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Can you suggest an increasing rational sequence converging to $\sqrt{2} $ in its closed form?

Can you make it from the recursion $x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}$? I dont think so, because, in that case $x_n$ should be $\sqrt{2} $ enough (Not interested). Further, how can we write the sequence $1,1.4,1.41,1.414,...$ in general form?

Riaz
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    What is meant by “$x_n$ should be $\sqrt{2}$ enough”? The sequence you give will decrease to the square root of two if initialized properly. You can make it increase to the square root of two via $x_{n+1}=ax_n +b/x_n$ for suitable $a,b$. – Michael Nov 26 '19 at 05:00

4 Answers4

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You are right, the sequence $a_0=1,a_1=1.4,a_2=1.41,a_3=1.414,...$ does increase to $\sqrt2$. The $n$-th term of the sequence, specifically, is $\sqrt2$ rounded down to $n$ decimal places. Note that, to round down, we use the floor function: $$\lfloor 1.9\rfloor=1.$$ Also note that if we multiply $\sqrt2$ by $10^n$ and round down we get an integer with $n$ digits which is less than $10^n\sqrt2$. Now, all we do is multiply our remaining integer by $10^{-n}$ to put the decimal point back were it's supposed to be. In symbols, $$a_n=10^{-n}\lfloor10^n\sqrt2\rfloor.$$ Note that for all $n$, $$1\le a_n<\sqrt2$$ and $a_n\le a_{n+1}$, so $\lim_{n\to\infty}a_n=\sqrt{2}$.

Yes, there are more efficient and correct methods for the calculation of square roots, but this is exactly what was asked for: an increasing rational sequence converging to $\sqrt2$.

clathratus
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    I don't think this is a good answer. It's not useful to actually compute $\sqrt{2}$, since the recursive formula contains $\sqrt{2}$. – MathematicsStudent1122 Nov 26 '19 at 05:01
  • This is a correct answer to the last question. I was writing a similar one, but feel this one misses pointing out that sequences that depend on the decimal representation of the number are not useful for most purposes. – Ross Millikan Nov 26 '19 at 05:02
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    @MathematicsStudent1122: OP specifically asked about this at the end. It does give a general form for the sequence, but does presume you know $\sqrt 2$. – Ross Millikan Nov 26 '19 at 05:03
  • @MathematicsStudent1122 see the edit – clathratus Nov 26 '19 at 05:05
  • What about the sequence $\frac{[\sqrt {2} n]}{n}$, where $[x]$ denotes the integer part of $x$, Is it increasing? – Riaz Nov 26 '19 at 05:21
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    @RIYASUDHEENT.K the sequence $x_n=[an]/n$ does increase and converge to $a$, but not very quickly. It is not strictly increasing. For example, with $a=\pi$, $$x_1=x_2=x_3=...=x_7=3.$$ It is not until $n=8$ that $x_n>3$. – clathratus Nov 26 '19 at 05:30
  • Could you pls show that $\frac{[an]}{n}$ is non decreasing to $a>0$? – Riaz Nov 26 '19 at 05:36
  • @RIYASUDHEENT.K We have that if $a>0$ then $a=m+\epsilon$ for some integer $m\ge0$ and $0\le \epsilon<1$. Set $x_n=[an]/a$. Then see that $ax_n=[n(m+\epsilon)]$, so $mn\le ax_n$. Thus $x_n$ does not decrease. – clathratus Nov 26 '19 at 05:46
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If we take $a_1=2$, and $a_{n+1}=\frac12(a_n+2/a_n)$ we get a famous decreasing sequence of rationals converging to $\sqrt2$. So instead take $b_n=2/a_n$. Now the $b_n$ increase and tend to $\sqrt2$. In fact $b_1=1$ and

$$b_{n+1}=\frac{4}{b_n+2/b_n}.$$

$$ $$ $$ $$

ngc1300
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Angina Seng
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Consider $f(x)=(x^2-2)(x-2)$. This has a root at $\sqrt{2}$, and $f'(\sqrt{2})<0$, and $f''(\sqrt{2})>0$. Therefore if you use Newton's root approximation method starting at a rational number just below $\sqrt{2}$, you will get a sequence of rational numbers converging to $\sqrt{2}$ from below.

The recursion is $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{(x_n^2-2)(x_n-2)}{3x_n^2-4x_n-2}$.

For example, $x_1=1$.

Then $x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{1}{-3}=\frac{4}{3}$.

Then $x_3=x_2-\frac{f(x_2)}{f'(x_2)}=\frac{4}{3}-\frac{4/27}{-2}=\frac{38}{27}$.

And $x_4=\cdots=\frac{23482}{16605}=1.41415236\ldots$ which already rounds to the same value as $\sqrt{2}$ at the ten-thousandths place.

And so on.

2'5 9'2
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  • @RIYASUDHEENT.K That was just an error from copy-pasting the first derivative statement. And having $f''(x)>0$ is actually what is needed. – 2'5 9'2 Nov 26 '19 at 05:04
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Why you think $x_{n+1}=\frac{1}{2}\left(x_n+\frac{2}{x_n}\right)$ will not work$?$ Hints: Show that $x_n^2\geq2$ (ignoring the initial term), and then use this to prove that $x_n-x_{n+1}\geq0$. Conclude that $\lim x_n=\sqrt{2}$

Rushabh Mehta
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emonHR
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