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Find the $1000$th digit after the decimal point of $\sqrt{n}$, where $n=\underbrace{11\dots1}_{1998 \text{ 1's}}$.

Obviously, $\underbrace{11\dots1}_{1998 \text{ 1's}}=\dfrac{1}{9}\left(9\cdot10^{1997}+9\cdot 10^{1996}+\dots+9\right),$ so we want to find $\left(\dfrac{10^{1998}-1}{9}\right).$ If only there was some way to convert this expansion into some closed form. I'm not sure if calculus would be useful. The problem asks for a single digit, so if we consider repeating digits, everything will be a lot easier. There seems to be a pattern in the decimal expansions of numbers consisting of only $1.$ For instance,

$$\sqrt{1}=1,$$ $$\sqrt{11}=3.3166247...$$ $$\sqrt{111}=10.5356537...$$ $$\sqrt{1111}=33.3316666...$$ $$\sqrt{11111}=105.408728...$$ $$\sqrt{111111}=333.333166...$$ $$\sqrt{1111111}=1054.09250...$$

Every term of the form $\displaystyle\sum_{n=0}^{2k+1}10^n$ has $k+1$ $3$'s at the beginning and $k+1$ 3's right after the decimal expansion, followed by one $1,$ and $2(k+1)$ $6$'s. Proving this would prove that the $1000$th digit is $1.$ This is the same as showing that $\sqrt{\left(\dfrac{10^{2m}-1}{9}\right)}=\dfrac{10^{m}-1}{3}+\dfrac{1}{3}-\dfrac{1}{6}\cdot 10^{-m}+\epsilon_m$ where $|\epsilon_m|<10^{-2m}.$

Edit: the previous question I asked was inspired by the current one, but the previous question seemed to have a rather unpleasant answer, so I changed it.

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    $\begingroup$ The exponent of seven is a largish odd integer $2M+1$. So you are looking at $7^M\sqrt{7}$. Unless there is a reason to suspect a trick, you need rather accurate information on both $\sqrt{7}$ and $7^M$. What is the source of this problem? That may give clues as to whether a trick exists :-) $\endgroup$
    – Jyrki Lahtonen
    Commented Oct 27, 2019 at 15:41
  • $\begingroup$ My guess is that there exists a "small" integer $n$ such that $7^{211^{67}}+n$ is a perfect square $m^2$. Then, $\sqrt{7^{211^{67}}}$ will either be slightly less than $m$ (in which case the $1000$-th digit is a $9$) or slightly greater than $m$ (in which case the $1000$-th digit is a $0$). Of course, you'll have to show that $|n| \lesssim 2 \cdot 10^{-1000} \cdot \sqrt{7^{211^{67}}}$. $\endgroup$
    – JimmyK4542
    Commented Oct 27, 2019 at 15:43
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    $\begingroup$ You keep changing your question. This is the third iteration I've seen. Please decide what you want to ask, and ask that. $\endgroup$
    – Arthur
    Commented Oct 27, 2019 at 15:52
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    $\begingroup$ Does that mean that you are just making up these problems as you go? $\endgroup$
    – Hagen von Eitzen
    Commented Oct 27, 2019 at 15:56
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    $\begingroup$ You completely changed the question with your edit! Please don't do this again. $\endgroup$
    – TonyK
    Commented Oct 27, 2019 at 16:39

2 Answers 2

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A little experimentation shows that if $n$ is the integer with $2m$ digits, where $m$ is an integer, all of them $1,$ then $\sqrt{n}$ has $m$ $3$'s, followed by a decimal point, $m$ $3$'s, a $1,$ and $2m$ $6$'s. If this was true, that would imply that the required digit is $1.$ Just substitute $2m=1998$ to verify.

Let $x=\dfrac{10^m-\frac{1}{2}10^{-m}}{3},m\in\mathbb{N}.$ Then $x$ has $m$ $3$'s, followed by a decimal point, followed by $m$ $3$'s, followed by one $1,$ followed by infinite $6$'s. So we just need to show that $(x-10^{-m-2})^2<n$ and $(x+10^{-m-2})^2>n,$ where $n=\dfrac{10^{2m}-1}{9}.$ This will show that the $(m+1)$st digit, or $1000$th digit, is indeed $1.$

$$(x-10^{-m-2})^2=\left(\dfrac{10^m}{3}-\dfrac{47}{300}10^{-m}\right)^2\\ =\dfrac{10^{2m}}{9}-\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\ =n-\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\ <n$$.

Similarly, $$(x+10^{-m-2})^2=\left(\dfrac{10^m}{3}+\dfrac{47}{300}10^{-m}\right)^2\\ =\dfrac{10^{2m}}{9}+\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\ =n+\dfrac{47}{450}+\dfrac{2209}{90\;000}10^{-2m}\\ >n$$.

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    $\begingroup$ @DavidK you're right. I'm going to fix that. $\endgroup$
    – Simon Fraser
    Commented Oct 29, 2019 at 4:33
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    $\begingroup$ now the proof should be true for all $n\in\mathbb{N}$ $\endgroup$
    – Simon Fraser
    Commented Oct 29, 2019 at 4:44
  • $\begingroup$ Yes, I think that fixed it. I think a couple of subtractions in the last part should be additions, but that seems just a cut-and-past error because the rest of that part is correct. $\endgroup$
    – David K
    Commented Oct 29, 2019 at 7:36
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It seems you actually are on the right track. Solving "smaller" problems of the "same kind" often pays off, and this is one of the times it does.

You already found that

$$ n = \frac{10^{1998}-1}{9} = \frac{10^{1998}}{9} - \frac19.$$

The binomial expansion for $(a+b)^{1/2}$ with $a = \frac{10^{2m}-1}{9}$ and $b = -\frac19$ gives us \begin{multline} \left(\frac{10^{2m}}{9} - \frac19\right)^{\!1/2} = \left(\frac{10^{2m}}{9}\right)^{\!1/2} + \frac12 \left(\frac{10^{2m}}{9}\right)^{\!-1/2} \left(-\frac19\right)\\ - \frac18 \left(\frac{10^{2m}}{9}\right)^{\!-3/2} \left(-\frac19\right)^{\!2} + \frac1{16} \left(\frac{10^{2m}}{9}\right)^{\!-5/2} \left(-\frac19\right)^{\!3} + \cdots.\end{multline}

Now try the following comparisons: \begin{align} \left(\frac{10^{2m}}{9}\right)^{\!1/2} && \text{vs.} &&& \dfrac{10^{m}-1}{3}+\dfrac13, \\ \end{align} \begin{align} \frac12 \left(\frac{10^{2m}}{9}\right)^{\!-1/2} \left(-\frac19\right) && \text{vs.} &&& - \dfrac16 10^{-m} \\ \end{align} \begin{align} - \frac18 \left(\frac{10^{2m}}{9}\right)^{\!-3/2} \left(-\frac19\right)^{\!2} + \frac1{16} \left(\frac{10^{2m}}{9}\right)^{\!-5/2} \left(-\frac19\right)^{\!3} + \cdots && \text{vs.} &&& 10^{-2m} \end{align}

You should be able to confirm the formula that you worked out from the pattern of digits in $\sqrt{11},$ $\sqrt{1111},$ and $\sqrt{111111}.$

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  • $\begingroup$ Note that the second degree term is subtracted, which depending on carries can matter no matter how many digits away it is. $\endgroup$
    – Arthur
    Commented Oct 27, 2019 at 17:38
  • $\begingroup$ @DavidK, are you implying that I should estimate the first 1000 digits of $\dfrac{10^{1998}}{9}$? And if so, would I need to use an $\epsilon$ to show this formally? Also, shouldn't the second term in your expansion be $+\dfrac{1}{2}\left(\dfrac{10^{1998}}{9}\right)^{-1/2}\left(\dfrac{1}{9}\right)$? $\endgroup$
    – user717371
    Commented Oct 27, 2019 at 18:28
  • $\begingroup$ @Arthur Subtraction can carry far to the left if you have a long string of zeros. Addition can carry a long way to the left if you have a long string of nines. So I think you have to watch out for those things either way. $\endgroup$
    – David K
    Commented Oct 27, 2019 at 19:18
  • $\begingroup$ @DavidK You're right. I had myself considered using Taylor before dividing by $3$, and in that case the subtraction is really relevant. $\endgroup$
    – Arthur
    Commented Oct 27, 2019 at 19:31
  • $\begingroup$ The second term in the expansion is negative because $b$ is negative. But I have rewritten the expansion to make this more explicit (the negative sign comes from $-\frac19$ and not from the coefficient $\frac12$). $\endgroup$
    – David K
    Commented Oct 27, 2019 at 20:04

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