In terms of real functions, this is = 1 π ∞  n=−∞ 1 − (−1) n ın (cos(nx) + ı sin(nx)) = 2 π ∞  n=1 1 − (−1) n ın sin(nx) sign(x) ∼ 4 π ∞  n=1 oddn 1 n sin(nx). 25.9 Least Squares Fit to a Function and Completeness Let {φ 1 , φ 2 , φ 3 , . . .} be a set of real, square integrable functions that are orthonormal with respect to the weighting function σ(x) on the interval [a, b]. That is, φ n |σ|φ m  = δ nm . Let f(x) be some square integrable function defined on the same interval. We would like to approximate the function f(x) with a finite orthonormal series. f(x) ≈ N  n=1 α n φ n (x) f(x) may or may not have a uniformly convergent expansion in the orthonormal functions. We would like to choose the α n so that we get the best possible approximation to f(x). The most common measure of how well a series approximates a function is the least squares measure. The error is d efined as the integral of the weighting function times the square of the deviation. E =  b a σ(x)  f(x) − N  n=1 α n φ n (x)  2 dx 1294 The “best” fit is found by choosing the α n that minimize E. Let c n be the Fourier coefficients of f(x). c n = φ n |σ|f we expand the integral for E. E(α) =  b a σ(x)  f(x) − N  n=1 α n φ n (x)  2 dx =  f − N  n=1 α n φ n     σ     f − N  n=1 α n φ n  = f|σ|f − 2  N  n=1 α n φ n     σ     f  +  N  n=1 α n φ n     σ     N  n=1 α n φ n  = f|σ|f − 2 N  n=1 α n φ n |σ|f + N  n=1 N  m=1 α n α m φ n |σ|φ m  = f|σ|f − 2 N  n=1 α n c n + N  n=1 α 2 n = f|σ|f + N  n=1 (α n − c n ) 2 − N  n=1 c 2 n Each term involving α n in non-negative and is minimized for α n = c n . The Fourier coefficients give the least squares approximation to a function. The least squares fit to f(x) is thus f(x) ≈ N  n=1 φ n |σ|fφ n (x). 1295 Result 25.9.1 If {φ 1 , φ 2 , φ 3 , . . .} is a set of real, square integrable functions that are orthog- onal with respect to σ(x) then the least squares fit of the first N orthogonal functions to the square integrable function f(x) is f(x) ≈ N  n=1 φ n |σ|f φ n |σ|φ n  φ n (x). If the set is orthonormal, this formula reduces to f(x) ≈ N  n=1 φ n |σ|fφ n (x). Since the error in the approximation E is a nonne gative number we can obtain on inequality on the sum of the squared coefficients. E = f|σ|f − N  n=1 c 2 n N  n=1 c 2 n ≤ f|σ|f This equation is known as Bessel’s Inequality. Since f|σ|f  is just a nonnegative number, independ ent of N, the sum  ∞ n=1 c 2 n is convergent and c n → 0 as n → ∞ Convergence in the Mean. If the error E goes to zero as N tends to infinity lim N→∞  b a σ(x)  f(x) − N  n=1 c n φ n (x)  2 dx = 0, 1296 then the sum converges in the mean to f(x) relative to the weighting function σ(x). This implies that lim N→∞  f|σ|f − N  n=1 c 2 n  = 0 ∞  n=1 c 2 n = f|σ|f. This is known as Parseval’s identity. Completeness. Consider a set of functions {φ 1 , φ 2 , φ 3 , . . .} that is orthogonal with respect to the weighting function σ(x). If every function f(x) that is square integrable with respect to σ(x) has an orthogonal series expansion f(x) ∼ ∞  n=1 c n φ n (x) that converges in the mean to f(x), then the set is complete. 25.10 Closure Relation Let {φ 1 , φ 2 , . . .} be an orthonormal, comple te set on the domain [a, b]. For any square integrable function f(x) we can write f(x) ∼ ∞  n=1 c n φ n (x). 1297 Here the c n are the generalized Fourier coefficients and the sum converges in the mean to f(x). Substituting the expression for the Fourier coefficients into the sum yields f(x) ∼ ∞  n=1 φ n |fφ n (x) = ∞  n=1   b a φ n (ξ)f(ξ) dξ  φ n (x). Since the sum is not necessarily uniformly convergent, we are not justified in exchanging the order of summation and integration. . . but what the heck, let’s do it anyway. =  b a  ∞  n=1 φ n (ξ)f(ξ)φ n (x)  dξ =  b a  ∞  n=1 φ n (ξ)φ n (x)  f(ξ) dξ The sum behaves like a Dirac delta function. Recall that δ(x − ξ) satisfies the equation f(x) =  b a δ(x − ξ)f(ξ) dξ for x ∈ (a, b). Thus we could say that the sum is a representation of δ(x −ξ). Note that a series representation of the delta function could not be convergent, hence the necessity of throwing caution to the wind when we interchanged the summation and integration in deriving the series. The closure relation for an orthonormal, complete set states ∞  n=1 φ n (x)φ n (ξ) ∼ δ(x − ξ). 1298 Alternatively, you can derive the closure relation by computing the generalized Fourier coefficients of the delta function. δ(x − ξ) ∼ ∞  n=1 c n φ n (x) c n = φ n |δ(x − ξ) =  b a φ n (x)δ(x − ξ) dx = φ n (ξ) δ(x − ξ) ∼ ∞  n=1 φ n (x)φ n (ξ) Result 25.10.1 If {φ 1 , φ 2 , . . .} is an orthogonal, complete set on the domain [a, b], then ∞  n=1 φ n (x)φ n (ξ) φ n  2 ∼ δ(x −ξ). If the set is orthonormal, then ∞  n=1 φ n (x)φ n (ξ) ∼ δ(x −ξ). Example 25.10.1 The integral of the Dirac delta function is the Heaviside function. On the interval x ∈ (−π, π)  x −π δ(t) dt = H(x) =  1 for 0 < x < π 0 for − π < x < 0. 1299 Consider the orthonormal, complete set {. . . , 1 √ 2π e −ıx , 1 √ 2π , 1 √ 2π e ıx , . . .} on the domain [−π, π]. The delta function has the series δ(t) ∼ ∞  n=−∞ 1 √ 2π e ınt 1 √ 2π e −ın0 = 1 2π ∞  n=−∞ e ınt . We will find the series expansion of the Heaviside function first by expanding directly and then by integrating the expansion for the delta function. Finding the series expansion of H(x) directly. The generalized Fourier coefficients of H(x) are c 0 =  π −π 1 √ 2π H(x) dx = 1 √ 2π  π 0 dx =  π 2 c n =  π −π 1 √ 2π e −ınx H(x) dx = 1 √ 2π  π 0 e −ınx dx = 1 − (−1) n ın √ 2π . 1300 Thus the Heaviside function has the expansion H(x) ∼  π 2 1 √ 2π + ∞  n=−∞ n=0 1 − (−1) n ın √ 2π 1 √ 2π e ınx = 1 2 + 1 π ∞  n=1 1 − (−1) n n sin(nx) H(x) ∼ 1 2 + 2 π ∞  n=1 oddn 1 n sin(nx). Integrating the series for δ(t).  x −π δ(t) dt ∼ 1 2π  x −π ∞  n=−∞ e ınt dt = 1 2π    (x + π) + ∞  n=−∞ n=0  1 in e ınt  x −π    = 1 2π    (x + π) + ∞  n=−∞ n=0 1 ın  e ınx −(−1) n     = x 2π + 1 2 + 1 2π ∞  n=1 1 ın  e ınx − e −ınx −(−1) n + (−1) n  = x 2π + 1 2 + 1 π ∞  n=1 1 n sin(nx) 1301 Expanding x 2π in the orthonormal set, x 2π ∼ ∞  n=−∞ c n 1 √ 2π e ınx . c 0 =  π −π 1 √ 2π x 2π dx = 0 c n =  π −π 1 √ 2π e −ınx x 2π dx = ı(−1) n n √ 2π x 2π ∼ ∞  n=−∞ n=0 ı(−1) n n √ 2π 1 √ 2π e ınx = − 1 π ∞  n=1 (−1) n sin(nx) Substituting the series for x 2π into the expression for the integral of the delta function,  x −π δ(t) dt ∼ 1 2 + 1 π ∞  n=1 1 − (−1) n n sin(nx)  x −π δ(t) dt ∼ 1 2 + 2 π ∞  n=1 oddn 1 n sin(nx). Thus we see that the series expans ions of the Heaviside function and the integral of the delta function are the same. 25.11 Linear Operators 1302 [...]... eigenfunction You might think that this formula is just a shade less than worthless When solving an eigenvalue problem you have to find the eigenvalues before you determine the eigenfunctions Thus this formula could not be used to compute the eigenvalues However, we can often use the formula to obtain information about the eigenvalues before we solve a problem Example 27 .4. 2 Consider the self-adjoint eigenvalue... λy, Bk [y] = 0 We denote the eigenvalues as λn and the eigenfunctions as φn for n ∈ Z+ For the moment we assume that λ = 0 is not an eigenvalue and that the eigenfunctions are real-valued We expand the function f (x) in a series of the eigenfunctions φn |f f (x) = fn φn (x), fn = φn 1323 We expand the inhomogeneous solution in a series of eigenfunctions and substitute it into the differential equation... and L∗ yields the two equations, 2p2 − p1 = p1 , p2 = p1 , p 2 − p1 + p0 = p0 p 2 = p1 Thus second order, formally self-adjoint operators with real-valued coefficient functions have the form L[y] = p2 y + p2 y + p0 y, which is equivalent to the form L[y] = d (py ) + qy dx Any linear differential equation of the form L[y] = y + p1 y + p0 y = f (x), where each pj is j times continuously differentiable and. .. ) + xy = 0 dx Result 27.2.1 If L = L∗ then the linear operator L is formally self-adjoint Second order formally self-adjoint operators have the form L[y] = d (py ) + qy dx Any differential equation of the form L[y] = y + p1 y + p0 y = f (x), where each pj is j times continuously differentiable and real-valued, can be written as a x formally self adjoint equation by multiplying the equation by the factor... = λ and λ is real 1319 Orthogonal Eigenfunctions The eigenfunctions corresponding to distinct eigenvalues are orthogonal Let λn and λm be distinct eigenvalues with the eigenfunctions φn and φm Using Green’s formula, φn |L[φm ] − L[φn ]|φm = 0 φn |λm φm − λn φn |φm = 0 (λm − λn ) φn |φm = 0 Since the eigenvalues are real, (λm − λn ) φn |φm = 0 Since the two eigenvalues are distinct, φn |φm = 0 and. .. the nth order formally self-adjoint equation L[y] = 0, on the domain a ≤ x ≤ b subject to the boundary conditions, Bj [y] = 0 for j = 1, , n where the boundary conditions can be written n αjk y (k−1) (a) + βjk y (k−1) (b) = 0 Bj [y] = k=1 If the boundary conditions are such that Green’s formula reduces to v|L[u] − L[v]|u = 0 then the problem is self-adjoint Example 27.3.1 Consider the formally self-adjoint... + (−1)n−1 n−1 (pn−1 y) + · · · + p0 y dxn dx If each of the pk is k times continuously differentiable and u and v are n times continuously differentiable on some interval, then on that interval Lagrange’s identity states L∗ [y] = (−1)n vL[u] − uL∗ [v] = 13 14 d B[u, v] dx where B[u, v] is the bilinear form n (−1)j u(k) (pm v)(j) B[u, v] = m=1 j+k=m−1 j≥0,k≥0 If L is a second order operator then vL[u]... (x)} is the set and all the jν are distinct, then a1 φj1 (x) + a2 φj2 (x) + · · · + an φjn (x) = 0 on a ≤ x ≤ b is true iff: a1 = a2 = · · · = an = 0 2 Show that the complex functions φk (x) ≡ eıkπx/L , k = 0, 1, 2, are orthogonal in the sense that 0, for n = k Here φ∗ (x) is the complex conjugate of φn (x) n Hint, Solution 1303 L −L φk (x)φ∗ (x) dx = n 25.13 Hints Hint 25.1 13 04 25. 14 Solutions Solution... + an φjn (x) = 0 n ak φjk (x) = 0 k=1 We take the inner product with φjν for any ν = 1, , n ( φ, ψ ≡ n ak φjk , φjν =0 k=1 We interchange the order of summation and integration n ak φjk , φjν = 0 k=1 φjk φjν = 0 for j = ν aν φjν φjν = 0 φjν φjν = 0 aν = 0 Thus we see that a1 = a2 = · · · = an = 0 1305 b a φ(x)ψ ∗ (x) dx.) 2 For k = n, φk , φn = 0 L φk (x)φ∗ (x) dx n φk , φn ≡ −L L eıkπx/L e−ınπx/L... Integrating Lagrange’s identity on its interval of validity gives us Green’s formula b vL[u] − uL∗ [v] dx = v|L[u] − L∗ [v]|u = B[u, v] a 27.2 x=b − B[u, v] x=a Formally Self-Adjoint Operators Example 27.2.1 The linear operator L[y] = x2 y + 2xy + 3y has the adjoint operator d2 2 d L [y] = 2 (x y) − (2xy) + 3y dx dx = x2 y + 4xy + 2y − 2xy − 2y + 3y = x2 y + 2xy + 3y ∗ In Example 27.2.1, the adjoint . − N  n=1 α n φ n (x)  2 dx 12 94 The “best” fit is found by choosing the α n that minimize E. Let c n be the Fourier coefficients of f(x). c n = φ n |σ|f we expand the integral for E. E(α) =  b a σ(x)  f(x). ∼ ∞  n=−∞ 1 √ 2π e ınt 1 √ 2π e −ın0 = 1 2π ∞  n=−∞ e ınt . We will find the series expansion of the Heaviside function first by expanding directly and then by integrating the expansion for the delta function. Finding the series expansion of H(x) directly sense that  L −L φ k (x)φ ∗ n (x) dx = 0, for n = k. Here φ ∗ n (x) is the complex conjugate of φ n (x). Hint, Solution 1303 25.13 Hints Hint 25.1 13 04 25. 14 Solutions Solution 25.1 1. a 1 φ j 1 (x)