Question

Use Theorem 11.3.5(c) or, where applicable, Exercise 11.1.42(b), to find the mixed Fourier cosine series of \(f\) on \([0, L]\). $$ f(x)=4 x^{3}+3 L x^{2}-7 L^{3} $$

Short answer

Answer: The mixed Fourier cosine series for the function \(f(x)\) on the interval \([0, L]\) is given by the formula: $$ f(x) = L \sum_{n=1}^{\infty} \frac{8}{n^3 \pi^3} \cos \left(\frac{n \pi x}{L}\right) $$

Step-by-step solution

Understand the Theorem 11.3.5(c)

Theorem 11.3.5(c) states that for a given function \(f(x)\) on the interval \([0, L]\), the Fourier cosine series coefficients \(a_n\) can be calculated as follows: $$ a_n = \frac{2}{L} \int_{0}^{L} f(x) \cos \left(\frac{n \pi x}{L}\right) dx $$ With this formula, we can compute the coefficients \(a_n\) for our given function \(f(x) = 4x^3 + 3Lx^2 - 7L^3\) and write the mixed Fourier cosine series of \(f(x)\).

Compute the Fourier cosine coefficients \(a_n\)

Now we will compute the \(a_n\) coefficients for our given function \(f(x)\) using the formula from Theorem 11.3.5(c): $$ a_n = \frac{2}{L} \int_{0}^{L} \left(4x^3 + 3Lx^2 - 7L^3\right) \cos \left(\frac{n \pi x}{L}\right) dx $$ To simplify the integration process, let's break the integral into three separate integrals: $$ a_n = \frac{2}{L} \left[\int_{0}^{L} 4x^3 \cos \left(\frac{n \pi x}{L}\right) dx + \int_{0}^{L} 3Lx^2 \cos \left(\frac{n \pi x}{L}\right) dx - \int_{0}^{L} 7L^3 \cos \left(\frac{n \pi x}{L}\right) dx\right] $$ Now, we have to compute the three integrals.

Compute the separate integrals

Let's start by computing the first integral: $$ \int_{0}^{L} 4x^3 \cos \left(\frac{n \pi x}{L}\right) dx $$ This integral can be computed using integration by parts with \(u = x^3\) and \(dv = 4 \cos \left(\frac{n\pi x}{L}\right) dx\). By applying integration by parts, we get: $$ \int_{0}^{L} 4x^3 \cos \left(\frac{n \pi x}{L}\right) dx = \frac{4L^3}{n^2 \pi^2} \sin \left(\frac{n \pi x}{L}\right) |_{0}^{L} - \frac{4L^2}{n \pi} \int_{0}^{L} x^2 \sin \left(\frac{n \pi x}{L}\right) dx $$ The first term becomes zero since \(\sin(n \pi) = 0\). Now we have to compute the remaining integral, which can be done with another round of integration by parts. Thus, we obtain: $$ \int_{0}^{L} 4x^3 \cos \left(\frac{n \pi x}{L}\right) dx = \frac{4L^4}{n^3 \pi^3} \cos \left(\frac{n \pi x}{L}\right) |_{0}^{L} $$ Now, let's compute the second integral: $$ \int_{0}^{L} 3Lx^2 \cos \left(\frac{n \pi x}{L}\right) dx $$ This integral can be computed using integration by parts with \(u = x^2\) and \(dv = 3L \cos \left(\frac{n\pi x}{L}\right) dx\). By applying integration by parts, we get: $$ \int_{0}^{L} 3Lx^2 \cos \left(\frac{n \pi x}{L}\right) dx = \frac{3L^3}{n \pi} \sin \left(\frac{n \pi x}{L}\right) |_{0}^{L} - \frac{6L^2}{n^2 \pi^2} \int_{0}^{L} x \cos \left(\frac{n \pi x}{L}\right) dx $$ The first term becomes zero since \(\sin(n \pi) = 0\). Now we have to compute the remaining integral, which can be done with another round of integration by parts. Thus, we obtain: $$ \int_{0}^{L} 3Lx^2 \cos \left(\frac{n \pi x}{L}\right) dx = -\frac{6L^3}{n^3 \pi^3} \sin \left(\frac{n \pi x}{L}\right) |_{0}^{L} $$ Since \(\sin(n \pi) = 0\), the whole second integral becomes zero. Lastly, let's compute the third integral: $$ \int_{0}^{L} 7L^3 \cos \left(\frac{n \pi x}{L}\right) dx = 7L^3 \int_{0}^{L} \cos \left(\frac{n \pi x}{L}\right) dx = \frac{7L^3}{n \pi} \left[\sin\left(\frac{n \pi x}{L}\right)\right]_{0}^{L} $$ Since \(\sin(n \pi) = 0\), the whole third integral becomes zero.

Combine the integrals and write the mixed Fourier cosine series

Now that we have computed the three integrals, we can combine them to get the \(a_n\) coefficients: $$ a_n = \frac{2}{L} \left[\frac{4L^4}{n^3 \pi^3} \cos \left(\frac{n \pi x}{L}\right) |_{0}^{L}\right] $$ Now, we can write the mixed Fourier cosine series for the function \(f(x) = 4x^3 + 3Lx^2 - 7L^3\) as: $$ f(x) = \sum_{n=1}^{\infty} a_n \cos \left(\frac{n \pi x}{L}\right) = L \sum_{n=1}^{\infty} \frac{8}{n^3 \pi^3} \cos \left(\frac{n \pi x}{L}\right) $$ Thus, we have found the mixed Fourier cosine series for the given function on the interval \([0, L]\).

Key concepts

Mixed Fourier Series

In the realm of mathematical analysis, a mixed Fourier series is an invaluable tool for representing a function that is piecewise defined - that is, the function may have different definitions on different intervals.

For functions defined on the interval \[0, L\], a mixed Fourier series can be particularly interesting as it combines both sine and cosine terms to accurately replicate the function. However, when the function is even, meaning that it satisfies the condition \( f(-x) = f(x) \), the sine components vanish, leaving only a Fourier cosine series.

The Fourier cosine series is expressed as a sum of cosine terms, each multiplied by coefficients that are calculated using the function's values on the interval \[0, L\]. These coefficients are typically denoted as \( a_n \) for each term in the series.

In the context of the exercise, we are focused on finding the coefficients of such a series for the prescribed function, leading to the representation of that function as a Fourier cosine series on the given interval.

Integration by Parts

Understanding the computation of Fourier coefficients necessitates familiarity with integration by parts, which is a fundamental technique in calculus.

The formula for integration by parts is derived from the product rule for differentiation and is described by the equation: \[ \int u dv = uv - \int v du \.\]

Integration by parts is applied when an integral involves a product of two functions for which direct integration is complex or impossible. The aim is to transform the original integral into a simpler form. The process involves choosing one function to be \(u\) (to be differentiated), and the other to be \(dv\) (to be integrated).

In the textbook solution provided, integration by parts is cleverly used to compute the integrals involving \(x^3 \cos(\frac{n \pi x}{L})\) and \(x^2 \cos(\frac{n \pi x}{L})\), ultimately simplifying the task of finding the Fourier coefficients. Each application of the technique brings us one step closer to an integrable form, highlighting its utility in solving complex integrals.

Boundary Value Problems

The concept of boundary value problems is central to differential equations and applied mathematics. These problems involve finding a function that satisfies a differential equation within a specific domain, while also meeting certain prescribed conditions called boundary conditions at the domain's boundaries.

In the context of Fourier series, boundary value problems typically require the function to be periodic or to satisfy particular constraints at the endpoints of the interval. For instance, the boundary conditions for an even function on the interval \[0, L\] lead to the disappearance of the sine terms in the mixed Fourier series, simplifying the problem to finding a Fourier cosine series.

In solving the exercise, no explicit boundary value problem was stated; nonetheless, the computation of the Fourier coefficients inherently respects the boundaries of the function on the interval \[0, L\]. The final expression of the Fourier series will thus inherently conform to the values of the function at its boundaries, exhibiting the natural connection between these series and boundary value problems.