**Area under curve 1/x ->
logs, and the area under 1/(1+x) -> logs and infinite series/****C****hapter
13**

Don graphed **
1/x** in *Mathematica *and found the following:

**The area under the curve 1/x from 1
to 3 is equal to the log _{e} 3 =~ 1.0986**

**The area under the curve 1/x from 3
to 6 is equal to the log _{e} 2 =~ 0.6931**

**The area under the curve 1/x from 1
to 6 is equal to the log _{e} 6 =~ 1.7917**

**The derivative of Log _{e}x
= 1/x -use the pencil approach, or **

**for a proof, see**

** http://www.5min.com/Video/Learn-about-Proof-ddx--ln-x--1x-99173876**

**and the
integral of 1/x = Log _{e}x**

You might try other areas, like A** _{1-4
}1/x + **A

Remember Ian's discovery in Ch. 3 **1/(1-x) **= **(1-x)**^{-}^{1}
= 1+ x + x^{2}** ^{
}**+ x

If we substitute ** ^{-}**x
in for x, we get

**Both of these series above converge
for ^{-}1<x<1**

Don graphed 1/(x+1) in *Mathematica *and found the following:

The graph of
y =
_{e}(1
+ a) and if we use the generalizations from the section above, this will
equal the integral of the infinite series 1
- a + a^{2} -
a^{3} +
a^{4} -
a^{5} +
... Taking the integral of each term separately, we get
A_{0} 1/(1+x)
= log_{-a}_{e}(1
+ a) = a
- a^{2}/2
+ a^{3}/3
- a^{4}/4
+ a^{5}/5
- a^{6}/6
+'
The natural log then, is the area under
the curve y=
1/x or y=
1/(1+x)
and it is the infinite series
you can't find the the loglog _{e}(1
+ a) = a
- a^{2}/2
+ a^{3}/3
- a^{4}/4
+ a^{5}/5
- a^{6}/6
+'
CAUTION: _{e}(1
+ 2) by |

putting 2 in for a in the series above because this series converges only
when a is between ^{-}1
and 1. _{e} 1**.**1
by putting **.**1 in for a in the series log_{e}(1
+ **.**1) = **.**1
- (**.**1)^{2}/2
+ (**.**1)** ^{3}**/3
- (

2 = 1.2*1.2/(0.8*0.9). Who else would think of writing 2 this way! Newton
then proceeded to get the log_{e} 2
by using the log identities **log(A*B) =
Log A + Log B** and **Log(C/D) = Log C - Log D**:

(check these identities that Kavi and Kevin found at logarithms )

then the log_{e}
2 = log_{e}(1
+ 0**.**2) + log_{e}(1
+ 0**.**2) - ((log_{e}(1
+ ** ^{-}**0

**Write a program **which will get this infinite series

**log _{e}(1
+ a) = a
- a^{2}/2
+ a^{3}/3
- a^{4}/4
+ a^{5}/5
- a^{6}/6
+'**

See
if you can find log_{e}
3 and log_{e}
4 and others using Newton's method and check these on a calculator.