Show That X2 Y2 X2 Y2 is Continuous

Daveyboy

Homework Statement

Show x^2 is continuous, on all reals, using a delta/epsilon argument.

Let E>0. I want to find a D s.t. whenever d(x,y)<D d(f(x),f(y))<E.
WLOG let x>y
|x^2-y^2|=x^2-y^2=(x-y)(x+y)<D(x+y)

I am trying to bound x+y, but can't figure out how.

Hurkyl

One problem is that you have the epsilon-delta statement wrong. What you wrote is the definition of "uniformly continuous", and squaring is not uniformly continuous.

Uniform continuity requires a delta that works for all values of y. Continuity only requires that, for each value of y, there exists a delta. If it helps, you should think of the delta you are choosing as a function of y.

Tobias Funke

I am trying to bound x+y, but can't figure out how.

Maybe writing x+y=x-y+2y will help?

HallsofIvy · Jul 17, 2010

You need to show: given any real number, a, then for any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|x^2- a^2|< \epsilon[/itex].

You might start by factoring [itex]|x^2- a^2|< |x-a||x+ a|[/itex].

Now, if x is close to a, so x- a is close to 0, how large can x+ a be?

Daveyboy

Now, if x is close to a, so x- a is close to 0, how large can x+ a be?

then x+a approaches 2a...

I feel like I should just take delta = sqrt(epsilon), and I'm fairly confident any delta less than that will suffice. I do not really understand how to show that though.

HallsofIvy · Jul 17, 2010

Saying x+ a "approaches" 2a is not enough. If [itex]|x- a|< \delta[/itex] then [itex]-\delta< x- a< \delta[/itex] so, adding 2a to each part, [itex]2a- \delta< x+ a< 2a+ \delta[/itex].

Now you can say [itex]|(x-a)(x+a)|= |x-a||x+a|< (a+\delta)|x- a|[/itex].

Mark44

Fixed your LaTeX.

Saying x+ a "approaches" 2a is not enough. If [itex]|x- a|< \delta[/itex] then [itex]-\delta< x- a< \delta[/itex] so, adding 2a to each part, [itex]2a- \delta< x+ a< 2a+ \delta[/itex].
Now you can say [itex]|(x-a)(x+a)|= |x-a||x+a|< (a+\delta)|x- a|[/itex].

Tip: Make sure that LaTeX expressions start and end with the same type of tag. [ tex] ... [ /tex] and [ itex] ... [ /itex]. You sometimes start the expression with [ itex] and end with [ /math]. I'm not sure that this Web site can render [ math] ... [ /math] expressions, but I am sure that you can't mix them.

HallsofIvy

Thanks, Mark44. I just need to learn to check my responses before going on!

Mark44

Thanks, Mark44. I just need to learn to check my responses before going on!

That's advice I'm trying to give myself, too. :approve:

Telemachus

How do this demonstrate the continuity? I mean, what you're saying is the same than [tex]\lim_{x \to c}{f(x)} = f(c)[/tex], but in delta epsilon notation? I don't get it :P

Daveyboy

shouldn't we conclude that

[itex]
|(x-a)(x+a)|= |x-a||x+a|< (2a+\delta)|x- a|
[/itex]
?

I still want to clearly define delta as a function of x, epsilon, and a (even though a is a constant)

I see the trick that was used to bound |x^2-a^2| and that was neat, but now I am confused.

If I solved
[itex]
(2a+\delta)|x- a|<\epsilon
[/itex]

for delta would that give me a correctly bounded delta?

Hurkyl

shouldn't we conclude that
[itex]
|(x-a)(x+a)|= |x-a||x+a|< (2a+\delta)|x- a|
[/itex]
?

I think you have a typo there....

I still want to clearly define delta as a function of x, epsilon, and a (even though a is a constant)

No, you want it as a function of epsilon and a. (Check your quantifiers -- delta is "chosen" before x enters the picture)

Johnson Jealifted

Show That X2 Y2 X2 Y2 is Continuous

Show x^2 is continuous.

Homework Statement

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