Question

Find the Fourier cosine series. $$ f(x)=x^{2} ; \quad[0, L] $$

Short answer

Answer: The Fourier cosine series of the function f(x) = x^2 in the interval [0, L] is F_n(x) = (1/3)L^2.

Step-by-step solution

Find a_0

To find the coefficient a_0, we need to integrate f(x) from 0 to L and multiply the result by 2/L: $$ a_0 = \frac{2}{L} \int_0^L x^2 dx $$ Now, we will compute the integral: $$ \int_0^L x^2 dx = \left[ \frac{1}{3}x^3 \right]_0^L = \frac{1}{3}L^3 $$ Thus, the value of a_0 is given by: $$ a_0 = \frac{2}{L} \cdot \frac{1}{3}L^3 = \frac{2}{3}L^2 $$

Find a_n

For finding the coefficient a_n, we need to integrate the product of f(x) and cosine term from 0 to L and multiply the result by 2/L: $$ a_n = \frac{2}{L} \int_0^L x^2 \cos\Big(\frac{n \pi x}{L}\Big) dx $$ To find the integral, we use integration by parts: Let u = x^2, dv = cos(nx\pi/L)dx Then du = 2xdx, v = (L/(n\pi))sin(nx\pi/L) Now we will compute the integral using integration by parts formula: $$ \int_0^L x^2 \cos\Big(\frac{n \pi x}{L}\Big) dx = \left[\frac{L}{n\pi}x^2\sin\Big(\frac{n \pi x}{L}\Big)\right]_0^L - \int_0^L\frac{L}{n\pi}\cdot 2x\sin\Big(\frac{n \pi x}{L}\Big)dx $$ The first term evaluates to 0 since the sine function becomes 0 at both ends. Now, we once again use integration by parts for the remaining term. Let u = 2x , dv = sin(nx\pi/L)dx Thus, du = 2dx, v = -(L/(n\pi))cos(nx\pi/L) Now, we compute the integral: $$ -\frac{2L}{n\pi}\int_0^L x\sin\Big(\frac{n \pi x}{L}\Big)dx = (-\frac{2L}{n\pi})\left[\frac{L}{n\pi}x\cos\Big(\frac{n \pi x}{L}\Big)\right]_0^L + \frac{2L^2}{(n\pi)^2}\int_0^L \cos\Big(\frac{n \pi x}{L}\Big)dx $$ Again, the first term evaluates to 0, and we have the final integral: $$ \frac{2L^2}{(n\pi)^2}\int_0^L \cos\Big(\frac{n \pi x}{L}\Big)dx = \frac{2L^2}{(n\pi)^2}\left[\frac{L}{n\pi}\sin\Big(\frac{n \pi x}{L}\Big)\right]_0^L = 0 $$ So, the coefficient a_n is given by: $$ a_n = \frac{2}{L} \cdot 0 = 0 $$

Fourier Cosine Series

Now, we have all the required coefficients, and we can write the Fourier cosine series for the given function: $$ F_n(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\Big(\frac{n \pi x}{L}\Big) = \frac{1}{2}\cdot \frac{2}{3}L^2 = \frac{1}{3}L^2 $$ Therefore, the Fourier cosine series of f(x) = x^2 in the interval [0, L] is given by: $$ F_n(x) = \frac{1}{3}L^2 $$

Key concepts

Integral Calculus

Integral calculus is a branch of calculus focused on the concept of the integral, which represents accumulation or the total effect of a quantity that changes over time or space. For example, if you want to find the area under a curve like a parabola, which is represented by a function like f(x) = x^2, you'd use an integral to do so.

To find the area under the parabola from 0 to an arbitrary length L, as is the case in our exercise, you calculate the definite integral of the function within those limits. The definite integral is visually represented by the area under the graph of the function between the two endpoints. To compute the definite integral of f(x) = x^2 from 0 to L, we use the fundamental theorem of calculus to find the antiderivative of f(x), evaluate this antiderivative at the upper limit L, subtract the evaluation at the lower limit 0, and obtain (1/3)L^3. This process simplifies finding the cumulative area under the curve, streamlining complex calculations that would otherwise be extremely difficult to carry out.

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation and is used to integrate the product of two functions. It’s especially handy when dealing with products of algebraic expressions and trigonometric or exponential functions, which seem tough to integrate on their own.

The formula for integration by parts is given by ∫u dv = uv − ∫v du. Here, you need to choose u and dv wisely, usually picking u to be a function that simplifies when differentiated and dv to be a function with an easily integrable derivative. In our exercise, to find the Fourier coefficient a_n, you apply integration by parts by letting u = x^2 and dv = cos(nπx/L)dx. After finding du and v, you substitute and evaluate the integral.

Performing integration by parts sometimes requires iteration; you may apply the technique more than once to simplify an integral to a solvable form. As shown in the solution, by repeating the integration by parts process, we're able to compute the required integral thanks to the properties of sine and cosine functions at multiples of π, which frequently reduce the evaluated terms to zero.

Series Representation of Functions

The series representation of functions is a powerful concept in mathematics that allows us to express complex functions as an infinite sum of simpler functions, often polynomials or trigonometric terms. This is particularly useful in many areas of science and engineering where solutions to problems can be approximated using series.

In the context of Fourier series, any periodic function can be written as a weighted sum of sines and cosines. These series take the form of a_0/2 + Σ(a_n cos(nπx/L) + b_n sin(nπx/L)), where Σ denotes the summation from n=1 to infinity, and a_0, a_n, and b_n are the Fourier coefficients. The computation of these coefficients involves integrating the product of the original function and the corresponding sine or cosine term, as we see in our exercise.

In our specific exercise, we've found a Fourier cosine series, which only includes cosine terms and represents even functions over [0, L]. Even in such seemingly simple outcomes as F_n(x) = (1/3)L^2, the process involves careful calculations of the coefficients, demonstrating the symmetrical nature of f(x) = x^2 over the given interval.