Homework 1 Solutions Math 587

Transcription

1 Homework 1 Solutions Math 587 1) Find positive functions f(x),g(x), continuous on (0, ) so that f = O(g(x)) as x 0, but lim x 0 f(x)/g(x) does not exist. One can take f(x) = sin(1/x) and g(x) = 1, for example. 2) Find all α so that f = O(ɛ α ) for the following: (a) f(ɛ) = ɛ tan(ɛ) (b) ln(ln(ɛ 1)) (a) Want to find all α such that lim ɛ tan(ɛ) ɛ <. Since lim α ɛ tan(ɛ) = 0 then for any α 0 have f(ɛ) = o(ɛ α ). Consider the case with α > 0. Taylor expanding tan(x) about x = 0 have so for ɛ near 0 have tan(x) = x x3 + O(x 5 ) ɛ tan(ɛ) ɛ α = ɛ 2 α + O(ɛ 4 α ) Thus, as long as α 2 have lim O(ɛ α ) for α 2. (b) Again want to find all α satisfying lim that lim ln(ln(1/ɛ)) = Thus, need α < 0 otherwise lim then ɛ tan(ɛ) ɛ α < and so ɛ tan(ɛ) = ln(ln(1/ɛ)) ɛ α ln(ln(1/ɛ)) ln( ln(ɛ)) lim ɛ α ɛ α So have ln(ln(1/ɛ)) = O(ɛ α ) for α < 0. <. First note ln(ln(1/ɛ)) ɛ α =. Suppose α < 0 1 ɛ ln(ɛ) αɛ α 1 1 αɛ α ln(ɛ) ɛ α α ln(ɛ) = 0 1

2 3) Suppose that f a 1 φ 1 (ɛ) + a 2 φ 2 (ɛ) + + a n φ n (ɛ), where a k 0. Show this is equivalent to m 1 f a k φ k (ɛ) a m φ m, m = 1,..., n In practical terms, this means that the m-th term in an asymptotic expansion can be found as an approximation to f minus the first m 1 terms in the approximation. By definition so that f m a k φ k (ɛ) = o(φ m (ɛ)) m 1 f a k φ k (ɛ) = a m φ m + o(φ m (ɛ)) = o(φ m 1 (ɛ)) where the last follows since φ k is well-ordered. 4) Find an infinite expansion of ln(1 + e ɛ 1 ) for ɛ 0. (Hint: factor out the dominant term inside the log first) ln(1 + e ɛ 1 ) = ln(e 1/ɛ (e 1/ɛ + 1)) = 1 ɛ + ln(1 + e 1/ɛ ) If ɛ << 1 then e 1/ɛ is very small so can Taylor expand ln(1 + x) about 0 to rewrite the log term in powers of e 1/ɛ. Continuing with the above have ln(1 + e 1/ɛ ) = 1 ɛ + ln(1 + e 1/ɛ ) = 1 ɛ + ( 1) n+1 n n=1 e n/ɛ 5) Find the two-term asymptotic expansion for the solutions of cos x = x/ɛ. Check your approximation for ɛ = 0.3 against a numerical solution (obtained by, for example, plotting cos x against x/ɛ). As ɛ 0 there is only one solution which is near 0. Trying the expansion x = x 1 ɛ + x 2 ɛ 3 + O(x 5 ) 2

3 and plugging it into the Taylor expansion of cos(x) at x = 0 have cos(x) = 1 x 2 /2 + O(x 4 ) = (x 1ɛ + x 2 ɛ 3 ) + O(ɛ 7 ) = x 1ɛ 2 O(ɛ 4 ) = 1 ɛ (x 1ɛ + x 2 ɛ 3 ) = x 1 + x 2 ɛ 2 Matching powers of ɛ have x 1 = 1, x 2 = 1/2. So two term expansion is x ɛ ɛ 3 /2 At ɛ = 0.3 the expansion has approximate solution x = The numerically obtained solution is x = ) (Bender Orszag 7.5) (a) Find expansion (at least to first order) for all roots of ɛx 3 +x 2 2x+ 1 = 0. (b) Consider ɛx 8 ɛ 2 x 6 + x 2 = 0 Clearly there is one O(1) root near x = 2. Looking for large roots which scale like ɛ α (where α > 0), show that there can be dominant balance between the first and last terms only, so the other root is (2/ɛ) 1/8. (c) Optional: do part (c) of this exercise. (a) When ɛ = 0 then x = 1 is a double root. So for ɛ small expect two roots each O(1). To find an appropriate expansion I tried x = 1 + x 1 ɛ β. Substituting this solution into the polynomial leads to the equation ɛ + 3x 1 ɛ β+1 + 3x 2 1ɛ 2β+1 x 3 1ɛ 3β x 1 ɛ 2β + x 2 1ɛ 2β 2 2x 1 ɛ 2β + 1 = ɛ + 3ɛ β+1 x 1 + 3x 2 1ɛ 2β+1 + x 3 1ɛ 3β+1 + x 2 1ɛ 2β = 0 Dominate balance can be obtained between the first and last term by setting β = 1/2. Then x 2 1 = 1 so have the two term approximation for the O(1) roots as x = 1 ± iɛ 1/2 To find leading order of the third root I tried x = ɛ α as a solution which leads to x 3 0ɛ 3α+1 + x 2 1ɛ 2α 2x 1 α + 1 = 0 3

4 Can get dominate balance between last 3 terms if α = 0 which leads to the O(1) roots. Can also achieve dominant balance if α = 1 in which case have x 1 = 1. So to leading order the third root is x = ɛ 1 To get a two term approximate I continued the expansion in powers of ɛ namely x = ɛ 1 + x 0. Substituting this into the polynomial and grouping powers of ɛ shows that x 0 = 1. Thus the two term approximation of the third root is x = ɛ 1 1 If ɛ > 0 then the roots of O(1) are complex whereas for ɛ < 0 the roots are real. (b) Supposing large root scales to leading order like ɛ α with α > 0 if we use the approximation x = x 0 ɛ α in the polynomial have x 8 1ɛ 8α+1 x 6 1ɛ 6α+2 + x 1 ɛ α 2 = 0 Now want to check for the possibility of dominant balance between various terms Case 1: First and Second Terms Then 8α + 1 = 6α + 2 thus α = 1/2. But want α > 0 so roots are large. Thus, cannot get dominant balance in this case. Case 2: First and Third Terms Then 8α + 1 = α so α = 1/7. In this case the first and third terms are O(ɛ 1/7 ) and the other terms are O(1) thus can have dominant balance between the first and third terms. Case 3: First and Last Terms Here α = 1/8 but in this case the first and last terms are O(1) but the third term is not O(ɛ 1/8 ) so don t have dominant balance between first and last terms. Case 4: Second and Third Terms In this case need α = 2/5 but then first terms is O(ɛ 11/5 ) while the second and third terms are O(ɛ 2/5 ). Case 5: Second of Last Terms Here α = 1/3 but while th second and last terms are O(1) the first term is now O(ɛ 5/3 ) so no dominant balance here. Case 6: Third and Last Terms This case requires α = 0 which will not find large roots. As a result dominant balance for roots which scale like x = ɛ α only occurs between the first and third terms. This balance is obtained when α = 1/7. In this case x 1 = 1 and so the large root 1/ɛ 1/7. 4

5 7) Let A, B be nonsingular n n matrices. Find a two term expansion of (A + ɛb) 1. (Hint: let C be equal to inverse so that C(A + ɛb) = I) Suppose C is of the form C = X 0 + ɛx 1 + O(ɛ 2 ) where X i are n n matrices. Substituting this into C(A + ɛb) = I and matching powers of ɛ have I = C(A + ɛb) = (X 0 + ɛx 1 + O(ɛ 2 ))(A + ɛb) = X 0 A + ɛ(x 1 A + X 0 B) + O(ɛ 2 ) So have X 0 A = I and X 1 A + X 0 B = 0 thus X 0 = A 1 X 1 A = X 0 B X 1 = X 0 BA 1 X 1 = A 1 BA 1 (O(1)) (O(ɛ)) And so C = (A + ɛb) 1 = A 1 ɛa 1 BA 1 + O(ɛ 2 ) 5