Question

Find the Fourier cosine series. $$ f(x)=\sin k x(k \neq \text { integer }) ; \quad[0, \pi] $$

Short answer

To find the Fourier cosine series for f(x) = sin(kx) in the interval [0, π], we first computed the Fourier coefficients a₀ and aₙ by integrating the given function multiplied by cos(nx). We found that a₀ = ( -1 / (πk) )(cos(kπ) - 1) and aₙ = (-1 / nπ)( [(cos((n+k)π) - 1] / (n+k) + [(cos((n-k)π) - 1] / (n-k) ). Finally, we put these coefficients into the Fourier cosine series formula, resulting in: f(x) = ( -1 / (πk) )[cos(kπ) - 1] + Σ((-1 / nπ)[(cos((n+k)π) - 1) / (n+k) + (cos((n-k)π) - 1) / (n-k)])cos(nx) for n = 1 to infinity.

Step-by-step solution

Understand Fourier cosine series formula

The Fourier cosine series of a function f(x) can be represented as: $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos{(nx)} $$ Where the Fourier coefficients (aₙ) are given by: $$ a_n = \frac{2}{\pi} \int_0^{\pi} f(x) \cos(nx)dx $$ In this case, f(x) = sin(kx) and we need to find the coefficients aₙ for n = 0, 1, 2, ...

Calculate a₀

First, let's calculate a₀, which is the average value of the function: $$ a_0 = \frac{1}{\pi}\int_0^\pi{\sin(kx)dx} $$ Integrate by substitution: Let u = kx, so du = kdx, and dx = (1/k)du. Change the limits of integration: When x = 0, u = k(0) = 0. When x = π, u = k(π) = kπ. Now, the integral becomes: $$ a_0 = \frac{1}{\pi k} \int_0^{k\pi}{\sin(u)du} $$ Now, integrate with respect to u: $$ a_0 = \frac{1}{\pi k}[-\cos(u)]_0^{k\pi} = \frac{-1}{\pi k}[\cos(k\pi) - 1] $$ Since k is not an integer, cos(kπ) ≠ 1, and a₀ ≠ 0.

Calculate aₙ for n ≥ 1

Now, let's find the coefficients aₙ for n ≥ 1: $$ a_n = \frac{2}{\pi}\int_0^{\pi}\sin(kx)\cos(nx)dx $$ Integrate by parts: Let: u = sin(kx), dv = cos(nx)dx. du = kcos(kx)dx, v = (1/n)sin(nx). Now, using integration by parts: $$ \int u\,dv = uv - \int v\,du $$ Here, we have: $$ a_n = \frac{2}{\pi}\left[\frac{1}{n}\sin(kx)\sin(nx) \bigg|_0^\pi - \int_0^\pi \frac{k}{n}\cos(kx)\sin(nx)dx\right] $$ The first term of the expression will be 0 because sin(0) = sin(nπ) = 0. Now we are left with the second term: $$ a_n = \frac{-2k}{n\pi}\int_0^\pi \cos(kx)\sin(nx)dx $$ Now, to compute this integral, we can use the product-to-sum formula: $$ \cos(kx)\sin(nx) = \frac{1}{2}[\sin((n+k)x) - \sin((n-k)x)] $$ Substitute this into the integral: $$ a_n = \frac{-k}{n\pi}\int_0^\pi [\sin((n+k)x) - \sin((n-k)x)]dx $$ Now, integrate with respect to x: $$ a_n = \frac{-k}{n\pi}\left[\left(\frac{-1}{n+k}\cos((n+k)x)\right) - \left(\frac{1}{n-k}\cos((n-k)x)\right)\right]_0^\pi = \frac{-1}{n\pi}\left[\frac{\cos((n+k)\pi) - 1}{n+k} + \frac{\cos((n-k)\pi) - 1}{n-k}\right] $$ Since k is not an integer, aₙ ≠ 0.

Combine results

Finally, put the calculated values of a₀ and aₙ into the Fourier cosine series formula: $$ f(x) = \frac{-1}{\pi k}[\cos(k\pi) - 1] + \sum_{n=1}^\infty\frac{-1}{n\pi}\left[\frac{\cos((n+k)\pi) - 1}{n+k} + \frac{\cos((n-k)\pi) - 1}{n-k}\right] \cos(nx) $$ This is the Fourier cosine series for the given function f(x) = sin(kx) where k is not an integer, in the interval [0, π].

Key concepts

Fourier Coefficients

Fourier coefficients are fundamental in determining the Fourier series, which represents a periodic function as a sum of sine and cosine functions. Specifically, the Fourier cosine series relies on coefficients to represent functions that are even around the origin. This is because an even function can be decomposed entirely into cosine terms, which themselves are even functions.

When working with Fourier coefficients, we focus on calculating a constant term, typically denoted as \( a_0 \), which represents the function's average value over its domain, and additional coefficients, \( a_n \), for each cosine term at different frequencies. The general formula for an \( n \)th Fourier coefficient in a cosine series is \( a_n = \frac{2}{\pi} \int_0^{\pi} f(x) \cos(nx)dx \). Calculating these coefficients for a non-integer variable, like in the exercise, requires careful integration, often resulting in a non-zero values for all terms.

Integration by Parts

Integration by parts is a useful technique in calculus, vital for solving integrals where the product rule for differentiation applies. The formula is derived from that product rule and can be written as \( \int u\,dv = uv - \int v\,du \).

When applying this to Fourier series, it helps us to handle products of sine and cosine functions encountered while finding Fourier coefficients. The selection of \( u \) and \( dv \) is crucial and typically selected to make the subsequent integration simpler. In our exercise, \( u = \sin(kx) \) and \( dv = \cos(nx)dx \) are appropriately chosen to facilitate integrating by parts. After applying the technique, students should take note of terms that vanish at the boundaries, which often simplifies the problem significantly.

Cosine Series Expansion

The cosine series expansion is a way to express even periodic functions as an infinite sum of cosine terms with distinct frequencies and amplitudes. The series takes the form \( f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos{(nx)} \).

For the case when \( f(x) = \sin(kx) \) with \( k \) not being an integer, this expansion highlights that even a non-even function may still have a cosine series representation on a specific interval, such as \( [0, \pi] \), by considering the function's symmetry with respect to the y-axis. Understanding this concept allows students to maneuver through mathematical challenges where periodic behavior is within a defined domain rather than across the entire x-axis.

Product-to-Sum Formulas

Product-to-sum formulas are trigonometric identities that transform the product of sine and cosine into a sum or difference of sines or cosines. These identities are extremely beneficial for integrating complex trigonometric products, particularly when calculating Fourier coefficients.

For the given exercise, utilizing the product-to-sum formula \( \cos(kx)\sin(nx) = \frac{1}{2}[\sin((n+k)x) - \sin((n-k)x)] \) simplifies what appears to be a daunting integral into a more manageable form. Once this reduction is achieved, integration becomes straightforward since the integral of sine and cosine functions over a period often results in zero or a constant value. This formula should be viewed as a powerful tool in the mathematician's toolkit, especially when dealing with trigonometric series expansions.