- #1

- 1,567

- 591

- TL;DR Summary
- About function that are continuous or differentiable at isolated points.

Many have probably seen an example of a function that is continuous at only one point, for example

##f:\mathbb{R}\rightarrow\mathbb{R}\hspace{5pt}:\hspace{5pt}f(x)=\left\{\begin{array}{cc}x, & \hspace{6pt}when\hspace{3pt}x\in\mathbb{Q} \\ -x, & \hspace{6pt}when\hspace{3pt}x\in\mathbb{R}\setminus\mathbb{Q}\end{array}\right.##

It is also possible to define a function for which the derivative defined as

##f'(x) = \underset{h\rightarrow 0}{\lim}\frac{f(x+h)-f(x)}{h}##

exists at only the point ##x=0##. One of the simplest examples is

##f:\mathbb{R}\rightarrow\mathbb{R}\hspace{5pt}:\hspace{5pt}f(x)=\left\{\begin{array}{cc}x^2 , & \hspace{6pt}when\hspace{3pt}x\in\mathbb{Q} \\ -x^2 , & \hspace{6pt}when\hspace{3pt}x\in\mathbb{R}\setminus\mathbb{Q}\end{array}\right.##

for which the derivative at the origin is zero. Extending this to higher derivatives is not as easy, because the second derivative is usually seen as the same finite difference applied to the first derivative:

##f''(x) = \underset{h\rightarrow 0}{\lim}\frac{f'(x+h)-f'(x)}{h}##

and it doesn't make much sense to start calculating this for a function that has only ##x=0## as its domain.

However, the second derivative of

##f:\mathbb{R}\rightarrow\mathbb{R}\hspace{5pt}:\hspace{5pt}f(x)=\left\{\begin{array}{cc}|x|^3 , & \hspace{6pt}when\hspace{3pt}x\in\mathbb{Q} \\ -|x|^3 , & \hspace{6pt}when\hspace{3pt}x\in\mathbb{R}\setminus\mathbb{Q}\end{array}\right.##

at ##x=0## could be calculated by the central finite difference

##f''(0) = \underset{h\rightarrow 0}{\lim}\frac{f(-h)-2f(0)+f(h)}{h}##

and found to have value zero.

A function with derivatives of arbitrary order being zero at ##x=0## and existing nowhere else could be defined by saying, e.g. ##f(0)=0## and ##f(x)=\pm\exp (-1/|x|)## when ##x\neq 0##.

Does the most conventional definition of 2nd and higher derivatives require that the derivative of previous order exist on some interval instead of just a single point?

##f:\mathbb{R}\rightarrow\mathbb{R}\hspace{5pt}:\hspace{5pt}f(x)=\left\{\begin{array}{cc}x, & \hspace{6pt}when\hspace{3pt}x\in\mathbb{Q} \\ -x, & \hspace{6pt}when\hspace{3pt}x\in\mathbb{R}\setminus\mathbb{Q}\end{array}\right.##

It is also possible to define a function for which the derivative defined as

##f'(x) = \underset{h\rightarrow 0}{\lim}\frac{f(x+h)-f(x)}{h}##

exists at only the point ##x=0##. One of the simplest examples is

##f:\mathbb{R}\rightarrow\mathbb{R}\hspace{5pt}:\hspace{5pt}f(x)=\left\{\begin{array}{cc}x^2 , & \hspace{6pt}when\hspace{3pt}x\in\mathbb{Q} \\ -x^2 , & \hspace{6pt}when\hspace{3pt}x\in\mathbb{R}\setminus\mathbb{Q}\end{array}\right.##

for which the derivative at the origin is zero. Extending this to higher derivatives is not as easy, because the second derivative is usually seen as the same finite difference applied to the first derivative:

##f''(x) = \underset{h\rightarrow 0}{\lim}\frac{f'(x+h)-f'(x)}{h}##

and it doesn't make much sense to start calculating this for a function that has only ##x=0## as its domain.

However, the second derivative of

##f:\mathbb{R}\rightarrow\mathbb{R}\hspace{5pt}:\hspace{5pt}f(x)=\left\{\begin{array}{cc}|x|^3 , & \hspace{6pt}when\hspace{3pt}x\in\mathbb{Q} \\ -|x|^3 , & \hspace{6pt}when\hspace{3pt}x\in\mathbb{R}\setminus\mathbb{Q}\end{array}\right.##

at ##x=0## could be calculated by the central finite difference

##f''(0) = \underset{h\rightarrow 0}{\lim}\frac{f(-h)-2f(0)+f(h)}{h}##

and found to have value zero.

A function with derivatives of arbitrary order being zero at ##x=0## and existing nowhere else could be defined by saying, e.g. ##f(0)=0## and ##f(x)=\pm\exp (-1/|x|)## when ##x\neq 0##.

Does the most conventional definition of 2nd and higher derivatives require that the derivative of previous order exist on some interval instead of just a single point?

Last edited: