0
$\begingroup$

I'm not going to go into detail why I am interested in the next iteration of these functions, but here they are:

1: 6/(x+1)

2: 8/(2^x)

3: 10/(?)

The question is, which one is next? I will say that the following lists will be found in with the equations shown above:

1: 6, 3, 2

2: 8, 4, 2

3: 10, 5, 2

@x=0-2 for each number

Notice how the first function uses multiplication, while the second uses exponentiation...... perhaps tetration will be used next?

Thanks.


Update:

I've been working with the current solution: x^2 + 1, but it doesn't seem to work for the overall parent formulæ that I am working with. The equations given above are indeed only part of a much larger equation, which does actually relate quite well to the real world. Like I've said, I think the correct solution will include tetration, because the operator number appears to grow by each equation. Maybe, just maybe, the integer operators won't cut it, and I'll need something in-between exponentiation and tetration...

$\endgroup$
  • $\begingroup$ Can you clarify how the lists of 3 numbers relate to the expressions? $\endgroup$ – CommonerG Jan 7 '16 at 23:14
  • $\begingroup$ @CommonerG Writing $f_1(x) = \frac 6 {x + 1}$, we get $f_1(0) = 6, f_1(1) = 3$ and $f_1(2) = 2$. Same for the second pair of sequences. $\endgroup$ – Jack M Jan 7 '16 at 23:28
2
$\begingroup$

I think I have worked out what you were trying to say:

  1. For $f_1(x)=\frac{6}{x+1}$ we get $f_1(0)=\color{red}{6},f_1(1)=\color{green}{3},f_1(2)=\color{blue}{2}$. Hence the pattern $\color{red}{6},\color{green}{3},\color{blue}{2}$
  2. For $f_2(x)=\frac{8}{2^x}$ we get $f_2(0)=\color{red}{8},f_2(1)=\color{green}{4},f_2(2)=\color{blue}{2}$. Hence the pattern $\color{red}{8},\color{green}{4},\color{blue}{2}$
  3. For $f_3(x)=\frac{10}{x^2+1}$ we get $f_3(0)=\color{red}{10},f_3(1)=\color{green}{5},f_3(2)=\color{blue}{2}$. Hence the pattern $\color{red}{10},\color{green}{5},\color{blue}{2}$

There are actually an infinite number of functions that can give you the desired values - are there any other restrictions here?

e.g. another solution to (3) would be:

  1. For $f_3(x)=\dfrac{10}{\dfrac{x^3}{3}+\dfrac{2x}{3}+1}$ we get $f_3(0)=\color{red}{10},f_3(1)=\color{green}{5},f_3(2)=\color{blue}{2}$. Hence the pattern $\color{red}{10},\color{green}{5},\color{blue}{2}$

One arbritary generalisation of this could be: $$ f_i(x)=\frac{2(i+2)(2^m-2^n)}{(i+1-2^n)x^m+(2^m-i-1)x^n+2^m-2^n} $$ where $i=1,2,\cdots$ and $m=1,2,\cdots$ and $n=1,2,\cdots$ (as long as $m\ne n$).

This gives you: $$\begin{align} f_i(0)&=2(i+2)\\ f_i(1)&=i+2\\ f_i(2)&=2 \end{align}$$

The simplest solution with this system is when $m=2$ and $n=1$ which yields: $$f_i(x)=\frac{4(i-1)}{(i-1)x^2+(3-i)x+2}$$ Resulting in: $$\begin{align} f_1(x)&=\frac{6}{x+1}\\ f_2(x)&=\frac{16}{x^2+x+2}\\ f_3(x)&=\frac{10}{x^2+1}\\ f_4(x)&=\frac{24}{3x^2-x+2}\\ \cdots \end{align}$$


Yet another solution but this time involving tetration is: $$ f_i(x)=\frac{2(i+2)(^i2-2^n-1)}{(i+1-2^n)2^{(^{(i-1)}x)}+(i^2-2-i)(1+x^n)} $$ where $i=1,2,\cdots$ and $n=1,2,\cdots$.

This again gives you: $$\begin{align} f_i(0)&=2(i+2)\\ f_i(1)&=i+2\\ f_i(2)&=2 \end{align}$$

The simplest solution with this system is when $n=1$ which yields: $$f_i(x)=\frac{2(i+2)(^i2-3)}{(i-1)2^{(^{(i-1)}x)}+(^i2-2-i)(1+x)}$$ Resulting in: $$\begin{align} f_1(x)&=\frac{6}{x+1}\\ f_2(x)&=\frac{8}{2^x}\\ f_3(x)&=\frac{130}{2^{1+x^x}+11(1+x)}\\ f_4(x)&=\frac{786396}{3.2^{x^{x^x}}+63330(1+x)}\\ \cdots \end{align}$$

$\endgroup$
  • $\begingroup$ I think you're on the right track, but I can tell you the equation begins to break down later on in my parent formulæ. At the first iteration above, it breaks down by half; by the second, it is unrecognizable broken. I am working on further restrictions, but what I am working on is still a bit secret, so I can't just explain what is wrong with the correction. I'll say this: I believe that f3(x) should involve tetration, otherwise the surrogate functions wind-up breaking down later-on. I'm still trying to grapple with this myself, but the more I try, the more impossible the transformation seems $\endgroup$ – Asta Jan 8 '16 at 2:32
  • $\begingroup$ Okay, I checked your second solution for the function, and after thorough analyisis, I can say that it doesn't work either, but instead of being off on the first iteration by a factor of 2, is instead off only by that of 5/3, so it is actually closer! This is pretty interesting... $\endgroup$ – Asta Jan 8 '16 at 3:12
  • $\begingroup$ I maybe be able to help more if you can divulge some of th additional restrictions on this without giving away whatever secret it is that you are trying to protect :) $\endgroup$ – Mufasa Jan 8 '16 at 14:14
  • $\begingroup$ Yeah, I'll go into a bit more detail (sorry for all of the sneakiness!): My function appears to rely on a system where the numerator of the function MUST be solved as well in the denominator and then multiplied by 2: Example: f2(x): (2(2^2+1))/(2^x+1). $\endgroup$ – Asta Jan 8 '16 at 18:38
  • $\begingroup$ Have you tried my latest version that includes tetration? $\endgroup$ – Mufasa Jan 8 '16 at 18:45
1
$\begingroup$

For the last one we want a denominator $f$ such that $f(0) = 1, f(1) = 2, f(2) = 5$. Thus $f(0) - 1 = 0, f(1) - 1 = 1, f(2) - 1 = 4$. Starting to look familiar? Just set $f(x) = x^2 + 1$.

$\endgroup$
  • $\begingroup$ This fix actually looked really great! but unfortunately the parent formulæ that I am working with didn't agree. Your answer is correct for the parameters I gave in the post, but it doesn't work with the results I should be getting. Something about maths is hiding and I can't seem to figure out which of the many solutions would be correct. Like I told Mufasa (above), I think the correct answer will involve tetration. No promises, just a hunch... $\endgroup$ – Asta Jan 8 '16 at 2:40
  • $\begingroup$ @Asta Well, we can hardly give satisfying answers if you don't post all the parameters. There are infinitely many functions that will generate the numbers 10, 5 and 2. You should post the full list of numbers. If it's a finite list, then I guarantee "the" answer won't involve tetration, and if it's infinite, you should explain where you're getting the numbers. $\endgroup$ – Jack M Jan 8 '16 at 8:25
  • $\begingroup$ @Asta If you do decide to do post more details though, you should definitely do so as a new question. $\endgroup$ – Jack M Jan 8 '16 at 8:27
1
$\begingroup$

Yet another solution: If the first number always goes up by $2$, and the second number by $1$, and the third number stays constant, then we can obtain a family of equations

$$ f_k(x) = \left[x-\left(2+\frac{k}{2}\right)\right]^2 +\left[2-\left(\frac{k}{2}\right)^2\right] $$

that produces these three-number sequences for $x = 0, 1, 2$, for $k = 0, 1, 2, \ldots$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.