Analytical function from numerable point set

[hilbert2]

Sometimes there are functions that are initially defined for only integer values of the argument, but can be extended to functions of real variable by some obvious way. An example of this is the factorial $n!$ which is extended to a gamma function by a convenient integral definition.

So, if I have an arbitrary function $f(x)$ for which I know its value for any $x\in\mathbb{N}$, is it possible to uniquely extend it to $x\in\mathbb{R}$ so that it becomes a $\mathcal{C}^\infty$ function?

Possible examples could be $f(n) = 1 + 2 + \dots + n$, or the "power tower" function $f(z,n) = z^{z^{\cdot^z}}$ with $n$ $z$:s in it.

Edit: For the first example it is straightforward to prove by induction that $f(n) = \frac{n(n+1)}{2}$ and then plug any value of $n$ in it, but for the second one it is not as obvious.

 

[mfb]

So, if I have an arbitrary function $f(x)$ for which I know its value for any $x\in\mathbb{N}$, is it possible to uniquely extend it to $x\in\mathbb{R}$ so that it becomes a $\mathcal{C}^\infty$ function?

You can always add something like sin(pi x), multiply by cos(2 pi x) and similar things. It can never be unique without additional requirements. As a prominent example, the gamma function can be uniquely defined by the factorials and by being logarithmic convex.

I wonder if, given f(z) for all $z\in\mathbb{N}$, there is always an entire function in the complex numbers.

 

[hilbert2]

Thanks. For a function with initially one real and one integer variable, $f(x,n)$, I think one approach could be to find the differential equations for which $f(x,1)$, $f(x,2)$, and so on, are solutions, and then try to decide a DEs that would naturally produce $f(x,1/2)$ etc. as one solution.

For instance, $f(x,n) = e^{nx}$ is a solution of $\frac{df(x,n)}{dx} = nf(x,n)$ with $f(0,n) = 1$. Sometimes fractional differentiation could also be of use if it can be shown that $f(x,n) = \frac{d^n g(x)}{dx^n}$ for some known function $g$.

Edit: Oh, looks like the extention of "power tower" to non-integer heights is discussed here in Wikipedia: https://en.wikipedia.org/wiki/Tetration#Extension_to_real_heights

 

[Svein]

Check out https://en.wikipedia.org/wiki/Lagrange_polynomial.

 

[hilbert2]

Check out https://en.wikipedia.org/wiki/Lagrange_polynomial.

Thanks. That seems to be for a function defined on an interval of finite length, though. This may cause some problems similar to how a Fourier series becomes a Fourier integral when you try to represent a function defined on whole $\mathbb{R}$ as a linear combination of $e^{ikx}$ terms.

 

[Svein]

Thanks. That seems to be for a function defined on an interval of finite length, though. This may cause some problems similar to how a Fourier series becomes a Fourier integral when you try to represent a function defined on whole $\mathbb{R}$ as a linear combination of $e^{ikx}$ terms.

OK. So: Assume a sequence {a_n} where $\sum_{n=1}^{\infty}\frac{1}{\vert a_{n} \vert}$ is convergent. Then $f(z)=z^{m}e^{g(z)}\prod_{n=1}^{\infty}(1-\frac{z}{a_n{}})e^{p_{n}(z)}$ is an entire function with a zero of order m at the origin and a zero at every a_n. Each p_n(z) is a polynomial of a finite degree and g(z) is an entire function.

Whew!

 

[Svein]

Comment to my post above: If $\sum_{n=1}^{\infty}\frac{1}{\vert a_{n} \vert}$ is convergent, all the p_n(z) ≡ 0 (they are not necessary). If $\sum_{n=1}^{\infty}\frac{1}{\vert a_{n} \vert}$ is not convergent, then the p_n(z) are necessary and different from 0.