Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version nyu B notes v1.5 12/10/84; site acf4.UUCP Path: utzoo!watmath!clyde!burl!ulysses!unc!mcnc!decvax!linus!philabs!cmcl2!acf4!percus From: percus@acf4.UUCP (Allon G. Percus) Newsgroups: net.math Subject: Re: REPEATED ROOTS Message-ID: <920006@acf4.UUCP> Date: Tue, 25-Jun-85 10:10:00 EDT Article-I.D.: acf4.920006 Posted: Tue Jun 25 10:10:00 1985 Date-Received: Sat, 29-Jun-85 00:36:49 EDT References: <15187@watmath.UUCP> Organization: New York University Lines: 91 > 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