Question

Use Theorem 11.3.5(a) to find the Fourier cosine series of \(f\) on \([0, L]\). $$ f(x)=x^{3}(3 x-4 L) $$

Short answer

In summary, the Fourier cosine series of the given function \(f(x) = x^3(3x - 4L)\) on the interval \([0, L]\) is: $$ f(x) = -\frac{17}{5}L^4 + \sum_{n=1}^{\infty} \left[\frac{6L^4}{n^2\pi^2} \cos\left(\frac{n\pi x^2}{2L}\right) - \frac{24L^3}{n^3\pi^3} \sin\left(\frac{n\pi x^2}{2L}\right) \cos\left(\frac{n\pi x}{L}\right) - \frac{8x^3L}{n^2\pi^2} \sin\left(\frac{n\pi x}{L}\right) \cos\left(\frac{n\pi x}{L}\right)\right] $$

Step-by-step solution

Find the Fourier cosine coefficients

To find the coefficients \(a_0\) and \(a_n\), we will use the given formulas: $$ a_0 = \frac{2}{L} \int_0^L f(x) dx $$ and $$ a_n = \frac{2}{L} \int_0^L f(x) \cos\left(\frac{n\pi x}{L}\right) dx $$ First, we will compute \(a_0\): $$ a_0 = \frac{2}{L} \int_0^L x^3(3x - 4L) dx $$ Now, find the antiderivative of the integrand: $$ \int x^3(3x - 4L) dx = \frac{3}{5}x^5 - 4Lx^4 + C $$ Evaluate the integral with the limits \(0\) and \(L\): $$ a_0 = \frac{2}{L} \left[\left(\frac{3}{5}L^5 - 4L^5\right) - \left(0\right)\right] = \frac{2}{L} \left(-\frac{17}{5}L^5\right) = -\frac{34}{5}L^4 $$ Next, we will compute \(a_n\): $$ a_n = \frac{2}{L} \int_0^L x^3(3x - 4L) \cos\left(\frac{n\pi x}{L}\right) dx $$ We can split the integral into two parts: $$ a_n = \frac{6}{L} \int_0^L x^4 \cos\left(\frac{n\pi x}{L}\right) dx - \frac{8}{L} \int_0^L x^3L \cos\left(\frac{n\pi x}{L}\right) dx $$ To solve these integrals, we can use integration by parts, but here we may use a library, like Wolfram, for example. The solution would then look like this: $$ a_n = \frac{6}{L}\left[\frac{L^5 \cos(\frac{n\pi x^2}{2L})}{n^2\pi^2} - \frac{4L^4 \sin(\frac{n\pi x^2}{2L})}{n\pi}\right]_0^L - \frac{8}{n\pi}\left[\frac{x^3L\sin(\frac{n\pi x}{L})}{n\pi}\right]_0^L $$ Now, we are ready to construct the Fourier cosine series.

Construct the Fourier cosine series

We can now write the Fourier cosine series of \(f(x) = x^3(3x - 4L)\) using the coefficients \(a_0\) and \(a_n\): $$ f(x) \approx \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\left(\frac{n\pi x}{L}\right) $$ Substitute the values for \(a_0\) and \(a_n\): $$ f(x) \approx -\frac{17}{5}L^4 + \sum_{n=1}^{\infty} \left[\frac{6}{L}\left(\frac{L^5 \cos(\frac{n\pi x^2}{2L})}{n^2\pi^2} - \frac{4L^4 \sin(\frac{n\pi x^2}{2L})}{n\pi}\right) - \frac{8}{n\pi}\left(\frac{x^3L\sin(\frac{n\pi x}{L})}{n\pi}\right)\right] \cos\left(\frac{n\pi x}{L}\right) $$ We can simplify the terms inside the summation: $$ f(x) = -\frac{17}{5}L^4 + \sum_{n=1}^{\infty} \left[\frac{6L^4}{n^2\pi^2} \cos\left(\frac{n\pi x^2}{2L}\right) - \frac{24L^3}{n^3\pi^3} \sin\left(\frac{n\pi x^2}{2L}\right) \cos\left(\frac{n\pi x}{L}\right) - \frac{8x^3L}{n^2\pi^2} \sin\left(\frac{n\pi x}{L}\right) \cos\left(\frac{n\pi x}{L}\right)\right] $$ And here is the Fourier cosine series of the given function \(f(x) = x^3(3x - 4L)\) on \([0, L]\).

Key concepts

Fourier coefficients

The Fourier coefficients are crucial in determining the Fourier cosine series. They capture the essence of the function to be expressed as a sum of cosine terms. This is what allows us to expand complicated functions into simpler trigonometric functions for analysis. The coefficients are denoted by \(a_0\) and \(a_n\).

• \(a_0\): Represents the average value of the function over the interval. It is given by \(\frac{2}{L} \int_0^L f(x) dx\).
• \(a_n\): These coefficients are found using the expression \(\frac{2}{L} \int_0^L f(x) \cos\left(\frac{n\pi x}{L}\right) dx\). They determine the contribution of each cosine term in the series.

Calculating these coefficients accurately is essential. For example, simplifies the complex function \(f(x) = x^3(3x - 4L)\) into a series that can be easily analyzed or used in applications such as signal processing.

integration by parts

Integration by parts is a mathematical technique used to solve integrals where ordinary integration becomes complex. It's based on the product rule for differentiation and is especially useful for integrating products of functions.
The formula is:
\[ \int u \cdot dv = uv - \int v \cdot du \] where \(u\) and \(dv\) are well-chosen parts of the original integral. This formula helps transform a challenging integral into a simpler one.

In the solution for the Fourier coefficients \(a_n\), integration by parts was necessary to make the difficult integrals involving \(x^4\) and \(\cos\left(\frac{n\pi x}{L}\right)\) more manageable. If calculations seem complex, computational tools can be invaluable in reaching accurate solutions efficiently.

cosine function

The cosine function, \(\cos(x)\), is a fundamental trigonometric function. It oscillates between -1 and 1 and is periodic, repeating every \(2\pi\). In Fourier series, the cosine function is a building block. It helps decompose complex functions on a fixed interval into simpler cosine terms.

• Cosine is even, meaning \(\cos(-x) = \cos(x)\), which simplifies many calculations in real-world applications.
• The frequency of the cosine in Fourier series is determined by \(\frac{n\pi x}{L}\). Its role is to capture the specific harmonics of the original function.

Hence, in constructing a Fourier cosine series, each function or system can be examined in terms of these elementary harmonics, making them easier to study or manipulate mathematically.

boundary value problems

Boundary value problems involve finding solutions to differential equations subject to specific conditions at the boundaries of the domain. These are often found in physics and engineering. Fourier cosine series are particularly advantageous for solving such problems.

Typically, when considering a system on \([0, L]\), the boundary conditions may require that the solution be zero at \(x = 0\) and \(x = L\). The even nature of cosine functions, paired with the symmetry of the interval, makes Fourier series a good fit.

• These boundary conditions help in determining appropriate series expansions, ensuring that solutions are not only accurate but also satisfy the physical system constraints.
• Cosine series expansions help in simplifying the complex boundary conditions, breaking them into more manageable algebraic relationships and thus providing a clear mathematical model for predicting behaviors.

Thus, Fourier series are invaluable in transforming and solving boundary value problems, providing clear insights and effective solutions.