The exponential function as geometric mean derivative of Fibonacci algebra.

Investigations into basic operations of geometric mean differentiation.

Geometric mean differentiation from first principles.

The idea of geometric mean differentiation is to extend conventional differentiation onto a more general statistical basis. Traditional differentiation is given a statistical interpretation as either an implicit arithmetic mean form of differentiation or harmonic mean form of differentiation.

My innovation to differentiation involves (usually an implicit) geometric mean form. There is also a difference from conventional differentiation, in the limit-taking process. Instead of taking a change in a functions independent variable towards a zero limit, geometric mean differentiation takes the change in the independent variable towards infinity, which being limitless is the opposite of a limit.

The geometric mean or average is calculated from two ends of a statistical range, by multiplying them and taking the square root. A GM of three items in a range is taken by multiplying them and finding their cube root. Four range items would be multiplied and their quartic root taken for the GM. And so on.

Logicly, there is the most limited case of taking the geometric mean, from only one item in a range, whose unitary root is taken, leaving the value of the item unchanged. This assumes that the item, or items, taken from a range are representative of the whole range, so the calculation of the average is accurate.

From this consideration of a variable as one item from a statistical range, I arrived at the assumption that conventional differentiation is implicitly an arithmetic mean differentiation or harmonic mean differentiation (as explained in my web page: Statistical basis of differentiation).

This page will try to give some more plausible reasons for developing geometric mean differentiation, by outlining what may be some of its basic operations.

The reader may bear in mind that this investigation is like the exploration of a maze. The best routes are not known yet. There may be false trails and dead-ends. The enterprise might not ultimately succeed, either from the limitations of the investigators basic idea or from the limitations of the investigator.

Take a simple function familiar in conventional differentiation from first principles and follow the alternative route of geometric mean differentiation.

Example:

1.
y = bx,

where y is the dependent variable, x is the independent variable, and b is a constant. The hash symbol, #, replaces the usual delta symbol, meaning "a change in" such as #y, #x. I call such a change in a variable, its "variation," #y or #x, etc.

As in conventional differentiation, equation 1 is manipulated into a ratio of changes: #y/#x. But GM differentiation soon departs from the traditional route, to put the equation in the form of the binomial theorem which can be infinitely expanded into an exponential function, which, like a geometric series, plots a non-linear graph or a curve that goes thru an accelerating change. The geometric mean is the average of such a curving range of values.

Consider a change in variable, y, as result of a change in variable, x, symbolicly as follows:

y + #y = b(x + #x)

= b.#x(1 + x/#x).

Therefore, #y = b.#x(1 + x/#x) - bx.

2.
#y/#x = b{(1 + x/#x) - x/#x}.

Most simply, GM differentiation seeks a relation like:

3.
#y/#x = (1 + x/#x).

Both sides of equation 3 can be raised to the power of #x, which is let approach infinity. This derives the required exponential function.
But is it justifiable to simplify the equation of #y/#x in this way?

Let:

4.
#y/#x = b{(1 + x/#x) - x/#x} = (1 + x/#x).

Then:

(1 + x/#x)(b - 1) = bx/#x.

Therefore:

(1 + x/#x) = bx/(b-1)#x.

That is if:

5.
b = (1 + x/#x).

Therefore:

6.
#y/#x = {(1 + x/#x) - (1 + x/#x)x/#x}.

Raising both sides to the power of #x:

7.
(#y/#x)^#x = {(1 + 2x/2#x) - (1 + x/#x)x/#x}^#x.

Let the powers, #x, approach infinity:

The final x/#x term on the right side of equation 7, is reduced to zero, which eliminates the second right side factor, which would otherwise be: e^x.

To signify the completion of this operation, the hash, #, signs are changed to the letter, d (as in conventional differentiation completion from first principles):

8.
(dy/dx)^dx = e^2x.

(The exponent, given by the letter, e, is an infinite number starting as 2.718...)

Equation 8 is the standard geometric mean derivative of equation 1.

The reason why the index is 2x is a result of a need for consistency. Numerator and denominator in the binomial factor are both multiplied by 2, because the denominator in the factor has to be the same as the index to the factor for purposes of turning this binomial factor, that expands into a binomial series, to expand into an exponential series instead, and whose series sum is the exponential function. (This is a standard textbook manipulation, shown on above cited web page).

In equation 1, the constant, b, has been given a definite value, so the original function can also be stated as:

9.
y = bx = (1 + x/#x)x.

The original equation 9 can be restated:

10.
x/#x + x - y = 0.

By the formula for solving quadratic equations:

x = {-1 (1 + 4y/#x)^1/2}/(2/#x).

Given y = #x, equation 9 could just as well have been solved by substituting y for #x.

If y = #x,

x/y = x/#x = {-1 5^1/2}/2.

Thus:
11.
x/#x = -.5 + 1.118 = .618.

Or:
x/#x = -.5 - 1.118 = -1.618.

And:
12.
#y/#x = (1 + x/#x) = 1 + .618 = 1.618 or #y/#x = 1 + -1.618 = -.618.

The change ratio, of #y/#x, thus equals the ratios, given here to three decimal places, that successive ratios, in the Fibonacci series, tend towards. So this is what the geometric mean derivative is of.
The exponential function is a geometric mean derivative of Fibonacci series algebra.
This is the main result of this investigation.

Given that y = #x and solving equation 9 for y, with the quadratic equations formula again. This comes out at:

y/x = (1 5^1/2)/2.

If the alternative solutions are considered as the limits of a range, then their average can be found.

If this range represented a constant change from one value to the next, like a velocity, then the arithmetic mean would be the appropriate way to find the most representative item in the range. (In this example the AM would be one half, the value equally between the alternatives, plus or minus square root of five divided by two.)

But since the Fibonacci series involves a non-constant change from one term in the series to the next, then the geometric mean applies. It multiplies the two alternative values and then takes the square root:

{(1 + 5^1/2)/2}{ (1 - 5^1/2)/2}^1/2 = (1/4 - 5/4)^1/2 = i.

Therefore, the geometric mean solution of y = bx is: (GM)y = ix.

Modifications of case 1.

To top

It may well be asked how geometric mean differentiation arrives at the simple exponential function, e^x.

This only requires a minor alteration to the coefficient b = (1 + x/#x), making variable, x, in the coefficient part of the equation, into a constant, in this case, zero.

This transforms equation 8 from:

(dy/dx)^dx = e^2x = e^x+x

to:
13.
(dy/dx)^dx = e^x+0 = e^x.

In fact, any other number than zero, can be similarly introduced. Indeed, the second x in the index could be an imaginary number, say: iy, for a GM derivation with a complex number in the index, e.g.: (dy/dx)^dx = e^x+iy.

Any desired coefficients to the variables can be bundled into over-all terms, simply making, say: fx + igy = X + iY.

My earlier attempts at geometric mean differentiation (including on previous web pages) simply treated coefficients as equal to one, to avoid an indeterminate result from raising the coefficient to a power taken to infinity.

This, however, confined my results to a special case, and prevented me from seeing the Fibonacci ratio meaning of the geometric mean anti-derivative to the exponential function.

Another modification is equation 14:

(dy/dx)^dx = e^-2x.

Its GM anti-derivative is a slight alteration of equation 1 to:
15.
y = 1/bx.

A GM derivative with a coefficient can be derived from equation 8:

(dy/dx)^dx = e^2x = e^x.e^x

modified to:

e^h.e^x,

where the first of the two exponents is derived from the constant coefficient, b, and can itself be given an arbitrary value, say: h. Examples above were h = 0 or h = iy.

Suppose a desired GM derivative is: 2e^x. Then this may be derived from: 2 = e^h. It is known that the converse of an exponential function is a natural logarithm (ln). So, h = ln2.

At high school and college we were taught and re-taught logarithm tables. (Yes, I'd already forgot by the time of my "further" education.) But my quarter of a century old scientific calculator tells me that: ln2 = .693 (to three decimal places).

An imaginary coefficient, i, to a GM derivative can be obtained by making h = iπ/2, because of the Euler formula: i = e^iπ/2.

Then:
(dy/dx)^dx = e^h.e^x

= e^iπ/2.e^x = ie^x.

Calculus texts provide trigonometric formulas such as:

16.
2.cos Q = e^iQ + e^-iQ.

The GM anti-derivative of the first of the two terms on the right is shown above to be:

y = bx. The second term on the right is the inverse of the first, so its GM anti-derivative should also be the inverse of the first, namely: 1/y = 1/bx.

Closely related is the formula:

17.
2i.sin Q = e^iQ - e^-iQ.

Adding equations 24 and 25:

18.
2.cos Q + 2i.sin Q = e^iQ + e^-iQ + e^iQ - e^-iQ

= 2e^iQ.

A complex number, in rectangular co-ordinates, z = x +iy, is given, in trigonometric form, by: r(cos Q + i.sin Q) = r.e^iQ, where r is for a circle radius. The GM anti-derivative of a complex number (assuming r = 1) is therefore given by case 1, to be: y = biQ.

The conjugate, of z, here labeled: Z = x -iy.

Therefore, the equation of a circle follows:

zZ = (x+iy)(x-iy) = x + y = r = r.e^iQ x r.e^-iQ = r.

The imaginary variables, +iy and -iy are the limits of a range about x. If the range exhibits a constant change in whatever values are in the range, then its average value is given by the arithmetic mean. If the range only has two values, which therefore must comprise just two ends to a discrete range, then this would seem to imply that the mid-value x, namely the arithmetic mean must be the most representative average.

However that may be, in this example, the range is not in uniform line, because real number, x, and imaginary number, iy, are in two dimensions at right angles. An object moving round a two-dimensional circle at a constant velocity, still is velocity in a changing direction, which means there is an aspect of acceleration or non-uniform motion to such motion.

Thus, the geometric mean is the representative average for this range. It is the square root of the multiple of the complex conjugates. In other words, the radius of a circle is the geometric mean of the range of its conjugate complex variables.

Case 2: y = b/x.

To top

Another example is equation 19:
y = b/x.

The process of geometric differentiation starts like conventional differentiation from first principles until arriving at the ratio of changes in the dependent and independent variables: #y/#x.

20.
y + #y = b/(x +#x).

Therefore:

#y = b/(x +#x) - b/x

= b(x - x - #x)/x(x +#x)= -#x.b/x(x +#x).

Therefore:

21.
#y/#x = -b/x(x +#x).

At this point, conventional differentiation would be completed by taking the change in the independent variable, #x, (which I call the "variation") towards a limit of zero.

That is to say: dy/dx = -b/x, where the hash, #, sign (usually delta sign in the textbooks) is changed to the letter, d, to signify the differentiation is completed.

But geometric mean differentiation takes the variation, #x, towards infinity.
As with conventional differentiation, the trick is to eliminate the variation from the right-hand side of the equation. In the case of GM differentiation, this can be done by putting the right side into binomial theorem form, where its power is made the variation, which is then taken towards infinity.

This process can be shown in an expansion of the binomial theorem to the binomial series. As the power is taken to infinity, the effect on the binomial series is to change it into an exponential series.
Hence, the result is an exponential function, which is the geometric mean derivative.

Transforming equation 21 to 22:

#y/#x = -b/#x.x(1 + x/#x).

So that the coefficient, -b/#x.x, won't become indeterminate, when both sides of equation 22 are raised to the power of #x, approaching infinity, the simplest option is to set the coefficient at unity. Then: b = -#x.x.

This is what my first explorations did, now superseded. Case 1, in the above section, introduced a more general method. The working below shows that this will entail the assumption that b = -#x. {Please note that for ease of writing (#x) = #x.} This is perhaps the more flexible condition, because b is only tied to the variation of x and not both the variable x and its variation, #x.

Letting:
23.
-b/#x.x = (1 + x/#x).

This is a slightly different value for term ,b, than in case 1. But the resulting quadratic equation solves for variable, x, such that the ratio of variations, #y/#x still equals a Fibonacci ratio, as in case 1.

The working is as follows:

-b/#x.x = ([#x + x]/#x).

Therefore:
-b = #x.x + x

Or:

x + #x.x + b = 0.

Using the quadratic equations formula to solve for x:

24.
x = {-#x (#x - 4b)}/2.

On the assumption that b = -#x:

x = #x(-1 5^1/2)/2.

Therefore:
x/#x = (-1 5^1/2)/2.

This is the Fibonacci ratios, .618 and -1.618.

1 + x/#x = 1 + (-1 5^1/2)/2

= (1 5^1/2)/2.

This is also the Fibonacci ratios, -.618 and 1.618.

Therefore:
25.
#y/#x = 1/(1 + x/#x) = 2/(1 5^1/2).

This is yet again the Fibonacci ratios, -1.618 and .618. (All to three decimal places.)

Therefore, for case 2, the coefficient identity, equation 23 satisfies the same condition as case 1, that the ratio of variations, #y/#x, is the Fibonacci series ratio.

The simplest coefficient condition: b = -#x.x does not satisfy this relation, without contriving to make b, in terms of x, in such a way that the equation becomes a Fibonacci quadratic equation.

Having gone to all that trouble to transform equation 22 to make it suitable for GM differentiation, this derives the rather limp equation 26:

#y/#x = (1 + x/#x)/(1 + x/#x) = 1.

This, in effect, reduces #y/#x to 1, which produces the obvious result of a GM derivative also equal to 1.
(Note, the Euler formula: e^i2πn = 1, where, n is any of the whole numbers: 0, 1, 2, 3,..)

However, following case 1, the variable x, in the numerator can be set at a constant, h, and modifications of this result produced for the GM derivative, along the lines in the preceding section.
For instance, if x becomes h = 0, in the numerator of equation 26, then its GM derivative becomes:

27.
(dy/dx)^dx = 1/e^x = e^-x.

(Case 2 could follow case 1, in letting b = 1 + x/#x, which then must equal -#x.x. This obliges a Fibonacci solution of the quadratic equation which makes x = i. As x is the independent variable, and a variable is supposed to be variable, this is a highly restrictive condition.)

Successive orders of geometric mean derivative.

To top

From case 1:

(dy/dx)^dx = e^2x.

Let:
28.
e^2x = e^i2π = 1.

Here comes the trick. As in case 1, above, it turned out that instead of reducing a coefficient to unity to get rid of an indeterminate result of raising to in infinite power, for GM differentiation, it made more sense to use the factor: (1 + x/#x).

So, we do the same trick here, for example, using the simplest exponential function, e^x.

Let:

e^x = e^i2πn = 1 + x/#x,

where the right side's second term, x/dx, can be considered to approach zero as #x approaches infinity. Consequently, #x may be redesignated dx.

This is, in effect, a complex number where, the imaginary part is zero.

The Euler formula is really a series signifying quarter turns of a circle, by a complex number, where either its real or imaginary part is zero:

e^i0π/2 = 1+0; e^i1π/2 = 0+i; e^i2π/2 = -1+0; e^i3π/2 = 0-i; e^i4π/2 = 1+0, and so on, repeatedly round the four compass points of a circle. The numbers, 0,1,2,3, etc in the index can be let equal to n.

Thus, GM differentiation or derivation of a GM derivative, that is a second-order GM differentiation would go like this:

29.
(dy/dx)^dx = e^x = e^i2π = (1 +x/#x).

Raising again to the power of #x:

30.
{(dy/dx)^dx}^#x = e^i2π.#x = (1 +x/#x)^#x.

Let #x approach infinity, for the completed second order derivative:

31.
{(dy/dx)^dx}^dx = e^x.

Thus, repeated GM differentiation of the simple exponential function leaves it unchanged.
(This also is the case with conventional differentiation of this function.)

An exponential function with an imaginary number in its index is called a circular function. The term, e^i2π.#x, where #x approaches infinity, may be considered as multiplying thru an infinite number of times. Indeed as a term, n = 0,1,2,3, etc, is already capable of this function, then #x might be identified with n, so that:

32.
e^i2πn = e^i2π.#x.

The Minkowski Interval in geometric mean velocity form.

To top

The uniform velocity in the standard equation, of the Minkowski Interval, may be changed to a geometric mean velocity.

Special relativity only coordinates the event measurements of observers in relative uniform motion in a straight line. But the innovation of a geometric mean velocity implies non-uniform motion.

33.
(I/t) = c - u.

Equation 32 of the Minkowski Interval, I, is the common space-time measurement shared by all local observers of an event, with their differing local times, t, and differing local observations of velocity, u.

Usually, the textbooks show the Minkowski Interval for two observers, with the second observer given time, t', and velocity, u', to show that they are different local time and velocity to the first observer.

And usually, the velocity, u, is shown in three dimensions, say: u, v, and w.

Equation 33 can be taken to mean either the velocity in just one dimension, supposing all observers were limited to that, or more realistically, the velocity is a three-dimensional vector, in other words, the hypotenuse equivalent to the three dimensions as given by Pythagoras theorem.

The Minkowski Interval is essentially Pythagoras theorem in four dimensions with the fourth dimension, involving the speed of light, c, having a different sign to the three spatial dimensions.

To both sides of equation 33, add: 2u - 2uc.

34.
2u - 2uc + (I/t) = c - u + 2u - 2uc

= (c - u) = (u - c).

Therefore:

{(I/t)}/(u-c) + 2u(u-c)/(u-c) = 1.

Therefore:

35.
{(I/t)}/(u-c) + 2u/(u-c) - 1 = 0.

This equation can be solved for 1/(u-c) according to the quadratic equations formula:

1/(u-c) = [-2u {4u - -4(I/t)}^1/2]/2(I/t)

= -u(t/I) t/I{(ut/I) +1}^1/2.

Therefore:

36.
{1/(u-c)}I/t = -ut/I {(ut/I) +1}^1/2.

Considering the plus or minus alternatives as the limits of a range, its average value can be found by taking their geometric mean velocity, U:

{1/(U-c)}I/t = <[-ut/I + {(ut/I) +1}^1/2][-ut/I - {(ut/I) +1}^1/2>^1/2

= [(ut/I) - {(ut/I) +1}]^1/2

= (-1)^1/2 = i.

Therefore:

37.
(U-c) = I/it.

Therefore, the geometric mean velocity form of the Minkowski Interval is:

38.
I = it(U-c).

Or, as the Minkowski Interval is usually squared:

39.
I = -{t(U-c)}.

Or, in comparison to equation 19, y = b/x:

40.
(it) = I/(U-c).

Substituting the equation 27 solution to equation 19, then the geometric mean derivative of equation 40 is:

41.
{[d(it)/d(U-c)]}^d(U-c)

= e^-(U-c).

The right side of equation 41 is the form of a normal distribution.
This is an equation of the squares of imaginary time as a function of motion or velocity. That is an equation of the inverse of acceleration squared.

Since the velocity, U, is a geometric mean, this might help to take the Interval out of the restricted framework of special relativity dealing with only relative uniform velocity.

Whatever its consequences, this characterisation is an improvement on my previous attempts to use geometric mean differentiation of the Interval into the normal distribution.

Closing remarks.

It's some years, and so many scribbled notes, since I first had the idea of geometric mean differentiation. The notion originated in forms of geometric mean I identified in special relativity formulas.
I thought this non-linear differentiation might be a direct route from special into general relativity. But as I've said before somewhere, I simply don't know enough about it.

I once did a check, from Wikipedia and elsewhere, on most of the mathematical terms indexed in The Road To Reality by Roger Penrose. I found they mostly seemed to relate to general relativity.

(Apart from the fact that I don't know about or understand any of them,) this made sense, as I'd read in a popular work that Penrose was responsible for making the mathematics of general relativity accessible to physicists.

All right, I'm not a mathematician or a physicist (Ive admitted that before, too). But this characterisation of geometric mean differentiation (written less than a year before pension age) yet may prove to be useful in many areas of research.


And summary.

To top

A few years ago, I suggested an alternate form of differentiation. Traditional differentiation is implicitly statistical, involving the averaging of an arithmetic mean or a harmonic mean from just one term in a potential range.

You could take the arithmetic mean or harmonic mean, as the case might be, from more than a one-term range and still get the standard derivatives.
Hence my suggestion that the same could be done with the geometric mean. Tho, in this case, the variation, to the independent variable, wouldn't be taken to a limit of zero but would be an unlimiting process, approaching infinity.

As with conventional differentiation, the knack of it is to get rid of the variation being taken to a limit, or an unlimit in the case of a GM differentiation.
This is achieved by a GM differentiation from first principles, that derives a binomial factor.

If the original equation is y = bx, then the form, b(1 + x/#x)^#x is derived where the hash, #, sign means the usual delta sign in first principles differentiation, and #x is taken to an unlimit of infinity.

A generated binomial series is changed to generating an exponential series which sums as an exponential function.
This derives an exponential function as GM derivative.

Like a one-trick pony, Ive only recently discovered the trick of repeating this trick to make further progress in understanding basic operations of a GM differentiation from first principles.

Formerly, there was the problem of what to do with coefficient, b, to prevent it becoming indeterminate, when it, too, would be raised to a power approaching infinity. Making b = 1 turns out to be only a special case of also making b = (1 + x/#x).

Thus repeating this form as a coefficient to the same binomial factor will result in an exponential function, e^2x as the GM derivative.
(You can make x, in the coefficient, some constant, thus creating a range of possible GM derivatives.)

Repeating the binomial factor in the coefficient turns the original equation, or GM anti-derivative into a Fibonacci quadratic equation. And it turns the ratio of variations, #y/#x, to be GM differentiated, into the Fibonacci ratio, that its infinite series approaches.

Further progress can be made by this one-trick pony repeating the trick a third time. Suppose an exponential function as GM derivative that is also equivalent to an Euler formula, like e^2iπn = 1.
This form is actually an implicit complex variable, 1 + i0. Similarly for the binomial form again (1 + x/#x).

And the number of circle revolutions, n, in the Euler formula, being potentially infinite, is equivalent to raising the formula to the power of the variation. #x, in a GM differentiation: n = #x.

This serves as the means of regenerating the exponential function as a second or successive order GM derivative. (In this particular case, conventional differentiation does the same thing a different way.)

That is the gist of making GM differentiation a viable set of operations from first principles.

Richard Lung
17; 19 june 2013.


To top

To home page