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As i defined E 1>x = (1>x)^(1>x) = 1>(x>x)

I define the C power,C of Complex Extended Own Power,as

C 1>x = (((1>x)^(1>x))^(1>x))^(1>x)......(1>x)times (1>x) in the power tower above base number(1>)......


From power math it can be derived that

C 1>x = (1>x) ^E (1>x)= 1>(x>(x>x)) because

from math we know (a^b)^c = a^(bc)

so in the power tower of C 100 = C 1>2 we get

100x100x100.....100times100 in the sum.....=100^100 = E 100

and in the power tower of C 1>x we get

(1>x)(1>x)(1>x)........(1>x)times..................= E 1>x


and thus C 1>x = (1>x)^(E 1>x)

and 1>x ^ ((1>x)^(1>x)) = 

(1>x)^(1>(x>x))=

10 to the power of (x((1>(x>x)))=

10 to the power of x>(x>x) or

1>(x>(x>x))


This is the mathematical proof,in the next part on this subject i will explore the patterns

of C and E powers.


A list demonstrates the patterns best as shown below;


1E 1>1 = 1>10                                                1C 1>1 = 1>(1>10)

1E 1>2 = 1>200                                              1C 1>2 = 1>(2>200)

1E 1>3 = 1>3000                                            1C 1>3 = 1>(3>3000)

1E 1>4 = 1>(4>4)                                           1C 1>4 = 1>(4>(4>4))

.....

1E 1>9 = 1>(9>9)                                           1C 1>9 = 1>(9>(9>9))

1E 1>10=                                                          1C 1>10=1>(1>((1>10)~1))

2E 1>1= 1>(1>11)                                           2C 1>1 = 1>(1>((1>((1>9)~8))~10))

2E 1>2 =1>(2>202)                                         2C 1>2 = 1>(2>((2>((2>197)~197))~(2>2)))

2E 1>3 =1>(3>3003)                                       2C 1>3 = 1>(3>((3>((3>2996)~2996))~(3>3)))

2E 1>4 =1>(4>((4>3)~4))                                2C 1>4 = 1>(4>((4>((4>(3~(9re3)~5))~(3~(9re3)~5)))~(4>4)))

.....

2E 1>9 =1>(9>((9>8)~9))

2E 1>10=

3E 1>1 = 1>(1>((1>9)~11))

3E 1>2 = 1>(2>((2>199)~202))

3E 1>3 = 1>(3>((3>2999)~3003))

3E 1>4 = 1>(4>((4>(3~(9re4)))~((4>3)~4)))

......

3E 1>9 = 1>(9>((9>(8~(9re9)))~((9>8)~9)))

3E 1>10 =

4E 1>1 = 1>(1>((1>(9re11))~(1>9)~11))

Now i will progress in E powers only with base number 1>1;

5E 1>1 = 1>(1>(                                                         (1>(9re((1>9)~11)))~  (1>(9re11))  ~(1>9)  ~11))

6E 1>1 = 1>(1>( (1>(9re((1>9re11)~(1>9)~11)))     ~(1>(9re((1>9)~11)))~   (1>(9re11))  ~(1>9)  ~11))    

7E 1>1 = 1>(1>( (1>(9re(                               (1>9re(1>9~11))~(1>9re11)~(1>9)~11))) ~repetition of the above))

8E 1>1 = 1>(1>( (1>(9re(    1>9re((1>9re11)~(1>9)~11)~...............

9E 1>1 = 1>(1>( (1>(9re(    1>9re((1>9re(1>9~11))~(1>9re11)~1>9)~11)~.........

etc. etc.


So we could say for

1E 1>1 = 1>1

2E 1>1 = 1>(1>1)

3E 1>1 = 1>(1>(                                                                                                                               (1>9re0)~1) )

4E 1>1 = 1>(1>(                                                                                                               (1>9re1)~(1>9re0)~1) )

5E 1>1 = 1>(1>(                                                                             (1>9re((1>9re0)~1))~(1>9re1)~(1>9re0)~1) )

no matter how far we go,the outcom of the number will be in the form of

        1>(1>(      (1>(9re(z)))~(1>(9re(y)))~...................~(1>(9re(11)))~(1>(9re(1)))~(1>(9re(0)))~1 ) )

where z ofcourse is not necesarily the 26th in sequence but could be any last sequence,the 5th,the 24th,the millionth etc.


When we aplly this listing on E powers of 1>2 base number the form of the number will only change in this way;

1>(2>(     (2>(1~(9re(z))))~(2>(1~(9re(y))))~....................~(2>(1~(9re(202))))~(2>(1~(9re(2))))~(2>(1))~2 ) )

Note that the z to a valeus here will be different than with 1>1 as base number,that can be seen at the tails of the number forms.


When we list the values from a to z say a is the last in sequence (the first from right on) from 4E on with base number 1>1 the list look as follows;

a=                                                                                                                                              (1>0)~1

b=                                                                                                                              (1>9re1)~(1>0)~1

c=                                                                                                          (1>(9re(a)))~(1>9re1)~(1>0)~1

d=                                                                                     (1>(9re(b)))~(1>(9re(a)))~(1>9re1)~(1>0)~1

e=                                                                  1>(9re(c)))~(1>(9re(b)))~(1>(9re(a)))~(1>9re1)~(1>0)~1

...........

(n)th letter=         (1>(9re((n-2)th letter)))~.....................~repetition of the above all down to ~(1>0)~1


The same principle goes for 1>2 or 1>3 or any base number.

only then the last term in the sequence the

(n)th letter for base number = 1>m will be

((m-1)>9re((n-2)th letter))).

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