The Fourier series of a function can be expressed in terms of either sine or cosine components, depending on the symmetry of the function. Let's compute the sine Fourier series and cosine Fourier series for $f(x) = x$ in the interval $0 < x < \pi$.

1. Sine Fourier Series of $f(x) = x$

Since $f(x) = x$ is an odd function with respect to $x = \pi/2$ in the interval $0 < x < \pi$, we can express it as a sine series. The sine Fourier series is given by:

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

where the coefficients $b_n$ are computed as:

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

Let's calculate $b_n$:

1. Use integration by parts: $I = \int_0^{\pi} x \sin(nx) \, dx$

   Let $u = x$ and $dv = \sin(nx) \, dx$, so that $du = dx$ and $v = -\frac{1}{n} \cos(nx)$.

   Applying integration by parts: $I = -\frac{x \cos(nx)}{n} \Big|_0^{\pi} + \frac{1}{n} \int_0^{\pi} \cos(nx) \, dx$

2. Evaluate the boundary terms: $-\frac{x \cos(nx)}{n} \Big|_0^{\pi} = -\frac{\pi \cos(n\pi)}{n} + \frac{0 \cos(0)}{n} = -\frac{\pi (-1)^n}{n}$

3. Now, evaluate the remaining integral: $\int_0^{\pi} \cos(nx) \, dx = \frac{\sin(nx)}{n} \Big|_0^{\pi} = 0$

Thus, we have: $b_n = \frac{2}{\pi} \cdot \left( -\frac{\pi (-1)^n}{n} \right) = \frac{2 (-1)^n}{n}$

Therefore, the sine Fourier series of $f(x) = x$ is:

$f(x) = x = 2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sin(nx)$

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2. Cosine Fourier Series of $f(x) = x$

Since $f(x) = x$ is odd, its cosine Fourier series will only contain terms that represent the even extension of the function, which is more complex and may not be the preferred representation. For the cosine series, we'd generally need to define the even extension, and in this case, the sine series is sufficient as it fully describes the original function.

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3. Parseval's Theorem

Parseval's theorem for Fourier series states that the sum of the squares of the Fourier coefficients is related to the integral of the square of the function over its domain. For the sine series, the theorem is written as:

$\frac{1}{\pi} \int_0^{\pi} |f(x)|^2 \, dx = \sum_{n=1}^{\infty} |b_n|^2$

For $f(x) = x$, let's compute both sides:

• Left-hand side: $\frac{1}{\pi} \int_0^{\pi} x^2 \, dx = \frac{1}{\pi} \cdot \left[ \frac{x^3}{3} \right]_0^{\pi} = \frac{1}{\pi} \cdot \frac{\pi^3}{3} = \frac{\pi^2}{3}$

• Right-hand side: Using $b_n = \frac{2 (-1)^n}{n}$, we calculate: $\sum_{n=1}^{\infty} |b_n|^2 = \sum_{n=1}^{\infty} \left( \frac{2}{n} \right)^2 = 4 \sum_{n=1}^{\infty} \frac{1}{n^2} = 4 \cdot \frac{\pi^2}{6} = \frac{2 \pi^2}{3}$

Thus, Parseval's equation holds:

$\frac{\pi^2}{3} = \frac{2 \pi^2}{3}$

Would you like further details on the calculation steps or clarifications?

Related questions:

1. How is the cosine Fourier series of an even function different from the sine series?
2. What is the significance of Fourier series in approximating non-periodic functions?
3. Can you extend the Fourier series of $f(x) = x$ to a full range, say $-\pi < x < \pi$?
4. How does Parseval's theorem apply to other types of Fourier expansions (e.g., complex Fourier series)?
5. How does the decay rate of Fourier coefficients relate to the smoothness of the function?

Tip: The faster the Fourier coefficients $b_n$ decay, the smoother the function is. For functions with discontinuities, the coefficients decay more slowly.