Using modulus to answer: "How close to $\pi$ do we need $x$ to be for $x^2$ to be within $10^{−100}$ of $\pi ^2$?"

Question

I'm studying the book: "The Princeton companion to mathematics", I'm in the part of fundamentals: continuity. There is an example I don't get:

How close to $\pi$ do we need $x$ to be for $x^2$ to be within $10^{−100}$ of $\pi ^2$? To answer this, we can use our earlier argument. Let $x = \pi + \delta$ again. Then $$x^2−\pi^2 = 2\delta \pi+\delta^2$$ and an easy calculation shows that this has modulus less than $10^{-100}$ if $\delta$ has modulus less than $10^{-101}$. So we will be all right if we take the first $101$ digits of $\pi$ after the decimal point.

I don't get which was the easy calculation. Can someone help me with that?

And, when it says modulus, is it talking about modular arithmetic?

$\delta< 10^{-101}\implies 2\delta \pi+\delta^2<7\times 10^{-101}+10^{-202}<10^{-100}$. — lulu, May 24 '18 at 17:52
"Modulus" here is a synonym for "absolute value". Nothing to do with modular arithmetic. — Nate Eldredge, May 24 '18 at 17:53

lhf · Accepted Answer · 2018-05-24T18:29:46.590

1

We have $\delta \pi + \delta^2 < \delta \pi + \delta$ if $\delta < 1$.

We want $2\delta \pi + \delta^2 < \varepsilon$.

So, it is enough that $2\delta \pi + \delta < \varepsilon$, or $\delta < \frac{\varepsilon}{2\pi + 1}$.

Now, $2\pi + 1 < 10$ and so it is enough that $\delta < \frac{\varepsilon}{10}$.

Putting this all together: $$ \delta < \frac{\varepsilon}{10} < 1 \implies x^2−\pi^2 = 2\delta \pi+\delta^2 < 2\delta \pi + \delta = (2\pi + 1)\delta < 10 \delta < \varepsilon $$

edited May 24 '18 at 18:29

answered May 24 '18 at 17:53

lhf

216,483

Do you mean "So, it is enough that $2\delta\pi+\delta\lt\color{#C00}{\epsilon}\dots$"? – robjohn May 24 '18 at 18:12
@robjohn, yes, thanks, fixed. – lhf May 24 '18 at 18:29

Using modulus to answer: "How close to $\pi$ do we need $x$ to be for $x^2$ to be within $10^{−100}$ of $\pi ^2$?"

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