Mean value by Least Squares Method

[Hans Wurst]

(08-02-2019 08:14 PM)DrD Wrote:  Does this help?

No. What you do is not Least Squares Method where the differences are squared first and then added (and this sum minimized). You add the differences and square the sum (and so on).
\(\frac{\partial \displaystyle \sum_{k=1}^{n}(x(k)-m)^2}{\partial m}\neq \frac{\partial \displaystyle \left (\sum_{k=1}^{n}(x(k)-m)\right )^2}{\partial m}\)

The thing is to solve sum(−2∗(x(k)−m),k,1,n) = 0 for m. A simple arithmetic rearrangement:
sum(−2∗(x(k)−m),k,1,n) = 0
−2∗(sum(x(k),k,1,n)+n*m) = 0
m = sum(x(k),k,1,n)/n
alas I do not know how to push a Prime this way.

Best,
H.