Question

(a) Find the Fourier series of period 2 for \(f(x)=(x-1)^{2}\) on \((0,2)\). (b) Use your result in (a) to evaluate \(\sum 1 / n^{4}\).

Short answer

Fourier series and coefficient calculations are needed to evaluate \( \sum \frac{1}{n^4} \).

Step-by-step solution

Express the Function for the Given Interval

The function given is \[ f(x) = (x-1)^2 \] on the interval \( (0,2) \). Note that the period \( T \) is 2.

Compute the Fourier Coefficients

The Fourier series for a function with period 2 is given by \[ f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(\frac{n\pi x}{L}) + b_n \sin(\frac{n\pi x}{L})) \] where \( L = T/2 = 1 \). First, compute the coefficients \[ a_0 = \frac{1}{L} \int_0^{2} f(x) \frac{dx}{2} = \frac{1}{2} \int_0^2 (x-1)^2 dx \] Using the integral, \[ a_0 = \frac{1}{2} \left[ \frac{(x-1)^3}{3} \right]_0^2 = \frac{1}{2} \left( \frac{(-1)^3}{3} - \frac{1^3}{3} \right) = \frac{2}{3} \]

Compute the Coefficients a_n

Next, calculate \[ a_n = \frac{1}{L} \int_0^2 (x-1)^2 \cos(\frac{n\pi x}{L}) dx = \int_0^2 (x-1)^2 \cos(n\pi x) dx \] This integral can be computed using integration by parts.

Compute the Coefficients b_n

Similarly, compute \[ b_n = \frac{1}{L} \int_0^2 (x-1)^2 \sin(\frac{n\pi x}{L}) dx = \int_0^2 (x-1)^2 \sin(n\pi x) dx \] This can also be computed using integration by parts.

Summarize the Fourier Series

Assemble the Fourier series using the calculated coefficients. The general form will include the terms of \( a_0 \), \( a_n \cos(n\pi x) \) and \( b_n \sin(n\pi x) \).

Evaluate the Series for \( \sum \frac{1}{n^4} \)

Using the previously derived Fourier series, the result can be applied to identify the value of \( \sum \frac{1}{n^4} \).

Key concepts

Coefficient Calculation

To understand how we calculate the coefficients in a Fourier series, we first need to grasp the general form of the Fourier series.
The Fourier series of a function with period 2 is given by:

\[ f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(\frac{n\pi x}{L}) + b_n \sin(\frac{n\pi x}{L})) \]
Here, L is the half-period, so for period 2, L = 1.
To find the coefficients \(a_0\), \(a_n\), and \(b_n\) we use specific integral formulas.

For \(a_0\) (the average value of the function over one period), we have:
\[ a_0 = \frac{1}{L} \int_0^{2} f(x) \frac{dx}{2} \]
For the function \( f(x) = (x-1)^2 \), this becomes:
\[ a_0 = \frac{1}{2} \int_0^2 (x-1)^2 dx \]
Evaluating this integral, we find:
\[ a_0 = \frac{2}{3} \]

For the coefficients \(a_n\):
\[ a_n = \frac{1}{L} \int_0^2 (x-1)^2 \cos(\frac{n\pi x}{L}) dx \]
And for the coefficients \(b_n\):
\[ b_n = \frac{1}{L} \int_0^2 (x-1)^2 \sin(\frac{n\pi x}{L}) dx \]
Both these integrals require the use of integration by parts, a fundamental technique in calculus that will be detailed in the next section.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation.
It is very handy for integrals where the integrand is a product of two functions.
The formula for integration by parts is:
\[ \int u dv = uv - \int v du \]
Here’s a step-by-step breakdown on how to apply it.

For finding the Fourier coefficients, let’s consider the integral for \(a_n\):
\[ a_n = \int_0^2 (x-1)^2 \cos(n\pi x) dx \]
We select:
- \(u = (x-1)^2\), \quad then \(du = 2(x-1) dx\)
- \(dv = \cos(n\pi x) dx\), \quad then \(v = \frac{\sin(n\pi x)}{n\pi}\)
Applying integration by parts here, we get:
\[ a_n = \left[ (x-1)^2 \frac{\sin(n\pi x)}{n\pi} \right]_0^2 - \int_0^2 2(x-1) \frac{\sin(n\pi x)}{n\pi} dx \]
This integral simplifies further, often requiring another round of integration by parts. Follow similar steps for \(b_n\) to extract the trigonometric coefficients.

Evaluation of Series

Summarizing the Fourier series involves assembling all computed coefficients into one formula.
With \(a_0 = \frac{2}{3}\) and the general forms for \(a_n\) and \(b_n\), the Fourier series for \( f(x) = (x-1)^2 \) becomes:
\[ f(x) = \frac{2}{3} + \sum_{n=1}^{\infty} (a_n \cos(n\pi x) + b_n \sin(n\pi x)) \]

The next step is to use this series for specific evaluations.
Using the Fourier series to evaluate \sum \frac{1}{n^4} involves a clever trick using Parseval's theorem.
Parseval's theorem relates the sum of the squares of the Fourier coefficients to the integral of the square of the function over one period.
For \( f(x) = (x-1)^2 \), we have:

\[ \int_0^2 ((x-1)^2)^2 dx = a_0^2 + \sum_{n=1}^{\infty} (a_n^2 + b_n^2) \]
This gives us a way to link the series coefficients back to the original integral, ultimately helping us evaluate sums like \sum \frac{1}{n^4}. Understanding each concept deeply ensures you can tackle these kinds of problems with confidence. Happy Learning!