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Let $n \in N$. Fix $m \in [-n,n]$. I am curious, how to bound from above the following expression $$ (n-m)^{\frac{n-m}{2}+1}(n+m)^{\frac{n+m+1}{2}}\leq \quad ? $$

Thank you.

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up vote 1 down vote accepted

$$ (n-m)^{\frac{n-m}{2}+1}(n+m)^{\frac{n+m+1}{2}} = n^{n+3/2}\exp\left(\frac{n-m+2}{2}\log(1+m/n)+\frac{n+m+1}{2}\log(1-m/n))\right) \leq n^{n+3/2}\exp\left(\frac{m(n-m+2)}{2n}-\frac{m(n+m+1)}{2n}\right) = n^{n+3/2} \exp\left(\frac{m-2m^2}{2n}\right) \leq n^{n+3/2} \exp\left(\frac{1}{16n}\right),$$

and from the Taylor series of $\log(1+x)$ and $\log(1-x)$ in a neighbourhood of zero it is not difficult to derive lower bounds, too.

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$(n-m)^{\frac{n-m}{2}+1}(n+m)^{\frac{n+m+1}{2}}\leq (2n)^{n+1}(2n)^{n+1/2}=(2n)^{2n+3/2}$

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Thank you, but I would like to get more accurate bound. I was trying to maximize this expression, but it turns out that there is no global maximum... Also, I've realized I did not mentioned that $m$ is fixed. Sorry. – Alex Jan 29 '13 at 0:06

I have had good luck tightening bounds by finding the form $(a-b)(a+b)$ which you already have. $$a^2 - b^2 \le a^2 - b^2 \le a^2$$ is a much better bound than the naive $$(a-b)^2 \le (a-b)(a+b) \le (a+b)^2$$ (with $b \lt a$)

So for your problem, I would take as many factors of $n^2 - m^2$ as I can. Which exponent is larger switches at $m=\frac12$ so treat each interval separately.

For $m \ge \frac 12$ we can break your expression as follows: $$ (n-m)^{\frac{n-m+2}2} (n+m)^{\frac{n-m+2}2} (n+m)^{m-\frac12} $$ which is $$ (n^2-m^2)^{\frac{n-m+2}2} (n+m)^{m-\frac12} $$ To simplify the powers I will only eliminate $m$ from the quantities in parentheses. The previous expression is less than: $$ (n^2)^{\frac{n-m+2}2} (2n)^{m-\frac12} $$ which equals $$ n^{n-m+2} 2^{m-\frac12} n^{m-\frac12} $$ and $$ 2^{m-\frac12} n^{n+\frac32} $$

You can complete the same steps for $m \le \frac12$ to obtain a similar expression, then combine these for the overall upper bound.

Edit: with the correction I just made it appears that both cases produce the same expression, so the result above should be the overall upper bound.

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