Buck Meta Chapter 4 Convergence

Buck Meta Chapter 4: Convergence PhL 1.19.15 This meta doc at 15 pages is one third the size of the 41 page raw notes as of 1/20/15. The book Chapter 4 is 63 pages long and is brimming chock full of theorems (33) and examples. Overview and the many meanings of "convergence". In Section 4.1 convergence refers to the convergence of infinite series of constants, and of the convergence of the sequence of partial sums associated such a series. Double series Σijaij are also considered. In Section 4.2 convergence refers to the convergence of a series of functions and to the sequences of partial sums associated with such a series of functions. In one sense (pointwise convergence), the functional series converges for some particular argument x with norm that of Em where f:En→Em. In another sense uniform convergence, the functional series converges over a set of x values D with the norm that of Lmax(D). I refer to this latter convergence as (uniform convergence)f. Both these functional convergence notions involve many functions usually called fn(x). In Section 4.3 Bucks apply the uniform convergence ideas to infinite function series Σnfn(x) which have the particular functions fn(x) = anxn. These are called power series. In Section 4.4 Bucks introduce another form of convergence I call (uniform convergence)p. It relates to a single function of two variables f(t,p) which function is continuous at t0 in t in a manner that is uniformly convergent for all p in set E as t→t0. [p = point] An example is the function f(t,x) = sin(xt)/ [t(1+x2)] which is continuous in t at t0 = 0, and as t→t0=0 this function exhibits (uniform convergence)p to F(x) = x/(1+x2) for all real values of x. The variable p or x I call a bystander variable and in one application this is the parameter in an "improper integral with a parameter". The main results of this section are a set of what I call "order interchange theorems" which involve limits and operators like ∂x and the integration operator I. These theorems always require that one or more functions exhibit (uniform convergence)p. For integrals, we want to take a limit of the parameter through the integral, or we may want to interchange the order of two integrals where one is improper. In the differential case, we want to differentiate the integral containing the parameter with respect to that parameter, and we would like to move the derivative operator through the integration and act directly on the integrand. Section 4.5 deals with a special function Γ(x) defined as an improper integral with parameter. Chapter 4: Convergence [ pp 158-220 ] 1 4.1 Infinite Series [158] // theorems 1 → 14 1 4.2 Uniform Convergence [180] // theorems 15 → 21 4 4.3 Power Series [ 196 ] // theorems 22→ 25 7 4.4 Improper Integrals with a Parameter [ 204 ] // theorems 26→29 8 4.5 The Gamma Function [ 213 ] // theorems 30→33 14 Chapter 4: Convergence [ pp 158-220 ] 4.1 Infinite Series [158] // theorems 1 → 14 Bucks develop many theorems here, most are tests for convergence of a series. Subjects include series rearrangements, alternating series, partial sums, comparison of sums with corresponding integrals. The final section has theorems about double sums, including some about the Cauchy product (zigzag order) of two regular series. The word series always means infinite series. Lots of examples in text and raw notes, but they are all omitted here. An infinite series involves terms {an} and partial sums {An}, both of which are sequences. These two sequences together compose the "infinite series" object. [160] If the {An} partial sum sequence converges, then one says "the series converges". So we may say, An → A series [ {an},{An} ] "converges" // else "diverges" Note that diverges does not imply An→ ∞ since you could have a jumping situation where An is finite (bounded), but won't stabilize. It is possible for two series to diverge but their sum converges, and an example is given. Cauchy Product of two series: Consider these two power series (from end of this chapter), (Σn=0∞ anxn) (Σm=0∞ bmxm) ≡ [ a0 + a1x + a2x2 + a3x3 + ...] [ b0 + b1x +b2x2 + b3x3 + ...] If you rearrange this series by increasing powers of x ( cnxn ) then the coefficients cn of the powers form the series called the Cauchy Product of the series Σan and Σbn . Theorem 1: [161] (Term Limit Test) If series Σan converges, then an → 0. Proof is totally trivial using the An. Contrapositive gives a very easy test for series convergence! [ Note: not true for analogous !Syntax Error, Icos(x2) dx = (1/4) , p 144 since integrand 0 as x→∞ ] Theorem 2: [161] (Comparison Test). If an ≤ bn and things non-negative, then Σbn conv Σan conv. Theorem is analogous to Comparison Test for integrals Theorem 14 p 140. Corollary: [161] (Two Series Ratio Test) If (an/bn) → L (positive and finite) with both an and bn non-negative, then when you consider Σan and Σbn, either both converge or both diverge. Analogous to Ratio Test for integrals p 140. Theorem 3. [161] (Ratio Comparison Test) Again we have both an and bn non-negative and an+1/an ≤ bn+1/bn for large n. This says that the A ratio is smaller than the B ratio in the limit. Theorem says: Σbn conv Σan conv. Proof is very simple. Theorem 4: [161] Consider Σn=0∞ xn. For |x| < 1, sum is 1/(1-x), otherwise series diverges (real x). Proof shown is very simple. Theorem 5. [161] (Bounce Ratio Test) Assume an > 0 and allow that the ratio an+1/an might bounce around for large n, so this ratio itself might not have a limit, but assume the ratio is bounded so lim sup ratio = L and lim inf ratio = l . That is to say, the ratio is at least bouncing around inside l ≤ an+1/an ≤ L. Obviously we have l ≤ L. Theorem says that L < 1 Σan converges and l > 1 Σan diverges, otherwise no conclusion. Corollary: [162] (No-Bounce Ratio Test) Now assume ratio an+1/an does not bounce around but has a simple limit r. Then in the previous theorem, we have L and l both equal r. Then r < 1 converge, r > 1 diverge, and r = 1 no conclusion. Theorem 6 [162] (The Root Test). Again an≥ 0 and assume lim sup (an)1/n = r. Then same conclusion as last theorem: r < 1 Σan converges and r > 1 Σan diverges. Theorem 7 [163] (Sum/Integral Compare Test) The sum 1 to ∞ of fn = f(n) and the integral of f(x) from 1 to ∞ both converge or both diverge if f is continuous and monotone decreasing (to 0, meaning f ≥ 0) on [1,∞). Corollary. [163] Series Σ1∞ (1/np) and Σ1∞ (1/n) / (ln n) p converge p > 1 else diverge. Theorem 8 [ p 163] (Raabe's Test). Assume an > 0 as usual, look at ratio rn ≡ an+1/an. Suppose rn ≤ 1 - p/n for p > 1 and large n. Then series converges. Theorem 9: [164] (Abs Value Test) If Σ |an| converges, so does Σan. Fact: It is possible for Σ|an| to diverge while Σan converges. In this case, Σan is conditionally convergent. If Σ|an| converges you have absolute convergence. This is the same idea as with integrals on page 141. The following two theorems involve products of series : Theorem 10 [165]: (very obscure) Here an→ 0, Σ|an+1- an| converges, Bn is bounded. Then the product sequence Σanbn converges. Says nothing about whether Σan or Σbn converges. [ compare Ch 3 Thm 16 ] Theorem 10A = Corollary 1: (very obscure) If monotone an→ 0 and Bn is bounded, then the product sequence Σanbn converges. Says nothing about whether Σan or Σbn converges. [ compare Ch 3 Thm 16A ] Corollary 2 [166] (Leibniz Alternating Series Test) : If an monotone → 0, then Σ (-1)n+1an converges. Notice that this does not seem to have a simple integral analog. But it's a bit like !Syntax Error, Isin(x) g(x) where the sign alternation is provided by sin(x), see Ch 3 Corollary 2 [144]. Theorem (Exercise 12 on p 179): For a convergent alternating series, if you stop at some term, the |error| is ≤ the |next term|. Also, the approach to the limit alternates between above and below. I have not proven this, but it gets used later on page 187. Proof seems would be very easy. Theorem 11: [169] Only if a series is absolutely convergent are you allowed to rearrange the series without changing the sum. Theorem A1 [170] Assume f is continuous, positive, and decreasing on [1,∞]. Then the difference between a sum and an integral from 1 to n is bounded on both sides this way: f(n) ≤ [ Σk=1n f(k) - !Syntax Error, If(x)dx] ≤ f(1). If we express Σ - ∫ as an error Cn, then Σ = ∫ + Cn where f(n) ≤ Cn ≤ f(1) Theorem A2. f(1) ≤ [ Σk=1n f(k) - !Syntax Error, If(x)dx] ≤ f(n) if f(n) increasing Theorem 12: The quantity [Σ - ∫- f(n)/2] is bounded above and below provided that f is positive and has negative curvature in the entire range 1 to ∞ (see p 171 figure). What this says is a ≤ [Σ1n f(k) - !Syntax Error, If(x)dx - f(n)/2] ≤ b a = f(1) - f(2)/2 b = f(1)/2 This really shows that Σ1n f(k) - f(n)/2 is a good fit to the integral. Double Series Stuff p 172 The convergence definition they choose on page 172 is Cauchy like in that the abs value shown has to → 0 as shown. In the general Cauchy you would have separate N and M, but here it is N and N as shown. They are keeping the matrix square NxN as they increase N to infinity. Theorem A: [173 top] Σij | aij| converges Σij aij converges [ like Theorem 9 p 164 ] If the abs sum converges, so does the other sum, series is absolutely convergent. Question: Look at Ch 3 meta below Thm 17. I would guess that the double sum is more like the double integral and so either both double sums above converge or both diverge. Is this true? Theorem B: [173] Σij aij is absolutely convergent you can rearrange arbitrarily and maintain the sum. This is like Theorem 11 p 169. Corollary C: [173] If Σij aij is absolutely convergent, you can rearrange as the zig-zag ordering which is in fact the Cauchy Product sequence as shown above. This theorem involves one double series. Theorem 13. If Σaj and Σbj are each absolutely convergent to A and B, then the zig zag ordered double sum Σcn converges to AB. This zig-zag ordering is in fact the Cauchy product defined earlier. Theorem 13A: [173] Σaj→ A and Σ|aj|→ A' (abs conv) => Σzzaibj → AB. and Σbi→ B but Σ|bj| diverges 13A is a little stronger than 13. It is called Theorem of Mertens and is in the Knopp book. Theorem 13B: [174] If both Σaj and Σbj are conditionally convergent, then Σcn may diverge. Theorem 14. [174] If aij ≥ 0, then columns-first sum converges rows-first sum converges The topic of double series is suddenly ended and here Bucks consider 5 examples of single series which contain a parameter. 4.2 Uniform Convergence [180] // theorems 15 → 21 Suddenly the topic is changing from the study of convergence of 1D and 2D series of constants, to the study of the convergence of 1D series of functions of one variable x. Again I suppress Examples. Since "uniform" has always been a problem for me, I copy all the background from raw notes here: Note: Uniform continuity for a function f(x) was treated in Section 2.3. Recall that you need a δ that works for all x in the domain D. Here we are talking uniform convergence of a sequence of functions or of a series whose terms are functions of x. Def: [180A] A series Σfn(x) converges pointwise to f(x) on domain D if it converges for each x in D. You might find that δ depends on x, and there is no specific δ that works for all x in D. Def: [180B] A sequence {fn(x)} converges pointwise to f(x) on domain D if it converges for each x in D. That is to say, for any given x in D, you can find N such that || fn(x) - F(x) || < ε for n > N. Again, you might find that N depends on x, so you really have N(x). Def: [182A] A sequence {fn(x)} converges uniformly to F(x) on domain D if it converges for all x in D "with the same N". That is, there is an N such that || fn(x) - F(x) ||D < ε for n> N and for all x in D. The point is that the same large integer N works for all x in D, you don't need N(x). Important point: Here || fn(x) - F(x) ||D is the max norm, sometimes called the infinity norm. I have much more on this later. Earlier definitions just have a normal Em geometric norm. Def: [183A] A series Σfn(x) converges uniformly iff the sequence of partial sums Fn(x) converges uniformly to F(x) as per the above definition. Def: [183B] A series Σfn(x) converges uniformly iff the tail of the series converges uniformly to 0. Can write this last as || Σk=n∞ fn(x)||D → 0. This is stated as if it were an alternate definition. Notice how the series convergence leans on the convergence of the partial sum sequence. Lots of examples are given. A good one is a series whose terms are fn(x) = n2x e-nx on [0,1]. This function has a peak at x = 1/n where fn(1/n) = n/e. Thus, as n→∞, || fn(x)||D → ∞ since this is the max norm and the max is going to ∞. Thus, not uniformly convergent on [0,1]. It is pointwise convergent since if you first pick some x, you can find N so n>N meets your ε rule (due to the e-nx). This is explained on page 182. It is always a slippery subject. Definition and Theorem (no number, top page 183) series is uniformly convergent partial sum sequence is uniformly convergent series is uniformly convergent series is pointwise convergent and | tailn |E → 0 def: [183] Cauchy Property for sequence: For any ε>0, you can find N such that | fn - fm |E < ε for n,m both > N . Theorem 15 (and precursor p 183) Sequence fn has Cauchy property sequence fn is uniformly convergent. Corollary [ p 184] : series is uniformly convergent || tailn,m ||E → 0 // Cauchy tail Theorem 16 (Weierstrass Comparison M-Test). [184] Consider a series Σkuk(x). Suppose you know that || uk(x) ||E ≤ Mk for all k (perhaps in an infinite tail) and suppose you also know that the series ΣkMk converges. Then Σkuk is uniformly convergent on E. Norm here is the max norm. Example: Consider fk(x) = k2x e-kx on [0,1] which we just saw has max = k/e and is not uniformly convergent on [0,1]. Then fk(x) ≤ k/e so you could try Mk = k/e, but alas, ΣkMk does not converge so test is a no go. Theorem 17: [184] Suppose fn(x) → F(x) uniformly on E. If the fn(x) are continuous, so is F. Suddenly the word "continuous" pops out of nowhere in this last. The premise here is that the sequence of functions shown converges uniformly to F(x). Restate this for a series: Theorem 17A: Suppose Σk=1n uk(x) → F(x) uniformly on E. If the uk(x) are continuous, so is F(x). [ Corollary 1 top of page 185] These both say that if a series or sequence converges uniformly to F(x) on [a,b], then F(x) must be continuous on [a,b] so it cannot be infinite at one of the endpoints (I think). Theorem 18: [185] Let E be closure of S which is open, like (0,1). Suppose fn converges uniformly on S, and the fn are continuous on E. Then fn converges uniformly on E. This says if uniformly convergent on (a,b), then so also on [a,b]. The notion of uniform convergence is strongly related to the notion of order interchange in various contexts. Here is our first order interchange theorem: Theorem 19. [186] (order interchange for sequence and integral in E2). The domain D of the integration is closed and bounded (compact) in E2. The sequence of functions {fn(x)} converges uniformly on D to some F(x), so limn→∞ fn(x) = F(x). Then: limn→∞ [ ∫D fn(x,y)dA] = ∫D [ limn→∞ fn(x,y)] dA which is ∫D F(x) dA This says you can take the n→∞ right through the integral if premise conditions are met. In other words, you can interchange the order of "sequence limit" and "integration" which itself is a certain limit. Corollary [186] : (order interchange for series and integral in E2) If uk(x) are continuous on D and if Σuk(x) → F(x) converges uniformly on D, then if you need to integrate F(x) over D, you may do so "term by term". This is just the above Theorem 19 restated as limn→∞ [ ∫D (Σk=1nuk(x,y) )dA] = ∫D [ limn→∞ (Σk=1nuk(x,y) )] dA which is ∫D F(x) dA term by term integration Theorem 20. [188] (Arzela-Osgood-Lebesgue) Assume fn(x) → F(x) pointwise on bounded D and fn and F are integrable on D. Suppose also that || fn(x) ||D ≤ M (a constant) for all n. Then limn→∞ [ ∫D fn(x) ] = ∫D [limn→∞ fn(x)] perhaps D is in En ? and you can do order interchange between sequence limit and integration. So this is a refinement of Theorem 19 which does not require uniform convergence, but requires something else. Comment: The above three items are treated where domain D lies in E2. Bucks are strangely silent about whether the theorems are valid for E1 or for general En. I presume all is OK for full En. The next theorem is stated for E1. Theorem 21 [188] (Series/Derivative Order Interchange) Suppose Σun(x) → F(x) pointwise on [a,b]. Suppose u'n(x) exists and is continuous on [a,b]. Suppose Σun'(x) → G(x) uniformly on [a,b]. Then G(x) = F'(x). That is to say, limn→∞ [Σk=1n ∂xun(x)] = ∂x[limn→∞ Σk=1nun(x)] G(x) = ∂x [F(x)] term by term ∂x Notice that this theorem requires F(x) pointwise converge, but F'(x) uniform converge, and more. Bucks consider on page 191 a certain Van Der Waerden function which is continuous but at the same time is nowhere differentiable. A grab bag item. Here is another: Tietze's Extension Theorem [193] You have E closed and bounded in En and f(p) [real] is continuous and bounded on E ( so |f(p)| ≤ M ). Then there is a way to extend f(p) to all of En (call it F(p)) such that F(p) = f(p) on E and such that F(p) is continuous and bounded on all of En. The proof constructs a solution to the problem. This is easy to see in one dimension, and plausible in 2D though not obvious there, not to mention higher dimensions. In 1D you just could just put linear tapering walls outside some [a,b] = E. This would not be differentiable at the walls, but it would be continuous and that is all that is required. 4.3 Power Series [ 196 ] // theorems 22→ 25 Theorem 22: [196] Power series Σanxn has some R radius of convergence so |x| < R gives convergence. You can find R from either the ratio test or root test, as shown top p 197. Could find R = 0 or R = ∞. You could restate this replacing x by g(x) so |g(x)| < R gives you some convergence domain in x. Corollary [197] You can generalize trivially to Σan(x-c)n. Theorem 23. [197] The power series Σanxn with R converges uniformly on |x| ≤ b < R. Theorem 24. [198] For |x| < R, you can differentiate f(x) = Σanxn term by term to get f'(x). The series f'(x) has the same R as f(x), and the same uniform convergence region. Corollary 1 [198] For f(x) = Σan(x-c)n, you can identify an = f(n)(c)/n!. You can keep diffing using Thm 24 and then just set x = c in the result to prove this corollary. Corollary 2 [198] Σan(x-c)n = Σbn(x-c)n in ball near x, then an = bn. This says that a function analytic near x = c can have only one power series there, and it is the Taylor series we studied earlier. Notice how the Bucks have set us up for easy treatment of the above theorems by previously discussing the notions of uniform convergence and order interchange for differentiation. Power series is just an application of all that previous stuff to series with terms fn(x) = anxn. Theorem 25 [199] (Abel) If f(x) = Σanxn has radius of convergence R, and if f(x) converges at x = R, then f(x) is uniformly convergent on [0,R]. Same idea if you do R → -R. Corollary [200]. If f(x) = Σanxn has radius of convergence R, and if f(x) converges at x = R to some value S, then you must get this same value S if you take the limx→R[Σanxn]. The section closes with various example, one of which is a series definition of ex called E(x). 4.4 Improper Integrals with a Parameter [ 204 ] // theorems 26→29 My notes begin with a long review of notions of continuity and convergence, each of the non-uniform and uniform varieties. Here is then the raw notes summary regarding convergence: 1. In the discussion of continuity and uniform continuity, we had limp→p0 f(p) = f(p0) using the Em norm saying that that | f(p) - f(p0)|Em < ε. We could then bring into the discussion a sequence of points pn and we could then say limpn→p0 f(pn) = f(p0) and we then have convergence of the sequence {f(pn)} which is a sequence of points in Em. This convergence is associated with the continuity of the function f(p). In the case that the same δ works for all p0, we had uniform continuity over E, and you could say that we have uniform convergence of f(pn) → f(p0) where pn → p0 where p0 is any point in E. In other words, we are saying that f(pn) → f(p0) converges to p0 sort of "uniformly" for any p0 in E. This use of the phrase "uniform convergence" is not used by the Bucks. You might call it (uniform convergence)p since it involves a sequence of points. Note that, although f(p) is a function, here we are really talking about sequences of points, { pn } and { f(pn) }. We are not talking sequences of function, only one function f is involved in the discussion. 2. In the above subsequent discussion of sequences of functions fn(p), we talked above about fn(p) converging pointwise to f(p) on D, which involved the En norm, and we talked about fn converging uniformly to f using the max norm over D. We say that fn converges uniformly to f over D, and this is the official definition of the phrase "uniform convergence". Maybe call this (uniform convergence)f since it involves a sequence of functions. The real distinction is this. In case 1 we are dealing with points in the space of points En with the En norm, whereas in case 2 we are dealing with functions in the space of functions called Lmax(D). The topic now changes toward consideration of functions of two variables in a certain sense. The functions are not like f(x,y) where (x,y) = r = a point in E2. In that case, we usually treat the two variables x and y as the single variable r. Rather, we consider here f(p,t) where p is going to be what I call a bystander parameter and where p in E and t in T are "unrelated variables". I found it useful in the raw notes to ponder how one might discuss the notion of continuity for functions of two somewhat unrelated variables f(p,t). The Bucks seemed a little abrupt on this transition to treatment of bystander parameters and a new use of the word "uniform convergence". I do this as a set of "upgrades" and shall copy the raw notes stuff here: Then consider this ε δ statement: 1. For any ε > 0 and for a specific p in E and for a specific t0 in T, we can find δ(t0,p) > 0 such that | f(t,p) - f(t0,p)| < ε when |t-t0| < δ. I would describe this as "continuity at a point t0 in T with a bystander parameter p in E". I would also say that f(t,p) "converges" to f(t0,p) as t → t0. This means there are sequences tn so that f(tn,p) "converges" to f(t0,p) as tn → t0. Our first "upgrade" of this concept would be: 2. For any ε > 0 and for all p in E and for a specific t0 in T, we can find δ(t0) > 0 such that | f(t,p) - f(t0,p)| < ε when |t-t0| < δ. I would describe this as "continuity at a point t0 in T which is uniform over E". I would also say that f(tn,p) (converges uniformly)p over E to f(t0,p) as tn → t0. Notice that there is only one function here, it is called f. A different upgrade of item 1 might be this: 3. For any ε > 0 and for a specific p in E and for all t0 in T, we can find δ(p) > 0 such that | f(t,p) - f(t0,p)| < ε when |t-t0| < δ. I would describe this as "uniform continuity over T with a bystander parameter p". One could also say we have (uniform convergence)p of f(tn,p) → f(t0,p) as tn → t0. The sequence involved is a sequence of points {tn} or { f(tn,p) }; only function is f. Now in the next item we combine both the above upgrades: 4. For any ε > 0 and for all p in E and for all t0 in T, we can find δ > 0 such that | f(t,p) - f(t0,p)| when |t-t0| < δ. I would describe this as "uniform continuity over T which is also uniform over E". This latter is a strange new use of the word "uniform". The statement involves being uniform in two different spaces at the same time, E and T. The "continuity" aspect only involves space T. One could also say we have (uniform convergence)p of f(tn,p) → f(t0,p) as tn → t0. The sequence involved is a sequence of points {tn} or { f(tn,p) }; only function is f. Notice that we still have the Em norm, not the max norm. We still have only one function. So even this double upgrade thing has nothing to do with (uniform convergence)f of a set of functions fn → f in the Hilbert Space of the max norm. Now can I connect what Bucks say on page 204 with one of the "upgrades" shown above? I first thought they were doing the complicated upgrade #3, but now I think it is upgrade #2 where we have ordinary continuity at the point t0 and where we are uniform in the bystander space p in E. From above: 2. For any ε > 0 and for all p in E and for a specific t0 in T, we can find δ(t0) > 0 such that | f(t,p) - f(t0,p)| < ε when |t-t0| < δ. I would describe this as "continuity at a point t0 in T which is uniform over E". I would also say that f(tn,p) (converges uniformly)p over E to f(t0,p) as tn → t0. Notice that there is only one function here, it is called f. Restate this as For any ε > 0 and for a for specific t0 (in some space T in En) one can find δ > 0 such that | f(t,p) - f(t0,p) |Em < ε when |t-t0|En < δ, where δ works for all p in E. In the Buck definition of "uniform convergence with a parameter", instead of saying one can find a δ where |t-t0| < δ (a neighborhood ball), they say one can find a neighborhood N around t0. For me, the distinction is very minor, though N is allowed to be more general in shape than a ball. I can restate the above ε,δ definition of "uniform convergence with a parameter" using their N : For any ε > 0 and for a for specific t0 (in some space T in En) one can find a neighborhood N around t0 such that | f(t,p) - f(t0,p) |Em < ε when t lies in N, where N works for all p in E. As noted above, it took me a long time to get happy with the above paragraph and to make the distinction between its implied (converges uniformly)p over E which differs from (converges uniformly)f which is a different kind of uniform convergence involving a function space rather than a point space. The first example of this concept is this: Example: Point p is now called x, it is the bystander variable and lies in some interval E. Consider f(t,x) = sin(xt)/ [t(1+x2)] and t0 = 0 is the specific point of interest in T f(t0,x) = f(0,x) = limt→0 {sin(xt)/ [t(1+x2)]} = x/(1+x2) ≡ F(x) Our goal of the example is to prove that limt→0 f(t,x) → F(x) in fact (converges uniformly)p for all real x. You always know it is not (converges uniformly)f if only a single function is involved. I have lots of raw notes on this example. But the main interest the Bucks have in this (converges uniformly)p of f(t,p) as t→t0 is in regard to integration and differentiation. We do integration first, then differentiation after that. In the first case, f(t,p) has the form g(t,p) =!Syntax Error, Idu f(p,u) where t is taken to be t0= ∞. Here the bystander parameter p is just some variable of the integrand function that is not integrated over in the integral. The big question here is whether you can take a limit limp→p0 through the integral or not. It is then really a case of order interchange of two limits, one being limp→p0 and the other being limt→∞ for the upper integration endpoint. A directly analogous situation would be when the function f(t,p) has the form g(t,p) = Σk=1t an(p) where t is taken to be t0= ∞ on the integer lattice In the first case t = r, whereas in the second case t = n. Several theorems are developed: Theorem 26 [206] (Comparison Test). If: f(p,u) is continuous in p over E, and in u on [c,∞) |f(p,u)| ≤ some g(u) for all p in E and for all u in [c,∞) !Syntax Error, Idu g(u) converges Then: limr→∞!Syntax Error, Idu f(p,u) (converges uniformly)p for all p in E to some F(p) ≡ !Syntax Error, Idu f(p,u) Theorem 27 [206] (Follow On). Assume the premises shown above so F(p) ≡ !Syntax Error, Idu f(p,u) converges uniformly on E. Then F(p) is continuous at p. Corollary 27 [206]. (order interchange of limits) Assume F(p) ≡ !Syntax Error, Idu f(p,u) converges uniformly on E. Theorem 27 then shows that limp→p0 F(p) = F(p0), which is the meaning of F(p) is continuous at p0. This says that limp→p0 [!Syntax Error, Idu f(p,u)] = !Syntax Error, Idu f(p0,u) =!Syntax Error, Idu { limp→p0 [ f(p,u)] } F(p) F(p0) and that in turn says you are allowed to move the limit through the integration in this case! The two limits interchanged are p→p0 and upper endpoint r → ∞. The text of the Corollary adds that p0 must be a cluster point in E. I think this is added to make sure that p is a continuous variable in some sense, and not perhaps some isolated lattice point in E. Example 1: Uses p = x in E1 and f(x,u) = x2/[1 + x2u2] and c = 1. Using the comparison test above, it is easily shown that !Syntax Error, Idu f(x,u) converges uniformly for all x, so [-∞,∞] ≡ E. Now take the point p0 to be x = ∞ which lies in E. Then the Corollary above says limx→∞ [!Syntax Error, Idu f(x,u)] = !Syntax Error, Idu f(∞,u) = !Syntax Error, Idu/u2 = 1. This same result is shown to obtain by just "doing the integral" naively and using l'Hospital, but here we are carefully justifying the process. Theorem 28 [207] ( order interchange of integrations ) If : f is continuous in u for [c,∞) and in x for [a,b] !Syntax Error, Idu f(x,u) converges uniformly for x in [a,b] then !Syntax Error, Idu [ !Syntax Error, I dx f(x,u) ] = !Syntax Error, Idx [ !Syntax Error, I du f(x,u) ] This is a generalization of Theorem 8 of Section 3.2 to the fancier case that one upper endpoint is infinite, which involves a limit limr→∞. Here is that Theorem 8 where the ∞ is the finite value d : ∫R f = [!Syntax Error, I!Syntax Error, Idx dy ] f(x,y) = !Syntax Error, Idx [ !Syntax Error, Idy f(x,y) ] = could do in other order It is probably true that !Syntax Error, Idy f(x,y) converges uniformly on x in [a,b] if f is continuous, so the second bullet item is automatically met for finite endpoint d. In Chapter 3 it was shown that the rectangle integral ∫R f(x,y) always exists and is well defined as long as f is continuous on the rectangle. This was Theorem 1 p 100, later generalized by Theorem 9 p 114 showing that you can do either single-variable integration first, and then do the other second. Since you can do the single integration first, it must converge for all values of the other variable, and that means uniform convergence on x in [a,b] which is the second bullet premise above. In Theorem 28 we only have the r→∞ limit, apart from the implied Riemann integration limits, so this theorem is not an "order interchange of limits" the way Corollary 27 is above. Bucks give a simple counterexample showing how lack of uniform convergence can result in a double integral giving different results when done in the two orders. Well, let's try one more time. Imagine that integration on a,b is an integral operator I, !Syntax Error, I dx f(x,u) = [I f](u) !Syntax Error, I dx h(x) = I h Then Thm 28 says (reversing the sides) I [ !Syntax Error, I du f(x,u) ] = !Syntax Error, Idu [I f](u) and this says we can move the integral operator I through the du integration, so that we have order interchange between the integral operator I and the integration on du which involves the r→∞ limit. The following theorem is the corresponding theorem where the integral operator I is replaced by the differential operator ∂x : Theorem 29 : [208] If: f is continuous in u for [c,∞) and in x for [a,b] !Syntax Error, Idu f(x,u) converges uniformly for x in [a,b] ∂xf is continuous in u for [c,∞) and in x for [a,b] !Syntax Error, Idu ∂xf(x,u) converges uniformly for x in [a,b] Then: ∂x [!Syntax Error, Idu f(x,u) ] = !Syntax Error, Idu [∂xf(x,u)] for any x in [a,b] The first two premises are the same as for Theorem 28, but we have to add the two corresponding premises regarding ∂xf. This theorem is stricter than Theorem 27 because differentiation is a less smooth operation that integration and requires these extra conditions. Example: Start with 1/x = !Syntax Error, Ie-xu du and apply ∂x to both sides and apply Thm 29 to get the new equation 1/x2 = !Syntax Error, Iu e-xu du . Can do this again and again. Lots of other examples treated in Buck and in my raw notes. 4.5 The Gamma Function [ 213 ] // theorems 30→33 An idea presented here is the definition of a function in terms of an integral of another function. The first example is log(x) as integral of 1/x. Second example is arctan(x) as an integral of 1/(1+x2), and how given that integral, you can derive various properties of the arctan function. But the big example is the Gamma function integral shown in (4-20). Cleverly, Bucks already computed this integral for integer x and showed it as the n! result as top page 214. This example is going to continue now for 4 pages. Theorem 30 [214] Γ(x+1) = xΓ(x) for any x> 0. This familiar fact is proved from the integral definition (4-20) using parts integration. By changing integration variables in (4-20) you can write down lots of other forms of the integral definition of Γ(x) and four such alternatives are given on mid p 214. Theorem 31 [214] Half integer formulas. We know Γ(1/2) just doing the integral, so Γ(n/2) can be done using Theorem 30's little formula. They then talk about "extending" the definition of Γ(x) to negative half integer values in the obvious manner. They don't mention the fact that really the parts integration provides an analytic continuation into a new convergence region, perhaps they have that in their appendix and I have seen it perhaps in Ahlfors or elsewhere. It then exposes the poles on the left. In this section the term pole does not appear since they have not talked at all about complex variables. Theorem 32. [216] Derivation of Stirling's Formula for Γ(x). The proof is a long one using a certain approximation method of breaking an integral into 3 terms and showing that only the middle one matters for large x. Two Lemmas are required along the way. I did some of it, but really it is not critical, nothing magic is done here. Theorem 33. [218] Beta B(p,q) is defined as a certain integral is shown equal to Γ(p)Γ(q)/Γ(p+q). The proof is pretty easy. It starts by writing Γ(p)Γ(q) as a double integral as shown in A. This is converted to polar coordinates, and further shuffling gives the desired result as shown p 219 B. I am quite familiar with "special functions" being defined by their integral representations as well as by their series, and of using things like connection formulas between the integrals to extend the domain of the function. For example F(a,b,c;z) hypergeometric, an old friend (now out of practice).