San José State University Department of Economics 

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This is an investigation of mathematical structures which contain within themselves a substructure which is identical to the structure itself. These are particularly easy structures to analyze. The term selfsimilar mathematical constructs is best explained by an illustration. Consider the infinite geometric series
Thus the infinite geometric series contains a subunit which is identical to itself.
Under the presumption that the series does converge to a number the last equation can be solved for S to give
Now, going back to the original definition of the series and noting that S is a function of x,
This must be the same as the derivative of S(x)=1/(1x); i.e., S'(x)=1/(1x)². Thus
The same procedure can be performed upon the above equation to give
The lefthand side (LHS) of the above equation can be expressed as
A repetition of the previous procedure yields
In terms of selfsimilarity the above expression may be derived from the equation S(x)=1+xS(x). One differentiation gives
A second derivation gives
S"(x) = S'(x) + S'(x) + xS"(x) = 2S'(x) + xS"(x)
so
S"(x) = 2S'(x)/(1x)
The general relation is
Consider the continued fraction
F = 1 + _x___________________________________ 1 + _x______________________________ 1 + _x_________________________ 1 + _x____________________ 1 + _x_______________ 1 + _x__________ 1 + _x____ 1 + …
Since this fraction continues on forever the expression under the first x is the same as F so the continued fraction can be written as
This means F must satisfy the equation
For x=1 these solutions reduce to F=1.6180 and 0.6180. There is no problem accepting +1.6180 as a value for F but 0.6180 is a puzzle even though it satisfies the equation F=1+1/F. These values however are values involved in the Golden Ratio. Their being solutions to F=1+1/F is confirmed as 1+1/1.618=1+0.6180=1.6180 and 1+1/(0.6180)=11.6180=0.6180.
Now consider the value of the continued fraction as a function of x. Differentiation produces
(To be continued.)
An infinite exponentiation is something which is raised to a power which is something raised to a power ad infinitum> Suppose
This equation might seem a puzzlement as to whether it has any solution other than the obvious one of a=1 and G=1. However a little manipulation turns it into a seemiongly trivial problem. The manipulation is to take the Gth root of both sides giving
Now if we want a value of a that gives G as a solution we need only take the Gth root of G and we have the answer. For example, for G=2, a=2^{½}=√2. Thus
Furthermore since (½)² = 1/4
To verify these relations consider an iterative scheme of the form
The results of the first 20 iterations are:
a=1/4  a=√2 
G  G 
1  1 
0.25  1.4142135623731 
0.707106781186548  1.63252691943815 
0.375214227246482  1.76083955588003 
0.59442699715242  1.84091086929101 
0.438651165125217  1.89271269682851 
0.544384414863917  1.9269997018471 
0.470162431706881  1.95003477380582 
0.52111552337359  1.96566488651732 
0.485575976841481  1.9763417544097 
0.510098600013408  1.98366839930382 
0.49304895344828  1.98871177341395 
0.504841387135777  1.99219088294706 
0.49665544235752  1.9945944507121 
0.502323653394958  1.99625666626586 
0.498391957558689  1.99740700114134 
0.501115853363732  1.9982034775087 
0.499227147304516  1.99875513308459 
0.500535987744584  1.99913731011939 
0.499628619597765  1.999402118325 
0.500257487555813  1.99958562293568 
0.499821555076828  1.99971279632964 
0.500123703895513  1.99980093549297 
0.499914262345387  1.99986202375778 
0.500059432345487  1.99990436444334 
0.499958806334303  1.9999337115821 
0.50002855408854  1.99995405289782 
0.499980208205761  1.99996815214924 
0.500013718814578  1.99997792487387 
0.499990490932778  1.9999846987471 
However everything is not as simple as the preceding. For one thing G^{1/C} does not have a single valued inverse.
Thus 4^{1/4}=2^{1/2} so the infinite exponentiation of 4^{1/4} does not converge to 4, instead it converges to 2.
Also the infinite exponentiation of the cube root of 3 ; i.e.,
should give 3 as a result but the iteration scheme does not converge to 3, instead it converges to a value of 2.478.... which is a lower value of G such that G^{1/G} is also equal to the cube root of 3.
The function G^{1/G} reaches its maximum for G=e, the base of the natural logarithms 2.71828.. At that value of G, G^{1/G} is equal to 1.44466786100977…, call it ζ. This is the maximum value of a for which infinite exponentiation has a finite value and
For details see Maximum.
Thus the maximum value an infinite exponentiation can converge to is e and the maximum base for the infinite exponentiation is ζ=1.44466786100977 so this is why the procedure worked for G=2 and G=1/2 but not for G=3.
(To be continued.)
The maximum of the function is found by finding the value of G such that the derivative is equal to zero. The derivative is found for a function in which its argument variable appears in more than one place is to get the derivative for the variable in each place it appears treating it as a constant in the other places.
For a(G)=G^{1/G} suppose the function is represented as G_{1}^{1/G2}, then
Setting this equal to zero
The maximum value of the function G^{1/G} is then e^{1/e}.
The Indian mathematician S. Ramanujan was interested in structures such as
which he called infinitely nested radicals.
Consider the simple case of such structures
This infinitely nested radical contains a substructure which is identical to the overall structure; i.e.,
H must satisfy the equation
For x=1, H must be such that
which has the solutions
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