Kvant Math Problem 254

Consider small cases of numbers of the form $0.

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Problem

Compute the value of $\sqrt{0{,}\smash{\underbrace{11111\ldots11111}_{100}}}\vphantom{\underbrace0_0}$ to an accuracy of

  1. 100 digits after the decimal point;
  2. 101 digits after the decimal point;
  3. 200 digits after the decimal point.

A. A. Egorov

Exploration

Consider small cases of numbers of the form $0.1$, $0.11$, $0.111$, $0.1111$, and so on. Denote by $S_n$ the number $0.\underbrace{111\ldots1}_{n}$ with $n$ ones after the decimal point. Compute their decimal expansions of the square roots. For $n=1$, $\sqrt{0.1} \approx 0.3162277660\dots$. For $n=2$, $0.11$, the square root is approximately $0.3316624793\dots$. For $n=3$, $0.111$, the square root is $0.3331666665\dots$. For $n=4$, $0.1111$, the square root is $0.3333166665\dots$.

From these examples, a pattern emerges: as $n$ increases, the decimal expansion begins with repeated digits $3$ followed by a slowly increasing sequence, eventually tending toward a sequence reminiscent of $0.\underbrace{3333\ldots}_{n}\dots$.

The crucial step is to find a representation of $0.\underbrace{111\ldots1}_{n}$ in a form suitable for series expansion or closed-form calculation of the square root. One natural approach is to express the number as a geometric series:

$0.\underbrace{111\ldots1}{n} = \sum{k=1}^{n} \frac{1}{10^k} = \frac{1 - 10^{-n}}{9}.$

Then the square root becomes

$\sqrt{0.\underbrace{111\ldots1}_{n}} = \sqrt{\frac{1 - 10^{-n}}{9}} = \frac{\sqrt{1 - 10^{-n}}}{3}.$

This shows the decimal expansion will be close to $\frac{1}{3} = 0.3333\dots$, with corrections due to the small term $10^{-n}$. To find the expansion to hundreds of decimal places, a series expansion using the binomial theorem or Newton iteration is required. The critical difficulty is controlling the error term to ensure that the decimal digits are correct to the desired precision.

Problem Understanding

The problem asks for the computation of the square root of the 100-digit repeating decimal $0.\underbrace{111\ldots1}_{100}$ to extremely high precision, specifically 100, 101, and 200 decimal digits. This is a Type C problem because it involves determining the precise numerical value of a function with maximal accuracy. The core difficulty lies in performing the expansion of $\sqrt{1 - 10^{-100}}$ to a sufficient number of terms so that the error is smaller than $10^{-200}$, which guarantees that all the required digits are correct. The answer will be extremely close to $\frac{1}{3}$, with small corrections starting around the 101st decimal digit.

Proof Architecture

Lemma 1. The number $0.\underbrace{111\ldots1}_{n}$ can be expressed as $\frac{1 - 10^{-n}}{9}$. This follows from summing the geometric series with ratio $\frac{1}{10}$.

Lemma 2. The square root can be expressed as $\sqrt{0.\underbrace{111\ldots1}_{n}} = \frac{1}{3}\sqrt{1 - 10^{-n}}$. This follows directly from Lemma 1 by factoring $1/9$.

Lemma 3. For small $x$, $\sqrt{1 - x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \dots$, an infinite binomial series convergent for $|x| < 1$. This is a standard binomial expansion.

Lemma 4. Truncating the series after $m$ terms yields an error smaller than $x^{m+1}$ times a factor of order 1, allowing explicit control of the decimal digits. The proof uses the alternating sign and decreasing magnitude of the terms.

The hardest step is Lemma 4 because the required precision demands careful estimation of the remainder term to guarantee no rounding affects the first 200 digits.

Solution

By Lemma 1, the number $0.\underbrace{111\ldots1}_{100}$ equals

$S_{100} = \frac{1 - 10^{-100}}{9}.$

By Lemma 2, its square root equals

$\sqrt{S_{100}} = \frac{1}{3} \sqrt{1 - 10^{-100}}.$

Denote $x = 10^{-100}$. Then $0 < x < 1$, and we can expand $\sqrt{1 - x}$ using the binomial theorem as

$\sqrt{1 - x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \frac{5 x^4}{128} - \frac{7 x^5}{256} - \dots.$

Thus

$\sqrt{S_{100}} = \frac{1}{3} \left(1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \frac{5 x^4}{128} - \dots \right).$

The term $x = 10^{-100}$, $x^2 = 10^{-200}$, and all higher powers are smaller. Therefore, truncating after the first term, we have

$\sqrt{S_{100}} = \frac{1}{3} - \frac{10^{-100}}{6} - \frac{10^{-200}}{24} - \dots$

To compute the first 100 decimal digits, the correction term $10^{-100}/6$ affects only the 101st digit and beyond. Therefore, for 100 digits we can write

$\sqrt{0.\underbrace{111\ldots1}{100}} = 0.\underbrace{3333\ldots3}{100}.$

For 101 digits, the term $-10^{-100}/6 = -0.000\ldots016666\dots$ contributes to the 101st decimal place. Dividing $1/6$ in decimal gives $0.1666666\dots$, hence the 101st digit decreases from $3$ to $2$, giving

$\sqrt{0.\underbrace{111\ldots1}{100}} = 0.\underbrace{3333\ldots3}{100}2.$

For 200 digits, the second term $10^{-200}/24 \approx 4.1666\dots \cdot 10^{-202}$ is smaller than $10^{-200}$, so it does not affect digits 1 through 200. Therefore the first 200 digits are

$\sqrt{0.\underbrace{111\ldots1}{100}} = 0.\underbrace{3333\ldots3}{100}2\underbrace{0000\ldots0}_{99}.$

In summary, the value is extremely close to $\frac{1}{3}$, with only the 101st digit adjusted by the small correction term from the binomial expansion.

Verification of Key Steps

The first delicate step is the expansion of $0.\underbrace{111\ldots1}_{100}$ as $\frac{1 - 10^{-100}}{9}$. Summing the geometric series with ratio $r = \frac{1}{10}$ gives

$\sum_{k=1}^{100} r^k = r \frac{1 - r^{100}}{1 - r} = \frac{1 - 10^{-100}}{9},$

confirming the formula.

The second delicate step is estimating the effect of the binomial expansion terms on the decimal digits. The first correction term is $10^{-100}/6 \approx 1.666\dots \cdot 10^{-101}$. Multiplying by $1/3$ gives $5.555\dots \cdot 10^{-102}$, which is smaller than $10^{-101}$, confirming that the first 100 digits are unaffected. The second term $10^{-200}/24$ is approximately $4.1666\dots \cdot 10^{-202}$, which is smaller than $10^{-200}$, confirming that truncation at the second term preserves 200-digit accuracy.

Alternative Approaches

An alternative method is to apply Newton-Raphson iteration for the square root: starting with an initial approximation, iterate $y_{k+1} = \frac{y_k + S_{100}/y_k}{2}$. Starting with $y_0 = 1/3$, each iteration approximately doubles the number of correct decimal digits. After seven or eight iterations, 200-digit precision can be achieved. This method is algorithmically efficient but requires careful high-precision arithmetic, whereas the binomial expansion provides a closed-form estimation with explicit control of error terms, making it preferable for exact decimal digit computation.