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Calculus Of One Real Variable
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7.1 |
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1. The Exponential Functions |
Note that the exponential function y
= bx
is different from the power function y = xb. For example,
for b = 2 and x = 3, we
have xb
= 32 = 9 and bx
= 23 = 8. In the power function
xb, the base x is variable and the exponent b is constant, while in
the exponential function bx,
the base b is constant and the
exponent x is variable. Remark that bx is called an exponential
function because the variable is in the exponent. Observe that the
variable x is the exponent and the
value bx
is the
exponential.
Definition
We know since high school what the exponential bx
means if x is a rational number. A
rational number is a number that is of
or can be written in the form of the ratio or fraction m/n, where m
and n are integers and n > 0. All the integers are rational,
since every integer k can
be written as k/1. Recall that the
definition of bm/n
is:
b3.1, b3.14, b3.141, b3.1415, b3.14159, ...,
Definition 1.1 – The Exponential Function
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The exponential function bx, where the base b is a positive constant different from 1, is defined for all real x as follows:
Thus the domain of bx is all x and its range is all y > 0. |
Remark 1.1
Since b
> 0 and b–x = 1/bx for any x > 0, we have
that bx > 0 for any x, positive or 0
or negative. For example, 2–3 = 1/23 =
1/8 > 0. That is, the domain of bx is all x and its range is all y > 0.
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2. The Number e |
Differentiability
Let's differentiate the exponential function bx. Note that so
far we don't yet know what (d/dx) bx
is. And we cannot claim that
it's xbx–1! Remember, bx is not a power function. We must go back to the
definition of the derivative. Let f (x) = bx.
First let's
differentiate it at x =
0:
exists. Of course this limit is the derivative of bx
at x = 0, and thus it's the
slope of the tangent to the curve y = bx
at its
y-intercept (0, 1), as b0
= 1. It's of the indeterminate form 0/0. If it exists, its value of course
depends on b.
Introducing e
Clearly things would simplify a lot if the above limit exists
and equals 1, because then the derivative of the function is the
function itself, a property that's simple plus interesting. So, if there's a
value of b, called e, such that:
That is, bx is a differentiable
function. This is an additional interesting and important property: the
existence of e assures that
bx
is differentiable. So our task of showing that bx is differentiable
reduces to showing that e exists.
We now summarize the interesting and amazing properties of e.
Properties Of e
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If there exists a positive number called e such that:
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Conversely, let's show that (4) implies (2): if (4) is
true, then there's a number e
such that (d/dx)
ex
= ex,
then (d/dx)
ex|x=0 =
ex|x=0 =
e0
= 1, then (2) is true, so (4) implies (2). Thus properties (2) and (4) are
equivalent. As a consequence, we can use
(4) to search for our desired e,
the number such that the derivative of the function ex is the
function ex itself.
But just one note before we proceed to search for e. If f (x) is a non-exponential function, is it
necessary that the properties
(d/dx)
f (x)|x=0 = 1 and (d/dx) f (x) = f (x) are equivalent? Let's examine some
examples. Let f (x) = x +
2. Then (d/dx)
f (x)
= 1 for all x, in particular (d/dx) f (x)|x=0 = 1. For this f
(x), the properties aren't
equivalent. Another example: let f (x) =
sin
x. Then (d/dx) f (x) = cos x
and (d/dx)
f (x)|x=0 = cos 0 = 1. For this f (x),
the properties aren't equivalent either. So the
answer is no. For non-exponential functions, the properties aren't necessarily
equivalent.
A Range In which e Lies If It Exists
To get a feeling of whether or not e
may exist, let's use its limit property to determine a range in which it lies.
Let's give h a
small value fairly close to 0, say h =
0.001, and find a value of b
that makes the ratio (bh
– 1)/h close to 1. We reach for a
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Fig. 2.1 Finding value of b that makes ratio (b0.001 – 1)/0.001 close to 1. |
calculator and proceed as shown in Fig. 2.1. Clearly e most likely exists and the range in which it lies is 2.70 < e < 2.72.
Search For e
As its derivative ex > 0 for all x, the function ex is strictly
increasing, so is one-to-one, thus has an inverse function. Let L(x) be
this inverse function. Let y = L(x),
so that x = ey. We have:
Consequently, if e
exists, then there's a function L(x) that's the inverse of ex
and that's an antiderivative of 1/x.
This gives us
an idea: find an antiderivative L(x) of 1/x,
then define a number called e
that will (1) lead us to establish that L(x) is the
inverse of ex,
and (2) be the desired e,
the number we search for, ie, such that (d/dx) ex = ex.
Now let's proceed to search for e,
using the idea described above. The process won't assume that e exists; of course it can't,
because it tries to show that e
exists. Instead, after finding an antiderivative of 1/x,
it'll define a number that exists by a
theorem. That number will be our desired e,
and hence will be called e
from the beginning. Then we'll have shown that our
desired e indeed exists.
Do We Already Have An Antiderivative Of 1/x?
Before we proceed to find an antiderivative of 1/x, let's ask ourselves if we already have
a function that's an antiderivative of
1/x. Since 1/x = x–1, let's examine the integer-power
functions, as displayed in the table in Fig. 2.2. In the f
(x) column, every
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Fig. 2.2 Examining Integer-Power Functions. |
integer power is present somewhere. However in the f '(x) column, the power x–1
= 1/x is “annihilated” by 0. So
1/x is absent.
Now we accept that so far we don't yet have an antiderivative of 1/x.
Finding An Antiderivative Of 1/x
As we want to find an antiderivative of 1/x, we of course have to consider the
function 1/x. So let's consider the
function y =
1/x for x
> 0, whose graph is shown in Fig. 2.3. The reason why we use x > 0 only is that the antiderivative
of 1/x that we
want will have to be the inverse function of ex, whose range is all y > 0, which implies that the domain
of that antiderivative is
all x > 0. For any u > 0, let A be the area of the plane region bounded
by the graph of y = 1/x, the x-axis,
the vertical line x =
1, and the vertical line x = u. The region is colored in Fig. 2.3.
Obviously, as u varies (while
remaining > 0), A
also varies.
That is, A
is a function of u. Thus it's only
natural to denote it A(u). Now we want to denote the variable of
the function A
by
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Fig. 2.3 Colored area A(u) is a function of u. |
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Fig. 2.4
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Fig. 2.5 L(x) = –A(x) if 0 < x < 1. |
the letter x,
instead of u, to conform to tradition.
In Fig. 2.4 we replace the letter x by t and the letter u
by x, to get the graph
of the function y = 1/t and the function A(x).
Note that the horizontal axis is re-named as the t-axis,
and that instead of the
letter t, we can utilize any other
letter that's normally utilized for variables, like u
or v, as long as it's not the
already-used x or
y.
However it’s a common practice to utilize the letter t.
Recall that areas are positive-or-0 quantities. Evidently,
A(x)
> 0 if x > 1 or 0 < x < 1 and A(1)
= 0. Define the function L(x)
as follows:
Clearly:
A1 < A < A2,
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Fig. 2.6 (d/dx) L(x) = 1/x. |
By the Squeeze Theorem in Section 1.1.2 Theorem 5.1 applied to one-sided limits we get:
Conclusion. It
follows that, as the right- and left-hand limits are equal to 1/x, we obtain:
We've found our function: an antiderivative of 1/x is that function L(x).
Defining e
We want to define e
utilizing the function L(x) in such a way that the inverse of L(x)
will be ex.
If y = L(x), then we should
have x = ey. Let k = L(e). Then e
= ek.
So k = 1. Thus we'll define e to be the number such that L(e) =
1.
We know that L(1) = 0. As seen in Fig. 2.7, it's clear that:
Now L(x) is differentiable for all x > 0, so it's continuous for all x > 0. As L'(x) = 1/x
> 0 for all x > 0, L(x) is
increasing. So
L(1) < 1 < L(4).
Thus, by the Intermediate-Value Theorem, there exists a number, let's call it e, where 1 < e
< 4, such that
L(e) =
1. See Fig. 2.8. That is, there exists a number e,
where 1 < e < 4, such that the area
of the region bounded by y =
1/x,
the x-axis, the vertical line x = 1, and the vertical line x = e
equals 1. The region is colored in Fig. 2.8.
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Fig. 2.7 L(4) > 1. |
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Fig. 2.8 L(e) = 1. |
It's Our Desired e
We're now going to show that the e
as defined above is our desired e.
Intuitively, as evidenced in Fig. 2.8, as x
increases
from the near-right of 0 toward infinity, L(x) is strictly increasing. Formally, L'(x) = 1/x
> 0 for all x > 0, so L(x) is
a strictly
increasing function. Thus it's one-to-one, and consequently it has an inverse
function. Let E(x)
be this inverse function.
First we show that E(x) = ex. This requires the
property E(tx)
= (E(t))x for any real t and x,
which in turn requires the
property x L(t) = L(t x), which we now show.
We have (d/dt)
(x
L(t)
– L(t x)) = x/t – (1/t x)(xt x–1) = x /t – x /t = 0
(it's the
derivative with respect to t, so
t x
is a power function). Consequently x L(t) – L(t x)
= C, a constant. At t = 1 we have C
=
x L(1) – L(1x) = x L(1) – L(1)
= x(0) – 0 = 0. This yields x L(t) = L(t x).
Now let y = E(tx).
Then L( y)
= L(E(tx)) = tx
= xt = xL(E(t)) = L((E(t))x). Hence E(tx) =
y = (E(t)) x, where the last
equation
is obtained because L is
one-to-one.
Since L(e) = 1, we have E(1)
= e. Then E(x) = E(1x) = (E(1))x = ex.
We've shown that E(x)
= ex.
It follows that ex
is the
inverse function of L(x). From now on we use ex instead
of E(x).
Next let's find the derivative of ex. Let y = ex. It follows that x = L( y). Differentiating this equation
implicitly with respect to x
we get:
Indeed the limit is 1. Thus we can use the earlier finding that 2.70 < e < 2.72.
Definition 2.1 - The Number e
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There exists a number called the Euler constant and denoted e, with 2.70 < e < 2.72, such that: The area of the plane region bounded by the curve y = 1/x, the x-axis, the vertical line x = 1, and the vertical line x = e equals 1 square unit. This number e has the property:
The existence of e guarantees that for any b > 0, bx is a differentiable function. |
Remarks 2.1
1. The process doesn't assume at all that e exists. It proves, with the help of the Intermediate-Value Theorem, that e exists.
2. The number e is called the Euler constant, named
after Leonhard Euler, an 18th-century Swiss mathematician. The name
“Euler” is pronounced “oiler”.
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3. Representation Of e As A Limit |
Let's calculate the value of:
for some larger and larger values of the positive integer n. First remark that as the exponent n gets larger and larger, the
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Fig. 3.1 Approximate Values Of (1 + (1/n))n For Some |
base 1 + 1/n gets smaller and smaller towards 1. Now using our calculator we have the table in Fig. 3.1. Apparently:
For the proof of the following theorem, we use L(x), which is the
inverse of ex, and the
property x L(t) = L(t x), as discussed
earlier under the heading It’s Our Desired e.
Theorem 3.1 – e As A Limit
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Proof
= 1.
Since L is differentiable (it has derivative 1/x), it's continuous. So we get:
EOP
The following theorem generalizes the above theorem to ex.
Theorem 3.2 – ex As A Limit
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For any real number x we have:
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Proof
= x.
Since L is differentiable (it has derivative 1/x), it's continuous. So we get:
EOP
Remark 3.1
Eq. [3.1] can be obtained from Eq. [3.2] by substituting x = 1 in Eq. [3.2]. Eq. [3.1] is a special case of Eq. [3.2].
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4. Approximations Of e |
Eq. 3.1, re-produced here for convenience:
can be used to compute approximations of e, as shown in the table in Fig. 4.1,
which, for convenience, is a re-production of
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Fig 4.1 Approximate Values Of (1 + (1/n))n For Some |
the table in Fig 3.1. For example,
the value 2.718 28, obtained when n = 1
000 000, can be used as an approximate value of e
accurate to 5 decimal places. Of course the larger n
is, the better the approximation is.
We'll see in Section 15.3 Part 3 that:
Just the first 10 terms of the series already give a more accurate
approximation of e than the 1 000 000th
term of the
sequence (1 + 1/n)n
does!! It can be proved that the series yields a better approximation of e than the sequence (1 + 1/n)n
does.
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5. The Natural Exponential Function |
The function ex is a particular exponential
function: it's the exponential function of base e.
The number e is very important in
mathematics and its applications. For this reason, ex is called the natural exponential function.
Definition 5.1 – The Natural Exponential Function
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The exponential function y = ex of base e is called the natural exponential function. For the sake of simplicity, it's often called just the exponential function. In contexts where there's no ambiguity, when an exponential function is referred to without specifying the base, it's understood to be the natural exponential function. |
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6. Laws Of Exponents |
We now establish the familiar properties of the natural
exponential function, properties associated with powers ex
of a fixed
base and a variable exponent. They're the laws of exponents.
The proof of the following theorem requires 2 additional
properties of L(x),
which are L(st)
= L(s)
+ L(t)
and L(s/t) = L(s) -
L(t),
which we now prove.
For L(st) = L(s) + L(t). We have (d/dt) (L(st) - L(t)) = (d/dt) L(st) - (d/dt) L(t) = (1/st)s – 1/t =
0. So L(st)
- L(t)
= C for
all t > 0, where C is some constant. In particular, for t = 1 we get L(s . 1) - L(1)
= C, or L(s) - 0 = C,
yielding C = L(s).
Thus L(st)
- L(t)
= L(s),
or L(st)
= L(s)
+ L(t).
Recall from under the heading It’s Our Desired e the property x L(t) = L(t x). Consequently L(1/t) = L(t -1) = -1L(t) = -L(t).
For L(s/t) = L(s) - L(t). We have L(s/t) = L(s(1/t)) = L(s) + L(1/t) = L(s) - L(t).
Theorem 6.1 - Laws Of Exponents
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a. exey = ex+y.
e. e 0 = 1. |
Proof
a. L(exey) = L(ex)
+ L(ey) = x + y = L(ex+y).
So exey = ex+y,
since L
is invertible and thus one-to-one.
b. L(e –x)
= -x = -L(ex)
= L(1/ex). Consequently
e–x = 1/ex.
c. L(ex/ey) = L(ex)
- L(ey) = x - y = L(ex-y). Hence ex/ey = ex-y.
d. L(ex) y = yL(ex) = yx = xy = L(exy).
It follows that (ex)y
= exy.
e. e 0 = e1–1 = e1/e1 = 1.
EOP
Example 6.1
Simplify the expression:
EOS
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7. Differentiation |
For formality and for now that e has been shown to exist, we re-state as a theorem with proof the fact that (d/dx) ex = ex.
Theorem 7.1 – Derivative Of ex
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The derivative of the natural exponential function ex is ex itself:
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Proof
Let y = ex. So x = L( y). Differentiating this equation
implicitly with respect to x we
get:
EOP
Remarks 7.1
1. Again note
that (d/dx)
ex
is not xex–1!! We have (d/dx) xt = txt–1, while (d/dx) ex
= ex.
The power function xt has the base
as the variable and the exponent as
the constant, while the exponential function ex has the base as the
constant and the
exponent as the variable.
2. Let y = ex. Then y '
= ex,
y '' = ex, y ''' = ex, y (4) = ex, y (5)
= ex,
etc. Since the (first) derivative of ex is ex
itself, the derivative
of any order of ex is ex
itself.
Example 7.1
Differentiate y = esin 3x.
Solution
dy/dx = esin 3x (d/dx) sin 3x = esin 3x (cos 3x)(d/dx) 3x = esin 3x (cos 3x)(3) = 3esin 3x cos 3x.
EOS
Example 7.2
Find the derivative of any order of eax+b as follows:
a. Find its
first three derivatives.
b. Guess a formula for its nth
derivative, where n is any positive
integer.
c. Prove your guess by using
mathematical induction.
Solution
Thus the formula is also true for n + 1. Consequently, by mathematical induction, it's true for all positive integers.
EOS
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8. Graph |
Inflection Points. Since there's no change of concavity, there's no inflection point.
We use a calculator to calculate the values of y = ex for some values of x, as displayed in Fig. 8.1. To find e in a calculator,
press the button e, or calculate e1,
or calculate (inverse of ln)(1), then store it in memory for use. We plot the
points from
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Fig. 8.1 Table Of Values For y = ex. |
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Fig. 8.2 Graphs Of y =
ex
And Its Tangent y = x + 1 At y-Intercept
(0, 1). The Slope |
Fig. 8.1 in Fig. 8.2, and join them, according to the properties
discussed above, by a smooth curve to obtain the graph of y
=
ex.
The slope of the tangent to the graph of y = ex at the y-intercept (0, 1) is y ’(0) = ex|x=0 = e0 = 1. So the equation of this
tangent is y = x
+ 1. The tangent is also sketched in Fig. 8.2.
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9. Fastness |
Refer to Fig. 9.1. When x is negative
and increases from near minus infinity to 0, ex increases only from
the near above of 0 to
1. But when x is positive and increases, ex
increases very fast. In fact, it increases faster and faster, as evidenced by
the fact
that its graph becomes steeper and steeper.
The limit:
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Fig. 9.1 Graphs Of y = ex. |
that this limit is infinity, meaning that ex approaches being infinitely larger than xn as x approaches infinity.
Let's compare the sizes of ex
and x1,000. Surely at x = 2 we have e2 < 21,000, and at x = 3 we still have e3 < 31,000. Now
consider
x = 10,000. Of course e10,000
has 10,000 factors e’s
while 10,0001,000 has only
1,000 factors 10,000’s. So, although e
< 10,000
by a large margin, it’s possible that e10,000 > 10,0001,000. With the help of our calculator we
get:
The proof of the following theorem makes use of this property of L(x):
Next we show that L(x) < xt/t
for all x > 0 and for any t > 0. We have, by the above
paragraph, L(xt) < xt.
Therefore, by the
discussion under the heading It’s Our Desired e, tL(x)
< xt.
So, since t > 0, we get L(x)
< xt/t.
Now we prove the above limit. By the above paragraph we have:
Theorem 9.1 - Fastness Of Growth Of ex
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For any positive integer n:
So, given any positive-integer power of x, when x
is positive and gets larger and larger, from some point on ex
grows faster |
Proof
EOP
Remarks 9.1
a. Again the
limit [10.1] means that although both ex and xn approach infinity as x does, ex approaches being infinitely larger
than xn.
b. For a given integer n
> 0, no matter how large it is, ex
may be smaller than xn
only for x in a bounded range, say for
x in
(1, b),
where b > 1 is finite. As x grows larger and larger away from b, ex
grows larger and larger than xn
does.
c. This “ fastness-of-growth” property of ex
has led to the popular “exponential” phrases such as “exponential growth” or
“increase exponentially”.
d. Eq. [9.1] can be written as:
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Problems & Solutions |
1. Simplify
the expression:
Solution
2. Find the derivative of any order of y = xeax as follows:
a. Find its
first three derivatives.
b. Guess a formula for its nth
derivative, where n is any positive
integer.
c. Prove your guess by using
mathematical induction.
Solution
a. y' = eax + axeax,
y'' = aeax + aeax + a2xeax = 2aeax + a2xeax,
y''' = 2a2eax + a2eax + a3xeax = 3a2eax + a3xeax.
b. y (n) = nan–1eax + anxeax.
c. As y' = eax + axeax, the formula
is true for n
= 1. Assume it's true for some positive integer n, so that y (n) = nan–1eax +
anxeax. Then:
y (n+1) = (d/dx) y (n)
= naneax + aneax + an+1xeax
= (n +
1)aneax + an+1xeax
= (n +
1)a(n+1)–1eax + an+1xeax.
Thus the formula is also true for n + 1. Consequently, by mathematical induction, it's true for all positive integers.
3. Find:
a. (d/dx) (Aeax cos bx + Beax sin bx).
Solution
a. (d/dx) (Aeax cos bx + Beax sin bx) =
Aaeax cos bx – Abeax sin bx + Baeax sin bx + Bbeax cos bx
= (Aa + Bb) eax cos bx +
(Ba
– Ab)
eax sin bx.
4. The function L(x)
defined in this section is the inverse of the natural exponential function ex
and is called the natural
logarithm and denoted ln x,
where “ln”
is pronounced like the word “lawn”. So ln x
is an antiderivative of 1/x, ln e = 1,
and ln xt = t ln x
for all x > 0 and for any t. Find the slope of the curve:
Solution
Differentiating the given equation implicitly with respect to x we get:
5. Prove that y ' = ky iff y
= Cekx, where y = f (x) is a function and k
and C are constants.
Hint: To show that y '
= ky implies y = Cekx,
you can show that y/ekx is a constant function.
Note: So a changing quantity has its
rate of change proportional to the quantity itself iff the quantity is an
exponential
function.
Solution
Suppose y ' = ky. Then:
So y/ekx = C, a constant. Thus y = Cekx.
Conversely, suppose y = Cekx. Then y ' = Ckekx = k(Cekx) = ky.
Consequently, y ' = ky iff y = Cekx.
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