Problem 17 - Advanced Engineering Mathematics

Question

An inner product defined on the vector space \(P_{2}\) of all polynomials of degree less than or equal to 2 , is given by $$ (p, q)=\int_{-1}^{1} p(x) q(x) d x $$ Use the Gram-Schmidt orthogonalization process to transform the given basis \(B\) for \(P_{2}\) into an orthogonal basis \(B^{\prime}\). $$ B=\left\\{1, x, x^{2}\right\\} $$

Step-by-Step Solution

Answer

Orthogonal basis for \( P_2 \) is \( \{1, x, x^2 - \frac{2}{3}\} \).

Step 1: Define the Problem and Basis

The goal is to transform the basis \( B = \{1, x, x^2\} \) for the vector space \( P_2 \) into an orthogonal basis \( B' \) using the inner product \((p, q) = \int_{-1}^{1} p(x) q(x) \, dx\).

Step 2: Start Gram-Schmidt Process with First Basis Vector

The first vector in the orthogonal basis is simply the first vector of the original basis: \( u_1 = 1 \). The Gram-Schmidt process starts with normalizing the first vector, but since normalization here is not necessary for orthogonality, we keep \( u_1 = 1 \).

Step 3: Apply Gram-Schmidt to Second Vector

To obtain the second orthogonal vector \( u_2 \), subtract the projection of \( x \) onto \( u_1 \) from \( x \). The projection is \( \text{proj}_{u_1}(x) = \frac{\int_{-1}^{1} x \, dx}{\int_{-1}^{1} 1^2 \, dx} \cdot 1 = 0 \). Thus, \( u_2 = x - 0 = x \).

Step 4: Apply Gram-Schmidt to Third Vector

Next, find \( u_3 \) by orthogonalizing \( x^2 \) against \( u_1 \) and \( u_2 \). Calculate the projection onto \( u_1 \):\[ \text{proj}_{u_1}(x^2) = \frac{\int_{-1}^{1} x^2 \, dx}{\int_{-1}^{1} 1^2 \, dx} \cdot 1 = \frac{2}{3} \cdot 1 = \frac{2}{3} \]Calculate the projection onto \( u_2 \):\[ \text{proj}_{u_2}(x^2) = \frac{\int_{-1}^{1} x^2 x \, dx}{\int_{-1}^{1} x^2 \, dx} \cdot x = \frac{0}{\frac{2}{3}} \cdot x = 0 \]Thus, \( u_3 = x^2 - \frac{2}{3} - 0 = x^2 - \frac{2}{3} \).

Step 5: Verify Orthogonality of New Basis

Verify that \( u_1 = 1 \), \( u_2 = x \), and \( u_3 = x^2 - \frac{2}{3} \) are orthogonal by checking that the inner products \((u_1, u_2)\), \((u_1, u_3)\), and \((u_2, u_3)\) are zero. Earlier steps ensure orthogonality, but verification includes checking integrals, which have already been computed.

Key Concepts

Inner Product Space

An Inner Product Space is a vector space equipped with an additional structure called the inner product. This inner product is a way to multiply vectors together in a manner that leads to a scalar, which often represents concepts like length or angle.
It establishes a method to compute the 'dot product' in more general spaces than just \mathbb{R}^n.

• The inner product is denoted as \(\langle p, q \rangle\) or (p, q), and assigns a real number to a pair of vectors, such as polynomials.
• In the exercise, the inner product is defined by an integral: \( (p, q) = \int_{-1}^{1} p(x) q(x) dx \).
• It satisfies properties of symmetry, linearity, and positive-definiteness.

These properties make inner product spaces essential for defining angles and lengths in abstract vector spaces, which helps in calculating orthogonal projections and lengths of vectors.

Orthogonal Basis

An orthogonal basis for a vector space is a basis composed of vectors that are all orthogonal, or perpendicular, to each other under the given inner product. This means the inner product of any pair of different basis vectors equals zero.

• Orthogonality simplifies many computations, like decomposing vectors uniquely and elegantly projecting vectors onto subspaces.
• In the Gram-Schmidt process used in this exercise, we transform a non-orthogonal basis \(B = \{1, x, x^2\}\) into an orthogonal basis \(B'\).
• Once orthogonal, it can be helpful to normalize each vector to a unit length, forming an orthonormal basis, although this is not required for orthogonality.

These properties make an orthogonal basis easier and more practical for things like simplifying calculations in vector spaces, particularly when solving linear equations.

Vector Spaces

Vector Spaces are abstract mathematical structures where vectors are the primary elements. These spaces are defined by a set of vectors and a set of operations, vector addition and scalar multiplication, that satisfy certain axioms, such as associativity, distributivity, and existence of an additive identity or a zero vector.

• Examples include the classic Euclidean space \mathbb{R}^n and spaces of functions like polynomials.
• The exercise's space \(P_2\) includes all polynomials with degrees less than or equal to two; its basis set \( \{1, x, x^2\} \) spans this space.
• Vector spaces provide a framework for various branches of mathematics and physics, helping define objects like lines, planes, and higher-dimensional spaces.

In vector spaces, we can perform linear algebra operations such as finding bases, calculating span, and determining linear independence.

Polynomial Vector Space

A Polynomial Vector Space is a vector space where the vectors are polynomials. The vector space discussed in the exercise is \(P_2\), consisting of polynomials with degrees up to 2.

• The basis set given, \(B = \{1, x, x^2\}\), represents polynomials up to degree 2.
• These polynomial vectors, under operations of addition and scalar multiplication, satisfy all vector space axioms, making \(P_2\) a vector space.
• Understanding polynomial vector spaces is crucial for computational fields like computer graphics and solving differential equations.

Through polynomial vector spaces, one can explore solutions to complex numerical problems where polynomial approximations are essential.