Question

Use Theorem \(11.3 .5(\mathbf{d})\) or where applicable, Exercise \(50(\mathbf{b}),\) to find the mixed Fourier sine series of the \(f\) on \([0, L]\). $$ f(x)=x^{2}(3 L-2 x) $$

Short answer

Answer: The mixed Fourier sine series for the given function is \(F_s(x) = \sum_{n=1}^{\infty} \frac{12(-1)^n}{n^3\pi^3} \sin\left(\frac{n\pi x}{L}\right)\).

Step-by-step solution

Write down the function and the interval

The function is given by \(f(x) = x^2(3L-2x)\), and we need to find its mixed Fourier sine series on the interval \([0, L]\).

Find the Fourier sine coefficients \(b_n\)

To find the Fourier sine coefficients \(b_n\), we will use the formula: $$ b_n = \frac{2}{L} \int_0^L x^2(3L-2x) \sin\left(\frac{n\pi x}{L}\right) dx $$ Now, we will compute the integral: $$ \int_0^L x^2(3L-2x) \sin\left(\frac{n\pi x}{L}\right) dx $$ We can use integration by parts to compute this integral. Let \(u = x^2(3L-2x)\) and \(dv =\sin\left(\frac{n\pi x}{L}\right) dx\). Then, \(du = (6Lx - 6x^2)dx\) and \(v= -\frac{L}{n\pi}\cos\left(\frac{n\pi x}{L}\right)\). Now we can rewrite our integral as: $$ -\frac{L}{n\pi}\int_0^L (6Lx - 6x^2)\cos\left(\frac{n\pi x}{L}\right) dx $$ We can split the integral and apply integration by parts one more time: $$ -\frac{L}{n\pi}\left[\int_0^L 6Lx\cos\left(\frac{n\pi x}{L}\right) dx - \int_0^L 6x^2\cos\left(\frac{n\pi x}{L}\right) dx\right] $$ Solving these integrals, we get: $$ b_n = \frac{12(-1)^n}{n^3\pi^3} $$

Sums of the series

Now we can write the mixed Fourier sine series for \(f(x)\): $$ F_s(x) = \sum_{n=1}^{\infty} \frac{12(-1)^n}{n^3\pi^3} \sin\left(\frac{n\pi x}{L}\right) $$ Therefore, the mixed Fourier sine series for the given function \(f(x) = x^2(3L-2x)\) on the interval \([0, L]\) is: $$ F_s(x) = \sum_{n=1}^{\infty} \frac{12(-1)^n}{n^3\pi^3} \sin\left(\frac{n\pi x}{L}\right) $$

Key concepts

Fourier Sine Coefficients

Understanding Fourier sine coefficients is essential when dealing with Fourier series in the context of odd periodic functions or functions defined on a finite interval. A Fourier sine series is a way to represent a function using a combination of sine functions. To find the coefficients, we integrate the product of the original function and a sine term that has increasing frequencies for all natural numbers.

The generic formula for the Fourier sine coefficients, denoted typically by \(b_n\), is given by \[ b_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) dx, \] where \( L \) is half the period of the function, \( n \) is an integer, and \( f(x) \) is the function we are trying to represent. These coefficients are a measure of the function's similarity to the sine functions at different frequencies.

Applying this to our exercise for the function \(f(x) = x^2(3L-2x)\), we compute the integral for each \( n \) to find the coefficients, which form the sine series that approximates the original function within its interval \( [0, L] \) as closely as possible.

Integration by Parts

Integration by parts is a powerful tool for evaluating integrals, especially when the integral consists of the product of two functions. It's based on the product rule for differentiation and is formally given by \[ \int u dv = uv - \int v du. \] To apply this technique, we cleverly choose \( u \) and \( dv \) from our integrand, then differentiate \( u \) to get \( du \) and integrate \( dv \) to find \( v \), applying the formula afterwards.

In the context of our example, we applied integration by parts to \( u = x^2(3L-2x) \) and \( dv =\sin\left(\frac{n\pi x}{L}\right) dx \) to find the Fourier sine coefficients. This helped us to break the problem into simpler parts that are easier to integrate. Remember that in some cases, applying integration by parts more than once is necessary to obtain an answer, as illustrated by our two separate integrals over \( x \cos\left(\frac{n\pi x}{L}\right) \) after the first application of this technique.

Orthogonal Functions

Orthogonality in functions is analogous to perpendicularity in vectors. Functions are considered orthogonal on an interval if their product integrated over that interval is zero. Mathematically, two functions \( f \) and \( g \) are orthogonal over the interval \( [a, b] \) if \[ \int_a^b f(x) g(x) dx = 0. \] This concept is integral to Fourier series, where sets of sine and cosine functions are orthogonal over specific intervals.

In our exercise, sine functions with different frequencies are orthogonal over the interval \( [0, L] \). Due to orthogonality, when computing the Fourier sine coefficients, the integral of the product of the original function and each sine term isolates the influence of that term in the approximation of the function. This is why the coefficient for each term in a Fourier series is a distinct constant, reflecting how much of that 'sine frequency' is present in the original function.