? operator

I was wondering (not for the first time) what comes next in the progression a^b, a\times b, a+ba?b
Probably this is a thing already. Anyway…
What is a?b ?
Well, my idea was that \ln ab=\ln a+\ln b, so \ln (a+b)=\ln a ? \ln b
So then a?b=\ln(e^a+a^b). Sounds reasonable. (and silly.)
Hmm so what’s the “I”, like a\times 1=a and a+0=a?
a?x=a \to \ln(e^a+e^x)=a \to e^x=0 \to x=\ln 0 \to x=-\infty
Hmm interesting hehe.
And a?a ?
2e^a=e^x\to x=\ln 2e^a =a+\ln 2
Hmm so a?0=\ln 2.. that’s weird.
Turns out that a?b is, for a<b, very close to b.
(And obviously for a>b, very close to a.)
It seems once a<\frac{4}{5}b, a?b is extremely close to b.
So it’s almost a max function! (just equals the higher value) How strange.
Except if a=b then a?a=a+\ln 2.
And a?0 is \ln(e^a+1)
e.g. 10?0 is 10.0000453988992\dots
1?0=1.31326168751822\dots
Well, I guess 10+0>10\times 0, so it’s only right that 10?0 be >10.


And going the other way, a+b\to a\times b\to a^b – why is a^b not symmetrical in a and b like + and \times? That doesn’t seem right.
So (you guessed it).. maybe call this \circ.

    \begin{align*} \ln(ab)&=\ln a+\ln b \\ \ln(a\circ b)&= \ln a \times \ln b \\ a\circ b &= e^{\ln a \ln b} \\ 2\circ 3 &\approx 2.1414?! \\ 100\circ 100 &= 1623081665.62\dots \\ &= e^{(\ln 100)^2} \\ \intertext{So, find I.} a\circ x &= a \\ e^{\ln a \ln x}&=a \\ \ln x \ln a&=\ln a \\ \ln x&=1\\ x &= e \to a\circ e=a\\  \text{and }e\circ e&=e \\ \text{hmm but } 1\circ \text{ anything } =e^{0}=1 \end{align*}

operator1


And the next one up? (and down?!)
Call the next one up \infty maybe.. a\infty b

    \begin{align*} \ln (ab)&=\ln a+\ln b \\ \ln(a\circ b)&=\ln a \ln b \\ \ln(a\infty b)&= \ln a \circ \ln b \\ a\infty b &= e^{\ln a \circ \ln b} \\ &= e^{e^{\ln \ln a \ln \ln b}} \\ 2\infty 3 &\approx 2.62772268\dots \\ 1000\infty 1000 &\approx 1.562581\times 10^{18} \\ &= e^{e^{\ln \ln 1000)^2}} \end{align*}

So, find I.

    \begin{align*} a\infty x &= a \\ e^{e^{\ln \ln a \ln \ln x}}&=a \\ \ln \ln x \ln \ln a&=\ln \ln a \\ \ln \ln x&=1\\ x &= e^e \to a\infty e^e=a\\  \text{and }e\infty e^e&=e^{e^{0}}=e \\ \text{hmm but} $e\infty$ \text{ anything }&=e \end{align*}

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