If $0<\delta<1$ and $x>1$ then what is the smallest $\ n$ such that $( x+\delta)^n - x^n \geq 1$ in terms of $\ x\ $ and $\ \delta\ ?$

If $\ 0<\delta<1\ $ and $\ x>1,\ $ then what is the smallest $\ n\in\mathbb{N}\ $ such that $\ \left( x+\delta\right)^n - x^n \geq 1,\ $ in terms of $\ x\ $ and $\ \delta\ ?\ $ Is there a known function $\ f:(x,\delta)\to \mathbb{N}\ ?$

I don't see how we can avoid using $\ n\ $ to answer this question - for example, any attempt with the Binomial expansion or the identity $\ (x-y)(x^{n-1} + x^{n-2} y + \ldots + x y^{n-2} + y^{n-1})\ $ involves $\ n.\ $ However, clearly such a function $\ f\ $ does exist - but maybe it has no nice form? But maybe there is a nice form with the ceiling/floor function, or splitting it into cases? But I don't see a straightforward way to do any of these.

Solution 1:

Not a complete answer. I give upper and lower bounds which seem to be good enough to determine the answer for "many" $x$ and $\delta$ but fail in certain situations.

Let $m$ be the smallest natural number with this property. Clearly, $m\ge 2$. Consider the function $f(t)=t^m$ and apply the mean value theorem on the interval $[x,x+\delta]$: $$\frac{(x+\delta)^m-x^m}{\delta}=ma^{m-1} $$ for some $a\in(x,x+\delta)$. Using $(x+\delta)^m-x^m\ge 1$ and $a<x+\delta$, we get $$ma^{m-1}\delta\ge 1\\ \Rightarrow m(x+\delta)^{m-1}\delta>1\\ \Leftrightarrow m(x+\delta)^m>\frac{x+\delta}{\delta}\\ \Leftrightarrow m\log(x+\delta)(x+\delta)^m>\frac{\log(x+\delta)(x+\delta)}{\delta} $$ Applying the principal branch of the Lambert W function (which is strictly increasing), we get $$m>\frac{W\left(\frac{\log(x+\delta)(x+\delta)}{\delta}\right)}{\log(x+\delta)} $$ You can get an upper bound by doing the same thing with the function $g(t)=t^{m-1}$. The difference is that we have, by the choice of $m$, $(x+\delta)^{m-1}-x^{m-1}<1$. We obtain $$\frac{W\left(\frac{(x+\delta)\log(x+\delta)}{\delta}\right)}{\log(x+\delta)}<m<\frac{W\left(\frac{x\log(x)}{\delta}\right)}{\log(x)}+1 $$ After some experimentation, it looks that this interval contains at most 2 integers in "many" cases for $x$ and $\delta$, so the answer for $m$ in those cases is either the floor of the upper bound or the ceiling of the lower bound. Specifically, that seems to be the case when $x\ge A$ and $1>\delta \ge B$ where $A>1$ and $B>0$ are appropriately chosen. However, if $(x,\delta)$ approaches $(1,0)$ in a certain way, then this interval becomes unbounded, so those bounds are useless for determining $m$.