Final answer:
To sketch f(x) from -2 ≤ x ≤ 2, graph the function f(x) = sin(πx) for 0 ≤ x ≤ 1 and then shift it to the left and right for other intervals of x. The Fourier series of f(x) = sin(πx) can be expressed as a0 + Σ [an*cos(nπx) + bn*sin(nπx)], where an = 0, bn = 2/nπ (for n odd), and bn = 0 (for n even).
Step-by-step explanation:
To sketch the function f(x) from -2 ≤ x ≤ 2, we can start by graphing the function f(x) = sin(πx) for 0 ≤ x ≤ 1, since this is the given function. This function is a sine function with a period of 1, which means it completes one full cycle between 0 and 1. So, we can plot the points (0,0), (1,0), (0.5,1), and (1.5, -1) to sketch the graph of f(x) for 0 ≤ x ≤ 1.
Next, to graph the function for -2 ≤ x ≤ -1, we can shift the graph of f(x) for 0 ≤ x ≤ 1 two units to the left. Similarly, to graph the function for 1 ≤ x ≤ 2, we can shift the graph of f(x) two units to the right. So, the graph of f(x) for -2 ≤ x ≤ 2 will look like a sine wave with peaks at -1 and 1, and the zero crossings at -2, -1, 0, 1, and 2.
The Fourier series of f(x) = sin(πx) is given by:
f(x) = a0 + Σ [an*cos(nπx) + bn*sin(nπx)], where n ranges from 1 to infinity,
a0 = 0,
an = 0 for all n,
bn = 2/nπ if n is odd, and
bn = 0 if n is even.