Finding the function that would describe this:

Question

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.

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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...

Answer 1

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 $\color{red}{10},\color{green}{5},\color{blue}{2}$

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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:

3. 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 $\color{red}{10},\color{green}{5},\color{blue}{2}$

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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: $$\begin{align} f_1(x)&=\frac{6}{x+1}\\ f_2(x)&=\frac{16}{x^2+x+2}\\ f_3(x)&=\frac{10}{x^2+1}\\ f_4(x)&=\frac{24}{3x^2-x+2}\\ \cdots \end{align}$$

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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: $$\begin{align} f_1(x)&=\frac{6}{x+1}\\ f_2(x)&=\frac{8}{2^x}\\ f_3(x)&=\frac{130}{2^{1+x^x}+11(1+x)}\\ f_4(x)&=\frac{786396}{3.2^{x^{x^x}}+63330(1+x)}\\ \cdots \end{align}$$

Answer 2

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 $f(x) = x^2 + 1$.

Answer 3

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 $k = 0, 1, 2, \ldots$