# Proof~Spivak Calculus: Problem 1.21

1. Feb 9, 2012

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I am a little stuck here. I was thinking of using the fact that if a<b and c<d then a+c<b+d to write:

|x-x0| + |y-y0| < ε / (2|y0|+1) + ε / (2|x0|+1) and then use the constraint |x-x0|<1 after,

but I am not sure how fruitful that would be....I first have to get at the heart of his "hint" in that I must relate xy-x0y0 to |x-x0| & |y-y0| first.....

Any thoughts?

2. Feb 9, 2012

### LCKurtz

Try subtracting and adding $xy_0$ in the expression $|xy-x_0y_0|$.

3. Feb 9, 2012

$\begin{array}{c} |xy-x_0y_0| &=& |xy + xy_0 - xy_0-x_0y_0| \\ &=& |x(y-y_0) + y_0(x-x_0) \end{array}$

Hmmm .... not sure what to do with this?

4. Feb 9, 2012

### LCKurtz

What happens as $x$ gets as close to $x_0$ and $y$ to $y_0$ as the problem suggests? You have the triangle inequality to use...

5. Feb 10, 2012

I'm sorry LC, but where is this suggested? The problem statement only implies that $-1 < x-x_0 < 1$ and says nothing about $y-y_0$.

Last edited: Feb 10, 2012
6. Feb 10, 2012

### LCKurtz

Read the "IF" in the first line of inequalities in your copy of problem 21 that you posted.

7. Feb 10, 2012

Hi LC Again, I addressed this in the post above. All that is said is that $-1 < x-x_0 <1$

Moreover, nothing is said about $|y-y_0|$

8. Feb 10, 2012

### LCKurtz

Are we in some kind of time and space warp here? Read the first two lines of the problem 21 you copied. It gives you assumptions about both $|x-x_0|$ and $|y-y_0|$.

9. Feb 10, 2012

Apparently we are. I actually logged out just to check that we are looking at the same image.

Line 1: $|x-x_0| < \min\left(\frac{\epsilon}{2|y_0|+1},1\right) \Rightarrow |x-x_0| < 1 \qquad \wedge \qquad |x-x_0| < \frac{\epsilon}{2|y_0|+1}$

AND

$|y-y_0| < \frac{\epsilon}{2|x_0|+1}$.

You seem to saying that I can use the fact that "$x$ is close to $x_0$" to do something. I have not even gotten to what that something is yet because I am

reluctant to accept your assertion that "$x$ is close to $x_0$." Like I said, I am *reluctant* and need convincing. What is it that is written above that makes you think this? All we know is that x is within "1" or $\frac{\epsilon}{2|y_0|+1}$ of $x_0$.

To me, that does not imply "close." What if we take the case where |x-x0|< 1 and x=0.1. The fact that x0 is within 1 of x does not see very close to me?

Maybe I am missing your point altogether. Can you be a little more explicit of what you are driving at?

Also: Thanks for your time and patience!

EDIT: ok. Wait a minute. Things are happening in my brain ...

I'll be back.

Edit 2: Yeah nope, nothing....I feel like batman in this right now :rofl:

Last edited: Feb 10, 2012
10. Feb 10, 2012

### LCKurtz

Have you used the triangle inequality on this yet? Remember what you are trying to do is make the left side < ε by overestimating it with something small. Do you see that getting $|x-x_0|$ and $|y-y_0|$ small will make it small if $x$ isn't too big? Can you keep the size of $x$ limited?

You are trying to get $x$ close enough to $x_0$ and $y$ close enough to $y_0$ to make $|xy-x_0y_0|<\epsilon$.

It is less than the minimum of those and if $\frac{\epsilon}{2|y_0|+1}$ is a tiny number less than 1 it will certainly be close. And if $\frac{\epsilon}{2|y_0|+1}> 1$ it will still be less than $\frac{\epsilon}{2|y_0|+1}$ .
After you use the triangle inequality suggested above, what happens if you put those overestimates in for $|x-x_0|$ and $|y-y_0|$?

11. Feb 10, 2012

Ok. So if I use the triangle inequality we have:

$|x(y-y_0) + y_o(x-x_0)| \le |x(y-y_0)| + |y_o(x-x_0)|$

which is bigger than the original expression. I don't know how that helps. Apparently I am too stupid for this subject.

12. Feb 10, 2012

### Broccoli21

You aren't too stupid. Just use what you know from the assumptions on that inequality that you just derived. The end goal is to chain together enough inequalities and get epsilon on the right side.

13. Feb 10, 2012

### LCKurtz

What did you start with that is equal to the left side that isn't in that equation? And you didn't answer the question in red above.

14. Feb 10, 2012

Hi LCKurtz I guess I might have been avoiding red question. I don't know what you mean "put in the overestimates." Do you mean to actually substitute $|x-x_0|$ with $\frac{\epsilon}{2|y_0|+1}$ ?

If so, I did not really know you could do that with inequalities. Is that what you meant though?

Here is what I have so far to summarize:

$$\begin{array}{c} |xy-x_0y_0| &=& |xy-x_0y_0 + xy_0 - xy_0| \qquad\qquad\qquad (1) \\ &=& |x(y-y0) + y_0(x-x_0)| \qquad\qquad\qquad (2) \\ &<& |x(y-y0)| + |y_0(x-x_0)| \qquad\qquad\qquad (3) \\ &<& \frac{|y_0|\epsilon(2|x_0|+1) + |x|\epsilon(2|y_0|+1)}{(2|y_0|+1)(2|x_0|+1)} \qquad(4) \end{array}$$

Edit OK. I see that I kind "substitute" them in by transitivity. Just had to convince myself. I added it in as line (4). It would seem that if I could prove that the RHS of (4) is less than $\epsilon$ I could finish. So now I need to see if:

$$\frac{|y_0|\epsilon(2|x_0|+1) + |x|\epsilon(2|y_0|+1)}{(2|y_0|+1)(2|x_0|+1)} \stackrel{?}{<} \epsilon \qquad\qquad(5)$$

Is this going in the right direction?

Last edited: Feb 10, 2012
15. Feb 10, 2012

### LCKurtz

You are getting the idea. Yes, the idea is to get it less than ε. But before you go from (3) to (4) write$$(3) \le |x||y-y_0| + |y_0||x-x_0|$$and then put in the overestimates and don't combine the fractions. Also, I didn't notice before but you have the parentheses in those denominators wrong. They are $2(|x_0|+1)$ and $2(|y_0|+1)$. Once you do that see if you can see why each fraction is less than $\epsilon/2$.

16. Feb 13, 2012

### LCKurtz

Where did you go Saladsamurai? Did you finish this? I'm curious what you did to finish.

17. Feb 13, 2012

Hi LCKurtz! I promise I am coming back to it! I had to just walk away from it for a bit to clear my head. I went back to some of the more basic problems in that section since they usually make the more challenging ones a little easier to grasp. So I am working through those now. I will be replying back to this thread, I promise!