# Construct a function with each derivative being non-differentiable at a distinct point

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!

• 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 $i!$... – Giuseppe Negro Sep 16 at 12:35

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 $$h are smooth, and the newly added term $$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).$$