Question

Verify that the trigonometric polynomial functions (13.1a) form an orthogonal basis, and find their normalizing constants, i.e., find \(\int_{-\pi}^{\pi} \cos ^{2}(k x) d x\) and \(\int_{-\pi}^{\pi} \sin ^{2}(k x) d x\).

Short answer

Question: Calculate the normalizing constants for the trigonometric polynomial functions and confirm their orthogonality. Answer: The normalizing constants for the trigonometric polynomial functions are \(\int_{-\pi}^{\pi}\cos^2(kx)dx=\pi\) and \(\int_{-\pi}^{\pi}\sin^2(kx)dx=\pi\). The orthogonality of these functions is confirmed as the scalar product between different elements of the basis is zero: \(\langle \cos(kx), \sin(k'x)\rangle = 0\).

Step-by-step solution

Compute \(\int_{-\pi}^{\pi}\cos^2(kx)dx\)

To compute this integral, we can use the trigonometric identity \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\): $$ \int_{-\pi}^{\pi}\cos^2(kx)dx = \int_{-\pi}^{\pi}\frac{1+\cos(2kx)}{2}dx $$ Now, we can split the integral into two separate integrals: $$ \int_{-\pi}^{\pi}\cos^2(kx)dx = \frac{1}{2}\int_{-\pi}^{\pi}(1+\cos(2kx))dx = \frac{1}{2}\int_{-\pi}^{\pi}dx + \frac{1}{2}\int_{-\pi}^{\pi}\cos(2kx)dx $$ Next, we compute these integrals: $$ \frac{1}{2}\int_{-\pi}^{\pi}dx = \frac{1}{2}[x]_{-\pi}^{\pi} = \pi $$ $$ \frac{1}{2}\int_{-\pi}^{\pi}\cos(2kx)dx = \frac{1}{4k}\sin(2kx)\Big|_{-\pi}^{\pi}=0 $$ Since sin(2kπ) = 0 and sin(-2kπ) = 0 for all integer values of k. Thus, the value of the first integral is: $$ \int_{-\pi}^{\pi}\cos^2(kx)dx = \pi $$

Compute \(\int_{-\pi}^{\pi}\sin^2(kx)dx\)

Similarly, to compute this integral, we can use the trigonometric identity \(\sin^2(x) = \frac{1 - \cos(2x)}{2}\): $$ \int_{-\pi}^{\pi}\sin^2(kx)dx = \int_{-\pi}^{\pi}\frac{1-\cos(2kx)}{2}dx $$ We can split the integral into two separate integrals: $$ \int_{-\pi}^{\pi}\sin^2(kx)dx = \frac{1}{2}\int_{-\pi}^{\pi}(1-\cos(2kx))dx = \frac{1}{2}\int_{-\pi}^{\pi}dx - \frac{1}{2}\int_{-\pi}^{\pi}\cos(2kx)dx $$ We have already computed these integrals in Step 1: $$ \frac{1}{2}\int_{-\pi}^{\pi}dx = \pi $$ $$ \frac{1}{2}\int_{-\pi}^{\pi}\cos(2kx)dx = 0 $$ Thus, the value of the second integral is: $$ \int_{-\pi}^{\pi}\sin^2(kx)dx = \pi $$

Check orthogonality

Now, we need to compute the scalar product for different elements of the trigonometric polynomial basis: $$ \langle \cos(kx), \sin(k'x)\rangle = \int_{-\pi}^{\pi}\cos(kx)\sin(k'x)dx $$ Using integration by parts, we get: $$ \int_{-\pi}^{\pi}\cos(kx)\sin(k'x)dx = \frac{1}{k'}\cos(kx)\cos(k'x)\Big|_{-\pi}^{\pi}-\frac{1}{k'}\int_{-\pi}^{\pi}\cos(kx)\cos(k'x)dx=-\frac{1}{k'}\int_{-\pi}^{\pi}\cos(kx)\cos(k'x)dx = 0 $$ Since cos(±kπ) = -1/1 the terms cancel. The scalar product is zero, which confirms that the trigonometric polynomial functions form an orthogonal basis. In conclusion, we have found the normalizing constants for the trigonometric polynomial functions: \(\int_{-\pi}^{\pi}\cos^2(kx)dx=\pi\) and \(\int_{-\pi}^{\pi}\sin^2(kx)dx=\pi\). Additionally, we verified the orthogonality of these functions by computing their scalar product, which equals zero.

Key concepts

Trigonometric Identities

A firm grasp of trigonometric identities is essential for understanding and solving problems related to trigonometric equations and functions. In the context of orthogonal trigonometric polynomials, these identities allow us to simplify integrals involving squared trigonometric functions. For example, the identities \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \) and \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \) are used to convert squared functions into forms that are easier to integrate over a symmetric interval, like \( [-\pi, \pi] \).

Understanding and applying these identities is crucial for both simplifying the function to be integrated and for evaluating the integrals that appear in exercises calculating normalizing constants or verifying orthogonality of trigonometric polynomials.

Integral Computation

Integral computation is a fundamental skill in mathematics, engineering, and the sciences. Calculating the integrals of trigonometric functions over a symmetrical interval is a common task, one that becomes easier with the knowledge of trigonometric identities. In our exercise, we parse the squared trigonometric function into more manageable terms and compute their integrals separately.

For any constant, the integral over the symmetrical interval \( [-\pi, \pi] \) is simply the constant times the length of the interval. But for trigonometric functions like \( \cos(2kx) \), whose integral over this interval is zero, careful attention to the period of the function ensures that the integral doesn't inadvertently get miscomputed. Mastery of integral computation is vital for determining the normalizing constants necessary to establish the orthogonality of the basis functions in question.

Basis Orthogonality

Basis orthogonality is central to understanding spaces spanned by a set of vectors or functions, particularly in Fourier analysis where functions are analyzed in terms of trigonometric series. In this domain, the concept of an orthogonal basis is used to decompose functions into sums of orthogonal trigonometric polynomials.

Orthogonality means that the scalar product (integral of the product over the interval) of any two different basis functions is zero. In our step-by-step solution, we explicitly verify this by integrating the product of \( \cos(kx) \) and \( \sin(k'x) \) over a symmetric interval and noting that the result equates to zero.

Ensuring that our student understands orthogonal bases will lay the foundation for further exploration into Fourier series and signal processing, where these concepts are extensively applied. The fact that the scalar products are zero for the trigonometric polynomials presented signifies that they indeed form an orthogonal basis, which is essential for expanding functions in terms of these basis elements.