The problem involves expanding the function $f(x) = 1 - x$ using Fourier series over the interval $0 < x < 2$. Let's break down each part step by step.

(a) Full-Range Fourier Series Expansion

The full-range Fourier series represents a function as a sum of sine and cosine terms over the interval $[-L, L]$, where the function is extended periodically. Here, $L = 1$, since the interval length is 2.

The general form of the full-range Fourier series is: $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos\left( \frac{n\pi x}{L} \right) + b_n \sin\left( \frac{n\pi x}{L} \right) \right)$ Where:

• $a_0 = \frac{1}{L} \int_0^{2L} f(x) \, dx$
• $a_n = \frac{1}{L} \int_0^{2L} f(x) \cos\left( \frac{n\pi x}{L} \right) \, dx$
• $b_n = \frac{1}{L} \int_0^{2L} f(x) \sin\left( \frac{n\pi x}{L} \right) \, dx$

Here $f(x) = 1 - x$, and $L = 1$, so the interval is $[0, 2]$.

We now compute the coefficients.

1. $a_0$ (DC term): $a_0 = \frac{1}{1} \int_0^2 (1 - x) \, dx = \int_0^2 (1 - x) \, dx$ $a_0 = \left[ x - \frac{x^2}{2} \right]_0^2 = (2 - 2) - (0) = 0$

2. $a_n$ (Cosine coefficients): $a_n = \int_0^2 (1 - x) \cos(n\pi x) \, dx$ For each $n$, we would need to solve this integral, which will depend on the value of $n$.

3. $b_n$ (Sine coefficients): $b_n = \int_0^2 (1 - x) \sin(n\pi x) \, dx$ This also depends on the value of $n$.

(b) Half-Range Cosine Series Expansion

In this case, we use the half-range cosine series since the function is even. The form is: $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\left( \frac{n\pi x}{L} \right)$ The coefficients are given by:

• $a_0 = \frac{2}{L} \int_0^L f(x) \, dx$
• $a_n = \frac{2}{L} \int_0^L f(x) \cos\left( \frac{n\pi x}{L} \right) \, dx$

For $L = 2$ and $f(x) = 1 - x$, we calculate the coefficients similarly to part (a).

(c) Half-Range Sine Series Expansion

For the half-range sine series (for odd extensions), the form is: $f(x) = \sum_{n=1}^{\infty} b_n \sin\left( \frac{n\pi x}{L} \right)$ The coefficients are given by:

• $b_n = \frac{2}{L} \int_0^L f(x) \sin\left( \frac{n\pi x}{L} \right) \, dx$

Again, substituting $L = 2$ and $f(x) = 1 - x$, we compute the sine series.

(d) Graphs

For the graphical part, you are required to:

1. Plot the original function $f(x) = 1 - x$.
2. Plot the periodic extensions of the function for the full-range Fourier series, half-range cosine series, and half-range sine series for $-8 \leq x \leq 8$.

This will illustrate how the Fourier series approximates the original function.

Would you like more details on how to compute each of the integrals for the Fourier coefficients? I can guide you through specific steps for any of the expansions if you'd like. Here are a few follow-up questions that might help:

1. Do you need a detailed breakdown of the integration for $a_n$ and $b_n$?
2. Would you like to see the calculations for specific values of $n$ in the full-range Fourier series?
3. Are you looking for the graphical representation of the Fourier series approximations?
4. Should I explain more about the symmetry considerations for the half-range expansions?
5. Do you want to focus on any specific part of the question first?

Tip: Fourier series can be tricky to compute manually for complex functions, but understanding the role of each term helps in interpreting the periodic behavior!