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# Surprising Rational solutions from a Real equation

Posted 10 years ago
 I like to study x^(1/x) as much as possible, because of two reasons. One is the MRB constant an alternating sum of all positive integers to the power of their reciprocal. The second is the following which is true only for a certain small range! A number, n, taken to the power of itself an infinite number of times found by solving for x in x^(1/x)==n. For example what is 2/10 to itself an infinite number of times? You could compute it through brute force by entering In[294]:= N[Nest[Power[2/10, #] &, 1, 100], 30] Out[294]= 0.469621922935881329117128363349,  Or, to get a much more accurate solution by using x^(1/x), you could enter In[295]:= FindRoot[x^(1/x) - 2/10, {x, .46}, WorkingPrecision -> 30] Out[295]= {x -> 0.469621922935610544117803066891}.  Nonetheless, here is what I wanted to say: Considering f(x)=x^(1/x) for the range [-1,0), when does the Im(f)==Re(f)? You might be surprised to find out, it is at rational values of x. Plot[Im[x^(1/x)] - Re[x^(1/x)], {x, -1, 0}]  Where on the graph does Im[x^(1/x)] - Re[x^(1/x)]==0? I used FindRoot to find a few of them:  In[222]:= FindRoot[{Im[x^(1/x)] == Re[x^(1/x)]}, {x, -.6}, WorkingPrecision -> 100] Out[222]= {x -> \ -0.5714285714285714285714285714285714285714285714285714285714285714285\ 714285714285714285714285714285714} In[299]:= N[-4/7, 100] Out[299]= \ -0.5714285714285714285714285714285714285714285714285714285714285714285\ 714285714285714285714285714285714 In[229]:= FindRoot[{Im[x^(1/x)] == Re[x^(1/x)]}, {x, -.4}, WorkingPrecision -> 100] Out[229]= {x -> \ -0.3636363636363636363636363636363636363636363636363636363636363636363\ 636363636363636363636363636363635} In[300]:= N[-4/11, 100] Out[300]= \ -0.3636363636363636363636363636363636363636363636363636363636363636363\ 636363636363636363636363636363636 In[233]:= FindRoot[{Im[x^(1/x)] == Re[x^(1/x)]}, {x, -.2}, WorkingPrecision -> 100] Out[233]= {x -> \ -0.2105263157894736842105263157894736842105263157894736842105263157894\ 736842105263157894736842105263158} In[301]:= N[-4/19, 100] Out[301]= \ -0.2105263157894736842105263157894736842105263157894736842105263157894\ 736842105263157894736842105263158  Notice, we found Im[x^(1/x)] == Re[x^(1/x)] at -4 over some (but not all) primes. In[269]:= FullSimplify[ Table[{p = -4/Prime[x], Im[p^(1/p)] == Re[p^(1/p)]}, {x, 1, 10}]] Out[269]= {{-2, False}, {-(4/3), True}, {-(4/5), False}, {-(4/7), True}, {-(4/11), True}, {-(4/13), False}, {-(4/17), False}, {-(4/19), True}, {-(4/23), True}, {-(4/29), False}}  I then found several of such primes: I haven't figured the pattern out for the primes, though. In[265]:= FullSimplify[ Table[{p = -4/Prime[x], Im[p^(1/p)] == Re[p^(1/p)]}, {x, {2, 4, 5, 8, 9, 11, 14, 15, 17, 19, 20, 22, 23, 27, 28, 31, 32, 34, 36, 38, 39, 41, 43, 46, 47, 48, 49, 52, 54, 56, 58, 61, 63, 64, 67, 69, 72, 73, 75, 76, 81, 83, 85, 86, 90, 91, 92, 93, 94, 95, 96, 99}}]] Out[265]= {{-(4/3), True}, {-(4/7), True}, {-(4/11), True}, {-(4/19), True}, {-(4/23), True}, {-(4/31), True}, {-(4/43), True}, {-(4/47), True}, {-(4/59), True}, {-(4/67), True}, {-(4/71), True}, {-(4/79), True}, {-(4/83), True}, {-(4/103), True}, {-(4/107), True}, {-(4/127), True}, {-(4/131), True}, {-(4/139), True}, {-(4/151), True}, {-(4/163), True}, {-(4/167), True}, {-(4/179), True}, {-(4/191), True}, {-(4/199), True}, {-(4/211), True}, {-(4/223), True}, {-(4/227), True}, {-(4/239), True}, {-(4/251), True}, {-(4/263), True}, {-(4/271), True}, {-(4/283), True}, {-(4/307), True}, {-(4/311), True}, {-(4/331), True}, {-(4/347), True}, {-(4/359), True}, {-(4/367), True}, {-(4/379), True}, {-(4/383), True}, {-(4/419), True}, {-(4/431), True}, {-(4/439), True}, {-(4/443), True}, {-(4/463), True}, {-(4/467), True}, {-(4/479), True}, {-(4/487), True}, {-(4/491), True}, {-(4/499), True}, {-(4/503), True}, {-(4/523), True}}  We need to figure out the pattern for {2, 4, 5, 8, 9, 11, 14, 15, 17, 19, 20, 22, 23, 27, 28, 31, 32, 34, 36, 38, 39, 41, 43, 46, 47, 48, 49, 52, 54, 56, 58, 61, 63, 64, 67, 69, 72, 73, 75, 76, 81, 83, 85, 86, 90, 91, 92, 93, 94, 95, 96, 99... . Then, whether we will have found all values for x where Im(f)==Re(f) in [-1,0),may still an open question. However, we will have found a countable infinitude of them that are rational! I posted a notebook with all of this information in it. Attachments:
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Posted 10 years ago
 More on x^(1/x) and 2 continued fractions for it are found in the wayback machine here
Posted 10 years ago
 This looks good too: R = .5; Plot3D[{Im[(a + I b)^(1/(a + I b))], Re[(b + I a)^(1/(b + I a))]}, {a, -R, R}, {b, -R, R}, Mesh -> None, PlotPoints -> 100, ExclusionsStyle -> Opacity[.5]] 
Posted 10 years ago
 Do not ask me what this is because I do not know but... would not fall solutions on intersection of surfaces? R = .5; Plot3D[{Im[(a + I b)^(1/(a + I b))], Re[(a + I b)^(1/(a + I b))]}, {a, -R, R}, {b, -R, R}, Mesh -> None, PlotPoints -> 100, ExclusionsStyle -> Opacity[.5]] Another way - looks a bit like Mandelbrot set: R = .01; ContourPlot[Im[(a + I b)^(1/(a + I b))] == Re[(a + I b)^(1/(a + I b))], {a, -R, R}, {b, -R, R}, PlotPoints -> 50, ContourStyle -> Directive[Thickness[0], Opacity[.5]]] 
Posted 10 years ago
 The roots of x^(1/x)==k are functions of Log and ProductLog. x /. Solve[(1/5)^x == x, x, Reals] (* Out[196]= {ProductLog[Log[5]]/Log[5]} *) To get a better sense of what goes on with the re/im of x^(1/x) when x is between -1 and 0, I'd recommend starting with ComplexExpand. Also simplify using that range. ss = Simplify[ComplexExpand[Re[x^(1/x)] - Im[x^(1/x)]], Assumptions -> -1 < x < 0] (* Out[199]= (-x)^(1/x) (Cos[\[Pi]/x] - Sin[\[Pi]/x]) *) Now solve for this vanishing and simplify the result.FullSimplify[x /. Solve[ss == 0 && -1 <= x < 0, x]] (* Out[202]= {ConditionalExpression[4/(1 + 8 C[1]), C[1] \[Element] Integers && C[1] <= -1], ConditionalExpression[4/(-3 + 8 C[1]), C[1] \[Element] Integers && C[1] <= -1]} *) So those denominators are comprised of two arithmetic sequences which, negating to put them in the more customary form, are 8*n-1 and 8*n+3. That is, {7,15,23,31,39,47,55,...} and {11,19,27,35,43,51,...}. Note that 3 is ruled out by the original restriction that -1<=x<0Anyway, here are a couple of immediate observations. One is that the denominators that "work" are not all prime. The other is that the primes that "fail" are those of the form 8n+1 (e.g. 17,41,...) and 8n+5 (so 13,29,...).
Posted 10 years ago
 In like manor, for the function g(x)=(-1)^x *x^(1/x) in (0,Infinity), Im[g[x]=Re[g[x] at rational values also! In[322]:= g[x_] = (-1)^x *x^(1/x) Out[322]= (-1)^x x^(1/x) In[324]:= Plot[Im[g[x]] - Re[g[x]], {x, 0, 10}]  In[326]:= FindRoot[Im[g[x]] - Re[g[x]], {x, {2, 3, 4, 5, 6, 7}}] Out[326]= {x -> {2.25, 3.25, 4.25, 5.25, 6.25, 7.25}} In[334]:= FullSimplify[ Table[{q = x + 1/4; Im[g[q]] == Re[g[q]]}, {x, 1, 100}]] Out[334]= {{True},..., {True}} 
Posted 10 years ago
 I found the pattern for pattern for {2, 4, 5, 8, 9, 11, 14, 15, 17, 19, 20, 22, 23, 27, 28, 31, 32, 34, 36, 38, 39, 41, 43, 46, 47, 48, 49, 52, 54, 56, 58, 61, 63, 64, 67, 69, 72, 73, 75, 76, 81, 83, 85, 86, 90, 91, 92, 93, 94, 95, 96, 99... Positions of 4k+3 primes, OEIS A080148. In[309]:= t = Flatten[Position[Prime[Range[1000]], _?(IntegerQ[(# - 3)/4] &)]]; In[317]:= Table[p = -4/Prime[t[[x]]]; If[FullSimplify[Im[p^(1/p)] == Re[p^(1/p)]] != True, "not right sequence", ""], {x, 1, 200}] Out[317]= {"",... ""}