Kdtyjb

You are given a function $f: mathbb{R}rightarrow mathbb{R}$. Every derivative $frac{d^n}{dx^n}(f(x)), ,n >0$ of the function is continuous.

Is there a function $g: mathbb{R}rightarrow mathbb{R}$, for which every derivative $frac{d^n}{dx^n}(g(x)), ,n >0$ is also continuous, such that:
$$forall xin[a,b]: , g(x) = f(x)land , exists x notin [a,b]: f(x) neq g(x),, a neq b$$

Thanks!

Define the real functions $f$ and $g$ thus:
$$
f(x) = begin{cases} expBig(-frac{1}{(x - 1)^2}Big) &text{if } x > 1 \
0 &text{if } x in [-1, 1] \
expBig(-frac{1}{(x + 1)^2}Big) &text{if } x < -1
end{cases}
$$ and
$g(x) = 0$. $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.

Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by applying L'Hôpital's rule inductively. Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
begin{align*}
lim_{x to 1^+}frac{df(x)}{dx} &= limlimits_{x to 1^+}frac{2expbig(- frac{1}{(x - 1)^2}big)}{(x - 1)^3} \
&= 2lim_{x to 1^+}frac{frac{1}{(x - 1)^3}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{this limit is of the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 2 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-3}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 2 lim_{x to 1^+}frac{-3(x - 1)^{-4}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= 3lim_{x to 1^+}frac{(x - 1)^{-1}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{again, this has the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 3 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-1}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 3 lim_{x to 1^+}frac{-(x - 1)^{-2}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= frac{3}{2} lim_{x to 1^+}frac{x - 1}{expbig(frac{1}{(x - 1)^2}big)} \
&= frac{3}{2} lim_{x to 1^+} Big[(x - 1)expBig(-frac{1}{(x - 1)^2}Big)Big] \
&= frac{3}{2} Big[lim_{x to 1^+} (x - 1)Big] Big[lim_{x to 1^+} expBig(-frac{1}{(x - 1)^2}Big)Big] = frac{3}{2} times 0 times 0 = 0
end{align*} That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_{x to 1+}frac{d^nf}{dx^n} = 0$ for all $n in mathbb{Z}_+!$ At all other points i.e. on $(-infty, -1) cup (-1, 1) cup (1, infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.

Bonus Fact:

Both $frac{d^n f(x)}{dx^n}$ and $frac{d^n g(x)}{dx^n}$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!

Taylors theorem implies that if two functions are the same in one interval, they must be the same everywhere. This is because when you look at one point in the interval, the nth derivatives of both will be equal. Thus, their Taylor series centered at that point will be equal. Then you can move away from the center and find the Taylor series of both centered around another point to get more information about the function, and they will still be equal. So anywhere you look, the two functions will be equal. (This applies for all analytic functions, not so much for piecewise functions)