Math 4318 : Real Analysis II Mid-Term Exam 1 14 February 2013

Transcription

1 Mth 4318 : Rel Anlysis II Mid-Tem Exm 1 14 Febuy 2013 Nme: Definitions: Tue/Flse: Poofs: Totl:

2 Definitions nd Sttements of Theoems 1. (2 points) Fo function f(x) defined on (, b) nd fo x 0 (, b) stte the definition of f (x 0 ). Solution: f (x 0 ) x x0 f(x) f(x 0 ) x x (2 points) Stte the Men Vlue Theoem fo function tht is continuous on [, b] nd diffeentible on (, b). Solution: Suppose tht f : [, b] R is continuous on [, b] nd diffeentible on (, b). Then thee exists c (, b) such tht (b )f (c) = f(b) f()

3 3. (3 points) Stte the Second Fundmentl Theoem of Clculus. Solution: Let f be integble on [, b]. Fo x [, b] let F (x) = x f(t)dt. Then F is continuous on [, b]. If f is continuous t x 0 in (, b), then F is diffeentible t x 0 nd F (x 0 ) = f(x 0 ). 4. (3 points) Stte one (of the mny) equivlent definitions fo function f to be integble on [, b]. Solution: A function f on [, b] is integble if nd only if fo ny ɛ > 0 thee exists ptition P of [, b] such tht U(f; P ) L(f; P ) < ɛ.

4 Tue o Flse (1 point ech) Fo Questions 1-5, let, b R with < b nd let f be function tht mps the intevl [, b] to R. 1. If f is diffeentible on (, b), then f is continuous on (, b). 2. If f is continuous on [, b], then f is integble on [, b]. 3. If f is monotonic on [, b], then f is integble on [, b]. 4. If f is integble on [, b], nd x [, b] then F (x) = x f(t)dt is continuous. 5. A bounded function f is integble on [, b] if nd only if thee exists n ɛ > 0 such tht fo ll ptition P of the intevl [, b] U(f; P ) L(f; P ) < ɛ. Solution: Flse 6. Evey continuous function on [, b] is diffeentible on (, b). Solution: Flse 7. If f nd g e integble on [, b], then fg is integble on [, b].

5 8. The uppe Dboux integl of f, U(f), is lwys less thn o equl to the lowe Dboux integl L(f). Solution: Flse 9. A function is Riemnn integble if nd only if it is Dboux integble. 10. If P nd Q e ptitions of [, b] nd P Q then L(f; P ) L(f; Q) U(f; Q) U(f; P ).

6 Poofs 1. (10 points) Let f be defined on R nd suppose tht f(x) f(y) (x y) 2 x, y R. Pove tht f is constnt function. Solution: It suffices to pove tht f (x) = 0 fo ll x R. Let ɛ > 0 be given. Then we need to show tht thee exists δ > 0 such tht f(x + h) f(x) 0 h = f(x + h) f(x) h < ɛ when h < δ But, by the popety of the function we hve tht f(x + h) f(x) h h2 h = h. So choose δ = ɛ, nd if h < δ we hve f(x + h) f(x) 0 h < ɛ, nd since ɛ > 0 ws bity we hve tht f (x) = 0.

7 2. (10 points) Suppose tht f is defined nd diffeentible fo evey x > 0 nd tht f (x) 0 s x. Set g(x) := f(x + 1) f(x). Pove tht g(x) 0 s x. Solution: Use MVT to wite g(x) = f(x + 1) f(x) = f (t x ) whee x < t x < x + 1. So s x then t x s well. Moe pecisely, let ɛ > 0 be given. Then select M such tht f (x) < ɛ fo ll x > M. Then if x > M, nd fo t x (x, x + 1) we hve tht f (t x ) < ɛ s well. Thus, fo ny x > M we hve tht nd so g(x) 0 s x. g(x) = f(x + 1) f(x) = f (t x ) < ɛ

8 3. (15 points) Two functions f : R R nd g : R R e equl up to ode n t the point if f( + h) g( + h) = 0. h 0 h n Suppose tht f(x), f (x),..., f (n) (x) exist nd e continuous t. Pove tht the function f nd the function n f (k) () g(x) = (x ) k k! e equl up to ode n t the point. k=0 Solution: Conside f( + h) g( + h) h n = f( + h) k=0 h n f (k) () k! h k As h 0 we hve tht this ppoches 0 since in the numeto we hve f() g() = 0 f() f() = 0. Since f () exists, nd g () exists, we by L Hopitl tht f( + h) g( + h) f( + h) h 0 h n h 0 k=0 h n k=1 f (k) () k! h k f (k) () (k 1)! hk f ( + h). h 0 nh n 1 Agin, evluting the numeto nd denominto s h 0 we find 0 since the numeto 0 now gives f () g () = f () f () = 0, nd cn gin pply L Hopitl s ule to see tht f( + h) g( + h) f( + h) h 0 h n h 0 f ( + h) h 0 k=0 h n k=1 nh n 1 k=2 f (k) () k! h k f (k) () (k 1)! hk f (k) () (k 2)! hk f ( + h). h 0 n(n 1)h n 2 By induction, this cn be epeted n times since we know tht f (k) () exists fo 0 k n, nd ive t f( + h) g( + h) f (n) ( + h) f (n) () h 0 h n h 0 n! = 0.

9 4. (15 points) A function f on [, b] is clled step-function if thee exists ptition P = { = x 0 < x 1 <... < x N = b} of [, b] nd constnts c j R such tht f(x) = c j fo ll x [x j 1, x j ]. () Pove tht step function is integble on [, b]. (b) Compute b f(x) dx when f is step function. Solution: Recll tht f is integble on [, b] if nd only if fo ny ɛ > 0 thee exists ptition P ɛ of [, b] such tht U(f; P ɛ ) L(f; P ɛ ) < ɛ. Let f be step function, nd let P be the ptition descibing the step function. Let ɛ > 0 be given, nd let P ɛ = P. Then note tht U(f; P ɛ ) = U(f; P ) = A simil computtion gives sup f(x)(x j x j 1 ) = x [x j 1,x j ] c j (x j x j 1 ). L(f; P ɛ ) = L(f; P ) = inf f(x)(x j x j 1 ) = c j (x j x j 1 ). x [x j 1,x j ] Thus, we hve U(f; P ɛ ) L(f; P ɛ ) = c j (x j x j 1 ) c j (x j x j 1 ) = 0 < ɛ. So f is integble on [, b].

10 5. (15 points) Let f be bounded function on [, b]. Suppose tht thee exist sequences {U n } nd {L n } of uppe nd lowe Dboux sums fo f such tht n (U n L n ) = 0. Show tht f is integble nd b f(x) dx n U n n L n. Solution: Let L n be the lowe Dboux sum, nd suppose tht it is ssocited to ptition P n, i.e, L n = L(f; P n ). Similly, let U n be the uppe Dboux sum nd suppose tht it ssocited to ptition P n, i.e. U n = U(f; P n ). Let Q n = P n P n. Note tht we hve U(f; Q n ) L(f; Q n ) U(f; P n) L(f; P n ) = U n L n Thus, given ɛ > 0, choosing n sufficenly lge will llow us to conclude tht thee exists ptition Q n so tht U(f; Q n ) L(f; Q n ) U(f; P n) L(f; P n ) = U n L n < ɛ. Thus, we hve tht f is integble. It suffices to show tht U n = b f(t)dt since we then hve L n (L n U n + U n ) (L n U n ) + U n = 0 + n n n n s well. Howeve, given ɛ > 0, select N so tht U n L n < ɛ. Note tht L n b f(t)dt U n, nd so b f(t)dt L n. Thus we hve U n b if n N. Theefoe, U n = b f(t)dt. f(t)dt U n L n < ɛ b f(t)dt

11 6. (15 points) Let 2 be n intege. Let f(x) = x defined on (0, ). Using the definition of deivtive, compute f (x 0 ) fo ny x 0 (0, ). Hint: b = ( b) 1 1 j b j. Solution: Using the hint, let = x nd b = x 0, then we hve Thus we hve nd so f (x 0 ) x x0 x x 0 = ( x ) 1 x 0 0. x x 0 x x 0 = x x 0 x x 0 ( 1 x x0 ) 1 0 ( 1 ) 1 0. Now since x, x 0 (0, ) we know tht the it bove is defined nd so x x 0 ( 1 0 ) 1 = ( x x 0 1 ) 1 0. Howeve, fo j = 0,..., 1, the function is continuous on (0, we hve 1 x x 0 0 = x j 0 = x = x 1 0. Using this we see tht f (x 0 ) = ) (x = x 1 0.