I'm just dabbling here.  Printable Version + Tetration Forum (https://math.eretrandre.org/tetrationforum) + Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) + Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) + 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+(31) = P((0+3),(0+1)) = P(3,0) = 2 0+(31) = 2 0+(31) = 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: 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 RE: 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(y1) 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 like the wikipedia hyperoperation definition since it shows the operator sequence. Why is "tet" the 4th operator?
