5
7. Our approach (based on Eq. 2229) consists of several steps. The first is to find an
approximate
value of
e
by taking differences between all the given data. The smallest
difference is between the fifth and sixth values:
18.08
×
10
–19
C – 16.48
×
10
– 19
C = 1.60
×
10
–19
C
which we denote
e
approx
. The goal at this point is to assign integers
n
using this
approximate value of
e
:
19
1
approx
19
2
approx
19
3
approx
19
4
approx
19
5
approx
6.563 10
C
datum1
4.10
4
8.204 10
C
datum2
5.13
5
11.50 10
C
datum3
7.19
7
13.13 10
C
datum4
8.21
8
16.48 10
C
datum5
10.30
10
n
e
n
e
n
e
n
e
n
e
−
−
−
−
−
×
=
¡
=
×
=
¡
=
×
=
¡
=
×
=
¡
=
×
=
¡
=
19
6
appeox
19
7
approx
19
8
approx
19
9
approx
18.08 10
C
datum6
11.30
11
19.71 10
C
datum7
12.32
12
22.89 10
C
datum8
14.31
14
26.13 10
C
datum9
16.33
16
n
e
n
e
n
e
n
e
−
−
−
−
×
=
¡
=
×
=
¡
=
×
=
¡
=
×
=
¡
=
Next, we construct a new data set (
e
1
,
e
2
,
e
3
…) by dividing the given data by the
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Fall '10 term at Kansas State University.
 Fall '10
 Rebello,NobelS

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