I'm just dabbling here. ChaoticMC Newbie Posts: 1 Threads: 1 Joined: Jul 2018 07/07/2018, 02:44 AM (This post was last modified: 07/07/2018, 02:46 AM by ChaoticMC. Edit Reason: Changed "Can" to "Could" at the very last line, to sound less ''pushy". ) 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? sheldonison Long Time Fellow Posts: 576 Threads: 22 Joined: Oct 2008 07/09/2018, 02:07 PM (This post was last modified: 07/09/2018, 08:23 PM by sheldonison.) (07/07/2018, 02:44 AM)ChaoticMC Wrote: 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...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 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    6561 - Sheldon Xorter Fellow Posts: 81 Threads: 26 Joined: Aug 2016 07/09/2018, 08:34 PM 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 Xorter Unizo sheldonison Long Time Fellow Posts: 576 Threads: 22 Joined: Oct 2008 07/10/2018, 10:34 PM (This post was last modified: 07/11/2018, 02:54 AM by sheldonison.) (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 $H_1(a,b)=a+b$ $H_2(a,b)=a\cdot b$ $H_3(a,b)=a\uparrow b=a^b$  $H_4(a,b)=a\uparrow\uparrow b=\text{tet}_a(b)$  $H_5(a,b)=a\uparrow\uparrow\uparrow b=\text{pent}_a(b)$ $\forall n>0\,H_{n}(a,b)=H_{n-1}\left(a,H_{n}(a,b-1)\right)$ Knuth's uparrow notation is great except when discussing the hyper operation sequence where uparrow of (0,-1,-2) donesn't make any sense.  Also, the recursive definition is limited to integer values of n, so I like $H_n(a,b)$. - Sheldon « Next Oldest | Next Newest »