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

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.

• 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 :-) Commented Oct 27, 2019 at 15:41
• 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 $|n| \lesssim 2 \cdot 10^{-1000} \cdot \sqrt{7^{211^{67}}}$. Commented Oct 27, 2019 at 15:43
• You keep changing your question. This is the third iteration I've seen. Please decide what you want to ask, and ask that. Commented Oct 27, 2019 at 15:52
• Does that mean that you are just making up these problems as you go? Commented Oct 27, 2019 at 15:56
• You completely changed the question with your edit! Please don't do this again. Commented Oct 27, 2019 at 16:39

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 $$(x-10^{-m-2})^2 and $$(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}\\ .

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

• @DavidK you're right. I'm going to fix that. Commented Oct 29, 2019 at 4:33
• now the proof should be true for all $n\in\mathbb{N}$ Commented Oct 29, 2019 at 4:44
• 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. Commented Oct 29, 2019 at 7:36

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.

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

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 $$\sqrt{111111}.$$

• Note that the second degree term is subtracted, which depending on carries can matter no matter how many digits away it is. Commented Oct 27, 2019 at 17:38
• @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 $+\dfrac{1}{2}\left(\dfrac{10^{1998}}{9}\right)^{-1/2}\left(\dfrac{1}{9}\right)$?
– user717371
Commented Oct 27, 2019 at 18:28
• @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. Commented Oct 27, 2019 at 19:18
• @DavidK You're right. I had myself considered using Taylor before dividing by $3$, and in that case the subtraction is really relevant. Commented Oct 27, 2019 at 19:31
• 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 $\frac12$). Commented Oct 27, 2019 at 20:04