1. **Problem:** Find the Fourier series representation of the function $$f(x) = x$$ for $$-\pi < x < \pi$$, assuming it is $$2\pi$$-periodic. 2. **Formula and rules:** The Fourier series of a $$2\pi$$-periodic function $$f(x)$$ is given by: $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(nx) + b_n \sin(nx)\right) $$ where $$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \, dx, \quad a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) \, dx, \quad b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) \, dx. $$ Since $$f(x) = x$$ is an odd function, all $$a_0$$ and $$a_n$$ coefficients will be zero because the product of an odd and even function is odd and integrates to zero over symmetric limits. 3. **Calculate coefficients:** - $$a_0 = \frac{1}{\pi} \int_{-\pi}^\pi x \, dx = 0$$ (odd function integral) - $$a_n = \frac{1}{\pi} \int_{-\pi}^\pi x \cos(nx) \, dx = 0$$ (odd function times even function is odd) - $$b_n = \frac{1}{\pi} \int_{-\pi}^\pi x \sin(nx) \, dx$$ (odd times odd is even, so this integral is nonzero) 4. **Evaluate $$b_n$$:** $$ b_n = \frac{1}{\pi} \int_{-\pi}^\pi x \sin(nx) \, dx = \frac{2}{\pi} \int_0^\pi x \sin(nx) \, dx $$ (using the function's odd symmetry) Integrate by parts: Let $$u = x$$, $$dv = \sin(nx) dx$$, then $$du = dx$$, $$v = -\frac{\cos(nx)}{n}$$. So, $$ \int_0^\pi x \sin(nx) \, dx = -\frac{x \cos(nx)}{n} \Big|_0^\pi + \frac{1}{n} \int_0^\pi \cos(nx) \, dx $$ Calculate each term: - $$-\frac{\pi \cos(n\pi)}{n} + 0 = -\frac{\pi (-1)^n}{n}$$ - $$\frac{1}{n} \int_0^\pi \cos(nx) \, dx = \frac{1}{n} \left[ \frac{\sin(nx)}{n} \right]_0^\pi = 0$$ Thus, $$ \int_0^\pi x \sin(nx) \, dx = -\frac{\pi (-1)^n}{n} $$ Therefore, $$ b_n = \frac{2}{\pi} \left(-\frac{\pi (-1)^n}{n}\right) = \frac{2 (-1)^{n+1}}{n} $$ 5. **Final Fourier series:** $$ f(x) = \sum_{n=1}^\infty \frac{2 (-1)^{n+1}}{n} \sin(nx) $$ This is the Fourier sine series representation of $$f(x) = x$$ on $$(-\pi, \pi)$$. **Answer:** $$ f(x) = 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sin(nx) $$