Solve the Fourier Series of these equation: f(x) = -x+y, limit = 0−2π

$$a_0 = \frac{1}{\pi} \int_{0}^{2\pi} (-x+4)\,dx$$ $$a_0 = \frac{1}{\pi} \left[ -\frac{x^2}{2} + 4x \right]_{0}^{2\pi}$$ $$a_0 = \frac{1}{\pi} \left( \left(-\frac{(2\pi)^2}{2} + 4(2\pi)\right) - \left(-\frac{0^2}{2} + 4(0)\right) \right)$$ $$a_0 = \frac{1}{\pi} \left( -\frac{4\pi^2}{2} + 8\pi \right)$$ $$a_0 = \frac{1}{\pi} (-2\pi^2 + 8\pi)$$ $$a_0 = -2\pi + 8$$

$$a_n = \frac{1}{\pi} \int_{0}^{2\pi} (-x+4) \cos(nx)\,dx$$

We use integration by parts, $\int u\,dv = uv - \int v\,du$. Let $u = -x+4$ and $dv = \cos(nx)\,dx$. Then $du = -dx$ and $v = \frac{1}{n}\sin(nx)$.

$$\int (-x+4) \cos(nx)\,dx = \left[ (-x+4)\frac{1}{n}\sin(nx) \right]_{0}^{2\pi} - \int_{0}^{2\pi} \frac{1}{n}\sin(nx)(-dx)$$ $$= \left[ \frac{-x+4}{n}\sin(nx) \right]_{0}^{2\pi} + \frac{1}{n}\int_{0}^{2\pi} \sin(nx)\,dx$$ $$= \left( \frac{-2\pi+4}{n}\sin(2n\pi) - \frac{4}{n}\sin(0) \right) + \frac{1}{n}\left[ -\frac{1}{n}\cos(nx) \right]_{0}^{2\pi}$$

Since $\sin(2n\pi) = 0$ and $\sin(0) = 0$:

$$= (0 - 0) - \frac{1}{n^2}[\cos(2n\pi) - \cos(0)]$$

Since $\cos(2n\pi) = 1$ and $\cos(0) = 1$:

$$= -\frac{1}{n^2}[1 - 1] = 0$$

Therefore, $a_n = \frac{1}{\pi}(0) = 0$ for $n \ge 1$.

$$b_n = \frac{1}{\pi} \int_{0}^{2\pi} (-x+4) \sin(nx)\,dx$$

We use integration by parts. Let $u = -x+4$ and $dv = \sin(nx)\,dx$. Then $du = -dx$ and $v = -\frac{1}{n}\cos(nx)$.

$$\int (-x+4) \sin(nx)\,dx = \left[ (-x+4)\left(-\frac{1}{n}\cos(nx)\right) \right]_{0}^{2\pi} - \int_{0}^{2\pi} \left(-\frac{1}{n}\cos(nx)\right)(-dx)$$ $$= \left[ -\frac{-x+4}{n}\cos(nx) \right]_{0}^{2\pi} - \frac{1}{n}\int_{0}^{2\pi} \cos(nx)\,dx$$ $$= \left( -\frac{-2\pi+4}{n}\cos(2n\pi) - \left(-\frac{4}{n}\cos(0)\right) \right) - \frac{1}{n}\left[ \frac{1}{n}\sin(nx) \right]_{0}^{2\pi}$$

Since $\cos(2n\pi) = 1$ and $\cos(0) = 1$:

$$= \left( -\frac{-2\pi+4}{n}(1) + \frac{4}{n}(1) \right) - \frac{1}{n^2}[\sin(2n\pi) - \sin(0)]$$

Since $\sin(2n\pi) = 0$ and $\sin(0) = 0$:

$$= \left( \frac{2\pi-4}{n} + \frac{4}{n} \right) - 0$$ $$= \frac{2\pi}{n}$$

Therefore, $b_n = \frac{1}{\pi} \left( \frac{2\pi}{n} \right) = \frac{2}{n}$ for $n \ge 1$.

Substitute the calculated coefficients into the Fourier series formula:

$$F(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))$$ $$F(x) = \frac{-2\pi+8}{2} + \sum_{n=1}^{\infty} \left( 0 \cdot \cos(nx) + \frac{2}{n} \sin(nx) \right)$$ $$F(x) = (4-\pi) + \sum_{n=1}^{\infty} \frac{2}{n} \sin(nx)$$

The Fourier series for $F(x) = -x+4$ on the interval $[0, 2\pi]$ is:

$$F(x) = (4-\pi) + 2 \sum_{n=1}^{\infty} \frac{\sin(nx)}{n}$$