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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

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