Question

Show that the set of functions \(\sin n x\) is not a complete set on \((-\pi, \pi)\) by trying to expand the function \(f(x)=1\) on \((-\pi, \pi)\) in terms of them.

Short answer

The set of functions \(\sin n x\) is not complete because \(f(x) = 1\) cannot be expanded in terms of them.

Step-by-step solution

Understand 'complete set'

A set of functions is said to be complete in a given function space if any function in that space can be expressed as a linear combination of those functions.

Express the problem

To show that \(\sin(n x)\) is not complete, try to expand \(f(x) = 1\) as a series of \(\sin(n x)\). This means looking for coefficients \(a_n\) such that \[\sum_{n=1}^{\infty} a_n \sin(n x) = 1 \]

Compute the Fourier coefficients

Find coefficients \(a_n\) using the formula for the Fourier series: \[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} 1 \sin(n x) dx\] where \(f(x) = 1\).

Evaluate the integrals

Calculate the integral \[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(n x) dx.\] Notice that \[a_n = \frac{1}{\pi} \left[ -\frac{1}{n} \cos(n x) \right]_{-\pi}^{\pi}.\]

Simplify the expression

Since \(\cos(n x) \) is periodic with period \2\pi\ and evaluates to the same value at \(\pm \pi\), we get \[\cos(n \pi) - \cos(-n \pi) = 0.\]

Conclusion of integral results

Thus, \(a_n = 0\) for all \(n\), implying that \[\sum_{n=1}^{\infty} a_n \sin(n x) = 0\] cannot be equal to \(1\).

Final conclusion

Since we cannot represent \(f(x)=1\) using only the set \(\sin(n x)\), this set of functions is not a complete set on \(-\pi, \pi\).

Key concepts

Fourier Coefficients

The Fourier coefficients play a crucial role in function expansion using Fourier series. To represent a function as a sum of sinusoidal components, the coefficients need to describe the amplitude of each sine or cosine function. In the given exercise, we are trying to express the constant function \(f(x) = 1\) using only sine functions \( \sin(nx) \). The formula to find these coefficients is given by:

\[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx\]

For our example, substituting \(f(x) = 1\) yields:

\[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(nx) \, dx\]

Calculating this integral will help us discover if we can find non-zero coefficients that would sum up to represent the function 1.

Integrals

Integrals are fundamental in determining the Fourier coefficients. Evaluating the integral \( a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(nx) \, dx \) allows us to understand the contribution of each sinusoidal function. The key to solving this integral relies on the periodic nature of sine and cosine functions.

We observe that the sine function integrated over a symmetric interval about zero always results in zero. Mathematically, integrating \( \sin(nx)\) over one period \( [-\pi, \pi ] \) gives zero because the positive and negative parts cancel each other out.

Thus, the computed integrals for this problem lead to:

\[a_n = \frac{1}{\pi} \, \left[ -\frac{1}{n} \cos(nx) \right]_{-\pi}^{\pi}\]

Evaluating this at the bounds results in zero since \( \cos(n\pi) \) and \( \cos(-n\pi) \) are equal, and their difference is zero.

Function Expansion

Function expansion in terms of a Fourier series involves expressing a function as an infinite sum of sines and cosines. However, for the function \(f(x) = 1\) over the interval \( (-\pi, \pi) \), trying to expand it solely using sine functions poses a problem.

The Fourier coefficients we computed turned out to be zero, leading to:

\[ \sum_{n=1}^{\infty} a_n \sin(nx) = 0 \]

This clearly cannot equal 1. The inability to find a set of non-zero coefficients \( a_n \) that represent the constant function 1 shows that we cannot accomplish a complete expansion using the set \( \sin(nx) \). Hence, the set \( \sin(nx) \) is not complete for this function.

Sinusoidal Functions

Sinusoidal functions, which include sines and cosines, are the core of Fourier series. They form the building blocks for representing more complex functions. These periodic functions have specific properties making them suitable for function expansion.

In the realm of Fourier series, we often combine both sine and cosine functions (or their complex exponentials) to achieve completeness.

For our exercise, focusing only on sine functions \( \sin(nx) \) was insufficient to expand a simple constant function like 1. This highlights an essential aspect: completeness often requires both sine and cosine terms to cover all possible functions in the interval.

The inability to represent the function \(f(x) = 1\) with just sine functions underscores the importance of comprehensiveness when choosing the basis functions for Fourier series expansion.