Chapter 5: Elementary Functions5.1 Infinite Series and ConvergenceThroughout this section, we will be dealing with numbers which may either be complex numbers or may be restricted to real numbers. Since the elementary functions will be defined via infinite series, we need to prove some criteria for checking for the convergence of such series. Definition 1: i. An infinite sequence is a function whose domain is the set of nonnegative integers. A infinite sequence is usually denoted as where is the value of the function at the nonnegative integer i. We say that this infinite sequence converges to a number a (or has limit a) if for every there is an such that for all . (Intuitively, if is as close as one wishes to a as soon as i is sufficiently large.) A sequence is said to be convergent if it converges to some number a; otherwise, it is said to be divergent. An infinite series is an infinite sum . The are called the terms of the infinite series. For every nonnegative integer n, the partial sum of the series is . The infinite series is said to converge to the number a, if the sequence of partial sums converges to a. A series is said to be convergent or conditionally convergent if it converges to some number; otherwise, it is said to be divergent. iii. An infinite series is said to be absolutely convergent if the series is convergent. It is important to distinguish between the sequence of terms of an infinite series and the sequence of partial sums of an infinite series. Proposition 1: If is convergent, then the sequence of its terms converges to zero. Proof: Suppose the series converges to a and that we are given an . There is an N > 0 such that satisfies for all k > N. Suppose that k > N + 1. then k  1 > N and so as well as . Since , one has by the triangle inequality: as desired. Example: i. The series is divergent ii. If is an infinite decimal, then the series converges to the value of the infinite decimal. iii. A geometric series converges to if . If , then the geometric series is divergent. Proposition 2: i. If the series converges to a and k is a nonnegative integer, then the series converges to . ii. Every absolutely convergent series is convergent. iii. If a series is absolutely convergent, then so is every series obtained from it by rearranging the order of its terms. Furthermore, reordering the terms does not affect the value to which the series converges. Proof: i. Let . Then the partial sums of the original series differ from the partial sums of the second series by b. The result is now a simple consequence of the definitions and the triangle inequality. ii. Suppose is absolutely convergent. Let the be the partial sums of this series. We know that the sum is convergent; by assertion i of this proposition, it follows that is also convergent for every nonnegative integer k. Further, one has for every n > k, and so we have in the interval . Now, it is easy to see that and that the lengths of these intervals approach zero. So, by Chapter 2, Corollary 8 of section 2.7.2 (or its complex analogue), we know that the intersection of all these intervals contains a single number and the series converges to this value. iii. For this assertion, it is enough to treat the case where the series consists of nonnegative terms. Then the partial sums are form a bounded increasing sequence. Furthermore, every partial sum of any rearranged series is bounded above by one of the partial sums of this original series. It follows that the partial sums of the rearranged series also converge. Proposition 3:(Ratio test) Consider an infinite series . Suppose there is a real number r with 0 < r < 1 and such that for all sufficiently large k. Then the series is absolutely convergent. Proof: Because convergence does not depend on the behavior of the first several terms, we can assume that the ratio condition holds for all nonnegative k. Further, one can assume all terms are nonnegative real numbers. But then we have But then the partial sum satisfies So, the partial sums form an bounded increasing sequence and therefore converge. Proposition 4: (Alternating Series Test) Let for be a sequence of real numbers which alternate in signs and such that their absolute values satisfy . If the approach 0, then the series is convergent. Furthermore, if the partial sums are and if the series converges to b, then we have for every nonnegative integer k. Proof: We can assume that the first term of the series is positive. Then it is easy to see that that So, the intervals are of the type in Corollary 8 of section 2.7.2. So, the partial sums converge to a value b contained in all the intervals. The final inequality follows from the fact that b is in the interval with endpoints and . Example 2: i. The ratio test shows that the series converges absolutely for all complex numbers z. This is the socalled exponential function exp(z). In particular, the value is the Euler constant e = 2.718283.... ii. The alternating series test shows that 1  1/2 + 1/3  1/4 + ... converges. It can be shown that this series is NOT absolutely convergent. iii. The alternating series exp(1) = 1  1/2! + 1/3!  1/4! + ... converges very rapidly because of the difference between the limiting value and the partial sum is no more than 1/(k+1)! in absolute value. In the next section, one will see that exp(1) = 1/e, and so one gets a very accurate estimate for the Euler constant e. Proposition 5: Let and be convergent series with partial sums and respectively Suppose that the series converge to numbers a and b respectively. Then
Proof: i. Let . We will see below that we need to choose an such that and . Assume that N is chosen so that for all n > N, one has and . Then for such n, one has:
ii. Apply the first assertion to the series obtained by replacing the terms and with their absolute values. One concludes that the sequence with terms are convergent. Now consider the series . Since all the terms are nonnegative, we really don't care about the order of the terms because if the series converges in one order, it has to converge in all orders. The series will converge if all the partial sums have a common upper bound. But, since the sequence with terms is convergent, we know that the partial sums do have a common bound. So, our series is absolutely convergent. But then the series . is also absolutely convergent. The value to which the series converges does not depend on the order of the terms. So, one might as well order them in an order whose partial sums would be the sequence of . But then the value of the series is ab by assertion i. 5.2 The Exponential and Logarithm FunctionsDefinition 2: The exponential function is defined for every complex number z to be the value of the series . For example, this means that exp(0) = 1. Proposition 6: For all complex numbers and , one has Proof: By Proposition 5, we know that is absolutely convergent and converges to . On the other hand, the inside sum is by the binomial theorem. So, the series is which completes the proof. Corollary 1: The exponential function is continuous at every complex number. Proof: By the addition theorem, one has for every complex number a: In particular, if the exponential function is continuous at 0, then it will be continuous everywhere. Let . Choose so that and . Then for all z with , one has So, the exponential function is indeed continuous at 0. Corollary 2: As a function of real numbers, the exponential function is an increasing function, i.e. exp(x) < exp(y) if x < y. Proof: By the addition theorem, it is enough to show that 1 = exp(0) < exp(y  x) if y  x > 0. But this is clear from the series since it starts out as 1 + (y  x) and all the remaining terms are positive. Corollary 3: Again restricting to real values, the range of the exponential function is the set of all positive real numbers. Clearly, the exponential function can be made arbitrarily large by taking x sufficiently large and positive. Since exp(x) = 1/exp(x) by the addition theorem, it follows that one can make the exponential function arbitrarily small by taking x negative and large in absolute value. Since the exponential function is continuous, the result now follows from the intermediate value theorem. It follows from Corollary 3 that the exponential function has an inverse function defined for all positive real numbers and with range the set of all real numbers. This function is denoted is called the logarithm function. Definition 3: Let a be a positive real number. Then define the power function for real numbers x. Corollary 4: For all positive real numbers a and b and real numbers c, one has
Proof: These are simple consequences of the definition of the logarithm function as the inverse of the exponential theorem. For example, Applying the logarithm to both sides gives the second assertion. Similarly, shows the third assertion. For the last assertion, suppose that . Then, since the exponential function is increasing, we have . But then, using the fact that the exponential function is the inverse of the logarithm function, we get . Taking the contrapositive, we see that we have shown that if , then . Corollary 5: For all real numbers a, b, r, and s with a and b positive, one has Proof: i. One has Assertion ii is proved similarly. iii. One has . So,
iv. One has . Proposition 7: The natural logarithm function is continuous at all x > 0. Proof: The hard part of the proof is to show that is continuous at x = 1. So, let's assume we have done that part. Let and let us then show that is continuous at . For this, let and, using the continuity at 1, we know that there is a such that provided that . Let . Suppose that x is any positive real number such that . Then and so . Since the left side of the inequality is just , we see that is continuous at . Now, let's try to see why is continuous at x = 1. Again, let . Let be any positive real number smaller than both 1 and . We will see below that the proper choice of is a positive number with . Suppose we choose such a and then choose any positive x with . There are two cases to consider:
