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I'm working through Rufus Isaacs's work on differential games and I need clarification on the notation used. Some context: The Value of the game is to be the minmax of the payoff which symbolically is

$$V(\mathbf{x})=\min\limits_{\phi(\mathbf{x})}\max\limits_{\psi(\mathbf{x})}\hspace{.2cm}(\text{payoff}).$$

The Lemma on Circular Vectorgrams is where I need help deciphering:

LEMMA 2.8.1. Let $u$, $v$ be any two numbers such that $$\rho=\sqrt{u^2+v^2}>0.$$ Then $$V(\mathbf{x})=\min\limits_{\phi(\mathbf{x})}\max\limits_{\psi(\mathbf{x})}\hspace{.2cm}(u\cos\phi+v\sin\phi).$$ is furnished by $\bar{\phi}$, where $$\cos\bar{\phi}=+[-]\frac{u}{\rho},\hspace{.4cm}\sin\bar{\phi}=+[-]\frac{v}{\rho}.$$ and the max[min] itself is $$+[-]\rho.$$

What does the author mean when there are these plus and minus signs by themselves?


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  • $\begingroup$ It's from his book "Differential Games: A Mathematical Theory with Applications to Warfare and Pursuit, Control and Optimization" $\endgroup$ – Peetrius Mar 23 at 22:16
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What Isaacs actually writes in his book is:

LEMMA 2.8.1. Let $u$, $v$ be any two numbers such that $$\rho=\sqrt{u^2+v^2}>0.$$ Then $$\max\limits_{\phi}\,[\min\limits_{\phi}]\hspace{.2cm}(u\cos\phi+v\sin\phi).$$ is furnished by $\bar{\phi}$, where $$\cos\bar{\phi}=+[-]\frac{u}{\rho},\hspace{.4cm}\sin\bar{\phi}=+[-]\frac{v}{\rho}.$$ and the max[min] itself is $$+[-]\rho.$$

All he is doing here is compressing the descriptions of, and soutions to, two separate optimisation problems into one version with alternative text given in square brackets. Ignoring the text in square brackets gives the description and solution of the maximisation problem, while substituting the text inside the square brackets for that which immediately precedes it gives the description and solution of the minimisation problem. Written out in full, the statement of the lemma would be:

LEMMA 2.8.1. Let $u$, $v$ be any two numbers such that $$\rho=\sqrt{u^2+v^2}>0.$$ Then $$\max\limits_{\phi}\hspace{.2cm}(u\cos\phi+v\sin\phi).$$ is furnished by $\bar{\phi}$, where $$\cos\bar{\phi}=+\frac{\vert u\vert}{\rho},\hspace{.4cm}\sin\bar{\phi}=+\frac{\vert v\vert}{\rho}.$$ and the max itself is $$+\rho\ ,$$ while $$\min\limits_{\phi}\hspace{.2cm}(u\cos\phi+v\sin\phi).$$ is furnished by $\tilde{\phi}$, where $$\cos\tilde{\phi}=-\frac{\vert u\vert}{\rho},\hspace{.4cm}\sin\tilde{\phi}=-\frac{\vert v\vert}{\rho}.$$ and the min itself is $$-\rho\ .$$

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  • $\begingroup$ Interesting. Now this makes more sense. Thank you. $\endgroup$ – Peetrius Mar 24 at 18:31

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