CTK Insights

02 Aug

Nested radicals

Elsewhere we proved that the sequence

(1) \sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}}\ldots

is convergent to 2. With this in mind, it is possible and meaningful to shorten the description of this fact to

  2 = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\ldots}}}}}

Expressions like that serve an example of infinite nested radicals. They are realized in terms of limits, if those exist, and are understood as the shorthand for the latter.

For a real s > 0,

(2) \sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s\ldots}}}}}

always exists. For s < 2, this follows immediately from the Monotone Convergence theorem because then sequence (2) is bounded by the sequence (1) termwise. For s > 2, the derivation is a little more complex.

The curious observation about (2) is that, like in (1), the infinite nested radical may equal an integer. In general, let

  S = \sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s\ldots}}}}}

Then, adding s to both sides,

  S + s = s + \sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s\ldots}}}}} = S^2.

Thus we obtain a relationship between s and S: s = S^{2}-S and using the quadratic formula also S = (1 \pm \sqrt{1+4s})/2. The negative number ought to be recognized as spurious and discarded. So, finally we have

  \sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s\ldots}}}}} = (1+\sqrt{1+4s})/2.

on the one hand and

(3) S = \sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s\ldots}}}}}, for s = S^{2}-S,

on the other. From (3) we get several examples:

  3 = \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6\ldots}}}}},
4 = \sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12\ldots}}}}},
5 = \sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20\ldots}}}}},
6 = \sqrt{30+\sqrt{30+\sqrt{30+\sqrt{30+\sqrt{30\ldots}}}}},

and so on.

An interesting quandary arises when we pick S = 1 which would require s = 0 and thus appear to imply

  1 = \sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0\ldots}}}}}

which, obviously could not be right. Indeed, the right-hand side represents the limit of the sequence

  \sqrt{0},\sqrt{0+\sqrt{0}},\sqrt{0+\sqrt{0+\sqrt{0}}}\ldots

with all terms zero. The sequence converges to 0, not 1! How does one explain that?

Answer →

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

How do you explain

  1 = \sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0\ldots}}}}}?

At one point we derived the relation S = (1 \pm \sqrt{1+4s})/2 and dismissed one solution with the sign minus as being negative and therefore unsuitable. However, for s = 0, the two solutions are (1 \pm 1)/2, i.e., 1 and 0. In this case, 1 is obviously spurious and should be dropped, leaving the correct identity

  0 = \sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0\ldots}}}}}.

References

  1. C. A. Pickover, A Passion for Mathematics, John Wiley & Sons, 2005, p. 96
  2. C.C. Clawson, Mathematical Mysteries, Plenum Press, 1996, pp. 140-144

4 Responses to “Nested radicals”

  1. 1
    Xamuel Says:

    Nice trick there with the spurious 1. I feel like this well deserves a place with all the other famous "1=0 contradictions" where the student is required to find the flaw.

  2. 2
    admin Says:

    Yes, I agree.

    In addition, this is a sort of interplay between algebra and analysis that is often disregarded, e.g., in questioning whether .999... = 1 or not. Several math notations include ellipses and have no meaning, except as limits. Once this is understood, algebra may be be fruitfully employed.

  3. 3
    Hubert Shutrick Says:

    The more obvious way to get S^2 = s + S is to just square the equation.

  4. 4
    admin Says:

    Hubert, hi.

    You are probably right, but should everything be always done the most obvious way? May there be a virtue in a little obfuscation so as to trigger a thought process. I am asking that because all my life - which probably means a few dozen times - I've been squaring the identity as the first step. This is the first time that I did it differently and here you at the right moment in the right place.

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