MATH 360 Exam 2 Wednesday, November 8, 2017.» b

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1 MATH 36 Exam Wedesday, November 8, 7 Name. Let f : ra, bs Ñ R be a cotiuous fuctio, ad suppose that for every cotiuous fuctio g : ra, bs Ñ R, Prove that fpxq for all x P ra, bs.» b a fpxqgpxq dx. We ca, i particular, choose g f, ad so we kow that the itegral of f is zero. If there were a poit x where fpx q the (because of the cotiuity of f ad hece of f ) there would be a δ such that if x x δ, the fpxq fpx q fpx q, which implies that fpxq fpx q for all x P px δ, x δq. But the» b fpxq dx a a» x δ which is a cotradictio. fpxq dx» x δ x δ fpxq dx» b x δ fpxq dx δfpx q δfpx q

2 . Let vptq be the velocity of a object for t P r, T s You kow that vptq is the derivative with respect to t of xptq, the positio of the object at time t. Assume that vptq for all t. (a) What is the average velocity with respect to t? (b) Show that xptq is a ivertible fuctio of t. What are its domai ad rage? (c) Sice x is a ivertible fuctio of t, we ca cosider v as a fuctio of x. Show that the average of v with respect to x is greater tha or equal to the average of v with respect to t. At a key poit, you will eed to use oe of those famous iequalities.

3 (a) The average velocity with respect to t is v t T. (b) Because dx vptq, we have that xptq is a (strictly) mootoically icreasig fuctio of t, dt i,e, xpt q xpt q if ad oly of t t (for t, t P r, T s). So xptq is a oe-to-oe map from r, T s to rxpq, xpt qs, ad thus ivertible (by the itermediate value theorem for every x P rxpq, xpt qs there is a uique t P r, T s such that xptq x ad we ca defie x px q t ). (c) The average value of v with respect to x is v x xpt q xpq» xpt q xpq vpxq dx. Make the chage of variables x xptq (so dx x ptq dt ) i the itegral (beig careful to chage the limits of itegratio as well) ad get v x» T vptq dt xpt q xpq vptq dt sice v is the derivative of x with respect to t. Now the temptatio to use Cauchy-Schwarz is almost overwhelmig with f ad g v, the iequality p ³ fgq p ³ f qp ³ g q becomes ad so v x vptq dt which is what we were tryig to show. T dt vptq dt T T» vptq dt. vpxq dx v t

4 3. Let fpxq ad gpxq be fuctios for which their Taylor series (cetered at x ) coverge to the fuctios themselves for x P r r, rs. (a) Show that the Taylor series for cfpxq coverges to cfpxq for x P r r, rs. (b) Show that the Taylor series for xfpxq coverges to xfpxq for x P r r, rs. (c) Show that the Taylor series for fpxq gpxq coverges to fpxq gpxq for x P r r, rs. (d) Suppose the Taylor series for ϕpxq is f f x f x where f, f ad f f f, so f is the th Fiboacci umber. What is ϕpxq? Let p pxq be the Taylor polyomial of degree (cetered at x ) for fpxq ad let q pxq be the Taylor polyomial for gpxq. For the first three parts, we kow that for each x P r r, rs, we have that p pxq Ñ fpxq ad q pxq Ñ gpxq as Ñ 8. For parts (a)-(c), stadard properties of limits of sequeces (limit of the sum is the sum of the limits, ad limit of a costat times a sequece is the costat times the limit of the sequece) shows that cp pxq Ñ cfpxq, xp pxq Ñ xfpxq ad p pxq q pxq Ñ fpxq gpxq. The oly issue that remais is to show that cp pxq, xp pxq ad p pxq q pxq actually are the th degree Taylor polyomials of cf pxq, xf pxq ad f pxq gpxq, respectively. (a) For apxq cfpxq, it is trivial that a pq pxq cf pq pxq, so cp pxq is the th degree Taylor polyomial of cf pxq. (b) For bpxq xfpxq, we have bpq ad b pq pxq xb pq pxq b p q pxq. Therefore b pq pq f p q pq, so the coefficiet of x i the Taylor polyomial of xfpxq is b p q pq{! f p q pq{p q!, which is the coefficiet of x i the Taylor polyomial of fpxq. Therefore the th degree Taylor polyomial of xfpxq is xp pxq. (c) Fially for cpxq fpxq gpxq it is agai trivial that c pq pxq f pq pxq g pq pxq so p pxq q pxq is the th degree Taylor polyomial of f pxq gpxq.

5 (d) If we let gpxq xfpxq x fpxq the the Taylor series for gpxq is (usig (b) ad (c) above) gpxq f x pf f qx pf f qx 3 pf 3 f qx 4 x f x f 3 x 3 f 4 x 4 fpxq x We coclude that xfpxq x fpxq fpxq x, or fpxq x x x.

6 4. Let fpxq # x for x 3 x for x Prove carefully (upper ad lower sums etc., ad mid the gap ) that f is itegrable o r, s ad evaluate the itegral. This is overly clever, but it s probably the easiest way to do it without ivokig bouded variatio or somethig: Write fpxq gpxq hpxq where gpxq # x for x x for x ad hpxq # for x for x The g is cotiuous, hece itegrable (or if you must:» gpxq dx lim ņ Ñ8 k g lim Ñ8 k lim Ñ8 ņ k k ņ k k where i the last sum i the first lie we split up as { ad oticed that the sum of pk q{ for k goig from to is exactly the same as the sum of k{ for k goig from to, so they cacel.) So g is itegrable ad its itegral is. For the itegral of hpxq, use partitios P t, {, {, u The clearly Lph, P q ad Uph, P q ad both approach as Ñ 8. So h is also itegrable ad its itegral is also. Therefore f, beig the sum of two itegrable fuctios, is itegrable ad its itegral is.

7 5. Let fpxq» x t 4 dt. Prove that f is uiformly cotiuous o all of R. Hit: Do ot attempt to evaluate the itegral!! Sice t 4 for all t P R, we have if x y δ the fpyq fpxq for all t P R. So give ε, choose δ ε ad so t4» y x t 4 dt» y x dt y x δ ε Sice the δ we chose does ot deped o x or y, fpxq is uiformly cotiuous (o all of R).

8 6. Suppose g : R Ñ R, with bouded derivative (say g M). Fix ε ad defie fpxq x εgpxq. Prove that f is oe-to-oe if ε is sufficietly small. Take ε {pmq. The εg pxq εg pxq satisfies, i other words, εg pxq εg pxq f pxq The importat thig is that f pxq, so by the mea value theorem, But the f pxq fpxq fpyq f pcq x y x y which shows that if x y the fpxq fpyq, i other words, f is oe-to-oe.

9 7, Prove that for every positive iteger, l. Recall the defiitio of the logarithm: so (otig that ) l lpxq» x t dt,» p q{ t dt. Sice {t is a strictly decreasig fuctio of t, we kow that But the which evaluates to or t» p q{ for all t satisfyig dt» p q{ l l t dt t» p q{ dt