Is Michael Spivak Wrong?

1. Jun 16, 2010

Buri

Is Michael Spivak Wrong!?

My text says,

In general, if ε > 0, to ensure that

|x²sin(1/x)| < ε,

we need only require that

|x| < ε and x ≠ 0, provided that ε ≤ 1. If we are given an ε which is greater than 1 (it might be, even thought it is "small" ε's which are of interest), then it does not suffice to require that |x| < ε, but it certainly suffices to require that |x| < 1 and x ≠ 0.

I don't see why if ε > 1 then it doesn't work. I've tried finding a counter example, but it always seem to work. Can someone please help me out?

EDIT:

x²sin(1/x) ≈ x when x → ∞ so it would seem that when x isn't very large then x²sin(1/x) will always be less then x and hence less then ε, even if ε were greater than 1.

BTW, this is Calculus by Spivak page 92-93.

2. Jun 16, 2010

rasmhop

Re: Is Michael Spivak Wrong!?

Try x = 6/pi.

Then $sin(1/x) = 1/2$, so $x^2\sin(1/x)=18/\pi^2$.

The reason why it doesn't necessarily work when $x>1$ is that we may have $|x^2| > \epsilon$ so we need sin to be suifficiently small, but x^2 grows towards infinity much faster than sin(1/x) grows towards 0.

It is true that we could still works with values of epsilon a little above 1, but we would need a different argument then.

3. Jun 16, 2010

Buri

Re: Is Michael Spivak Wrong!?

So what would be wrong with this proof then?

sin x < x for x > 0, so then
|sin x| < |x| for x ≠ 0, so in particular we have

|sin x| < |x| for nonzero x in the interval (-1,1)

This means that

|sin(1/x)| < |1/x| for |x| > 1

which implies

|x^2 sin(1/x)| < |x| for |x| > 1

This should guarantee then that if ε > 1 and 1 < |x| < ε, then
|x^2 sin(1/x)| < |x| < ε

4. Jun 16, 2010

Landau

Re: Is Michael Spivak Wrong!?

Yes, you are totally correct.

In fact, for all x in R, we have |x^2 sin(1/x)|<=|x|, so we can take delta=epsilon.

I think what Spivak meant, was that in this chapter you know almost nothing about the sine function. He only uses the very rough estimate |sin x|<= 1 for all x, which everyone knows and which is clear from the picture of the unit circle. If you want to extend the argument for epsilon >1, you will have to come up with a better estimate of |sin x| (or |sin(1/x)|): for example you are using the inequality
|sin x|<=|x|. I think Spivak does not want to assume you to know this fact, because the sine has not been introduced formally.

5. Jun 16, 2010

Buri

Re: Is Michael Spivak Wrong!?

I'm confused now. Landau, you're telling me I'm right, but rasmhop has provided the counter-example x = 6/&pi;. So I'm massively confused now...

6. Jun 16, 2010

rasmhop

Re: Is Michael Spivak Wrong!?

That isn't really a counterexample I responded too hastily. Taking $x=6/\pi$ we get:
$$x^2\sin(1/x) > 1$$
but obviously not
$$x^2\sin(1/x) > x$$
(for instance by the argument you provided). So my counterexample doesn't work.

7. Jun 16, 2010

Buri

Re: Is Michael Spivak Wrong!?

Wait a second, x = 6/&pi; doesn't work!! lol

8. Jun 16, 2010

Buri

Re: Is Michael Spivak Wrong!?

Well thanks for helping me guys, this had been bothering me lol So thanks a lot for the help!

9. Jun 16, 2010

Landau

Re: Is Michael Spivak Wrong!?

Yes, you were right. You're welcome :)

10. Jun 16, 2010

Office_Shredder

Staff Emeritus
Re: Is Michael Spivak Wrong!?

$$|sin(\frac{1}{x})|<\frac{1}{x}$$ (since |sin(x)|<|x|) so it's certainly always true that $$|x^2sin(\frac{1}{x})|<|x|$$

It might be that the way he proves it is far more convenient to assume that x is small, since a proof of sin(x)<x is non-trivial from first principles

11. Jun 19, 2010

xeno_gear

Re: Is Michael Spivak Wrong!?

interesting...