Question

(a) Sketch several periods of the function $f(x)$ of period $2\pi$ which is equal to $x$ on $-\pi <x<\pi .$ Expand $f(x)$ in a sine-cosine Fourier series and in a complex exponential Fourier series. (b) Sketch several periods of the function $f(x)$ of period $2\pi$ which is equal to $x$ on 0 $<x<2\pi$. Expand $f(x)$ in a sine-cosine Fourier series and in a complex exponential Fourier series. Note that this is not the same function or the same series as (a).

Short answer

For (a): Fourier series is $\sum _{n=1}^{\mathrm{\infty }}\frac{2(-1{)}^{n+1}}{n}\sin (nx)$ and $c_n=\frac{1}{in}$ for complex exponential. For (b): Fourier series is $\pi -\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n}\sin (nx)$ and $c_n=\frac{1}{in}$ for complex exponential.

Step-by-step solution

Understand the problem

We need to sketch the function and find the Fourier series expansions for two different cases: (a) when the function is equal to $x$ on $-\pi <x<\pi$ and (b) when the function is equal to $x$ on $0<x<2\pi$, both with a period of $2\pi$.

Sketch the function for (a)

For case (a), sketch $f(x)=x$ over one period from $-\pi$ to $\pi$. Then repeat this pattern to cover several periods.

Compute the Fourier series for (a)

The Fourier series expansion of a periodic function $f(x)$ with period $2\pi$ is given by: $$f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}(a_n\cos (nx)+b_n\sin (nx))$$ First, find the coefficients: $$a_0=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }xdx=0$$ $$a_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }x\cos (nx)dx=0$$ $$b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }x\sin (nx)dx=\frac{2(-1{)}^{n+1}}{n}$$ Thus, the Fourier sine-cosine series is: $$f(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{2(-1{)}^{n+1}}{n}\sin (nx)$$ For the complex exponential series: $$f(x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n{e}^{inx}$$ where $$c_n=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }x{e}^{-inx}dx=\{\begin{array}{lll}\frac{1}{in}& \text{if }neq00& \text{if }n=0\end{array}$$

Sketch the function for (b)

For case (b), sketch $f(x)=x$ over one period from $0$ to $2\pi$. Then repeat this pattern to cover several periods.

Compute the Fourier series for (b)

The Fourier series expansion of the function in case (b) is similar to case (a) but with different integration limits: First, find the coefficients: $$a_0=\frac{1}{2\pi }{\int }_0^{2\pi }xdx=\pi$$$$a_n=\frac{1}{\pi }{\int }_0^{2\pi }x\cos (nx)dx=0$$ $$b_n=\frac{1}{\pi }{\int }_0^{2\pi }x\sin (nx)dx=\{\begin{array}{lll}-\frac{2}{n}& \text{if }n\text{ is odd}0& \text{if}n\text{ is even}\end{array}$$ Thus, the Fourier sine-cosine series is: $$f(x)=\pi -\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n}\sin (nx)$$ For the complex exponential series: $$f(x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n{e}^{inx}$$ where $$c_n=\frac{1}{2\pi }{\int }_0^{2\pi }x{e}^{-inx}dx=\{\begin{array}{lll}\frac{1}{in}& \text{if }neq00& \text{if }n=0\end{array}$$

Key concepts

Fourier sine-cosine series

A Fourier sine-cosine series is a way to represent a periodic function as an infinite sum of sine and cosine functions. This is particularly useful because sine and cosine functions form an orthogonal basis, meaning they can represent any periodic function accurately over a specified interval. For a periodic function with period $2\pi$, the general form of the Fourier series is:

$$f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}\text{(}{a}_n\cos (nx)+b_n\sin (nx)\text{)}$$

Here:
- $a_0$ is the average value of the function over one period.
- $a_n$ and $b_n$ are the Fourier coefficients that determine the amplitude of the cosine and sine components, respectively.

In the exercise, the function $f(x)=x$ on both intervals is expanded into such series, albeit producing different series for the different intervals. The sine-cosine series for these functions are as follows:

For case (a) on $-\pi <x<\pi$, the series is: $$f(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{2(-1{)}^{n+1}}{n}\sin (nx)$$

For case (b) on $0<x<2\pi$, the series becomes: $$f(x)=\pi -\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n}\sin (nx)$$

complex exponential Fourier series

The complex exponential Fourier series is another method to represent periodic functions, this time using complex exponentials of the form $e^{inx}$. The advantage of this method is that it simplifies the mathematics, combining sine and cosine terms into a single exponential term. The general form is:

$$f(x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n{e}^{inx}$$

The coefficients $c_n$ are crucial to this representation and are calculated as:

$$c_n=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(x)e^{-inx}dx$$

In our exercise, the coefficients were computed for both cases (a) and (b):
Case (a): $$c_n=\frac{1}{in}\text{for}n\text{eq}0,0\text{for}n=0$$
Case (b): Similarly, $$c_n=\frac{1}{in}\text{for}n\text{eq}0,0\text{for}n=0$$

In both cases, the series covers all integer values of $n$, allowing us to represent the function in a compact and elegant form.

periodic functions

Periodic functions are those that repeat their values in regular intervals or periods. The given function $f(x)=x$ is periodic with a period of $2\pi$ in both problem instances. This means that:

$$f(x+2\pi )=f(x)$$

In the exercise, we examine the behavior of $f(x)=x$ over different intervals to form a clear periodic pattern:
- Case (a): The function is defined on $-\pi <x<\pi$, meaning it starts at $-\pi$, increases linearly to \pi, and then repeats this pattern.
- Case (b): Defined on $0<x<2\pi$, the function starts at 0, increases linearly to $2\pi$, and then repeats.

Understanding these intervals and how the function behaves over each helps in computing the Fourier series. By analyzing such periodic behavior, we can efficiently break down more complex functions into manageable periodic components.

coefficients calculation

Calculating Fourier coefficients is essential for constructing Fourier series, both sine-cosine and complex exponential. The coefficients $a_0$, $a_n$, and $b_n$ for the sine-cosine series are given by integrals over one period:

$$a_0=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(x)dx$$
$$a_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\cos (nx)dx$$
$$b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\sin (nx)dx$$

For the complex exponential series, the coefficients $c_n$ are calculated as:

$$c_n=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(x)e^{-inx}dx$$

In our given problems, these integrals were solved step-by-step for specific periodic intervals:
- For case (a): The integral for $b_n$ for $f(x)=x$ on $-\pi <x<\pi$ results in non-zero values only for $b_n$, leading to a pure sine series.
- For case (b): The calculation shifts to intervals from $0$ to $2\pi$, giving different coefficients yet a similar sine-cosine expansion.

Each coefficient tells us about the contribution of a particular sine or cosine wave (or the exponential term) of specific frequencies to the overall function.