Math221: HW# 7 solutions

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1 Math22: HW# 7 solutions Andy Royston November 7, let x = e u. Then ln x = u, x2 = e 2u, and dx = e 2u du. Furthermore, when x =, u, and when x =, u =. Hence x 2 ln x) 3 dx = e 2u u 3 e u du) = e 3u u 3 du. ) In the last step I used the minus sign to flip the limits of integration. Note that, if we want to apply the definition of the Gamma function to do an integral, it is necessary that the limits of the integral go from to. If they don t, it s not a Gamma function. Also, it s necessary that we have e t in the integrand, where t is the integration variable. In this case, we do not, so we need to make another change of variables. Let t = 3u du = dt/3. Note that the limits do not change: e 3u u 3 du = e t t 3 dt = 3 3 e t t dt = Γ). 2) 3 In this case we can simplify the result, using Gamman) = n )!, for n N. Thus x 2 ln ) 3 dx = 3! x 3 = ).3.6 The initial conditions are x) =, ẋ) =. The Lagrangian is The Euler-Lagrange equation is d dt L = T V = 2 mẋ2 m ln x. ) 2 ) L L ẋ x = d dt mẋ) m 2x ) = ẍ + 2x =. 5)

2 This is the second order differential equation that we need to solve. It is of the type d2 x = F x), dt 2 with F x) =. See formulas 7.) and 7.5), in section 7 of chapter 8. It can be integrated 2x once if we multiply each side by ẋ. The result is 2ẋ2 dx = F x)dx = 2x + c = 2 ln x + c, or ẋ2 = ln x + c. 6) This gives us ẋ. We still need to integrate again to get xt). First, however, use the initial conditions to solve for the integration constant c. At time t = we know that x =, ẋ =. Plugging these in, Then = ln + c = + c c =. 7) ẋ 2 = ln x = ln x ẋ = ± ln x dx ln x ln x = dt. 8) Note that since we know x <, ln x is negative, so ln x = ln is positive. Furthermore, we chose x the minus sign from taking the square root because we know that the particle s velocity is negative; it s traveling from x = to the origin. Now, let the time T be the time it takes for the particle to reach the origin. Then T = T dt = dx ln x = dx. 9) ln x To do the x integral, make the change of variable x = e u, so that ln x = u, dx = e u du. Also, x = u =, x = u =. Then T =.5. We want to prove dx ln x = = e u du u = e u u /2 du e u u /2 du = Γ 2 ) = π. ) 3 2n ) 2n)! Γn + /2) = π = π. ) 2 n n n! Use the fundamental property Γp + ) = pγp) for any p >. Then 2

3 Then just keep going: Γn + /2) = Γn /2) + ) = n /2)Γn /2). 2) n /2)Γn /2) = n /2)Γn 3/2) + ) = n /2)n 3/2)Γn 3/2) = n /2)n 3/2)Γn 5/2) + ) = n /2)n 3/2)n 5/2)Γn 5/2) =.... 3) Now, since n is some fixed integer, this process must come to an end. Eventually, I have to get down to Γ/2). Then at this point, I can t go any further. Γp + ) = pγp) only holds for p >. However, I do know that Γ/2) = π. Thus, Γn + /2) = n /2)n 3/2)n 5/2) 3/2)/2)Γ 2 ) n /2)n 3/2)n 5/2) 3/2)/2) π. ) Now, observe that this is a product of precisely n terms multiplying the π. Multiply and divide by 2 n. In the numerator, distribute a factor of 2 to each term. Then Γn + 2n )2n 3)2n 5) 3 ) = π, or 2 2 n Γn + 3 2n ) ) = π, 5) 2 2 n which is the first equality. For the second equality, note that the numerator above is a product of all odd numbers from to 2n. Let us multiply and divide by the product of all even numbers from 2 to 2n. Then the top will just be the product of all integers from to 2n, otherwise known as 2n)!: 3 2n ) 2 3 2n 2)2n )2n) π = 2 n 2 n 2 2n) 2n)! = π. 6) 2 n 2 2n) Now the product of even numbers from 2 to 2n) has n terms. Pull out a 2 from each: Hence, noting 2 n 2 n = 2 2n = 2 2 ) n = n, 2 2n) = 2 n 2 n) = 2 n n!. 7) 3 2n ) 2n)! 2n)! π = π = π, 8) 2 n 2 n 2 n n! n n! which is the second equality. 3

4 .7.2 We have sin 3 x cos xdx = 2 2 sin 3/2 x cos /2 xdx. 9) Let 2p = 3/2 p = 5/ and 2q = /2 q = 3/ and compare with formula 6.) of the text. Then 2 2 Writing this in terms of Gamma functions, sin 3/2 x cos /2 xdx = 2 B5, 3 ). 2) 2 B5, 3 ) = Γ 5)Γ 3) 2 Γ 5 + 3) = Γ 5)Γ 3) 2 Γ2) = 2 Γ5 )Γ3 ), 2) where in the last equality I used Γ2) =! =. Now there is some trickery that can be done. Write Then Γ 5 ) = Γ + ) = Γ ). 22) 2 Γ5 )Γ3 ) = 8 Γ )Γ3 ). 23) The reason I did this is because now we have it in the form Γp)Γ p), where p = /. We can use the mystical result, equation 5.): Γp)Γ p) = Hence, the whole result is π sin πp) Γ )Γ3 ) = π sin π/) = π / 2 = π 2. 2) 2 B5, 3 ) = 8 π ).7.7 Observe that if p = /, q = /2. Hence dθ = sin θ 2 2 sin 2p θ) cos 2q θ)dθ, 26) dθ = sin θ 2 B, 2 ) = Γ )Γ ) 2 π Γ 2 Γ 3) = ) 2 Γ 3 27) ). I don t know any way to simplify this further, and neither does Steven Wolfram, so this is the final answer. Numerically, it turns out to be

5 π Γ ) 2 Γ ) ).7.9 First we show Bn, n) = Bn, ) 2. 29) 22n Start with the trig expression for Bn, n) and use the trig identity as suggested: Bn, n) = 2 = 2 sin 2n θ) cos 2n θ)dθ = 2 2 sin 2θ))2n dθ = 2 2 2n sin θ cos θ) 2n dθ sin 2θ)) 2n dθ. 3) Now let s change variables, φ = 2θ. Then dθ = dφ/2, and note that the upper limit changes: when θ = π/2, φ = π. Thus, 2 2 2n sin 2θ)) 2n dθ = 2 2n π sin φ) 2n dφ = 2 2n π sin φ) 2n cos φ) 2 /2 dφ. 3) Note in the last step that cos φ) 2 /2 = cos φ) =, so I didn t do anything. Now, this looks like a Beta function except that the upper limit is wrong. But observe that sin φ is symmetric about π/2 from to π. The area under the graph from to π/2 equals the area under the graph from π/2 to π. Hence, this is also true of sin φ) 2n. Then the integral from to π is just twice the integral from to π/2. The result follows: Bn, n) = 2 2n π sin φ) 2n cos φ) 2 /2 dφ = 2 2n 2 sin φ) 2n cos φ) 2 /2 dφ = 2 Bn, ). 32) 2n 2 From this result, we can derive the duplication formula for the Gamma function. Simply use the formula Bp, q) = Γp)Γq) Γp + q) on each side of the above result. On the left side, 33) Bn, n) = Γn)Γn). 3) Γ2n) 5

6 On the right side Now set them equal and get the result: 2 2n Bn, 2 ) = 2 2n Γn)Γ 2 ) Γn + /2) = π 2 2n Γn) Γn + /2). 35) Γn)Γn) Γ2n) = π 2 2n Γn) Γn + /2) Γ2n) = 22n Γn)Γn + /2) π. 36) Finally, let us check this formula for n = /. On the left side, Γ2n) Γ2 ) = Γ/2) = π. 37) On the right side, we will need the result Γp)Γ p) = π/ sin πp). We have So the formula checks out. 2 2 / Γ/)Γ3/) = π π 2π sin π/) = 2π π 2 = π. 38).. Use Stirling s formula in the numerator and denominator. In the limit n, it becomes exact...5 We have 2n)! n lim n 2 2n n!) = lim 2n) 2n e 2n 2π2n) n 2 n 2 2n n n e n 2πn) 2 2 2n n 2n e 2n 2n π = lim n 2 2n n 2n e 2n 2πn = lim =. 39) n π π lim n n + /2) n+/2 e n+/2) 2πn + /2) = lim n nnn e n 2πn n + /2 = lim e /2 n n 6 Γn ) nγn + ) = lim Γn + /2) + /2) n nγn + ) n + /2 e /2 n n+/2 n ) n+/2+/2 = e /2 lim + n 2n )n+. ) = lim n n + /2) n+/2

7 For the last limit, put it in a more standard form: lim + n 2n )n+ = lim + n 2n )n lim + n 2n ) = lim + n = lim n + 2n )2n = 2n )n lim + n 2n )2n. ) Here I pulled off one factor of + /2n), and took its limit, which is. Then I introduced the square root so that the remaining limit has the form + x) x, 2) lim x just let x = 2n). This limit is something we all supposedly learned in high school... The answer is e. If you just quote this or look it up, that s fine by me, but it s actually pretty easy to prove. The trick is write it as an exponential of a log: lim x + x) x = lim x eln +/x)x) = lim x e x ln +/x). 3) Now bring the limit in the exponential; you can do this bring the limit inside the function with any continuous function. That s what it really means to be continuous). Then observe that, using the Taylor series ln + ɛ) = ɛ + Oɛ 2 ), for ɛ = /x small, lim x ln + /x) = lim x/x + x x O/x)2 )) = lim + O/x)) =. ) x Therefore, the thing in the exponential goes to, and we get + x) x = e. 5) Plugging this result into our problem, lim x lim n Γn ) nγn + ) = e /2 lim n + 2n )n+ = e /2 e =. 6) 2..2 Let s first solve the equation by pedestrian methods: y = 3x 2 y dy y = 3x2 dx ln y = x 3 + c 7 yx) = Ae x3. 7)

8 Now for the power series solution, assume yx) = a n x n = a + a x + a 2 x 2 + a 3 x 3 + 8) n= for some coefficients a n that we need to solve for. Then 3x 2 y = 3a x 2 + 3a x 3 + 3a 2 x + + 3a n 2 x n +. 9) The derivative is y = a + 2a 2 x + 3a 3 x n + )a n+ x n +. 5) Now write out y 3x 2 y =, grouping powers of x. y 3x 2 y = a + 2a 2 x + 3a 3 3a )x 2 + a 3a )x 3 + +n + )a n+ 3a n 2 )x n + =. 5) Now here s why we did this. The above equation has to hold for all x. It s just the original differential equation, with our guess solution plugged in). Since the x n are linearly independent functions for each n =,, 2,..., the only way the whole thing will be zero is if the coefficient of every power of x is zero. Thus, we get a list of equations: a =, 2a 2 =, 3a 3 a ) =, a 3a ) =, 5a 5 3a 2 ) =, 6a 6 3a 3 ) =,... n + )a n+ 3a n 2 ) =, 52) In particular, we learn that a = a 2 =, but a 3 = a. Then a a = and a 5 a 2 =, but a 6 = 2 a 3 = 2 a. This pattern continues; the general term tells us that a coefficient a n is always connected to three back, a n 3. We will have a 7 = a 8 =, but 9a 9 = 3a 6 a 9 = 3 a 6 = 3 2 a 3 = You can now guess the pattern: a. 53) 8

9 a 3m = mm ) 2 a, a 3m = a 3m 2 =, m N. 5) Thus we have determined all of the coefficients in terms of a single undetermined one, a. This is expected. We have one undetermined coefficient because we started with a first order equation. Let us now write down the solution: yx) = a n x n = a + n= a 3m x 3m = a + a m= = a m= m= m! x3m Hence we get the correct result, with a serving as A, the arbitrary constant. m! x3 ) m = a e x3. 55) 2..7 The differential equation is x 2 y 3xy + 3y =. This is a Cauchy type equation. To get the solution, try y = x k. Then x 2 kk )x k 2 3xkx k + 3x k = kk ) 3k + 3)x k = k 2 k + 3 = k 3)k ) = k =, 3. 56) Therefore the general solution is yx) = Ax + Bx 3. 57) Let s try it by power series. y = a + a x + a 2 x 2 + a 3 x 3 + a x + y = a + 2a 2 x + 3a 3 x 2 + a x 3 +, y = 2a a 3 x + 3a x a 5 x ) Then x 2 y = 2a 2 x a 3 x 3 + 3a x + + nn )a n x n +, 3xy = 3a x + 3)2a 2 x 2 + 3)3a 3 x )na n x n +. 59) Now write down the differential equation term by term in x: x 2 y 3xy + 3y = 3a )a x )a 2 x )a 3 x nn ) 3n + 3)a n x n + = 6)

10 Every coefficient must vanish. Look at the general term, which holds for all n: nn ) 3n + 3)a n = n 3)n )a n =, n =,, 2,.... 6) If n =, 3 then this is satisfied, and there are no restrictions on a, a 3. However, for all other n, we must require a n =. Hence, the solution is yx) = a n x n = a x + a 3 x 3. 62) n= This is the solution we had before with a = A, a 3 = B The equation is x 2 + 2x)y 2x + )y + 2y =. I actually don t see a way to solve this, except by power series. So let s find the power series solution first and then check the solutions we get. y, y, and y are as in equation 58) above. All the terms we need are x 2 y = 2 a 2 x a 3 x 3 + 3a x +, 2xy = 2)2 a 2 x + 2)3 2a 3 x 2 + 2) 3a x 3 +, 2xy = 2)a x + 2)2a 2 x 2 + 2)3a 3 x 3 + 2)a x +, 2y = 2)a + 2)2a 2 x + 2)3a 3 x 2 + 2)a x 3 +, 2y = 2a + 2a x + 2a 2 x 2 + 2a 3 x 3 + 2a x +. 63) Adding everything up, grouping in powers of x, 2a + 2a ) + 2)2 a 2 2a + 2a 2 ) + 2a )x +2 a 2 + 2)3 2a 3 22a 2 + 3a 3 ) + 2a 2 )x a 3 + 2) 3a 23a 3 + a ) + 2a 3 )x nn )a n + 2)n + )na n+ 2na n + n + )a n+ ) + 2a n )x n + =. 6) The general term actually gives the correct form for all n =,, 2,.... Hence, we require nn )a n + 2)n + )na n+ 2na n + n + )a n+ ) + 2a n ) =, n =,,... nn ) 2n + 2)a n + 2nn + ) 2n + ))a n+ = n 2 3n + 2)a n + 2n + )n )a n+ = n 2)n )a n + 2n + )n )a n+ =. 65) Now, as long as n, we can divide out the n ) and obtain a n+ = n 2) 2n + ) a n, n. 66)

11 For n = we have a = a. For n =, we learn nothing about the a s. Equation 65) is automatically satisfied. Now, for n = 2, the equation says a 3 = a 2 =. Then, from this, a = 2 =, a 5 a =, ) That is, all of the a n, starting with a 3 are. Note that we never learn anything about a 2. Summarizing, Hence, yx) is given by a = a = arbitrary const. yx) = a 2 = arbitrary const. a n =, n 3. 68) a n x n = a + x) + a 2 x 2, 69) n= where a, a 2 are arbitrary constants. You can plug in each of these functions, + x) and x 2, and verify that they are indeed solutions to the differential equation Legendre s equation is x 2 )y 2xy + ll + )y =. The general series solution is +a [ x y = a [ l )l + 2) x 3 + 3! ] x + ] ll + ) x 2 ll + )l 2)l + 3) + 2!! l )l + 2)l 3)l + ) x 5 + 5!. 7) When l is an integer, one of these series will terminate after a finite number of terms. These special solutions are the Legendre polynomials P l x). To find them, just see which series is going to stop, and write down all the nonzero terms. For P 2 x), look at the top series. At the third term, and in all subsequent terms, there is the factor l 2), which will vanish. Therefore we are interested in the solution yx) = a ) 22 + ) x 2 = a 3x 2 ). 7) 2! In order to make it the Legendre polynomial P 2 x), we fix the constant a by demanding P 2 ) =. This gives Hence, = a 3) a = 2. 72)

12 P 2 x) = 2 3x2 ). 73) For the next one, l = 3, look at the lower series. The third term, and all terms afterwards, have the factor l 3). Therefore, we are interested in Normalize by P 3 ) = : yx) = a x ) 3 )3 + 2) x 3 = a x 5 3! 3 x3 ). 7) Thus = a 5/3) = a 2 3 a = ) P 3 x) = x3 x) = 2 5x3 3x). 76) Finally, for l =, the top series will terminate at the fourth term. The fourth term is given by ll + )l 2)l + 3)l )l + 5) x 6. 77) 6! You can see this by the recursion relation, formula 2.6) of the text, or just observe the pattern. The point is that it, and every term after it, contain the factor l ), and so vanish. So write Normalizing, yx) = a + ) x 2 2! ) + ) 2) + 3) x! = a x x ). 78) Hence, = a ) = a ) = 8 3 a a = ) P x) = 3 8 x x ) = 8 3 3x2 + 35x ). 8) 2

13 2.2.2 Observe that in the previous problem, we always used the top series for l even and the bottom series for l odd. This will always be the case because the general term in the top series has structure while the general term in the bottom series has structure ll 2)l ) l + )l + 3), 8) l )l 3)l 5) l + 2)l + ). 82) So if l is even, and remember, the Legendre polynomials are only defined for l ), then the top series will eventually terminate, while the bottom does not. On the other hand, if l is odd, the bottom series will be the one that terminates. Now, the top series contains only even powers of x, so it is an even function. The bottom series contains only odd powers of x, so it is an odd function. Therefore we learn in general P l x) = P l x), l even; P l x) = P l x), l odd. 83) Now apply this result to x =. There we know, by definition really), that P l ) =, for any l, even or odd. Hence, = P l ) = P l ), l even; = P l ) = P l ), l odd. 8) This may be summarized as P l ) = ) l. 85) 3