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### A Thought…

There’s a place for all God’s creatures. Right next to the potatoes and gravy. — [Traveler]

I did not fight and claw my way up to the top of the food chain to eat plants. — dflak, Non Sequitur comic letters

## Benderama equation

Posted on 2022-May-19 at 11:50:00 (last update on 2022-Jun-11 at 22:13:00) by Phil

In the 2011 (Season 6) Futurama episode titled Benderama, Bender “borrowed” Professor Farnsworth’s Banach-Tarski Dupla-Shrinker to make copies of himself. It is stated that two copies will be made, each 60% the size of the original. A certain amount of unspecified “matter” (e.g., Zoidberg’s wedding album) is needed to feed the machine (which Bender has hidden within his Compartment of Mystery). When the Professor realizes that Bender is making copies, and each copy is also making copies, he announces that the entire Earth will be consumed in the process and the total mass of $$n$$ generations of Benders will be:

$$M = \sum^\infty_{n=0} {2^n \cdot \left[{M_0} \over {2^n (n+1)}\right]}~~.$$

Now, there are a few issues here (besides the very existence of such a duplicator in the first place…). First of all, the “60%” figure is a bit ambiguous — is it 60% of the mass, 60% of the volume, or 60% of the height (also width and depth)? From the pictures (animation), it would appear that it’s 60% of the height (linear dimensions), meaning $$0.6^3 = 0.216$$ of the volume, and (assuming constant density) the mass. Also, we only see two copies being made at a time — no limit on usage is mentioned, but it appears that the duplicator gives up after being used once (to produce two smaller copies). The Big Question: is this equation correct, and is it divergent (goes to infinity), as claimed?

Futurama was, in part, written by three PhD’s (including applied mathematician Ken Keeler, chemist Bill Odenkirk, and computer scientist Jeff Westbrook), not to mention a number of others with graduate STEM degrees, so I was disappointed to see that the formula (upon which the plot depends), appears to be wrong. This equation does not appear to account for the 60% scale factor (see this discussion, among others). If you pull out the $$M_0$$ and cancel the two $$2^n$$ terms, you get:

$$M = M_0 \sum^\infty_{n=0} {1 \over {n+1}}~~,$$

and indeed, the series is divergent ($$M$$ goes to $$\infty$$). However, wouldn’t the correct formula be something closer to:

$$M = M_0 \sum^\infty_{n=0}~{2^n \cdot 0.6^{3n}}~~?$$

Each generation is 60% (linear dimensions) of the previous one, with 21.6% of the volume (and, presumably, mass). There are two copies, so $$2^n$$ Bender copies at generation $$n$$, each of mass $$M_0 \cdot 0.6^{3n}$$. I think that’s the correct formula, but I am unable to prove that it converges or diverges!

So, I wrote a little Perl program to calculate the result for various scale values:

use strict; use warnings; my $scale = .7925; my$MAXn = 1000; my $term; my$sum; my $n;$sum = 0.0; print "1/(n+1)\n"; for ($n=0;$n<$MAXn;$n++) { $term = 1.0/($n+1); $sum +=$term; print "+ $term =$sum\n"; } $sum = 0.0; print "2**n*($scale)**3n\n"; for ($n=0;$n<$MAXn;$n++) { $term = 2**$n * $scale**(3*$n); $sum +=$term; print "+ $term =$sum\n"; }

My results are that it converges up to a \$scale of about 0.792+, and diverges beyond that. Can anyone come up with a precise answer?

• .1 converges to 1.002
• .5 converges to 4/3
• .6 converges to 1.76 (this is the scale stated in the show)
• .75 converges to 6.4
• .775 converges to 14.5
• .78 converges to 19.65
• .79 converges to 67.5
• .792 converges to 156ish
• .7925 may converge to 218.4ish
• .793 may converge to 352ish
• .795 diverges
• .8 diverges
• .9 diverges

So, it looks to me that (using the 60% scale factor), even an infinite number of ever-smaller Benders will only consume less than one additional Bender in mass… nothing to worry about (at least, the entire planet will not be consumed in the process).

There are some other, non-mathematical physics issues, such as what happens when the Benders shrink down to atomic size — what are they made of? Do their atoms keep shrinking? Beyond that, is there some absolute smallest-size limit, such as the Planck Length? Stay tuned, or just suspend your disbelief and enjoy the show!

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