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by: Cassidy Grimes
Cassidy Grimes

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This 24 page Class Notes was uploaded by Cassidy Grimes on Monday October 26, 2015. The Class Notes belongs to MATH 555 at University of South Carolina - Columbia taught by Staff in Fall. Since its upload, it has received 17 views. For similar materials see /class/229550/math-555-university-of-south-carolina-columbia in Mathematics (M) at University of South Carolina - Columbia.

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httpWWW mathsceduNsharpleymathS55Lect uresRSiFacw html Analysis 11 RiemannStieltjes Integration Additional Results Theorem Suppose that f and x are both continuous and nondecreasing then r b r b J f ddfbxbfaxa J ocdf Proof First we observe both integrals exist by our previous results Let P be a partition PIX02altX1ltltXi1ltXiltltbIXn and set Ed Xi1 Eji Xi then RP f0 fb0 b f20063 RP 0 f This follows from the summation by partsquot equation 1391 1391 2 u vjvj1 unvn u0vO 2 vj1ujuj1 j1 j1 We set uJ fXJ and v 0LXJ and chose a partition P so that both the left hand side is arbitrarily close to lab f doc and the RiemannStieltj es sum on the right hand side is arbitrarily close to lab OLdf L Theorem If x is monotone increasing on ab and f is bounded then f is RiemannStieltjes integrable with respect to x onab if and only if f oc is Riemann integrable In this case I b f doc I b foc dx a Pf Suppose 8 gt 0 is arbitrary Since OL is Riemann integrable pick a partition P such that UP0L LP0L lt 8 For each subinterval Ii XiXi1 by the Mean Value Theorem there is a ti in the interval so that Adi x ti Axi If si is arbitrary in Ii then 1 of4 04242000 1055 PM httpWWW mathsceduNsharpleymathS55Lect uresRSiFacw html 2 mi mi 2 mg m Axi l l s 2 W low was Axi UPfoc l S llfllm UPOc LPoc UPf oc S 8 UPf0c By varying si this implies UPf0 UPf 0639 S Hill 3 Using the same inequality this estimate also holds for Ul f OL UPfoc A similar argument establishes the corresponding estimate for lower sums Theorem Suppose that f has range mM and I is continuous on mM If f is Riemann Stieltjes integrable with respect to x then F bof is Riemann Stiletjes integrable with respect to x Pf Let 8 gt 0 be arbitary Since I is uniformly continuous there is a 5 gt 0 which we may as well assume is smaller than 8 such that u Vlt8 if uV lt 5 For this 5 pick a partition P so that UPfx LPfoc lt 82 J mi as the sup and inf of F over same subinterval Let G be de ned as the index set of good intervals where M m lt 5 and B as the remainder where M m 2 5 Forj e G we have Mfgmlt lt 8 while forj e B there holds Ej E B AOL lt 5 The last estimate follows since 5 Ej E B AOL S EJ E B Mimi AOL S 52 Now we can estimate the difference between the upper and lower RiemannStieltjes sums of F Let M mJ denote sup and inf of f respectively over the subinterval Xj1XJ Similarly define Milt 2 of4 04242000 1055 PM httpWWW mathsceduNsharpleymathS55Lect uresRSiFacw html UPF0LLPF0L s 2 MJmJA0Lj 2 MjmjAOLj j E G j E B s e ocboca 2 Film 2 A09 j E B s e 0Lb0La 2 an 5 s c e D Corollary If f is RiemannStieltjes integrable with respect to x then so is f2 Further if g is also RiemannStieltjes integrable with respect to x then the product f g is as well Pf Apply the above theorem with My y2 to establish the rst statement To establish the second use the identity fg fg2 f2 g22 and note that each term of the sum on the right hand side is RiemannStieltjes integrable with respect to 1 0L Corollary If f is RiemannStieltjes integrable with respect to x then so is Pf Apply the above theorem with My E Defn A function y is said to be of bounded variation if for any partition P of the interval ab varabCY suppartitionsP KP is finite where tPy 2 1n Theorem A function y is of bounded variation if and only if it can be decomposed as a difference 7 300 of two monotone nondecreasing functions Pf First define uJr maXu0 and u39 minu0 then u 1fr u39 and u uJr u39 For a partition P of aX define PPX Ej1nAerr 3 of4 04242000 1055 PM httpWWW mathsceduNsharpleymathS55Lect uresRSiFacw html and nPx ZJ 1n It is clear that Bx suppmi ons P of 81x pPx and 0Lx suppamtions P of ax nPx are nondecreasing functions Finish by showing that y BOL D Note Using this decomposition one may now extend both the definition of the Riemann Stieltjes integral and its properties from y monotone nondecreasing to ybeing a function of bounded variation Jab fdy Jab de Jab fdoc All the properties given in the previous lectures have their obvious analogues We may also easily extend to the case of complex and vector valued functions with corresponding results Robert Sharpley March 4 I 998 4 of4 04242000 1055 PM lof4 httpWWW mathsceduNsharpleymathSSSLect39uresRSJNTRO html ANALYSIS 11 Introduction to RiemannStieltjes Integration Defn A collection of n1 distinct points of the interval ab PIX02altX1ltltXi1ltXiltltbIXn is called a partition of the interval In this case we de ne the norm of the partition by HP max Axi lSiSn where Axi Xi Xi1 is the length of the ith subinterval Xi1Xi For a nondecreasing function x on ab define AOLi ocxi ocxi1 Defn Suppose that f is a bounded function on ab and x is nondecreasing For a given partition P we define the RiemannStiely39es upper sum of a function f with respect to 06 by n UPf0 I 2 Mi A08 11 where denotes the supremum of f over each of the subintervals Xi1 Xi Similarly we define the RiemannSlieltjes lower sum by n LPf0 I 2 mi A08 il where here mi denotes the infimum of f over each of the subintervals Xi1 Xi Since n1i S Mi and AOLi is nonnegative we observe that 04242000 1053 PM 20f4 httpWWW mathsceduNsharpleyma iSSSLect uresRSJNTRO html LPf0c s UPf0c for any partition P Defn Suppose P1 P2 are both partitions of ab then P2 is called a re nement of P1 denoted by P1 S P2 if as sets P1 g P2 Note If P1 S P2 it follows that P2 S P1 since each of the subintervals formed by P2 is contained in a subinterval which arises from P1 Lemma If P1 3 P2 then LP1f 0c 3 LP2f 0c and UP2foc s UP1foc Pf Suppose first that PI is a partition of ab and that P2 is the partition obtained from P1 by adding an additional point z The general case follows by induction adding one point at a time In particular we let P12X02altX1ltltXi1ltXiltltbIXn and P22X02altX1ltltXi1ltZltXiltltbIXn for some fixed i We focus on the upper sum for these two partitions noting that the inequality for the lower sums follows similarly Observe that n UP1foc 2 M Au jl and 04242000 1053 PM httpWWW mathsceduNsharpleymathSSSLect39uresRSJNTRO html il n UP2f0L 2 MJ AOLJ M OLZXXi1 M 0LXi0Lz 2 MJ AOL J39 1 il where M sup fX I Xi1 S X S 2 and MN sup fX I z S X S Xi It then follows that UP2foc S UP1foc since Defn If P1 and P2 are arbitrary partitions of ab then the common re nement of P1 and P2 is the formal union of the two Corollary Suppose P1 and P2 are arbitrary partitions of ab then LP1foc s UP2foc Pf Let P be the common refinement of P1 and P2 then LP1foc s LPf0c s UPf0c s UP2foc 3 Defn The lower RiemannStiely39es integral of f with respect to 06 over ab is defined to be b L fX d 0L sup LPfoc all partitionsP of ab Similarly the upper RiemannStiely39es integral of f with respect to a over ab is de ned to be b U fX d 0LX inf UPfoc all partitions of ab By the de nitions of least upper bound and greatest lower bound it is evident that for any function f there holds 3 of4 04242000 1053 PM httpWWW mathsceduNsharpleymathSSSLect39uresRSJNTRO html r b r b L J fX d OLX S U J fX d OLX a Defn A function f is RiemannStieltjes integrable over ab if the upper and lower Riemann Stieltjes integrals coincide We denote this common value by lab fX d XX Examples 1 Obviously if XX X then the Riemann Stieltjes integral reduces to the Riemann integral of f 2 lab fX d XX fX0 if f is continuous at X0 and x is de ned to be the step function which is one for X larger than X0 and zero otherwise Robert Sharpley Feb 23 I 998 4 of4 04242000 1053 PM MATH 5557041 NOTES LEBESGUE S THEOREM ON RIEMANN INTEGRABLE FUNCTIONS In this note we will present a self contained proof of Lebesgue s characterization of Riemann integrable functions on ab as those bounded functions which are continuous ae on ab The text contains a proof which uses the theory of the Lebesgue integral Other proofs exist which use either the notions of semi continuous functions or the modulus of continuity of a function Our proof is modelled after one which uses the theory of the Lebesgue integral but only uses the concepts of measure and content zero Recall that E C R has Lebesgue measure zero if for all 6 gt 0 there exists a countable collection In of open intervals with E C Uff In such that Z lIn lt 6 where lIn denotes the length of the interval In If we replace in this definition the countable collection of intervals by a nite collection of of open intervals then we say that E has content zero Obviously a set of content zero has measure zero Lemma 1 A countable union of sets of measure zero has measure zero Proof Let En C R have measure zero and put E UnEn Let E gt 0 Then for each n there exist a countable collection Imk il of open intervals such that En C UzilImk and Zk lIn7k lt Now Imkh n is again a countable union of open intervals and E C Ukamk such that Zk lIn7k lt 6 Hence E has measure zero D 7 Lemma 2 Let 0 S f a b gt R be a Riemann integrable function with fab f 0 Then for all c gt 0 the set as E a b 2 c has content zero Proof Let c gt 0 and denote by E the set as 6 ab 2 0 Let E gt 0 Then there exists a partition 73 30 33 of a b such that HUD f lt 6 c where 17 f denotes the Riemann upper sum corresponding to 73 ie we f Z Mimi 2391 where Mi supfc a Ei1ci Denote I E rel1 31 0 lfi E I then Mi 2 0 Hence we have 60 gtU73f 2 ZMZAri 2 02ml iEI ieI From this it follows that Zia Mag1 Zia Ami lt 6 Since E is covered by lag1 i E I it follows that E has content zero D We say that a property P holds almost everywhere abbreviated by ae on a b if the set as E a b P fails for x has measure zero 1 Corollary 3 Let 0 S f a b gt R be a Riemann integrable function with fab f 0 Then f is zero ae on ab Proof The set as 6 ab 7 0 Uffd c 6 ab 2 i which is by the above lemma a countable union of sets of content zero and has thus measure zero B Let 73 90 337 be a partition of a b Let Mi supfc a E ml491 and mi inffc a E ml491 Denote by b the upper function for f corresponding to 73 and by 11 the corresponding lower function ie i1 and Zmixlmiilymi r 21 It is easy to see that b and 11 are Riemann integrable and that f b 17 f and ff LJ 03 Moreover S S on ab Recall that the upper Riemann integral is given by Lbf infb73f 73 partition of ab and that the lower Riemann integral is given by I f sup 73 f 73 partition of a By de nition f is Riemann integrable if the lower integral of f equals the upper integral of f Theorem 4 Lebesgue A bounded function f 1 b gt R is Riemann integrable if and only if it is continuous ae on a b Proof Assume rst that f is Riemann integrable on a b Let 7 be a sequence of partitions of a b with P C 73 and such that the mesh PM gt 0 as k gt 00 Let bk the upper function for f corresponding to 73k and by wk the corresponding lower function Then T S l for all a 6 ab and f wk T fabf and f bk f f Let 933 limknoo and limknoo for a E a b It follows now that 3 933 3 S S for a 6 ab Hence we have b b b b b b b kSgSgSfShShS k Letting k gt 00 we conclude that g and h are Riemann integrable and that f g f h f f As h 2 9 it follows from Corollary 3 that g h ae Hence the set E as 6 ab 933 7 U Uk P has measure zero We claim that f is continuous on ab Let 330 6 ab E and let 6 gt 0 Then 9330 hco implies that there exists k E N such that bkc0 7 wkr0 lt 6 Now bk 7 wk is 2 constant in a neighborhood of 60 since 60 7 Hence there exists 3 gt 0 such that 711kx kx0711kx0for all x 7 x0 lt 5 For x 7 x0 lt 3 we now ave 6 lt 111M930 klt930 S 33 i 930 S klt330gt 11k930lt Ea which shows that f is continuous at 30 This completes the proof that f is continu ous except for a set of measure zero Assume now that f is continuous on 1 b E where E has measure zero Let E gt 0 and M such that S M on 1 b Then 7 3 2M for all x y E 1 b Since E has measure zero there exists open intervals 112 such that E C UWL and Z In lt W For all x E 1b E there exists an open interval Jc with x E Jx such that 7 3 Mia for all 242 6 Jm 11 since f is continuous at such 3 Now 1k U Jm x E 1 b is an open cover of 11 so by compactness of 1b there exists a finite cover 1329 U Ug jgg xi 6 1b of 1b Let 73 1 t0 tN b be the partition of 1 b determined by those endpoints of 1521 and Jwi xi 6 1 bE which are inside 1b For each 1 S j S N the interval trhtj is contained in some k or some Jmi Let J j trhtj C k for some Then we have that N we f 7 m f 7 2mm supfltcgt 7 y 2 m e tights 31 ZAt 2M ZAt m jeJ 1 lt 2Mltb7ame Hence f is Riemann integrable D Exercise 1 Prove that a set E has content zero if and only if there exists a closed bounded interval 1 b containing E such that XE is Riemann integrable on 1 b and has Riemann integral zero Exercise 2 Prove that a set has zero content if and only if its closure is a bounded set with measure zero Exercise 5 Give an example of a bounded set with measure zero which does not have content zero CONVEX FUNCTIONS GIVE INEQUALITIES PROF MARIA GIRARDI 1 BASICS Notation 11 Throughout this presentation 0 ab 1 C R where ab 6 RUjoo o p I a R is a function 0 p E 17 00 and its conjugate ewponent p E 17 00 is de ned by i 1 1 De nition 12 p I a R is conve ltgt Ly 61 t6 01 W l 01 S tltp1tlt y 11 Picture 1 1 1 1 1 a z tz 17ty y b Date 7 June 2001 2 HOLDER7S INEQUALITY Lemma 21 Let p I a R be convex Then i1 i1 i1 TL ti 139 7 ti 139 1 6 I 7 ti 6 071 gt p S w 22 211 211 Proof 21 Use 11 and induction 22 Let ti 221 tjl lti and apply 21 I Recall Geometric Arithmetic Mean lnequality GM 3 AM n ln 1 2120 nEN gt S Using convex functions7 we can generalize the GM AM inequality Proposition 22 Generalized GM AM inequality 9520 te01 221 g 23 i1 i1 i1 Proof Let p 07 00 a R be x 7 lnx Then p is convex gt 7 ln ti 3 i1 gt 2 ln S lnlt i1 Now exponentiate both sides of 24 An immediate corollary follows now Corollary 23 Youngjs lnequality p igt0 1ltpltoo gt zlzgginLi 25 p Proof Apply 23 to I 1 961 zz 2 5 5 p I Recall For a sequence L1 x17 zn from R n 117 Halwally Z i1 Theorem 24 Holder7s Inequality in 17 For sequences L1 and L1 from R lli39yi1lll1 S Lilla Lilla that is 117 n n 117 n 2 W g z 2 26 i1 i1 i1 When p 2 10 7 note that 26 is just the Cauchy Schwarz Inequality Proof WLOG Neither nor is 0 WLOG 1 for if not then h yi z ll963939ll 1 7 y39i lly3939ll 1 1 llzjjllzp l l p l lHljb39llzp l l I By Youngjs Inequality ml to 1 p 1 p gl iyz l E 2 p 7 7 5 llillzp 17 lHlil llzp 139 i1 p Here are two exercises of Generalized Holder7s Inequalities in 17 k 1 111773971 k I n k I zz Human 71 11 1 711 J 3 739 7L ExerCISe For sequences ah1 from R and Z n n 1 1 7 1 ExerCISe For sequences 11 and 11 from R and a a i E imi 211am went nyil f iiipz Hint 1 1 171173 122173 Recall For a nice function f I a R7 117 ML WM dz Theorem 25 Holder7s Inequality in LP For nice functions g I a R7 HfgHL1 S Han HgHLp that is7 1 1 112 1f969961dz U MW V 19W 1 I I I Proof WLOG Neither HfHLF nor HgHL is 0 or 00 WLOG HfHLF 1 HgHL for if p p not so f f 7 ML By Youngjs lnequality WM 19W 1 p 1 1 11fzgz1dz p l i p ML p Ham 7 1 t f g i my LP 1 Ham 1 Here are two exercises of Generalized Holder7s Inequalities in LP Exercise For nice functions fj I a R and 21 1 1 7 k k Hfj HimML 11 L1 11 Exercise For nice functions g I a R and 1 i pig HfgHLp3 Hm ML 4 3 JENSEN7S INEQUALITY Lemma 31 Let p I a R be convex Let 7y72 E I with z lt y lt 2 Then gamma 3 2ltP S lt2gt7 ltyr 31 yes 27x 27y Picture Proof Key idea 27 7x yy ziz ziz y m 17tz z y S tltpx1tltpz Think of what this says in the picture the needed algebra then follows easily I 5 Proposition 32 Let p I a R be convex and 0 6 I Then is continuous at 0 ltp 2 LO 311mm m07 W exists ltP 0 311mm mg W exists ltp Lltzogt ltp 96o Proof 1 Find a lt 11 lt 12 lt 0 lt b2 lt b1 lt b If x E a27b2 zo7 then a a1 12 z x0 x b2 b1 b by 31 A 39 ltM02 7 ltMal S ltMo 7 WC S WM 7 lt7092 B 12701 oii 1in and so M g maxAB oii 274 Consider a 1 1 0 i z b By 31 ltMo 7 ltMl lt ltMo 7 M551 lt WET 7 ltMo lt Wu 7 ltMo x07xl 7 x07il 7 E77z0 7 x77z0 and so ltMo 7 ltMl 1 WW 7 ltMo 1 lt 1 32 5210 0 7 m 4960 7 ltP950 lt30 95 7 0 I Observation 33 Let p I a R be convex and x0 6 I Let ltPL950 S m S ltP950 Consider the line 195 7719c 950 ltM950 through the point my 0 Then lz S x V E I 33 A line y lz through the point 07 p x0 that satis es 33 is called a supporting line of y ltp at 0 Draw yourself a picture to see the choice of terrninolgy here Thus Observation 33 says that convex functions always have supporting lines Picture Proof By 32 ltMo 7 ltMl m lt Wm 7 ltMo 39 x0 7 x1 7 7 x 7 x0 Thus mltz 7 goo we 7 we W e I zo Theorem 34 Jensen7s Inequality Let p R a R be convex and f I a R be integrable If I 017 then l ltp mom f96 dz If I a7b has nite length7 then 2133 3253 Remark Theorem 34 may be thought of as a continuous version of Lemma 21 34 35 Proof 34 Let L f dx x0 and ltpLz0 S m S ltp r0 Then by Observation 33 mx 7 x0 xo 3 x Vs E R Thus mf96 7 960 We S f 96 V96 6 1 Now integrate both sides of 36 over I 01 35 follows applying 34 to 36 Completeness in Metric Spaces lofl httpwwwmathsced nmpletene html ANALYSIS II Metric Spaces Completeness Defn Suppose Xd is a metric space A sequence Xn is said to a be Cauchy sequence in Xd if for each 5 gt 0 there is a natural number N such that dXn Xm lt e if N 3 nm Proposition Convergent sequences are Cauchy Lemma Cauchy sequences are bounded but not necessarily convergent Pf Consider X as the interval 01 with the absolute value as norm The sequence 31 with x ln is Cauchy but not convergent in X To show that each Cauchy sequence is bounded apply the definition withE l to obtain an N such that dxn xm lt 1 if N 3 nm LetR max 1 dxN x1 dxN x2 dxN xN1 then Kn is contained in B2RXN Defn If Xd is a metric space for which each Cauchy sequence converges then Xd is said to be a complete metric space Lemma If a subsequence of a Cauchy sequence converges then the sequence is itself convergent to the same limit Proposition If C is a closed subset of a complete metric space Xd then C is a complete metric space with the restricted metric Examples R C Rk Ck are all complete metric spaces Pf Use the fact that convergence of a sequence in each of the spaces C Rk Ck is equivalent to convergence in each coordinateof the sequence This follows since the sup norm is equivalent to the Euclidean norm Theorem Let Cab denote the normed linear space of continuous functions on the interval ab equipped as before with the supnorm then Cab is complete Pf Let fn be a Cauchy sequence in Cab m Establish a pointwise limit for fnand call this function f 0 m Next show that the function f is continuous on ab Hint Use an E 3 argument Step 2 Prove thatH fn fH Robert Sharpley Feb I I 998 04242000 1029 PM Compactness in Metric Spaces lof2 httpWWW mathsc d nmpzot html ANALYSIS II Metric Spaces Compactness Defn A collection of open sets is said to be an open cover for a set A if the union of the collection contains A A subset of an open cover whose union also contains the set A is called a subcover of the original cover A cover is called nite if it has finitely many members Defn A set K in a metric space Xd is said to be compact if each open cover of K has a finite subcover Theorem Each compact set K in a metric space is closed and bounded Proposition Each closed subset of a compact set is also compact Theorem HeineBorel Theorem from last term Each closed and bounded interval ab is a compact subset of the real numbers Pf Let C be an open cover for ab and consider the set A x l ax has an open cover from C Note that A is not empty since a belongs to A Let c lub A It is enough to show that c gt b since if x1 belongs to A and a S x 5 x1 then x belongs to A Suppose instead that c 5 b then there must be some 00 in C such that c belongs to 00 But 00 is open so there exists 5 gt 0 so that N5c is contained in 00 Since c is the least upper bound for A then there is an x in A such that cd ltx 5c But x belongs to A so there are members 01 On of C Whose union covers ax The collection 01 On covers a c62 Contradiction since c is the least upper bound for the set A Corolla1v Each closed and bounded set of real numbers is compact Theorem If a set A is compact in a metric space X and f X gt Y is continuous then fA is compact in Y Corolla1v If f X gt Y is continuous and X is compact then f is a bounded function Corolla1v If f X gt R is continuous and X is compact then f attains its extremal values Theorem Suppose that f ab gt K is onetoone onto and continuous then f1 is continuous Pf Let 0 ab be relatively open then f13910 fO Let C be the complement in ab of 0 then C is closed and hence compact Therefore fC is compact in K and conseqently it is closed Its complement in K must then be relatively open That complement however is fO Compactness Characterization Theorem Suppose that K is a subset of a metric space X then the 04242000 1030 PM Compactness in Metric Spaces httpWWW mathsc d 39 nmpzot html following are equivalent 1 K is compact 2 each infinite subset of K has a limit point in K 3 each sequence from K has a subsequence that converges in K K Click here for the details of the proofs Corollarv Each closed and bounded set K in Rk or Ck is compact Pf Use the sequential convergence criterium and consider projections into each coordinate Recall that convergence in Rk is equivalent to convergence in each coordinate Defn A set K in a metric space X is said to be totally bounded if for each 5 gt 0 there are a finite number of open balls with radius 6 which cover K Here the centers of the balls and the total number will depend in general on 5 Theorem A set K in a metric space is compact if and only if it is complete and totally bounded Homework Robert Sharpley Feb 14 1 998 2 of2 04242000 1030 PM Special Functions log and exp httpWWWmathsceduNsharpleyma iSSSLecturesSpecFunctions html ANALYSIS 11 Special Functions Defn The logarithm function is defined by LX lt dt xgt0 Properties Ll 0 L X lX L strictly increasing and continuous 0 LX1 X2 LX1 LXZ Pf Use additivity of integration over subintervals and change of variables M o LXn nLX lim XWOLX 00 Pf Use induction with the previous result 0 LlX LX lnn X gt 0 LX oo Pf Notice that Lx UK 0 DomL K RangeL R Pf Since L is continuous by the intermediate value theorem each real number must be in the range Defn The real number e is defined as the unique number such that Le 1 This number is unique since L is 11 which follows since L is strictly increasing 0 e gt2 Pf Use the fact that L strictly increases Le l gt LZ L X lX2 lt 0 L decreases and L is concave down Defn The logarithm function with base agt0 is defined by logaX LXLa This function satisfies all the properties of logarithm above but with logaa l Defn The exponential function is defined as the inverse function of LX ie EX L391X Properties 1 of2 04242000 1017 PM Special Functions log and exp 20f2 http WWW mathsceduNsharpleymathSSSLecturesSpecFunctions html EO l EX is a strictly increasing function with domain R and range R E is continuous and differentiable with E X EX Pf Use the formula to compute derivatives of inverse functions g x lfgx if gf391 EX1 X2 EXl EXZ EmX EXm Pf Use the fact that Ex y if and only if L07 x Eln e1 n Emn emm for each rational mn Note This allows us to extend eX to all real numbers as EX by using the fact that the rationals are dense and E is continuous Pf Let y Eln then y En ln e So y is the nth root of e The second statement follows as before by induction E X EXgtO so E strictly increases and E is concave up Set yj Exj jl2 Defn The exponential function aX EX La has all the anticipated properties and is the inverse function of loga X Robert Sharpley Jan 12 1998 04242000 10 17 PM


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