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Sunday, September 21, 2014

1st-Year-Maths-Notes CH #04

 
Logarithm and exponential functions
4.Logarithm
We have seen that, for
a
=
1,
 
x
a
dx
=
x
a
+1
a
+ 1
a
=
1
,
and we still have to define this integral for
a
=
1. For this, we introducthe logarithmasln
x
=
 
x
1
duu,
so that the logarithm gives the surface area between the function 1
/u
and the horizontalaxis, from 1 to
>
0. The real logarithm is not defined for
<
0, sincthe correspondingsurface area would be infinite. The number
e
is defined by ln
e
= 1 and
e
2
.
718281828.
Properties
:
We have seen that the integrals
 
1
dx/x
and
 
10
dx/x
both diverge, such thatlim
x
→∞
ln
x
+
anlim
x
0
ln
x
=
−∞
From the definition of the logarithm, one can see thatln(
ab
=
 
ab
1
duu
=
 
a
1
duu
+
 
aba
duu
ln
a
+
 
b
1
dvv
ln
a
+ ln
b,
(6)where we make the change of variable
u
=
av
.
One can also see thatln(
x
a
=
 
x
a
1
duu
=
 
x
1
advv
=
a
ln
x,
(7)where we make the change of variable
u
=
v
a
.
We have in general, for any differentiable function
,
 
dx
(
x
)
(
x
)ln
|
(
x
)
|
+
c,
where
c
is a constant18
 
Logarithm in base
a
The logarithm in base
a
is defined aslog
a
(
x
=ln
x
ln
a,
and is equal to 1 when
x
=
a
. Note that ln
x
= log
e
(
x
).
Integral of the logarithm
Tcalculate
 
ln
dx
, one uses an integration by parts:
 
ln
dx
=
 
(
x
)
ln
dx
=
x
ln
x
 
x
(ln
x
)
dx
=
x
ln
x
x
+
c,
where
c
is a constant.
Limits
When 
x
+
:
We show here the important limitlim
x
+
ln
xx
a
0
>
(8)which means that any (positive-) power law goes quicker to infinity than the loga-rithm, when
x
+
.
Proof 
For any
u
1 and for any
a >
0, we have1
u
1
u
1
a/
2
.
Integrating this inequality from 1 to
x
leads to0
<
ln
x
2
a
(
x
a/
2
1)
<
2
ax
a/
2
.
Dividing by
x
a
gives the expected result:0
<
ln
xx
a
2
ax
a/
2
when
x
+
.
When 
x
0
:
Another important limit to know islim
x
0
x
a
ln
x
0
a >
0
,
(9)which means that any (positive-) power law kills the divergence of the logarithm at
x
0.
Proof 
For any
u
satisfying 0
u
1 and any
a >
0, we have1
u
1
u
1+
a/
2
.
Integrating this inequality from
x
to 1, we obtain0
<
ln
x
2
a
(
x
a/
2
1)
<
2
ax
a/
2
.
Multiplying by
x
a
gives the expected result:0
x
a
|
ln
x
|
2
ax
a/
2
when
x
0
.
19
 
4.Exponential
The exponential is defined as the inverse function of the logarithm:
y
ln
x
x
= exp
y
=
e
y
From property (6), if we note
u
ln
a
and
v
ln
b
, we haveexp(
u
+
v
) = (exp
u
)
×
(exp
v
)
,
and from property (7), if we note
y
ln
x
, we have
exp
y
a
= exp(
ay
)
.
Derivative of the exponential
one can differentiate the definition exp(ln
x
=
x
, which,using the chain rule, leads to exp
(ln
x
=
x
. We therefore conclude that the derivative of the exponential is the exponential itself:exp
x
= exp
x.
Exponential of base
a
This function is defined as
a
x
= exp(
x
ln
a
)
,
which is consistent with the properties of the logarithm and the exponential. It’s derivativeis then(
a
x
)
= (ln
a
)
a
x
.
One can also define the function
x
x
, with derivative(
x
x
)
=
ddx
exp(
x
ln
x
)
= (1 + ln
x
)
x
x
.
Limits
From the limit (8), if we note
y
ln
x
and
b
1
/a >
0, we havelim
y
+
exp
yy
b
+
,
and the exponential goes to infinity quicker than any power law.
From the limit (9), if we note
y
=
|
ln
x
|
and
b
1
/a >
0, we havelim
y
+
y
b
exp(
y
0
,
and the decreasing exponential kills the divergence of any power law.20
 
4.Hyperbolic functions
The hyperbolic functions are defined as
hyperbolic cosine: cosh
x
(
e
x
+
e
x
)
/
2;
hyperbolic sine: sinh
x
(
e
x
e
x
)
/
2;
hyperbolic tangent: tanh
x
= sinh
x/
cosh
x
;
hyperbolic cotangent: coth
x
= cosh
x/
sinh
x
,and their derivatives are given bycosh
x
sinh
x
sinh
x
cosh
x
tanh
x
1
tanh
2
x
coth
x
1
coth
2
x
It can easily be seen that, from their definition, the functions cosh and sinh satisfy, for all
x
,cosh
2
x
sinh
2
x
1
.
Also, it can be easily checked thatcosh(2
x
cosh
2
(
x
) + sinh
2
(
x
)sinh(2
x
2sinh(
x
)cosh(
x
)21

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