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For positive integers $ b_1,b_2,...,b_T $, I am trying to solve the following optimization problem: $$ \min_{\substack{p_1,\ldots,p_T\\ p_1+\cdots+p_T=1\\}} \sum_{i=1}^T {p_i} ^ {b_i} .$$

The solution I would like to find is an expression for $ p_i $ in terms of $ b_i $. For example, something like $ p_i \sim 1/b_i $.

I have tried a few different things, including experimenting with the simple case where $ T = 2 $, but I have been having some trouble. If anyone can tell me how to solve this, give me any guidance, or even tell me whether this problem is tractable, I would really appreciate it. Thanks!

Update:

I have now tried using Lagrange multipliers to solve this optimization problem. However, I'm not entirely sure that this is correct.

The Lagrangian is:

$$ L(p,\lambda) = \sum_{i} {p_i}^{b_i} - \lambda \bigl(\sum_{i}{p_i -1} \bigr) $$

Next I calculate the gradient

$$ \frac{dL}{dp_i} = b_i p_i^{b_i - 1} - \lambda$$ $$ \frac{dL}{d\lambda} = \sum_i p_i - 1 $$

I set these to zero to calculate the fixed points. But lambda is only in my first equation? I end up with:

$$ p_i = \bigl( \frac{\lambda}{b_i} \bigr)^{\frac{1}{(b_i - 1)}}$$

So then, to solve for lambda, I should substitute this expression in for the 2nd equation:

$$ \sum_i p_i = \sum_i \bigl ( \frac{\lambda}{b_i} \bigr ) ^{\frac{1}{(b_i - 1)}} = 1 $$

adn bps
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  • Thanks for the help! I have tried using Lagrange multipliers. Would be interested to see if I'm on the right track. – adn bps Aug 23 '17 at 19:32
  • Not quite. The x numbers are integers and not changing. (Call them b's; it looks better.) You have the constrained variables being the ones you use for the derivative, and you want to look for when the gradient of your sum function is the same as the normal to the plane. Gerhard "Or Something Along Those Lines" Paseman, 2017.08.23. –  Aug 23 '17 at 21:37
  • Thanks again! I updated my response. It seems that I can solve for lambda using my second equation. Does this look correct? – adn bps Aug 23 '17 at 22:04
  • That looks better, and close to what I would expect. (I have not verified it.) I recommend using Joseph's example below, and small (perhaps two or three variable) examples using Wolfram Alpha, to check your calculations and build your intuition. You might not get an explicit functional answer, so numerically verifying is important. This is not the forum for this discussion, but writing up your experience (including especially the mistakes) on a blog would help other students learning Lagrange multipliers. Gerhard "And Become The Education Multiplier" Paseman, 2017.08.23. –  Aug 23 '17 at 22:19
  • Yes, your Lagrangian approach is correct. Solve that numerically for $\lambda$ and you're there. – Michael Grant Aug 25 '17 at 22:31
  • Note that you need to separate the cases where $b_i = 1$ from the cases when $b_i > 1$. A more careful analysis is required. – Alex Shtoff Aug 27 '17 at 09:51

2 Answers2

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I'll focus on solving for $\lambda$ in your last equation.

$$\sum_{i}\left(\frac{\lambda}{b_i}\right)^\frac{1}{b_i-1}=1\quad\equiv\quad\sum_{i}\left(\frac{1}{b_i}\right)^\frac{1}{b_i-1}\left(\lambda\right)^\frac{1}{b_i-1}=1\quad \equiv:\quad\sum_i c_i\ \sqrt[b_i-1]{\lambda}=1$$

which clearly shows, that the problem amounts to removing the radicals; that can be done by the nice method of Dr. Vogler, which is described on Stackexchange https://mathoverflow.net/questions/177765/rewrite-sum-of-radicals-equation-as-polynomial-equation.

After the radicals have been removed, the problem has been converted into the easier task of solving a polynomial equation; that problem can then be fed into a symbolic or numeric solver.

Manfred Weis
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  • Hi Manfred, thanks for the suggestion! Actually, I already know the $ b_i $ variables (I should have mentioned this in my post). So I just substitute these into the equation, and I only need to solve for lambda. – adn bps Aug 25 '17 at 00:26
  • @adnbps I concluded that from the formulation that you are trying to find the $p_i$ – Manfred Weis Aug 25 '17 at 04:34
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If it would help to have an explicit solution on hand to test hypotheses, solvers have no difficulty with individual instances. For example, $$ \min \; p_1^2 + p_2^3 + p_3^2 + p_4^4 $$ leads to $$ (p_1,\, p_2,\, p_3,\, p_4) \approx (0.14058,\, 0.30614,\, 0.14058,\, 0.41270) \;. $$

Joseph O'Rourke
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