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 x+x+x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}} for x=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 6+6+6+.\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}. Let k=6+6+6+.k=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}. You may notice that you are adding kk to 66 and taking the square root of this. I'll demonstrate this with some parentheses. 6+6+6+=6+(6+6+)=6+(6+6+)=\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 6+6+6+=6+(6+6+).\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=6+k.k=\sqrt{6+k}. Let's try solving that. k=6+kk2=6+kk2k6=0(k3)(k+2)=0k=3 or k=2 \begin{aligned} 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{aligned} We can eliminate the k=2k=-2 solution because the nested radical obviously is non-negative. Thus, we can say that 6+6+6+=3.\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=\boxed{3}.


Now we can use a similar line of logic to find a general form for x+x+x+.\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}. Let that equal y.y. We will make use of the same procedure, but using the quadratic formula to solve for y.y. y=x+x+x+y=x+yy2=x+yy2yx=0y=1±1+4x2 \begin{aligned} 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{aligned} Since we are dealing with real numbers, the value under the radical cannot be negative (except for x=0x=0.) Thus, we can say that x+x+x+=1+1+4x2.\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 1a+1a+\dfrac{1}{a+\frac{1}{a+\ddots}} as well as continued fractions where the numerator is not necessarily 1.1.

First, the obligatory demonstration of such a fraction. 12+12+=12+(12+)k=12+kk2+2k=1k2+2k1=0k=2±4+42k=1±2 \begin{aligned} \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{aligned} Again, we can throw out the negative solution and say that 12+12+=1+2.\dfrac{1}{2+\frac{1}{2+\ddots}}=-1+\sqrt{2}.

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

y=cx+cx+y=cx+yy2+xy=cy2+xyc=0y=x+x2+4c2 \begin{aligned} 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{aligned}


So a summary of what we have proved here:

  • x+x+x+=1+1+4x2\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}=\dfrac{1+\sqrt{1+4x}}{2}
  • cx+cx+=x+x2+4c2\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+12x+12x+.x+\dfrac{1}{2x+\frac{1}{2x+\ddots}}.

  2. This calculus problem

  3. Find all values of nn for which n+n+n+\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.
4 years, 6 months ago

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Some more nested radicals

x+xx+=1+4x32\displaystyle \sqrt{x + \sqrt{x - \sqrt{x + \ldots }}} = \dfrac{1 + \sqrt{4x -3}}{2}

xx+x=1+4x32\displaystyle \sqrt{x - \sqrt{x + \sqrt{x - \ldots }}} = \dfrac{-1 + \sqrt{4x - 3}}{2}

Krishna Sharma - 4 years, 6 months ago

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If you want to practice some problems on Nested Radicals ,
you may refer to my set Nested Radicals

Brilliant Member - 4 years, 6 months ago

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Thanks dude, I'll keep your tips it in mind while solving such questions ¨\ddot\smile

A Brilliant Member - 4 years, 6 months ago

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This is excellent. Thanks Trevor.

Silas Hundt Staff - 4 years, 6 months ago

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3.Let n+n+=x\sqrt{n+\sqrt{n+\sqrt{\cdots}}}=x.Then we can write it as n+x=x\sqrt{n+x}=x as you have just shown.Now solving: n+x=x2x2xn=0x=1±(1)24(1)(n)2(1)x=1±4n+12n+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: x=1+4n+12\boxed{x=\frac{-1+\sqrt{4n+1}}{2}} By observation we conclude that 4n+14n+1 must be a perfect square odd number.That's as far as I got.Perhaps someone could complete this.

Abdur Rehman Zahid - 4 years, 6 months ago

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My workings are similar to yours. I think your application of the quadratic equation is wrong though. If I am mistaken please correct me!

x = [1 +sqrt(1+4n)] / 2

x is an integer when 1 + 4n is a perfect square (and positive). 1 + 4n is an odd perfect square, therefore its square root is an odd integer.

Let (y + 1)^2 = 1 + 4n y^2 + 2y + 1 = 1+ 4n 4n = y^2 + 2y n = [y^2 + 2y] / 4 = y^2/4 + y/2 y is an even integer.

Thus n is an integer, whereby for all even integers y: n = (y^2)/4 + y/2

Jia En - 2 years, 4 months ago

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Thank you :)

Muhammad Rasel Parvej - 4 years, 5 months ago

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