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
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 :
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
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 :
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.