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The Elementary Transcendental Functions

Module by: Lawrence Baggett. E-mail the author

Summary: This module dives deeper into power series functions, examining their properties, which coefficients in a power series give special properties, which have converging tendencies, and when they are defined everywhere. The exponential function is defined.

Having introduced a class of new functions (power series functions), we might well expect that some of these will have interesting and unexpected properties. So, which sets of coefficients might give us an exotic new function? Unfortunately, at this point in our development, we haven't much insight into this question. It is true, see (Reference), that most power series functions that we naturally write down have finite radii of convergence. Such functions may well be new and fascinating, but as a first example, we would prefer to consider a power series function that is defined everywhere, i.e., one with an infinite radius of convergence. Again revisiting (Reference), let us consider the coefficients an=1/n!.an=1/n!. This may seem a bit ad hoc, but let's have a look.

Definition 1:

Define a power series function, denoted by exp, as follows:

exp ( z ) = n = 0 z n n ! . exp ( z ) = n = 0 z n n ! .
(1)

We will call this function, with 20-20 hindsight, the exponential function.

What do we know about this function, apart from the fact that it is defined for all complex numbers? We certainly do not know that it has anything to do with the function ez;ez; that will come in the next chapter. We do know what the number ee is, but we do not know how to raise that number to a complex exponent.

All of the exponential function's coefficients are positive, and so by part (d) of (Reference) exp is not a rational function; it really is something new. It is natural to consider the even and odd parts expeexpe and expoexpo of this new function. And then, considering the constructions in (Reference), to introduce the alternating versions expeaexpea and expoaexpoa of them.

Definition 2:

Define two power series functions cosh (hyperbolic cosine) and sinh (hyperbolic sine) by

cosh ( z ) = exp ( z ) + exp ( - z ) 2 and sinh ( z ) = exp ( z ) - exp ( z ) 2 , cosh ( z ) = exp ( z ) + exp ( - z ) 2 and sinh ( z ) = exp ( z ) - exp ( z ) 2 ,
(2)

and two other power series functions cos (cosine) and sin (sine) by

cos ( z ) = cosh ( i z ) = exp ( i z ) + exp ( - i z ) 2 cos ( z ) = cosh ( i z ) = exp ( i z ) + exp ( - i z ) 2
(3)

and

sin ( z ) = - i sinh ( i z ) = exp ( i z ) - exp ( - i z ) 2 i . sin ( z ) = - i sinh ( i z ) = exp ( i z ) - exp ( - i z ) 2 i .
(4)

The five functions just defined are called the elementary transcendental functions, the sinhsinh and coshcosh functions are called the basic hyperbolic functions, and the sine and cosine functions are called the basic trigonometric or circular functions. The connections between the hyperbolic functions and hyperbolic geometry, and the connection between the trigonometric functions and circles and triangles, will only emerge in the next chapter. From the very definitions, however, we can see a connection between the hyperbolic functions and the trigonometric functions. It's something like interchanging the roles of the real and imaginary axes. This is probably worth some more thought.

Exercise 1

  1. Verify the following equations:
    exp(z)=n=0znn!=1+z+z22!+z33!+...+zkk!+...,=cosh(z)+sinh(z).exp(z)=n=0znn!=1+z+z22!+z33!+...+zkk!+...,=cosh(z)+sinh(z).
    (5)
    sin(z)=z-z33!+z55!-z77!+...+(-1)kz2k+1(2k+1)!+...=k=0(-1)kz2k+1(2k+1)!,sin(z)=z-z33!+z55!-z77!+...+(-1)kz2k+1(2k+1)!+...=k=0(-1)kz2k+1(2k+1)!,
    (6)
    cos(z)=1-z22!+z44!-z66!+...+(-1)kz2k(2k)!+...=k=0(-1)kz2k(2k)!,cos(z)=1-z22!+z44!-z66!+...+(-1)kz2k(2k)!+...=k=0(-1)kz2k(2k)!,
    (7)
    sinh(z)=k=0z2k+1(2k+1)!,sinh(z)=k=0z2k+1(2k+1)!,
    (8)
    and
    cosh(z)=k=0z2k(2k)!.cosh(z)=k=0z2k(2k)!.
    (9)
    (These expressions for the elementary transcendental functions are perhaps the more familiar ones from a calculus course.)
  2. Compute the radii of convergence for the elementary transcendental functions. HINT: Do not use the Cauchy-Hadamard formula. Just figure out for which zz's the functions are defined.
  3. Verify that exp(0)=1,exp(0)=1,sin(0)=sinh(0)=0,sin(0)=sinh(0)=0, and cos(0)=cosh(0)=1.cos(0)=cosh(0)=1.
  4. Prove that all five of the elementary transcendental functions are not rational functions.
  5. Can you explain why sin2(z)+cos2(z)1?sin2(z)+cos2(z)1? What about the “ Addition Formula”
    sin(z+w)=sin(z)cos(w)+cos(z)sin(w).sin(z+w)=sin(z)cos(w)+cos(z)sin(w).
    (10)

Exercise 2

  1. Show that the elementary transcendental functions map real numbers to real numbers. That is, as functions of a real variable, they are real-valued functions.
  2. Show that the exponential function exp is not bounded above. Show in fact that, for each nonnegative integer n,n,exp(x)/xnexp(x)/xn is unbounded. Can you show that exp(x)=ex?exp(x)=ex? What, in fact, does exex mean if xx is an irrational or complex number?

At this point, we probably need a little fanfare!

Theorem 1: THEOREM 3.14159 (Definition of ππ)

There exists a smallest positive number xx for which sin(x)=0.sin(x)=0. We will denote this distinguished number xx by the symbol π.π.

Proof

First we observe that sin(1)sin(1) is positive. Indeed, the infinite series for sin(1)sin(1) is alternating. It follows from the alternating series test (Theorem 2.18) that sin(1)>1-1/6=5/6.sin(1)>1-1/6=5/6.

Next, again using the alternating series test, we observe that sin(4)<0.sin(4)<0. Indeed,

sin ( 4 ) < 4 - 4 3 3 ! + 4 5 5 ! - 4 7 7 ! + 4 9 9 ! - 0 . 4553 < 0 . sin ( 4 ) < 4 - 4 3 3 ! + 4 5 5 ! - 4 7 7 ! + 4 9 9 ! - 0 . 4553 < 0 .
(11)

Hence, by the intermediate value theorem, there must exist a number cc between 1 and 4 such that sin(c)=0.sin(c)=0. So, there is at least one positive number xx such that sin(x)=0.sin(x)=0. However, we must show that there is a smallest positive number satisfying this equation.

Let AA be the set of all x>0x>0 for which sin(x)=0.sin(x)=0. Then AA is a nonempty set of real numbers that is bounded below. Define π=infA.π=infA. We need to prove that sin(π)=0,sin(π)=0, and that π>0.π>0. Clearly then it will be the smallest positive number xx for which sin(x)=0.sin(x)=0.

By (Reference), there exists a sequence {xk}{xk} of elements of AA such that π=limxk.π=limxk. Since sinsin is continuous at π,π, it follows that sin(π)=limsin(xk)=lim0=0.sin(π)=limsin(xk)=lim0=0. Finally, if ππ were equal to 0, then by the Identity Theorem, Theorem 3.14, we would have that sinx=0sinx=0 for all x.x. Since this is clearly not the case, we must have that π>0.π>0.

Hence, ππ is the smallest (minimum) positive number xx for which sin(x)=0.sin(x)=0.

As hinted at earlier, the connection between this number ππ and circles is not at all evident at the moment. For instance, you probably will not be able to answer the questions in the next exercise.

Exercise 3

  1. Can you see why sin(x+2π)sin(x)?sin(x+2π)sin(x)? That is, is it obvious that sinsin is a periodic function?
  2. Can you prove that cos(π)=-1?cos(π)=-1?

REMARK Defining ππ to be the smallest positive zero of the sine function may strike many people as very much “out of the blue.” However, the zeroes of a function are often important numbers. For instance, a zero of the function x2-2x2-2 is a square root of 2, and that number we know was exztremely important to the Greeks as they began the study of what real numbers are. A zero of the function z2+1z2+1 is something whose square is -1, i.e., negative. The idea of a square being negative was implausible at first, but is fundamental now, so that the zero of this particular function is critical for our understanding to numbers. Very likely, then, the zeroes of any “new” function will be worth studying. For instance, we will soon see that, perhaps disappointingly, there are no zeroes for the exponential function: exp(z)exp(z) is never 0. Maybe it's even more interesting then that there are zeroes of the sine function.

The next theorem establishes some familiar facts about the trigonometric functions.

Theorem 2

  1.  exp(iz)=cos(z)+isin(z)exp(iz)=cos(z)+isin(z) for all zC.zC.
  2. Let {zk}{zk} be a sequence of complex numbers that converges to 0. Then
    limsin(zk)zk=0.limsin(zk)zk=0.
    (12)
  3. Let {zk}{zk} be a sequence of complex numbers that converges to 0. Then
    lim1-cos(zk)zk2=12.lim1-cos(zk)zk2=12.
    (13)

Exercise 4

Prove Theorem 2.

HINT: For parts (2) and (3), use (Reference).

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