Answer Exercise 3.4.27 using the weighted inner product ⟨f, g⟩=∫-1^1 f(x) g(x)(1+x) d x

Name: Problem 28, Chapter 3: Applied Linear Algebra (Undergraduate Texts in Mathematics)
Uploaded: 2022-02-08T10:04:53-08:00
Duration: 8 min 44 s
Channel: Ahmad Reda
Description: Answer Exercise 3.4.27 using the weighted inner product ⟨f, g⟩=∫-1^1 f(x) g(x)(1+x) d x

Answer Exercise 3.4.27 using the weighted inner product ⟨f,...

Key Concepts

Projection onto Subspaces

Projection involves finding the closest element in a subspace to a given element using the inner product structure. This technique is used to approximate a function by elements of a subspace (such as polynomials) in a least-squares sense, ensuring the error is orthogonal to the subspace, which minimizes the overall error in the approximation.

Orthogonality

Orthogonality in an inner product space indicates that two elements have an inner product of zero. This concept is central to many techniques in linear algebra and numerical methods, such as decomposing functions into sums of orthogonal components or constructing bases in function spaces, which lead to simplified computations and analysis.

Weighted Inner Product

A weighted inner product extends the standard inner product by incorporating a weight function into the integration process. This method modifies the contribution of different parts of the domain, which is crucial in many applications such as probability, approximation theory, and spectral methods, where certain regions are given more emphasis than others.

Inner Product Space

An inner product space is a vector space equipped with an inner product that defines notions like length, angle, and orthogonality. In the context of functions, this often involves integration over a domain with or without a weight function, providing a framework for analyzing convergence, projection, and decomposition in spaces of functions.

Transcript

00:02 In problem 41, the inner product between two functions is defined as the definite integral from minus 1 to 1 for the multiplication of these two functions.

00:11 Here, the two functions are defined as x and g and e use of out of x.

00:19 For party a, it's required to find the value of the inner product between f and g.

00:28 It equals by definition the definite integral from minus 1 to 1 for 1.

00:33 Because it's f of x multiplied by e to the power of x because it's re of x d x we can evaluate this integral by birds then it equals u multiplied by v minus the integral from minus 1 to 1 for v d u by defining u as x and dv as as e as e to power of x d x then v is just e to the word of x by integrating from minus 1 to 1 and u is defined as x then it equals x multiplied by e to the 1 of x integrates from minus 1 to 1 minus sorry we substitute from minus one to one minus integral from minus one to one for v which is defined as e to the bar of x d u d u is d x let's substitute for the first term we substitute by x equals 1 it's e multiplied by sorry it's e minus we substitute by the second term it's minus 1 e to the bar of minus x minus 1 minus the integral of e to the bar of x is e to the bar of x we substitute from minus 1 to 1 then it's e plus e to the bar of minus 1 minus 1 minus e minus e to the bar of minus 1 we can find that it's 2 divided by e this is for part a let's move to part b it's required to find the magnitude of the function f by definition it's the square root of the inner product between f and itself then it's the square root of the definite integral from minus 1 to 1 for f for f which is x multiplied by f d x then it's the square root of the integral of x squared the integral of x squared is x squared is x cubed divided by 3 we substitute for minus 1 to 1.

04:01 It equals the square root we substitute by x equals 1 first minus minus 1 third.

04:10 It's square root of 6 divided by 3...

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