determination of Fourier coefficients
Suppose that the real function f may be presented as sum of the Fourier series:
f(x)=a02+∞∑m=0(amcosmx+bmsinmx) | (1) |
Therefore, f is periodic with period 2π. For expressing the Fourier coefficients am and bm
with the function itself, we first multiply the series (1) by cosnx (n∈ℤ) and integrate from -π to π. Supposing that we can integrate termwise, we may write
∫π-πf(x)cosnxdx=a02∫π-πcosnxdx+∞∑m=0(am∫π-πcosmxcosnxdx+bm∫π-πsinmxcosnxdx). | (2) |
When n=0, the equation (2) reads
∫π-πf(x)𝑑x=a02⋅2π=πa0, | (3) |
since in the sum of the right hand side, only the first addend is distinct from zero.
When n is a positive integer, we use the product formulas of the trigonometric identities, getting
∫π-πcosmxcosnxdx=12∫π-π[cos(m-n)x+cos(m+n)x]𝑑x, |
∫π-πsinmxcosnxdx=12∫π-π[sin(m-n)x+sin(m+n)x]𝑑x. |
The latter expression vanishes always, since the sine is an odd function. If m≠n, the former equals zero because the antiderivative consists of sine terms which vanish at multiples of π; only in the case m=n we obtain from it a non-zero result π. Then (2) reads
∫π-πf(x)cosnxdx=πan | (4) |
to which we can include as a special case the equation (3).
By multiplying (1) by sinnx and integrating termwise, one obtains similarly
∫π-πf(x)sinnxdx=πbn. | (5) |
The equations (4) and (5) imply the formulas
an=1π∫π-πf(x)cosnxdx |
and
for finding the values of the Fourier coefficients of .
Title | determination of Fourier coefficients |
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Canonical name | DeterminationOfFourierCoefficients |
Date of creation | 2013-03-22 18:22:47 |
Last modified on | 2013-03-22 18:22:47 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 7 |
Author | pahio (2872) |
Entry type | Derivation |
Classification | msc 26A42 |
Classification | msc 42A16 |
Synonym | calculation of Fourier coefficients |
Related topic | UniquenessOfFourierExpansion |
Related topic | FourierSineAndCosineSeries |
Related topic | OrthogonalityOfChebyshevPolynomials |