## Futurama infinite series

 Quote by blade123 Actually, I think it would have been cooler if they would have used that series which actually works...
It would have been, though as I see it, they couldn't have because they set up the situation in such a way that the mass of the Benders did not diverge. It was key to the plot that the Benders would eventually devour the Earth, but that is not what would happen if each Bender were a 60% scale copy of the previous Bender. So, rather than change the rules of the game (for instance, if the copies were 80% scale versions of the original...), they just threw some nonsense on the board. The series that they show does diverge, it just seems a rather poor model for the situation they describe.

xander
 $M_{0}$= Mass of initial bender $2^{n}$ Number of benders in nth generation 60%* Mass of bender in n-1 generation= Mass of individual nth generation bender EG. $M_{1}$=.6$M_{0}$ $M_{2}$=(.6)(.6)$M_{0}$ $M_{3}$=(.6)(.6)(.6)$M_{0}$ In general $M_{N}$=$.6^{n}$$M_{0}$=Mass of nth generation bender $2^{n}$$.6^{n}$$M_{0}$=$1.2^{n}$$M_{0}$=mass of nth generation $\sum_{n=0}^{ \infty}$=$1.2^{n}$$M_{0}$=mass of total system This is a non convergent geometric series with a rate of 1.2
 I was thinking about the scaling factor like you did in your post when I hastily proposed an alternative series a few posts above. (Darn my late nights!) As one poster previously pointed out the mass should scale with the cube of the height. So it would be the sum as n goes from 0 to infinity of [((.6)^3n)Mnaught(2^n)] <=> the sum as n goes form 0 to infinity of [(.432^n)Mnaught] So the series is a convergent power series with r=.432 So it really hinges on if you do or don't cube the mass scaling factor. Is there any reason not to cube the scaling factor? After careful consideration, I'm convinced it must be cubed to be correct, but my thinking could be flawed. Zach
 I just brought this up with my calculus teacher. He said the episode would make much more sense if instead of 2 copies, the machine would create 5 60% copies. It would give r=1.08 and thus would be divergent. At 2 60% copies, the series is convergent.
 shouldn't series be $$\sum_{n=0}^{\infty}M_{0}(1.2)^{n}$$ each bender duplicates in each generation ....

 Quote by zetafunction shouldn't series be $$\sum_{n=0}^{\infty}M_{0}(1.2)^{n}$$ each bender duplicates in each generation ....
You seem to be assuming that the mass of each generation is 60% the mass of the previous generation. As previously noted in this topic, it is probably better to assume that the 60% scale refers to the linear scale of the Benders, implying that the mass of an (n+1)st generation Bender should be 21.6% of an nth generation Bender.

xander

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