How's this epsilon-delta proof?

These proofs are killing me! But I'm finally starting to get a grasp of them. So I just wanted to check if this proof is sane and correct since it is different from the ones usually given: all of the proofs I found use the minimum function,

\min{(x,y)}

, one way or another; this one doesn't.
Anyways here is the limit:

\displaystyle \lim_{z \rightarrow z_0}z^2=z_0^2

.
Proof:

Suppose

\epsilon >0

. Let

\delta=\sqrt{|z_0|^2+\epsilon}-|z_0|

and let

z

be any complex number, where

0<|z-z_0|<\delta

. Then:

|z^2-z_0^2|=|z-z_0||z+z_0|<\delta |z+z_0|=\delta |z-z_0+2z_0|

This part is straight forward, and it follows from

|z-z_0|<\delta

.

\delta|z-z_0+2z_0|\leq\delta(|z-z_0|+|2z_0|)\leq\delta(\delta+2|z_0|)

This part uses the triangle inequality. The equality here will not hold for

|z^2-z_0^2|

since it is strictly less than

\delta|z-z_0+2z_0|

.

[br]\mid z^2-z_0^2 \mid<\delta(\delta+2|z_0|)

[br]\mid z^2-z_0^2 \mid<(\sqrt{|z_0|^2+\epsilon}-|z_0|)(\sqrt{|z_0|^2+\epsilon}+|z_0|)=\epsilon[br]

Substitution.

DONE!

So how was that?

(edited 8 years ago)

Reply 1

BlueSam3

17

In line 1, you assume that

|z_0|^2 - \varepsilon > 0

. This isn't necessarily true: what happens if

z_0 = 0

? You also assume that

\sqrt{|z_0|^2 - \varepsilon} > |z_0|

, which is never true.

You've got a few inequalities that you're claming are strict that aren't.

The whole thing badly needs some explanation and adjustment to make it clearer.

Reply 2

username1539513

21

Original post by gagafacea1

These proofs are killing me! But I'm finally starting to get a grasp of them. So I just wanted to check if this proof is sane and correct since it is different from the ones usually given: all of the proofs I found use the minimum function, <a rel="nofollow" href="javascript:void(0)"><img src="http://www.thestudentroom.co.uk/latexrender/pictures/c9/c9b09d247b505b567e9a89e678bb82ba.png" width="79" height="21" style="border: 0px; margin: 0px; padding: 0px; vertical-align: -5px;" alt="\min{(x,y)}" title="\min{(x,y)}" onclick="newWindow=window.open('http://www.thestudentroom.co.uk/latexrender/latexcode.php?code=%5Cmin%7B%28x%2Cy%29%7D','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100');" /></a>, one way or another; this one doesn't.
Anyways here is the limit:<a rel="nofollow" href="javascript:void(0)"><img src="http://www.thestudentroom.co.uk/latexrender/pictures/bd/bddac2347e38ad96a3677589470fc512.png" width="99" height="31" style="border: 0px; margin: 0px; padding: 0px; vertical-align: -15px;" alt="\displaystyle \lim_{z \rightarrow z_0}z^2=z_0^2" title="\displaystyle \lim_{z \rightarrow z_0}z^2=z_0^2" onclick="newWindow=window.open('http://www.thestudentroom.co.uk/latexrender/latexcode.php?code=%5Cdisplaystyle+%5Clim_%7Bz+%5Crightarrow+z_0%7Dz%5E2%3Dz_0%5E2','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100');" /></a>.
Proof:

Suppose <a rel="nofollow" href="javascript:void(0)"><img src="http://www.thestudentroom.co.uk/latexrender/pictures/93/93e50e2f00ace9894edc61ad0fe45bb3.png" width="43" height="15" style="border: 0px; margin: 0px; padding: 0px; vertical-align: -1px;" alt="\epsilon >0" title="\epsilon >0" onclick="newWindow=window.open('http://www.thestudentroom.co.uk/latexrender/latexcode.php?code=%5Cepsilon+%3E0','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100');" /></a>. Let <a rel="nofollow" href="javascript:void(0)"><img src="http://www.thestudentroom.co.uk/latexrender/pictures/bb/bb4e87881d7233dac12d878343cd727e.png" width="175" height="25" style="border: 0px; margin: 0px; padding: 0px; vertical-align: -6px;" alt="\delta=\sqrt{|z_0|^2-\epsilon}-|z_0|" title="\delta=\sqrt{|z_0|^2-\epsilon}-|z_0|" onclick="newWindow=window.open('http://www.thestudentroom.co.uk/latexrender/latexcode.php?code=%5Cdelta%3D%5Csqrt%7B%7Cz_0%7C%5E2-%5Cepsilon%7D-%7Cz_0%7C','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100');" /></a> and let <a rel="nofollow" href="javascript:void(0)"><img src="http://www.thestudentroom.co.uk/latexrender/pictures/9d/9d5ed8764b4f74e3b56fbe649cf91c3a.png" width="10" height="10" style="border: 0px; margin: 0px; padding: 0px;" alt="z" title="z" onclick="newWindow=window.open('http://www.thestudentroom.co.uk/latexrender/latexcode.php?code=z','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100');" /></a> be any complex number, where <a rel="nofollow" href="javascript:void(0)"><img src="http://www.thestudentroom.co.uk/latexrender/pictures/ab/abd73f685648c305bf62adb1d0a38b42.png" width="134" height="20" style="border: 0px; margin: 0px; padding: 0px; vertical-align: -5px;" alt="0<|z-z_0|<\delta" title="0<|z-z_0|<\delta" onclick="newWindow=window.open('http://www.thestudentroom.co.uk/latexrender/latexcode.php?code=0%3C%7Cz-z_0%7C%3C%5Cdelta','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100');" /></a>. Then:

<a rel="nofollow" href="javascript:void(0)"><img src="http://www.thestudentroom.co.uk/latexrender/pictures/52/520b54988803a430641bc9d4931c01ab.png" width="466" height="21" style="border: 0px; margin: 0px; padding: 0px; vertical-align: -5px;" alt="|z^2-z_0^2|=|z-z_0||z+z_0|<\delta |z+z_0|=\delta |z-z_0+2z_0|" title="|z^2-z_0^2|=|z-z_0||z+z_0|<\delta |z+z_0|=\delta |z-z_0+2z_0|" onclick="newWindow=window.open('http://www.thestudentroom.co.uk/latexrender/latexcode.php?code=%7Cz%5E2-z_0%5E2%7C%3D%7Cz-z_0%7C%7Cz%2Bz_0%7C%3C%5Cdelta+%7Cz%2Bz_0%7C%3D%5Cdelta+%7Cz-z_0%2B2z_0%7C','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100');" /></a>

<a rel="nofollow" href="javascript:void(0)"><img src="http://www.thestudentroom.co.uk/latexrender/pictures/48/4813a42030fdf5acc6c6bbe624b05b15.png" width="418" height="21" style="border: 0px; margin: 0px; padding: 0px; vertical-align: -5px;" alt="\delta|z-z_0+2z_0|<\delta(|z-z_0|+|2z_0|)<\delta(\delta+2|z_0|)" title="\delta|z-z_0+2z_0|<\delta(|z-z_0|+|2z_0|)<\delta(\delta+2|z_0|)" onclick="newWindow=window.open('http://www.thestudentroom.co.uk/latexrender/latexcode.php?code=%5Cdelta%7Cz-z_0%2B2z_0%7C%3C%5Cdelta%28%7Cz-z_0%7C%2B%7C2z_0%7C%29%3C%5Cdelta%28%5Cdelta%2B2%7Cz_0%7C%29','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100');" /></a>

<a rel="nofollow" href="javascript:void(0)"><img src="http://www.thestudentroom.co.uk/latexrender/pictures/32/32d6b5ff49249e0d6fd964929facc9db.png" width="198" height="21" style="border: 0px; margin: 0px; padding: 0px; vertical-align: -5px;" alt="
\mid z^2-z_0^2 \mid<\delta(\delta+2|z_0|)" title="
\mid z^2-z_0^2 \mid<\delta(\delta+2|z_0|)" onclick="newWindow=window.open('http://www.thestudentroom.co.uk/latexrender/latexcode.php?code=%0D%0A%5Cmid+z%5E2-z_0%5E2+%5Cmid%3C%5Cdelta%28%5Cdelta%2B2%7Cz_0%7C%29','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100');" /></a>

<a rel="nofollow" href="javascript:void(0)"><img src="http://www.thestudentroom.co.uk/latexrender/pictures/50/50c3a26bb9859da917c4fb903b51a707.png" width="448" height="25" style="border: 0px; margin: 0px; padding: 0px; vertical-align: -6px;" alt="
\mid z^2-z_0^2 \mid<(\sqrt{|z_0|^2-\epsilon}-|z_0|)(\sqrt{|z_0|^2-\epsilon}+|z_0|)=\epsilon
" title="
\mid z^2-z_0^2 \mid<(\sqrt{|z_0|^2-\epsilon}-|z_0|)(\sqrt{|z_0|^2-\epsilon}+|z_0|)=\epsilon
" onclick="newWindow=window.open('http://www.thestudentroom.co.uk/latexrender/latexcode.php?code=%0D%0A%5Cmid+z%5E2-z_0%5E2+%5Cmid%3C%28%5Csqrt%7B%7Cz_0%7C%5E2-%5Cepsilon%7D-%7Cz_0%7C%29%28%5Csqrt%7B%7Cz_0%7C%5E2-%5Cepsilon%7D%2B%7Cz_0%7C%29%3D%5Cepsilon%0D%0A','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100');" /></a>

DONE!

So how was that?

[video="youtube;ld2nWfIap2k"]https://www.youtube.com/watch?v=ld2nWfIap2k[/video]

I apologise in advance for not being able to help here but I couldn't resist

Reply 3

username1599987

OP

12

Original post by BlueSam3

In line 1, you assume that

|z_0|^2 - \varepsilon > 0

. This isn't necessarily true: what happens if

z_0 = 0

? You also assume that

\sqrt{|z_0|^2 - \varepsilon} > |z_0|

, which is never true.

You've got a few inequalities that you're claming are strict that aren't.

The whole thing badly needs some explanation and adjustment to make it clearer.

That was a typo, it should have been a plus not a minus epsilon. I always do this when using LaTeX!
Oh and you mean the second line, where I used the triangle inequality? You're right but that shouldn't affect it much, since

\delta|z-z_0+2z_0|

is strictly bigger than

|z^2-z_0^2|

. I edited the post now, thanks for pointing those out.

(edited 8 years ago)

OP

Bump!

Original post by gagafacea1

Bump!

I found it a bit hard to read, so only scanned it. You can simplify it anyway with something like this (I'll use

x,a

rather than

z,z_0

)

With

|x-a| < \delta

we have

|x^2-a^2| = |x-a||x+a| < \delta|x+a|

But if

\delta < 1

then:

|x+a| =|x-a+2a| \le |x-a| +2|a| < \delta +2|a| < 1+2|a|

So

|x^2-a^2| < \delta(1+2|a|) = \epsilon

if

\delta = \frac{\epsilon}{1+2|a|}

So having chosen

\epsilon

choose

\delta = \text{min}\{1, \frac{\epsilon}{1+2|a|}\}

and the argument above will work.

Reply 6

ghostwalker

17

Original post by gagafacea1

Bump!

Looks fine to me, though that's not the ordering I'd expect to see in an exam. You've clearly worked out a possible function for delta initially.

Reply 7

BlueSam3

17

Original post by ghostwalker

Looks fine to me, though that's not the ordering I'd expect to see in an exam. You've clearly worked out a possible function for delta initially.

That's how I'd do it: I'd just write "

\delta =

", leave it blank, then figure out what I needed as I went.

Reply 8

Sourestdeeds

8

What level is this? Im hoping to do a maths degree but this looks crazy! hehe

Reply 9

DFranklin

18

Original post by Sourestdeeds

What level is this? Im hoping to do a maths degree but this looks crazy! hehe

1st year undergrad (at a decent university, at any rate).

Reply 10

Zacken

22

Original post by Sourestdeeds

What level is this? Im hoping to do a maths degree but this looks crazy! hehe

First year (basic) analysis work.

Reply 11

Sourestdeeds

8

Original post by DFranklin

1st year undergrad (at a decent university, at any rate).

Oh cool

Have my hopes on Nottingham Uni. Seeing this makes me feel better about teaching myself FP2. Knowing that what I'm doing now isn't that hard in comparison is good!

Reply 12

username1599987

OP

12

Original post by atsruser

I found it a bit hard to read, so only scanned it. You can simplify it anyway with something like this (I'll use

x,a

rather than

z,z_0

)

But...but...but that's how they write it in the books? And I don't see how yours is simpler than mine to be honest. Not saying you're wrong or anything, but as I said, I didn't want to use that min function.

Original post by ghostwalker

Looks fine to me, though that's not the ordering I'd expect to see in an exam. You've clearly worked out a possible function for delta initially.

Of course I have done some scratch work, but from what I've read in multiple places, when you present the proof, you're supposed to present it in the order I did. Though maybe not in actual exams?

(edited 8 years ago)

Reply 13

DFranklin

18

Original post by gagafacea1

Of course I have done some scratch work, but from what I've read in multiple places, when you present the proof, you're supposed to present it in the order I did. Though maybe not in actual exams?

"Supposed" is a bit strong. Suppose you want to show f is continuous at a. There's nothing wrong with starting off with the standard "take

\epsilon > 0

", and then simply the assumption

\delta > 0

. You then start to put bounds on |f(x) - f(a)| given

|x-a| < \delta

, piling extra conditions on to

\delta

as necessary until you can finally say triumphantly: "if

\delta

satisfies all these conditions then

|f(x) - f(a)| <\epsilon

whenever

|x-a| < \delta

, therefore f is continuous at a".

Reply 14

username1599987

OP

12

Original post by DFranklin

"Supposed" is a bit strong. Suppose you want to show f is continuous at a. There's nothing wrong with starting off with the standard "take

\epsilon > 0

", and then simply the assumption

\delta > 0

. You then start to put bounds on |f(x) - f(a)| given

|x-a| < \delta

, piling extra conditions on to

\delta

as necessary until you can finally say triumphantly: "if

\delta

satisfies all these conditions then

|f(x) - f(a)| <\epsilon

whenever

|x-a| < \delta

, therefore f is continuous at a".

Oh, I did not know that. Because many sources present these proofs in the way I did, so I just assumed it's the norm. Thanks!

Reply 15

atsruser

11

Original post by gagafacea1

But...but...but that's how they write it in the books? And I don't see how yours is simpler than mine to be honest. Not saying you're wrong or anything, but as I said, I didn't want to use that min function.

I'm not sure what you mean by "that's how they write it in the books" - with

z

? If so, that's usually complex variable notation (which is fine here, but I'm not sure if you were intending it to be a proof about complex variables).

As for simplicity, yeah, maybe you're right. It's easier to understand an argument that you've written down yourself, so I guess that may be a draw.

And as for the min business, I clearly skimmed your post a bit too quickly, and didn't see that. Sorry. A lot of these proofs do it that way though. BTW, your penultimate line looked to me like you would need delta to be a function of epsilon which is itself a function of delta, but the clever choice of delta fixes that. The min approach avoids that in a different way.

Of course I have done some scratch work, but from what I've read in multiple places, when you present the proof, you're supposed to present it in the order I did. Though maybe not in actual exams?

A lot of books present in in this "rabbit out of the hat" way (which is dismal pedagogy IMHO), but in an exam, you won't be able to do that easily, as you'll find it tricky to start off with an appropriate choice of delta.

Reply 16

username1599987

OP

12

Original post by atsruser

I'm not sure what you mean by "that's how they write it in the books" - with

z

?

That was a reply to your "It's hard to read" comment. Meaning in the books, the proofs skip a lot of those "straight-forward" steps, so they become hard to read. But I see now that I'm not the only one who finds them hard to read, because I thought that all you experts can read those like reading a poem! lol

How's this epsilon-delta proof?

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