find the half-range sine series of f(x) = x² in (0, π).


To find the half-range sine series for the function \( f(x) = x^2 \) on the interval \( (0, \pi) \), we use the following form for the half-range sine series expansion:

\[
f(x) = \sum_{n=1}^{\infty} b_n \sin(nx)
\]

Where the Fourier sine coefficients \( b_n \) are given by the formula:

\[
b_n = \frac{2}{\pi} \int_0^\pi f(x) \sin(nx) \, dx
\]

For the given function \( f(x) = x^2 \), we need to calculate the coefficients \( b_n \).

### Step 1: Compute the Fourier sine coefficients

We substitute \( f(x) = x^2 \) into the formula for \( b_n \):

\[
b_n = \frac{2}{\pi} \int_0^\pi x^2 \sin(nx) \, dx
\]

To solve this integral, we use integration by parts. Let's perform the integration step-by-step:

1. Let \( u = x^2 \) and \( dv = \sin(nx) \, dx \).
2. Then, \( du = 2x \, dx \) and \( v = -\frac{1}{n} \cos(nx) \).

Now, applying integration by parts:

\[
\int_0^\pi x^2 \sin(nx) \, dx = \left[ -\frac{x^2}{n} \cos(nx) \right]_0^\pi + \int_0^\pi \frac{2x}{n} \cos(nx) \, dx
\]

At the boundaries:

\[
\left[ -\frac{x^2}{n} \cos(nx) \right]_0^\pi = -\frac{\pi^2}{n} \cos(n\pi) + 0 = \frac{\pi^2}{n} (-1)^n
\]

Next, we calculate the remaining integral:

\[
\int_0^\pi \frac{2x}{n} \cos(nx) \, dx
\]

We apply integration by parts again to this new integral, where we let \( u = x \) and \( dv = \cos(nx) \, dx \):

1. \( du = dx \) and \( v = \frac{1}{n} \sin(nx) \).

Now, we integrate by parts:

\[
\int_0^\pi \frac{2x}{n} \cos(nx) \, dx = \left[ \frac{2x}{n^2} \sin(nx) \right]_0^\pi - \int_0^\pi \frac{2}{n^2} \sin(nx) \, dx
\]

At the boundaries:

\[
\left[ \frac{2x}{n^2} \sin(nx) \right]_0^\pi = 0
\]

The remaining integral is:

\[
- \int_0^\pi \frac{2}{n^2} \sin(nx) \, dx = -\frac{2}{n^2} \left[ -\cos(nx) \right]_0^\pi
\]

Evaluating:

\[
- \frac{2}{n^2} \left[ -\cos(n\pi) + 1 \right] = \frac{2}{n^2} \left(1 - (-1)^n \right)
\]

Now, putting everything together, the Fourier sine coefficient \( b_n \) becomes:

\[
b_n = \frac{2}{\pi} \left( \frac{\pi^2}{n} (-1)^n + \frac{2}{n^2} \left(1 - (-1)^n \right) \right)
\]

### Step 2: Final Expression for the Half-Range Sine Series

The half-range sine series for \( f(x) = x^2 \) on the interval \( (0, \pi) \) is:

\[
f(x) = \sum_{n=1}^{\infty} b_n \sin(nx)
\]

Where the Fourier sine coefficients \( b_n \) are calculated above.

Would you like a detailed breakdown of these coefficients or further explanation on any part? 

### Related Questions:
1. How does the half-range sine series differ from a full Fourier series?
2. What is the general method for finding the Fourier coefficients for any piecewise continuous function?
3. How can we interpret the convergence of the Fourier sine series for a given function?
4. Can you find the half-range cosine series for a different function on the same interval?
5. How do boundary conditions affect the form of the Fourier series?

### Tip:
The sine series is particularly useful for functions defined on intervals starting from zero, as it naturally satisfies the boundary condition \( f(0) = 0 \) without needing any modifications.

find the half-range sine series of f(x) = x² in (0, π).

Learn how to find the half-range sine series for the function f(x) = x² over the interval (0, π). This detailed solution covers the calculation of Fourier sine coefficients, integration by parts, and the final sine series expansion. Suitable for advanced students studying Fourier analysis or related topics in calculus.

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