Exercises 3.6
In order to approximate the solution
φ
(
x
)=
e
x
at the point
x
=1w
i
th
N
steps, we take
h
=(
x
−
x
0
)
/N
=1
/N
,andso
N
=1
/h
. Then the above formula becomes
y
N
=
±
1+
h/
2
1
−
h/
2
²
N
y
0
=
±
1+
h/
2
1
−
h/
2
²
N
=
±
1+
h/
2
1
−
h/
2
²
1
/h
and hence
e
=
φ
(1)
≈
y
N
=
±
1+
h/
2
1
−
h/
2
²
1
/h
.
Substituting
h
=10
−
k
,
k
=0
,
1
,
2
,
3, and 4, we Fll in Table 3A.
Table 3–A
: Approximations
±
1+
h/
2
1+
h/
2
²
1
/h
to
e
≈
2
.
718281828
...
.
h
Approximation
Error
1
3
0.281718172
10
−
1
2.720551414
0.002269586
10
−
2
2.718304481
0.000022653
10
−
3
2.718282055
0.000000227
10
−
4
2.718281831
0.000000003
These approximations are better then those in Tables 3.4 and 3.5 of the text.
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details