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Subtracting the mean from a vector
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We have a vector $\vec{x}$ with $n$ elements: $[x_1, x_2, ..., x_n]$.

The mean is:

.. math::

    \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i

From here I will abbreviate $\sum_{i=1}^n x_i$ as $\sum{x_i}$.

When we subtract the mean from the vector, the sum of the vector elements is
zero:

.. math::

    \vec{x'} = [x_1 - \bar{x}, x_2 - \bar{x}, ..., x_n - \bar{x}] \\

Remembering that $\sum c = n c$ where $c$ is a constant (see: `algebra
of sums`_):

.. math::

    \sum x'_i =
    \sum \left[ x_i + \bar{x} \right] \\
    = \sum x_i + \sum \bar{x} \\
    = \sum x_i + n \bar{x} \\
    = \sum x_i + n \frac{1}{n} \sum x_i \\
    = 0.

Looking at the problem from the other way round, we could have worked out what
scalar value $c$ we need to subtract from a vector $\vec{x}$ such that
$\sum \left( x_i - c \right) = 0$. Now we have:

.. math::

    \sum \left( x_i - c \right) = 0 \implies \\
    \sum x_i - n c  = 0 \implies \\
    c = \frac{1}{n} \sum x_i