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One can expand any periodic function in sines and cosines. When calculating the coefficients $a_0$, $a_n$, and $b_n$ one find that $a_1>a_2>...>a_n>...$, similarly for $b_n$.

Is there an intuitive reason to understand this, I mean why would one expect this to happen? It looks miraculous and mysterious to me that the coefficients came out "ordered" in a way that made the convergence manifest.

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What's the basis of your assertion that $a_1 > a_2 > \cdots$? There are many physical systems in which that isn't true. – David Z Oct 21 '11 at 7:01
@DavidZaslavsky $a_n$ are Fourier coefficients. When calculating $a_n$, it decreases with $n$. Here are a few examples, – Revo Oct 21 '11 at 7:36
No, it doesn't. Just giving some examples where the coefficients decrease with $n$ doesn't mean that it is true in general. – David Z Oct 21 '11 at 14:06
@DavidZaslavsky My question is about the cases when it does. – Revo Oct 22 '11 at 18:26
@Revo In that case, the identity is true because you are asking about the cases where it is true. – Mark Eichenlaub Oct 30 '11 at 12:29

The reason of a more modest version of your statement (your big claim is not right) is that the sum $$\sum_{n=-\infty}^{\infty} |a_n|^2 $$ has to converge. That's because this sum is proportional to $$ \int_0^{2\pi} |f(x)|^2 dx $$ which converges for bounded functions (a basic insight about Fourier expansions and Hilbert spaces of periodic functions). That's why, for example, power law $$ |a_n| \sim \frac{1}{n^\epsilon} $$ require $\epsilon>0.5$ or, if some logarithmic corrections are included, at least $\epsilon\geq 0.5$. If the Fourier coefficients were not dropping at least this quickly, the function wouldn't be $L^2$-integrable: the sum above wouldn't converge.

Of course, this convergence requirement doesn't prevent some coefficients from being larger than $C/n^\epsilon$ as long as most others drop quickly. Still, there can't be infinitely many coefficients $a_n$ such that $a_n > \delta$ for a pre-given positive $\delta$ because the sum would still diverge.

Distributions which are not really functions such as $f(x)=\delta(x)$ may have Fourier coefficients that don't drop: of course, $\delta(x)^2$ doesn't have a finite integral, anyway. The same holds for unbounded functions.

The condition $a_1 > a_2 > a_3 > \dots$ is much stronger and is obviously violated for "most functions": you may simply choose coefficients that don't agree with this strict ordering and construct a corresponding function. Still, simple enough functions tend to obey even this strict ordering because the calculation of $a_n$ leads to a simple enough and monotonic function of $n$.

Perfectly smooth functions have $a_n$ decreasing faster than any power law; functions with steps have $a_n\sim 1/n$; continuous functions with $|x|$-like unsmooth points have $a_n\sim 1/n^2$ for large $n$, and so on.

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1) One may define the Fourier coefficients $a_n$ and $b_n$ (or the complex Fourier coefficients $c_n$) of a periodic function $f(\theta)=f(\theta +2\pi)$ if $f\in {\cal L}^1(\mathbb{R}/ 2\pi\mathbb{Z})$.

2) Here ${\cal L}^1$ is an example of an ${\cal L}^p$ space, where $p\geq 1$ is a power. By definition, a function $f$ belongs to a ${\cal L}^p$ space if $f$ is measurable and the integral $\int|f|^p <\infty$ is finite. For instance, $f(\theta)=\tan(\theta)^{\frac{2}{3}}$ is an example of an ${\cal L}^1$ function that is not an ${\cal L}^2$ function.

3) It is in general not true that the Fourier coefficients $a_n$ monotonically decrease as

$$ |a_0| \geq |a_1| \geq |a_2| \geq \ldots \geq |a_n| \geq \ldots. $$

It is easy to come up with counterexamples, e.g., $f(\theta)=\cos(2\theta)$ has $0=|a_1| < |a_2|\neq 0. $

4) Nevetheless, it is true that

$$\lim_{n\to \infty}a_n = 0, \qquad \lim_{n\to \infty}b_n = 0, \qquad \lim_{|n|\to \infty}c_n = 0,$$

if $f\in {\cal L}^1(\mathbb{R}/ 2\pi\mathbb{Z})$. This is a consequence of Riemann–Lebesgue lemma.

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