Calculus Of One Real Variable
By Pheng Kim Ving
Chapter 7: The Exponential And Logarithmic Functions
Section 7.8: Transcendency Of The Exponential And Logarithmic Functions

 

7.8
Transcendency Of The Exponential And Logarithmic
Functions

 

 

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1. Transcendency Of The Exponential Functions

 

For the definition of transcendental functions, see Section 6.3.1 Definition_3.1. In this section we’re going to show that the
exponential, logarithmic, hyperbolic, and inverse hyperbolic functions are transcendental. Like the function sin x as seen in
Section 6.3.1 Eq. [3.3], these functions have their own infinite expansions in x, which can be used to compute their
approximate values.

 

Theorem 1.1 - Transcendency Of The Natural Exponential Function

 

The natural exponential function ex is a transcendental function.

 

 

Proof

 

 

for all x, where qn(x) isn't identically 0. Dividing both sides of Eq. [2] by ex we get:

 

EOP

 

Theorem 1.2 - Transcendency Of The General Exponential Function

 

The general exponential function bx is a transcendental function.

 

 

Proof

 

where the ak,i's are the real coefficients and mk is the degree of pk. So:

 

EOP

 

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2. Transcendency Of The Logarithmic Functions

 

In Section 6.3.2 Theorem 2.1, it's shown that a function is transcendental iff its inverse function is transcendental. So the
natural logarithm function
ln x, which is the inverse of the transcendental natural exponential function ex, is transcendental.

 

There's also a way to prove directly that the function ln x is transcendental, ie, by utilizing the definition of transcendental
functions. We're going to do it, believing that it may be of interest to some students, especially math majors.

 

The Natural Logarithm Function Isn’t A Rational Function

 

The direct proof of the transcendency of the natural logarithm function that we're going to present employs the property that
this function isn’t a rational function. Recall that a function f (x) is said to be a rational function  if it is or can be written in the
form of the ratio p(x)/q(x), where p(x) and q(x) are polynomials in the real variable x with real coefficients and q(x) isn’t
identically 0.

 

We say that polynomial D(x) (the dividend) is divisible by the polynomial d(x) (the divisor) or that d(x) divides D(x) if the
remainder of the division of D(x) by d(x) is 0, ie, if D(x)/d(x) = Q(x) (the quotient) or D(x) = d(x)Q(x), where Q(x) is a
polynomial, ie, if d(x) is a factor of D(x). If D(x) has a non-0 constant term, then D2(x) has a non-0 constant term. If D2(x) is
divisible by the polynomial x, then its constant term is 0:

 

D2(x) = xQ(x) = anxn + an–1xn–1 + an–2xn–2 + ... + a2x2 + a1x,

 

assuming Q(x) = anxn–1 + an–1xn–2 + an–2xn–3 + … + a2x + a1. So the constant term of D(x) must also be 0, for otherwise the
constant term of D2(x) would be non-0. Thus D(x) is also divisible by x.

 

Lemma 2.1 - The Natural Logarithm Function Isn’t A Rational Function

 

The natural logarithm function ln x isn’t a rational function.

 

 

Proof
Assume that ln x is a rational function, so that ln x = p(x)/q(x), where p(x) and q(x) are polynomials and q(x) isn't 0 for any
x > 0 (the function ln x is defined for all x > 0, so p(x)/q(x) must be defined for all x > 0). Cancel all the common factors of
p(x) and q(x) if any, and re-use the letters p and q, so that p(x) and q(x) have no common factors other than 1. We’ll show
that p(x) and q(x) have x as their common factor, which is a contradiction. Differentiating both sides of ln x = p(x)/q(x) we
have:

 

 

EOP

 

Transcendency Of The Natural Logarithm Function

 

We now prove that the natural logarithm function ln x is transcendental.

 

Theorem 2.1 - Transcendency Of The Natural Logarithm Function

 

The natural logarithm function ln x is a transcendental function.

 

 

Proof


 

 

 

 

The last equation shows that ln x is a rational function, which is impossible by Lemma 2.1. It follows that the coefficient of
lnn–1 x in Eq. [4] isn't identically 0.

 

We've shown that Eq. [4] with the (ln x)-degree being n – 1 satisfies the assumption that ln x is an algebraic function. This
contradicts the fact that n is the minimum. Therefore ln x must be a transcendental function.
EOP

 

Transcendency Of The General Logarithmic Function

 

Recall that for any constant b positive and different from 1, we have logb x = ( ln x)/ ln b = (1/ ln b) ln x. Of course ln b is a
non-0 constant, and so is 1/ ln b. As seen in Section 6.3.1 Theorem 3.1, for any non-0 constant c, a function f (x) is
transcendental iff c f (x)  is transcendental. Thus, as ln x is transcendental, logb x is also transcendental.

 

Corollary 2.1 - Transcendency Of General Logarithmic Function

 

The general logarithmic function logb x is a transcendental function.

 

 

It can also be proved directly that the function logb x is transcendental, as carried out in problem & solution 3.

 

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3. Transcendency Of The Hyperbolic Functions And Their Inverses

 

Since the hyperbolic functions are defined in terms of the natural exponential function which is transcendental, it's only natural
to expect that they may be transcendental too. We're going to show that they indeed are.

 

Theorem 3.1 - Transcendency Of Sine And Cosine Hyperbolic Functions

 

The sine and cosine hyperbolic functions sinh x and cosh x are transcendental functions.

 

 

Proof


 

 

Substitute these expansions into Eq. [1]. Then for each term, multiply out its pk(x) and expansion. Factor out each factor emx
that occurs twice or more. The power emx with the greatest m is enx, and the power emx with the least m is e–nx. There are
2n + 1 terms (m goes from n to 0 to – n). Eq. [1] now becomes:

 

 

 

{2.1} Section 6.3.1 Theorem 3.1.

EOP

 

Theorem 3.2 - Transcendency Of Tangent Hyperbolic Function

 

The tangent hyperbolic function tanh x is a transcendental function.

 

 

Proof
We have:

 

 

{2.2} Section 6.3.1 Problem & Solution 5.

{2.3} Section 6.3.2 Theorem 1.1.
{2.4} Section 6.3.1 Theorem 3.1.
EOP

 

Corollary 3.1 – Transcendency Of The Reciprocal Hyperbolic Functions

 

The cotangent, secant, and cosecant hyperbolic functions coth x, sech x, and csch x are transcendental functions.

 

 

This corollary is a direct consequence of Theorems 3.1 and 3.2 above and Section 6.3.2 Theorem 1.1.

 

Corollary 3.2 – Transcendency Of The Inverse Hyperbolic Functions

 

The inverse sine, cosine, tangent, cotangent, secant, and cosecant hyperbolic functions sinh–1 x, cosh–1 x,
tanh–1 x, coth–1 x, sech–1 x, and csch–1 x are transcendental functions.

 

 

This corollary is a direct consequence of Theorems 3.1 and 3.2 and Corollary 3.1 above and Section 6.3.2 Theorem 2.1.

 

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Problems & Solutions

 

1. Is the function:

 

    

 

    algebraic or transcendental? Justify your answer.

 

Solution

 

It's algebraic, because it's a polynomial.

 

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2. Is it possible that:

 

   

 

    for all x > 0? Explain, without doing any numerical calculations to compare the 2 sides.

 

Solution

 

No, because ln x is a transcendental function while the expression on the right-hand side is a finite combination of algebraic
operations applied to polynomials in x and thus is an algebraic function.

 

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3. Prove directly (ie, by using the definition of a transcendental function) that the general logarithmic function logb x is
    transcendental.

 

Solution

 

 

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4. Prove that the cosine hyperbolic function cosh x is a transcendental function. Hint: For any non-0 constant c, a function f  is
    transcendental iff the function cf  is transcendental.

 

Solution

 

 

 

Substitute these expansions into Eq. [1]. Then for each term, multiply out its pk(x) and its expansion. Factor out each factor
emx that occurs twice or more. The power emx with the greatest m is enx and the power emx with the least m is e-nx. There are
2n + 1 terms (m goes down from n to 0 to -n). Eq. [1] now becomes:

 

 

 

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5. Prove directly (ie, by using the definition of a transcendental function) that the cotangent hyperbolic function coth x is a  
    transcendental function. Hint: if f (x) is transcendental then 1 + f (x), cf (x), where c is any non-0 constant, and 1/ f (x) are
    transcendental.

 

Solution

 

We have:

 

 

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