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

LIES MY CALCULATOR AND COMPUTER TOLD ME ■ 7
SOLUTIONS
1. The computer results are from Maple, with Digits:=16. The last column shows the values of the sixth-degree
Taylor polynomial for () =csc 2 2 near =0. Note that the second arrangement of Taylor’s polynomial is
easier to use with a calculator:
6 () = 1 3 + 1 15 2 + 2
289 4 + 1
675 6 = 1
675 2 + 2
289
2 + 1 15
2 + 1 3
() calculator () computer () Taylor
01 033400107 03340010596845 033400106
001 0333341 033334000010 033334000
0001 03334 0333333400 033333340
00001 034 03333334 033333333
000001 20 033334 033333333
0000001 100 or 200 0334 033333333
00000001 10,000 or 20,000 04 033333333
000000001 1,000,000 0 033333333
We see that the calculator results start to deteriorate seriously at =00001,andforsmaller, they are entirely
meaningless. The different results “100 or 200” etc. depended on whether we calculated [(sin ) 2 ] 1 or
[(sin ) 1 ] 2 . With Maple, the result is off by more than 10% when =00000001 (compare with the calculator
result!) A detailed analysis reveals that the values of the function are always greater than 1 , but the computer
3
eventually gives results less than 1 3 .
The polynomial 6 () was obtained by patient simplication of the expression for (), starting with
sin 2 () = 1 (2)2
(1 cos 2),wherecos 2 =1 + (2)4 ··· (2)10 +
2 12 (). Consequently, the exact
2! 4!
10!
value of the limit is 6(0) = 1 3
. It can also be obtained by several applications of l’Hospital’s Rule to the expression
() = 2 sin 2
2 sin 2
with intermediate simplications.
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.
3. From () =
25
(We may assume 0; Why?), we have ln () =25ln ln(10001) and
(10001)
0 ()
() = 25 ln(10001). This derivative, as well as the derivative 0 () itself, is positive for
0 0 =
( 0 )=
25
249,971015, and negative for 0. Hence the maximum value of () is
ln(10001)
25
0
, a number too large to be calculated directly. Using decimal logarithms,
(10001) 0
log 10 ( 0 ) 12408987757,sothat( 0 ) 1229922 × 10 124 . The actual value of the limit is lim () =0;
it would be wasteful and inelegant to use l’Hospital’s Rule twenty-ve times since we can transform () into
25
() =
, and the inside expression needs just one application of l’Hospital’s Rule to give 0.
(10001) 25
Stewart: Calculus: Early Transcendentals, Seventh Edition. ISBN: 0538497904. (c) 2012 Brooks/Cole. All rights reserved.