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Multivariable Calculus 9th Edition by James Stewart solution manual

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( ◦ )() + ( ◦ )() = sin + sin = 2sin. The two sides are not equal, so the given statement is false.
19. Let   be the statement that 7  − 1 is divisible by 6
•  1 is true because 7 1 − 1 = 6 is divisible by 6.
• Assume   is true, that is, 7  − 1 is divisible by 6. In other words, 7  − 1 = 6 for some positive integer . Then
7 +1 − 1 = 7  · 7 − 1 = (6 + 1) · 7 − 1 = 42 + 6 = 6(7 + 1), which is divisible by 6, so  +1 is true.
• Therefore, by mathematical induction, 7  − 1 is divisible by 6 for every positive integer .
20. Let   be the statement that 1 + 3 + 5 + ··· + (2 − 1) =  2 .
•  1 is true because [2(1) − 1] = 1 = 1 2 .
• Assume   is true, that is, 1 + 3 + 5 + ··· + (2 − 1) =  2 . Then
1 + 3 + 5 + ··· + (2 − 1) + [2( + 1) − 1] = 1 + 3 + 5 + ··· + (2 − 1) + (2 + 1) =  2 + (2 + 1) = ( + 1) 2
which shows that  +1 is true.
• Therefore, by mathematical induction, 1 + 3 + 5 + ··· + (2 − 1) =  2 for every positive integer .
21.  0 () =  2 and  +1 () =  0 (  ()) for  = 012.
 1 () =  0 ( 0 ()) =  0   2  =
  2  2
=  4 ,  2 () =  0 ( 1 ()) =  0 ( 4 ) = ( 4 ) 2 =  8 ,
 3 () =  0 ( 2 ()) =  0 ( 8 ) = ( 8 ) 2 =  16 ,.Thus, a general formula is   () =  2
+1 .
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
CHAPTER 1 PRINCIPLESOF PROBLEMSOLVING
¤
73
22. (a)  0 () = 1(2 − ) and  +1 =  0 ◦   for  = 012.
 1 () =  0

1
2 − 

=
1
2 −
1
2 − 
=
2 − 
2(2 − ) − 1
=
2 − 
3 − 2 ,
 2 () =  0

2 − 
3 − 2

=
1
2 −
2 − 
3 − 2
=
3 − 2
2(3 − 2) − (2 − )
=
3 − 2
4 − 3 ,
 3 () =  0
 3 − 2
4 − 3

=
1
2 −
3 − 2
4 − 3
=
4 − 3
2(4 − 3) − (3 − 2)
=
4 − 3
5 − 4 
Thus, we conjecture that the general formula is   () =
 + 1 − 
 + 2 − ( + 1) .
To prove this, we use the Principle of Mathematical Induction. We have already verified that   is true for  = 1.
Assume that the formula is true for  = ; that is,   () =
 + 1 − 
 + 2 − ( + 1) . Then
 +1 () = ( 0 ◦   )() =  0 (  ()) =  0

 + 1 − 
 + 2 − ( + 1)

=
1
2 −
 + 1 − 
 + 2 − ( + 1)
=
 + 2 − ( + 1)
2[ + 2 − ( + 1)] − ( + 1 − )
=
 + 2 − ( + 1)
 + 3 − ( + 2)
This shows that the formula for   is true for  =  + 1. Therefore, by mathematical induction, the formula is true for all
positive integers .
(b) From the graph, we can make several observations:
• The values at each fixed  =  keep increasing as  increases.
• The vertical asymptote gets closer to  = 1 as  increases.
• The horizontal asymptote gets closer to  = 1
as  increases.
• The ­intercept for  +1 is the value of the
vertical asymptote for   .
• The ­intercept for   is the value of the
horizontal asymptote for  +1 .
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
 

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