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Show that \(f\) and \(g\) are orthogonal in the inner product space \(C[a, b]\) with the inner product $$\langle f, g\rangle=\int_{a}^{b} f(x) g(x) d x$$ $$C[-\pi / 2, \pi / 2], \quad f(x)=\cos x, \quad g(x)=\sin x$$

Short Answer

Expert verified
The functions \(f(x) = cos(x)\) and \(g(x) = sin(x)\) are orthogonal in the inner product space \(C[-\pi / 2, \pi / 2]\).

Step by step solution

01

- Identify the functions and the inner product

The functions given are \(f(x) = cos(x)\) and \(g(x) = sin(x)\). The inner product in the space \(C[-\pi / 2, \pi / 2]\) is defined by the integral \(\int_{-\pi/2}^{+\pi/2} f(x) g(x) dx\). To prove the orthogonality, this inner product must be equal to zero.
02

- Calculate the integral

Substitute \(f(x)\) and \(g(x)\) into the integral, we get \(\int_{-\pi/2}^{+\pi/2} cos(x)*sin(x) dx\). This integral can be computed either by direct integration or using integration by parts.
03

- Evaluate the integral

Calculate the integral, we get \(\int_{-\pi/2}^{+\pi/2} cos(x)*sin(x) dx = 0 \). This is because the integral of \(cos(x)*sin(x)\) over the interval \([- \pi/2 , + \pi/2]\) results in zero.
04

- Draw the conclusion

Since the value of the integral, which defines the inner product of the two functions, equals zero, \(f(x)\) and \(g(x)\) are orthogonal in the inner product space \(C[-\pi / 2, \pi / 2]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inner Product Space
An inner product space is a mathematical structure that extends the concept of the dot product from vector spaces to function spaces. It's a crucial concept when exploring orthogonality in functions. The inner product of two functions, say \( f(x) \) and \( g(x) \), over an interval \([a, b]\), is typically defined by the integral:\[\langle f, g \rangle = \int_{a}^{b} f(x) g(x)\, dx\]This integral evaluates the "overlap" or "interaction" between the two functions. In function spaces, similar to dot products in vector spaces, when the inner product is zero, the functions are orthogonal. Here, the exercise requires us to check orthogonality of \( f(x) = \cos x \) and \( g(x) = \sin x \) within the space \( C[-\pi/2, \pi/2] \).
This means we need to compute the integral of their product over the given interval and check if it equals zero.
  • If the integral (inner product) equals zero, \(f(x)\) and \(g(x)\) are orthogonal.
  • This property is essential because orthogonal functions simplify complex problems, especially in Fourier series and solving differential equations.
Integration by Parts
Integration by parts is a valuable technique used to evaluate integrals where direct integration is complex. It stems from the product rule of differentiation and is expressed as:\[\int u\, dv = uv - \int v\, du\]In our exercise, to compute \(\int_{-\pi/2}^{\pi/2} \cos(x) \sin(x) dx\), integration by parts is a useful method if the integral is not immediately obvious. The selection of functions \( u \) and \( dv \) in this technique is critical. In this case:
  • Choosing \( u = \cos(x) \) and \( dv = \sin(x) dx \) would lead to more manageable derivative and integral calculations.
Using integration by parts can simplify integration processes, which is beneficial when dealing with trigonometric or logarithmic functions. While the direct computation in some cases might seem easier, understanding and applying integration by parts is an invaluable skill in various mathematical contexts.
Trigonometric Functions
Trigonometric functions, like \( \cos(x) \) and \( \sin(x) \), play a pivotal role in many areas of mathematics and physics. These functions represent periodic oscillations and have properties that make them orthogonal over certain intervals.The orthogonality of \( \cos(x) \) and \( \sin(x) \) on the interval \([-\pi/2, \pi/2]\) arises because:
  • Their product, integrated over this symmetrical interval around zero, results in zero.
  • This orthogonality simplifies Fourier analysis and signal processing.
Understanding these functions and their relationships as oscillatory solutions helps solve various real-world problems involving waves, vibrations, and cycles. Moreover, their orthogonality property is utilized in solving differential equations and in expressing functions as an infinite series in terms of sines and cosines.

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Most popular questions from this chapter

(a) find the least squares approximation \(g(x)=a_{0}+a_{1} x\) of the function \(f,\) and (b) use a graphing utility to graph \(f\) and \(g\) in the same viewing window. $$f(x)=e^{-2 x}, \quad 0 \leq x \leq 1$$

Let \(W\) be a subspace of the inner product space \(V\). Prove that the set \(W^{\perp}=\\{\mathbf{v} \in V:\langle\mathbf{v}, \mathbf{w}\rangle= 0 \text { for all } \mathbf{w} \in W\\}\) is a subspace of \(V\) Getting Started: To prove that \(W^{\perp}\) is a subspace of \(V,\) you must show that \(W^{\perp}\) is nonempty and that the closure conditions for a subspace hold (Theorem 4.5 ). (i) Find a vector in \(W^{\perp}\) to conclude that it is nonempty. (ii) To show the closure of \(W^{\perp}\) under addition, you need to show that \(\left\langle\mathbf{v}_{1}+\mathbf{v}_{2}, \mathbf{w}\right\rangle= 0\) for all \(\mathbf{w} \in W\) and for any \(\mathbf{v}_{1}, \mathbf{v}_{2} \in W^{\perp} .\) Use the properties of inner products and the fact that \(\left\langle\mathbf{v}_{1}, \mathbf{w}\right\rangle\) and \(\left\langle\mathbf{v}_{2}, \mathbf{w}\right\rangle\) are both zero to show this. (iii) To show closure under multiplication by a scalar, proceed as in part (ii). Use the properties of inner products and the condition of belonging to \(W^{\perp}\).

Find (a) proj, u and (b) proj, v. Use the Euclidean inner product. $$\mathbf{u}=(0,1,3,-6), \quad \mathbf{v}=(-1,1,2,2)$$

Prove that \(\mathbf{u} \times(\mathbf{v}+\mathbf{w})=(\mathbf{u} \times \mathbf{v})+(\mathbf{u} \times \mathbf{w})\)

Find proj, \(u,\) (b) find proja \(v\), and (c) sketch a graph of both proj, u and proj.v. Use the Euclidean inner product. $$\mathbf{u}=(-1,3), \quad \mathbf{v}=(4,4)$$

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