I can figure out cases when \$xgg y\$ or \$ygg 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.

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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 + frac12 epsilon^2 + O(epsilon^3) ight) ight> = ln 2 + frac12 epsilon + frac18 epsilon^2 + O(epsilon^3),\$\$ and so for \$x, y gg 0\$ we have\$\$ln(1 + e^x + e^y) approx ln2 + frac12 (x + y) + frac18 (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 o x\$ (i.e., \$epsilon o 0\$) this is just \$log 2 + x\$, recovering vadim123"s observation in his comment.

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edited Mar 7, 2015 at 1:21
answered Mar 6, 2015 at 5:49

Travis WillseTravis Willse
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We will break this into 3 parts.

Part \$1\$ provides derivation of an series expansion for \$log(1+e^x+e^y)\$ in terms of powers of \$e^-left(x-y ight)\$, where it is assumed that \$e^x>1+e^y\$. This is equivalent to assuming that \$x-y>logleft(1+e^-y ight)\$

In Part \$2\$, we develop a series converges rapidly when \$x-y\$ is adequately small. This development is motivated since the series in Part 1 converges slowly when \$x-y\$ is small.

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Part \$3\$ provides upper and lower bounds that are global (i.e., We impose no conditions on \$x\$ or \$y\$.).

PART \$1\$:

Without loss of generality, assume \$x>y\$. Let \$M=e^x\$ and \$m=e^y\$ and further assume that \$m+10\$.

Part \$2\$ considers the case in which \$x-y\$ is "small."

PART \$2\$:

To address the case for which \$0>1\$ so that \$w=e^-x

PART \$3\$:We may also obtain crude bounds as follows. For the upper bound we have

\$\$eginalign1+e^x+e^y&=(e^x+e^y)left(1+frac1e^x+e^y ight)\\&y\$, this upper bound becomes

\$\$logleft(1+e^x+e^y ight)lelog(2)+x+frac12e^-y\$\$

For the lower bound, we have

\$\$eginalign1+e^x+e^y&=(e^x+e^y)left(1+frac1e^x+e^y ight)\\&ge2sqrt(e^xe^y)left(1+frac1e^x+e^y ight)endalign\$\$

Now use (i) the series for \$log(1+x)\$ and (ii) the inequality for the geometric and arithmetic means to reveal

\$\$eginalignlog(1+e^x+e^y)&ge log(2)+frac12left(x+y ight)+frac1e^x+e^y-frac12left(frac1e^x+e^y ight)^2\\&gelogleft(2 ight)+frac12left(x+y ight)+frac12minleft(e^-x,e^-y ight)-frac18 maxleft(e^-2x,e^-2y ight)endalign\$\$

Under the assumption that \$x>y\$, this lower bound becomes

\$\$log(1+e^x+e^y)gelogleft(2 ight)+frac12left(x+y ight)+frac12e^-x-frac18 e^-2y\$\$