When we have a tetration \( \\[15pt]
{y_{x}=\,^xa} \) with his base "a" between \( \\[25pt]
{1< a \leq e^{\frac{1}{e}}} \), it tends to an asymptote value \( \\[15pt]
{y_{\infty}} \) such that \( \\[25pt]
{a^{y_{\infty}}={y_{\infty}}} \).
But that depends on the definition \( \\[15pt]
{^0a=1} \). If it were defined \( \\[15pt]
{^0a=y_{\infty}} \), we would get an horizontal line (drawn with red dashes in the graphic down).
The interesting part is when we define \( \\[15pt]
{^0a} \) a bit larger than \( \\[15pt]
{y_{\infty}} \), the tetration also converges to the same asymptote to the right, and another asymptote to the left, at the limit \( \\[15pt]
{y_{-\infty}} \), defined by \( \\[25pt]
{a^{y_{-\infty}}={y_{-\infty}}} \).
We get a Z-shaped tetration function (drawn in green), contained between \( \\[15pt]
{y_{-\infty}} \) and \( \\[15pt]
{y_{\infty}} \), which I call the Z curve.
![[Image: a8V4PJV.gif?1]](http://i.imgur.com/a8V4PJV.gif?1)
The problem is that the Z curve is not uniquely defined. It depends on the value choosen for \( \\[15pt]
{^0a} \). Any value between \( \\[20pt]
{{y_{\infty}} < ^0a < {y_{-\infty}}} \) is valid, and the only difference this choice make, is the horizontal displacement on the curve. I drew the Z curve matching the origin with his inflection point (roughly).
The upper asymptote is also a non trivial function, and there is also another non trivial possible upper branch, obtained by choosing °a>-oo (drawn in blue line).
The question is, what would be convenient values for the definition of \( \\[15pt]
{^0a} \) for the green and blue branches?
If we take \( \\[15pt]
{a=\sqrt{2}} \), \( \\[15pt]
{^0a} \) may be multivalued at x=0: \( \\[15pt]
{^0(\sqrt{2})=(1,2,\; 2<y_3<4 \; , 4 , \; 4<y_5)} \).
The inflection point in the Z curve is near to 3. Maybe that base has all integer values for °a? (°1,41421356 = (1,2,3,4,5) )
![[Image: lynxfBI.jpg?1]](http://i.imgur.com/lynxfBI.jpg?1)
Note how the blue branch resembles the function \( \\[15pt]
{e^x} \).
{y_{x}=\,^xa} \) with his base "a" between \( \\[25pt]
{1< a \leq e^{\frac{1}{e}}} \), it tends to an asymptote value \( \\[15pt]
{y_{\infty}} \) such that \( \\[25pt]
{a^{y_{\infty}}={y_{\infty}}} \).
But that depends on the definition \( \\[15pt]
{^0a=1} \). If it were defined \( \\[15pt]
{^0a=y_{\infty}} \), we would get an horizontal line (drawn with red dashes in the graphic down).
The interesting part is when we define \( \\[15pt]
{^0a} \) a bit larger than \( \\[15pt]
{y_{\infty}} \), the tetration also converges to the same asymptote to the right, and another asymptote to the left, at the limit \( \\[15pt]
{y_{-\infty}} \), defined by \( \\[25pt]
{a^{y_{-\infty}}={y_{-\infty}}} \).
We get a Z-shaped tetration function (drawn in green), contained between \( \\[15pt]
{y_{-\infty}} \) and \( \\[15pt]
{y_{\infty}} \), which I call the Z curve.
The problem is that the Z curve is not uniquely defined. It depends on the value choosen for \( \\[15pt]
{^0a} \). Any value between \( \\[20pt]
{{y_{\infty}} < ^0a < {y_{-\infty}}} \) is valid, and the only difference this choice make, is the horizontal displacement on the curve. I drew the Z curve matching the origin with his inflection point (roughly).
The upper asymptote is also a non trivial function, and there is also another non trivial possible upper branch, obtained by choosing °a>-oo (drawn in blue line).
The question is, what would be convenient values for the definition of \( \\[15pt]
{^0a} \) for the green and blue branches?
If we take \( \\[15pt]
{a=\sqrt{2}} \), \( \\[15pt]
{^0a} \) may be multivalued at x=0: \( \\[15pt]
{^0(\sqrt{2})=(1,2,\; 2<y_3<4 \; , 4 , \; 4<y_5)} \).
The inflection point in the Z curve is near to 3. Maybe that base has all integer values for °a? (°1,41421356 = (1,2,3,4,5) )
Note how the blue branch resembles the function \( \\[15pt]
{e^x} \).
I have the result, but I do not yet know how to get it.