Sum of Deviations: Proving $\sum_{i=1}^Nv_i(v_i - \langle v \rangle) = 0$

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The discussion focuses on proving that the sum of deviations from the average of a set of measurements equals zero. The average value of N measurements is defined, and the deviation for each measurement is expressed as (v_i - ⟨v⟩). Participants clarify that the correct expression for the sum of deviations is ∑(v_i - ⟨v⟩), and they emphasize the need to expand this summation to demonstrate that it evaluates to zero. The conversation highlights the mathematical steps necessary to achieve this proof, ensuring clarity on what is required to show that the sum of all deviations is indeed zero. Understanding this concept is crucial for grasping the properties of averages in statistical analysis.
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Homework Statement


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
$$\sum_{i=1}^Nv_i(v_i - \langle v \rangle) = 0$$

Homework Equations


##?##

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.
 
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kubaanglin said:

Homework Statement


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
$$\sum_{i=1}^Nv_i(v_i - \langle v \rangle) = 0$$
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##.
kubaanglin said:

Homework Equations


##?##

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.
Simply expand the summation.
 
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$$((v_1 - \langle v \rangle) + (v_2 - \langle v \rangle) + \cdots (v_N - \langle v \rangle))$$
$$((v_1 - \frac {v_1 + v_2 + \cdots v_N}{N}) + (v_2 - \frac {v_1 + v_2 + \cdots v_N}{N}) + \cdots (v_N - \frac {v_1 + v_2 + \cdots v_N}{N}))$$
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.
 
kubaanglin said:
$$((v_1 - \langle v \rangle) + (v_2 - \langle v \rangle) + \cdots (v_N - \langle v \rangle))$$
$$((v_1 - \frac {v_1 + v_2 + \cdots v_N}{N}) + (v_2 - \frac {v_1 + v_2 + \cdots v_N}{N}) + \cdots (v_N - \frac {v_1 + v_2 + \cdots v_N}{N}))$$
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.

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##.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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