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This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

392 pickover/pickover.18.p


Title: Cliff Puzzle 18: Difficult Nested Roots
From: cliff@watson.ibm.com

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover

* * *

Consider the following nested set of square roots.

.eq ? = sqrt <1 + G sqrt <1+(G+1) sqrt < 1 + ... >>>

Here, G indicates "Googol" or 10**100.
The "<" and ">" symbols indicate where the beginning and ends of the
the nested roots.

1. What is the value for in this infinite set of nested roots.
2. What is the next term under the root?

Hint:
In 1911, the famous mathematical prodigy Srinivasa Ramanujan posed the
following question (#298) in a new mathematical journal called the
:Journal of the Indian Mathematical Society.

.eq ? = sqrt <1 + 2 sqrt <1+3 sqrt <1 + ... >>>

pickover/pickover.18.s

-------------------------

In article <1992Nov11.221749.129578@watson.ibm.com> you write:
: Title: Cliff Puzzle 18: Difficult Nested Roots
: From: cliff@watson.ibm.com
: Consider the following nested set of square roots.
:
: ? = sqrt <1 + G sqrt <1+(G+1) sqrt < 1 + ... >>>
:
: Here, G indicates "Googol" or 10**100.
: The "<" and ">" symbols indicate where the beginning and ends of the
: the nested roots.
:
: 1. What is the value for in this infinite set of nested roots.
: 2. What is the next term under the root?
: Hint:
: In 1911, a twenty-three-year-old Indian clerk named Srinivasa Ramanujan
: posed the following question (#298) in a new mathematical journal called
: the Journal of the Indian Mathematical Society.
:
: ? = sqrt <1 + 2 sqrt <1+3 sqrt <1 + ... >>>
:
Doing a n-depth thing-a-ding on this.....

n=1   v=1
2     1.732
3     2.236
4     2.5598
5     2.7551
6     2.867
....
20    2.99999376
....

so I expect that the sum is actually 3. Or in the general case when the
2 (or the G from above) is replaced by m, then the evaluation of the series
is m+1. This CAN be shown as follows....

m+1 = sqrt(1+m sqrt(1+(m+1)*sqrt(....))
m^2 + 2m +1 = 1 + m *sqrt(1 + (m+1)*sqrt(...))
m^2 + 2m = m*sqrt(1+(m+1)*sqrt(...))
m+2 = sqrt(1+(m+1)*sqrt(1+(m+2)*sqrt(...))

Thus if m+1 is then sum when the series is based off m, then m+2 is then
sum when the series is based off m+1. Since this works for m=2 (as shown
above), then it must work for all whole numbers (mathematical induction is
such a wonderful thing...)

Therefore, the sum with m=G is G+1.

The next term, as show above, is (1+(m+2)*sqrt(1+....))

--
Michael Neylon   aka Masem the Great and Almighty Thermodynamics GOD!
      //         | Senior, Chemical Engineering, Univ. of Toledo
  \\ // Only the |   Summer Intern, NASA Lewis Research Center
\  \X/   AMIGA!  |         mneylon@jupiter.cse.utoledo.edu           /
 --------+ How do YOU spell 'potato'?  How 'bout 'lousy'? +----------
    "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L

 

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