θ d2 +r 2 r dr d (d 2 +R 2 ) 1/2. (1)

Transcription

1 Lecture 33 Applications of Taylor and Maclaurin polynomials (cont d) Leading-order behaviours and limits using Taylor series Example: We return to the problem of the gravitational force exerted on a point mass by a circular disc. (Week 5 Tutorial - the discussion follows Lecture 12). Let D denote a circular disc of radius R on the xy-plane centered at (0,0). This disc represents a thin plate with constant density ρ 0 (mass/unit area). (Therefore the mass of the plate is M = ρ 0 πr 2. A point mass m is situated at the point on the z-axis at a distance d from the center, i.e., at coordinate (0,0,d) as sketched below. We shall compute the total gravitational force F = Fj exerted on m by the disc. m d θ d2 +r 2 r dr y x The magnitude of the total vertical force F was found to be F = 2πGmρ 0 [1 (Details are presented in the tutorial discussion.) ] d (d 2 +R 2 ) 1/2. (1) Note that as d, the second term in the square brackets approaches 1, which implies that F 0, as expected. But let us now refine this result, using an binomial expansion, to show how F tends to 306

2 0 as d. We ll first rewrite the result in (1) as follows, F = 2πGmρ 0 1 ( = 2πGmρ 0 [1 1 ) 1/2 1+ R2 d 2 (1+ R2 d 2 ) 1/2 ]. (2) Now let x = R2 d 2 so that as d, x 0+. We ll now employ the binomial series for k = 1 2, (1+x) 1/2 = x+ ( 1 2 )( 3 2 ) x 2 + 2! Now with an eye to Eq. (2), rearrange the above slightly, Now employ this result in Eq. (2) to yield = x+ 3 8 x2 +. (3) 1 (1+x) 1/2 = 1 2 x 3 8 x x = 1 2 R 2 d 2. (4) 1 R 2 F 2πGmρ 0 2 d 2 = Gm (ρ 0πR 2 ) d 2 = GmM d 2, (5) where M = ρ 0 πr 2 is the mass of the circular disc. This result shows that for d sufficiently large, i.e., x = R2 1, the magnitude F of the force is well approximated by considering the circular disc D as d2 a point mass M situated at its center. For the sake of interest, let us now employ the first two terms of the binomial approximation in (4, i.e., 1 (1+x) 1/2 = 1 2 x 3 8 x x 3 8 x2 = R 2 d 2 3 R 4 8 d 4. (6)

3 We now employ this result in Eq. (2) to yield [ 1 R 2 ] F 2πGmρ 0 2 d 2 3R4 8d 4 = Gm (ρ 0πR 2 [ ) d R 2 ] 4 d 2. = GMm d 2 [ R 2 ] d 2. (7) Once again, this approximation is made in the case that d is much larger than R. Regardless of the actual form of second term in the square brackets, we note that it is a negative correction we are subtracting something positive from the leading term GMm d 2. The reason that this correction is negative is that the leading term is based on the approximation that the entire mass M of the disc is concentrated at the center point (0,0,0). For d <, if there is any correction to be made, it must be negative. The reason is that the actual mass of the disc is not concentrated at the center point it is distributed over points that are situated at distances larger than d from the point mass m. As such, the magnitude of the gravitational force exerted by the disc of total mass M must be less than that of the force exerted by a point mass M at the center of the disc. This is precisely what the correction is telling us. The important point from this exercise is that if one is obtaining corrections to leading order behaviour, it is a good idea to check whether the sign of the correction makes sense or at least why the correction has the sign that it does. Example: This problem is taken from Stewart s textbook, Section 11.11, Exercises, Question 33. An electric dipole consists of two electric charges of equal magnitude and opposite sign separated by a nonzero distance. If the charges are q and q are located at a distance d from each other, then the electric field E at the point P in the figure below is, up to a multiplicative constant, Eî, where E = q D 2 + ( q) (D +q) 2 = q D 2 q (D +q) 2. (8) (The unit vector î points in the positive x-direction, to the right.) This implies that if a charge Q were situated at P, the net force on Q would be F = EQi, where EQ = qq D 2 qq (D +q) 2. (9) 308

4 P q q D d If q and Q have the same sign, i.e., both positive or negative, then EQ > 0, representing a repulsive force acting to the left. If q and Q have opposite signs, then EQ < 0, representing an attractive force acting to the right. Of course, as the distance D, the magnitude E 0. If only the charge q were present, then 1 E decays as D2, in accordance with the inverse squared law of classical electrostatic forces. But let us see how E behaves with respect to D as D when both charges are present. We ll write the second term as follows, q (D +q) 2 = q D 2( 1+ d D ) 2 = q D 2 ( 1+ d D) 2. (10) As D, x = d D brackets. Recall that so that 0. We ll employ the binomial series expansion for k = 2 on the term in (1+x) k = 1+kx+ k(k 1) x 2 +, (11) 2 (1+x) 2 = 1 2x+3x 2 +. (12) Substitution into the final term in (10) yields, Now substitute this result into Eq. (8), ( q D 2 1+ d ) 2 = q ) dd d2 (1 2 D D 2 +3 D 2 + E = q D 2 = q D 2 2qd D 3 + 3qd2 +. (13) D4 q (D +q) 2 = q D 2 [ q D 2 2qd D 3 + 3qd2 D 4 + = 2qd D 3 3qd2 +. (14) D3 309 ]

5 As such, for large D, the field strength is well approximated as E 2qd > 0. (15) D3 1 It is still a repulsive field, but it decays as D 3 instead of 1 D2. The presence of the charge q situated at distance D + d from P produces a slight cancellation not a total one since it is slightly farther away that q. Note that if we switched the positions of the two charges, then we obtain the negative of the above result so that i.e., an attractive field which decays as E 2qd < 0, (16) D3 1 D 3. Finally, in the case that both charges were identical, i.e., q, then from Eq. (14) the total field would be E = q D 2 + The dominant behaviour of the field strength is q (D +q) 2 = q D 2 + [ q D 2 2qd D 3 + 3qd2 D 4 + = 2qd D 2 2qd +. (17) D3 ] E 2qd > 0, (18) D2 which is the field produced by a charges 2q situated at a distance D from P. But the first correction to this leading behaviour produces a slight decrease since the second charge is situated at a slightly larger distance D +d away. For very large distances D, in which case d D behaviour is a good approximation. is negligible, the dominant 310

6 Example: Now consider two dipoles separated by a distance D as shown in the figure below. q q q q d D d Since the two charges q of same sign are closest to each other, we expect that there will be a net repulsive force between the dipoles. But let s see if we can determine the leading order behaviour of this force for large D. First of all, we compute the magnitude F of the net force. There are four contributions: 1. repulsive force between q and q situated D apart, 2. two attractive forces between q and q situated D +d apart, 3. repulsive force between q and q situated D +2D apart. Note that we are disregarding the forces between the two charges comprising the same dipole. The total magnitude of these two forces is constant since d is kept constant. We are interested in the forces which are D-dependent. Up to a multiplicative constant, the total force F is F = q2 D 2 2q2 (D +d) 2 + q 2 (D +2d) 2. (19) We ll perform the same operations on the last two terms as we did in the previous example, F = q2 D 2 2q 2 D 2 (1+ d D )2 + q 2 = q2 D 2 2q2 D 2 ( 1+ d D D 2 (1+ 2d ) 2 + q2 D 2 D )2 ( 1+ 2d D We expanded the middle term in brackets earlier - let s do it again: ( 1+ d ) 2 = 1 2 d d2 +3 D D D 2 We ll also expand the final term in brackets, ( 1+ 2d ) 2 = 1 2 2d D D +34d2 D 2 ) 2. (20) = 1 2d D + 3d2 +. (21) D2 = 1 4d D + 12d2 +. (22) D2 311

7 We now insert these results into (20), F = q2 D 2 2q2 D 2 ( 1 2d D + 3d2 D 3 + We now collect like terms in inverse powers of D: )+ q2 D 2 ( 1 4d ) D + 12d2 D 2 +. (23) F = 1 D 2[q2 2q 2 +q 2 ]+ 1 D 3[4q2 d 4q 2 d+ ]+ 1 D 4[ 6q2 d 2 +12q 2 d 2 + ]. (24) Notice that the coefficients of the terms in D 2 and D 3 are zero! The first non-zero term in the expansion of F is We conclude that the leading order behaviour of F as D is F = 6q2 d 2 D 4 +. (25) F 6q2 d 2 D 4 (26) This might seem to be a rather suprising result the repulsive force decays even slower than the 1 D 3 point-dipole interaction of the previous example. Example: An attractive dipole-dipole interaction. As mentioned in class, a more energetically favourable situation involving two dipoles would be the one sketched in the figure below. The charge q from one dipole would be attracted to the charge q of the other dipole nearest to it. q q q q d D d In this case, the net force (once again ignoring the forces between charges in the same dipole) will be given by F = q2 D 2 + 2q2 (D +d) 2 q 2 (D +2d) 2. (27) This actually turns out to be the negation of the force in Eq. (19) for the previous problem. As such, we can immediately conclude that the leading behaviour as D will be the following attractive force, F 6q2 d 2 D 4. (28) Theabove areexamples ofmanyproblemsinwhichonewishestodeterminethebehaviourof aphysical quantity, e.g., field strength, for large values of some variable, e.g., distance, field strength itself. The 312

8 important idea is that if you wish to know how something behaves as x, you ll want to express it in terms of a variable that involves some inverse power of x, e.g., = 1. We ll see this idea in the x next mathematical problem. Example: Find provided that the limit exists. lim [ x x 2 x 4 +4x 2 ], (29) x Note that this is an indeterminate form. The dominant part of the first term is the x 4 term in the integrand which becomes x 2 after taking the square root. The same can be said for the second term. So to leading order, we have x 2 x 2 = 0. But what else happens? To answer this, we ll start with the first term and rewrite it as follows, x x 2 = x x 2. (30) Now define y = 100. As x, y 0. For x sufficiently large, so that y < 1, we can employ the x2 binomial expansion, and resubstiute for x: Finally, multiply by x 2 : Do the same with the second term, (1+y) 1/2 = y 1 8 y2 +, (31) ( ) 1/2 x 2 = x x 4 +. (32) ( x ) 1/2 x 2 = x x 2 +. (33) x 4 +4x 2 = x x 2. (34) Now define = 4 and use the binomial expansion in (31) to obtain, x2 ( 1+ 4 ) 1/2 x 2 = 1+ 2 x (35) x4 Multiply by x 2 : x ( ) 1/2 x 2 = x (36) x2 313

9 Now combine the results of Eqs. (33) and (36) to return to the original problem, x x 2 x 4 +4x 2 = [ x ] x 2 + [x x 2 + ] = x 2 +. (37) Taking limits, we obtain [ lim x x 2 ] x 4 +4x 2 n [ = lim ] n x 2 + = 48. (38) Note: This limit could also have been obtained by other methods, e.g., multiplying by x x 2 + x 4 +4x 2 x x 2 + = 1. (39) x 4 +4x2 That being said, the series approach possesses a kind of elegance. 314

10 APPENDIX: Material covered in Monday, March 27, 2017 Tutorial Another problem involving Taylor series approximation to a function f(x) over an interval [a, b] In Lecture 32 (Week 11), we considered the problem of approximating the function f(x) = e x over the interval [ 2,2] (and then [ 10,10]) using Maclaurin polynomials T n (x). In particular, we wanted to find the minimum value of n such that the approximation e x T n (x) does not exceed 10 2 in magnitude at any point x [ 2,2]. Here, we consider the problem of approximating f(x) = sin x with Maclauring polynomials. We look for the minimum value of n such that the approximation sinx T n (x) does not exceed (i) 10 2 and (ii) 10 3 for any x in the interval [0,2π]. Recall that the Maclaurin series for sinx is sinx = x x3 3! + x5 5! = ( 1) n x2n+1 (2n+1)!. (40) n=0 We should also keep in mind that this series is rather unique in that it is composed only of odd powers of x. As such, we need only consider Maclaurin polynomials of odd degree. The first four such polynomials are T 1 (x) = x, T 3 (x) = x x3 3!, T 5(x) = x x3 3! + x5 5! T 7 (x) = x x3 3! + x5 5! x7 7!. (41) (Technically, the second-degree Taylor polyonomial T 2 (x) is identical to T 1 (x) since the coefficient of x 2 is zero, etc..) We are therefore considering approximations to sinx of the form, sinx T 2n+1 (x) x [0,2π]. (42) Associated with this approximation is the remainder term, R 2n+1 (x) = sinx T 2n+1 (x). (43) Recall that this comes from the equation, sinx = T 2n+1 (x)+r 2n+1 (x). (44) 315

11 We now wish to find n such that R 2n+1 (x) 10 2 for all x [0,2π]. (45) We shall use Taylor s Inequality, where n is now replaced by 2n + 1 since we are considering only odd-numbered indices. Here, Taylor s Inequality will be as follows: If f 2n+2 (x) M 2n+2 for all x [0,2π], then R 2n+1 (x) M 2n+1 (2n+2)! x 2n+1 x [0,2π]. (46) As we have seen before, we must provide bounds M 2n+2 for the derivatives f (2n+2) (x) which are valid for all x [0,2π]. This is easy for the function f(x) = sinx since its derivatives are either sinx, sinx, cosx or cosx. Therefore, for all k 0, f (k) (x) 1 for all x R. (47) Taylor s Inequality becomes, R 2n+1 (x) 1 (2n+2)! x 2n+2, x [0,2π]. (48) The largest value that the term x 2n+1 can achieve over [0,2π] is (2π) 2n+1. As such, we look for n such that R 2n+1 (x) 1 (2n+2)! (2π)2n (49) This will guarantee that the error at any point x [0,2π] is not greater than Let us define A 2n+1 = After a little computation, we find that 1 (2n+2)! (2π)2n+2, n 0. (50) A , A 19 = , (51) so 2n+1 = 19 satisfies the inequality in (49). Thus, we conclude that the approximation, sinx T 19 (x), (52) will satisfy the requirement that the error will never exceed 10 2 in magnitude. In fact, since Taylor s Inequality generally overestimates the error, we expect that the Maclaurin polynomial T 17 (x) may also provide approximations with errors below To check this, we examine the approximations to sinx by T 17 (x) and T 19 (x) at the point in [0,2π] farthest away from the center point x = 0, namely, 316

12 x = 2π, where we expect the error to be the largest. Of course, sin(2π) = 0 so we can easily see how close the approximations are to zero. Numerically, we find: T 17 (2π) = , Error = T 19 (2π) = , Error = (53) We see that the error of the approximation to sin(2π) = 0 yielded by T 19 is well below the desired value of The error of the approximation yielded by T 17 is very slightly over the desired value. If we now wish that the error of the approximations does not exceed 10 3 over [0,2π], we must find n such that A 2n (54) We find that A A (55) As such, we conclude that the magnitude of the error in the approximation, sin T 21) (x), (56) will not exceed 10 3 for all x [0,2π]. Once again, we examine the approximations at x = 2π: T 19 (2π) = , Error = T 21 (2π) = , Error = (57) The approximation yielded by T 21 certainly satisfies the requirement, as predicted. But the approximation yielded by T 19 is almost there as well. 317

13 Use of Taylor series in Special Relativity The material followed the discussion in Stewart s textbook, Example 3, Section 11.11, Page 778. Very briefly, in Einstein s theory of special relativity, the mass of an object moving with velocity v < c is m = m 0 1 v 2 /c 2, (58) where m 0 is the rest mass of the object at rest and c is the speed of light. The kinetic energy K of the object is the difference between its total energy and its energy at rest, i.e., K = mc 2 m 0 c 2. (59) When v is very small compared to c, i.e., v c 1, then to a very good approximation, K = 1 2 m 0v 2, (60) the kinetic energy of the particle according to classical Newtonian mechanics. To derive this result, we substitute (58) into (59), K = m 0 c 2 1 v 2 /c 2 m 0c 2 [ ( ) 1/2 = m 0 c 2 1 1] v2 c 2. (61) We ll employ the binomial expansion for the term in round brackets, setting x = v 2 /c 2 : Inserting this result into (61), (1+x) 1/2 = 1+( 1 2 )x+ ( 1 2 )( 3 2 ) x 2 +. (62) 2! [ K = m 0 c v 2 2 c v 4 ] 8 c [ 1 = m 0 c 2 v 2 2 c v 4 ] 8 c 4 +. (63) When v is much smaller than c, i.e., v 1, then all terms after the first are very small compared to c the first term. If we neglect them, we obtain the approximation, ( 1 K m 0 c 2 v 2 ) 2 c 2 = 1 2 m 0v 2, (64) which is the classical Newtonian mechanical kinetic energy. In Stewart, it is shown that for velocities v 100 m/s, the error in using the Newtonian expression for kinetic energy is at most ( )m

14 Lecture 34 Applications of Taylor/Maclaurin series and approximations (cont d) The pendulum problem and its linearization Consider a pendulum comprised of a mass m suspended from the ceiling by a thin and inflexible rod of length l (also assumed to be massless, at least compared to m), as pictured below. P θ l 0 m The angle θ represents the angle of displacement of the mass from its equilibrium position θ = 0. The mass m is acted upon by the gravitational force F = mgk. It is constrained to move over a circular curve of radius l centered at the point where the rod meets the ceiling. We ll proceed stepby-step to derive the equation of motion of the mass m. This is, in fact, a very appropriate subject to examine at this final stage of the course, since it also requires the use of parametric curves. First of all, let x(t) = (x(t),y(t)), t 0, denote the position of the mass m at time t. We ll let (0, 0) denote the equilibrium position, i.e., the lowest point of the circular curve. From the figure, x(t) = lsinθ(t), y(t) = l(1 cosθ(t)). (65) Note that the angle θ(t) is time-dependent it is responsible for the variations in x(t) and y(t). The associated velocity vector is v(t) = x (t) = (x (t),y (t)) = (lθ (t)cosθ(t),lθ (t)sinθ(t)) = lθ (t)(cosθ(t),sinθ(t)),. (66) The speed associated with the trajectory of the mass m is given by v(t) = l θ (t). (67) 319

15 The acceleration vector is obtained by differentiating v(t): a(t) = v (t) = (lθ (t)cosθ(t) l(θ (t)) 2 sinθ(t),lθ (t)sinθ(t)+l(θ (t)) 2 cosθ(t)) = lθ (t)(cosθ(t),sinθ(t))+l(θ (t)) 2 ( sinθ(t),cosθ(t)) = a T (t)+a R (t). (68) Here, a T (t) = lθ (t)(cosθ(t),sinθ(t)) = lθ (t)ˆt(t) (69) is the tangential acceleration vector since it is a multiple of the unit tangent vector ˆT(t) = (cosθ(t),sinθ(t). (70) Furthermore, a R (t) = l(θ (t)) 2 ( sinθ(t),cosθ(t)) = l(θ (t)) 2 ˆN(t) (71) is the radial acceleration vector since it is a multiple of the unit radial or normal vector ˆN(t) = ( sinθ(t),cosθ(t)) (72) which is orthogonal to the unit tangent vector. These two unit vectors are sketched in the figure below. P θ l ˆN(t) 0 ˆT(t) As mentioned earlier, the mass m is acted upon by the gravitational force F = mgk. It is convenient to decompose F in terms of its components in the radial and tangential directions, F R and F T, respectively, as shown in the figure below. We have that F R = mg cosθ(t) ˆN(t), F T = mg sinθ(t) ˆT(t). (73) 320

16 P θ l ˆN(t) ˆT(t) 0 F θ F T F R It is only the tangential component F T that contributes to the motion of the mass along the circular arc. The radial component F R is balanced by the tension in the string which acts in the opposite direction with equal magnitude (to keep the mass m moving in a circular trajectory). We now apply Newton s Second Law of Motion with regard to the tangential motion of the mass: tangential force = mass tangential acceleration. Mathematically, F T = ma T. (74) At any time t, this force/acceleration relationship will hold in the tangential direction T(t) which, of course, changes in time. Mathematically, we have mg sinθ(t) ˆT(t) = mlθ (t) ˆT(t). (75) Equating components, we have mlθ (t) = mg sinθ(t). (76) We ll divide by ml and rearrange to produce the following differential equation in θ(t), d 2 θ dt 2 + g sinθ = 0. (77) l This second order nonlinear DE in the function θ(t) is known as the pendulum equation. The solution to the pendulum equation is complicated, involving nonlinear elliptic functions. From your first-year courses in Physics, you are most probably familiar with a standard approach to simplify 321

17 this problem. For small oscillations, i.e., θ << 1, we use the linear approximation to sinθ, i.e., sinθ T 1 (θ) = θ (78) so that the pendulum is converted to the following linear differential equation, d 2 θ dt 2 + g θ = 0, (79) l which is also known as the linear oscillator DE or harmonic oscillator DE. Earlier in this course, we saw that the general solution to this DE is θ(t) = C 1 cosω 0 t+c 2 sinω 0 t, ω 0 = g l. (80) For example, in the special case that the mass m is pulled to angle θ 0 and released from rest, which translates to the initial conditions, then the behaviour of θ(t) for t 0 is given by θ(0) = θ 0 θ (0) = 0, (81) θ(t) = θ 0 cosω 0 (t). (82) Note that the general solution in (80) can be written in phase-shifted form, θ(t) = Acos(ω 0 t φ). (83) This reveals one of the most characteristic properties of the linear harmonic oscillator: All solutions θ(t), as expressed in Eq. (80) have the same frequency ω 0 hence period T = 2π ω 0. The frequency/period is independent of the amplitude of the oscillation. Of course, we have to recall that the linear oscillator DE is an approximation and is only valid for small oscillations, i.e., oscillations with low amplitude. Of course, the approximation sinθ θ is obtained by using only the first term in the Taylor series expansion of sinθ. If we include another term in the Taylor series expansion, we obtain the following nonlinear DE, d 2 θ dt 2 + g l θ g 6l θ3 = 0, (84) which is known as the Duffing nonlinear oscillator. One of the most important differences between this, and other, nonlinear oscillators is that the frequency/period of oscillation depends on the amplitude. This is the case for the pendulum equation as well. 322

18 Energy formulation of pendulum problem The treatment discussed above is an F = ma analysis of the pendulum problem. It is also instructive to examine this problem in terms of conservative forces and energy. First of all, the forces involved in this problem are conservative we are ignoring air or other resistance. As discussed earlier, the tangential force, F T = mg sinθ ˆT, (85) is responsible for the motion. We have to be a bit careful when computing the potential energy function V, which we shall consider to be a function of the angle θ. We ll use the equilibrium point, θ = 0, as the reference point. The work done against the force in moving the mass m from θ = 0 to some θ > 0 must be computed as an integral over the arclength of the circular curve from θ = 0 to θ. The arclength of this curve is s = lθ. As such, the infinitesimal arclength is ds = ldθ. The potential energy function is then θ V(θ) = = mg = mgl 0 θ F T (φ)ds 0 θ 0 (sinφ)ldφ sinφdφ = mgl(1 cosθ), π θ π. (86) From Eq. (67), the kinetic energy of the mass m at a time t is given by K = 1 2 m v(t) 2 = 1 2 ml2 (θ (t)) 2. (87) Combining these results, the total mechanical energy E(t) of the mass m is given by E(t) = 1 2 ml2 (θ (t)) 2 +mgl(1 cosθ(t)). (88) Since the force F T is conservative, we expect that the energy E(t) of a particular trajectory is constant in time. We can confirm this by computing E (t): E (t) = ml 2 θ θ +mglθ sinθ = ml 2 θ [ θ + g l sinθ ] = ml 2 θ 0. (89) 323

19 The final line follows from the pendulum DE in (77) which, in turn, comes from F = ma. (Recall the formal proof of the conservation of energy to show that E (t) = 0.) The importance of Linearization in Physics Linearization, as was done in the approximation sinθ θ, is perhaps one of the most important approximative tools in Physics. You will see it employed in a wide variety of applications. Indeed, another application was given earlier in this course in Assignment No. 8, a bonus question was devoted to the family of so-called Lennard-Jones (n, 6) potential energies for diatomic molecules. In many cases, the interaction between the two atoms in a diatomic molecule A-B is well modelled by the so-called Lennard-Jones (n, 6) potential energy function, having the form Here, r denotes the separation between the nuclei of atoms A and B and n > 6. The first term in Eq. (90) models the repulsion of the nuclei and the second term models the attractive force caused by complicated electronic effects. The net result of adding these terms is a potential well of depth D with minimum situated at r = r e, the equilibrium separation. This potential well is sketched qualitatively at the right. V(r) = C n r n C 6 r 6. (90) D r e repulsive potential V (r) net potential attractive potential One of the most useful LJ potentials is the (12,6) potential, where n = 12, which is written in the following convenient mathematical form, [ (r0 ) 12 ( r0 ) ] 6 V(r) = 4D. (91) r r Note that V(r 0 ) = 0. It is not difficult to show that the potential minimum is located at r = r e = 2 1/6 r 0 and that the well depth is D. For small oscillations about equilibrium, i.e., r close to r e, the potential function V(r) may be r 324

20 approximated by a harmonic oscillator-like potential: V(r) V(r e )+ 1 2 V (r e )(r r e ) 2 = D k(r r e) 2, (92) where k = 72D r 2 e may be viewed as the Hooke s constant for the linear oscillation about r e. (93) This approximation is very important in the study of diatomic molecules, but from the perspective of quantum mechanics. The discrete energy levels of a harmonic oscillator potential such as the one in 92 can be determined in closed form. Perhaps the most important property of such energy levels is that they are equally spaced the spacing depends on k. For many molecules, especially those with deep potentials, i.e., high D-values, the harmonic oscillator approximation is very good for the first few energy levels and can match the energy-level spacings observed in molecular spectroscopy (where a photon of energy E equal to the difference in two vibrational energy levels E 1 and E 2 is emitted/absorbed when the molecule transforms from one vibrational state to the other. 325

21 A final sample problem involving approximation by Taylor polyomials and estimation of error From Stewart: Exercises, Section 11.11, Question 15, p Use Taylor s Inequality to estimate the accuracy of the approximation f(x) = x 2/3 T 3 (x) (94) where the center point of T 3 (x) is a = 1 and the interval is I = [0.8,1.2]. Solution: We first compute f(1) = 1, followed by the necessary derivatives, f (x) = 2 3 x 1/3 f (1) = 2 ( )( 3 2 f (x) = 1 ) x 4/3 f (1) = ( )( 2 f (x) = 1 )( 4 ) x 7/3 f (1) = ( )( 2 f (4) (x) = 1 )( 4 )( 7 ) x 10/3. (95) We can now compute the Taylor polynomial T 3 (x), T 3 (x) = f(1)+f (1)(x 1)+ 1 2 f (1)(x 1) ! (x 1)3 = (x 1) 1 9 (x 1) (x 1)3. (96) We must now estimate the error in the approximation f(x) T 3 (x) over the interval I = [0.8,1.2] using Taylor s Inequality with n = 3, R 3 (x) M 4 4! x 1 4 (97) where M 4 is an upper bound to the derivative f (4) (x) on I. From the equation for f (4) (x) above, f (4) (x) = x 10/3. (98) The function x 10/3 is a decreasing function which implies that the maximum value of f (4) (x) is achieved at the left endpoint x = 0.8, i.e., f (4) (x) (0.8) 10/3 = M 4. (99) 326

22 The maximum value that x 1 4 can achieve for x [0.8,1.2] is (0.2) 4. Substituting these results into (97), we have the estimate R 3 (x) (0.8) 10/3 (0.2) (100) This is the maximum error of the approximation x 2/3 T 3 (x) on the interval I = [0.8,1.2]. END OF COURSE 327