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 Uniqueness Criterion for Tetration UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 08/13/2007, 01:33 PM jaydfox Wrote:I'm not sure I follow why your solution is $C^\infty$ The function is continuous, of course: $\begin{eqnarray} \lim_{y \to {\small 1^{-}}} F(x,y, 0) & = & x^y \\ & = & x^1 \\ \vspace{10} \\ & = & x \\ \vspace{15} \\ \lim_{y \to {\small 1^{+}}} F(x,y-1, 1) & = & x^{x^{y-1}} \\ & = & x^{x^0} \\ \vspace{5} \\ & = & x^1 \\ \vspace{10} \\ & = & x \end{eqnarray}$ And for base e, the first derivative is continuous as well: $\begin{eqnarray} \lim_{y \to {\small 1^{-}}} D_y F(x,y, 0) & = & D_y x^y \\ & = & x^y \ln(x) \\ & = & x^1 \ln(x) \\ & = & x \ln(x) \\ \vspace{10} \\ \lim_{y \to {\small 1^{+}}} D_y F(x,y-1, 1) & = & D_y \left(x^{x^{y-1}}\right) \\ & = & \left(x^{x^{y-1}}\right) \ln(x) D_x (x^{y-1}) \\ & = & \left(x^{x^{y-1}}\right) \left(x^{y-1}\right) \left(\ln(x)\right)^2 \\ & = & \left(x^{x^0}\right) (x^0) \left(\ln(x)\right)^2 \\ & = & x \left(\ln(x)\right)^2 \end{eqnarray}$ However, even at the second derivative, we begin to see problems: $\begin{eqnarray} \lim_{y \to {\small 1^{-}}} D_y^2 F(x,y, 0) & = & D_y^2 x^y \\ & = & D_y \left( x^y\right) \ln(x) \\ & = & x^y \left(\ln(x)\right)^2 \\ & = & x^1 \left(\ln(x)\right)^2 \\ & = & x \left(\ln(x)\right)^2 \\ \vspace{10} \\ \lim_{y \to {\small 1^{+}}} D_y^2 F(x,y-1, 1) & = & D_y^2 \left(x^{x^{y-1}}\right) \\ & = & D_y \left( \left(x^{x^{y-1}}\right)\left( x^{y-1}\right)\right) \left(\ln(x)\right)^2 \\ & = & \left(\left( \left[D_y \left(x^{x^{y-1}}\right)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left[D_y \left( x^{y-1}\right)\right]\right)\right) \left(\ln(x)\right)^2 \\ & = & \left(\left( \left[x^{x^{y-1}} \ln(x) D_y \left(x^{y-1}\right)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left( x^{y-1}\right) \ln(x) \right)\right) \left(\ln(x)\right)^2 \\ & = & \left(\left( \left[x^{x^{y-1}} \ln(x) \left(x^{y-1}\right)\ln(x)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left( x^{y-1}\right) \ln(x) \right)\right) \left(\ln(x)\right)^2 \\ & = & \left( \left(x^{x^{y-1}}\right) \left(x^{y-1}\right) \left(\ln(x)\right)^3 \right) \left(\left(x^{y-1}\right)\ln(x) + 1\right) \\ & = & \left( \left(x^{x^0}\right) \left(x^0\right) \left(\ln(x)\right)^3 \right) \left(\left(x^0\right)\ln(x) + 1\right) \\ & = & \left( \left(x^1\right) \left(1\right) \left(\ln(x)\right)^3 \right) \left(\left(1\right)\ln(x) + 1\right) \\ & = & \left( x \left(\ln(x)\right)^3 \right) \left(\ln(x) + 1\right) \\ \end{eqnarray}$ For base e, we get a second derivative of 1 from the left, but 2 from the right. Due to the product rule, the second and higher derivatives will get quite complex, preventing a simple solution. I see. I guess you are right. Thanks for the correction. « Next Oldest | Next Newest »

 Messages In This Thread Uniqueness Criterion for Tetration - by jaydfox - 08/09/2007, 07:01 AM RE: Uniqueness Criterion for Tetration - by bo198214 - 08/09/2007, 09:05 AM RE: Uniqueness Criterion for Tetration - by jaydfox - 08/09/2007, 03:58 PM RE: Uniqueness Criterion for Tetration - by bo198214 - 08/10/2007, 06:19 PM RE: Uniqueness Criterion for Tetration - by UVIR - 08/12/2007, 10:52 PM RE: Uniqueness Criterion for Tetration - by jaydfox - 08/13/2007, 02:58 AM RE: Uniqueness Criterion for Tetration - by UVIR - 08/13/2007, 01:33 PM RE: Uniqueness Criterion for Tetration - by tommy1729 - 04/27/2014, 08:17 PM RE: Uniqueness Criterion for Tetration - by Gottfried - 05/01/2014, 09:07 PM RE: Uniqueness Criterion for Tetration - by tommy1729 - 05/01/2014, 10:21 PM

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