*http://I have already pointed out that you do not understand the term exponential.*I wrote the book on exponentials.

I provided the first explicit formula in the history of mathematics for the exponential function.

**GLOBAL COSINE/ARCCOSINE/EXPONENTIAL/LOGARITHM/ARCTANGENT FUNCTIONS**The hypothenuse is labeled as c (which unites points A and C), side a is located on the x axis (which unites points A and B), and we also have side b. Angle θ is located between sides c and a (cos θ = a/c).

Point D will be the intersection of the circle with the positive x axis.

We first calculate the value of segment CD, in terms of a, b and c: (2c

^{2} - 2ac)

^{1/2}We then succesively bisect the chord CD, and each hypothenuse thus obtained (if we divide CD in half the midpoint will be E, and the intersection of the segment AE with the circle will be labeled as F; then we calculate this new hypothenuse CF in terms of the values obtained earlier, and so on, aiming to get as close to the value of s [arc length of CD] as possible], into smaller and smaller equal segments, calculating each succesive value in terms of a, b and c, in order to obtain a very close approximation of s (the arc length between points C and D); since s = rθ, where r = c = 1, by acquiring an exceptional figure for s, we correspondingly then get the value of θ.

Letting c = 1, we finally obtain:

**COS θ = 1/2 x (({ [( (2 - θ**^{2}/2^{N})^{2} - 2)^{2}...]^{2} - 2}^{2} - 2)) (n/2 + 1 evaluations)

COS^{-1} θ = 2^{n} x {2 - ((2 + {2 + [2 + (2 + 2θ)^{1/2}]^{1/2}}^{1/2}...))}^{1/2} (n + 1 evaluations)

The cosine formula is a GLOBAL formula; by contrast the Maclaurin cosine series is a local formula:

My global cosine formula is the SUM of the Maclaurin cosine expansion.

We know that the Maclaurin hyperbolic cosine expansion is:

cosh x = 1 + x

^{2}/2! + x

^{4}/4! + ...

Therefore, by just changing the sign in the global cosine formula, we obtain immediately the GLOBAL hyperbolic cosine formula:

**COSH V = 1/2 x (([(({[(2 + V**^{2}/2^{n})^{2} - 2]^{2}} - 2))^{2}...-2]^{2} - 2)) (n/2 + 1 evalutions)This is the global hyperbolic cosine formula which is the sum of the corresponding local Maclaurin power series expansion.

We then immediately obtain the GLOBAL natural logarithm formula:

**LN V = 2**^{n} x ((-2 + {2 + [2 + (2 + 1/V + V)^{1/2}]^{1/2}...}^{1/2}))^{1/2} (n+1 evaluations)By summing the nested continued square root function, we finally obtain:

**LN V = 2**^{n} x (V^{1/2n+1} - 1/V^{1/2n+1})This is the first explicit global formula for the natural logarithm, which can be used immediately to find LN V without resorting to logarithm tables, or calculators which feature the logarithm key: all we need is a calculator which has the four basic operations and the square root key. It links algebraic functions with elementary and higher transcendental functions.

For a first approximation,

LN V = 2

^{n} x (V

^{1/2n} - 1)

First results appear for n = 8 to 12, all the remaining digits for n = 19 and higher...

Example: x = 100,000 LN x = 11.5129255

with n = 20 the first approximation is LN x = 11.512445 (e

^{11.512445} = 100,001.958)

We also can get the corresponding arctangent formula:

**ARCTAN V = 2**^{n} x ((2- {2+ [2+ (2+ 2{1/(1+ V^{2})}^{1/2})^{1/2}]...^{1/2}}))^{1/2} (n+1 parentheses to be evaluated)ERROR ANALYSIS

Here is the Maclaurin expansion for e

^{x}:

Let us obtain a remainder form for the Maclaurin expansion for e

^{x} (Lagrange remainder):

R

_{n}(x) = f

^{(n+1)}(c)[x

^{n+1}]/(n+1)! , where c is between 0 and x

f

^{(n+1)}(c) = e

^{c}An approximation is said to be accurate to n decimal places if the magnitude of the error is less than 0.5 x 10

^{-n}.

e

^{1} to four decimal place accuracy:

R

_{n} = e

^{c}/(n+1)!

since c<1, then e

^{c} < e

since e<3, then R

_{n} < 3/(n+1)!

then, for n=8 we will obtain four decimal place accuracy.

Local formulas are difficult to use because of their very slow convergence.

By contrast, my formula is a GLOBAL formula, which rapidly converges to the result, even for large x.

**COSH v = (e**^{v} + e^{-v})/2 =~ 1/2e^{v} = 1/2 x (({[( ( (2 + v^{2}/2^{n})^{2}) -2)^{2}] -2)^{2} ...-2}^{2} -2)) (n/2 +1 evaluations)

We can turn this formula into an exact formula for e

^{v} by simply substituting y for e

^{v}, and then solve the quadratic equation for y.

One might ask, could you not use Taylor expansions to obtain cosh 10 (as an example)? No, because you would need some other value to start with, cosh 9.5 or cosh 9.8 or cosh 10.3, to apply Taylor series.

With my global formula, no such approximations are needed, we can start directly with the value v = 10.

My formula also has a built-in remainder approximation estimation: the term v

^{2}/2

^{n}.

That is, we can estimate the accuracy from the very start: this is the power of a global formula.

The higher the value of n, the better the approximation that we will obtain.

Example:

COSH 10 = 11013.233

10

^{2}/2

^{20} = 0.00009536

Using the global hyperbolic cosine formula with n = 20, we get: 11012.762