1.2 Fourier Series on (-pi,pi).pdf

Fourier series from - pi to pi

1. Chapter 9
Fourier Series

‘The reader is familiar with the Maclaurin's expansion of a function f(x) in a
series of integral powers of x. Series ofthis kind are called power series. In
the present chapter, we consider expansions of f(x) in series containing sine
and cosine functions. Such series are called Trigonometric Fourier Series.

9.1 Trigonometric Fourier Series

Consider a real-valued function f(x) which obeys the following conditions, known as the
Dirichler' conditions:

(1) FG) is defined in an open or closed interval (a, a+21) withthe property that f{x+ 29 = f(x),
so that f(x) isa periodie function with period 2.

€) Sta) is continuous or f(x) has only a finite number of simple (finite) discontinuities in the
interval (a, + 20).

(6) f(x) has no or only a finite number of maxima or minima in the interval (a, + 2).

Alo et
ae
227 | 104 10)
ont [ sereos(M) cda, form = 1,23, o
a
ant seo) ast o
‘Then the infinite series
2 Sroncon(E) 4 Soi @

529

2. 9. Fourier Series

led the Trigonometric Fourier Series or just the Fourier Series forthe function f(x) in (or
cover) the interval (a+ 21). Also, the real numbers ap, came and Med yy
are called the Trigonometric Fourier Coeficiems or just the Fourier Coefficients of f(x) The
formulas (1), (2), (3) are referred to as Euler formulas.

Alcan be proved (- the proof is beyond the scope of this ext) thatthe series (4) has its sum
equal to FF f(x) is continuous at x. Then we write

Eno rl): a

‘The series on the right hand side of (5) is usually referred to as the Fourier series expan-
‘sion or just the Fourier expansion of f(x) in (over) the interval (a, a + 20)

ne

Note: If f(x) is discontinuous at an interior point x of the interval (a, a +20) then the sum Of
the series (4) would be

1

FU
where OZ fee, h>0 and O h>0
Remark: By vue ofthe definition of the Fourier coins, follows that if a and are
constants, then, over a specified interval,

Fourier expansion of {af(x) + 8 x(x)

= (a x Fourier expansion of /(x)] + 18 x Fourier expansion of g(x)|,

Particular Cases

‘The following are some important particular cases.

Case 1: Suppose a = 0. Then (a, a + 21) = (0,21), and formulas (1) to (3) become

©
m
by 8)

With ay and dp. by, n = 1,2,:==, given by formulas (6) 0 (8). the right hand side of
expression (5) is the Fourier expansion of f(x) over (in) the interval (0,20).

3. 9.1. Trigonometric Fourier Series si

Case 2: Suppose a =
we get

1. = a Then (aa + 21) = 40.21). Setting 1 = x in formulas (6) 10 (8),

fry dx 0)

ES

£ f toycocmás, torn = cio)
à
£ [ronde orne 1.23, ay
Aso, in this ease, expression ($) becomes
me P ,
fo) =F + Yi arcosme+ Dh sin ne a

ot ot
With ay and dy, ba. n = 1,2,---. given by formulas (9) to (11), the right hand side of
expression (12) is the Fourier expansion of f(x) over (in) the interval (0, x).

Case 3: Suppose a = 1. Then (a,a +21) =

(-1.D, and formulas (1) t0 (3) become

e
wet f rode a
an

1
tno 4 [ seosin() ads for n= 1.2.3, us

With ay and au, ba, n= 1,2,:-+, given by formulas (13) to (15), the right hand side
of expression (5) is the Fourier expansion of f(x) over (in) the interval (-1, 0.

Case 4: Suppose a
(15), we get

Lam, Then (aya + 20) = (~n.m). Setting I = x in formulas (13) 0

4. 9. Fourier Series

an

a 2 [rca fra 123:

ed

Also, in this case, expression (5) becomes identical with expression (12); that is,

a9

With ao and au. br, given by formulas (16) to (18), the right hand side
of expression (19) isthe Fourier expansion of f(x) over (in) the interval (7,7).

Expansions of Even and Odd functions

“A function Jc) defined in the interval (4,1) is said to be (i) an even function if f(x) =
fix) and (ii) an odd function if fi-x) = FO for al x in the interval (-L.D. For example,
os (nz!Dx is an even function and sin (nn/D) is an odd function in the interval (4.0. Ie is
easily verified that (i) the products of two even functions or two odd functions are even, and
(ii the product of an even function and an odd function is odd

IF Ga is an even function in the interval (LD then f(x)c0s (ux/ Ds s an even function and
sin Ce am dá function in the interval. In this cas, formulas (13) 10 (15) become”

we} f roe e
i
2 ar
ae FS mereos(F) ses, forn = 1,23 en
a

, for n = 12.3, e»

“Thus for an even function f(x) defined in the interval (-1,D, ao and a, are given by for-
mulas (20) and (21) and by are all identically zero. Accordingly, the Fourier expansion of an
‘even function f(x) over the interval (-1,1) does not contain sine functions. This is tu in the
panicular case of = us well.

roda à joe po

eat f fonds =
2 o e

5. 9.1.1. Fourier Expansion over the interval (~7.) sa

If fix) is an odd function in the interval (1.1), then f(x) cos (nz/D) x isan odd function and
fi) sin (nx JD x is an even function in this interval. In this case, formulas (13) 10 (15) become
a = 0 es

0, for n = 1. as

as

j Hadsin 2) ede, or

‘Thus, for an odd function f(x) defined inthe interval (-1.D. ap and ay ar all zero, and by
are given by formula (25). Accordingly, the Fourier expansion of an odd function f(x) over the
interval (-1,D does not contain cosine functions. This is true in the particular case of! = mas
well
Important Note: It must be emphasised that formulas (20)-(22) and (23)-(25) are valid (for
even and odd functions respectively) only when the interval over which the Fourier expansion
is desired is ofthe form (4.0.

9.1.1 Fourier Expansion over the interval (-1,71)

Below we illustrate, through examples, the Fourier expansion of a function f(x) of period 2
over the interval (rx). In this case, expressions (16)-(19) of Secion 9.1 are employed. That
is. in this case, the Fourier expansion of f(x) is given by

10 B+ Zmcnneı San a
where loa o
£ f srcosmeds, forn = 1.2.3 o

ña)

4 [nosinnzan forn= 1,

If #44) is an even function in (-a,m) then by = 0 and expressions (2) and (3) become

2 fra 2
a= | fords ea
o

6. su 9. Fourier Series

2
[ro cosnx dx, form = 1,23, Ga

IF fx) is an odd function in (3,7). then ay = 0, a, = 0, and expression (4) becomes

by

2
2 [rs nxdx, forn = 1,2,3, a)
5

Po AA

Deduce that

of the function f(x) = x over the interval (1.1).

> Here, f(x) = fix), Hence the given f(x) is an odd function in the interval (1, 7).
‘Therefore, ao Oforn = 1,2,---, and the Fourier expansion of f(x) over (=. 7) is
Kor Iren m o
wee
Taking f(x) = x in (i), we get
522 fran más 2 St! ©
= dE incosmn +0] -i{
Using isin, a
FM sme A,

=
‘This is the Fourier expansion of f(x) = x over the interval (~,).
sng the method of Integration by pars
“The geveralized Rue for Integration by Partsreads: fur
where au, a successive derivatives of u, ad
‘This Role frequently employed in problems on Fourier Seren,

7. 9.1.1. Fourier Expansion over the interval (1. ) ss

Deduction
Taking x = 7/2 in the expansion (

Find the Fourier series for the function f(x) = = in the interval - € x & 7
Deduce the following:

Psi feos
oh © FE

> Forthe given function, f(x) = (47 = x = fa), Hence the given f(x) isan even function

in he interval (1.1. Therefore alo by = O nd the Fourier series for f(x) over (=)
en a
pue 6S oem o

Her.
à 2 [ro 2 f tare? Bl = w

x n e
Using (i) and (ii) in i, we get the Fourier series for f(x) = + over the intel

asisnas
im
Deductions
(a) Putting x =
w

8. 536 9. Fourier Series

(0) Taking x = 0 in iv), we get
e y cont
one or ao = 0
34 IE me
(©) Adding he results (4) and (vi) we get
la E
le E .

DERI Fie he rier series ere function f(x) = lat ré Er
Hence deduce that

» Hon 103 =12 2 2 JU goa ane tn te a
(rm). Therefore, all of b, = 0 and the Fourier series for f(x) = |x| for =x < x < xis
Wa PD men nx 0)
tn
2
2 ana fa en Oma weitesten

r 2
und ay =2 AS
3

A

Using these in (i), we obtain the required Fourier series as

pans
E

1,251
7 Ed m IP eos

Ir Wwersıco

a acta nnsinef TE

9. Fourier Expansion over the interval (~n,) 537

‘iy

Deduetion
Putting x = in the expansion

). we get

sos (2n = Dr
EM

or a .
Expand fs given by
fo -nexeo
fix
i for Oe nen,
where > Dis constant, ina Fourier series for=n < x en.
Deduce that byt >
Rise

> First, ve note that, forthe given /G0.
ak for =n (<0, ie forocier
a ER ui mere

=f)

Therefore, the given /(x) is an add function over (~77). Its Fouries series expansion is

10

dE f reina dem? fia dr

2 canst O rn
O EA

10. 538 9, Fourier Series

Using this in (i), we get

aaa
jive El Em aro Sse Sin

‘This isthe required expansion of the given (0)

Deduction
For x = 1/2, we have f(x) = & and expression ii) becomes
ak ok te
CEE EE 0

IE 0210 ic Fourier expansion of the faction fx defined by

Team. 1 <x<0

1-Qx/m, Osan

Law

Deduce that

DEA

> First, we note that, for the given f(x).

fon

1- um) for ESO, ie forOsase

Harn) for0s-nsm iefr-msıs0
=f

“Thus, the given f(x) isan even function in the given interval (=n, m). Accordingly, all of by
are zero, and

ports? ffr- Eee image aus

¿Je
e

sem Zi Fit co) = = CD

mi

q.

11. 9.1.1. Fourier Expansion over the interval (1.1) 39

Hence the Fourier expansion of the given f(x) over (mm) is

+ Dance EY cosy

fa

Deduction

For x = 0. we have f(x) = /(0) = 1, and the above expansion becomes

}

Ne (2

DRM oi he Fourier series expansion of fs) = sina, where a snot an integer
over he interval =.)

> First we observe that, for f(x) = sinax,

SEDE sina(-20 = ~sinax = -f(0.

‘Therefore, f(x) = sinax in as odd function in the interval (-,m). Therefore, all of a, = 0
and the Fourier series expansion of f(x) = sin ax over (x, 1) is

Gin asin mo de

(cos (a=mx=cos (a+ nx} de

sin a-mı _ sin e | A sen)

A nées munie
ATEN ses ar )

; i men
UT June anne EU ue

12. so 9. Fourier Series

Using this in (1). we get

This isthe required expansion
IE Evo 109 = TRG ma Founer series over he interval CR)

> Forthe given f(x), we note that
Sen T0 = Vicos = fin)

‘Therefore. the given f(x) is an even function in the interval (xx) and its Fourier expan-
sion over this interval is.

o

Gi

Fria en amin ud

12h | cosin= 1/20 [

CANTES A

cos (n+ BR = 1, cost = 12m 1
we 12)

13. 9-11. Fourier Expansion over the interval (-.m)

si
A 1 af
a (nei CONTI a)

‘Substituting for an and a, fom (i) and ii) in (i), we obtain the, required Fourier expansion
as

26 VIE cosnr
ee a
DERI) A in air series oer the mer
con = eos =f,

AG) isan even function and the Fourier series for fix) over (

nm

mis

Sia) = leon st= o

cn

A [E Joona}

2 din

J'istn + sets 113 Finsine De sear nar

Forn # 1, this yields

a. al (A any. fen, sen) |

dates Um o |, (Ron on
ee ee

le = 1Xcos nn /2) ~ (m + 1)costnm/2) cos (nr mn
* mt

Roe tht con + Vin = (conn con w/2) € (sin sin 2) =O
(6081) > On (0.4/2) and (cos 1) < On (472.0)

14. sa 9. Fourier Series

For n = 1, we get from (ii)

- 1 lb snare a] [
¿ona [host o
Using (ii), (v) and (iv) in (i), we get
coe
‘as the required expansion. *

o Evene 39 = sain sa fourier series in he intervl (n,n) Hence deduce
the following:

Tet 2,2

mes 2,2.

> First, we observe that, for f(x)

SED = (sin (2)

ME sin x) = sin x = f(x),

‘Therefore, f(x) = xsin xis an even function in the interval (1,1). Therefore, all of by
and the Fourier Series expasion of (x) over (=r, 1) is

ae
1093 Foco o
Here
2
2 [roue 2 finas Las d--smu=Fne2 Gi)
and e622 ro cosmede=2 f rsinscosm ds cio
3 ’
This gives

ain

ober

Feunzourerat fan a

(0)

15. 9.1.1. Fourier Expansion over the interval (~7,7) 543

Forn > 2, we obtain, from (ii),

NECE Lada De tn De)
met nat Creer ey

Le lo
O ce mal ft, E ul A
la en) ®

‘Using (i), Civ) and (v) in (i), we obtain the Fourier series for f(x) = xsinx as

mn
2 the eta) ils
fate
ri os eign
ifs
36-) + .

DR our ve rier expansion ofthe function 160 = ax + BE. where and 8

‘are non-zero constants, over the interval -n $ x 5%.

> Inthe given interval, the given f(x) is neither odd nor even. Therefore, the Fourier expansion
of the given f(x) is (See expression (1))

ape =D +P acotnes
a 2

basinnx o

We find that (See expressions (2)-(4))

ao L finas

16. 54 9. Fourier Series

for

“I O A

oi
«2 f rovosm

D as in Example 2 mi)

frowns thf ‘nal

It, as in Example 1. ww)

if [rico

“le

With the Fourier coefficients given by in), the required Fourier expansion (i reads
er
+43 cosme +20 SC ions .

MRR et the Fourier Series forthe función f(x) = à + 3° over the interval

1 exsn

ara PE

Hence deduce the following:
11

CEA

w

17. 9.1.1. Fourier Expansion over the interval (7,1) 545

For x = x, expression (i) reduces to.

2
mr ) Gi)
For x =, expression (i) reduces to
2: A
a+ a tn dv)
4
‘Adding (ii) and (iv), we get zu?
m
or E w
‘Adding (ii) and (v), we get
or Gi)
All the required results are thus deduced. .

Expand fix) = € as a Fourier series over the interval (=.
Hence deduce he Fourie expansions of, coshax and sinhax over the interval (2)

Also, prove the following:

eee ir 2ecy
© Fane ar (y ‘tes Fre
> Here, the given JG is neither even nor odd in the given interval (2.5). Its Fourier
coefficients are

{-acos nx + nsinnal], o

acosa + nina) - e(-acos nx — nsinne)]

5 = awash + bin a “asin b= boos
Rec fe cond da = SASHES DAME yg Fe nb de nu

18. 9. Fourier Series

acos nr

co AY ay
- (ét = em) = sinhar,
mat +n) en" Ka
and 2 [er anna | " -asinnx- ncosnallt, o

ra lO Ca sine = cos) = (asian = neo]
n 60m 2n(-1

AT (et = erat) N ah am
Maren 1 nee

Hence, the Fourier series expansion for f(x) = €

+ D ax cos ne Sy sin nx
2 a

ar a 7 aon

_Sinh ar | 2sinh ar & a(-1y" 2sinh ar & ny
ce He FS conn tS

2si

me 5 PM ecosme+ asians o

Deductions
Changing a to ~a in the result (i), we get the Fourier series expansion for 2% over the
interval (1,1) as

sishar _ 2sinhar & CI

ET Na pocos encino) 0)

From expressions (i) and (ii), we get the following Fourier expansions of coshax and
sinh ax over the interval (1,1)

cosh ax = ye 10%

sinhax 5(é 0%

19. 9.1.1. Fourier Expansion over the interval (5,1) 547

This yields
.
for -rex<0
Sfx) = i =
sinx fr O<xen,
1 sinx 2 = cos2mx
od IO tT Za
Deduce the following:
eee 1 1 1,1
m Basti 2 ®) 1.3 3:5"3.7
> For the given /(x), the Fourier coefficients are
A 2 4 1
m + [oa 1 fos fase) «econo 0)
= H +
1 (roc ncdc= À f snacoancen
4 [ nocos neax= 4 f sinxeo
E 3
if 3 ji
a fesse vo w

Form + 1, this gives

ac [pete De ODT, Lfi-emint ie 1m Im
le n= a

ft ly u I) qu eur)

1

0)

20. 548 9. Fourier Series

Further,

sin xsinnx dx

a

1
24 rasta puce td w

Forn # 1, this gives.

sin(n— Dx _ sin On + Dx

n+l 2 ®

Gin

fuse +I NN

ren 3 and (vi)

Wii)

is the first of he required results.
Deductions

(a) For x = 0, we have f(x) = 0. In this case, (vii) yields

1
OS

mi

21. 9.1.1. Fourier Expansion over the interval (-.m) 549

E CPL RE
"Lana 13 3857

as required,

Obtain the Fourier expansion of

m -n<x<0
re
x O<x<m

> Forthe given function, the Fourier coefficients are

ant frou-1|f one fu
un [roms | fonemanı froma
| œuf} wen.

and y= 4 fé 1 flaca + fran]

i

1-2-4]

dfn E

in)

al ns

| cosmo 1 cosnx | sinnx

! p
see

22. 550 9. Fourier Series

‘Therefore, the Fourier expansion of the given /() is

EN cs A ings

Deduction
From the definition of the given
© isa point of discontinuity of f(a), with 0"

(o). we noe that f(x) jumps from = to at x = 0. There-
and /(0*) = 0. Accordingly. a

or .
Obtain the Fourier series expansion of the function f(x) given below over
En.
[x in —n<x<0
Si) = 10 in O<x<n/2

ka in mexen

> For the given f(x), the Fourier coefficients over are

el ere
ALT}

(OC ete nr 0) =

3
0)

(See Noi in page 530.

23. 9.1.1. Fourier Expansion over the interval (a,

nn .

cn cate

=

ss

a)