Some confusing functions

A collection of weird counterexamples in analysis ...

An example is a powerful thing. It can challenge our intuition and prevent us from making claims which are plausible but false.

In this issue, I’ll present several examples in real analysis that challenge our intuition. Enjoy!

A function which does not equal its Taylor series

In calculus, we often learn about Taylor series. Many of our favorite functions like trigonometric functions, polynomials, etc. can be expressed as Taylor series.

But just because a function is infinitely differentiable, does not mean that it equals its Taylor series. Consider the function below:

\(f(x)=\begin{cases} e^{-1/x} &\text{if }x>0 \\ 0& \text{if }x\leq0. \end{cases}\)

It looks like this near the origin:

This function is smooth at the origin.

But all of its derivatives are zero at the origin, so its Taylor series at x=0 is identically zero!

Therefore this function does not equal its Taylor series.

A convergent series which is not absolutely convergent

It is a fact from calculus that the alternating harmonic series converges:

\(\sum_{n=1}^{\infty}\frac{(−1)^{n+1}}{n}=\dfrac{1}{1}−\dfrac{1}{2}+\dfrac{1}{3}−⋯=\ln⁡(2).\)

However, the series is not absolutely convergent because the harmonic series diverges:

\(\sum_{n=1}^{\infty}\frac{1}{n}=\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+⋯=\infty.\)

A one-to-one correspondence between two intervals that is nowhere monotonic

Consider the function f(x) defined on the interval [0,1] as follows:

\(f(x) = \begin{cases} x &\text{if }x \in \mathbb{Q} \\ 1-x &\text{if }x \not\in \mathbb{Q}. \end{cases}\)

Then there is no subinterval of [0, 1] on which this function is monotonic, since the rationals are dense in the reals. The image of f is again the interval [0, 1], and f is one-to-one.

A power series convergent at only one point

Consider the power series defined by

\(\sum_{n=1}^{\infty} n! \,\,x^n.\)

This power series only converges at x=0; it diverges if x is nonzero.

A bounded sequence which diverges

You might think that a bounded sequence has to converge to something. But this is false.

The simplest example of a bounded divergent sequence is the sequence

\(0,1,0,1,0,1,0,…,\)

which alternates between 0 and 1.

If you’d like to see more, check out the book "Counterexamples in Analysis" by Gelbaum and Olmsted.

Problem of the Week

Here is this week’s challenge problem:

Each of the numbers 1 to 10⁶ is repeatedly replaced by its digital sum until we reach 10⁶ one-digit numbers. Will these have more 1’s or 2’s?

Got a solution to the challenge problem? Submit it here.

Solution from last week

See here for the solution to last week’s problem. (Thank you to Olivier Massicot from Champaign, IL, whose solution is being featured this week.)

Thanks for reading and happy learning! Until next time,

Adithya