Infinite Series. 1 Introduction. 2 General discussion on convergence

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1 Infinite Series 1 Introduction I will only cover a few topics in this lecture, choosing to discuss those which I have used over the years. The text covers substantially more material and is available for reference if you need this information later in your professional practice of physics. Become familiar with what is included in Chapter 5, however I expect you to know in detail only that information covered in this written lecture and presented in class. A useful way of representing a function is by an infinite series, particularly when a numerical value of the function is required. We have previously seen that a series a function can be obtained using the Laurant expansion about a non-singular point. Thus; f(z) = a n = 1 2π i a n (z z ) n dz f(z) (z z ) n In the case that f(z) is analytic, the evaluation of the integral results in a Taylor expansion of the function, f(z), about the point, z. In the lecture, we mostly, but not exclusively, use real values of z. Therefore, within the radius of convergence (distance from z to the first singular point) the series converges. In subsequent sections, we look at tests to determine convergence. Remember as stressed previously, the function may exist whether on not the series representation converges. 2 General discussion on convergence Suppose a function is represented by a series; f(x) = a n x n The values of the expansion coefficients can be obtained by integration about a point x. This yields the terms; f(x) = f(x ) + (x x ) f (x + (x x ) 2 2! R n represents the n th term. R n = x a x a f (n) (x) d n x f (x ) + + (x x ) n 1 (n 1)! + R n + 1

2 Using the mean value theorem; R n = (x x ) n n! f (n) (η) a η x When lim n R n = then the Taylor series results. f(x) = (x x ) n n! f (n) (x ) 3 Example We represent the function f(x) = (1 + x) m by series. Insert in the Taylor expansion in the last section with expansion about x =. m(m 1) f(x) = 1 + mx + + 2! ( ) f(x) = m! n!(m n)! xn = M m x n n n ( ) m In the above is the binomial coefficient. We note that this series converges for all x, n as indeed it should since it is everywhere analytic. Of course the series terminates after m terms. In general a series does not terminate and converges only up to, but not including, the radius of convergence. As another example we find the series representation of ln(1 + x) expanded about x =. The radius of convergence is x < 1. f(x) = ln(1) + (x 1) dx d (x 1)2 ln(1 + x) + 2! f(x) = x x 2 /2 + x 3 /3 x 4 /4 d 2 dx 2 ln(1 + x) + 4 Types of series Unless the series is absolutely convergent (to be defined below), it must be handled carefully to determine if it converges, and is therefore a valid representation of a function. Consider the alternating series; S = ( 1) n =

3 The result of the sum is 1 or depending on how the series is summed. To explore this further, expand the function f(x) = x about x =. This is; f(x) = 1 x + x = ( x) n Now the series does not converge if x = 1 as there is a singularity at x = 1, however the value of the function is 1/2. Another example is the geometric series; S N = N = 1 + r + r r n This series can be evaluated as follows; rs N = N S N = 1 rn+1 1 r In the limit as N if r < 1; S = 1 1 r r n+1 = r + r 2 + r r N+1 The singular point at r = 1 is obvious. A power series should be evaluated to determine its convergence. The series above is convergent inside the unit circle in the complex plane. However, we know that we can use a Laurant expansion, say about the point x = 3, to obtain a series representation for x > 1. f(x) = 1 1 (z 3) = (1/2) + (Z 3)/4 (Z 3)2 /8 + (z 3)n The general term is a n = 2 n+1 and this can be shown to converge by the ratio test described in a next section. The radius of convergence for this expansion is a circle of radius 2 about the point z = 3 5 Tests for convergence A description of natural phenomenon must remain finite. However, the mathematical representation of models can diverge indicating that these models, or their representations, are not correct, at least within certain parameter spaces. Thus for series representations we need tests to determine if the series, S = a n, leads to finite results. A series converges if; 3

4 S N a n < ǫ In the above, ǫ is any small number and we find an N such that the sum over N terms approaches the value of the function, S. However, this definition is not useful for testing series convergence. 5.1 Comparison test If term by term a series is less than those of a series known to converge, then the series converges. However, if term by term a series is larger than any series which converges, then the series does not converge. 5.2 Ratio Test If a term of the series, a n, is the n th element of a sum, then; lim n a < 1 Convergent n+1 a n = > 1 Divergent = 1 Indeterminant Consider the harmonic series, 1/n. R = n n + 1 The ratio has the value of 1 as limn. Thus the test fails. In this case the series is not convergent. 5.3 Integral test Suppose f(x) is a continuously decreasing function. We replace f(n) = a n. Then; Thus; S N = N 1 n=1 f(n) = N 1 a n n=1 dxf(x) < a n < 1 dxf(x) + a 1 This is shown in Figure 1 where the area of the 2 histograms bounds the area under the function. 4

5 f(x) f( 1 ) f( 2) f( 3 ) f( 4 ) f( 5 ) f( 6 ) x Figure 1: A geometric figure illustrating the intergral trest for convergence of a series As an example, consider the Riemann Zeta function. This series is defined by; ζ(p) = n p n=1 Define f(x) = x p 1 dxx p = ( x p+1 p + 1 p 1 ln(x) p = 1 ) This series converges for p < 1, diverges logarithmically if p = 1 and diverges if p > Raabe s test a n If a n > and the ratio R = n[ a 1] for all n > N then a n converges if R > 1 and n+1 diverges if R < Gauss test If a n > for all n and; a n a n+1 = 1 + h/n + B(n)/n 2 5

6 where B(n) is bounded for n. Then a n converges if h > 1 and diverges for h 1. As an example, in Legendre s equation the recursion relation is; r = a 2n+2 a 2n = 2n(2n + 1) l(l + 1) (2n + 1)(2n + 2) Then for x = 1 the ratio of the terms in the series as n and n l is; (2n + 1)(2n + 2) 2n(2n + 1) = 2n + 1 2n = 1 + 1/n Using Gauss test, the series does not converge. 5.6 Cauchy test Determine if the the radius of convergence is R = lim n a n a n+1 A finite radius of convergence occurs if R is finite. In this case let x = zr f(z) = b n z n b n = a n R n Whether the series converges or diverges at z = e iφ the character of the function (singular or analytic) is undetermined. For example; z = ( z) n converges for z < 1, and diverges for Z = 1 but z hand; is analytic at z = 1. On the other z dw ln(1 w) = 1 + (1 z)[ln(1 z) 1] This is singular at z = 1 but the integral of the series z n+1 n(n + 1) The derivative z n /n does not converge at z = 1. converges for z = Absolute convergence A series, a n, is absolutely convergent if we can replace a n by a n and the series converges. The order of the terms in the sum does not matter for an absolutely convergent series. 6

7 5.8 Conditional convergence If the series a n converges but a n diverges, the series is conditionally convergent. As an example, the series; ( 1) n 1 n 1 = 1 1/2 + 1/3 1/4 converges by the Leibnitz criterion. This criterion is applied to a series; ( 1) n+1 a n a n > n=1 Let a n be monotonicaly decreasing for sufficiently large n and lim n a n =. Then the series converges. As an example, ( 1) n 1 1/n = 1 1/2 + 1/3 1/4 n=1 The above is convergent but the series is not absolutely convergent. Note that the ratio test yields a n+1 a = n n n Uniform convergence Given the series S = a n, consider the partial sum S N = N a n. If the series, S is convergent and S n goes uniformly to its limit, the the series is uniformly convergent. If each term in the series is analytic. If the series converges uniformly then the sum is analytic. 5.1 Alternating series The series considered in the last subsection is an alternating series. If an alternating series converges absolutely it has absolute convergence. If the series converges but is not absolutely convergent, it is conditionally convergent. A conditionally convergent series can be made convergent to any desired value or it can be divergent. Rearranging terms in the previous example; 1 1/2 + 1/3 1/4 = 1 (1/2 1/3) (1/4 1/5) The result is obviously < 1. But also upon re-arrangement; 7

8 (1 + 1/3 + 1/5) 1/2 + (1/7 + 1/9 + 1/11 + 1/13 + 1/15) 1/4 = 3/2 6 Example Consider the hypergeometric series; F(α, β, γ; x) = 1 + αβ 1! γ x+ α(α + 1)β(β + 1) 2! γ(γ + 1) x 2 + Look at the ratio R of terms. The general term has the form; a n = R = (α + n 1)! (β + n 1)! (γ 1)! n! (γ = n 1)! (α 1)! (β 1)! n(γ + n) x (α + n)(β + n) = (1/x) (n γ + n 2 (αβ + n(α + β) + n 2 For n R (1/x) which converges if x < 1. For x = 1 use the Gauss test so that convergence occurs if γ > (α + β) x n 7 Series of functions We can define the convergence of a series of functions in order to represent another function in a similar way to a power series. S(x) = U n = lim n S n (x) This converges uniformly if; S(x) S n (x) < ǫ For all n N Note several items. 1. If the individual terms of the series are continuous, the series is continuous. 2. If the series is continuous the series may be integrated term by term. 3. If the series is continuous the derivative of the series may be taken term by term. 8

9 First develop the Taylor expansion; x a a dxf (n) (x) = f (n 1) (x) f (n 1) (a) x x [ dx] dx = f (n 2) (x) f (n 2) (a) (x a)f (n 1) (a) a Then we fit any boundary condition by a series of eigenfunctions. Thus represent a function by; F(z) = a n ψ n (x) This represention is to be valid between x 1 and x 2. The function set ψ n must be complete, i.e it must cover all dimensions within the function space. Completeness is defined as; x 2 lim m x 1 [F(z) m a n ψ n (z)]r(z) dz The above represents convergence in the mean, and this type of convergence can represent discontinuous functions. As an example, consider the Fourier represention of the sawtrooth function shown in Figure 2. The series is given by; f(x) 1/2 2π/k 4π/k 6π/k 1/2 π/k 3π/k 5π/k x Figure 2: The sawtooth function (k/2π)x 1/2 f(x) = a n sin(nkx) The coefficients are obtained using the orthogonality of the harmonic functions. 2π/k dxf(x) sin(mkx) = 2π/k a n an δ mn (π/k) = a m (π/k) dx sin(nkx) sin(mkx) = Thus; 9

10 a m = 2π/k dxf(x) sin(mkx) The sawtooth function is f(x) = (k/2π)x 1/2 and is repeated every x = k/2π. The first few coefficients are then; a = a 1 = 1/π a 2 = 1/2π a 3 = 1/3π a 4 = 1/4π a n = 1/nπ The fit to the function is shown in Figure 3. Note that the function is discontinuous and the fit represents the function at these discontinuities by their mean value. This type of fit requires eigenfunctions which satisfy the boundary conditions of the geometry. For the example, here the boundary conditions were set by choosing the value of the sin function (eigenfunction) to vanish every x = 2π/k. It is useful to have eigenfunctions which are orthogonal and normalized, although this is not necessary in general. We will study eigenfunctions later. Figure 3: The fit to the sawtooth function showing the terms in the series and the convergence to the mean value at a discontiunity 1

11 8 Elliptic integrals Elliptic functions are periodic and have some similarities to harmonic functions. We introduce them here by considering the motion of a pendulum as illustrated in Figure 4. The force equation using Newton s laws, after resolving forces along, and perpendicular to, the massless string, results in the equations; θ T M Mg Figure 4: The forces describing the movement of a pendulum M[ d2 r dt 2 r ( dθ dt )2 ] = Mg cos(θ) T M[2 dr dt dθ dt + r d2 θ dt 2 ] = Mg sin(θ) Assume that r does not change so that r = L. The above equations reduce to; ML( dθ dt )2 = T Mg cos(θ) ML d2 θ dt 2 = Mg sin(θ) T = Mg cos(θ) + ML( dθ dt )2 d 2 θ dt 2 = (g/l) sin(θ) For small oscillations, sin(θ) θ and the solution is; θ = θ cos( g/l(t t )) For large amplitudes we must solve the differential equation exactly. Thus convert the differential equation in θ into; 11

12 ( dθ dt )2 = (2g/L)(cos(θ) cos(θ )) We have used the initial condition that dθ dt = when θ = θ. The above equation can be put into the form; dt = L/2g dθ cos(θ) cos(θ ). The period, P, is twice the time for the pendulum to swing from θ to +θ. P = 2 L/2g θ θ dθ cos(θ) cos(θ ) Make a change of variable sin(θ/2) = sin(θ /2) sin(ψ) to obtain the expression; P = 2 L/g π/2 π/2 A power expansion then gives; P = 2 L/g π/2 π/2 dψ 1 sin2 (θ /2) sin 2 (ψ) dψ [1 + 1/2 sin 2 (θ /2) sin 2 (ψ) + 3/8 sin 4 (θ /2) sin 4 (ψ) + ] P = P [1 + 1/4 sin 2 (θ /2) + 9/64 sin 4 (θ /2) + ] However, we can consider the solution in terms of a representation of other periodic functions. Mathematically we expect to find a form, f(z + a) = f(z) having period a. The harmonic functions have period 2π. Looking in the complex plane the exponential e z has a period along the y axis of 2π i, as well as a period along the real axis. The equation for y is d 2 y dt 2 = y ( dy dt )2 = 1 y 2 the latter equation has solution; t = y du 1 u 2 t = sin 1 (y) This solution is 1/4 of a period, so the equation is written; 1 τ = 4 du 1 u 2 12

13 This is similar to the equation for the pendulum. We ask the question if it is posssible to have a function with two periods. Obviously this cannot occur in the same direction, but in the complex plane we have 2 dimensions, so we seek a solution with different periods along the real and imaginary axes. Thus we propose a solution of the form; ( du dz )2 = (1 u 2 )(1 k 2 y 2 )) This equation is guessed by considering the zero s (poles)of the function we seek. The equation may be inverted to represent the period, P. 1 P = 4K = 4 du (1 u2 )(1 k 2 u 2 ) There is also a period along the imaginary axis. P = 2K = 2 1/k 1 du (1 u2 )(1 k 2 u 2 ) Here K is an elliptic function of the 1 st kind. Generally the complete elliptic integral of the 1 st kind is expressed as; K(m) = 1 dt 1 (1 t2 )(1 mt 2 ) m 1 The complete elliptic integral of the 2 nd kind has the form; 1 E(m) = dt 1 mt 2 1 t 2 m 1 These integrals have series expansions which are uniformly convergent. K(m) = (π/2)[1 + (2n 1)!! [ ] 2n!! 2 m n ] n=1 E(m) = (π/2)[1 (2n 1)!! [ ] 2n!! 2 2n mn 1 ] n=1 9 Asymptotic series In the last lecture we discussed asymptotic series. As a reminder, assume a function, f(z), is represented by the expansion as z ; f(z) = φ(z)[a + A 1 /z + a 2 /z 2 + ] 13

14 In the above φ(z) is a known function and the series represents f(z)/φ(z) asymptotically. This means that for a given number of terms in the series, the value approaches the true value of the function as z. For each term the error will be on the order of 1/z n+1. The series, however, diverges, so one must take only the first few terms. As an example, take the exponential integral. Ei( x) = x dt e t /t This integral is developed by a series of partial integrations. Ei( x) = e x /x[1 1/x+2!/x 2 3!/x 3 + +( 1) n n!/x n ]+( 1) n+1 n! Test for convergence of the series using the ratio test. a n+1 a n = n/x lim n = x dt e t /t n+2 Table 1: A demonstation of the asymptotic series showing convergence properties 4e 4 Ei( 4) = n Value of the n th term Sum including the n th term Upper bound to error The series diverges. Two successive terms are equal when n = x, so this is approximately the optimum value. Table 1 shows how this works out for Ei( 4). The development of an asymptotic series is most often obtained using the method of steepest descents. Some properties of asymptotic series are; 1. An asymptotic series is not unique. 2. Asymptotic series may be added or multiplied term by term by another function. 3. Asymptotic series may be integrated term by term. 14

15 4. Asymptotic series may NOT be differentiated to obtain an asymptotic series for the differential unless the differential has an asymptotic expansion. 1 Interpolation Often we have the value of a function at a number of discrete points, but there is not an analytic expression for the function. In many cases it is useful to obtain a series representation of the function, for example a power series, in order to find values of the function at any point. Obviously the series will be valid over a finite range of the function. Obtaining a value in this range is called interpolation. If the value we seek lies outside the range over which the series was fit to the function, then the extraction of the value is called extrapolation. One must be VERY careful using extrapolation. Any 2 points uniquely determines a line, and 3 points a parabola. We will wish to find the value of a function at a point P using a polynomial of order N or some other function by fitting to the known values at specific points. Most often this is done my minimizing the error defined by the closeness of the fitted point to the value of the function at that point. The general procedure, at least if the distribution of the errors is Normal, is to minimize the value of the χ 2 of the distribution. χ 2 = N n=1 [G(A m, x n ) Data n ] 2 σ 2 n In the above, N represents the number of known points, G(A m, x n ) is the value of the function to be fit to the data points having A m adjustable parameters, and σ n is the error in the data point Data n. The value of χ 2 is minimized by varying the parameters A m. This gives a set of m equations; χ 2 A m = Usually this parameter space is large and the coupled equations must be solved numerically. If the χ 2 distribution obtained is assumed to be Normal, the goodness of the representation is determined by the normalized value of χ 2. All this will be discussed next semester. Finally, a smooth function is more accurately interpolated by a high order polynomial. However, it is better to fit the function over the full range rather than fit several pieces. A discontinuous function is best fit with a low order polynomial over continuous, discrete pieces, Figure 5. A spline fit provides continuity of derivatives. It fits a polynomial to the points, so that the fitted function is cointinuous through the 2 nd derivative. Most modern fitting programs have the ability to do spline fitting. 15

16 Figure 5: Representative accuracy using different orders of polynomials for various types of functions 16