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Suppose $y>1$ is some approximation to $x=\sqrt{2}+1$. Give a brief reason (not a proof) why one should expect $(1/y)+2$ to be a closer approximation to $x$ than $y$ is.

After testing this out for a bit, it looks like we can let $y_{n+1}=\frac{1}{y_n}+2$ and $\lim_{n\to\infty}y_n=\sqrt{2}+1$, but this does not give me any intuitive idea as to why $y_{n+1}$ should be a better approximation to $x$ than $y_n$ is.

Can anyone give a brief reason for this improvement in aproximation, especially a more "intuitive" one than simple numerical data?

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    $\begingroup$ A side note: If $x=\sqrt{2}+1$, then $x=(1/x)+2$ (rationalizing the denominator for $1/(\sqrt{2}+1)$ makes this more apparent). $\endgroup$ – interrogative Feb 15 '16 at 17:15
  • $\begingroup$ A good intuition is to think that if your initial $y_n$ is in $(1-\sqrt(2),1+\sqrt(2))$, then the $y_n's$ increase towards $x$, and therefore $|y_{n+1}-x|<|y_n-x|$. You can see this by setting up $y_{n+1}>y_{n}=1/y_n+2>y_n$ which is quadratic and easily solvable. The same argument applies for $y_{n+1}<y_n$ Hope this helps! $\endgroup$ – Fede Poncio Feb 15 '16 at 17:35
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Let$ y<1+\sqrt{2}$,

So $1+\sqrt{2} - y=D$ (say).

$1/y > \sqrt{2}-1$

And $1/y+2>1+\sqrt{2}$

So let $1/y+2-(1+\sqrt{2}) =d.$

So we have to prove d < D.

Let D-d>0.

Simplifying you get : 2$\sqrt{2} >y+1/y $

Which is true as $1-\sqrt{2}<y<1+\sqrt{2}$.

You can do similarly for $y>1+ \sqrt{2}$

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  • $\begingroup$ I think this is the most natural/easiest way to approach this problem. One thing though: I think you left out one important piece at the very end..."which is true as $y<1+\sqrt{2}$." It really should be "which is true as $1<y<\sqrt{2}+1$" because $2\sqrt{2}>y+1/y$ holds for all $y\in(\sqrt{2}-1,\sqrt{2}+1)$. $\endgroup$ – interrogative Feb 15 '16 at 17:59
  • $\begingroup$ @interrogative Yes you are right. $\endgroup$ – Sudhanshu Feb 15 '16 at 18:22
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Let $y=\sqrt2+1+\epsilon=x+\epsilon$ with $\epsilon\to 0$ so $$\frac1y=\frac1{\sqrt 2+1+\epsilon}=\frac{\sqrt{2}-(1+\epsilon)}{2-(1+\epsilon)^2}=\frac{\sqrt2-1-\epsilon}{1-2\epsilon-\epsilon^2}$$ so that $$z=\frac1y+2= \frac{\sqrt2+1-5\epsilon-2\epsilon^2}{1-2\epsilon-\epsilon^2}\sim(\sqrt2+1-5\epsilon-2\epsilon^2)(1+2\epsilon)\sim x+(2\sqrt 2-3)\epsilon$$ and then $$|z-x|\sim(3-2\sqrt2)|\epsilon|<|z-y|\sim(4-2\sqrt2)|\epsilon|$$

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Let $y_n=\frac{\sqrt{2}+1}{1+\delta}$, with $\delta$ small (not necessarily positive or nonzero, but surely not $-1$), then \begin{align}y_{n+1}&=\frac1{y_n}+2\\ &=\frac{1+\delta}{\sqrt{2}+1}+2\\ &=(1+\delta)\frac{\sqrt{2}-1}{(\sqrt{2}-1)(\sqrt{2}+1)}+2\\ &=(1+\delta)(\sqrt{2}-1)+2\\ &=\sqrt{2}+1+\delta(\sqrt{2}-1) \end{align} We get $|y_n-(\sqrt{2}+1)|=|\frac{\delta}{\delta+1}|(\sqrt{2}+1)$ and $|y_{n+1}-(\sqrt{2}+1)|=|\delta|(\sqrt{2}-1)$ so what we want is $$\left|\frac{\delta}{\delta+1}\right|(\sqrt{2}+1)>|\delta|(\sqrt{2}-1)$$ and this is true since $$\left|\frac{1}{\delta+1}\right|(\sqrt{2}+1)^2>1$$ for small enough $\delta$ (we're good if $|\delta|<2+2\sqrt{2}$). And since this is true (if you're starting with a good enough approximation, that is. It'll still work for larger starting values, but you'll have to show that $\delta$ gets small enough), we'll get better and better approximations, since the errors get smaller and smaller.

Note that this was secretly induction. I was unable to do a base case, since you didn't provide $y_0$ or $y_1$.

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