Andrew Yanthar-Wasilik

26 December 2016 AD, Feast of St Stephen

I was able to derive general formulas for multiplication, division, powers and logarithms for the Transcendental Constants.

Addition and subtraction are more difficult to derive, it was done only partially.

Transcendental Constants have their own unique way of calculations, i.e. they use what I called

Index Mathematics.

It means, that indexes (subscripts) of the given constants are used to calculate new values

of multiplication, division, powers, and logarithms, possibly integrals and derivatives.

I will start from simple examples, so it is easier to understand, and then derive the general formulas.

1. For example, multiplication of two constants can be described as follows:

Cm× Cn= (C(m + n)⁄2)2 (Eqn. 1)

So, in concrete example lets say

Cm = C8= π = 3.141 592 654...

and

Cn= C7= e = 2.718 281 828...

then

C8× C7 = π  ×  e = 3.141 592 654...  ×  2.718 281 828... = 8.539 734 223...

now,

(C(m + n)⁄2)2 = (C(8 + 7⁄2))2 = (C(15⁄2))2 = (C7.5)2

using formula Eqn. 11 from “Book 1 - Transcendental Constants - Introduction”

we can calculate any value of constant with real index, as follows:

FT(x) = (C0)× (π/e)x (Eqn. 2)

FT(7.5) = (0.986 976 350...)  ×  (1.155 727 350...)7.5 = 2.922 282 364...

squaring that we get

(2.922 282 364...)2= 8.539 734 216...

relative error is

ε = -0.000 000 001

ie minimal error (if at all) - calculations are done on a hand-held calculator.

2. Adding powers to the formula for multiplication of two constants (Eqn. 1) gives:

(Cm)p×  (Cn)q = [C(p × m + q × np + q)](p + q) (Eqn. 3)

Let us use the previous example with some added powers:

(C8)1⁄4×  (C7)3 = (C(0.25 × 8 + 3 × 7⁄0.25 + 3))0.25 + 3

Left Hand Side equals to

= 26.740 585 61...

and Right Hand Side ( again, using Eqn. 2) is

= (C23⁄3.25)3.25 = (2.748 713 730)3.25 = 26.740 585 57...

relative error

ε = 0.000 000 001

3. The general formula for multiplication for any powers and any number of factors.

(Cm)p× (Cn)q× (Co)r× ... × (Cx)z= (Eqn.4)

= [C(p × m + q × n + r × o + ... + z × x)⁄(p + q + r + ... + z)](p + q + r + ... + z)= (Eqn. 4a)

= (C(p × m + q × n + r × o + ... + z × x))× (C0)(p + q + r + ... + z − 1)(Eqn.4b)

In (Eqn. 4a) p + q + r +...+ z  ≠ 0, so the (Eqn. 4b) is much more robust.

Example for the last formula for three factors and three powers

(C8)1⁄4×  (C7) − 3× (C0 = 0.986976350...)2= 0.064 568 027...

The second part of general formula gives

(C(0.25 × 8 − 3 × 7 + 2 × 0⁄0.25 − 3 + 2))(0.25 − 3 + 2)= (C25.3333) − 0.75=

using Eqn.2 to calculate C25.3333we get

= (38.604 978 32...) − 0.75= 0.064 568 027...

Third part of general formula gives

(C(0.25 × 8 − 3 × 7 + 2 × 0))  × (C0)(0.25 − 3 + 2 − 1)=

= (C − 19)× (C0) − 1.75=

= (0.063 103 627...)  × (1.023 206 273) = 0.064 568 027...

So, all three results are the same.

4.The general formula for the logarithm of the products and power.

There is not much to it. Taking logarithms of the Equation 4, we get:

ln[(Cm)p × (Cn)q × (Co)r × … × (Cx)z] = (Eqn. 5)

= p × ln(Cm)+ q × ln(Cn)+ r × ln(Co)+… + z × ln(Cx) = (Eqn. 5a)

= (p + q + r + … + z)× ln(C(p × m + q × n + r × o + ... + z × xp + q + r + ... + z)) = (Eqn. 5b)

= ln(C(p × m + q × n + r × o + ... + z × x))+ [(p + q + r + ... + z) − 1]× ln(C0)(Eqn.5c)

Again, in the (Eqn. 5b) p + q + r +...+ z  ≠ 0

5. Division of two constants.

CMCn= (C( − M + n)) − 1⁄(C0) − 1(Eqn. 6)

eg., C8C7= πe = 1.155 727 350...

Now, using (Eqn. 6)

C8C7= πe = (C( − 8 + 7)) − 1⁄(C0) − 1=

= (C − 1) − 1⁄(C0) − 1= (0.853 987 189...) − 1÷ (0.986 976 350...) − 1 =

= 1.155 727 350...

The values of constants are from the Blog section “Table of Transcendental Constants...”

Same results.

6. Division of two constants with powers.

(CM)P⁄(Cn)q = (Eq. 7)

= (C( − P × M + q × n⁄ − P + q))(P − q) = (Eqn. 7a)

= (C( − P × M + q × n)) − 1⁄(C0)( − 1 − (P − q))(Eqn. 7b)

eg., from (Eqn. 7): (C8)2.5⁄(C7) − 0.5= (π)2.5⁄(e) − 0.5= 28.841 770 89...

from (Eqn. 7a): (C( − 2.5 × 8 − 0.5 × 7⁄ − 2.5 − 0.5))(2.5 − ( − 0.5))=

= (C( − 20 − 3.5⁄ − 3))3 = (C7.8333)3

Using (Eqn. 2) to caclulate this result:

TF(7.8333) = (C0)× (πe)7.8333 = 3.066 718 931...

(C7.8333)3 = (3.066 718 931...)3 = 28.841 770 86...

From (Eqn. 7b):

(C( − 2.5 × 8 − 0.5 × 7)) − 1⁄(C0)( − 1 − (2.5 + 0.5))=

= (C − 23.5) − 1⁄(C0) − 4=

Using (Eqn. 2) to calculate this result:

TF(-23.5) = (0.986976350...) × (πe) − 23.5 = 0.032 900 694...

Now:

(C − 23.5) − 1⁄(C0) − 4= (0.032 900 694...) − 1÷ (0.986976350...) − 4=

(30.394 495 37...) ÷ (1.053 835 963...) = 28.841 770 86...

i.e. same results.

7. General formula for division with any number of factors and any powers.

((CM)P×  (CN)Q×  (CO)R×  ...  ×  (CX)Z) ÷((Cm)p× (Cn)q× (Co)r× ...  × (Cx)z) = (Eqn. 8)

= (C(( − P × M − Q × N − R × O − ... − Z × X) + (p × m + q × n + r × o + ... + z × x)⁄( − P − Q − R − ... − Z) + (p + q + r + ... + z)))[(P + Q + R + ... + Z) − (p + q + r + ... + z)]= (Eqn.8a)

= (C(( − P × M − Q × N − R × O − ... − Z × X) + (p × m + q × n + r × o + ... + z × x))) − 1⁄(C0)[ − 1 − ((P + Q + R + ... + Z) − (p + q + r + ... + z))](Eqn. 8b)

(Eqn. 8a) has a limitation:

( − P − Q − R − ... − Z)+ (p + q + r + ... + z) ≠ 0

8. Logarithms.

Taking logarithm on both sides we get a similar equation to the (Eqn. 5a, 5b and 5c).

It is too tedious to write it here.

In all these Index Math formulas, Constant C0 = 0.986 976 350... seems to be of utmost importance,

as if all the other constants can be calculated with this particular Constant C0 plus C8 = π and C7 = e.

Constant C0(representing quantum scale, the smallest limit) together with Constant C16 representing the cosmological scale

i.e. the largest limit, are connected with the Constant CHT, i.e. constant of Holy Trinity (see theology in “About” section),

as follows:

C0 = 0.986 976 359...

C16 = 9.999 838 797...

CHT = 0.0986 960 440...

Now,

C0 /10= CHT⁄(C16⁄10)(Eqn. 9a)

or

CHT = (C0 /10) × (C16⁄10)(Eqn. 9b)

substituting numerical values in (Eqn. 9b)

0.0986 960 440... = 0.0986 976 350...  ×  0.999 983 880...