Fourier Series Convergence Analysis at x=0

This problem requires us to understand how a Fourier series behaves at a point of jump discontinuity and relate this behavior to specific mathematical series identities.

Function Definition and Interval

The function f(x) is defined on the interval [‐π, π] as follows:

• f(x) = 0 for ‐π < x ≤ 0
• f(x) = π - x for 0 < x < π

Understanding Convergence at Jump Discontinuities

A fundamental property of Fourier series is that at points where a function has a jump discontinuity, the series converges to the average of the left-hand and right-hand limits of the function at that point. This is a direct consequence of the convergence criteria for Fourier series.

Calculating Limits at the Point x = 0

We need to find the limits of f(x) as x approaches 0:

• Left-hand limit: For x ≤ 0, f(x) = 0. $$ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 0 = 0 $$
• Right-hand limit: For 0 < x < π, f(x) = π - x. $$ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (\pi - x) = \pi - 0 = \pi $$

Determining the Convergence Value of the Fourier Series

At x = 0, the Fourier series converges to the average of these limits:

Thus, the Fourier series converges to &frac{\pi}{2} at x = 0.

Evaluating the Provided Fourier Series at x = 0

The question provides the specific Fourier series expansion for f(x):

Let's substitute x = 0 into this series. We use the known values cos(0) = 1 and sin(0) = 0.

• The cosine terms contribute: $$ \frac{2}{\pi }\left[ {\frac{{\cos 0}}{1^2} + \frac{{\cos 0}}{3^2} + \ldots } \right] = \frac{2}{\pi }\left[ {\frac{1}{1^2} + \frac{1}{3^2} + \ldots } \right] $$ This sum of reciprocals of squares of odd integers can be written as: $$ \frac{2}{\pi } \sum_{n=1, n \text{ odd}}^\infty \frac{1}{n^2} = \frac{2}{\pi } \sum_{n=1}^\infty \frac{1}{(2n-1)^2} $$
• The sine terms contribute: $$ \left[ {\frac{{\sin 0}}{1} + \frac{{\sin 0}}{2} + \frac{{\sin 0}}{3} + \ldots } \right] = [0 + 0 + 0 + \ldots] = 0 $$

Substituting these results back into the Fourier series expression at x = 0, denoted as S(0):

Connecting Fourier Convergence to Series Sum Identities

We found that the Fourier series converges to S(0) = &frac{\pi}{2} at x = 0. Equating this with the evaluated series expression:

Now, we solve this equation for the summation term, which represents the convergence result being asked for:

To isolate the sum, we multiply both sides by &frac{\pi}{2}:

This calculation demonstrates that the convergence of the given Fourier series at x = 0 yields the specific mathematical identity \(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{\left( {2n - 1} \right)^2}} = \frac{{{\pi ^2}}}{8}\).

Matching the Result with Provided Options

Let's compare our derived identity with the options given in the question:

• Option 1: \(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{{n^2}}} = \frac{{{\pi ^2}}}{6}\)
• Option 2: \(\mathop \sum \limits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{n^2}}} = \frac{{{\pi ^2}}}{{12}}\)
• Option 3: \(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{\left( {2n - 1} \right)^2}} = \frac{{{\pi ^2}}}{8}\)
• Option 4: \(\mathop \sum \limits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{2n - 1}} = \frac{\pi }{4}\)

The result we obtained, \(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{\left( {2n - 1} \right)^2}} = \frac{{{\pi ^2}}}{8}\), directly corresponds to Option 3.