My Math Forum Differentiation of a (discrete) series

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 March 13th, 2017, 03:39 PM #1 Senior Member     Joined: Oct 2014 From: I don't know... Posts: 220 Thanks: 26 Math Focus: Calculus Differentiation of a (discrete) series Hello, I have a discrete series: $\displaystyle f(n) = \frac{1}{\sigma^2} \sum_{n = 0}^{N-1}(x(n) - A) \qquad (1)$ and a result which states that if this series is equal to 0 yields to: $\displaystyle A = \frac{1}{N}\sum_{n = 0}^{N-1}x(n) \qquad (2)$ I understand this is a mean value. Since I am would like to so good handling series I have a doubt resolving by myself... I understand that if $\displaystyle A$ does not depend on $\displaystyle n$, if I equate the equation (1) to 0 and multiply the denominator $\displaystyle \sigma^2$ with it, I obtain: $\displaystyle A = \sum_{n = 0}^{N-1}x(n) \qquad (3)$ If my reasoning is correct, what about the term $\displaystyle \frac{1}{N}$ at the denominator of (2)? Perhaps the answer is in my own question when I write that the equation (2) is a mean value, but what is the mathematical step to obtain it? I don't understand it, or maybe I am confused. Could someone give a hint? Thank you in advance, szz Last edited by skipjack; March 14th, 2017 at 12:46 AM.
 March 13th, 2017, 04:24 PM #2 Global Moderator   Joined: May 2007 Posts: 6,259 Thanks: 509 $\displaystyle \sum_{n=0}^{N-1}(x(n)-A)=\sum_{n=0}^{N-1}x(n)-NA$ Thanks from szz Last edited by skipjack; March 14th, 2017 at 12:45 AM.

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