Fourier Series ad their Applicatios The fuctios, cos x, si x, cos x, si x, are orthogoal over (, ). m cos mx cos xdx = m = m = = cos mx si xdx = for all m, { m si mx si xdx = m = I fact the fuctios satisfy these relatios over ay iterval (α, α + ). Assumig that f(x), defied ad itegrable i (, ), has a expasio. a + (a cos mx + b si x) uiformly coverget over (, ) f(x)dx = a f(x) cos xdx = a therefore a + ib = f(x) si xdx = b f(x)e ix dx These coefficiets exist irrespective of whether or ot the series coverges ad is equal to f(x), ad they are called the Fourier coefficiets. Sufficiet Coditios for covergece f(x) f(ξ) a) If f(x) is differetiable at ξ (or if m, such that x ξ < m, xɛ(ξ h, ξ + h)) the the fourier series coverges at ξ to f(ξ). b) If f(x) is mootoic i ξ < x < ξ + h ad i ξ h < x < ξ for some h >, the the fourier series coverges at ξ to the value {f(ξ ) + f(ξ + )}.
Geeral Rage The rage a x b is stadardised by substitutig X = ( ) x a+b ) ( b a the < X <. The series a + a cos X + b si X becomes ( ) ( ) x (a + b) x (a + b) a + a cos + b si b a b a Periodicity of f(x) We suppose that f(x) is represeted by the series (whe cgt.) for all x, hece sice the sum fuctio of the series is periodic, with period, we have f(x + ) = f(x) which defies f(x) outside the origial rage. Fourier Series for t ( < t < ) First cosider the idetity m + e ix = e(m+)ix e ix i si ) ( x x + i si m + = e(m+ )ix e ix ) x ( cos x i si x) = cos ( m + i si x Hece for x real (x, ±, ) takig real ad imagiary parts: m Re : + cos x = + si ( m + ) x si x or m + si ( m + ) x si x () cos x = Im : cot m x + si x = cos ( m + ) x si x () Itegrate () ad () from x to m (): ( x) si x = si ( m + ) t x si t dt (3) [ ( (): log si )] [ m ] t cos t + = cos ( m + ) t x x si t dt (4) x Now suppose δ x δ.
The usig the Riema Lebesgue theorem, we have, lettig m i (3) ad (4) ( x) = si x (5) log si x ( ) cos x + = Therefore log si x = cos x (6) Alterative Proof of (5) ad (6) ξ dt log( ξ) = t where we take a cut alog the positive real axis i the t-plae from to. The brach of log( ξ) chose is that which is real whe ξ is real, ad is oe-valued i the cut plae. I particular this vaishes at ξ = t = + t + + tm + tm t ξ dt m therefore t = ξ ξ + t m t dt where the path is take alog the radius ξ. DIAGRAM For all t o the radius through ξ t si θ Re(ξ) > otherwise Hece i all cases t si δ whe δ arg ξ δ ( < δ < ) ξ t m Therefore t dt = r (pe iθ ) m e iθ dp t r p m si δ dp = r m+ si δ m + r (m + ) si δ ξ m Hece lim t m dt t δ arg ξ δ r Whe r = the covergece is uiform with respect to θ. Hece we have ξ log( ξ) = Where the series coverges o ξ = except at ξ =, uiformly i δ arg ξ δ 3
log( ξ) = log ξ + i arg( ξ) = log ( si ) ( θ i θ ) Takig real ad imagiary parts gives θ = si θ log ( si ) θ cos θ = Covergece beig uiform i δ θ δ. < θ < Fourier Expasio of the Beroulli polyomials i t. Values of the Beroulli umbers. Put x = t i ( x) = si x Therefore t = si t si t = () < t < si t Therefore P (t) = P (t) = P (t) P () = Therefore P (t) = t P (s)ds The series () coverges uiformly i ɛ t ɛ t t si s P (s)ds = ɛ ɛ ds ɛ t ɛ cos ɛ cos t = () The series o the right coverges absolutely ad uiformly sice cos(t) () () ad coverges. t cos t Hece P (s)ds = t () (usig cotiuity) Hece P (t) = P cos t + () () 4
P = () = () S Next P 3(t) = P (t) P si t Therefore P 3 (t) = () 3 ad geerally we have P m (t) P m = ( ) m cos t m= () m P m+ (t) = ( ) m si t () m+ S m m= P m = ( ) m () m We also have φ m (t) = P m (t) m =, 3, m! B m (m)! = ( )m P m For k it ca be show that S k + (S k ) Therefore S k = + o(k) Also P m+ (t) ( ) m si t () m+ P m (t) ( ) m ( cos t) () m Fourier Series of the Square Wave We have ( x) = si x Write y = x y = si y ( ) si x x = ( ) si x Write f(x) = ( ) < x < < y < < x < 5
Graph of f(x) is show by solid lies. Graph of f(x + ) is show by broke lies. Graph of f(x) f(x + ) is show by dotted lies. The fourier series of f(x) f(x + ) is the si x ( ) ( ) si x ( ) si( + )x = 4 = + { 4 si( + )x + < x < = = + < x < Fid coefficiets by direct itegratio. Gibbs Pheomeo Write S m (x) = 4 m si( + )x = + d m 4 dx S si(m + )x m(x) = cos( + )x = = si x (m + ) This vaishes i o < x < at x = m + m + S m (x) is symmetrical about. Hece cosider the value of S m for < x <, ad i particular at x =, the first max. m ( + ) 4 S m+ si(m + )t m = dt m + si t s Put t = the we have m + 6
( ) 4 si = si sds m + (m + ) si s = ( ) sis m+ s φ s ds m + where φ(u) = u si u. s Now φ(u) φ(δ) u δ < ad m + m + ( ) ( ) So φ φ s m + m + si s i s s si s ( Hece s ds si s s φ s ( ) m + si s φ m + s ds ( ) Sice lim φ = we have m m + ( ) si s lim m s φ s si s ds = m + s ds ( ) Hece lim S m = m m + ) ds si s ds.79 > s Dirichlet s Formula (sufficiet coditios for covergece) Assume that f(x) is bouded ad itegrable over [, ], ad f(x + ) = f(x) Write S m (x) = m a + (a cos x + b si x) = [ m ] f(t) + (cos t cos x + si t si x) dt = [ m ] f(t) + cos (t x) dt = f(t) si ( ) m + (t x) si dt (t x) = = = x x f(x + s) si ( ) m + s si s ds f(x + s) si ( ) m + s si s ds by periodicity [f(x + t) + f(x t)] si ( ) m + t si t dt as si ( ) m + t si t is eve. 7
Sice m + cos x = si ( ) m + x si x, = si ( m + ) x si x dx Therefore [f(x + ) + f(x )] = [f(x + ) + f(x )] si ( m + si t Therefore S m (x) [f(x + ) + f(x )] = [f(x + t) f(x + )] si ( ) m + t si t dt + [f(x t) f(x )] si ( ) m + t si t dt () Whe f(x + ) = f(x ) = f(x) () becomes S m (x) f(x) = [f(x + t) + f(x t) f(x)] si ( ) m + t si t dt (a) The itegrals appearig i () ad (a) are all of the form b φ(t) si λtdt where a =, b =, λ = m + a f(x + t) f(x + ) f(x t) f(x ) φ(t) = si t, si t, or f(x + t) + f(x t) f(x) si t Hece if φ(t) is bouded ad itegrable over [, ], the by the Riema Lebesgue theorem, φ(t) si λt as λ. I fact i the above cases φ(t) is bouded ad itegrable over [h, ] h > ad so the covergece depeds oly o the behaviour of the fuctio i a sufficietly small iterval [, h]. Itegratio of a Fourier Series If f(x) is bouded ad itegrable i [, ] ad F (x) = ) t ( f(t) ) a dt Where a = f is the costat term i the Fourier series for f, the F (x) has a Fourier series, coverget everywhere to F (x), obtaied by itegratig the Fourier series for f(x) a term by term. [This holds eve if the Fourier series for f does ot coverge.] 8
F (x) is a absolutely cotiuous fuctio ad hece possesses a Fourier series covergig everywhere to F (x). F (x) = A + (A cos x + B si x) Assumig that f is cotiuous o (, ) esures the existece of F (x), ad A = F (x) cos xdx =,, = [ ] si x F (x) F (x) si xdx = ( f(x) ) a si xdx [F () = F () = ] = f(x) si xdx = b Similarly B = a Therefore F (x) = A + b cos x + a si x Puttig x = gives (a b ) = ( ) Therefore F (x) = = x a si x + b (( ) cos x) {a cos t + b si t}dt Differetiatio of a Fourier Series This is ot always valid. si x e.g. = ( x) x d si x = cos x which does ot coverge. dx Sufficiet Coditios If f(x) is cotiuous ad f (x) exists except at a fiite umber of poits, ad both f(x) ad f (x) have Fourier series which coverge, the the series for f (x) is obtaied by term by term differetiatio of the Fourier series for f(x). i.e. f(x) = a + a cos x + b si x 9
[f (x + ) + f (x )] = b cos x a si x [This is really just the same as the result for itegratio, with slightly weaker coditios.] Half-Rage Series Let f(x) be bouded ad itegrable i [, ] () Cosie Series defie f c (x) = { f(x) x f( x) x The f c (x) is a eve fuctio, which has a Fourier series i which b a = = () Sie Series defie f s (x) = f c (x) cos xdx f c (x) cos xdx = { f(x) < x < f( x) < x < If f() f s is discotiuous at. If f() f s is discotiuous at. f(x) cos xdx The f s (x) is a odd fuctio, ad has a Fourier series i which a b = = f s (x) si xdx f s (x) si xdx = f(x) si xdx Order of magitude of Fourier coefficiets a ib = f(x)e ix dx = c () Suppose f(x) ad all its derivatives are bouded ad cotiuous i (, α ), (α, α ) (α k, ) Write c m = f (m) (x)e ix dx
αr+ = k f (m) (x)e ix dx () α r Itegratig () by parts gives [ k c = c = f(x) ] αr+ αr+ i e ix f (x)e ix dx = i = i r= [ k α r + i f(α r + )e iαr f(α r+ )e iα r+ + r= α r [ f( + )e i f( )e i αr+ α r f (x)e ix dx ] k + {f(α r + ) f(α r )}e iαr + c () r= f( + ) = f( + ) by periodicity Therefore f( + )e i f( )e i = [f( + ) f( )]e i = [f(α k+ + ) f(α k+ )]e iα k+ Hece we have c () = c() i Write J (m) + i = k+ r= k+ r= {f(α r + ) f(α r )}e iαr {f (m) (α r + ) f (m) (α r )}e iαr Therefore c () = c() i + J () i Similarly c () = c() i + J () i Therefore c () = J () i + J () (i) + c() (i) Sice c () = f (x)e ix dx is bouded, if J () = for all m the c = c () is O ( ) as If also J () = for all m the c = c () is O ( ) as 3 I particular if f, f (), f (r) are cotiuous but f (r+) is ot cotiuous the c = O ( I fact J () ) as r+ vaishes oly if f is cotiuous for if we write k+ j r e iαr. f(α r + ) f(α r ) = j r the J () = If J () = for m the r= ]
k+ r= j r e iαr = = m, m +, Takig = m, m +,, m + k, we write e iαr = z r z m z m z m k+ j Therefore.. = z m+k z m+k zk+ m+k j k+ The determiat of the matrix is (z z z k+ ) m z z k+ = (z z z k+ ) m r z s ). z k zk+ k r>s(z z r z s for r s Therefore the determiat is o zero. Therefore j = j = = j k+ = Therefore f is cotiuous. Parseval s Theorem If f(x) is bouded ad itegrable i (, ) the (f(x)) dx = a + (a + b ). [Note that this is true eve though f(x) does ot equal the sum of its Fourier series.] If we assume that f(x) is cotiuous ad the Fourier series coverges to f(x), f (x) = [ ] f(x) a + (a cos x + b si x) dx Sice uiformity of covergece is ot affected by multiplyig by f(x) we ca itegrate term by term RHS= a f(x)dx + a f(x) cos xdx + b f(x) si xdx = a + a + b