Any tight lower bound for $ (1+\frac{w}{d})^{\frac{d}{w} }$, where $d \gg w$ and both are constants? $\lim_{d \to \infty} (1+\frac{w}{d})^{\frac{d}{w} } = e$. But, what if the number of bits used to encode $d$ is polynomial in length. In this model, infinity can't be encoded. However, $d$ is polynomialy  much larger than $w$.
Is there any tight lower bound, a closed form function $f(d)$ such that
$$ f(d) \le (1+\frac{w}{d})^{\frac{d}{w} }$$
 A: Let $x = d/w$. Since $d \ll w$, we have $x \ll 1$. Therefore the Taylor expansion of $(1+x)^{1/x}$ should offer a good approximation to the function:
$$ (1+x)^{1/x} = e \left[1 - \frac{1}{2} x + \frac{11}{24} x^2 - \frac{7}{16} x^3 + \frac{2447}{5760} x^4 - \frac{959}{2304} x^5 + O(x^6) \right]. $$
In particular, we have the following lower bounds:
$$ e \left[1 - \frac{1}{2} x\right], e \left[1 - \frac{1}{2} x + \frac{11}{24} x^2 - \frac{7}{16} x^3\right], \ldots$$
If you're interested in the entire Taylor expansion, have a look here, here and here.
A: For every fixed positive $d$, $(1+w/d)^{d/w}\to1$ when $w\to+\infty$ while $(1+w/d)^{d/w}\gt1$ for every positive $w$. Hence the best lower bound of $(1+w/d)^{d/w}$ valid for every positive $w$ (or only for every $w$ large enough) is $f(d)=1$.
A: By expanding the left-hand side as a power series one can show that
$$
\left(1+\frac{1}{x}\right)^x > e - \frac{e}{2x}
$$
for $x > 0$, where the approximation gets better as $x$ gets larger.  By setting $x = d/w$ we get
$$
\left(1 + \frac{w}{d}\right)^{d/w} > e - \frac{ew}{2d}
$$
for $d/w > 0$.
