1

It's Problem 22. from Chapter 1. I'm given:

$y_0 \neq 0$

$|y - y_0| < \frac{|y_0|}{2}$

$|y - y_0| < \frac{\epsilon|y_0|^2}{2}$

and I must use them to prove that:

$y \neq 0$

$|\frac{1}{y} - \frac{1}{y_0}| < \epsilon$

I haven't really ever done proof based problems before this book, so I'm having a little hard time.I can do the problems once I get the general direction, but I'm not sure how exactly to start.Can someone offer a hint where to start?I'm also confused by this:

$|y - y_0| < \frac{\epsilon|y_0|^2}{2} , |y - y_0| \geq 0 => \frac{\epsilon|y_0|^2}{2} > 0 => \epsilon > 0, y_0 > 0$

(you can just use the $\frac{|y_0|}{2}$ part, but the 2nd part also tells you about $\epsilon$)

However $y_0 \neq 0$ is already granted at the start, even tho it's obviously the first thing you notice, so does it mean the author is hinting at something with this?

darkradeon
  • 297

2 Answers2

1

PROOF:

Step 1: $|y| > \frac{|y_{0}|}{2}$, since $|y|-|y_{0}| < |y-y_{0}| < \frac{|y_{0}|}{2}$.

Step 2: $|\frac{y - y_{0}}{y\cdot y_{0}}|<\frac{|y-y_{0}|}{(\frac{|y_0|^{2}}{2})} < \varepsilon\cdot \frac{(\frac{|y_{0}|^{2}}{2})}{(\frac{|y_{0}|^{2}}{2})} = \varepsilon$.

I hope this helps.

Christopher K
  • 3,383
  • @DEIMOS, does this answer help? – Christopher K Nov 15 '13 at 23:05
  • Not sure how you make it to be greater, if I make $|y - y_0|$ to equal $|y + (y - y - y_0)|$ and to get $|y| + |y - y - y_0| < \frac{|y_0|}{2}$ ...but I'm not sure if I'm properly using the triangle inequality – darkradeon Nov 15 '13 at 23:12
  • There is something called the inverse triangle inequality. $|y|-|y_{0}|\leq|y-y_{0}|$. Now why is this? $|(y-y_{0})+y_{0}|\leq |y-y_{0}| + |y_{0}|$. So... – Christopher K Nov 15 '13 at 23:15
  • As for your conclusion, it is clearly wrong. Just note that you have "shown" that $|y| < \frac{-|y_{0}|}{2} < 0$, which is nonsense. – Christopher K Nov 15 '13 at 23:17
  • @DEIMOS, I hope my edited version (with a sketch of the proof) helps. Please see above. – Christopher K Nov 15 '13 at 23:52
  • Ok I know I'm really gonna sound stupid now, but with $|y| - |y_0| \leq |y - y_0| < \frac{|y_0|}{2}$ I can get that $|y| - |y_0| < \frac{|y_0|}{2} => |y| < \frac{3|y_0|}{2}$ but not quite the $|y| > \frac{|y_0|}{2} that I need.Sorry if the question looks too basic. – darkradeon Nov 16 '13 at 00:01
  • Not stupid at all. You agree that $|y-y_{0}| = |y_{0}-y|$, right? Then $|y_{0}| - |y| < \frac{|y_{0}|}{2}$. So... – Christopher K Nov 16 '13 at 00:06
  • oh yeah (facepalm) I forgot to use that property, now it's all clear thanks – darkradeon Nov 16 '13 at 00:15
1

You are given that $\varepsilon > 0$ and $|y - y_0| < \min(|y_0|/2,\varepsilon|y_0|^2/2)$, and asked to prove that $\varepsilon$ is greater than $|1/y - 1/y_0|$. A useful first step would be to transform $|1/y - 1/y_0|$ into an expression that's easier to work with, given the assumptions. To this end, note that $$\left|\frac{1}{y} - \frac{1}{y_0}\right| = \left|\frac{y_0 - y}{y\cdot y_0}\right| = \frac{|y-y_0|}{|y|\cdot|y_0|} = \frac{1}{|y|} \cdot \frac{1}{|y_0|} \cdot |y - y_0|$$ We already have a bound on $|y - y_0|$ involving $\varepsilon$, namely $|y - y_0| < \varepsilon|y_0|^2/2$. Thus, it remains to find an appropriate bound on $1/|y|$. We note that if $|y - y_0| < \varepsilon|y_0|^2/2$, then in order to prove the desired result, we require that $1/|y| < 2/|y_0|$. I will leave this proof as an exercise, but note that you should use the fact that $|y - y_0| < |y_0|/2$ in conjunction with the reverse triangle inequality and the fact that $a > b$ implies $a^{-1} < b^{-1}$ for all positive real $a$ and $b$. I hope this answer was helpful.

emi
  • 1,640