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Futurama infinite series 
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#1
Jun2411, 03:19 AM

P: 30

So tonight there was a new episode of Futurama. The professor created something that could create 2 copies of something. Bender got ahold of it, and started duplicating himself.
Bender duplicated himself once, giving 3 benders. Then those 2 duplicates duplicated themselves, giving 7 benders and so on ad infinitum. During the episode, they gave the infinite series as 2^{n}[m_{0}/(2^{n}(n+1)] Which I don't think works. The 2^{n} cancel out and you're left with gibberish. Even leaving it unsimplified doesn't work. Do I have a poor understanding of infinite series or is it a mistake? 


#2
Jun2411, 03:48 AM

P: 138

I'm a little confused by exactly what you mean when you say it does not work.
The expression simplifies to the mass of the original Bender times the harmonic series. Since the harmonic series diverges, the total mass of Benders must increase without bound. This is exactly what the Planet Express crew was concerned about! 


#3
Jun2411, 04:57 AM

P: 9

I noticed an inconsistency as well. Bender states that his first two replicas were "60% scale replicas." That suggests a series of the form:
n from zero to infinity, M(n) = (n+1)((.6)^n)(MNaught) Such a series should accurately describe the total mass of all benders at a given generation n. Clearly, M(n)>0 for all n in Z. So, apply the ratio test. M(n+1)/M(n)= ((n+2)(.6^(n+1))(MNaught)) / ((n+1)(.6^(n))(MNaught)) = .6((n+2)/(n+1)) Taking the limit as n goes to infinity, Lim N > infinity [.6((n+2)/(n+1))] = .6 < 1 Hence the series converges. I think my math is correct, but if anyone sees an error let me know. I'm rusty on this stuff. If I'm correct, that's a pretty big plot hole. :) 


#4
Jun2411, 11:44 AM

P: 4

Futurama infinite series
zach12_2: If I understand what you are trying to do, there are a couple of problems. First, if the firstgeneration miniBenders are 60% scale duplicates of the original, then they should mass [itex]0.6^3=0.216[/itex], or about 22%, what the original masses. Of course, this will make your series converge even faster. However, and this may be the more important issue, you need to multiply by the number of Benders in each generation. If the original Bender is generation 0, then in the [itex]i[/itex]th generation, there are [itex]2^i[/itex] Benders.
My approach to modeling the situation would be as follows (be aware that there are probably mistakes): First, I'm going to define a couple of variables: Let [itex]m_i[/itex] be the mass of an individual Bender in the [itex]i[/itex]th generation, and let [itex]n_i[/itex] be the number of Benders in the [itex]i[/itex]th generation. In general, [itex]m_i = 0.6^{3i}m_0[/itex], and [itex]n_i=2^i[/itex]. Thus the total mass of the ith generation is [itex]m_in_i = 0.6^{3i}m_02^i[/itex]. Hence we are interested in [itex]\sum_{i=0}^{\infty} 0.6^{3i}m_02^i[/itex]. Of course, like your result, this also converges (though if you don't cube the scaling factor, i.e. you assume that the mass of the [itex]i+1[/itex]st generation Benders is 60% the mass of the [itex]i[/itex]th generation Benders, then you end up with a divergent series). I'm still not quite sure what the series in the show was meant to represent, though as noted above, it definitely diverges. xander 


#5
Jun2411, 01:07 PM

P: 30

I definitely agree, it does diverge.
I might not have interpreted "m" correctly. I thought it stood for number of benders. It makes much more sense that it stands for mass, hence the m ahahaha. However I still don't think it works. At n=o, you get Mo At n=1, you get Mo+Mo/2 or 3Mo/2 If the second generation is 60% scale replicas. It would be 100% + 2*60% or 220% I don't understand why they put the 2^n in the denominator and outside of the fraction, wouldn't it just simplify to Mo/(n+1) 


#6
Jun2411, 01:58 PM

P: 138

In my opinion, we shouldn't worry about whether or not the show is consistent. Just look at the Benders rearranging atoms only from water to make alcohol. There are obviously two things wrong with that, but I enjoyed the episode anyways! 


#7
Jun2411, 02:02 PM

P: 9

The professor said that the series represents the mass of successive generations of benders.
Now since the machine that duplicates objects creates two of every object it duplicates if follows that for a given generation of benders, n, there are Sum from 0 to n {2^n} benders. Whoops. That's what I get for doing math at 4AM, lol. So I was certainly incorrect. I understand the errors I made except for the scaling factor in the mass. Why, exactly, does the mass scale this way? Why would it not scale linearly as I assumed in my series? (My initial reasoning is since bender is 3 dimensional, we scale by .6 in all 3 planes, is that correct?) Thanks for you help. 


#8
Jun2411, 02:07 PM

P: 4

The [itex]2^n[/itex] in the numerator is the number of miniBenders in generation [itex]n[/itex]. This implies that [itex]\frac{m_0}{2^n(n+1)}[/itex] is meant to represent the mass of each of the miniBenders in generation [itex]n[/itex]. This is the term that I am having trouble understanding. As to the [itex]2^n[/itex] in the numerator and denominator cancelingyes, they do cancel each other out, at which point we easily see that the remaining series is the mass of the original Bender times the harmonic series (which diverges). The animators or writers of Futurama probably left the terms in to make the sum look more complicated, or for some other aesthetic reason. That being said, I clearly agree with your reasoning. xander 


#9
Jun2411, 02:15 PM

P: 9

"Mass is directly proportional to the cube of the scale."
Ah, first year physics is coming back to me now... I feel silly for having forgot that. I've been doing too much pure math I suppose. Thanks, again for clarifying. 


#10
Jun2411, 02:16 PM

P: 138




#11
Jun2511, 12:21 AM

P: 30

I guess I'm not understanding the harmonic series
Let's simplify it to Mo/n+1 At Generation 0, you should get Mo right? Mo/(0+1)=Mo At generation 1, you should get Mo/(1+1) or Mo/2. Add that to the previous generation and you get 3Mo/2 or 1.5Mo Now according to xhenderson, the first generation should be 143.2% or 1.432Mo. Clearly not 1.5Mo Does it work out later on? I guess I don't know enough about infinite series, but does a series have to work every step of the way? Or can a series only work only at infinity/other determined number? 


#12
Jun2511, 01:00 AM

P: 138

[itex]m_{0}+\frac{m_{0}}{2}+\frac{m_{0}}{3}+\frac{m_{0}}{4}+ ...[/itex] or a bit more simplified: [itex]m_{0}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ ...\right)[/itex] The expression in the parenthesis is the harmonic series. As stated earlier, since the harmonic series diverges, the mass of all the Benders must increase without bound. 


#13
Jun2511, 01:13 AM

P: 30

Oh for sure, the harmonic series diverges. But not all divergent series are the same. So you think that the series given in the show doesn't work (isn't consistent with the circumstances)?
I understand the series, I just didn't understand how it worked for the circumstances. If THAT series isn't consistent, what would be a more consistent series? 


#14
Jun2511, 01:27 AM

P: 138

If each Bender is 60% the size of the original then the total mass of all Benders should be given by: [itex]M = \sum^{\infty}_{n=0}m_{0}\left(\frac{54}{125}\right)^{n}[/itex] 


#15
Jun2511, 02:01 AM

P: 30

Wooo! Thank you.



#16
Jun2511, 02:02 AM

P: 30

Actually, I think it would have been cooler if they would have used that series which actually works...



#17
Jun2511, 02:30 AM

P: 138




#18
Jun2511, 10:06 AM

P: 4

xander 


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