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How to solve "nested" problems

Something that I see a lot at math competitions or on Brilliant is a problem asking to find the value of \(\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}\) for \(x=\) some value. I'm going to talk about how to solve that "nested radical" as well as "nested" fractions.

First I'm going to talk about how to solve a problem such as finding the value of \(\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}.\) Let \(k=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}.\) You may notice that you are adding \(k\) to \(6\) and taking the square root of this. I'll demonstrate this with some parentheses. \[\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=\sqrt{6+\left(\sqrt{6+\sqrt{6+\ldots}}\right)}=\sqrt{6+\sqrt{\left(6+\sqrt{6+\ldots}\right)}}=\ldots\] For simplicity, we're just going to use \(\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=\sqrt{6+\left(\sqrt{6+\sqrt{6+\ldots}}\right)}.\) You may notice that this is the same thing as saying that \(k=\sqrt{6+k}.\) Let's try solving that. \[ \begin{align} k&=\sqrt{6+k}\\ k^2&=6+k\\ k^2-k-6&=0\\ (k-3)(k+2)&=0\\ k=3\text{ or }k=-2\\ \end{align} \] We can eliminate the \(k=-2\) solution because the nested radical obviously is non-negative. Thus, we can say that \(\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=\boxed{3}.\)

Now we can use a similar line of logic to find a general form for \(\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}.\) Let that equal \(y.\) We will make use of the same procedure, but using the quadratic formula to solve for \(y.\) \[ \begin{align} y&=\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}\\ y&=\sqrt{x+y}\\ y^2&=x+y\\ y^2-y-x&=0\\ y&=\dfrac{1\pm\sqrt{1+4x}}{2}\\ \end{align} \] Since we are dealing with real numbers, the value under the radical cannot be negative (except for \(x=0\).) Thus, we can say that \(\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}=\dfrac{1+\sqrt{1+4x}}{2}.\)

Now let's use the same logic to derive a formula for nested fractions of the form \(\dfrac{1}{a+\frac{1}{a+\ddots}}\) as well as continued fractions where the numerator is not necessarily \(1.\)

First, the obligatory demonstration of such a fraction. \[ \begin{align} \dfrac{1}{2+\frac{1}{2+\ddots}}&=\dfrac{1}{2+\left(\frac{1}{2+\ddots}\right)}\\ k&=\dfrac{1}{2+k}\\ k^2+2k&=1\\ k^2+2k-1&=0\\ k&=\dfrac{-2\pm\sqrt{4+4}}{2}\\ k&=-1\pm\sqrt{2} \end{align} \] Again, we can throw out the negative solution and say that \(\dfrac{1}{2+\frac{1}{2+\ddots}}=-1+\sqrt{2}.\)

Moving on, we can derive a general formula for finding \(\dfrac{c}{x+\frac{c}{x+\ddots}},\) allowing this expression to be equal to \(y.\)

\[ \begin{align} y&=\dfrac{c}{x+\frac{c}{x+\ddots}}\\ y&=\dfrac{c}{x+y}\\ y^2+xy&=c\\ y^2+xy-c&=0\\ y&=\dfrac{-x+\sqrt{x^2+4c}}{2} \end{align} \]

So a summary of what we have proved here:

  • \(\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}=\dfrac{1+\sqrt{1+4x}}{2}\)
  • \(\dfrac{c}{x+\frac{c}{x+\ddots}}=\dfrac{-x+\sqrt{x^2+4c}}{2}\)

If you want some practice problems, here's two.

  1. Find a closed form for \(x+\dfrac{1}{2x+\frac{1}{2x+\ddots}}.\)

  2. This calculus problem

  3. Find all values of \(n\) for which \(\sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}\) is an integer.

I hope this helps you out! Please feel free to share what you think in the comments. Thank you very much!

Note by Trevor B.
2 years, 1 month ago

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Some more nested radicals

\(\displaystyle \sqrt{x + \sqrt{x - \sqrt{x + \ldots }}} = \dfrac{1 + \sqrt{4x -3}}{2}\)

\(\displaystyle \sqrt{x - \sqrt{x + \sqrt{x - \ldots }}} = \dfrac{-1 + \sqrt{4x - 3}}{2}\) Krishna Sharma · 2 years, 1 month ago

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If you want to practice some problems on Nested Radicals ,
you may refer to my set Nested Radicals Pranay Singh · 2 years, 1 month ago

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Thank you :) Muhammad Rasel Parvej · 2 years, 1 month ago

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3.Let \(\sqrt{n+\sqrt{n+\sqrt{\cdots}}}=x\).Then we can write it as \(\sqrt{n+x}=x\) as you have just shown.Now solving: \[n+x=x^2\\ x^2-x-n=0\\ x=\frac{-1\pm\sqrt{(-1)^2-4(1)(-n)}}{2(1)}\\ x=\frac{-1\pm\sqrt{4n+1}}{2}\] As the nested radical is positive,so we discard the negative square root.And we are left with: \[\boxed{x=\frac{-1+\sqrt{4n+1}}{2}}\] By observation we conclude that \(4n+1\) must be a perfect square odd number.That's as far as I got.Perhaps someone could complete this. Abdur Rehman Zahid · 2 years, 1 month ago

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This is excellent. Thanks Trevor. Silas Hundt Staff · 2 years, 1 month ago

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Thanks dude, I'll keep your tips it in mind while solving such questions \(\ddot\smile\) Azhaghu Roopesh M · 2 years, 1 month ago

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