Problem 1 Kaplan, p. 436: 2c,i

Transcription

1 Mathematical Methods I - Fall 03 Homework Solutions Page Problem Kaplan, p 36: c,i Find the first three nonzero terms of the following Taylor series: c) ln + x) about x = 0 i) arctanh x about x = 0 The definition of a Taylor Series is: Part c fx) = n! x n x x 0 ) n ) x=x0 Using the definition from Equation for c), the summary in Table below is produced n x n x n x=x0 n! x n x=x0 n! x n x=x0 x x o ) n 0 log + x) log+x) +x x) log+x) +x) x 3 6 +x) + log+x) 3 +x) 6 x 3 3 log+x) +x) +x) Table : Development of Taylor Series for ln x + ) x Summing the elements of the final column, the three term approximation is: Part i f x x 3 + x Using the definition from Equation for c), the summary in Table below is produced n x n x n x=x0 n! x n x=x0 n! x n x=x0 x x o ) n 0 arctanh x x x x x ) x 3 x ) + 3 x ) 3 8x 3 x ) + x x ) x 5 x ) + 88x 5 x ) + x ) 3 5 Table : Development of Taylor Series for arctanh x x 3 3 x 5 5 Summing the elements of the final column, the three term approximation is: f x + x3 3 + x5 5

2 Mathematical Methods I - Fall 03 Homework Solutions Page Problem Kaplan, p 69: 5 Show that Satisfies Bessel s equation of order 0 in the form J 0 x) = Computing the fist and second derivatives of Equation as J 0x) = J 0 x) = n= ) n x n n!) ) xy + y + xy = 0 3) n ) n x n /n!) nn ) ) n x n /n!) n= Remapping these indices of the summation to n = 0 J 0x) = n + ) ) ) n x n+ /n + )!) J 0 x) = n + )n + ) ) n x ) n /n + )!) ) Placing the definitions from Equations and, into Equation 3 and combining like terms, [{ n + )n + ) ) + n + ) ) } + n + ) n ] x ) n+ /n + )! = 0 5) For this to be true for the summation it must hold for all values of x and n, therefore n + )n + ) ) + n + ) ) + n + ) = 0 6) And upon expanding Equation 6, we see that this is true and therefore the series from Equation satisfies Equation 3

3 Mathematical Methods I - Fall 03 Homework Solutions Page 3 Problem 3 Course notes, a Solve as a series in x for x > 0 about the point x = 0: x y xy + x + )y = 0 Examining Equation 7, and recognizing that is of the form y) = 7) y) = 0 P x)y + Qx)y + Rx)y = 0 Since P 0) = 0, is not an ordinary point Examining xqx)/p x) = and x Qx)/P x) = x + are both analytic at x = 0 therefore, this is a regular singular point problem Therefore there exist a solution for y is of the form: y = a n x n+r 8) Plugging this assumed form into the differential equation and simplifying yields, [ ] [ ] [ ] [ ] a n n + r)n + r )x n+r + a n n + r)x n+r + a n x n+r+ + a n x n+r = 0 Combining like powers of x into a single summation and simplifying the result [ ] [ ] a n [n + r) 3n + r) + ]x n+r + a n x n+r+ = 0 Removing the n = 0 term from the first summation and remapping the index back to zero, [ ] [ ] a 0 r 3r + )x r + a n+ [n + r + ) 3n + r + ) + ]x n+r+ + a n x n+r+ = 0 Combining the two summations a 0 r 3r + )x r + [a n + a n+ {n + r + ) 3n + r + ) + }]x n+r+ = 0 The term outside the summation gives us the indicial equation r 3r + = 0 Therefore r = 3± 5 Using one of these values, the term outside of the summation is zero and therefore the coefficient of x n+r+ inside the summation must be zero for all values of x and n Therefore, For r = 3+ 5 a n + a n+ n + r + ) 3n + r + ) + ) = 0 a n a n+ = n + r + ) 3n + r + ) + a n a n+ = n + + 5)n + ) = a 0 n + )! n+ i + 5) Therefore, the first solution to Equation 7 is, y = a 0 x 3+ 5) + 5 x x + ) ) 5 9 x 3 + i=

4 Mathematical Methods I - Fall 03 Homework Solutions Page For r = 3 5 b n b n+ = n + 5)n + ) = b 0 n + )! n+ i 5) Therefore, the second solution to Equation 7 is, y = b 0 x 3 5) + x ) ) ) x )x3 + Combining the two solutions, y = y +y, then considering the boundary conditions a 0 = and b 0 = Therefore, we can plot the solution below i= Exact Term Approximation 0 05 Figure : Exact and approximate solution to x y xy + x + )y = 0

5 Mathematical Methods I - Fall 03 Homework Solutions Page 5 Problem Course notes, Find two-term expansions for each of the roots of: where λ is large Multiplying out the terms: x )x + 3)x 3λ) + = 0 x 3 + 3λ)x 6λ + 3)x + + 9λ) = 0 Dividing through by λ, and substituting in a small number ɛ for λ, ɛx 3 + ɛ 3)x 6 + 3ɛ)x + ɛ + 9) = 0 9) Examining this problem, we see that as ɛ 0, a solution is lost Therefore, we must change variables, we can perform the transformation, X = x ɛ α 0) Using the transformation from Equation 0 on Equation 9, X 3 ɛ 3α+ 3X ɛ α + X ɛ α+ 3Xɛ + )ɛ α + ɛ + 9 = 0 ) The two highest order terms of X are in the same order of ɛ if + 3α = α, Therefore, we demand that α = With this, Equation becomes, Let x be expressed as a sum in ɛ, X 3 + X ɛ 3) 3Xɛɛ + ) + ɛ ɛ + 9) = 0 ) X = X n ɛ n X 0 + X ɛ + X ɛ + X 3 ɛ 3 3) Therefore, separating on orders of ɛ and recognizing that since the expansion is 0 each term must also equal 0, the equations become X 3 0 3X 0 = 0 3X X 0 + X 0 6X X 0 6X 0 = 0 3X X 0 + 3X X 0 + X X 0 6X X 0 3X 0 3X 6X + 9 = 0 X 3 + X + 6X 0 X X 6X X 3X + X 0 X 6X + 3X 0 X 3 6X 0 X 3 + = 0 Sequentially solving these equations until we have two terms for each solution, we find X 0, X, X ) = 0, 3, ) X 0, X, X ) = 0,, ) X 0, X, X, X 3 ) = 3, 0, 0, ) 9 ) Then combining the values from Equation and the initial approximation from Equation 3, we find, Therefore X ɛ ɛ 3ɛ, + ɛ or 3 ɛ3 9 x ɛ 3, ɛ +, or 3 ɛ ɛ 9 x λ 3, +, or 3λ λ 9λ

6 Mathematical Methods I - Fall 03 Homework Solutions Page 6 Problem 5 Course notes, b Find all solutions through Oɛ ), where ɛ is a small parameter, and compare with the exact result for ɛ = 00 ɛx + ɛ + )x ɛ)x 5x = 0 5) Starting this problem, we see that as ɛ 0, a solution is lost Therefore, we must change variables, we can perform the transformation, X = x ɛ α 6) Using the transformation from Equation 6 on Equation 5, ɛ α+ X + ɛ 3α+ + ɛ 3α )X 3 + 7ɛ α ɛ α+ )X 5ɛ α X = 0 7) The two highest order terms of X are in the same order of ɛ if α + = 3α, Therefore, we demand that α = With this, Equation 7 becomes, ɛ 3 X + ɛ + ɛ 3 )X 3 + 7ɛ ɛ )X 5ɛ X = 0 8) In order to solve this problem, we start by assuming that X can be written in the form, X = X n ɛ n = X 0 + X ɛ + X ɛ + X 3 ɛ 3 + 9) Then placing the expansion from Equation 9 into Equation 8, we find, ɛ 3 X 0 + X ɛ + X ɛ + X 3 ɛ 3 + ) + ɛ + ɛ 3 )X 0 + X ɛ + X ɛ + X 3 ɛ 3 + ) 3 + 0) 7ɛ ɛ )X 0 + X ɛ + X ɛ + X 3 ɛ 3 + ) 5ɛ X 0 + X ɛ + X ɛ + X 3 ɛ 3 + ) = 0 Expanding this and separating by powers of ɛ and recognizing that if the sum of all terms is 0 then each order of ɛ must be 0 as well Examining these equations, ɛ 0 X 0 X 3 0 = 0 ) ɛ 8X X 3 0 X 3 0 6X X 0 7X 0 = 0 ɛ 8X X 3 0 X X 0 X X 0 6X X 0 + X 0 6X X 0 X X 0 + 5X 0 = 0 ɛ 3 8X 3 X 3 0 X X X 0 X X 0 6X 3 X 0 8X 3 X 0 X X 0 + X X 0 X X X 0 X X 0 X 3 7X + 5X + = 0 Solving Equations we see, X 0, X, X, X 3 ) = 0,, 3 ) 5, 33, 0, 875,, ), 0,, 378 5, 75 ),, , 7, 7 ) Therefore using the definition from Equation 9, the data from Equation and the transformation defined in Equation 6, the Oɛ ) solutions are, X = ɛ 3 5 ɛ ɛ3 x = ɛ 3 5 ɛ X = ɛ ɛ 378 ɛ3 x = 378 ɛ ɛ X = ɛ 5 ɛ ɛ3 x = ɛ 5 ɛ + X = + 3 ɛ + 7 ɛ + 7 ɛ3 x = 7 ɛ + 7 ɛ + 3 ɛ Now, in the case of ɛ = 00, our estimate of the roots of Equation 5 would be While the exact solution is, x = 06757, , , 989 x = 06783, , , 9887 The Oɛ ) approximation provides an excellent approximation of the roots, the maximum relative error is )

7 Mathematical Methods I - Fall 03 Homework Solutions Page 7 Problem 6 Course notes, Find three terms of a solution of x + ɛ cos x + ɛ) = π 3) where ɛ is a small parameter For ɛ = 0, compare the best asymptotic solution with the exact solution In order to solve this problem, we start by assuming that x can be written in the form, x = x n ɛ n = x 0 + x ɛ + x ɛ + x 3 ɛ 3 + ) Then placing the approximation from into Equation 3, we find, x 0 + x ɛ + x ɛ + x 3 ɛ 3 + ) + ɛ cos x 0 + x ɛ + x ɛ + x 3 ɛ 3 + ) + ɛ) = π 5) Performing a Taylor series expansion in ɛ about ɛ = 0, splitting this up by powers of ɛ, ɛ 0 : x 0 π = 0 6) ɛ : x + cosx 0 ) = 0 ɛ : x x + ) sinx 0 ) = 0 ɛ 3 : x 3 x sinx 0 ) x + ) cosx 0 ) = 0 Solving for the terms in Equation 6, we find x 0 = π, x = 0, x =, x 3 = Therefore the three term asymptotic solution of Equation 3 is, x = π + ɛ + ɛ 3 In the case of ɛ = 0, this yields x Appx = 6668, while the exact solution is x Exact = 6658

8 Mathematical Methods I - Fall 03 Homework Solutions Page 8 Problem 7 Course notes, 6 The solution of the matrix equation A x = y can be written as x = A y Find the perturbation solution of where ɛ is a small parameter Assuming x is of the form x = A + ɛb) x = y, 7) ɛ n x n = x 0 + ɛx + ɛ x + 8) This means that Equation 7 can be written as, A + ɛb) x 0 + ɛx + ɛ x + ) = y 9) Distributing through the dot product and grouping by powers of ɛ, Equation 9 becomes, A x 0 y) + ɛb x 0 + A x ) + ɛ B x + A x ) + = 0 30) Since the summation of the terms is zero, each power of ɛ must be as well therefore, Solving each line for the unknown x, Back substituting, we find that in general, A x 0 = y B x 0 + A x = 0 B x + A x = 0 x 0 = A y x = A B x 0 x = A B x x n = A B A ) n y 3) Therefore using the definition of x from Equation 8 and the values of x n from Equation 3, the perturbation solution of Equation 7 is, [ x = A ɛb A ) ] n y

9 Mathematical Methods I - Fall 03 Homework Solutions Page 9 Problem 8 Course notes, 7 Find all solutions of ɛx + x = 0 3) approximately, if ɛ is small and positive If ɛ = 000, compare the exact solution obtained numerically with the asymptotic solution Note as ɛ 0, the equation becomes singular Let Using the transformation from Equation 33 on Equation 3, X = x ɛ α 33) ɛ α+ X + ɛ α X = 0 3) The two highest order terms of X are in the same order of ɛ if α + = α, Therefore, we demand that α = 3 With this, Equation 3 becomes, ɛ 3 X + ɛ 3 X = 0 35) In order to solve this problem, we start by assuming that X can be written in the form, X = Then placing the expansion from Equation 36 into Equation 35, we find, X n ɛ n 3 = X0 + X ɛ 3 + X ɛ 3 + X3 ɛ + 36) X 0 + X ɛ 3 + X ɛ 3 + X3 ɛ + ) + X 0 + X ɛ 3 + X ɛ 3 + X3 ɛ + ) ɛ 3 = 0 37) Expanding this and separating by powers of ɛ and recognizing that if the sum of all terms is 0 then each order of ɛ must be 0 as well Examining these equations, ɛ 0 X 0 + X 0 = 0 38) ɛ 3 X X X = 0 ɛ 3 X X X X 0 + X = 0 ɛ X 3 X X X X 0 + X 3 X 0 + X 3 = 0 Solving Equations 38 we see, X 0, X, X, X 3 ) =, 3, 8 ) 9, 60, 39) 8 0,, 0, 0), 3, 3, 8 3 ) 60 3 ),, 9 8 ) /3), ) 8 60) )/3 ) /3, 3, 8 9 Therefore using the definition from Equation 36, the data from Equation 39 and the transformation defined in Equation 33, the Oɛ) solutions are, X = 8ɛ/ ɛ 60ɛ 8 X = 3 ɛ X = 8 3 ) ɛ / ɛ ɛ X = 8 ) /3) ɛ / ɛ 60 8 )/3 ɛ ) /3

10 Mathematical Methods I - Fall 03 Homework Solutions Page 0 Performing the inverse transformation to convert X to x, x = 60ɛ/ ɛ 9 3 ɛ 3 x = 8 3 ) 9 ɛ /3 3 x = 3 ɛ ɛ ɛ x = 7 + ) /3) ɛ /3 5 3 ɛ 60 ) /3 ɛ 8 ) /3 8 3 ɛ Using this approximation when ɛ = 000, we find the approximate roots are and the exact roots are x = x = x 3 = i x = i x = 0593 x = 989 x 3 = i x = i

11 Mathematical Methods I - Fall 03 Homework Solutions Page Problem 9 Course notes, 8 Obtain the first two terms of an approximate solution to ẍ ɛ)ẋ + x = 0, with 0) x0) = + ɛ), ẋ0) = 3 + ɛ), for small ɛ Compare with the exact solution graphically in the range 0 t for a) ɛ = 0, b) ɛ = 05 and c) ɛ = 05 Letting x be of the form x = x 0 + ɛx + Therefore the second derivative takes the form, ẍ = ẋ 0 + ɛẋ + and the second derivative takes the form, ẍ = ẍ 0 + ɛẍ + Equation 0 then becomes, ẍ 0 + ɛẍ + ) ɛ)ẋ 0 + ɛẋ + ) + x 0 + ɛx + ) = 0, with x0) = + ɛ), ẋ0) = 3 + ɛ) Combining the sums with the same power of ɛ, and recognizing that all orders of ɛ are linearly independent, since the of all powers of ɛ must be zero, each order of ɛ sums to zero, ie, ɛ 0 : x 0t) + 3x 0t) + x 0 t) = 0, x 0 0) =, x 0) = 3 ɛ : x t) + 3x t) + x t) + 3x 0 = 0, x 0) =, x t) = 6 Solving these sequentially note that since the initial conditions are of mixed order of ɛ they are separated as well), x 0 = e t e t + ) x = e t 6t + e t 3t + ) + ) We find the exact and the two term approximate solution to Equation 0 to be, x Appx = e t 6tɛ + e t 3tɛ + ɛ + ) + ɛ + ) Plotting the cases of ɛ = 0, ɛ = 05 and ɛ = 05

12 Mathematical Methods I - Fall 03 Homework Solutions Page x 0 Exact 5 Approximation t Figure : = 0 x 5 0 Exact Approximation t Figure 3: = 05 x 30 5 Exact 0 Approximation Figure : = 05 0 t

13 Mathematical Methods I - Fall 03 Homework Solutions Page 3 Problem 0 Course notes, 58 Find the solution of the transcendental equation near x = π for small positive ɛ If we substitute x = x 0 + ɛx + ɛ x + into Equation, we find sin x = ɛ cos x, ) sin x 0 + ɛx + ɛ x + ) = ɛ cos x 0 + ɛx + ɛ x + ) ) Performing a Taylor Series on Equation about ɛ = 0 and collecting by powers of ɛ, ɛ 0 : sinx 0 ) = 0 3) ɛ : cosx 0 )x cosx 0 ) = 0 ɛ : sinx 0)x + sinx 0 )x + cosx 0 )x = 0 ɛ 3 : 6 cosx 0)x 3 + cosx 0 )x sinx 0 )x x + sinx 0 )x + cosx 0 )x 3 = 0 Solving Equation 3, we see x 0 = π, x =, x = 0, x 3 = 6 Therefore x = π ɛ + 6 ɛ3 +