Growth of the "cube of square root" function - MathOverflow

Growth of the "cube of square root" function

2010-03-15T06:17:46Z

Hello all, this question is a variant (and probably a more difficult one) of a (promptly answered ) question that I asked here, at http://mathoverflow.net/questions/18054/is-it-true-that-all-the-irrational-power-functions-are-almost-polynomial.

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 $n=100000$.

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.

Answer by S. Carnahan for Growth of the "cube of square root" function

2010-03-15T07:44:43Z

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

$f(n) = t^3 + (3/2)tr + \lfloor (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)r^4/t^5 - \dots \rfloor$
$f(n+1) = t^3 + (3/2)t(r+1) + \lfloor (3/8)(r+1)^2/t - (1/16)(r+1)^3/t^3 + (3/128)(r+1)^4/t^5 - \dots \rfloor$
$f(n+2) = t^3 + (3/2)t(r+2) + \lfloor (3/8)(r+2)^2/t - (1/16)(r+2)^3/t^3 + (3/128)(r+2)^4/t^5 - \dots \rfloor$.

$g(n)$ then has no contributions from the first two terms of each series, since they cancel. Therefore:

$g(n) = \lfloor (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)(r+1)^4/t^5 - \dots \rfloor +$ $\qquad + 2\lfloor (3/8)r^2/t + (3/4)r/t - (1/16)(r+1)^3/t^3 + (3/8t) + (3/128)(r+2)^4/t^5 - \dots \rfloor +$ $\qquad + \lfloor (3/8)r^2/t + (3/2)r/t - (1/16)(r+2)^3/t^3 + (3/2t) + (3/128)(r+2)^4/t^5 - \dots \rfloor$.

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.

Answer by Sergei Ivanov for Growth of the "cube of square root" function

2010-03-15T11:36:05Z

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.

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 $\frac{const}{\sqrt{n}}$).

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 $(n+1)^{3/2}-n^{3/2}<m$ and $(n+2)^{3/2}-(n+1)^{3/2}>m$ . Observe that $$ \frac{3}{2}\sqrt{n}<(n+1)^{3/2} - n^{3/2} < \frac{3}{2}\sqrt{n+1} $$ (the bounds are just the bounds for the derivative of $x^{3/2}$ on $[n,n+1]$. Therefore $$ \frac{3}{2}\sqrt{n} < m < \frac{3}{2}\sqrt{n+2} $$ or, equivalently, $$ n < \frac49 m^2 < n+2. $$ 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 $|r|<1$. From Taylor expansion $$ f(x+r) = f(x) +r f'(x) +\frac12 r^2 f''(x+r_1), \ \ 0<r_1<r, $$ 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 $0<\delta<\frac1{4m}$. This cannot be close to an integer because it is close (from above) to a fraction with denominator 27.