robertj@garfield.UUCP (Robert Janes) (06/18/85)
This is a small problem which has bothered me for some time and I and I would like to hear some comments from various folks out there in netland. The problem arises in considering the expression: EQ 1 sqrt( 2 + sqrt( 2 + sqrt( 2 + sqrt( 2 + sqrt( 2 + ....))))) = x where sqrt is taken to mean the standard positive square root of a number and there are an "infinite" number of these square roots nested one inside another as shown. The problem is what is x ? ( METHOD I) The first solution which came to mind arose from the observation, which I now suspect to be invalid, that if we strip off the first radical ( that is we square both sides of the equation ) we get: EQ 2 2 + sqrt( 2 + sqrt( 2 + sqrt( 2 + ....)))) = x^2 which of course ( here's were I get a bit suspicious ) can be rewritten as EQ 3 2 + x = x^2 which can then be rearranged and solved as follows EQ 3a x^2 - x - 2 = 0 EQ 3b ( x - 2 ) ( x + 1 ) = 0 EQ 3c x = 2, x = -1 :DONE: QUESTIONS: 1) Is EQ 3 a legitimate expression to derive from EQ 2 ? 2) If so, why are there two different answers produced, at first glance I would expect the equation to generate a quadratic with repeated roots as the positive square root is well defined. 3) If EQ 3 is legitimate are these equations legitimate ? EQ 4 4 + 4( sqrt( 2+ sqrt( 2 + sqrt( 2 + .... )))) + 2 + sqrt( 2 +..))) = x^4 ( square of equation 2 ) EQ 5 6 + 5x = x^4 EQ 6 x^4 - 5x - 6 = 0 which has solutions EQ 7 x = -1 x = 2 x = -1/2 + i*sqrt( 11 ) x = -1/2 - i*sqrt( 11 ) question 4) If the answer to 3 is yes then can we continue this process indefinitely and if so: a) if we are considering the 2^n power will all the solutions of 2^n-1 power be included ? b) are there infinitely many solutions ? METHOD 2 This is a very indirect approach to the problem: let a1 = pi/4 then cos( a1 ) = sqrt( 2 )/2 let ai = a(i-1)/2 then by induction we can show that cos( ai ) = sqrt( 2 + sqrt( 2 + ....)))))..)/2 where there are i sqrt's but in the limit as i->infinity ai = 0 hence cos( ai ) = 1 hence 2 = sqrt( 2 + sqrt( 2 +sqrt( 2 + .....))))))) :DONE: QUESTIONS: question 1) Is this a legitimate method ? question 2) Are there any other such constructions which will yield similar results using trignometric functions question 3) Can a similar series be constructed for each of the solutions given by METHOD 1 ? _______________________________________________________________________________ I would appreciate any comments on the above and any corrections appropos algebra or trigonometry. I suspect that since we are dealing with infinite sequences that the results shown in section 1 are somewhat bogus as they make that great leap into defining certain things to be x when it may not be rigourously certain to be true. As you may have gathered I have a certain preference for METHOD 2 as it assumes very little in comparison to METHOD I and seems somewhat more elegant. Robert Janes Memorial University of Newfoundland
csc@watmath.UUCP (Computer Sci Club) (06/20/85)
>EQ 1 > > sqrt( 2 + sqrt( 2 + sqrt( 2 + sqrt( 2 + sqrt( 2 + ....))))) = x > > The problem is what is x ? First we must define the problem rigorously. Let a(1) = sqrt(2) a(2) = sqrt( 2 + sqrt(2)) a(3) = sqrt( 2 + sqrt(2 + sqrt(2))) a(n) = sqrt( 2 + sqrt(2 + ... +sqrt(2)))...) (n sqrt's) Let x = lim a(n) n->inf Show that the limit (x) exists, and find its value. We have a(1) = sqrt(2) (i) a(n+1) = sqrt( 2 + a(n)) (ii) Therefore if a(n) < 2 then a(n+1) < sqrt (2 +2) = 2 but a(1) < 2 Therefore by induction a(n) < 2 for all n (iii) by (iii) a(n)**2 = a(n) * a(n) < 2 * a(n) < a(n) + a(n) < 2 + a(n) Therefore by (ii) a(n+1) > a(n) for all n (iv) The a(n)'s form an increasing sequence bounded above (by 2), so the sequence must have a limit. (i.e. x exists) Furthermore x must be greater than a(1) = sqrt(2) so: sqrt(2) < x <= 2. Now a(n+1) = sqrt(2 + sqrt(a(n)) a(n+1)**2 = 2 + a(n) Taking the limit as n goes to infinity of each side (this step is now justified as we know the limit exists) we obtain lim a(n+1)**2 = 2 + lim a(n) n->inf n->inf 2 x = 2 + x (v) The solutions of (v) are x = 2 and x = -1. But as x > sqrt(2) we must have x=2. Note that although x must be a solution of equation (v) it is not true that any solution to (v) is equal to x. -1 is called an extraneous root, that is an extra solution brought about by squaring an equation. Consider a simple example the equation a=b. If we square both sides we obtain a**2=b**2 with solutions a=b and a = -b . When solving an equation by squaring both sides one must always be on the lookout for extraneous roots. (as indicated in the original posting if we take both sides of the equation to the forth power we obtain even more extraneous roots). Therefore METHOD 1 is valid but needs more details to be rigorous METHOD 2 is equivalent to noting that a(n) = 2 cos [ pi/(2**(n+1)) ] (vi) Therefore lim a(n) = lim 2 cos [ pi/(2**(n+1)) ] n->inf n->inf = 2 cos(0) = 2 This method is perfectly valid (although to make it rigorous one would have to prove the identity (vi) ). It is not however as general as method 1 the basic ideas of which can be applied to a very large class of similar problems. William Hughes
percus@acf4.UUCP (Allon G. Percus) (06/25/85)
> The first solution which came to mind arose from the observation, > which I now suspect to be invalid, that if we strip off the first radical > ( that is we square both sides of the equation ) we get: > > EQ 2 > 2 + sqrt( 2 + sqrt( 2 + sqrt( 2 + ....)))) = x^2 > > > which of course ( here's were I get a bit suspicious ) can be rewritten as > > EQ 3 > > 2 + x = x^2 > > which can then be rearranged and solved as follows > > EQ 3a > x^2 - x - 2 = 0 > > EQ 3b > ( x - 2 ) ( x + 1 ) = 0 > > EQ 3c > x = 2, x = -1 > QUESTIONS: > > 1) Is EQ 3 a legitimate expression to derive from EQ 2 ? Absolutely. You're just manipulating an identity. > 2) If so, why are there two different answers produced, at first glance > I would expect the equation to generate a quadratic with repeated > roots as the positive square root is well defined. There is only one answer produced. You specified that "sqrt" means the "positive square root", so, plugging x=-1 into it: sqrt(2+sqrt(2+sqrt(2+...)))=-1 sqrt(2+x)=-1 The positive square root of 2+x can most certainly NOT be -1!!! Therefore, x=-1 must be discarded as an extraneous root. > 3) If EQ 3 is legitimate are these equations legitimate ? > > EQ 4 > 4 + 4( sqrt( 2+ sqrt( 2 + sqrt( 2 + .... )))) + 2 + sqrt( 2 +..))) > = x^4 > > ( square of equation 2 ) > > EQ 5 > 6 + 5x = x^4 > > EQ 6 > > x^4 - 5x - 6 = 0 > > which has solutions > > EQ 7 > x = -1 > x = 2 > x = -1/2 + i*sqrt( 11 ) > x = -1/2 - i*sqrt( 11 ) Equations 4, 5, and 6 are absolutely correct. However, in #7, we must again discard the negative root as extraneous, and the same can be shown to hold for the complex roots. > 4) If the answer to 3 is yes then can we continue this process > indefinitely and if so: > a) if we are considering the 2^n power will all the solutions > of 2^n-1 power be included ? > b) are there infinitely many solutions ? If you go on to 2^n, as n -> inf, you'll get an infinite number of roots, but again, you'll have to discard all but one as extraneous: except for x=2, all roots will either be negative or complex. If I have some time later on, I'll provide a semi-rigorous proof of this. As for your first part of the question, I'll have to think about it. METHOD 2, I agree, is somewhat more elegant than METHOD 1, but I'm afraid I can't look at any more now. I find this a VERY interesting problem. A. G. Percus (ARPA) percus@acf4 (NYU) percus.acf4 (UUCP) ...!ihnp4!cmcl2!acf4!percus
meister@linus.UUCP (Phillip W. Servita) (06/27/85)
In article <920006@acf4.UUCP> percus@acf4.UUCP (Allon G. Percus) writes: >> The first solution which came to mind arose from the observation, >> which I now suspect to be invalid, that if we strip off the first radical >> ( that is we square both sides of the equation ) we get: >> >> EQ 2 >> 2 + sqrt( 2 + sqrt( 2 + sqrt( 2 + ....)))) = x^2 by now everybody knows the problem we are talking about here. when i was back in high school, i discovered this by accident. however, i took it a little further than people seem to be taking it here. take the general function: y = f(x) = sqrt(x + sqrt(x + sqrt(x + sqrt(x + ...)))) hence, 2 y = x + sqrt(x + sqrt(x + sqrt(x + sqrt(x + ...)))) or: 2 2 y = x + y , so y - y - x = 0, with solutions easily given by the quadratic formula: y = (1 +/- sqrt(1 + 4x))/2 for x = 2, this gives values of 2 and -1, giving rise to the extraneous root discussion. what i noticed, is that: TRY PLUGGING IN -1/4 for X! this gives a repeated root of 1/2! hence sqrt(-1/4 + sqrt(-1/4 + ...)) = 1/2. this one just defies all intuition. enjoy. -the venn buddhist
percus@acf4.UUCP (Allon G. Percus) (06/28/85)
> ...what i noticed, is that: TRY PLUGGING IN -1/4 for X! > this gives a repeated root of 1/2! hence > > sqrt(-1/4 + sqrt(-1/4 + ...)) = 1/2. this one just defies all > intuition. enjoy. > > -the venn buddhist Hmm...Very strange indeed. Obviously it's true, but very unexpected... A. G. Percus (ARPA) percus@acf4 (NYU) percus.acf4 (UUCP) ...!ihnp4!cmcl2!acf4!percus
sinclair@aero.ARPA (William S. Sinclair) (09/19/85)
Dear Robert:
What you really have is the recursion formula:
x(n+1)=2+sqrt(x(n))
If you assume x(n+1)=x(n) [the limit case] then
x=2+sqrt(x)
Which gives x=4 or 1. THe second value assumes that you use the negative root.
Bill Sinclair 213/647-1753