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. \(\int_{0}^{\pi} \sin m x \cos n x d x=0\)

Short answer

Question: Prove that the following orthogonality relations hold true for integers \(m\) and \(n\) such that \(m \neq n\): 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. \(\int_{0}^{\pi} \sin m x \cos n x d x = 0\) Answer: a. Using the product-to-sum trigonometric identity and evaluating the integral, we find that \(\int_{0}^{\pi} \sin m x \sin n x d x = 0\). b. Similarly, using the product-to-sum trigonometric identity and evaluating the integral, we find that \(\int_{0}^{\pi} \cos m x \cos n x d x = 0\). c. Using integration by parts and knowing the integral from part a, we find that \(\int_{0}^{\pi} \sin m x \cos n x d x = 0\). Thus, all three orthogonality relations are proven true for integers \(m\) and \(n\) with \(m \neq n\).

Step-by-step solution

a. Proving \(\int_{0}^{\pi} \sin m x \sin n x d x=0\)

To prove this orthogonality relation, we will use the product-to-sum trigonometric identity: \(\sin A \sin B = \frac{1}{2} \left[ \cos (A - B) - \cos (A + B) \right]\). Applying this identity to our integral, we get: \(\int_{0}^{\pi} \sin m x \sin n x d x = \frac{1}{2}\int_{0}^{\pi} \left[\cos ((m-n) x) - \cos ((m+n) x)\right] d x\) Now, let's integrate each term separately: \(\frac{1}{2}\int_{0}^{\pi} \cos ((m-n) x) d x - \frac{1}{2}\int_{0}^{\pi} \cos ((m+n) x) d x\) The anti-derivatives of \(\cos((m-n)x)\) and \(\cos((m+n)x)\) are given by \(\frac{1}{m-n}\sin((m-n)x)\) and \(\frac{1}{m+n}\sin((m+n)x)\), respectively. \(\frac{1}{2}\left[\frac{1}{m-n}\sin((m-n)x)\Big|_{0}^{\pi} - \frac{1}{m+n}\sin((m+n)x)\Big|_{0}^{\pi}\right]\) For integer values of \(m\) and \(n\), where \(m\neq n\), both \(\sin((m-n)x)\) and \(\sin((m+n)x)\) are equal to \(0\) at the limits \(x=0\) and \(x=\pi\). Therefore, this expression simplifies to \(0\). Thus, \(\int_{0}^{\pi} \sin m x \sin n x d x = 0\) has been proven.

b. Proving \(\int_{0}^{\pi} \cos m x \cos n x d x=0\)

Similar to a, we will use the product-to-sum trigonometric identity: \(\cos A \cos B = \frac{1}{2} \left[ \cos (A - B) + \cos (A + B) \right]\). Applying this identity to our integral, we get: \(\int_{0}^{\pi} \cos m x \cos n x d x = \frac{1}{2}\int_{0}^{\pi} \left[\cos ((m-n) x) + \cos ((m+n) x)\right] d x\) Now, let's integrate each term separately: \(\frac{1}{2}\int_{0}^{\pi} \cos ((m-n) x) d x + \frac{1}{2}\int_{0}^{\pi} \cos ((m+n) x) d x\) The anti-derivatives of \(\cos((m-n)x)\) and \(\cos((m+n)x)\) are given by \(\frac{1}{m-n}\sin((m-n)x)\) and \(\frac{1}{m+n}\sin((m+n)x)\), respectively. \(\frac{1}{2}\left[\frac{1}{m-n}\sin((m-n)x)\Big|_{0}^{\pi} + \frac{1}{m+n}\sin((m+n)x)\Big|_{0}^{\pi}\right]\) For integer values of \(m\) and \(n\), where \(m\neq n\), both \(\sin((m-n)x)\) and \(\sin((m+n)x)\) are equal to \(0\) at the limits \(x=0\) and \(x=\pi\). Therefore, this expression simplifies to \(0\). Thus, \(\int_{0}^{\pi} \cos m x \cos n x d x = 0\) has been proven.

c. Proving \(\int_{0}^{\pi} \sin m x \cos n x d x=0\)

To prove this orthogonality relation, we will use integration by parts, with \(u = \sin(mx)\) and \(dv = \cos(nx)dx\). Then, \(du = m\cos(mx)dx\) and \(v = \frac{1}{n}\sin(nx)\). Using integration by parts, we get: \(\int_{0}^{\pi} \sin m x \cos n x d x = uv\Big|_{0}^{\pi} - \int_{0}^{\pi} v du\) Now, let's calculate \(uv\Big|_{0}^{\pi}\) and \(\int_{0}^{\pi} v du\) separately: \(uv\Big|_{0}^{\pi} = \left[\frac{1}{n}\sin(mx)\sin(nx)\right]\Big|_{0}^{\pi}\) Since \(m\) and \(n\) are integers, and \(x=0\) or \(x=\pi\), both values of \(\sin(mx)\) and \(\sin(nx)\) are \(0\). So, \(uv\Big|_{0}^{\pi} = 0\). Now, calculating the other part: \(-\int_{0}^{\pi} v du = -\int_{0}^{\pi} \frac{m}{n}\cos(mx)\sin(nx) dx\) This integral is the same as part a when \(m \neq n\). So we know that this integral is also equal to \(0\). Therefore, \(\int_{0}^{\pi} \sin m x \cos n x d x = 0\) has been proven.

Key concepts

Orthogonality Relations

Orthogonality relations are crucial in the study of Fourier series because they allow for the decomposition of functions into sine and cosine components. Imagine orthogonality as the idea that two functions are "at right angles" to each other in a functional sense, meaning their dot product, or inner product, is zero. When dealing with trigonometric functions, this often translates to the integral of their product over a specific interval being zero.
For example, as seen in the exercise, the integrals \( \int_{0}^{\pi} \sin m x \sin n x \ dx \), \( \int_{0}^{\pi} \cos m x \cos n x \ dx \), and \( \int_{0}^{\pi} \sin m x \cos n x \ dx \) are all zero when \(m eq n\). This property is fundamental in Fourier series, allowing us to isolate and determine coefficients for each component of a periodic signal. Orthogonality in this context ensures that each frequency component is uniquely determined.

Trigonometric Identities

Trigonometric identities simplify complex expressions and integrals involving sine and cosine functions. They are powerful tools in proving orthogonality relations as they transform products of trigonometric functions into sums or differences, which are often easier to integrate.
A key identity used in the provided solution is the product-to-sum identity:

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

By applying these identities, the integrals were rewritten as a sum of cosine terms, whose antiderivatives are straightforward. Understanding and using these identities are key to simplifying and solving complex trigonometric integrals, making them accessible and manageable.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It is primarily used to integrate products of functions. In calculus, it's a handy tool for converting complex integrals into simpler ones, especially when dealing with transcendental functions like trigonometric expressions.
The formula for integration by parts is:

• \( \int u \ dv = uv - \int v \ du \)

In the third part of the exercise, this technique was used to integrate \( \sin(m x) \cos(n x) \). By choosing \( u = \sin(mx) \) and \( dv = \cos(nx) dx \), we apply the integration by parts formula to solve the integral. This approach reveals how integration by parts can strategically reduce the complexity of evaluating certain integrals, especially when direct integration is cumbersome. Understanding when and how to apply this technique is crucial for solving advanced calculus problems effectively.