Math1202 Supplementary Notes

The Logarithm and Exponential Functions We begin with the following question: If a > 0, what is ax ? We define a1 = a,

a2 = a · a,

a3 = a · a · a,

...

and more generally, an = |a · a{z· · · a} n times

for positive integers n. We also define a0 = 1

a−n =

and

1 for n in N. an

1

To define ax for rational x, we first let a n be the unique b > 0 such that bn = a. This b is the nth root of a √ and is also denoted by n a. It is the image of the inverse function of f (x) = xn

for all x ≥ 0

at the point a. It can also be obtained directly from the Completeness Axiom as sup{x ≥ 0| xn < a}. Then we define m

1

1

a n = (am ) n = a n


m

and

m

a− n =

1 m an

for every natural numbers m, n. Thus ax is defined for every rational number x. The following index laws hold for every rational numbers x and y: ax+y = ax · ay

and

y

axy = (ax ) .

Can we define ax for irrational numbers x so that the above index laws still hold? For example, what √

should be a 2 ? One approach is to interpolate a nice curve through the points (x, ax ), for all rational numbers x, and define ax for irrational numbers x according to the values on the curve. We observe that if the function f : R → R satisfies f (x + y) = f (x)f (y)

for every x, y in Q,

and that f (1) 6= 0, then f (x) = f (1)x for every rational number x. Proof: Note that f (0) = f (0 + 0) = f (0)2 so that f (0) = 0 or 1. If f (0) = 0, then f (x) = f (x + 0) = f (x)f (0) = 0 for every x. In particular, f (1) = 0. Hence f (0) = 1. For every positive integer n, f (n) = f (1 + · · · + 1) = f (1) · · · f (1) = f (1)n . and 1 = f (0) = f (n + (−n)) = f (n)f (−n) so that f (−n) =

1 1 = = f (1)−n . f (n) f (1)n 1

(∗)


n As f n1 =f
m m f n = f (1) n .

1 n

+ ··· +

1 n




= f (1), we have f

1 n




1

= f (1) n . Finally, it is straightforward to prove that

Hence we have to find a solution of (∗) such that f (1) = a. If such an f is further assumed to be differentiable, then f (x + h) − f (x) f (x)(f (h) − 1) f ′ (x) = lim = lim = cf (x), h→0 h→0 h h for c = f ′ (0). In other words, f is a solution of the differential equation f ′ (x) = cf (x)

for every x.

It turns out that it is easier to find the inverse function g = f −1 . Intuitively, g = loga , as loga y is the unique x such that ax = y. Now g = f −1 is also differentiable and g ′ (y) =

1 1 1 = = . f ′ (g(y)) cf (g(y)) cy

As g(1) = 0, g(y) = g(y) − g(1) =

Z

y

g ′ (t) dt =

1

1 c

Z

y

1

1 dt. t

When c = 1, we get the natural logarithm function, log, given by log x =

Z

x

1

1 dt t

for every x > 0.

It is clear that if x < 1, log x =

Z

1

x

1 dt = − t

Z

1

x

1 dt < 0, t

log 1 = 0, and log x > 0 if x > 1. Furthermore we have i. log is differentiable and log′ x = 1/x; ii. log is strictly increasing and is hence injective; iii. log xy = log x + log y; iv. log(x/y) = log x − log y; v. log xn = n log x for every integer n. Proof: iii. Fix any y > 0 and define h(x) = log(xy). The function h is differentiable, and h′ (x) = y log′ (xy) = y ·

1 1 = = log′ x. xy x

Hence log(xy) = h(x) = log x + c for some constant c. Putting x = 1, we get c = log y and the result follows.

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iv. It follows from the equations log x = log



x ·y y



= log

x + log y. y

v. It can be proved by Mathematical Induction that log xn = n log x

for every natural number n.

It also follows from (iv) that  n 1 1 = n log = −n log x. = log x x

−n

log x

To recover f = g −1 , note that lim log x = +∞

x→+∞

and

lim log x = −∞.

x→0+

Proof: Observe that log 2n = n log 2,

log 2−n = −n log 2,

where

log 2 > 0.

Hence the image of log (which is the domain of f ) is R. Let exp = log−1 . It is called the exponential function and is defined on R into the set of all positive numbers. We have the following properties: i. exp is differentiable and exp′ x = exp x; ii. exp is strictly increasing; iii. exp (x + y) = exp x · exp y. Proof: i. For all number x, exp′ x = (log−1 )′ x =

1 1 = = log−1 x = exp x. log′ (log−1 x) 1/ log−1 x

iii. Suppose u = exp x and v = exp y. Then x + y = log u + log v = log uv. This means exp(x + y) = uv = exp x exp y.

Let e = exp 1. Then e is characterized by Z

1

2

Re

1 1 t

dt = 1. As

1 dt < 1 · (2 − 1) = 1, t

and Z

1

4

1 1 1 1 13 dt > · (2 − 1) + · (2 − 1) + · (2 − 1) = , t 2 3 4 12 3

we get 2 < e < 4. By some more delicate computation, we can show that e ≈ 2.71828. From property (iii) above, exp x = (exp 1)x = ex . for every rational number x. Define ex = exp x for every x ∈ R. It follows that ex+y = ex · ey for every x, y ∈ R. In order to define ax , observe that if a > 0, then ax = (elog a )x = ex log a , for every rational number x. As ex log a is defined for all x, we can use it to define ax . Definition: If a > 0, define for every real number x, ax = ex log a . Note that the requirement a > 0 is necessary in order that log a is defined. The following properties holds: i. ax is differentiable and (ax )′ = ax log a; ii. ax is strictly increasing; y

iii. axy = (ax ) ; iv. a1 = a; v. ax+y = ax · ay . Proof: i. Recall that ax = ex log a . Hence by the Chain Rule, (ax )′ = log a · ex log a = log a · ax . iii. (ax )y = ey log a = ey log(e x

x log a

)

= ey(x log a) = exy log a = axy .

iv. a1 = e1 log a = elog a = a. v. ax+y = e(x+y) log a = ex log a+y log a = ex log a · ey log a = ax · ab . The function loga is the inverse function of ax . It can also be expressed in terms of log as follows. If y = loga x, then x = ay = ey log a . So log x = y log a, and y=

log x . log a

4

It follows that (loga x)′ =

1 . x log a

Let a in R. Define f : (0, ∞) → R by f (x) = xa

for every x > 0.

Then f ′ (x) = axa−1

for every x > 0.

Proof: Write f (x) = xa = ea log x , then f ′ (x) =

a a log x a ·e = · xa = axa−1 . x x

Exercise: If f is differentiable and f ′ (x) = f (x)

for all x,

then there is a number c such that f (x) = cex

for all x.

Solution: Let g(x) =

f (x) . ex

Then g ′ (x) =

ex f ′ (x) − f (x)ex = 0. ex

Hence there is a number c such that g(x) =

f (x) = c. ex

Reference Michael Spivak, Calculus, W.A. Benjamin, Menlo Park, 1967.

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