Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I was basically exploring the function:

$\displaystyle {\text{frac}(x)}$ which is the fractional part function and I noticed that it has a nice fourier series definition which is:

${\text{frac(x)}}$ = $\frac{1}{2}$ - $\frac{1}{\pi}$ $\sum\limits_{j=1}^{\infty}{\frac{\sin(2{\pi}jx)}{j}}$

When looking at the function:


I notice it exhibits a wave behavior which suggests a wave function with asymptotes (ex: tangent, secant, cosecant, etc...) might be able to create a series definition for this function. Is this a good intuition? If it is, how do I go about finding that series definition?

share|cite|improve this question
replace taylor expansion with a harmonic analysis tag – frogeyedpeas Apr 28 '13 at 21:16

You are dealing with the function $f(x)=1/x$ on the interval $(0,1)$, extended to the line as $1$-periodic function. The definition of Fourier coefficients involves an integral with $f$, and this function $f$ is not integrable. ($\int_0^1 \frac{1}{x}\,dx$ diverges.) So, the short answer is: $\frac{1}{\operatorname{frac}(x)}$ does not have a Fourier series.

But if you insist on having a series of some sort, you can expand $\log x$ into Fourier series $$\ln x = \sum_{n=0}^\infty (a_n\cos 2\pi n x+ b_n\sin 2\pi n x) \tag1$$ and then formally (i.e., without justification) differentiate (1), obtaining $$\frac{1}{\operatorname{frac}(x)} \sim \sum_{n=1}^\infty (2\pi n b_n\cos 2\pi n x - 2\pi n a_n\sin 2\pi n x) \tag2$$ (I write $\sim$ instead of $=$ because there is no guarantee that the series on the right of (2) converges; in fact it does not.)

The coefficients $a_n,b_n$ involve trigonometric integrals $\operatorname{Si}$ and $\operatorname{Ci}$, as well as the Euler constant $\gamma $. With Maple I get $$a_0=-1,\quad a_n=-\frac{\operatorname{Si}(2\pi n)}{\pi n}\ \text{ for } n\ge 1 \tag3$$ and $$b_n=\frac{\operatorname{Ci}(2\pi n)-\ln(2 \pi n)-\gamma }{\pi n}\ \text{ for } n\ge 1 \tag4$$ Therefore, (2) takes the form $$\frac{1}{\operatorname{frac}(x)} \sim \sum_{n=1}^\infty 2 \bigg((\operatorname{Ci}(2\pi n)-\ln(2 \pi n)-\gamma ) \cos 2\pi n x + \operatorname{Si}(2\pi n) \sin 2\pi n x\bigg) \tag5$$ Unfortunately, this divergent series appears to be useless. The partial sums do not look much like the left side of (5). Here are the 50th partial sums: for logarithm on top, for its derivative at the bottom. The approximation to logarithm is of okay quality (except for the severe Gibbs phenomenon at $1^-$), but its wobbliness becomes a disaster when the derivative is taken.

share|cite|improve this answer
what about using tangents and cotangents for approximation? – frogeyedpeas Apr 28 '13 at 22:57
If given the argument npix where n is an integer they appear to have the correct form for converging – frogeyedpeas Apr 28 '13 at 22:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.