Levels of addition...

[alkatran]

Well, I've been thinking.

An addition is a series of multiplications, an exponential is a series of multiplications, whatever is a series of... well you get the point.

Now, with the way we do it now, we would need an infinite amount of symbols for this... so I was thinking I'd just represent it like this:

4:1;5 = 4 + 5
4:2;5 = 4*5

etc. (addition should be 0, maybe?)

ANYWAYS, that's not my point. My point is, if we have X as a level:

4:x;5 = 20

How do we solve x?? (it's obviously "multiply" there, but you know)

What if X is a real number? (non-whole) what if it's negative? imaginary? *pensive*

[jemidiah]

Well, there are many different operators you could make that would accept 4 and 5 and give you twenty. If you had only a limited number to choose from, you'd have to plug in each operator and see if it worked.

Back to the original thoughts, there's actually something called Sigma notation for the repeated additions. This thread explains it and its notation, along with the multiplicative version. You could actually represent the multiplicative one using a series of summations. Anyhow, what's wrong with what we alread use?

[alkatran]

This isn't changing the system we use, it's adding to it.

I've though about it.. addition is 1, subtraction is -1, etc, so:
x:y;z = t goes to x = t:-y;z

my question is: Has anyone ever looked into this before?? What if I have 1:0.5;1?

i already know about sums, trust me.. I was coming up with the equations for summing geometrical and arithmatical (french math, sorry if terms are wrong) series before we were taught them... or even began the section for that matter. (My technique.. was actually better than the books since it allowed for a different exponent than 0 as the first element).
...Sorry if that came off as arrogant.

[jemidiah]

You know, it just hit me what you're doing. Hmmm...... this will require further thought......

[alkatran]

There we go, don't think too hard about it. I tried to make some equations for it once... but It's so hard to test them because you need to go into series of exponentials and stuff.

BTW: is 3^3^3 = (3^3)^3 or 3^(3^3)
(PEDMAS says left to right ((3^3)^3) but it seems like most people do it the other way around)

[jemidiah]

Google says 3^(3^3): http://www.google.com/search?num=20&...E3&btnG=Search

[alkatran]

My friends TI-83 says it's (3^3)^3 but my TI-89 says it's 3^(3^3). Just great. lol

As a consequence, my equation solvimicating func could be invalid. lol

[jemidiah]

It strikes me to look into the properties of such a function instead of trying to find the function itself.

4+5=5+4
If x(p,q) = a+b Then x(b,a) = x(a,b)
4*5 = 5*4
If x(p,q) = a*b Then x(b,a) = x(a,b)
4^5 <> 5^4
If x(p,q) = a^b Then x(b,a) <> x(a,b)

Well, there's something breaking down in these relationships. Is this chaotic or orderly? Let's look.

4+5 = (1+1+1+1)+(1+1+1+1+1)
1 added to itself 4 times and 1 to itself 5 times.
4*5 = 4+4+4+4+4
4 added to itself 5 times.
(1 added to itself 4 times) 5 times.
4^5 = 4*4*4*4*4
4 times itself 5 times.
(((((1 added to itself 4 times) 4 times) 4 times) 4 times) 4 times)

It seems chaotic to me, though I may have missed something about it.

If it truly is chaotic, then there is (by definition) no extending pattern. Otherwise, there is an answer out there.

[alkatran]

I was thinking along the lines of...

for X:Y;X
X Y=1,2,3,4,5,6 etc
1-> 2,1,1,1,1,1,1,1,1,1
2-> 4,4,4,4,4,4,4,4,4,4
3-> 6,9,27,19683, holy ****
4-> 8,16,256,(256^4)^4

The thing I find weird is that two is ALWAYS 4. That would be my basis for any equation.

[twanvl]

I have tought about this in the past too, x:.5;y is interesting, hadn't tought about that one. The inverse could be described in different ways:
a = x:1;y = x+y => x = a-y = a:-1;y, y=a:-1;x
That would make division -2
x:-2;y = x/y
But for 3, two inverses are posible, roots and logarithms
a = x:3;y
x = a:3;(1/y)
y = a:-3;x ?

how about: Every step there is one more argument:
[x] = increment(x)
[x,y] = add(x,y) = y times inc(x)
[x,y,z] = mult(x,y,z) = z times x:=add(x,y)
[x,y,z,a] = power(x,y,z,a) = a times x:=mult(x,y,z,a)
etc.
That would make:
[x] = [x,1,1,1,1,1,...]
a+b = [a,b]
a*b = [a,a,b]
a^b = [a,a,a,b]
a:b;c = [(b times a),c]

I also noticed that somtimes you get a converging series, for example for 1<=x<=e:
lim a->Inf : (x:3;-x):4;a = x

[alkatran]

Great post twanvl, I'm still trying to understand what you wrote down! (It's hard to read with this system, hehe).

how about : Every step there is one more argument:
[x] = increment(x)
[x,y] = add(x,y) = y times inc(x)
[x,y,z] = mult(x,y,z) = z times x:=add(x,y)
[x,y,z,a] = power(x,y,z,a) = a times x:=mult(x,y,z,a)
etc.
That would make:
[x] = [x,1,1,1,1,1,...]
a+b = [a,b]
a*b = [a,a,b]
a^b = [a,a,a,b]
a:b;c = [(b times a),c]

[x,y,z] -> from what I understand this means "X, Y times, Z times" right? But don't we want an exponenet... I don't exactly understand the system, could you explain more?

Until I understand.. I'll stick with x:y;z..



Anyways, as for your other point, that two inverses were possible:

X:Y;Z = T -> X = T:-Y;Z
X:Y;Z = T ->
2:1;-3 = -1 = 2-3 -> -3 = -1 -2 = -1:-1:2

Z = T:-Y;Z ?? (for addition)

4*3 = 12
4:2;3 = 12
isolate 4
12:-2;3 = 4

isolate 3

12/4 = 3 so
12:-2;4 = 3

A:B;C = D then C = D:-B;C and A = D:B;C
Well, it works so far, but exponents should mess this up...


2^4 = 16
2:3;4 = 16

isolate 2

4th root of 16
16:-3;4 = 2

isolate 4
square root of 16
16:-3;2 = 4

...hold it.. that worked?? trying other numbers..

2^3 = 8
2:3;3 = 8
8:-3;3 = 2 cubic root of 8
8:-3;2 = square root of 8 is NOT 3...

Buuuuut.. maybe we need to go a level lower.. like with exponents we have to divide instead of subtract.. if you get what I mean?
Of course, that brings up the question, what's level 0? I'd say X:0;Y = X.

[alkatran]

Thought: What if instead of changing Y (in X:Y;Z) to -Y, we do 1/Y for exponentials? ... I really have to stop thinking about this.

[alkatran]

if X:Y;Z = T then X = T:-Y;Z
We've proven this. The problem comes when we try to isolate Z.

if X:1;Z = T then Z = T:-1;X
1+2 = 3, 3-2 = 1 true

if X:2;Z = T then Z = T:-2;X
2*3 = 6, 6/2 = 3 true

if X:3;Z = T then Z = T:-3;X ?
2^3 = 8, 8^(1/2) != 3 false

However.. log(8)/log(3) is 2.. so, the question is.. how do we know which one we want?? The flaws in the system begin to show... It feels so... primitive.

*idea!* what if we said that:

X:Y;Z = T, Z = T:-1/Y;X

1 goes to -1, 2 to goes to -.5, 3 goes to -.33333 etc.... hmmm...

[jemidiah]

Well, we haven't actually proved that the inverses are anything, we're just using the notation of -X.... is the inverse of Y...... (ish )

Actually, why would

2^3 = 8, 8^(1/2) != 3 false

Need to equal three? Using the logarithms, you've actually gotten close to finding the roots (Log(8)/log(2) = 3). I'd say that, if -3 is the inverse of 3, that X:-3;Y = X^(1/Y) = Yth root of x.


Based on your previous post, I found some equations relating X:1;X and X: (1-3);X. Here they are:

X:1;X = ((X:1;X)/2)*2
X:2;X = ((X:1;X)/2)^2
X:3;X = ((X:1;X)/2)^((X:1;X)/2)

Now, I see that the operation done to (X:1;X) is one operation greater than the one we're finding. IE: Take the first one. You multiply by 2, and multiplication is the 2nd operation. In the second one, you are using exponents, operation 3. In the third one, you are using the ((X:1;X)/2) exponent, which would presumably be operation 4. By the way, how did you get this table:

for X:Y;X
X Y=1,2,3,4,5,6 etc
1-> 2,1,1,1,1,1,1,1,1,1
2-> 4,4,4,4,4,4,4,4,4,4
3-> 6,9,27,19683, holy ****
4-> 8,16,256,(256^4)^4

For operations greater than 3?

[alkatran]

I used my head and my computer's calculator.

(3^3)^3 = 27^3 = 27*27*27 = 19683
((4^4)^4)^4) = (256^4)^4

As for 1, and 2. Just think about it logicly. operations based on 2 always happen twice. 1 always happens once (except for +1)

4 times itself 5 times.
(((((1 added to itself 4 times) 4 times) 4 times) 4 times) 4 times)

It seems chaotic to me, though I may have missed something about it.

If it truly is chaotic, then there is (by definition) no extending pattern. Otherwise, there is an answer out there.

I will dispute that. It isn't chaotic because we can represent it as X:Y;Z (where Y is changing). A similar example would be saying 2^x is chaotic because it can't be represented in an xth degree polynomial.

[sql_lall]

Interesting...

I was also thing about these things before, but more along these lines:

O(x) is the xth level of operand:
O(1) = '+'
O(2) = '*'

BUT with the recursion:
log(A O(n) B) = log(A) O(n-1) log(B)

any ideas?

[twanvl]

sql_lall, although the idea looks good there are some problems at higher levels. The biggest problem is that O(3) is dependent upon the base of the logarithm:
O(3) = @ (got to use some symbol...)
log(a @ b) = log(a) * log(b)
log(a @ b)/log(base) = log(a) * log(b) / log(base)²
log(a @ b) = log(a) * log(b) / log(base)

The second problem is what this means? If we take for example a base 2 logarithm:
a@b = 2^(₂log(a) * ₂log(b)) = (2^₂log(a))^₂log(b) = a^₂log(b)
This system is lacking a power operator, that would require an inverse for the logarithm, which is the power operator itself.

For completeness, let's also look at O(4):
a O(4) b = 2^(log(a)^log(log(b))) = ?

But it does make for an interesting series.

[alkatran]

The symbol used for higher levels go lik this:
+
*
^
^^
^^^
^^^^
^^^^^

etc...


I like this O(x) system though, it's much 'cleaner' than mine because it resembles what we use now.

log(a @ b) = log(a) * log(b)
log(a @ b)/log(base) = log(a) * log(b) / log(base)^2
log(a @ b) = log(a) * log(b) / log(base)

How do you come up with the second line? It looks like you just arbitralily multiplied by log(base) on one side of the equation?? (you know.. come to think of it.. log(a^b) = log(a)*log(b), log(b^a) = log(b)*log(a), log(a^b)=log(b^a), a^b = b^a....
You sure you're doing your math right?????? Isn't it log(a ^ b) simplifies to log(a) * b???

[twanvl]

That second line is based on the equation:
_blog x = log x / log b

[alkatran]

log(A O(n) B) = log(A) O(n-1) log(B)

Isn't that log(A O(n) B) = log(A) O(n-1) B?

log(100) = 2 = log(10^2) = log(10) * 2 = 1*2 = 2


This means that:

log(A + B) = log(A) O(0) B
*base 2 log
Log(2+2) = log(2) O(0) 2
2 = 1 0(0) 2

...multiplication? 2/1? hmmm

[alkatran]

o(0) = @

Log(B + A) = Log(B) @ A

*base 10 log
Log(4 + 6) = 4 @ 6
1 = 4 @ 6 = 6 @ 4

*base 3 log
Log(5+4) = log(5) @ 4
2 = log(5) @ 4 = log(4) @ 5

Wait a sec.. I think I see something..

4 = 100
5 = 101
If we (not or) this (both must be false to give true)

100
101
010 = 2


4 = 100, 6 = 110
100
110
001 = 1 (HOLY ****!)



I'm not sure, but it looks like:

Log(A) @ B = Log(B) @ A = not (A or B)

*edit* just realized those weren't the same logs


I know what I'll be doing all day at school on my calculator...

[alkatran]

Just realized that:

log(a*b) = log(a) + log(b)
log(a^b) = log(a) * b (not log(b))

this means the rule doesn't apply, and I wasted some time