Posts

Showing posts from August 6, 2018

Complex set of linear equations

Image
Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite I'm trying to find $beta$'s which solve the following problem: $sum Vi beta i = V$ Or that at least minimize ($sum Vi beta i - V)$ where $Vi$ are vectors. Additionally, there are a few other constraints/properties for these numbers. $sum beta i =1 $ and $beta i gt 0$ Also, the sum of the vector components is also 1. Another formulation is imagine you have a population of Things. For each one of these, you can apply some random input (from a set of 6 inputs, not necessarily equally likely), and you get some output that depends solely on the input, given the Thing. I need to find the best distribution of inputs so that, if I apply them to the entire population, my output match some control totals that are pre-established. In this case though, I don't need to exactly match everything, but find the solution which gives the closest (or close enough) set of outputs to my controls. I'm not really sure whe...

Which functions satisfy $f^n(x) = f(x)^n$ for some $n ge 2$?

Image
Clash Royale CLAN TAG #URR8PPP up vote 30 down vote favorite 13 Let $n$ be an integer greater than $1$. The notation $f^n$ is notoriously ambiguous: it means either the $n$-th iterate of $f$ or its $n$-th power. I was wondering when the two interpretations are in fact the same. In other words, if we write $f^n(x)$ for $f(f(dotsb f(x) dotsb))$ and $f(x)^n$ for $f(x) cdot f(x) dotsb f(x)$: For which functions $f colon I to mathbb R$ is $f^n(x) = f(x)^n$ for all $x in I$? So far I have been able to show that: The constant functions $f(x) = 0$ and $f(x) = 1$ satisfy the condition for all $n$ and $f(x) = -1$ satisfies the condition for all odd $n$. Also, if $f$ satisfies the condition for $n$, then the only fixed points of $f$ can be either $0$, $1$, or if $n$ is odd also $-1$. The squaring function $f(x) = x^2$ satisfies the condition for $n = 2$ and is essentially the only non-constant differentiable function to do so. Indeed, $$f(f(x)) = f(x)^2 implies f'(f(x)) f'(x...