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 Calculus Calculus Math Forum

 March 13th, 2017, 03:39 PM #1 Senior Member   Joined: Oct 2014 From: EU Posts: 224 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: 6,807 Thanks: 716 $\displaystyle \sum_{n=0}^{N-1}(x(n)-A)=\sum_{n=0}^{N-1}x(n)-NA$ Thanks from szz Last edited by skipjack; March 14th, 2017 at 12:45 AM. Tags differentiation, discrete, series Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post redglitter Calculus 5 September 3rd, 2014 07:24 AM Jhenrique Calculus 1 March 1st, 2014 11:47 AM rohit99 Calculus 1 January 26th, 2012 07:08 PM wnvl Calculus 4 November 25th, 2011 12:44 PM dha1973 Calculus 4 December 7th, 2009 06:53 PM

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