# Is it possible to solve this equation system (containing summations)?

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!

• All you can say is that $1-C$ is between $\pm\sqrt{(1-A)(1-B)}$. – Lord Shark the Unknown Aug 9 at 6:15
• 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 $0 \leq C \leq \sqrt{(1-A) (1-B)}$ – Anti Aug 9 at 6:22
• $1-A$ 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. – Lord Shark the Unknown Aug 9 at 6:25

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 $$C$$.
• That's pity. Nevertheless, thanks a lot! BTW - $x_i , y_i \in [0,1]$ (just edited my post) – Anti Aug 9 at 6:17
$$C= (1/2)* [3-A-B-∑(x_i+y_i )^2 ]$$