Physics 116A Solutions to Homework Set #2 Winter 2012

Transcription

1 Physics 6A Solutions to Homework Set #2 Winter 22. Boas, problem. 23. Transform the series 3 n (n+ (+ n determine the interval of convergence to a power series and First we want to make the replacement u = +, which allows us to read this as a power series, yielding 3 n (n+. Now use the ratio test. The term a (u n n+ = 3n+ (n++ u n+ so the ratio a n+ a n = 3(n+ which is less than if u > 3. This translates to > 2 or u(n+2 < -4. But we need to check these end points. Note that at u = 3, the sum itself can be written as (n +, which clearly diverges, for u=-3, the series can be written as ( n (n +, which also diverges as the terms do not tend to. So the correct limit is > 2 or < Boas, p. 29, problem.3 4. Write the Maclaurin series for / + in form using the binomial coefficient notation. Then find a formula for the binomial coefficients in terms of n. Using the binomial coefficient notation we can write this compactly as: ( p ( + p = n, <. n Setting p = /2, it follows that: ( /2 = n, <, + n where we have noted that the series is conditionally convergent at = (as it is an alternating series. Using ( ( p p p(p (p n + =, =, for n =, 2,..., p, n n! it follows that for n =, 2,..., p, ( /2 = ( ( 3( 5 ( n n n! where = ( n 3 5 (2n 2 n n! = ( n (2n!! (2n!!, ( (2n!! = 3 5 (2n, (2n!! = 2 n n! = n.

2 It is convenient to define (!! =!! =, in which case eq. ( also applies to the case of n =. Hence, we can write: + = ( n (2n!! (2n!! n, <. (2 3. Boas, p. 32, problem.3 3. Find the first few terms of the Maclaurin series for the function e t2 dt. Find the general term of the Maclaurin series and write the series in summation form. Using the Maclaurin series for e y, where y t 2, it follows that ( t 2 n e t2 = n! = ( n t2n n!. Integrating from t = to t =, and using one obtains e t2 dt = t 2n dt = 2n+ 2n +, The first few terms of the series are given by: e t2 dt = ( n 2n+ (2n + n!. 3 3! ! 7 7 3! + O(9. 4. Boas, p. 4, problem.5. Use the Maclaurin series to evaluate sin 2 lim. 2 Use the trig identity sin 2 ( = 2 ( cos(2, which implies sin2 (2 = 2 ( cos(4 from the Maclaurin series for cos( you obtain cos = ( n 2n (2n! O( 6. Taking only the leading two terms, because we are worried about the limit as goes to, and the higher order terms will have larger values of, and will thus be irrelevant in the limit (the reason for taking a second term will become apparent in a moment. So cos(4 2 (42. Then the limit can be epressed as (remember the etra constant from the conversion of sin 2 to cos: 2

3 lim ( ( 2 2 (42. 2 The constants cancel, and we have a limit of 4 (62 2 which is Boas, p. 4, problem.5 23 (b. Find the following limit using Maclaurin series. ( (b lim cos 2 sin 2, (b In this case, it is convenient to rewrite the limit as ( lim 2 cos 2 sin 2. (3 First consider the Maclaurin epansions for sin sin cos = and cos, = ( n 2n (2n +! = O( 4, (4 ( n 2n (2n! = O( 4. (5 Squaring the epansion given in eq. (4 yields sin 2 2 = [ O( 4 ] 2 = O( 4. To invert this series, we employ the geometric series, y = y n = + y + O(y 2, y <. (6 Choosing y = 3 2 O( 4, and noting that O(y 2 = O( 4, it follows that Combining eqs. (5 and (7 yields Hence, lim 2 sin 2 = O( 4 = O( 4. (7 2 cos sin 2 = [ O( 4 ][ O( 4 ] = O( 4. ( 2 cos [ 2 sin 2 = lim O( 4 ] [ = lim + 6 O(2 ] =, 6 3

4 since lim O( 2 =. Additional Information (not required for full credit - To find the behavior as, we have to keep higher order terms in the epansion. Repeating the above derivation but working to one higher order in the series epansions, sin = ( n 2n (2n +! = O( 6, (8 cos = ( n 2n (2n! O( 6. (9 Squaring the epansion given in eq. (8 yields sin 2 2 = [ O( 6 ] 2 = ( O( 6 = O( 6. To invert this series, we again refer to the geometric series [cf. eq. (6], y = + y + y2 + O(y 3. Identifying y = O( 6 and noting that O(y 3 = O( 6, it follows that: 2 sin 2 = O( 6 = + ( ( O( 6 = ( Hence, eqs. (9 and ( yield 4 + O( 6 = O( 6. ( 2 cos sin 2 = [ O( 6 ][ O( 6 ] = + ( ( O( 6 = O( 6. Thus, the behavior as is: ( 2 cos 2 sin 2 = [ ] O( 6 = 6[ O( 4 ]. For an alternative method for solving this problem, see the Appendi at the end of these solutions. 6. Boas, p. 5, problem Find a series epansion for the relativistic equation of energy E = mc 2 ( v2 c

5 Using the series epansion around a velocity v=, we have an energy derivative: E (v = mc 2 ( v2 ( 3 2 c 2 2 (2v v2 = mv( c2 which is when v=. So we epand to a second term: 3 c 2 2 E (v = mc 2 ( 3 v2 5 4v 2 v2 3 ( 4 c 2 2 ( + m( c4 c 2 2 which gives us a value of m when v=. Thus using the series epansion we obtain: E mc mv2 + O(v 4 interestingly, we note that this first term is the classical kinetic energy! Only now with higher order corrections. Another way to say this is that the energy calculation in special relativity reduces to classical mechanics (with the addition of a rest energy, so long as the velocity is small compared to the speed of light (i.e. v <<. c 7. Boas, p. 5, problem For the comple number z = 3 + i, determine, y, r, and θ, where z = + iy = r(cos θ + i sin θ = re iθ. Plot the number in the comple plane and label it in five ways as in Figure 3.3 on p. 48 of Boas. Also, plot the comple conjugate of z. Given z = 3 + i, we have (, y = ( 3,, and r = 2 + y 2 = 2. Writing z = r(cos θ + i sin θ, we identify cos θ = 3/2 and sin θ = /2, which corresponds to an angle θ = 5π/6 = 5. In the comple plane, we can draw the point z and its comple conjugate point z: (, y = ( 3, 3 + i (r, θ = (2, 5 6 π 2(cos 5 6 π + i sin 5 6 π 2e 5iπ/6 (, y = ( 3, 3 i (r, θ = (2, 5 6 π 2(cos 5 6 π i sin 5 6 π 2e 5iπ/6 z z y 5π/6 8. Boas, p. 53, problem 2.5. Simplify the fraction 3i 7 i+4 to either + iy or reiθ form. In this case, it is much easier to get the formula into +iy form, so we will do this. 5

6 We want to multiply by the comple conjugate of the denominator, which is 4-i. This yields 25+9i 25, which gives us + i Boas, p. 53, problem Find the absolute value of 2 + 3i i. First, we write the number in the form of a+bi. Then we compute a+bi = a 2 + b 2. ( ( 2 + 3i 2 + 3i + i i = = 2 + 3i2 + 5i = ( + 5i. 2 i + i + Hence, 2 + 3i i = 4 ( + 25 = Boas, p. 53, problem Solve +iy = i for all possible values of the real iy numbers, y. Cross multiply to obtain +iy = -i(-iy which simplifies to +iy = -i-y. Taking only the real terms in this equation yields =-y and iy=-i, both of which point to the solution y = -. Boas, p. 53, problem Describe geometrically the set of points in the comple plane satisfying the equation z + + z = 8 Remember we are just asked to describe the set of points (not solve for it - note that the first term describes points which are a distance z away from the point (, while the second describes points that are a distance z away from (-, - so this is the definition of a ellipse. Note that because the foci lie in the plane, so will the semi-major ais. Plugging in for z (since y will be in the plane, we obtain = 8 which gives a semi-major ais of 4. Alternatively, we can do this algebraically. First, move the z term to the other side of the equation. z + 2 = 64 z 2 which yields ( y 2 = 64 ( 2 + y 2 6 ( 2 + y y y 2 = 6 ( 2 + y 2 canceling terms leads to: 6

7 squaring again yields: 4 64 = 6 ( 2 + y = 256(( 2 + y = y 2 canceling terms again leads to: y 2 = y2 = which is the equation of an ellipse with major ais of 4 and foci at (,, (-,. 7

8 An alternative method for solving problem 7 [Boas, problem.5-23(b] Problem: Find the behavior of as. f( 2 cos sin 2, ( The original strategy presented in the solutions above was to insert the epansions for sin and cos into eq. ( and manipulate the resulting epression. You may be wondering why I did not simply compute the Taylor series for the function given in eq. ( directly. Let us try to do this. It is easy to check that this function satisfies f( = f(, since cos( = cos and sin( = sin. Moreover, f( has a finite limit as. Using sin 2 lim =, 2 it follows that lim f( is a finite constant. Combining these two properties of f(, we conclude that f( possesses a Taylor series about = of the form: f( = c n n = c + c O( 4, where c n = n! f (n (. That is, f( = f( implies that c n = for all odd values of n. Thus, all we need to do is to compute c and c 2. I challenge you to try this yourself by hand. You will quickly realize that the required computation is quite intense. Each of the two computations requires you to compute the limit as of some rather nasty looking epression that involves powers of sin in the denominators. To compute these limits, you will need to employ a method similar to the one we have presented in our original solution. At the end of the day, it will be quite apparent that the original method is far simpler (and less prone to error. If you are unsure of this argument, go ahead and check eplicitly that c = c 3 =. 8