Problem 16 - Differential Equations and Linear Algebra

Question

Show that the given functions in \(C^{0}[-1,1]\) are orthogonal, and use them to construct an orthonormal set of functions in \(C^{0}[-1,1]\) $$f_{1}(x)=2 x, f_{2}(x)=1+2 x^{2}, f_{3}(x)=x^{3}-\frac{3}{5} x$$

Step-by-Step Solution

Answer

The given functions \(f_1(x)=2x, f_2(x)=1+2x^2\), and \(f_3(x)=x^3-\frac{3}{5}x\) are orthogonal in the interval \([-1, 1]\) as their inner products are all zero. The orthonormal set of functions is then given by \[\hat{f_1}(x) = \frac{3}{2}x,\ \hat{f_2}(x) = \frac{3}{\sqrt{14}}(1 + 2x^2),\ \hat{f_3}(x) = \frac{5}{2}(x^3 - \frac{3}{5}x)\]

Step 1: Compute the inner product between \(f_1(x)\) and \(f_2(x)\)

Let's compute the integral: \[\langle f_1, f_2 \rangle = \int_{-1}^1 (2x)(1 + 2x^2)dx\] \[ = 2\int_{-1}^1 (x + 2x^3)dx\] \[ = 2\left[\frac{1}{2}x^2 + \frac{1}{2}x^4\right]_{-1}^1\] \[ = 2((\frac{1}{2} + \frac{1}{2}) - (-\frac{1}{2} - \frac{1}{2})) = 0\] So, \(f_1(x)\) and \(f_2(x)\) are orthogonal.

Step 2: Compute the inner product between \(f_1(x)\) and \(f_3(x)\)

Let's compute the integral: \[\langle f_1, f_3 \rangle = \int_{-1}^1 (2x)(x^3 - \frac{3}{5}x)dx\] \[ = 2\int_{-1}^1 (x^4 - \frac{3}{5}x^2)dx\] \[ = 2\left[\frac{1}{5}x^5 - \frac{1}{5}x^3\right]_{-1}^1\] \[ = 2((\frac{1}{5} - \frac{1}{5}) - (-\frac{1}{5} + \frac{1}{5})) = 0\] So, \(f_1(x)\) and \(f_3(x)\) are orthogonal.

Step 3: Compute the inner product between \(f_2(x)\) and \(f_3(x)\)

Let's compute the integral: \[\langle f_2, f_3 \rangle = \int_{-1}^1 (1 + 2x^2)(x^3 - \frac{3}{5}x)dx\] \[ = \int_{-1}^1 (x^3 + 2x^5 - \frac{3}{5}x - \frac{6}{5}x^3)dx\] \[ = \int_{-1}^1 (-\frac{6}{5}x^3 + 2x^5 - \frac{3}{5}x)dx\] \[ = \left[-\frac{3}{10}x^4 + \frac{2}{6}x^6 - \frac{3}{10}x^2\right]_{-1}^1\] \[ = ((-\frac{3}{10} + \frac{2}{6} - \frac{3}{10}) - (\frac{3}{10} + \frac{2}{6} + \frac{3}{10})) = 0\] So, \(f_2(x)\) and \(f_3(x)\) are orthogonal. Now that we have shown that all three functions are orthogonal, let's calculate their norms and divide each function by its norm to obtain the orthonormal set.

Step 4: Compute the norms of the functions

The norms of the functions are given by the square root of the inner product of the function with itself: \[\|f\| = \sqrt{\langle f, f \rangle}\] \[\|f_1\| = \sqrt{\langle f_1, f_1 \rangle} = \sqrt{\int_{-1}^1 (2x)^2 dx} = \sqrt{\frac{4}{3}}\] \[\|f_2\| = \sqrt{\langle f_2, f_2 \rangle} = \sqrt{\int_{-1}^1 (1 + 2x^2)^2 dx} = \sqrt{\frac{14}{3}}\] \[\|f_3\| = \sqrt{\langle f_3, f_3 \rangle} = \sqrt{\int_{-1}^1 (x^3 - \frac{3}{5}x)^2 dx} = \sqrt{\frac{4}{35}}\]

Step 5: Construct an orthonormal set

Now, let's normalize each function by dividing each function by its norm: \[\hat{f_1}(x) = \frac{f_1(x)}{\|f_1\|} = \frac{2x}{\sqrt{\frac{4}{3}}} = \frac{3}{2}x\] \[\hat{f_2}(x) = \frac{f_2(x)}{\|f_2\|} = \frac{1 + 2x^2}{\sqrt{\frac{14}{3}}} = \frac{3}{\sqrt{14}}(1 + 2x^2)\] \[\hat{f_3}(x) = \frac{f_3(x)}{\|f_3\|} = \frac{x^3 - \frac{3}{5}x}{\sqrt{\frac{4}{35}}} = \frac{5}{2}(x^3 - \frac{3}{5}x)\] The orthonormal set of functions in \(C^{0}[-1,1]\) is: \[\hat{f_1}(x) = \frac{3}{2}x,\ \hat{f_2}(x) = \frac{3}{\sqrt{14}}(1 + 2x^2),\ \hat{f_3}(x) = \frac{5}{2}(x^3 - \frac{3}{5}x)\]

Key Concepts

Inner Product

In mathematics, the concept of an "inner product" is crucial for understanding the geometry of function spaces. An inner product is a way of multiplying two functions that results in a scalar. This scalar gives us information about the functions, such as their orthogonality. For the given functions, we use integrals to calculate the inner products.Let's take the inner product of functions \(f_1(x)\) and \(f_2(x)\) as an example:

• We calculate \(\langle f_1, f_2 \rangle = \int_{-1}^1 (2x)(1 + 2x^2)dx\).
• By solving the integral, the result is 0, which indicates that \(f_1(x)\) and \(f_2(x)\) are orthogonal.

This procedure is repeated for other pairs of functions. If the inner product between any two functions is zero, it confirms their orthogonality. This is a key element when proving that a set of functions is orthogonal.

Orthonormal Set

An "orthonormal set" is a collection of functions that are both orthogonal and normalized. A function is considered normalized if its norm is equal to 1. When building an orthonormal set, we start with an orthogonal set and follow through the normalization process.
For example, consider three functions \(f_1(x)\), \(f_2(x)\), and \(f_3(x)\) that are shown to be orthogonal through their inner products with each other yielding zero.Next, we need to normalize each function:

• Calculate the norm \(\|f\|\) for each function by taking the square root of the inner product of the function with itself.
• Normalize each function by dividing it by its norm.

This procedure gives us an orthonormal set, where not only does each function have a norm of 1 but also every pair is orthogonal.

Function Norm

The "function norm" is a measure of the "size" or "length" of a function. Like the length of a vector in regular Euclidean space, the norm of a function is derived from an inner product. Specifically, the norm is the square root of the inner product of the function with itself.For example, to find the norm of \(f_1(x) = 2x\):

• Compute the inner product \(\langle f_1, f_1 \rangle = \int_{-1}^1 (2x)^2 dx\).
• The result of this computation is \(\frac{4}{3}\), and so the norm is \(\sqrt{\frac{4}{3}}\).

Understanding function norms is essential, especially when normalizing functions to create orthonormal sets. This process ensures that each function in the set maintains a standard "length" of one.

Orthogonal Set

An "orthogonal set" of functions is one where each pair of functions is orthogonal. Orthogonality, in this context, means that the inner product between any two different functions in the set is zero. This property implies the functions are completely independent in terms of data information.In the exercise, the functions \(f_1(x), f_2(x),\) and \(f_3(x)\) have been shown to form an orthogonal set by checking these conditions:

• The calculation of inner products \(\langle f_1, f_2 \rangle\), \(\langle f_1, f_3 \rangle\), and \(\langle f_2, f_3 \rangle\) all result in zero.

This indicates none of the functions can be expressed as a combination of the others. Establishing an orthogonal set is a critical step towards forming an orthonormal set by normalizing the functions.