Can Absolute Value Solve the Problem of Convergence for Inverse Functions?

[Julio1]

Show that if $y_0\ne 0$ and $|y-y_0|<\min\left(\dfrac{|y_0|}{2},\dfrac{\varepsilon|y_0|^2}{2}\right)$, then $y_0\ne 0$ and $\left|\dfrac{1}{y}-\dfrac{1}{y_0}\right|<\varepsilon.$

Spoiler

Hello !, in this case what is the minimum? For example, $\min\left(\dfrac{|y_0|}{2},\dfrac{\varepsilon|y_0|^2}{2}\right)=\dfrac{|y_0|}{2}$?

 

[Prove It]

Show that if $y_0\ne 0$ and $|y-y_0|<\min\left(\dfrac{|y_0|}{2},\dfrac{\varepsilon|y_0|^2}{2}\right)$, then $y_0\ne 0$ and $\left|\dfrac{1}{y}-\dfrac{1}{y_0}\right|<\varepsilon.$

Spoiler

Hello !, in this case what is the minimum? For example, $\min\left(\dfrac{|y_0|}{2},\dfrac{\varepsilon|y_0|^2}{2}\right)=\dfrac{|y_0|}{2}$?

We don't know what is the minimum, that is the point. But we know that one of them is, so the other one will be no less than this value, and thus won't affect any of the inequalities.

If $\displaystyle \begin{align*} \left| y - y_0 \right| < \frac{ \left| y_0 \right| }{2} \end{align*}$, and this is the minimum, then it must also be less than $\displaystyle \begin{align*} \frac{ \epsilon \left| y_0 \right| ^2 }{2} \end{align*}$ and vice versa.

 

[Euge]

Show that if $y_0\ne 0$ and $|y-y_0|<\min\left(\dfrac{|y_0|}{2},\dfrac{\varepsilon|y_0|^2}{2}\right)$, then $y_0\ne 0$ and $\left|\dfrac{1}{y}-\dfrac{1}{y_0}\right|<\varepsilon.$

Spoiler

Hello !, in this case what is the minimum? For example, $\min\left(\dfrac{|y_0|}{2},\dfrac{\varepsilon|y_0|^2}{2}\right)=\dfrac{|y_0|}{2}$?

Hi Julio,

To prove the result, assume the hypothesis and manipulate the expression $|\frac1{y} - \frac1{y_0}|$ to a get a term involving $|y - y_0|$ as follows:

$(*) \displaystyle \left|\frac1{y} - \frac1{y_0}\right| = \left|\frac{y_0 - y}{yy_0}\right| = \frac{|y - y_0|}{|yy_0|}$.

These steps will now be justified. Since

$\displaystyle |y - y_0| < \min\left(\frac{|y_0|}{2}, \frac{\varepsilon |y_0|^2}{2}\right)$,

in particular

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

This inequality implies $y \neq 0$. Otherwise, $|y_0| < \frac{|y_0|}{2}$, which is contradiction. Thus $yy_0 \neq 0$ and the equations in $(*)$ are valid. Again by assumption,

$\displaystyle |y - y_0| < \frac{\varepsilon |y_0|^2}{2}$.

Further, by the triangle inequality, $|y - y_0| < \frac{|y_0|}{2}$ implies $|y| \ge |y_0| - |y - y_0| > \frac{|y_0|}{2}$. Hence

$\displaystyle \frac{|y - y_0|}{|yy_0|} < \frac{\varepsilon |y_0|^2}{2} \frac{2}{|y_0|^2} = \varepsilon$.

 

[Julio1]

Hi Julio,

To prove the result, assume the hypothesis and manipulate the expression $|\frac1{y} - \frac1{y_0}|$ to a get a term involving $|y - y_0|$ as follows:

$(*) \displaystyle \left|\frac1{y} - \frac1{y_0}\right| = \left|\frac{y_0 - y}{yy_0}\right| = \frac{|y - y_0|}{|yy_0|}$.

These steps will now be justified. Since

$\displaystyle |y - y_0| < \min\left(\frac{|y_0|}{2}, \frac{\varepsilon |y_0|^2}{2}\right)$,

in particular

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

This inequality implies $y \neq 0$. Otherwise, $|y_0| < \frac{|y_0|}{2}$, which is contradiction. Thus $yy_0 \neq 0$ and the equations in $(*)$ are valid. Again by assumption,

$\displaystyle |y - y_0| < \frac{\varepsilon |y_0|^2}{2}$.

Further, by the triangle inequality, $|y - y_0| < \frac{|y_0|}{2}$ implies $|y| \ge |y_0| - |y - y_0| > \frac{|y_0|}{2}$. Hence

$\displaystyle \frac{|y - y_0|}{|yy_0|} < \frac{\varepsilon |y_0|^2}{2} \frac{2}{|y_0|^2} = \varepsilon$.

Thanks :)

But how it's conclude that $|y|\ge |y_0|-|y-y_0|>\dfrac{|y_0|}{2}$? Why $|y|\ge \dfrac{|y_0|}{2}$?

 

[Euge]

Thanks :)

But how it's conclude that $|y|\ge |y_0|-|y-y_0|>\dfrac{|y_0|}{2}$? Why $|y|\ge \dfrac{|y_0|}{2}$?

By the triangle inequality,

$\displaystyle |y_0| = |y + (y_0 - y)| \le |y| + |y_0 - y| = |y| + |y - y_0|$.

Hence, $|y| > |y_0| - |y - y_0|$. Since $|y - y_0| < \frac{|y_0|}{2}$,

$|y_0| - |y - y_0| \ge |y_0| - \frac{|y_0|}{2} = \frac{|y_0|}{2}$.

Therefore, $|y| > \frac{|y_0|}{2}$.

 

[Julio1]

By the triangle inequality,

$\displaystyle |y_0| = |y + (y_0 - y)| \le |y| + |y_0 - y| = |y| + |y - y_0|$.

Hence, $|y| > |y_0| - |y - y_0|$. Since $|y - y_0| < \frac{|y_0|}{2}$,

$|y_0| - |y - y_0| \ge |y_0| - \frac{|y_0|}{2} = \frac{|y_0|}{2}$.

Therefore, $|y| > \frac{|y_0|}{2}$.

Thanks :) Now its clear the problem. The condition $|y-y_0|<\dfrac{|y_0|}{2}$ is for see that $y\ne 0.$ And the condition $|y-y_0|<\dfrac{\varepsilon|y_0|^2}{2}$ is for show that $\left|\dfrac{1}{y}-\dfrac{1}{y_0}\right|<\varepsilon.$

Bye and Thanks !