San José State University
Department of Economics

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Thayer Watkins
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 Self-Similar Mathematical Constructs

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 self-similar mathematical constructs is best explained by an illustration. Consider the infinite geometric series

#### S = 1 + x + x² + x³ + … which can be factored to give S = 1 + x[1 + x + x² + x³ + … ] which is equivalent to S = 1 + xS

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 Σj=0∞ ((j+2)!/j!)xj = 2/(1-x)³.

A repetition of the previous procedure yields

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

## 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 = aaa…which can be represented as G = aG

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

#### G1/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=2½=√2. Thus

#### √2√2√2… = 2

Furthermore since (½)² = 1/4

#### ¼¼¼… = 1/2

To verify these relations consider an iterative scheme of the form

#### Gn = aGn-1

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 G1/C does not have a single valued inverse.

Thus 41/4=21/2 so the infinite exponentiation of 41/4 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 G1/G is also equal to the cube root of 3.

The function G1/G reaches its maximum for G=e, the base of the natural logarithms 2.71828.. At that value of G, G1/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

#### ζζζ… = 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 G1/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)=G1/G suppose the function is represented as G11/G2, then

#### da/dG1 = (1/G2)G11/G1-1 and da/G2 = d(eln(G1)(1/G2)/dG2 = ln(G1)eln(G1)(1/G2)(−1/G2²) and therefore da/dG = G(1/G -2) − ln(G)G1/G/G²

Setting this equal to zero

#### G(1/G -2) − ln(G)G1/G/G² = 0 and dividing G1/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 G1/G is then e1/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.

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