CRAIM GRADUATE SEMINAR ALGEBRA PROJECT DATA BASE AND HOMEWORK WORK DOCUMENT III ALL YOU EVER WANTED TO KNOW ABOUT RADICALS AND EXPONENTS BUT WERE AFRAID TO ASK ´ ´ ´ OMAR HERNANDEZ RODR´IGUEZ, JORGE M. LOPEZ FERNANDEZ, ´ AND AILEEN VELAZQUEZ ESTRELLA
Contents Part 1. On the definition of powers and logarithms 1. Roots and rational exponents 2. The Intermediate Value Theorem 2.2. The role of the IVT in the definition of powers 3. Rational exponents: the rules step by step 3.1. ax for x ∈ N 3.2. ax for x ∈ Z 3.3. ax for x ∈ Q 3.4. ax for x ∈ R Part 4. 5. 6. 7. 8. 9. 10. 11.
2. Addendum: analytics of exponential and logarithmic functions Continuity of the exponential function An axiomatic definition of the exponential function Euler’s approach to the exponential logarithmic functions The logarithmic function The parameter c and the base of the natural logarithm e: some conclusions Exercises for all students Exercises for students of mathematics A Exercises for students of mathematics B
2 2 2 3 3 3 4 5 7 8 8 9 10 12 13 13 14 15
Date: November 24, 2015. 2010 Mathematics Subject Classification. Primary. Key words and phrases. Realistic Mathematics Education, algebra education, operational definition, curricular goals.
Hern´andez, L´opez & Vel´azquez, p. 2, References
16
Abstract. This work presents an exposition of the rules governing the existence of radicals and rational exponents and their algebraic rules. A step by step presentation is given, based on the intermediate value property of continuous functions. The evolution of the knowledge base is carried on in steps from the natural numbers to the integers, to the rational numbers and finally to the real numbers. The second part of the work presents a development of exponents and logarithms close to the presentation of Euler. Some exercises of general interest as well as material of a foundational mathematical nature are also presented.
Part 1. On the definition of powers and logarithms 1. Roots and rational exponents The discussion about radicals and rational exponents, if we take it from the very first principles, has to be centered around a topological result. This result is the the Intermediate Value Theorem (IVT henceforth). We first discuss the role it plays in the definition of powers. 2. The Intermediate Value Theorem Theorem 1. (IVT) Any continuous function f : [a, b] → R (with −∞ < a < b < ∞) with f (a)f (b) < 0 (that is, having opposite signs at the end points of the interval) must cross the x axis, that is, for some θ ∈ (a, b), f (θ) = 0. Remarks 2.1. i. Note that the condition f (a).f (b) < 0 says in an elegant fashion that y = f (x) has values with different parity at the endpoints of interval [a, b]. ii. A metaphorical statement of this result states that a car on a trip to be completed within the bounds of two contiguous municipalities of a city, starting in one municipality and ending in the other, must cross the dividing line between the municipalities. iii. Note that the conclusion states that the point θ lies in the interior (a, b) of [a, b], and this is clear since f (θ) = 0 and neither f (a) or f (b) can be zero (by hypothesis). iv. The IVT is a theorem of a mathematical discipline called topology and in that context the result has many interesting and useful equivalent statements and versions thereof. Specialized to the real numbers the statement is, basically, as we have stated it. v. The IVT was first proved by Bernard Bolzano in 1817. Today, as it happens for many important mathematical results, there are various proofs of this result. vi. The IVT is intimately related to completeness of the real number line. In fact, an Archimedean ordered field is complete if and only for all real numbers a, b with a < b and for all continuous functions f : [a, b] → R such that f (a) · f (b) < 0, there exists a θ ∈ (a, b) such that f (θ) = 0. (see Exercise 13.)
Seminario Graduado CRAIM p. 3 vii. Suppose I is an non empty interval of the real line. A function f : I → R has the intermediate value property (IVP) provided for each u, v ∈ I and every β ∈ R such that f (u) < β < f (v) or f (v) < β < f (u), then there is some θ ∈ (u, v) such that f (θ) = β. A function f : I → R has the IVP if and only if f [I] ⊆ R is an interval; see Exercise 11. Bolzano gave examples of non continuous functions on intervals I with the IVP. This says that the IVP does not characterize continuous functions on intervals. viii. An important theorem of analysis states that if f : [a, b] → R is a derivative1, then f has the IVP. It should be stressed that f itself need not be continuous. 2.2. The role of the IVT in the definition of powers. We need the IVT to prove the following: Theorem 2. Let a > 0 and let n > 1 be an integer. Then there is a unique real number √ θ > 0 such that θn = a. We say that θ is the principal n-th root of a and write θ = n a or 1 θ = an Proof. Define the function f (x) = xn − a for all x > 0. We will show that f has a a positive and a negative value in the interval of definition (0, ∞). Choose x1 ∈ R such that 0 < x1 < min{a, 1} Then x1 < 1 and x1 < a. Then f (x1 ) = xn1 − x1 = x1 (x1n−1 − 1) < 0 (since x1 < 1, we have x1n−1 < 1). On the other hand we can choose x2 ∈ R such that x2 > max{a, 1}. Then x2 > 1 and x2 > a and it is clear that x2 > 0 and f (x2 ) = xn2 − a > xn2 − x2 > 0 (since x2 > a and x2 > 1). So we have found a = x1 and b = x2 so that f (a)f (b) < 0. But by the standard theorems on limits f is continuous: If (xk )∞ k=1 is a sequence in (0, ∞) converging n to x ∈ (0, ∞), then limk→∞ f (xk ) = limk→∞ (xk − a) = xn − a = f (x). By the IVT there is a θ ∈ (a, b) such that f (θ) = θn − a = 0, that is θn = a. 3. Rational exponents: the rules step by step 3.1. ax for x ∈ N. Step 1. Definition of natural exponents i If a ∈ R and a > 0 we define a0 = 1, and a
n+1
n
= a · a, for all n ∈ N.
(3.1) (3.2)
ii The previous definition is an example of an inductive definition. Essentially, the process defines the first exponent a0 as 1 by (3.1) and if an is defined for some n ∈ N, then (3.2) defines the next exponent that follows, an+1 . The important thing to remember is that even though we have chosen to define an by steps, by going from one exponent to the next, beginning with the first one, this produces a sequence an defined for all n ∈ N. There is a subtlety here that is related to the old dichotomy in Aristotelian philosophy related to the potential infinity as opposed to the actual infinity. We want to define the domain of an exponential function n 7→ an , by first defining it for {0} and by adding one more element to 1That
is, for some F : [a, b] → R, F 0 (x) = f (x) for all x ∈ [a, b] (at the endpoints of the interval derivatives are interpreted as one sided); F is said a primitive function for f .
Hern´andez, L´opez & Vel´azquez, p. 4, the domain each time we have the function defined in a given domain. In our case, we want to add n + 1 to the set {0, 1, 2, · · · n} once we know the latter is part of the domain of the function. The whole point of this discussion is that the step by step definition ends up capturing the whole of the set of natural numbers N as the domain of the function. This is the gist of the discussion in Stoll [9, 1961, Theorem 1.1, p. 61] or in Halmos [6, 1960,§12 p. 48] to which we refer the interested reader. Step 2. Rules for handling natural exponents: These are the basic rules for handling exponents. In this discussion a and b are positive real numbers and n, m ∈ N. i. an am = an+m . ii. (an)m = anm . n n iii. ab = abn . iv. 1 n a n−m = a m a 1 am−n
if n = m if m < n if n < m.
Proof. This is a proof by straight induction; we illustrate with the proof of 2i. Let n ∈ N and consider the set S = {m ∈ N | an · am = an+m }. We must proof that 0 ∈ S and that for all k ∈ N, if k ∈ S then k + 1 ∈ S. Clearly, since an+0 = an and an a0 = an · 1 = an , it follows that 0 ∈ S. Suppose that k ∈ S. Then an ak = an+k and it follows that an ak+1 = an (ak a) = (an ak )a = a(n+k) a = a(n+k)+1 = an+(k+1) ; here we used the inductive definition of exponents, the hypothesis that k ∈ S and the associative law of addition. All this goes to show that k + 1 ∈ S whenever k ∈ S. By the principle of mathematical induction, S = N. We leave the other proofs as informal exercises for the interested reader.
3.2. ax for x ∈ Z. i Note that for n ∈ N, a−n = a1n . All the laws of exponents contained in Step 2i, Step 2ii and Step 2iii hold for integer exponents. This is all straightforward to prove. For instance lets consider 2iii: an am = an+m . We consider cases using the fact that Z = N ∪ −N; here Z stands for the set of integers and we employ the notation −N = {−k | k ∈ N}. If n and m are natural numbers, we know the rule is valid. Suppose otherwise, for instance, that n ∈ N and m ∈ / N. Then m < 0 and m = −|m| where |m| ∈ N. Hence,
Seminario Graduado CRAIM p. 5
an am = an
1 a|m|
=
an = an−|m| |m| a an−|m| if n ≥ |m| = 1 if n < |m| |m|−n a n+m if n ≥ |m| = a 1 if n < |m| a−n−m = an+m
1 = am for all m ∈ Z (not just for m ∈ N. The other We used the fact that a−m rules go as expected and they can be proved by similar arguments or, more laboriously by considering cases (we really considered cases, but in a very efficient way). We state the result: Here a and b are as before but n, m ∈ Z.
i. an am = an+m . ii. (an )m = anm . a n an = n. b b n a iv. m = an−m . a
iii.
Compare these relations with Step 2i–Step 2iv.
3.3. ax for x ∈ Q.
√ 1 i Remember that for a real number a > 0 and n ∈ N with n > 0, a n or n a stand for the unique positive real number θ that satisfies θn = a. Here we need a consistency theorem anticipating the definition of rational exponents: Theorem 3. For all n, m ∈ N with n 6= 0 and for all a ∈ R with a > 0 we have: 1
1
(a n )m = (am ) n .
(3.3)
Proof. We use the following property for exponents in N with n 6= 0: Principle 1. If x > y > 0 are real numbers then xn > y n for all n ∈ N with n > 0. This is readily shown by a proof by mathematical induction. We leave the straightforward details to the reader.
Hern´andez, L´opez & Vel´azquez, p. 6, Observe that 1 m n 1 mn 1 nm 1 n m = an = an = an = am , and (3.4) an m n1 n a = am . (3.5) 1 m n m n1 n 1 1 m By (3.4) and (3.2) we have that a n = a . By Principle 1 a n = am n as was to be proved. (1) Using the previous work now we can define rational exponents: n
Definition 1. If a ∈ R, a > 0 and n, m ∈ N with m 6= 0, we define a m as the value 1 n of a m . Now we can prove n
Corollary 1. With a, n, m as in the Theorem, a m defines well a rational exponent. n In other words, given a as indicated the value of a n is independent of the way we n write m as the quotient of two natural numbers. p 1 1 1 1 √ 1 1 p m m ) p = (a p ) m = a mp = a pm ; in radical notation First we show that (a a = p √ √ √ m p mp pm a= ap= a.p √ √ √ √ p m Notice that a, m p a, mp a, and pm a all stand for the unique positive real number that satisfies the equation xmp = a. Proof. Follows directly from the Theorem by using the properties of exponents in N. n For the last statement let n, m, r, s ∈ N with m, s 6= 0 and m = rs . Let xr be the n reduced expression of m , so that for some k ∈ N we have n = kr and m = ks. The reader is invited to justify the following equations: n
1
a m = [a m ]n 1
= [a ks )]kr 1
1
= [({a s } k )k ]r 1
= [a s ]r r
= as . i. Extension to negative rational exponents This extension is subsumed in the following Theorem: Theorem 4. If a and b are positive real numbers and r, s ∈ Q (here Q stands for the number system of all rational numbers). i. ar as = ar+s . ii. (ar )s = ars . iii.
a r ar = r. b b
Seminario Graduado CRAIM p. 7 iv.
ar = ar−s . as
Proof. As in the proof for the laws of exponents for integers, it suffices to prove the laws for positive rational numbers. As an illustration we prove rule iii for this case, asuming n and s = pq , the corresponding rule for integers has been proved. So suppose that r = m where n, m, p, q ∈ N and m, q 6= 0. We invite the reader to justify each line of the following proof, keeping in mind the definition of roots and the properties of natural exponents. n p (ar )s = a m q 1 n pq = am 1 p n = am q 1 1 np = am q 1
= anp mq np
= a mq = ars . When one or more of r or s is negative, we could use an argument similar to the extension of exponents from N to Z and it will work accordingly. 3.4. ax for x ∈ R. (1) (Last step) Extension of exponents from Q to R Here there are two approaches and we just sketch them for the reader. One approach uses the fact that Q is dense in R so that for each x ∈ R there is a sequence of rational rn numbers (rn )∞ n=1 converging to x. It can be shown that the sequence (a ) is a Cauchy x sequence and it converges to a number we denote by a . Furthermore the convergence does not depend on the sequence (rn )∞ n=1 , the sequence chosen that converges to x,k that is to say, if (sn )∞ also converges to x, then limn→∞ arn = limn→∞ asn . The n=1 details of the presentation are not devoid of effort and technicalities and depends on knowledge of calculus. As an example we give the proof that ax ay = ax+y supposing ∞ the rule is valid for exponents x, y in Q. Choose sequences (rn )∞ n=1 and (sn )n=1 such that rn → x (that is limn→∞ rn = x) and (sn )∞ n=1 such that sn → y. Then, rn + sn → x + y and it follows that ax+y = lim arn +sn n→∞
= lim (arn · asn ) n→∞
= lim arn · lim asn n→∞ x y
n→∞
=a a . where we have used the rules for exponents for rational exponents, and the continuity of addition and multiplication on R. The ease of the proof is typical once we have worked out the initial details of the existence of the limits and the wellness of the definitions that ensue. Another approach that also uses the density of the rationals
Hern´andez, L´opez & Vel´azquez, p. 8, in the real line but no limits, is described here. Because of the need to handle inequalities it is best to start with positive real exponents. So if x is one such, we define ax = sup{ar | r ∈ Q, and r < x}.
(3.6)
Here, as suspected, sup stands for the least upper bound. If we set K = {ar | r ∈ Q, and r < x} then, if r1 and r2 are rational numbers with r1 < x < r2 , then it is clear that ar1 ∈ K and K is bounded above by ar2 (here a > 1 has been assumed so that x 7→ ax is an increasing function). The proof of the same exponent law presented before can be easily proved using (3.6). Let x, y ∈ R. We will use the property of bounded non empty sets of positive real numbers A, B as follows: sup(A · B) = sup A · sup B; here A · B = {a · b | a ∈ A and b ∈ B}. In fact if we denote the set in (3.6) by Kx , and by Ky and by Kxy for the corresponding sets for y and xy respectively, then, ax ay = sup Kx · sup Ky , that is sup(Kx · Ky ) = sup{ar · as | r, s ∈ Q and 0 < r < x, 0 < s < y} = sup{at | t ∈ Q and 0 < t < x + y} = ax+y The equivalence of the second and third statements uses the fact that if t ∈ Q and 0 < t < x + y, then we can write t = r + s where r, s are positive rational numbers with r < x and s < y. Part 2. Addendum: analytics of exponential and logarithmic functions In this section we consider matters related to the differentiation and integration of the exponential function and its inverse, the logarithm. We will present our own version of the theory which is, in our view, a modern rendition of Euler’s own treatment Euler [3, 1774]. This section has been written, mainly, for the mathematics students in the CRAIM Seminar. 4. Continuity of the exponential function If a ∈ R, a > 1 and x ∈ R we define ax by (3.6). We will need the following Proposition 1. If A ⊆ R is a non empty set of non zero numbers, bounded above and below, then 1 sup A = inf ; A here 1 1 = a∈A . A a Proof. We remind the reader that if A is a bounded non empty set then there are sequences ∞ (pn )∞ n=1 and (qn )n=1 in A such that pn → inf A and qn → sup A. In our case, there is a ∞ sequence (rn )n=1 such that rn → sup A. Then r1n → sup1 A . But r1n ≥ inf A1 for each index n. 1 1 It follows that sup1 A ≥ inf A1 . Conversely, if (rn )∞ n=1 is a sequence in A such that rn → inf A , then, since r1n ≥ sup1 A , it follows that inf A1 ≥ sup1 A . Hence, inf A1 = sup1 A .
Seminario Graduado CRAIM p. 9 Corollary 2. Let a > 1. Then, for all x ∈ R, sup{ar | r ∈ Q and r < x} = inf{ar | r ∈ Q and r < x}. Proof. Note that ax = sup{ar | r ∈ Q and r < x} 1 = 1 inf | r ∈ Q and r < x ar 1 = inf {a−r | r ∈ Q and − x < −r} 1 = ; r inf {a | r ∈ Q and − x < r}
(4.1) (4.2)
(4.3) (4.4)
here the Theorem was used in obtaining (4.2). It follows that ax · inf{ar | r ∈ Q and − x < r} = 1. From the last relation we conclude that for all x ∈ R, inf{ar | r ∈ Q and − x < r} = a−x . Exchanging −x for x, we get the desired result.
Theorem 5. The function Ea (x) = ax is continuous from R onto (0, ∞). Proof. (Recall we have been assuming all along that a > 1; there are analogous results for bases a < 1, but we leave that as an informal exercise for the interested reader. In fact, since 1 for all x ∈ R and a < 1 if and only if a1 > 1, the results are readily extended to the ax = a−x case a < 1 by the techniques that have already been discussed.) Let x ∈ R. We show Ea is continuous at x. Let > 0. Then, by Corollary 2, there are r, s ∈ Q such that r < x < s and ax − < ar < ax < as < ax + . Let δ be a number such that 0 < δ < min{x − r, s − x}. Then it is clear that if u ∈ R and |u − x| < δ, then u ∈ (x − δ, x + δ) ⊆ (r, s). Since x 7→ ax is strictly increasing, ar < au < as and ax − < ar < au < as < ax + , that is to say, |au − ax | < . This completes the proof. 5. An axiomatic definition of the exponential function Definition 2. . An exponential function is a continuous non constant function E with domain R such that for every pair of real numbers x and y we have E(x + y) = E(x)E(y). Remarks 5.1. Here are some properties about an exponential function: i. If E is an exponential function as in Definition 2, then E(x) 6= 0 for all x ∈ R. In fact if E(u) = 0 for some u ∈ R, then E(x) = E[(x − u) + u] = E(x − u)E(u) = 0 and E would be a constant function.
Hern´andez, L´opez & Vel´azquez, p. 10, ii. Note that if E is an exponential function then E(x) > 0 for all x ∈ R. In fact, E(x) = E( 21 x + 12 x) = E( 12 x)2 > 0. iii. If E is an exponential function then, a straightforward induction argument shows E(nx) = E(x)n for every n ∈ N and all x ∈ R. iv. The function x 7→ ax is an exponential function for a > 1, in view of Theorem 5. Similar arguments show that x 7→ ax is also continuous for a < 1, thus it is also an exponential function. v. If E is an exponential function then a = E(1) 6= 1. In fact, if a = E(1) = 1, then, by induction E(n) = an = 1 for all n ∈ N. By standard arguments seen already, this implies n that E(n) = an = 1 for all n ∈ Z. If r = m with m > 0, then E(r)m= E(mr) = E(n) = an n and E(r) = a m = 1. By continuity E(x) = 1 for all x ∈ R contradicting that E is not a constant function. vi. Two exponential fuctions E1 and E2 such that E1 (1) = E2 (1) are equal, that is to say, E1 (x) = E2 (x) for all x ∈ R. In fact, by hypothesis, if a = E1 (1) = E2 (1), then E1 (r) = ar = E2 (r) for all r ∈ Q. By continuity, E1 (x) = ar = E2 (x) for all x ∈ R. vii. If E is an exponential function, then E[R] = (0, ∞). We already know that E[R] ⊆ (0, ∞) In view of the IVT, to finish the proof, it suffices to observe that limn→∞ E(n) = limn→∞ E(1)n = ∞ and limn→∞ E(1)−n = 0 (we are assuming that E(1) > 1; the case E(1) < 1 is analogous). Proposition 2. If E is an exponential function, then E is differentiable at each x ∈ R is and only if it is differentiable at x = 0. Proof. Notice that for any x ∈ R E(x + h) − E(x) E(h) − 1 = E(x) and h h E(h) − 1 E(x + h) − E(x) lim = E(x) lim , h→0 h→0 h h that is, whenever one of the limits exist so does the other. Of course, the limit on the right hand side of the equality is E 0 (0) if it exists. So the result is clear. Remark 5.2. The last limit relation in the proof says, of course, that E 0 (x) = E 0 (0)E(x) when the derivatives exist. 6. Euler’s approach to the exponential logarithmic functions In what follows we present an approach to the exponential and logarithm functions, very close to Euler’s own approach (without infinitesimals). Euler’s original presentation can be found in Bruce [3, 2013, P. 184] and commentated in Dunham [2, 1999, p. 25]. From reading Dunham’s nice exegesis of Euler’s arguments, one cannot help but belive that it can all be made formally correct when framed in Robinson’s version of non standard calculus. Euler’s argument is extremely ingenious and all students of mathematics should at some
Seminario Graduado CRAIM p. 11 point look at such a reflection of his genius. Since we are not interested here to make any detailed incursion into non standard analysis, we will present our own reading of Euler’s presentation. As shown in Bruce [3, 2013, P. 184] Euler brings to the limelight right at the beginning the famous infinite series expansion for exponential functions. In this area there are various presentations, among them, a very clean and elegant R x 1 presentation starting with the definition of the natural logarithm function as x 7→ 1 u du, (x ∈ (0, ∞)) and, from there, powers are then defined. Such an approach appears, for example, in Apostol [1]. By the way, (ibid) presents integration before differentiation. We think that the presentation of logarithms before exponents is counter productive for student understanding, although very ingenious and creative. In our view, this is an example of what today is referred to as a “didactic inversion”. Freudenthal [5, 1973, p. 327] has some additional criticisms for this approach to the logarithm and the relevance of the integral as part of the problem of defining the logarithm. We start by assuming the following: invite the reader to follow our presentation which is closely related to Euler’s own presentation. Principle 2. Every exponential function is differentiable at x = 0, that is limx→0 exists and we call the value of the limit c.
E(x)−1 x
By the Principle 2 and Proposition 2 it follows that E is differentiable everywhere. We are now ready to prove our main result. Theorem 6. Let E be an exponential function as in Definition 2 with E 0 (1) = c, Then E(x) =
∞ X (cx)n n=0
n!
for each x ∈ R.
(6.1)
Remark 6.1. Note that, as is true of all convergent power series, convergence is uniform in every bounded interval of the real line. Note also that any exponential function is infinitely differentiable, and since exponential functions are unique, continuity in Definition 2 guarantees, eventually, infinite differentiability. Proof. We will suppose that E is an exponential function and that it has a derivative at x = 0. We have seen that if E has a derivative at all x ∈ R, then E 0 (x) = cE(x). Then, applying this to x = 0, an induction argument shows E (n) (0) = cn E(0) = cn for all n ∈ N. Now we examine the Taylor polynomials of degree n − 1 around x = 0 (to generate a Maclaurin series) and the reminder Rn (x) for each n ∈ N. The Taylor polynomial of degree n − 1 about x = 0 is given by (Ross, et all [7, 2013, p. 250]) n−1 X E (k) (0) k=0
k!
k
x =
n−1 X (cx)k k=0
k!
,
(6.2)
and the remainder is: Rn (x) = E (n) (θ) = θn
xn , n!
xn n! (6.3)
Hern´andez, L´opez & Vel´azquez, p. 12, where θ lies between x and 0. Hence, |Rn | = |Rn+1 |/|Rn | we get,
|θx|n . n!
If we consider the consecutive ratios
|θx|n+1 |θx| (n + 1)! = → 0 when n → ∞. n |θx| n+1 n! Since the limit of the ratios is less than 1 (Ross, et all [7, 2013, 9.11, p. 55]), it follows that Rn (x) → 0 for each x ∈ R; see see Exercise 18 in Exercises 11. To finish the proof we must show that power series expansion is actually an exponential (after all, we assumed differentiability). For this we use the multiplication of power series by means of the Cauchy product using Merten’s theorem and Cauchy products; Stromberg [10, 2015, p. 76]. If x, y ∈ R, then X X ∞ ∞ yn xn · E(x) · E(y) = n! n! k=0 k=0 =
p ∞ X X xp−k y k (p − k)! k! p=0 k=0
∞ X 1 = (x + y)p p! p=0
= E(x + y).
7. The logarithmic function The logartithmic function is the inverse of the exponential function. That is, the logarithmic function is the function L : (0, ∞) → R such that L ◦ E = idR and E ◦ L = id(0,∞) ; here idA is the identity function on the set A. The relation between the derivatives of inverse functions of differentiable functions (Ross et all [7, 2013, p. 237]) gives, L0 (x) =
1 E 0 (L(x))
1 cE(L(x) 1 = . cx =
(7.1) (7.2)
The following are properties of the function L that can be deduced readily from those of the function E. i limx→∞ L(x) = ∞ ii limx→∞ L(x) = 0 iii L is strictly increasing.
Seminario Graduado CRAIM p. 13 iv L is a isomorphism of the multiplicative group (0, ∞) onto the additive group R. This means among other things that that L(uv) = L(u) + L(v) for all u, v ∈ (0, ∞).
8. The parameter c and the base of the natural logarithm e: some conclusions Now we choose c = 1 and the expression for exponential function becomes, E(x) =
∞ X xn n=0
n!
.
(8.1)
Since L0 (x) = x1 for all x ∈ (0, ∞) by our choice of c, it follows from the fundamental theorem of calculus that, Z x 1 L(x) = L(x) − L(1) = dt. (8.2) 1 t Rx 1 Rx 1 Since L(1) = 0, this says L(x) = dt. But, since dt is a divergent integral (see t 1 1 t Rx 1 Exercise 19), that is, limx→∞ 1 t dt = ∞, and the integral is a continuous function of x, we can invoque the IVT to conclude that there is some value e > 1 such that Z e 1 dt = 1. (8.3) 1 t In particular, L(e) = 1 and E(1) = e. By arguments seen before we conclude that E(x) = ex for all x ∈ R. We can conclude by (8.1) and (8.3) that x
e =
∞ X xn n=0
n!
.
(8.4)
When we take the basis of logarithms as e, we write ln for L. We conclude this exposition with two little jewels discovered by Euler: Theorem 7. 1 lim 1 + x) x = e. and x→0+ n 1 lim 1 + = e. n→∞ n Proof. See Excercise 19.
(8.5) (8.6)
Exercises 1. 9. Exercises for all students (1) Find an interval where the equation 5x4 − 7x2 − 1 = 0 has a solution.
Hern´andez, L´opez & Vel´azquez, p. 14, (2) What are the least number of solutions that the equation f (x) = 0 can have in the interval [25, 40] if y = f (x) is a continuous function, f (25) = 51, f (40) = 13 and f (a) = 56 for some a ∈ (25, 40)? (3) How many solutions does the following equation have in the interval [−2, 2]: 24x + 8 = 0. 5x2 − 15 (4) Show that the equation cos x = x has a root between 0 and 1. (5) Show that p(x) = 2x3 − 5x2 − 10x + 5 has a root somewhere between −1 and 2. (6) Show that there is some u with 0 < u < 2 such that u2 + cos(πu) = 4. (7) Show that ln(x) = e?x has a root between 1 and 2. (8) Show that the equation arctan x = 1 − x has at least one real solution. (9) Can you place two areas on the plane and draw a single line through both of them so that the line divides both areas in half? What does this have to do with the IVT? (10) Prove that the function f (x) = x2 − 4x + 2 intersects the x-axis on the interval [0, 2]. Can the same be said on the same question for the function: f (x) = 2x−3 ? x−1
10. Exercises for students of mathematics A Part A: The intermediate value theorem (11) An interval in R is a non empty set I such that for every x, y, z ∈ R, if x, y ∈ I and x < z < y, then z ∈ I. Show that if I ⊆ R is an interval, then I is of the form (−∞, ∞), (a, ∞), [a, ∞), (−∞, b), (−∞, b] for some real numbers a, b or [a, b), (a, b], (a, b], or (a, b) for real numbers a, b with a < b. (12) Prove that if I ⊆ R is an interval, then f : I → R has the IVP if and only if f [I] is an interval of R. (13) Show that an Archimedean ordered field (F, +, ·, ≤) is complete if and only if the IVT is a valid statement in F; see Exercise 11. [A possible approach to this problem: First argue that in view of what we have proved, it is enough to prove the following: There are a, b ∈ F such that a < b, and a continuous function f : [a, b] → F such that f (a) · f (b) < 0 and f (x) 6= 0 for all x ∈ [a, b]. Furthermore, it is enough to show that there is a continuous f : F → F such that for some distinct a, b ∈ F we have f (a) · f (b) < 0 and f (x) 6= 0 for all x in the interval determined by a and b (you can restrict this function to the interval determined by a and b). So, suppose F is not complete. Then there is a nonempty set S with an upper bound and no least upper
Seminario Graduado CRAIM p. 15 bound. Let B the set of bounds of S and let A = F \ B. Define f = the characteristic or indicator function of A). Prove that:
1 2
− χA (χA is
i. Both A and B are non empty. ii. S ⊆ A and for all a ∈ A, f (a) = − 12 < 0. iii. S has more than one element, in fact, for every s ∈ S there is s0 ∈ S such that s < s0 . (in fact, it has an infinite number of elements) and S ⊆ A. iv. If a ∈ A and b ∈ B then a < b. In particular f (b) =
1 2
> 0 for all b ∈ B.
v. If a ∈ A, there is some interval(u, v) ⊆ A such that a ∈ (u, v) ⊆ A and we have that f is constant on (u, v). vi. If b ∈ B, there is some interval(u, v) ⊆ B such that b ∈ (u, v) ⊆ B and we have that f is constant on (u, v). vii. Show that every function which is “locally constant” as in 13iv and 13v is continuous. Now put everything together and finish this up.] viii. (Intermediate Value Property for Derivatives) [In this problem we use the notation I(x, y) for the open interval having as endpoints the distinct real numbers x an y.] Let I ⊆ R be a non empty interval and let f : I → R be a differentiable function on I (derivatives are taken as unilateral at the endpoints of I if any). If a, b ∈ I are points for which f (a) and f (b) are distinct, and if α ∈ I(f (a), f (b)), then for some θ ∈ I(a, b), we have f 0 (θ) = α. Note that f 0 is not assumed continuous. [Suggestions: Note that if f (a) and f (b) are distinct real numbers, then so are a and b. For definiteness assume a < b and f 0 (a) < α < f 0 (b). Other cases are treated similarly. Define φ(x) = f (x) − αx for all x ∈ [a, b]. Then φ is differentiable on [a, b], φ0 (a) < 0 < φ0 (b). Since φ0 (a) < 0, there are values x to the right of a and close to a such that φ(x) < φ(a). Similarly, argue that there are points y to the left of b and close to b such that φ(y) < φ(b). By the Extreme Value Theorem, in this case, φ has a minimum in some iterior point of [a, b], say θ. By calculus φ0 (θ) = 0. This says f 0 (θ) = α.]
11. Exercises for students of mathematics B Part B: Adendum Exponential and logarithmic functions ∞ (14) If A is a bounded non empty set show that there are sequences (pn )∞ n=1 and (qn )n=1 in A (that is to say the sequence elements are numbers in A) such that pn → inf A and qn → sup A.
Hern´andez, L´opez & Vel´azquez, p. 16, (15) Let x, y, s ∈ R and suppose s < x + y. Show that there are p, q ∈ R such that p < x, q < y and s = p + q. (16) Show that the function x 7→ ax is strictly increasing if a > 1 and prove all the laws of exponents for this fucntion. (17) Knowing that the function x 7→ ax (x ∈ R) is continuous for a > 1, show that the function x 7→ ax (x ∈ R) is continuous for a < 1. an+1 (18) If (an )∞ n=1 is a sequence of positive real numbers and limn→∞ an < 1. Show that n limm→∞ an = 0. Conclude that limn→∞ an! = 0 for each a ∈ R with a > 0. [Suggestion: Show that for some θ ∈ (0, 1) and some index N , ap < θp−N an for all p > N .]
(19) Show that for x > 1, Z
x
n
X1 1 dt > , k 1 t k=2 Rx 1 where n = bxc. Conclude that limx→∞ 1 t dt = ∞. item Prove Theorem 7. References [1] T. Apostol Calculus Vol. 1: One-Variable Calculus with an Introduction to Linear Albebra 2nd Edition, Wyley, 1991 [2] W. Dunham The Master of Us All Dolciani Mathematical Expositions No. 22 The Mathematical Association of America, 1999 [3] L. Euler Introduction in Analysin Infinitorum, Vols. 1 and 2 Tranlated and annotated by Ian Bruce Euler’s Archive, 2013 [4] H. Frudenthal Didactical Phenomenology of Mathematical Structures Kluwer Academic Publishers Dordrecht, 1999. [5] H. Frudenthal Mathematics as an Educational Task D. Reidel Publishing Commpany Dordrecht, 1973. [6] Naive Set Theory The University Series in Undergraduate Mathematics D. Van Nostrand Company, Inc., New York, 1960 [7] K. Ross & J. L´ opez Elementary analysis: the theory of calculus Springer (2013) [8] An Introduction to Classical Real Analysis Originally published by Wadsworth, Belmont, 1981, published with corrections by the American Mathematical Society Chelsea Publishing, Providence, Rhode Island, 2015 [9] R. Stoll Set Theory and Logic Dover Publications Inc, 1961 [10] An Introduction to Classical Real Analysis Originally published by Wadsworth, Belmont, 1981, published with corrections by the American Mathematical Society Chelsea Publishing, Providence, Rhode Island [11] H. Wu Fractions, decimals and rational numbers. Notes for education researchers and K-8 teachers University of California at Berkeley 2008, revised 2014. [12] H. Wu. How to prepare students for algebra American Educator. p. 1-7. Summer 2001. Department of Graduate Studies, Faculty of Education, University of Puerto Rico, R´ıo Piedras E-mail address:
[email protected] Department of Mathematics, University of Puerto Rico, R´ıo Piedras E-mail address:
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