Chapter 7: Problem 17
Show that 1 is orthogonal to \(\cos k x\) and \(\sin k x\) in \(\mathscr{C}[-\pi, \pi]\) for \(k \geq 1\)
Short Answer
Expert verified
1 is orthogonal to both cos(kx) and sin(kx) on [-π, π] for k ≥ 1.
Step by step solution
01
Understand Orthogonality
Orthogonality in the context of functions means that the inner product (integral) of the two functions over a specified interval is zero. We need to show that the integral of 1 times each of the functions \(\cos kx\) and \(\sin kx\) over \([-\pi, \pi]\) is zero.
02
Set Up the Integrals
Set up the integrals for the functions involved. The integrals we need to evaluate are: \[\int_{-\pi}^{\pi} 1 \cdot \cos(kx) \, dx\] and \[\int_{-\pi}^{\pi} 1 \cdot \sin(kx) \, dx\].
03
Evaluate the Integral with Cosine
Evaluate \(\int_{-\pi}^{\pi} \cos(kx) \, dx\). The antiderivative of \(\cos(kx)\) is \(\frac{1}{k}\sin(kx)\). Evaluate it from \(-\pi\) to \(\pi\):\[ \left. \frac{1}{k} \sin(kx) \right|_{-\pi}^{\pi} = \frac{1}{k} \sin(k\pi) - \frac{1}{k} \sin(-k\pi) = 0 - 0 = 0.\]This is because \(\sin(n\pi) = 0\) for any integer \(n\).
04
Evaluate the Integral with Sine
Evaluate \(\int_{-\pi}^{\pi} \sin(kx) \, dx\). The antiderivative of \(\sin(kx)\) is \(-\frac{1}{k}\cos(kx)\). Evaluate it from \(-\pi\) to \(\pi\):\[ \left. -\frac{1}{k} \cos(kx) \right|_{-\pi}^{\pi} = -\frac{1}{k}(\cos(k\pi) - \cos(-k\pi)) = 0.\]This follows because \(\cos(n\pi) = (-1)^n\) and thus \(\cos(k\pi) = \cos(-k\pi)\).
05
Conclude Orthogonality
Since both integrals are zero, the function 1 is orthogonal to \(\cos(kx)\) and \(\sin(kx)\) in the interval \([-\pi, \pi]\) for \(k \geq 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inner Product
The inner product is a central concept used to determine orthogonality between functions. In mathematical terms, it is a process that allows us to "multiply" functions together and integrate the result over a specific interval. More specifically, if two functions are said to be orthogonal on a particular interval, their inner product (or integral over that interval) must equal zero. In this exercise, the functions are 1, \(\cos(kx)\), and \(\sin(kx)\), and the interval of interest is \([-\pi, \pi]\). By showing that the inner product of 1 with each trigonometric function over this interval is zero, we confirm their orthogonality.
Steps to calculate an inner product:
Steps to calculate an inner product:
- Multiply the two functions together.
- Integrate the resulting function over the given interval.
Cosine Function
The cosine function, \(\cos(kx)\), is a fundamental trigonometric function that describes a wave pattern. Trigonometric functions, such as cosine, are periodic and have values between -1 and 1. For any integer \(k\), the function oscillates based on the frequency determined by \(k\).
Evaluating Integrals Involving Cosine:
Evaluating Integrals Involving Cosine:
- The integral of \(\cos(kx)\) is \(\frac{1}{k}\sin(kx)\).
- Substitute the boundaries of the interval into the antiderivative.
- For orthogonality, check if the result simplifies to zero.
Sine Function
Just like the cosine function, the sine function \(\sin(kx)\) is a crucial component of trigonometric operations. It also exhibits periodic behavior and ranges from -1 to 1. For different integer values of \(k\), it determines how often the wave oscillates within a given interval.
The Process of Evaluating the Sine Integral:
The Process of Evaluating the Sine Integral:
- The integral of \(\sin(kx)\) is \(-\frac{1}{k}\cos(kx)\).
- Insert the limits of integration into the antiderivative.
- The result simplifies to zero if the sine function is orthogonal to a constant function over the interval.
Integrals
Integrals form an essential part of calculus and are used to calculate a variety of things including the area under a curve. In the context of orthogonality, integrals are employed to determine the inner product of two functions.
Understanding Integrals for Orthogonality:
Understanding Integrals for Orthogonality:
- Set up the correct integral for the functions involved.
- Calculate or find the antiderivative of the integrand.
- Evaluate it using the boundaries of the interval.
Antiderivative
An antiderivative, or primitive function, is the reverse process of differentiation. When given a function, finding its antiderivative allows you to express the integral of that function. Antiderivatives play a key role in calculating definite integrals, especially in determining the area under a curve or, as in this case, verifying orthogonality.
Steps to Identify and Use Antiderivatives:
Steps to Identify and Use Antiderivatives:
- Recognize the function to integrate.
- Determine the antiderivative of the function.
- Use it to find the value of the definite integral across an interval.