Question

At x=59, the Fourier sine series of f(x)={

2,0≤x<1
4,1≤x≤3
​ } on [0,3] converges to (a) 3 (b) -3 (c) 4 (d) -4 (e) -2

Answer 1

The Fourier sine series for the given f(x) converges to 1.5, which is not an option provided. There seems to be a typo as 59 is outside the interval [0, 3]. Options (a)-(e) are incorrect.

Step-by-step explanation:

The Fourier sine series of a function f(x) on a given interval converges to the average of the function's values at the endpoints of the interval in question. In this case, the function f(x) = 2 for 0 ≤ x < 1 and f(x) = 1 for 1 ≤ x ≤ 3. The series will converge to the average of f(x) values at x = 0 and x = 3. At x = 0, f(x) = 2, and at x = 3, f(x) = 1. Thus, the series converges to (2 + 1) / 2 = 1.5, but since the options provided do not include 1.5, this indicates there may be a typo or misunderstanding in the question itself. The function's Fourier series won't evaluate to any of the options given at x = 59 because 59 lies outside the defined interval [0, 3].