Question

1 a)

1) Sketch from (-3,3) and find the Fourier Series of f(x)= f(x+2) = f(x) xif -1 < x < 0 -X if 0 < x < 1 크 a) Apply the Fourier Convergence theorem to your result with an appropriate value of x to evaluate the sum: 1 (2n – 1)2 n=1

Answer 1

© Answed. Givendolo 1. 70 sketch from 63,3] and to find the Fouricor Sedics of f(x) = if Isaco 2 -1 if OLX21 If f(x) is bounded then the Foudica Sedies is NIT (2x -Q-b) Ź hot & {ay COS MITT (20-0-b)) ton. sin (6.a) (6-0) On. ns! HO where an 2 s f(x). Cos b-a a so NI (2X-0-b) b-a dx, n21, 00 = 2 S 2002X ba b MT (2x-a-b) fla). sin dx. bra bo b-a The propedties O.de b)*(+8)+ funny when acxcb (2) { f (at) + f(b)} at X=Q 08 b, and is pediddic with Period (b-a).

3 X @). TO Opply the Foubicor convergence theorem. From cuotono con be written as b 002 f(x) dx = J FLU dx b-a -! jºx.dx + * ¢{-x) dx - [*]* - (6->) - 6-1) - OD x-a-b ni (2X+1-1) PYT צוות ba 1+1

flex) as (M14) dix an : b-a a s' Ixl cosntx dx What, I COS 11.X. is an eve) function, -2 J'x cos NITO da 6 - 2. Sinonix) 1. (-Cosnta) 2²12 TI 0 1 x Sin (nn) + cos(n) N²72 Cos (779)-1 22 an =-2 (6-on-1) 272 4. when n'is odd integer 0²72 when in is even integer 0 b For bn 2 Jof(a) sin nit (31-0-b) dx b-o (6-0) - JOUS: 77 (1772) dx Ibn -D

The Foudico Sedies of the given function is 00 9 00 + E ancos 11134 + Š bn Sinntil. n=1 n=1 =-+ CQ, COSTIX, + 03 COS317% + 050055774 +. ...) (02.CO5211%. + Qy Cosufix +06COS 67TX + - ...). + +0 : Dos! br=0 un COSTX, +4. C0531731 +4 U12 (3²7² + 4 . CD 5571+.... (5) 12 At the point when x=tl is the sum of the series ^ [F (+1+0) + f(1-02 -Ź [4) + (-1)] When - DV -KX c', the Partial sum of the sedies is - * { fbx-3}+{{1+ { f (x) + f(x)? .: fuis continuous f). is to ((را از

flu) = **+ Co COSTIX. + (DS3X + COS 511 ... M3 1 : 00S (29-DFIX TV 2 Now Putting aro on both sides we have f0- + + nije