Exercises 1.4
For approximation of
φ
(
t
)a
tth
epo
in
t
t
=1w
i
th
N
= 20 steps, we take the step size
h
=(1
−
t
0
)
/
20 = 0
.
05. Thus, the recursive formulas for Euler’s method are
t
n
+1
=
t
n
+0
.
05
,
x
n
+1
=
x
n
+0
.
05
(
1+
x
2
n
)
.
Applying these formulas with
n
=0
,
1
,...,
19, we obtain
x
1
=
x
0
+0
.
05
(
1+
x
2
0
)
=0
.
05
,
x
2
=
x
1
+0
.
05
(
1+
x
2
1
)
=0
.
05 + 0
.
05
(
1+0
.
05
2
)
=0
.
100125
,
x
3
=
x
2
+0
.
05
(
1+
x
2
2
)
=0
.
100125 + 0
.
05
(
1+0
.
100125
2
)
≈
0
.
150626
,
.
.
.
x
19
=
x
18
+0
.
05
(
1+
x
2
18
)
≈
1
.
328148
,
φ
(1)
≈
x
20
=
x
19
+0
.
05
(
1+
x
2
19
)
=1
.
328148 + 0
.
05
(
1+1
.
328148
2
)
≈
1
.
466347
,
which is a good enough approximation to
φ
(1) = tan 1
≈
1
.
557408.
13.
From Problem 12,
y
n
=(1+1
/n
)
n
and so lim
n
→∞
[(
e
−
y
n
)
/
(1
/n
)] is a 0
/
0 indeterminant. If
we let
h
=1
/n
in
y
n
and use L’Hospital’s rule, we get
lim
n
→∞
e

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