Guest Posted June 6, 2010 Report Share Posted June 6, 2010 if N = the set of real numbers find X XN = X(N-1) + X(N-2} I'm pretty sure this works for any value of N and only one value for X Please give reason for the answer and the method used(this will be very important to me because I got it by observation) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 6, 2010 Report Share Posted June 6, 2010 I have an answer to the problem, but not yet worked out for all powers, although it does work for N=1 and N=2 Putting N=2 gives X^2 = X + 1 Solving for X gives X = (1+ sqrt 5)/2 Interestingly, this is the limit of the ratio of consecutive terms in the Fibonacci sequence, the golden ratio. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 6, 2010 Report Share Posted June 6, 2010 By iteration, if X^2 = X^1 +X^0, and we multiply both sides by X, then we get X^3 = X^2 + X^1, and so it is true for any N X^N = X^(N-1) + X^ (N-2) :) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 6, 2010 Report Share Posted June 6, 2010 (edited) If you divide through by X^(N-2) you get X^2 = X + 1 Which is a simple quadratic with roots of (1+SQRT(5))/2 and (1-SQRT(5))/2 and doesn't depend on N Edited June 6, 2010 by dannchen Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 6, 2010 Report Share Posted June 6, 2010 Whoops, I missed a minus sign in my first reply. Dannchen is right Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 6, 2010 Report Share Posted June 6, 2010 Try this one. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 6, 2010 Report Share Posted June 6, 2010 X^(N) = X^ (N-1) + X^(N-2) N Log X = (N-1) Log X + (N-2) Log X N = N - 1 + N - 2 N = 2N - 3 N = 3 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 7, 2010 Report Share Posted June 7, 2010 For positive N, X^N = X^(N-1)+X^(N-2) X^(N+2) = X^(N+1) + X^N X^N(X^2+X+1)=0 Either X = 0 or X = (1+-Sqrt(5))/2, by the quadratic equation. For negative N, X = (1+-Sqrt(5))/2, only. Quote Link to comment Share on other sites More sharing options...
0 random7 Posted June 7, 2010 Report Share Posted June 7, 2010 (edited) if N = the set of real numbers find X XN = X(N-1) + X(N-2} I'm pretty sure this works for any value of N and only one value for X Please give reason for the answer and the method used(this will be very important to me because I got it by observation) XN = XN-1 + XN-2 XN - XN-1 - XN-2 = 0 XN-2 (X2 - X - 1) = 0 Now if X2 - X - 1 = 0 then: X = (1 +- sqrt(5))/2 for all N an element of the reals. If XN-2=0 then: X = 0 for all N an element of the reals except N = 0, 1, 2. This proves that there are three potential answers for X, with X = 0 only if N != 0, 1, 2. EDIT: After posting this at the same time as marsupialsoup, I realised from their response that N cannot be less than or equal to 2 for the X = 0 solution. That is, if X = 0, N > 2. Edited June 7, 2010 by random7 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 7, 2010 Report Share Posted June 7, 2010 Just substitute XN-2 for t and you'll get: t³=t²+t One answer is 0 but that only works for N>2 Other answer is like everyone else answered: t²-t-1=0 (1±sqrt(5))/2 Which I believe is the Golden Ratio. http://en.wikipedia.org/wiki/Golden_ratio Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 7, 2010 Report Share Posted June 7, 2010 For positive N, X^N = X^(N-1)+X^(N-2) X^(N+2) = X^(N+1) + X^N X^N(X^2+X+1)=0 Either X = 0 or X = (1+-Sqrt(5))/2, by the quadratic equation. For negative N, X = (1+-Sqrt(5))/2, only. My mistake, I multiplied both sides of the equation by 0. The two cases are N>2 and N<=2, for N being real, X should be positive. This is a problem with polynomials over the real field; sometimes there are no roots... Quote Link to comment Share on other sites More sharing options...
0 Quantum.Mechanic Posted June 7, 2010 Report Share Posted June 7, 2010 X^(N) = X^ (N-1) + X^(N-2) N Log X = (N-1) Log X + (N-2) Log X N = N - 1 + N - 2 N = 2N - 3 N = 3 The log of a sum is not the sum of the logs. The problem stated that this worked for all N, but only 1 X. Where's your X? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 7, 2010 Report Share Posted June 7, 2010 ^ You mean it should have been in step 2 instead of: N Log X = (N-1) Log X + (N-2) Log X Should have been N Log X = Log(X^ (N-1) + X^(N-2)) Just in case anyone like me didn't understand... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 7, 2010 Report Share Posted June 7, 2010 Here's a thought: I strikes me that there is a simple solution based on the concept that there is one real number that stays the same whether you add it to itself or multiply itself by itself... You could project that through simple identity definitions, but I may be underthinking this. Quote Link to comment Share on other sites More sharing options...
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if N = the set of real numbers
find X
XN = X(N-1) + X(N-2}
I'm pretty sure this works for any value of N and only one value for X
Please give reason for the answer and the method used(this will be very important to me because I got it by observation)
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