is it possible to find a function knowing all of its derivatives at $x=0$?

Question

Let's say I have a series of real values $y_0,y_1,y_2\cdots$. My question is if it's always possible to find (at least one) $C^\infty$ real function such that \begin{equation} f^{(n)}(0)=y_n \end{equation} and in the affirmative case, how. It is a kind of "reverse taylor" problem... any hints?

Answer 1

I am going to address a complementary issue: the fact that if a solution to this problem exists, it is not unique.

This is due to the existence of (Laurent) Schwartz function defined by

$\varphi(x)=\exp(-\tfrac{1}{x^2})$ for $x \neq 0$ and $\varphi(0) = 0$, which is such that $\forall n\in\mathbb{N}, \ \varphi^{(n)}(0)=0$.

Thus, if one has a function $f$ such that $f^{(n)}(0)=y_n,$ then $f+k\varphi$ is also a solution.

Remark: Using this function, one can define more general so-called "bump functions" (https://en.wikipedia.org/wiki/Bump_function).

Graphical representation of function $\varphi$: the very flat behavior in $0$ accounts for the fact that $0$ has an infinite order of multiplicity. One notes also a horizontal asymptote with equation $y=e^0=1$

Answer 2

Yes, it is possible to find a function that satisfies the condition $f^{\left(n\right)}\left(0\right)=y_n$ when the sequence $\left<y_n\right>$ is given, though infinite such functions exist.

One way of finding a possible function would be to a polynomial of degree $n$. Consider $$p\left(x\right)=a_0+a_1x+a_2x^2 +\cdots+a_nx^n$$

Now, let $$f\left(x\right)=p\left(x\right)e^x$$ Then, we have $$\begin{array}\\ f\left(0\right)&=&a_0&=&y_0\\ f^1\left(0\right)&=&a_0+a_1&=&y_1\\ f^2\left(0\right)&=&a_0+2a_1+2a_2&=&y_2\\ &\vdots& \end{array}$$

We get $n+1$ equations for the $n+1$ variables $\left<a_0,\,a_1,\,\dots,\,a_n\right>$, and simply solve the equations to get the values of the coefficients and hence the function. A computer algebra system can be used if the value of $n$ is large.

But, the function so obtained is clearly not the only possible function. To generate more functions, we can simply increase the degree of the polynomial $p$ to an arbitrary value. Since the coefficients of powers of $x$ higher than $n$ do not occur in our set of equations, we may assign any arbitrary value to them. Still then, we cannot cover all possible definitions of $f$.

NOTE: We could simply choose $f$ to be a polynomial, but since $f$ must be $C^\infty$, therefore we must choose such a function which has derivatives of all orders everywhere in its domain.