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.
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 $y$.).
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."
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
$$eginalign1+e^x+e^y&=(e^x+e^y)left(1+frac1e^x+e^y ight)\\&y$, this upper bound becomes
For the lower bound, we have
$$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$$