View
7
Download
0
Category
Preview:
Citation preview
Natural Logarithm as an IntegralMATH 161 Calculus I
J. Robert Buchanan
Department of Mathematics
Summer 2019
Background
I Previously we had worked with functions ln x and ex
numerically and hypothesized the general properties ofthese two functions.
I Now that we have encountered the Fundamental Theoremof Calculus we can treat these two functions rigorously.
I We will see that the rigorous treatment will lead to thesame properties we had hypothesized earlier.
Definition of ln x
DefinitionFor x > 0 we define the natural logarithm function as
ln x =
∫ x
1
1t
dt .
Note:
ddx
[ln x ] =ddx
[∫ x
1
1t
dt]
=1x
by the FTC, Part II.
Definition of ln x
DefinitionFor x > 0 we define the natural logarithm function as
ln x =
∫ x
1
1t
dt .
Note:
ddx
[ln x ] =ddx
[∫ x
1
1t
dt]
=1x
by the FTC, Part II.
Properties of the Natural Logarithm
TheoremFor any a, b > 0 and any rational number r ,
1. ln 1 = 02. ln(a b) = ln a + ln b
3. ln(a
b
)= ln a− ln b
4. ln(ar ) = r ln a.
Proof (1 of 2)
1.
ln 1 =
∫ 1
1
1t
dt = 0
2.
ln(ab) =
∫ ab
1
1t
dt =
∫ a
1
1t
dt +
∫ ab
a
1t
dt
= ln a +
∫ ab
a
1t
dt
Let a u = t then a du = dt .
ln(ab) = ln a +
∫ b
1
1u
du = ln a + ln b
Proof (2 of 2)
4.
ddx
[ln(x r )] =1x r
ddx
[x r ] =1x r
[rx r−1
]=
rx
ddx
[r ln x ] =rx
The two functions have the same derivative and thus bythe MVT
ln(x r ) = r ln x + Cln(1r ) = r ln 1 + C
ln 1 = r ln 1 + C0 = r(0) + C0 = C
which implies ln(x r ) = r ln x .
Approximating Values of the Natural LogarithmUse a Riemann sum to approximate ln 3 =
∫ 3
1
1t
dt .
Let n = 100 and use midpoint evaluation points.
∆t =b − a
n=
3− 1100
=1
50
ti = a +
(i − 1
2
)∆t = 1 +
(i − 1
2
)1
50
Therefore
ln 3 =
∫ 3
1
1t
dt ≈n∑
i=1
f (ti)∆t
=100∑i=1
[1
1 +(i − 1
2
) 150
]150
≈ 1.0986.
Approximating Values of the Natural LogarithmUse a Riemann sum to approximate ln 3 =
∫ 3
1
1t
dt .
Let n = 100 and use midpoint evaluation points.
∆t =b − a
n=
3− 1100
=1
50
ti = a +
(i − 1
2
)∆t = 1 +
(i − 1
2
)1
50
Therefore
ln 3 =
∫ 3
1
1t
dt ≈n∑
i=1
f (ti)∆t
=100∑i=1
[1
1 +(i − 1
2
) 150
]150
≈ 1.0986.
The Transcendental Number e
DefinitionWe define e to be the real number such that ln e = 1.
Use Newton’s method to approximate e by solving the equation
ln x − 1 = 0.
The Transcendental Number e
DefinitionWe define e to be the real number such that ln e = 1.
Use Newton’s method to approximate e by solving the equation
ln x − 1 = 0.
Approximation
Let f (x) = ln x − 1 and x0 = 2 and use the Newton’s Methodformula:
xn = xn−1 −f (xn−1)
f ′(xn−1)= xn−1 −
ln xn−1 − 11
xn−1
.
Then
n xn0 2.01 2.613712 2.716243 2.718284 2.71828
Approximation
Let f (x) = ln x − 1 and x0 = 2 and use the Newton’s Methodformula:
xn = xn−1 −f (xn−1)
f ′(xn−1)= xn−1 −
ln xn−1 − 11
xn−1
.
Then
n xn0 2.01 2.613712 2.716243 2.718284 2.71828
The Exponential Function
Sinceddx
[ln x ] > 0 then the natural logarithm function isinvertible on (0,∞).
DefinitionFor any real number x we define y = ex to be the number forwhich
ln y = ln(ex ) = x .
Thus we have
ln(ex ) = x for −∞ < x <∞,eln x = x for x > 0.
The Exponential Function
Sinceddx
[ln x ] > 0 then the natural logarithm function isinvertible on (0,∞).
DefinitionFor any real number x we define y = ex to be the number forwhich
ln y = ln(ex ) = x .
Thus we have
ln(ex ) = x for −∞ < x <∞,eln x = x for x > 0.
Properties of the Exponential Function
TheoremFor any real numbers r and s and any rational number t,
1. er es = er+s
2.er
es = er−s
3. (er )t = ert
Proof
ln(er es) = ln(er ) + ln(es)
= r ln e + s ln e= (r + s) ln e= ln(er+s)
Since ln x is a one-to-one function, then er es = er+s.
Derivative of the Exponential Function
We have defined y = ex if and only if ln y = x .
ddx
[ln y ] =ddx
[x ]
1y
dydx
= 1
dydx
= y
ddx
[ex ] = ex
General Exponential Functions
If y = ax where a > 0 and a 6= 1 then
ln y = ln(ax )
= x ln aeln y = ex ln a
y = ax = ex ln a.
Thus
ddx
[ax ] = (ln a)ax∫ax dx =
ax
ln a+ C
General Logarithmic Functions
If a > 0 and a 6= 1 then x = ay if and only if loga x = y .
Question: how can we work with logarithms with generalbases?
ln x = ln(ay )
ln x = y ln aln xln a
= y
ln xln a
= loga x
General Logarithmic Functions
If a > 0 and a 6= 1 then x = ay if and only if loga x = y .
Question: how can we work with logarithms with generalbases?
ln x = ln(ay )
ln x = y ln aln xln a
= y
ln xln a
= loga x
General Logarithmic Functions
If a > 0 and a 6= 1 then x = ay if and only if loga x = y .
Question: how can we work with logarithms with generalbases?
ln x = ln(ay )
ln x = y ln aln xln a
= y
ln xln a
= loga x
Derivative of General Logarithmic Functions
ddx
[loga x ] =ddx
[ln xln a
]=
1ln a
ddx
[ln x ]
=1
ln a
[1x
]=
1x ln a
Homework
I Read Section 4.8I Exercises: 1–39 odd
Recommended