Problem Set 1 October 30, 2017

Transcription

1 1. e π can be calculated from e x = x n. But that s not a good approximation method. The n! reason is that π is not small compared to 1. If you want O(0.1) accuracy, then the last term you need to include must satisfy xn < 0.1 (why?). What is the minimum n required to achieve this n! accuracy? A better approach is to expand e 3+x near x = 0. However, you will need to know e How many terms do you need to get O(0.1) accuracy now? Solution. The Taylor expansion of e x can be truncated at order N with an error of order N + 1 e x = N x n n! + O(xN+1 ) (1) where the leading term in O(x N+1 x ) is N+1. if we want the accuracy of the truncated (N+1)! Taylor expansion to be of the order 0.1 (i.e. error be of order less than O(0.1)), we should have x N (N)! < 0.1. In our case, this translates into finding N s.t. π N N! < 0.1. () Since π 3, we can see that the first term satisfying eq. is corresponding to N = 9 ( π 0.3! but π9 0.0). 9! However, if we expand e π as e x+3, we ll get [ N ] [ e π = e 3 x n N ] n! + x n O(xN+1 ) 0.1 n! + O(xN+1 ). (3) So, we need (0.1 xn ) < 0.1 where x You will see that in this case only 4 terms N! (N = 0,, 3) is sufficient to reach the desired accuracy since ( ) The first few terms of e x = 1 x + x + usually provide pretty good approximations. So what s wrong with ] e x = [x x + x3 + = (4) Solution. 0 This should be clear even with your high school knowledge! The first thing to notice is that this expansion is designed to work for x 1 and it is definitely not a good idea to use it in order to calculate the value of the function as x (especially when we are truncating the expansion after the third term). Another way of looking at it is that in eq. 4 we are adding and subtracting infinities which is vague and cannot be manipulated in the form written here. 1 0

2 3. For each of the following functions find the maximum n Z such that f(x) = O(x n ) as x 0. (In some cases no such n exists, and when it does f(x) = Θ(x n ) may be false.) (i) f(x) = e x Solution. Whenever a function has a Taylor expansion, f(x) is of the same order as the first term appearing in the expansion. In the case of e x we have e x = O(1) and n = 0. (ii) f(x) = sin(x) x e x = Solution. The Taylor expansion of sin(x) can be written as sin(x) = x n n! = 1 + x + x +. (5) ( 1) n x n+1 (n + 1)! = x x3 + x5 5! +. (6) Using eq. 6, the first term appearing in the Taylor expansion of f(x) = sin(x) x will be x 3. Therefore, sin(x) x = O(x3 ) and n = 3. (iii) f(x) = log(x) Solution. As x 0, log(x) slower than any negative power of x, i.e. log(x) = o(x 1 ) = O(x 1 ) and n = 1. (iv) f(x) = x p log(x) lim x 0 log(x) x 1 = lim x 0 xlog(x) = 0. (7) Solution. As is stated in eq. 7, lim x 0 xlog(x) = 0. Therefore, lim x 0 x p log(x)/x (p 1) = 0, and n = p 1. (v) f(x) = (1 cos(x))/tan 1 (x) Solution. (1 cos(x))/tan 1 (x) has a Taylor expansion of the form Then, it is obvious from eq. that n = 1. (vi) f(x) = e 1/x 1 cos(x) tan 1 (x) = x + x3 41x O(x7 ). ()

3 Solution. lim x 0 e 1 x 1 = 0 for all n Z n Z ; e x xn = o(x n ). (9) n Z s.t. e 1 x = Θ(x n ). (vii) f(x) = x Solution. the maximum n is 0. x lim = 0 for n 0. (10) x 0 xn (viii) f(x) = x x = e x log x Solution. As we have already seen in eq. 7 lim x log x = 0. x 0 So, we have Using eq. 11, we see that Therefore, maximum n = 0. lim x 0 lim x 0 xx = 1. (11) x x = 0 for all n 0. (1) xn (ix) f(x) = 1 x ( 1 x 1)( 1 x ) ( 1 x 1 x + 1) Solution. For simplicity let us only work with values of x = 1/N for N N; then, f(x) = N! and the limit x 0 is equivalent to the limit N. Using Stirling s formula for N!, we have lim ln N! = N ln N N. N But, as N gets bigger and bigger we can even ignore the second term. So Using eq. 13, we can see that no such n Z exists s.t. lim ln N! N ln N N! N N. (13) N N N lim N N <. n 4. Find the first four coefficients a 0,1,,3 of the Taylor expansion sin 1 (x) = a nx n by plugging it in sin(sin 1 x) = x and using the Taylor series of sin x. Repeat the calculation starting from sin 1 (sin x) = x. 3

4 Solution. sin x has a Taylor expansion of the form sin x = ( 1) n x(n+1) (n + 1)! = x x3 + O(x5 ). (14) Therefore, assuming that sin 1 x has a Taylor expansion of the form a nx n and only keeping terms up to order 3, we get x = (a 0 + a 1 x + a x + a 3 x 3 ) 1 (a 0 + a 1 x + a x + a 3 x 3 ) 3 = (a 0 a3 0 ) + (a 1 3a 0a 1 )x So, we have +(a 3a 0a 1 a 0 a3 0 = a 0 (1 a 0 ) = 0, a 1 1 a 0a 1 = 1, a 1a 0a 1 1 a 0a = 0, a 3 1 a3 1 1 a 0a 3 = 0. 3a 0a )x + (a 3 a3 1 3a 0a 3 a 0 = 0, ±() 1, a 1 = (1 a 0 ) 1, a = 1 a 0a 3 1, a 3 = 1 a4 1. ) + O(x 4 ). It seems that we are in trouble! depending on the a 0 that we choose from the three possible options, we will get different Taylor expansions for sin 1 x. What comes to our rescue, as always, is the initial value; we want sin 1 (0) = 0, so we should choose a 0 = 0. Therefore (15) (16) sin 1 x = x + x3 + O(x4 ). (17) 5. Find the first three coefficients a 0,1, of the Taylor expansion log(1+x) = a nx n by plugging it in exp(log(1 + x)) = 1 + x and using the Taylor series of e x. Repeat the calculation starting from log(e x ) = x. Solution. We will proceed in the same manner as we did in the problem (a 0 + a 1 x + a x ) + (a 0 + a 1 x + a x ) + = 1 + x (a a 0) + (a 1 + a 0 a 1 )x a 0 = 0,, a 1 = (1 + a 0 ) 1, a = 1 a3 1. +(a + 1 a 1 + a 0 a )x = x. Using the same reasoning as in problem 4, we demand a 0 to be equal to 0. Therefore (1) (19) log(1 + x) = x 1 x + O(x 3 ). (0) 6. Find the first three non-zero terms in the power series of the following expressions. Don t use Taylor s formula for the expansion coefficients; rather manipulate series you already know. 4

5 (i) sin x/(1 x) as x 0. Solution. Let us assume that sin x/(1 x) has a power series expansion of the form We also know that Using eq. 1 and eq., we get x x3 + x5 5! + O(x 7 ) (1 x) = sin x = sin x (1 x) = a n x n. (1) ( 1) m x(m+1) (m + 1)!. () m=0 a n x n x x3 + x5 5! + O(x7 ) = a n x n a n x n+1. (3) eq. 3 can be solved for a n s by comparing the coefficients of x n s order by order x x3 + O(x5 ) = a 0 + (a 1 a 0 )x + (a a 1 )x + (a 3 a )x 3 + O(x 4 ) a 0 = 0, a 1 = 1, a = 1, a 3 = 5 6,. (4) Therefore sin x (1 x) = x + x x3 + O(x 4 ). (5) (ii) sin x/(1 x) as x π/. Solution. Let us introduce y x π/. sin x/(1 x) can be written is terms of y as sin( π + y) [1 ( π + y)] = cos y (1 π) y = a n y n. (6) Then, substituting the power series expansion of cos x in eq. 6, we get 1 + y! + y4 4! + O(y6 ) = (1 π )a 0 + (a n+1 (1 π )a n)y n+1. (7) a 0 = (1 π ) 1, a 1 = 1, a = (1 π ),. sin x (1 x) = (1 π ) 1 + (x π ) + (1 π )(x π ) + O(x 3 ). () 5

6 (iii) (x + 1)/(x 1) as x 1. Solution. We will proceed as we did in the previous part: we introduce y x + 1 and write (x + 1)/(x 1) in terms of y y y = a n y n y = a 0 + (a n a n+1 )y n+1. (9) a 0 = 0, a 1 = 1, a = 1, 4 a 3 = 1,. a n = n,. (x + 1) (x 1) = (x + 1) (x + 1) 4 (x + 1)3 (x + 1)n n. (30) (iv) e x log( + x/) as x 0. Solution. we start with rewriting e x log( + x/) as e x log( + x ) = ex log[(1 + x 4 )] = ex [log + log(1 + x )]. (31) 4 Plugging the standard Taylor expansions of exponential and logarithm functions into eq. 31, we get the final answer e x log( + x x ) = log (1 + x + + O(x3 )) + (1 + x + x + O(x3 ))( x 4 + x 16 + O(x3 )) = log + (log + 1 )x + (log )x + O(x 3 ). (3) 7. Visualize the solution to the equation x = ɛe x by plotting the intersection between the line y = x/ɛ and the curve y = e x. Now solve it perturbatively by writing x = a nɛ n and equating powers of ɛ on both sides. Find a 0,1,. Show that dx/dɛ > 0, so that as you increase ɛ the location of x moves to the right. Does this agree with what you have on your plot? Solution. Figure 1 Shows the intersection of the line y = x/ɛ and the curve y = e x for three different values of ɛ. In order to solve x = ɛe x perturbatively, we write x = a nɛ n and equating powers of ɛ on both sides a 0 + a 1 ɛ + a ɛ + O(ɛ 3 ) = ɛ[1 (a 0 + a 1 ɛ + a ɛ + O(ɛ 3 )) + (a 0 + a 1 ɛ + a ɛ + O(ɛ 3 )) + ]. (33) 6

7 Figure 1: We can visualize the solution to the equation x = ɛe x by plotting the intersection between the line y = x/ɛ and the curve y = e x. The red curve represents y = e x and the green, orange, and blue lines represent y = x/ɛ for ɛ = 0.01, 0.1, 1 respectively. As it is clear in this figure, as ɛ gets bigger the location of intersection (i.e. the solution to the equation) moves further to the right (x gets bigger). keeping only terms up to order of ɛ, we get a 0 + a 1 ɛ + a ɛ + O(ɛ 3 ) = (1 a a 0)ɛ + (a 1 a 0 a 1 )ɛ + O(ɛ 3 ). (34) a 0 = 0, a 1 = 1, a = 1. Also, note that taking derivative of equation 35 w.r.t. ɛ we get x = ɛ ɛ + O(ɛ 3 ). (35) d dɛ x = 1 ɛ + O(ɛ ), (36) for ɛ small enough so that keeping only two terms makes sense, we can see that this derivative is positive which is exactly the behaviour we saw in Fig. 1.. The equation of motion of a pendulum of length l is θ(t) + ω sin θ(t) = θ, where ω = g/l. For small oscillations, and with initial conditions θ(0) = θ 0, θ(0) = 0, this has solution θ(t) = θ 0 cos ωt, because sin θ θ. (i) Now we want to solve the original equation perturbatively. parameter of the problem? Plug What is the small expansion θ(t) = θ 0 cos ωt + T (t)θ 0 + T 3 (t)θ O(θ 4 0). (37) into the equation and obtain T,3 (t). Be careful about the initial conditions; we want θ(0) = θ 0 and θ(0) = 0 to still hold. 7

8 Solution. By now you should know the procedure! We start with Taylor expanding sin θ and then plug in eq. 37 d dt (θ 0 cos ωt + T (t)θ 0 + T 3 (t)θ O(θ 4 0)) + ω [(θ 0 cos ωt + T (t)θ 0 + T 3 (t)θ O(θ 4 0)) 1 (θ 0 cos ωt + T (t)θ 0 + T 3 (t)θ O(θ 4 0)) 3 ] = 0. (3) We need only to keep terms up to order θ 3 0 and also we know that the differential equation for the first order in θ 0 is already solved by design. So { [ T (t) + ω T (t)]θ0 = 0, [ T 3 (t) + ω T 3 (t) ω cos 3 ωt]θ 3 0 = 0. { T (t) + ω T (t) = 0, T 3 (t) + ω T 3 (t) ω cos 3 ωt = 0. (39) Solving the differential equation for T (t) is easy; we already know that T will be of the form T (t) = c 1 cos ωt + c sin ωt. (40) In order to solve the differential equation for T 3 (t) we note that the final solution will be the sum of complementary solution (the solution of the homogeneous part i.e. T3 (t)+ω T 3 (t) = 0) and the particular solution. The complementary solution is just the same as the solution of T (t) i.e. T 3 (t) = c 31 cos ωt + c 3 sin ωt + particular solution. (41) In order to solve for the particular solution, we first simplify the equation using trigonometric identities and then solve the differential equation with the method of undetermined coefficients T 3 (t) + ω T 3 (t) = ω ω cos ωt + cos 3ωt. (4) 4 Then, the particular solution will be the sum of two particular solutions of { T3 (t) + ω T 3 (t) = ω cos ωt, T 3 (t) + ω T 3 (t) = ω cos 3ωt. (43) 4 The particular solution of T 3 (t) + ω T 3 (t) = ω cos ωt is of the form (T 3 ) p1 = t(b 1 cos ωt + b sin ωt), (44) where (b 1 cos ωt + b sin ωt) was multiplied by t to account for cos ωt in the complementary solution. The particular solution to T 3 (t) + ω T 3 (t) = ω cos 3ωt is of the form 4 Adding eq. 44 and eq. 45, we get (T 3 ) p = b 3 cos 3ωt + b 4 sin 3ωt. (45) (T 3 ) p = b 1 t cos ωt + b t sin ωt + b 3 cos 3ωt + b 4 sin 3ωt. (46)

9 Taking the second derivative and plugging it in the original equation we can calculate the coefficients b 1,,3,4. T 3 (t) = c 31 cos ωt + c 3 sin ωt + ωt 1 sin ωt cos 3ωt. (47) Combining eq. 37, eq. 40, and eq. 47, we get θ(t) = θ 0 cos ωt + [c 1 cos ωt + c sin ωt] θ0 [ + c 31 cos ωt + c 3 sin ωt + ωt ] 1 sin ωt cos 3ωt θ0 3 + O(θ ). (4) Imposing the initial conditions θ(0) = θ 0 and θ(0) = 0, we get the final answer θ(t) = θ 0 cos ωt + [ ] 1 ωt 1 cos ωt + sin ωt cos 3ωt θ0 3 + O(θ ). (49) (ii) As time t increases, the correction you found grows, until it becomes greater than the leading term θ 0 cos ωt. This is where our approximation breaks down and we have to include higher order terms. At what time does this happen? Solution. In order to find the time that this approximation breaks down we have to find the time that correction term gets bigger than the leading term. Therefore, we should find the time t c for which the amplitude of the term θ3 0 ωt 16 sin ωt gets big enough to be equal to θ 0 (we don t care about the two other terms because they are just oscillating and their amplitude does not change with time) θ 3 0ωt c 16 = θ 0 t c = 16 θ 0ω. (50) 10. Solve θ(t) + ω sin θ(t) = 0 iteratively. To do so, write the equation as θ(t) + ω θ(t) = ω [θ(t) sin θ(t)]. This way, all linear terms are on the left and all non-linear terms are on the right (similar to x = f(x) that we solved in the class). Then plug θ 1 (t) = θ 0 cos ωt as your first guess in the right hand side and compute θ (t) from θ (t) + ω θ (t) = ω [θ 1 (t) sin θ 1 (t)].keep only terms that are at most O(θ 3 0) and compare with the result of problem. Solution. We will Taylor expand sin θ and will only keep terms up to order θ 3 0 θ (t) + ω θ (t) = ω [θ 1 (t) θ 1 (t) + 1 θ3 1 + O(θ0)] 5 θ (t) + ω θ (t) = θ3 0ω cos 3 ωt. (51) Which is exactly the differential equation we solved for T 3 (t) in problem (remember that T (t) vanished when we imposed the initial conditions). these two methods lead to the same answer. 9