prove Taylor of $R(a)$ converges $R$ but its sum equals $R(a)$ for $a$ in interval.Which interval? Pls I'm glad to give an idea or hint?:

$T(a)=\begin{cases} 0 & a\leq 0\\ e^{-1/a} & a>0 \end{cases}$

$Z(a)=\begin{cases}0 & a\geq 1\\ (1+a) e^{-1/(1-a)} & a<1 \end{cases}$

$p=\int_{-1}^{1}Z(x)dx$

$Y(a)=\frac{1}{p}\int_{-1}^{a}Z(x)dx$

$R(a)=Y(a-1)Y(2-a)$

(Why all four functions are in $C^{\infty}(R)$)

My real question is that –>

how to prove the Taylor series of $R(a)$ centered at zero converges to $R$ however its sum equals to $R(a)$ for $a$ in a much smaller interval. What is the interval?

Please Help me for how to solve this problem? It is so difficult. I dont know how to solve. Thank you:)

Solutions Collecting From Web of "prove Taylor of $R(a)$ converges $R$ but its sum equals $R(a)$ for $a$ in interval.Which interval? Pls I'm glad to give an idea or hint?:"

First part: show by induction that
$$
T^{(n)}(x)=\frac{P_n(x)}{x^{2n}}e^{-1/x} \qquad\forall x>0
$$
whith $P_n$ a polynomial.

It follows that
$$
\lim_{0^-}T^{(n)}(x)=\lim_{0^+}T^{(n)}(x)=0
$$
for all $n$.

So $T$, which is obviously $C^\infty$ on $\mathbb{R}^*$, is also $C^\infty$ at $0$.

Now $Z,Y,R$ are built from $T$ in such a way that they are also $C^\infty$.

Second part: Note that
$$
1=Y(1)=Y(x)\qquad \forall x\geq 1.
$$
It follows that
$$
R(x)=Y(x-1)\qquad \forall x\leq 1.
$$

Now observe that the complex function
$$
f(z)=(1+z)e^{-\frac{1}{1-z}}
$$
is holomorphic on $\mathbb{C}\setminus\{1\}$.

In particular, $f$ is analytic on $(-3,1)$, so $Z$ is analytic on $(-3,1)$. Both are equal to their Taylor series on $(-3,1)$, by restriction of the holomorphic function $f$ to the reals.

Therefore the antiderivative
$$
g:x\longmapsto \int_{-1}^xZ(t)dt
$$
is analytic and equal to its Taylor series at $0$ on $(-3,1)$, so
$$
g(-1+x)=\sum_{n\geq 0}\frac{g^{(n)}(-1)}{n!}x^n \qquad \forall x\in (-2,2).
$$

Now a fortiori
$$
R(x)=Y(x-1)=\frac{1}{p}g(-1+x) \qquad \forall x\in (-1,1)
$$
is analytic and equal to its Taylor series at $0$ on $(-1,1)$.

Recall the radius of convergence of the Taylor series of a holomorphic function at $z_0$ is the distance between $z_0$ and the complement of the domain of analyticity. It follows that the radius of convergence of the Taylor series above is $2$.

Also, $Y(x-1)$ is equal to its Taylor series at $0$ on $(-2,2)$.
So it is equal to
$$
Y(2-x)Y(x-1)=R(x) \qquad \forall x\in (-2,1]
$$
since $Y(2-x)=1$ there.

And on $(1,2)$, we have $Y(2-x)\neq 1$ and $Y(x-1)\neq 0$ so the Taylor series does not equal $R(x)$ there.

For the first one we obviously have any order derivative of $T(a)$ is differentiable for $a < 0$ and $a > 0$ so you are left with the case $a = 0$. For that you can just get a formula for the $n^\text{th}$ derivative of the exponential part of the function and show that the limit as $a \to 0$ is $0$. So $T \in C^\infty$.

$Z(a)$ can be written as $Z(a) = (1 + a)T(1 – a)$ which shows $Z \in C^\infty$.

$Y$ is defined as an integral of a continuous function, so it must be differentiable and its derivative is $Y'(a) = \frac{1}{p}Z(a)$ whose higher order derivatives are obviously differentiable and so $Y \in C^\infty$.

For $R$, any order derivatives are just a sum of products of differentiable functions, so $R \in C^\infty$.