{"id":1084,"date":"2013-06-22T06:10:09","date_gmt":"2013-06-22T10:10:09","guid":{"rendered":"http:\/\/allegro.mit.edu\/~zong\/wpress\/?p=1084"},"modified":"2013-06-22T19:43:23","modified_gmt":"2013-06-22T23:43:23","slug":"tetration","status":"publish","type":"post","link":"https:\/\/blog.yhuang.org\/?p=1084","title":{"rendered":"tetration"},"content":{"rendered":"

Here is a problem from fakalin. Find the error in the following proof:<\/p>\n

Let \\(x^{x^{x\\cdots}} = 2\\). Then \\(x^{x^{x\\cdots}} = x^{x^{x^{x\\cdots}}} = 2\\). Substituting, we get \\(x^2=2\\) and \\(x=\\sqrt{2}\\). Now let \\(x^{x^{x\\cdots}} = 4\\). Similarly, \\(x^{x^{x^{x\\cdots}}} = 4\\), \\(x^4=4\\) and \\(x=\\sqrt{2}\\). But then \\(2=x^{x^{x\\cdots}}=4\\) and \\(2=4\\).
\n
\nObviously 2 cannot equal 4.<\/p>\n

The problem seems to occur with solving for \\(x\\) by inverting the function \\(f(x)=x^{x^{x\\cdots}}\\) via the recursion \\(x^{f(x)}=f(x)\\), \\(x=f(x)^{1\/f(x)}\\). While we can set \\(f(x)\\) to any value \\(s\\) to solve \\(x=s^{1\/s}\\), we mustn’t neglect also the condition that \\(s=f(x)\\) should have a solution.<\/p>\n

Here we plot \\(x=s^{1\/s}\\) for \\(s\\) from 0 to 5. (Let us only consider \\(x\\ge 0\\), where everything is real.) Both \\(s=2\\) and \\(s=4\\) give \\(x=\\sqrt{2}\\), but only the former satisfies \\(s=f(x)\\), while no \\(s\\) to the right of the peak has a solution. The reason is that \\(f(x)\\) for \\(x>1\\) has to be increasing by the monotonicity of exponentiation in each of its two arguments (\\(a^b\\) increasing in both \\(a,b\\) if \\(a,b>1\\)), and so the proper inverse (where it is defined) can only be taken from the left branch of the plot. The peak occurs at \\(s=e\\), \\(x=K=e^{1\/e}\\). For every \\(x>K\\), \\(f(x)\\) diverges to \\(\\infty\\). For \\(1< x\\le K[\/latex], [latex]\\overbrace{x^{x^x\\cdots}}^n[\/latex] is increasing in the tetration depth [latex]n[\/latex] and has an upper-bound ([latex]e[\/latex]), so [latex]f(x)[\/latex] is a well defined limit which is exactly [latex]s[\/latex].\n\nThe other half is more complicated. When [latex]x=0[\/latex], [latex]0^s=0[\/latex] for all [latex]s>0\\) and \\(0^s=1\\) for \\(s=0\\), so there is no solution to \\(s=f(x)\\). This is a consequence of where \\(a,b<1[\/latex], [latex]a^b[\/latex] is increasing in [latex]a[\/latex] while decreasing in [latex]b[\/latex]. From [latex]x<1[\/latex], we get [latex]x^x>x\\), \\(x^{x^x} < x^x[\/latex], and so on in this alternating fashion. We also get from [latex]x^x < 1[\/latex] that [latex]x^{x^x} > x\\), so in fact we have two non-crossing monotonic subsequences — the odd-depth tetrations increasing and the even-depth ones decreasing — that must therefore both converge. \\(f(x)\\) is well defined iff they converge to the same value \\(s\\). Suppose one subsequence has limit \\(s_1\\) and the other \\(s_2\\). Either of the cases \\(s=s_1\\) or \\(s=s_2\\) satisfies \\(x^{x^s}=s\\), so where there is a unique solution for a given \\(x\\), the limits must equal and \\(f(x)\\) must be well defined. From the plot (no proof) this occurs from \\((1,1)\\) down to the intersection at \\(s=1\/e\\), \\(x=L=(1\/e)^e\\). So for \\(L \\le x < 1[\/latex], [latex]f(x)[\/latex] is well defined. For all [latex]x < L[\/latex] though, multiple solutions for [latex]s[\/latex] actually correspond to the curved branch of this [latex](s_1,s_2)[\/latex] plot, which shows solutions to [latex]x^{s_1}=s_2[\/latex] and [latex]x^{s_2}=s_1[\/latex], i.e., [latex]s_1^{1\/s_2}=s_2^{1\/s_1}[\/latex], so [latex]f(x)[\/latex] is not defined there. (*)\n\nIn summary, [latex]f(x)[\/latex] is well defined between [latex]1\/e[\/latex] and [latex]e[\/latex] for [latex](1\/e)^e\\le x \\le e^{1\/e}[\/latex], goes to [latex]\\infty[\/latex] over the range, and flips between two values under the range. Since [latex]x^{x^{x\\cdots}}=4[\/latex] can never happen, it is a vacuous assumption that can logically lead to any conclusion.\n\n\n

\n

(*) Rather than looking for the limit of a sequence of tetrations, [latex]f(x)\\) can be formally defined in the lower range \\(\\)0 < x < L[\/latex] by the recursion [latex]x^{f(x)}=f(x)[\/latex] directly: in the second plot, the branch from [latex](1,1)[\/latex] is exactly this equilibrium (i.e., that branch is the first plot), but in this lower range the equilibrium is unstable and therefore not reachable by a finite tetration of [latex]x[\/latex].\n<\/p>\n","protected":false},"excerpt":{"rendered":"

Here is a problem from fakalin. Find the error in the following proof: Let . Then . Substituting, we get and . Now let . Similarly, , and . But then and .<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[3000,3001,394],"class_list":["post-1084","post","type-post","status-publish","format-standard","hentry","category-uncategorized","tag-exponentiation","tag-monotonicity","tag-proof"],"_links":{"self":[{"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=\/wp\/v2\/posts\/1084","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1084"}],"version-history":[{"count":0,"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=\/wp\/v2\/posts\/1084\/revisions"}],"wp:attachment":[{"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1084"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1084"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1084"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}