**Problem** A real sequence satisfies for all , where . Let be the length of the th interval. Show that there exists a unique such that for all .

*Remark * This is a problem that arises from nonlinear quantization in AAC. There, MDCT coefficient is power-compressed to before linear quantization. Then we face the problem of choosing boundaries for the th quantization interval where all are quantized to and de-quantized to , such that the de-quantization noise is minimized statistically. Of course, the solution to this problem depends on the underlying probability distribution of . In case of uniform distribution, the de-quantized value should be at the center of the th interval, or . With this restriction, all the boundaries are uniquely determined by . But the length of the interval , as revealed by numerical experiments, almost always undulates and at the same time, asymptotically increases to as ( ), for a randomly chosen . Then, is there a to make the length of the interval monotonically increase? The answer is yes, and this is uniquely determined by , as shown in the following.

*Proof* Let us denote the difference of th and th intervals’ lengths by . We claim that for all . If so, to ensure for all , then we need to show that there exists a unique such that both and exist and are non-negative. First, by the relation , we have . Furthermore, let , then

$latex \begin{array}{rcl} \Gamma_{n+2}-\Gamma_n

& = & 2(n+3)^\alpha+2(n+2)^\alpha-2(n+1)^\alpha-2n^\alpha-4T_{n+3}+4T_{n+1} \\

& = & 2(n+3)^\alpha-6(n+2)^\alpha+6(n+1)^\alpha-2n^\alpha \\

& = & 2(f(n+2)-2f(n+1)+f(n)) \\

& = & 2f”(\eta_n), \end{array}$

where and the last equation is due to recursive application of Rolle’s theorem (first about , then about about ). Since for when , we have thus . Next, consider the limit of :

$latex \begin{array}{rcl} \Gamma_{2n}+\Gamma_{2n+1}

& = & \Delta_{n+2}-\Delta_n \\

& = & T_{n+3}-T_{n+2}-T_{n+1}+T_n \\

& = & T_{n+3}+T_{n+2}-2T_{n+2}-2T_{n+1}+T_{n+1}+T_n \\

& = & 2[f(n+1)-f(n)] \\

& \xrightarrow{n\to\infty} & 0, \end{array}$

where the last equation is due to for . Thus, the desired , if it ever exits, must ensure . Conversely, if with some , , then for all .

Now, if suffices to show there exists a unique such that . Let , then . We claim that converges as . In that case, we shall have

$latex \begin{array}{rcl} \Gamma_{2n}

& = & 2(2n+1)^\alpha+2(2n)^\alpha-4T_{2n+1} \\

& = & 2(2n+1)^\alpha+2(2n)^\alpha-8S_{2n+1}-4T_1 \\

& = & 2(2n+1)^\alpha+2(2n)^\alpha-4(2n+\tfrac12)^\alpha+4\xi_n-4T_1 \\

& = & 2[g(2n+\tfrac12)-g(2n)]+4\xi_n-4T_1 \\

&\xrightarrow{n\to\infty} & 4\xi_n-4T_1, \end{array}$

where and the last equation above follows from when . Therefore, the unique will be . It remains to show that does converge.

Let us investigate the difference between and :

$latex \begin{array}{rcl} \xi_{n+1}-\xi_n

& = & (2n+\tfrac52)^\alpha-(2n+1)^\alpha-2[(2n+2)^\alpha-(2n+1)^\alpha] \\

& = & f(2n+\tfrac32)+f(2n+\tfrac12)-2f(2n+1) \\

& = & \tfrac14\alpha(\alpha-1)(\alpha-2)\theta_n^{\alpha-3}, \end{array}$

where and the last equation is also due to recursive application of Rolle’s theorem. (First about , then about , and finally about .) Therefore, we have , which implies that converges.

This completes the proof.