



 
Simon, I doubt very much that you will find what you are looking for in fact I very much expect that there is no solution in terms of elementary functions. I did experiment with some graphing software and, at least for small values of n, 1≤ n ≤ 20, I saw that the behaviour depended on n. For n = 2, 3, 5, 7, 11, 13, 17 and 19 it looked like there was no solution in (0, 1). For n = 4 and 9 the only solution is (0, 1) seems to be x = 1/2. For n = 6, 8, 10, 14, 15 and 16 it looks like there is one solution in (0, 1/2) and for n = 12, 18 and 20 there seem to be 2 solutions in (0, 1/2). You didn't say how you want to use this but one possibility is to approximate the function f(x) = cos^{2}(π × n^{x}) + cos^{2}(π × n^{1x})  2 using a power series. I am no expert on this and I don't even have a suggestion of the type of series to try. I'm sorry I'm not much help, Well, RD showed me wrong in a very nice way. Harley Simon, It's not as hard as you may have thought. Assuming we're looking only at real numbers here, cos^{2}(alpha) cannot be outside [0,1] for any alpha. (1) So what can you say about cos^{2}(π × n^{x}) and cos^{2}(π × n^{1x})? (2) Then what can you say about cos(π × n^{x}) and cos(π × n^{1x}) ? (3) And what does this tell you about n^{x} and n^{1x}? (4) How are n^{x} and n^{1x} related? (Hint: what is their product?) (5) You should now be able to see why there are (eg) two solutions in (0, 0.5) for n=12, and to write them as ratios of logarithms. By the way, are solutions in (0, 0.5) really sufficient? If n is a perfect square, do you really not want the solution x=0.5)? Good Hunting!  


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