I'm looking for two exponents cancelling each other. Let $0 < a < 1$ be the first and $b > 1$ be the other. If I calculate $x^b$ and then raise the result to the power $a$, getting the original $x$ is easy enough. $a$ must be $\frac{1}{b}$.

Let's generalise a bit. Is there an exponent $a$ that gives the original $x$ after raising $x^b$ to the power $n$ times? If so, it must satisfy the following equation:


The solution of this equation is:

$$a=\left(\frac{\ln x}{b\ln x}\right)^\frac{1}{n}=\frac{1}{b}^\frac{1}{n}$$

Now I'm applying this to a machine learning algorithm that I'd want to behave 'nicely'. The algorithm is to predict actions based on learned behavior. There are $n$ available actions. The weight of an individual action is 0.5 for a start. Now only doing one action over and over again reinforces that cell and punishes others, but the 'nice' behavior would be that if all actions are performed once, all weights remain at 0.5. In addition, the more actions available, the system ought to adjust the weights slower. So:


$$a=\left(\frac{\ln x}{\frac{n+1}{n}\cdot\ln x}\right)^\frac{1}{n-1}=\left(\frac{n}{n+1}\right)^\frac{1}{n-1}$$


The weights don't stay the same. Granted - they are equal after a round, but all decrease. So I'm lost. The problem must be with my reasoning above, but I can't find where. Any insight?

Read the picture as a state machine, state transition from row number to column number. So the first row should be all 0.5's. $n=4$ and one round was performed.



1 Answer 1


Turns out I had the concepts of reward and punishment backwards. For values between 0 and 1 an exponent greater than 1 obviously punishes a weight. In the question I was finding a repeat punishment for a reward.

A suitable reward could be $\frac{n}{n+1}$ which leads to a punishment of:


Correct weights


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