# Need mathematical function for “adding” 0.5 and 0.5 and getting 0.4

I'm looking for a mathematical function that would have the following attributes:

• Reasonably smooth -- continuous to the second or third derivative, say, for values greater than zero.
• Given two values 1.0 and 1.0 it produces 1.0
• Given two values less than 1.0 (but greater than 0), such as 0.5 and 0.5 it produces a number less than either number but greater than the product of the two numbers (eg, 0.4).
• Similarly, for two values greater than 1.0 it produces a number greater than either but less than their products (less important).
• ADDED: Given one value 1.0 and another not, should produce the second value -- f(1.0, N) = N.
• Ideally (not a hard requirement), the function is associative and commutative.
• Ideally (not a hard requirement), there is a "knob" one can turn to adjust the "strength" of the function, in terms of, eg, whether f(0.5,0.5) = 0.4 or instead 0.3.

Eg, I could simply use multiplication, where 0.5 $\cdot$ 0.5 = 0.25, but that results in a number (to be used as a weighting factor) that is too small.

After some experimentation (in a "toy" Java test program):

Math.exp(-Math.pow(Math.pow(Math.abs(Math.log(A)), fudge) + Math.pow(Math.abs(Math.log(B)), fudge), 1.0/fudge));


$$f(A,B)=\exp(-((|\ln A|^{\text{fudge}}+|\ln B|^{\text{fudge}})^{(1/{\text{fudge})}}))$$

comes pretty close, where "fudge" is roughly 2.0, and the inputs are <= 1. However, it obviously doesn't work right for values of A and B > 1.0, and my crude attempts to extend it didn't produce a very smooth function. [I found out later that the function was reasonably smooth, only Excel was plotting it strangely due to the way I generated the input file.]

(exp, pow, abs, and log are all the mathematical functions you'd expect from their names.)

So: Is there an obvious mathematical formula that provides the desired characteristics?

Background:

This function is used to combine "adjustment factors" used to correct for the interdependence of observations in a Bayesian inference. In each step of the Bayesian calculations, a multiplier consisting of the conditional probability divided by the marginal probability is generated. That multiplier is "adjusted" by raising it to the power of the combined (using the sought-after formula) "adjustment factors" reaching it from previous terms in the equation.

(I'm not here to have this adjustment scheme critiqued or debated, I'm just giving this for background.)

I plotted, in Excel, the results (for fudge = 2.0, alpha = 1.0) from Didier Piau's scheme, close as I could understand it. (Sorry, Didier, if it's not correct.) It came out looking the same as the kludge I had cooked up earlier, though that earlier version occupied about 3 times as much code.

Since I don't understand Excel plotting very well the X axis is screwed up -- it represents B values times 10. The Y axis is the function value. (The curve fitting may be wrong in spots -- it's Excel's default.)

It certainly looks a bit peculiar, but I can't see anything specific that's wrong with it.

Musing: It kinda seems to me like there would be a name for this sort of function. It's kind of a "mean", or maybe a "product". I tried looking up "logarithmic mean", but that turned out to be something else.

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I am curious what do you want for f(0.5, 1.5)? when one argument is lower than 1 and the other one higher than 1? – Alice Aug 9 '11 at 16:58
@Alice -- Generally I'd expect a value between the two input values in that case. – Daniel R Hicks Aug 9 '11 at 17:07
Taking the absolute values of the logs of A and B means your output will be the same for A and 1/A. – Ross Millikan Aug 9 '11 at 17:28
@Ross Millikan -- Correct, which is why it doesn't extend nicely above 1.0. – Daniel R Hicks Aug 9 '11 at 17:51
@Daniel, nice simulations (and nice question, by the way). – Did Aug 10 '11 at 1:28

A solution is to fix a positive parameter $a$ and to define $x*y$ for every positive $x$ and $y$, as the unique positive number $z$ such that $$\log(z)|\log(z)|^a=\log(x)|\log(x)|^a+\log(y)|\log(y)|^a.$$ One sees that:

Commutativity, associativity, the fact that $1*x=x*1=x$ for every $x$ are obvious. Furthermore, for every $y>1$, $x*y>x$ and for every $y<1$, $x*y<x$, and $(1/x)*(1/y)=1/(x*y)$, hence all there remains to check is that $x*y<xy$ for $x$ and $y$ greater than $1$. This reads as the fact that for every $x>1$ and $y>1$, $$\log(x)|\log(x)|^a+\log(y)|\log(y)|^a<\log(xy)|\log(xy)|^a.$$ But $$\log(xy)|\log(xy)|^a=\log(x)|\log(xy)|^a+\log(y)|\log(xy)|^a,$$ and $$|\log(x)|^a<|\log(xy)|^a,\qquad |\log(y)|^a<|\log(xy)|^a,$$ and the logarithms are positive, hence the inequality holds.

Note that $x*x=x^b$ for every $x$, with $b=2^{1/(1+a)}$, hence $b$ may be any exponent between $1$ and $2$, for example, if $a=.7$, $b=1.5034$ and $.5*.5=.3527$.

More generally, one could define $z$ as the unique solution of the equation $u(z)=u(x)u(y)$ for a given increasing positive function $u$ such that $u(1/x)=1/u(x)$ for every positive $x$, and such that $u(x)u(y)<u(xy)$ for every $x$ and $y$ greater than $1$.

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This is a very nice solution! However, it has infinite gradient at the points $xy = 1$, because the right-hand side of your first equation crosses zero and you have to take the $(1+a)$th root of that to get $\log z$. – Rahul Aug 9 '11 at 20:32
@Rahul: yes, as may be guessed from the Excel plot the OP added to the question. – Did Aug 17 '11 at 21:59
I later realized that one can easily get rid of that issue by replacing $\log x \lvert \log x \rvert^a$ by some other increasing function of $\log x$ that has positive slope everywhere. Now I'd call it unconditionally a very nice solution :) – Rahul Aug 17 '11 at 22:28

How about $f(a,b)=(ab)^p$, where $p$ is chosen less than $1$ for the strength you want. Maybe $3/4$ would suit your needs.

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Not associative. (In case it's relevant.) – anon Aug 9 '11 at 17:14
Darn -- left out the requirement that f(1.0, N) = N. (I've added that above -- sorry.) – Daniel R Hicks Aug 9 '11 at 17:15