Question

Find the mixed Fourier cosine series. $$ f(x)=\sin x ; \quad[0, \pi] $$

Short answer

Question: Find the mixed Fourier cosine series representation of f(x) = sin(x) on the interval [0, π]. Answer: The mixed Fourier cosine series representation for f(x) = sin(x) on the interval [0, π] is given by: $$ f(x) = \sum_{n=1}^{\infty} \left[ \frac{1}{\pi} \left( \frac{(-1)^{n+1} - (-1)^{n-1}}{n^2 - 1} \right) \right] \cos(nx) $$

Step-by-step solution

Determine the a_n Coefficient Formula

The formula for the a_n coefficients for the mixed Fourier cosine series is: $$ a_n = \frac{2}{L} \int_{0}^{L} f(x) \cos \left(\frac{n\pi x}{L}\right) dx $$ For this problem, since we are dealing with the interval [0, π], we will have L = π. So, the a_n coefficient for this problem will be: $$ a_n = \frac{2}{\pi} \int_{0}^{\pi} \sin x \cos \left(\frac{n\pi x}{\pi}\right) dx $$

Calculate the a_n Coefficients

To calculate the a_n coefficients, we need to solve the definite integral for a_n: $$ a_n = \frac{2}{\pi} \int_{0}^{\pi} \sin x \cos(nx) dx $$ To solve this, let's use the sine-cosine product-to-sum identity, sin(α)cos(β) = ½[sin(α+β) - sin(α-β)]: $$ a_n = \frac{1}{\pi} \int_{0}^{\pi} [\sin((n+1)x) - \sin((n-1)x)] dx $$ Now, integrate and evaluate as a definite integral: $$ a_n = \frac{1}{\pi} \left[-\frac{1}{n+1}\cos((n+1)x) + \frac{1}{n-1}\cos((n-1)x) \right]_{0}^{\pi} $$ Evaluating the integral at the limits, we get: $$ a_n = \frac{1}{\pi} \left[-\frac{1}{n+1}(-1)^{n+1} + \frac{1}{n-1}(-1)^{n-1}- \left( \frac{1}{n+1} - \frac{1}{n-1} \right) \right] $$

Write Down the Mixed Fourier Cosine Series Formula

Now that we have the a_n coefficients, we can write down the mixed Fourier cosine series formula: $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx) $$ Substitute a_n from Step 2: $$ f(x) = \sum_{n=1}^{\infty} \left[ \frac{1}{\pi} \left(-\frac{1}{n+1}(-1)^{n+1} + \frac{1}{n-1}(-1)^{n-1}- \left( \frac{1}{n+1} - \frac{1}{n-1} \right)\right)\right] \cos(nx) $$

Simplify the Mixed Fourier Cosine Series Formula

We can further simplify the expression in the series: $$ f(x) = \sum_{n=1}^{\infty} \left[ \frac{1}{\pi} \left( \frac{(-1)^{n+1} - (-1)^{n-1}}{n^2 - 1} \right) \right] \cos(nx) $$ This is the mixed Fourier cosine series representation for the given function f(x) = sin(x) on the interval [0, π].

Key concepts

Mixed Fourier Cosine Series

The mixed Fourier cosine series is a way to represent a function using a series of cosine terms. It incorporates aspects of both sine and cosine components but emphasizes cosine terms to match the boundary conditions of the problem. This series is particularly useful in specific intervals, such as [0, π], where functions may exhibit particular symmetries or periodicities.

While most Fourier series deal with full cycles, mixed Fourier series allow you to work over half-ranges, simplifying calculations when symmetry can be leveraged. For functions like sine that naturally fit these half-range intervals, it provides a concise representation that can handle oscillatory behavior efficiently. The series is generally written as:

• \( f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx) \)

This format captures the essence of the function through the sum of cosine harmonics, each influenced by its respective coefficient \( a_n \).

Fourier Coefficients

Fourier coefficients are the backbone of any Fourier series, determining the amplitude of each cosine (or sine) component in the function's representation. For the mixed Fourier cosine series, the formula for calculating these coefficients \( a_n \) is:
\[a_n = \frac{2}{L} \int_{0}^{L} f(x) \cos \left(\frac{n\pi x}{L}\right) dx\]

In the exercise at hand, we're considering \( f(x) = \sin(x) \) over the interval [0, π], so \( L = π \). The formula simplifies to:
\[a_n = \frac{2}{\pi} \int_{0}^{\pi} \sin x \cos(nx) dx\]

Each coefficient \( a_n \) is calculated by integrating the product of the function and the relevant cosine term over the specified interval. These coefficients tell us how much of each cosine harmonic is needed to closely approximate the original function.

Sine-Cosine Product Identity

The sine-cosine product identity is a crucial tool in transforming complex Fourier integrals into simpler forms. This identity converts products of sine and cosine into sums of sines, making integration more straightforward. The identity is:

• \( \sin(\alpha)\cos(\beta) = \frac{1}{2}[\sin(\alpha+\beta) - \sin(\alpha-\beta)] \)

In step 2 of the solution, we use this identity to simplify the integral:
\[ a_n = \frac{1}{\pi} \int_{0}^{\pi} [\sin((n+1)x) - \sin((n-1)x)] dx \]

By applying this identity, we effectively break the problem down, reducing the complexity involved in solving the integral. As the two resulting sine terms are integrated separately, this approach simplifies the calculation process, enabling a simpler path to find \( a_n \).

Definite Integral Calculation

Calculating definite integrals is a key part of determining Fourier coefficients, particularly when transforming complex expressions into a usable form. Definite integration gives the precise accumulation on an interval and is used here to evaluate the contributions of different components to the series.

For the integral we need to solve in this problem:
\[a_n = \frac{1}{\pi} \left[ -\frac{1}{n+1}\cos((n+1)x) + \frac{1}{n-1}\cos((n-1)x) \right]_{0}^{\pi}\]

We start by evaluating each term at the bounds 0 and π. Usually, trigonometric identities and properties are employed at this stage to compute these values efficiently. Considering how cosine behaves over these limits helps exploit symmetry and properties like \( \cos(n\pi) = (-1)^n \), further simplifying our expressions.

These calculations combine to explain how each Fourier coefficient \( a_n \) is determined, which in turn contributes to the accuracy and completeness of the mixed Fourier cosine series representation.