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FSU STA 5446 - Homework

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18 100B Fall 2002 Homework 7 Was due by Noon Tuesday November 5 This was a bit of a stinker Rudin 1 Chapter 5 Problem 12 Solution In x 0 x 3 x3 so is in nitely di erentiable being a polynomial and has derivative 3x2 Similarly in x 0 x 3 x3 is again a polynomial and has derivative 3x2 The limit lim 0 t 0 f 0 f t lim t 3 t 0 0 t 0 0 t so f is di erentiable at 0 and f x 3x x everywhere As already noted this is di erentiable in x 0 and has derivative 6 x The limit 1 f 0 f t lim 3 t 0 0 t 0 0 t 0 0 t lim again exists so f x 6 x exists everywhere Finally the third derivative exists for x 0 and is f 3 x 6 sgn x sgn x 1 as x 0 or x 0 The limit of f 0 f t f 0 f t 6 sgn t 0 t 0 t does not exist as 0 t 0 so f 3 0 does not exist 2 Chapter 5 Problem 14 Solution By assumption f x is convex and di erentiable on a b Thus f tx 1 t y tf x 1 t f y t 0 1 x y a b For any three points x y z a b the di erence quotient satis es f x f y f x f z f y f z x y x z y z as shown last week Letting y x in the rst inequality and using the di erentiability of f shows that f x f x f z f y f z x z y z where x y z are again any points satisfying x y z Now letting y z we conclude that f x f z if x z Conversely suppose f x is monotonically increasing on a b Using the mean value theorem if x z then f z f x z x f q for some x q x z so f x f z f f z For three points x z y this z x gives f z f x f z f y z x z y y z and setting t y x so z tx 1 t y this is precisely f tx 1 t y tf x 1 t f y t 0 1 which is convexity 1 2 If f x exists for all x a b and f 0 then f x is increasing and so f is convex Conversely if f is convex then f is increasing and hence f 0 3 Chapter 5 Problem 15 I should have said not to do the last part since I have not talked much about di erentiation of vector valued functions Solution The question is quite as clear as should be you are supposed to assume that M0 and M2 are nite Following the hint recall that Taylor s theorem shows that 2h 2 f 2 for some x x 2h which can be written f x 2h f x 2hf x f x 1 f x 2h f x hf 2h Thus 1 f x 2h f x h f 2h and so with M0 an upper bound for f and M2 and upper bound for f f x M0 h 0 x a h Taking the supremum over x for each h 0 we nd f x hM2 M0 h 0 h We can assume M0 M2 0 since if M2 0 then f is linear and M0 is in nite If M0 0 then f 0 The right side is di erentiable in h with derivative M2 h 2 M0 This vanishes when h M0 M2 0 substituting this gives M1 2 M0 M2 M12 4M0 M2 M1 hM2 For the given f x we see that f x 2x2 1 x2 1 x2 1 4x 4x x2 1 2 1 x 0 0 x 1 x 0 0 x also exists at 0 where it has the value 0 Then f x also exists at 0 taking the value 4 and 4 1 x 0 f x 4 1 x2 0 x x2 1 3 Now f 0 in x 0 and f 0 for x 0 so sup f x M0 1 3 Similarly f 0 in x 1 and f 0 in x 1 so f takes its maximum value at x 1 and since it is positive for x 0 its minimum is 4 so M1 sup f x 4 Finally then M2 sup f 4 since in x 0 it decreases to its zero at x 1 and for x 1 f 4x2 x2 1 3 4 Thus equality can occur Yes the result is true for vector valued functions for the usual Euclidean norms Let f f1 f2 fk be a function with values in Rk Thus the assumption is that each of the components satis es the assumptions of the question and we set Mi sup f i x x a with the Euclidean norm Now suppose that a a1 ak Rk is a constant vector We can apply the result above to g x a f x a1 f1 x ak fk x We see then that for any x a g x 4 sup g sup g 4 a 2 M0 M2 Now we can set a f x for a given x and divide by a factor of f x 2 and so conclude that f x 2 4M0 M2 Taking the supremum over x now gives the vector valued result 4 Chapter 6 Problem 2 Solution Since f is continuous it is Riemann integrable and f 0 either f 0 or there exists an interval of positive length l 0 in a b on which f x c 0 Then there exists a partition with the end points of this interval as two of its points such that L P f lc 0 b Since a f dx L P f for any partition this implies b f dx 0 must imply f 0 a Or you could use the fundamental theorem of calculus b a f dx 0 so 5 Chapter 6 Problem 4 Solution For any partition P we have n xi xi 1 sup f U P f L P f i 1 xi 1 xi xi 1 xi inf xi 1 xi f Now any interval of non zero length contains both rational and irrational points so the di erence of sup f and inf f is always one It follows that U P f L P f b a so the function cannot be Riemann integrable


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