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Futurama infinite series |
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| Jun25-11, 10:06 AM | #18 |
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Futurama infinite seriesxander |
| Jun26-11, 11:47 PM | #19 |
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[itex]M_{0}[/itex]= Mass of initial bender
[itex]2^{n}[/itex] Number of benders in nth generation 60%* Mass of bender in n-1 generation= Mass of individual nth generation bender EG. [itex]M_{1}[/itex]=.6[itex]M_{0}[/itex] [itex]M_{2}[/itex]=(.6)(.6)[itex]M_{0}[/itex] [itex]M_{3}[/itex]=(.6)(.6)(.6)[itex]M_{0}[/itex] In general [itex]M_{N}[/itex]=[itex].6^{n}[/itex][itex]M_{0}[/itex]=Mass of nth generation bender [itex]2^{n}[/itex][itex].6^{n}[/itex][itex]M_{0}[/itex]=[itex]1.2^{n}[/itex][itex]M_{0}[/itex]=mass of nth generation [itex]\sum_{n=0}^{ \infty}[/itex]=[itex]1.2^{n}[/itex][itex]M_{0}[/itex]=mass of total system This is a non convergent geometric series with a rate of 1.2 |
| Jun27-11, 08:10 AM | #20 |
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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 |
| Jun27-11, 03:24 PM | #21 |
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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. |
| Jun29-11, 06:42 AM | #22 |
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shouldn't series be [tex] \sum_{n=0}^{\infty}M_{0}(1.2)^{n} [/tex]
each bender duplicates in each generation .... |
| Jun29-11, 11:10 AM | #23 |
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xander |
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