Find the half-range (i) Cosine series and (ii) sine series for
f(x)= x
2

We are asked to find the **half-range Fourier series** for the function \( f(x) = x^2 \) over the interval \( 0 \leq x \leq L \). There are two types of half-range series: the cosine series (which represents the function as an even extension) and the sine series (which represents the function as an odd extension).

### (i) Half-Range **Cosine Series** for \( f(x) = x^2 \)

The half-range cosine series assumes that the function is extended to be **even** on the interval \( -L \leq x \leq L \). The cosine series is given by:

\[
f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\left(\frac{n \pi x}{L}\right)
\]

Where the Fourier coefficients are:

\[
a_0 = \frac{2}{L} \int_0^L f(x) \, dx
\]
\[
a_n = \frac{2}{L} \int_0^L f(x) \cos\left( \frac{n \pi x}{L} \right) dx \quad \text{for} \quad n \geq 1
\]

For \( f(x) = x^2 \):

#### 1. Calculation of \( a_0 \):
\[
a_0 = \frac{2}{L} \int_0^L x^2 \, dx = \frac{2}{L} \left[ \frac{x^3}{3} \right]_0^L = \frac{2}{L} \cdot \frac{L^3}{3} = \frac{2L^2}{3}
\]

#### 2. Calculation of \( a_n \):
\[
a_n = \frac{2}{L} \int_0^L x^2 \cos\left( \frac{n \pi x}{L} \right) dx
\]

This integral can be solved using integration by parts, or referring to standard Fourier integral tables. The result for \( a_n \) is:

\[
a_n = \frac{4L^2}{n^2 \pi^2} (-1)^n
\]

Thus, the half-range cosine series is:

\[
f(x) = \frac{L^2}{3} + \sum_{n=1}^{\infty} \frac{4L^2 (-1)^n}{n^2 \pi^2} \cos\left( \frac{n \pi x}{L} \right)
\]

---

### (ii) Half-Range **Sine Series** for \( f(x) = x^2 \)

The half-range sine series assumes that the function is extended to be **odd** on the interval \( -L \leq x \leq L \). The sine series is given by:

\[
f(x) \sim \sum_{n=1}^{\infty} b_n \sin\left( \frac{n \pi x}{L} \right)
\]

Where the Fourier sine coefficients \( b_n \) are:

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

For \( f(x) = x^2 \):

#### 1. Calculation of \( b_n \):
\[
b_n = \frac{2}{L} \int_0^L x^2 \sin\left( \frac{n \pi x}{L} \right) dx
\]

Again, this can be computed using integration by parts or standard integral tables. The result for \( b_n \) is:

\[
b_n = \frac{-4L^2}{n^2 \pi^2}
\]

Thus, the half-range sine series is:

\[
f(x) = \sum_{n=1}^{\infty} \frac{-4L^2}{n^2 \pi^2} \sin\left( \frac{n \pi x}{L} \right)
\]

---

Would you like to see more details on the integral steps or specific examples?

### 5 Follow-up Questions:
1. How does the choice of interval affect the cosine and sine series expansion?
2. Why does the cosine series represent an even extension of the function?
3. What is the physical interpretation of a half-range Fourier series?
4. How would the series change if \( f(x) \) were a different polynomial?
5. Can we derive the sine and cosine series for a non-polynomial function?

### Tip:
For solving Fourier integrals like those in this problem, integration by parts is often essential. Keep handy integral tables for faster results!

Find the half-range (i) Cosine series and (ii) sine series for f(x)= x^2

Explore the half-range Fourier series for the quadratic function f(x) = x^2 over the interval [0, L]. Learn how to derive the cosine and sine series expansions, including detailed steps for calculating the Fourier coefficients. This comprehensive guide is ideal for undergraduate students studying advanced mathematics.

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