Continuous function that can be written as two distinct infinite polynomials I want to find a function $f:[0,1]\rightarrow \mathbb{R}$, where I can write
$$f(x)=\sum_{n=0}^\infty c_nx^n$$
and such $(c_n)_n$ is not unique.
I can exclude $f$ that is infinitely differentiable at $0$ since $f$ would have unique Maclaurin series,
but then I don't know which function would be equal to two distinct infinite polynomials.
 A: Let's assume you have sucha function, and you can write
$$
\sum_{n=0}^\infty a_n x^n = f(x) = \sum_{n=0}^\infty b_n x^n
$$
Then, subtract the RHS from the LHS to get
$$
0 = \sum_{n=0}^\infty (a_n - b_n) x^n, \quad \forall x \in (0,1).
$$
Can you prove this is only possible if $a_n=b_n$ for all $n \in \mathbb{N}$?
A: You don't actually say that your series converge on $[0,1]$, so we are free to use series that converge on smaller domains centered at $0$.  (We don't explicitly use this freedom below, but it is an area of freedom that one can exploit in this setting.)  You don't say that your series are produced by the recipe for Taylor series, so in fact we could just write down any two series and claim these are the two "Jason Dil" series for this function.  But let's assume you intend to use Taylor's recipe for projecting a function onto the space of analytic functions (centered at $0$).
Let $f$ be some function with domain $[0,1]$,  \begin{align*}
f_1 &= \sum_{n=0}^\infty c_n x^n \text{, and}  \\
f_2 &= \sum_{n=0}^\infty d_n x^n \text{,}
\end{align*}
with $(c_n)_n \neq (d_n)_n$, where $f_1$ and $f_2$ each converge on at least one point.  Let $S_1$ be the radius of agreement of $f_1$ with $f$,
$$  S_1 = \sup \{u : \left. f \right|_{[0,u)} = \left. f_1 \right|_{[0,u)} \}  \text{,}  $$
and $S_2$ be the radius of agreement of $f_2$ with $f$.  Let $S = \min \{S_1, S_2\}$.
(Note that $S_i$ is less than or equal to the radius of convergence of $f_i$ for $i \in \{1,2\}$.  That equality is not assured can be seen most easily by expanding our context to $\Bbb{C}$.  Start with an entire $f$.  Outside of the strip $-1 < \Im(x) < 1$, replace $f$ with $f+1$.  The radius of agreement with this modified function cannot exceed $1$, but the radius of convergence of the power series is $\infty$.)
If $S > 0$, since $f_1 = f_2$, on $0 \leq x < S$, we have
$$  f_1 - f_2 = 0,  \quad 0 \leq x < S  \text{,}  $$
or, what is the same thing,
$$  \sum_{n=0}^\infty (c_n - d_n) x^n = 0,  \quad 0 \leq x < S  \text{.}  $$
By the identity theorem for power series (see this for a statement and some discussion),
$c_n = d_n$ for all $n$.
If $S = 0$, then at least one of $f_1$ and $f_2$ has radius of agreement $0$.  WLOG, assume $f_1$ has radius of agreement $0$ (otherwise, swap the labels "$f_1$" and "$f_2$").  This means $f_1$ only agrees with $f$ when $x = 0$ and therefore has $c_0 = f_1(0) = f(0)$.  Since $f_2$ at least agrees with $f$ on $\{0\}$, we also have $d_0 = f_2(0) = f(0)$.  So $d_0 = c_0$ is forced.  But then every other coefficient is completely unconstrained.
So, for your desired scenario to happen, let $f_2$ be a Taylor series of $f$ at $0$ (using one-sided derivatives at $0$) agreeing with your function on some interval (possibly of zero width) and let $f_1$ be any Maclaurin series (different from $f_2$) with zero radius of convergence and same constant coefficient as $f_2$ (for example, $c_0 = f(0)$ and $c_n = n!$ for $n > 0$).  This is perhaps unsatisfying as these two series only agree at one point -- but we have seen above that if the series agree on an interval, they are identical, so one point is the largest interval of agreement for two series having different coefficients.
Perhaps such an $f_1$ seems strange.  Here is an example.  Observe $\lim_{x \rightarrow 0} \mathrm{e}^{-1/x^2} = 0$ and define the continuous function
$$  f(x) =  \begin{cases} \mathrm{e}^{-1/x^2} ,& x \neq 0  \\
0 ,& x = 0 \end{cases}  \text{.}  $$
Since $f(0) = f'(0) = f''(0) = \cdots = 0$, the Maclaurin series of $f$ is $0$.  It agrees with $f$ only at $x = 0$, even though the series has infinite radius of convergence.
A: This is not possible. $f$ can be extended to a map defined on $(-1/2, 1/2)$ expandable into a power series on that interval. Therefore $f$ is indefinitely differentiable and $c_n = \frac{f^{(n)}(0)}{n!}$.
