An Infinitely Differentiable Extension for hyper4

(The presentation that follows is a light introduction. A more austere article is Manuscript 2).

While looking for an analytic continuation of the hyper4 operator, one is tempted to search for series expansions for nx. Having such an expansion at hand may offer additional insight into what an analytic yx can look like, for y in R.

Instead of searching for an expansion of nx, it is often convenient to look for an expansion for n(ex) first. We summarize the results found in A Series Expansion for m(ex)":

For m = 1:

m(ex) = ex, of course.

For m > 1:

m(ex) = nexpx.GIF

with:

amn.GIF.

Having those coefficients at hand, we can now construct a Coo extension of the hyper4 operator as follows:

We first extend the coefficients above to conform with the hyper4 definition, i.e. 0x=1, so we define:
am,n={1, if n=m=0,
            0, if m=0 and n=/=0,
            1/n!, if m=1,
             Sum(j*am,n-j*am-1,j-1,j=1..n)/n, otherwise}     (6.1)

We will use the well-known Coo function,

f(x)={exp(-1/x) if x>0,
          0, if x<=0}

Elementary calculus shows that f(1-x2) is symmetric about the origin and there it attains its maximum, f(1)=1/e. We first change the support of f to be the interval [-1/2,1/2], so we consider the function

phi(x)=f(1/4-x2)={exp(4/(4x2-1)), if |x|<1/2,
                            0, otherwise}

phi(x) is shown below.

phi.GIF

Set Am=Int(phi(t-m-1/2),t=m-1..m). Since phi(x-m-1/2) is simply a right translation of phi, it follows that for all m,n in N, Am=An. We first normalize phi with respect to its integral, so we set:

psim(x)=(am,n-am-1,n)phi(x-m-1/2)/Am and finally,

Definition:
For all m in N, n in N U {0}, x>=0 and with initial values for am,n as in recursion (6.1):
alphan(x)={1, if n=0,
               alpha_n.GIF, if n<>0}

If x=m>0, then since all the psim are non-zero on disjoint subsets, alphan(m)=am,n, while if x>=n, then alphan(x)=an,n. The next figure is the graph of alpha3(x).

alpha3.GIF

Lemma 1:
If x>=0, alphan(x) is infinitely differentiable with respect to x.
Proof:
If x>= n>0, alphan(x)=an,n = constant, so derivatives of all orders exist and are 0. If x<n, then the existence of the p-th derivative of alpha depends on the existence of the p-1-th derivative of psim(x), which is simply a right translation of phi, for which derivatives of all orders exist and the Lemma follows.

We are ready for the extension. With alphan(x) as above and y>=0,
y(ez)= ext2.GIF (6.2)

If we fix z and call F(y)=y(ez), then for y in N, the above function satisfies the functional equation, F(y+1)=(ez)F(y). We have to prove convergence.

Lemma 2:
If y>=0, then Sk(z)=Sum(alphan(y)*zn, n=0..k) converges uniformly on compact subsets of C.
Proof:
If y=0 then Sk(z)=1-> 1. Fix y>0 and z in U subset C, U compact. Then y in [m-1,m) for some m in N, and there |alphan(y)|<=am,n, for all n in N, therefore for all z in U and each y>0, |alphan(y)zn|<=am,n|z|n=Mn and Sum(Mn,n=0..+oo)=m(e|z|) by the analysis of coefficients above, so by the Weierstrass M-test, the series Sk(z) converges (absolutely and) uniformly on compact subsets and the Lemma follows.

If y=0, then y(ez)=1 as required by the definition of hyper4, while if y=m in N, then  y(ez) coincides with the corresponding expansion for the tower  m(ez) in the analysis above. Therefore  y(ez) interpolates all finite towers of iterates of ez. The important question now is if this interpolation is not only continuous, but Coo with respect to y. We are ready for the second result.

Lemma 3:
If y>=0, then y(ez) is infinitely differentiable with respect to y.
Proof:
Since Sk(z) converges uniformly on compact subsets, we can differentiate term by term. But dp/dyp{alphan(y)} exists in the domain of alphan, for all p>=1 by Lemma 1, therefore dp/dyp{y(ez)} also exists for y>= 0 for all p>= 1 and the Lemma follows.

We can now define a corresponding Coo function that interpolates between all the finite power iterates of z, as
yz=ext3.GIF


Afterthoughts:

It is interesting to note that if we consider limy->+ooy(ez), then the limit of the summation becomes a summation of limits, and limy->+oo alphan(y)=an,n, therefore,
limy->+ooy(ez)=infex.GIF,
which is the series expansion of W(-z)/(-z), where W is Lambert's W function. The last series converges for |z|<=1/e, as shown in the article for "A Series Expansion for m(ez)".

Because the radius of convergence of the above series is 1/e, the behavior of any numerical computations for the extension  y(ez) that depend on the coefficients (6.1) for convergence, will most likely be bad past this radius of convergence. Indeed, we can set up some Maple code to calculate  y(ez) for |z|<=1/e, but past these bounds (or disk in the complex case) any code will likely produce garbage. And this is to be expected. If convergence of the series expansions of the iterates  y(ez) was everywhere nice, the limit function would be entire, but it's not, although all finite iterates are entire.

In general, calculating  y(ez) (resp.  yz) even when |z|<1/e (resp. |Log(z)|<=1/e), is quite difficult, because it depends on explicitly defining alphan(y). The alphan functions are all defined piecewise and as n grows larger, they become harder and harder to define explicitly. A numerical approximation for y(e-0.34) using just 6 terms in the extension (n=5), coupled with the extension of the article "A Continuous Extension for the hyper4 Operator", is shown in the next figure, with y in [0,5], where it is known that limy->+ooy(e-0.34)=W(0.34)/0.34 ~ 0.76973.

twoexts.GIF

The above extension is obsolete as it suffers from the fact that the functional equation of tetration is only satisfied at the naturals and 0 (As noted by Andrew Robbins). For what appears to be a C solution which doesn't suffer thus, check Andrew Robbins' solution. Another (possibly Cn) solution is given by Robert Munafo.


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