between addition and multiplication

[martin]

Hey Martin, we are not talking about the function f(x) = a [4] x but about the function g(x) = a [x] b.

Right, I didn't remember correctly - but it had to do with considering only an interval for which arithmetic progression turns into geometric progression and then extrapolating to obtain further values.
I've worked on so many different ideas during the last few years, so I didn't dig into it any further. (The most proliferate idea being a real number Y for which floor(Y*p#) always produces primes for p=prime>1 with a probability of at least 1-10^-560)

[nuninho1980]

This has probably been thought of before, but here goes anyway. I was thinking about the "sesqui" operation intermediate between adding and multiplying; I'll write "@" here. Obviously a @ b should lie between a+b and ab. Maybe we should take the mean. But which one, arithmetic or geometric? Since one applies to addition and the other to multiplication, why not take both? Then we'll take the mean of these two. But which mean? Again, take both; the proposed value for the sesqui-operation is then the limit of this process when iterated many times.

In fact the two values converge quite quickly and for 10-digit precision we usually have convergence within 3 or 4 iterations. Here are some values for a @ a:

(...)

I discovered this forum after asking a question recently on sci.math. It looks like people here have been thinking about the same thing: I asked if the next operation after exponentiation should require new numbers, the way that addition/subtraction, multiplication/division, exponentiation/root-taking/logarithms lead from the counting numbers to negative, real and complex numbers respectively.

I'm very interesting about operation "@". but what's your code (pari/gp, maple...)?

[lloyd]

So I tried this out on the weekend to see what it looked like, and it works very well. Between addition and multiplication I used the "weighted arithmetic-geometric mean" that I described earlier in the post. Between multiplication and exponentiation I used a similar weighted function I call the "weighted exponential-geometric mean". Here is the C code; I lazily had it iterate 6 times instead of calling itself recursively and ending when the two means were close enough. I calculate m = a * b and n = a ^ b, I specify phi with some value between 1 and 2 and I call this function:

float wgxm(float m, float n, float phi)
{
int i;
float p,q;
phi--;
for (i=1;i<6;i++) {
p = pow(n/m,phi) * m;
q = pow(m, pow(log(n)/log(m),phi));
m = p;
n = q;
}
return p;
}

The attached graph shows the functions y = x {phi} 3, where phi varies from 0 to 2 by stepping 0.2. The light blue lines are at integer values of phi, i.e. y=x+3, y=3x and y=x^3. The yellow lines are the values calculated by the function above (and the corresponding one between addition and multiplication. The x axis goes from 1 to 20 and the y axis goes from 0 to 800. The grid lines are every unit in the x direction, and every 40 units in the y direction.

Something needs to be done to clean up the exponential/geometric interpolation function a little for it to work with values close to x=1 (the log of values close to 0 is negative). I don't think the formula works quite right whenever x^3 is smaller than x*3. The second attached graph is a close-up of the first, where x goes from 1 to 2 and y goes from 1 to 9, with gridlines in both directions every 0.25 (meaning the stretching/scaling is only by a factor of 2).

I'll have to see how the wgxm function looks if I interpolate between 3x and 3^x next, instead of x^3. Let me know if you think any of this looks nice.

Surely, though, {0.25} should be weighted 3/4s towards the arithmetic mean, and 1/4 towards the geometric mean. Ah but is the weighting carried out arithmetically or geometrically? Apply a 3/4 arithmetic : 1/4 geometric weighting there too! And take the limiting case again.

In other words, for a {0.25} b, with a<b,

m1 of a and b = a + (b-a)*0.25 (0.25 of the way between a and b, judged arithmetically)
m2 of a and b = a * (b/a)^0.25 (0.25 of the way between a and b, judged geometrically)

Now plug m1 and m2 into a and b, and iterate until you get something stable (i.e. m1 = m2 to whatever degree of precision you need)

This seems like a good generalisation to real values between 0 and 1!

Yes! I like that a lot.

[lloyd]

I'm very interesting about operation "@". but what's your code (pari/gp, maple...)?

Here's the C code. Feed this function three values:

m is a+b (don't give it m=0)
n is a*b
phi is a value between 0 and 1.

Sorry, the formatting (indenting) will disappear on this forum.

float wagm(float m, float n, float phi)
{
int i;
float p,q;
for (i=1;i<6;i++)
{
p = phi * (n-m) + m;
q = pow(n/m,phi) * m;
m = p;
n = q;
}
return p;
}

See my other post for a graph and for the extension to operations between multiplication and exponentiation.

[nuninho1980]

thank you! but did you use "microsoft visual c++" or "turbo c++". or what's other program "C"?

[lloyd]

thank you! but did you use "microsoft visual c++" or "turbo c++". or what's other program "C"?

uh, it's just plain vanilla C using the pow function from the math header math.h. It'll probably work in c++ too, it's extremely basic code.

[JmsNxn]

This is fucking beautiful! I've been working at this for three years now and you beat me to the punch! LOL!

Do you have a closed form expression for it yet? I have so many ideas involving rational operators.

And one thing that MUST be true is

x {1.5} 2 = x {0.5} x

If that's not true then the system doesn't meet requirements. But it wouldn't be hard to improvise a system since 0 <= q <= 1, {q} is defined.

Also, I wonder if we could try to implement the laws of logarithmic semi operators into this system:

q:log(x) = exp^[-q](x)

q:log(x {q} y) = q:log(x) + q:log(y)
q:log(x {1+q} y) = q:log(x) * y

And if we implement this we can feasibly solve for a new variation of tetration.

Since
x {0.5} y = arithmetic/geometric limiting algo
= -0.5:log(0.5:log(x) + 0.5:log(y))
= sexp(slog(sexp(slog(x)-0.5) + sexp(slog(y)-0.5)) + 0.5)

Holy jesus, yes!

And also, I was wondering what our identities are?, if S(q) is the identity function and x {q} S(q) = x, S(1) = 1, and S(0) = 0 obvi, but what is S(q)?

I found S(0.5) = 0.7019920407 for 2 {0.5} S(0.5) = 2. I used 1000 cycles so my numbers are probably more accurate.
However, sadly, S(0.5) for 3 {0.5} S(0.5) = 3 is a different number and therefore {q} has no identity. This is very sad indeed.

[tommy1729]

if i understand this confusing thread well

x {0.5} x = x !

in fact x {y} x = x

at least following from

( quote )

In other words, for a {0.25} b, with a<b,

m1 of a and b = a + (b-a)*0.25 (0.25 of the way between a and b, judged arithmetically)
m2 of a and b = a * (b/a)^0.25 (0.25 of the way between a and b, judged geometrically)

Now plug m1 and m2 into a and b, and iterate until you get something stable (i.e. m1 = m2 to whatever degree of precision you need)

( end quote )

unless you guys started using different ideas since the time of the quote.

furthermore i think this has nothing to do with * tetration * and see it more like an idea inspired by Gauss Aritmetic-Geometric Mean.

If this was not my favorite forum and due to the lack of good math forums in general imho , i might not have read it in the first place ( OP was already unaware of Gauss AGM Mean ) although it might get intresting soon ...

i would be more carefull to associate this immediately with a new * slog *.

i think JmsNxn is a bit overenthousiastic.

maybe its me , but the only intresting thing i can see at the moment is :

does this ' new ' mean have a closed form similar to gauss his result ?

( i assume its analytic ?? for complex z_i : z1 {z2} z3 ? )

furthermore , i have no idea why JmsNxn thinks x {1.5} 2 = x {0.5} x
???

tommy1729

[JmsNxn]

if i understand this confusing thread well

x {0.5} x = x !

in fact x {y} x = x

at least following from

( quote )

In other words, for a {0.25} b, with a<b,

m1 of a and b = a + (b-a)*0.25 (0.25 of the way between a and b, judged arithmetically)
m2 of a and b = a * (b/a)^0.25 (0.25 of the way between a and b, judged geometrically)

Now plug m1 and m2 into a and b, and iterate until you get something stable (i.e. m1 = m2 to whatever degree of precision you need)

( end quote )

unless you guys started using different ideas since the time of the quote.

furthermore i think this has nothing to do with * tetration * and see it more like an idea inspired by Gauss Aritmetic-Geometric Mean.

If this was not my favorite forum and due to the lack of good math forums in general imho , i might not have read it in the first place ( OP was already unaware of Gauss AGM Mean ) although it might get intresting soon ...

i would be more carefull to associate this immediately with a new * slog *.

i think JmsNxn is a bit overenthousiastic.

maybe its me , but the only intresting thing i can see at the moment is :

does this ' new ' mean have a closed form similar to gauss his result ?

( i assume its analytic ?? for complex z_i : z1 {z2} z3 ? )

furthermore , i have no idea why JmsNxn thinks x {1.5} 2 = x {0.5} x
???

tommy1729

I thought the identity function would have been stable. And if it was, it would've been possible to create a variant of tetration. Solving semi-operators can solve tetration.

and
x {1.5} 2 = x {0.5} x
the same way x * 2 = x + x,
and x ^ 2 = x * x
The law of recursion must hold.

[lloyd]

Tommy, the relevance to tetration is that tetration is an attempt to extend the sequence a{1}b, a{2}b, a{3}b... (where these are respectively addition, multiplication, exponentiation) to a{4}b (tetration) -- but another way to extend the sequence is to find the functions where phi (which I use for the value value in curly brackets) is inbetween the integer values we already know. Your guesses that x {0.5} x = x!, and x{y}x = x are wrong and do not follow from the definitions. It's true there is no closed form for these intermediate operations as I defined them yet, and it's true that I didn't know Gauss had invented the arithmetic-geometric mean, if that matters to you. But look at the graphics I posted and tell me you don't think the numerical solutions using these weighted functions look good as interpolated curves.

Finding closed forms will be very difficult, as already the closed form for Gauss's mean involves some tough steps.

if i understand this confusing thread well

x {0.5} x = x !

in fact x {y} x = x

at least following from

( quote )

In other words, for a {0.25} b, with a<b,

m1 of a and b = a + (b-a)*0.25 (0.25 of the way between a and b, judged arithmetically)
m2 of a and b = a * (b/a)^0.25 (0.25 of the way between a and b, judged geometrically)

Now plug m1 and m2 into a and b, and iterate until you get something stable (i.e. m1 = m2 to whatever degree of precision you need)

( end quote )

unless you guys started using different ideas since the time of the quote.

furthermore i think this has nothing to do with * tetration * and see it more like an idea inspired by Gauss Aritmetic-Geometric Mean.

If this was not my favorite forum and due to the lack of good math forums in general imho , i might not have read it in the first place ( OP was already unaware of Gauss AGM Mean ) although it might get intresting soon ...

i would be more carefull to associate this immediately with a new * slog *.

i think JmsNxn is a bit overenthousiastic.

maybe its me , but the only intresting thing i can see at the moment is :

does this ' new ' mean have a closed form similar to gauss his result ?

( i assume its analytic ?? for complex z_i : z1 {z2} z3 ? )

furthermore , i have no idea why JmsNxn thinks x {1.5} 2 = x {0.5} x
???

tommy1729