f(*)$ from the knowledge of a discrete set of value $f_{k}$ . I started supposing that $f(*)$ is a complex variable function which is analytic in the whole complex plane, i.e. that $f(*)$ is an entire function, so that we can write it as Taylor expansion around $z=0$ …

$f(z)=\sum_{n=0}^{\infty}a_{n}\cdot z^{n}$ (1)

We consider $f(*)$ ‘sampled’ on the real axis at $z=k$ with $k = 0 , 1 , 2 , \dots$ , call $f_{k}=f(k)$ the ‘samples’ and use them to construct the following real variable function…

$\varphi(t)=\sum_{k=0}^{\infty} (-1)^{k}\cdot f_{k}\cdot \frac{t^{k}}{k!}= \sum_{k=0}^{\infty} (-1)^{k}\cdot \frac{t^{k}}{k!}\cdot \sum_{n=0}^{\infty}a_{n}\cdot k^{n} = \sum_{n=0}^{\infty}a_{n}\cdot \varphi_{n}(t)$ (2)

In order to find the $\varphi_{n}(*)$ let’s write (2) ‘in extenso’…

$\varphi(t)= 1\cdot \sum_{n=0}^{\infty}a_{n}\cdot 0^{n} -$

$- t\cdot \sum_{n=0}^{\infty} a_{n}\cdot 1^{n} +$

$+ \frac {t^{2}}{2}\cdot \sum_{n=0}^{\infty}a_{n}\cdot 2^{n} -$

$- \frac{t^{3}}{6}\cdot \sum_{n=0}^{\infty}a_{n}\cdot 3^{n} + \dots =$

$= a_{0}\cdot (1-t+\frac{t^{2}}{2} - \frac{t^{3}}{6} +\dots)+$

$+ a_{1}\cdot (-t+t^{2} - \frac{t^{3}}{2} +\dots) +$

$+ a_{2}\cdot (-t+ 2\cdot t^{2} - \frac{3\cdot t^{3}}{2} +\dots) +$

$+ a_{3}\cdot (-t+ 4\cdot t^{2} - \frac {9\cdot t^{3}}{2} + \dots) + \dots$ (3)

Comparing (2) and (3) it is easy to observe that the $\varphi_{n}(t)$ obey to the recursion formula…

$\varphi_{0}(t)= e^{-t}$

$\varphi_{n+1}(t)= t\cdot \frac {d}{dt} \varphi_{n}(t)$ (4)

At this point some question for you…

a) what I have done till now is correct or not?…
b) what can we say more about the $\varphi(t)$ and the $\varphi_{n}(t)$ ?…

Kind regards

$\chi$ $\sigma$

**CaptainBlack** · March 27th 2009, 08:17 AM

Originally Posted by chisigma

Ladies and Gentleman…
… at the end of this thread…

http://www.mathhelpforum.com/math-he...variables.html

… Opalg kindly invited me to ask you some questions regarding a problem about which I’m working now. The problem is to examine the possibility to full reconstruction of a function $f(*)$ from the knowledge of a discrete set of value $f_{k}$ . I started supposing that $f(*)$ is a complex variable function which is analytic in the whole complex plane, i.e. that $f(*)$ is an entire function, so that we can write it as Taylor expansion around $z=0$ …

$f(z)=\sum_{n=0}^{\infty}a_{n}\cdot z^{n}$ (1)

We consider $f(*)$ ‘sampled’ on the real axis at $z=k$ with $k = 0 , 1 , 2 , \dots$ , call $f_{k}=f(k)$ the ‘samples’ and use them to construct the following real variable function…

$\varphi(t)=\sum_{k=0}^{\infty} (-1)^{k}\cdot f_{k}\cdot \frac{t^{k}}{k!}= \sum_{k=0}^{\infty} (-1)^{k}\cdot \frac{t^{k}}{k!}\cdot \sum_{n=0}^{\infty}a_{n}\cdot k^{n} = \sum_{n=0}^{\infty}a_{n}\cdot \varphi_{n}(t)$ (2)

In order to find the $\varphi_{n}(*)$ let’s write (2) ‘in extenso’…

$\varphi(t)= 1\cdot \sum_{n=0}^{\infty}a_{n}\cdot 0^{n} -$

$- t\cdot \sum_{n=0}^{\infty} a_{n}\cdot 1^{n} +$

$+ \frac {t^{2}}{2}\cdot \sum_{n=0}^{\infty}a_{n}\cdot 2^{n} -$

$- \frac{t^{3}}{6}\cdot \sum_{n=0}^{\infty}a_{n}\cdot 3^{n} + \dots =$

$= a_{0}\cdot (1-t+\frac{t^{2}}{2} - \frac{t^{3}}{6} +\dots)+$

$+ a_{1}\cdot (-t+t^{2} - \frac{t^{3}}{2} +\dots) +$

$+ a_{2}\cdot (-t+ 2\cdot t^{2} - \frac{3\cdot t^{3}}{2} +\dots) +$

$+ a_{3}\cdot (-t+ 4\cdot t^{2} - \frac {9\cdot t^{3}}{2} + \dots) + \dots$ (3)

Comparing (2) and (3) it is easy to observe that the $\varphi_{n}(t)$ obey to the recursion formula…

$\varphi_{0}(t)= e^{-t}$

$\varphi_{n+1}(t)= t\cdot \frac {d}{dt} \varphi_{n}(t)$ (4)

At this point some question for you…

a) what I have done till now is correct or not?…
b) what can we say more about the $\varphi(t)$ and the $\varphi_{n}(t)$ ?…

Kind regards

$\chi$ $\sigma$

Shannon's theorem?

CB

**chisigma** · March 27th 2009, 11:55 AM

Originally Posted by CaptainBlack

Shannon's theorem?

CB

The analysis is just in the initial stage... but it is not impossible we arrive to a result that has much in common with the Shannon's theorem and also the Nyquist's theorem...

... in any case i do hope that Captain Black, Opalg, Perfect Hacker and other 'theorists' will give great contribution

...

Kind regards

$\chi$ $\sigma$

**chisigma** · March 30th 2009, 12:27 AM

In my previous post it has been demonstrated that, given an entire function $f(z)= \sum_{n=0}^{\infty}a_{n}\cdot z^{n}$ ‘sampled’ at $z=k, k=0,1,\dots$ , a real function $\varphi(t)$ can be constructed from it as…

$\varphi(t)=\sum_{k=0}^{\infty} (-1)^{k}\cdot f_{k}\cdot \frac{t^{k}}{k!}= \sum_{n=0}^{\infty}a_{n}\cdot \varphi_{n}(t)$ (1)

… where $f_{k}=f(k)$ and the $\varphi_{n}(*)$ can be obtained in recursive way as…

$\varphi_{0}(t)= e^{-t}$

$\varphi_{n+1}(t)= t\cdot \frac{d}{dt} \varphi_{n}(t)$ (2)

Using the definition od ‘exponential generating function’ (Exponential Generating Function -- from Wolfram MathWorld), one can also define $\varphi (t)$ as the ‘exponential generating function’ of the real sequence $s_{k}= (-1)^{k}\cdot f(k)$ . In order to proceed we have to establish some important details. For example: at what conditions the series (1) does converge?… It is easy to demonstrate that (1) converges if exists a real constant $a>1$ for which is…

$\forall z , |f(z)|<a^{|z|}$ (3)

… so that ‘almost all’ the $f(*)$ do have the corresponding $\varphi(*)$ . The second step is to find an explicit expression for the $\varphi(*)$ . Appling the recursive relations (2) it is not difficult to arrive to demonstrate that…

$\varphi_{n}(t)= \sum_{k=0}^{\infty}(-1)^{k}\cdot k^{n}\cdot \frac{t^{k}}{k!}$ (4)

… so that each $\varphi_{n}(*)$ is the ‘exponential generating function’ of the sequence $e_{k , n}= (-1)^{k}\cdot k^{n}$ . Because the problem I'm 'attacking' is [for me of couse...] 'a little hard', I do hope somebody answers to a little question: is correct what I have done till now?...

Kind regards

$\chi$ $\sigma$

**chisigma** · March 31st 2009, 04:54 AM

It seems that for you what I’ve produced till now is ‘all right’, so that we can proceed in the next step… but before proceeding let’s have a little ‘pause of careful consideration’. In the XVIII° century the German mathematician Leonhard Euler dedicated great attention to the function…

$\Gamma(z)= \int_{0}^{\infty} t^{z-1}\cdot e^{-t}\cdot dt$ (1)

… so that it is commonly called ‘Euler’s gamma function’. For reason you easily understand sometime is better use this little different form of the function…

$\gamma(z)= \int_{0}^{\infty} t^{z}\cdot e^{-t}\cdot dt$ (2)

Of course is $\gamma(z)=\Gamma(z+1)$ and the two function have similar properties. For $z=k , k=0,1,\dots$ is $\gamma(k)=k!$ [and it is very comfortable for my to avoid confusion in remembering if it is $\Gamma(k)=(k-1)!$ or $\Gamma(k)=(k+1)!$

…]. As for the $\Gamma(*)$ there are the ‘forward’ and ‘backward’ identities…

$\gamma(z+1)= (z+1)\cdot \gamma(z)$

$\gamma(z-1)=\frac{\gamma(z)}{z}$ (3)

Strictly connected to the function $\gamma(*)$ is its reciprocal that we call $\eta(*)$ …

$\eta(z)=\frac{1}{\gamma(z)}$ (4)

… which is an entire function which is alternatively defined as ‘infinite product’…

$\eta(z)= e^{\gamma\cdot z}\cdot \prod_{n=1}^{\infty}(1+\frac{z}{n})\cdot e^{- \frac{z}{n}}$ (5)

... where $\gamma$ is the so called 'Euler's constant' and is...

$\gamma = \lim_{k\rightarrow \infty} \sum_{n=1}^{k}\frac{1}{n} - \ln k = .577215664901\dots$ (6)

Taking into account that is…

$sinc(z)= \frac{\sin (\pi z)}{(\pi z)}= \prod_{n=1}^{\infty}(1 - \frac{z^{2}}{n^{2}})$ (7)

… from (5) you can derive the nice identity…

$\eta(z)\cdot \eta(-z)= sinc(z)= \frac{\sin (\pi z)}{(\pi z)}$ (8)

All that will be very useful in next posts…

Kind regards

$\chi$ $\sigma$

**chisigma** · April 1st 2009, 02:03 AM

Today we will proceed in the next step… a very important step!…

In the previous post I’ve demonstrated that, given an entire complex variable function $f(z)=\sum_{n=0}^{\infty}a_{n}\cdot z^{n}$ defined for $z=k, k=0,1,\dots$ , we can construct a real variable function $\varphi(*)$ defined as…

$\varphi(t)= \sum_{k=0}^{\infty} (-1)^{k}\cdot f(k)\cdot \frac{t^{k}}{k!}= \sum_{n=0}^{\infty}a_{n}\cdot \varphi_{n}(t)$ (1)

… where…

$\varphi_{0}= e^{-t}$

$\varphi_{n+1}(t)= t\cdot \frac{d}{dt} \varphi_{n}(t)$ (2)

Now let’s consider the integral…

$\int_{0}^{\infty}t^{z-1}\cdot \varphi(t)\cdot dt= \sum_{n=0}^{\infty}a_{n}\cdot \int_{0}^{\infty} t^{z-1}\cdot \varphi_{n}(t)\cdot dt$ (3)

... and start computing the integral at second member for $n=0$ . It is...

$\int_{0}^{\infty}t^{z-1}\cdot \varphi_{0}(t)\cdot dt= \Gamma(z)= \gamma (z-1)$ (4)

... where $\Gamma(*)$ is the 'Euler's gamma function' and $\gamma(*)$ is the 'modified gamma function' we have defined in last post. For the others index $n$ you have to take in mind the (2) and use them to compute the following indefinite integral...

$\int t^{z-1}\cdot \varphi_{n}(t)\cdot dt= \int t^{z}\cdot \frac{d}{dt} \varphi_{n-1}(t)\cdot dt= t^{z}\cdot\varphi_{n-1}(t) - z \cdot \int t^{z-1}\cdot \varphi_{n-1}(t)\cdot dt$ (5)

Integrating the (5) in the interval $[0,\infty]$ we obtain first…

$\int_{0}^{\infty}t^{z-1}\cdot \varphi_{n}(t)\cdot dt= -z\cdot \int_{0}^{\infty}t^{z-1}\cdot \varphi_{n-1}(t)\cdot dt$ (6)

… and after n successive steps…

$\int_{0}^{\infty}t^{z-1}\cdot \varphi_{n}(t)\cdot dt = (-z)^{n}\cdot \int_{0}^{\infty} t^{z-1}\cdot \varphi_{0}(t)\cdot dt= (-z)^{n}\cdot \gamma (z-1)$ (7)

Now if we insert the (7) into the (1) the result is the following…

$\int_{0}^{\infty}t^{z-1}\cdot \varphi(t)\cdot dt = \gamma (z-1)\cdot \sum_{n=0}^{\infty}a_{n}\cdot (-z)^{n}= \gamma(z-1)\cdot f(-z)$ (8)

… or, that is the same, the following…

$f(-z)= \eta(z-1)\cdot \int_{0}^{\infty}t^{z-1}\cdot \varphi(t)\cdot dt= \eta(z-1)\cdot \int_{0}^{\infty}t^{z-1}\cdot \sum_{k=0}^{\infty}(-1)^{k}\cdot f(k)\cdot \frac{t^{k}}{k!} \cdot dt$ (9)

The (9) is quite interesting because it allow us, given the ‘samples’ $f(k)$ of an entire function $f(*)$ in $k=0,1,\dots$ , to compute $f(*)$ in the whole left half-plane of the domain of the complex variable z… not bad!

…

The familiar question for you: is all that correct or not?…

Kind regards

$\chi$ $\sigma$

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