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16. Use the epsilon-delta definition to verify that
g(x,y)=x^2y^2 is continuous at (-1,2).
Definition of continuity:
A function f is continuous at x0 if, given any eps>0, there is a
delta>0 so that
if |x-x0|
Applying the definition to this problem: vector_x=,
vector_x0=<-1,2>.
We need to verify:
IF: sqrt((x+1)^2+(y-2)^2)<delta, then |(x^2*y^2)-(-1^2*2^2)|<epsilon
sqrt((x+1)^2+(y-2)^2)<delta: Since both sqrt((x+1)^2) and
sqrt((y-2)^2) are greater than or equal to zero, if the restriction is
satisfied, then each of the parts satisfies it. So then
|x+1|<delta and |y-2|<delta are true also (because
(sqrt((x+1)^2)=|x+1| and sqrt((y-2)^2)=|y-2|). We can look for delta
restrictions on |x+1| and |y-2| separately.
|(x^2*y^2)-(-1^2*2^2)|<epsilon =>*1
In order to control the changes of x&y individually, we can rewrite
this equation as:
|((x^2*y^2)-(x^2*2^2))[part1]+((x^2*2^2)-(-1^2*2^2))[part2]|<epsilon.
We are allowed to do this because: -(x^2*2^2)+( x^2*2^2)=0. Therefore,
we are not actually adding anything to the equation). For part 1 of
the equation, only the y-value is changing, while in part 2 only the
x-value is changing.
|(x^2*y^2)-(x^2*2^2)+(x^2*2^2)-(-1^2*2^2)|<epsilon
According to the triangle inequality:
|(x^2*y^2)-(x^2*2^2)+(x^2*2^2)-(-1^2*2^2)|<=|(x^2*y^2)-(x^2*2^2)|+|(x^2*2^2)-(-1^2*2^2)|
Since I have just split the equation into 2
pieces, I know that each piece makes up some fraction of the original
formula *1. Therefore, if I can verify that
|(x^2*y^2)-(x^2*2^2)|<epsilon/2 and if I can verify that
|(x^2*2^2)-(-1^2*2^2)|<epsilon/2, then I will be done. (Adding these 2
equations will again give *1.)
|(x^2*y^2)-(x^2*2^2)|<epsilon/2
Factoring out x^2 gives:
|x^2|*|y*2-4|<epsilon/2 or |x^2|*|y+2|*|y-2|<epsilon/2
I want this inequality to be true. Suppose I know that |x|<=2 and
|y|<=3. If these conditions are true, then I know that the maximum
value of |x^2|=4 and the maximum value of |y+2|=5. Therefore, the
equation, solved for |y-2|, now becomes: |y-2|<epsilon/40.
Factoring out a |2^2| gives: 4|(x^2)-1|<epsilon/2, so |(x+1)*(x-1)|<epsilon/8.
Using the same estimates that I suggested before, I know that
the maximum value of |x-1| is 1. Plugging this in to the equation I now have
|x+1|<epsilon/8.
Now I collect the conditions that must be satisfied for this inequality
to apply to the more general condition. I need:
|x|<=2 and |y|<=3, |x+1|<epsilon/8, |y-2|<epsilon/40.
What I am able to specify is sqrt(|x+1|^2+|y-2|^2)<delta.
I now choose delta to be the minimum of eps/8 and eps/40 and 2 and
3. I just really need the minimum of epsilon/40 and 2.
Collecting all the restrictions on the variables shows that:
If |(x,y)-(-1,2)|<delta, then |f(x,y)-f(-1,2)|<eps,
where delta is the minimum of epsilon/40 and 2.
Sent by Jason Tokayer on Sun, 13 Oct 2002 10:29:19
Special added bonus!
Here is the substance of several e-mail messages I wrote to a student
explaining another example. Maybe what's written here will be
helpful to some people.
let's work through a 1 variable example. the algebra will be
irritating, but, what the heck. i believe that the limit of x^3 as
x--> 2 is 8. so i should be able to make |x^3-8| small by making
|x-2| small. but how could i verify that? (yes, i could graph things,
etc., but this is really preparation for a more than 1 variable
example, so i would prefer to stick with algebraic techniques.)
the connection between x^3-8 and x-2 is "simple". i hope i have it
right:
(x^3-8)=(x^3-2^3)=(x-2)(x^2+2x+4)
(you aren't watching me, thank goodness ... i am painfully checking
what i just typed ... yes, it IS correct.)
suppose i wanted |x^3-8| to be less than, say, 1/(10,000). (i am
picking a specific number just to have something specific to work
with). well, if i look at the equation above, all i seem to need to do
is make |x-2| really really small. but i do need to deal with the
other factor. is IS possible that x-2 could be small while the other
stuff (x^2+2x+4) could be really big, so the product wouldn't be
small. well, i need some sort of control. so right now, i will insist
that |x-2|<1. (again, the "1" is pulled out of the air, but i do need
something.) if |x-2|<1, then x must be between 1 and 3, i think. and
then i will look at x^2+2x+4: the pieces are all positive, so if i
want to estimate the largest possible size i just need to find the
largest size of each piece. and the pieces are increasing. so, in this
case (which is "easy") i can just plug in 3 for x. and i have the
following logical implication:
IF |x-2|<1, THEN |x^2+2x+4|<19
(i put x=3 in the quadratic, i hope correctly.)
well then, if |x-2|<1, i know that |x^3-8|= |x-2|*|x^2+2x+4|< 19|x-2|.
if i want this less than 1/(10,000), i should choose
|x-2|<1/(190,000).
o.k. now put it together. if i want |x^3-8| to be less than some
epsilon (the "output error") then i'd better make |x-2| less than
... let's see: i need epsilon/19 AND i also need it to be less than 1
(otherwise i can't control |x^2+2x+4| enough to get the estimate
|x^2+2x+4|<19 which i need.
so there seem to be two possible restrictions: |x-2|<epsilon/19 AND
|x-2|<1. the traditional name for the restriction on |x-2| is
delta. so if i say that delta is the MINIMUM of 1 and epsilon/19, this
logical statement is correct:
IF |x-2|<delta, THEN |x^3-8|<epsilon.
i think there's very little "higher math" in this, but certainly some
intricate logic.
well, what we did here was a rather straightforward computation. in
more complicated examples, especially in more than 1 variable, there
may be more restrictions needed.
let me try a 2 variable example and see if it helps. look at xy^2 as
(x,y)-->(4,5). i am confident that xy^2 will get close to 100. (i
think that is 4*5^2.) but i would like to control the closeness. in
fact, i would like to know restrictions to put on x and y so that i am
GUARANTEED that |xy^2-100|<1/(10,000). i will use the techniques i
showed in class yesterday. again, this willbe tedious, but what i'm
doing has these advantages: 1. it is essentially elementary (no fancy
stuff needed!). and 2. it works -- it gives an effective answer.
so now we start:
|xy^2-100|=|xy^2-4*5^2|=|xy^2-4y^2+4y^2-4*5^2|
here in this step i have replaced "0" by "-4y^2+4y^2" and the aim is
to make a 2 variable change into a succession of 1 variable
changes. now i will use the triangle inequality in order to make
things easier to handle:
|xy^2-4y^2+4y^2-4*5^2|<=|xy^2-4y^2|+|4y^2-4*5^2|
i use <= for "less than or equal to" because the ascii character set
doesn't have the appropriate thing.
there are two pieces. i need to keep my eye on the final goal, which
is to get things less than 1/(10,000). since there are two pieces, i
know that if i can guarantee each of the pieces is less than
1/(20,000), i will have won. (yes, i could split things up
differently, but i just want to "win" here.)
so i want |4y^2-4*5^2|<1/(20,000) and i will do this by "controlling"
|y-5|. here is the algebra:
|4y^2-4*5^2|=4|y^2-5^2|=4|y-5|*|y+5|.
now i need to control the size of |y+5|. i will make an assumption
about |y-5|. just to be a bit different, let me try |y-5|<.5 (yeah, i
could have used 1, but let me do this in an effort to convince you
that other choices work.) if |y-5|<.5, i know that -.5<y-5<.5 so
4.5<y<5.5 and so 9.5<y+5<10.5 and so |y+5|<10.5. so:
IF |y-5|<.5, THEN |4y^2-4*5^2|=4|y-5|*|y+5|<4*|y-5|*10.5=42|y-5|.
but i want this to be less than 1/(20,000). i can achieve this if i
know |y-5|<1/(840,000). the 840,00 is the product of 42 and
20,000. (hey, now you may begin to see why people invented algebra in
the first place: the numbers are getting weirder and weirder.)
i am NOT done yet. i still must make the other term "small". that is,
i still want |xy^2-4y^2|<1/(20,000). so let's see:
|xy^2-4y^2|=y^2|x-4|
now the multiplier is y^2. but, thank goodness, i already have some
control over y because of the previous work. i know that 4.5<y<5.5, so
20.25<y^2<30.25 (in another window i just used a calculator. i was NOT
too intelligent when i chose that darn .5 instead of 1. but now you
may see why people make simple choices like 1 instead of things like
.5).
i now know:
if |y-5|<.5, then |xy^2-4y^2|=y^2|x-4|<30.25|x-4|.
this is less than 1/(20,000) if i require
|x-4|<1/(30.25*20,000)=1/(605,000).
sigh. the numbers are getting more and more ludicrous.
o.k. let me sum up the EASY EXAMPLE (which i think i have managed to
make complicated enough!).
IF |x-4|<1/(605,000) and IF |y-5|<.5 and IF |y-5|<1/(840,000) THEN
|xy^2-100|<1/(10,000).
this is too complicated. why not look carefully at the numbers. the
"tightest" restriction is with the smallest number, in this case
1/(840,000). so a simpler statement is the following:
IF the distance from (x,y) to (4,5) is less than 1/(840,000), THEN
|xy^2-100|<1/(10,000) is guaranteed.
notice please that this may not be "best possible" -- i did NOT ask
for that. what i mean is that maybe a looser restriction may serve
just as well to control the output. i wanted a simple way to get a
simple restriction on the input error to guarantee the output
error.
in this example, if instead of 1/(10,000) i had written epsilon, what
would i need to get the output restriction? let's see: if i stick with
the darn |y-5|<.5, i get that i need |y-5|<epsilon/84 and i also need
|x-4|<epsilon/60.5. there are 3 restrictions. now if i make delta
equal to the minimum of .5 AND epsilon/84 AND epsilon/60.5 (stupid
choice, that original .5, darn it!) then i will know:
IF the distance from (x,y) to (4,5) is less than this delta described
above, THEN |xy^2-100|<epsilon is guaranteed.
We need to control |y-5| and |x-4|. Specifically, we need several
assumptions to get the sequences of inequalities to work.
So: |x-4|<epsilon/60.5, |y-5|<epsilon/84 and |y-5|<.5 are all
needed.
It would be nice to "package" this better, so that anyone who actually
wants to use it could easily verify that all three (or four or five or
sixty-eight, however many in the application being considered)
inequalities are correct.
CLAIM: if sqrt((x-4)^2+(y-5)^2) < K then |x-4|<K and |y-5|<K.
Why is this correct? Well, take sqrt((x-4)^2+(y-5)^2) < K and square
it. Then (x-4)^2+(y-5)^2 < K^2 of course. Both of the terms on the
left-hand side of the inequality are non-negative. Certainly each of
them is < K^2, otherwise the stated inequality would be wrong. SO we
know (x-4)^2<K^2. Take the square root. Then sqrt((x-4)^2)<K. But
sqrt((x-4)^2) is actually |x-4|. So |x-4|<K. The same is true for the
other term, so |y-5|<K.
If we use the CLAIM, then
|x-4|<epsilon/60.5, |y-5|<epsilon/84 and |y-5|<.5
will be verified if sqrt((x-4)^2+(y-5)^2) < K where K is selected so
that K is the smallest of the numbers epsilon/60.5, epsilon/84, and .5
and that suggests that we just take K to be the minimum of those three
numbers.
The standard name for K is delta. Another reason that people like to
use sqrt((x-4)^2+(y-5)^2) to measure "error" is that it corresponds
to distance between (x,y) and (4,5), so that there is a picture (?) of
what it supposedly measures.
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