Question

Find the complex Fourier series of \(f(t)=\frac{1}{1-\frac{1}{2} e^{-i t}}\) on the interval \([-\pi, \pi]\).

Short answer

The complex Fourier series is \(f(t) = \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n e^{-int}\).

Step-by-step solution

Identify the periodicity

The function given is \(f(t) = \frac{1}{1 - \frac{1}{2} e^{-it}}\). This function can be interpreted as being defined over the interval \([-\pi, \pi]\), which is a typical interval for Fourier series calculations. However, we must check if it's periodic on this interval. As it is not inherently periodic, it will be treated as periodic with period \(2\pi\).

Express as a Geometric Series

Note that \(\frac{1}{1 - \frac{1}{2} e^{-it}}\) resembles a geometric series. The sum of the series \(\frac{1}{1-x}\) is \(1 + x + x^2 + x^3 + \ldots\) for \(|x|<1\). Here, \(x = \frac{1}{2}e^{-it}\). Hence, \(\frac{1}{1 - \frac{1}{2} e^{-it}} = \sum_{n=0}^{\infty} \left(\frac{1}{2} e^{-it}\right)^n\).

Simplify the Series Terms

Substitute \(x = \frac{1}{2} e^{-it}\) into the series, we have \(\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n e^{-int}\). This series is convergent since \(\left|\frac{1}{2} e^{-it}\right| < 1\).

Write the Complex Fourier Series

The complex Fourier series for \(f(t)\) is given by \(f(t) = \sum_{n=-\infty}^{\infty} c_n e^{int}\). Observing the derived series \(\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n e^{-int}\), it matches the structure of \(\sum_{n=-\infty}^{\infty} c_n e^{int}\), with non-zero terms only for \(n \geq 0\). Thus, \(c_n = \left(\frac{1}{2}\right)^n\) for \(n \geq 0\) and \(c_n = 0\) for \(n < 0\).

Summarize the Complex Fourier Coefficients

Summarize the coefficients as \(c_n = \left(\frac{1}{2}\right)^n\) for \(n \geq 0\) and \(c_n = 0\) for \(n < 0\). This concludes the derivation of the complex Fourier series for the given function.

Key concepts

Function Periodicity

At the heart of working with complex Fourier series is understanding across which interval the function is periodic. Periodicity refers to how a function repeats its values over regular intervals. Take for example, the function \(f(t)=\frac{1}{1-\frac{1}{2} e^{-i t}}\). Even though functions like these may not seem periodic initially, for the Fourier series application, they are often *treated as periodic*. This means that you impose a repeat pattern over the regular intervals.For the function given, the interval \([ -\pi, \pi ]\) is a common choice when dealing with complex Fourier series. By setting the function to repeat over this interval, you can analyze its behavior and expand it in terms of sinusoidal components that repeat every \(2\pi\). Essentially, regardless of the function's original behavior, you impose a periodic nature where the function's values repeat, creating a bridge for Fourier series representation.

Geometric Series

A geometric series is a powerful tool in mathematics, and it's particularly handy when working with functions like \(\frac{1}{1-x}\). This form already presents itself as a geometric series, summed up as \(1 + x + x^2 + x^3 + \ldots\), valid for \(|x|<1\). These series form when you keep multiplying by a constant ratio, which in this case is \(x\).For the given exercise, the function resembles a geometric series with \(x = \frac{1}{2}e^{-it}\). Recognizing this structure allows for expressing the function as a sum of its components: \(\frac{1}{1 - \frac{1}{2} e^{-it}} = \sum_{n=0}^{\infty} \left(\frac{1}{2} e^{-it}\right)^n\). This thoughtful approach breaks down complex functions into manageable parts by using the simple idea of repetitive multiplication.

Fourier Coefficients

Fourier coefficients are central to building the Fourier series representation of a function. These coefficients act as weights that determine how much of each frequency component, in terms of complex exponentials, is present in the original function.Looking at our example, the identified Fourier series expresses itself as: \(f(t) = \sum_{n=-\infty}^{\infty} c_n e^{int}\). From the analysis: only terms where \(n\geq 0\) have non-zero coefficients: \(c_n = \left(\frac{1}{2}\right)^n\). For \(n < 0\), the coefficients are zero, \(c_n = 0\).This reveals how Fourier coefficients assemble into the overarching series, guiding the decomposition of a function into its fundamental frequencies. By carefully determining these coefficients, you accurately capture the periodic nature of the function.

Convergence of Series

The convergence of a series tells you when it's reasonable to substitute the series in place of its original function. This means looking at the conditions under which an infinite series converges to a finite, well-defined value.In the context of our function, convergence is ensured because \(|\frac{1}{2} e^{-it}|<1\). This inequality maintains that each term in the series \(\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n e^{-int}\) becomes smaller as \(n\) increases, allowing the series to sum to a specific value reliably.Ensuring convergence is crucial, as it underpins the legitimacy of using infinite series to approximate periodic functions without fear of divergence or instability in calculations. This inherent stability supports effective function representation in Fourier analysis.