[PDF] [PDF] Lecture 14: Half Range Fourier Series: even and odd - UBC Math

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[PDF] Lecture 14: Half Range Fourier Series: even and odd - UBC Math

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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 Extensions

14.1 Even and Odd Functions

Even:f(x) =f(x)

Odd:f(x) =f(x)

14.1.1Integrals of Even and Odd Functions

L

Lf(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) E

1(x)E2(x)!even.

(3) O

1(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) 2

Check: LetE(x) =1

2 [f(x) +f(x)]. ThenE(x) =1 2 [f(x) +f(x)]=E(x) even. Similarly let

O(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 L

Lf(x)cos|

{z even( nx L dx=2 L L 0 f(x)cos(nx L dx (14.7) b n=1 L L

Lf(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 L

Lf(x)cos(nx

L {z odddx= 0 (14.10) b n=1 L L

Lf(x)sin(nx

L {z evendx=2 L L 0 f(x)sin(nx L dx

Therefore

f(x) =1∑ n=1b nsin(nx L ;bn=2 L L

0f(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 L

1∑

n=1b nsin(nx L

Fourier Series3

where a n=2 L L 01 2 [f(x) +f(x)]cos(nx L dx=1 L L

Lf(x)cos(nx

L dx b n=2 L L 01 2 [f(x)f(x)]sin(nx L dx=1 L L

Lf(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.

Cosine Series:

f(x) =a0 2 +1∑ n=1a ncos(nx L (14.12) a n=2 L L 0 f(x)cos(nx L dx: (14.13)

Sine Series:

f(x) =1∑ n=1b nsin(nx L (14.14) b n=2 L L 0 f(x)sin(nx L dx: (14.15)

Example 14.1

Expandf(x) =x,0< x <2in a half-range (a) Sine Series, (b) Cosine Series. (a) Sine Series: (L=2) b n=2 L L 0 f(t)sinn tdt (14.16) 2 0 tsinn 2 tdt (14.17) =tcosn 2 t n 2 2 0 +2 n 2 0 cosn 2 tdt (14.18) =4 n cos(n) +(2 n 2 sin(n 2 t) 2 0 (14.19) =4 n (1)n (14.20)

Therefore

f(t) =4 1 n=1(1)n+1 n sin(n 2 t) (14.21) 4 f(1) = 1 =4 1 n=1(1)n+1 n sin(n 2 (14.22) therefore 4 = 11 3 +1 5 1 7 (14.23) (b) Cosine Series: (L=2) a 0=2 2 2 0 tdt=t2 2 2 0 = 2 (14.24) a n=2 0 tcosn 2 tdt=(2 n tsin↗n 2 t2 0(2 n

2∫

0 sinn 2 tdt (2 n 2 cosn 2 t 2 0 =4 n

22fcosn1g

(14.25)

Therefore

f(t) = 1 +4 21
n=1[ (1)n1] n 2cosn 2 t (14.26) = 18 21
n=0cos(2n+ 1) 2 t=(2n+ 1)2: (14.27) The cosine series converges faster than Sine Series. f(2) = 2 = 1 +8 21
n=01 (2n+ 1)2;2 8 = 1 +1 3 2+1 5 2+

Example 14.2

Periodic Extension: Assume thatf(x) =x,0< x <2represents one full period of the function so thatf(x+ 2) =f(x).2L= 2)L= 1. a 0=1 L L

Lf(x)dx=1

1f(x)dx=2

0 xdx=x2 2 2 0 = 2 (14.28) sincef(x+ 2) =f(2). (14.29)

Fourier Series5

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