Desarrolla en serie de Fourier:. Desarrolla en serie de Fourier:

Presentación del tema: "Desarrolla en serie de Fourier:. Desarrolla en serie de Fourier:"— Transcripción de la presentación:

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2 Desarrolla en serie de Fourier:

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4 La función f es continua en (−, ) excepto en x = 0. Así su
serie de Fourier converge en x = 0 a: La serie es una extensión periodica de la función f. Las discontinuidades en x = 0, 2, 4, … convergen a: Y las discontinuidades en x = , 3, … convergen a:

5 Secuencia de sumas parciales y su representación gráfica

6 Evaluar donde el contorno C
es el círculo |z|= 2.

7 A Question In part (b) of Example 2 in Sec.19.3, we showed that the Laurent series of f(z) = 1/z(z – 1) valid for 1 < |z| is The point z = 0 is an isolated singularity of f and the Laurent series contains an infinite number of terms involving negative inter powers of z. Does it mean that z = 0 is an essential singularity?

8 The answer is “NO”. Since the interested Laurent series is the one with the domain 0 < |z| < 1. From part (a) of that example, we have Thus z = 0 is a simple pole for 0 < |z| < 1.

9 C is positively oriented circle | z – 2| = 1.
Integrand is analytic everywhere except z=2 and z=0. Find Laurent series of f(z) in the disk 0 < | z – 2 | < 2 Residue of f at the isolated singular point z0

10 ALTERNATIVE METHOD Residue of f at the isolated singular point 2 is the coefficient of 1/(z–2). Solve for A, B, C, D, E by setting coefficients of z, z2, z3, z4 equal to 0. A + E = 0 (A(z – 2) – Az)(z – 2)3 = – 2A(z – 2)3 D – 2A = 0 (– 2A (z – 2) + 2A z)(z – 2)2 = 4A (z – 2)2 4A + C = 0 (– 4A z + 4A (z – 2)) (z – 2) = –8A(z – 2) B – 8A =0 8A z – 8A (z – 2) = 16 A = 1 A = 1/16, E = – 1/16

11 Partial Fraction Expansion Review

12 C is positively oriented circle | z – 2| = 1.
Integrand is analytic everywhere except z=2 and z=0. Find Laurent series of f(z) in the disk 0 < | z – 2 | < 2 Residue of f at the isolated singular point z0 b1 All terms have positive exponents