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Differentiation, Local Behavior: The Exponential and Logarithm Functions

Module by: Lawrence Baggett. E-mail the author

Summary: We derive next the elementary properties of the exponential and logarithmic functions. Of course, by “exponential function,” we mean the power series function exp.exp. And, as yet, we have not even defined a logarithm function.

We derive next the elementary properties of the exponential and logarithmic functions. Of course, by “exponential function,” we mean the power series function exp.exp. And, as yet, we have not even defined a logarithm function.

Exercise 1

  1. Define a complex-valued function f:CCf:CC by f(z)=exp(z)exp(-z).f(z)=exp(z)exp(-z). Prove that f(z)=1f(z)=1 for all zC.zC.
  2. Conclude from part (a) that the exponential function is never 0, and that exp(-z)=1/exp(z).exp(-z)=1/exp(z).
  3. Show that the exponential function is always positive on R,R, and that limx-exp(x)=0.limx-exp(x)=0.
  4. Prove that expexp is continuous and 1-1 from (-,)(-,) onto (0,).(0,).
  5. Show that the exponential function is not 1-1 on C.C.
  6. Use parts b and e to show that the Mean Value Theorem is not in any way valid for complex-valued functions of a complex variable.

Using part (d) of the preceding exercise, we make the following important definition.

Definition 1:

We call the inverse exp-1exp-1 of the restriction of the exponential function to RR the (natural) logarithm function, and we denote this function by ln.ln.

The properties of the exponential and logarithm functions are strongly tied to the simplest kinds of differential equations. The connection is suggested by the fact, we have already observed, that exp'=exp.exp'=exp. The next theorem, corollary, and exercises make these remarks more precise.

Theorem 1

Suppose f:CCf:CC is differentiable everywhere and satisfies the differential equation f'=af,f'=af, where aa is a complex number. Then f(z)=cexp(az),f(z)=cexp(az), where c=f(0).c=f(0).

Proof

Consider the function h(z)=f(z)/exp(az).h(z)=f(z)/exp(az). Using the Quotient Formula, we have that

h ' ( z ) = exp ( a z ) f ' ( z ) - a exp ( a z ) f ( z ) [ exp ( a z ) ] 2 = exp ( a z ) ( f ' ( z ) - a f ( z ) ) [ exp ( z ) ] 2 = 0 . h ' ( z ) = exp ( a z ) f ' ( z ) - a exp ( a z ) f ( z ) [ exp ( a z ) ] 2 = exp ( a z ) ( f ' ( z ) - a f ( z ) ) [ exp ( z ) ] 2 = 0 .
(1)

Hence, there exists a complex number cc such that h(z)=ch(z)=c for all z.z. Therefore, f(z)=cexp(az)f(z)=cexp(az) for all z.z. Setting z=0z=0 gives f(0)=c,f(0)=c, as desired.

Corollary 1: Law of Exponents

For all complex numbers zz and w,w,exp(z+w)=exp(z)exp(w).exp(z+w)=exp(z)exp(w).

Proof

Fix w,w, define f(z)=exp(z+w),f(z)=exp(z+w), and apply the preceding theorem. We have f'(z)=exp(z+w)=f(z),f'(z)=exp(z+w)=f(z), so we get

exp ( z + w ) = f ( z ) = f ( 0 ) exp ( z ) = exp ( w ) exp ( z ) . exp ( z + w ) = f ( z ) = f ( 0 ) exp ( z ) = exp ( w ) exp ( z ) .
(2)

Exercise 2

  1. If nn is a positive integer and zz is any complex number, show that exp(nz)=(exp(z))n.exp(nz)=(exp(z))n.
  2. If rr is a rational number and xx is any real number, show that exp(rx)=(exp(x))r.exp(rx)=(exp(x))r.

Exercise 3

  1. Show that lnln is continuous and 1-1 from (0,)(0,) onto R.R.
  2. Prove that the logarithm function lnln is differentiable at each point y(0,)y(0,) and that ln'(y)=1/y.ln'(y)=1/y. HINT: Write y=exp(c)y=exp(c) and use Theorem 4.10.
  3. Derive the first law of logarithms: ln(xy)=ln(x)+ln(y).ln(xy)=ln(x)+ln(y).
  4. Derive the second law of logarithms: That is, if rr is a rational number and xx is a positive real number, show that ln(xr)=rln(x).ln(xr)=rln(x).

We are about to make the connection between the number ee and the exponential function. The next theorem is the first step.

Theorem 2

ln(1)=0ln(1)=0 and ln(e)=1.ln(e)=1.

Proof

If we write 1=exp(t),1=exp(t), then t=ln(1).t=ln(1). But exp(0)=1,exp(0)=1, so that ln(1)=0,ln(1)=0, which establishes the first assertion.

Recall that

e = lim n ( 1 + 1 n ) n . e = lim n ( 1 + 1 n ) n .
(3)

Therefore,

ln ( e ) = ln ( lim n ( 1 + 1 n ) n ) = lim n ln ( ( 1 + 1 n ) n ) = lim n n ln ( 1 + 1 n ) = lim n ln ( 1 + 1 n ) 1 n = lim n ln ( 1 + 1 n ) - ln ( 1 ) 1 n = ln ' ( 1 ) = 1 / 1 = 1 . ln ( e ) = ln ( lim n ( 1 + 1 n ) n ) = lim n ln ( ( 1 + 1 n ) n ) = lim n n ln ( 1 + 1 n ) = lim n ln ( 1 + 1 n ) 1 n = lim n ln ( 1 + 1 n ) - ln ( 1 ) 1 n = ln ' ( 1 ) = 1 / 1 = 1 .
(4)

This establishes the second assertion of the theorem.

Exercise 4

  1. Prove that
    e=n=01n!.e=n=01n!.
    (5)
    HINT: Use the fact that the logarithm function is 1-1.
  2. For rr a rational number, show that exp(r)=er.exp(r)=er.
  3. If aa is a positive number and r=p/qr=p/q is a rational number, show that
    ar=exp(rln(a)).ar=exp(rln(a)).
    (6)
  4. Prove that ee is irrational. HINT: Let pn/qnpn/qn be the nnth partial sum of the series in part (a). Show that qnn!,qnn!, and that limqn(e-pn/qn)=0.limqn(e-pn/qn)=0. Then use (Reference).

We have finally reached a point in our development where we can make sense of raising any positive number to an arbitrary complex exponent. Of course this includes raising positive numbers to irrational powers. We make our general definition based on part (c) of the preceding exercise.

Definition 2:

For aa a positive real number and zz an arbitrary complex number, define azaz by

a z = exp ( z ln ( a ) ) . a z = exp ( z ln ( a ) ) .
(7)

REMARK The point is that our old understanding of what arar means, where a>0a>0 and rr is a rational number, coincides with the function exp(rln(a)).exp(rln(a)). So, this new definition of azaz coincides and is consistent with our old definition. And, it now allows us to raies a positive number aa to an arbitrary complex exponent.

REMARK Let the bugles sound!! Now, having made all the appropriate definitions and derived all the relevant theorems, we can finally prove that eiπ=-1.eiπ=-1. From the definition above, we see that if a=e,a=e, then we have ez=exp(z).ez=exp(z). Then, from part (c) of (Reference), we have what we want:

e i π = - 1 . e i π = - 1 .
(8)

Exercise 5

  1. Prove that, for all complex numbers zz and w,w,ez+w=ezew.ez+w=ezew.
  2. If xx is a real number and zz is any complex number, show that
    (ex)z=exz.(ex)z=exz.
    (9)
  3. Let aa be a fixed positive number, and define a function f:CCf:CC by f(z)=az.f(z)=az. Show that ff is differentiable at every zCzC and that f'(z)=ln(a)az.f'(z)=ln(a)az.
  4. Prove the general laws of exponents: If aa and bb are positive real numbers and zz and ww are complex numbers,
    az+w=azaw,az+w=azaw,
    (10)
    azbz=(ab)z,azbz=(ab)z,
    (11)
    and, if xx is real,
    axw=(ax)w.axw=(ax)w.
    (12)
  5. If yy is a real number, show that |eiy|=1.|eiy|=1. If z=x+iyz=x+iy is a complex number, show that |ez|=ex.|ez|=ex.
  6. Let α=a+biα=a+bi be a complex number, and define a function f:(0,)Cf:(0,)C by f(x)=xα=eαln(x).f(x)=xα=eαln(x). Prove that ff is differentiable at each point xx of (0,)(0,) and that f'(x)=αxα-1.f'(x)=αxα-1.
  7. Let α=a+biα=a+bi be as in part (f). For x>0,x>0, show that |xα|=xa.|xα|=xa.

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