How to solve this product recursion?

Question

The recursion is given by $$T(N)=2((\log N)T(N^{3/8})T(N^{1/4}))^2$$ $$T(N)=1\mbox{ if }M<1.$$ Is there a good upper bound?

Answer 1

No, there are no upper bounds on this function. In fact, either there are infinite families of functions satisfying this equation, or there are no solutions at all. Define the equivalence relation $\sim$ on real numbers by $x\sim y$ if there is exist $n,m \in \Bbb N$ such that $x^n = y^m$. The equivalence classes of this relation are countable, so there must be an uncountable number of such classes. And it is easy to verify that an infinite number of distinct equivalence classes contain natural numbers.

Yet $T$ can be defined independently on each equivalence class, as $N, N^{3/8}, N^{1/4}$ are all equivalent to each other. (In fact, there are certainly subclasses of these classes that can be defined independently - I just went with the easiest relation that worked, here.) For each independent set, you can choose an arbitrary $x$ for the set, choose an arbitrary value for $T(x)$ and figure out what values of $T(y)$ for $y$ in the same set are compatible with it (you will likely find other arbitrary choices are required to fully define $T$ even on that set).

As a result of being able to choose $T(x)$ arbitrarily for at least one $x$ in an infinite number of classes, $T$ does not satisfy any definite upper bound. You might be able to make your choices so that a bound exists, but you can also make choices that defy any bound.