Analytical function from numerable point set

In summary, there are functions that can be extended from integer values to real values in a straightforward way, such as the factorial function extended to a gamma function. However, for arbitrary functions, there is no unique way to extend them to real values without additional requirements. Some possible methods for extension include using differential equations or Lagrange polynomials. Additionally, for certain sequences, there exists an entire function that can be constructed using a combination of exponential and polynomial terms.
  • #1
hilbert2
Science Advisor
Insights Author
Gold Member
1,598
605
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.
 
Last edited:
Mathematics news on Phys.org
  • #2
hilbert2 said:
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.
 
  • #3
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
 
  • #5
Svein said:

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.
 
  • #6
hilbert2 said:
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 {an} where [itex] \sum_{n=1}^{\infty}\frac{1}{\vert a_{n} \vert}[/itex] is convergent. Then [itex]f(z)=z^{m}e^{g(z)}\prod_{n=1}^{\infty}(1-\frac{z}{a_n{}})e^{p_{n}(z)}[/itex] is an entire function with a zero of order m at the origin and a zero at every an. Each pn(z) is a polynomial of a finite degree and g(z) is an entire function.

Whew!
 
  • #7
Comment to my post above: If [itex]\sum_{n=1}^{\infty}\frac{1}{\vert a_{n} \vert} [/itex] is convergent, all the pn(z) ≡ 0 (they are not necessary). If [itex]\sum_{n=1}^{\infty}\frac{1}{\vert a_{n} \vert} [/itex] is not convergent, then the pn(z) are necessary and different from 0.
 

Related to Analytical function from numerable point set

1. What is an analytical function?

An analytical function is a mathematical function that can be expressed as a finite combination of basic algebraic operations, such as addition, subtraction, multiplication, division, and exponentiation, along with well-defined functions like trigonometric, logarithmic, and exponential functions.

2. What is a numerable point set?

A numerable point set is a collection of points that can be counted and listed in a one-to-one correspondence with the positive integers, meaning that every point in the set can be assigned a unique whole number.

3. How is an analytical function from numerable point set different from other functions?

An analytical function from numerable point set is different from other functions in that it is a continuous and differentiable function that can be expressed as a finite combination of basic algebraic operations and well-defined functions, while also being defined on a numerable point set. This makes it a powerful tool for analyzing and understanding complex data sets.

4. What are some applications of analytical functions from numerable point sets?

Analytical functions from numerable point sets have a wide range of applications in various fields, including mathematics, physics, engineering, and data analysis. They are often used to model and analyze real-world phenomena and make predictions based on collected data.

5. What are some common techniques used to study analytical functions from numerable point sets?

Some common techniques used to study analytical functions from numerable point sets include calculus, numerical analysis, and computer programming. These techniques help to analyze the behavior and properties of these functions, as well as to visualize and interpret the data they represent.

Similar threads

Replies
4
Views
664
Replies
5
Views
896
  • General Math
Replies
12
Views
2K
Replies
2
Views
1K
Replies
13
Views
1K
Replies
7
Views
1K
Replies
5
Views
1K
  • General Math
Replies
16
Views
2K
Replies
6
Views
1K
Replies
1
Views
2K
Back
Top