
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: $\displaystyle f(n) = \frac{1}{\sigma^2} \sum_{n = 0}^{N1}(x(n)  A) \qquad (1)$ and a result which states that if this series is equal to 0 yields to: $\displaystyle A = \frac{1}{N}\sum_{n = 0}^{N1}x(n) \qquad (2)$ 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: $\displaystyle A = \sum_{n = 0}^{N1}x(n) \qquad (3)$ 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: 6,213 Thanks: 491 
$\displaystyle \sum_{n=0}^{N1}(x(n)A)=\sum_{n=0}^{N1}x(n)NA$
Last edited by skipjack; March 14th, 2017 at 12:45 AM. 

Tags 
differentiation, discrete, series 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Taylor Series and Differentiation  redglitter  Calculus  5  September 3rd, 2014 07:24 AM 
Taylor series in terms of discrete derivative  Jhenrique  Calculus  1  March 1st, 2014 12:47 PM 
calculus, sequences, series , differentiation  rohit99  Calculus  1  January 26th, 2012 08:08 PM 
Proof series using differentiation  wnvl  Calculus  4  November 25th, 2011 01:44 PM 
Differentiation + series problem  dha1973  Calculus  4  December 7th, 2009 07:53 PM 