* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Real Exponents
Survey
Document related concepts
History of logarithms wikipedia , lookup
Location arithmetic wikipedia , lookup
Georg Cantor's first set theory article wikipedia , lookup
Infinitesimal wikipedia , lookup
Series (mathematics) wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Large numbers wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
Hyperreal number wikipedia , lookup
Non-standard calculus wikipedia , lookup
Collatz conjecture wikipedia , lookup
Proofs of Fermat's little theorem wikipedia , lookup
Real number wikipedia , lookup
P-adic number wikipedia , lookup
Approximations of π wikipedia , lookup
Transcript
Real Exponents We begin this note by recounting the standard definition of bx for real x by approximating x by rational numbers for a base b > 0. For simplicity, we focus on 2x . We investigate what happens if we try to do the same with a negative base b. We discuss of the following fact: the natural domain of (−2)x consists of rational numbers x = p/q with p an integer and q an odd integer. Using base-3 numbers, we show how to approximate any real number by numbers in the domain of (−2)x . 1. Defining 2x for Real x If x is an integer, then it is clear what 2x is. For example, 20 = 1, 22 = 2(2) = 4, 2−1 = 21 . For any rational number x = p/q, where p is an integer and q is a nonzero integer, we can also define 2x by √ p 1 q 2x = 2 q = (2p ) q = 2p . Exercise: Use a calculator to plot the points on the graph y = 2x for x = 0, 0.1, 0.2, 0.3, · · · , 0.9, 1. Notice that these points look as if they are part of a continuous graph. What happens if we plot more points (for example, for x = 0.15)? √ From here, we may wonder how to define 2 2 , for example. Perhaps the most obvious way to approximate an irrational number is by its decimal expansion. Use a calculator to compute √ 2 = 1.414213 · · · . √ Then the successive decimal approximations to 2 are 1, 1.4, 1.41, etc., and we may compute x= 1 1.4 1.41 1.414 1.4142 1.41421 1.414213 x 2 ∼ 2 2.63902 2.65737 2.66475 2.66512 2.66514 2.66514 So the values of 2x for these values of decimal approximations√of x approach a limit, which is about 2.66514. Then we can define 2 2 as √ 2 this limit so that 2 ∼ 2.66514. [Rigorously, we could say that the sequence of approximations on the bottom row is monotone increasing √ (since 2x is an increasing function for rational x). Also, since 2 < 2, the sequence of approximations are all less than 22 = 4. A basic 1 2 property of the real numbers says that any bounded monotone sequence √ 2 must have a limit. In this case that limit is our 2 .] 2. What About (−2)x ? The first step in trying to define (−2)x goes the same way. If x is an integer, then (−2)x is clearly defined so that (−2)2 = (−2)(−2) = 4. √ 1 We run into trouble with rational exponents though. (−2) 2 = −2 p is not a real number, and similarly, we cannot define (−2) q if q is even. We can define (−2)x for many rational numbers x though. So the domain of (−2)x is all rational numbers x = p/q so that q is odd. Exercise: Use a calculator to plot the points on the graph y = (−2)x for x = 0, 91 , 29 , · · · , 89 , 1. Notice that these points do not seem to make a part of a continuous graph. Also notice that we can choose x on a 1 2 finer scale: Between 0 and 91 for example, we can choose 27 and 27 in x the domain of (−2) . Plot some more points such as these. Describe what you see. The graphical evidence in the previous exercise makes it clear that we cannot expect to extend (−2)x to a continuous function for all real x. We can also look at this problem in terms of approximating a real number x by values in the domain of (−2)x . We don’t have all rational numbers to work with, so it may seem that we can’t approximate every real number by numbers in the √ domain of x (−2) . Think about what we did with the real number 2 above. We didn’t approximate it by just any √ rational numbers—we used the decimal expansion to approximate 2. Let’s see√ if these particular rational numbers—the decimal approximations of 2—will work. (−2)1 = −2 is clearly OK, and we can make sense out of 14 7 (−2)1.4 = (−2) 10 = (−2) 5 ∼ −2.63902 √ as well. The next approximation of 2 is impossible, however: 141 (−2)1.41 = (−2) 100 . The denominator of the exponent is even and we can’t factor to make it odd. [Question: Is there any nonterminating decimal number so that each of the decimal approximations can be reduced to the form pq , for p an integer and q an odd integer? Produce one or prove such a number cannot exist. 3 Answer: It cannot exist. In fact, such a number can only have nonzero digits in up to three places after the decimal place. ] So it appears we are out of luck. The decimal approximations do not work either. The problem is that we get powers of 10 (which are even) in the denominator of the exponent, and so we cannot have a real number as an answer. The problem with this approach is then 10, the base for our decimal number system. If we take an odd base, say 3, then we don’t have this problem. Let 1.213 be written in base 3. In other words 2 1 16 + 2 = , 3 3 9 which is clearly in the domain of (−2)x . So let’s take a number x which is not in the domain of (−2)x , and use its base-3 approximation to see what happens to the values of (−2)x . We don’t even need to try an irrational number; any number not in the domain of (−2)x will work—e.g. 12 . So write 12 in base 3. 1.213 = 1 + 1 2 = 0.111111 · · ·3 Let’s see what happens to (−2)x for these approximations. x = 03 = 0 (−2)x ∼ 1 0.13 0.113 0.1113 0.11113 0.111113 0.1111113 1 3 4 9 13 27 40 81 121 243 364 729 −1.25992 1.36079 −1.39618 1.40818 −1.41220 1.41354 So notice that the approximations (−2)x alternate in sign and don’t settle down to a limit. (Their absolute values do converge to a limit; what is it?) Project: In base 3, √ 2 ∼ 1.10201123 . √ Use the first several base-3 approximations of 2 to work out a similar table to the one above. Based on the base-3 digits, when does the sign of (−2)x change? Why is this the case? Solution. If x is a number with a terminating expression in base 3, then we can represent p x= n 3 for an integer p (n is the number of places after the decimal point in the base-3 expression of x). p (−2) 3n > 0 for p even, p (−2) 3n < 0 for p odd. 4 p is even or odd depending on whether the number of 1’s in the base-3 representation of x is even or odd. (Why?) So the sign will change only when the current digit is a 1. Example: Are there then numbers with nonterminating base-3 representations so that the limit computed as above exists. A few interesting ones to try are 41 = 0.020202 · · ·3 and 1 = 0.222222 · · ·3 . Are the limits computed in this way appropriate? Remark. There is a natural way to define (−2)x as a continuous function, but not as a real-valued function. If we define ln(−2) = ln 2 + iπ, we can define x (−2)x = eln(−2) = = = = ex ln(−2) ex ln 2+iπx ex ln 2 (cos πx + i sin πx) 2x (cos xπ + i sin xπ). Notice that in this formulation (−2)x is real only if x is an integer. This is the price you have to pay if you want to make (−2)x into a continuous function.