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Chapter 1
Plugging these values into the recursive equations above yields
t
2
=0
.
25 + 0
.
25 = 0
.
5a
n
d
x
2
.
25 + 0
.
25
·
[1 + 0
.
25
·
sin(0
.
0625)] = 0
.
503904
.
Continuing in this way gives
t
3
.
75
and
x
3
.
503904 + 0
.
25
·
[1 + 0
.
5
·
sin(0
.
251952)] = 0
.
785066
,
t
4
=1
.
00
and
φ
(1)
≈
x
4
.
13920
.
For
N
=8,wehave
h
/
8=0
.
125 . Thus, the recursive formulas become
t
n
+1
=
t
n
+0
.
125
and
x
n
+1
=
x
n
.
125
·
[1 +
t
n
sin(
t
n
x
n
)]
.
Using these formulas and starting with
t
0
=0and
x
0
= 0, we can ±ll in Table 1A. From this
we see that
φ
(1)
≈
x
8
.
19157, which is rounded to ±ve decimal places.
Table 1–A
: Euler’s method approximations for the solution of
x
0
=1+
t
sin(
tx
),
x
(0) = 0,
at
t
= 1 with 8 steps (
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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