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

S = 1/(1-x)
 

Now, going back to the original definition of the series and noting that S is a function of x,

S(x) = 1 + x + x² + x³ + …
which can be differentiated with respect to x to give
S'(x) = 1 + 2x + 3x² + 4x³ + …
 

This must be the same as the derivative of S(x)=1/(1-x); i.e., S'(x)=1/(1-x)². Thus

1 + 2x + 3x² + 4x³ + … = 1/(1-x)².
 

The same procedure can be performed upon the above equation to give

2 + 3*2x + 4*3x² + 5*4x³ + … = 2/(1-x)³
 

The left-hand side (LHS) of the above equation can be expressed as

(2!/0!)x⁰ + (3!/1!)x¹ + (4!/2!)x² + (5!/3!)x³ + … = 2/(1-x)³
Or, in a different notation
Σ_j=0^∞ ((j+2)!/j!)x^j = 2/(1-x)³.
 

A repetition of the previous procedure yields

Σ_j=0^∞ ((j+3)!/j!)x^j = 2*3/(1-x)⁴.
and more generally
Σ_j=0^∞ ((j+n)!/j!)x^j = (n-1)!/(1-x)ⁿ.
 

In terms of self-similarity the above expression may be derived from the equation S(x)=1+xS(x). One differentiation gives

S'(x) = S(x) + xS'(x)
so
S'(x) = S(x)/(1-x)
 

A second derivation gives S"(x) = S'(x) + S'(x) + xS"(x) = 2S'(x) + xS"(x)
so
S"(x) = 2S'(x)/(1-x)
 

The general relation is

dⁿS/dxⁿ = n(d^n-1S/dx^n-1)/(1-x)
 

Continued Fractions

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

F = 1 + x/F
 

This means F must satisfy the equation

F² = F + x
or, equivalently
F² − F -x = 0
which has the solutions
F = (1 ± (1 + 4x)^½)/2
 

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)=1-1.6180=-0.6180.

Now consider the value of the continued fraction as a function of x. Differentiation produces

F'(x) = 1/F(x) - (x/F²)F'(x)
and hence
F'(x) = 1/F(x)/[1+x/F²] = 1/[F(x) + x/F(x)]
and since x/F = F-1
F'(x) = 1/[2F(x)-1]
 

(To be continued.)

Infinite Exponentiation

An infinite exponentiation is something which is raised to a power which is something raised to a power ad infinitum> Suppose

G = a^aa…
which can be represented as
G = a^G
 

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 G-th root of both sides giving

G^1/G = a
 

Now if we want a value of a that gives G as a solution we need only take the G-th root of G and we have the answer. For example, for G=2, a=2^½=√2. Thus

√2^√2√2… = 2
 

Furthermore since (½)² = 1/4

¼^¼¼… = 1/2
 

To verify these relations consider an iterative scheme of the form

G_n = a^G_n-1
 

The results of the first 20 iterations are:

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.,

³√^3√…
 

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

ζ^ζζ… = e
 

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 Value of G^1/G

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 G₁^1/G₂, then

da/dG₁ = (1/G₂)G₁^1/G₁-1
and
da/G₂ = d(e^{ln(G₁)(1/G₂)}/dG₂ = ln(G₁)e^{ln(G₁)(1/G₂)}(−1/G₂²)
and therefore da/dG = G^{(1/G -2)} − ln(G)G^1/G/G²
 

Setting this equal to zero

G^{(1/G -2)} − ln(G)G^1/G/G² = 0
and dividing G^1/G yields
1/G² − ln(G)/G² = 0
multiplying by G²
further reduces it to
ln(G) = 1
which means
G = e
 

The maximum value of the function G^1/G is then e^1/e.

Infinitely Nested Radicals

The Indian mathematician S. Ramanujan was interested in structures such as

√ 1 + 2√1 + 3√ …

 

which he called infinitely nested radicals.

Consider the simple case of such structures

√ 1 + x√1 + x√ …

This infinitely nested radical contains a substructure which is identical to the overall structure; i.e.,

√1 + xH

= H
 

H must satisfy the equation

H² = 1 + xH
 

For x=1, H must be such that

H² −H − 1 = 0
 

which has the solutions

H = (1 ± (1 + 4)^½/2 = 1.6180 and −0.6180,
values related to the Golden Ratio.