# Sum of deviations

1. Oct 3, 2016

### kubaanglin

1. The problem statement, all variables and given/known data
The average value of N measurements of a quantity $v_i$ is defined as
$$\langle v \rangle \equiv \frac {1}{N} \sum_{i=1}^Nv_i = \frac {1}{N}(v_1 + v_2 + \cdots v_N)$$
The deviation of any given measurement $v_i$ from the average is of course $(v_i - \langle v \rangle)$. Show mathematically that the sum of all the deviations is zero; i.e. show that
$$\sum_{i=1}^Nv_i(v_i - \langle v \rangle) = 0$$
2. Relevant equations
$?$
3. The attempt at a solution
I understand that this is simply describing an average, but I am not sure how to express this mathematically. It makes sense to me that the sum of the deviations would be zero.

2. Oct 3, 2016

### Staff: Mentor

Your formula above is incorrect, as it has an extra $v_i$.
The sum of the deviations is
$$\sum_{i = 1}^N (v_i - \bar{v})$$
Here $\bar{v}$ is the mean of the measurements $v_i$.
Simply expand the summation.

Last edited: Oct 4, 2016
3. Oct 4, 2016

### kubaanglin

$$((v_1 - \langle v \rangle) + (v_2 - \langle v \rangle) + \cdots (v_N - \langle v \rangle))$$
$$((v_1 - \frac {v_1 + v_2 + \cdots v_N}{N}) + (v_2 - \frac {v_1 + v_2 + \cdots v_N}{N}) + \cdots (v_N - \frac {v_1 + v_2 + \cdots v_N}{N}))$$
Is this sufficient to "show mathematically" that the sum of all the deviations is zero? I am just not sure what I am being asked to do.

4. Oct 4, 2016

### Ray Vickson

You are being asked to show that the summation you wrote above evaluates to $0$ for any possible inputs $v_1, v_2, \ldots, v_N$.