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Unformatted text preview: gral, thus,
"✓ ◆
#Z
✓ ◆9
⇣ n ⌘9
1
1
19
2
x10
lim
+
+ ··· +
=
x9 dx =
n!1 n
n
n
n
10
0
10. (a) Set up a Newton’s method algorithm to approximate
(b) Now, starting with x1 = p
5 Z 1 x9 dx, 0 1 = 1/10
0 2. 1, ﬁnd x2 , the second iteration approximation of p
5 2. Solution: p
(a) Let x = 5 2, then this implies x5 +2 = 0, and we wish to ﬁnd x, that is, we wish to use Newton’s
Method to approximate a root of the equation x5 + 2 = 0, so let f (x) = x5 + 2 then f 0 (x) = 5x4 and
so
f ( xn )
x5 + 2
5x5 (x5 + 2)
4x 5 2
n
n
xn+1 = xn
= xn
=n
= n4
f 0 ( xn )
5x 4
5x4
5x n
n
n
so the resulting algorithm is xn+1 =
(b) Starting with x1 = 4x5 2
n
.
5x4
n 1 we have x2 = 4x 5 2
42
1
=
=
4
5( 1)4
5x 1 6
5 11. Find a positive number such that the sum of the number and its reciprocal is as small as possible.
Solution:
We wish to ﬁnd a number x > 0 such that the sum S = x + 1/x is minimized. Note that
x2 1
S 0 (x) = 1 1/x2 =
and so the critical points are x = ±1. Note...
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This note was uploaded on 02/04/2014 for the course APPM 1350 taught by Professor Segur,harv during the Fall '07 term at Colorado.
 Fall '07
 SEGUR,HARV

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