(Compiled 4 August 2017) In this lecture we consider the Fourier Expansions for Even and Odd functions, which give rise to cosine and sine half range Fourier
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Introductory lecture notes on Partial Differential Equations - c⃝Anthony Peirce. Not to be copied, used, or revised without explicit written permission from the copyright owner.1 Lecture 14: Half Range Fourier Series: even and odd functions (Compiled 4 August 2017)
In this lecture we consider the Fourier Expansions for Even and Odd functions, which give rise to cosine and sine half
range Fourier Expansions. If we are only given values of a functionf(x) over half of the range [0;L], we can dene two
different extensions offto the full range [L;L], which yield distinct Fourier Expansions. The even extension gives rise
to a half range cosine series, while the odd extension gives rise to a half range sine series. Key Concepts:Even and Odd Functions; Half Range Fourier Expansions; Even and Odd Extensions14.1 Even and Odd Functions
Even:f(x) =f(x)
Odd:f(x) =f(x)
14.1.1Integrals of Even and Odd Functions
LLf(x)dx=0
Lf(x)dx+L
0 f(x)dx (14.1) L 0[ f(x) +f(x)]dx (14.2) 8 :2L∫0f(x)dx feven
0fodd:
(14.3) Notes: LetE(x) represent an even function andO(x) an odd function. (1) Iff(x) =E(x)O(x) thenf(x) =E(x)O(x) =E(x)O(x) =f(x))fis odd. (2) E1(x)E2(x)!even.
(3) O1(x)O2(x)!even.
(4) Any function can be expressed as a sum of an even part and an odd part: f(x) =1 2 [f(x) +f(x)] {z even part+ 1 2 [f(x)f(x)] {z odd part: (14.4) 2Check: LetE(x) =1
2 [f(x) +f(x)]. ThenE(x) =1 2 [f(x) +f(x)]=E(x) even. Similarly letO(x) =1
2 [f(x)f(x)] (14.5)O(x) =1
2 [f(x)f(x)]=O(x) odd: (14.6)14.2 Consequences of the Even/Odd Property for Fourier Series
(I) Letf(x) be Even-Cosine Series: a n=1 L LLf(x)cos|
{z even( nx L dx=2 L L 0 f(x)cos(nx L dx (14.7) b n=1 L LLf(x)sin(nx
L {z odddx= 0: (14.8)Therefore
f(x) =a0 2 +1∑ n=1a ncos(nx L ;an=2 L L 0 f(x)cos(nx L dx: (14.9) (II) Letf(x) be Odd-Sine Series: a n=1 L LLf(x)cos(nx
L {z odddx= 0 (14.10) b n=1 L LLf(x)sin(nx
L {z evendx=2 L L 0 f(x)sin(nx L dxTherefore
f(x) =1∑ n=1b nsin(nx L ;bn=2 L L0f(x)sin(nx
L dx:(III) Since any function can be written as the sum of an even and odd part, we can interpret the cos and sin series
as even/odd: f(x) =evenodd 1 2 [f(x) +f(x)]+1 2 [f(x)f(x)] (14.11) a 0 2 +1∑ n=1a ncos(nx L1∑
n=1b nsin(nx LFourier Series3
where a n=2 L L 01 2 [f(x) +f(x)]cos(nx L dx=1 L LLf(x)cos(nx
L dx b n=2 L L 01 2 [f(x)f(x)]sin(nx L dx=1 L LLf(x)sin(nx
L dx:14.3 Half-Range Expansions
If we are given a functionf(x) on an interval [0;L] and we want to representfby a Fourier Series we have two
choices - a Cosine Series or a Sine Series.