Spivak's Calculus - Chapter 1 Question 23

Question

In Spivak's Calculus Chapter 1 Question 23:

Replace the question marks in the following statement ing $\varepsilon, x_0$ and $y_0$ so that the conclusion will be true: If $y_0\neq 0$ and $$|y-y_0|<? \qquad\text{and}\qquad |x-x_0|<?$$ then $y\neq 0$ and $$ \bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg|<\varepsilon.$$

The answer in its Solution Manual is $$|x-x_0|<\min\bigg(\frac{\varepsilon}{2(1/|y_0|+1)},1 \bigg)$$ and $$|y-y_0|<\min\bigg( \frac{|y_0|}{2},\frac{\varepsilon|y_0|^2}{4(|x_0|+1)} \bigg).$$ The latter is same as my solution, but the first is a bit different. So I wonder if $$|x-x_0|<\min\bigg(\frac{\varepsilon |y_0|}{4},1 \bigg)$$, is the solution still be true? Thanks in advance.

Answer 1

$|x-x_0|<\min\bigg(\frac{\varepsilon |y_0|}{4},1 \bigg)$ is wrong, here's why:

We have:

\begin{align*} \bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg| &\leq \frac{\left | x \right |\left | y_{0}-y \right |+\left | y \right |\left | x_{0}-x \right |}{\left | yy_{0} \right |}\\ &\leq \frac{\left | x-x_{0} \right |\left | y-y_{0} \right |}{\left | yy_{0} \right |}+\frac{\left | x_{0} \right |\left | y_{0}-y \right |}{\left | yy_{0} \right |}+\frac{\left | y \right |\left | x_{0}-x \right |}{\left | yy_{0} \right |} \end{align*}

where I used: $\left | x \right |\leq \left | x-x_{0} \right |+\left | x_{0} \right |$

Now if you consider the first term, i.e. $\frac{\left | x-x_{0} \right |\left | y-y_{0} \right |}{\left | yy_{0} \right |}$ alone and try to get it to be smaller than $\frac{\varepsilon }{2}$ you would end up with:

$\frac{\left | x-x_{0} \right |\left | y-y_{0} \right |}{\left | yy_{0} \right |}\leq \frac{\left | x-x_{0} \right |\frac{y_{0}}{2}}{\frac{\left | y_{0} \right |^{2}}{2}}= \frac{2\left | x-x_{0} \right |}{\left | y_{0} \right |}< \frac{\varepsilon }{2}$ giving the result you found which is $|x-x_0|<\frac{\varepsilon |y_0|}{4}$.

But looking at the decomposition above you realise this is wrong because you need to consider:

$\frac{\left | x-x_{0} \right |}{\left | yy_{0} \right |}(\left | y-y_{0} \right |+\left | y \right |)$ instead of $\frac{\left | x-x_{0} \right |\left | y-y_{0} \right |}{\left | yy_{0} \right |}$

The correct answer for the first is the one stated by the manual as: $\left | \frac{x}{y}-\frac{x_{0}}{y_{0}} \right |=\left | \frac{x}{y}-\frac{x}{y_{0}}+\frac{x}{y_{0}}-\frac{x_{0}}{y_{0}} \right |\leq \left | x \right |\left | \frac{1}{y}-\frac{1}{y_{0}} \right |+\left |\frac{1}{y_{0}} \right |\left | x-x_{0} \right |$

So: $\left |\frac{1}{y_{0}} \right |\left | x-x_{0} \right |\leq \frac{\frac{1}{\left |y_{0} \right |}\varepsilon }{2(\frac{1}{\left |y_{0} \right |}+1)}\leq \frac{\varepsilon }{2}$

Answer 2

To solve question 23 you use the inequalities from questions 21 and 22, which I give here with their proofs.

Qu. 21) If $$|x-x_0|<\min\bigg(\frac{\varepsilon}{2(|y_0|+1)}, 1 \bigg)\quad\text{and}\quad |y-y_0|< \frac{\varepsilon}{2(|x_0|+1)} \tag{1}$$ then $|xy-x_0y_0|<\varepsilon$.

Proof:

Since $|x-x_0|<1$ we have $|x|-|x_0|<|x-x_0|<1\Longrightarrow |x|<|x_0|+1$, and so \begin{align*} |xy-x_0y_0|&=|x(y-y_0)-y_0(x_0-x)|\\ &\le|x||y-y_0|+|y_0||x_0-x|\\ &< (1+ |x_0|)\frac{\varepsilon}{2(|x_0|+1)}+|y_0|\frac{\varepsilon}{2(|y_0|+1)}\\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{align*}

Qu. 22) If $y_0\neq0$ and $$|y-y_0|<\min\bigg(\frac{|y_0|}{2}, \frac{\varepsilon |y_0|^2}{2} \bigg), \tag{2}$$ then $y\neq0$ and $$\bigg| \frac{1}{y}-\frac{1}{y_0}\bigg|<\varepsilon.$$

Proof:

$|y_0|-|y|\le|y-y_0|<\frac{|y_0|}{2}\Longrightarrow |y|>\frac{|y_0|}{2}$. Since $y\neq0$ we have $$\frac{1}{|y|}<\frac{2}{|y_0|}.$$ Then $$\bigg| \frac{1}{y}-\frac{1}{y_0}\bigg|=\frac{|y-y_0|}{|y||y_0|}<\frac{1}{|y|}\cdot\frac{2}{|y_0|}\cdot\frac{\varepsilon |y_0|^2}{2}=\varepsilon.$$

Qu. 23) If $y_0\neq 0$ and $$|y-y_0|<\min\bigg( \frac{|y_0|}{2},\frac{\varepsilon|y_0|^2}{4(|x_0|+1)} \bigg) \qquad\text{and}\qquad |x-x_0|<\min\bigg(\frac{\varepsilon}{2(1/|y_0|+1)},1 \bigg)$$ then $y\neq 0$ and $$ \bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg|<\varepsilon.$$

Proof: We note that $$\bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg|=\bigg| x\cdot\frac{1}{y}-x_0\cdot\frac{1}{y_0}\bigg|$$ which is just $|xy-x_0y_0|$ of question 21 but with $y$, and $y_0$ replaced by $\frac{1}{y}$, and $\frac{1}{y_0}$ respectively, where $y\neq0$. Hence we require, by inequality (1) that $$|x-x_0|<\min\bigg(\frac{\varepsilon}{2(1/|y_0|+1)}, 1 \bigg)\quad\text{and}\quad \bigg|\frac{1}{y}-\frac{1}{y_0}\bigg| < \frac{\varepsilon}{2(|x_0|+1)}. \tag{3}$$

Now in question 22, inequality (2) allows $\big| \frac{1}{y}-\frac{1}{y_0}\big|<\varepsilon$. However we must replace the $\varepsilon$ of inequality (2) with $\frac{\varepsilon}{2(|x_0|+1)}$ of (3) to give:

$$|y-y_0| <\min\bigg(\frac{|y_0|}{2}, \frac{\varepsilon |y_0|^2}{4(|x_0|+1)} \bigg).$$

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To see why $$|x-x_0|<\min\bigg(\frac{\varepsilon |y_0|}{4},1 \bigg)$$ is the wrong choice, we rework question 22: \begin{align*} \bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg|&= \frac{\left|xy_0-x_0y\right|}{\left|yy_0\right|}\\ &=\frac{ |x (y_0-y)- y(x_0-x)|}{|y||y_0|}\\ &\leq\frac{ | x ||y-y_0|}{|y||y_0|}+\frac{|x-x_0|}{|y_0|}\\ &\leq\frac{ (1+| x _0|)|y-y_0|}{|y||y_0|}+\frac{|x-x_0|}{|y_0|}\\ &<\frac{ (1+| x _0|)}{|y||y_0|}\cdot\frac{\varepsilon |y_0|^2}{4(|x_0|+1)}+\frac{|x-x_0|}{|y_0|}\tag{$\star$}\\ &<\frac{2}{|y_0|}\cdot\frac{ 1}{|y_0|}\cdot\frac{\varepsilon |y_0|^2}{4}+\frac{|x-x_0|}{|y_0|}\\ &=\frac{\varepsilon}{2}+\frac{|x-x_0|}{|y_0|}. \end{align*} Where we have used $$\frac{1}{|y|}<\frac{2}{|y_0|}$$ from question 22 in the penultimate inequality. Now if $|x-x_0|<\min\big(\frac{\varepsilon |y_0|}{4},1 \big)$, we should have $$\bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg|<\frac{\varepsilon}{2}+\frac{\varepsilon}{4}=\frac{3\varepsilon}{4}.$$ Now here could be a source of confusion because for sure $\frac{3\varepsilon}{4}<\varepsilon$, but that is not what is required. What is required is for $\big| \frac{x}{y}-\frac{x_0}{y_0}\big|<\varepsilon$, and as such we need to end up with two numbers $\varepsilon_1$ and $\varepsilon_2$, say, which add up to $\varepsilon$, which is some fixed number $>0$. The point is that when you assert something is less than $\varepsilon$ in the proof, you don't know if it is less than $\frac{3\varepsilon}{4}$, indeed, it could be greater than $\frac{3\varepsilon}{4}$, just as long as it is $<\varepsilon$.

The answer given in (3) works since $$\frac{|x-x_{0}|}{|y_0|} <\frac{\varepsilon/|y_0|}{2(1/|y_0|+1)}< \frac{\varepsilon }{2},$$ which yields the desired $\varepsilon$ when plugged into $(\star)$.