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Construct a function $f_1:\mathbb{R}\to\mathbb{R}$ with the following properties or show that no such function exists:

$1.$ $f_1$ is differentiable everywhere except one point $x_1.$

$2.$ Define $f_2 : \mathbb{R}\setminus\{x_1\} \to \mathbb{R}$ as $f_2(x) := $ derivative of $f_1$ at $x.$ This $f_2$ must be differentiable everywhere in its domain except one point $x_2.$

$3.$ Define $f_3 : \mathbb{R}\setminus\{x_1,\;x_2\} \to \mathbb{R}$ as $f_3(x) := $ derivative of $f_2$ at $x.$ This $f_3$ must be differentiable everywhere in its domain except one point $x_3.$

$\vdots$

$n.$ Define $f_n : \mathbb{R}\setminus\{x_1, \cdots, x_{n-1}\} \to \mathbb{R}$ as $f_n(x) := $ derivative of $f_{n-1}$ at $x.$ This $f_n$ must be differentiable everywhere in its domain except one point $x_n.$

$\vdots$

(Note that we do not stop at any $n.$)

I found this question in a collection of extra questions for my Calculus course.

I started out by trying something along the lines of $f(x) = \lim_{n\to\infty}\sum_{i=1}^n (x-i)^i|x-i|$, but the function itself isn't defined anywhere and I couldn't figure out how to fix it with minimum effort.

So next, I tried out something that might actually be defined somewhere such as $$f(x) = \lim_{n\to\infty}\sum_{i=1}^n \frac{\left(\frac{2}{\pi}\arctan(x-i)\right)^i\left(\frac{2}{\pi}\arctan|x-i|\right)}{(i+1)!}$$

which is defined for $x \in \mathbb{R}$, but I wasn't able to prove continuity or differentiability. Intuitively, I feel that since it's a sum of continuous functions, it should be continuous, but I'm not sure whether this intuition is correct because it's an infinite sum.

I asked the person whose website I found the question on (another student), and he said that he wasn't sure whether such a function was even possible.

Any help would be appreciated!

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    $\begingroup$ You were on the right track with your first attempt (but replace $x-i$ by $x-x_i$, where $(x_i)$ is the given sequence of points). It is true that the series needs not converge at any point, but that can be remedied easily by multiplying every summand by a sequence of numbers that decreases sufficiently fast, such as $i!$... $\endgroup$ – Giuseppe Negro Sep 16 at 12:35
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A different solution is the following 'lazy exponential' - there are easier solutions (maybe look up bump functions), but I like delay ODEs. set \begin{align}x\in(-\infty,0]&\implies f(x):=1,\\ x\in(0,1] &\implies f(x) := 1+x, \\ x\in (1,2] &\implies f(x) := 1+x + \frac{(x-1)^2}{2!},\\ x\in(2,3] & \implies f(x) := 1+x + \frac{(x-1)^2}{2!} + \frac{(x-2)^3}{3!}, \end{align} and in general $$x\in(n,n+1]\implies f(x) := \sum_{k=0}^{n+1} \frac{(x-k+1)^k}{k!}. $$

If you differentiate, you find for $x\in (n,n+1)$, where $n>1$:

$$ f'(x) = \sum_{k=1}^{n+1} \frac{(x-k+1)^{k-1}}{(k-1)!} =\sum_{j=0}^{n} \frac{(x-j)^{j}}{j!}= f(x-1)$$ so to the right of 1, it solves a delay ODE with initial data prescribed on $x\in(0,1]$ above. $f'$ is clearly discontinuous at $0$, but $$\left.\frac{d}{dx}\frac{(x-1)^2}{2!}\right|_{x=1} = 0 $$ so the derivative is continuous at $x=1$. In general, for any integer $n\ge 2$, near $x=n-1$, all the terms $\frac{(x-h+1)^h}{h!}$ for $h<n$ are smooth, and the newly added term $T_n$, $$ T_n(x) := \begin{cases} \frac{(x-n+1)^n}{n!} & x>n-1,\\ 0 & x\le n-1\end{cases}$$ is $C^1$. Conclusion - $$f \in C^0(\mathbb R)\cap C^1(\mathbb R\setminus \{0\}).$$

To finish, we use the delay ODE, which says that differentiating is the same as translating the function to the right by one. Thus for $x\in \mathbb (0,\infty)\setminus \mathbb N$, $i\in\mathbb N$, $$ f^{(i+1)}(x+i) = f'(x).$$ So the discontinuity of $f^{(i+1)}$ at $x=i-1$, and the continuity at integers $x=\tilde i > i-1$ follows directly from the dis/continuity of $f'$ at $0,1,2,\dots$. We conclude $$ f \in C^0(\mathbb R)\cap \left(\bigcap_{k=1}^\infty C^k(\mathbb R\setminus{\{0,1,\dots,k-1\}})\right).$$

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  • $\begingroup$ Hm, nice solution! Thanks! $\endgroup$ – Amit Rajaraman Sep 17 at 5:53
  • $\begingroup$ @Epiksalad You're welcome; there were some typos, and Ive fleshed out the argument (hopefully correctly) $\endgroup$ – Calvin Khor Sep 17 at 9:16

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