# Futurama infinite series

1. Jun 24, 2011

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

2n[m0/(2n(n+1)]

Which I don't think works. The 2n 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. Jun 24, 2011

### Dschumanji

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! :tongue:

Last edited: Jun 24, 2011
3. Jun 24, 2011

### zach12_2

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. :)

Last edited: Jun 24, 2011
4. Jun 24, 2011

### xhenderson

zach12_2: If I understand what you are trying to do, there are a couple of problems. First, if the first-generation mini-Benders are 60% scale duplicates of the original, then they should mass $0.6^3=0.216$, 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 $i$th generation, there are $2^i$ 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 $m_i$ be the mass of an individual Bender in the $i$th generation, and let $n_i$ be the number of Benders in the $i$th generation.

In general, $m_i = 0.6^{3i}m_0$, and $n_i=2^i$. Thus the total mass of the ith generation is $m_in_i = 0.6^{3i}m_02^i$. Hence we are interested in $\sum_{i=0}^{\infty} 0.6^{3i}m_02^i$. 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 $i+1$st generation Benders is 60% the mass of the $i$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. Jun 24, 2011

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. Jun 24, 2011

### Dschumanji

It was most likely to make the infinite series look more complicated. Most people would not be able to easily recognize it in the second that it was flashed (I know I didn't recognize it). I think by doing this it is supposed to make the audience feel exactly like Fry.

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. Jun 24, 2011

### zach12_2

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. Jun 24, 2011

### xhenderson

I'm note sure where you are getting these terms, or perhaps I am not understanding what they represent. Can you explain your thinking, please?

Again, mass doesn't scale in that fashion. Mass is directly proportional to the cube of the scale. To make things a bit simpler, suppose that you have a 1 kg cube with uniform density that measures 1 m on a side. If we scale the cube by 60%, we get a new cube that measures 0.6 units on a side. Mass is density times volume, and the volume of a cube is the cube of the side length. Thus the mass of the smaller cube is $(0.6{\rm\ m})^3\cdot 1\ {\rm kg}/{\rm m}^3 = 0.216\ {\rm kg}$. Thus in the second generation, there is one full-sized Bender that masses 100%, and two mini-Benders which mass 21.6% each, for a total of 143.2% the mass of the original Bender.

I assume that this has something to do with how the formula was generated. Every term of the sum should basically have two terms: the number of mini-Benders in a particular generation, and the mass of each of those mini-Benders.

The $2^n$ in the numerator is the number of mini-Benders in generation $n$. This implies that $\frac{m_0}{2^n(n+1)}$ is meant to represent the mass of each of the mini-Benders in generation $n$. This is the term that I am having trouble understanding.

As to the $2^n$ in the numerator and denominator canceling---yes, 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.

Because most of their viewers would instantly recognize $\sum_{n=0}^\infty \frac{m_0}{n+1}$ as a harmonic series. :P

That being said, I clearly agree with your reasoning.

xander

9. Jun 24, 2011

### zach12_2

"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. Jun 24, 2011

### Dschumanji

Yeah, yeah, I know. :tongue:

11. Jun 25, 2011

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. Jun 25, 2011

### Dschumanji

I don't think it is possible to reconcile the infinite series given in the show and the comment about the smaller Benders being 60% the size of the original. Only focusing on the infinite series that is given, you can show the following:

$m_{0}+\frac{m_{0}}{2}+\frac{m_{0}}{3}+\frac{m_{0}}{4}+ ...$

or a bit more simplified:

$m_{0}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ ...\right)$

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. Jun 25, 2011

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. Jun 25, 2011

### Dschumanji

The series does not represent the mass of all the Benders if each of the smaller Benders are 60% the size of the original.

If each Bender is 60% the size of the original then the total mass of all Benders should be given by:

$M = \sum^{\infty}_{n=0}m_{0}\left(\frac{54}{125}\right)^{n}$

15. Jun 25, 2011

Wooo! Thank you.

16. Jun 25, 2011

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

17. Jun 25, 2011

### Dschumanji

This wouldn't be very exciting for the story line unless the total sum of all Benders equaled the mass of the Earth. Bender would have to be pretty massive.

18. Jun 25, 2011

### xhenderson

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

19. Jun 26, 2011

### Goodderf

$M_{0}$= 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​

20. Jun 27, 2011

### zach12_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