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I have difficulty with understanding V. Zorich's (Mathematical Analysis, p.85 of English edition) proof of a
$$\lim\limits_{n\to\infty}\frac{x_n}{y_n}=\frac{A}{B}, \quad y_n\neq0,\quad B\neq0$$
Proof: Using estimate $$\tag1 \lvert \frac{A}{B}-\frac{x_n}{y_n}\rvert \leq\frac{\lvert x_n\rvert\Delta(y_n)+\lvert y_n\rvert \Delta(x_n)}{y^2_n}\cdot \frac{1}{1-\delta(y_n)}$$ where $$\tag2\delta(y_n)=\frac{\Delta(y_n)}{\lvert y_n\rvert}$$

For a given $\varepsilon>0$ we find numbers N' and N'' such that
$$\tag3\forall n>N' \quad \left(\Delta(x_n)<min \left\{1,\frac{\varepsilon \lvert B\rvert}{8}\right\}\right)$$

$$\tag4\forall n>N' \quad \left(\Delta(y_n)<min \left \{\frac{|B|}{4},\frac{\varepsilon \lvert B^2\rvert}{16|A|+1}\right\}\right)$$

Then for $n>max\{N',N''\}$ we shall have
$$\tag5\lvert x_n\rvert<\lvert A\rvert+\Delta(x_n)< \lvert A\rvert+1$$ $$\tag6\lvert y_n\rvert >B-\Delta (y_n)>\lvert B\rvert-\frac{\lvert B\rvert}{4}>\frac{\lvert B\rvert}{2}$$ $$\tag7 1-\delta(y_n)>\frac12$$ and therefore
$$\lvert x_n \rvert\cdot\frac{1}{y^2}\Delta (y_n)<(|A|+1)\cdot\frac{4}{B^2}\cdot\frac{\varepsilon\cdot B^2}{16(|A|+1)}=\frac{\varepsilon}{4}\tag8$$ $$\lvert \frac{1}{y_n}\rvert\Delta(x_n)<\frac{2}{|B|}\cdot{\frac{\varepsilon |B|}{8}=\frac{\varepsilon}{4}}$$ $$\tag9 0<\frac{1}{1-\delta(y_n)}<2$$ and consequently $$\tag{10}\lvert\frac{A}{B}-\frac{x_n}{y_n}\rvert<\varepsilon$$ when n>N
Questions: begin with (3) and (4). It is obvios that minimal elements have been choosen to give to give in sum $\varepsilon$, or as we see in (8) to be equal $\frac{\varepsilon}{4}$ but how they were constructed? It seems that we first note, see (8) that $|x_n|<|A+1|$ then we see from (6) that $y>\frac{|B|}{2}$ and part of (8) makes seance. Can we then construct such $\Delta(y_n)$ to give us what we want in (8)?

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  • $\begingroup$ Let $\varepsilon>0$ be given, and let $\Delta(y_n)=|B-y_n|$. Since $\frac{\varepsilon \cdot B^2}{16(|A|+1)}>0$ there exists $N \in \mathbb{N}$ such that $\Delta(y_n)<\frac{\varepsilon \cdot B^2}{16(|A|+1)}$ whenever $n \geq N.$ $\endgroup$ – Matt A Pelto Sep 30 '14 at 21:51
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I presume that $\Delta(x_n)$ is defined as $|x_n - A|$ and $\Delta(y_n)$ as $|y_n - B|$.

There is an error in this proof, where $N'$ is defined. $\Delta(y_n)$ should be taken to be less than $\dfrac{\varepsilon B^2}{16(|A| + 1)}$, not $\dfrac{\varepsilon B^2}{16}$.

In the definition of $N'$, the parts about $1$ and $|B|/4$ were just there so that there would be a fixed upper bound on $|x_n|$ and a fixed lower bound on $|y_n|$, as well as so that $\Delta(y_n)$ would not exceed a fixed fraction of $|y_n|$, here chosen to be $1/2$, but which could have been any fixed positive number less than 1. In order for the last part to occur, the upper bound on $\Delta(y_n)$ needed to be expressed as a fraction of $|B|$.

Once these things had been decided, the given upper bound of $2$ was determined by the upper bound of $1/2$ on $\delta(y_n)$. At that point, the numbers $\varepsilon/4$ and $\varepsilon/4$ were chosen by taking ($\varepsilon$ divided by the bound of $2$) and splitting that into two parts. After that, the numbers $\dfrac{\varepsilon |B|}{8}$ and $\dfrac{\varepsilon B^2}{16(|A| + 1)}$ would have been found by working backwards from $\varepsilon/4$ and $\varepsilon/4$ together with the known upper bounds on $|x_n|$, $1/|y_n|$ and $1/y_n^2$.

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