SOLITU Plus:  JMM Math Notes (from years past), posted 2015

Series Calculus [mathematical induction] evolution

for N squared, cubed, et cetera:

Exponential Powers as a Series

Here I will explain the principles and the notation I will use to denote or “notate” a series formula or series expression of a mathematical reality.  The notation I am using here is not necessarily the same as you will find in other math books, but it is similar, and my intention was to use a method of notation that is clear.

First, a sequence or series can be said to be comprised of separate elements or terms.  Each element or term follows, or grows out of, the previous element or term, by the computation of a particular formula.  We are going to start with the most simple formula.  Later they will get more complex.  But remember!  Every formula will be reduced or de-constructed to a series of additions.

The symbol for the elements of the series will be the capital letter K.  The first element will be K sub zero (K0).  The second will be K sub one (K1) and so on.  In any of the formulas, we may include in our formula a value that has an immediate relationship with the previous element.  We refer to each element as K sub en (Kn), where the n stands for the number of the element ( 0 to ¥).  That symbol that looks something like a number 8 taking a rest or ¥ is a symbol for infinity.  Obviously, K1 is the first element and K2 is the second element and so on.  We may have to fudge a little on the K0 element.  Sometimes we think of K0 as the first element.  However, in nature, there is no K0 element that we can identify, because that would mean the element that existed before the first element, and that is like saying we know what existed before the beginning of physical reality.  Therefore, the K0 element is usually going to be equal to zero.  Only if there is some good reason will the K0 element be equal to some value other than zero.  When we include in our formula one of those references to a previous element, we will use the notation K sub en minus one (K(n-1) ).  That symbol (K(n-1) ) means the value which is the previous element.  If that is not perfectly clear, I will try to make it perfectly clear shortly.

For any sequence or series we will use the label of the capital letter S followed by a number.  The first will be S1 and the second S2 and so on, followed by a colon, and then by the formula.  Let’s do S1, which denotes the series of one number multiplied by another, or n x M.

S1:       Kn =  K(n-1)  + M,  K0 = 0 ,      (n x M)            (n = 0 --> ¥ ) to infinity.

Here’s what it means, quite simply, in English.  This is sequence or series number S1.  The formula for each element is that we take the previous element and add M to it.  The capital M is the value of the number we have assigned to M in the formula n x M.  Therefore, if we are denoting the sequence for multiplying the number 7, then M = 7.  And, the value of n goes from 1 to ¥.  We don’t really need that "(n = 0 --> ¥ )" all the time, so I will often leave it out when it does not appear to be necessary.  If I excluded it from this first sequence formula, then S1 would have been as follows:

S1:       Kn =  K(n-1)  + M,  K0 = 0 ,      (n x M)

In common math notation, it is the "n" that goes from 1 to ¥, so that each element of the sequence (k1, k2, k3, ...kn) is labeled with the "subscript" n number.  Notice that we have to fudge right off on that first element.  We have to say that K0  is equal to zero because if we don’t we have a silly problem that arises from the fiction of minus reality.  In mathematics we use the convention of minus or negative numbers.  I will discuss this further later as a fiction of consciousness, just as fractions are sometimes a fiction of consciousness, or a fiction of our calculating consciousness.  In this case of S1,

if we plug in Kn as K0 our K0 element will be:

K(n-1)  + M        which means K (0-1) + 7  =  what?  0 or 7 or - 7 ?

Actually, if we made K(-1) equal to -7, which has some logic to it, then K0 would be equal to zero, because +7 added to -7 equals zero.  In a sense it really doesn’t matter, because as I said before we are not going to try to establish the value of any element that is K (0-1) which means K(-1) because we are not trying to use a notation for “minus reality.”  We could say that K(-1) is equal to minus 7.  Or, we could just say that K0 is equal to zero.  We have to have K0 equal to zero so that the formula works for K1 , where K1 is equal to 7 x 1 which is equal to 0 + 7.

Look at S1 again and you will see not only does it yield the sequence of every whole number times 7, but it also reveals that to complete this calculation we need only use addition.  The simplest reality of mathematical calculation is that multiplication is addition.

S1:       Kn =  K(n-1)  + M,  K0 = 0 ,      (n x M)

Use your calculator if you wish to assure yourself how simple S1 is.  Don’t be freaked out by the simplicity of this beginning.  The formulas will get more complex and interesting very soon.

n = 1                            n = 2                            n = 3                            n = 4

K1 = 0 + 7 = 7             K2 = 7 + 7 = 14           K3 = 14 + 7 = 21         K4 = 21 + 7 = 28

K1 = 1 x 7 = 7             K2 = 2 x 7 = 14           K3 = 3 x 7 = 21           K4 = 4 x 7 = 28

The elements of S1 when M = 7 are:  7, 14, 21, 28, 35 ...

The elements of S1 when M = 8 are:  8, 16, 24, 32, 40 ...      et cetera.

Although that probably seemed a bit boring, it was important to explain the system in its most simple form in the beginning.  And, we will re-use this formula ( n x M) many times in the more complex formulas that are coming up right now.

The second sequence is N squared (N2), which means of course N x N.  Instead of the usual statement that N2 means multiplying N by itself, we have an arithmetic formula that yields the square of N for each element n.  An arithmetic formula (arithMETic with the emphasis on “met” not on “rith”) means even to conventional mathematicians a formula that involves only addition.  I have done this to show that it can be done, and to show that we can obtain the sequence of squares using a simple addition formula.

The mathematical symbols follow, which I believe you will easily understand now without further explanation.

S2:       Kn =  K(n-1)  + (n - 1) + (n - 1) + 1,      K0 = 0              (N x N)            (n = 1 to ¥)

n = 1:              K1 =  K0 + (1 - 1) + (1 - 1) + 1 = 1

n = 2:              K2 =  K1 + (2 - 1) + (2 - 1) + 1 = 4

n = 3:               K3 =  K2 + (3 - 1) + (3 - 1) + 1 = 9

n = 4:               K4 =  K3 + (4 - 1) + (4 - 1) + 1 = 16

n = 5:               K5 =  K4 + (5 - 1) + (5 - 1) + 1 = 25                et cetera

The sequence formula for N cubed (N3) is a little more creative, and rather interesting I think.  The "M" here has a different meaning from before.

S3:       (N x N x N)

S3:       Kn =  K(n-1)  + ( Mn x 6 ) + 1,              K0 = 0      M1 = 0         Mn = M(n-1) + (n -1)                                         K0 = 0                   M1 = 0          M(-1) = dinglefritz

n = 1:               K1 =  K0 + ( 0 x 6 ) + 1 = 1                 then M2 = 0 + 1 = 1

n = 2:               K2 =  K1 + ( 1 x 6 ) + 1 = 8                 then M3 = 1 + 2 = 3

n = 3:               K3 =  K2 + ( 3 x 6 ) + 1 = 27               then M4 = 3 + 3 = 6

n = 4:               K4 =  K3 + ( 6 x 6 ) + 1 = 64               then M5 = 6 + 4 = 10

n = 5:               K5 =  K4 + ( 10 x 6 ) + 1 = 125           then M6 = 10 + 5 = 15

Further explanation:  Our "then M2 = 0 + 1 = 1" means:

M2 = M1 + the value of (n - 1) = 1      where n = 2

or    M2 = 0 + the value of (2 - 1) = 1

M3 = M2 + the value of (n - 1) = 3      where n = 3

or    M3 = 1 + the value of (3 - 1) = 3

Note that, as is always the case, a multiplication operation can be deconstructed to addition.  So, our N = 5 term, also denotable as K5, can be stretched out as:

K5 = 64 + ( 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 ) + 1 = 125

This means, if it is not boringly obvious, that we have an arithMETic formula that consistently yields the cube of N for each N element in the sequence, using only the previous element and addition.  In this sense the sequence is “evolutionary.”  Each subsequent element grows out of the previous element, or “evolves.”  If you write a computer program to apply this formula, as I did, then, the answer to any question in the form of “What is the cube of X?” is automatically there, like an item on a grocery list, where n = X.  All “cube” problems are resolved forever through this formula.  Again, this may not seem Earth shaking yet, but I may be shaking the Earth a bit soon when I start showing that the same can be done for any mathematical problem.  And, does this simpler process solve the same problems as "the calculus" attributed to Isaac Newton?

I hope you have a scientific calculator handy so that you can check out these formulas.  You don’t need to write computer programs to see that these sequential formulas work.  It is satisfactory to check them out by just looking at the formula and using a hand-held scientific calculator.

Now we can do N to the fourth power (N4)simply by multiplying N2 x N2.  Remember the rule you learned in high school math:  Xn x Xm = X(n + m) .  This means that  N4 equals N2 x N2 and 22 x 23 equals 25 .

S4:       is like S2 times S2       (N x N x N x N)

S4:       Kn =  [K(n-1)  + (n - 1) + (n - 1) + 1] x [K(n-1)  + (n - 1) + (n - 1) + 1] ,

K0 = 0              (N x N x N x N)

n = 1:               K1 = [ 1 ] x [ 1 ] = 1                 ( 1 x 1 x 1 x 1 = 1 = 14 )

n = 2:               K2 = [ 4 ] x [ 4 ] = 16               ( 2 x 2 x 2 x 2 = 16 = 24 )

n = 3:               K3 = [ 9 ] x [ 9 ] = 81               ( 3 x 3 x 3 x 3 = 81 = 34 )

n = 4:               K4 = [ 16 ] x [ 16 ] = 256         ( 4 x 4 x 4 x 4 = 256 = 44 )

n = 5:               K5 = [ 25 ] x [ 25 ] = 625         ( 5 x 5 x 5 x 5 = 625 = 54 )

S5:       is like S2 times S3       (N x N x N x N x N)

S5:       Kn =  [K(n-1)  + (n - 1) + (n - 1) + 1] x [K(n-1)  + ( Mn x 6 ) + 1] ,

Mn = M(n-1) + (n -1)

K0 = 0              M1 = 0             (N x N x N x N x N)

n = 1:               K1 = [ 1 ] x [ 1 ] = 1                 = 15

n = 2:               K2 = [ 4 ] x [ 8 ] = 32               = 25      same as 8 added 4 times

n = 3:               K3 = [ 9 ] x [ 27 ] = 243           = 35      same as 27 added 9 times

n = 4:               K4 = [ 16 ] x [ 64 ] = 1024       = 45      same as 64 added 16 times

n = 5:               K5 = [ 25 ] x [ 125 ] = 3125     = 55      same as 125 added 25 times

If you are brilliant, you have already noticed, or strongly suspected, that since we have sequences for the exponential powers 2, 3, 4 and 5, it follows that we can write sequences for any exponential power because of the rule:  Xn x Xm = X(n + m) .

To get N6 we write out what is essentially S3 x S3.  For N7 we have S3 x S4.  For N8 we can use S4 x S4 or S3 x S5.  Using this same simple procedure, we can thereafter find a combination, or several combinations, which will yield Nx by application of a series formula, using addition as the only necessary mathematical operation.

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