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.

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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