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<h1>The Nested Radical Constant</h1>
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<span> Posted on Tue 19 November 2019 in <a href="https://newptcai.github.io/category/math.html" style="font-style: italic">math</a>
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<p>My colleagues at Uppsala University regularly give some puzzles to everyone. We got the following
one for this week. I like it because it sort of shows how computer can help with this type of
problems.</p>
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<h2 id="The-problem">The problem<a class="anchor-link" href="#The-problem">¶</a></h2><p>Let $a_n = 1+\sqrt{2+\sqrt{3+\dots+\sqrt{n+\sqrt{n+1}}}}$. Show that $a_n$ is bounded</p>
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<h2 id="The-experimental-solution">The experimental solution<a class="anchor-link" href="#The-experimental-solution">¶</a></h2>
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<p>Let's try some numeric experiment first.</p>
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<div class=" highlight hl-ipython2"><pre><span></span><span class="k">def</span> <span class="nf">a</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="n">prec</span><span class="o">=</span><span class="mi">53</span><span class="p">,</span> <span class="n">numeric</span><span class="o">=</span><span class="kc">False</span><span class="p">):</span>
<span class="n">n0</span> <span class="o">=</span> <span class="n">n</span>
<span class="n">value</span> <span class="o">=</span> <span class="n">n</span><span class="o">+</span><span class="mi">1</span>
<span class="k">while</span> <span class="n">n0</span> <span class="o">></span><span class="mi">0</span><span class="p">:</span>
<span class="n">value</span> <span class="o">=</span> <span class="p">(</span><span class="n">n0</span><span class="o">+</span><span class="n">sqrt</span><span class="p">(</span><span class="n">value</span><span class="p">))</span>
<span class="k">if</span> <span class="n">numeric</span><span class="p">:</span>
<span class="n">value</span> <span class="o">=</span> <span class="n">value</span><span class="o">.</span><span class="n">n</span><span class="p">(</span><span class="n">prec</span><span class="o">=</span><span class="n">prec</span><span class="p">)</span>
<span class="n">n0</span> <span class="o">-=</span> <span class="mi">1</span>
<span class="k">return</span> <span class="n">value</span>
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<p>The sequence</p>
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<div class=" highlight hl-ipython2"><pre><span></span><span class="p">[</span><span class="n">a</span><span class="p">(</span><span class="n">n</span><span class="p">)</span> <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">6</span><span class="p">)]</span>
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<pre>[sqrt(2) + 1,
sqrt(sqrt(3) + 2) + 1,
sqrt(sqrt(5) + 2) + 1,
sqrt(sqrt(sqrt(sqrt(5) + 4) + 3) + 2) + 1,
sqrt(sqrt(sqrt(sqrt(sqrt(6) + 5) + 4) + 3) + 2) + 1]</pre>
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<p>The numeric value</p>
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<div class=" highlight hl-ipython2"><pre><span></span><span class="p">[</span><span class="n">a</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="n">numeric</span><span class="o">=</span><span class="kc">True</span><span class="p">)</span> <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">10</span><span class="p">)]</span>
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<pre>[2.41421356237309,
2.93185165257814,
3.05817102727149,
3.08437362526552,
3.08930271111850,
3.09016194475189,
3.09030226595168,
3.09032389883267,
3.09032706620193]</pre>
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<p>This does look convergent. What is the limit?</p>
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<div class=" highlight hl-ipython2"><pre><span></span><span class="n">a</span><span class="p">(</span><span class="mi">100000</span><span class="p">,</span> <span class="n">numeric</span><span class="o">=</span><span class="kc">True</span><span class="p">,</span> <span class="n">prec</span><span class="o">=</span><span class="mi">100</span><span class="p">)</span>
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<pre>3.0903275767905763591925445067</pre>
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<p>If you check this number on <a href="https://www.wolframalpha.com/input/?i=3.0903275767905763591925445067">WolframAlpha</a>, you will see that it's the square of the <a href="http://mathworld.wolfram.com/NestedRadicalConstant.html">nest-radical-constant</a>, which is defined by the limit of
$$
b_n = \sqrt{1+\sqrt{2+\dots \sqrt{n}}}
$$
And you will find this <a href="http://oeis.org/A072449">OEIS page</a> for it. This page tells you that,
$$
\sqrt{c_1+\sqrt{c_2+\dots \sqrt{c_n}}}
$$
converges if and only if
$$
\limsup_{n \to \infty} \frac{\log(c_n)}{2^n} < \infty
$$
In other words, we take $c_n = e^{2^n}$ and we will still get convergent. This gives us a lot of liberty to upper bound the original problem.</p>
<p>The easiest choice is actually just take $c_n = 2^{2^n} \ge n$. Let see what we get.</p>
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<div class=" highlight hl-ipython2"><pre><span></span><span class="k">def</span> <span class="nf">a1</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="n">prec</span><span class="o">=</span><span class="mi">53</span><span class="p">,</span> <span class="n">numeric</span><span class="o">=</span><span class="kc">False</span><span class="p">):</span>
<span class="n">n0</span> <span class="o">=</span> <span class="n">n</span>
<span class="n">value</span> <span class="o">=</span> <span class="mi">2</span><span class="o">^</span><span class="p">(</span><span class="mi">2</span><span class="o">^</span><span class="p">(</span><span class="n">n</span><span class="o">+</span><span class="mi">1</span><span class="p">))</span>
<span class="k">while</span> <span class="n">n0</span> <span class="o">></span><span class="mi">0</span><span class="p">:</span>
<span class="n">value</span> <span class="o">=</span> <span class="p">(</span><span class="mi">2</span><span class="o">^</span><span class="p">(</span><span class="mi">2</span><span class="o">^</span><span class="n">n0</span><span class="p">))</span><span class="o">+</span><span class="n">sqrt</span><span class="p">(</span><span class="n">value</span><span class="p">)</span>
<span class="k">if</span> <span class="n">numeric</span><span class="p">:</span>
<span class="n">value</span> <span class="o">=</span> <span class="n">value</span><span class="o">.</span><span class="n">n</span><span class="p">(</span><span class="n">prec</span><span class="o">=</span><span class="n">prec</span><span class="p">)</span>
<span class="n">n0</span> <span class="o">-=</span> <span class="mi">1</span>
<span class="k">return</span> <span class="n">sqrt</span><span class="p">(</span><span class="n">value</span><span class="p">)</span>
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<div class=" highlight hl-ipython2"><pre><span></span><span class="n">a1l</span><span class="o">=</span><span class="p">[</span><span class="n">a1</span><span class="p">(</span><span class="n">n</span><span class="p">)</span> <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="mi">6</span><span class="p">)]</span>
<span class="n">a1l</span>
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<pre>[2*sqrt(2),
2*sqrt(sqrt(2) + 1),
2*sqrt(sqrt(sqrt(2) + 1) + 1),
2*sqrt(sqrt(sqrt(sqrt(2) + 1) + 1) + 1),
2*sqrt(sqrt(sqrt(sqrt(sqrt(2) + 1) + 1) + 1) + 1)]</pre>
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<p>So we just need to show this sequence converges</p>
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<div class=" highlight hl-ipython2"><pre><span></span><span class="n">a1l</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="o">/</span><span class="mi">2</span> <span class="o">*</span> <span class="n">v</span> <span class="k">for</span> <span class="n">v</span> <span class="ow">in</span> <span class="n">a1l</span><span class="p">]</span>
<span class="n">a1l</span>
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<pre>[sqrt(2),
sqrt(sqrt(2) + 1),
sqrt(sqrt(sqrt(2) + 1) + 1),
sqrt(sqrt(sqrt(sqrt(2) + 1) + 1) + 1),
sqrt(sqrt(sqrt(sqrt(sqrt(2) + 1) + 1) + 1) + 1)]</pre>
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<p>And this is quite obvious if you look at this picture</p>
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<div class=" highlight hl-ipython2"><pre><span></span><span class="n">var</span><span class="p">(</span><span class="s1">'x'</span><span class="p">)</span>
<span class="n">plot</span><span class="p">([</span><span class="n">sqrt</span><span class="p">(</span><span class="mi">1</span><span class="o">+</span><span class="n">x</span><span class="p">),</span><span class="n">x</span><span class="p">],</span> <span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">),</span> <span class="n">legend_label</span><span class="o">=</span><span class="p">[</span><span class="sa">r</span><span class="s2">"$\sqrt{1+x}$"</span><span class="p">,</span><span class="sa">r</span><span class="s2">"$x$"</span><span class="p">],</span> <span class="n">aspect_ratio</span><span class="o">=</span><span class="mi">1</span><span class="p">)</span>
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<img 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<p>This means if we let $f(x)=\sqrt{1+x}$ and $f_n(x)$ be $f$ applied to $x$ for $n$ times, then $f_n(x)$ converges to the fixed point of $f(x)$, regardless of where you start. And this fixed point is</p>
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<div class="prompt input_prompt">In [9]:</div>
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<div class=" highlight hl-ipython2"><pre><span></span><span class="n">sol</span> <span class="o">=</span> <span class="n">solve</span><span class="p">(</span><span class="mi">1</span><span class="o">+</span><span class="n">x</span><span class="o">==</span><span class="n">x</span><span class="o">^</span><span class="mi">2</span><span class="p">,</span><span class="n">x</span><span class="p">)</span>
<span class="n">show</span><span class="p">(</span><span class="n">sol</span><span class="p">[</span><span class="mi">1</span><span class="p">])</span>
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<html><script type="math/tex; mode=display">\newcommand{\Bold}[1]{\mathbf{#1}}x = \frac{1}{2} \, \sqrt{5} + \frac{1}{2}</script></html>
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<p>This concludes the proof.</p>
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