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+--- Thread: I'm just dabbling here. (/showthread.php?tid=1195)
I'm just dabbling here. - ChaoticMC - 07/07/2018
I'm only a new user, so I don't know what to do here. I'm only an amateur too.
Anyways, here's how we define hyperoperations.
H(b,x,0) = S(b)
H(b,0,1) = b
H(b,1,r≥2) = b
H(b,x+1,r+1) = H(H(b,x,r+1),b,r)
Alright, so we want the negative hyperoperations.
So let's denote S(b) as S(b,x). This time with 2 inputs.
So, it has to obey the rule that H(b,b,r) = H(b,2,r+1)...right?
Well, if S(b,b) did equal b+2, then would there something wrong?
S(b,b) = b+2 = S(b,b+1), nothing wrong with that.
However, 0+1 = S(0+0,0) = S(0,0) = 2. This is a contradiction.
This is the only instance, however, where this is wrong. Why? Because 0+0=0. It's the identity of addition.
In fact, even if the identity was 2, then 2+3 = S(2+2,2) = S(2,2) = 2. If the identity was 3, then
3+4 = S(3+3,3) = S(3,3) = 2.
-Wait! If the identity was 1... Then 1+2 = 2...which would actually be true. So, in order for S(b,x) to be functional, then the identity of addition has to be 1.
But wait! [Note: Let's denote the inverse of S as P]
0+(3-1) = P((0+3),(0+1)) = P(3,0) = 2
0+(3-1) = 2
0+(3-1) = 0+3 = 3
2=3
And this is when the identity of addition is 1.
So, no matter which identity we choose, there will always be a contradiction. So, that's it. It's just 0.
Meaning that S(b,b) ≠ b+2.
Meaning that S(b,x) = S(b).
Alright, well this does change a bit about the negative hyperoperations, but not completely. Anyways.
Let's denote H(b,x,-1) as ®.
So, S(b,x) just equals S(b) for all x.
So, S(b,S(x)) = S(b,x)®b = S(b)®b
S(b)®b = S(b)
And...this is just it. This is the solution to the hyperoperation rank -1.
We can't go any farther.
Could somebody check my work?
RE: I'm just dabbling here. - sheldonison - 07/09/2018
(07/07/2018, 02:44 AM)ChaoticMC Wrote: here's how we define hyperoperations.I would try this definition from wikipedia:
H(b,x,0) = S(b)
H(b,0,1) = b
H(b,1,r≥2) = b
H(b,x+1,r+1) = H(H(b,x,r+1),b,r)
Alright, so we want the negative hyperoperations...
https://en.wikipedia.org/wiki/Hyperoperation
The standard definition then follows, where a is the base
H0(a,b)=b+1 independent of the value of a
H1(a,b)=a+b where H1(a,0)=a
H2(a,b)=a*b where H2(a,0)=0
H3(a,b)=a^b where for n>=3, H(a,0)=1, and H(a,1)=a
With that definition, surprisingly H{-1}(b)=H0(b)=b+1; this works for H{-1,-2,-3.....}(b)=b+1 as well. Here is the table for a=3.
Code:
a=3 b+1 b+a b*a a^b a^^b
b h0(b) h1(b) h2(b) h3(b) h4(b)
====================================
0 1 3 0 1 1
1 2 4 3 3 3
2 3 5 6 9 27
3 4 6 9 27 7625597484987
4 5 7 12 81
5 6 8 15 243
6 7 9 18 729
7 8 10 21 2187
8 9 11 24 6561RE: I'm just dabbling here. - Xorter - 07/09/2018
Hi!
I agree with Sheldon almost at all. If I were you, I would use the following notations:
Hn(a,b) = a[n]b = H(a,n,b) or the Knuth's uparrows, the second is simplest as I think.
If you want iterate an operator, just use the following functional power formula:
y [z+1] x = (y [z] x)^o(y-1) o y
I suppose that z can be negative, zero, rational, real or complex, too.
But if you follow Sheldon's zeration formula, it will not lead to addition, so I think that is wrong.
But Sheldon has a lot of successful pari/gp programme and ideas for the realizations of from the tetration to heptation, but hard to interpolate with rational or real numbers.
Good luck!
Xorter
RE: I'm just dabbling here. - sheldonison - 07/10/2018
(07/09/2018, 08:34 PM)Xorter Wrote: Hi!
I agree with Sheldon almost at all. If I were you, I would use the following notations:
Hn(a,b) = a[n]b = H(a,n,b) or the Knuth's uparrows, the second is simplest as I think...
I like the wikipedia hyperoperation definition since it shows the operator sequence. Why is "tet" the 4th operator?
- \( H_0(b)=1+b\; \) successor, which is a unary operation
- \( H_1(a,b)=a+b \)
- \( H_2(a,b)=a\cdot b \)
- \( H_3(a,b)=a\uparrow b=a^b \)
- \( H_4(a,b)=a\uparrow\uparrow b=\text{tet}_a(b) \)
- \( H_5(a,b)=a\uparrow\uparrow\uparrow b=\text{pent}_a(b) \)
- \( \forall n>0\,H_{n}(a,b)=H_{n-1}\left(a,H_{n}(a,b-1)\right) \)