Approximating $\ln(1+\exp(x)+\exp(y))$ Ok, so I know that
$\ln(1+e^x)\approx x$ when $x$ is large.
But what about
$\ln(1+e^x+e^y)$ when both $x$ and $y$ are large?
I can figure out cases when $x\gg y$ or $y\gg x$ since that simplifies to the earlier case. But, not the case when both are large. Are there any approximations in this example?

One can certainly write for large $x, y$ that $$\ln(1 + e^x + e^y) \approx \ln(e^x + e^y) = \ln(e^x) \ln(1 + e^{y - x}) = x + \ln(1 + e^{y - x}),$$ which quickly leads to the given formulas when $x$ and $y$ are far apart.
In the limit where they are close, that is when $\epsilon := y - x$ is close to zero, we have that
$$\ln(1 + e^{\epsilon}) = \ln \left[1 + \left(1 + \epsilon + \frac{1}{2} \epsilon^2 + O(\epsilon^3)\right)\right] = \ln 2 + \frac{1}{2} \epsilon + \frac{1}{8} \epsilon^2 + O(\epsilon^3),$$ and so for $x, y \gg 0$ we have
$$\ln(1 + e^x + e^y) \approx \ln2 + \frac{1}{2} (x + y) + \frac{1}{8} (x - y)^2 + O((x - y)^3).$$ (It turns out that the part of the approximation that is polynomial in $(x - y)$ is even, and in particular we can replace $O((x - y)^3)$ with $O((x - y)^4)$.) In the limit $y \to x$ (i.e., $\epsilon \to 0$) this is just $\log 2 + x$, recovering vadim123's observation in his comment.
