How to optimize $p + p^\delta + (1 - p)^\delta$ For an exercise in a book I'm reading I have come across the expression $p + p^\delta + (1 - p)^\delta$ and I need to come up with a value for $p \in [0, 1]$ which minimizes this. With the aid of a computer I've determined that $p \approx \frac{\log \delta}{\delta}$, but I'm not sure how to prove this.
I'm pretty new to this whole bounding and optimization business so I'm guessing I'm not seeing some obvious bound. Thanks.
 A: Let's call the expression you need to minimize $E(p)$ and its optimal value $p^*$. We immediately see that $p^* < 1/2$, and thus $1-p > p$.
Now, $p^*$ will have to satisfy
$$
\left.\frac{dE}{dp}\right|_{p^*} = 1 + \delta\left[{p^*}^{\delta-1} - (1-p^*)^{\delta-1}\right] = 0
$$
This isn't super helpful at the moment because it's not solvable analytically. Still, we can get an idea of the behavior of $p$ by looking at certain limits. Let's start with large $\delta$. In this case, since $1-p^* > p^*$, $(1-p^*)^{\delta-1} \gg {p^*}^{\delta -1}$, giving $\delta(1-p^*)^{\delta-1} \simeq 1$, which can be solved for
$$
p^*\simeq 1 - \delta^{1/(1-\delta)}
$$
As it happens, this is asymptotically equal to $\ln\delta/\delta$, but converges much more quickly to the answer for large $\delta$. That said, $\ln\delta/\delta$ is still pretty good here.
For $\delta$ close to 1, a series expansion in $\delta$ gives $(\delta - 1)\ln[(1-p)/p]\simeq 1$, giving
$$
p^*\simeq [1 +e^{1/(\delta-1)}]^{-1}
$$
This is actually not even asymptotically equal to the answer, as the series expansion in $\delta$ ends up being inconsistent. However, it is a much better approximation than $\ln \delta/\delta$, and more accurate for all $\delta < 2.84$. 
A: If $\lambda \in \mathbb{N} \land \lambda > 0$, then taking the derivative and setting it to zero and solving for $p$ gives the value of $p$ where the expression is minimized. $$\frac{d}{dp} (p+p^\lambda+(1-p)^\lambda)=1 + \lambda p^{\lambda-1} - \lambda(1-p)^{\lambda-1}$$
Setting to zero: $1 + \lambda p^{\lambda-1} = \lambda(1-p)^{\lambda-1}$.
Trying out a few values of $\lambda$:
2 yields linear equation: $1+ 2p= 2(1-p) \Rightarrow p=\frac{1}{4}$
3 yields linear equation: $1 + 3p^2 = 3(1-p)^2 \Rightarrow p=\frac{1}{3}$
4 yields cubic equation: $1 + 4p^3 = 4(1-p)^3 \Rightarrow p \approx .3389$
The degree of the equation to solve for p is: $\lfloor\lambda/2\rfloor$
A: I am assuming that δ is a positive real number.  As noted previously, δ=2 yields an answer of ¼, while δ=3 yields ⅓.  Neither of those is equal to ln(δ)/δ, but as δ gets larger, then the approximation to ln(δ)/δ appears to get closer.
For natural number δ, the polynomial x + x^δ + (1-x)^δ is always of even degree.  It's derivative, f', has the property that f'(0)=1-p and f'(1)=1+p, so f' always crosses the x-axis in the interval [0,1].  An examination of the second derivative, f", shows that it is always positive in the interval [0,1], which tells us that f is convex upward facing in the interval [0,1].  We can therefore deduce that f has a unique local minimum in the interval [0,1], which will be located at the root of f' in the interval [0,1].
It appears that the root of f'in that interval approaches ln(δ)/δ.  This would be a very curious result, since ln(δ)/(δ) is also the asymptotic limit of the 1/PI(δ), the inverse function of the number of prime numbers less than δ.  
