The discussion revolves around the derivation of the finite geometric series formula, specifically focusing on the rationale behind multiplying by "r" in the derivation process and the validity of cancelling terms in the resulting equations. Participants explore the logical foundations of the operations involved in the derivation.
Participants generally agree on the validity of the operations discussed, but there is no consensus on a comprehensive explanation of the rationale behind them. The discussion remains exploratory with multiple viewpoints presented.
Some assumptions regarding the operations and their implications may not be fully articulated, and the discussion does not resolve the deeper logical foundations of the derivation.
Terrell said:what is the rationale of multiplying "r" to the second line of series? why does cancelling those terms give us a VALID, sound, logical answer? please help. here's a video of the procedure
Samy_A said:The "rationale" is that it works.
Why wouldn't that operation be valid?
What he basically does is:
##a=b+c+d+e##
##f=\ \ \ \ \ \ c+d+e+g##
then, subtracting the second equation from the first gives ##a-f=b-g##
The other terms cancel each other.
i get that it works, but the closest logical explanation i have for myself is 1 - (r^n) divided by 1 - r is the polynomial 1 + r + r^2 + ... + r^(n-1) + r^n. which is identical to the finite geometric series. Thus, distributing (1 - r) to the polynomial gives us 1*S -r*S = S - rS.Samy_A said:The "rationale" is that it works.
Why wouldn't that operation be valid?
What he basically does is:
##a=b+c+d+e##
##f=\ \ \ \ \ \ c+d+e+g##
then, subtracting the second equation from the first gives ##a-f=b-g##
The other terms cancel each other.
I see.Terrell said:i get that it works, but the closest logical explanation i have for myself is 1 - (r^n) divided by 1 - r is the polynomial 1 + r + r^2 + ... + r^(n-1) + r^n. which is identical to the finite geometric series. Thus, distributing (1 - r) to the polynomial gives us 1*S -r*S = S - rS.