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March 13th, 2017, 03:39 PM   #1
szz
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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.
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March 13th, 2017, 04:24 PM   #2
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$\displaystyle \sum_{n=0}^{N-1}(x(n)-A)=\sum_{n=0}^{N-1}x(n)-NA$
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Last edited by skipjack; March 14th, 2017 at 12:45 AM.
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