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I am working on a problem crop harvesting/sowing problem. A beet seed will yield 1 beet. Beets left in the ground will produce 3 seeds per beet to be replanted in the next season. I'm curious if there's a direct way to solve this like you can for an arithmetic series or something? What would this math area be called so that I can study further?

The problem is that if I want to sow enough seeds so that I can harvest $d$ beets, then I won't have any beets in the ground to produce seeds for the next season. I now need to plant enough beets to account for the extra beets needed to seed the next season. I played with this and got $d + d/3$. This gets me closer but now I have to account for the beets that result from $d/3$. This keeps getting closer but doesn't solve the problem since now I have to account for the beets, the additional beets for seeds and now the beets for that addition, and so on.

I noticed that this is really just repeatedly dividing the result of the previous result. This is where I started going with the math but I don't know what branch/topic of math this would be called to learn more about it. I simply can't stop accumulating partial results at hard coded amount of attempts as illustrated here:

$$ d + {{d + {{d + {d\over 3}}\over 3}}\over 3} $$

Depending on the size of $d$ this will start to lose seeds.

I didn't give up entirely though! I don't know how to continue the with the math so I came up with a recursive solution in code:

def seeds_needed need
  seeds = need / 3.0

  if seeds < 1
    return 1   # because I can't use a partial seed :)
  else
    return seeds + seeds_needed(seeds)
  end
en

This function gets me as close as I can. I pass it the number of beets I wish to harvest each year and it will return to me the number of seeds that I need to plant so that I can leave the extra beats in the ground to become seeds and start the process all over again.

Note: I do normalize the results later to account for the seeds being multiples of 3 per plant.

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  • $\begingroup$ Your function looks like a geometric series. $\endgroup$
    – Michael Burr
    Apr 13, 2021 at 15:01
  • $\begingroup$ A broad classification would be "algebra", in the sense of tag algebra-precalculus. Evaluating the expression relates to its form as an iterate formed starting from a seed. Some interesting problems about function iteration can be asked. $\endgroup$
    – hardmath
    Apr 13, 2021 at 15:03
  • $\begingroup$ @samhw, did you happen to approach 1.5 d? $\endgroup$
    – Subbota
    Apr 13, 2021 at 15:15
  • $\begingroup$ @Subbota yes, but I don't know why or how. $\endgroup$
    – samhw
    Apr 13, 2021 at 16:12

2 Answers 2

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It’s 2am, I want sleep, my Mathjax is really bad. So I tried handwriting it. Sorry. I do hope I understood correctly and that my math holds. enter image description here

Second line, I rearranged the given equation, equal to total seeds needed. Second to third line, I expanded the brackets. It was a bit brief the working. One could easily jump straight to the $d + d/3 + d/3^2...$ bit just by looking at the given equation, but it is what it is, this might help you see/understand it better maybe.

I think the topic you are interested in is geometric progressions and sums. https://www.varsitytutors.com/hotmath/hotmath_help/topics/geometric-series Something like this. Common ratio is $T_2/T_1 $in a series, to see a sort of relationship, or more generally $T_n / T_{n-1}$, to see the relationship between a term and the next. Hopefully this helps, and again sorry for the sloppy work.

Edit: It has occurred to me that T.S = $ d + d/3 + d/3^2 ... + d/3^{y-1}$ not T.S = $d + d/3 + d/3^2 ... + d/3^y$ as I wrote. As y—> infinity, this shouldn’t affect it, but if you’re looking at a specific year, that isn’t too large it might have an impact.

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    $\begingroup$ Excellent. This is exactly where I was trying to go but couldn't figure it like you have here. I will refresh my understanding of geometric series! $\endgroup$
    – samhw
    Apr 13, 2021 at 16:14
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    $\begingroup$ +1. Better than mine. $\endgroup$
    – Yuki.F
    Apr 14, 2021 at 15:35
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Try let $u = d + (d + (d + \cdots)/3)/3$, then because this series is infinite, $u = d + u/3$ and you should get $u$ quickly.

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  • $\begingroup$ I'm not sure an infinite series is meant by the OP. It seems possible that the real problem is determining how many iterations are required, e.g. to reach some goal such as a total number of beets. $\endgroup$
    – hardmath
    Apr 13, 2021 at 15:07
  • $\begingroup$ Correct. Either determining the number of iterations or whether there's a direct answer as you'd use in an arithmetic series. $\endgroup$
    – samhw
    Apr 13, 2021 at 16:11

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