FOURIER SERIES Math 21b, Spring 07Smooth functions f(x) on [−π, π] form a linear space X. There is an inner product in X defined byhf, gi =1πRπ−πf(x)g(x) dxIt allows to define angles, length, projections etc in the space X as we did in finite dimensions.THE FOURIER BASIS.THEOREM. The functions {cos(nx), sin(nx), 1/√2 } form an orthonormal bas is in X.Proof. To check linear independence a few integrals nee d to be computed. For all n, m ≥ 1, with n 6= m youhave to show:h1√2,1√2i = 1hcos(nx), cos(nx)i = 1, hcos(nx), cos(mx)i = 0hsin(nx), sin(nx)i = 1, hsin(nx), sin(mx)i = 0hsin(nx), cos(mx)i = 0hsin(nx), 1/√2i = 0hcos(nx), 1/√2i = 0To verify the above integrals in the homework, the fo llowing trigonometric identities are useful:2 co s(nx) co s(my) = cos(nx − my) + cos(nx + my)2 sin(nx) sin(my) = cos(nx − my) − cos(nx + my)2 sin(nx) cos(my) = sin(nx + my) + s in(nx − my)FOURIER COEFFICIENTS. The Fourier coe fficients of f are defined asa0= hf, 1/√2i =1πRπ−πf(x)/√2 dxan= hf, cos(nt)i =1πRπ−πf(x) cos(nx) dxbn= hf, sin(nt)i =1πRπ−πf(x) sin(nx) dxFOURIER SERIES. The Fourier representation of a smooth function f is the identityf(x) =a0√2+P∞k=1akcos(kx) +P∞k=1bksin(kx)We take it for granted that the series converges and that the identity holds for all x.ODD AND EVEN FUNCTIONS. Here is some advise which can save time when computing Fourier series:If f is odd: f(x) = −f(−x), then f has a sin series.If f is even: f(x) = f(−x), then f has a cos series.If you integrate an odd function over [−π, π] you get 0.The product of two odd functions is even, the product between an even and an odd function is odd.EXAMPLE 1. Let f (x) = x on [−π, π]. This is an odd function (f (−x)+f (x) = 0) so that it has a sin series: withbn=1πRπ−πx sin(nx) dx =−1π(x cos(nx)/n + sin(nx)/n2|π−π) = 2(−1)n+1/n, we get x =P∞n=12(−1)n+1nsin(nx).For exampleπ2= 2(11−13+15−17...)is a formula of Leibnitz.EXAMPLE 2. Let f (x) = cos(x) + 1/7 co s(5x). This trigonometric polynomial is already the Fourier series.The nonzero coefficients are a1= 1, a5= 1/7.EXAMPLE 3. Let f (x) = 1 o n [−π/2, π/ 2] and f(x) = 0 else. This is an even function f(−x) − f (x) = 0 sothat it has a cos series: with a0= 1/(√2), an=1πRπ/2−π/21 co s(nx) dx =sin(nx)πn|π/2−π/2=2(−1)mπ(2m+1)if n = 2m + 1 isodd and 0 e lse. So, the series isf(x) =12+2π(cos(x)1−cos(3x)3+cos(5x)5− ...)Remark. The function in Example 3 is not smooth but Fourier theory still works. What happened at thediscontinuity point π/2? The Fourier series gives 0. Diplomatically, it has chosen the point in the middle of thelimits from the right and the limit from the left.FOURIER APPROXIMATION. For a smooth function f, theFourier series of f converges to f . The Fourier coefficients arethe coordinates of f in the Fourier basis.The function fn(x) =Pnk=1aksin(kx) is called a Fourier ap-proximation o f f. The picture to the right shows an approxima-tion of a piecewise continuous even function in example 3).-3 -2 -1 1 2 3-0.20.20.40.60.811.2THE PARSEVAL EQUALITY. When e valuating the square of the length of f with the square of the length ofthe series, we get||f||2= a20+P∞k=1a2k+ b2k.EXAMPLE. We have seen in ex ample 1 that f (x) = x = 2(sin(x) − sin(2x)/2 + sin(3x)/3 − sin(4x)/4 + ...Because the Fourier coefficients are bk= 2(−1)k+1/k, we have 4(1 + 1 /4 + 1/9 + ...) =1πRπ−πx2dx = 2π2/3 andso11+14+19+116+125+ ... =π26Isn’t it fantastic that we can sum up the reciprocal squares? This formula has b e e n obtained already by Leonar dEuler, who celebrated his 300’th birthday two weeks ago, on April 15.HOMEWORK: (this homework is due either Wednesday or Thursday. Remember that there is a lab due onThursday or Friday).1. Verify tha t the functions cos(nx), sin(nx), 1/√2 form an orthonormal family.2. Find the Fourier series of the function f(x) = |x|.3. Find the Fourier series of the function cos2(x) + 5 sin(x) + 5. You may find the double angle formulacos2(x) =cos(2x)+12useful.4. Find the Fourier series of the function f(x) = |sin(x)|.5. In the previous problem 4) you should have gotten a seriesf(x) =2π−4πcos(2x)22− 1+cos(4x)42− 1+cos(6x)62− 1+ ...Use Parseval’s identity to find the value of1(22− 1)2+1(42− 1)2+1(62− 1)2+ ···HEAT AND WAVE EQUATION Math 21b, Spring 07FUNCTIONS OF TWO VARIABLES. We consider functions f(x, t) which are fo r fixed t a piecewise smoothfunction in x. Analogously as we studied the motion of a vector ~v(t), we are now interested in the motion of afunction f in time t. While the governing equation for a vector was an ordinary differential equatio n ˙x = Ax(ODE), the describing equation is now be a partial differential equation (PDE)˙f = T (f). The functionf(x, t) could denote the temperature of a stick at a position x at time t or the displacement of a stringat the position x at time t. The motion of these dynamical systems will be easy to describe in the orthonormalFourier basis 1/√2, sin(nx), cos(nx) treated in an earlier lecture.PARTIAL DERIVATIVES. We write fx(x, t) and ft(x, t) for the partial derivatives with resp e c t to x or t.The notation fxx(x, t) means that we differentiate twice with re spect to x.Example: for f(x, t) = cos(x + 4t2), we have• fx(x, t) = −sin(x + 4t2)• ft(x, t) = −8t sin(x + 4t2).• fxx(x, t) = −cos(x + 4t2).One also uses the notation∂f (x,y)∂xfor the partial derivative with respect to x. Tired of all the ”par tial derivativesigns”, we always write fx(x, t) for the partial derivative with respect to x and ft(x, t) for the partial derivativewith respect to t.PARTIAL DIFFERENTIAL EQUATIONS. A partial differential e quation is an eq uation for an unknownfunction f(x, t) in which different partial derivatives occur.• ft(x, t) + fx(x, t) = 0 with f (x, 0) = sin(x) has asolution f(x, t) = sin(x − t).• ftt(x, t) − fxx(x, t) = 0 with f(x, 0 ) = sin(x) andft(x, 0) = 0 has a s olution f (x, t) = (sin(x − t) +sin(x + t))/2.THE HEAT EQUATION. The temperature distribution f(x, t) in a metal bar [0, π] satisfies the heat equationft(x, t) = µfxx(x, t)This partial differential equation tells that the r ate of change of the tempera tur e at x is proportional to thesecond space derivative of f(x, t) at x. The function f (x, t) is assumed to be ze ro at both ends of the bar andf(x) = f (x, t) is a given initial temperature distribution. The constant µ depends on the
View Full Document
Unlocking...