# Shannon Entropy Minimization

The Shannon Entropy for an observation is given by $-x \log_2(x)$. Why is the maximum entropy achieved at $x = \frac{1}{e}$, and not at $x = 0$? Could someone provide a logical explanation that justifies the mathematics?

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There is no entropy for "an observation". And I wonder how you got that it's maximized at $x=1/e$ (?) –  leonbloy Apr 3 at 15:52
It is true that the maximum of $-xlog_2(x)$ is at $x=1/e$, as you can see by differentiating/graphing. But I don't see any relevance of that to entropy. –  Martin Leslie Apr 3 at 16:19

Shannon Entropy, given a discrete probability distribution with probabilities $p_1,p_2,\dots,p_n$ is $$-\sum_i p_i\log_2(p_i).$$
This is not equal to the quantity you gave. If $n=1$ then we must have $p_1=1$ and then the Shannon Entropy is 0.
A relevant example is to figure out what $p_1$ maximizes Shannon entropy when $n=2$. This can be thought of as what coin (not necessarily fair) gives you the most information when you flip it.