For polynomials $p(x)=x^n$ and $q(x)=x^n-x^{n-1}$, let $z(x)=\min_{a\in\mathbb{R}}\sqrt{(x-a)^2+\big(p(x)-q(a)\big)^2}$ be the distance between the point $\langle x,p(x)\rangle$ and the nearest point on the graph of $q(\cdot)$. I've noticed that it looks like this distance approaches a finite limit; i.e., $\displaystyle\lim_{x\to\infty}z(x)=c$ for some $c$. Is this true and, if so, how can I find $c$? https://www.desmos.com/calculator/foqvlfqyhh
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Do you really want that minus sign after $(x-a)^2$? Seems to me that a distance would have a plus here, but maybe you've got some other idea. – Matthew Conroy Jun 13 '17 at 01:49
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thanks for catching that! – Jacob Claassen Jun 13 '17 at 01:55
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1It is not clear to me what you are asking. If $p(x) \neq q(a)$, then $z>0$. Is there something else you want to show? – Matthew Conroy Jun 13 '17 at 02:00
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well, I'm not really sure how to describe it mathematically, and I'm hella tired right now, but basically, the two functions stay together but don't ever have $0$ distance between them(like how $x^2$, and $x^2-1$ do as they approach an infinity) and they never have infinite distance. – Jacob Claassen Jun 13 '17 at 02:07
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I don't know what you mean by "they never have infinite distance". Any pair of points have a finite distance between them. Are you familiar with the concept of limits? Perhaps the question you want to ask involves limits. – Matthew Conroy Jun 13 '17 at 02:14
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I'm not sure how to state it. I'm comfortable with limits, just not applications. Maybe $0<\lim_{x\to\infty}\min(z)<\infty$? – Jacob Claassen Jun 13 '17 at 02:21
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Let us continue this discussion in chat. – Matthew Conroy Jun 13 '17 at 02:22
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1I've offered a pretty substantial edit for clarity; please let me know if you think this matches the spirit of your initial question. (It's simpler to ask about the limit on one side, and the limit on the other will easily fall out of the same analysis.) – Steven Stadnicki Jun 13 '17 at 23:47
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I agree, that's much better, thank you! – Jacob Claassen Jun 13 '17 at 23:51
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While finding the exact value of the limit is a little bit trickier, it's straightforward to show that such a limit exists: we can show that the value $x'$ with $q(x')=p(x)$ is within a bounded distance of $x$, and this clearly provides an upper bound on the distance-between-graphs requested here. Indeed, just let $w=x+1$; then $q(w)=(x+1)^n-(x+1)^{n-1}$ $=x^n+nx^{n-1}+O(x^{n-2})-(x^{n-1}+O(x^{n-2}))$ $=x^n+(n-1)x^{n-1}+O(x^{n-2})$. Since $n\gt 1$, this will be $\gt x^n$ for all sufficiently large $x$. But since $q(x)\lt p(x)$ and $q(w)\gt p(x)$, there's some $\zeta\in[x,w]$ with $q(\zeta)=p(x)$. But this means that $z(x) \leq \sqrt{(\zeta-x)^2+(q(\zeta)-p(x))^2}$ $=\sqrt{(\zeta-x)^2}=|\zeta-x|\leq|w-x|=1$ and so $z(x)\leq 1$ for all sufficiently large $x$.
In fact, if we set $w=x+t$ here (with $t$ to be determined) and run the same analysis, we get $q(w) = x^n+ntx^{n-1}-x^{n-1}+O(x^{n-2})$, so for $t\approx \frac1n$ (plus smaller terms) we already have $q(x+t)\approx p(x)$. I suspect that you'll find the exact value of your constant $c$ is comparable to the horizontal distance, $\frac1n+O(n^{-2})$, but it might be hard to get a precise handle on the higher-order terms.
Steven Stadnicki
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