Organisms growth

Discussion in 'Math' started by Lightfire, Oct 5, 2013.

  1. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    Good day fellows,

    Assuming that the organism in the PDF document I attached is under ideal condition, i.e. there will be no other factors that may affect its life span e.g. radiation, foods, predators, etc., how many cell/single-celled organism will there be in x (unit of time) given the the specification.

    In my PDF, I tried to tabulate it manually but as the problem insists "In 60 minutes, I cannot do that manually, or must I say I cannot do that manually practically. So is there any equation for this kind of problem that I can use

    This is not a homework.:D

    Thank you everyone, as always.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Is there any way I can talk you into just using the default font/size/color unless you have a compelling reason not to? It makes it hard to provide detailed feedback because we have to deal with the text formatting when we split up your post to quote different pieces of it and this is wasted time because it adds nothing to the information content.
     
  3. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    Sure. But what do you mean split when quoting? You can just quote it.

    But can you help me please?
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Yes, you can derive a general formula that will give you the total population after a certain amount of time has passed. The basic approach is to develop what is known as a recurrance relation. In this case, let's pick some moment in time, T, and see how many organisms will be alive at T+1s.

    You specification is a bit murky. It says that it can ONLY reproduce in its 30th second of life. That means that it can reproduce one time. Yet you then say that if can reproduce until its death, which implies that it can START reproducing in its 30th second of life.

    According to the rules, all of the organisms alive at T-60s will not survive to T+1s. In addition, all organsims alive from T-60s to T-30s will reproduce during this second (so organisms alive at T-60s will both reproduce and die before T+1s, which seems consistent with the rules).

    Thus

    N[T+1s] = N[T] + Ʃ (N[T-60s] ... N[T-30s]) - N[T-60s]

    Simplifying this a bit:

    N[T+1s] = N[T] + Ʃ (N[T-59s] ... N[T-30s])

    This is your recurrence relation and this alone makes it easy to set up a spreadsheet to calculate the answer.

    What you want to do at this point is start working this toward a closed-form solution. You can do this by applying the recurrence relation to walk the terms involving prior states back toward T=0s, since you have the initial conditions that

    N[T] = 0 for T < 0s
    N[T=0s] = 1

    N[T+1s] = N[T] + Ʃ (N[T-59s] ... N[T-30s])
    N[T] = N[T-1s] + Ʃ (N[T-60s] ... N[T-31s])

    Therefore

    N[T+1s] = N[T-1s] + N[T-60s] + 2Ʃ (N[T-59s]) ... N[T-30s]) + N[T-31s]

    As you go through a few levels of this, you look for patterns that you can use to simplify your expression. In another level or two you will start seeing one of the patterns emerge here.
     
    killivolt likes this.
  5. WBahn

    Moderator

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    Look at my responses to some of your other posts. I frequently break up the quoted material (and often only quote part of the post) so that I can insert responses to the difference pieces within the direct context of the part of the post I am responding to. Many people do that.
     
  6. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    What is t+1s?

    Why is "all of the organisms alive at T-60s will not survive to T+1s"?

    Why is "all organsims alive from T-60s to T-30s will reproduce during this second (so organisms alive at T-60s will both reproduce and die before T+1s, which seems consistent with the rules)"?

    Sorry if I have to format my text. But can't you just delete the portion that you don't want. But nevertheless, I'm sorry if that somewhat offends you.

    THANKS!!!
     
  7. WBahn

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    Mar 31, 2012
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    T+1s is simply 1 second after time T. So if T is 30 minutes, then

    N[T] is the number of organisms alive at time T=30minutes and N[T+1s] is the number of organisms alive one second later at time 30:01

    Because they would then be 61 seconds old, which violates the rules, right?

    Because any organism that was alive 30 seconds ago or longer is old enough to reproduce.

    Why do you HAVE to format your text?

    I am happy to delete portions of a quote that are not relevant to the response. But I am far less willing to deal with the formatting. If I only want to quote the top of the post, I have to preserve the formatting (specifically the closing tags) at the end of the post. Or I can go through and delete all of the formatting, but sometimes the formatting that is there is useful in conveying information and should be kept. My tolerance in dealing with non-informative formatting and separating it from informative formatting is limited.

    It's not a matter of it offending me. It's a matter of it making it harder for someone that you are asking for free assistance to provide that assistance. Now, you are under no obligation to make it easy for me to help you. But I am under no obligation to help you if it is more difficult to do so than I am willing to put up with.

    You're welcome.

    See what I mean by splitting up your quote so that I can insert responses at appropriate points. If you had formatted this post as you normally do, I would have either had to strip out that formatting or replicate the open/close tags in each quoted segment.
     
  8. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    Truth to be told: I am a bit slow. I really can't follow. I understand the sigma sign as a summation. recurrence relation? I know I know the definition of it but I can't understand it.

    So wouldn't you mind giving an example? Say at 60 minutes (3600 seconds) how many organisms?

    Thank you very much for your time and effort.
     
    killivolt likes this.
  9. killivolt

    Active Member

    Jan 10, 2010
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    Last edited: Oct 6, 2013
  10. Lightfire

    Thread Starter Well-Known Member

    Oct 5, 2010
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    I already know about sigma notation. But as you may see on WBahn's post, there is no lower and upper limit in the sigma sign.....

    Pls. pls. give me an example so I can understand even better (sorry if i demand too much)

    thanks......
     
  11. killivolt

    Active Member

    Jan 10, 2010
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    I just thought till WBahn, maybe some light reading would help, cause I sure can't.

    (Edit: Don't you hate it, when people don't finish there...)

    I will be following the post though. Good Luck.

    kv
     
    Last edited: Oct 7, 2013
  12. WBahn

    Moderator

    Mar 31, 2012
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    First, a "recurrence relation" is simply a relation that defines the present value of a function in terms of values of that same function at other times (generally, but not always, at times in the past).

    As for the form that I used in my post, the limits indicated by the values on either side of the "...". This was done due to the limitations of putting them in text. So to be more explicit:

    This is the same as:

    <br />
N[T+1s] \, = \, N[T] \, + \, \sum_{k=0}^{30s} N[T-60s+k] \, - \, N[T-60s]<br />

    Now, technically we shouldn't have an index variable that implies a continuous domain (which is implied since time can take on values of 1s or 1.254s). Instead, we should have an integer index that represents samples taken at, usually, regular intervals. This index is often called 'n' (just as the continuous time variable is generally called 't') and where t = nTo and, in turn, To is the sampling period which is, in our case, To=1s.

    Thus we have

    <br />
N[n+1] \, = \, N[n] \, + \, \sum_{k=0}^{30} N[n-60+k] \, - \, N[n-60]<br />

    Although the rules of summation notation say that only the first term after the summation sign is included in the summation, this is more clearly indicated by either

    or

    <br />
N[n+1] \, = \, N[n] \, + \, \sum_{k=0}^{30} \( N[n-60+k] \) \, - \, N[n-60]<br />

    or probably even better

    <br />
N[n+1] \, = \, N[n] \, - \, N[n-60] \, + \, \sum_{k=0}^{30} N[n-60+k]<br />

    We can simplify this (but it may or may not be helpful in the long run) to:

    <br />
N[n+1] \, = \, N[n] \, - \, N[n-60] \, + \, \sum_{k=30}^{60} N[n-k]<br />

    or

    <br />
N[n+1] \, = \, N[n] \, - \, N[n-60] \, + \, \sum_{k=0}^{30} N[n-(k+30)]<br />

    You should satisfy yourself that all three of these are the same (barring mistakes on my part).

    In each of these last three cases, we are just saying that the number of organisms alive 1 second from now will be the number of organisms alive now, minus the organisms that are going to die in the next second, plus the organsism that are going to be produced in the next second.
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    Let's use a simpler example so that we can work through it completely.

    Let's assume that each organism lives for 6 seconds and can reproduce after it has lived for 3 seconds. That means that our recurrence relation is

    <br />
N[n+1] \, = \, N[n] \, - \, N[n-6] \, + \, \sum_{k=0}^{3} N[n-(k+3)]<br />

    Note that this reduces to

    <br />
N[n+1] \, = \, N[n] \, - \, N[n-6] \, + \, \sum_{k=0}^{2} N[n-(k+3)] \, + \, N[n-6]<br />

    or

    <br />
N[n+1] \, = \, N[n] \, + \, \sum_{k=0}^{2} N[n-(k+3)]<br />

    When you think about it, all this merely reflects is that, in their last second of life, each organism is simply replacing itself with its offspring.

    Where we go from here involves a bit of art -- by that I mean that recurrence relations do not lend themselves to cookie-cutter recipes when trying to reduce them to closed form.

    Frankly, I haven't done much in this regard for several years, so I will have to ponder this a bit. My first couple thoughts didn't seem too promising and I don't want to lead you down a rabbit hole. Perhaps someone else with deeper (or at least more recent) experience can guide you a bit further. Otherwise, I will try to play with it a bit and come back to it when I can.
     
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