Solutions
BLOCK 3 : PARTIAL DERIVATIVES Solutions
Block 3: P a r t i a l D e r i v a t i v e s
1. U s e p o l a r c o o r d i n a t e s t o o b t a i n w =
2 r s i n 8 (r cos 8) 2
r
2 s i n 8 c o s 8 i f r # 0.
therefore,
3.
h" (r)
+
1
T h e r e f o r e , w = s i n 28
lim
w =
(x,y)+(O, O )
hl (r).
1i m
(r,e)+(o,eo)
(r # 0 ) and
w = s i n 28 0 and t h i s depends
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
3.1.1(L)
While it was more n a t u r a l , i n t e r m s of an e x t e n s i o n of lower
dimensional c a s e s , t o i n v e n t t h e E u c l i d e a n m e t r i c , t h e p o i n t i s
t h a t a metric i s d e f i n e d s o l e l y by t h e p r o p e r t i e s mentioned i n
t h i s exercise.
S i n c e , from a l o g i c a l p o i n t o f view, w e u s e n o t h i n g
b u t i n e s c a p a b l e consequences o f t h e s e p r o p e r t i e s , i t f o l l o w s
t h a t any m e t r i c t h a t h a s t h e s e same p r o p e r t i e s may be used i n o u r
study.
I n p a r t i c u l a r , w e want t o show i n t h i s e x e r c i s e t h a t t h e
Minkowski metric h a s t h e s e p r o p e r t i e s .
Thus,
,...
a . S i n c e llxll
- = m a x { l x l l , . . . , l x n I } t h e f a c t t h a t each l x i l , ( i = l
,n),
i s a t l e a s t a s g r e a t a s z e r o g u a r a n t e e s t h a t t h e maximum of t h e
x i ' s i s a l s o a t l e a s t a s g r e a t a s zero.
T h i s proves t h a t
IkII
i f i t happens t h a t
a c t u a l l y e q u a l s 0, t h e n by d e f i n i t i o n ,
t h e maximum / x i ] i s a l s o z e r o . But no 1x.I can be less t h a n z e r o ;
NOW,
1 t h u s , t h e maximum lxil
being zero guarantees t h a t a l l / x i / equal
z e r o o r t h a t a l l xi = 0 , and t h i s means t h a t
Ik+YII\< ]lXll + Ikll,
w e see from t h e d e f i n i t i o n
o f t h e Minkowski m e t r i c t h a t t h i s i s e q u i v a l e n t t o proving t h a t
A s f o r checking t h a t
Unless w e a r e b o t h e r e d by t h e f a i r l y a b s t r a c t n o t a t i o n ,
Namely, l e t Ixk+Ykl d e n o t e t h e maximum
W e t h e n have
xn+yn
a l m o s t t r i v i a l t o prove.
of
I xl+yl I ,..., and I
(1) i s
I.
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.1 (L) c o n t i n u e d
and, s i n c e xk and yk a r e r e a l numbers, w e a l r e a d y know from p a r t 1
o f o u r c o u r s e t h a t ixk+YkI
Finally.
s
s i n c e lxk14 max{lxll
lxkl
+ IYkl.
,....l x n l }
Hence.
and l y k l s maxtlyll
,.., l y n l } ,
w e have from ( 2 ) t h a t
I n f a c t , i f t h e x ' s and y ' s a l l have t h e same s i g n , e q u a l i t y can
h o l d i f and o n l y i f
x and y "match" ( i . e . ,
which means t h a t t h e maximum components of have t h e same s u b s c r i p t ) .
For example, i f x = ( 2 , 4 , 1 ) and y = ( 3 , 2 , 5 ) , t h e maximum components
do n o t match ( i n q it i s t h e second component, 4 , and i n y i t i s
t h e t h i r d component, 5 )
I n t h i s c a s e x+y = ( 5 , 6 , 6 ) , whence
.
Ik(1+ [yll = 4+5
hand i f x = (2,5,1)
Thus nx+yll
<
Ilx+yll = 6 ; w h i l e
= 9.
On t h e o t h e r
and Y = (3,7,2)
components m a t c h ] ,
so that
Ik+yII = 1x11 + Ikll.
rate,
[where t h e maximum
Ik+YII = max{5,12,3} = 12 w h i l e llscll +
F i n a l l y , w e must show t h a t
A t any
l l ~ l l + 11y11.
Ikll
= 5+7 = 1 2 ,
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.1 (L) c o n t i n u e d
b. From a mechanical p o i n t o f view, t h i s i s h a r d l y a l e a r n i n g e x e r c i s e .
I n f a c t , i t i s more l i k e a t r i v i a l d r i l l e x e r c i s e .
x = (2,4,1) while y = (4,4,5).
that -
We a r e g i v e n
( W e p i c k e d n = 3 s o t h a t you
c o u l d u s e your g e o m e t r i c i n t u i t i o n i f you s o d e s i r e d ; t h a t i s , w i t h
-f
n = 3 w e may t h i n k o f x a s being 2 1
+
t
41
+
-+
k, e t c ) . Then, -x
* =~ ( 2 ) ( 4 )
+
( 4 ) (4)
+
(1)( 5 ) = 29
On t h e o t h e r hand, o u r d e f i n i t i o n of t h e Minkowski metric y i e l d s
that
llxll
- = max{2,4,11 = 4 and llyll
= max{4,4,51 = 5 1k11 1k11 =
From ( 2 ) and ( 3 ) we have t h a t
r e s u l t w i t h (11, w e have t h a t
*N o t i c e
that i f
c>O
i s any c o n s t a n t ,
max{cx l , . . . c x n I = c max{xl,
s i n c e c "magnifies" each x
20, and comparing t h i s
then
...,xnl i n t h e same r a t i o . i Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 1: Functions of More Than One Variable
3.1.1 ( L ) continued
Equation ( 4 ) v i o l a t e s a key p r o p e r t y of a d o t product, e s p e c i a l l y
a s w e know it i n t h e 2- and 3-dimensional cases.
I n o t h e r words, t h e
Minkowski m e t r i c does n o t obey t h e s t r u c t u r a l p r o p e r t y t h a t i s s o
o f t e n used i n d o t products (namely, Ix*yl 6 1 1 ~ 1 1
while t h e
IkII)
Euclidean m e t r i c does have t h i s s t r u c t u r a l p r o p e r t y ( a s w e have
proven i n o u r supplementary n o t e s ) .
This e x p l a i n s why most t e x t s stress t h e E u c l i d e a n - m e t r i c over t h e
Minkowski m e t r i c , even though a s we s h a l l s e e i n t h e n e x t few
e x e r c i s e s , t h e Minkowski m e t r i c i s most h e l p f u l i n our d i s c u s s i o n
of l i m i t s i n t h e case of r e a l f u n c t i o n s of n-tuples,
expecially
with n>3.
In our l a t e r work ( e s p e c i a l l y Block 7) we w i l l do a g r e a t d e a l of
work with t h e g e n e r a l i z e d n o t i o n of a d o t product and i n t h i s
c o n t e x t t h e Euclidean m e t r i c has d e s i r e d p r o p e r t i e s n o t shared
by t h e Minkowski m e t r i c .
From t h e p o i n t of view of o u r "game," however, t h e Minkowski
m e t r i c i s on a p a r with t h e Euclidean m e t r i c i n t h e type of
i n v e s t i g a t i o n we a r e c u r r e n t l y undertaking where t h e d o t product
of two n - t u p l e s i s n o t an i s s u e .
lim
W e have x+a f
(x) =
L1
lim
g (5) = L2.
x+a
- -
and
Given E>O, w e must f i n d 6>0 such t h a t 0 < Ik-all
- < 6
+
I [ f ( x ) + g ( ~ )-l
[ L ~ + LI ~< I E -
J u s t a s i n t h e s c a l a r case, we observe t h a t
I [f(x)+g($l -
[L +L ] I =
.1 2
I
[£(XI-L1l
+ [9(g-L21 t
-
6 I f ( x ) - ~ ~+ l l g ( x ) - ~ ~ I
BY d e f i n i t i o n of L1 and L 2 , we can f i n d
and d 2 such t h a t
Solutions
Block 3 : P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.2
continued
J
Thus i f 6 6 min{61,62)
i t f o l l o w s from ( 2 ) t h a t
Combining ( 3 ) w i t h (1) y i e l d s t h e d e s i r e d r e s u l t , namely
0 <
1lz-zll
< 6
+
1 [ f ( z ) + g ( z )1 -
[L1+L21
I
< E
Note :
I n t h i s e x e r c i s e , w e d i d n o t have t o s p e c i f y whether i t was t h e E u c l i d e a n m e t r i c o r t h e Minkowski m e t r i c t h a t was being used.
The p o i n t i s t h a t o u r proof used o n l y t h o s e p r o p e r t i e s t h a t were t r u e f o r e i t h e r metric.
v a l u e of
E
To be s u r e , t h e a c t u a l v a l u e of 6 f o r a g i v e n might depend on which m e t r i c was used, b u t t h e e x i s t e n c e of 6 d i d n o t . On t h e o t h e r hand, t h e r e may be a tendency t o t h i n k i n terms o f t h e
Minkowski m e t r i c when w e s a y " l e t 6 = min{61,62)." The p o i n t i s
t h a t w h i l e t h i s p a r t of o u r l o g i c a l s t r a t e g y i s t h e same whether
i t i s t h e E u c l i d e a n m e t r i c o r t h e Minkowski m e t r i c which i s b e i n g
used, what i s t r u e i s t h a t t h e m o t i v a t i o n behind t h e s t r a t e g y i s
t h e same a s t h a t which i n s p i r e d t h e " i n v e n t i o n " o f t h e Minkowski
m e t r i c . Namely, i n b o t h i n s t a n c e s , t h e i d e a i s t h a t t o make s u r e
t h a t something i s s u f f i c i e n t l y s m a l l , w e make s u r e t h a t i t s maximum
dimension i s s u f f i c i e n t l y s m a l l , r e g a r d l e s s of how we d e f i n e " d i m e n s i ~ n . ~
One f i n a l word t h a t may b e of i n t e r e s t i s t h a t i n o u r 1-dimensional
s t u d y o f c a l c u l u s , t h a t i s , when w e were s t u d y i n g l i m i t s of f u n c t i o n s
o f a s i n g l e r e a l v a r i a b l e , t h e r e was no d i s t i n c t i o n between t h e
E u c l i d e a n and t h e Minkowski m e t r i c .
The i n t e r e s t i n g p o i n t i s t h a t
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
i n t h e s p e c i a l c a s e of n = l , t h e two m e t r i c s a r e i d e n t i c a l .
f o r a r e a l number, x ,
Ix(
Namely,
a n d 0 a r e t h e same (and, o b v i o u s l y , f o r
t h e c a s e of o n l y one component, 1x1 = m a x { l x l } ) . That i s , e i t h e r
t h e Minkowski o r t h e Euclidean m e t r i c would have l e d t o t h e same
d e f i n i t i o n of magnitude i n t h e c a s e o f a 1-dimensional s p a c e .
3.1.3(L)
a.
W e must f i n d 6 such t h a t
AS s e e n i n o u r supplementary n o t e s ,
Ix +y -311 < € i f
and 2
TO make l x -41 <
2 (i.e.,
E
5
we take i n t o consideration t h a t x i s near
i s small)
T h a t i s , w e may assume t h a t
2-5 < x < 2
+ 5
.
where 1 5 1 < 1
In p a r t i c u l a r , then,
1 < x < 3
and t h e r e f o r e
3 < x+2 < 5
Thus, near x=2,
o r , from (1)
From ( 2 ) it i s e a s i l y s e e n t h a t Ix-21
< 10 i m p l i e s t h a t
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.3 (L) c o n t i n u e d
r e f e r r e d t o i n o u r n o t e s may be t a k e n t o be
Thus, t h e
6.I n
o t h e r words,
A s f o r making ly3-271<
5 , w e observe t h a t
2
y3-27 = (y-3) (y +3y+9)
Hence 1 y3-27 1
=
1 y-31 1 y 2
+3y+9 1
2
S i n c e w e know t h a t y i s n e a r 3 , it f o l l o w s t h a t y +3y+9 i s n e a r 27
2
(because we a l r e a d y Know t h a t y +3y+9 i s a continuous f u n c t i o n of y ) .
I n p a r t i c u l a r , w e s h a l l assume y i s c l o s e enough t o 3 s o t h a t
2
y2+3y+9 (= 1 y +3y+9
< 28.
1)
With t h i s i n mind,
( 4 ) becomes
From (51, i t f o l l o w s immediately t h a t i f ly-31
3
E
I n o t h e r words
ly -271 < p.
and
56
<
then i s t h e 6 2 of o u r n o t e s .
E
Since E i s p o s i t i v e , it i s c l e a r t h a t 56 <
6 = min{b 1' 62 w e have
E
-
.
Hence, i f
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 1: Functions of More Than One V a r i a b l e
3.1.3 (L) continued
I n summary, using t h e Minkowski m e t r i c , given E > O choose 6 =
Then
E
E
x
[and by o u r remarks i n t h e supplementary n o t e s , t h e v a l u e of 6 a l s o
s u f f i c e s i f we a r e using t h e Euclidean m e t r i c , a s we s h a l l review
in part (b)l.
b.
P i c t o r i a l l y , w e have
2 3
Since Ix t y -311 i s t r u e f o r a l l (x,y) i n r e c t a n g l e ABCD, it i s , i n
particular,
with r a d i u s
.
t r u e f o r a l l (x,y) i n t h e c i r c l e c e n t e r e d a t (2,3)
The equation of t h i s c i r c l e i s
&- .
In any e v e n t
Solutions
Block 3 : P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.3 (L) c o n t i n u e d
Hence,
o
\I
€
< ~ ~ ( x , ~ ) - ( 2 <~ 3 )
+
1 ~ ~ + ~ ~ -< 3s 1 1
-
i s t r u e even i f l l ( x f y ) ( 2 , 3 )
W e have x = (x1.x2 ,x3 , x 4 )
,
11
now d e n o t e s t h e Euclidean m e t r i c .
-
3
= (1,1,1,1) and f (x)=x12+2x2+x3 +x4
W e must prove t h a t
Since 1 = (1,1,1,1), t h e d e f i n i t i o n of f t e l l s u s t h a t
2
f (1 ) = f ( l , l , l , l ) = 1 + 2 ( 1 ) + l3 + l2 = 5. Thus, w e s e e t h a t
f (x) e x i s t s a t 1 and i t s v a l u e i s 5. Next w e must show t h a t
Equation ( 2 ) , by d e f i n i t i o n , now i m p l i e s t h a t f o r a g i v e n E > O w e
must f i n d 6>0 such t h a t
If we
Up t o now w e have n o t s p e c i f i e d what m e t r i c w e a r e u s i n g .
s p e c i f y t h e Minkowski m e t r i c ( n o t i n g t h a t t h e same 6 w i l l a l s o
work f o r t h e E u c l i d e a n m e t r i c )
Find 6>0 such t h a t
,
( 3 ) becomes
2
.
Solutions
Block 3 : P a r t i a l D e r i v a t i v e s
Unit 1: Functions of More Than One Variable
3.1.4 (L) continued
We rewrite 1x12+2x2+x33+x42-51 i n t h e more s u g g e s t i v e form
( s i n c e t h i s i s how we g o t 5 i n t h e f i r s t p l a c e ) , where upon we
may conclude t h a t
3
From ( 5 ) , lx12+2x2+x3 +xp2-51 w i l l be l e s s than
a s soon a s
E
T h i s , i n t u r n , w i l l happen a s soon a s each of o u r f o u r summands
i s l e s s than
i.That3i s ,
2
x1 -1 , 2 1 x2-1
w e must have:
I
I
1 , I x3
-1
w e need only make each of t h e numbers
I , and 1 x12-1 1
E
less than 5. I n o t h e r words
The key now i s that each of t h e f o u r i n e q u a l i t i e s i n ( 6 ) involves
only s c a l a r s .
lim
x2+l X2 =
lim
3 =
x3+l X3
lim
2
x4+l 4
=
I n p a r t i c u l a r , s i n c e we know t h a t
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s of More Than One V a r i a b l e
3.1.4(L)
continued
i t f o l l o w s t h a t w e can f i n d 61,
6 2 r cT3
,
and 6 4 such t h a t
I
Hence, i f w e l e t 6 = min~61162163164}Iw e may conclude from ( 7 ) t h a t
i f lxi=ll <6 ( i = 1 , 2 , 3 , 4 )
then
and from ( 5 ) t h i s g u a r a n t e e s t h a t
I n o t h e r words, from t h e d e f i n i t i o n o f t h e Minkowski metric,
which was p r e c i s e l y what had t o be shown.
A major aim o f t h i s e x e r c i s e i s t o have you become convinced t h a t
t h e i d e a of a m e t r i c a p p l i e s t o h i g h e r dimensional c a s e s ( i n o u r
example, we used i t i n t h e 4-dimensional c a s e ) , and t h a t t h e
Minkowski m e t r i c a l l o w s us t o f i n d an n-dimensional d i s t a n c e i n
terms o f n 1-dimensional d i s t a n c e s .
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 1: Functions of More Than One Variable
Given E > O w e wish t o determine 6 , i n terms of
E
such t h a t
We may t a k e equation (6) of t h e previous e x e r c i s e a s our s t a r t i n g
point.
Namely, n e a r xl = 1, xl + 1 i s near 2.
assume I xl+l 1 = xl+l < 3. Then
Hence, i f w e make s u r e t h a t Ixl-11
I n summary, w i t h 61 =
0 < lxl-11
< 61
-+
<
I n p a r t i c u l a r , w e may
E
, we
obtain
E
E! , we
2
lxl -11 <
have
E
and (1) of Equation ( 6 ) i n t h e previous e x e r c i s e i s obeyed.
E
Next, w i t h r e s p e c t t o (2) of Equation ( 6 ) . w e want ix2-11 < 3,
b u t now it i s t r i v i a l t o s e e t h a t ti2 =
allows us t o say
i,
A s f o r ( 3 ) of Equation ( 6 ) , we f i r s t w r i t e
when x3 i s n e a r 1, x '+x3+l i s n e a r 3.
3
s m a l l enough s o that
w e have
1 xg2+x3+l1
Again, we pick our i n t e r v a l
= x3 '+x3+1 < 4 .
Once t h i s i s done
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.5
continued
E
, we
and i f w e now make s u r e t h a t Ix3-11 <
3
E
Ix3 -1 < 4 )
=
,E and 3 i s obeyed.
have
1
I n summary, i f w e l e t
=
E
then
F i n a l l y , w e t r e a t ( 4 ) of Equation ( 6 ) by o b s e r v i n g t h a t n e a r 1,
x 4 + l i s n e a r 2, hence, f o r a s u f f i c i e n t l y s m a l l neighborhood of
1, w e may conclude t h a t 1x4+11 = x 4 + l < 3 .
so that i f
1 x4-1 1
< E
c a s e w e may l e t g 4 =
, t h e n 1 x4 2-11
1E2
<
aE
Then.
.
That is, i n t h i s
.
E
E
E
E
L e t t i n g 6 = min{61,62,63.64),
w e have 6 = En,
gr
and
s i n c e c>O, t h e f r a c t i o n w i t h t h e g r e a t e s t denominator w i l l be
t h e minimum.
Hence, w e may l e t
6 = =
E
.
That i s
The key p o i n t i s t h a t 6 was found by l o o k i n g a t f o u r 1-dimensional
l i m i t s , and i t i s i n t h i s s e n s e t h a t h i g h e r dimensional problems
a r e a s r e a l a s t h e lower d i m e n s i o n a l ones. Moreover, f o r e v e r y
-x such
(Note:
-
E
t h a t IIx-lll
-- < 16
'
(f(x ) - 5 1 < E.
I f you have no q u a n t i t a t i v e f e e l i n g f o r 6 =
E
in this
e x e r c i s e , it might be worthwhile t o c a r r y t h e computational d e t a i l s
a b i t further.
Namely,
E 6 = 16 d e s c r i b e s t h e domain
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s of More Than One V a r i a b l e
3.1.5
continued
i.e.,
a "4-dimensional cube"
Assuming t h a t E > O i s s u f f i c i e n t l y s m a l l t o i n s u r e t h a t a l l numbers
i n t h e above i n e q u a l i t i e s a r e p o s i t i v e , w e have
Theref o r e
i s a lower bound f o r
i n t h i s case, while
i s an upper bound.
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.5
continued
S i m p l i f y i n g w e f i n d t h a t x12+2x2+x33+x42 i s between
and
Our c l a i m now i s t h a t f o r
E
s u f f i c i e n t l y s m a l l t h e numbers named
by (1) and ( 2 ) be between 5-E and 5 + ~ . For example, assuming t h a t
~ < 1w ,e have
E
<
E~
< c3, whereupon
Theref o r e ,
[ A c t u a l l y , by choosing 6 t o be t h e minimum of 61, ti2, 6 3 ,
and 6 4
2481
w e narrowed t h e i n t e r v a l and t h i s i s why
E occurred r a t h e r t h a n something c l o s e r i n v a l u e t o E ]
I n any e v e n t ,
.
( 3 ) shows t h a t
3.1.6 ( L )
a . Many o f t h e same q u e s t i o n s t h a t could have been asked a b o u t l i m i t s
i n o u r s t u d y of c a l c u l u s of a s i n g l e v a r i a b l e can a l s o be asked now.
Among o t h e r t h i n g s , t h e form 0/0 e n t e r s i n t o t h i n g s h e r e j u s t a s
before.
P a r t ( a ) of t h i s e x e r c i s e i s meant t o emphasize t h a t
f ( x , y ) c a n n o t be c o n t i n u o u s a t (0,O) s i n c e it i s n o t even d e f i n e d t h e r e .
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 1: Functions of More Than One V a r i a b l e
3.1.6 (L) continued
I n s t i l l o t h e r words, i f f were continuous a t ( 0 , 0 ) , i t would
mean t h a t
lim (x,yHO,O) f (x,y) b u t t h i s i s impossible s i n c e f (0,O) i s n o t d e f i n e d ( 0 / 0 ) .
b. Even though f i s n o t d e f i n e d a t ( 0 , 0 ) , it does n o t p r e v e n t our
w r i t i n g 1i m
f ( x , y ) s i n c e then w e a r e i n t e r e s t e d i n what
(x,yXO,O)
happens t o f a s (x,y) g e t s a r b i t r a r i l y c l o s e t o (0,O) without
e q u a l i n g (0,O).
The problem i s t h a t t h e l i m i t might (and i n
t h i s c a s e , it does) depend on j u s t how ( x , y ) approaches (0,O)
.
For example, suppose both x and y a r e n o t e q u a l t o 0 and we l e t
(x, y ) approach (0,O) along t h e i n d i c a t e d p a t h :
(Figure 1)
I n t e r m s of t h e i n d i c a t e d p a t h , w e f i r s t hold y c o n s t a n t and l e t
x approach 0. As soon a s o u r path reaches t h e y-axis we hold x
c o n s t a n t (x=O) and l e t y approach 0. Computing t h e l i m i t t h i s
way means
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s of More Than One V a r i a b l e
3.1.6 (L) c o n t i n u e d
N o t i c e t h a t (1) i s a s p e c i a l way i n which w e may w r i t e
one
That i s ,
(1) i n d i c a t e s b u t
o f t h e i n f i n i t e l y many p a t h s by
which ( x , y ) may approach (0,O). E v a l u a t i n g (1) y i e l d s
2 2
l i m [ l i m x2-y2
y+o x+O
x +Y
]
- l i m (-1) = -1
Y+O
On t h e o t h e r hand, t h e p a t h
suggests
l i m x 2 -y 2
x+o
lim
[Y+O x2+y2
A comparison of
]
- lim
-
x+O
2
x - 1 x2 ( 2 ) and ( 3 ) shows t h a t t h e v a l u e o i
i n t h i s example depends on t h e p a t h by which w e approach ( 0 , O ) ;
s i n c e even f o r t h e two s p e c i a l p a t h s w e c o n s i d e r e d , w e g o t two
d i f f e r e n t answers.
The p o i n t i s t h a t when we s a y
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 1: Functions of More Than One Variable
3.1.6 ( L ) continued
w e mean t h e l i m i t e x i s t s and i s L f o r every p o s s i b l e p a t h by which
,...,
x+a.
-
( I n h i g h e r dimensions it means t h a t when w e make )xl-all
and I xn-an 1 s m a l l , o u r numerical answer must be independent of
t h i s o r d e r i n which we make t h e s e q u a n t i t i e s s m a l l , and t h i s , i n
t u r n , says t h a t our answer i s uniquely determined once xk i s
s u f f i c i e n t l y c l o s e t o ak f o r k = 1,
n).
...,
From a more p o s i t i v e p o i n t of view, i f w e a r e t o l d t h a t
lim
( x , y ) + ( a , b )f (x,y)
e x i s t s , then we can be s u r e t h a t i t s v a l u e , among o t h e r ways, i s
given by e i t h e r
The main c a u t i o n i s t h a t u n l e s s we know t h e l i m i t e x i s t s , t h e s e
two d i f f e r e n t l i m i t s may y i e l d d i f f e r e n t answers, a s i n t h e case
of t h e p r e s e n t e x e r c i s e .
c . Here we t r y t o show what goes wrong n e a r t h e o r i g i n by u t i l i z i n g
polar coordinates.
(Among o t h e r t h i n g s , t h i s shows us t h a t
p o l a r c o o r d i n a t e s have a purpose q u i t e a p a r t from c e n t r a l f o r c e
fields: )
L e t t i n g x = r cos 8 and y = r s i n 8 we s e e t h a t
*~
x +Y
2
2
2
= r (cos + s i n 8)
2
r
2
= cos 8
-
= cos 28 2
sin 8
provided r f 0
provided r f 0
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s of More Than One V a r i a b l e
3.1.6 (L) c o n t i n u e d
I n o t h e r words, computing
limcos
r+o
i s e q u i v a l e n t t o computing
20.
The p o i n t i s t h a t t h e v a l u e of c o s 20 depands o n l y on 8
IT
-
For example, i f w e choose t h e r a y 8 = g r then c o s 28 = 2'
IT
i f ( x , y ) + ( O, O ) a l o n g t h e r a y 9 =
then
n o t r.
Therefore,
'TT
A s a check, n o t i c e t h a t i n C a r t e s i a n c o o r d i n a t e s t h e r a y 8 = -i;
becomes t h e h a l f - l i n e
tan
1T
1
= -
43
$=
w e have y =
IT
t a n i; i n t h e f i r s t q u a d r a n t .
X
on
t h i s ray.
Thus,
J?j
Therefore,
lim
Pictorially,
2
2
1
,
a s predicted.
Since
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 1: Functions of More Than One V a r i a b l e
3.1.6 ( L ) continued
I f (x,y)#(O,O), f (x,y) i s c o n s t a n t on each r a y 0=8
b u t t h e v a l u e of 0'
t h e c o n s t a n t depends on O O .
Namely, t h e c o n s t a n t i s cos 200; i.e., f o r each p o i n t ( x , y ) on t h e l i n e 0=00, f (x,y) = cos 200. -
[ ( 0 , 0 ) i s excluded from c o n s i d e r a t i o n on 8 = 0 0 1 s i n c e .'.
f (x,y) =
x
.'.
~
-
~
only i f
~
(x,y)#(O , O )
I. x +Y (Figure 2)
I n summary, here, we have a r a t h e r e x c e p t i o n a l example i n which
w e can make f ( x , y ) t a k e on any value i n [-1,lI
merely by choosing
t h e a p p r o p r i a t e p a t h by which w e allow (x,y)+(O,O).
3.1.7
a.
We have
L e t t i n g x = r cos 0 and y = r s i n 0 , we have t h a t
f (x,Y) = f (r cos 8, r s i n 0 ) =
2 r cos 8 (r s i n 8)
2
r
= 2 s i n 8 cos 8, provided r
theref ore
f ( x , y ) = s i n 2 0,
provided r # 0
# 0
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.7
continued
therefore
1i m lim s i n 28 = s i n 200
(x,y)-+(O,O)f ( x , y ) = r+O 88
'
S i n c e t h e p a t h i s determined by
eO,
( 2 ) shows t h a t d i f f e r e n t p a t h s
lead t o d i f f e r e n t values f o r the l i m i t .
T h i s w i l l be e x p l o r e d
i n more d e t a i l i n t h e remaining p a r t s o f t h i s e x e r c i s e and a l s o
i n the next exercise.
b.
If
e0
=
a'A ,
(2) y i e l d s
1i m
f(x,y) = s i n 2
(x,y)+(O,O) A s a check O0 =
($) =
aIT c o r r e s p o n d s
sin T = 1
2
.
t o t h e r a y y=x, ~ 2 0 . I n t h i s e v e n t ,
w e have
2
f ( x , ~ )=
2x
X +X
--
1
( i f x#Ol T h e r e f o r e , on y=x, x30, w e have lim (x,y)+lO,O)
f(x,y) = 1 which checks w i t h ( 3 ) .
c.
Along t h e r a y 8=0, ( 2 ) y i e l d s
lim
f ( x , y ) = s i n 2 (0) = 0
(x,y)+(O,O) A s a check t h i s i s t h e r a y y=O, xbO.
In t h i s event Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 1: Functions of More Than One V a r i a b l e
-
3.1.7
- -
continued
theref ore
When 8 =
71
,
28=r
and s i n 28 = 0.
Therefore, on t h e r a y x=O,
theref ore,
lim
(x,y)+(O,O)
%.- +y
=
o
i n this c a s e a l s o .
Comparing (b) and (c) we see t h a t while l i m
f (x,y) = 0
(x,y)+(O , O )
i f (x,y)+(O,O) e i t h e r along t h e x-axis o r t h e y-axis,
1i m
IT
f ( x , y ) = 1 along t h e l i n e of approach 8 = T .
(x,y)+tO,O) This shows
lim
t h a t (1) ( x , y ) + ( O , O )f (x,y) does n o t e x i s t s i n c e i t s v a l u e depends
on t h e p a t h by which (x,y)+(O,O) , and a s an a s i d e ,
( 2 ) merely
knowing t h a t f ( x , ~ )approaches t h e same l i m i t a s (x,y) approaches
(0,O) along e i t h e r t h e x- o r t h e y-axis i s n o t enough t o say t h a t
this common v a l u e i s l i m
f (x,y).
(x,y)+(O,O)
3.1.8(L) I n a way, this e x e r c i s e i s a c o r o l l a r y t o t h e previous one, b u t i t i s of enough s i g n i f i c a n c e i n terms of what comes n e x t i n o u r course t h a t w e wanted t o p r e s e n t the i d e a here. Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s of More Than, One V a r i a b l e
3.1.8 (L) c o n t i n u e d
I n o u r l a t e r work, e x p e c i a l l y w i t h 2 - t u p l e s ,
w e s h a l l s e l e c t two
p a t h s ( f o u r , c o u n t i n g s e n s e ) by which ( x , y ) w i l l approach ( a , b ) ,
and w e s h a l l t r y t o do a l m o s t e v e r y t h i n g e l s e i n terms of what
happens a l o n g t h e s e two p a t h s .
The p a t h s w i l l b e (1) w e h o l d
y c o n s t a n t and l e t x approach a , t h e n w e h o l d x c o n s t a n t and l e t
y approach b;
( 2 ) w e h o l d x c o n s t a n t and l e t y approach b , t h e n
w e h o l d y c o n s t a n t and l e t x approach a .
These two p a t h s a r e
i n d i c a t e d below.
hY
( a ,b)
9
X
Now, a s w e s a i d i n E x e r c i s e s 3.1.6
and 3.1.7,
the l i m i t of f(x,y)
a s ( x , y ) approaches ( a , b ) may be d i f f e r e n t along t h e s e two p a t h s .
The p o i n t i s t h a t it s h o u l d n o t be s u r p r i s i n g t h a t t h e l i m i t
depends on t h e p a t h a s w e l l a s t h e p o i n t of e v a l u a t i o n of t h e
l i m i t ; y e t , a s w e g e t f u r t h e r i n t o t h e c o u r s e , we s h a l l become
more and more p r e o c c u p i e d w i t h t h e two s p e c i a l p a t h s d i s c u s s e d
above. What w i l l happen i s t h a t w e s h a l l have, i n g e n e r a l ,
s u f f i c i e n t l y smooth f u n c t i o n s s o t h a t what happens along t h e s e
two s p e c i a l p a t h s w i l l govern what happens along any o t h e r p a t h ,
b u t b a r r i n g t h i s " s u f f i c i e n t smoothness," there i s no g u a r a n t e e
t h a t what happens a l o n g t h e s e two s p e c i a l p a t h s i s enough t o t e l l
u s what happens a l o n g e v e r y p a t h .
T h i s e x e r c i s e i s meant a s a forewarner; it i s d e s i g n e d t o make
s u r e t h a t w e u n d e r s t a n d t h a t when w e s a y "independent of t h e
p a t h " w e mean more t h a n j u s t e i t h e r t h e h o r i z o n t a l o r v e r t i c a l
directions.
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 1: Functions of More Than One V a r i a b l e
3.1.8 (L) continued
Based on t h e f a c t t h a t f a s d e f i n e d i n t h e previous e x e r c i s e was
n o t d e f i n e d a t ( 0 , 0 ) , coupled with t h e f a c t t h a t t h e l i m i t of f
a s ( x , y ) approached (0,O) along t h e r a y s 8=0 and 8 =
IT was 0,
w e a r e motivated t o d e f i n e a new f u n c t i o n g by:
r
I n o t h e r words, t h e g of t h i s e x e r c i s e and t h e f of t h e previous
e x e r c i s e a r e i d e n t i c a l provided t h a t ( x , y ) # ( 0 , 0 ) , b u t a t (0,O)
i s d e f i n e d , and, i n p a r t i c u l a r , g(O,O)=O.
f i s n o t d e f i n e d while g -
-
a . We a l r e a d y know from t h e previous e x e r c i s e t h a t g i s n o t cont i n u o u s a t ( 0 , 0 ) , s i n c e among o t h e r t h i n g s we showed t h a t i f
IT
( x , y ) approached (0,O) along t h e r a y 8 = -4 , then
lim
(x.~)'(o.o)
= 1
*.
+Y
Notice t h a t t h e meaning of " l i m i t " i n (2) t e l l s us t h a t
(x,y)# ( O , O ) ,
Hence,
and under t h i s c o n d i t i o n ,
(1) t e l l s us t h a t
(2) may be replaced by
lim
g ( x , y ) = 1 # g(0,O)
(x,y)+(O,O)
,
s i n c e g(O,O) = 0 .
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.8 (L) c o n t i n u e d
Thus, a t l e a s t a l o n g t h e p a t h d e f i n e d by 8 =
aIT
Equation ( 4 ) t e l l s u s t h a t g i s n o t c o n t i n u o u s a t (0,O) s i n c e i f
i t were
lim
(x,y)-t(Q,O)
g ( x , y ) = g(0,O)
a l o n g e v e r y p a t h c o n n e c t i n g t o (0,O)
b.
.
What w e want t o show h e r e i s t h a t
i f t h e approach i s a l o n g e i t h e r a x i s .
I f w e approach (0,O) a l o n g t h e x - a x i s ,
y=O f o r t h e e n t i r e p a t h .
Thus we o b t a i n , a l o n g t h i s p a t h ,
lim
(x,y)+(O,O)
lim
-
lim
4%
+Y
x-to
y=o
lim
x-to
x
*
0
7
I n terms o f p o l a r
* R e c a l l t h a t x+O i n c l u d e s b o t h X+O+ and X+O-.
c o o r d i n a t e s , we must c h e c k t h a t t h e l i m i t s e x i s t and a r e e q u a l f o r
t h e two r a y s 8=0 and @ = I T . In t h i s c a s e , s i n c e g ( x , y ) i s s i n 2 8
f o r ( x , y ) # ( 0 , 0 ) , b u t 8=0 and IT make g ( x , y ) = 0 a s ( 5 ) i n d i c a t e s .
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 1: F u n c t i o n s of More Than One V a r i a b l e
3.1.8 ( L ) continued
S i m i l a r l y , approaching (0,O) a l o n g t h e y-axis y i e l d s
lim
(x,y)+(O I01
-
xI-" 0
Y+O
- lim
Y+O
*
x +Y
0
y2 T h i s example shows once and f o r a l l t h a t i f
lim
g ( x , y ) = g(0,O)
(x,y)+(O,O)
a s ( x , y ) +(0 , O )
along e i t h e r a x i s , w e cannot g u a r a n t e e t h a t g i s
continuous a t (0,O).
( T r i v i a l l y , Of c o u r s e , i f w e know t h a t g i s
continuous a t (0,O) t h e n i n p a r t i c u l a r ( 6 ) must be t r u e . T h a t i s ,
once w e know t h a t 1i m + ( O , O ) g ( ~ I y
= )g(0,O) f o r e v e r y p a t h ,
( ~ I Y )
t h e n w e know t h a t i t ' s t r u e f o r a p a r t i c u l a r p a t h ; i f , however,
w e know what t h e l i m i t i s f o r some p a r t i c u l a r p a t h , w e cannot
n e c e s s a r i l y conclude what it i s f o r e v e r y p a t h . )
Our main aim h e r e i s t o show how much more complicated f u n c t i o n s
of s e v e r a l v a r i a b l e s a r e t h a n f u n c t i o n s of a s i n g l e v a r i a b l e .
W e have d e l i b e r a t e l y chosen t h e c a s e n=2 h e r e s o t h a t (1) w e can
a g a i n c a p i t a l i z e on any g e o m e t r i c i n t e r p r e t a t i o n s and ( 2 ) w e can
show t h a t t h e c o m p l i c a t i o n s do n o t r e q u i r e t h a t we have more t h a n
t h e " u s u a l " 3-dimensional space.
The proof i s n o t r e a l l y t o o d i f f i c u l t once w e g e t t h e hang of
things.
( I f you have d i f f i c u l t y v i s u a l i z i n g o u r approach, perhaps
i f , a s w e proceed, you r e f e r t o F i g u r e s 1 and 2 , t h i n g s w i l l s e e m
To b e g i n w i t h , i f f happens t o be a c o n s t a n t f u n c t i o n
clearer.)
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.9 (L) c o n t i n u e d
( i . e . , f (xl,yl)
= f ( x 2 , y 2 ) f o r e v e r y p a i r of p o i n t s (xl,yl)
and (x2 ,y 2 ) i n t h e domain of f ) , t h e a s s e r t i o n o f t h e theorem
f o l l o w s t r i v i a l l y , f o r i n t h i s c a s e t h e r e i s a number c such t h a t
f ( x , y ) = c f o r a l l ( x , y ) i n t h e domain of f .
Thus, t h e i n t e r e s t i n g c a s e i s when f i s n o t a c o n s t a n t f u n c t i o n .
Assuming, t h e r e f o r e , t h a t f i s n o t a c o n s t a n t f u n c t i o n , t h e r e a r e
a t l e a s t two p o i n t s P (xl, yl) and Q ( x 2 ,y 2 ) f o r which f (PI # f ( Q )
.
Now p i c k any number, s u b j e c t o n l y t o t h e c o n d i t i o n t h a t it l i e
between f ( P ) and f ( Q ) .
T h a t t h e r e a r e such numbers i s g u a r a n t e e d
L e t u s d e s i g n a t e one such number
by t h e f a c t t h a t f (P) # f ( Q ) .
by m.
L e t C d e n o t e any c u r v e t h a t c o n n e c t s P and Q.
Since f
i s c o n t i n u o u s , t h e r e must be some p o i n t on C , s a y X , f o r which
f ( X ) = m, s i n c e a continuous f u n c t i o n t a k e s on a l l i t s i n t e r m e d i a t e
values.
Of c o u r s e , w h i l e you a r e l i k e l y t o a c c e p t t h i s l a s t s t a t e m e n t a s
b e i n g t r u e ( o r a t l e a s t , a s b e i n g r e a s o n a b l e ) , t h e f a c t remains
t h a t w e proved i t o n l y i n t h e c a s e of a s i n g l e r e a l v a r i a b l e .
T h i s proof can be extended t o t h e c a s e o f f u n c t i o n s of s e v e r a l
r e a l v a r i a b l e s , b u t it i s beyond o u r p r e s e n t purposes t o become
While t h e i n t e r e s t e d
s o computationally involved a t t h i s t i m e .
r e a d e r s h o u l d f e e l f r e e t o prove t h i s r e s u l t on h i s own i f he s o
d e s i r e s , w e w i l l use a geometrical interpretation.
Recall t h a t
Now suppose
i n t h i s c a s e t h e domain of f i s t h e e n t i r e xy-place.
t h a t f i s c o n t i n u o u s . T h i s means, p i c t o r i a l l y , t h a t t h e s u r f a c e
z = f ( x , y ) i s unbroken. Again, t o add c o n c r e t e n e s s t o o u r d i s c u s s i o n , l e t u s r e p l a c e such l i t e r a l p o i n t s a s P and Q by t h e more
s p e c i f i c p o i n t s , s a y , P ( 1 , 2 ) and Q ( 3 , 4 ) . Suppose, w i t h o u t l o s s of
g e n e r a l i t y , t h a t f ( 1 , 2 ) = 6 w h i l e £ ( 3 , 4 ) = 9. Now l e t C be any
The p a t h g e n e r a t e d by a l i n e
c o n t i n u o u s c u r v e t h a t j o i n s P t o Q.
(Recall
p a r a l l e l t o t h e z - a x i s running a l o n g C g i v e s us a c y l i n d e r .
t h a t m a t h e m a t i c a l l y a c y l i n d e r i s o b t a i n e d by t r a c i n g along any
c u r v e i n t h e p l a n e w i t h a l i n e p e r p e n d i c u l a r t o t h e c u r v e ) , and
t h i s c y l i n d e r i n t e r s e c t s t h e s u r f a c e z = f ( x , y ) t o form an unbroken ( s p a c e ) c u r v e .
A l l we a r e now s a y i n g i s t h a t s i n c e
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s of More Than One V a r i a b l e
3.1.9 (L) c o n t i n u e d
t h i s s p a c e c u r v e i s unbroken, and i t p a s s e s through t h e h e i g h t s 6 and 9 , i t must p a s s through e v e r y h e i g h t between 6 and 9. S t a t e d i n terms o f p l a n e s , w e a r e s a y i n g t h a t o u r s p a c e c u r v e originates
z = 9.
i n t h e p l a n e z = 6 and t e r m i n a t e s i n t h e p l a n e S i n c e t h e c u r v e i s unbroken, i t must i n t e r s e c t e v e r y p l a n e z = k where 6 < k < 9. Thus, s t i l l u s i n g o u r p a r t i c u l a r example, s i n c e 6 s 8
< 9, there is a point P
on t h e c o n t i n u o u s c u r v e C1 which c o n n e c t s 1
( 1 , 2 ) and ( 3 , 4 ) f o r which £ ( P I ) = 8. N o t i c e t h a t C1 was
any c u r v e t h a t connected ( 1 , 2
and ( 3 , 4 )
.
The p o i n t i s t h a t t h e r e a r e i n f i n i t e l y many such continuous c u r v e s which have no p o i n t s o t h e r t h a n ( 1 , 2 ) and ( 3 , 4 ) i n common. For each of t h e c u r v e s C n , t h e r e i s a p o i n t Pn f o r which f ( P n ) = 8 . S i n c e t h e c u r v e s i n t e r s e c t o n l y a t ( 1 , 2 ) and ( 3 , 4 ) t h e p o i n t on C (no Pn can j
j
be e i t h e r ( 1 , 2 ) o r ( 3 , 4 ) because f ( P n ) = 8 w h i l e f ( 1 , 2 ) = 6
and f ( 3 , 4 ) = 9 ) . Thus, w e have i n f i n i t e l y many d i f f e r e n t p o i n t s
f o r which f ( p l ) = f ( P 2 ) = . . . = f ( P n ) =
= 8.
P1, P 2 , . . . , Pn,
An a t t e m p t t o summarize t h e s e r e s u l t s p i c t o r i a l l y i s made below:
Pi on Ci
c a n n o t be t h e same a s t h e p o i n t P
....
...
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.9 (L) c o n t i n u e d
1. Cn i s a c o n t i n u o u s c u r v e
i n t h e xy-plane c o n n e c t i n g P ( 1 , 2 ) and Q ( 3 , 4 ) . 2. The ( r i g h t ) c y l i n d e r
a l o n g Cn i n t e r s e c t s t h
surface z=f(x,y)
along t h e space
curve P'Q'.
3.
P' i s i n t h e p l a n e
2=6 s i n c e i t s
coordinates are (1,2,6)
Q ' i s i n 2=9 s i n c e i t s
coordinates a r e (3,4,9).
4.
The p l a n e 2=8 i n t e r sects P'Q' a t , a t least,
one p o i n t Qn s i n c e
P ' Q ' i s a continuous
curve.
Y
The p r o j e c t i o n of Qn i n t o t h e xy-plane
y e i l d s Pn on Cn whereupon f (Pn) = 8.
X
( F i g u r e 1)
The i m p o r t a n t t h i n g i n F i g u r e 1 i s t h a t Cn was any continuous c u r v e
There a r e i n f i n i t e l y many such c u r v e s w i t h
t h a t j o i n s P t o Q.
I n Figure 2 four
no p o i n t s o f i n t e r s e c t i o n o t h e r t h a n P and Q.
such c u r v e s C1,
C2, Cj,
C 4 a r e drawn.
By t h e construction i n
F i g u r e 1, w e may o b t a i n P1 on C1, P2 on C 2 , P3 on C 3 , P4 on C 4
such t h a t f (Pn) = 8 f o r e v e r y n = 1 , 2 , 3 , 4 .
Solut.ions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 1: Functions of More Than One V a r i a b l e
3.1.9 (L) continued
(Figure 2)
The p o i n t w e now wish t o make a c t u a l l y begins with t h e s o l u t i o n of
t h i s exercise.
We have now shown t h a t i f f:E2+E i s any continuous
f u n c t i o n , then it i s impossible f o r f t o be 1-1. I n f a c t , i n t h e
p a r t i c u l a r i l l u s t r a t i o n we used, t h e r e were i n f i n i t e l y many p o i n t s
i n t h e domain of f a t which f was 8.
What t h i s e x e r c i s e shows i s t h a t i f t h e mapping from n-dimensional
space (with n > l ) i n t o 1-dimensional space i s continuous, then f
cannot be 1-1. R e s t a t e d more formally:
I f n > l and f : E ~ + E i s continuous then f i s n o t 1-1, and, i n f a c t ,
t h e r e can be i n f i n i t e l y many elements i n En which have t h e same
image i n E.
(Notice t h a t our argument holds i f t h e domain of f i s any s u b s e t
of En.
That i s , a l l t h a t o u r argument r e q u i r e d was t h a t
f ( P ) # f ( Q )n o m a t t e r how c l o s e t o each o t h e r P and Q might be.)
Thus, w e should begin t o f e e l t h a t while much of t h e theory f o r
f u n c t i o n s of n - t u p l e s i s a g e n e r a l i z a t i o n of t h e r e s u l t s f o r t h e
1-dimensional c a s e , t h e r e a r e enough new "wrinklesn when n > l t o
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s o f More Than One V a r i a b l e
3.1.9 (L) c o n t i n u e d
remind u s t o be on guard.
Among o t h e r t h i n g s , t h e r e a r e i n f i n i t e l y
many c o n t i n u o u s f u n c t i o n s £:E'+E
which a r e 1-1 (namely t h o s e
whose g r a p h s a r e e i t h e r monotonically r i s i n g o r monotonically
falling)
.
By now, w e hope t h a t i t i s v i r t u a l l y s e l f - e v i d e n t t o you t h a t n=4
was chosen simply f o r c o n c r e t e n e s s and t h a t o u r approach e x t e n d s
v e r b a t i m t o any v a l u e of n.
TO prove t h a t "=" i s an e q u i v a l e n c e r e l a t i o n w e must show t h a t
-
4
(1)
5
(2)
a = b + b = a , a a n d b i n E
= a f o r e a c h asE
4
(3) g = b a n d-b = c + a = c
( I ) , w e have 5 = (al,a2 , a 3 , a 4 ) where al,a2 , a 3 , a 4 a r e a l l
r e a l numbers. By t h e p r o p e r t i e s o f e q u a l i t y f o r r e a l numbers, we
AS f o r
know t h a t
al = a l ,
a2 - a 2 , a 3 = a3 , and a4 = a4.
Hence, by d e f i n i t i o n of v e c t o r ( n - t u p l e ) e q u a l i t y , w e have
which e s t a b l i s h e s t h e v a l i d i t y of
(1).
( N o t i c e how (1) was e s t a b l i s h e d f o r
p r o p e r t y o f e q u a l i t y f o r r e a l numbers).
by v i r t u e of t h e corresponding
Solutions Block 3: Partial Derivatives Unit 1: Functions of More Than One Variable 3.1.10 continued AS for (21, let a = (al,a2,a3,a
by definition, a = b
- means
)
al = bl, a2 = b2, a3 = b3, and a4
and b
=
=
(bl,b2,b3,b4).
Then,
b4
But, for real numbers, equality is symmetric.
conclude that
Hence, we may
bl = al, b2 - a2, b3 = a3, and b4 = a4
whence, by definition,
or b=5
which establishes the validity of (2). Finally, to establish the validity of (3), we let a = (al,a2,a3,a4),
-
b
= (bl,b2,b3,b4) and c = (c1,c2,c3,c4).
Then, from a = b
- we conclude that al = bl, a2 = b2, a3 = b3 and a4 = b4, while from
b1 = c
b
= c we conclude that b2 = c2, b3 = c3, and b4
=
C4
Since for real numbers, a = b and b = c+a = c, we may conclude
from (1) and (2) that
al = cl, a2 = c2, a3
=
c3 and a4 = c4
.
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 1: F u n c t i o n s of More Than One V a r i a b l e
3.1.10
continued
Hence
Notice how t h i s proof augments o u r remark i n t h e supplementary
n o t e s t h a t t h e f a c t t h a t = i s an e q u i v a l e n c e r e l a t i o n on E~ i s
induced by t h e f a c t t h a t v e c t o r s may be viewed a s n - t u p l e s ,
each o f whose components i s r e a l , and e q u a l i t y i s an e q u i v a l e n c e
r e l a t i o n on t h e r e a l numbers.
3.1.11
(Note:
The t e c h n i q u e used i n t h i s s p e c i f i c e x e r c i s e i s t h e one
used whenever w e d e c i d e t o view v e c t o r s i n terms of t h e i r components)
.
We have, by d e f i n i t i o n of t h e a r i t h m e t i c of E
4
,
that
Hence, by o u r d e f i n i t i o n o f d o t p r o d u c t
a
-
(b+c) =
(al,a2 ,a 3 l a 4 )
S i n c e t h e a s , bk I
s
(bl+clIb2+c2 ,b3+c3 , b 4 + c 4 )
and c k l s (k=112,3,4) a r e numbers, we know
t h a t (1) may be r e w r i t t e n a s
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 1: Functions of More Than One Variable
~
-
3.1.11 continued
which, i n t u r n , i s
which proves t h e r e q u i r e d r e s u l t .
Again, n o t i c e how we e f f e c t i v e l y reduce t h e study of E~ t o f o u r
s e p a r a t e s t u d i e s of E1 ( i . e . , t h e f o u r components of any v e c t o r
4
[n-tuple] i n E ). That i s , t h e v e c t o r r e s u l t i s induced by t h e
corresponding s c a l a r r e s u l t .
MIT OpenCourseWare
http://ocw.mit.edu
Resource: Calculus Revisited: Multivariable Calculus
Prof. Herbert Gross
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