Download Real Exponents

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.