# Fourier Series/1 over -1 to 0, Cosine of Pi x over 0 to 1/Mistake

## Mistake

By formulae $(3)$ the Fourier coefficients are:
$a_0 = \ds \int_{-1}^0 \rd x + \int_0^1 \rd x = 1$
$a_n = \ds \int_{-1}^0 \map \cos {n \pi x} \rd x + \int_0^1 \map \cos {n \pi x} \map \cos {\pi x} \rd x = 0$
$b_n = \ds \int_{-1}^0 \map \sin {n \pi x} \rd x + \int_0^1 \map \sin {n \pi x} \map \cos {\pi x} \rd x$
$= \dfrac {\map \cos {-n \pi} - 1} {n \pi} + \dfrac 1 {2 \pi} \set {\dfrac {1 - \map \cos {n + 1} \pi} {n + 1} + \dfrac {1 - \map \cos {n - 1} \pi} {n - 1} }$

## Correction

The author appears to have omitted to note that the term $a_1$ is non-vanishing:

 $\ds a_1$ $=$ $\ds \int_0^1 \cos \pi x \cos \pi x \rd x$ $\ds$ $=$ $\ds \int_0^1 \cos^2 \pi x \rd x$ $\ds$ $=$ $\ds \intlimits {\frac x 2 + \frac {\sin 2 \pi x} {4 \pi} } 0 1$ Primitive of $\cos^2 \pi x$ $\ds$ $=$ $\ds \paren {\frac 1 2 + \frac {\sin 2 \pi} {4 \pi} } - \paren {\frac 0 2 + \frac {\sin 0} {4 \pi} }$ $\ds$ $=$ $\ds \paren {\frac 1 2 + 0} - \paren {0 + 0}$ Sine of Multiple of Pi $\ds$ $=$ $\ds \frac 1 2$

This is the graph of $\map f x$ and its expansion to $b_7$ according to the correct analysis:

In comparison, this is the corresponding graph as a result of I.N. Sneddon's analysis, that is, without the $a_1$ term.

The missing $a_1 \cos \pi x$ component has been presented in red, to illustrate the contribution it makes: