Growth of the "cube of square root" function - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:08:27Z http://mathoverflow.net/feeds/question/18246 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18246/growth-of-the-cube-of-square-root-function Growth of the "cube of square root" function Ewan Delanoy 2010-03-15T06:17:46Z 2010-03-15T12:58:21Z <p>Hello all, this question is a variant (and probably a more difficult one) of a (promptly answered ) question that I asked here, at <a href="http://mathoverflow.net/questions/18054/is-it-true-that-all-the-irrational-power-functions-are-almost-polynomial" rel="nofollow">http://mathoverflow.net/questions/18054/is-it-true-that-all-the-irrational-power-functions-are-almost-polynomial</a>.</p> <p>For $n\geq 1$, let $f(n)$ denote the "integer part" (largest integer below) $n^{\frac{3}{2}}$, and let $g(n)=f(n+2)-2f(n+1)+f(n)$. Question : Is it true that $g(n)$ is always in $\lbrace -1,0,1\rbrace$ (excepting the initial value $g(1)=2$) ? I checked this up to $n=100000$.</p> <p>It is not too hard to check that if $t$ is large enough compared to $r$ (say $t\geq \frac{3r^2+1}{4}$), $f(t^2+r)$ is exactly $t^3+\frac{3rt}{2}$ (or $t^3+\frac{3rt-1}{2}$ if $t$ and $r$ are both odd ) and similarly $f(t^2-r)$ is exactly $t^3-\frac{3rt}{2}$ (or $t^3-\frac{3rt+1}{2}$ if $t$ and $r$ are both odd ). So we already know that the answer is "yes" for most of the numbers.</p> http://mathoverflow.net/questions/18246/growth-of-the-cube-of-square-root-function/18255#18255 Answer by S. Carnahan for Growth of the "cube of square root" function S. Carnahan 2010-03-15T07:44:43Z 2010-03-15T07:44:43Z <p>It's not too hard to put a bound on the size of second differences (since without the truncation, they are bounded above by a constant times $n^{-1/2}$), but getting the bound down to one seems delicate. It looks like it can be done with mindless brute force, though. I won't write all of the cases, but here is a start. Write $n = t^2 + r$, for integers $t,r$ satisfying $|r| \leq t$. The binomial theorem implies $n^{3/2} = t^3 + (3/2)tr + (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)r^4/t^5 - \dots$. I'll look at the case where $t$ is even. Then</p> <ul> <li>$f(n) = t^3 + (3/2)tr + \lfloor (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)r^4/t^5 - \dots \rfloor$</li> <li>$f(n+1) = t^3 + (3/2)t(r+1) + \lfloor (3/8)(r+1)^2/t - (1/16)(r+1)^3/t^3 + (3/128)(r+1)^4/t^5 - \dots \rfloor$</li> <li>$f(n+2) = t^3 + (3/2)t(r+2) + \lfloor (3/8)(r+2)^2/t - (1/16)(r+2)^3/t^3 + (3/128)(r+2)^4/t^5 - \dots \rfloor$.</li> </ul> <p>$g(n)$ then has no contributions from the first two terms of each series, since they cancel. Therefore:</p> <p>$g(n) = \lfloor (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)(r+1)^4/t^5 - \dots \rfloor +$ $\qquad + 2\lfloor (3/8)r^2/t + (3/4)r/t - (1/16)(r+1)^3/t^3 + (3/8t) + (3/128)(r+2)^4/t^5 - \dots \rfloor +$ $\qquad + \lfloor (3/8)r^2/t + (3/2)r/t - (1/16)(r+2)^3/t^3 + (3/2t) + (3/128)(r+2)^4/t^5 - \dots \rfloor$.</p> <p>At this point, you can break the analysis into more cases involving the fractional part of $(3/8)r^2/t$ and the size of $r/t$, and then invoke some estimates about the remaining parts of the sum.</p> http://mathoverflow.net/questions/18246/growth-of-the-cube-of-square-root-function/18260#18260 Answer by Sergei Ivanov for Growth of the "cube of square root" function Sergei Ivanov 2010-03-15T11:36:05Z 2010-03-15T12:58:21Z <p>Here is a proof that $|g(n)|\le 1$ for all but finitely many $n$. You can extract an explicit bound for $n$ from the argument and check the smaller values by hand.</p> <p>If $f(n)=n^{3/2}$ without the floor, then $g(n)\sim \frac{3}{4\sqrt n}$, so it is positive and tends to 0. When you replace $n^{3/2}$ by its floor, $g(n)$ changes by at most 2, hence the only chance for failure is to have $g(n)=2$ when the fractional parts of $n^{3/2}$ and $(n+2)^{3/2}$ are very small and the fractional part of $(n+1)^{3/2}$ is very close to 1 (the difference is less than $\frac{const}{\sqrt{n}}$).</p> <p>Let $a,b,c$ denote the nearest integers to $n^{3/2}$, $(n+1)^{3/2}$ and $(n+2)^{3/2}$. Then $c-2b+a=0$ because it is an integer very close to $(n+2)^{3/2}-2(n+1)^{3/2}+n^{3/2}$. Denote $m=c-b=b-a$. Then <code>$(n+1)^{3/2}-n^{3/2}&lt;m$</code> and <code>$(n+2)^{3/2}-(n+1)^{3/2}&gt;m$</code>. Observe that $$\frac{3}{2}\sqrt{n}&lt;(n+1)^{3/2} - n^{3/2} &lt; \frac{3}{2}\sqrt{n+1}$$ (the bounds are just the bounds for the derivative of $x^{3/2}$ on $[n,n+1]$. Therefore $$\frac{3}{2}\sqrt{n} &lt; m &lt; \frac{3}{2}\sqrt{n+2}$$ or, equivalently, $$n &lt; \frac49 m^2 &lt; n+2.$$ If $m$ is a multiple of 3, this inequality implies that $n+1=\frac49 m^2=(\frac23m)^2$, then $(n+1)^{3/2}=(\frac23m)^3$ is an integer and not slightly smaller than an integer as it should be. If $m$ is not divisible by 3, then $$n+1 = \frac49 m^2 + r$$ where $r$ is a fraction with denominator 9 and $|r|&lt;1$. From Taylor expansion <code>$$f(x+r) = f(x) +r f'(x) +\frac12 r^2 f''(x+r_1), \ \ 0&lt;r_1&lt;r,$$</code> for $f(x)=x^{3/2}$, we have $$(n+1)^{3/2} = (\frac49 m^2 + r)^{3/2} = \frac8{27}m^3 + mr + \delta$$ where $0&lt;\delta&lt;\frac1{4m}$. This cannot be close to an integer because it is close (from above) to a fraction with denominator 27.</p>