Question

Find the Fourier series of fon the given interval. Rx) sin(x). - XCO OSX 00 Hx) = + Glve the number to which the Fourier series converges at a point of discontinuity of /. (1ff is continuous on the given interval, enter CONTINUOUS.)

Answer 1

solution: f(x) O-T<<<0 sinx Os* T nel Is defined L+2T T -] O I 27 1 27 - 7 Let the fourier series of f be f(x)= ao + Ž an cos (2014) + Ź on sim(2012) your function on the interval (-2, 0)[0,7) That means the length of the interval is Now. Ao = t i f(x) dx tai 0.0x + sina dhe ] .605x1" ] [ 1+1] -t and, an = { f(a) cosantx) dx [ - coylanting) te] : [s sinx cos ne ose ] att fisin(n+1)x+ sin co-n) x ) mm. -cos(n+1) cos(n-1) (n-1) "] cos(n.) (n+1) (n-1) to [ come COS (n +)* (+) (n-1)(n+1) } $(x) sin(272) de sina sin (nx) de . O dx + À 27 o 자 + o COS(n+) -in- (n+1 2 COS (n-1) (11) Now, bn 2 27 S at sa sina sinna de 2 "F[eos (1-1)x - - cos (n+1)x] te sin (n-1)=1 ***/*] (n-1) an I sin(n-1)+ - Sin inti)* ). (as sinna 0, n. 1,2,3, Ts, 2, 3.

So, f(a)= ☆ + Ž HAI 2 (n-1)/n. los (1) cos(n+ (n-1) (no) o sinna. N2 + 2* eos (n-1) cos (n+1)* rese I (nt) (--)(n+) The fourier soures converges to the [ f(x+) + f(x-)] at the point of at which of is discontinuous te, at the point o. (Ans)

Hope this solution will help you, thank you!