I'm studying book about digital filter by Richard Hamming. And there is exercise to get odd and even expansion of g(x)=x where x is from 0 to $\pi$. I understood even expansion, but can't get into odd expansion: $\pi$ - $2$*(${\rm sin} x$ + $\frac{1}{2}$${\rm sin} 2x$ + $\frac{1}{3}$${\rm sin} 3x$...)

Can anybody explain odd expansion of this function g(x)=x?

Thanks in advance.

UPDATE: Let me be more clear. The exersise is: show that g(x) = x has two expansions (0 < x < $\pi$)

$x$ =$\begin{cases}\pi - 2*(\sin(x)+\sin(2x)/2+\sin(3x)/3+...) \\\pi/2 - 4/\pi*(\cos(x)+\cos(3x)/3^2+\cos(5x)/5^2+...)\end{cases}$

I merely understand cosine expansion, but I don't understand sine expansion...

  • 2
    $\begingroup$ Perhaps you can include what Richard Hamming means by odd and even expansion for the benefit of those readers who do not have access to the book? $\endgroup$ Nov 23, 2013 at 14:48
  • $\begingroup$ According to book: to get cosine expansion you extend the function about origin as an even function g(x)=(-x); and for only sines as an odd function g(x)=-g(-x). $\endgroup$
    – Sharov
    Nov 23, 2013 at 18:34
  • $\begingroup$ There is something wrong. If the odd expansion of $g(x) =\begin{cases}x,&0 \leq x \leq \pi,\\0,&\text{otherwise,}\end{cases}$ is the odd function $g(x) =\begin{cases}x,& |x| \leq \pi,\\0,&\text{otherwise,}\end{cases}$, then its Fourier series cannot have a DC term: odd functions have a DC term $0$, not $\pi$ as you have it. On the other hand, the even function $p(x) =\begin{cases}|x|,&-\pi \leq x \leq \pi,\\0,&\text{otherwise,}\end{cases}$ does indeed have a DC term of $\pi$ but the coefficients of the $\sin$ functions are all $0$. Please check that you have copied things correctly $\endgroup$ Nov 23, 2013 at 21:23
  • $\begingroup$ That is why I'm asking this question... I guess in order to make function odd, we extend it with $\pi$ - x (or x - $\pi$) on interval from -$\pi$ to 0. There is another excersice where g(x) = x needed to be expanded on [0,$\pi$) he notes that if you substract $\pi$ from x you get odd function. $\endgroup$
    – Sharov
    Nov 23, 2013 at 22:26

1 Answer 1


We want to find a representation in terms of sines and cosines for the function $g(x)$ which has value $x$ for $x \in [0,\pi]$. We don't really care if this representation does not work well outside the interval $[0,\pi]$, but inside the interval, the representation better work perfectly. This seems to be crying out for a Fourier series approach, and so that's what we will use. The standard approach begins with functions that are periodic with period $2\pi$ and so let us extend the definition of $g(x)$ to cover one period $[0,2\pi]$.

  • If we choose $g(x)$ to have value $2\pi-x$ for $x \in [\pi, 2\pi]$, then, when repeated periodically, this extended $g(x)$ is an even function of $x$ with average value $\pi/2$. Furthermore, as is the case with all even functions, the sine terms all have $0$ coefficients. Thus, the Fourier series works out to have only cosines in it, and the DC term is $\pi/2$.

  • If we choose $g(x)$ to have value $x$ for $x \in [\pi, 2\pi]$, then when repeated periodically, this extended $g(x)$ is not an odd function of $x$, but nonetheless it is true that all the cosine terms in the Fourier series turn out to be $0$ except the DC term which has value $\pi$.

  • $\begingroup$ Thank you so much, Dilip! Very clear! The last question: why extended $g(x)$ is not an odd function? I guess it is truly odd function... Nevertheless, thank you very very much. $\endgroup$
    – Sharov
    Nov 27, 2013 at 19:20
  • $\begingroup$ @Sharov If $g(x) = x, 0 \leq x < 2\pi$ is extended periodically with period $2\pi$, then for $-2\pi \leq x < 0$, the extended $g(x)$ equals $2\pi+x$, having value $0$ when $x = -2\pi$ and approaching $2\pi$ as $x$ approaches $0$. So, the extended $g(x)$ is not an odd function. If you subtract off the DC value $\pi$ and shift the function to the right by $\pi$, then it is an odd function, being given by $x$ for $-\pi < x < \pi$ $\endgroup$ Nov 27, 2013 at 20:04

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