Strange equality involving the sum of squares of a finite set in Euclidean vector space

49 Views Asked by At

Conjecture. Let $Q$ be a finite subset of a finite-dimensional Euclidean vector space $V$ and let $p \in V$. Then $$\sum_{q \in Q} ||q - p||^2 = \sum_{q \in Q} \left(||q - \mathbb{E}[Q]||^2 + ||\mathbb{E}[Q] - p||^2 \right)$$

Note that for an individual $q \in Q$ it is not true in general that $||q - p||^2 = ||q - \mathbb{E}[Q]||^2 + ||\mathbb{E}[Q] - p||^2$. I tried many numerical examples and it always worked.

Can anyone prove or disprove this conjecture?

1

There are 1 best solutions below

2
On BEST ANSWER

number the $q_i$ from $1$ to $n$ Then name $T = \sum q_i^2 = \sum q_i \cdot q_i $ which is scalar

Then $S = \sum q_i $ which is a vector with $E = S/n$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

$$ \sum (p - q_i)^2 = T - 2 p \cdot S + n p^2 $$

$$ \sum ( p - \frac{1}{n}S)^2 = n p^2 - 2 p \cdot S + \frac{1}{n} S^2 $$

$$ \sum ( q_i - \frac{1}{n} S)^2 = T - \frac{1}{n} S^2 $$

and the second plus third line equals the first