Approximating $x=\sqrt{2}+1$

Question

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?

Answer 1

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 $y>1+ \sqrt{2}$

Answer 2

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 $$|z-x|\sim(3-2\sqrt2)|\epsilon|<|z-y|\sim(4-2\sqrt2)|\epsilon|$$

Answer 3

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 $y_1$.