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Math Help - Does a 'general extrapolation formula' exist?...

  1. #1
    MHF Contributor chisigma's Avatar
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    Does a 'general extrapolation formula' exist?...

    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
    Last edited by chisigma; March 27th 2009 at 07:00 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by chisigma View Post
    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
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    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
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  4. #4
    MHF Contributor chisigma's Avatar
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    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


    Last edited by chisigma; April 7th 2009 at 01:07 AM.
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  5. #5
    MHF Contributor chisigma's Avatar
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    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
    Last edited by chisigma; March 31st 2009 at 07:59 AM.
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  6. #6
    MHF Contributor chisigma's Avatar
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    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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