Answer book for calculus 3rd (spivak, 1994) PDF

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ANSWER BOOK FOR CALCULUS Third Edition Michael Spivak Publish or Perish, Inc. HOUSTON, TEXAS ANSWER BOOK FOR CALCULUS Third Edition Copyright © 1984, 1994 by Michael Spivak All rights reserved Manufactured in the United States of America ISBN 0-914098-90-X CHAPTER 1 1. (ii) (x - y)(x + y) = [x + (-...

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ANSWER BOOK FOR CALCULUS Third Edition

Michael Spivak

Publish or Perish, Inc. HOUSTON, TEXAS

ANSWER BOOK FOR CALCULUS Third Edition Copyright © 1984, 1994 by Michael Spivak All rights reserved

Manufactured in the United States of America ISBN 0-914098-90-X

CHAPTER 1 1.

(ii) (x

-

y)(x + y)

[x + (-y)](x +

=

y)

x(x + y) +

=

=x(x+y)-[y(x+y)]=x2+xy-

=x2+xy-y

(iv) (x--y)(x

=x2-y2.

)=x(x +xy+y

+xy+y

(-y)(x + y) [yx+y2

)-[y(x

=x3+x2y+xy

-[yx

+ y+y B +xy2+y3]=x3-y3.

(v) (x

y)(xn-1 4

-

n-2,

4 =

.

.

xn y xn

n-2

n-1 n-2

...

2 n-2

¿n-1 n-2

[xn-1

-

=

n--2

[y(xn-1

-

=

n-1

4 yn-2 4 x(xn-1 4 n-2

.

2

#

•·

n-ip

n-1 ·

X

n-1

n

yn

_

Using the notation of Chapter 2, this proof can be written as follows: n-1

n-1

(x

-

y)

n-1

j n-1-j

xiyn-1-j

-

jyn-1-j

__

j=0

j=0

j=0 n-1

n-2 =

j+1yn--i-j

xn

jyn-j

_

j=0

j=1 n-2

n-2 =

n

j+1yn-1-j

xn 4

k+1pn-(k+1)

__

n

k=0

(lettingk =

xn

_

=

j

-

1)

yn

A formal proof requires such a scheme, in which the expression

(x"¯ I

+ xn-2, y

n-1 -·

-

+ xyn-2 4 yn-1) is replaced by the inductively defined symbol

2 j=0

xiy"¯ I¯ J.

Along the way we have used several other manipulations which can, if necessary, be justifiedby inductive arguments. 3.

(iv) (a/b)(c/d) (ac)/(bd).

=

(ab--1)(cd¯ I)

=

(ac)(b-Id¯ ')

1

=

(ac)(bd)-1 (by (iii))

=

2

Chapter 1 ab¯ I

(vi) If

=

cd¯ ',

(ab¯ I)bd

then

(ad)d¯ Ib¯

(cd I)bd,

=

(bc)d¯ Ib¯ ',

bc, then or ab¯ ! b2, so by Problem 1(iii), a b or a = then a2 a/b b/a 1 and if a then a/b b/a ad

=

=

-b.

=

=

-b,

=

=

-1.

=

=

or ad =

bc. Conversely, if

=

If ab I Conversely, if a cd

1.

= =

ba¯ ', b, then

=

4.

(ii) All x. (iv) x > 3 or x < 1. 2 or x < [-1 Ã + Á (vi) x > [--1 ] 2. (viii) All x, since x2 + x + 1 [x + (1/2)]2+ 3/4. (x) x > Ä or x < Ã. (xii) x < 1. (xiv) x > 1 or x < -

=

-1.

5.

(-b) is in P. a is in P, so (iv) b a is in P and c is in P, so c(b a) bc ac is in P. (vi) If a > 1, then a > 0, so a2 > a 1, by part (iv). (viii) If a 0 or c 0, then ac 0, but bd > 0, so ac < bd. Otherwise we have ac < bc < bd by applying part (iv) twice. (x) If a < b were false, then either a b or a > b. But if a b, then a2 = b2, and if a > b >_ 0, then a2 > b2, by part (ix).

(ii) b

-a

-

-

-

=

-

-

-

=

=

=

=

=

y and Problem 5(viii) we have x2 < y2 [asin Problem 5(ix)]. Then from 0 5 x < y and x2 < y2 we have x3 < y3. We can continue in this way to prove that x" < yn for n = 2, 3, (a rigorous proof uses induction, covered in

6.

(a) From 0 5

x

<

.

.

.

the next chapter).

< y, then x" < y" by part (a). If x < y 5 0, then 0 5 so n by and < this that is odd) hence part (a); means (sincen yn (sincen is odd). Thus, in all x" < yn. Finally, if x < 0 5 y, then xn < 0 cases, if x < y, then xn « yn

(b) If 0 5 (-y)n

x

-y

<

-y"

_

_<

(c) This follows immediately from part (b),since x while y

<

x would imply that y"

(d) Similarly, if >_

x, y

-x,

-xn

0 and xn

<

<

y would imply that xn

,

yn,

x".

n is even, then using part (a) instead of part (b) we see that if yn, then x = y. Moreover, if x, y 5 0 and x" yn, then -

_

0 and (-x)" (-y)", so again x y. The only other possibility is that one of x and y is positive, the other negative. In this case x and are both n, negative. Moreover xn positive or both since n is even, so it follows from y the previous cases that x -x,

-y

>_

=

=

-y

_

-y.

=

Chapter 1 7. If a

b, then

<

b+b 2

a+b

a+a a= 2ab, If 0

<

<

a

<

Ñ

by Problem 5(x).

-

a2 + 2ab + b2

>

4ab,

(a + b)2 > 4ab,

2Ñ.

Moreover, for all a, b we have (a b)2 à 0, and thus (a + b)2 4ab, which implies that a + b > 2Ñ for a, b à 0.

so a + b >

-

>

8. Two applications of P'12 show that if a < b and c < d, then a+c < b+c < b+d, so a + c < b + d by P'11. In particular, if 0 < b and 0 < d, then 0 < b + d, which < 0 > 0; for if proves Pl l. It follows, in addition, that if a < 0, then were true, then 0 = a + (-a) < 0, contradicting P'10. Consequently, any number 0, a > 0, a < 0, the last being a satisfies precisely one of the conditions a equivalent to > 0. This proves P10. Finally, P'13 shows that if 0 < a and 0 < c, then 0 < ac, which proves P12. -a

-a

=

-a

9.

(ii) la|+ |b| |a + bl. (iv) x2 2x y + y2. -

-

10.

(li) x-1

ifx>1;

1-x

if05xd1;

1+x

if-15x50;

-1

-1.

-

x

if x á

(iv) aifa>0;

3a if a 5 0. -5

11.

(ii)

<

x

<

11.

2 (thedistance from x to 1 plus the distance from x to 2 equals 2). 1 precisely when 1 5 x

(iv) x

<

1 or x

(vi) No x. (vili) If x

x2 + x

-

5

>

-2,

> =

1 or x < then the condition becomes (x 1) (x + 2) 0, for which the solutions are + B)|2 and -

(-1

(-1

-

=

3, or

)|2.

4

Chapter 1 -2,

Since the first is > 1 and the second is < both are solutions to the equation x)(x + 2) < x < 1 the condition becomes (1 3 |x 1| |x + 2| = 3. For x2 which has no solutions. + x + 1 0, or -2

-

·

=

-

=

12.

(ii) [1/x\ \xl \(1/x) x\ (by(i)) |1] 1, so ll/x| =

-

(iv) |x

-

y

=

=

-

|x+(-y)|

=

|x|+ |-y|=

5

=

1|lxl.

|x|+ |y|.

(vi) Interchanging x and y in part (v) gives |y| with part (v) yields |(|x| |yl)| 5 lx y|.

|x 5 |x

-

-

y|. Combining this

-

-

13.Ifxsy,then|y-x|=y-x,sox+y+|y-x|=x+y+y-x=2y, which is 2 max(x, y). Interchanging x and y proves the formula when x à y, and the same type of argument works for min(x, y). Also max(x, y, z)

max(x, max(y, z))

=

y+z+|y-z| 2

x+

y+z+|y-z| 2

+

-x

2

|y-zl+y+z+2x+|y+z+|y--z|-2x|

=

4 -(-a)

14. (a) If a 2 0, then lal = a proved for a 5 0 by replacing a by

=

=

]-a|, since

-a

5 0. The equality is

-a.

(b) If |a i b, then clearly b à 0. Now |a| 5 b means that a 5 b if surely a 5 b if a 5 0. Similarly, |a| 5 b means 5 b, and hence -a

5 a if a >_ 0. So 5 a 5 b, then |a|

-b

Conversely, if a 5 0. 5 a 5

|a| and -(|a

b| 5 ja|+

|a +

15. If x

¢

=

5 a 5 b. b if a 2 0, while a

|a]

5 b if

-a

=

-|b|

-|a|

so

-b

-b

-b

a 5 0, and surely

(c) From

a 2 0, and 5 a, if

|b| it follows that + [bj)5 a+b 5 |a + |b|, 5 b5

|b|.

y, then x +xy+y

=

x

3

-y

3 .

-y

x

Problem 6(b) shows that the quotient on the right is always positive (sincex3 y3 > 0 if x y > 0 and x3 y3 < 0 if x y < 0). Moreover, if x y ¢ 0, then X2 3x2 > 0. The other inequality is proved similarly, using the # Xy # y2 factorization for x' y'. -

-

-

=

-

-

=

Chapter 1 16.

(a) If

then xy

=

X2

0, so x

2

0 or y

=

=

=

(X

i y)

=

X2

5

+ 2xy + y2,

0. If

x3+yS=(x+y)3=X3+3x2y+3xy2+y3,

then3xy(x+y)=0,sox=0ory=0orx=-y.

(b) The first equation implies that 4x2 + 8xy + 4y2

>_

0.

Suppose that we also had 4x2+6xy+4y2

_ 0. If neither x nor y is 0, this means that we must have 2xy > 0; but this implies that 4x2 + 6xy + y2 > 0, a contradiction.

Moreover, it is clear that if one of x and y is 0, but not the other, then we also have 4x2 + 6xy + 4y2 > 0.

(c) If

X4

4

(X

=

# y)4

=

X4

+ 4x3y + 6x2y2 + 4xy3 + y',

then 0=4x3y+6x2y*+4xy*

=xy(4x

+6xy+4y

), 0. But by part (b), the last

0, or 4x2 + 6xy + 4y2 0 or y implies that x and y are both 0. Thus we must always have x so x

=

=

=

=

0 or y

=

equation

0.

(d) If x' + y'

y)'

(x+

=

=

x' + 5x4y + 10x3y2 + 10x2

3

+ 5xy' + y',

then 0

=

=

5x4y + 10x2y + 10x y* + 5xy4 5xy(x3 + 2x2y + 2 y2 + y*¾

=0or

soxy

x'+2x2y+2xy

+y*=0.

Subtracting this equation from

(x+y)'=x3+3x2y+3xy2+y' we obtain

(x+y)'=x2y+xy2=xy(x+y). So either x+ y 0 or (x+ y)2 xy; the latter condition implies that x2+xy+ 0, so x 0 or y 0 by Problem 15. Thus x = 0 or y = 0 or x =

=

-y.

=

=

=

y2

-·

6

Chapter 1 17. (a) Since 2

3 x-4

2x2-3x+4=2

=2

( (

3

x--

the smallest possible value is 23/8, when

9 8

+4-23 +-, 8

4

(x 3/4)2 -

0, or x

=

3/4.

=

(b) We have x2-3x+2y

+4y+2=

+2(y+1)2-

x-

-9/4,

-1.

so the smallest possible value is

(c) For each

,

when x

3/2 and y

=

=

y we have

x2+4xy+5y2-4x-6y+7=x2+4(y-1)x+Sy2-6y+7

[x + 2(y [x + 2(y

=

=

so the smallest possible value is 2, when y

(a) is a straightforward

18.

1)]2 + Sy2

6y + 7 1)]2 + (y + 1)2 + 2,

-

-

-1

-

-

4(y

-

1)

-2(y

and x

=

--·

=

1) = 4.

check.

(b) We have b x+2

X2+bx+c=

*

(

b2 +

c---

b2 Ec-----, 4

4

butc-b2/4>0,sox2+bx+c>0forallx. y2 for c: we have b2 4C = y2 4y2 < 0 for 0 for all x, if y ¢ 0 (andsurely x2 + xy + y2 > 0 for all

(c) Apply part (b) with y for b and y x

¢ 0, so x2 + xy ¢ 0 if y 0).

+ y2 >

-

-

=

(d) a

must satisfy

(ay)2 4y2 -

<

0, or a2

(e) Since x2+bc+c=

b x+2

(

and since x2 + bx + c has the value c is c b2/4. Since

-

<

4, or

2

|œ\< 2. b2

+

c--

>c--,

4

b2/4 when x

4 -b|2,

the minimum value

=

-

ax2+bx+c=a

x2+b

a

b2

a

Chapter1 the minimum value is

(c

19.

(a) The proofs

when xi

b2

b2 =c--.

-----

a

a

4a2

4a

lyi and x2

=

7

Ay2, or yi

=

ward. If there is no such A, then the equation 2 À2 2 #X2Ï2) #

A(X1Ïl

_

(X12

=

y2 2)

4 y

0, are straightfor-

=

0

=

has no solution A, so by Problem 18(a) we must have 2

2(xi yi + x2y2) (712+ 72 ) which yields the Schwarz inequality.

(b) We have 2xy

5 x2 + y2, since 0 5

xi +

X22

-

<

(X12

712 + 722 <

#

(XI2

¯

xi2 + x22

=

x2

0

2xy + y*. Thus

-

yi' X22)

+

(Ï12Ÿ Ï22)

X22

2x171

(2)

(x y)2

<

xi'

2xi yi

(1)

4(xi2 + yi2) (yi2 + y22)

--

yi2 + y22

'

Ï22

+ x22)

+

(712 722)

'

addition yields

2(xiyi + x2y2) x:2 + x22

(c) The

Ï12 Ÿ Ï22

equality is a straightforward computation. Since

(xiy2

--

x2yi)2 à 0, the

Schwarz inequality follows immediately.

(d) The proof in part (a) already yields the desired result. In part (b),equality holds only if it holds in (1)and (2). Since 2xy only when 0 (x y)2, i.e., x y, this means that =

2

x2

=

-

x¿

y¿

xi2 + x22 so we can choose A =

=

y

2

X22

x12 +

for x

+ y22 2

=

1, 2,

2

In part (c),equality holds only when xi y2 x2yi = 0. One possibility is yi 0. If yi ¢ 0, then xi (xi/yi)yi and also xi (xi/ys)y2;similarly, if y2 y2 0, then A x2|y2· =

-

=

=

=

=

=

20, 21, 22. See Chapter 5.

23. According to Problem 21, we have x

--

xo|

<

min

|x/y

(

--

xo/yo|

<

€

2(1||yo| + 1)

,

e if 1

8

Chapter 1

and 1

1

s

2(|xoj + 1) and the latter is true, according to Problem 22, if

'

yo

y

slyol2

(lyol

|y-yo|

<

an

=

=--=1

g

1 and n 2. Now suppose that the assertion is true for 3. Then it is true, in particular, for n 1 and n 2, so

the assertion is true for n all k

2

=

=

-

-

an-i + an_2

1+Ê

n-2

)( 2

1+

1-Ä

(1+Ê ¯

n-1

2

1-d

2

2

2

"

Á

2

1+

-

"¯ '(1+B

2

2

1+

2

2

(1+B

n-2

1-Ä

2 *

(1-4 2

n

"¯ '{1-4

(

2

2

n-1

Chapter 2

16

21. x¿

=

(a) As before, the proof is trivial if all y¿ 0 or if there is some number =

ly¿ for all i. Otherwise, O<

(ly¿ x¿) -

i=1 =

y¿2

A

21

-

x¿2,

x¿y¿ +

i=1

i=1

i=1

so Problem 1-18 again gives the result.

(b) Using 2xy

<

x2 + y2 witli x

x¿

=

n

y¿

=

y

,

n

X¿2

i=1

2

i=1

we obtain

(1)

<

+

.

X¿

2

x¿2 i=1

2

x¿2

2x¿y¿

i=1

i=1

i=1

Adding we obtain

i2x¿y¿ i=l i=1

<

...