Transcendental Functions and Initial Value
Problems: A Different Approach to
Calculus II
Byungchul Cha
Byungchul Cha ([email protected]) teaches at
Muhlenberg College in Allentown, Pennsylvania, after
working for three years at Hendrix College. He received his
B.S. from the Korea Advanced Institute of Science and
Technology, and his Ph.D. from Johns Hopkins University.
His area of specialization is number theory. When he is not
teaching calculus, he enjoys engaging his young son in
doing other mathematics, such as counting from 1 to 10
and drawing octagons.
Imagine that all of our knowledge of elementary transcendental functions has been
somehow lost. Using the differential equations that these functions satisfy, how can
we rebuild them?
This material is taken from a Calculus II curriculum developed and used at Hendrix
College for a number of years. This article consists of several snapshots of our approach, and gives a coherent exposition of the materials on transcendental functions.
We present two theorems on initial value problems (IVP) and coupled initial value
problems (CIVP) that simply assert the existence and uniqueness of their solutions.
After this, we study various initial value problems that are satisfied by the functions
exp t, ln t, sin t, cos t, and arcsin t. Our theorems guarantee that the solutions to these
initial value problems exist and are unique. Therefore, we can define these functions as
the solutions. By using our fundamental theorems as tools, we derive important properties of these functions. We conclude with a brief description of how we integrate
these materials with other traditional calculus topics to make a coherent Calculus II
course.
The idea of using differential equations to define transcendental functions appears
in some calculus textbooks. For example, Callahan et al. [4] influenced the development of the Hendrix curriculum, and, in particular, our exposition on the exponential
function overlaps with Chapter 4 in [4] substantially. However, we include it here for
completeness.
Initial value problems
Recall that an initial value problem (IVP) consists of a differential equation and an
initial condition. In this article, we assume that all of our differential equations are of
the type
dy
= F(t, y),
dt
(1)
where both the partial derivatives of F(t, y) exist and are continuous. This helps us
avoid some technical difficulties. We now state, without proof, our fundamental theorem of initial value problems (see, for example, [3]), which will serve as the cornerstone for our exposition.
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Theorem 1 (Fundamental theorem of IVPs).
the form
dy
= F(t, y) and
dt
Every initial value problem of
y(t0 ) = y0
has a unique solution y = f (t) on an open interval containing t0 .
Intuitively, this theorem means this: At each point in the t y-plane, equation (1)
specifies the slope of the line tangent to a solution curve passing through the point,
and therefore defines a direction field (see Figure 1). When a particle is released at the
point (t0 , y0 ), the Fundamental Theorem states that the particle must travel along the
solution curve determined by this direction field, at least for a short period of time. One
consequence of the theorem is that there cannot be two solution curves on the direction
field that meet one another. If two such curves did exist, the intersection point could
be taken as an initial condition, and this would define an IVP possessing two different
solutions, contradicting the theorem.
y
y = f (t)
(t 0 ; y 0 )
t
Figure 1. Fundamental theorem of IVP.
The exponential function. Consider the initial value problem
dy
=y
dt
with
y(0) = 1.
(2)
Theorem 1 asserts there is a unique solution, which we define to be y = exp t. (Later,
we define other transcendental functions this way, and each defining IVP will appear
in a box as above.) Starting from this definition, we prove the exponential laws: for all
real numbers x and y,
•
•
•
•
exp(x + y) = exp x exp y
exp(−x) = 1/ exp x
exp(x − y) = exp x/ exp y
exp(x y) = (exp x) y .
To prove these, we use the following strategy. We construct an appropriate IVP and
show that each side of an equality is a solution to this IVP. Then, it follows from the
uniqueness in Theorem 1 that the equality holds. We will call this an IVP-style proof.
To prove exp(x + y) = exp x exp y, we take f 1 (t) := exp(t + a) and f 2 (t) :=
exp t exp a (treating a as a constant). Because we have defined exp t as the solution to
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(2), we can differentiate f 1 (t) directly to get
d
d
( f 1 (t)) =
(exp(t + a)) = exp(t + a) = f 1 (t),
dt
dt
using the chain rule. In addition, we see that f 1 (0) = exp a. This says that f 1 (t) is a
solution to the IVP
dy
=y
dt
with
y(0) = exp a.
(3)
Also, we verify by direct computation that
d
( f 2 (t)) = exp t exp a = f 2 (t) and
dt
f 2 (0) = exp 0 exp a = exp a,
using (2) again. So, f 2 (t) is also a solution to (3). By uniqueness, we conclude that
f 1 (t) = f 2 (t). So, exp(x + y) = exp x exp y for all x and y.
The other exponential laws can be shown to hold by similar arguments. For example, we let f 1 (t) := exp(at) and f 2 (t) := (exp t)a , and show that both f 1 (t) and
f 2 (t) are solutions to the IVP
dy
= ay
dt
with
y(0) = 1.
Again, by uniqueness, f 1 (t) = f 2 (t).
Note that, in differentiating this f 2 (t), we have used the power rule (x a ) = ax a−1
for any real number a. For a nonnegative integer a, the power rule can be easily established using the binomial theorem, and implicit differentiation extends it for rational a.
Usually, it is not until an advanced calculus course that a rigorous definition of x a for
an irrational exponent a is given, so that the expression x a , for a fixed x > 0, gives a
continuous function of a. If we accept this fact, then our proof for a rational exponent
a is sufficient to imply exp(at) = (exp t)a for all real a because both sides represent,
for a fixed t, continuous functions of a that agree on all rational numbers. (See an interesting exposition by Anselone and Lee [1] regarding diffentiability of exponential
functions.) Alternatively, one can take the equation (exp t)a := exp(at) as a definition
of x a for any real a for a fixed x = exp t. This is possible because the range of exp t can
be shown to be all positive reals (see below). Our earlier proof for rational a then says
that this definition is consistent with the conventional definition of x a for rational a.
As a corollary of the exponential laws, we argue that the range of y = exp t is all
positive reals. Observe that the line y = 0 is a solution to the differential equation
dy/dt = y. Therefore, the graph of y = exp t must always stay above the t-axis since
otherwise it would intersect y = 0. Hence (exp t) = exp t > 0 for all t. Thus y =
exp t is a strictly increasing function. Take > 0. Then exp > 1 because exp 0 = 1.
Now, we use the fourth exponential law to obtain exp(n · ) = (exp )n , which becomes arbitrarily large as n increases. From the intermediate value theorem, we conclude that the range of exp t contains (1, ∞). This, in turn, implies that the range
contains the interval (0, 1) as well, thanks to the second exponential law exp(−x) =
1/ exp x. This completes the proof that the range of exp t is the set of all positive real
numbers.
There is an issue that we have overlooked in our discussion of y = exp t. Theorem 1
guarantees the existence of a solution to (2), but only locally. In particular, exp t has not
been shown to be well-defined on the entire t-line yet. For example, a solution of the
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IVP dy/dt = y 2 and y(0) = 1 is y = 1/(1 − t), which is undefined at t = 1. In fact,
all solutions of dy/dt = y 2 blow up at some value of t; that is, they possess vertical
asymptotes (see Figure 2). After taking a calculus course based on the material in this
article, one student, Jared Williams, asked when in general solutions of differential
equations blow up, and eventually, this question resulted in an undergraduate research
paper [5]. In particular, his theorem [5] implies that y = exp t does not blow up, and
is well-defined for all real t.
y
t
Figure 2. Direction field of y = y 2 and its solutions.
The logarithmic function. Before we define the logarithmic function, we study
how two IVPs are related when their solutions are inverses of each other. Suppose that
y1 (t) and y2 (t) satisfy
⎧
⎧
⎨ dy2 = F (t, y ),
⎨ dy1 = F (t, y ),
1
1
2
2
dt
dt
and
(4)
⎩
⎩
y1 (a) = b
y2 (b) = a.
Further, assume that they are inverses of each other. A standard result from implicit
differentiation (see, for example, [6, p. 235]) is that, if f is a one-to-one differentiable
function whose inverse is also differentiable, then
1
( f −1 ) (x) =
f
(
f
−1 (x))
.
Therefore, the expressions F1 and F2 are related by
F2 (t, y) =
1
.
F1 (y, t)
(5)
In words, we can obtain F2 from F1 by interchanging t and y, and flipping.
Now we define the logarithmic function. Since we expect ln t to be the inverse of
exp t, equation (5) suggests that we define y = ln t as a solution to the IVP
dy
1
=
dt
t
with
y(1) = 0.
(6)
du
u
(7)
Its integral version
t
ln t =
1
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is nothing but a standard definition of ln t used in many calculus textbooks (see, for
example, [6]). It is immediate from (7) that the domain of ln t is (0, ∞) because the
integrand 1/u is continuous on this interval. In particular, we see that the domain of
ln t is the same as the range of exp t.
Here is an IVP-style proof of the fact that exp t and ln t are indeed inverse functions
of each other. Define f 1 (t) := (exp ◦ ln)(t) and f 2 (t) := t, for all t > 0. Note that f 1 (t)
is defined for t > 0, because ln t is defined for t > 0. We differentiate f 1 (t) directly
using (2), (6), and the chain rule, and obtain
exp(ln t)
f 1 (t)
d f 1 (t)
=
=
.
dt
t
t
Further, f 1 (1) = 1 from (6) and (2). This shows that f 1 (t) is a solution to the IVP
dy/dt = y/t with y(1) = 1. On the other hand, it is easy to verify that f 2 (t) is a solution to the same IVP. Therefore, f 1 (t) = f 2 (t). That is, (exp ◦ ln)(t) = t for t > 0.
Similarly, define f 3 (t) := (ln ◦ exp)(t) and f 4 (t) := t for all t. Since the range of exp t
is the positive reals, f 3 (t) is defined for all t. Further, we can show by direct computation that both f 3 (t) and f 4 (t) are solutions to the IVP dy/dt = 1 and y(0) = 0. For
example, by (2), (6), and the chain rule, we get
1
d f 3 (t)
=
· exp t = 1.
dt
exp t
The remaining computations are straightforward, with the result that exp t and ln t are
inverses.
Coupled initial value problems
A coupled initial value problem (CIVP) consists of a pair of initial value problems
⎧
dx
⎪
⎪
= F1 (t, x, y),
⎨
x(t0 ) = x0 ,
dt
(8)
and
⎪
dy
y(t0 ) = y0 .
⎪
⎩
= F2 (t, x, y)
dt
As we did with (1), we assume that all of the partial derivatives of F1 (t, x, y) and
F2 (t, x, y) exist and are continuous in order to avoid some technical difficulties. We
state a theorem similar to Theorem 1 that asserts the existence and uniqueness of a pair
of solutions under similar conditions [2, p. 205].
Theorem 2 (Fundamental theorem of CIVPs). Every coupled initial value
problem of the form (8) has a unique pair of solutions x = x(t) and y = y(t) on an
open interval containing t0 .
Sine and cosine functions. We define x(t) := cos t and y(t) := sin t to be the
unique solutions to the CIVP
⎧
dx
⎪
⎪
= −y,
⎨
dt
⎪
dy
⎪
⎩
=x
dt
292
and
x(0) = 1,
y(0) = 0.
(9)
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With this definition, there is an IVP-style proof for the fundamental identity
sin2 t + cos2 t = 1.
(10)
Let f (t) := cos2 t + sin2 t. Then it is not difficult to check that f (t) = 0 and f (0) = 1
by (9) and the chain rule. Hence, Y = f (t) is a solution to the IVP
dY
=0
dt
and
Y (0) = 1.
On the other hand, the constant function Y = 1 is also a solution to this IVP. This
proves (10) by the uniqueness statement of Theorem 1.
Next, we sketch a proof of the addition formulas
sin(t + a) = sin t cos a + cos t sin a
(11)
cos(t + a) = cos t cos a − sin t sin a
using what we will call a CIVP-style proof. Define the pair of functions f 1 (t) :=
sin(t + a) and f 2 (t) := cos(t + a), and another pair f 3 (t) := sin t cos a + cos t sin a
and f 4 (t) := cos t cos a − sin t sin a. Then, a direct computation using (9) and the
chain rule shows that both pairs are solutions to the CIVP
⎧
dy1
⎪
⎪
= y2 ,
⎨
y1 (0) = sin a,
dt
and
⎪
y2 (0) = cos a.
dy
⎪
⎩ 2 = −y1
dt
Therefore, the two pairs are the same by the uniqueness statement of Theorem 2. The
addition formulas (11) are thus proved.
Next, we consider a direction field which is determined by the differential equations
(9) in t x y-space (see Figure 3(a)). The solution curve starts at the point (t0 , x0 , y0 ) =
(0, 1, 0). If we project this direction field onto the x y-plane, we obtain a new direction
field (Figure 3(b)) on the x y-plane associated with dy/d x. We leave it to readers to
y
y
1
1
(cos t, sin t)
θ= t
1
t
(a)
x
x
1
(b)
Figure 3. Direction field of sine and cosine in the t x y-space and its projection onto the x yplane.
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check, using the chain rule and (9), that
x
dy
=− .
dx
y
Also, the projection of the solution curve lies on the unit circle, as evidenced by the
identity (10).
Recall that the angle θ in polar coordinates of the point (x, y) is given by θ =
tan−1 (y/x). A little calculation shows that dθ/dt = 1. Further,
θ(0) = tan
−1
y(0)
x(0)
= 0.
So, we get θ(t) = t. In other words, a particle released at the point (1, 0) in this direction field will move along the circle at a constant angular speed 1 radian per unit time.
This links our CIVP approach with the usual definitions of the sine and cosine.
Arcsine function The last function that we consider here is the arcsine function
y = arcsin t. It is defined to be a solution to the IVP
1
dy
=√
dt
1 − t2
with
y(0) = 0.
(12)
This IVP is motivated by the fact that y = sin t satisfies the initial condition y(0) = 0
and the differential equation
dy
= 1 − y2,
dt
(13)
(swap t and y, and flip!) on a small interval containing t = 0, over which y = sin t is
one-to-one. Since y (0) > 0, such an interval does exist, and (13) follows from (9) and
(10).
This IVP definition is, by part one of the fundamental theorem of calculus [6,
p. 396], equivalent to the integral version
t
arcsin t =
0
du
.
√
1 − u2
(14)
√
Since the integrand 1/ 1 − u 2 is a continuous function of u on (−1, 1), the domain
of arcsin t contains the open interval (−1, 1). However, we can show that both
lim
t→1−
0
t
du
√
1 − u2
and
t
lim
t→−1+
0
du
√
1 − u2
exist, so arcsin t can be continuously extended to the closed interval [−1, 1]. We prove
the convergence of the improper integrals as follows. It is not difficult to see that
d 1
,
−u 1 − u 2 = −2 1 − u 2 + √
du
1 − u2
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which, in turn, implies that
du
= −u 1 −
√
1 − u2
u2
+2
1 − u 2 du.
From this, we deduce that
arcsin(1) = lim
t→1−
0
t
du
=2
√
1 − u2
1
1 − u 2 du =
0
π
,
2
(15)
1√
since 0 1 − u 2 du is the area of a quarter circle of radius 1. The computation of
arcsin(−1) = −π/2 is similar.
Equation (15) has many important consequences. We briefly describe some of them
here as exercises for the readers.
(a) Show that the function arcsin t gives a one-to-one correspondence from [−1, 1]
to [−π/2, π/2]. Hint: (The equation (14) says arcsin t has a positive derivative,
from which the injectivity of arcsin t follows. For surjectivity, use the intermediate value theorem.)
(b) Prove that sin(arcsin t) = t for all t ∈ [−1, 1]. Together with (a), this implies
that sin t and arcsin t are inverses of each other. (Hint: instead of proving just
this, show even more: for all t ∈ [−1, 1],
sin(arcsin t) = t,
and
cos(arcsin t) =
1 − t 2.
This can be done by a CIVP-style proof with this CIVP:
⎧ dy
y2
1
⎪
⎪
⎨ dt = √1 − t 2 ,
⎪
dy
y
⎪
⎩ 2 = −√ 1
dt
1 − t2
and
y1 (0) = 0,
y2 (0) = 1.)
(c) Define arccos t := π/2 − arcsin t. Then deduce results similar to those in (b)
for cos t.
(d) From (15) and (b), we get sin(π/2) = 1. Combine this with (10) and (11) to
recover many other properties of the sine and cosine functions, such as periodicity, odd/evenness, and so on.
IVPs as teaching tools for calculus II
The materials we outlined in this article have been successfully used in the second
calculus course at Hendrix College for many years. We mix these with other standard
materials, including the following:
•
•
•
Qualitative, numerical, and analytic tools—graphical analysis, Euler’s method, and
separation of variables ([6, §9.1–§9.5])—for studying differential equations.
Integration techniques, such as partial fractions and trigonometric substitutions.
Sequences and series.
VOL. 38, NO. 4, SEPTEMBER 2007 THE COLLEGE MATHEMATICS JOURNAL
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The emphasis on differential equations provides meaningful contexts for various
integration techniques. Also, the IVP- and CIVP-style proofs require students to use
many familiar ideas from Calculus I—the chain rule, the fundamental theorem of calculus, and so on—in an unfamiliar setting. Although not mentioned in this article, our
approach to IVPs lends itself nicely to an overview on how to obtain power series
expansions of transcendental functions via Picard iteration.
Acknowledgments. Many of my colleagues at Hendrix College contributed to this approach for Calculus II,
including Bob Eslinger, Dwayne Collins, David Sutherland, and Duff Campbell. This article has been greatly
improved by suggestions made by Carl Burch, an anonymous referee, and the editor.
References
1. P. Anselone and J. Lee, Differentiability of Exponential Functions, College Mathematics Journal 36 (2005)
388–393.
2. P. Blanchard, R. Devaney, and G. Hall, Differential Equations, 3rd ed., Brooks/Cole, 2005.
3. W. Boyce and R. DiPrima, Elementary Differential Equations, 8th ed., Wiley, 2004.
4. J. Callahan, D. Cox, K. Hoffman, D. O’Shea, H. Pollatsek, and L. Senechal, Calculus in Context, Freeman,
1995.
5. D. Campbell and J. Williams, Exploring Finite-Time Blow-Up, PME Journal 11 (2003) 423–428.
6. J. Stewart, Calculus, Early Transcendentals, 5th ed., Brooks/Cole, 2003.
Proof Without Words: A Graph Theoretic Summation of the First n
Integers
n
i=
i=1
n+1
2
For example, n = 5:
1
6
2
5
3
4
1
1
1
1
1
6
2
6
2
6
2
6
2
6
2
5
3
5
3
5
3
5
3
5
3
4
4
4
4
4
For a related result, see Mathematics Magazine 80 (2007) 182.
Joe DeMaio
Joey Tyson
Kennesaw State University
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