Find the complex form of the Fourier Series expansion of
$f (x) = \cos a x, - π < x < π$

Question

Find the complex form of the Fourier Series expansion of  f(x)=cos⁡ax, −π&amp;lt;x&amp;lt;π

Charles Benedict · Accepted Answer

Expand cos⁡(ax)=12(eiax+e−iax)  Then  fn=12π∫−ππcos⁡(ax)e−∈xdx  =12π12∫−ππ(eiax+e−iax)e−∈xdx  =14π∫−π(ei(a−n)x+e−a(a+n)x)dx  This is a straightforward integration. Take care when a is an integer (in which case the Fourier series is trivially obtained from the above expansion).  If a∈Z, then the formula for cos⁡ gives fn=12σ|n||a|  fn=14πi(−1)n(eiaπ+e−iaπa−n−e−iaπ−eiaπa+n)  =14πi(−1)n(eiaπ−e−aπ)(1a−n+1a+n)  =12π(−1)nsin⁡(aπ)2aa2−n2  =1π(−1)n+1sin⁡(aπ)an2−a2

Jacob Homer · Accepted Answer

cn=12π∫−ππcos⁡axe−∈xdx  We make integration by parts twice to  I=∫−ππcos⁡axe−∈xdx=incos⁡axe∈x∣−ππ+ain∫−ππsin⁡axe−∈xdx  =−an2sin⁡axe−∈x+a2n2I  I think you can handle it from here.

karton · Accepted Answer

Complex form of Fourier series is given by
$f (x) \sim \sum_{n = - \infty}^{\infty} c_{n} (f) e^{i n x}$
You need to find Fourier coefficients
$c_{n} (f) = \frac{1}{2 π} \int_{- π}^{π} f (x) e^{- i n x} d x$
Thus, simply integrate by parts (twice)
$c_{n} (f) = \frac{1}{2 π} \int_{- π}^{π} f (x) e^{- i n x} d x$
$= \frac{1}{2 π} \int_{- π}^{π} \cos (a x) e^{- i n x} d x .$

Find the complex form of the Fourier Series expansion of f(x)=\cos

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