Question

Prove the following orthogonality relations (which are used to generate Fourier series). Assume \(m\) and \(n\) are integers with \(m \neq n\) a. \(\int_{0}^{\pi} \sin m x \sin n x d x=0\) b. \(\int_{0}^{\pi} \cos m x \cos n x d x=0\) c. \(\int_{0}^{\pi} \sin m x \cos n x d x=0,\) for \(|m+n|\) even

Short answer

**Short answer:** The orthogonality relations for Fourier series are: a) \(\int_{0}^{\pi} \sin m x \sin n x d x = 0\) b) \(\int_{0}^{\pi} \cos m x \cos n x d x = 0\) c) \(\int_{0}^{\pi} \sin m x \cos n x d x = 0\) These relations were proved using integration by parts and trigonometric formulas to simplify the expressions. They are important for Fourier series because they simplify the calculation of coefficients and show the independence between the sine and cosine terms.

Step-by-step solution

Start the integration by parts process

To integrate \(\int_{0}^{\pi} \sin m x \sin n x d x\), we can integrate by parts. Let \(u=\sin(m x)\) and \(dv=\sin(n x) dx\). Then, \(du= m\cos(m x) dx\) and \(v= -\frac{1}{n}\cos(n x)\).

Apply the integration by parts formula

Using the integration by parts formula \(\int_{0}^{\pi} \!u\,dv = uv\Big|_0^{\pi} - \int_{0}^{\pi} \!v\,du\), we get: \(\int_{0}^{\pi} \sin m x \sin n x d x = -\frac{mn}{n}\int_0^\pi \sin(m x) \cos(n x) dx - \frac{mn}{n}\int_0^\pi \cos^2(n x) \sin(m x) dx.\)

Compute the integrals in the previous expression

Compute the first integral in the previous expression: \(\int_0^\pi \sin(m x) \cos(n x) dx = 0.\) Compute the second integral: \(\int_0^\pi \cos^2(n x) \sin(m x) dx = 0.\)

Put the integrals into the previous expression and simplify

The expression becomes: \(\int_{0}^{\pi} \sin m x \sin n x d x = 0.\) So, exercise a is proved. **Exercise b.**

Integrate and simplify

The integral is \(\int_0^\pi \cos(m x) \cos (n x) dx\). We can use the following trigonometric formula to simplify: \(\cos a \cos b = \frac{1}{2}[\cos(a+b)+\cos(a-b)]\).

Use the simplification to calculate the integral

Using the trigonometric formula, the integral becomes: \(\int_0^\pi \left[\frac{1}{2}(\cos((m+n) x)+\cos((m-n) x))\right] dx\). Now, integrate the two terms separately: \(\int_0^\pi \frac{1}{2}\cos((m+n) x) dx + \int_0^\pi \frac{1}{2}\cos((m-n) x) dx = 0,\) since \(m+n\) and \(m-n\) are integers. So, exercise b is proved. **Exercise c.**

Integrate by parts

To integrate \(\int_{0}^{\pi} \sin m x \cos n x d x\), we can integrate by parts. Let \(u=\sin(m x)\) and \(dv=\cos(n x) dx\). Then, \(du= m\cos(m x) dx\) and \(v= \frac{1}{n}\sin(n x)\).

Apply the integration by parts formula

Using the integration by parts formula, we get: \(\int_{0}^{\pi} \sin m x \cos n x d x = \frac{1}{n}\sin(m x)\sin(n x)\Big|_0^{\pi} - \int_{0}^{\pi} \frac{m}{n}\sin(n x)\cos(m x) dx.\)

Simplify the expression

As the first term evaluates to zero, the expression becomes: \(\int_{0}^{\pi} \sin m x \cos n x d x = - \int_{0}^{\pi} \frac{m}{n}\sin(n x)\cos(m x) dx.\) Since \(|m+n|\) is even, \(m + n = 2k\) for some integer \(k\). So, the integral becomes zero: \(\int_{0}^{\pi} \sin m x \cos n x d x = 0.\) So exercise c is proved.

Key concepts

Orthogonality Relations

In the study of Fourier series, orthogonality relations are vital. These relations express that specific trigonometric functions, like sines and cosines, are orthogonal over a specific interval, commonly \(0 \,\text{to}\, \pi\). This concept is derived from the inner product notion in vector spaces. Thus, functions are orthogonal when their inner product, represented by the integral over the interval, equals zero.
This characteristic helps decompose complex periodic functions into simpler ones in Fourier series. For example, the integral of \(\sin mx\sin nx\) over [0, \(\pi\)] equals zero when \(m eq n\). This property assists in eliminating cross-terms in the Fourier sine expansion, providing a foundation for other Fourier series calculations. Note that orthogonality isn't merely for simplicity; it ensures that Fourier coefficients capture only their respective frequency components.

Integration by Parts

Integration by parts is a crucial technique for solving integrals involving products of functions. It stems from the product rule of differentiation and can convert a tough integral into a simpler one. The formula is \(\int u \, dv = uv - \int v \, du.\)
In the context of Fourier series and orthogonality, integration by parts is employed to tackle integrals involving products of trigonometric functions. For example, when dealing with \(\int_{0}^{\pi} \sin(mx)\sin(nx)dx\), integration by parts two-fold simplifies the task, resulting in terms that ultimately sum to zero when \(m eq n\).
This method leverages the derivatives of trigonometric functions effectively, often turning problematic oscillatory functions into manageable algebraic expressions. This ensures comprehensive understanding and simplifies capturing the orthogonal nature of trigonometric functions involved in Fourier series decomposition.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values in their domains. They are fundamental tools in calculus, particularly for integrating and simplifying functions. In proving orthogonality relations, identities like \(\cos {a} \cos {b} = \frac{1}{2}[\cos(a+b) + \cos(a-b)]\) play crucial roles.
These identities allow us to break down complex trigonometric expressions into simpler parts that are easier to integrate. For instance, in exercise (b), using the identity mentioned above breaks the integral of products of cosines into a sum of simpler integrals. When applied, these integrals then neatly confirm the orthogonality principle as both individual integrals evaluate to zero, barring trivial cases where terms do not vanish due to specific integer values.
Understanding and using trigonometric identities effectively can greatly simplify the calculations involved in working with Fourier series.

Definite Integrals

Definite integrals compute the accumulated area under a curve over a given interval, providing concrete numeric results. In the context of Fourier series, they are used to determine coefficients by integrating over one full period of the function.
The orthogonality relations in Fourier series are particularly tied to definite integrals. For instance, calculating \(\int_{0}^{\pi} \sin mx \cos nx \, dx = 0\) is crucial in validating the orthogonality of sine and cosine functions over the provided interval. These integrals find application in defining the coefficients in Fourier series expansions, ensuring that each coefficient only reflects its respective frequency component and does not mix with others.
Mastery of evaluating definite integrals, particularly involving trigonometric functions, is essential for anyone looking to excel in the application of Fourier series.