Question

In this problem we indicate certain similarities between three dimensional geometric vectors and Fourier series. (a) Let \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3}\) be a set of mutually orthogonal vectors in three dimensions and let \(\mathbf{u}\) be any three-dimensional vector. Show that $$\mathbf{u}=a_{1} \mathbf{v}_{1}+a_{2} \mathbf{v}_{2}+a_{3} \mathbf{v}_{3}$$ where $$a_{i}=\frac{\mathbf{u} \cdot \mathbf{v}_{i}}{\mathbf{v}_{i} \cdot \mathbf{v}_{i}}, \quad i=1,2,3$$ Show that \(a_{i}\) can be interpreted as the projection of \(\mathbf{u}\) in the direction of \(\mathbf{v}_{i}\) divided by the length of \(\mathbf{v}_{i}\). (b) Define the inner product \((u, v)\) by $$(u, v)=\int_{-L}^{L} u(x) v(x) d x$$ Also let $$\begin{array}{ll}{\phi_{x}(x)=\cos (n \pi x / L),} & {n=0,1,2, \ldots} \\ {\psi_{n}(x)=\sin (n \pi x / L),} & {n=1,2, \ldots}\end{array}$$ Show that Eq. ( 10 ) can be written in the form $$\left(f, \phi_{n}\right)=\frac{a_{0}}{2}\left(\phi_{0}, \phi_{n}\right)+\sum_{m=1}^{\infty} a_{m}\left(\phi_{m}, \phi_{n}\right)+\sum_{m=1}^{\infty} b_{m}\left(\psi_{m}, \phi_{m}\right)$$ (c) Use Eq. (v) and the corresponding equation for \(\left(f, \psi_{n}\right)\) together with the orthogonality relations to show that $$a_{n}=\frac{\left(f, \phi_{n}\right)}{\left(\phi_{n}, \phi_{n}\right)}, \quad n=0,1,2, \ldots ; \quad b_{n}=\frac{\left(f, \psi_{n}\right)}{\left(\psi_{n}, \psi_{n}\right)}, \quad n=1,2, \ldots$$ Note the resemblance between Eqs. (vi) and Eq. (ii). The functions \(\phi_{x}\) and \(\psi_{x}\) play a role for functions similar to that of the orthogonal vectors \(v_{1}, v_{2},\) and \(v_{3}\) in three-dimensional

Short answer

Question: Express a 3D vector in terms of three mutually orthogonal vectors and relate this to expressing a function in terms of the Fourier series. Answer: To express a 3D vector \(\mathbf{u}\) in terms of three mutually orthogonal vectors \(\mathbf{v}_{1}\), \(\mathbf{v}_{2}\), and \(\mathbf{v}_{3}\), we find the coefficients \(a_i\), where \(\mathbf{u} = a_{1}\mathbf{v}_{1} + a_{2}\mathbf{v}_{2} + a_{3}\mathbf{v}_{3}\). These coefficients can be found using the formula: $$a_i = \frac{\mathbf{u}\cdot\mathbf{v}_{i}}{\mathbf{v}_{i}\cdot\mathbf{v}_{i}}, \quad i=1,2,3$$ In a similar manner, a function \(f(x)\) can be expressed in terms of its Fourier series, with coefficients \(a_n\) and \(b_n\), representing the function in terms of an orthogonal basis of sine and cosine functions. The coefficients are found using the orthogonality relations for the Fourier series, given by: $$a_n = \frac{(f, \phi_n)}{(\phi_n, \phi_n)}, \quad n=0,1,2,\ldots$$ $$b_n = \frac{(f, \psi_n)}{(\psi_n, \psi_n)}, \quad n=1,2,\ldots$$ This demonstrates that both 3D vectors and functions can be expressed using orthogonal bases, with the coefficients found using similar projection techniques.

Step-by-step solution

Part (a): Representation of \(\mathbf{u}\) in Mutually Orthogonal Vectors

Given a set of mutually orthogonal vectors, \(\mathbf{v}_{1}\), \(\mathbf{v}_{2}\), and \(\mathbf{v}_{3}\), and a vector \(\mathbf{u}\), our goal is to express \(\mathbf{u}\) as a linear combination of the given mutually orthogonal vectors: $$\mathbf{u} = a_{1}\mathbf{v}_{1} + a_{2}\mathbf{v}_{2} + a_{3}\mathbf{v}_{3}$$ We know that for mutually orthogonal vectors, their dot product is zero: \(\mathbf{v}_{i}\cdot\mathbf{v}_{j}=0\), for \(i\neq j\). To find the coefficient \(a_i\), dot the equation above by \(\mathbf{v}_{i}\). This gives: $$\mathbf{u}\cdot\mathbf{v}_{i} = a_{1}(\mathbf{v}_{1}\cdot\mathbf{v}_{i}) + a_{2}(\mathbf{v}_{2}\cdot\mathbf{v}_{i}) + a_{3}(\mathbf{v}_{3}\cdot\mathbf{v}_{i})$$ Since \(\mathbf{v}_{i}\) is orthogonal to the other two vectors, their dot products will vanish. Only the term with \(\mathbf{v}_{i}\cdot\mathbf{v}_{i}\) remains, leading to: $$a_i = \frac{\mathbf{u}\cdot\mathbf{v}_{i}}{\mathbf{v}_{i}\cdot\mathbf{v}_{i}}, \quad i=1,2,3$$ To interpret \(a_{i}\) as a projection of \(\mathbf{u}\) in the direction of \(\mathbf{v}_{i}\) divided by the length of \(\mathbf{v}_{i}\), recall that the projection of \(\mathbf{u}\) onto \(\mathbf{v}_{i}\) is given by: $$\text{proj}_{\mathbf{v}_i}(\mathbf{u})=\frac{\mathbf{u}\cdot\mathbf{v}_{i}}{\|\mathbf{v}_{i}\|}\frac{\mathbf{v}_{i}}{\|\mathbf{v}_{i}\|}$$ Dividing the projection by the length of \(\mathbf{v}_{i}\), we get: $$\frac{\text{proj}_{\mathbf{v}_i}(\mathbf{u})}{\|\mathbf{v}_i\|}=\frac{\mathbf{u}\cdot\mathbf{v}_{i}}{\|\mathbf{v}_i\|^2}=a_{i},$$ where \(\|\mathbf{v}_i\|^2=\mathbf{v}_{i}\cdot\mathbf{v}_{i}\).

Part (b): Writing a function in terms of Fourier Series

Define the inner product \((u, v)\) of two functions as $$(u, v) = \int_{-L}^L u(x) v(x) dx$$ The Fourier series basis functions are defined as $$\phi_{n}(x) = \cos\left(\frac{n\pi x}{L}\right) ,\quad n = 0, 1, 2, \ldots$$ $$\psi_{n}(x) = \sin\left(\frac{n\pi x}{L}\right) ,\quad n = 1, 2, \ldots$$ To show that the equation given in the exercise can be written in the form: $$\left(f, \phi_{n}\right)=\frac{a_{0}}{2}\left(\phi_{0}, \phi_{n}\right)+\sum_{m=1}^{\infty} a_{m}\left(\phi_{m}, \phi_{n}\right)+\sum_{m=1}^{\infty} b_{m}\left(\psi_{m}, \phi_{m}\right)$$ Let's write the Fourier series representation of \(f(x)\): $$f(x) = \frac{a_{0}}{2} + \sum_{m=1}^{\infty} a_{m} \phi_{m}(x) + \sum_{m=1}^{\infty} b_{m} \psi_{m}(x)$$ Now, take the inner product of \(f\) with \(\phi_n\): $$(f, \phi_n) = \left(\frac{a_{0}}{2}\phi_{0} + \sum_{m=1}^{\infty} a_{m} \phi_{m} + \sum_{m=1}^{\infty} b_{m} \psi_{m}, \phi_{n}\right)$$ Distributing the inner product and using the linearity property, we have: $$\left(f, \phi_{n}\right)=\frac{a_{0}}{2}\left(\phi_{0}, \phi_{n}\right)+\sum_{m=1}^{\infty} a_{m}\left(\phi_{m}, \phi_{n}\right)+\sum_{m=1}^{\infty} b_{m}\left(\psi_{m}, \phi_{n}\right)$$

Part (c): Finding Fourier coefficients using orthogonality relations

The orthogonality relations are given by: $$(\phi_m, \phi_n) = \begin{cases} 0, & m \neq n; \\ L, & m = n; \\ L/2, & m = n = 0\end{cases} \quad (\psi_m, \psi_n) = \begin{cases} 0, & m \neq n; \\ L/2, & m = n \end{cases}$$ From part (b), we have: $$\left(f, \phi_{n}\right)=\frac{a_{0}}{2}\left(\phi_{0}, \phi_{n}\right)+\sum_{m=1}^{\infty} a_{m}\left(\phi_{m}, \phi_{n}\right)+\sum_{m=1}^{\infty} b_{m}\left(\psi_{m}, \phi_{n}\right)$$ Using the orthogonality relations and noticing that the last term vanishes as the dot product of sine and cosine is zero, we have: $$a_n = \frac{(f, \phi_n)}{(\phi_n, \phi_n)}, \quad n=0,1,2,\ldots$$ Similarly, taking the inner product of \(f\) with \(\psi_n\), we get: $$b_n = \frac{(f, \psi_n)}{(\psi_n, \psi_n)}, \quad n=1,2,\ldots$$ Now, we can see the resemblance between these coefficients and the coefficients we got in part (a), which highlights that Fourier series can be seen as an expansion in a suitable orthogonal basis in a similar way as for three-dimensional vectors.

Key concepts

Orthogonal Vectors

When we talk about orthogonal vectors, we're discussing a fundamental concept in mathematics and physics. Imagine you're looking at the corner of a room, where two walls meet at a right angle. In a similar fashion, two vectors in a three-dimensional space are orthogonal to each other if they meet at a right angle, meaning they have no component in common along each other's direction. This is paramount for understanding Fourier series because they rely on the principle that certain function sets are 'orthogonal' in function space, just like perpendicular walls in a room.

In our exercise, we deemed vectors \( \mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3} \) as mutually orthogonal, indicating no two share a common direction. This enables a unique representation of any other vector in that space as a sum of scales of these orthogonal vectors. Here's why that's important: it simplifies calculations, just like breaking down a complex motion into movements along the x, y, and z axes. Each coefficient \( a_{i} \) then represents how much of \( \mathbf{v}_{i} \) is 'in' the vector \( \mathbf{u} \) we are trying to describe.

Vector Projection

The vector projection comes into play when we want to measure how much one vector contributes to the direction of another. It's like casting a shadow; one vector projects onto the other, and the length of this 'shadow' tells us how much influence one vector has in the direction of the other.

In the context of our problem, the projection of \( \mathbf{u} \) onto \( \mathbf{v}_{i} \) is tantamount to finding its component along \( \mathbf{v}_{i} \). This is given by the dot product of \( \mathbf{u} \) and \( \mathbf{v}_{i} \) divided by the magnitude of \( \mathbf{v}_{i} \) squared. Technically, we're scaling \( \mathbf{v}_{i} \) by the factor of \( \mathbf{u} \) directed in \( \mathbf{v}_{i} \)’s direction. This directed scaling is the coefficient \( a_{i} \) in our orthogonal vector representation, and it's vital for constructing Fourier series, which are projections of functions onto a set of orthogonal basis functions.

Inner Product

Understanding the inner product can be quite intuitive if we envision it as a way of quantifying the 'overlap' between two entities. In a more grounded sense, for vectors, it blends the concepts of magnitude with the cosine of the angle between them—it's their dot product. When we extend the idea to functions in the context of Fourier series, the inner product involves integrating the product of the two functions over a certain interval.

In this exercise, we define the inner product for functions as an integral of their product over the interval from \( -L \) to \( L \). This mathematical operation probes similarities between the functions across the interval, which is crucial for Fourier analysis. By doing so, we can establish coefficients in the Fourier series that correspond to the strength of each basis function's presence in the function we're examining, similar to how \( a_{i} \) represents how much of \( \mathbf{v}_{i} \) is in \( \mathbf{u} \) in the three-dimensional vector scenario.