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The below listed 3 equations are given:

$$A = 1-\sum_{i=1}^n x_i^2,$$ $$B = 1-\sum_{i=1}^n y_i^2,$$ and $$C = 1-\sum_{i=1}^n x_i y_i$$

with $x_i, y_i \in [0,1]$ and $\sum_{i=1}^n x_i = \sum_{i=1}^n y_i = 1$.

Is it possible to write $C$ as a function of $A$ and $B$, i.e. as $C(A,B)$?

Every help is appreciated! Many thanks in advance!

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  • $\begingroup$ All you can say is that $1-C$ is between $\pm\sqrt{(1-A)(1-B)}$. $\endgroup$ Aug 9, 2019 at 6:15
  • $\begingroup$ Thank you. Since I added info about $x_i$ and $y_i$ (both are in the interval $[0,1]$), I now see at least that $0 \leq C \leq \sqrt{(1-A) (1-B)}$ $\endgroup$
    – Anti
    Aug 9, 2019 at 6:22
  • $\begingroup$ $1-A$ is the square of the length of the vector $x=(x_1,\ldots,x_n)$. Similarly, $1-B$ is the square of the length of $y$. Then $1-C$ is the dot product of $x$ and $y$ and so equals $\sqrt{(1-A)(1-B)}\cos t$ where $t$ is the angle between them. In general, knowing the lengths of these vectors won't tell you the angle between them. $\endgroup$ Aug 9, 2019 at 6:25

2 Answers 2

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No. Consider $x_i = y_i = 1$ vs $x_i = -y_i = 1$.

In both cases, $A = B = 1 - n$. In the first case $C = 1 - n$ and in the second $C = 1 + n$. Therefore knowing $A$ and $B$ does not tell you $C$.

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  • $\begingroup$ That's pity. Nevertheless, thanks a lot! BTW - $x_i , y_i \in [0,1]$ (just edited my post) $\endgroup$
    – Anti
    Aug 9, 2019 at 6:17
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We can write

$$C= (1/2)* [3-A-B-∑(x_i+y_i )^2 ]$$

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  • $\begingroup$ That's cool, thx! $\endgroup$
    – Anti
    Aug 9, 2019 at 6:38
  • $\begingroup$ mostly welcome! $\endgroup$ Aug 9, 2019 at 7:19

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