Differential Equations - Fourier Series: Cosine

Differential Equations - Fourier Series: Cosine - ii

Example 2 Problem Statement.

$f (x) = L - x$ on $0 \leq x \leq L$
$f (x) = x^{3}$ on $0 \leq x \leq L$
$f (x) = {\begin{array}{l} \frac{L}{2} & if 0 \leq x \leq \frac{L}{2} \\ x - \frac{L}{2} & if \frac{L}{2} \leq x \leq L \end{array}$

f (x) = L - x

0 \leq x \leq L

Here is the even extension of this function.

\begin{aligned} g (x) & = {\begin{array}{l} f (x) & if 0 \leq x \leq L \\ f (- x) & if - L \leq x \leq 0 \end{array} \\ = {\begin{array}{l} L - x & if 0 \leq x \leq L \\ L + x & if - L \leq x \leq 0 \end{array} \end{aligned}

Here is the graph of both the original function and its even extension. Note that we’ve put the “extension” in with a dashed line to make it clear the portion of the function that is being added to allow us to get the even extension

f (x) = x^{3}

0 \leq x \leq L

The even extension of this function is,

\begin{aligned} g (x) & = {\begin{array}{l} f (x) & if 0 \leq x \leq L \\ f (- x) & if - L \leq x \leq 0 \end{array} \\ = {\begin{array}{l} x^{3} & if 0 \leq x \leq L \\ - x^{3} & if - L \leq x \leq 0 \end{array} \end{aligned}

The sketch of the function and the even extension is,

f (x) = {\begin{array}{l} \frac{L}{2} & if 0 \leq x \leq \frac{L}{2} \\ x - \frac{L}{2} & if \frac{L}{2} \leq x \leq L \end{array}

Here is the even extension of this function,

\begin{aligned} g (x) & = {\begin{array}{l} f (x) & if 0 \leq x \leq L \\ f (- x) & if - L \leq x \leq 0 \end{array} \\ = {\begin{array}{l} x - \frac{L}{2} & if \frac{L}{2} \leq x \leq L \\ \frac{L}{2} & if 0 \leq x \leq \frac{L}{2} \\ \frac{L}{2} & if - \frac{L}{2} \leq x \leq 0 \\ - x - \frac{L}{2} & if - L \leq x \leq - \frac{L}{2} \end{array} \end{aligned}

The sketch of the function and the even extension is,

Okay, let’s now think about how we can use the even extension of a function to find the Fourier cosine series of any function

f (x)

0 \leq x \leq L

So, given a function

f (x)

we’ll let

g (x)

be the even extension as defined above. Now,

g (x)

is an even function on

- L \leq x \leq L

and so we can write down its Fourier cosine series. This is,

g (x) = \sum_{n = 0}^{\infty} A_{n} \cos (\frac{n π x}{L}) A_{n} = {\begin{array}{l} \frac{1}{L} \int_{0}^{L} f (x) d x & n = 0 \\ \frac{2}{L} \int_{0}^{L} f (x) \cos (\frac{n π x}{L}) d x & n \neq 0 \end{array}

and note that we’ll use the second form of the integrals to compute the constants.

Now, because we know that on

0 \leq x \leq L

we have

f (x) = g (x)

and so the Fourier cosine series of

f (x)

0 \leq x \leq L

is also given by,

f (x) = \sum_{n = 0}^{\infty} A_{n} \cos (\frac{n π x}{L}) A_{n} = {\begin{array}{l} \frac{1}{L} \int_{0}^{L} f (x) d x & n = 0 \\ \frac{2}{L} \int_{0}^{L} f (x) \cos (\frac{n π x}{L}) d x & n \neq 0 \end{array}

Let’s take a look at a couple of examples.

Example 3 Find the Fourier cosine series for

f (x) = L - x

0 \leq x \leq L

All we need to do is compute the coefficients so here is the work for that,

A_{0} = \frac{1}{L} \int_{0}^{L} f (x) d x = \frac{1}{L} \int_{0}^{L} L - x d x = \frac{L}{2}

\begin{aligned} A_{n} & = \frac{2}{L} \int_{0}^{L} f (x) \cos (\frac{n π x}{L}) d x = \frac{2}{L} \int_{0}^{L} (L - x) \cos (\frac{n π x}{L}) d x \\ = \frac{2}{L} {(\frac{L}{n^{2} π^{2}}) (n π (L - x) \sin (\frac{n π x}{L}) - L \cos (\frac{n π x}{L})) |}_{0}^{L} \\ = \frac{2}{L} (\frac{L}{n^{2} π^{2}}) (- L \cos (n π) + L) = \frac{2 L}{n^{2} π^{2}} (1 + {(- 1)}^{n + 1}) n = 1, 2, 3, \dots \end{aligned}

The Fourier cosine series is then,

f (x) = \frac{L}{2} + \sum_{n = 1}^{\infty} \frac{2 L}{n^{2} π^{2}} (1 + {(- 1)}^{n + 1}) \cos (\frac{n π x}{L})

Note that as we did with the first example in this section we stripped out the

A_{0}

term before we plugged in the coefficients.

Next, let’s find the Fourier cosine series of an odd function. Note that this is doable because we are really finding the Fourier cosine series of the even extension of the function.

Comments

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