Fourier Series Evaluation:
We evaluate the expression:
$$\frac{2}{\pi n^2}\Big[(-1)^n - 1\Big]\cos(nx)$$
For \( n = 1, 2, 3, 4, \dots \) until we have three non-zero terms.
Note: If \( n \) is even, \([(-1)^n - 1] = (1 - 1) = 0\), so only odd \( n \) contribute.
For odd \( n \):
$$[(-1)^n - 1] = (-1 - 1) = -2$$
So the term becomes:
$$\frac{2}{\pi n^2}(-2)\cos(nx) = -\frac{4}{\pi n^2}\cos(nx)$$
First three non-zero terms (n = 1, 3, 5):
$$-\frac{4}{\pi}\cos(x) \;-\; \frac{4}{9\pi}\cos(3x) \;-\; \frac{4}{25\pi}\cos(5x) + \dots$$
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