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An unknown function $y=f(x)$ is infinitely differentiable everywhere on $\mathbb{R}$.

Known the values of every points $ (x,y) $ where $0<x<1 $ ,

Could you please show whether there exist a unique solution of $y=f(x)$ ?

If not, could you please show that there exist infinite number of solution of $y=f(x)$ ?


My approach:

$$y=f(x)=\Sigma a_nx^{n}$$

we have infinite number of data point $(x_i,y_i)$ to solve for $a_n$, through a system of infinite number of linear equations:

Xa = y

Where a is single column, infinite matrix $(a_1,a_2,...,a_n, ...)$

y is a single column, infinite matrix $(y_1,y_2,...,y_m, ...)$

X is a $\mathbb{R}^2$ infinite matrix

\begin{pmatrix} x_1^{1} & x_1^{2} & ... & x_1^{n} & ...\\ x_2^{1} & x_2^{2} & ... &x_2^n & ... \\ ... \\ x_m^{1} \\ ...\end{pmatrix}

Using definitions from paper 'SEQUENCE SPACES AND INVERSE OF AN INFINITE MATRIX', we could evaluate the inverse of an infinite matrix.

Then

a = $ \textbf{X}^{-1}$ y

This way we solved all $a_n$, and the solution to f(x) found.

However my intuition told me that there should be infinite number of solutions available for f(x).


What branch of Math shall I explore to solve this kind of problems?

{Math Newbie and lover here. Thanks for your help!!}


Case 2, instead of known the values of every points $ (x,y) $ where $0<x<1 $,

we know the values of every points $ (x,y) $ where $x \in \mathbb{N} $


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Your intuition is correct, there are infinitely many possible solutions. Indeed, you may multiply any given solution by an everywhere differentiable function that equals 1 on $0 <x <1$ but not everywhere (*) to obtain a new and different solution.

But your results are still correct, in that there is a unique solution of the form $\sum a_n x^n$ (actually, there could also be no solution of this form). To learn more about solutions of the this form, and why they are unique if they exist, you may want to pick up a book on 'Complex analysis'.

EDIT: In your case 2, there actually exist multiple solutions. Consider $ sin(2\pi x) $ and the constant zero function.

(*) Such functions exist, Compactly supported infinitely differentiable function constant on an interval

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  • $\begingroup$ OMG I love your username $\endgroup$ – High GPA May 7 '17 at 7:25
  • $\begingroup$ Wrong about what? BTW, I now also comment on your case 2 $\endgroup$ – Bananach May 7 '17 at 21:22
  • $\begingroup$ Please correct me if I am wrong: Since there are infinitely number of solutions for y=f(x), and all f(x) are infinitely differentiable. That is, all the solutions f(x) could be written as the form of ∑$a_nx^x$. Therefore, the system of linear equations $\textbf{Xa=y}$ must produce either infinite sets of solutions for $\textbf{a}$, or no solutions at all. $\endgroup$ – High GPA May 7 '17 at 21:35
  • $\begingroup$ Please put more effort into your writing. A lack of English language knowledge cannot excuse sentences without verbs, statements out of nowhere, broken math markup, and complete lack of explanation what you are even trying to do. Are you trying to summarize my answer and ask me if I agree with the summary? $\endgroup$ – Bananach May 7 '17 at 21:43
  • $\begingroup$ I have the impression you believe that any function that is infinitely often differentiable can be written as $\sum a_n x^n $. This is not true $\endgroup$ – Bananach May 7 '17 at 21:46

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