We have been going places in the car of calculus for years, but this analysis course is about how the car actually works.

Transcription

1 Analysis I We have been going places in the car of calculus for years, but this analysis course is about how the car actually works. Copier s Message These notes may contain errors. In fact, they almost certainly do since they were just copied down by me during lectures and everyone makes mistakes when they do that. The fact that I had to type pretty fast to keep up with the lecturer didn t help. So obviously don t rely on these notes. If you do spot mistakes, I m only too happy to fix them if you me at mdj27@cam.ac.uk with a message about them. Messages of gratitude, chocolates and job offers will also be gratefully received. Whatever you do, don t start using these notes instead of going to the lectures, because the lecturers don t just write (and these notes are, or should be, a copy of what went on the blackboard) they talk as well, and they will explain the concepts and processes much, much better than these notes will. Also beware of using these notes at the expense of copying the stuff down yourself during lectures it really makes you concentrate and stops your mind wandering if you re having to write the material down all the time. However, hopefully these notes should help in the following ways; you can catch up on material from the odd lecture you re too ill/drunk/lazy to go to; you can find out in advance what s coming up next time (if you re that sort of person) and the general structure of the course; you can compare them with your current notes if you re worried you ve copied something down wrong or if you write so badly you can t read your own handwriting. Although if there is a difference, it might not be your notes that are wrong! These notes were taken from the course lectured by Dr Paternain in Lent If you get a different lecturer (increasingly likely as time goes on) the stuff may be rearranged or the concepts may be introduced in a different order, but hopefully the material should be pretty much the same. If they start to mess around with what goes in what course, you may have to start consulting the notes from other courses. And I won t be updating these notes (beyond fixing mistakes) I ll be far too busy trying not to fail my second/third/ th year courses. Good luck Mark Jackson Schedules These are the schedules for the year 2009/10, i.e. everything in these notes that was examinable in that year. The numbers in brackets after each topic give the subsection of these notes where that topic may be found, to help you look stuff up quickly. Limits and convergence (1) Sequences (1.1) and series (1.3) in and. Sums, products and quotients (1.1.2). Absolute convergence (1.5); absolute convergence implies convergence (1.5.1). The Bolzano-Weierstrass theorem (1.1.3) and applications (the General Principle of Convergence) (1.2). Comparison (1.4.1) and ratio (1.4.2) tests, alternating series test (1.4.5). Continuity (2) Continuity of real- and complex-valued functions defined on subsets of and (2.1). The intermediate value theorem (2.3). A continuous function on a closed bounded interval is bounded and attains its bounds (2.4).

2 Differentiability (3) Differentiability of functions from to (3.1.1). Derivative of sums and products (3.1.2). The chain rule (3.1.3). Derivative of the inverse function (3.1.2). Rolle s theorem (3.2.1); the mean value theorem (3.2.2). One-dimensional version of the inverse function theorem (3.3). Taylor s theorem from to ; Lagrange s form of the remainder (3.5.1). Complex differentiation (3.6). Taylor s theorem from to (statement only). Power series (4) Complex power series and radius of convergence (4.1). Exponential (4.3), trigonometric (4.4) and hyperbolic (4.5) functions, and relations between them. *Direct proof of the differentiability of a power series within its circle of convergence* (4.2). Integration (5) Definition and basic properties of the Riemann integral (5.1, 5.3). A non-integrable function (5.1.2). Integrability of monotonic functions (5.2.2). Integrability of piecewise-continuous functions (5.2.4). The fundamental theorem of calculus (5.4.1). Differentiation of indefinite integrals (5.4.2). Integration by parts (5.4.3). The integral form of the remainder in Taylor s theorem (5.5.1). Improper integrals (5.6). Contents 1. Limits and Convergence Sequences Cauchy sequences Series Convergence tests Absolute convergence Continuity Definitions and basic results Limits of functions Intermediate value theorem Cofu clobi ibaatib Inverse functions Differentiability Basic calculus The mean value theorem Inverse rule (Inverse function theorem) Cauchy s mean value theorem Taylor s theorem Some comments on differentiability of functions Power series Radius of convergence Differentiability of power series The standard functions: exponentials Trigonometric functions Hyperbolic functions Integration Riemann integration Integrability of monotonic and continuous functions Elementary properties of the integral Integration rules and tools... 33

3 5.5 Taylor s theorem revisited Infinite integrals (Improper integrals) Limits and Convergence 1.1 Sequences Review from Numbers and Sets Let be a sequence of real numbers. Definition. as if given, Note. in most cases. We say that is increasing if, that is decreasing if, that is strictly increasing if, and that is monotonic if it is either increasing or decreasing The Fundamental Axiom and some properties Fundamental Axiom of the Real Numbers. Suppose, and, and is increasing. Then as. In other words, an increasing sequence bounded above converges. Notes. 1) We could have phrased the Axiom with decreasing sequences bounded below. 2) The Axiom is equivalent to the fact that every non-empty set of real numbers bounded above has a least upper bound (supremum). Lemma 1.1. (i) The limit is unique. That is, if and, then. (ii) If as and, then as (subsequences converge to the same limit). (iii) (iv) If and, then (v) If and, then. (vi) If, and, then. (vii) If and, then. Proof. (i),. Similarly,,. Using the triangle inequality, for, we obtain for. If, just take so which is absurd. Thus. (ii),. For,. Thus, that is, as. (v) Definitions as before. By triangle inequality, If, then Lemma 1.2. as.

4 Proof. The sequence is decreasing and bounded below. The Fundamental Axiom. Now we claim that. by Lemma 1.1(iv). But is a subsequence of. By Lemma 1.1(ii),. By uniqueness of limit, Remark. If we were instead considering sequences of complex numbers,, we could give essentially the same definition of limit; as if given,, where here stands for the modulus of a complex number. All properties of Lemma 1.1 carry over to (we have no order in ). except the last one The Bolzano-Weierstrass Theorem Theorem 1.3 (Bolzano-Weierstrass Theorem); If and, then we can find and as. (Every bounded sequence has a convergent subsequence). Remark. We say nothing about uniqueness of. E.g. has and. Proof. Let and. Then either 1) for infinitely many values of, or for infinitely many values of. In case 1), set In case 2), set. Proceed inductively to obtain sequences and such that the following holds; (i) (ii) (iii) for infinitely many values of Then is a bounded increasing sequence and is a bounded decreasing sequence. By the Fundamental Axiom, and. By property (iii), passing to the limit, Having selected such that, select such that. We can always do this because for infinitely many values of. In other words,. Since and,. This argument is called the bisection method or lion hunting. 1.2 Cauchy sequences Definition. A sequence is a Cauchy sequence if given,. As before, Lemma 1.4. A convergent sequence is a Cauchy sequence. Proof. and. This means,.. So if we have. As an application of the Bolzano-Weierstrass theorem, we now show that the converse is true; Lemma 1.5. A Cauchy sequence is convergent. Proof. First we ll prove that if is a Cauchy sequence, then it is bounded. Cauchy means that,. Choose. Then. In particular,.

5 Take for. For this choice of,.. Now by the Bolzano-Weierstrass theorem, has a convergent subsequence. Now we show that in fact.. Choose large enough so that (since Cauchy, ) and (since ). Thus. Thus, for sequences of real numbers, convergence and Cauchy property are equivalent. This is called the General Principle of Convergence. 1.3 Series Convergent and divergent series We begin with some generalities. Definition. or. We say that converges to if the sequence of partial sums tends to as. In that case we wrire. If does not converge, we say that diverges. Remark. Any question about series is really a question about the sequence of partial sums. Lemma 1.6 (i) If and converge, then so does where. (ii) Suppose. Then either and both converge or both diverge (i.e. initial terms do not matter). Proof. (i) If and, then clearly by Convergent series and decreasing terms Lemma 1.7. If the series converges, then. Proof.. If the sum to infinity converges, then. Therefore. The converse is not true! Example. Consider the series Then, but the series diverges. This can be shown as follows; so. So if converges, then, but since, in the limit we get which is a contradiction Geometric series Consider

6 If, If, the series clearly diverges. First note that. So, if, In conclusion, converges iff, and if then. 1.4 Convergence tests The first four tests in this section are for series of positive terms Comparison test The most powerful test for convergence comes out straight from the fundamental axiom. Theorem 1.8 (Comparison test). Suppose. Then if converges, so does. Proof. Let Now, and the same for, so and are increasing sequences. If, then, so it follows that if converges, then and thus, since, then is bounded above. By the fundamental axiom, has a limit. Remark. Since initial terms do not matter for convergence, in the theorem it is enough to assume that Root and ratio tests We will now derive two applications of the comparison test, the root and ratio tests. Theorem 1.9 (Root test, Cauchy). If, suppose as. Then if, converges, and if, diverges. Remark. If, the test does not give any information. In fact, later on, we ll see examples with which are convergent and others which are divergent. Proof. Assume that with. Take such that. Then means that given,,, or in other words. Now take, so that. But since the geometric series converges for, the comparison test tells us that converges. Suppose now that with. By the definition of a limit,,. Thus ; in particular, does not tend to. So by Lemma 1.7, diverges. Example. Consider Since, by the root test, converges. Theorem 1.10 (Ratio test). Suppose and. If, then converges. If, then diverges. Remark. Again, the test is inconclusive if.

7 Proof. Suppose with. Take. As in the root test, let, so that, we have. Now write Using the above result we get for. In other words, there is a constant (independent of ) such that for. Since, the series converges, and by the comparison test, also converges. Now suppose with. From the definition of limit,,. This is saying that the sequence increases after. In particular,. But this clearly implies that, and thus must diverge Some inconclusive examples Examples. (i) We know diverges, but both tests are inconclusive. From the root test, because because. [This can be proved as follows. Let with, so by the binomial expansion. Thus Hence.] Applying the ratio test gives so also. (ii) Try both tests on Ratio test; Root test; So for this series it also holds that. However, we can show that the series converges. Thus the series converges, and by the comparison test, converges.

8 This shows that if and/or we can t conclude anything. (iii) By the ratio test, We conclude from the ratio test that the series converges Cauchy condensation test Theorem 1.11 (Cauchy condensation test). Let be a decreasing sequence of positive terms. Then converges iff converges. Proof. Note that for and. Suppose first that converges. Then, which by is, i.e. Hence Multiply by to get because converges. Since the sequence of partial sums is bounded, converges. Suppose now, conversely, that converges. Then Thus because converges. Therefore is bounded in, so converges. QED. Example. Consider for ; if, then. Now is decreasing, so let s investigate.

9 which is a geometric series, so converges iff converges iff.. By the condensation test, Alternating series test Theorem 1.12 (Alternating series test). If and, then converges. Example. converges, because and is decreasing. Proof., so In both expressions, all terms are since is decreasing. Thus is an increasing sequence bounded above. Now by the fundamental axiom,. But now. Since,. Since, given,,. And since, given,,. Therefore,. Hence converges. 1.5 Absolute convergence Definition. Take or. If is convergent, then the sequence is called absolutely convergent. Note. is convergent but not absolutely convergent Absolute convergence implies convergence Theorem If is absolutely convergent, then it is convergent. Proof. Suppose first that. Introduce Note that,, and. Since converges, by the comparison test and also converge, and since, must also converge. If, write and note that. So, if converges, by the comparison test, and converge. In other words, and converge absolutely, therefore and converge. Since, converges. Remark. There is an alternative quick proof of this using Cauchy sequences. First define For, So convergent Cauchy is Cauchy is convergent.

10 1.5.2 Conditional convergence and weird sums Definition. If converges but does not, then we say that is conditionally convergent. The word conditionally is used because in this case, the sum to which the series converges is conditional on the order in which the terms are taken. E.g. converge to different limits ( and respectively). A rearrangement of is, a bijection, taking. Theorem If is absolutely convergent, then every series consisting of the same term in any order (i.e. a rearrangement) has the same sum. Proof. We prove this for ; the extension to is an exercise. Let be a rearrangement of. Let. Suppose we are given a fixed. Then such that contains every term in, so ( ). By the fundamental axiom, with. But by symmetry, If has any sign, consider and from the proof of 1.13, and,,. Since converges, both and converge, and now use the case to conclude that and, and the result follows. QED. 2. Continuity 2.1 Definitions and basic results Two definitions of continuity where and (also applies of course to ). Usually will be some kind of interval. E.g. Definition 1. For, we say is continuous at if given any sequence, then. Definition 2. For, we say is continuous at if given, such that if and, then. ( - definition). Proof that the two definitions are equivalent; D2 D1; We know that given, such that if and, then. Take with, then it is required to prove that. Since,,. But this implies, i.e.. D1 D2; Suppose D2 not true. Then we can find with and. Choose, thus there is with and., but. So does not tend to, which contradicts D1.

11 2.1.2 Sums, products, multiples and reciprocals Proposition 2.1. Let, and with both continuous at. Then,, for any constant are also continuous at. In addition, if, then is also continuous at. Proof. This is a direct consequence of Definition 1 and Lemma 1.1 (about sequences). For example, to show that is continuous at, we take. Since and are continuous at, and. Lemma 1.1 implies ; that is, continuous at. Similarly with the other claims. Consequence. is clearly continuous. By Proposition 2.1, any polynomial is continuous at every point of. Quotients of polynomials (rational functions) are continuous at every point where the denominator does not vanish. Terminology. We say that is continuous in if it is continuous at every point in Compositions We now look at compositions. Theorem 2.2. Let and be two functions such that ( and are subsets of ). Suppose is continuous at and is continuous at, then is continuous at. Proof. Take with. Then it is required to prove that. Since is continuous at,. Call. Since is continuous at,. Remark. Of course, one can prove this with the - definition Examples Examples. (1) (assuming continuous). At, is continuous by 2.1 and 2.2. However is not continuous at. Take so. Also, but. Thus does not tend to zero. (2) For, is continuous (same as before). Now let. from definition of. But so so is continuous at. 2.2 Limits of functions. We d like to make sense of even when may not be in. E.g.

12 but. Similarly if e.g.. Definition.,. Take and assume that there exists a sequence,,. (Note the point may be in but need not be in.) We say that (or as ), if given, such that whenever and, then. Remarks. 1) iff for every sequence,, we have. The proof of this is exactly as the proof that Definition 1 Definition 2 from the last section. 2) If, then iff is continuous at. There is nothing to prove here really, it follows straight from the definitions. This limit enjoys the properties which one would expect, i.e. (i) the limit is unique (ii) if as and as, then, and if. We will start by looking at continuous. 2.3 Intermediate value theorem Theorem 2.3 (Intermediate value theorem). Let. Then takes every value which lies between and. Proof. Without loss of generality, assume. Take and let. Now, because, so is bounded above (by, in fact). Thus has a supremum,. If, then obviously holds. If, use the following argument. (Alternatively, by continuity of at,.) Given a positive integer, is not an upper bound for, otherwise it would contradict the definition of supremum. Then such that, so. Now. So as, so by continuity of,. Thus. Note that, otherwise, but. Thus, for all sufficiently large,, and. Again by continuity of,. Since,. In the limit,. Hence. Application. We can show the existence of the th root of a positive number, where is a positive integer. Look at,, which is continuous. We look at in. Now. By the intermediate value theorem,, thus. This means that has a positive th root. In fact, this th root is unique, since if is another positive th root with, then without loss of generality, so, so which is a contradiction. So definitely exists. 2.4 Cofu clobi ibaatib If continuous, Theorem 2.4. such that. Theorem 2.5. such that. In words, a continuous function on a closed bounded interval is bounded and attains its bounds. Note. on is not bounded. Proof (2.4). Suppose the statement is not true; then for any positive integer,. By Bolzano-Weierstrass, has a convergent subsequence, with

13 , which is a contradiction.. Now since is continuous, so, but Proof (2.5). Let. Then by 2.4. Now, so by the definition of supremum, such that. By Bolzano-Weierstrass, has a convergent subseequence, and. But with. Let, then, so by continuity of. Similarly and the infimum. Alternative proof. Let. Now consider. Work by contradiction and suppose in. Now is continuous, since is. By 2.4 applied to,. Since is positive,. Then, i.e., so that is an upper bound for the set. This contradicts the definition of, since. 2.5 Inverse functions Definition. is said to be increasing if and strictly increasing if. Similarly for decreasing and strictly decreasing. A function is monotone if it s either increasing or decreasing. Theorem 2.6. Let be continuous and strictly increasing. Let and. Then is bijective, and the inverse is also continuous and strictly increasing. Proof. is injective; If then. Otherwise, if,, or if,. ( strictly increasing.) is surjective; take. By the intermediate value theorem,. So is bijective and has an inverse. is strictly increasing; take,,. If, then since is increasing,, i.e.. is continuous; given, let and. Then and for any. Take For this, if then. This was for, but a similar argument gives continuity at the end points also. QED. 3. Differentiability 3.1 Basic calculus Differentiability Let be a function. Mostly we ll be looking at the case where is an interval on and is real-valued. Take,,. Definition. is differentiable at with derivative if

14 is differentiable on if it is differentiable at every point in. Remarks. (i) We could also write with. (ii) Consider Then. Also. Thus an equivalent way of saying that is differentiable at with derivative is to say that there is a function such that, with. is a linear function in. Other ways of writing the same thing;, with as. Also with. (iii) If is differentiable at then it is continuous at ; suppose with. Then i.e. is continuous at. Example. Take,. If, If, then is differentiable, but. But at, is not differentiable. Does exist? is continuous at, therefore Constants, sums, products and reciprocals Proposition 3.1. (i) If, then is differentiable at with. (ii) If differentiable at, then so is, and. (iii) If differentiable at, then so is, and. (iv) If is differentiable at and, then is differentiable at and Remark. From (iii) and (iv) we get Proof. (i) (ii)

15 (iii) Let. Then As, and this becomes. (iv) Let. Then By continuity, this Examples. (i). Then (ii), with a positive integer. Write it as. Then by induction. (iii), for a positive integer ( ). Using 3.1(iv) we get All polynomials and rational functions are differentiable Compositions the chain rule Theorem 3.2 (Chain rule). Let which is differentiable at, and, and is differentiable at. Then. Example.,,. Then, so. Proof. differentiable at means Similarly, differentiable at means We substitute to obtain Let. Then

16 We need to prove that. Define and. Then are continuous at respectively. But now is continuous at because it is products, sums and compositions of continuous functions. Thus. QED. 3.2 The mean value theorem Up to now. everything has worked for functions.. Now we look in more detail at the case of Rolle s theorem We first prove the following basic existence result; Theorem 3.3 (Rolle s theorem). If continuous and differentiable on, and, then Proof. Let be the maximum value of in and be the minimum. (By Theorem 2.5, the values and are achieved.) Let. If, then is constant. Otherwise, or. Suppose (the proof for is similar). We know by Theorem 2.5 that. We ll prove that. Suppose first. Since is differentiable at, we can write with. Since, for all sufficiently small. If in addition we take, then. Absurd because is the maximum of. If, the same argument shows that there are points at left of for which is strictly bigger than, which is again absurd. Thus The mean value theorem Theorem 3.4 (Mean value theorem). If continuous and differentiable on, then Remark. We can rewrite this as follows; letting, where Proof. We consider the auxiliary function. Let s choose. Then. So So, for this choice of, satisfies the hypothesis of Rolle s theorem. Hence. But, so Corollaries We have the following important corollary; Corollary 3.5. continuous and differentiable. (i) If, then is strictly increasing. (ii) If, then is increasing. (iii) If, then is constant. Proof. (i) with. The mean value theorem for If then. (ii) If, same argument gives for,. (iii) Consider on the interval and apply mean value theorem to get, for

17 3.3 Inverse rule (Inverse function theorem) Statement and proof Theorem 3.6. continuous and differentiable in with Let and. The function is a bijection and is continuous on and differentiable on with Proof. Since, then by Corollary 3.5 we know that is strictly increasing. By Theorem 2.6, exists and is continuous. Let. We need to prove that is differentiable with If (small) then there is a unique such that. Note;. Writing we get Note; if, then ( continuous at ) Differentiating rational powers of Example., a positive integer, ( ). Then and if. The inverse rule is differentiable in and Example., any integer and a positive integer. We find using the chain rule, as ; In other words, if where is any rational number, then. Later on we ll define for real and discover that Some lead up to Taylor s theorem with continuous on and differentiable on. The Mean Value Theorem for and for. If, We ll see that we can choose. (Taylor s theorem).

18 3.4 Cauchy s mean value theorem Theorem 3.7 (Cauchy s mean value theorem) Suppose differentiable on. Then such that Proof. Consider the function with are continuous and Then is continuous on and differentiable on. (It is just a product and sum of functions with the same property.) because in either case, 2 columns are equal. By Rolle s theorem. Now expand the determinant and differentiate to see that gives exactly what we want. Example. Find Apply CMVT on. Then Let, then (L Hôpital s rule) 3.5 Taylor s theorem Taylor s theorem with Lagrange s form of the remainder Theorem 3.8 (Taylor s theorem with Lagrange s remainder). Suppose and its derivatives up to order are continuous on, and exists for. Then where. Notes. 1) is Lagrange s form for the remainder. 2) For this is the mean value theorem; Proof. Consider for is chosen so that. Then We ll apply Rolle s theorem times. We apply it to first, to get an such that. Then Apply Rolle to in to get such that. Note that in fact from the definition of we see that We keep applying Rolle to get at which for.

19 Now note. Therefore. Write for to get. Now put this value of in the definition of and the statement is exactly the statement of the theorem. QED Taylor s theorem with Cauchy s form of the remainder Theorem 3.9 (Taylor s theorem with Cauchy s form of the remainder). Let satisfy the same hypothesis as in Theorem 3.8 and in addition suppose (this is just to simplify matters). Then where Proof. Define, for, Set Then (from definition of ). Rolle s theorem applied to gives such that. Now we compute Rearranging Set Note. Setting gives which is Lagrange s form of the remainder Taylor series Needs as

20 Remarks. 1) is just to simplify matters 2) The same result holds in an interval. 3) is the Maclaurin expansion Application to the binomial series Application. Binomial series;. Then. Claim; for, where is the generalised binomial coefficient. If is a positive integer, then for. Let s prove that for the series is absolutely convergent. Ratio test; So, the series converges absolutely for. In particular,, i.e. as for. Now we study the remainders in the Taylor theorem. missing line We look at Lagrange s form; If, if if. Let to conclude that. For, the argument breaks, so Cauchy to the rescue! Note that Careful, may depend on. If,. If,. In any case you get

21 where is a constant that depends on and but not. Therefore as. 3.6 Some comments on differentiability of functions Standard properties work for both functions are differentiable. Example., and ; chain rule, the sums and products of differentiable So does not exist!,, is differentiable with real glasses Complex differentiable functions are called holomorphic. 4. Power series We will look at series of the form 4.1 Radius of convergence Convergence of power series Lemma 4.1. If converges, and, then converges absolutely. Proof. Since converges, as. In particular, there is a constant such that. since so. Thus the geometric series converges. By comparison, converges, i.e. Theorem 4.2. A power series either (i) converges absolutely for all, or converges absolutely. (ii) converges absolutely for all inside a circle and diverges for all outside it, or (iii) converges for only. Definition. The circle is called the circle of convergence, and the radius of convergence. In case (i) we agree that, and in case (iii) we agree that. Proof. Let and converges. Now. If, then by Lemma 4.1,. If is unbounded, then and we have case (i).

22 If is bounded, there exists a finite supremum for which we call. If, we ll prove that if, then converges absolutely. Choose such that. Then converges, and by Lemma 4.1, converges absolutely. Finally we show that if, then diverges. Take such that. If converges, again by Lemma 4.1, converges. But this contradicts the definition of as supremum of. Thus diverges Computing the radius of convergence The next lemma is useful in computing ; Lemma 4.3. If as, then. Proof. By the ratio test we have absolute convergence if, i.e.. And if, then and does not tend to zero. Therefore. Remark. By the root test, if then. Examples. (geometric series). We know, but note that if, the series diverges, since which does not tend to. as, but what if? If, diverges (Example Sheet 1). Abel s test;,, is bounded bounded in. Therefore converges for and. has but converges for every.

23 , but on, we have divergence. Conclusion In general nothing can be said at. 4.2 Differentiability of power series Theorem 4.4. Suppose has radius of convergence, so that for with. Then is differentiable and Remark. Iterate this theorem, to get that were a polynomial. Proof (non-examinable). We ll use two auxiliary lemmas. can be differentiated infinitely many times as if it Lemma 4.5. If has radius of convergence, then so do and. Lemma 4.6. (i) (ii). Proof of 4.4. By Lemma 4.5, since has radius of convergence, it defines a function for with. Then we would like to show that This implies that is differentiable with. By Lemma 4.6,. Take. If, we get By Lemma 4.5, we know that converges to some number. Therefore Proof of 4.5. has radius of convergence. Then has radius of convergence and it is required to prove that. Take with. Choose. Since, as. In particular,.

24 But converges. Ratio test; By comparison, converges absolutely. Therefore. But in fact we have equality because So by comparison if converges absolutely, so does. Therefore. What we proved also implies that also has radius of convergence of. Proof of 4.6. (i) (ii) by the binomial theorem. Therefore Example. We saw last time that it has radius of convergence. Then with defined to be The theorem we just proved tells us right away that is differentiable and 4.3 The standard functions: exponentials General remark. Let be differentiable. If, then is constant. Proof. Let,. Write, By the chain rule,. But we also see (Corollary 3.5) constant. Therefore since The exponential function Claim. for. Proof.,

25 By the general remark above, is constant. From the definition of,, so Now at, so the claim is proved. Now we restrict and prove: Theorem 4.7. (i) is differentiable everywhere. (ii) (iii) (iv) is strictly increasing (v) as, as (vi) is a bijection. Proof. (i) and (ii) have been done. (iii) Clearly from the definition of, if and (iv) is strictly increasing by theorem we proved before (v) From definition of, for, so as,. Also, as. (vi) is strictly increasing so is injective. For surjectivity pick. Since as and as,. The intermediate value theorem says that. Remark. (vi) and (ii) are saying that The logarithmic function is a group isomorphism. Since is a bijection, such that and. Theorem 4.8. (i) is a bijection and, (ii) is differentiable and. (iii) Proof. (i) obvious. (ii) is differentiable by the inverse rule (Theorem 3.6). (iii) By groups, is a homomorphism functions Definition. Let, and be any positive number. Then. E.g.. Theorem 4.9. Suppose and. Then

26 (i) (ii) (iii) Proof. (i) (ii) (iii) Let be a positive integer. Then Let be a positive integer, then what is We can now set for. This definition agrees with the one given before for. We can now prove that and. Define the real number as Then, but, so. can be rewritten as. Then for, by the chain rule. Let. Then by the chain rule. 4.4 Trigonometric functions We define Both have infinite radius of convergence (check).,. By Theorem 4.4 they re differentiable, and, Relation to exponential function Now and, so

27 Now from definitions, and. So. Therefore It is also obvious that and. Now we get, using, that for. Also, for. If, from which it follows that and. Warning: (or ) are not bounded for, since for, with, Periodicity of the trigonometric functions Proposition There is a smallest positive number (where ) such that. Proof. If,. Then for and (since and ). Therefore for. But for is a strictly decreasing function in. We ll prove that,. Then by the intermediate value theorem, with. where each term is Therefore. Now where all bracketed terms are for. For, Therefore. Corollary Proof. But we know.

28 Definition. We define Theorem (i). Finally, (ii) (iii) Proof. Immediate from addition formulae and Note. This implies periodicity of! 4.5 Hyperbolic functions Definition. From this we get,. We can check (exercise) that and, and identities such as. Two final remarks; 1) The other trigonometric functions (, etc) are defined in the usual way (e.g. ). 2) Exponentials beat powers, i.e. as for. To see this, take a positive integer such that and observe (from the definition of as a power series) that 5. Integration 5.1 Riemann integration bounded, i.e Dissections Definition. A dissection (or partition) of the interval is a finite subset of which contains and. We ll write it as We define the upper and lower sums of to, and, as with respect

29 Clearly Lemma 5.1. If and are dissections with, then Proof. Suppose has one extra point, let s say for some. Note that Then If now has more than one extra point, just do this argument for each extra point. ; we already noted this. The proof for is similar to the one for the upper sums. QED. Lemma 5.2. If and are any two dissections, then This is a key lemma for integration. Proof., so by Lemma 5.1,. QED Definitions of integrals and integrability Definition. The upper integral is defined as The lower integral is Remark. These numbers are well-defined because and. is bounded, so Definition. We say that a bounded function is (Riemann) integrable if. In this case we write (or just ). Remark., thanks to Lemma 5.2. and Example. (Dirichlet) A function given by For any, because every interval contains a rational number. However, for any partition. Therefore and so is not integrable. 5.2 Integrability of monotonic and continuous functions Riemann s theorem We now prove the following useful criterion for integrability;

30 Theorem 5.3 (Riemann s theorem). A bounded function. Proof. Let s assume first that is integrable. is integrable iff given Given, by definition of,. By definition of,. Consider. Then by Lemma 5.2, because on account of being integrable. To prove the converse, assume that given,. Then by definitions of and. Since this is true for all. Therefore is integrable. QED Integrability of monotonic functions We now prove that monotonic functions are integrable and continuous functions are integrable. Remark. Monotonic and continuous functions on are bounded. Theorem 5.4. If monotonic, then is integrable. Proof. Assume is increasing (same proof if is decreasing). Take to be any partition of. Then But is increasing, so Thus Now consider, for a positive integer, For this When, take large enough so that Therefore by Theorem 5.3, is integrable. QED Uniform continuity Lemma 5.5 (Uniform continuity). If is continuous, then given and if then. Remark. Continuity at a point, let s say, means that given,. The lemma is saying that we can choose a which works for every!

31 Proof. Suppose claim is not true. Then there exists such that for every, we can find with but. Take, to get with but. By Bolzano- Weierstrass,. Then. Let, then, so. On the other hand,. continuous, therefore,. Then, passing to the limit,, but. Absurd. QED Integrability of continuous functions Theorem 5.6. If continuous, then is integrable. Proof. Consider the partition with points for a positive integer. Then Let be given, and consider the given by Lemma 5.5. Choose large enough such that. Then for any, Lemma 5.5 tells us that. Therefore By Riemann s criterion, Example. is integrable. We claim that is integrable and Take any partition, then since every interval contains irrational numbers where is zero. Clearly. To prove the claim, it suffices to show that given theorem and the fact that ). (by Riemann s Let be given, and take an integer with. Then let. Then and, so. So is a finite set with cardinality.. Choose a partition such that the points belong to the intervals with length less than. First sup is and second sup is (outside ). So. QED.

32 5.3 Elementary properties of the integral bounded and integrable on then (1) If on, then. (2) is integrable over and. (3) For any constant, is integrable and. (4) is integrable and. (5) The product is integrable. (6) If except at finitely many points in, then is integrable and. (7) If, is integrable over and and. Proof (1). If, then integral is upper sum, so; for any. Now take infimum over all, so Proof (2). Observe Now take any two partitions by Lemma 5.2. Now keep fixed and take infimum on all s. Then Now take inf over all since integrable. A similar argument for the lower sums gives Putting everything together we get, since, that. Proof (3). Exercise. Proof (4). Consider. Then we claim that is integrable. (this needs checking). For any partition, By Riemann, since is integrable,. By Riemann again and, is integrable. Now note that since. So ; since and are integrable by (2) and (3), is integrable.

33 follows from (1) and the fact that. QED. Proof (5). First we ll prove that if is integrable, then is integrable. Suppose first that. Since is integrable, given, a partition of such that. Let since. bounded, thus where is integrable, by Riemann s theorem. If is now arbitrary (i.e. just integrable) then is integrable by property (4). But since is integrable. To prove that is integrable, just write. Now integrable and integrable and integrable, and we re done. Proof (6). Let. Then for every except perhaps a finite set (just from the definition of the integral) must be integrable with. But is integrable with. Proof (7). Exercise. Convention. If, define Agree that. 5.4 Integration rules and tools How do we compute? The fundamental theorem of calculus Let be a bounded Riemann integrable function. For define Theorem 5.7. Proof. For is continuous.

34 by properties of integral, assuming. is bounded so. Thus In fact for any,. Let, therefore Theorem 5.8 (Fundamental Theorem of Calculus). If in addition is continuous, then is differentiable and. Proof. For,, we d like to make small as. We write it as since Now Thus So we have proved that Now if, then by continuity of, Integration is the inverse of differentiation Corollary 5.9. If is continuous on, then Proof. From 5.8 we know that is constant. Therefore. Therefore This gives a way of computing if we know a primitive (anti-derivative) for, that is, a function such that. Primitives of continuous functions always exist (5.8) and any two primitives differ by a constant. such that constant Integration by parts Corollary Suppose and exist and are continuous on. Then

35 Proof. Product rule gives. Therefore and the rest follows immediately Integration by substitution Corollary 5.11 (Integration by substitution). Let with,, and assume that exists and is continuous on. Let be continuous. Then Set, to get a more familiar version. Proof. Set Let with taking values in. Then by the chain rule, 5.5 Taylor s theorem revisited Taylor s theorem with integral form of the remainder We now revisit Taylor s theorem and find an integral form for the remainder. Theorem 5.12 (Taylor s theorem with remainder an integral). Let, then be continuous for where Remark. Note that here we assume continuity of and not just mere existence (as in our previous versions), so this theorem is a little weaker, but just fine for most practical purposes. Proof. Let s make the substitution, so that. Then By integration by parts this becomes That is,

36 Integrate by parts times to get where the last term is. QED Link to Cauchy s form This integral form gives back Cauchy s and Lagrange s forms. Previous remarks. continuous, then The mean value theorem applied to. Thus gives (mean value theorem for integrals). Let s apply this to which is Cauchy s form of the remainder Link to Lagrange s form To get Lagrange s form we use the following proposition. Suppose, then for some. Then which is Lagrange s form of the remainder. Now we prove the proposition we just used, namely that if, then such that continuous and If you have Proof. Let then you get the mean value theorem. By the Fundamental Theorem of Calculus, and.

37 Cauchy s mean value theorem (3.7) applied to and gives Since, we simplify to get the desired result. 5.6 Infinite integrals (Improper integrals) We d like to make sense of Definition Suppose we have such that for every, is bounded and integrable on. Definition. If we say that exists and that its value is. In this case we also say that converges. If does not tend to a limit, we say that diverges. Example. If, the integral Let, then we say that the limit is finite only if. If, Relation between improper integrals and series Notice the similarity with the series then Remark. If, for and for constant, then if converges, converges and This is the analogue of the comparison test. Proof. The function is an increasing function. Thus if,

38 If converges, then is bounded above, and so we can consider We claim By definition of supremum, given, such that. If, then i.e., i.e. Remark. For series, we know that if converges, then. However, it is not true that if converges, then as. Example. converges, so converges. But, so as The integral test Theorem 5.13 (Integral test). Let be a positive decreasing function. Then 1) both converge or both diverge. 2) As, tends to a limit as, such that. Examples. (1) We just say that converges iff. We can apply the integral test to deduce that 2) converges iff. Let. Then

39 As, this goes to infinity, so the sum diverges. Proof (1). Note that since is decreasing, it is Riemann integrable in every interval. For, since decreasing. Integrate between and to get Adding, Suppose converges, then is bounded in. By, is bounded, and therefore it must converge by the fundamental axiom (as it is increasing in ). Suppose converges, then is bounded and by is bounded, thus since is increasing, it follows that Proof (2). Let converges. Therefore is decreasing. Also from,. Thus is a bounded decreasing sequence, and therefore it must converge to a limit with. QED Euler s constant Corollary 5.14 (Euler s constant). as with. Proof. Let and apply the theorem. Remarks. 1) It is still unknown whether is rational or irrational! 2) 3) Proof that. Define

40 for. We ll derive estimate for. Let and in the fact from the last section that for,,. for. So But can be easily computed by integration by parts. If you do the calculation, you get by definition of Integrals which are improper at both ends So far we have looked at but we can also make perfect sense of We can also make sense of. Definition. converges if converges and converges. Also if and, then we let. This defintion is independent of ; so the sum will not change, it will always be equal to. This is not quite the same as saying that Example. exists.

41 but does not exist, because neither or converge. Example. on. Let. and the same for general functions. This is an improper integral of the second kind.