James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

8 ■ LIES MY CALCULATOR AND COMPUTER TOLD ME
5. For () =lnln with [ ], =10 9 ,and =10 9 +1,weneed 0 () = 1
ln , 00 ln +1
() =
2 (ln ) . 2
(a) 0 () 0 (),where 0 () 48254942434 × 10 11 , 0 () 48254942383 × 10 11 .
(b) Let’s estimate 0 () 0 () =( ) 00 ( 1 )= 00 ( 1 ).Since 00 increases (its absolute value decreases), we
have | 0 () 0 ()| | 00 ()| 50583 × 10 20 .
7. (a) The 11-digit calculator value of 192 sin is 62820639018, while the value (on the same device) of before
96
rationalization is 6282063885,whichis168 × 10 8 less than the trigonometric result.
96
(b) =
2+ 3 · 2+ 2+
3 · 2+ 2+ 2+
3 · 2+ 2+ 2+ 2+ 3
but of course we can avoid repetitious calculations by storing intermediate results in a memory:
1 = 2+ 3, 2 = 2+ 1 , 3 = 2+ 2 , 4 = 96
2+ 3 , andso =
1 2 3 4
According to this formula, a calculator gives 62820639016, which is within 2 × 10 10 of the trigonometric
result. With Digits:=16;,Maplegives 6282063901781030 before rationalization (off the trig result by
about 11 × 10 14 )and 6282063901781018 after rationalization (error of about 17 × 10 15 ), a gain of
about one digit of accuracy for rationalizing. If we sets Digits:=100;, the difference between Maple’s
calculation of 192 sin 96 and the radical is only about 4 × 1099 .
9. (a) Let =
27 + 13
729 2 + 108 3 and =
1
2
1
2
27 729 2 + 108 3 13.
Then 3 + 3 =27 and = 1 4 [7292 (729 2 + 108 3 )] 13 = 3. Substitute into the formula
+ =
3 + 3
3
where we replace by
2 + 2 :
= 1 273
( + ) =
3
1
2
27 + 729 2 +108 3 23
+3 +9
2
1
2
27 + 729 2 +108 3 23
which almost yields the given formula; since replacing by results in replacing by , a simple discussion
of the cases 0 and 0 allows us to replace by || in the denominator, so that it involves only positive
numbers. The problems mentioned in the introduction to this exercise have disappeared.
(b) A direct attack works best here. To save space, let =2+ 5, so we can rationalize, using 1 =2+ 5
and 1 =4(check it!):
=
4
23 +1+ · 13 13
23 13 = 4(13 13 )
= 13 13
13 1
and we cube the expression for : 3 = 3 13 +3 13 1 =4 3,
3 +3 4=( 1)( 2 + +4)=0, so that the only real root is =1. A check using the formula
from part (a): =3, = 4, so729 2 + 108 3 =14,580 = 54 2 × 5, and
=
36
54 + 27
5
23
+9+81
54 + 27
5
23
, which simplies to the given form.
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.
Stewart: Calculus: Early Transcendentals, Seventh Edition. ISBN: 0538497904. (c) 2012 Brooks/Cole. All rights reserved.