Question of "min" function from Spivak

[Amad27]

Hi,

Suppose you want to prove |x - a||x + a| < \epsilon

You know

|x - a| < (2|a| + 1)
You need to prove

|x + a| < \frac{\epsilon}{2|a| + 1}

So that

|x - a||x + a| < \epsilon

Why does Michael Spivak do this:

He says you have to prove --> |x + a| < min(1, \frac{\epsilon}{2|a| + 1}) in order to finally prove, |x + a||x - a| < \epsilon

Why do we need the "min" function there?
Amad27 - The closing itex tag starts with /, not \. I fixed them all for you. - Mark44

Thanks!

 

[pasmith]

If we assume |x - a| &lt; 1 then a - 1 &lt; x &lt; a + 1 so that 2a - 1 &lt; x + a &lt; 2a + 1 \leq 2|a| + 1. That in turn means that <br /> |x - a||x + a| &lt; |x - a|(2|a| + 1) and we can ensure |x - a||x + a| &lt; \epsilon by requiring that <br /> |x - a|(2|a| + 1) &lt; \epsilon, so that <br /> |x - a| &lt; \frac{\epsilon}{2|a| + 1}. However the preceding argument assumed that |x - a| &lt; 1, so |x - a| must satisfy that constraint as well. Hence we must have <br /> |x - a| &lt; \min\left\{1, \frac{\epsilon}{2|a| + 1}\right\}.

 

[Amad27]

If we assume |x−a|<1|x - a| < 1 then a−1<x<a+1a - 1 < x < a + 1 so that 2a−1<x+a<2a+1≤2|a|+12a - 1 < x + a < 2a + 1 \leq 2|a| + 1. That in turn means that

Hello @pasmith, I don't understand;

On WHAT basis and WHY are you assuming |x - a| &lt; 1??

 

[Mark44]

If we assume |x−a|<1|x - a| < 1 then a−1<x<a+1a - 1 < x < a + 1 so that 2a−1<x+a<2a+1≤2|a|+12a - 1 < x + a < 2a + 1 \leq 2|a| + 1. That in turn means that

Hello @pasmith, I don't understand;
On WHAT basis and WHY are you assuming |x - a| &lt; 1??

Amad, be careful when you quote someone. @pasmith did not write "|x−a|<1|x - a| < 1" - what he wrote was "If we assume |x−a| < 1 then ..."

He made this assumption, and if it turns out that this assumption is incorrect, all that is accounted for in the last line of his post where he says that |s - a| will be less than the minimum of 1 and the expression involving $\epsilon$.

 

[Amad27]

He made this assumption, and if it turns out that this assumption is incorrect, all that is accounted for in the last line of his post where he says that |s - a| will be less than the minimum of 1 and the expression involving ϵ\epsilon.

Hi,

What does min(1, \frac{\epsilon}{2|a| + 1|})

Actually mean? I don't understand this "min" idea??

 

[Mark44]

Hi,

What does min(1, \frac{\epsilon}{2|a| + 1|})

Actually mean? I don't understand this "min" idea??

It's very simple - it means the minimum, or smaller, of the two values.

 

[Amad27]

@Mark44, why can't we directly prove:

|x - a| &lt; \frac{\epsilon}{2|a| + 1|}

Why is that 1 required.

Why does it matter if |x - a| < min of something?

Thanks!

 

[Mark44]

@Mark44, why can't we directly prove:

|x - a| &lt; \frac{\epsilon}{2|a| + 1|}

Why is that 1 required.

Why does it matter if |x - a| < min of something?

Thanks!

The goal is to prove that |x - a||x + a| < $\epsilon$

You need to get bounds on the value of |x + a|. He could just as easily assumed that |x - a| < 1/2 or |x - a| < 3 or whatever.

 

[Amad27]

You need to get bounds on the value of |x + a|. He could just as easily assumed that |x - a| < 1/2 or |x - a| < 3 or whatever.

Oh,

So its we know that

\frac{\epsilon}{2|a| + 1} &lt; 1?

If \frac{\epsilon}{2|a| + 1} &lt; 1 is minimum of course.

 

[Mark44]

Oh,

So its we know that

\frac{\epsilon}{2|a| + 1} &lt; 1?

If \frac{\epsilon}{2|a| + 1} &lt; 1 is minimum of course.

It all depends on the value of $\epsilon$. For some values $\frac{\epsilon}{2|a| + 1}$ will be smaller than 1, but for suitably large values of $\epsilon$, 1 will be larger.

 

[Amad27]

It all depends on the value of ϵ\epsilon. For some values ϵ2|a|+1\frac{\epsilon}{2|a| + 1} will be smaller than 1, but for suitably large values of ϵ\epsilon, 1 will be larger.

Right, since it depends for every \epsilon

But why would he randomly choose one (1) as a part of the minimum function? Why not two(2) or three(3)?

So, if 1 is the minimum then how does it prove

|x-a||x+a| &lt; \epsilon?

 

[HallsofIvy]

The "\delta" was derived under the assumption that |a- 1|< 1 (so that -1< a- 1< 1 and then 1< a+ 1< 3). In order for the proof to work, both |a- 1|&lt; \delta and |a- 1|&lt; 1 must be true. That will be the case if it is less than the smaller of the two.

 

[Amad27]

"δ\delta" was derived under the assumption that |a- 1|< 1 (so that -1< a- 1< 1 and then 1< a+ 1< 3). In order for the proof to work, both

|a−1|<δ​

|a- 1|< \delta and

|a−1|<1​

|a- 1|< 1 must be true. That will be the case if it is less than the smaller of the two.

But it has work for all real numbers,

Not just |x-a| < 1 what about |x-a| > 1 ?

 

[Mark44]

But it has work for all real numbers,

Not just |x-a| < 1 what about |x-a| > 1 ?

We don't care about that. You're missing the big picture here. The goal is to see how close x needs to be to a so that |x² - a²| < $\epsilon$.
We don't care about any x values that are relatively distant from a.

 

[Amad27]

We don't care about that. You're missing the big picture here. The goal is to see how close x needs to be to a so that |x2 - a2| < ϵ\epsilon.
We don't care about any x values that are relatively distant from a.

I really don't understand this concept.

The definition says you have to prove there is some \delta such that |x - a| &lt; \delta not that |x - a| &lt; \delta &lt; 1

Also, there is no proper definition of "relatively distant." Relatively close could mean 10000000000000. Why are you making assumptions anyway?

 

[Char. Limit]

Because if you don't make assumptions, you'll never get anywhere.

 

[Amad27]

Because if you don't make assumptions, you'll never get anywhere.

I don't believe so; The goal I think is to bound the number |x - a| so that it doesn't go overboard. Because when looking at limits you generally look at values close to x = a.

 

[Char. Limit]

I don't believe so; The goal I think is to bound the number |x - a| so that it doesn't go overboard. Because when looking at limits you generally look at values close to x = a.

And that right there, that's the assumption. We assume that x is close to a - namely, that |x - a| < 1. We can safely assume this, so we do, and the rest of the proof follows from such.

 

[pasmith]

But it has work for all real numbers,

Not just |x-a| < 1 what about |x-a| > 1 ?

We don't care about |x - a | \geq 1.

To conclude that x^2 - a^2 is continuous at a you need to prove:

Statement 1
For all \epsilon &gt; 0 there exists a \delta &gt; 0 such that for all x, if |x - a| &lt; \delta then |x^2 - a^2| &lt; \epsilon.

What Spivak invites you to prove is a stronger statement:

Statement 2
For all \epsilon &gt; 0 there exists a \delta &gt; 0 such that \delta &lt; 1 and for all x, if |x - a| &lt; \delta then |x^2 - a^2| &lt; \epsilon.

Note that Statement 2 implies Statement 1. Of course one could substitute any R &gt; 0 for '1' in Statement 2 and the result would still imply Statement 1, but 1 is the most natural choice.

 

[Mark44]

Also, there is no proper definition of "relatively distant." Relatively close could mean 10000000000000.

Really? In what context could 10 trillion be considered relatively close?

Saying that "relatively close" could mean 10 trillion comes off in my mind as trolling. Since the question has been asked and answered, I am closing this thread.