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Fourier Cosine Series Examples

January 7, 2011

It is an remarkable fact that (almost) any function can be expressed as an infinite sum of cosines, theFourier cosine series. For a functionf(x)defined onx2[0;p], one can writef(x)as f(x) =a02 k=1a kcos(kx) for some coefficientsak. We can compute thea`very simply: for any given`, we inte- gratebothsidesagainstcos(`x). Thisworksbecauseoforthogonality:Rp

0cos(kx)cos(`x)dx

can easily be shown to be zero unlessk=`(just do the integral). Plugging the above sum intoRp

0f(x)cos(`x)dxtherefore gives zero fork6=`andRp

0cos2(`x) =p=2 for

k=`, resulting in the equation a `=2p Z p

0f(x)cos(`x)dx:

Fourier claimed (without proof) in 1822 thatanyfunctionf(x)can be expanded in terms of cosines in this way, even discontinuous functions. This turned out to be false for various badly behavedf(x), and controversy over the exact conditions for conver- gence of the Fourier series lasted for well over a century, until the question was finally settled by Carleson (1966) and Hunt (1968): any functionf(x)whereRjf(x)j1+edx is finite for somee>0 has a Fourier series that convergesalmost everywheretof(x) [except possibly at isolated points of discontinuities]. At points wheref(x)has a jump discontinuity, the Fourier series converges to the midpoint of the jump. So, as long as one does not care about crazy divergent functions or the function value exactly at points of discontinuity (which usually has no practical significance), Fourier"s remark- able claim is essentially true.

Example

To illustrate the convergence of the cosine series, let"s consider an example. Let"s try f(x) =x, which seems impossible to expand in cosines because cosines all have zero slope atx=0 whereasf0(0) =1. Nevertheless, it has a convergent cosine series that can be computed via integration by parts: a k=2p Z p

0xcos(kx)dx=2pkxsin(kx)p

0 2pkZ p

0sin(kx)dx=(

0keven

4pk2kodd:

1

00.20.40.60.810

0.5 1 1.5 2 2.5 3 3.5 x/πone term (up to a 0)

00.20.40.60.810

0.5 1 1.5 2 2.5 3 3.5 two terms (up to a1) x/π

00.20.40.60.810

0.5 1 1.5 2 2.5 3 3.5 three terms (up to a3) x/π

00.20.40.60.810

0.5 1 1.5 2 2.5 3 3.5 five terms (up to a7) x/πFigure 1: Fourier cosine series (blue lines) for the functionf(x) =x(dashed black lines), truncated to a finite number of terms (from 1 to 5), showing that the series indeed converges everywhere tof(x). We divided by 0 fork=0 in the above integral, however, so we have to computea0 separately:a0=2p R p

0xdx=p. The resulting cosine-series expansion is plotted in

figure 1, truncated to 1, 2, 3, or 5 terms in the series. Already by just five terms, you can see that the cosine series is getting quite close tof(x) =x. Mathematically, we can see that the series coefficientsakdecrease as 1=k2asymptotically, so higher-frequency terms have smaller and smaller contributions. 2

General convergence rate

Actually, this 1=k2decline ofakis typical for any functionf(x)that does not have zero slope atx=0 andx=plike the cosine functions, as can be seen via integration by parts: a k=2p Z p

0f(x)cos(kx)dx=2pkf(x)sin(kx)p

0 2pkZ p

0f0(x)sin(kx)dx

2pk2f0(x)cos(kx)p

0 2pk2Z p

0f00(x)cos(kx)dx

=2pk2[f0(0)f0(p)]2pk3f00(x)sin(kx)p 0 +2pk3Z p

0f000(x)sin(kx)dx

=2pk2[f0(0)f0(p)]+2pk4[f000(0)f000(p)]; where theisfor evenkand+for oddk. Thus, we can see that, unlessf(x)has zero first derivative at the boundaries,akdecreases as 1=k2asymptotically. Iff(x)has zero first derivative, thenakdecreases as 1=k4unlessf(x)has zerothirdderivative at the boundaries (like cosine). Iff(x)has zero first and third derivatives, thenak decreases like 1=k6unlessf(x)has zero fifth derivative, and so on. (Of course, in all of the above we assumed thatf(x)was infinitely differentiable in the interior of the integration region.) Ifallof the odd derivatives off(x)are zero at the endpoints, then a kdecreases asymptotically faster thananypolynomial in 1=k- typically in this case, a kdecreases exponentially fast.11 Technically, to getakdecreasing exponentially fast (or occasionally faster), we needf(x)to have zero

odd derivatives at the endpoints and be an "analytic" function (i.e., having a convergent Taylor series) in a

neighborhood of[0;p]in the complexxplane . Analyzing this properly requires complex analysis (18.04),

however. 3quotesdbs_dbs17.pdfusesText_23