# Course 161/2S3, Tutorial, Hilary Term, 2006 Use Simpson’s rule to evaluate • 5x

```Course 161/2S3, Tutorial, Hilary Term, 2006
• Use Simpson’s rule to evaluate
Z
5
5x5 dx
−
with n = 4.
Solution n = 4 ⇒ h = (1 − 0)/4 = 0.25.
h
(f (a) + 4f (a + h) + 2f (a + 2h) + 4f (a + 3h) + f (b))
3
0.25 5 ∗ 05 + 4 ∗ 5 ∗ 0.255 + 2 ∗ 5 ∗ 0.55 + 4 ∗ 5 ∗ 0.755 + 5 ∗ 15
=
3
= 0.83984375
S =
The analytic result is 5/6.
What value of n is needed to solve this problem to an accuracy of 10−12
(approx machine precision).
(b − a)5 1
max(f 0000 )x=0,1 &lt; 10−12
180 N 4
1 1
53024 &lt; 10−12
180 N 4
N 4 &gt; 55.05
N = 56 will give the answer with error &lt; 10−12.
1
• Given the first-order differential equation
dy
= −y(x)
dx
and initial condition y(0) = 1. Determine y(x), x = 2 for a step size
h = 0.5 using the Euler method.
yn+1 = yn + hf (x, y)
= yn + h(−y(x))
For y0 = 1, h = 0.5
y1
y2
y3
y4
=
=
=
=
0.5
0.25
0.125
0.0625
At x = 2 we have y = 0.0625
Note the analytic solution to this equation is y(x) = e−x . See how your
answer compares to the analytic solution.
At x = 3, e−3 = 0.1353.
2
```