Hfrhyu

You are given a function $f: mathbbRrightarrow mathbbR$. Every derivative $fracd^ndx^n(f(x)), ,n >0$ of the function is continuous.

Is there a function $g: mathbbRrightarrow mathbbR$, for which every derivative $fracd^ndx^n(g(x)), ,n >0$ 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!

Edit:

I asked the question because I intuitively wondered if there would be functions, which could behave like "the same" in a given interval, but then behave differently so that they start diverging or at least stopped being the same the anymore. The answer to this question gave me a bigger understanding of real analysis.

If you would like to know which made me think about such a problem:
Although this is a vague formulation, generally, this question asks if two (completely) different things can develop (themselves) to be exactly equal in at least one part of there whole existence...

Define the real functions $f$ and $g$ thus:
$$
f(x) = begincases expBig(-frac1(x - 1)^2Big) &textif x > 1 \
0 &textif x in [-1, 1] \
expBig(-frac1(x + 1)^2Big) &textif x < -1
endcases
$$ 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 (Edit: A more elementary demonstration is given below based on the exponential series). Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
beginalign*
lim_x to 1^+fracdf(x)dx &= limlimits_x to 1^+frac2expbig(- frac1(x - 1)^2big)(x - 1)^3 \
&= 2lim_x to 1^+fracfrac1(x - 1)^3expbig(frac1(x - 1)^2big) quadtextthis limit is of the form fracinftyinfty text so L'Hôpital applies \
&= 2 lim_x to 1^+fracfracddx(x - 1)^-3fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 2 lim_x to 1^+frac-3(x - 1)^-4-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= 3lim_x to 1^+frac(x - 1)^-1expbig(frac1(x - 1)^2big) quadtextagain, this has the form fracinftyinfty text so L'Hôpital applies \
&= 3 lim_x to 1^+fracfracddx(x - 1)^-1fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 3 lim_x to 1^+frac-(x - 1)^-2-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= frac32 lim_x to 1^+fracx - 1expbig(frac1(x - 1)^2big) \
&= frac32 lim_x to 1^+ Big[(x - 1)expBig(-frac1(x - 1)^2Big)Big] \
&= frac32 Big[lim_x to 1^+ (x - 1)Big] Big[lim_x to 1^+ expBig(-frac1(x - 1)^2Big)Big] = frac32 times 0 times 0 = 0
endalign* That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_x to 1+fracd^nfdx^n = 0$ for all $n in mathbbZ_+!$ 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 $fracd^n f(x)dx^n$ and $fracd^n g(x)dx^n$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!

Edit:

A more elementary way of seeing $lim_x to 1^+fracdf(x)dx = 0$ is as follows:
beginalign*
lim_x to 1^+fracdf(x)dx &= 2lim_x to 1^+fracfrac1(x - 1)^3expbig(frac1(x - 1)^2big) \
&= 2lim_y to +inftyfracy^3expbig(y^2big) quadtext changing variables to y = frac1x - 1 \
&= 2lim_y to +inftyfracy^3sumlimits_n = 0^infty fracy^2nn! \
&= 2lim_y to +inftyfracy^31 + y^2 + fracy^42 + sumlimits_n = 3^infty fracy^2nn! \
&= 2lim_y to +inftyfrac1frac1y^3 + frac1y + fracy2 + sumlimits_n = 3^infty fracy^2n - 3n! quadtext dividing top and bottom by y^3 \
&= 2 times 0 = 0
endalign* Essentially, the top term $1$, which remains constant, is a sitting duck. Meanwhile the bottom term blows up to infinity because in the bottom, only two terms ($frac1y^3$ and $frac1y$) go to zero as $y$ becomes large. The rest of the terms ($fracy2$ and $fracy^2n - 3n!$ for $n in mathbbZ_+ geq 3$) have at least a $y$ in the numerator and since no cancellation happens (all terms are positive), their sum becomes arbitrarily large and the overall ratio becomes arbitrarily small.

Sure. In fact, there is a whole class of functions which not only exist, but are specifically made to do something that effectively implies exactly what you are looking for: they are called bump functions, and are defined as smooth (differentiable everywhere, arbitrarily many times) functions that have compact support, meaning (almost) that they are zero everywhere except on a compact set, which for the real numbers as domain basically means a closed, bounded (i.e. contained within an interval) subset thereof, such as a closed interval. This compact set where they are nonzero is called the "support". The trick is to exploit the "everywhere else zero"-ness, as that gives you what you're after.

Namely, any two different (i.e. not equal) bump functions with the same supporting interval $[a, b]$, will be smooth, zero on any interval outside this interval, and yet different, because they differ on such interval. More generally, given any two different bump functions, period, you just have to find an interval outside both of their support sets, which is always possible because they are both bounded.

A simple example of such a bump function is

$$mathrmbump: mathbbR rightarrow mathbbR, mathrmbump(x) := begincases e^-frac11 - x^2, mboxwhen $x$ is in $(-1, 1)$\ 0, mboxotherwise endcases$$

Then consider just $mathrmbump(x)$ and a nonzero multiple thereof, say, $mathrmbump2(x) := 2 cdot mathrmbump(x)$. We now have $mathrmbump(x) = mathrmbump2(x)$ when, say, $x in [10, 11]$, since they are both zero there. Yet, they are ostensibly not equal when $x$ is in $(-1, 1)$.

ADD: It appears to have been asked as to how one can do this without explicitly constructing the bump function. The above is just to show (not completely thoroughly) that bump functions exist. Indeed, we can do so as well. Let now $mathrmbump(x)$ be a general bump function. Let its support set be $mathrmsupp[mathrmbump]$. That is,

$$mathrmsupp[mathrmbump] := mathrmclleft( x in mathbbR : mathrmbump(x) ne 0 right)$$

(n.b. "cl" means to take the closure; basically this includes all "endpoints" of regions in which it is nonzero, even if it is zero at those endpoints - e.g. the support of the just-given-explicitly bump function is $[-1, 1]$, not $(-1, 1)$. This is a bit of technicality that was wrapped earlier when I said "almost" in "meaning (almost)" above.)

Since the support set is bounded and closed, it has a maximum and minimum (largest and smallest element): assign $M := mathrmmax mathrmsupp[mathrmbump]$. Now consider the interval $mathrmext ival := [M+1, M+2]$. If $x in mathrmext ival$, then it is clearly not in the support set, but rather to the right of it. Thus $mathrmbump(x) = 0$ there. Now set $mathrmbump2(x) := 2 cdot mathrmbump(x)$ as before (if you want even more generality, just replace $2$ with an arbitrary vertical rescaling coefficient $a$ that is not $0$ or $1$). Congrats, you now have two bump functions that are unequal but equal on the external interval $mathrmext ival$.