Full Fourier series

mark

Compute the full Fourier series of $f(x) = x-x^2$ over the interval $[0,1/2]$. How does your result differ from what we saw in class?

Zach

The full Fourier series is generally:

$$ f(x)~ \frac{a_0}{2} + \sum_{n=1}^{\infty} a_ncos(\frac{n\pi x}{L}) + b_nsin(\frac{n\pi x}{L})$$

First find the value of $a_0$ for $f(x) = x - x^2$ over the interval $[0,\frac{1}{2}]$

$$ a_0 = \frac{2}{L} \int_{0}^{L} f(x)\,dx $$

$\implies a_0 = 4 \int_{0}^{\frac{1}{2}}(x - x^2)\,dx$ = $4 \int_{0}^{\frac{1}{2}} x\,dx$ - $\int_{0}^{\frac{1}{2}} x^2\,dx$

$$\implies 4[\frac{1}{2} x^2 \biggr\rvert_{0}^{\frac{1}{2}} - \frac{1}{3} x^3 \biggr\rvert_{0}^{\frac{1}{2}}] = 4[\frac{1}{2} (\frac{1}{2})^2 - \frac{1}{3} (\frac{1}{2})^3]$$

$$\implies 4[\frac{1}{8} - \frac{1}{24}] = 4(\frac{1}{12})$$

Therefore $a_0 = \frac{1}{3}$ and plugging this value into the general equation gives $a_0 = \frac{1}{3}*\frac{1}{2} = \frac{1}{6}$

Next, find the value of $a_n$ for $f(x)$

$a_n = \frac{2}{L}$ $\int_{0}^{L} f(x)cos(\frac{n\pi x}{L}\,dx$ = $4\int_{0}^{\frac{1}{2}} (x - x^2)cos(\frac{2\pi nx)\,dx$

Can split into integrals of $x$ and $x^2$

$4 \int_{0}^{\frac{1}{2}} xcos(2\pi nx)\,dx$

Using integration by parts, with $u = x, du = dx, v = \frac{sin(2\pi nx)}{2\pi n}, dv = cos{2\pi nx}dx$

$$\implies 4[ \frac{xsin(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} - \frac{1}{2\pi n} \int_{0}^{\frac{1}{2}}sin(2\pi nx)\,dx ]$$

$$\implies \frac{1}{(\pi n)^2} cos(2\pi nx) \bigg\rvert_{0}^{\frac{1}{2}} = \frac{1]{(\pi n)^2} [(-1)^n - 1] = \frac{(-1)^n -1}{\pi^2n^2}$$

Now, plugging in $x^2$ gives:

$4 \int_{0}^{\frac{1}{2}} x^2cos(2\pi nx)\,dx$

Again use integration by parts, with $u = x^2, du = 2xdx, v = \frac{sin(2\pi nx)}{2\pi n}, dv = cos(2\pi nx)dx$

$$\implies 4[ \frac{(x^2)sin(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} - \frac{1}{\pi n} \int_{0}^{\frac{1}{2}} xsin(2\pi nx)\,dx ]$$

The first term is equal to 0, and integration by parts must be used to solve the integral, with $u = x, du = dx, v = \frac{-cos(2\pi nx)}{2\pi n}, dv = sin(2\pi nx)dx$

$\implies \frac{-4}{\pi n} \frac{-xcos(2\pi nx)}{2\pin} \bigg\rvert_{0}^{\frac{1}[2}}$ + $\frac{1}{2\pi n} \int_{0}^{\frac{1}{2}} cos(2\pi nx)\,dx$

$$\implies \frac{4}{\pi n}(\frac{(-1)^n}{4\pi n}) = \frac{(-1)^n}{\pi^2n^2}$$

$$\implies a_n = \frac{(-1)^n - 1}{\pi^2n^2} - \frac{(-1)^n}{\pi^2n^2} = \frac{(-1)}{\pi^2n^2}$$

Now move onto $b_n$ and again split the integrals of $x$ and $x^2$

$$ b_n = \frac{2}{L] \[ \int_{0}{L} f(x)sin(\frac{\pi nx}{L})\,dx \]$$

$b_n = 4[ \int_{0}^{\frac{1}{2}} xsin(2\pi nx)\,dx$ + $\int_{0}^{\frac{1}{2}} x^2sin(2\pi nx)\,dx ]$

Again use integration by parts for the integral of $x$, with $u = x, du = dx, v = \frac{-cos(2\pi nx)}{2\pi n}, dv = sin(2\pi nx)dx$

$\implies 4[\frac{-xcos(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}}$ + $\frac{1}{2\pi n} \int_{0}^{\frac{1}{2}} cos(2\pi nx)\,dx$

$$\implies 4[-\frac{(-1)^n}{4\pi n}] = -\frac{(-1)^n}{\pi n}$$

Use integration by parts for the integral of $x^2$, with $u = x^2, du = 2xdx, v = \frac{-cos(2\pi nx)}{2\pi n}, dv = sin(2\pi nx)dx$

$$\implies 4[\frac{-x^2cos(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} + \frac{1}[\pi n} \int_{0}^{\frac{1}[2}} xcos(2\pi nx)\,dx $$

$$\implies 4[-\frac{(-1)^n}{8\pi n}] = -\frac{(-1)^n}{2\pi n}$$

$$\implies b_n = -\frac{(-1)^n}{\pi n} + \frac{(-1)^n}{2\pi n} = -\frac{(-1)^n}{2\pi n}$$

Therefore, the full Fourier series for $x - x^2$ over $[0,\frac{1]{2}]$ is

$$\frac{1}{6} + \sum_{n=1}^{\infty} \frac{(-1)}{\pi^2n^2}cos(2\pi nx) - \frac{(-1)^n}{2\pi n}sin(2\pi nx)$$

This differs from the Fourier series we saw in class as it contains both the Fourier Sine series and Fourier Cosine series

Zach

I am having trouble figuring out why the these remaining lines won't compile. I have used most of the editing window to fix other things, just can't seem to find what's wrong.

Zach

The full Fourier Series is generally:

$$f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} a_ncos(\frac{n\pi x}{L}) + b_nsin(\frac{n\pi x}{L})$$

First find the value of $a_0$ for $f(x) = x - x^2$ over the interval $[0,L]$

$$a_0 = \frac{2}{L} \int_{0}^{L} f(x)\,dx$$

$$\implies a_0 = 4 \int_{0}^{\frac{1}{2}}(x - x^2)\,dx = 4\int_{0}^{\frac{1}{2}} x\,dx - \int_{0}^{\frac{1}{2}} x^2\,dx$$

$$\implies 4[\frac{1}{2} x^2 \biggr\rvert_{0}^{\frac{1}{2}} - \frac{1}{3} x^3 \biggr\rvert_{0}^{\frac{1}{2}}] = 4[\frac{1}{2} (\frac{1}{2})^2 - \frac{1}{3} (\frac{1}{2})^3]$$

$$\implies 4[\frac{1}{8} - \frac{1}{24}] = 4(\frac{1}{12})$$

Therefore $a_0 = \frac{1}{3}$ and plugging in this value into the general equation gives $\frac{a_0}{2} = \frac{1}{6}$

Next, find the value for $a_n$ for $f(x)$

$$\int_{0}^{L} f(x)cos(\frac{n\pi x}{L})\,dx = 4\int_{0}^{\frac{1}{2}} (x - x^2)cos(2\pi nx)\,dx$$

Can split into integrals of $x$ and $x^2$

$$4\int_{0}^{\frac{1}{2}}xcos(2\pi nx)\,dx$$

Using integration by parts, with 

$u = x, du = dx, v = \frac{sin(2\pi nx)}{2\pi n}, dv = cos(2\pi nx)dx$

$$\implies 4[ \frac{xsin(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} - \frac{1}{2\pi n} \int_{0}^{\frac{1}{2}}sin(2\pi nx)\,dx ]$$

$$\implies \frac{1}{(\pi n)^2} cos(2\pi nx) \bigg\rvert_{0}^{\frac{1}{2}} = \frac{1}{(\pi n)^2} [(-1)^n - 1] = \frac{(-1)^n -1}{\pi^2n^2}$$

Now, plugging in $x^2$ gives

$$4 \int_{0}^{\frac{1}{2}} x^2cos(2\pi nx)\,dx$$

Again using integration by parts, with 

$u = x^2, du = 2xdx, v = \frac{sin(2\pi nx)}{2\pi n}, dv = cos(2\pi nx)dx$

$$\implies 4[ \frac{(x^2)sin(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} - \frac{1}{\pi n} \int_{0}^{\frac{1}{2}} xsin(2\pi nx)\,dx ]$$

The first term is equal to $0$, and integration by parts must be used to solve the integral with

$u = x, du = dx, v = \frac{-cos(2\pi nx)}{2\pi n}, dv = sin(2\pi nx)dx$

$$\implies \frac{-4}{\pi n} \frac{-xcos(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} + \frac{1}{2\pi n} \int_{0}^{\frac{1}{2}} cos(2\pi nx)\,dx$$

$$\implies \frac{4}{\pi n}(\frac{(-1)^n}{4\pi n}) = \frac{(-1)^n}{\pi^2n^2}$$

$$\implies a_n = \frac{(-1)^n - 1}{\pi^2n^2} - \frac{(-1)^n}{\pi^2n^2} = \frac{(-1)}{\pi^2n^2}$$

Now move onto $b_n$ and again split the integrals of $x$ and $x^2$

$$b_n = \frac{2}{L} \int_{0}^{L} f(x)sin(\frac{\pi nx}{L})\,dx $$

$$b_n = 4[\int_{0}^{\frac{1}{2}} xsin(2\pi nx)\,dx - \int_{0}^{\frac{1}{2}} x^2sin(2\pi nx)\,dx ]$$

Use integration by parts for the integral of $x$, with

$u = x, du = dx, v = \frac{-cos(2\pi nx)}{2\pi n}, dv = sin(2\pi nx)dx$

$$\implies 4[\frac{-xcos(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} + \frac{1}{2\pi n} \int_{0}^{\frac{1}{2}} cos(2\pi nx)\,dx ]$$

$$\implies 4[-\frac{(-1)^n}{4\pi n}] = -\frac{(-1)^n}{\pi n}$$

Use integration by parts for the integral of $x^2$, with 

$u = x^2, du = 2xdx, v = \frac{-cos(2\pi nx)}{2\pi n}, dv = sin(2\pi nx)dx$

$$\implies 4[\frac{-x^2cos(2\pi nx)}{2\pi n} \bigg\rvert_{0}^{\frac{1}{2}} + \frac{1}{\pi n} \int_{0}^{\frac{1}{2}} xcos(2\pi nx)\,dx ]$$

$$\implies 4[-\frac{(-1)^n}{8\pi n}] = -\frac{(-1)^n}{2\pi n}$$

$$\implies b_n = -\frac{(-1)^n}{\pi n} + \frac{(-1)^n}{2\pi n} = -\frac{(-1)^n}{2\pi n}$$

Therefore, the full Fourier Series for $f(x) = x - x^2$ over $[0,\frac{1}{2}]$ is:

$$\frac{1}{6} + \sum_{n=1}^{\infty} \frac{(-1)}{\pi^2n^2}cos(2\pi nx) - \frac{(-1)^n}{2\pi n}sin(2\pi nx)$$

This differs from the Fourier series we saw in class as it contains both the Fourier Sine series and the Fourier Cosine series.

Zach

Fixed it using Overleaf, made it easy to see places where I had typed [ instead of {.

mark

@Zach Holy Moly!! 💪