My Math Forum Differentiation of a (discrete) series

 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.

 Thread Tools Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top