Topics i Fourier Aaysis-I 1 M.T.Nair Departmet of Mathematics, IIT Madras Cotets 1 Fourier Series 1.1 Motivatio through heat equatio.............................. 1. Fourier Series of -Periodic fuctios........................... 5 1.3 Fourier Series for Eve ad Odd Fuctios........................ 8 1.4 Sie ad Cosie Series Expasios............................. 11 1.5 Fourier Series of -Periodic Fuctios........................... 13 1.6 Fourier Series o Arbitrary Itervas............................ 14 1.7 Exercises........................................... 15 1 Lectures for the course MA 68, Juy-November 14.
1 Fourier Series 1.1 Motivatio through heat equatio The cosideratio of Fourier Series ca be traced back to the situatio which Fourier ecoutered i the begiig of ast cetury whie sovig heat equatio: Cosider a thi metaic wire of egth. Suppose a iitia temperature is suppied to it, ad suppose the temperature at both the ed poits kept at. The oe woud ike to kow the temperature at each poit of the strig at a particuar time. Let us represet the strig as a iterva [, ]. Let u(x, t) be the temperature at the poit x [, ] at time t. It is kow that u(, ) satisfies the partia equatio u t = c u, < x <, (1.1) t where c > is the heat coductivity of the materia. Sice the temperature at both the ed poits kept at, we have u(, t) = = u(, t), t >. (1.) Let the iitia temperature at the poit x be f()x), i.e., u(x, ) = f(x), x. (1.3) Exercise 1.1. Equatio (1.1) satisfyig (1.) ad (1.3) caot have more tha oe soutio. I order to fid u(x, t), we use a procedure caed method of separatio of variabes. I this method, we assume first that u(x, t) is of the form: u(x, t) = φ(x)ψ(t). The we have Hece, from (1.1), Hece, Hece, u t = φ(x)ψ (t), u t = φ (x)ψ(t). φ(x)ψ (t) = c φ (x)ψ(t). ψ (t) c ψ(t) = φ (x) = K, cost. φ(x) ψ (t) = Kc ψ(t), φ (x) = Kφ(x). (1.4) Let us cosider differet cases: Case(i): K = : I this case, φ (x) = so that φ is of the form φ(x) = ax + b.
By (1.), φ()ψ(t) = = φ()ψ(t). bψ(t) = = (a + b)ψ(t). Thus, we arrive at either ψ = or φ =. Case(ii): K > : I this case, K = α for some α. The we have φ (x) α φ(x) = so that φ is of the form φ(x) = ae αx + be αx. Agai, by (1.), φ()ψ(t) = = φ()ψ(t) so that (a + b)ψ(t) = = (ae α + be α )ψ(t). This agai ead to u(x, t) =. Case(iii): K < : I this case, K = α for some α. The we have φ (x) + α φ(x) = so that φ is of the form φ(x) = a cos αx + b si αx. By (1.), φ()ψ(t) = = φ()ψ(t) so that if ψ, we obtai a = ad b si α =. Assumig b (otherwise φ = ), we have α =, Z. Thus, α / : Z}. Now, from (1.4), with α / : Z}. Hece, ψ (t) = α c ψ(t) ψ(t) = ae α c t, ad hece, u is of the form u(x, t) = ae α c t si αx, α / : Z}. Thus, for each Z, u (x, t) = a e λ ct si λ x, λ :=, with a R satisfies (1.1) ad (1.). But, this u eed ot satisfy (1.3), uess f(x) = a si(x/) for some Z. If f is of the form for some k N, the we see that k f(x) = a si(x/) (1.5) k u(x, t) := a e λ ct si(x/) 3
satisfies (1.1), (1.) ad (1.3). What ca we say if f is, i some sese, arbitrary? The cosideratio of the fuctios of the form i (1.5) suggests the foowig query: If f is of the form f(x) = a si(x/), ca we say that u(x, t) := a e λ ct si(x/) is a soutio of (1.1) satisfyig (1.) ad (1.3) with appropriate otio of covergece? As a first step, et us assume that f is of the form f(x) = a si( x ). (1.6) Assume further that, term by term itegratio of the above series is possibe. The, we have f(x) si( mx )dx = a si( x ) si(mx )dx. Sice ad we obtai si( x ) si(mx )dx = for m, si ( x )dx = a = 1 cos ( x Note that, if f has the form as i (1.6), the f is -periodic, i.e., ad f is a odd fuctio, i.e., If f is as i (1.6), we we may defie u(x, t) := ) dx =, f(x) si( x )dx. (1.7) f(x + ) = f(x) for a x R f( x) = f(x) for a x R. a e λ ct si(x/), λ := /. Assumig that the above series is coverget ad ca be differetiated term by term, we see that u(, ) is a soutio of (1.1) satisfyig (1.) ad (1.3). Exercise 1.. Show that each λ := / is a eigevaue of the operator eigevector si λ x. d dx wit correspodig 4
1. Fourier Series of -Periodic fuctios I the ast sectio, we assumed that the fuctio f ca be represeted as f(x) = If =, the the above series takes the form a si( mx ). f(x) = A si x, A := f(x) si xdx. The above series is a specia case of the Fourier series that we are goig to itroduce. Let us cosider a few defiitios. Defiitio 1.3. A fuctio of the form k c + (a cos x + b si x). where c, a, b R, is caed a trigoometric poyomia, ad a series of the form c + (a cos x + b si x) with c, a, b R is caed a trigoometric series. Note that a trigoometric poyomia is a specia case of a trigoometric series. We observe that trigoometric poyomias are -periodic o R, i.e., if f(x) is a trigoometric poyomia, the f(x + ) = f(x) x R. From this, we ca ifer that, if the trigoometric series c + (a cos x + b si x) coverges at a poit x R, the it has to coverge at x + as we; ad hece at x + for a itegers. This shows that we ca restrict the discussio of covergece of a trigoometric series to a iterva of egth. Hece, we caot expect to have a trigoometric series expasio for a fuctio f : R R if it is ot a -periodic fuctio. We kow that a coverget trigoometric series is -periodic. What about the coverse? Suppose that f is a -periodic fuctio. Is it possibe to represet f as a trigoometric series? 5
Suppose, for a momet, that we ca write f(x) = c + (a cos x + b si x) for a x R. The what shoud be the coefficiets c, a, b? To aswer this questio, et us further assume that f is itegrabe o [, ] ad the series ca be itegrated term by term. For istace if the above series is uiformy coverget to f i [, ], the term by term itegratio is possibe. By Weierstrass test,we have the foowig resut: If = ( a + b ) coverges, the c + (a cos x + b si x) is a domiated series o R ad hece it is uiformy coverget. For, m N }, we observe the foowig orthogoaity reatios: cos x cos mxdx = si x si mxdx = cos x si mxdx =., if m, if = m,, if = m =,, if m, if = m, Thus, uder the assumptio that f is itegrabe o [, ] ad the series ca be itegrated term by term, we obtai a = 1 c = 1 f(x)dx, f(x) cos xdx, b = 1 f(x) si xdx. Defiitio 1.4. The Fourier series of a -periodic fuctio f is the trigoometric series where a = 1 a (a cos x + b si x), f(x) cos xdx ad b = 1 f(x) si xdx ad this fact is writte as f(x) a (a cos x + b si x). The umbers a ad b are caed the Fourier coefficiets of f. 6
If f is a trigoometric poyomia, the its Fourier series is itsef. Writig cos x = 1 [eix + e ix ], si x = 1 i [eix e ix ], we have Thus, writig we have a cos x + b si x = a [eix + e ix ] + b i [eix e ix ] ( a = + b ) ( e ix a + i b ) e ix. i a + c := a + b i, c := a b i, (a cos x + b si x) = Z c e ix. Now, suppose f(x) = Z c e ix with c C, ad this series ca be itegrated term by term. The, we have But, Hece, f(x)e imx dx = Z e i( m) dx = c m, i.e., e i( m) dx = c e i( m)x dx. if m = m, if m m. c = 1 f(x)e ix dx, Z. The foowig theorem show that there is a arge cass of fuctios which ca be represeted by their Fourier series (see Bhatia [1]). We sha come back to this theorem at a ater stage. THEOREM 1.5. (Dirichet s theorem) Suppose f : R R is a -periodic fuctio which is piecewise differetiabe o (, ). The the Fourier series of f coverges, ad the imit fuctio f(x) is give by f(x) = f(x) if f is cotiuous at x, 1 [f(x ) + f(x+)] if f is ot cotiuous at x. I Theorem 1.5 we used the termioogy piecewise differetiabe as per the foowig defiitio. Defiitio 1.6. A fuctio f : [a, b] R is said to be piecewise differetiabe if f exists ad is piecewise cotiuous o [a, b] except possiby at a fiite umber of poits. 7
Remark 1.7. It is kow that there are cotiuous fuctios f defied o [, ] whose Fourier series does ot coverge poitwise to f. Its proof reies o UBP (see []). We sha cosider this at a ater occasio. Athough each term ad the partia sums of a Fourier series are ifiitey differetiabe, the sum fuctio eed ot be eve cotiuous at certai poits. This fact is iustrated by the foowig exampe., x, Exampe 1.8. Let f(x) = By Dirichet s theorem (Theorem 1.5), the Fourier 1, < x. series of f coverges to f(x) for every x, ad at the poit, the series coverges to 1/. Note that ad for N, b = 1 Thus, Fourier series of f is I particuar, for x = /, a = 1 cos xdx = 1, =,,, si xdx = 1 [ ] 1 cos = 1 [ ] 1 ( 1) = 1 = 1 + = 1 + which eads to the Madhava Niakaṅtha series = si[( + 1)/] ( + 1) si( + 1)x. ( + 1) = 1 + =, ( 1) ( + 1) odd,, eve. 4 = ( 1) ( + 1). = 1.3 Fourier Series for Eve ad Odd Fuctios The foowig ca be verified easiy: Suppose f is a eve fuctio, i.e., f( x) = f(x) x X. The f(x) cos x is a eve fuctio ad f(x) si x is a odd fuctio, so that a = 1 f(x) cos xdx = b = 1 f(x) si xdx. f(x) cos xdx, 8
Suppose f is a odd fuctio, i.e., f( x) = f(x) x X. The f(x) cos x is a odd fuctio ad f(x) si x is a eve fuctio, so that a = 1 f(x) cos xdx =, b = 1 f(x) si xdx = Thus, we have the foowig: f(x) si xdx. (1) Suppose f is a eve fuctio. The the Fourier series of f is I particuar, respectivey. a a cos x with a := f(x) cos xdx. f() = a a, f() = a ( 1) a, = () Suppose f is a odd fuctio. The the Fourier series of f is b si x with b := I particuar, f(/) = ( 1) b +1. Exampe 1.9. Cosider the fuctio f defied by = f(x) si xdx, f(x) = x, x [, ]. I this case, f is a eve fuctio. Hece, the Fourier series is with a a cos x, x [, ] a = x dx = ad for = 1,,..., a = x cos xdx = [ si x x = [ cos x ] = [ ( 1) ] 1 ] } si x dx 9
Thus, so that a =, a +1 = x 4 = Takig x = (usig Dirichet s theorem), we obtai Exampe 1.1. Let f(x) = x, series is with b = Thus the Fourier series is = 4, = 1,,... ( + 1) cos( + 1)x ( + 1), x [, ]. 8 = 1 ( + 1). = x [, ]. I this case, f is a odd fuctio. Hece, the Fourier b si x, x [, ] x si x dx = cos } [ x = ( 1)+1. ( 1) +1 si x. cos x ] + } cos x dx I particuar (usig Dirichet s theorem), with x = / we obtai the Madhava-Nīakaṅtha series 4 = ( 1) +1 si = = ( 1) + 1. Exampe 1.11. Let f(x) = Fourier series is with Thus b = 1, x <, 1, x. b si x, I this case, f is a odd fuctio. Hece, the si x dx = (1 cos ) = [1 ( 1) ]. f(x) 4 = si( + 1)x. + 1 Takig x = /, agai we obtai the Madhava-Nīakaṅtha series 4 = ( 1) + 1. = 1
Exampe 1.1. Let f(x) = x, a x [, ]. Sice f is a eve fuctio, its Fourier series is a cos x, x [, ], a = It ca be see that a = /3, ad a = ( 1) 4/. Thus x 3 + 4 ( 1) cos x, x [, ]. Takig x = ad x = (usig Dirichet s theorem), we have respectivey. 1 = ( 1) +1, 6 = 1 x cos x dx. 1.4 Sie ad Cosie Series Expasios Suppose a fuctio f is defied o [, ]. By extedig it to [, ] so that the exteded fuctio is a odd fuctio, we obtai Forier sie series of f, ad by extedig it to [, ] so that the exteded fuctio is a eve fuctio, we obtai Fourier cosie series of f. The odd extesio ad eve extesio of f, deoted by f odd ad f eve are defied by f odd (x) = f(x) if x <, f( x) if x <,, f eve (x) = f(x) if x <, f( x) if x <, respectivey. Therefore, f(x) = f odd (x) b si x, x [, ] ad f(x) = f eve (x) a a cos x, x [, ] with a = f(x) cos x dx, b = f(x) si x dx. Exampe 1.13. Let f(x) = x, x [, ]. The eve extesio of f is itsef. Its odd extesio is: x, if x <, f odd (x) =, x, if x <. Hece, f(x) = f odd (x) b si x, x [, ], 11
with b = Note that [ cos x x [ cos x si x x dx = x Thus, x si x dx = [ cos x x ] ] cos = ] + = ( 1)+1, } cos x x dx. si x [ cos x ] dx = = [ ( 1) 1 b = [ ( 1)+1 ( 1) ]} 1 + = ( 1)+1 + 4 [ ( 1) ] 1. ]. Exampe 1.14. Let f(x) = x, x [, ]. Its odd extesio is itsef, ad f eve (x) = x, x [, ]. From Exampes 1.1 ad 1.9, we obtai ( 1) +1 x si x, x [, ] ad x 4 Exampe 1.15. Let us fid the sie series expasio of the fuctio, if x < /, f(x) = 1, if / x <. The sie series of f is give by where Note that b 1 = b = / f(x) = cos( + 1)x ( + 1), x [, ]. b si x, si x dx = [ cos x ] ( 1) ad b = [( 1) 1] = x [, ], / = [ cos / cos if odd, if eve. ]. Thus, for x [, ], we have si x si x si 3x si 5x si(4 3)x f(x) + + + + 1 1 3 5 4 3 si(4 )x si(4 1)x si(4 + 1)x + + +. 4 4 1 4 + 1 1
1.5 Fourier Series of -Periodic Fuctios Suppose f is a T -periodic fuctio. We may write T =. The we may cosider the chage of variabe t = x/ so that the fuctio f(x) := f(t/), as a fuctio of t is -periodic. Hece, its Fourier series is a (a cos t + b si t) where I particuar, a = 1 b = 1 ( t f ( t f ) cos tdt = 1 ) si tdt = 1 f eve = b = ad a = f odd = a = ad b = f(x) cos x dx, f(x) si x dx. f(x) cos x dx, f(x) si x dx. Exampe 1.16. Let f(x) = 1 x, 1 x 1. Takig = 1, we have ad Now, Hece, Thus, 1 a = 1 x cos x dx = a = 1 1 (1 x ) cos x dx = b = 1 1 1 (1 x ) si x dx =. (1 x ) cos x dx 1 [ si x cos x dx = =, [ ] 1 si x 1 si x [ cos x x dx = ] 1 (1 x ) cos x dx = [1 ( 1) ] = f(x) = 4 cos x. + 1 ] 1 = ( 1) 1., eve, 4/, odd. 13
1.6 Fourier Series o Arbitrary Itervas Suppose a fuctio f is defied i a iterva [a, b]. We ca obtai Fourier expasio of it o [a, b] as foows: Method 1: Let us cosider a chage of variabe as y = x a+b. Let ϕ(y) := f(x) = f(y + a + b ), where y with = (b a)/. We ca exted ϕ as a -periodic fuctio ad obtai its Fourier series as where Thus, where a = 1 a = 1 ϕ(y) a (a cos y + b si ) y ϕ(y) cos y dy, b = 1 ϕ(y) si y dy. f(x) a (a cos y + b si ) y with = (b a)/ ad y = x a+b. f(x) cos y dx, b = 1 f(x) si y dx Method : Cosiderig the chage of variabe as y = x a ad := b a, we defie ϕ(y) := f(x) = f(y + a) where y <. We ca exted ϕ as a -periodic fuctio i ay maer ad obtai its Fourier series. Here are two specific cases: (a) For y [, ], defie f e (y) = ϕf( y). Thus f e o [, ] is a eve fuctio. I this case, where = (b a)/ ad ϕ(y) a a cos y a = 1 ϕ(y) cos y dy. (b) For y [, ], defie f o (y) = ϕ( y). Thus f o o [, ] is a odd fuctio. I this case, where ϕ(y) b = 1 b si y ϕ(y) si y dy. From the series of ϕ we ca recover the correspodig series of f o [a, b] by writig y = x a. 14
1.7 Exercises The foowig are take from [3]. 1. Fid the Fourier series of the - period fuctio f such that: 1, (a) f(x) = x <, < x < 3. (b) f(x) = (c) f(x) = x, x, x < < x < 3. 1 + x, x 1 x, x. (d) f(x) = x 4, x.. Usig the Fourier series i Exercise 1, fid the sum of the foowig series: (a) 1 1 3 + 1 5 1 7 +..., (b) 1 + 1 4 + 1 9 + 1 16 +.... (c) 1 1 4 + 1 9 1 16 +..., (d) 1 + 1 3 + 1 5 + 1 7 +.... si x, x 4 3. If f(x) = cos x, 4 x < f(x) 8 cos 4, the show that [ si x 1.3 + si 3x 5.7 ] si 1x + 9.11 +.... 4. Show that for < x < 1, x x = 8 [ ] si x si 3x si 5x 1 3 + 3 3 + 5 3 +.... 5. Show that for < x <, 6. Show that for < x <, ad fid the sum of the series si x + si 3x 3 + si 5x 5 +... = 4. x si x = 1 1 cos x cos x + cos 3x cos 4x +..., 1.3.4 3.5 1 1.3 1 3.5 + 1 5.7 1 7.9 +.... 7. Show that for x, [ ] x( x) = cos x 6 cos 4x cos 6x 1 + + 3 +..., x( x) = 8 [ ] si x si 3x si 5x 1 3 + 3 3 + 5 3 +.... 15
8. Assumig that the Fourier series of f coverges uiformy o [, ), show that 9. Usig Exercises 7 ad 8 show that 1 (a) 4 = 4 9, (b) ( 1) 1 (c) 1 6 = 6 945 (d) 1 [f(x)] dx = a (a + b ). = 1 ( 1) 1 ( 1) 3 = 3 3 1. Write dow the Fourier series of f(x) = x for x [1, ) so that it coverges to 1/ at x = 1. Refereces [1] R. Bhatia, Fourier Series, TRIM series, [] Nair, M. T. () (Fourth Prit: 14): Fuctioa Aaysis: A First Course, New Dehi: Pritice-Ha of Idia. [3] Nair, M. T. (14): Cacuus of Oe Variabe, Ae Books, Pvt Ltd. [4] M.T. Nair, Measure ad Itegratio, Notes for the MSc. Course, Ja-may, 14. [5] R. Radha ad S. Thagaveu (13): Fourier Aaysis, NPTEL Notes, IIT Madras. 16