Solutions to Homework 2

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Claribel Gilmore
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1 Solutions to Homewor Due Tuesday, July 6,. Chapter. Problem solution. If the series for ln+z and ln z both converge, +z then we can find the series for ln z by term-by-term subtraction of the two series: + z ln = ln + z ln z z = z z + z z + z z z z = z + z + z An easy computation shows that if + x = + z/ z, then z = x/x +. For example if you want to use this series to find ln.5, then put z = /5. Problem 9 solution. a We enclose in the + st pair of parenthesis the terms with + digits none of which are 9 in the denominator. Then there are 8 choices for the first digit the numbers 8 and 9 choices for the remaining digits the numbers 8. This gives a total of = 8 9 possiblities, and so there are 8 9 terms in the + st pair of parenthesis. b In the + st pair of parenthesis there are 8 9 terms, each of which is less than the first term, /. So the sum enclosed in the + st pair of parenthesis is less than 8 9 /, and so the series is bounded by

2 c The above can be written as a geometric series: = 8 9 = 8 This shows that if we remove from the harmonic series all terms which contain a 9 in the denominator then the resulting series converges! Problem solution. a Let be a positive integer. Rationalizing the denominator gives = = Since + >, we have + + > and so + = < + + This gives the first inequality. For the second inequality note > + so = > + Putting these two inequalities together gives + < < b < < < <. n n < < n n n n + n < n < n n

3 Adding these equations and simplifying gives n + < n < n. Since = and = 6.555, putting n = in the above gives the estimate 6.7 < < 6.5. c Mathematica says = 6.8. d Since 9 + = 6.55 and 9 = 65.55, putting n =,,, = 9 in part b gives 6 < < 65,,, Mathematica says the true value of the sum above is about 6. e No. If the infinite series was bounded by a number M, then for all positive integers n we would have This is false for all n M + M. n + < M. Problem solution. Write γ x n = + 5 All the numbers inside the parenthesis are positive, so Similarly if we write γ x n = + γ x n > then all the terms in parenthesis are negative so γ x n <

4 To estimate the sums by integrals, note that on the interval, + we have /x > / + and /x > / +. Thus + x= /x dx > / + and < < + + x dx = n x dx = n x dx x dx. Similarly on the interval, + we have / > + /x dx and so > + dx x = dx n x. Putting these together gives the inequalities in the exercise. To prove Equation.5: it is easy to see that Equation.5 is equivalent to n > γ x n > n 6n. We have shown that n dx x > γ x n > dx n x dx n x. Evaluating the definite integrals gives the result. Problem 9 solution. Putting x = in Equation.5 gives + + +! + + n! e + + +! + + n! + e n!. This gives immediately the first inequality in Equation.55. To prove the second inequality, let a = + + +! + + n!. It remains to show that We have e n! n! a. e a + e n!. Doing a little algebra gives the result.

5 Problem solution. The difference between the two bounds in problem 9 is using the same notation as in the solution of problem 9 n! a n! = a. n! Using Mathematica we see that this is less than is n is at least. Chapter. Problem solution. We can assume α and β are both nonnegative since only their squares enter into the differential equation. Equating coefficients of r λ on both sides of Equations.59.6 gives λ λ + λ β =. It immediately follows that λ = ±β. For the physically interesting solution we tae the positive sign, λ = β >. Equating coefficients of r λ and replacing λ by β gives ββ a + β + a β a = This gives a β + =, or a = since β, we have β +. This gives Equation.6. Equating lie powers of r λ +j and replacing λ by β gives β + j β + ja j + β + ja j β a j + α a j =. Doing a little algebra gives Equation.6. From Equations.6.6 we have a = α a = β + = α β + a = α a = β + α α = β +!β + β + a 5 = a 6 = 6 β + α!β + β + α 6 =!β + β + β + α This gives Equation.65. 5

6 Problem solution. Equation.65 is not true for all functions. What you can prove is that if f is an infinitely differentiable function and n is a non-negative integer, then x ft dt = xfx x! f x+ x! f x+ + n xn n! f n x+ n+ x t n n! f n t dt. This is proved using induction. The case n = and n = is given in the exercise. If we assume the result is true for n, then applying integration by parts to the integral gives n+ x t n n! f n t dt = xn+ n+ n +! f n x + n+ which gives the result for n +. x t n+ n +! f n+ t dt Problem 9 solution. For the first part, assume n is a positive integer. Then the following inequalities are equivalent: n 99 > n > 99 > n n >. For the second part, note the hypothesis n α+ > implies α n+ = n α. If α n + x/n >, then α n + x > n n α x > n n x > α + x. Now n α + >, so x > also and n > α + x x = α + x. 6

7 Conversely, if x > and n > α + / x, then n > n + x x n x > α + x α n + x n >. If α + < : assume n is a positive integer. The inequality is equivalent to α + n x > n. Since α+ <, we have α+ n < and α+ n = n α = n+ α+. Then n + α + x > n n x > α + x If x >, then this holds for any positive integer n. If x = and n is a positive integer, then this holds only if α =. If x <, then n < α + x. Problem solution. The problem is that c depends on n also. As n increases, c might go to zero fast enough that / + c n+ does not go to zero. For example if c = /n + then lim n + n+ n+ = lim + n+ = e. n n + Problem 5 solution. For fx = ln + x, we have and so f n x = n n! + x n D n, x = n + c n x n n for some c between and x. Problem 6 solution. Put x = /99 and estimate + c n. Then D n, /99 /99n. n We must minimize F n = /99n n as a function of n. Computing F n we see the only critical point is at n = / ln/99 = Tae n =. 7

Review (11.1) 1. A sequence is an infinite list of numbers {a n } n=1 = a 1, a 2, a 3, The sequence is said to converge if lim

Review (11.1) 1. A sequence is an infinite list of numbers {a n } n=1 = a 1, a 2, a 3, The sequence is said to converge if lim Announcements: Note that we have taking the sections of Chapter, out of order, doing section. first, and then the rest. Section. is motivation for the rest of the chapter. Do the homework questions from