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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:

$\displaystyle{\frac{1}{\text{frac}(x)}}$

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?

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replace taylor expansion with a harmonic analysis tag –  frogeyedpeas Apr 28 '13 at 21:16

1 Answer 1

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.

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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

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